Soil Mechanics & Foundation Engineering by K R Arora_6th Edition

SOIL MECHANICS AND FOUNDATION ENGINEERING Dr. K.R. Arora

SOIL MECHANICS AND

FOUNDATIONENGThlliEmNG [ IN SI UNITS 1

Dr. K.R. ARORA /l,E. (Civil); ME. (/10m",): Ph. O. (lrrf)) F.I.£.; M.I.C.S.. FISDT; MIWRS

Fonner Professor and Head {)f Civil Engg. Depanmcnl Engineering College. KOTA (Raj.)

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First Edition, Second Edition,

Third Edition,

Fourth Edition,

1987 1989 1992 1997

(Revised and Enlarged) Reprint. 2000 Fifth Edition, 2000 Reprint, 200 I Reprint, 2002 Sixlh Edition, 2003 (Revised and Enlarged) Reprint, 2004

o

K.R, ARORA

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ISBN, 81-8014-028-8

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ACKNOWLEDGEMENT Fig. No. 32.2

page 839 and Fig. No. 32.7 on pagt: 848 of !.hIs publication have been reproduced wiLh permission of 81S, from IS: 1893 (Part 1)-2002 to ~hich reference IS invited forJrurther details. It is desirable Oil

thai for more complete details, reference be made only (0 the lalest version of this standard. which is available from Bureau of Indian Standards, Manak Shawan, New Delhi.

PREFACE TO THE SIXTH EDITION In this edition, the text has been revist..xJ and updated. A new chapler on 'Geotechnical Earthquake Enginccring' has been includt:d to introduce the readers to the recent developments. The importance of geo(Cehnical aspeclS of earthquake engineering has considerably increased in recent years, especially after !.be Bhuj Earthquake of 2001. On the suggestions received from readers, this chapter has bLocn included in this text book. TIle uulhor heartily Ihanks his wife Mrs Raui Arora and son-in-law Dr. B.P. Suncja, Lecturer (Selection scale) in Civil Enginecring, Engineering College, Kota for the assist
-Dr. K.R. ARORA

PREFACE TO THE FOURTH EDITION The basic aim of the four!.h edition of Soil MecIJanics and FOllndlltion Engineering is the ~e as lbat of the earlier thrL"C editions. namely. to present Ihe fundllmentals of the subjcct in a Simplified manner. III this edition, a number of improvements and additions have been incOrl)Orated to make the text more useful. A large number of multiple·choice questions and objective type questions (wilh answers) have been added at the end of each chapter, Chapter 30 gives the detailed procL>dure for conducting nineteen common laboratory experiments. Olapter 31 covers !.he basic principles of Rock Mechanics. Appendix A gives the glossary of common terms for ready reference. SelCCted rcferel ccs and a list of relevant publications of Bureau of Indian Standards are given at the end for furlher study. It is gratifying that the book has heen appreciated by students, teachers and practising engineers throughout the country. TI1C book has established itself as a useful text in most of the enginccring colleges and technical institutions of the country. The author is grateful to !.he teachers and students who have sem !.heir comments, suggestions and letters of appreciation. 111e author thanks his colleagues Dr. R. C. Mishra and Sh. N. P. Kaushik for Lheir help in proof reading. 111e author also thanks his wile. Mrs. Rani Arora for her assistance in the revision of the book. The help received from Shri Bhagwan SlIwroop Sharma, Draughtsman, in improving the diagrams is appreciated . EffoTlS made by !.he publishcr Sh. N. C. Jain and his sons Sh. Ajay Kumar Jain and Sh. Atul Kumar Jain for bringing out this edition in a short lime and in a good form are appreciated. 10 spite of every care taken to ensure accuracy. some errors might have crept in. The au!.hor will be grateful 10 the readers for bringing such errors, if any. to his notice. Suggestions for the improvements of the text wiU be gratefully acknowledged. KOTA (Raj.) -Dr. K.R. ARORA February 26, 1997

PREFACE TO THE FIRST EDITION Soil mcdtanics and Foundation engineering (gcolcchniClI engineering) is a [asl developing discipline of civil engineering. Considerable work: has been done in [he field in the last 6 dcc.'ldes. A student finds it difTiOJII to have access to the latest literature in the field. The author b.1S tried to collect the material from various sources and [0 prescnt in the form of a lext. The text bas been divided into twO parts. The first pan dc.'lls with the fundamentals of soil mcchanics. The second pout dc.lIs with earth rCUlining structures and foundat ion engineering. 'nle subject matter has been presented in a logical :lntl org:mi.scd manner such liwi it may be laken up serially without llny loss of continuity. ' :hc book covers the syllabi of undergraduate courses inn Soil Mechanics appearing for various competitive examinations and AMlE will :llsa find the text useful. A large number of ch•• rts and tables have been included to make the text useful (or'pmctising engineers. lbc author is grateful to Prof. Alam Singh of Jodhpur University who introduced the subjcct to him about 3 decades ago as a student :1t M.B.M. Engineering College, Jodhpur. lbe author is indebted to Prof. A. Varadarnjan of nT, Delhi, who helped him in understanding some of lhe intricate problems during his doctoral programme. The author thanks the faculty of Geotechnical Division of liT, Delhi. for the help extended. '[be author al<;o thanks his fellow research scholars, Dr. K.K, Gupta, Dr. D. Shankcriah, Dr. T.S. Rekhi, Dr. 8.S. Salija, and Dr. R.N. Shahi for the fruitful discussions. Ihe autbor is grateful to Prof. A.V. Ramanujam. Principal, Engineering College, Kola for constant encouragement. 'Ibe author thanks his colleagues at Engineering COllege, Kota, especially Sh. Amin Uddin, Drnughtsman. 'Il1e author also thanks his wife Mrs. Rani Arora who helped in proof reading and other works related with this tex!. 'Ille help received from his daughter Sangeela Arora and son Sanjeev Arora is also acknowledged. In spite of every care Inken to cnsure acx:uracy. somc errors might have crept in. The author will be grateful to readers for bringing such errors to his notice. Suggestions for improvement of the text wilt be acknowledged wilh lhanks. KOTA(Raj.)

January 4,1981

-K.II.AROUA

NOTATIONS The notations have been explained wherever they appear. The following notations have been more commonly used. Pore p~ure parameter '" Actlvjtyofsoils

A ..

A,,= Arcaofvoids A"". Angstrom

A .. ::: Air conlcnt Qv = CoefficicnI of compressibility B= I'orepressureparamelcr c.. ::: Compression index a Coefficient of curvature

e,,:

Uniformity Cocfficient

= Coefficient of elastic uniform compression c:: Unit cohesion c' ... EffeCtive unil cohesion

e.. ::: Apparent u;>hcsion Cv'" Coefficient of consolidation DlO'" Effectivesize D,= Foundation depth Dr = RelalivedcnsilY E:: Modulus of elasticity c o: Voidralio FS::: Factor of safely

f:c Friction G '" g::: h= I", Ip =

Specific gravity of panides

Acceleration due to gravity Hydraulichcad Momcnl.of inertia Plasticityindcx i"" Hydraulic gradient ::: Angleofsurcharge

K" Cocfficicntofabsolutc permeability Ko = Coefficient of eanh pressure al .."

K a ", Coefficient of UClive pressure Kp '" Coefficient of passive pressure k '" Coefficient of pcrmwbility

.. k, = kp '" N ..

Coefficient 0( subgmde reaction Coefficient of subgrade reaction erefficient of percolation Numberofblows(SP1)

'" Perocntfincr .s Normal romponent n:: Porosity IS.," Percentage air voids p .. Forre

w......

Weigh t of water W,,,, WeighlofsoHds Wq ", Wotertablcfactor Wy '" Water Illble factor IV '" Water content M= Mass,lotalmass Mw: M
P,,; Activeprcssureforce Pp z: Passive pressure force p= Pressure p"", Activeprcssure Pp'" PlL'iSiveprcssure pit:. Horizontal pressure Q= force,Load "' Totlliquantityofwatcr Q.. '" Allowable load

(w)

/'

CONVERSION FACTORS (a) MKS to SI Units F~

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Multiply by

kgf gmf

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9.81)( 10-3

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9.81

kN

k~flcm2 2 kg£Jcm

kN/m2

9.81 98.1

N/mm2

9.81 x 10- 2

gmflcm 2

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98.1

2 Ilm

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9.81

kgfiln 3

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kNlm' kN/m)

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gl""Jtcm)

kgflrn

Nl m N·m

9.81

kgf_seclm2

N_s/m2

9.81

11 .. 9.81 kN 1 kgf/cm2 .. 98.1 kN / m2 1 kgf/cm2 .. 0.0981 N/mrn2 1 gmflcm2 .. 98.1 N/m2 I I/m2 .. 9.81 kN/ml

1 legUm) .. 0.00981 kN/ mJ

Illm J .. 9.8L kN/ml 1 gm£lcml I k,grlm 1 kgf-m 1 kgC_sec/m 2

9.81

kg£.m

Equivalence 1 kgf .. 9.81 N lsmf .. O.00981N

9.81

.. 9.1:U kN/mJ .. 9.81 N/ m ... 9.81 N-m

.. 9.81 N-s/m2

(b) Sl to MKS Units

From

To

Multiply by

Equivalellcl:

N N

kgf

0.102

grnf too"' kgfJan2

102.0 0 .102

IN -O.101kgf IN _102gmf

kN kN/m2 N/mm2

2 kgf/an

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0.102

2 kN/m

vm'

0.102

1

lIeN .. 0.102/

O.l02x 10- 1 10.2 Ie

10-1

1 kN/m2 .. 0.0102 legf/cm 2 1 Nk,vnl .. 10.2 kgf/an1

1 N/ml .. 0.0102 gmf/cm2

kN/inl

vm'

0.102

1 kN/m 2 .. O.102Vm 2 1 kN/mJ .. 102.0 kgf/m3 1 kN/m3 _ 0.10211m3

kN/ml

grnf/m'

0.102

lkN/m3 _ O.102grnf/an3

N/m N·m

kgf/m kgf-m

0.102 0.102

N-s/m2.

ktJ«I;/m2.

0.102

kN/ml

kgf/ml

0.102)(

Note: 1 poise '" 0.1 N-tlm2. _ l.02 x 1O-~ kgf-seclm2 2 1 bar .. 100 ItN/m

tal

1 N/m - 0.102 kgflm 1 N-m - 0.102 kgf-m 1 N--s/m2. _ 0.102 kgr~m2.

(vii)

CONTENTS Chapter' 1.

Page No.

PART I. FUNDAMENTALS OF SOIL MECHANICS Introduction

3 -12

1.1. Definition of soil, 1; 1.2. Definition of soil mechanics, 2; 1.3. Definition of Soil Engineering ond Geotechnical Engioecring, 1; 104. Scope of soil Engineering, 2; 1.5. Origin of Soils, 4; 1.6. Fonnution of Soils, 5; 1.7. Transportation of Soils, 6; 1.8. Major Soil Deposits of India, 7; 1.9. Comparison of Soils with a:her materials, 8; 1.10. Umltation.s of Soil Engineering 8; 1.11. Thrminology ofdiffeR:n1 types of soils, 9; 1.12. Cohesive and CohesionJess Soils, IU; 1.1:\. Brief History of Soil Engineering, li; Problems, 11.

2.

Basic DerrniUons and Simple Tests

13 - 44

21. Introduction, 13; 2.2 Volurnetrjc Relationships. 14; 2.3 WIlter content, 15; 204. Units, 1; 2.5 Volume Mass Relationship, 16; 26. VoluriJe..Weight Relationships, 17, 2.7.lnter-relalion between Mass and Weight Units, 18; 2.8. Specific Gravity of Solids, 19; 2.9. Three-Phase Diagram inn Terms of Void ratio, 10; 210. Three-Phase Oiagrom in Terms of Porosify, 22; 211. Expressions for Mass Density in Terms of WJter Cantant, 23; 2.12. Expression fa- mass density in tenns of water rontent, 24; 2.13. Relationship between Dry Mass Density and Percentage Air \bids, 25; 2.14. Water Content Determination, 26; 2.15. Specific Gravity Determinatlon, JO; 2.16. Measurement of Mass Density, 32; 2.11. ~ennination of Void Ratio, Porosity and Degree of Saturation, 36; illustrative Examples, 37; Problems, 42.

3.

Particle Size Analysis

4S - 68

3.1. Introduction, 45; 3.2 Mechanical Analysis. 46; 33. Sieve Analysis, 46; 3.4. Stokes' Ulw, 47; 3.5. Preparation of suspeMion for sedimentation analysis, 49; 3.6. Theory of Sedimentation, 50; 3.7. Pipette Method, 51; 3.8. Hydrometer Method, 52; 3.9. Relationship Between Percentage Fiocr and Hydrometer Reading, SS; 3.10. Limitation of Sedimentation Analysis, 57; 3.11. Combined Sieve and ScdimentllIion Analysis, 57; 3.12 Panicle Size Distribution Curve, 57; 3.13. Uses of Particle Size Distribution Curve, 59: 3.14. Shape of Partideo>, 59; 3.15. Relative Density, 60; 3.16. Determination of Relative Density, 61; lIIustrative Examples. 62; Problems, 66.

4.

Plasticity Cbaracterlstlcs of Solis

69 - K8

4.1. Plaslicity of Soils, 69; 4.2. Consistency limits, 69; 43. Uquid Limit, 70; 4..4. Cone Pcoclromctcr Method, 73; 4.5. Plastic Limit, 73; 4.6. Shrinkage limit, 74; 4.7. Alternative Method for determination of shrintage limit, 75; 4.8. Shrinkage Parameters, 76; 4.9. Plasticity, Uquidity and Consistency Indexes, 78: 4.10. Flow Index, 78; 4.11. Toughness Index, 79: 4.12 Mea<>urement of Consistency, 80; 4.13. Sensitivit) 80; 4.14. Thixotropy, 81; 4.15. Activity of Soils, 81; 4.16. Uses or consistency Limits, 82; Illustrative Examples,83; Problems, 87.

5.

SoD Classification

89 -106

S.1. Introduction, 89: 5.2. Pllrtide Size Oassification, 89; 5.3. Thxtural Oassification, 91; SA. AASlrfO OassHication System, 92; 5.5. Unified soil Oassifiallion System, 72; 5.6. Compari:-;on of AASlim and USC systems, 95: 5_7. Indian Standar.d Oassifiemion System, 98; 5.8. Boundary O[l$ificrltion, 99; 5.9. Field Identification of Soils, 101; 5.10. General ClJarnderiSlics of Soils or Different Groups. 103; lII~trBtive Examples, 103; Problems, 105.

6.

Clay Mineralogy and SOU StructUR

107 -119

6.1. Introduction, 107; 6.2. Gravitational and Surface (oroes, 107; 6.3. Primary %lcnce Bonds, 108; 6.4. Hydrogen Bond, 109; 6.5. Seo::todary \1aImoe Bonds, uo; 6.6. BasIc Structural Units of Oay Minerals,

(viii) 111; 6.7. lsomot:phous Substitution, 112; 6.8. Kaolinite Mineral , 112; 6.9. Mo ntmorillonite Mineral , 112; 6.10. Illite Mineral, 113; 6.11. Electrical charges on clay minerals, 113; 6.12.11ase E;(change Capm.i ty, 114; 6.13. Diffuse Double Layer, 114; 6 14. Adsorbed Wl11 er, 116; 6.15. Soil Structurcs.1l6, ProbJem~ , 118.

7.

Capillary Water

120 - 133

7.1. Types of Soil Water, 120; 7.2. Surface Tension, 120; 7.3. Capillary Rise in Small DiameterTubcs, 121 ; 7.4. Capillary Thnsion, 122; 7.5. Capill3fY Rise in Sroils, 123; 7.6. Soil Suctio n, '12S; 7.7. Capillary Potential, 125: 7.8. Capillary Thnsion During Drying,.(l(.SoiIs, 126; 7.9. Factors Affecting Soil Suction, 126; 7.10. Measurement of Soil Suction, 127; 7.11:'~~t H.c~e, 128; 7: 12. Fra;t Doil , 129; 7.13.

~~::~~~~~! ~~~~~~~'; ~~~;. 7~~jl~:~n~p~::~ndi~~~~~:rn~~~\:!~:r~~~'ri~~k~~~~!~~r3i~; 8.

Penneabllily ~r Soil 8.1. Introduction, 134; 8.2. Hydroulic Hcad, 134; 8 .3. Darcy's Law, 135; 8.4. Validi ty of Darcy's

134-162 Low, 136;

8.5. Determination of Coeffi cient of Permeabili ty, 136; 8.6. ConSlant Head Penncabilily Test, 137; 8.7. Variable·Head Permeability Test, 138; 8.8. Seepage Velocity, 140; '8.9. General Expression for Laminar Flow, 141; 8.10. Laminar Flow through Porous Media, 142: 8.11. Factors affccting Permeability of Soils,

143; 8.12. Coefficient of Absolute Permeability, 145; 8.13. Pumping Out Tests, 146; 8.14. Pumping in Thsts, 148; 8.15. Coefficient ofpcrmeability by Indirect Methods, 151; 8.16. Caei.llarity- Permeability Test,

152; 8.17. Permeability of Stratifi ed Soil Deposits, 154; l11ustralive Examples, 156; Problems, 160.

9.

Seepage Analysis

163 - J 88

9.1. Introduction, 183: 9.2. l:lplooe's equation 164; 93. Stream and Potential Functions, 165; 9.4. Owacteristics of Row Net, 167; 9.5. Graphical Method, 168; 9.6. Electrical Analogy Methcxl, 168; 9.7. Soil Models, 171; 9.8. Plastic Models, 172; 9.9. Flow Net by Solution of Laplocc's Equation, 172; 9.10 flow Net in Eanh Dams with tI lIorizonml Filler, 173; 9.11. Seepage through Eanh Dam with Sloping Discharge face, 175; 9.12. Seepage through Eanh Dam with Discharge angle less than 30\ 176: 9.13. Seepage through Eanh Dam with Discharge angle greater than 30°, 177; 9.14. Uses of Flow Net, '178; 9.15. flow Net for Anisolropic Soils, 180: 9.16. Coefficient of Penncability:in an Inclined Direaion, 182; 9.17. flow Net in a Non-homogellOOus Soil Mass, 182; Ill ustra tive Examples. 184; Problems. 185.

10. Effective Stress Principle

189-217

10.1. Introduction, 189; 10.2. Erfective Stress Principle, 189; 10.3. Nature of Effective Slrcs.~ 190; 10.4.

Effect of water Table fluctuations on Effcctive Stress, 192; 10.5. Effective Stress in a Soil Ma.,,-" under Hydrostatic Conditions, 193; 10.6. Increase in effective Stresses due 10 surcharge, 195; 10.7. Effective Stresses in Soils saturated by Capillary Action, 195; 10.S. Seepage Pressure, 197; 10.9. Force Equilibrium in Seepage Problems, 198; 10.10. Effective Stresses under Steady Seepage Conditions, 200; 10.11. Quick Sand Condition 201; 10.12. Seepage Pressure Approach for Quick Qlndition, 203; to.13. [creel of Surdlarge on Quick Conditions, 203; 10.14. Failures of Hydrnulic Suucturcs by Piping, 204; 10.15. Prevention of Piping Failures, 206; 10.16. Design ofGroded Filter, 207; 10.17. Effective Stress in Panially Saturaled Soils, 209; Illustrative Examples, 210; Problems, US.

11. Slresses Due 10 Applied Loads

218 - 255

11.1 InlrOOudion, 218; 11.2. Suess·Slroi n Paramelers, 218; 11.3. Geostatic Stresses, 219; 11.4. Venical Stresses Due to Concentrated Loads, 221; 11.S. Horizontal and Shear Snesses Due to Concentrated Loads, IZ3; 11 .6. Isobar Diagram, 225; 11.7. Vertical StftSS Distribution on 3 Horizontal Plane, 225; 11 .8. lnfluence Diagram, 226; 11.9. Venical Stress Distribution on a Venical Plane, 227; 11.10. Vertical Stresses Due lo a Une Lond, 227; 11.11 . Venic:al Stresses Under a Strip Load, 229; 11 .12. Maximum Shenr Strcsses at a Point Under a Strip Load, 232; 11 .13. Venical Stresses Under a Circular Area, 233; 11 .14 . Vcr1ical Stress Under Comer of a Rectangulor Area, 234; 11.15. Venical Stress al any Poin t Under a Rectllngulur Area, 236; 11 .16. Newmark's InfluenceChurts, 237: 11.17. Comparison ofStrc.o;scs Due 10 l..ood
fix)

12. Consolidation of Soils

256- 305

12.1. Introduction, 256: 12.2. IrIIllal. Primary and Secondary Consolidation. 257; 12.3. Spring Analogy for Pnmary ConsulktLtlon. 257: 12.4. Behnviour of Satumtec.l Soils Under Press ure. 258: 12.5. Consolidntion 'res!. 259: 12.6. Dctenlllll:Ltmn u! VOid Rmio at Various Load Increments. 261: 12.7. COl1solid:uion Test Result~. 263: 12.!). Ba."lc Dottinitions. 265: 12.9. Terzaghi's Theory of Consolidation. 267: 12. 10. Solution 0 1 fllL~ I C DlffcrelltlHl Equatllm. 271 : 12.1 1. Determination of Coefficient of Consolidation. 277: 12. 12. Preconsolidatlon Pressure. 280: 12. 13. Causes of Preconsoliti:llion in Soil s.l8 1: 12. 14. Finol Settlement of ;1 Soil DepoSli in the Fn:ld. 28 1: 12.15. Time Sell[emcnt Curve. 283: [2.16. Field Consolidation Curve. 2X4: 12.17. Secondary Co nsnliu.llion. 2115: 12.18. 3-D Consolidation Equation in Cartesinn Coordinates. 287: 12.IY. 3-D Consolidation Equation in Cylindrical Co-ordi nates. 289: 12.20. Sand Dmin~. 291: 12.21. Effect or L:ller.ll Stmin ml C()nsohdlltion. 294: IIlustrmivc Exn!llpl~, 295; Problems. 302.

13. Shear Strength

306-356

1]. 1. Im roduClitin. 30h: 13.2. StrC-'is Sy~tcm with Prindp.11 Planl!s P:lr.lllel to the Coordinate Axcs, 306: 13.3. Mohr's Circle. 3d7: Il4. Pri nc ipal planes mclinl!d to the coordinate axis. 308; 13.5. Stress system with Vertical and Horimntu! Plnnl!s not Principal Plnnc.~. 309: 13.6. Import::lIlt Characteristics of Mohr's Circle. 311 : 11.7. Muhr·Cou lomb TIleory. 3 12: 13.8. Revised Muhr· Coulomb equation. 313: 13.9. Different Typc~ of tc~ t s nnd Dminnge Condi(ion~. 3 13; 13. 10. Mode o f Application of SheH Force 314: 13 .11. Direct Shear Test. 314: 13. 12. Presentation ()t" Results or D I ~cCI ShearT..::s(' 316: 13. 13. Merits alld Demerits of DirCl.:"t SheOlr Tc~t. 3 1H: 13. [4. Triaxml Compression Applirmus, 318: [3. 15. Trillx ia! Tests on Cohc!<.i\lc Soils. 321; IJ.16. Triaxia l 'reSiS on Cohesiunlc:ss Sui Is. 322: 13.17. Merits ::md Demerits of Tri.lxinl Tcs!. -'2-': [J. [1I . Cmnput;l1 ion o f various Pnmmeters. 324: 13.19. Presentatio n of Results of Triaxial Te~ts. 325 I J.20. Elfcct ofCunsolidation Pre.~~urc o n Undraim.'d Strength 328: 13.2\. Relationship Betwecn Unur.lincd Shear Strength and Effective O\lerburden Pressure. 329:: 13.22. Unconfined Compression Te:.t. 330: 13.2-'. Vanl! Shear Test, 332: 13.24. Pore Pressure Parameters. 333: 13.25. Mohr·Coolo mb Fai lure Cntl!nun. 337: 13.26. Mo(lillt.:d F.u[ure en\lelope. 338: 13.27. Stress Path. 339: 13.28. Shear Slro;:ngth o f Partially Satur.Jte(/ Soils, 341; 13.29. H\lo rslev's Strength TIleory. 342: 13.30. Liquet":lo;,:tion of S:mds. 343: 13.3 1. Shear Characteristics of Co hesionles.' Soils. 144: [3.32. Shear Charncteristics of Cohesive Sui Is. 345: U.3J. Ch"ire of Test Conditiuns and Shear Pamlllcters. 347 • Ill ustrative EX;lll\ple~. 347: Problelll~. 353.

14. Compaction of Soils

357 -375

14. 1. Introduction. 357: 14.2. S1andani Proc1or Te~t. 358: 14.3. Modified Procto r Tesi. 360: 14.4. Compaction of Sands. 361 : 14.5. Jodhpu r Mini CompaclllfTc~l. 362; 14 .6. Harvard Mini;Jture Compaction Tc.~t. 362: 14.7. Ahbot Cump;u:llon TC~I. 362: 14.S. Fal1or~ Affccting Compaction. 362; 14.9. EITel'! of CompaCIIOI1 on PrOJ>CrllO;:~ of Soih. 364: 14.10. Methods of C(Illlpaction Used in Field. 366: 14. 11. PI'lcement Water Content. 367: 14. 12. Relative COmp;Jl1ion. 368: 14. 13. Compaction Control. 368; 14. 14. .. ,broll m;n il)n Method. 36?: 14. 15. Te ml Probe Method, 370: [4. 16. Compaction by Pounding. 370: 14.17. Cnmpa':1I011 by Explosl\·e.,. 37 1: 14.1B. Prccomprcssion. 37 1: 14.19. Compaction Piles, 371 : 14.20. Suitability of Various i\.1t:thod~ uf Compaction. 371: lllustrati ve Exa mples. 372; Problems. 374.

15. Soil Stab ilisation

376 - 390

15. 1. Introduction. 37(, : 15 .2. Medwnical St;lbi!isntion. 376; 15.:1. Cement Stabi lisation. 377: [5.4 Lime Stabilisation. 3811: 15.5. Bituminous $t;lhilisalion. 31B ; 15.6. Chemical Stabilisatiun. 3H2: 15.7. TIlcrm;1 Stabili~ation. 383: 15.8. Electrical St;lbilisation. 384: 15.9. Stabilisation by grouting. 384: 15. 10 Stabilis;Jtion by C<.'utuxtilc :lnu Fnbrics. 3115: 15.1 1. Reinforced Eolrth. 3M7: Prob lem~. 3S?

16. Drainage, DC·w:ltcri ng and Wclls

391 - 414

16.1. Int roductiun. 391 : 16.2. Interceptor Ditches. 39 1; 16.3. Single Stage Well Points. 392: 16.4. Mult"i.$I:J£c Well POIOIS, 393: 16.5. Vacuum Well Points. 393; 16.6. Shallow Well System. 394: 16.7. Deep Well System. 394: 16.8. Hori zontal Wd ls. 394: 16.9. Electl1}-Osmosis. 39-1: 16. 10. Permanent Drainage After Con~tnlctil)n. 395: 16.1 1. Design of Dewatering Sy.~ tcm s. 396: 16. 12. Discharge from :I Fully Penetrating Slu\. 396: 16.1]. Di sc harge from a Partially Penctr.lling Slot, 399: 16. 14. Discharge in a Slot from Bolh sides. 400 : 16. 15. Well Hydraulics. 4() 1: 16.[6. Tem1.~ Uscd in We ll Hydraulics, 402; 16.17. Discharge From a Fully P..::netnull1g WeI [. 403: [6. 18. Disc harge From a Paniall y Pc netrnting Well, 404: 16. 19. IllIerf..::rcnce among Wells, 4115: 16.20. Spherical Flow in a We ll. 407: 16.2 1. Discharge Froman Open Well. 407; 16.22. Advt':rse Eff..:cts of Dramage. 44.19; Ill ustrative Examples. 4O!J; Problems, 412.

(r)

PART II. EARTH RETAINING STRUCTURES AND FOUNDATION ENGINEERING 17. Site Investigations

415 - 439

17.1. Introduction, 415; 17.2. Planning a Sub-Surface Explor.lIion )rogrnmmes, 416; 17.3. Slagcs in

Sub-surface Explorations, 416; 17.4. Reconnaissance, 417; 175. Depth of Exploration, 417; 17.6. Lnternl Exlent of E"plorlLlion, 419; 17.7. Open Excavation Methods of Explomtion, 420; 17.8. Borings for Exploration, 420j 17.9. Auger Doring, 420; 17.10. Wnsh Boring, 420; 17.11. Rotary Drilling, 422; 17.12. Percussion Drilling, 42Z; 17.13. Core Drilling. 422; 17.14. Types of Soil Samples, 423; 17.15. Design Fealuws AfJa:ting the Sample Disturbance, 423; 17.16. Split- Spoon Samplers, 424; 17.17. Sa"aper-Buckel Sampler, 425; 17.18. Shelby Tubes and Thin Walled Samplers, 425; 17.19. PiSlon Samplers, 426; 17.20. Denison Sampler, 426; 17.21. lIand-Carvcd Samples, 426; 17.22. Slandard Pcnclmtion Test, 427; 17.2.1. Cone Pcnclrarion T~SlS, 429; 17.24. Tn-sHu Vane Shear Test, 431; 17.25 . In-situ Tcst Using a Pressure Mctcr, 431; 17.26. Observation of Ground W:Jtcr 111ble 432; 11.27. Geophysical Mcthods, 433; 17.28. Sdsmic Mcthods, 433; 17.29. Elcctrical Resistivity Methods, 435; 17.30. Sub-Soillnvcstigmion RcporlS, 437; Problems, 438.

18.. Stability or Slopes

440-477

18.1. Introduction, 440; 18.2. l3asis of Analysis, 441; 18.3. Different Factors of S3fety, 441; 18.4. Types of Slope. Failures, 442; 18.5. Stability oron Infinite Slope of Cohesionlcss Soils, 444; 18.6. Stability An.:lIysis of nn Infinite Slope of Cohesive Soils, 446; 18.7. W(:dgc Failure, 447; 18.8. Culmann's Method, 448; 18.9. '" .. 0 Analysis, 450; 18.10. FriCtion Circle Method, 4s(); 18.1 L SI.1bility Chans, 453; 18.12. Swedish Cirde Method, 455: 18.13. Stability of Slope Under Steady Seepage Condition, 460; 18.14. Stability of Slope Under Sudden During ConstM., ion, 461; 18.15. Stability of Slopes During Construction, 462; 18.16. Bishop's Simplified Method, 46..1; 18.17. Other Mcthods of Analysis, 466; 18.18. Improving Stability of Slopes, 467; IIlUSlrutive Examples, 467: Problems, 475

19. Earth Pressure Theories

478 - 516

19.1. Introduction, 478; 19.2. Diffcrcnltypcs of uterol Earth Pressure, 478; 193. Earth Pressure at Rest. 480: 19.4. Rankine's Earth Pressure Theory, 481: 19.5. Runkine's Earth Pressure when the Surf:Jce is Inclined, 485; 19.6. Itnnkinc's Earth Pressure in Cohesive Soils, 491; 19.7. Coulomb's Wedgc Thcory, 494: 19.8. Coulomb's Active Pressure in Cohcsionless Soils, 494: 19.9. Rehbann's Construction for Active Pn.'SSurc, 497; 19.10. Culmnnn's ConstruClion for Active Pressure, SOl; 19.11. Coulomb's Active Earth !'ressure for Cohesive Soils, S02; 19.12. Trial Wedge Methoo, 503; 19.13. Coulomb's Passive Earth Pressure for Cohesionlcss Soil, S()4; 19.14. Passive Pressure By Ihe Friclion Circle Method, 50S; 19.15. Determination of ShCllr Strength Parameters, 507; Illustrative Examples, 508; Problems, 515.

20. Design of Ret:lining Walls and Bulkheads

517 - 549

21. Braced Cuts and Coffer Dams

550 - 569

W .1. InlrOOUClion, 517: 20.2. l)'pcs of Retaining Walls, 517; 20.3. Pri~iples of the Design of retaining Walls, 517; 20.4 . Gravity Rctaining Walls, 520; 20.5. Cantilevcr Rctaining walls, 52J ; 20.6. Counterfo rt Retaining Walls, 523; ZO.7. Other Modes of Failure of Retaining Walls, 524; 20.8. Drainage from the Backfill, 525; 20.9. 'Iypcs ofshcel pile Walls, 526; 20.10. Free Cantilever shcct pile, 527; 20.11 . Cantilever Sheet Pile in Cohesionlcss Soils, 528; 20.12. Cantil~..... cr Sheet Pile Penetrating Clay, 530; 20.13. Anchored Sheet Pile with Free Earth support, 532; 20.14. Rowe's Moment Reduction Curves, 53-1; W.15 . Anchored Shcct Pile with fixed Eartb Support, 535; 20. 16. Design of AnchOl'S, 536; lIIustrntive exa.mples, 53Sj Problcm,s 547. 21.1 . Introduction, 5S(); 21 .2. Lateral Earth Pressure on Shccting.<:. 551 . 213. Different 'I'ypes of Sheeting and Bracing Systems, 553; 21.4. OcsiJ!n of Various Components of nracing, 554; 21.5. Types of Coffer Dams, 556; 21.6. Design of Ccllulm- Coffer dams on Rock, 559; 21.7. Design of Cellular Coffer dams on Soil, 562; II1US1ldtive Example, 564; Problems, 568.

22: Shal'ts, Tunnels and Underground

Condlli~

570 - 586

22.1. Stresses in Soil in the Vicinity of Vertical Shaft, 570; 22.2 Stresses in Soil around Tunnels, 57.1; 22.3. Construction of Ellnh Tunnels, 572: 22.4. Arching in Soils, 573; 22.5. Types of Unde!grOlmd Conduits,

(xi)

575; 22.6, Ditch conduits. 575; 22.7. Positive Projecting Conduil~. 577; 22.8. Negative Projecting Conduits, 580: 22.9. Im perfect Ditch Condui!. 582; 22. 10. Tunndcd Conduits. 51:12: 22.11. Loads on Conduits Due 10 Surface Loads, 583: 22.1 2. COnSlmCI10n of Conduits. 583; Illustrative ElIamp1cs. 584: Problems. 585.

23. Bearing Capacity of Shallow Foundations

587 -635

23.1. Introducti on. 5~7: 23.2. Basic Definitions, 581: 23.3. GI1IS~ and Net fooling Pressure. SKS: 23.4. Rankine's Anllly!>is. 5~1: 23.5. HO!!Clllog1cl' and l c r/.!i£hi's An3lysis, 591; 23.6. Prandt]'s Anal ys is. 592: 23.7. li: rzag hi's bearing Capacity 1110(1)'. 593: 23.8. Types of ShCltr Fail ures. 596: 23.9 . Ultimate BC3ring CapllclIY in casc of Local Shear Failure. 597: 23.10. Effect of Wmcr lanle on Beanng Cllp,n:i ty. 600: 23. 11 . Beming CIIJlllcity of Square and Circulnr Footings, 601 : 23. 12. Mcyemof's BCllring Cap:u:ity Theory. 602: 23. 13. Hansen 's Bcaring ClIpacilY 1l100ry, 60.1: 23.14. VClIic's Be:ui ng Capacity Theory, 605: 23.15. IS Code Method 606: 23.1(1. Skcmpton 's Analysis for CoheSive Suils, 607; 23.17. IS Code Method for Cohcloive Soil. 608: 23. 18. Heave of the Buttom of the Cut in Clay. 60N: 23. 19. Foundations on Layered C lny. 6111: 23.20. Bt,tring Capa,,;ity fru m Standard Penetration lest. 6H1: 23.21. El:centne:tll y Loaded r,()und:u io ns. 611 : 23.22. SeU lemcnt of FoumJations. 612: 23.23. Loads for Sett lement An:llysis. 613: 23.24. Immediat e Scll!cmcllt ofCohc$iw Soils. 613; 23.25. Immedi:lIC SeUlemeot ofCohesionlcss Soils. 614; 23.26. Consolid.Ltion SClllcmcnt in ClllYS. 6 15: 2.l27. Sel1lement of foundations on CoheslOn lcss Soils, 616: 23.28. Accuracy of foundation Settlement Prl.-diction. 617: 23.29. Artuwablc ScUlcmenl. 617; 23.30. Allowable Soil Pressure for Cohcloionlcss Soils. 618: 23.31. Allowahle Soil Prcs~ ure ror Cohesive Soils. 621 : 23.32. Presu mptive Bcaring C:1P:1Clly. 621: 23.33 . Plate LO:ld Test. 621; 23.34. Housel's Method for destgn o f Foundation. 625; lIIusmtuve Ex:unplcs. 625 : Problems. 625.

,- ,

.-.

24. Design of Shallow Foundations

636 - 670

24.1. Types of SlmllolV fou ndations. 636: 24.2. Depth u f Footings. 637; 24.3. Foundation Loading, 639; 24.4. Principle of Design of Footings. 640: 24.5. Proport ioning FOO1ings for E
t.

25, Pile Foundations

1

671-705

25. 1. Introduction. 671 ; 25.2. Necessity uf Pile ruuIl(Jntiun. 671 ; 25.3. Cla~silication of Piles. 672; 25.4. Pile Driving, 674: 25.5. Conmllction o f Bored Piles, 675: 25.6. Driven Cast-in-situ Concrete Piles. 676 25.7. Lo,ld CarrYlllg Ca pacity of Piles. 677; 25 .!:\. Stallc Methods for Driven Piles in Sand. 677: 25.9. Static Method f()r Driven Piles in SllIUr.'lIt:d Clay. 681 : 25.10. Stalic Method tor Bored Piles. 683; 25. 11. Factor of Safet y. 684: 25.12. Negative Skin Friction. 684 25. 13. Dynamic Fommillc, 685; 25. 14. Wave Equation A naJ Y~t~. 61:17: 25. 15. In-loitu penetr.'llion tests for Pile capllcity, 688: 25. 16. Pile Load Tcst. 688: 25.17. Other tYJ>cs uf Pile Luad IcSt. 690: 25. 18. Gmup Aclion of Piles. 690 25.19. Pile Groups in Sand aod gr.'lve1. 691 : 25.20. Pile G roups in day. 692: 25.21 . Seulcment of Pile Groups. 692: 25.22 Sharing of Loads in It Pil e Group. 694 25.23. Tcn~ioll PiJc ~. 694; 25.24. Laterally Lunded Piles . 696; lIIustrativc Examples. 697; Problems. 70....

26. Drilled Piers and CaL..sons

706 -721

26.1. Introduction. 7('11',: 26.2. Drilled Piers. 706: 26.3 . Construction of Drilled Piers 708; 26.4. Advnlll~ge.~ and Dis.1dvllntngcs of Drilled Piers. 709: 26.5. Dcsigll o f opcn Cllbson~. 710; 26.6. Construction of open caissons. 713: 26.7. Pneumali,,; Caissons. 714: 26.8 . Con ~ lru cli() n of PneLimatic Caissons. 715: 26.9. Advllnt
27. Well Foundations

722-754

27. 1. Introduction, 722; 27.2. Dil"ferent Shapc.\ IlfWells. 72.3 : 27J. Gri p Length. 723: 27.4. Forces ACling

('\'ii)

00 the Well Fououmion. 724: 27.5. Tel7.aghi's Analysis, 725: 27.6. B;mcrjee and Gangopadhyay's r\nalysis. 728: 27.7. Si lllplilicu Antlly~is lor Heavy Welts, 733: 27.8. IRe method, 734: 27.9. Individual Components of the welt. 739: 27,10. Sinking of Wells, 742: 27.1 I. Mca~urc,~ for Rectification o f Tilts nnd Shins, 744: IJl U.,tr,lIl\·C Examplc!>. 746: Pmbkms. 754.

28. Machine Foundations

755-772

28. 1. Introduction. 755: 28.2. 'TYpes of Machine Foundations. 755: 28.3. Bllsic Definitions. 756; 2~.4. Degrcc of Frc ...'. 766: 2S. 11 . Reinforcement and Con~truction Dcrails. 767: 28. 12. Weight of Found:lt iun. 767: 2tU3. Vibration IsolatlU n and Control. 767; l1lustrJtive EX:llllples. 76H ; Problems. 771.

29. Pavement Design

773 -787

29.1 Typc~ of PavemcnT~. 773; 21).2. Bask Requirements of P:lvemCnls . 175: 29.3. Functions of Different Components of a Pave ment. 774: 29.4. Fm:tors Affecting Pnvement Design, 775: 29.5. California Bcaring Rutio T~'st. 775: 29.6. Design of Flexihle Pavcmcnts. 777; 2<;.7. GroUI' Index Mcthod. 777 29.8. CBR MCIJlOd. 17M: 2Y.'J. Culifornla Resiswnce Value Method 778; 29. 10. MeLeod Mo.: thod. 779: 29. I I. Triaxial T..::st Method. 7HO: 21). 12. Blirmister's Metbud. 780: 29. 13. Coefficient oj 'iubgrade Reaction, 781 : 29. 14. Westergaard's Analysis . 782: 29. 15. Temperature ~trcsscs in Rigid Pn"emcnh. 784: 29.16. Combined Stressc.~ In Rigid P:lVclllellts. 785: ltIuSlrative EX;lmplcs. 785: Problems. 786:

30. Laboratory Experiments

788 - 816

30.1. To determine Ihe watcr cOlltelil of a sample hy ovendrying met hod. 788: ~O.2. To determine tb e water content of a soil hy pyonomcter method. 789: 30.3. To determ ine the !>pt.'Cilic gravity of M)lids by the dcnslIY holl!c l11elhOpccilic gravity of solids by pycnomcter method. 79J : 30.5. To determine th e dry den.~ity of the soil by core cutter method. 792: 30.6. To dt.'tcrmioe the in.situ dry density by the sand repilicement method. 793; 30.7. To determ ine Ihe dry densi ty of ;1 soil by water-(lisplacclllent method. 795: 3O.S. To determine the particle sil.e dlst ributi(1O of a soil by sieving, 796: 30.9. To dCh!nnmc the p:trt icle size distri but ion by the hydrometer m...1hOO. 797: 30.10. To determine the hqmd Illllit of II ~()iJ !>pcclll1Cn. MOO; 30. 11 . To delennine the pla~tlc limit of a ~oil specimen. 801 ; 30.12. To detemline the .\ Imnkngc limit of a spc!Clmen of the rernouldt:d soil, 802: 30. 13. To determine the pcrm..-ahiJity of a !toil spt.'Clmcn by the constant· head pcnneamctcr. 804; 30.14. To determinc the permeahi lity o f II ~()!I specimcn by th..: vanable head pcrmc:l1netcr. X05: 30. 15. To detemline the conslJlkl;ltroll chal',l!;teri~tic~ of or soil spedmen. 807; 30. 16. To detcnnioe the shear parametcrs of a sandy soi l by direct ~hcar le~t. X09: 30.17. To dO:lenmne th e unconlined eomprc.~sivc stren gth of a cohesive soi t. 811 : JO. It\. Tu dctcnnmc the compaction Ch;lr:tClcristjc of a soil specime n by Proctor's test. S12: 30. 19. To detemlinc the Culi forrlra Bcnring Ratio (CBR) of a soil specimen. 813.

31. Introduction to Rock Mechanics

817 - 837

3 1. 1. Introduetkm, 8 17: 3 1.2. Geologic,ll Classification o/' Rocks, 1:117: 3 I .3. 9,lsic Tenninolagy. 818: 3 1.4. Index Properties of Rocks. H19: 31.5. Uni t weight (ar ma~s density), 819: 31.6. Porosity. H20; 31.7. Permeability, H20: 3 1.8. Point loud strength. 821: 31.9. Slaking and Durahility. H22: 3 UO. Sanic Velocity, 823; 4 1.1 I. Cli..~silicmian of Rock.~ for Engineering pmperties. 824: 31.12. Strength c1assifiention of Intac t Rocks, K27: 3 1.13 . LH borlltary tests lilr determination of strength of Rocb, 1:128: 31.14. Stre.~s.strain curve~. K29: 3 1.15. Modes of Failure of Rocks. 1'131; 31.16. Mohr-Coulomb Criterion lor Rocks. 832: 31.17. Shear Strength of Rocks. K33: 31. 18. H
\I

iii)

32. Gt!ott!chnical Eurlhquakt! Enginct!l"ing

838 - 863

32.1. Introduction. 838: 32.2. H i~IOI)' of Earthq uakes in India. 838: J2 ..l Seisml\: Zonc~ of India. 840: 32.4. Magnitude of :111 Earthquake. 840: 32.5. Intensity of Earth(IUnkcs. 842: 32.6. EO·I."CI of Ground motion on Smll:ture~. S44; 32.7. Gcnernl Principles of Earthquake-Resistant design. 1«46: 32.8. ~Ii SeismiC coefficient. 848j 32.9. Dc~ign Seismic forces. 849j 32.10. Site.Spccific Respunse ~pcclrn :H50: 32 , J l. Hazards due to Earthquakes. 851; 32.12. Liquefaction Phenomenon. 852: 32.13. P:lctors t\1!1.'ClIn~ Liqucfnctlon. 854; 32. 14. A s~ss mc nt of Susceptihility ofn Soil 10 Liqucl",\ction. 854: n. ls. Preventio n nl Liquefoction. S57: Illustrative EXHll1pJes. 858; Problems. 861: Selected References. 863

]

I.

Appt!ndix

A-GloSS~lrY

of Common Terms

Appendix B-Miscellaneous Objective-Type Questio ns

864 - 868 869 - 876

References

877-881

Publications of Bureau of Indian Standards

882- 883

Index

884- 886

PART-I

FUNDAMENTALS OF SOIL M'ECHANICS

1 Introduction 1.1. DEFINmON m' SOIL The word 'soil' is derived from the btin wort! so/iI/ill whic.:h. according 10 Webster's dictionary. means the upper layer of the earth thai may be dug or plowooj spccilically. the loose surface material of the earth in which plants grow. lhe above definition of soil is used in the field of agronomy where the main concern is in the use of soil for raising crops. In geology, eanh's crust is assumed to consist of unconsolidated sediments, called mantle or regolith, overlying rocks. 111C (enn 'soil' is used for the upper layer of mantic which can support plants. 'Ine matcrj~ll which is called soil by the agronomist or the geologist is known as lOp soil in geotechnical engineering or soil enginccring. lhe top soil c.onwins a large quantity of organic matter and is nOt suitable as a construCtiOn material or as a foundation for structures. The top soil is removal from the earth's surface before the construction of structures. Ollie (erm 'soil' in. soil engineering is defined as an unconsolidated material. romJXlSCd of solkl particles, proouccd by the disintegrntion of rocks. The void space between the particles may contain air, water or both. The solid particles may contain organic matter. The soil particles can be separated by such mechanical means as agit..1tion in water. A nalural aggregate of mineral particles bonded by strong and pennancnt cohesive forces is called 'rode'. It is an indurated material that requires drilling, wedging or blasting for its removal from the earth's surface. Since the Icons weak and strong have different interpretations, the boundary between soU and rock is rather arbitrary. In case of a partially disintegrated rock, it is extremely difIicult to locate th~ boundary between soil and rock. Fig. 1.1 shows a cros.c;.seCliorr through the canh's surface, indicating the nomenclature used in geology,

-r.J.S~f ,.

Manll e

Grp uqd

sUrfgce..

(regolith )

S oil

1

~RO'k

Rock

(a)

Ground surfacrl

Nomandalura in Grlology

(b) Nomt.nclalure in Soil Engintaring

Fig. 1.1. Nomendature.

SOIL MECHANICS AND FOUNDATION ENGINEERING

and in l Soil Engineering. It may be noted that the material which is called mantle (regolith) in geology is known:as soil in Soil Engineering.

1.2. DEFINITION OF SOIL MECHANICS The tenn 'soil mechanics' was coined by Dr. Karl Terzaghi in 1925 when his book Erdballmecllanic on the subjcct was published in Genn:m. According to Terz.:1ghi, 'Soil mechanics is the appliCltion of the laws of mechanics and hydraulics to cnginccring problems dealing with sediments and other unconsolidated accumulations of solid particles produced by the mechanical and chemical disintegration of rock, regmdlcss of whether or not they contain an admixture of organic constituents'. Soil mechanics is, therefore, a branch of mechanics which dC.1is with the action of forces on soil and with the flow of water in soil. The soil consists of discrete solid pmtic1es which arc neither strongly bonded as in solids nor they nrc as free as p::!rtic1cs of lluids. Consequently, the behaviour of soil is somewhat intermediate between tiM of a solid and a nuid. It is not; therefore, surprising th:1I soil mechanics draws hctlvily from solid mechanics and fluid mechanics. As the soil is inherently a IXlrIiculate system. soil mcch:mics is also caBcd paniell/me mechanics. Rock mechanics is the science de:.lling with thc mechanics of rocks.

1.3. DEFINITION OF SOIL ENGINEERlNG ANI) GEOTECHNICAL ENGlNEERING

Soil engineering in :m appUed science dealing with the applic
Load --Column _Column Ground level

Ground Level J/ .

5

0

i I. ~ooting

Soit

So i I (a) Shallow foundation

i\ra 51ratum (b) Pile foundation Fis. 1.2.

DiITel'l:ntlypts ofrOLlI\li-llions.

INTRODUcnON

(1) Foundations-Every civil engineering structure, whether it is a building. a bridge, or a dam, is founded on or below the surface of the earth. Foundations are required to transmit the load of the structure to soil safely and efficiently. A foundation is termed shallow foundation when it transmits the load to upper strata of earth. A foundation is called deep foundation when the load is transmitted to strata at considcl1lble depth below the ground surface (Fig. 1.2). Pile foundation is a type of deep foundation. Foundation engineering is an importana branch of soil engineering. (2) Retaining Structures-When sufficient space is not availnble for a mass of soil to spread and form a safe slope. a structure is required to rct"lin the soil. An earth retaining structure is also required to keep the

Dredge

level Soil

Earth pressure'

(a) R«.taining wall (b) Shut pitt. Fig. 1.3. Retaining structures.

soil at different levels on its either side. The retaining structure may be a rigid retaining wall or a sheet pile bulkhead which is relatively flexible (Fig. 13). Soil engineering gives the theories of earth pressure 00 retaining structures. (J) Stability of Slopes-If soil surface is not horizontal. there is a oomp:ment of weight of the soil which

~ay Soil

~bilnkm.nt slope

(a)

Soil

Excavation slopq; (b) Fig. 1.4. Slopes in (Q) filling and (b) cutting.

tends to move it downward and thus causes instability of slope. The slopes may be natural or man-made Fig. 1.4 shows slopes in filling and culting. Soil engineering provides the methods for checking the stability of slopes. (4) Underground Structures-The design and construction of underground structures, such as tunnels, sbafts, and oonduits, require evaluation of forces exerted by the soil on these structures. These forces are discussed in soil engineering. Fig. 1.5 shows a tunnel oonstructed below the ground surface and a oonduit laid below the ground surfaCe. .

o

SOIL MECHANICS ANI) FOUNDATION ENGINEERING

~ -:."

..

..•...

.~". ~

'

.

-:-

--

·,-"' ,~~ -:;',"~e::z1t""'"' ~E~~~n~:~~~: ,~~h:~~~:~~;:"~~:)~~: i:.:; .;~.:: L·.. i; (b1 Conduit

(al lunn/l.l

hard crust placed on soil (subgrnde) Cor the purpose of providing a smooth and strong

~~~~:~~~~ ~~~ue~g:i~::~~I~ ~~~~;s ~ns~~~~:

. . ... _ . -.' - '.' , .', " - , ,' ',', ', ... : _. ~ ':. :•. _. : . . _: '. -.' ..:

:

Sa

se

:.!:.,: ub base

Subgrade

(50i~)

in soil engineering. Fig. 1.6. Pavement del:tlls. (6) Eurth Dam-Earth dams arc huge structures in which soil is used as a construction material (Fig. 1.7). The earth dams arc bu ill for cfc::lling water reservoirs. Since the failure of an earth dam may cause widespread catastrophe, extreme care is taken in its design and construction. It requires a thorough knowledge

of soil enginccring.

Sh~ l\

(Pervious so il ) Fig. 1.7. Earth Dam.

(7) Miscellaneous

Soil .P roblems-The geotechnical engineer has sometimes to tackle miscellaneous

problems related with soil, such as soil heave, soil subsidence, frost heave, shrinkage and swelling. of soils.

Soil engineering provides an in-depth study of such problems. 1.5. aruCIN OF SOILS Soils arc formed by we.1lilering of rocks due to mechanical disintegration or chemical deoomIXlsition. When a rock surface gets exposed to tllmOSphere for an appreciable time, it disintegrates or decomposes inlO small particles and thus the soils are fanned. Soil may be considered as an incidental material obtained from the geologic cycle which goes on oontinuously in naturc. lhe geologic cyde consists of eros.ion, transportation, deposition and upheaval of soil (Fig. 1.8). Exposed rocks are eroded :md dcgraded by various physical and chemical processes. TIle products of era>ion are picked up _ _ __ _ __ _ by agencies of transportation, such as water and wind. and arc ~rosion Tran~\a\ion carried to new locations where they are deposited. This shilling Uph~aval Deposi ti on of the material disturbs Fig. 1.8. Gc~c Cydc.

INTRODucnON

the equilibrium of forces on the earth and causes large scale earth movemcnts and upheavals. 1l1is process results in further CX(Xl')'Ure of rocks and Ihe geologic·cydc gelS repeated. If the soil stays at the place of its formation just above the parent rock, it is kllOwn as residual soil or sedentary soil. When the soil has been deposited at a place away from the place of its origin, it is called a transported soil. The engineering properties of residual soils vmy considernbly from the top layer to the bollom layer. Residua! soils Iwve a grndual trnnsition from relalively fine material near the surface to large frJgments of stones al greater depth. 'nle properties of the bottom layer resemble that of the parent rock in many respects. The thickness of the rcsidu::li soil fonnation is generally limited to a few metres. The enginccring properties of transported soils arc entirely different from the properties of the rock at the place of deposition. Deposits of transported soils are quite thick and are usually uniform. Moot of the soil deposits with which a geotechnical engineer has to deal arc transported soils. 1.6. FORMATION OF SOILS As mentioned above, soils are formed by either (A) physical disintcrgration or (0) chemical decomposition of rocks. A. IJhysicul Disintcgrntion-Physical disintegmtiOO or mech:mic.ll weathering of rocks occurs due to the following physical proc'CSScs : (1) Temperature changes-Different minerals of:J rock huve different coefficients of thennal cxprlOsion. Unequal cXlxmsion and contraction of these minerllis occur due 10 temperature changes. When the slresses induced due to such changes arc repe"lIcd many times, the particles gcl dctached from the rocks and the soils arc formed. (2) Wedging action of Ice-Water in the pores and minute crncks of rocks gets frozen in very cold climates. As the volume of icc formed is more than that of water, expansion occurs. Rocks get broken into pieces when large stresses develop in the cracks due to wedging action of the icc formed. (3) Spreading of roots of phm1s-As the roots of trees and shrubs grow in the cracks and fISSUres of the rocks, forces act on the rock. The segments of the rock arc forced apart and disintegration of rocks occurs. (4) Abrasion-As water, wind :Jnd glaciers move over the surface of rock, abrasion :Jnd scouring takes place. It results in the formation of soil. In all the processes of physical diSintegration, there is no change in the chemical composition. 1llc soil formed has the properties of the parent rock. Coarse grained soils, such as grnvel and sand, 3re fonned by the process of physical disintegration. B. Chemical Decomposition-When chemical decomposition or chemical weathering of rocks takes place, original rock minerals arc transformed into new minerals by chemica] reaction.<>. The soils (onned do not have the properties of the parenl" rock. The following chemical proc:csses generally OCOJr in nature. (1) Hydration-In hydmtion, water combines with the rock minerals and results in the formation of a new chemicnl compound. loe chemical reaction causes a dmnge in volume and decomposition of rock into small particles. (2) Carbonation-It is a type of chcmical decomposition in which carbon dioxide in the atmosphere combines with water to form carbonic xid. Ibe c.lrbonic acid reacts chemically with rocks and causes their decomposition. (3) Oxidation--Oxidation occurs when oxygen ions combine with minerals in rocks. Oxidation results in decomposition of rocks. Oxidmion of rocks is somewhat similar to rusting of steel. (4) Solutlon-Somc of the rock minernls fonn a solution with water when they get dissolved in water. Chemical reaction t:Jkes place in the solution and the soils are formed. (5) Hydrolysis-It is a chemical process in which water gets dissociated into W and Olr ions. The hydrogen cal ions replnc:c the metallic ions such as calcium, sodium :Jnd potassium in rock minerals and soils are formed with a new chemical dccompa:>ition. Chemical dccomposit.ion of rocks results in form:Jtion of clay minerals. These clay minerals impart plastic properties to soils. Oayey soils are fonned by chemical decomposition.


1.7. TRANSPORTATION OF SOIlS The soils formed at a place may be transported to other places be agents of trarL<;portion, such as water, wind. ice and gravity. (1) Wllter transported Soils- Flowing water is one of the most important agents of transportation of soils. Swill running water cnrries a large quantity of soil either in suspension or by rolling along (he bcd. Water erodes the hills and deposits the soils in the valleys. The size of the soil particles carried by w:Jter depends upon the velocity. 1bc swift water can carry the particles of large size such as boulders and gravels. With a dca'casc in velocity, the coarse particles get deposited. The [mer particles are carried further downstream and are deposited when the velocity reduces. A delta is fanned when the velocity slows down to almost zero al the confluence with a receiving body of sliU water, such as a lake, a sea or an oct.1n (Fig. 1.9). riginal

ground

, Eroded ') _ .... grou nd-./" ........ ,

Still · walen Ag. 1.9. Alluvial Deposits.

All type of soils amied and deposited by water are known as alluvial deposits. Deposits made in lakes are called lacustrine deposits. Sudl deposits are laminated or varved in layers. Marine deposits are formed when the flowing water carries soils to ocean or sea. (2) Wind transported Solls-Soil particles are transported by winds. The particle size of the soil depends upon the velocity o[ wind. 'The finer partiCles are amied far away from the place of the [ormation. A dust storm gives a visual evidence of the soil part icles carried by wind. Soils deposited by wind are known as aeolian deposits. Large sand dunes are fanned by winds. Sand dunes occur in arid regions and on the leeward side of sea with sandy beaches. Loess is a sill deposit made by wind. These deposits have low density and high compressibility. The bearing capacity of such soils is very low. The permeability in the vertical dire<.:tioo is large. (3) Glncier-Deposited SoiJs..---.Glaciers are large masses of ice facmed by the oompadion of snow. As the glaciers grow and move, they carry with them soils varying in size [rom fine grained to huge boulders. Soils get mixed with the ice and are transported far away from their original position. Drift is a general term used for the deposits made by glaciers directly or indirealy. Deposits direct.ly made by melting of glaciers are called till. Termina l morcl ln e "

.'

.. Gr ound moraine Fig. 1.10. Glader Deposited Soils.

-.

,.

INTRODUcnON

During their advancement, glociers tr.msport soils. At the lenninus, a melting glacier drops the material in the fonn of ridges, known as terminal moraine (Fig. ] .10). '1l1e land which was once covered by glaciers and on which till has been deposited after melting is called ground moraine. lbe soil carried by the melting water from the front of a glacier is termed out-wash. Glaciofluvial deposits arc fanned by glaciers. The material is moved by glaciers and subsequently deposited by streams of melling water. These deposits have stratification. Deposits of glacial till arc generally well-graded and can be compacted to a high dry density. lbcse have generally high shearing strength. (4) Gravity-deposited soil.<;-Soils C<'ln be transported through short distances under the action of gravity. Rock fragments and soil masses collected at the foot of the cliffs or steep slopes had fallen from higher elevation under the action of the gravitational force . Colluvial soils, such as talus, have been dcposited by the gravity. Talus consists of irreguJar, coarse particles. It is a good source of broken rock pieces and coarse-grained soils for many engineering works. (5) Soils tr"ansporled by combined IIction-Somelimes, two or morc agenrs of transportation aCI jointly and tr.lnsport the soil. For example, a soil portiele may fall under gravity and may be carried by wind to a for off place. It might by picked up again by flowing waler and deposited. A glacier may carry it still further. 1.8. MAJOR SOIL DEPOSITS OF INOlA The soil deposits of India may be classified in the following five major groups : (1) Alluvial Deposils-A large part of north india is oovered with alluvial deposits. lhe thickness of alluvium in the Indo-Gangctic and Drnhmputra flood plains varies from a few mctn:s to more than one hundred metres. Even in the pcninsul:lr India, ll11uvi'll deposits occur at some places. The distinct characteristics of alluvial deposits is the existence of alternming layers of sand, silt and clay. The thickness of each layer depends uiX>n the local terrain and the nature of floods in the rivers causing deposition. The deposits are generally of low density and are liable to liquefaction in earthquake-prone areas. (2) Black Cotton Soils-A large part of cenlral India and a portion of South India is oovered with black cotton soils. These soils are residual deposits fonned from basalt or trap rocks. The soils are quite suitable for growing collon. Black cotton soils are clays of high plasticity. 'Ihey contain essentiaUy the clay mineral montmorillonite. The soils have high shrinkage and sweUing eharncteristics. The shearing strength of the soils is extremely low. The soils are highly compressible and have very low bearing capacity. It is extremely diffiadt to work with such soils. (3) Lateritic Soils-Lateritic soils arc formed by decomposition of rock. removal of bases and silica, and accumulation of iron oxide and -aluminium oxide. The presence of iron oxide gives these soils the characteristic red or pink colour. Thcsc are residual soils, formed from basalt. Lateritic soils exist in the central. southern and c..1stem India. The lateritic soils are soft and can be cut with a chisel when wet. However, these harden with lime. A hard crust of gravel size particles, known as laterite, exists ncor the ground surface. The plasticity of the lateritic soils decreases with depth as they approach the parent rock. These soils, especially thaie which contain iron oxide, have relatively high specific gravity. (4) Desert Soils-A large part of Rajasthan and adjoining states is covered with sand dunes. In this area, arid conditions exist, with practically lillie mineaU. Dune sand is uniform in gradation. lhe size of the particles is in the range of fine sand. The sand is non-plastic and highly pervious. As the sand is gcncnltly in loose condition. it requires dcnsi[ic.1tion 10 increase its strength. (5) Marine Deposits-Marine depooilS arc mainly confined along a narrow belt ncar the coast. In the south-west coost of India, there are thick layers of sand above deep deposits of soft marine clays. The marine deposits have very low shearing strength and are highly oomprcssible. They contain a large amount of organiC mailer. The marine days are soft and highly plastic.

SOIL MECllANICS AND FOUNDl\nON ENGINEERING

J.9. COMPARISON OF SOILS WITH OTHER MATERIALS Soil is a highly complex material. It differs from conventional structural IT'3teriaLs, such as steel and concrete. (1) Steel is a m~mufactured material the properties of which are accurately controlled. The properties of concrete are also controlled to some extent during its preparation. Soil is a material which ha<> been subjected to vagaries of nature, without any control. Conscquenlly, soil is a highly heterogeneous and unpredictable material. (2) The properties of a soil change not only from one place to the other but also at the place with depth. '1l1C properties also change with a Change in the environmental, loading and drainage conditions. lbc properties of a soil depend not only on its type but also on the conditions under which it exists. (3) The main engineering properties of steel and concrete are modulus of elasticity and tensile and compressive strength. Most of the design work can be done if these properties are known or determined. However, the engincering properties of soils ucpend upon a number of f;:lclors and it is not possible to characterise them by two or three parameters. ElabUroItc h.:,:.;ting is required to dctennine the characteristics of the soil before design can be donc. (4) Because of huge qu:mtilics of soils involved. it is not econom ically feasible to tnmsport the soils from other places like steel or ·concrete. Soils rlre gCl1cr:llly used in the conditions in which they exist. (5) Whereas steel and concrcte C'ln be inspected bcfore usc, soils for foundmions are at great depth and not open to inspection. lhe ~Imples of the soil Ulken from the bore holes are generally disturbed rind do nOI represent the lrue in-situ conditions.

1.10. LIMITATIONS OF SOIL ENGINEEIUNG Soil engineering is not an exact science. Because of the nature and the variability of soils, sweeping assumptions are made in the derivation of equations. '[he solution obtained in most cases are for an idealised, hypothetical material, which may not truly represent the actual soil A good engineering judgment is required fOf the interpretation of the results. In f'let, each problem in soil engineering is a unique problem because the soils at two places arc seldom identical. The following limitations must be kept in mind when tackling problems relate
INTRODUCfION

1.11. TERMINOLOGY OF DIFFERENT TYPES OF SOILS A geotechnical engineer should be well versed with the nomenclature and tenninology of different types of soils. The foUowing list gives the names and salient characteristics of different types of soils, arranged in alphabetical order. (1) Bentonite-It is a type of clay with a very high percentage of clay mineml montmorillonite. It is a highly plns!ic clay, resulting from the decomposition of volc:lOic ash. It is highly water absorbent and hao; high shrinkage and swelling charuaeristics. (2) Black Cotton Soil-It is a residual soil containing a high percentage of the C1.1y mineral montmorillonite. It hao; very low bearing capacity and high swelling and shrinkage properties. (3) Boulders-Boulders arc rock fragments of large size, more than 300 mm in size. (4) Calcareous soils-These soils contain a largc quantity of calcium carbonatc. Such soils effervesce when tCSled with weak hydrochlOriC acid. (5) Caliche-It is a type of soil which cOntains gravel, sand and s ilt. 111c panicles are cemented by calcium carbonatc. (6) Cluy-It consists of microscopiC and sub-microscopic panicles derived from the chemical dccompooition of rocks. It contains a large quantity of clay mincl1lis. It can be made plastic by adjusting the water content. It exhibits considerable strength when dry. Clay is a finc-grained soil. It is a chocsive soil. The particle size is less than 0.002 mm. Drgunie eluy cont
10

SOIL MECHANICS AND FOUNDATION ENGIN,EERING

(19) Loess-It is a wind blown deposit of siJL II is generally of uniform gradation, with the particle size between 0.01 to 0.05 mm. It consists of quartz and feldspar particles, cemented with calcium carbonate or iron oxide. When wet, it becomcs soft and compressible because cementing action is loot. A loess deposit has a loose structure with numerous roo! holes which produce vertical cleavage. The permeability in the vertical direction is generally much greater than thaI in the horizontal direction. (20) Marl-It is a stiff, marine calcareous clay of greenish colour. (21) Moorum-ll1c word moorulII is derived from a Tamil word, meaning powdered rock. It consists of small pieces of disintegrated rock Of shale, with or without boulders. (22) Muck-It denotes a mixture of fmc soil particles and highly deoomposed organiC matter. It is black in colour and of extremely soft consistency. It caonot be used for engineering works. The organic matter is in an advanced stage of decomposition. (23) Peat-It is an organic soil having fibrous aggregates of macroscopic and microscopic particles. It is fonned from veget.'ll matter under conditions of excess moisture, such as in swamllS. It is highly compressible and not suitable for foundations. (24) Sund-It is a coarse-grained soil, having particle size between 0.075 mm to 4.75 mm. The particles are visible to naked eye. The soil is cobesionless and pervious. (25) Silt-It is a fine-grained soil, with particle size between 0.002 mm and 0.075' mm. The particles are not visible to naked eyes. Inorganic silt consists of bulky, equidimensional grains of quartz. It has little or no plasticity, and is cohesionless. Organic silt contains an admixture of orgily dcnsified by compaction. Till is also known as boulder-clay. (27) Top soils-Top soils are surface soils that support plants. They contain a large quantity of organic matter and nrc not suitable for foundations. (28) Tuft-It is a fine-grained soil composed of very small particles ejected from volcanoes during its explosion and deposited by wind or water. (29) Thndru-It is a mat of peat and shrubby vegetation that oovers clayey subsoil in arctic regions. The deeper layers are permanently frozen and are called permafrost. lbe surface deposit is the active layer which alternately freezes and thaws. (30) Varved clays-These are Sedimentary deposits consisting of alternate thin layers of silt and clay. The thickness of each layer seldom exceeds 1 cm. These clays are the results of deposition in lakes during perioos of alternately high and low waters. [Note. For glossary of technical terms, sec APPENDIX A].

1.12. COHESIVE AND COHESIONLESS SOILS Soils in which tbe adsorbed water and particle attraction act such that it defonns plastically at varying water contents are known as cohesive soils or clays. This cohesive property is due to presence of clay minerals in soils. Therefore, the term cohesive soil is used synonymously for clayey soils. The soils composed of bulky grains are cohesionlcss regardless of the fineness of the particles. The rock flour is cohesionless even when it hac; the particle size smaller than 21l size. Non-pla'ltic s ilts and coarsegrnined soils are oohcsionlcss. [Nofe. 1 Il = 1 micron = 1O~ m = 10-3 mmJ. Many soils are mixture of bulky grains and clay minerals and exhibit some degree of plasticity with varying water content. Such soils are termed cohesive if the plasticity effect is significant; otherwise, cobesionless, Obviously, there is no sharp dividing line between cohcsionless and cohesive soils. However, it is sometimes convenient to divide the soil into above two groups.

INTRODUCfION

II

111e term cohesive-soil is used for clays and plastic silt, and the term cohcsionlcss-soil, for non·plastic silts. sands and gravel

1.13. BRIEF mSTORY OF SOIL ENGINEERING

1

According to the author, the history of soil engineering can be divided into three periods, as described below: (1) Ancient to Mediey,,1 perlod-Man's first contact with soil was when he placed his foot on the earth. In ancient times, soil was used as a construction material for building huge earth mounds for religious purposes, burial places and dwellings. Caves were built in soit 10 live in. ExceUent pavements were construded in Egypt and India much before the OI.ristian era. Some earth dams have been storing water in India for more than 2000 years. Remnants of various underground waler structures. such as aqueducts. tunnels and large drains. found in the excavation at the sites of early civilisation at Mohenjodaro and l-Iarrappa in the Indian subcontinent indicate the use of soil a.<; foundation and construction material. Egyptian used caissons for /Jeep foundations j::vcn 2000 D.C. I hmging gClrden at Babylon (Iraq) was also built during that period. The city of D.1bylon was built on fills above the adjoining flood plains. During Roman times, heavy structures, such as bridges, aqueducts, harbours and buildings, were built. Some of these works are in existence even today. After the collapse of the Roman Empire, tbe construction activities declined. However, some heavy city walls and forts were built from the strategic considerations. Cathedrals. casLJes and campaniles (bell towers) were also constructed. lbe famous tower of !lisa. known as the leaning tower of Pisa, was also built. The tower has leaned on one side because of the diITerentiai sctllement of its base. The famous Rialto Bridge was constructed in Venice (Italy) in the seventeenth century. Leonardo da Vinci constructed a number of structures in France during the same perioo. The famous London Bridge in England was also built. The mausoleum Thj Mahal at Agra (India) was constructed by the emperor Shah Jehan to commemorate his favourite wifc Mumtaz Mahal. It is built on masonry cylindrical wclls sunk into the soil at close intcrvals. 11 is certain that early builders. while constructing such huge structures, encountcred and successfully tackled many challenging problems. However, no record in available about the methods adopted. No scientific study seems to have been made. The builders were guided by the knowledge and experience passed down from generation to generation. (2) Period of Early Developments-The eighteenth century caD be considered as the real beginning of soil engineering when early developments in soil engineering look place. In 1773, a French engineer Coulomb gave a thcory of earth pressure on retaining walts. 1be theory is used by the gcotechniall engineers even today (chapter 19). Coulomb also introduced the concept thill the shearing resistance of soil consists of two components, namely, the cohesion compunent ~md the rric.1ion component (ch.1plcr 13). Culmann gave a geneI"dl gT'dphical solution for the earth pressure in 1866. Ibmkine. in 1857, published a theory on earth pressure considering the plastic equilibrium of the earth mass. In 1874, Rehbann gave a graphical method for computaHon of earth pressure based on Coulomb's theory. Darcy gave the law of the permeability of soils in 1856. Darcy's law is used for the computation of seepage through soils (chapters 8 and 9). In the same year, Stokes gave tbe law for the velocity o[ fall of solid particles through fluids. The law is used [or determining the particle size, as disoJssed in chapter 3. Q-Mohr gave the rupture theory for soils in 1871. He also gave a graphical method of representation of slresses, popularly known as Mohr's circle. II is extremely useful for delerminalion of stresses 00 inclined planes (Chapter 13). Boussinesq, in 1885, gave the theory of stress distribution in a semi·infmile, homogeneous, isotropic, elastic medium due to an externally applied load. The theory is used for detennination of stresses in soils due to loads, as discussed in Chapter 11. . In 1908, Marston gave the theory for the load carried by underground conduits (chapter 22). Atlerberg. in 1911, suggested SOQl~ simple tests for characterizing consistency of cohesive soils. The

roiL MECHANICS AND FOUNDATION ENGINEERING

12

limits, commonly known as Altcrbcrg's limits, are useful for identification and classification of soils, as discussed in chaplers 4 and 5. Swedish Geotechnical Commission of the Siale Railways of Sweden appointed a committee headed by Prof. Fellcnius in 1913 \0 study the st.'lbility of slopes. The commillee gDvC the Swedish circle method for checking the stability of slopes, dcsaibcd in ch.'lptcr 18. In 1916, Petterson gllvc the friction circle method for the stability of slopes. (3) Modem Era-The modem em of Soil Engineering I;Icgan in 1925. with the publicaliOl) of the book E,dballmechanic by KJolri TCL,taghi. The contribution made by Tcrzaghi in lhe development of soil engineering is immense. He is fittingly called the father of soil mechanics. For the first time, he adopted a scientific approach in the study of soil mechnnics. His theory of consolidation of soils (chapler 12) and the effective stress principle (chapler 10) gave a new direction. ProcIor did pioneering work on compaction of soils in 1933. ~ discussed in chapter 14. Taylor made major contributions on consolidation of soils, shear strength of clays and the stability of slopes. Casagmnde made significant contributions on classification of soils, seepage through earth masses and consolidation. Skempton did pioneering work on the pore pressures, effective stress, bearing capacity and the stability of slopes. Meyerhof gave the theories for the bearing capacity of shallow and deep foundatioos. Hvorslcv did commendable work on subsurface exploration and on shear strength of remouldcd clays. The above list is far from complete. Many other distinguished geotechnical engineers have made a mark on the development of soil engineering. Because of space limitation, their mention could not be made in the above list.

A. Oescripllve 1.1. DefiDC the term 'soil', 'soil mcchaniu;' and soil engineering. What are limillltions of soil engineering? 1.2. Whot is geologic eycle ? Expl;)jn the phenomena of formation and ltaosporUition of soils. 13. What arc the major soil deposits of India? Explain their characteristics. 1.4. Write D bricf history of soil engineering.

n.

Multiple·Choice Questions 1. Colluvial soils (talus) are transported by: (a) Water (b) Wind (e) Grovity (d) Ice 2. Water-tronsponed soils are termed: (a) Aeoline (b) Alluvial (e) Colluvial (d)1i1l 3. Glacier-dcpositcd soils are called: (a) Talus (b) Loess (e) Drin (d) None of above 4. Cohesionlcss soils ate fonned due to: (a) Oxidation (b) Hydration (e) Physical disintegration Cd) Chemical decomposition 5.. When the prcxluCiS of rock wC
2 Basic Definitions and Simple Tests 2.1. INTRODUcnON A soil mass consists of solid particles which form a jXlrous structure. The voids in the soil mass may be filled with air. with water or partly Air with air and partly wiLh water. In general.., a soil mass consists of solid particles, water and air. The three Wat/i!f constituents are blended together to form a complex material (Fig. 2.1. a). However, for OJnvcnicncc, aU Solid the solid particles are segregated and placed in the lower layer of the three-phase diagram (Fig. 2.1b). Ukewise, water and air particles are

placed separately. as shown. The 3-phase diagram is alSo known as Block diagram.

(a)

(b)

It may be noted that the three constituents

cannot

be

actually

Fig. 2.1. Conlititueflts of Soil.

segregated, as shown. A 3-phase diagram is :10 llrtince ll.<>ed for easy understanding Dnd convenience in

cairuIalion.

Although the soil is a three-phase system, it becomes a two-phase system in the following two cases: (1)

It

::f2r~~~~~~~Eli~i;"~:~ T~

saturated, there is no air phase (Fig. 2.2b). It i, the
~T~~~:Cl~;~::; lV

The phase diagram is a simple, diagrnmmetic representation of a real soil. It is extremely useful for

studying the various tenns used in soil engineering and their interrelationships.

\10

.

tI rtr I-::-:;;~;: :-:-:-:-:-:-:::- - T tI

T

Mo"O

Vw

-= -: -=- =-:

- ~- =

v

1Ms 1" 1L

1 Vs

v,

- - - - - - -

I

{ol Dry soil

(o)Soluroled soil Fig. 22. Two-phRse diagrams.

Mw

11" ",


14

r ~,,~.,.:,:,~~c ~ T 11 '=''''''~o:' f

In a 3-phase diagram, it is conventional to write volumes 00 the left side and the mass on the right side (Fig. 2.3 0). The t~otal volume of , gwen soil m"j.in designatal as V. equal to the sum of ' nvolume h e of solids (V,~ "e

Air

f

Mo=O

"" .. --.-.------.-.. T

'4:J

"" T

Air

'No:0

----- --.--

T

J "" fI 11 ~" 11 1'

(0)

Fig. 2_1. 'I1m:c-phasc lIiagram.

(b)

the volume of water (V...) ilnd the volume of air (V,,). '11m volume of voids (V,.) is equal (0 the sum of the

volumes of water and air. lbe lotal mass of the soil mass is represented as M. lllC mass of air (MIJ) is very small and is neglected. lbcrcfOfc. the lotlll mass of the soil is equal to the mass of solids (M,) and the mass of water (M..,). Fig. 2.3b shows the 3-phase diagram in which the weights are written on the right side. 2.2. VOLUME'I1UC RELATIONSHIPS 'Jbe following five volumetric relationships are widely used ip soil engineering. (1) Void Rutio (e)-ll is defmed as the ratio of the volume of voids to the volume of solids. Thus < -

i

..

(2.1)

The void ratio is expressed as a decimal, such as 0.4, 0.5, etc. For coarse-grained roils, the void ratio is gcncr.llly smaller than that for fine-grained soils. For some soils, it may have a value even greater than unity. (2) l'orosity (n)-It is defined as the ratio of the volume of voids to the total volume. Thus

n·

~

... (2.2)

Poror;ity is gcneraUy expressed as percentage. However, in equations. it is used as a ratio. For example, a porosity ' of 50% will be used as 0.5 in equations. The porosity of a soil cannot exceed 100% as it woukl mean V~ is greater than V, which is absurd. 10 fact, it will have a much smaller value. Porosity is aJso known as percentage voids. Doth porosity and void ralio are mea'iurcs of the denseness (or loosencs..'9 of soils. As the soil becomes more and more dense, their values dc<'T~sc. The lenn porosity is more oommunly used in other disciplines such as agricultural enginccring. In soil engineering. lhe term void mHo i"i more popular. It is more convenient to use void ratio Ihan porosity. When the volume of a soil mass changes., only the numerator (i.e. V~) in the void ratio changes and the denominator (i.e. V,) remains constant. However, if the lenn porosity is used, both the numerator and the denominator change and it becomes inconvenient. An inter-relationship can be found between the void ratio and the porosity as under.

From Eq. 2.2,

1

V

V" + V,

ii·~·-V;-

!.1+!.!...:!:...! n

or

<

n _ -
e

... (0)

... (23)

BASIC DEFINmONS AND SIMPLE TESTS

Also, from Eq. (a),

15

~ _;; _ 1 _ l~n e .. 1

:n

..

(2.4)

In Eqs. (2.3) and (2.4), the porosity should be expressed as a ratio (and not pcrentagc). (3) Degree of Saturation (5)-The degree of saturation (S) is the ratio of the volume of water to the volume of voids.

s-~

Thus

,.. (2.5)

V" The degree of 5.1luralion is generally expressed as a percentage. It is equal to zero when the soil is absolutely dry and 100% when the soil is fully saturated. In expressions, the degree of saturation is used as a decimal. In some texts, the degree of saturation is expressed as S,. (4) Percentuge Ai.- voids (n,,)-It is the ralio of the volume of air to the tolal volume. Vo

111us

na"

As the name indicates, it is represented

(5) Air Content Thus

.

(Q~)-Air

V

.. .(2.6)

as a percentage.

oontent is defined as the ratio of the volume of air to the volume of voids.

Vo

ar -

... (2.7)

-v::-

Air content is usually expressed as a percentage. Both air content and the percentage air voids are zero when the soil is saturated (V" = 0). An inter-relationship between the percentage air voids and the air oontent can be obtained .

~

From Eq. 2.6,

n" -

V"

V -

v:Va )( VVv

or n" - n Q c ... (2.8) [Note_ In literature, the ratio V" IV is also·c.111cd air content hy some authors. However. in this lext, this ralio would be lenned percentage nir voids ..nd nOI air contentJ.

2.3. WATER CONTENT The water content (w) is defined as the ratio of the mass of water to the mass of solids.

w_~ ... (2.9) M, The water content is also known as the moisture conlent (m). 11 is expressed as a percentage, but used as a decimal in computation. The water content of the fine-grained soils, such as silts and clays, is generally more than that of the coarse grained soils, such as gravels and sands. The water cootent of some of the fme-gained soils may be even more than 100%, which indicates that more than 50% of the total mass is that of water. The water content of a soil is an important property. The characteristics of a soil, especially a fine-grained soil, change to a marked degree with a variation of its water content. In geology and some other disciplines, the water content is defined as the ratio of the mass of water to the total mass. Some of tbe instruments, such as moisture tesler, also give the water content as a ratio of the total mass. In this text water content (w) will be taken as given by Eq. 2.9, unless mentioned otherwise. The symbol m' shall be used in this texl for the water content based on the total wet mass. Thus /II' -

¥f)(

100

... (2.10)

Note. Certain quantities, as defined above, are expressed as a ratio and certain other quantities, as a

SOIl. MECIIANICS AND fOUNDATION ENGINEERING

16

percentage. To avoid confusion. it is a<.Ivis
= tOl

gm

1 mcgagrammc (Mg) = 106 gm = 103 kg Likewise, 1 millinewton (mN) = 10--3 newton (N) 1 lcilonewton (kN) = 103 N 1 meg
2.5. VOLUME· MASS RElA'110NSIIWS '(be volume-mass n;l;ltiornhip ure in tenns or mass density. 'Ibe rna..... of soil per unit volume is known as mass density. In soil cngin\:cring. the fullowing 5 dilTerent muss densities arc usct.I. (1) Bulk Muss Den.. Uy-l11e bulk mao;s density (p) is defined m the total mass (M) per unit lotal volume (Y). Thus, from Fig. 23 (a), M

... (2.11) P -V The bulk mass densily is also known as the wei mass density or simply bulk density or density. It is expressed in kg/ml, gm/ml or Mg/ml. Obviously. 1 Mg/m 3 1000 kg/m l 1 gm/ml

=

=

(2) Dry Mass DensUy-The dry mass density (p.,) is defined as the mass of solids per unit lotal volume.

Thus M,

Pd-

V

... (2.12)

As the soil may shrink during drying. the mass density may not be equal to the bulk mass density of the soil in the dried condition. '(be lotal volume is measured before drying.

The dry mas... density is also known as the dry density. The dry mass density is used to express the denseness of the soil. A high value of. dry mass density indicates that the soil is in a compact condition. (3) Saturated Mass DensJty-The saturated mass density (PS
M_

P,. - I I

... (2.13)

(4) Submerged Muss liel\.~Uy-When Ihe soil cxisL" beluw water, it is in II submerged condition. Wheo a volume V of soil is Submerged in water, it displaces an equal volume of water. Thus the net mass of soil when submerged is reduced (Fig. 2.4 (o)}. The submerged mass density (p') of the soil is defined as the submerged rna<>s ~ unit of total volume. Thus

BASIC DEFINITIONS AND SIMPLE TESrS

The

submerged

dcnsily

:

-

17

a:~br ~"':._= _:-_=_=_ 1_ ~= __-_____ ~~~14) Tr ----------- 11 TT ------------1 1 I I

::V;:I~C~:yi:nt::~t;:~it~:;b)~IsoJ FIg. 2.4 (a) shows a sOli m~ submerged under water. The soil solids which have a volume of V, arc buoyed up by Ihe walec. The uplhrusl

~ equal 10 Ihe

mass of water diplaced by the solids. Thus

1 v,

v, G

I

~

M,

1

Vs

11

u:VsJ'w

U _ V,P",

I

U:Vs'6w (b)

(o)

Therefore,

Ws

1

~

v, G'W

Fig. 2.4. Submerged mass.

- V, Gp ... - VsP ...

From Eq. 2.14,

,

P

V;p.(G-l)

- - -v- -

... (2.15)

Alternatively, we can also consider the equilibrium of the entire volume downward mass, including lhe mass of water in the voids, is given by

M.

In this case. the total

M'Ol - M,+V~ p ... The total upward thrust, including that on the water in voids, is given by U _ Vp.

Therefore, the submerged mass is given by M.•"h

From Eq. 2.14,

p'

= (M,,' + = (M., +

V" p".) - V p".

V,. ~".) - V p",

=_ M_ ,,,,_~_v_P_".

or Using Eq. 2.13

p' -

r'aI -

p...

...(2.16)

The submerged density p' is roughly one-half of the saturated density.

(5) Mass Density of Sollds-1be mass density of solids (p,) is equal to the ratio of thc mass of solids to the volume of solids. Thus

M,

p, -

V,

... (2.17)

2.6. VOLUME-WEIGHT RELATIONSHIP The volume-weight relDtionships are in terms of unit weights. The weight of soil per unit volume is known as unit weight (or specific weight). In soil engineering, the following five different unit weights are used in various computations. (1) Bulk Unit Weight-The bulk unit weight (y) is defined as the total weight per unit total volume [Fig. 2.3 (b)] . Thu, W ... [2.11(0)] "t - V The bulk unit weight is alSo known as the total unit weight ("tl)' or the wet unit weight. In 51 units, it is expressed as N/m 3 or kN/ m3. In some texts. the bulk unit weight is expressed as "tb or "tr

SOIL MECHANICS AND rOUHDATION ENGINEERING

18

(2) Dry Unit Wdght-The dry unit weight (Yd) is defined as the weight of solids per unit total volume. Thus

W,

'fd""-Y

... (2.12(a)J

(3) Sutur-lled Unit Wcight-The saturated unit weight (llol1') is the bulk unit weight when the soil is fully saturated. lYr ....

Thus

y,« -

II

... (2.13(a)J

(4) Submerged Unit Weight-When Ihe soil exists below water. it is in a submerged condition. A buoyant Corce acts on the soil solids. According to Archimedes' principle, the buoyant [orce is equal to the wcighi of water displaced by Ihe solids. The net mass of the solids is reduced. The reduced mass is known as the submerged mass or the buoyant mass.

lltc submerged unit weight (y') of the soil is defined as the submerged weight per unit of total volume.

Tbus

, lVslIh y.--y

... (2.14(a)l

Fig. 2.4 (b) shows a soil mass submerged under water. lbc soil solids which have a volume of V, are buoyed up by the water. The buoyant force (U) is equal to the weight of wuter displaced by the solids. U - Viy ... The weight of water in the voids has a zero weight in water, as the weight of water and the buoyant force just balance c.'lch other. When submerged, all voids can be assumed 10 be filled with water. lltercforc,

w. ....... w,-u

From Eq. 2.14.

- V,Gy. - V, y. - V,y.(G - 1) • V,y.(G - 1) Y ---V--

. .. [2.15(a)J

We can also consider the equilibrium of the entire volume (Y). The lotal downward force, including the wight of water in the voids, is given by W..'" .. W, + V" Y... The tOial upward force, including that on the water in voids, is given by U .. Vy", Therefore, the Submerged unit weight is given by W,uh = (W~ + V,.y",).- Vy"

( ~, + V,. ~,,) - Vy". = ~.." ~ VYw

From Eq. 2.14,

i

Using Eq. 2.13

Y' - YI(;I( - Y...

=

... [2.16(a)]

llte submerged unit weight is roughly one-half of the saturated unit weight. In literature, the submerged uni! weight is also frequently expressed as 'fsub' For convenience, the submerged uni! weight wiD be expressed as y' in Ihis tex\. (S) Unit weight of Soil SolicJs.-The unit weight oC solids (Y.) is equal!o the mtio of the weight of solids to the volume of solids. Thus W, 't, -

V;

... (2.17(a)J

2.7. INTER-RELA"HON BETWEEN \\lASS AND WEIGln' UNITS In Sect. 2.5, the mass-volume relationships have been developed. The corresponding weight-volume relationships arc given in Sect. 2.6. The reader should carefully understand the difference between the two units and should be able to convert the mass densities to the unit weights and vice-llCrsa. The mass and weight are related by Newton's second law of motion, viz, Force :: Mass )( Acceleration

BASIC DEFINI110NS AND SIMPLE TESTS

19

When a force of one newton (N) is applied to a mass of one kilogrammc (kg), the acceleration is 1 mlsec2. The weight of 1 kg mass of material on the surface of earth is 9.81 N hecausc the acceleration due to gravity (g) is 9.81 mlsec". Thus we can ('{)Overt the mass in kg into weight in N by multiplying it by g. In otberwards, W = Mg. Because the unit weight '( is expressed as 1VIV and the mass density (p) as MIV. the two quantities can be related as

y-*- Y -pg

Thus unit weight in Nlml = mass density in kglm l For example, for water Pw is 1000 kglm 3. Therefore,

1000

'(W -

)C

)C

9.81

9.81 _ 9810 Nlm 3

=:

9.81 kNlml _ 10 kNlm 3

Sometimes, the mass density is expressed in Mgfm 3 or glml. The corresponding unit weight in kNlm 3 is equal to 9.81 p. For example, for water Pw is 1 Mg/m3 or 1 glml. The corresponding unit weight is 9.8l kN/ml. Likewise. mass density of 1600 kglm l corresponds to a unit weight of 1600 x 9.81 N/ml = 15696 Nlml '" 15.696 kNlm 3. In the reverse order, a unit weight of 18 kNlml corresponds to a mass density of 1800019.81 l = 1834.62 kglm . It will not be OUI of place to give a passing reference to the MKS unils still prelevant in some fields . In MKS units, the weight is expressed in kilogram me force (kgf). It is equal to the force exerted on a mass of 1 kg due to gravity. As the same force is also equal to 9.81 N, we have 1 kgf= 9.81 N unit

~e~h7~~nw~t~/~s~~ ~;;,~ra~~f;h~ =rr!~~~~:l~~ i~e;:o:i~ ~~;i~~;i~;9~~r ~~~le, Likewise,

1 kgf/ cm 2 _ 104 kgflm2 _ 104

X

9.81 N/m2 _ 98.1 kN/m 2 _ 98.1 k Pa

The unit kgf is not used in this text. Measurement of Mass The mass of a quantity of matter is determined with a weighing balance. lbe weights which have previously been used in MKS units are also used in SI units for measurement of mass. In other words, the weight of I kgf is called as the mass of I kg. The quantity of matter which weighs I kgf in MKS units will have a mass of I kg in SI units. Of course, the weight of that quantity of matter will be 9.81 N. Thus the weighing balances and weights which were previously used for determining the weights in kgf are used to determine the mass in kg. 2.8. SPECIFIC GRAVITY OF SOLIDS The specific gravity of solid particles (G) is defined as the ratio of the mass of a given volume of solids to the mass of an equal volume of water at 4°C. Thus, the specific gravity is given by G -

Ii:

..

(2.18)

The mass d~nsity of water pw at 4°C is one gmlm l, 1000 kglm 3 or 1 Mg/ml. [Note. In some texts, the specific gravity is represented as Gs .] The specific gravity of solids (or most natural soils falls in the general range of 2.65 to 2.80; the smaller values are for the coarse-grained soil... Table 2.1 gives the average values for different soils. It may be mentioned that the specific gravity of different panicles in a soil mass may not be the same. Whenever the specific gravity of a soil mass ~ indicated, it is the average value of all the solid particles present in the soil mass. SpecifiC gravity of solids is an important parameter. It is used for determination of void ratio and particle size.

roll MECHANICS AND FOUNDATION ENGINEERING

Table 2.1. Typlatl Values of G Specific Gravity

Soil Type

Grovel

2..65-2.68 2..65-2.68 2..66-2.70 2..66-210 2.68-280 Variable, may fall below 2.00

So""

Silty Sands

Sill InocganicQays Organic Soils

In addition to thc standard tcrm of specific gravity as defined, thc following two tcnns related with the specific gravity are also occasionally used. (1) Muss Specific Gravity (G",~1t is defined as the ratio of lhe mass density of the soil to the ma
.1J ill

Z.9. THREE· PHASE DIAGRAM IN TERMS OF VOID RATIO

ill.

--

The relationships developed in the preceding sections are independent of the actual dimensions of the soil

~~OI~:C ~r:~:~~~

. ..L

::;.:~.:";;~ ~,:.::: ~~e~":n7:~~ l'W~t>S,I~O ::~:~::-.:-~ ::-: "'tS"So}'W the volume of solids is I+e also equal to the height of solids. Fig. 2.5 (a) shows the phase diagram with volume of solids V. equal to unity. Since the void ratio is equal to the ratio of the volume of

1l' l+e

1

w Ms"GP

1

fa) Fig. 2.S.

-L

~I~ l'Wt"S. :.=.:.:.-:-:;.---::..:.:.. w",t"S.lW '

"'l"G'W (b)

Three.phase d iagram in terms of void ratio.

BASIC DEFlNmONS AijO SIMPLE TESTS

21

voids to the volume of solids, the volume of voids in Fig. 2.5 (0) becomes equal to e. The total volume ('\I) is obviously equal to \1 + e). 1be volume of air is shown bye" and the volume of water, bye.... The volumes are shown on the left side and the mrresponding mass on the right side in Fig. 2.5 (a). 1be volumetric relationships developed in Sect. 2.2 can be written direaly in tenns of void ratio as under: Poru;ity,

n ..

~ .. ~

Degree of saturation,

s ..

~_~

V. e The volume of water (V...) is shown as Se in Fig. 2.5(0). Obviously, the volume of air (V,,) is equal to (e - Se) = e(1 - S). Therefore,

percentage air voids, n" ..

~

.. e

and air content, a" ..

~

..

~1 +- eS)

¥ .

(1 - S)

Various mass densities discussed in Sect. 2.4 can be expressed in terms of the void rotio from Fig. 2.5 (a).

From Eq. 2.11,

p ..

M

M~+M...

V .. I-:;e ..

Gp ... + Sep ... --1-.-e-

(G + Se)p ... p - -l-.-eFrom Eq. 2.12 From Eq. 2.13,

M,

".(2.21)

Gp...

".(2.22)

Pd-V-~

P,...

M_

----y-

As the degree of saturation for a saturated soil is l.0 (i.e. 100%), Eq. 2.21 gives

P,... From Eq. 2.16

or

(G. e)p.

".(2.23)

.,~

(G. ')P. P .. p- - P..... -1-.-.- - P. ,

,

(G - 1)

P"~P ... In case the soil is not fully saturated, the submerged mass density is given by p' .. P - P...

From Eq. 2.21

p' .. (G 1+

:e;

".(2.24)

P ... _ p ...

(G • Se) p. - (1 • e) p.

1 • e , [(G - 1)- e(l - S)] P. p .. 1 + e

Eq. 2.25 reduces to Eq. 2.24 when the soil is fully saturated (S

".(2.25)

= 1.0).

Equations In Weight Units Eqs. 2.21 to 2.25 can be expressed in terms of weights. Equations can be derived comidering the vOlume-weight phase diagrams [Fig. 2.5 (b)] or simply by multiplying both sides of the equations by g and remembering thaI 1 .. pg. Thus

22


(G + Se)l.., y - -1-+-.-

Eq. 2.21 be
[2.21(a)1

Gy..,

Eq. 2.22 becomes

'td"

I-:;:-; (G +

Eq. 2.23 becomes

... [2.12 (a)1

ely.

'fJQI-~

•

. .. [2.23 (a)]

(G - I)

Eq. 2.24 becomes

'f -

Eq. 2.25 be
[(G - 1)-.(1 - s)]y. 'f .. 1 + e

--y-:;:-e Y...

... [2.24 (a)]

•

... [2.25 (a)1

In geotcchnical engineering, unit wcighlS are generally expres...oo in IcN/ml. The unit weight of waler l (Y ...) is 9.81 kN/m , which is sometimes taken as 10 kN/ml, for convenience.

It may be mentioned once again that mass density in glm! can be converted into unit weight in kN/ml by multiplying it by 9.81. For water, p... .. 1 glm!. For soils, if p .. 2 glm!.

"t.., .. 9.81 kN/ml y .. 19.62 kN/ml

2.10. THREE.PIIASE DIAGRAM IN TERMS OF POROSfrv Fig. 2.6(a) shows the three-phase diagrom in which the total volume is taken as unity. According 10 the dcfinilion of porosity n,

(0)

Fig. 2.6. Ph.1se-dillgram in terms of porssity.

(b)

n-~_~_vv Thcremre, the volume of voids in shown as n.

Obviously, the volume of solids is (1 - n).

Void ratio,

From Eq. 2.11 ,

or From Eq. 2.12,

n • - ~ P _

(same as Eq. 2.4)

%- _ M~

+ :",p", _ G p",(l - ;) + Sn p",

p - IG (I - n) + Sn] P.

Pd _

*" _

. .. (2.26)

GP",~l-n)

Pd - Gp",(l-n)

... (2.27)

23

BASIC DEFINITIONS AND SI¥I'LE TESrs

From Eq. 2.13

Ps... _ M~
or

P,. - [G(I-a) + alp.

From Eq. 2.15

_

G p",(l - In) + n p ...

. .. (2.28)

p' _ V'P.~-I) _ (I - a)p•. (G - I)

or p' - (G - 1)(1 - a) p. . .. (2.29) It mily be mentioned that Eqs. (2.26) to (2.29) in terms of porosity can also be derived from Eqs. (2.21) to (2.25) dircaly by substituting e - nl(l -n). This is left an exercise for the readers. Equations In tenns of Weight units Eqs. 2.26 to 2.29 can be written in terms of unit weights as under. Eq. 2.26 becomes y - [G(I-a) + SalY.

Eq. 2.27 becomes Eq. 2.28 becomes Eq. 2.28 becomes

.[2.26(a)1 ... [2.27(a)1 ... [2.28(a)1

y" - Gy.(I-a) '($'" - [G (1 - n) + n] ,(", y' _ (G - 1)(1 - n) y.

. .. [2.29(a)J

2.11 . RELATIONSIIH' nETWEEN THE VOID RATIO AND THE WATER CONTENT

kl'; !~I'; 1 1

An extremely useful relationship between the void ratio (e) and the water content (w) can be developed

as under.

Fig. 2.7 (a) shows the three-phase diagram.

t VW

t t ------------ t

Mw~Vwfw

-------------

------.-----~

------------ Mw~S'Yw

l'w

IYw

T~WG

(bl

Fig. 2.7. Three-phase ddiagram .

M M;

From Eq. 2.9

w -

From Eq. 2.5,

V... •

SV~

and from Sq. 2.18,

£!..

G

w _ V... P... V,PS

Therefore,

p.

W·

orps·Gp...

SV, V.. G

But Vv IV, "" void ratio (e). Therefore. oK)

ore -

S

... (2.30)

Z4

SOIL MECIIANICS AND FOUNDATION ENGlNEERING

For a fully salurnlcd soil, S = 1.0, and

e - "" ... (2.31) Alternatively Eqs. (2.30) and (2.31) can be derived using the 3·phasc diagram in terms of the void ratio [Mg. 2.7 (b»). w _ ~

From Eq. 2.9,

or

w -

Sep ...

M~

or w - G P...

~

or e·

i

,.

(same as Eq. 2.30)

II may be noted that it is morc convenient to work with 3-phasc diagram in (enns of void £"'ollio. The reader is advised to use 3-phnse diagram in terms of void ralio as far $ possible. 2.12. EX)'RESSIONS }"OR MASS DENSITY IN TERMS OF WATER CONTENT The e.'{prcssions for mass density c.1n be written in terms of water content by writing the void mtio in

terms of water content using Eq. 2.30.

(G • Se) p. p - -I-.-e-

From Eq. 2.21

(G • "") p. p - 1 • ("")IS If the soil is fully saturated, S ::: 1.0, and Eq. 2.32 becomes

or

(1 + w)Gp .... P - 1 • ("")IS

. .. (2.32)

(1 + w)Gp",

Pl"-~

... (2.33)

From Eq. 2.16, the submerged density is given by

P.../>-

(1 + w)Gp",

Pl"-P",·~

-P...

(G - I)p. Psub-~

or

...(2.34)

Eq.2.34 can also be obtnined directly from Eq. 2.24 by substituting e • wG. Gp.

From Eq. 2.22,

PJ -

or

Gp. p, - 1 • (""IS)

t;e

...(2.35)

From Eqs. 2.32 and 2.35,

... (236) Eq. 2.36 is nn extremely useful equation for determination of the dry density from the bulk density and vice versa. For a given water content w, a soil becomes saturated when S = 1.0 in Eq. 2.35. The dry density of the

soil in such a condition can be represented as

(Pd)zQt • 1 ~P~

.

... (237)

(p.dsQt is called saturated dry density.

The reader should carefully nOle tt).e difference between {P)SGI and (PJ),..,. In the first case, the water conlent of a partially saturated is inqeased so tha: all the voids are filled with water, whereus in the second case, the water content is kepi oonsLant and the air voids are removed by compaction so tlwt all tbe remaining voids are saturated with water. lbe Jailer condition is only hypothetical as it is not fcasibfe 10 remove all the air voids. Equations In terms

or Weight

Units

.Eqs. 2.32 to 2.37 can be written in terms of unit weighls as under.

BASIC DEFINITIONS AND SIMI'LE TESTS

. ..[2.32(a)1

Eq. 2.32 becomes Eq. 2.33 becomes

.. .\2.33(a)1 (G - I)r.

Eq. 2.34 becomes

. .. [2.34(a)[

'tlub-~

Gr.

Eq. 2.35 becomes

. .. [2.35(a)[

'td - 1 + (MIlS)

'td-~

Eq. 2.36 beaxnes Eq. 2.37 becomes

('td)'<6 - 1

.[2.36(0)1

~ 't:«,

... [2.37(a)1

2.13. RE1ATIONSIIII' BETWEEN OUY MASS DENSITY AND PERCENTAGE AIU VOIDS In the study o[ compaction of soils (Chapter 14), a relationship between the dry mass density and the percentage air voids is required. The relationship can be developed from the 3·phase diagram shown in Fig. 2.8 (a).

ill t"·, 1111 (0)

lb)

Fig. 2.8. Three-phase diagram

Now

v ..

V, + V ... + V"

l-~+Yv+~ Bul

V.

v-n"

CEq. 2.6)

Therefore

.f

(1 - n,,) ..

~

(1 _ n,,) ..

M, /~GP ...) + M",:pw

+

_ k. Gp,.,

+ (wM,)/pw V


26

... (2.38) When the soil becomes fully saturated at a given water rootent,

(Pd)". - 0 - 1 A little refieaioo win show that

(Pd)/I~

II" -

0, and Eq. (2.38) can be written as

~P:o

_ 0 and (P.d.... represent the same oondil ion.

In lenns of unit weights Eq. 2.38 becomes [Fig. 2.8 (b)]

(1 - a.)Gy.

. .. [2.38 (a)]

Yd - - - 1-.-e-

Table 2.2 gives a summary of the various relationships. The reader should make these equations as a pan of his soil engineering vocabulary. Thble 2.2. Dasic Relationships S. No.

Eq. No.

Relatiollship in Mass Density

Relationship in Ullil weigh/

1.

2.3

n ... ef(l + e)

2.

2.4

e ... n/(1 - 1/)

" ... eI(1 +e) e ... n/(I-II)

3.

2.8

4.

2.21

S.

2.22

6.

2.23

7.

2.24

8.

230

(0 +Se) pw

p..

9.

2.36

10.

238

Pd-~

'I"

Yd-~ I+< ,/Id---

.~ p... 1 +e

,---

e ... wG/S pol ... p/(l + w)

(0 +e)y... 1 +<

,

1+..0

(0-1)'1'" I+<

e .. wG/S Yd " y/(l +

(l-a.IGe_ _

(0+&)1'" 1 +e

1 " (G+e)pw 1 +e

p,......

pJ-

(Note. p...... 1(0) kg/m 3

I +e

'Id"

w)

~ I +WG

1.0 glml. ,/",'" 9810 N/m2 ... 9.81 kN/mJ _ 10 kN/mll

2.14. WATER CONTENT DETERMINATION The waler content of 8 soil is an important parameter that controls its behaviour. It is a quantitative measure of the wetness of a soil mass. 'Ibe water content of a soil can be determine
BASIC DI!flNnlONS AND SIMPLE 'f'ESTS

27

The soil sample is taken in a smaU. non-corridible, ainighl container. The mass of the sample and that of the container are obtained using an aex:urate weighing balance. According to IS : 2720 (pan 1I}-1973, the mass of the sample should be taken to an accuracy of 0.04 per cent. The quantity of the sample to be taken for the test depends upon the gradation and the maximum size of the panicles and the degree of wetness of the soil. The drier the soil. tbe more shall be the quantity of the specimen. Table 2.3 gives the minimum quantity of soil specimen to be taken for the test. The soil sample in the container is then dried in an oven at a temperature of 110° ::t SoC for 24 hams. The temperature range selected is suitable for most of the soils. The temperature lower than 110° ::t 5°C may not cause oomplete evaporation of water and a temperature higher than this temperature may c.'1use the breaking down of the crystalline structure of the soil panicles and laiS of chemic.'111y bound. st ructural water. However, oven-drying at 110° ::t 5°C does not give reliable resulLS for soils oontaining gypsum or other minerals having loooely bound waler of hydration. This temperature is aL~ not suitable for soils containing significant amount of organic matter. for all such soils, a temperature of 60° to 80°C is recommended. At higher tempcraturt; gypsum loses its waler of crystaUine and the organic soils tend to decompose and get oxidized. 'lhble 2.3. Minimum Quantity of Soil for Water Content Detenninatlon S. No. l.

,

2.

3.

~.

4.

s. 6.

Size of Particles more tilan 90% passing 425--micron IS sieve 2mm IS sieve 4.75 mmlSsieve lOmm IS sieve 20 rom IS sieve 40 mm IS sieve

Millimum Qualltity (gm) 25 50 200 300 SOO 1000

The drying pcriod of 24 hours has been rccommemled for normal soils, as it has been found that this period is sufficient to cause complete evaporation of water. lbc sample is dried till it attains a constant mass. The soil may be deemed to be dry when the difference in successive wcighings of the cooled sample docs nol exceed about 0.1 percent of the original mass. The soils oontaining gypsum and organic matter may require drying for a period longer Ihan 24 hours. The water content of the soil sample is caiCUl.1tcd from the following equation. w ..

where

~_M2-MJ)(lOO

M, M)-MI M 1 - mass of container, with lid M2 - mass of container, lid and wet soil M) - mass of container, lid and dry soil

... (2.39)

The water content of the soil is reponed to two Significant figures. (Refer to Chapter 30, Sect. 30.1 for the laboratory experiment) (2) Torsion Balance Method. lbe infra-red lamp and torsion balance moisture meter is used for rapid and accurate determination of the water content. The equipment has two main parts: (I) the infra-red lamp, and (il) the torsion balance. The infra-red mdiation is provided by a 250 W t.1mp built in the torsion balance for use with an alternating current 220--230 V, 50 cycles, single- phase main supply (IS : 2720 (part

0)--1973]. As the moisture meter is generally calibrated for 25 gm of soil, the maximum size of particle in the specimen shall be k!ss than 2 mm. The sample is kept in a suitable container so that its water content is not affected by ambient cooclitions. lbe torque is applied to one end of the torsion wire by means of a calibrated drum to balance the loss of weight of the sample as it dries out under infrared lamp. A thermometer is


provided for recording the drying temperature which is kept at 110° :!: SoC. Provision is made to adjust Ihc input vOltage to the infra·red lamp to conlrol the beat for drying of the specimen. The weighing mechanism, known as a torsion balance, has a built· in magnetic damper which reduces pan vibration<> for quick drying. TIle balance scale (drum) is divided in terms of moisture content (m') based on wet mass. lbe water mnlent (w), based on the dry mass, can be determined from the value of m' as under.

m'_~_~

From Eq. 2.10,

M

~ _ Ms

;w

M, +M",

M

", _

~

+ 1

w.--L

... [2.40(0)]

1 - m'

If wand m' are expressed as percentage,

w -

IOC,n~ m'

)( 100

... [2.4O(b)]

The time required for the test depends upon the type of the soil and the quantity of water present. It takes about 15 to 30 minutes. Since drying and weighing occur Simultaneously, the method is useful for soils which quickly rc.absorb moisture after drying. (3) Pycnomeler melhod. A pycnometer is a glass jar of about I litre capacity and filled with a brass conic.Ji cap by means of a SCf'C\HYPC cover (Fig. 2.9). The cap has a smaU hole of 6 mm diameter at its apex. A rubber or fibre washer is placed between the cap and the jar to prevent leakage. There is a mark on Ibe cap and also on the jar. The cap is screwed down to the same mark ...-Brass top such thai the volume of the pycnometer used in calculations remains constant. The pycnometer method for the - type c.over determination of water content can be used only if the specific gravity of solid (G) particles is known. A sample of we' soil, about 200 to 400 g, is taken in the pyalOmeter and weighed. Water is then added to the soil in the pycnometer to make it about hllif full. The mntents are GlilSS jar thoroughly mixed using a glass rod to remove the entrapped air. More and more water is added and stirring process continued till (he pycnometer is fiUed flush with the hole in the conical cap. The pycnometer is wiped dry and weighed. The pycnometer is then completely emptied. It is washed thoroughly and filled with water, flush with the lOp hole. 1bc pycnometer is wiped dry and weighed. Fig. 2.9 PyCllomelcr.

)~~~~~-screw

Let

MI

-

mass of pycnom
M2 - mass of pycnometer + wet soil M) - mass of pycnometer + wet soil + water M4 - mass of pycnometer filled with water only. Obviously, the mass M4 is equal to mass M) minus the mass of solids Ms plus the mass of an equal volume of water (see Fig. 2.10).

Thus

M,

M4 - M) - Ms + (G P...) . P... M4 - M)- Ms +

~

BASIC DEF1NmONS AND SIMI'LE TESrS

Fig. 2.10. Pycnometer Method Derivation.

- M, - M, ( I -

Mz - eM) - M4 )

or Now, mass of wet soil

= M2

G

(M j

-

~ 1)

- Mi

Therefore. mass of water M.., - (M2 - M i ) From Eq. 2.9.

b)

• (

w -

it )(

-

M.) ( G~I )

100

_ [(M, - M (Q.::..!.) _ I] (M,-M,) G I)

x 100

... (2.41)

This method for the detcnnination of the water OJOtCDt is quite suitable for roarse-grained soils from which (he entrapped air can be easily removed. If a vacuum pump is available, the PYOlometcr can be connected 10 II for about 10 to 20 minutes to remove the entrapped air. 11lc rubber tUbing of the pump shoukl be held tightly with the pYOlometcr 10 preveDt leakage. (Refer to Dlapter 30. Sect. 30.2 for the laboratory experimcot) (4) Sand Bath Method. Sand balh method is a field method for the determination of water content. The method is ropid, but not very accurate. A sand bath is a large, open vessel oontaining sand filled to a depth of 3 em or more. The soil sample is taken in a troy. The sample is crumbled and placed loosely in the tray. A few pieces of white paper are also placed on the sample. The tray is weighed and the mass of wet sample i£ obtained. The tray is then placed on the sand-bath. The sand bath is heated over a stove. Drying takes about .20 to 60 minutes, depending upon the type of soil. During heating, the specimen is tumed with a palette knife. Overheating of soil should be avoided. The white paper turns brown when overheating occurs. The drying should be continued till the sample attains a constant mass. When drying is oomplete, the tray is removed from the sand bath. cooled and weighed. ]be water content is determined using Eq. 239. (5) Alcohol Method. The soil sample is taken in an evaporating dish. urge lumps of soil, if any. should be broken and crumbled. The mass of the wet sample is taken. The sample is then mixed with methylated spirit (alcohol). The quantity of methylated spirit required is about one millilitre for every gram of soil. The methylated spirit and the soil should be turned several times, with a palette knife, to make the mixture uniform. The methylated spirit is then ignited. The mixture is stirred with a spatula or a knife when ignition ~ talciog place. After the methylated spirit bas bumt away completely, the dish 'is allowed 10 be cooled, and the mass of the dry soil obtained. 1bc metbod takes about 10 minutes. Methylated spirit is extremely volatile. Care shall be taken to prevent fire. 1be method cannot be used if the soil contains a large proportion of clay, organic maller, gypsum or any other caJcareous materiaL The method is quite rapid, but not very accurate.

SOIL MECHANICS AND FOUNDATION ENINEERING

30

(6) Calcium Cllrbide Method. This method of the dctcnninalion of water contenl makes use of the fact

that when water reacts wiLh calcium carbide

(c., C:z). acetylene gas (Cz Hi) L<.;

produced.

Cay

+ 2H 20 - CzH 2 + Ca (Ollh The water rooteol of the soil is determined indirectly from the pr
of

- SO,tt

51 pet casing

Stezezl cdsing

8

A caps-ute

Hydrog~n aloms of, ' , waiN in soil

-

Oetector

Fig.2. 1l .

in a steel casing A, placed in a bore hole as shown in Fig. 2.11. The steel casing has a small opening on its one side through which rays can come out. A detector is placed inside another steel casing B, which also has an opening facing that in casing A. Neutrons are emitted by the radio-active material. The hydrogen atoms in water of the soil cause scattering of neutrons. As these neutrons strike with the hydrogen atoms,they lose energy. The loss of energy is proportional to 'he quantity of water present in the soil. The detector is calibrated to givc directly the water content The mcthod is extremely useful for tbe determination of water cootcnl of a soil in the in-situ conditions. The methcx:l should be very carefully used, as it m3Y lead to radiation problems if proper shielding precautions are not taken. 2.15. SPECIHC GRAVny DETERMINATION The specific gravity of solid particles is determined in the laboratory using the following mcthods: (2) Pycnometer method (3) Measuring flask mcthod (1) Density boule method (4) Ga<> jar mcthod (5) Shrinkage limit mcthod. The last method of determining thc specific gravity of solid particles from thc shrinkage limit is discussed in Sect. 4.6.

BASIC DEFINmONS AND SIMPLE TESrs

31

(I) Density Bottle Method. TIle specific gravity of solid particles can be determined in a laboratory using a density botlle filled with a stopper having a hole (Fig. 2.12). The density bottle of 50 ml capacity is generally used [IS : 2720 (Pan

II) 1980]. The density bottle is cleaned and dried at a temperature of 105° to 110°C and cooled. 'The mass of the bottle. including that of stopper. is taken. About 5·10 g of oven dry sample of soil is taken in the bottle and weighed. If the sample contains particles of large size, it shall be ground to pass a 2·mm sieve before the test. Distilled water is then added to cover the sample. The soil is allowed to soak water for about 2 hOurs. More water is added until the bottle is half full. Air entrapped in the soil is expelled by applying a V3aJum pressure of about 55 em of mercury for about one hour in a vacuum dcssicalor. Alternatively. the entrapped air can be removed by genqe heating. More water is added to the bottle to make it full. 111e slopper is inserted in the bottle and its mass is taken. The temperature is also recorded. The bottle is emptied. washed and then refilled with di'itilled water. The bottle Fig. 212 Density bottle. must be filled to the; same mark as in the previous case. The mass of the botLle filled with water is taken. The temperature should be the same as before.

Let

MI

.. massofemptybottle M2 - mass of bottle and dry soil M) _ mass of bottle, soil and water

M4 - mass of bottle filled with water. If the mass of solids M. is subtracted from M) and replaced by the mass of walcr equal to the volume of solid. the mass M4 is obtained.

Thus

M4 - M) - MI + aM.

M. ( 1 -

h) -

P.

(P...)

M) - M4

M. -M2 -Ml

8uI

(M, - M,) ( 1 -

Therefore

~

~)

_ M, - M.

(M, - M,) _ (M, - M,) - (M, - M.) G-

M2 -Mj

... (2.42)

(M2 - M I) - (M) - M 4 )

0-

Alternatively,

M, M. +M4

M)

... [2.42(a))

Eq. 2.42 gives the specific gravity of solids at the temperature at which the test was condUdcd. SpecifiC gravity of solids is generally reported at 2-rC (IS: 2720-11) or at 4°C. The speciHc gravity al 27°C and 4"C can be dc!con ined from thc following equations. G TI and

where Gv

G4

=sp sr. of particles at 27",

_ G )( specific gravity of water al tOC , specific gravity of watcr at 27°C

... (2.43)

G,)( specificgravityofw3leratl°C

... (2.44)

-

G4

=sp. gr. of partiCles at 4°C, G, =sp. gr. of particles al t"C


32

Table 2.4 give.,> the specific gravity of water al different temperatures. The specific gravity of solids is reported as the average of the two tests, to the nearest 0,01. provided the difference between lhe lwO lests docs not differ by 0.03. Kerosene is a better wetting agent than water and is sometimes used in place of water. If Gk is the specific gravity oC kerosene at the test temperature. Eq. 2.42 becomes

G.

(M, - MdG. (M, - M I ) - (M, - M,)

. .. (2.45)

Table 2.4. Specific Gravity of Water I·e

sp. gr.

I·C

sp. gr.

I"C

sp. gr.

0.9999

11

12

0.99% 0.9995

21

0.9999

13

0.9994

31 32 33

0.9954 0.9951 0.9947

14 15 16 17 I• 19 20

0.9993 0.9991 0.9990 0.9988 0.9986 0.9984 0.9982

0.9980 0.9978 0.9976 0.9973 0.9971 0.9968 0.9965 0.9963 0.9%0

l"C

sp. gf.

1 2

3 4 5 6 7

••

10

1.0000 1.0000 1.0000 1.0000 0.9999 0.9999 0.9998

0.9997

22 23 24 25 26 27 2J3 29 30

0.9957

3.

0.9944 0.9941

35 36 37 38

0.9937 0.9934 0.9930 0.9926 0.9922

3. 40

Sometimes, other liquids, such as paramo, alcohol and benzene. arc also used. Density bottle method is suitable for fine-grained soils, with more than 90% passing 2 mm-IS sieve. However the method can also be used for medium and coarse-grained soils if they are pulverised such that the particles pass 2 mm-IS sieve. (See Chapter 30, Sect. 30.3 for the laboratory experiment). (2) Pycnometer Method. The method is similar to the density boute method. As the capacity of the pYOlometer is larger, about 200-300 g of oven-dry soil is required for the test. The method can be used for all types of soils, bul is more suitable for medium-grained soils, with morc than 90% passing a 20 mm IS sieve and for ~rse-grained soils with more Ihan 90% passing a 40 mm IS sieve. (See Chapter 30, Sect. 30.4 for the laboratory experiment). (3) Measuring Flask Method. A mea'iuring nask is of 250 ml (or 500 ml) capacity. with a graduation mark at Ihat leveL It is fitted with an adaptor for connecting it to a vacuum line for removing entrapped air. The method is similar 10 the density bottle method. About 80--100 g of oven dry soil is required in Ihis case. The method is suitable for fine-grained and medium grained1soits. Rubber bung (4) Gus Jar Method. In this method. a ga.. jar of about I litre capacity is used. The jar is fitled with a rubber bung (Fig. 2.13). The gas jar serves as a pycnometer. The method is similar to the pycnometer method.

2.16. MEASUREMENT OF MASS DENSITY

The bulk mass density of a soil sample, as per Eq. 2.11, is the·mass per unit volume. Allhough lhe mass of a soil sample can be determined to a high degree of precision, it is rather difficult to determine the volume of the sample accurately. The methods discussed below basically differ in the prOCedure for the measurement of the volume. Once the bulk mass density has been detennincd. the dry mass density is found using Eq. 2.36. Thus

, LItre glass jar

Soil

Fig. 2.13. Gas Jar.

BASIC DEFlNn10NS AND SIMPLE 1T:SrS

33

p • Mand V The volume of the specimen used in various tcsts can be computed from the measured dimensions. as Ihey have regular shapes, such as a cylinder or a cube. Ilowever, precise measurements arc not possible. If the sample is made in a container of known dimensions. much more accurate measurements arc possible. The following methods are genemlly used for the detennination of mass density. (1) Water Displacement Method (2) Submerged mass density Method (3) Core Cutler Method (4) Smld Replacement Method. (5) Water Balloon Method (6) Radi:ltion Method. The methods are discussed below. 1lIc first two methods arc laboratory methods and the !'CSt, field methods. (I) Water Dl~placement Method. The volume of the specimen js dClcnmned in Ihis method by waler displacement, As the soil mass disintegrates when it comes in contact with water, the sample is cooted with paraffin wax to make it impervious. A Icst specimen is trimmed to more or less a regular shape and weighed. It is then coated Valva with a trun lay.er of .paraffin wax by dipping it. in molten ~ wax. The specimen IS allowed to cool and weighed. 1llc Mtasurrng . =difference between the two observations is equal 10 the mass of the paraffin. 'llie waxed specimen is then immersed in a waterdisplacement container shown in Fig. 2.14. Thc volume of the specimen is equal 10 the volume of WOlter which comes out of the outflow lube. The actual volume of the soil Fig. 2.14. WIlICr di~placemcnl cont.,incr. specimen is less th3I1 the volume of the waxed specimen. The volume of the wax is determined from the mass of the wax peeled orr from the specimen afler the test and the mass density of wax. Now

V _ V, _ (Al, - M)

... (2.46)

p, V = volume of specimen, V, = volume of waxed specimen, M, = mass of waxed specimen, M = mass of specimen, Pp = mass density of paraffin (approximately 0.998 gm/ml). A representative sample of the soil is laken from the middle of specimen for the walcr content detennination. Once the mass, volume and the water content of the specimen have been determined, the bulk density and the dry density arc found from Eqs. 2.11 and 2.36, respectively.

where

(See Chapter 30, Sect. 30.7 for the laboratory experiment). (2) Submerged Muss Den~ity method. lltis method is ba<>ed on Archimedes' principle that when a body is submerged in water, the reduction in mass is equal to the mass of the volume of water displaced. The sample is first trimmed and weighed and then it is immersed in moiten wax and again weighed, as in tbe water displacement method. The specimen is then placed in the cmdlc of special type balance. The cradle dips in the water contained in Ibe bucket placed just below. Tne npparent mnss of the waxed specimen in water is delennincd. The volume of the specimen is determined as below: V .. (101, - 101 1) _ (M, - M) ... (2.47)

P..·

where M,

Pp

= mass of waxed speCimen. M = mass of specimen,


34

M 1 = mass of waxed specimen in water, Pp = mass density of wUJt, p... = mass density of water. Eq. 2.47 can be derived, using Archimedes' principle. Ml - M, - U - M, - VIP... V, _ M, - M\

P. Substituting this value V, in Eq. 2.46. we gel Eq. 2.47. This method is suitable for finc-gruincd soils. (3) Core Cutler Method. It is a field method for determination of mass density. A core cutter consists of an open, cylindrical barrel, with a hardened, sharp cutting edge (Fig. 2.15). A dolly is placed over the cutter and it is rammed into the soil. lne dolly is required to prevent burring of the edges of the cutter. 1nc cutter containing the soil is taken oul of the ground. Any soil extruding above the edges of the culler is removed. The mass of the cutter filled with soil is taken. A representative

sample is taken for water content dctermin.'llion. lbc volume of the soil is equal to the internal volume of the cutter, whidl can be detcnnincd from the dimensions of tbe cutter or by filling the cutler with water ~nd finding the mass of water.

... (2.41)

Bulk mass density,

r-----' ~lmm---..,.j I

I

T

Cutter __ where M 2 :: mass of culter, with soil, 13 0 rr.m M I = mass of empty cutter, V:: intCITh'l1 volume of cutter. lhe method is quite suitable for son, fine grnined soils. It cannot be used for stoney, graven), soils. The method is practicable only at the places where the surface of the soil is exposed and the cutter con be easily driven. Fig. 2.15. Core.Culler with dolly. (See Chapter 30, Sect. 30.5 for the experiment).

I i

1

(4) Sand Replacement Method. Fig. 2.16 shows a sand-pouring cylinder, which has a pouring cone at its base. TIle cylinder shown is placed with its base at the ground level. There is a shutter between the cylinder and the rone. The cylinder is firsl calibrated to delennine the mass density of sand. For good results, the $and used should be uniform, dry and clean, passing a 600 micron sieve and rctuined on a 300 micron sieve. (0) Callbrntlon of appurotus-The cylinder is filled with sand and weighed. A calibrating oontainer is then placed below the pouring cylinder and the shutler is opened. The sand fills the calibrating container and the cone. The shutter is closed, and the mass of the cylinder is again laken. lbe ma5S of Ihe sand in the container and the cone is equal to the dirl'crencc or the two observations. The pouring cylinder is again filled 10 the initial mass. The sand is allowed 10 run 001 of the cylinder, equal to the volume of the calibrating cootaincr and the shutler is closed. The cylinder is then placed over a pt.!in surface and the shutler is opened. 'Ihe sand runs Oul of the cylinder and fills the cone. The shutler is closed when no further :novement of sand takcs place. 1nc t.)'linder is removed and the sand filling the rone is collected and weighed (Mi). "he mass density of the sand is dctennincd as under:

P. ..

All - M2 -M) V t

where M 1 = initial mass of cylinder with sand,

... (2.49)

BASIC DEFINITIONS AND SIMPLE TFSrs M2

M3

= mass of sand in cone only, = mass of cylinder after pouring sand into the cone

and the container, Vc = Volume of the container. Note. Mass of sand in both the oontainer and cone is M I

-

M3-

(b) Measurement or Volume or lIole-A tray with a central hole is placed on the preparoo ground surface which has been cleaned and properly levelled. A hole about tOO mm diameter and 150 mm deep is excavated in the ground, using the hole in the tray as a pattern. The soil removed is carefully collected and weighed. The sand pouring cylinder is then placed over the excavated hole as shown in Fig. 2.16. The shutler is opened and the sand is filled in the cone and Ihe hole. When the sand Slops running out, the shutler is closed. Tbc cylinder is removed and weighed. 111c volume of the hole is determined from Ihe mass of sand filled in the hole and the unit mass density of sand.

Volume of hole where M I =

.. M I

-

M. - M2

p,

...(2.50)

mass

of cylinder and sand before pouring into the hole, M2 = mass of sand in cone only, M. = mass of cylinder after pouring sand into the hole, P, = mass density of sand, as found from calibration.

The bulk mao;s density of the in-situ soil is determined from the mass of soil excavated and the volume of the hole. Fig. 2.16. Sand Replacement method. The method is widely used for soils of various particle sizes, from fine-grained to coa~grained. For accurate results, the height of sand column in the cylinder is kept approximately the same as that in the calibration test. The depth of the hole should also be equal to the depth of calibrating container. (See Chapter 30, Sect. 30.6 for the experiment). 5. Rubber Balloon Method. The volume of the hole in this method is determined using a rubber balloon' or by filling water in the hole after covering it with a pl...1Stic sheet. The rubber balloon method is explained below. The apparatus consists of a density plate and a graduated cylinder, made of lucite, encloocd in an airtight aluminium case (Fig. 2.17). 11te cylinder is partly filled with water. There is an opening in the bottom of the case, which is sealed by a rubber balloon. The balloon can be pulled up into the cylinder or may be pushed down through the bottom. A pump is attached to the cylinder for this purpose. When the pressure is applied, balloon comes out the aluminium case through the hole in the density piate. When a vacuum is applied, the balloon is pulled up into the cylinder. For determination of the volume of the hole. the density plate is placed on the levelled ground. The cylinder is then placed over the plate. The pressure is applied to the balloon. The balloon deflates against the surface of the soil. The .volume of water in the cylinder is ooserved. The cylinder is removed from the base plate. 1be soil is taken out

Hand pump

W<"lltr

Otn:;lly

ba,lloon

pial!!:

~~..!r~~'='=rlG::.~"d Holf. in ground

SOil MECHANICS AND FOUNDA110N ENINEERING

from the hole through the opening in the base plate. All loose material is removed. llle soil removed is collected and weighed. The cylinder is
2.17. DETERMINATION O}O'

vom

RATIO, l'OnOSITY AND DEGREE 0.' SATURATION

The void ratio of a soil s.1mp!e is a measure of its den'lcncss. It is one of the important parameters of soils. Engineering properties of soils depend upon void mtio 10 a large extent. The void mHo is determined in the labordtory indirectly from the dry mass density. From I3q. 2.22.

e _ Gp•. _ 1 .. . (2.51) p, The methods for determin:ltion of the spccilic gravity of solids G and the dry density Pd have been discussed in the preceding sections. For a saturated soil. the void ratio is determined using Eq. 2.31, e .. ~. This method is a very convenient and accurate method. as the water content of a soil can be determined quite easily and acaJrnlcly. The specific gravity of soil (G) can also be determined in the laboratory. Once the void mlio hns been detennined. other volumetric relationships such as porosity and degree of saturation can be determined using Eqs. 2.3 and 2.30, respectively. Percentage air voids are determined indirectly, using Eq. 2.38, (l-n.)Gp. Pd"~

n" .. I - :;'" (1 +

~)

... (2.52)

Eq. 2.52 can be reduced to the following form . PJw n,,_I1 _ _

r.·

... (2.53)

Ibis is len as an exercise for the r~uJers. Table 2.5 gives typical values of void ' rl.ltio, porosity. dry density. and dry unit weight of dilTerent soils in loosest and densest conditions.

BASIC DEFINlllONS AND SIMPLE TESTS

37

Thble 2.5. lypical Values or Void Ratio lind Dry Denl;ily lind Dry Unll'i: Weights

S.No.

Soil type

Gravel

Slale oj soil """",,

Void Ratio

PorosilY

0.60 0.30

'"

2.

Coarse sand.

"''''''', """",,

3.

Medium sand Unifonn, fine

"""",,

4.

",' Coorse silt

S.

Fine silt

"',"'" """",, "'''''', Softest

O.

Lean Clay

Softest IIDrdCSI

7.

fm clay

Sortesl

2.20

Ilnrd(.'$l

OAO

0.85 0.4 1.0 0.45 1.00 0.4 1.20

IIDrcicsl

(kglm

OAO

J

)

''''"

23 42 2. 40 29 50 31 SO 29 55 29 69 29

0.75 035

Densest

Dry defLSity

('!o)

Dry unit weight

(kNlm 10

2000 1S00 1900

'"

1400 1900

14 I.

1300 1800

13 I. 13 I. 13 I. 10

1300 1900 1300 1900 1000 2000

J

)

IS I.

'"

ILLUSTRATIVE EXAMPLES

1II1J.~lruti't'e EXlIIIlpie 2 .1. 71u: mass of a clwnk of moist soil is 20 kg, and its volume is 0.011 ml. After drying in an oven, the mass reduces 10 16.5 kg. Determine the water content, the density of moist soil, tile dry density. void rario, porosity and the degree of saluration. Take G = 2.70. Solulion. Mass or water. AI... = 20.0 - 16.50 = 3.50 kg F'rom Eq. 2.9, water content, F'rom Eq. 2.11, the wet mass

dt~nsity

From Eq. 2.12, the dry density, From Eq. 2.22.

or

w ..

;6~5~

.. 0.2121 (21.21%)

p ..

O.~ 1

.. 1818.18 kg/Oil

Pd ..

~.~~l

.. 1500.0 kg/OIl

Gp l+e .. ""

p,

e .. 2.701;~OOO _ I .. 0.80

From Eq. 2.3,

n ..

~

..

From Eq. 2.30,

S ..

~

O.212~.;'

..

~::

.. 0.444(44.44%) 2.70 .. 0.7158 (71.58%)

IIIustratl't'e Example 2.2. A ~Qil specimen has a water content of }O% and a wet unit weighl of 20

kN/nl If the specific gravity oj solids is 2.70, determine the dry unit weight, void ratio, and the degree of samra/;OIL Take 't ... = 10 /eN/m . Solution. From Eq. 236 (a), From Eq. 2.22 (a), From Eq. 230,

'td -

~

- 1

;°

. .. 18.18 kN/m.l 01

1 + e _ G 't"" .. 2.70 x 10 .. 1.49 Yd 18.18

S -

7-

or e _ 0.49

0.1 0~4~·70 - 0.551 (55.1 %)

38

SOIL MECHANICS AND FOUNDATION ENINEERIN(

Illustrative Example 2.3. A sample of dry soil 'Weighs 68 gm. Find the volum~ of voids if t.he tOla volume of the sample is 40 ml and the specific gravity of Solids is 2.65. Also determine the void ratio. Solution. From Eq. 2.12,

Pd ..

*" . ~ . M,

1.70 gm/ml

68

Volume of solids,

V~ .. Gp", .. ~ .. 2S.66ml

Volume of voids,

V~ .. V - V, .. 40.00 - 25.66 .. 14.34 ml

e-~-~::':-O.s6

From Eq. 2.1,

Illustrative E1UIrnple 2.4. A moist soil sample weighs 3.52 N. After drying in an oven, its weight is redl~ced to 2.9 N. The specific gravity o/solids and the mass specific gravity are, respectively, 2.65 and 1.85. Determine the water content, void ralio, porosity and the degree of saLUration. Take "t ... = 10 leN/mJ,

Solution.

= 3.52-2.90 = 0.62 N

Weight of water

~~

From Eq. 2.9, w
w ..

From Eq. 2.19,

'I .. Gm y", .. 1..85 )( 10 .. 18.5 IcN/m l

From Eq. 2.36 a,

Yd ..

From Eq. 2.22 a,

.. 0.2138 (21.38%)

Ifw .. 1 + 1~';138 .. 1524 kN/ ml

I + e .. Gy", .. 2.65 x 10 .. 1.74 Yd 15.24

e .. 0.74 From

Eq. 23,

From Eq. 2.30,

n ..

~

S.

7 _0.21~7:

.. 1

~'~74

.. 0.4253 (42.53%)

2.65 _ 0.7656(76.56%)

illustrative Example 2.5. A soil has a porosity of 40%, the SpecIfIC gravity of solids of 2.65 and a WQter content of 12%. Determine the mass of water reqllired to be added to 100 m) o/tltis .foil for /ull saturation. Solution. Let us take unit volume of solids, i.e. V, .. 1.0 ml. From Eq. 2.9,

Mass of solids., mass of water, Volume of water

From Eq. 2.4, From Eq. 2.1, volume of voids,

M, .. G P ..... 2.65 x 1000 .. 2650 kg

M", .. 0.12 x 2650 .. 318 kg ..

13~

.. 0.318m l

e .. l:n"

1.00~~.40"

V~ .. e V, .. 0.667

l(

0.667

1.0 .. 0.667013

Therefore, volume of air, .. 0.667 _ 0318 .. 0.349ml Volume of additiOllal water for full saturation = 0.349 ml Total volume of soi~ V .. V, + Vv .. 1.0 + 0.667 .. 1.667 01 3 Volume of water required for 100 013 of soil ..

~:!:~ x 100 .. 20.94 013

Mass of water required = 20940 kg. Illustrative Example 2.6. A sample 0/ saturated soil has a water content of 25 percent and a bulle unit 1 weight of 20 kN/m . Determine dry density, void ratio and specific gravity of solid particles.

BASIC DEFINITIONS AND SIMPLE TESfS

39

What would be the bulk uni, weight of the same soil at the same void ratio hut at a degree of saturation of 80% ? Ta/ce y", = 10 leN/mJ.

Solution. From Eq. 2.33 (a),

'is" ... 1 ~ Y::v (1 + w) 20 ... G ;

From Eq. 2.22

;oo.~ +)( 0(,25)

or G ..

2.67

e ... M-G _ 0.25 )( 2.67 .. 0.67

From Eq. 2.30, laking S = 1.0, (a~

Yd'" IG

In the scoond case, as e :; 0.67 and S

;"'e . . 2i6: ~.;~

= 0.80,

.. 15.99 kN/m

J

Eq. 2.21 (a) gives

y ... (G 1++S:h w

_

(2.67 +

~.~ ~.~;,67)

)C

10 .. 19.20 kN/mJ

lIIustrallve EXllmple 2.7. A sample of clay was coated wl'tll paraffin wax and its mass, including the mass of wax, was found to be 697.5 gm. The sample was immersed in water and the volume of the water displaced was found to be 355 1111. The mass of the sample wit/JO/d wax was 690.0 gill, and the water content of the representative specimen was 18%. Determine the bllik densil){ dry density, void ratio and the degree of saturation. The specific gravity of . the solids WQS 2.70 and that of tite wax was 0.89. . Solution. Mass of wax ... 697,5 - 690,0 ... 7.5 gm

O.~~O 1.0

VOlume of wax

...

Volume of soil

... 355.0 - 8.43 ... 346.57 mt

Bulk density

From Eq. 2.36, dry density

From Eq. 2.22, From Eq. 2.30,

...

~:!7

... 1 1+ e ...

... 8.43 ml

.. 1.99 gm/ml

!'~18

... 1.69 gm/ml

27~,:9t.O ...

1.60

Of

e ... 0.60

S _ ~ _ O.l80.~2.70 _ 0.81 (81%)

illustrative Example 2.8. (a) During a lesl for water content determination on a soil sample by pycnometer, the following observations were recorded (1) (2) (3) (4)

Mass of wet soil sample Mass of pycnometer with soil and filled with water Mass of pycnometer filled with water only Specific gravity of solids

= 1000 gm

·2000 gm = 1480 gm ·2.67

Determine the water content. (b) If the b/Jlk density of the soil

Solution. (a) From Eq. 2.41,

is 2.05 gm/ml, determine the degree of saturation. w - [

(M,-M1)

(M3 _ M4 ) '

(a-I) ----a- -

1

1)( 100

- [ (20001000 _ 1480) x (2.67-1.0) ~ - 1

1x 100 -

20'· ._%

40

SOIL MECHANICS ANI,) FOUNDATION ENINEERING

(b) From &j. 2.36.

Pol -

From Eq. 2.22,

1 + e ..

Now

5 ,.,

t!-;;.

1

+20~02.8

2.6~.;O 1.0 .. 1.57

.. 1.70gm/ ml

or

t! ..

0.57

~ '" O.20~.5; 2.67 .. 0.950 .. 95.0%

IIhl!drnCive Exumple 2.9. The mass of an empty gas jQl' was 0.498 kg. Wilen completely filled with water, its mass was 1.528 kg. An oven-dried sample of soil of IIIOSS 0.198 kg was placed in the jar and water was added to fif/the jar and irs mass was found to be 1.653 kg. Determine the specific gravity of panicles. M2 - M[ G .. (M -M ) (M)

Solulloo. From Eq. 2.42,

2

or

M

I

(~:~~~

G - 0.198

) 4

1.528) - 2.71

IIIustruUve Exumple 2.10. In a compaction test on a soil, the 1II1ISS of wei soil when compacted in the mould was 1.855 kg. The water content of the soil was 16%. If the VQ11III/c of the /IIould was 0.945 litres, determine the dry density, void ratio, degree of samra/ion and percentage air voids. Take G '" 2.68. Solution.

Bulk density

p '"

From Eq. 236.

Pd ..

From Eq. 2.22.

1+ e -

From Eq. 2.30,

S ..

From Eq. 2.38.

p, -

0.94~·~510::J

~

-

- 1962.96kg/ m)

/~~~6

2.~69;.~~

3

- 1692.21 kg/ m

.. 1.584

or e .. 0.584

"'~ - o. I ~;si·68 _

0 .7342 _ 73.42%

(l-n.)G·p. 1 + wG

1 - nQ _ 1692.21 ;~l x+ l~ )( 2.(8) .. .0.9022

or

n. - 0.0978 (9.78%) Illustrative Exumple 2 .U . A compacted cylindrical specimen, 50 111m dia and 100 111111 length, is to be prepared from oven-dry soil. If the specimen is required to have a waler contenl of 15% and the percentage air voids of 20%, calClilate the //lass of the soil and water required for the preparation of tlte sample. Take G = 2.69. Solution. Let M, be the mass of solids in kg. Mass of water,

V .. ~ .. _ _M_,_ _ .. ~ m) # G P... 2.69)( 1000 2690

Volume of water,

1.'... _

Total volume of sample, From Eq. 2.6, volume of air, Now

.. wM# _ 0.15 M#

Volume of SOUds.

O'I~'

_ 0.15 )( 10-) M, m)

V .. x /4 )( (0.05)1 x 0.10 .. 196.35 x 10-4 m 3 VB - 0.20 )( V - 0.2 )( 196.35 )( 10-6 _ 39.27 )( 10-4 m 3

BASlC DEFINmONS AND SIMPLE TESTS

2~

41

+ 0.15 x 10-3 M, + 39.27 x 10-6 _ 196.35 )( 10-6 M, _ 0.301 kg

Mass of water .. 0 .15 x 0.301 _ 0.045 kg llJustrnUve Example 2.12. A borrow area soil has a lIatural water comem of 10% and a bulk density of 1.80 Mg/l,r. The soil is used for an embankment to be compacted at 18% moisture content to a dry density of 1.85 Mg/m J• Determine the amount of water to be added to 1.0 m J of borrow soil. flow nJllI1Y cubic metres of excavation is required for I nl of compacted embankment ?

Solution. Borrow area soil.

Pd"

Unit weight

~:~

.. 1.636 glml

_ 1.636 )( 9.81 _ 16.05 kN ml

U us consider 1 m1 of borrow soil. W, z: Dry weighllm)

= 16.05 kN 1 = 1.605

W ... = Weight of water/m

In embankment,

W...1

"

kN

0.18 )( 16.05 _ 2.889kN

Wnter to be added Weight of dry soil in embnnkment/m

1

- 2.889 - 1.605 - 1.284 kN _ 1.85 )( 9.81 ,. 18.15 kN

1 Volume of soil rcquired/m of embankment _

\86.~~

_1.131 m l

Illustrative Example 2.13. There are two borrow areas A and B which have soils with void ratios of 0.80 and 0.70, respectively. The inplace water content is 20%, and 15%, respect;IIC[Y. The fill at the end of constmction will have a total voillme of 10,000 /II), bulk density of 2 Mg/mJ and a placement water content of 22%. Determine the volllllle of the soil required to be excavated from both arcas. G = 2.65. If the cost of excavmion of soil and trlUl.Sportation is Rs. 200/· per 100 nI for area A and Rs. 220/. per 100 mJ for area A, which of the borrow area is more economical?

2~6: ~. ~OO

Solution. Borrow area A.

p" ==

Let us consider I m.l of borrow soil.

W, .. Dry weightlm 1 .. 14.44 kN

In embankment.

Pd - 1 +2 . .. 1.639 glml 0 22

Weight of dry soil per m3 Volume of soil required

=:

1.47 g/ml (14.44 kN/mJ)

( .. 16.08 kN/m)

= 16.08 kN ..

!=:~

.

1.114 m

l

Cost of soil _ Rs 200/100 )( 1.114 .. Rs 2.23 per m1

Pd ..

Borrow area B.

2.~.~ 1.0

.. 1.559 g/ml (15.29 kN/ml)

W, .. dry weight/m 1

..

15.29 kN

In embankment, weight of dry soil = 16.08 leN Volume of soil required

..

~

.. 1.052 m1

Cost of soil .. Rs 220/100 x 1.052 .. Rs 2.31 per ml. Borrow area A is more economical.


42

PROBLEMS A. NumeriCllI :U. (D) Deline the [elTI1S void ralio, specific gravit), of particles, degree of saturation and dry densit)'. (b) Develop a relationship between the void rolio, water cootem, specific gravity of particles and the degree of saturation. Z.Z. (0) Describe ovcn.(lrying method for the delenninmion of waler oooten! of a soil sample in a laboratory. l (b) A sample of wei soil has a volume of 0.0192 m and a mass of 32 kg. When the sample is dried oul in an oven, its mass reduces to 28.S kg. Determine (I) Bulle. density. (il) Wllter rontcnl, (;il) Dry density, (iv)

:I~~~~ dcngty, ([~~:,dl:~6jV~):~;i~;); ~:: ~~:)~~~~·.5~a~~:~;sr,:~ ;4~i~~:~~~~ l

2.3. (a) A $lmple of saturated soil hOlS a water content of 2."S percent and a bulk unil weight 0£20 kN/m , Determine the dry unit weight, void ratio and the specific gravity of solids. (b) What would be the bulk unit weighL of the soil in en) if it is compacted LO the same void ratio but hos I) degree of saLUration of 90% ? (Ans. 16 kNIm\ 0.667, 2.667 19.60 kNlmll 2.4. A sample of soil has a volume of 65 ml and weighs 0.96 N. After oomplete drying, its weight reduces 10 0.78.'i N. If the specific gravity of solid particles is 2.65, determine the degree of saturation. [Ans.51%J 2.5. A saturated soil sample has 0. water content of 40%. If the specific gravity of solids is 2.67, dctennine lhe void ratio, saturated denSity, and submerged density. [An!i. 1.07 i 1807 kg/m l i 807 kg/mll 2.6. (a) Define the terms void ratio, dry density, submerged density and mass specific gravity. (b) Derive on expression for bulk density in tenTIS of its water content, void ralio, specific gravity of solids and density ofwatet. l 2.7. A partially saturated sample of a soil has a density of 1950 kg/m and a water content of 21%. If the specific gravity of solids is 2.65, ClIlculate the degree of saturation and void ratio. If the sample subsequently gets saturoted, determine its saturated density. (Ans. 86%; 0.645 ; 2003 kglmlJ 2.S. A sample of soil has a volume of 1 litre and lL wcight of 17.5 N. The specific gruvity of the solids is 2.68. If the dry unit weight of the soil is 14.8 leN/ml, determine (a) water content, (b) void ratio, (e) porosity, (d) saturated unit weight, (e) submerged density and (j) degree of saturation. [Ans. 18.2% : 0.811 : 44.8% ; 19.28 kN/ml, 9.28 kN/ml and 60.2%1 2.9. A fully saturated day sample has a mass of 130 gm and hos a volume of 64 anl . The sample mass is 105 gm nfler oven drying. Assuming thaI the volume docs not change during drying, dC1ennine the following; (,) specific gravity of soil solids. (il) void ratio, (iii) porosity, (iv) dry density. [Ans. 269 ; 0.64 i 39% and 1.641 gm/cn?] 2.10. Prove thnt the water content (w) of a p3nially saturated soil can be expressed as \I' -

1 - (011010) (0",/5) _ 1

where Gm "" mass specific gravity, G "" specific gravity of solids and S '" degree of salUralion. 2,11 (a) Prove that the degree of saturation of 8 panially saturtlled soil ClIn be expressed os S _ ::--''-----;-

~(l+W)-t

where p .. bulk densilY, G .. specific gravity of solids nnd w water content. (b) A eyliodrical specimen oC soil is 7.50 em long and 3.75 em in diameter and has a

mMS of 175 gm. If the water content is 18 percent and the specific gravity of solids is 2.68, detennine the degree of saluration. What 'NOuld be the error in the degree of saturation if there has been an error of 1 mm in measuring the length ? (An&. 96.7%, 4.62%) 1.12. A pycnometer having a mass of 600 gm was used in the following measuremenls of three samples of soil. Sample No.1 was ovendricdi sample no. 2 wos partially saturated and sample no. 3 was Cully saturated. The bulk density of the sample no. 2 was 2.05 gmfml. Sample No. 1 No.2 No.3 Muss of samples (gm) 960 970 1000 Mass of sample + water. pycnometer (gm) 2080 2050 2010

43

BASIC DEFTNI1l0NS AND SIMPLE TESTS

If the mass of pycnometer when filled with water only was 1475 gm, dClermine the specifie gravity of solids. (b) Also determine the water content and void ratio of samples no. 2 and 3, and the degree of saturation of sample no. 2. IAns. 2.70; 6.3%, 0.40; 11.70: 0.32 and 41.85%1

2.13. An undisturbed specimen of clay was tested in a laboratory and the following results were obtaine
'" 210 gm '" 175 gm '" 2.70

What was the totuJ volume of the original undiswrtx:d spccimcn ns..c;uming that the specimen was 50% !Illturatcd ? (Ans. 134.8 ml] 2.14. A soil deposit to be used for construction of an eanh embankment has an average dry density of 1.62 gmJmI . If the compacted embankment is to havc an average dry densi ty of 1.72 gmlmI, determine the volume of soil to be ex:cavated for 1000 m) of embankment. The water content of the soil in the bonow pit is 10%. lAos. 1.06] x 10) mll

2.15. Determine the specific gravity of solids from the following observations: (i) Mass of dry sample '" 0.395 kg (ij) Mass of pycnometer full of water '"' 1.755 kg (iii) Mass of pycnometer containing soil and full of watet ::::I 2.005 kg.

IAns.2.72J

2.16. A sample of clay having a mass of 675 gm was coaled with paramn wax:. 1be combined mass of the clay and the wax was found to be 682 gm. The volume was found by immersion in water as 345 mt. The sample was then broken open and the water content and the specific gravity of solids were found 10 be 15% and 2.70, respeaively. calculate the bulk density of soil, its void ratio, and degree of saturation. Thke specific gravity of wax: as 0.89. {Ans. 2.002 gmlml, 0.551 and 735%J 2.17. In order to determine the bulk density of a soil insi tu, 4.7 kg of soil was e."~tractcd from a hole al the surface of the soil. The hole required 3.65 kg of loose dry s:lnd for its filling. If il takcs 6.75 kg of the SlIme sand to fin a calibrating can of 4.5 lilre capacity, dl!termine the bulk density of the soil. [An.... 1932 kglm)l 2.18. A litre capadty cullcr of mass I kg WIlS pu.<;hed into an emban~cnt under construction and the mass of the culler with soil was found to be 2.865 kg. If the sample had wnter content of 11 %, determine the void ratio of the soil in embankment. G:: 2.67. rAm•. 0.59J

8. Descriptive and Objective lYpes 2.19 What is a block diagrom ? WhDl is its use ? 2.20. Differentiate between : (a) (b) (c) (d) (e)

Percentage air voids and air content, Void ratio and porosity. Specific gravity of solids and mass specific gravity. Watcr content based on solid material and that based on total mass. Saturated density and bulk density.

2.21. How do you determine the void ratio of a soil? 2.22. Discuss various methods for detcrmination of water content in a laboratory.

2.23. Describe a method for dctermination of the specific-gravity of solids of fine.grained soils. 2.24. How would you determi ne the bulk: density of a soil specimen in a laboratory ? 2.2S. Discuss various methods for the determination of bulk density of a soil in field. 2.26. Slllte whether the following statements are true or false (a) The water content of a soil can be more than 100%. (b) The porosity of a soil can be more than 100% (e) The specific gravity of particles of coarse-grained is seldom greater than 2.70. (d) Thc submerged density is about onc·half of the SlltUrnted density. (e) For dcterminmion of water coment of all types of soils, the oven temperature Is 1000

:t

5°C.

fAns. True (a),(c), (d)J

2.27. (a) Which of the following relation is nOi correct ?

44

SOIL. MECHANICS AND FOUNDATION ENINEERJNG

(i) c = J

:1/

(;,) 11 = - ' -

1-,

(iii)PJ=~

(il')

P'

(G,-:)t,·

=

(Ans. (;1)]

I

(h) Which of the (o llowing S1;lICl11cnls is wrong '!

(n The void rml0 of u snlunucd soil can ~ determined from its wmer COntent. (il) The dry density is 1thc bu lk density of soil in dried condition. (iii ) 100% .5>iltumtioo linc lind zero percent air void lines are identicaL

IAns.(ii»)

C. Multiple-Choice Questions 1. TIle waler L-on lenl of ;\ highly organic soil i~ dctcrmmed in tin o~'e n III II temperature of: (ti) lOSoC (b) 800C Ce) 60 0 e (dJ 27°C 2. Pycnometer method I'M water conte nt dclCmlin:llion i~ more suitan le for: ((I)

Clny

(b)

Ie) Sand

Loess

(If) Silt

3. The gas formed by lhe rem,lion 01' calcium carbide with water is: (a) Carboy dhlXldc (b) Sulphur dioxide (e) Ethane (dJ Acetylene 4. The rmin of the volume til' voids to the total volume of soil is: (a) Voids r.ltlO (b) Degree of saturlllion (e) Ai r content «(I) Porosity 5. Dry density of soil is equal to the: (lI) Mass of solids to Ihe volume of solids. (h) Mass of solids to th e tot al vo lume of soil. Ie) Density of soi l in the dried condition. (tI) No ne of the above. 6. The most accurate method for th e determination of water content in the laboratory is: «(/) Sand hm h method. (b) Oven-dryi ng melh·od. Ie) Pycnometer method. (d) Calcium carbide method. 7. A soil ha~ a bulk. density of 1.80 g}cm"J a~ a ~llter content of 5%. If the void r:llio remai ns constant then the

~:)lk2.:n;:':fr 8.

a

water

L'On!cnt

o f 10%

~~; ~S8

glcm3

(e) 1.82 glcmJ (tl) 1.95 glcm) In a wet soil mas!>, air occupies one·sixth of ils vol ume and WilIer occupies one-third of its volume. The void ratio of the soil is (n) 0.25 (b) 0.50 ~) 1 .5 0

(0)1.00

9. A soil sample has a specific gravity of 2.60 and a void rat,io of 0 .78. The water contenl required to fu lly saturale the soil at that vuid nltio will be ta) 20% ....{-b")30% (el40% (tl) 60% [_I .~~~1~ ~~~~~~~~L~a~

3 Particle Size Analysis 3.1. INTRODUC!lON (u) Engineering Propertles-lhc main engineering properties of soils are penncabilily, comprcs.<;ibility. and shear strength. Pcnncability indicates the facility with which water can flow through soils. It is requiroo for estimation of seepage discharge through earth m~. Compressibility is related with the deformations produced in soils when they are subjected to compressive loads. Compression chanlClCrislics arc required for computation of the settlements of Structures founded on soils. ShC..lf strength of a soil is ils ability to resist Shc.1r stresses. l11c shear strength determines the stability of slopes. bearing capacity of soils and the earth pressure on retaining structures. Engineering properties of soils are discussed in latter Ch..1pICrs. (b) Index Properties-The tests required [or determination of engineering properties arc generally

elaborate and time-consuming. Sometimes, the gcotechnical engineer is interested to h'lve some rough assessment of the enginccring properties without conducting elaborate testS. This is possible if index properties are determined. The properties of soils which are not of primary interest to the geotechnical engineer but which are indicative of the engineering properties are caned index properties. Simple tests which are required to determine the index properties are known as classification tests. The soils arc cJ:tSSified and identified based on the index properties. as discussed in Chapter 5. The main index properties of coarsegrained soils ace panicle size and the relative density. which are described in this chapter. for finc-grained soils, the main index propcnics are Ancrberg's limits and the consistency (chapter 4). The index properties arc sometimes divided into two categories. (I) Properties of individual particles. and (2) Properties of the soil mass. also known as aggregate properties. The properties of individual particles can be dctennined from a remouldcd. disturbed sample. These depend upon the individu.,l grains and are independent of the manner of soil formation. 1llc soil aggregate properties depend upon the mode of soil fonnmion, soil history and soil structure. lbese properties should be determined from undisturbed samples or preferably from in-situ tests. lbe most important properties of the individual particles of coarse- grained soils arc the particle size distribution and grain shape. The aggregate property of the coarse-grained soils of great prnctical importance is its relative density. lbe index properties give some infonnation about the engineering properties. It is IaciUy assumed that soils with like index properties have identical engineering properties. However, the correlation between index properties and engineering properties is not perfe,,;. A liberal factor of safety should be provided if the design is b.ascd only on index properties. Ocsign of large. imponant struau[CS should be done only aRer ddenninalion of engineering properties. 3.2. MECHANICAL ANALYSIS The mechanic.1i analysis. also known as par/icle size annfysis, is a method of scp.1ralion of soils into different fr.lctions b.1SCd on the panicle size. It expresses quantitatively the proportion". by mass. of various sizes of particles present in :l soil. It is shown grtlphically on (I p
SOIL MECIIANICS AND fOUNDATION ENGINEERING

The mechanical analysis is done in two stagcs : (1) Sieve Analysis. (2) Sedimentation Analysis. 1nc first analysis is meant for coarse-grained soils (particle Si7.c greater Ulan 75 micron) which can easily pass through a set of sievcs. 'Ine second analysis is used for fine-grained soils (size smaller than 75 microns). Sedimentation analysis is also known £IS wet lJJJQlysis. As a soil mass may contain the pm1iC\cs of both types of soils, a combined analysis comprising both sieve analysis and sedimentation analysis may be required for such soils. Particle size smaller than 0.2 micron cannot be determined by the sedimentation method. These can be determined by an electron microscope or by X-ray diffraction techniques. However, such analysis is of lillie practical importance in soil engineering. 3.3. SIEVE ANALYSIS lbe soil is sieved through a sct of sieves. Sieves are generally made of spun brass and phosphor bronz (or stainless steel) sieve clolh. According to IS : 1498--1970. the sieves are designated by the size of square 3 opening, in mm or microns (1 micron 10-6 m 10- mm). Sieves of various sizes ranging from 80 mm to 75 microns arc available. '(he diameter of the sieve is generally between 1510 20 em. As mentioned before, the sieve analysis is done for coarse-grained soils. 1nc coarse-grained soils can be further sub-divided into gravel fmction (sizc > 4.75 mm) and sand fraction (751' < size < 4.75 mm), where Greek leiter I' is used to represent microo. A set of coarse sieves, consisting of the sieves of size 80 mm, 40 mm, 20 mm, 10 mm and 4.75 mm, is required for the gravel fmction. 'Ille second set of sieves, ronsisting of the sieves of size 2 mm, I mm, 600 ",. 415 1',212 ",. ISO I' and 75 "', is used for sieving minus 4.75 mm fraction. However. all the sieves may not be required for a particular soil. The selection of the required number of sieves is done to obtain a good particle size distribution curve. The sieves are stacked one over the other, with decreasing size from the top to the bottom. Thus the sieve of the largest opening is kept at the top. A lid or co..-er is placed at the top of the largest sieve. A receiver, known as pan, which has no opening, is placed at the bottom of the smallest sieve. (a) Dry Sieve Analysis-The soil sample is taken in suitable quantity. as given in Table 3.1, The larger the particle size, the greater is the quantity of soil required. The soil should be oven-dry. It should not contain any lump. If necessary, it should be pulverized. If the soil contains organic matter, it can be taken air-dry inste..'1d of oven dry. The sample is sieved through a 4.75 mm [S sieve. loe portion retained on the sieve is the gravel fraction or plus 4.75 mm material. The gravel fraction is sieved through the set of coarse sieves manually or using a mechanical shaker. Hand sieving is nonnally done. The weight of soil retained on each sieve is obtained. 2·0mm The minus 4.75 mm fraction is sieved through the set of fine sieves. '·Omm The sample is placed in the top sieve and the set of sieves is kept on a mechanical shaker (Fig. 3.1) and the machine is started. Nonnally, 10 GOOr minutes of shaking is sufficient for most soils. The mass of soil retained (. 25,.. on each sieve and on pan is obtained to the nearest 0.1 gm. The mass of the retained soil is checked against the original mass. 212 rDry sieve analysis is suitable for c:ohesionlcss soils, with little or no fines. If the sand is sieved in wet conditions. the surface tension may 150rcause a slight increase in the size of the particles and the particles smaller 7S ~ than the aperture size may be retained on the sieve and. the results would be crroneol.1';. Pan Thble 3.1. Quantity of Soil for Sieve Anulysls

=

=

Maximum Size

Quall/ily (kg)

SOmm ZOmrn

60 6.5 0.5

4.75 mm

lSi ...

,ha'"

I

Fig. 3.1. StackingoC Sieves.

47

PARnCLE S17.£ ANAlYSIS

(b) Wet Sieve Anulysis-Ir the soil contains a substuntial quantity (say. more than 5%) of fine particles, a wet sieve analysis is required. All lumps arc broken into individual purticlcs. A representative soil sample in the required quantity is taken, using a rimer. and dried in an oven. Tbe dried sample is taken in a tray and soaked with water. If denocculalion is required. sodium hex.:,meta-phosphate, at the mte of 2 g per litre of water, is added. lbc sample is stirred and left for a soaking period of at leas( one hour. '!be slurry is then sieved through a 4.75 mm IS sieve, and washed with a jet of water. 1lle material retained on the sieve is the gravel fraction. It is dried in an oven, and sieved through SCI of ~ sieves. 'llie material passing through 4.75 mm !iieve is sieved through a 75 1.1. sieve. The material is washed until tile wash water becomes clear. 'Ibe material retained on the 75 1.1. sieve is collected and dried in an oven. It is then sieved through the sel of fine sieves of the size 2 mm, 1 mm, 600 1.1., 425 1.1., 212 ~ 150 lA, and 75 IA. The material retained on each sieve is oollCCled and weighed. The material that would have been retained on pan is equal to the tOlal mass of soil minus the sum of the masses of material retained on all sieves.

Computation of I'ercentage Finer For determination of the p.orticle si ...c distribution (:urve, percentage of particles finer than a p..or1icular size is required. This om be found as under: Let us consider the case when the sieving has been done through seven sieves, no I (coarsest) to no. 7 (fincst). Let the mass of soil retained on the....e sievC-I; be respectively. M I , M2 ... ,M7 , and the mass of soil retained on the JXln (receiver) be Mil' The sum of all these masses is, obviously. equal to the tottll mass of samplcM. Eltprcsscd as percentage. the materials retained on the sieves and pan are

and

PI -

~

P1"

~-

)( 100

and

100

P2 -

o/J )(

100, etc.

Pa ..

o/J )(

100

The cumulative percentage (q of material retained on any sieve is equal to the sum of the percentage of soil retained on the sieve and that retained on aU sieves coarser than that sieve. Therefore, C1

-

PI

C2 .. PI + P2 C, - PI + P2 + ... + P1 The percentage fmcr (N) than any sieve size is obtained by subtracting the cumulative percentage retained on the sieve from 100%>.

N2 .. lOO-Cz , etc. lbus, N t - lOO-C\; N, _ 100-C, and It may be noted that the dimension of the soil particle that controls whether a particle shall pass through 3 sieve opening is the intermediate dimension (width) of the particle.For eltample, a particle with dimensions 3 mm )( 2 mm )( I mm shall pass through a sieve of size 2 mm if il is assumed that the particle is aligned such that the largest dimension is oormal to the plane of sieve opening and is at right angles to the side of the square. (See Chapter 30, Sect. 30.8 for the laboratory eltperiment) 3.4. STOKES' LAW

Soil particles finer than 75 1.1. size cannot be sieved. The particle size distribution of such soils is detennined by sedimentation analysis. The analysis is based on Stokes' law, which gives the terminal velocily of a small sphere settling in a fluid of infinite elttenl. When a small sphere sculcs in a Ouid, its velocity firs! increases under the aaion of gravity, but the drag force oomes into action, and retards the velocity. After an initial adjustment period, steady conditions are attained and the velocity beoomes oonstant. The velocity


48

attained is known as terminal velocity. The expression for leonina! velocity can be obtained from the equilibrium of the particle. The drag force, F D • experienced by a sphere of radius r when it falls through a fluid of viscosity" is

given by ... (a) where v is the velocity. The other two forces acting on the sphere arc the weight (W) of the sphere and the buoyant force (U).

.?

W .4/3

y, • 4/3

.?

(p,g)

... (b)

'?(P.g)

... (e)

where 1, is the unit weight of the material of sphere

and U. 4/3 .? y.' 4/3 From equilibrium of [orces in vertical direction.

W .. U + PD 4/3lt?-y... 4/31try ... + 6 llTlrv 4/31t,3 gp, .. 4/31t,}gp ... + 61tTJTV

2 V-

9

,>

":;:J(p,-p",)g

, • .l...

gd'(G-I)p. . .. (3.1) 18 ~ where D is the diameter of the sphere, G is the specific gravity of the material of sphere, and g is the ;)cceieration due to gravity. If a spherical particle falls Ihrough a height Ht! centimeters in t minutes,

v .. He an/sec 60,

.. .(3.2)

From Eqs. (3.1) and (3.2), 11,

I

gd'(G-I)P.

60t - 18 --~--

D-V

. .. (3.3)

0.3'l xlie g(G-I)p.

x,

... [3.4(a)J

D_M-{if;., where M is a facto" equal to

... [3.4(b)J

[g (~'=r) P.]"

in which 11 is the viscosity in poise (dyne- sec/em1- g _ 981 em/sc2, and p.., is in gm/ml. D is in cenlimeters. Table 3.2 gives the values of the rocfficienl of viscosity 'l for water at different temperatures. The values of the factor M can be computed and Ulbulated for different temperatures. For example, for G = 2.67 and T:: 20°C, and taking p.., z:: 1.0 gm/ml, and 11 .. 10.09 X 10-3 poise. g .. 981 cro/see-2, M = [0.3 x 10.09 x IO-J]'h = 136 981 x 1.67 x 1.0 .

X

JO-J

An approximate expression for diameter D of the panicle can be obtained from Eq. 3.1.'

PARTICLE ·SIZE ANALYSIS

,.c 0 1 2 3 4 5 6 7 8 9

49

lhble 3.2 Coefficienl of viscoslly (Vulues in millipol-.e)

"

17.94 17.32 16.74 16.19 15.68 15.19 14.73 14.29 13.87 13.48

,·C

,.C

"

13.10 12.74 12.39 12.06 11.75 11.45 11.16 10.88

10 11 12 13 14 15

16 17 18 19

'" 21 22 23

24

25 26 1:1 28 29

10.60

1034

or water TJ

,.C

10.09

,.

9.84 9.61 9.38 9.16 8.95 8.75 855 8.36 8.18

31 32 33 3. 35 36 37 38 39

"

"

8.00

7.83 7.67 7.51 7.36 7.21 7.06 6.92 6.79

6."

2

\Note. 1 millipoise = 0.1 mN-stm ] v _

l... 18

981 x rr(2.67-1.0) x 1.00 10.09 x Wool

rr

or v • 9020 where v is the velocity in em/sec and D is the d iameter in em . If v is expressed in mm/sec and D in mm, v _ 902d

... [3.5(a)]

... [3.5(b)]

If v is expressed in em/sec and D in mm,

... [3.5(c)] v _ 90.2 d Table 3.3 gives the lime required for the scUlcmenl of ~rticles of different sizes through a height of 100 mm. Thble 3.3. nme of Settlemenl for 100 mm lIeight S.No.

Diameter (mm)

Time

1. 2. 3.

0.075 0.02

19.72 sec

.. 5.

3.s. PREPARATION OF SUSI'ENSION

0.006

0.002 0.001

4.62 min 51.36 mm 7.70 hr JO.81 hr

FOR SEDIMENTATION ANALYSIS

About 50 g of oven-dried soil is weighed accurately and transferred to an evaporating dish. Th have proper dispersion of soil, about 100 ml of a dispersion solution is added to the evaporating dish to covcr the soil. IS ; 2nD-Part IV recommends the use of dispersion solution obtained after adding 33g of sodium hex am eta-phosphate and 7g of sodium carbonate to distilled water to make one litre of solution. After the dispersing solution has been added 10 soil, the mixlure is wanned gently for about 10 minutes. The contents of the evaporating dish are then transfcrred to the cup of a mochanical stirrer. Distilled water is added to make the cup about three-fourth full. TIle suspension is stirred for about 15 minutes. However, the stirring period is more for clayey soils. The suspei'!Sion is then washed through a 75 \.l sieve, using jets of distilled water. The portion of the suspension which has passed through the sieve is used for sedimentation analysis. 'The specimen is washed into a 1000 ml jar and enough watcr is added to make 1000 ml of suspension. If the soil cont~ins organic mallcr and calcium compounds, il should be pretreated before adding the dispersing agent. This is done is two stnges. (1) 1bc soil is taken in a beaker and first treated with a 20 volume hydrogen peroxide solution to remove the organic matter, at tbe rate of about 100 ml of hydrogen peroxldc for 100 gm of soil. The mixture is

so


wanned to a temperature nor. exceeding 60°C. Hydrogen peroxide causes oxidation of organic maHer and gas is Ubernled. When no more gas comes out. the mixture is boiled to decompose the remaining hydrogen peroxide. The mixture is then cooled. (2) Calcium compounds in the soil arc removed by adding 0.2 N hydrochloric acid at the rate of 100 ml for every 100 g of soil. When the reaction is oomplete, the mixture is filtered. The filtrate is washed with distilled water until it is free from the acid. The damp soil on the filler is placed in a evaporating dish and dried in an oven to constant mass.

3.6. THEORY OF SEDIMENTATION AI the commenIXmenl of the sedimentation, the soil particles arc unifoonly dispersed throughout the th~ same at all depths. After a lime period, at a particular depth, only those particles remain which have nol settled. Because all particles of the same size have the same velocity, the particles of a given size, if they exist at any level, are in the same concentration as at the beginning of sedimentation. In other words., all particles smaller than a particular size D will be present at a depth 1I~ in the same degree of concentration as at the beginning. All panicles larger than the size D would have settled below that depth. For illustration, let us assume that the soil is composed of particles of only three sizes, which have terminal velocities in the ratio of 1:2:3. The three types of panicles. two at each level. are shown in the kfi

suspenSion, and the concentration of particles of different sizes is

Level A A

T h

Level B B

t

level C- C

t

.1. 00

S

2.

10

Jb

~~

13

14

15

7

2

~.

0 0

h

Levctl 00

+ h

Levilli E E

1.

V3 = 3 V1 V2 = 2 V,

(.)

5

7

00

10.J3.

S'g~g 6:9~~:5

(b)

Fig. 3.2. Settlement of particles.

column, middle column and the right column in Fig., 3.2 (a). At the beginning of the sedimentation, the concentration of particles is the same at all levels. After some time, the particles take the position as shown in Fig. 3.2 (b). The particles of the smallest size have settled to a depth h, those of the intermediate size and the largcst size to 2h and 3h, respectively. At lever B-B, only the particles of the smallest size exist, and the concenlratjon of these particles is the same as at the beginning, viz. 2 particles. At level C-C, the concentration of the particles of the smallest and intermediate sizes is the same as at beginning. Likewise, at level D-D, the particles of all the three sizes exist with the same concentration. If mD is the mass of parCdes per ml of.suspcnsion at depth fie after time t, and m, is the mass of partida. per ml of suspension at the beginning of sedimentation. the percentage finer than the size D is given by

N. !!'!.Q x 100

m.

The particle size D is detennined using Eq. 3.4(a).

... (3.6)

PARTICLE SIZE ANALYSIS

51

3.7. PIPETI'E ME11IOD In this method, 500 ml of soil suspension is required. The procedure for preparation of 1000 ml of suspension has been discussed in Sect. 3.5. All the quantities required for 1000 ml of suspension are halved to get a 500 ml of suspension. The suspension is taken in a sedimentation tube. Fig. 3 .3 shows a 10 '" _----<--+-- Bulb funnel ml capacity pipette used for extnlction of the (distilled water) . sample. The pipette is fitted with a suction inlet. Suction---,_,-,,~,,"

(a) Calibration or Pipette For delenninalion of the volume of pipette. it is calibrated before usc. For calibration. the ncyale Wash ouHet of the pipette is immersed in distilled water. The stop cock: S is closed. The three-way stop cock T is opened, and the water is sucked up into the pipette until it rises in safety bulb. The stop cock T is then Pipette closed and the pipette is taken oul. The stop cock T is now turned the other way round to connect it to the wash outlet to drain the excess water from the safety bulb. The stop cock is the:) turned thc other way round to discharge the water comained in the pipette into a glass weighing bottle. The mass of water in the botUe in grams is equal to the volume of the pipette in ml.

Safety bulb

Stop cock

Scale

Sliding carriage

(b) Sedimentation Thst

The sedimentation tube containing the suspension is placed in a constanHempcrature bath Sedimentation tube at 2rC for about one hour. The lube is then removed from the bath, a rubber bung is placed on its top to close the mouth. 'Ibe tube is inverted endover-end a number of times to cause thorough mixing. The bung is removed and the tube is again Fig. 33. Pipette Method placed in the constant temperature bath kept just below the tip of the pipette. The instant when the tube is placed in the constant temperature bath is taken as the beginning of the sedimentation. 1be slap watch is slarted to record the time. The pipette is gradually lowered into the suspension in thc sedimentation tube. The samples are taken from a depth of 100 mm below the surface of thc suspension. The first sample is taken after 2 minutes of the start of sedimentation. The pipette is lowered slowly about 20 seconds before the sample is due to be taken. More samples are taken after 4, 8, 15 and 30 minutes, and 1, 2, 4, 8 and 24 hours. Exact time at which the sample is taken is noted from the stop watch. The procedure for taking samples is as follows. The stop cock Tis opene..::l and tbe suspension is drawn into the pipette until it is full of suspension. The time taken for actual sampling is about 20 seconds. The stop cock T is then closed, and any surplus suspension dawn up in the safety bulb is drained away through the wash outlet. The safety bulb is flushed out with distilled water stored in the bulb funnel. The stop cock is again turned and the soil sample in the pipette is transferred to a weighing bottlc. Distilled water is used to tmnsfer any solid particles adhering to the inside. The samples taken are dried in an ovcn at 105-110°C for 24 hours to obtain the mass of solids per ml. As the solids also contain dispersing agent. a correction in the mass of solids is required. If m is the ma5S of dispersing agent per ml of suspension, tbe weight of soil solids ml is given by

per

... (3.7)


52

where "'n'

= mass of solids/ml as

obtained from the sample. /liD = actual mass of soil solids/ml. '[he percentage finer than 'any size D can be obtained using Eq. 3.6.

Merits and Demerits of the PipeUe Method The pipette method is a standard laboratory method for the particle size analysis of finc-grained soils. It is a very accurate method. However, the apparatus is quite delicate and expensive. It requires a very sensitive ~c~::g:n~~'::~tor quick particle size analysis, the hydrometer method, described in the following section,

3.8. liYDROMETER METllOD A hydrometer is an instrument used for the determination of the specific gravity of liquids. As the specific gravity of the soil suspension depends upon the particle Si7.c, a hydrometer can be used for the particle size analysis. A spccialtypc of hydrometer with a long stem (neck) is used. 'Ille stem is marked from lop to bottom, generally in Ihe range of 0.995 to 1.030 (Fig. 3.4). AI the lime of commenement of 0·995

\·000 1.005

I

_

-,

=0 = '5

Stem

B

--- B B

-.l."t1

TA

1 B

H,

H+~

L

~

A

h

Bulb

I

L

TlH TA

1

(0)

Fig. 3.4. Hydrometer Method

(0)

sedimentation, the specific gravity of suspension is uniform at all depths. When the sedimentation takes place, thee larger particles settle more deept:r than the smaller oncs. This results in non-uniform specific gravity of Ihe suspension at different depths. The.lower layers of the suspension have specific gravity greater than thai of the upper layers. Casagrande has shown that the hydrometer measures the specific gravity of suspension at a point indicat~d by the centre of the immersed volume. If the volume of the stcm is neglected. the centre of the immersed volume of the hydrometer is the same as the centre of the bulb. Thus, the hydrometer gives the specific gravity of the suspension at the centre of the bulb.


"

(a) Calibration of hydrometer

To determine the depth al which the specific gravity is measured, calibration of the hydrometer is done. The volume of the hydrometer, V", is fimt determined by immeming it in a graduated cylinder partly filled with water and noting down the volume due to the rise in water level The volume of the hydrometer can also be determined indirectly from its mass. The volume of hydrometer in ml is approximately equal 10 the mass of hydrometer in grdms, assuming that the specifK: gravity of hydrometer is unity. The depth of any layer A·A from the free surface 8-0 is lhe effective depth at which the specific gravity is mca')ured by the hydrometer ((Fig. 3.4 (b)]. As soon as the hydrometer is inserted in the jar, the layer of suspension whieh was at level A· A rises to the level A' -A', and that at level B· B rise to the level B' - B'. TIle effective depth He is given by

Ile -

(II +~) - ~ + ~

... (a)

H = depth from the free surface B' - B' to the lowest mark on the stem, h = height of bulb, V" = volume of hydrometer, A "" cross·sectional area of jar. In Eq. (a), it has been assumed that the rise in suspension level from A -A to A' -A' at the centre of the bulb is cqu.11 to half the total rise due to the volume of the hydrometer. where

Thus

lie .. H +

i(

h -

~)

... (3.8)

lbe markings on the hydrometer stem give the specific gravity of the suspension at the centre of the bulb. The hydrometer readings are recorded after subtracting unity from the value of tlle specific gravity and multiplying the remaining 1 BO digit by 1000. Thus, a specific gravity of 1.015 is represented by a hydrometer 'SO reading of (1.015 - 1.000) x 1000 15. The graduations on the right side 1I. 0

R,.

~~t~: ~~~ :;~t~~: . depth He depends upon the hydrometer reading R", a calibration chart can be obtained between the hydro· meter reading Rh and the elIective depth He. For deter· mination of the effective depth He from Eq. (3.8), an

~ k

:~ ~

120

W '00_1,,5--..,----:,---:',,-0---:''''5---:2'''0-- '''25'---''::-;30 Hydrome.hr re.ading (R h ) fiB· 3.5. Calibration Chart.

accurate scale is used to determine the height h and the depth H to various graduations. Fig. 3.5 shows a typical calibration charI. As the sedimentation progresses, the specific gravity of the suspension decreases and the hydrometer goes deeper and deeper, and the effective depth increases. The hydrometer reading of course, decreases (Fig. 3.6).

R".

(b) Test Procedure

Exactly 1000 ml of suspension is prepared as explained in Sect. 3.5. After stirring, the suspension is washed into a 1000 ml'jar and willer is added 10 il to bring the level to 100(} ml mark. 1bc suspension is


54

mixed thoroughly by placing a bung (or the palm of a hand) on the open end of the

r

jar and turning it upside down and back a few times. The jar is then placed on a table, and a stop walch is started. The hydrometer is inserted in the suspension and the first reading is laken aOcr minute of the

r

t

commencement of the sedimentation. Further readings

are laken after one minute, two minutes and four minutes of the commencement of tbe sedimentation. The hydrometer is then

t: 12 (b) t4

>3>t:z>t,

t = 13 (el

Fig. 3.6. Downward movemenT or hydromeTer. removed from the jar and rinsed with distilled water and floated in a comparison cylinder oonlaining distilled water with the dispersing agent added to the same concentration a<> in the soil suspension. Further readings are taken after 8, 15 and 30 minutes and t, 2, 4, 8 and 24 hours reckoned from the beginning of sedimentation. For each of these reading'>, the hydrometer is inserted about 20 seconds before the reading. lbc hydrometer is taken out after the reading and floated in the comparison cylinder.

(c) Corrections of Hydrometer Reading

The hydrometer reading;; are corrected as under: (I) Meniscus correclion--Sincc the suspension is opaque, the observations are taken at the top of the meniscus. The meniscus correction is equal to the reading between the top of the meniscus and the level of the suspension. As the marking on the stem increases downward, the COfT'Cdion is positive. The meniscus correction (elll) is determined from the reading;; at the top and bottom of meniscus in the comparison cylinder. The meniscus correction is constant for a hydrometer. If Rio' is the hydrometer reading of the suspension at a partio.l1ar time. corrected hydrometer Rio reading is given by

the

Rio - Rio' + C", .. .(3.9) The corrected hydrometer reading (Rio) is required. for determining the effective depth from the calibration chart (Fig. 3.5). (it) Temperature correction-The hydrometer is generally calibrated at 27°C. If the tempernture of the suspension is different from 2rc. a temperature correction (Cf ) is required for the hydrometer reading. IT the temperature is more than 27°C. the suspension is lighter. and tbe actual reading will be less than the corrected reading. The temperature correction is positive. On the other hand, if the temperature is less than 27°C, the temperature corrcction is negative. The temperature corrcction is obtained from the chans supplied by the manufacturer. (iit) Dispersion agent Con-ection-Addition of the dispersing agent 10 the soil specimen causes aod increase in the specific gravity of the suspension. Therefore. the dispersing agent correction is always negative. The dispersing agent correction (C~ can be determined by noting the hydrometer reading in clear water and again in the same water after adding the dispersing agent. Thus, the corrected reading R can be obtained from the observed reading Rio' as under.


... (3.10)

ComJ:lQ!lite Correction-Inslcad of finding the corrections individually, it is convenienl to find one composile correction. The composite correction (C) is the algebraic sum of all the corrections. Thus,

n.R,.C

.. .(3.11)

The composite correction is found directly from the readings taken in a comparison cylinder, which has. distilled waler and the dispersing agenl in the same concentrntion. and has the same temperature. As the hydrometer has been calibrated at 27°C to indicate a specific gravity of 1.(X)J, the difference between the reading taken at the top of meniscus and 1.(X)J is in magnitude equal 10 the composite oorrcct,ion. The negative of the hydrometer reading in the comparison cylinder is equal LO the composite oorrection. The composile correction can be positive or negative. For example. if the hydrometer reading is +2 (i.e. 1.002), the correction is -2, and if the reading is -3 (Le. 0.997), the rorrection is +3. The composite correction is found before the start of the test and at every 30 minute interval.

3.9. RELATION BETWEEN PERCENTAGE FINER AND IIYDROMETER READING The corrected hydrometer reading R can be related to the percentage finer N than any size D as under: Let M$ be the mass of dry soil in a sLl'>pension of volume V. At the commencement of the sedimentation, the soil-water suspension is uniform, and. therefore, the mass of solids per unit voluQe of suspension at any depth is M,!V. The initial density of suspension is given by M$ + mass of water in suspension p;' V

or

p,'.

~

+ massofwater!volumeofsuspension.

. .. (a)

The mass of water per unit volume of suspension can be detennined from the volume of water per unit volume of suspension. as explained below. Mass of solids/volume of suspension

Volume of solidslvolume of suspension

M,

·v M,

• V(G P.)

Volume of water/Volume of suspension

_l_~

Mass of water/volume of suspension

• [I -

From Eq. (a),

V(Gp.)

M,

Pi - V + -P ...

or

V(~'P.) 1P. [

M,

I - V(G P...)

M, +-y

(1-

1P...

I)

0

(G-I)

M, p;.p..,+-y----c;-

... (3.12)

If MD is tbe mass of solids in volume V at that depth after time t, Eq. 3.12 gives the density of suspcru;ioo at that depth as

MD po. P... + V

(G-I) ----a-

... (3,13)

,.


From Eq. 3.6, the percentage liner N than any size is given by

N- !!!Q)( 100 Ills

N ' m,

"'0 - """"iOO where /liD" MolV and III, . AI/ V lbereforc, Eq. 3.13 becomes

p_p~,+~~(G~l)

... (3.14)

P-Pw" ~; (G~l) N .

(~) G- I

As the hydrometer reading R is cqUll1 to (P - P..,)

As Ills- AI/ V,

It

(p - Pw) x 100

... (3.15)

Ills

1000. Eq. 3.15 can

be

written as

N _

(~) . ~

It

J...

x 100

... (3.16)

N .

(~) . ~

x

~

x 100

... (3.17(Q)J

G- 1 G- 1

1000 1000

N.. (G~ t) . k

x

m,

M,

100

... (3.17(b)J

where M, is the mass of the solids in a volume Vof 1000 ml. The particle size D is determined using Eq. 3.4, laking the value of efTeaive depth He from the calibration curve for the hydrometer reading RI!' (See Chapter 30, Sect. 30-9 for Ihe laboratory experiment) 3.10. LIMITATION OF SEDIMENTATION ANALYSIS The sedimentation analysis docs not give correct values of the particle size and the percentage [iner due to the following limitations. (I) The sedimentation analysis gives the panicle size in terms of equivalent diameter. which is less than the particle size given by sieve analysis. The soil particles arc not spherical. The equivalent diameter is close to the lhickness (smallest dimension) rather than the length or width. (The equivalent diameter is the diameter of the sphere which falL.. with the same velocity as the actual particle.) (2) As the specific gravity of solids for different panicles is different. the use of an average value of G in Eq. 3.17 (b) is a source of error. However, as the variation of the values of G is small, the erroe is negligible. (3) Stokes' law is applicable only when the liquid is infinite. The presence of walls of the jar affects the results to some extent. (4) In Stokes' law. it has been assumed that only one sphere settles. nnd there is no interference from other spheres. In the scdimentntipn an<11ysis. as many panicles sellie simultaneously. there is some interference. However. the effca of errors mentioned in paras (3) and (4) is negligibly small if the mass of dry soil used per 1000 ml of suspension is not more lhan 5Og. (5) The sedimentation analysis cannot be used for particlt:S larger than 0.2 mm as turbulent conditions develop and Stokes' law is not applicable.


"

(6) TIle sedimentation method is not applicable for particles smaller than O.2~ because Brownian movement takes place and the particles do not setUe as per Stokes' law. (7) The sedimentalion method cannot be used for chalky soils, because of the removal of the calcium carbonate of chalky soils in the pretre.1tement by hydrochloric acid. Despite above limitations,the sedimentation analysis is used for detennination of the particle size" of fine-grained soils. '[be particle sizes of such soils is not of much practical significance and, therefore. even approximate analysis is good enough. The index properties of such soils are plac;.ticity characteristics and not the panicle size. The main use of the sedimentation analysis is to detennine the clay content (particles less than 2 f.1 size) in a soil mass. 3.11. COMBINED SIEVE AND SEDIMENTATION ANALYSIS

It the soil mass consists of particles of both coarse-grained and fioe-grained soils. a combined analysis is done. The slurry of the soil is made as mentioned in the wet sieve analysis (Sect. 33). 1be slurry is sieved through a 4.75 mm IS sieve. The material retained on the sieve is oven-dried and a coarse-sieve analysis is done. The material retained on a 75 fA. IS sieve is also oven-dried and the sieve is analysis is done using the set of fine sieves. The suspension passing the 75 fA. sieve is mixed with a deflocaJlating agent, if oot already done. The hydrometer test is performed on the suspension, as explained in Sect. 3.B. The percentage finer than any size can be calculated on the basis of the original mass of soil taken for the combined analysis. 3.12. PARTICLE SIZE DISTRIBUTION CURVE The particle size distribulion rurve, also known as a gradation curve, represents the distribulion of particles of different sizes in the soil mass. The percentage finer N Ihan a given size is ploUed as ordinate (on natural scale) and Ihe panicle size ac;. abscissa (on log scale). In Fig. 3.7 (0), the particle size decreases from

~100~ ~ 80

~ 60

g

40

~

20

C

&01.0

0.1

0.01

- - Par ticle

0.001

0.0001

size (mm)

(a)

~1DO~ E 80 1l>60 ~ 40 ~

&

20 0 .001 0

0.01

0.10 Particle (b)

1.0

size (mm)-

Fig. 3.7. Pnrlide Size Curve.

16.0


.18

leO 10 right, whereas in Fig. 3.7 (b), the particle size increases from left \0 right. Both the methods are prclevant. The reader should carefully observe the horizontal scale of the particle size distribution curve. In this lexl, tbe particle size distribution is shown as in Fig. 3.7 (b), i.e., the particle sizc i~ from left to , righI, which is also the usual convention. The semi-log plol for the particle size distribution, as shown in Fig. 3.7, has lhe following advantages over nalural plots. (1) The soils of equal uniformity exhibit the same shape, irrespective of the adual particle si1.c. (2) A<; the range of the particle sizes is very large, for better representation. a log scale is required.

Grading of Soils-The distribution of particles of differcOi sizes in a soil mass is called grading. The grading of soils can be determined from the particle size distribution curves. Fig. 3.8 shows the patlicle size distribution curves of different soils. 100

9
~

~Fio,

70

V

rained

"so,0..,.

Am

50

I~

7lJ

/

uop grad

./

/0,

j-cCXJrsegrOined

;f1fo-

.o.61'mm

y.--- L

.I

I I

f

'" I 30

ill I VI II

/

"'c:.!~1l9~

I

60

~O

I

UO

~O.oo'mm __ -f""" A, _______ 0.01

1

0·1

1.0

1M

Porlicle size 0 (mm)-_ Fig. 3.8. Grading 05 Soils.

A curve with a hump, such as curve A, represents the soil in which some of the intermediate size particles are missing. Such a soils is called gap-graded or skip-graded A flat S-oJrve, such as curve B. represents a soil which contains tbe particles of different sizes in good proportion. Such a soil is called a wellvgraded (or uniformly graded) soil. A steep curve, like C, indicates a soil oontaining the particles of almost the same size. Such soils are known as unIform soils. The particle size distribution curve also reveals whether a soil is coarse..grained or fine-grained. ro general, a curve situated higher up and to the left (curve D) indicates a relatively fine-grained soil, whereas a curve situated 10 lhe right (rurve E) indicates a coarse-grained soil. The uniformity of a soil is expressed qualilatively by a term known as uniformity coefficient., Cu. given

by

Dro

C" -

... (3.18)

1>;;:

where D6fJ = particle size such that 60% of the soil is finer than this size, and DIO = particle size such that 10% of the soil is finer than this size. D IO size is also known as the effective size. In Fig. 3.8, Dw and DIO (or the soil B are, respectively. 0.08 m.m and 0.004 mm. Therefore, Cu 0.0810.004 20 The larger the numerical value of Cu. Ihe'more is Ihe range of particles. Soils wilh a value of C u less

=

=

,.


tban 2 are uniform soils. Sands with a value of C" of 6 or . more, are wcll·graded. Gravels with a value of CIl of 4 or more are weU·graded. The general shape of the particle size distribution curve is described by another coefficient lrnown as the coefficient of curvature (Cc) or the coefficient of gradation (Cg ).

(D",l' Cc • D(IJ x DIO

... (3.19)

where D)(J is the particle size corresponding 10 30% finer. For a well-graded soil, the value of the coefficient of curvature lies between 1 and 3. It may be noted that the gap grading of the soil cannot be detected by C" only. The value of C c is also required to detect. it. For the soil shown by curve B in Fig. 3.9. the particle size D:Jo is 0.025 mm. Therefore, Cc

-

o.:·~~~~

.

1.95

3.13. USES OF PARTICLE SIZE DISTRIBUTION CURVE The particle size distribution curve is extremely useful for coarse-grained soils. As the behavior of fine-grained soils (minus 75 IA) depends upon the plasticity characteristic and not on the particle size, its use for fine- grained soils is limited. (1) The particle size dislribulioo curve is used in the cla<>sification of coarse·grained soils (sec chapter 5). (2) The coefficient of permeability of a coarse-grained soil depends to a large extent on the size of the particles. An approximate value of the coefficient of permeabi1ily can be determined from the particle size as discussed in chapter 8. (3) The particle size is used to know the susceptibility of a soil to frost action. (4) The particle size distribution curve is required for the design of drainage filters. (5) The particle size distribution provides an index to the shear strength of the soil. Generally. a well·graded. compacted sand has high shear strength. (6) The compressibility of a soil can also be judged from its particle size distribution curve. A unifonn soil is more compressible than a well-graded soil. (7) The particle size distribution curve is useful in soil stabilisation and for tbe design of pavements. (8) The particle size distribution curve may indicate the mode of depositioo of a soil. For example, a gap·graded soil indicates deposition by two different agencies. (9) The particle size distribution curve of a residual soil may indicate the age of the soil deposit. 'With increasing age, the average particle size deaeases because of weathering. The particle size distribution curve whidl is initially wavy becomes smooth and regular with age. 3.14. SHAPE OF PARTICLES The engineering properties of soils, especially coarse.graincd soils, depend upon the shape of particles. As it is more difficult to measure the shape lhan the size. the shape of the particles does not get the required attention. When the length, width and thickness of the particles are of same order of magnitude, the particles are known to have a bulky Shape. Cohesionless soils have bulky particles. As sta~ed in chapter 1, bulky particles are formed by physical disintegration of rocks. Rock flour, which has the size of the particles in the range of fme-grained soils, behaves like rohesionless soils because its particles are bulky. Soils rontaining bulky grains behave like a heap of loose bricks or broken stone pieres. Such soils can suppoc1 heavy. loads in static conditions. However, when vibration I.!lkes place, large settlements can occur. Cohesive, clayey soils have particles which are thin and flaky, like 8 sheet of paper. Soils composed of flaky


60

particles arc highly compressible. These soils deform easily under SIHtic lo.'K1s, like dry leaves or loose papers in a b~kcl subjected to a pressure. However, such soils arc relatively morc stable when subjected to vibrations. The shape of tbe coarse-grained soils can be described in terms of sphericity, flatness or angularity. Sphericity (S) of the particle is defined as S. D,IL wbere D.. is equivalent diameter of tbe particle assuming It to be a sphere, given by D.. - (6V/a)Vl, where V is the volume of the particle and L is the length of the particle. The particles with a high value of sphericity (more roundness) are easy [0 manipulate in construction and their tendency to fracture is low. Flatness (/') and elongation (E) are defined as as F-BIT and E-LI8 where L. Band T are. respectively. length. widlh and thickness. The higher the value of the flatness or the elongation. the morc is the tendency of the soil to fracture. loe angularity (R) of a particle is defined as R.. average radius of comers and edges radius of maximum inscribed circle Depending upon angUlarity. the panicles are qualitative ly divided into 5 shapes (Fig. 3.9).

00000 AnguLar

Subangular

Subrounded

Fi.g 3.9. Different shapes of

Rounded

Will[ rounded

p~rtidC5

The angularity of particles has great influence on the behavior of marse-grained soils. The particles with a high value of angularity lend to resist the displacement, but have more tendency for fracturing. On the o ther hand, the particles with low angularity (more roundness) do not crush easily under loads. but have low resistance to displacements as they have a tendency to roll. In general. the angular particles have good engineering properties, such as shear strength.

3.15. RELATIVE DENSllY The most important index aggregate propeny of a cohesionlcss soil is iLS relative density. 1lle engineering properties of a mass of cohesionless soil depend to a large extenl on its relative density (D,). also known as

density index (Iv). The relative density is defined

liS

... (3.20) where emu = maximum void ratio of the soil in the loosest condition. emin = minimum void ratio of the soil in the densest condition. e = void ratio in the naturaL state. The relative density of 3 soil gives a more clear idea of the denseness than does the void ratio. 1Wo types of sands having the same void ratio may have entirely different state of denseness and engineering properties. However, if the two sands have the same relative density. they usually behave in identical manner. . 11lC relative density of a soil indicates how it woukl behave under loads. If the deposit is dense, it can take heavy loads with very little settlements. Depending upon the relative density, the soils are generally divided into 5 categories (Thble 3.3).

PARl'ICLE SIZE ANALYSIS

61

Table 3.3. Denseness or Soils Very

Dellselless

Dense 85

Dr(%)

[0

100

3.16. DETERMINATION OF RElATIVE DENSITY Fig;. 3.10 (0). (b) and (c) show the soil in the densest, natural and loosest statcs_ As it is difJicull to measure the void ratio directly, Eq. 3.20 cannot be used. However, it is oonvenicnt to express the void ratio in terms of dry density (p.,).

(e)

( b)

(a)

Fig. 3.10

e.Gpw_1 Pd Representing the dry density in the loosest, densest and natural oon
GP __ ) _ (GP __ ) I I ( Pm,n Pd D,- (GP__ ) _ (GP _ _ ) Pm;n

I

Pmruc

I

D,- Pmu ( P.-Pm;, ) ... (3.21) Pd Pm:", - Pm;n Eq.3.21 is used to determine the relative density of an in-silu deposit. The methods for the determination of the dry density (Pd) have been discussed in ch:tpter 2. The methods for the determination of p""", and PlI'Iin are described below. For determination of the minimum dry density. a representative, oven dry sample of soil is taken. The sample is then pulverised and sieved through the required sieve. The minimum dry density is found by pouring the dry soil in a mould using a pouring device (IS : 272Q-Pal1 XlV). The spout of the pouring device is so adjusted that the height of free fall is always 25mm. lbe rna<;S and the volume of the soil deposited are found, and the dry density of the soil is determined as under. Mmln

PlIlin·

where

Mmln

V",

-v;:-

... (3.22)

= mass of dry soil.

= volume of soil deposited

in the mould.


62

The maximum dry density is detennined either by the dry method or the weI method. In the dry method, the mould is filled with thoroughly mixed oven-dry soil. A surcharge load is placed on the soil surface, and the mould is fixed to a vibrntor deck. The specimen is vibrated for 8 minutes. 'Ibe mass and volume of the soil in the compacted state are found. The m3:ltlmum dry density is given by

Pmu-

M.~ --v;:-

where Mm11% = mass of dry soil and V... = volume of mould. The maximum dry density of a soil can be determined also in the saturated state. In this method. the mould is filled with wet soil and water is added till a small quantity of free water accumulates on the free surface of the soil. During and just after filling the mould. vibrntion is done for a total of six minutes. Water appearing on the surface of soil is removed. A surcharge mass is placed on the soil and the mould is vibrated again for 8 minutes. The volume (\1,;.) of the soil is determined. Ibe mass Mmnx of the soil is determined after oven drying the sample. Note. If the sand is vibrated under more severe conditions, it may have a relative density of more than 100%. ILLUSTRATIVE EXAMPLES

lliustrallve Example 3.1. The results lOtal

0/ a

sieve analysis

0/ a

soil are given below:

mass 0/ sanlple = 900 gm. Pan

IS Sieve

Mnssofsoil

75

retained (gm)

Draw the partick size distribution cun.oe and hence determine the uniformity coefficient and lhe coeffICient 0/ curvature. Solution. The calQJlatioos for percentage finer N than different sizes are shown (fable 3.1).

Tuble E-3.1 IS S;""

M~,

retained

Percenlage retailled

-~xlOO (1)

(2)

20mm 10 4.75 20 1.0 0.6 425 • 2121' 150 •

3S 10" 40

75.

Pan 1: _ 9OO.0gm

80 ISO ISO 140 115 55 3S 25 7S

(3) 3.89 4.44 &89 16.67 16.67 15.56 12.78 6.11 3.89 278 &32

shown in Fig. E-3.1. From plot, DfIJ" 1.55 mm; D3iJ = 0.53 mm; 0'0 = 0.115 mm

The particle size distribution ruNe is

Cumu/QJive perr:nuage mained

Pereenlage Finer(N) = 100 - (4)

(4)

(5)

3.89 833

96.11 91.67 82.78

17.22 33.89 5056 66.12 78.90

BS-.01 88.90 91.68 100.00

66.11 49.44 33.88 21.10 14.99 11.10 &32


63

"Xl

~~----~----~--~~~--~

/1

7O~----------+------------+--__L-;'-----+______~~ r

60

---------

---------

1:1

3 ~~----------+_----------~/~I--------_+------~~ ~

f

~~--------~--------~+-+--------+------~

3

-~-------- - - - - - - --II---+---L-----+----+---1

J1

~

,

I

I

gk,,~________~~~,L'O~~~.O~.'~"~mm~-L~~o~:~~:fu~IO~60~·~,·~"~m~m--~,O~.O------~,"~.O~ Fig. E-3.1

C _

From Eq. 3.19,

~i2l!

Porticle

e....

From &j. 3.18

~

D60 D

10

0.115 .. 13.48

(D,,)2

Doo

(mm) - - - - _

1.55

..

X

D 10

_ ~ _ 1.58 1.55 X 0.115

Illustrative Example 3.2. The following observations we~ IiJJren during a pipette analysis for the determination of particle size distribution of a soil sample. (a) Depth below the water surface at which the sample was taJcen = 100 mm (b) Capacity of pipette = 10 ml (c) Mass of sample when dried = 03 gm Cd) Tune of talcing sample = 7 minutes after tM start. (e) \-illume of soil suspension in the sedimentation tube = 500 mi. if) Dry mass of soil used in making suspension = 25 gm. Determine the e'"IOrdinate of the point on the particle size distribution curve corresponding to above ooservlUWns. 2.70 and TJ 10.09 miIlipoise. p ... = 1 gmlml Take G

=

=

Solution. From Eq. 3.4 (0),

From Eq. 3.6,

D ..

Yg(G - I)p.

D ..

YO.30981x x10.09 x 10-3 x 10 .. 000161 em (2.70-1) x 7 .

N_

~

0.30 TJ H,

x 100 ..

x I

~%:

x 100-60%

The coordinates of the point on the particie-size distribution curve are (0.0161 mm. 60%). U1ustratlve Example 3.3. A dry sample of mass 50 gm is mixed with distilled water 10 p~pare a

SOlL MECHANICS AND FOUNDATION BNGlNEERING

64

suspension of 1000 ml for hydrometer analysis. The reading of the hydrometer taken after 5 minutes was 25 and the depth of die centre of the bulb below the water surface when the hydrometer was in the jar was 150 "VII. The vollmll! of me hydrometer was 62 1111 and lhe area of cross-section of the jar was 55 cm 2. Assuming G :: 2.68 and'l1 = 9.81 miflipoise, determine lhe coordinates of the point corresponding IfJ above observation. Solullon. 1be depth between levels B' -B' and A' -A' in Fig. 3.4, is given to be 150 mm. The effective deplh between B - B and A - A is given by. 11,,- 15.0-.!f+

.. 15.0 From Eq. 3.4 (a),

D_

2

~- 15.0-~

~2S5

.. 14.43600

YO.3)( g(G-1)r

11 11"

.. YO.3981x 9.81(2.68-1.0) x 10'-3 x 14.436 .. 00023 5 . x

From Eq. 3.17 (a1

N - (G

~ 1)

2.68)

.. ( 1.68

x (

x

x

k )(1~

1000 --.so x

.. 0023 an

•

rnm

) x 100

25

or N- 79.76

1000 )( 100

The coordinates of the point on the particle-size distribution curve are (0.023 mm, 79.76%). lIIustrative Example 3.4. A soil has a dry del1sity 0/1.816 gm/ml in the MturaJ corulition. When 410 gm 0/ the soil was poUTed il1 a vessel in a very loose stale, its volume was 290 mi. The same soil when vibrated and compacted was found to have a volwlle of 215 mI. Determine the relative density. Solution. From Eq. 3.22,

Pmin"

M V. ..

410

From Eq. 3.23,

Pnwc _

M~

~!~

From Eq. 3.21,

m ;"

..

ill

.

1.907 gm/ml

Dr"' p""", ( Pd-Pmin ) Pd

or

290 - 1.414 gm/ml

X

100

Pm;\X - Pmin

.. 1.007 (1.816-1.414) 100 1.816 1.907 -1414 )( ~_~Q.

Illustrative Example 3.5. A test lor the relative density 0/ soil il1 place was performed by digging a small hole in ule soil. The volume of ule hole was 400 ml and ule moist weight O/Ihe excovated soil was 9 N. A/ter oven drying, the weight was 7.8 N. 0/ the dried soil, 4 N was poured into a vessel in a very loose state, and its volume was found to be 270 mi. The same weight 0/ soil when vibrated and tamped had a volume of 200 mI. Determine the relative density. Solution.

w..

9.07~:O&J

Yd-~" (ld)min"

~~

.. 0.1538

O.0195N/ml .. 19.5kN/m3

- 0.0148 N/ml .. 14.8 kN/m

3

.,


(yJIlWI -

~

.. 0.02 N/ml .. 20.0 kN/m

3

From Eq. 3.21, subslituting y for p Ihroughout,

20.0 (195 - 14.8) Dr .. 19.5 20.0- 14.8 x 100 .. 92.70% Dlustratlve Example 3.6. A sample of sand has a volume of 1()(){) ml in its natural state. Its mininwm voIwne when compacted is 840 mi. When gently poured in a measuring cylinde1i its marimum volume is 1370 mi. Determine the relative density. SoluCion. Let M be the dry mass (in gm) of the sample.

'Iberefore,

PrMII ..

D, _

From Eq. 3.21,

::0;

1~0;

Pmin"

P." ltix,

Pm~ ( P.-P"'o ) Pd

Pma:. -Pmln

MI840 (MIIOCJO-MI1370) .. MIlOOO MI840 MI1370 x 100 .. 0.6981 x 100 .. 69.81%.

filuslratlYe: Example 3.7. In order to find the relaJive density of a sand, a mould of volume 1000 ml was used When the sand was dynamically compacted in the mould, its mass was 2.10 kg, whereas when the sand was poured in loosely, its mass was 1.635 kg. If the in·situ density of the soil was 1.50 Mglm J• calculate the relative density. G = 2.70. Assume thot the sand is saturated.

(p_l.... _ 2.1~~ 103

SolutJon.

(Pmin)...... As Pd" 1.50 Mg/m

3

_

1.6J~~ 103

150 glml,

.. 2.1 glm!

.. 1.635 g/ml

e .. G P", _ 1 .. 2.70 _ 1 .. 0.80 Pd 1.5

Now

or ernln .. 0545

likewise

or

1.635 ..

2.70 + .~)

(

~

x 1.0

or emu: ..

From Eq. 3.20

..

11.~~ ~ g~5

x 100 .. n.47%

1.677


66

PROBLEMS A. Numerical 3.1. One kg of soil was sieved through a sel of 8 sieves. with the size 4.75 rom, 2.0 mm, 600 Il, 425 I-t. 300..... 212f.t, ISOlA and 75j.l. The mass of soil retained on these sieves was found to be 50, 78, 90, 150, 160, 132, 148 and 179 gm, respectively. Determine the percentage finer than the corresponding sizes. (AIlS. 995, 87.S , 78.2, 63.2, 47.2, 34.0, 19.2 and 1.]] 3.2. Prove Ihal the particle diameter and the terminal velocity of panicle are related as

v_9020d where

I'

= velocity in an/sec,

D :, diameter in em Oearly stale the various assumptions made. 33. Determine the maximum void rmio for II sand compa;ed of grains of spherical shapes. (1I1nt Consider a cubical box of size al, where d is the diameter of sphere. The nunDer of pm in the box is 81 IAns. 0.911 3.4. The minimum and the maximum dry density of a sand were found to be 1.50 and 1.70 gmlml. CalculDte the dry density corresponding \0 relative densities of 50% and 75%. fAns. 1.594 gmt1; 1.645 gmIm1] 3 3.5. An undisturbed sample of fine sand has II dry unit weight of 18 kN/m . At the maximum density. the void ratio is 035, and that at the minimum defL~ity, 0.90. ()ctermine the relative density of the undisturbed soil. G = 2.65. [Ans. 77.82%] 3.6. A coarse-grained soil is oompocted to a wet density of 2Mglm3 lit II WilIer coolenl of 15%. Determine the relative density of the wmpoctcd sand. given emu _ 0.85 and em;n _ 0.40 and G _ 2.67. fAns. 70%] 3 .7. How long would it take for 11 particle of soil 0.002 an in diameter 10 settle from the surface to the bottom of 2 the pond 15 m deep? Tllke G '" 2.60 and TJ '" 1.0 x 10-S gmf_seclcm _ [Ans. 11.72 hours] 3.8 A sample of soil of moss 40 gm is dispersed in 1000 mI of water. How long after the commencement of scdimentntion should the hydrometer reading be IIIken in order to estimate the percentage of particles less than 0.002 mm effective dillJ1)Cter ? 1be centre of the bulb is at an effective depth of 20 em below the surface of water. Thke G ;; 2.70, TJ '" 0.01 poise. tAns. 14.99 hoursJ 3_9_ In a sedimentation test, 25 gm of soil was dispersed in 1000 mI of water (TJ '" 0.01 poise). Doe hour after the commencement of sedimentation, 25 ml of the suspension Wll') IIIken by means of a pipette from a depth of 10 em. The mass of solid pDrticles oblllined on drying was 0.09 g. Determine (a) 1be largest size of the particle remai ning in suspension al a depth of 10 an after one hour of the beginni ng of sedimentation. (b) The percenlllge or particles finer than Ihis size in the original suspension. (e) Tbe lime interval from tile commencement, after which the largest particle remaining in suspension al 10 an depth is one-half of this size. (Hint. Volume of suspension;; 1009.3 ml) [Ans. 0.0055 rnntj 14.53%; 4 bours] 3.10 The results of a sedimcnllltion test of a SIlmple P.!lSSing distribution. Use approxima te formula v =- 9100 0 2.

ObservaviOfl

TIme

75~

sieve are given below. Determine the grain size .

Depth

Mass of soil in 25 ml sample

all depth 10 em 10 em 5em

2Sgm

No.

L

23. 4. 5.

Z
San

15 gm

lOgm 5 gm 0.5 gm

{Ans. Percentage finer than 0.075 mm, 0.0428 mm, 0 .0191 mm 0.0095 mm and 0.0017 nun, respectively, 100%,60%, 40%, 20% and 2%J. 3.11 In a lesl 10 grn of fine-grained soil of specific gravity 2.70 was dispersed 10 make 500 mI of suspension. A

PAlmCLE SIZE ANALYSIS

67

sample of volume JO mI was taken by means 0( a pipette 9t a depth of 100 mm, 50 minutes after the comrnenrement of sedimentation. The sample was dried in an oven. If the dry "taSS of the soil was 0.03 gm. calculate the larga;t size of the particle remaining in the suspension at a depth of 100 mm and the percentage of particles liner than this size in the original soil. 11 " 0.01 poise. IAns. 0.006 mm; 15%) 3.12. Ouring a scdirnentalion test for grain size analysis. the corncted hydrometer reading in a 1000 ml uniform soil suspension al the cornmenoemem of sedimentation is 1.028. After 30 minutes, the corrected hydrometer reading is 1.012, and the COCTesponding effective depth is 105 em. Determine (I) the IOtal mass of solids dispersed in 1000 mI of suspension, (;1) lbe portide size mrresponding 10 the 30 minute reading. and (iii) the percentage fiDef than this size. TIIke G " 2.67 and 11 ,,0.01 poise. . (Aos. 44.77 gm; 0.00796 mm; 42.86%) 3.13. A dry soil sample is 49 8m in mass. It is composed of the following: Particle size (mm) Mass (8m)

0.05

0.02 20

0.01 18

O.OOt

The sample is mixed with enough water 10 make a uniform suspension of 1000 ml. Detennine

(I) The largest particle size at a depth of JO em after 5 minutes of the commenocment of sedimentation and the specifie gravity of the suspension al that time III thut depth. (i/) The time required for 1111 the pDrliclcs to scllie belcr.v 10 an depth. Thke G .. 2.70 lind 11 '" 9.81 millipoise. [Ans. mm; 1.014; 1.06 )( 10-5 seconds}

om

3.14, An air-dry soil sample weighing 2S kg was sieved in a laboratory. The results are given below. 15 Sieve (mm) Mass rela;IIed

(.g)

0.08

Draw the grain size distribution curve and delenniile the coefficient of curvalure and the uniformity coefficient. IAns. 1.15; 259J

3.15. A 1000 rnI suspension containing 30 gm of dry soil ~ prepared for a hydrometer analysis. If the temperalUfe is the same as that at which it was allibrated, what whouJd be the hydrometer reading al the instant of commencement of sedimenl.8tion ? Take G " 2.70. IAns. 1.019)

B. Descriptive and Objedlve 'TYpe 3.16. 3.17. 3.18. 3.19.

What do you understand about index properties~? Whal is their importance?

How would you determine the perrentlge finer than different sieve sizes in the laboratory ?

What are the main index properties of a COIlISC-grained soil? How are these determined? Differentiate between the dry sieve analysis and the wet sieve analysis. Why the wet sieve anlllysis is required? Stnte Stokes' low. What is its use in the scdimenllliion mcthOO of analysis? Whlll are its limitations? Compare the pipette method imd the hydrometer meiOOd. Why the hydrometer method is more popular? State the various corrections required for a hydrometer reading. How these corrections ore determined? What is particle size distribution curve? What is its use in soil engineering? 3.24. What is relative density? How is it determined? What is ilS imJX)tlllnoc for a ooarse-grnined soil? 3.25. What do you understllnd by allibrotion of a hydrometer? How is it done? 3.26. State whether the following statements are true or false.

3.20. 3,Z1. 3.22:. 3.23.

(a) The sill size pAnicles can be seen by unaided (nllked) eye. (b) The sieve annlysis gives tbe largest dimension of the soil particle. (c) The wei sieve analysis gives slightly larger size than that by the dry sieve analysis. (d) The reading:; on a hydrometer inaease in upward directioo. (e) The sedimentation analysis is useful for al\ soil panicles smaller than 75", size. (j) The rock Hour even of clay size panides is non- plastic. (g) A gap-graded soil is also allied 8 uniform soil. (h) A well-graded soil contains particles of one size.

[...... nu.(c)'(J)]

.,

SOIL MECHAN ICS AND FOUNDATION ENGINEERING

C. Multiple Choice Questions I. Ill' Stokes ' law. the termin:!1 velocity of the particle is (a) Proportional 10 the mehtls of the panicle. (b) l'roportional 10 the squ:trc 01 the radius of particle. (c) Inverscly prOllOrliOMII \0 the ~qllarc of the radius of particle. Id) None

or the above,

2. Stoke's law docs nol hold good if Ihe (a)

Greater than 0.2 mill

~izc

of particles is

(b) (ellS

lhan 0.2 ).1m

(c) Neither (a) Nur (b)

(til Bo\h (al and (b) 3. Prelrcalcrncnl ur sOil [0 rcnl<'l\'C the orgamc m3l1cr by oxidation is done with ((I) Sodium hexametaphosphate (b) Oxygen (c) Hydrogen peroxide (tl) Hydrochloric acid 4. The particie-slIe diwibulion curve with a hump is obtained for a (ll) Unifonl1 soil (b) Well-gmded ~oil l(.') Gap-gmdecl soil (1/) Pourly-graded soil 5. For a well· graded sllnd. the coefficient of eUTV
(Ans. l. (b). 2. (d). 3. (e) . 4. (e). 5. (b). 6. (b). 7. (b). 8. (e). 9. (b)J

4 Plasticity Characteristics of Soils 4.l. PlASTICITY OF SOILS The plasticity of a soil is its ability to undergo deformation without cractking or fracturing. A plastic soil can be moulded into various shapes when it is weI. Plasticity is an impol1ant index property of Hoe-grained

soils, especially clayey soils. Plasticity in soils is due to presence of clay minerals. The clay particles carry a negative charge on their surfaces, as disrussed in chapter 6. The water molecules are dipolar (dipoles) and are attracted towards the clay surface. The phenomenon is known as adsorption (not absorption) of water, and the water SO attracted to the clay surface is called adsorbed water. Plasticity of the soil is due to adsorbed water. The clay particles are separated by layers of adsorbed water which allow them to slip over one another. When the soil is subjected to deformations, the particles do nol return to their original pa'>itions, with the result that the defonnations are plastic (irreversible). As the water content of the soil is reduced, the plasticity of the soil is reduced. Ultimately, the soil becomes dry when the particles are cemented together as a solid

mass. The presence of adsorbed water is necessary to impart plasticity characteristics to a soil. 1be soil does not become plastic when it is mixed with a non-polarizing liquid, such a<; kerosene or paraffin oil. These liquids do not have electromagnetic properties to react with clay mincrals. The soil becomes plastic only when it has clay" minerals. If the soils contains only non-clay minerals, such as quartz, it would not become plastic whatever may be the fmcness of soil. Whcn such soils are ground to very fine size, these cannot be rolled into threads. Rode. flour, which contains very fine particles of non-clay particles. does not become plastic. This chapter deals with plasticiiy characteristics and consistency of fine.grained soils.

43. CONSISTENCY LIMITS The consistency of a fine-grained soil is the physical state in which it exists. It is used to denote the degree of finnness of a soil. Consistency of a soil is indicated by such tenns as soft, firm or hard. In 1911, a Swedish agriculture engineer Atterberg mentioned that a fme-grained soil can exist in four states, namely, liquid, plastic, semi-solid or solid state. The water contents at which the soil changes from one Slat~ to the other are known as consistency limits or Atterberg's limits. The water content alone is not an adequatl! index property of a soil. AI the same water oontent, one soil may be relatively soft, whereas another soil may be hard. However, the soils with the same consistency limits behave somewhat in a similar manner. Thus consistency limits are very important index properties of finegrained soils A soil containing high water content is in a liquid state. It offers no shearing resistance and can flow lik.e liquids. It has no resistance to shear deformation and, therefore, the shear strength is equal to zero. As the water content is reduced, the soil becomes stiffer and starts developing resistance to shear dcfonnation. At


70

some particular water contenl. the soil becomes plastic (Fig. 4.1). l11e water content at which the soil chang~ from the liquid state to the plastic Slale is known as liquid limit (ll, w,), In other words, the liquid limit ~ the water content at which the soil ceases 10 be liquid. The soil in the plastic stale can be moulded into various shapes. As the water content is reduced, tht plasticity of the soil decreascs. Ultimately, the soil passes from the pla<>lic state to tbe semi~so1id state whet it stops behaving as a plastic. It crocks when moulded. The water content at which the soil become! semi-solid is known as the plastic limit (PL, wp ). In other words tbe plastic limit is the water content at wbicll the soil just fails to behave plastically. The numerical difference between the liquid limit and the plastic limit is known as plasticity inde" (PI,I, ).

lbus PI - U - PL 'The soil remains plastic when lhe water content is between the liquid limit and the plastic limit. Th( plasticity index is an imponant index property of fine-grained soils. When the water content is reduced below the plastic limit, the soil attains a semi-solid state. The SOL cracks when moulded. In the semi-solid stale, the volume of the soil decreases with a deaea
~;:~ ~st::)~ as the shrinkage

.5

~

t---~--,'~

Below the shrinkage limit, the soil does not remain Solid stale S!2:mi solid state : Plastic stah! Liqui saturated. Air enters the voids stale of the soil. However, because of capillary tension developed, Ws wp' w the volume of the soil docs not L change. Thus, the shrinkage (S L) ( PLl ( LL i limit is the water content at Wal!2:r content _ _ which the soil slaps shrinking FIg. 4 1 Different states of SoIl further and attaincs a constnnt volume. The shrinkage limit may also be defined as the lowest water content at which the soil is fully saturated. Fig. 4.1 shows sodden changes in the states of the soil at different consistency limits. Actually• .the tronsition between different states is gradual. 'The consistency limits are detennined rather arbitrarily, as explained in the following sections. [Note. In liquid Slate, the soil is like soup; in plas6c Slate, like soft buller; in seml·solid state, like cheese; and in solid Slate like hard candy.)

4.3. UQUID LIMIT As defined above, the liquid limit is the water oootent at wbich the soil chaoges from the liquid state 10 the plastic state. At the liquid limit. the clay is practically like a liquid, but possesses a small shearing strength. The shearing strength at that stage is the smallest value that can be measured in the Lllboratory. The liquid limit of soil depends upon the clay mineral present. The stronger the surface charge and the thinner the particle, the greater will be the amount of adsorbed water and, therefore. the higher will be the liquid limit.

".

G

PLAsrJCITY CHARACTER1S1.·'CS OF SOILS

is

lbe liquid Umit is delennincd in the laborntory cilhec by Casagrande's apparatus or by cone penetration method. The latter is discussed in Sect. 4.4. The device used in

faU

Casagmnde's method consists of a brass cup

which drops through a height of t em on a Groove Sample hard base when opcmlcd by the handle (Fig. 4.2). The device is opernted by turning the handle which raises the cup and lets it drop on the rubber base. The height of drop is adjusted with the help of adjusting screws. Rubber About 120 gm of an air.-dricd sample block passing through 425 J.I. IS sieve is laken in a dish and mixed with dlstillcd water to form a Fig. 4.2. Liquid Umil App.1rntus. unifonn paste. A portion of this paste is placed in the cup of the liquid limit devia:, and lhe surface is smoothened and a levelled with a spatula to a maximum depth of I em. A groove is CUI through tbe sample

along the symmetrical axis of the rup. preferably in one stroke. using a standard grooving tool. IS : 272.G---Part V recommends two types of grooving tools : (1) Casagrande lOOt. (2) ASfM toot. The Casagrande tool C\JIS a groove of width 2 mm al the bouom, II mm at the top and 8 mm deep. The ASTM 1001 cuts a groove of width 2 mm at the bottom, 13.6 mm 31 Ihe lop and 10 mm deep (Fig. 4.3). The Casagrande 1001 is recommended for normal fine.grained soils. whereas the ASTM 1001 is recommended for sandy, fioc grained soils, in which the Casagrande 1001 tends to tear the soil in the groove. After the soil pal has been cut by a proper grooving 1001. the handle is turned at a rate of 2 revolutioos per second until the two parts of the soil sample come into contact al the bottom of the groove along a distance of 12 mm. The groove should close by a now of the soil, and not by slippage between the soil and the cup. When the groove closes by a flow, it indicates the failure of slopes formed on the two sides of the groove.

Cd) DIVIDeD SOIL

CAKE BEFORE

(el SOIL CAKE

TEST

Fig. 4.3. Details of Appa11l11a and Tools.

AFTER

TEST

72


'{be soil in the cup is again mixed, and the tcst is repealed until two COflSeOJtivc tests give the same number of blows. About 15 gm of soil near the closed groove is taken for water content determination. The soil in the cup is tr.msfemxllo the dish containing the soil p8Sleatld mixed thoroughly after adding more

water. The soil sample is again taken in the cup of the Uquid limit device and the lest is repeated. The liquid limit

:~U:i~~~u~~~~ya~:~:

35,--------.-__-._,--.-"-.-rn

now when the device is given

25 blows. As it isdifficull to gel exactly 25 blows for the sample flow, the test is conducted at

~

30

10

different water contents so as to gel blows in the range of 10 to

25 ~

40. A plot is made between the

Wilier content as ordinate and

:lCn:~~~~~I~~ ~~ot'°~ o! ~'5' ~:.!~-------------approximately a straight line. _

I

~

•

The plOI is known as flow ~ curve. The liquid limit is B obtained, from the plot, COI'l"Qiponding to 25 blows (Fig. 10 4.4). The liquid limit is ~ expressed as the nearest whole number. The rappings in the liquid limit device cause smaU shear-

/i}.r-J

~n:u~~ 0: :it=r;e~a~ O,'~----j,c-~'-::"5 --!.:-,;-,:c,to";,:'o---il1Oc-;!:,,-.,!cO-C,!nO-'50h\60~BO!,L!"" as the water contenl when the Numbt'~

soil has shear strength just sufficient to withstand the

01

blows

(N) _ _

Ag. 4.4. Flow Curve.

shearing stresses induced in 25 blows. 'P.le shear strength of the soil at liquid limit is about 2.7 .kN/m2.

One-poInt Method The above procedure for detennining the liquid limit requires the test 10 be repealC
... [4.1(0)] = water content of the soil when the groove closes in N blows. n = an index, as given below. According 10 IS : 272D-V, for soils with liqUid limit less than 50%, the value of n is equal to 0.092 and for soils with liquid limit greater than 50%, the value of n c 0.12. The acocpted range for N is 15 to 35 for soils with liquid limit less than 50% and 20 to 30 for soils with liquid limit more than 50%. Alternalively, ... [4.I(b)] 1.3215 _ 0.23 iogloN Eq. 4.1 (a) can be written a<; where

wN

W, -

where C is the correction fador.

...(4.2)

PlASTICITY CHARACfERlsrlCS OF SOILS

The value of the factor is approximately 0.98 for N = 20 and 1.02 for N (See Chapter 30, Sect. 30.10 for the laboratory experiment)

= 30.

"

4.4. CONE PENETROMETER METIIOD The liquid limit of a soil can also be detennin(X! by Cone Penetrometer (IS : 2nO-V). It oonsists of a stainless steel cone having an apex angle of 30 0 ;t; 10 and a length of 35 mm. The cone is fixed al the lower end of a sliding rod which is fiued with a disc at its lop (Fig. 4.5). The total mass of the cone, Sliding rod and the disc is 80 g ;t; 0.05 g. The soil sample is prepared as in the case of the Casagrande method. The soil pat is placed in a cup of 50 mm internal diameter and 50 mm height. The cup is filled with the sample, taking care so as . Clomp not to entrap air. Excess soil is removed and the surface of the soil is levelled up. The cup is placed below the cone, and the cone is gradually lowered so as to just touch the surface of the soil in the cup. The graduated scale is adjusted to zero. The cone is released, and allowed to penetrate the soil for 30 seconds. 100 water content at which the penetrotion is 25 mm is the liquid limit. Since it is difficult to obtain the penetration of 25 mm exactly, liquid limit is detennined from the equation given below. W,. Wy

+ 0.01 (25 - y) (Wy + 15)

...(4.3)

Fig.

4.5. Cone Penetrometer

where y (in mm) is the penetration when the water content is wy • and

w, = liquid limit.

Eq. 4.3 is applicable provided the depth of penetration y is betweeo 20 to 30 mm. IT the penetration is oot in this range, the soil in the cup is taken out, and the water content adjusled 10 get the required penetration. A chart can also be drawn for direct determination for the liquid limit from the observed value of y and The shear strength of soil at liquid limit, as determined by tbis method, is about 1.76 kN/m2 which occurs when the penetration is 25 mm. The cone penetrometer method has several advaotages over the casagrande method. (1) It is easier to perform. (2) The method is applicable to a wide range of soils. (3) The results are reliable. and do nol depend upon the judgment of the operator.

w,..

4.5. PLASTIC LIMIT Plastic limil is the water content below which the soil stops behaving as a plastic material. II begins to crumble when rolled into a thread of soil of 3 mm diameter. AI this water content, the soil loses its plasticity and passes to a semi-solid state. For determination of the plastic limit of a soil, it is air-dried and sieved through a 425 .... IS sieve. About 30 gm of soil is taken in an evaporating dish. It is mixed thoroughly with distilled water till it becomes plastic and can be easily moulded with fingers. About 10 gm of the plastic soil mass is. takeo in one band and a ball is formed. 'The ball is rolled with fmgers 00 a glass plate 10 form a soil thread of uniform diameter (Fig. 4.6). The rate of rolling is kept about 80 to 90 strokes per minute. If the diameter of thread becomes smaller than 3 mm, without aack formation, it shows that the water content is more than the plastic limit. The soil is kneaded further. 1ltis results io the redudion of the waler content, as some water is evaporated due to the heat of the hand. 'The soil is re-rol1ed

SOIL MECHANICS AND FOUNDATION ENGINHERINO

74

and the procedure repeated lill

lhe thread aumbles. The water content at which the soil can be rolled into a lhrtad of approximately 3 mm in diameter without crumbling is known as the plastic limit (PL or wp). The test is repeated, taking a fresh sample each time. The plastic limil is taken as the average of three values. The plastic limit is reported 10 the nearest whole number. The shear strength at the pJastic limi\ is about 100 limes that al the liquid limit. (See Olapter 30, Sect. 30.11 for the laboratory experiment)

Fig. 4.6. Determination of P\alic limit.

4.6. SHRINKAGE UMrf Shrinkage limit is the smallest water content at which the soil is saturated. It is also defined as the maximum water cooleOI at which a reduction of water content will oot cause a decrease in the volume of the soil mass. In other words, at this water content, tbe shrinkage ceases. An expression for the shrinkage limit can be obtained as given below Fig. 4.7 (a) shows the block diagram of a soil sample when it is fully saturated and has the water content

p'T~ii-:rr;~~~-~~-'-~""""-

1%}tiII11'~ll' Stage 1 (e)

Stoge II r

Stage II

(c)

(b)

Fig. 4.7. Stages ror Derivation of Shrinkage Umil.

the oondition when the soil sample bas been ovendried. The total volume V] in Fig. 4.7 (c) is the same as the lotal volume V1 in Fig. 4.7 (b). The throe figures indicate, respectively, stage I, II and m. Let M~ be the mass of solids. Mass of water in stage I - Ml -M, loss of mass of water from stage I to stage II - (VI p ... Mass of water in stage n - (MJ - M,) - (VI - Vi) p... From definition, shrinkage limit '" water content in stage II (MI - M,) - (VI - V,)P. w, • M, ... (4.5)

Vv

P[ASI'lCITY CHARACfERI5nCS OF SOILS

or

75

(V, - V:z)

w. - wI - ~ P...

•..(4.6)

where wI represents the water content in stage [. For determination of the shrinkage limit in the laboratory. about 50 gm of soil passing a 425 Il sieve is laken and mixed with distilled water to make a aeamy paste. The waler content (wI) of the soil is kept greater then the liquid limit. A cirallar shrinkage dish, made of porcelain or stainless steel and having a diameter 30 to 40 mm and a height of 15 mm, is taken. The shrinkage dish has a flat bottom and has its intemal comers well rounded. The capacity of the shrinkage dish is first determined by fllling it with mercury. The shrinkage dish Is placed in a large porcelain evaporating dish and filled with mcccury. Excess merrury is removed by pressing a plain glass plate fumly over the top of the shrinkage dish. The mass of mercury is the shrinkage dish is obtained by transferring the mcccw-y into a mercury weighing dish. The capacity of the shrinkage dish in ml is equal to the mass of mercury in gm divided by the specific gravity of mercury (usually, taken as 13.6). The imide surface of the empty shrinkage dish is mated with a Ihin layer of vaseline or silicon grease. The mass of empty shrinkage dish is obtained aa:urately. 111e soil sample is placed in the shrinkage dish, about one-third its capacity. The dish is tapped on a firm surface to ensure that no air is entrapped. More soil is added and the tapping continued till the dish is completely filled with soil. The excess soil is removed by striking off the top surface with a straight edge. The mass of the shrinkage dish with soil is taken to obtain lbc mass (Mt,) of the soil. 1be volume of the soil VI is equal 10 the capacity of the dish. The soil in the shrinkage dish is allowed to dry in air unlil the oolour of the soil pal turns light. It is then dried in a oven. The mass of the shrinkage dish with dry soil is taken to obtain the mass of dry soil M •. For determination of the volume of the dry pat, a glass OJP, about 50 mm diameter and 25 mm height, r. taken and placed in a large dish. The OJp is filled with mercury. 'The excess mercury is removed by pn=ssing a glass plate with three prongs firmly over the top of the cup. Any mercury adhering on the side of the alp is wiped off, and the OJp full of mercury is transferred to another large dish. The dry pat of the soil is removed from the shrinkage dish, and placed on tbe surface of the mercury in the OJp and submerged inlO il by pressing il with the gl
also equal to V". The shrinkage limit of the soil is detenniOC(l, using Eq. 45, from the measured values of VI' V2 ,M1 andM•. (See Olapter 30, Sea. 30.12 for the laboratory experiment).

4.7. ALTERNATIVE ME11IOD FOR DETERMINATION OF SIIRINKAGE LIMIT The shrinkage limit of a soil can be determined by an alternative method if the specific gravity of solid particles (G) is known or is determined separately. An expression for shrinkage limit in terms of the specific gravity of solids can be developed from Fig. 4.7 (b). At that stage, the water 'oonteOI is al the shrinkage limit, given by,

(V, - V,)P. w·---.-M where V. is the volume of solids.

. •.(4.7)


76

Eq. 4.7 can be written as

W, -

V, V,] p... [Ii; Gp ... (V,)

.. _ [V'P. _ ,

M,

l]

... (4.8)

G

.. . M, Now, from the defimtlon of the dry mass densIty, Pd"

v;-

Therefore,

w, ..

(~ - ~ )

... (4.9)

Eq. 4.8 can be used for the delennination of the shrinkage limit, as explained below. A smooth, round-edge(! pal of wet soil is made in a shrinkage dish. It is then dried in an oven and cooled in a dcssicalor. Any dust on the sample is brushed off. The dry mass Ms of the sample is delennined. The volume Vz of the dry soil pal is obtained by placing it in a glass cup and delcnnining the displacement of mercury, as discussed in Sea. 4.6.

Determination of Specific Gravity of 80Uds

rr.om

Shrlnkage Urnit

L Method-The specific gravity of solids (G) can be delennined using Eq. 4.8 if the Shrinkage limit has already been determined.

(V'P,.l~,)

... [4.10(.)J G .., Sometimes, Eq. 4.9 is written in tenns of mass specific gravity (G".) in dried slale. Thking Girl" p/p""

From Eq. 4.8,

G -

lI(G.\ _..,

...

[4.10(b)]

n.

Method-The observations made in the shrinkage limit test, as desaibcd in Section 4.6. can be used to determine the approximate value of G. The volume of solids (V~) is stage III (Fig. 4.7(c)] V, _

.!!:.... Gp.

...(0)

Also, the volume of solid can be detennined from the volume VI in Fig. 4.7(a) (stage I) as V~

- VI - volume of water

V _ VI _ (M} - M,)

, From Eqs. (.) and (b),

... (b)

_ (MI-M,)

Po

Gp.

~ or

p•

.!!:.... _ VI

- VIP ... - (MI-M,)

G _

M, VIP ... -(M1 - M,)

.. .(4.11)

1be methods for determination of Vh MI and M, t-.sve already been discussed in Sect. 4.6. 4.8. SBRlNKAGE PARAME:I'ERS

(~p)

The following parameters related with shrinlcage limit are frequently used in soil engineering. (1) Shrinkage Index-The shrinkage index (I~) is the numerical difference between the plastic limit and the sbrinkage limit (w,).

I, - wp - w~ ... (4.1:l) (2) ShrInkage Ralio-The shrikage ratio (SR) is dermed as the ratio of a given volume change, expressed as a percentage of dry volume, to tbe corresponding change in water cootent.

PlASTICITY CHARAcrERlsrtCS OF SOILS

17

SR _ (V, - V,)/V, x 100 wl-Wz

... (4.13)

= volwne of soil mass at water content WI V 2 = volume of soil mass at water content w2 Vd = volume of dry soil mass.

where VI

When the volume V2 is at the shrinkage limit. SR .. (VI - Vd)/Vd )( 100

. .. (4.14)

WI-w.

Another expression for shrinkage ratio (SR) can be found from Eq. 4.13, by expressing the water rontent

(V,- V,)P.

WI-W2"~

SR_~

Therefore,

VdP",

SR ..

~

. .. (4.15)

.. G.

Thus the shrikage rntio is equal to the mass gravity of the soil in dry state (Gift). From &po 4.9 and 4.15, tbe shrinkage limit.

w_(-"-_.!.) •

S.R.

••.[4.15(0)J

G

(3) Volumetric Shrinkage-The volumetric shrinkage (VS). or volumetric change, is defined as tbe change in volume expressed as a percentage of the dry volume when the water mnlen! is reduced from a given value of the shrinkage limit. Thll'> \IS..

V,-V,) )( (----v;-

100

... (4.16)

But From Eq. 4.14, [(VI - Vd)/Vd) )( 100 .. SR (wI- w,)

Therefore.

\IS .. SR (wI - W,)

... (4.17)

(4) Linear Shrinkage-Unear shrinkage (IS) is defined as the change in kngth divided by the initiaJ length when the water content is reduced to the shrinkage limit. It is expressed as a percentage. and reported to the nearest whole number.

Thus

LS ..

(Initiallen~h - final length ) )( 100

... (4.18)

Iml13llength

The linear shrinkage can be detennined in a laboratory (IS : 2720-Part XX). A soil sample about ISO gm in mass and passing through a 425" sieve is taken in a dish. It is mixed with distilled water 10 fonn a smooth paste at 8 water content greater than the liquid limit. 1be sample is placed in a brass mould, 140 mm long and with a semi-circular sealon of 25 mm diameter. The sample is allowed to dry slowly first in air and tben in an oven. The sample is oooled ana its fmal length measured. The. linear shrinkage is calculated using the following equal:ion. LS _ [1 _

Le~~h of oven-dry ~ple ImUallength ofspecunen

1)( 100

... (4.19)

In Eq. 4.19, it has been assumed that the length of the spedmen in oven-dried state is the same as that at the shrinkage limit.


78

1be linear shrinkage may also be obtained from the volwnetric shrinkage (VS) as under. [S -

HXl[ 1 - ( I'S 1.00100

f1

... (4.20)

The linear shrinkage is related 10 the plasticity index (Ip). as under: [S_ 2.13

x (LS)

... (4.21)

4.9. PLASTICITY, LIQUIDITY AND CONSISTENCY INDEXES (1) PlastkJty Index-Plasticity index (II' or PI) is the range of water rooteDt over which the soil remains in the plastiC stale. It is equal to the difI~ between the liquid (w,) and the plastic limit (w,,). Thus,

W,

When eilher WI or

wI'

I... -wI' canDOl be dctennined. the soil is

... (4.22) noo~plastic

(NP). When the plastic limit is greater

Lban the liquid limit, the plasticity index is reported as zero (and not negative).

(2) Liquidity Index-Uquidity index (I, or LI) is defined a<>

II wbere w

T )(

100

... (4.23)

= water content of the soil in

Datuml condition. The liquidity iodex or a soil indicates the nearness of its water content to its liquid limit. When the soil is at its liquid limit, its liquidity index is 100% and it behaves as a liquid. When lhe soil is at the plastic limit. its liquidity index is zero. Negative values of the liquidity index indicate a water content smaller than the plastic limit. The soil is then in a hard (dessicated) state. The liquidity index is also known as Water-Plasticity ratio. (3) Consistency lndex-Consislency index (Ie. Cf) is defined as

Ie - W//:W )( 100

... (4.24)

where W = water cootents of the soil in natural rondition. The ronsistency index indicates the ronsistency (finn ness) of a soil. It shows the nearness of the water content of the soil to its plastic limit. A soil with a ronsistency index of zero is at the liquid limit. It is extremely soft and ha<> negligible shear strength. 00 the other hand, a soil at a water moteot equal to the plastiC limit bas a amc;istency index of 100%, indicating that the soil is relatively firm. A cono;istency index of greater than 100% shows that the soil is relatively strong, as it is the semi-solid state. A negative value of consistency index is also possible, which indicates that the water content is greater than the liquid limit. The consistency index is also known are relalive consistency. It is worth noting that the sum total of the liquidity index and the consistency index is always equal to 100%, indicating that a soil having a high value of liquidity index has a low value of consistency index and vice-versa. 4.10. FLOW INDEX Flow index (I,) is the slope of the flow CUIVe obtained between the number of blows and the water content in Casagrande's method of determination of the liquid limit (Fig. 4.4). Thus

I, -

log:' W~Nl)

••.[4.25(a)J

W -

-I,log,oOO + C

..•[4.25(b)J

where N, = Dumber of blows required al water rontenl of w,_ and N1. = number of blow required at water amlent of Wz. Eq. 4.25 (a) can be written in the general form

PLASIlCITY CHARACfERI5nCS OF SOILS

The flow index can be dctcnnincd from the flow curve from any two points. For convenience, the number of blows N) and HI are taken corresponding 10 ooe log cycle, i.e. N)INI - 10. In that case,

70

60

so

... [4.25(c)]

~-------------~-

It may"be mentioned tbat the number of blows actually GO observed in tests are in a narrow range, nonnally in the range of 20 10 30 and lhe of )0 NyN I - 10 can be obtained ~ only after extrapolation of the ~ plot. -:1he flow index is tbe rate at

ratio

=~:::~ ~n~= ii~:: 8

I

;If(2)::: "i -~ I 109!.0INzIN1)

content. Fig. 4.9 shows the flow curves of two soils (1) and (2). ~

:r~:iljj~~~ ::t:;~~~~ possesses

shear Strength as rompared to soil (l}-with a flatter slope. In order to decrea<>e the waler rootent by the same amount, the soil with a steeper slope takes a smaller number of blows, and, therefore, has lower shear strength.

and

15

~,-~2~ __ -

- __ -

- -

-r- - - - - ____

lower

I I t

10

:",

4.11. TOUGHNESS INDEX Toughness index (I,) of a soil is defined as the ratio of the plasticity index (Ip) and the flow index (I,.)

10 Number 01

100 blo'WS ( N )

fiB- 4.9, Row IndexCi

!£.

...

I, _ (4.26) I, Toughness index of a soil is a measure of the shearing Stralgth of the soil at the plastic limit. This can be proved as under: Let us assume that the flow curve is a straight Une between the Uquid limit and the plastic limit. As the shearing resistance of the soil is direcUy proportional to .the number of blows in Casagrande's devi~ k SI _ NI ... (a) aDd k S, _ H, ... (b) Thus

where HI ::: number of blows at the liquid limit when the shear strenglb is SI Np ::: number of blows at th~ plastic limit when the shear strellgth is Sp k::: constan.l.


80

From Eq. 4.25 (a), taking wl ..

W,

and Nt .. 1.0, 1 I, .. 10;1 0 WI - I, JaglO N,

ZN;I) .. :~:~,

W, ..

... (e)

Ukewise, fcr the plastic limit. ... (d) From

Eqs. (e) and (d). w, - Wp

.. -

1,1oglO(N,INp )

Substituting the value of (N/INp) from Eqs. (a) and (b).

W, -

wp .. -I,loglO (S,ISp) .. 1,log(SpfS,) ... (e) Since the shearing strength of aU soils at the liquid limit is almost roostanl and C(jual to 2.7 kN/m2, Eq. (e) can be written

as. laking w/ -wp

..

IF' Ip - If log (S,I2.7)

or

If .. 10gIO (Sp) _ log%,7) log 10 (Sp) .. I, + C

or

.. (4.27)

where C is constant equal 10 loiJ~·7) (::: 0.431). Eq. (4.27) proves that the shear strength al plastic limit depe~ upon the toughness index. The value of the toughness index of most soils lies between 0 to 3.0. A value of toughness index less than unity indicates thaI the soil is friable at the pia..lic limit.

4.12. MEASUREMENT OF CONSISTENCY

Consistency of a soil. a
Table 4.1 gives the uoconfined compressive strength of soils of different oonsislency.

Table 4.1. Consistency In lerms of Consistency Index and Unconfined Compressive Strength (q.,) S.HO.

Consistency

Consistency

i_

Unconfined compressive

(%)

strength~q,J

CharQcteristics o/soil

(kN/m

Yay soCt Soft

0-25

< 2S ltN/m

2.

~50

~50

3.

Medium (Firm)

50-75

50-100

1.

4 ..

5. 6.

Stiff Vel)'

stiff

"Old

75-100

100-200

> 100

200-400

> 100

>400

Fist can be pressed inlO soil Thumb can be pressed ioto

"'I

Thumb can be pressed with

1"=""

Thumb can be pressed wilh great difficulty The be readily indented with thumb nail The soil be indented with difficulty by thumb nail

"'I ao. ao.

pL\SfICn-V CHARACTERlsrlCS OF SOILS

'81

4.13. SENSmVITY A cohesive soil in its natural state of occurrence has a certain structure (see chapter 6). When the structure is disturbed, the soil becomes remoulded. and its engirieering properties dlange considerably. Sensitivity (S,) of a soil indicates its weakening due to remoulding. It is defined as the ratio of the undisturbed strength to the remoulded strength at the same water content. S • (q,,). . .. (4.28) , (q.), where (q,,).. = unconfined compressive strength of undisturbed clay (q..). unconfined compressive strength of remoulded clay. Depending upon sensitivity, the soils can be classified into six types, as given in Table 4.2.

=

Table 4.2. Classification

or Soils

based on SensitIvity

S.No.

Sellsitivity

Soil Type

1. 2. 3.

< 1.00 1.0-2.0 2.0-4.00 4.0-8.00 8.0-16.0 > 16.0

Insensitive Little sensitive Moderately sensitive Sen.'!itive EXIra sensitive

4.

5. 6.

Quick

For most days, sensitivity lies between 2 and 4. Clays considered sensitive have S, values between 4 and 8. In C$e of sensitive clays, remoulding causes a large reduction in strength. Quick clays are unstable. These tum into slurry when remoulded. High sensitivity in clays is due to a weU-developed flocculent structure which is disturbed when the soil is remoulded. High sensitivity may also be due to leaching of soft glacial clays deposited in salt water and subsequenUy uplifted. Extra-sensitive day, such as clays of Mexico city, are generally derived from the decomposition of volcanic ash.

4.13. mIXOTROPY The word Thixotropy is derived from two words : tl!ixis meaning touch, and tropo. meaning to change. Therefore, thixotropy means any dlange that occurs by touch. The loss of strength· of a soil due to remoulding is partly due to change in the soil structure and partly due to disturbance caused to water thplecules in the adsorbed layer. Some of lhese changes are reversible. If a remoulded soil is allowed to staM, 'filhout loss of water, it may regain some of its lost strength. In soil engineering, this gain in strength of ute soil with passage of time after it has been remoulded is called thixotropy. It is mainly due 10 a gradual itprientation of molecules of water in the adsorbed water layer and due to re-establisbment of chemical equilibfi!.im.

driV~~~!ro~~~.~ 1!s°~l=tQ;::~~rt:':is~~~~enc!=~n~i:~t~~~mi~:ica~::noc:wp~:! sbear strength will be regained after the pile hm been driven and left in place for some time.

4.14. ACIlVITY OF SOllS Activity (A) of a soil is the ratio of the plaslicity index and the percentage of clay fraction (minus 2,",

sjze). Thus ... (4.29) wbere lp = plasticity index, F = clay fraction. The clay fraction F is percentage finer than 21-1 size. The amount of water is a soil mass depends upon .the type of clay mineral present. Activity is a measure


82

of the water·holding capacity of cl.•'1yey soils. The changes in the volume of a clayey soil during swelling or shrinkage depend upon the activity.

A number of samples of a particular soil arc taken and their plasticity index and clay fraction determined. If a plot is obtained between the clay fradion (as abscissa) and the plastit.ity index (as ordinate). it is observed that all the points for a particular soil lie on a straight line (Fig. 4.10).

eo

I ~

60 1.0

~

(1)

"

n:

Kaol '\ ni\e

(Z) II\lte

20

(3)

Mon\omorillonitl;?;

40 Clay fra c t'lon (m i nus 2 r-) Fig. 4.10. ActiYity of Soils.

The slope of the line gives the activity of soil. The steeper the slope, the greater is the activity. TIle lines with different slopes are obtained for different soils. The soils containing the clay mineral montmorillonite have very high activity (A > 4). The soil containing the mineral kaolinite are least active (A < 1). whereas the soils oontaining the mineral illite arc moderately active (A = 1 to 2). Depending upon activity, the soils are classified into three types (Table 4.3). Tobie 4.3 Clas.<;ification of Soils Based on Activity S.

No. 1. 2. 3.

Activity A < 0.75 A::: 0 .75 to 1.25 A> 1.25

Soil type Inactive Normal Active

Activity gives information about the type and effect of clay mineral in a soil. The following two points are worth noting: (1) For a soil of specific origin, the activity is constant. 1be plasticity index increases as .the amount of clay fraction increases. (2) Highly active minerals, such as montmorillonite,. can produce a large increase in the plasticity index even when present in small quantity. 4.15. USES OF CONSISTENCY LIMITS The consistency limits are detemlined fo r remoulded soils. However the Shrinkage limit can also be obtained for the undisturbed sample. Since the actual behavior of a soil depends upon its natural structure, the consistency limits do not give complete information about the in-situ soils. lbey give at best a rough estimate about the behaviour of in-situ soils. . Although it is not possible 10 interprete the consistency limits and other plasticity characteristics in fundamental terms, yet these parameters are of great practical use as index properties of [ine-grained soils. The engineering propenies of such soils can be empirically related to these index properties as under. (1) It has been found that both the liquid and plastic limits depend upon the type and amount of clay in


84

From Eq. 4.25 (a), for

N2

Ii;"

w .. 10. 100

If .. wI - W:l .. 55.0 - 42.0 .. l3.0% llIustrative Example 4.2. A soil has a liquid limit of 25% and a j1uw index 0/ 12.5%. If the plastic limit is 15%, determine the plasticity index and me toughness index. 11 the water content of the soil in its natural condition in the .field is 20%, find lhe liquidity index and the relative consistency. ;f! ~Iution. From Eq. 4.22,lp" W, - wp" 25-15 .. 10% ~SO w~;1.9.1.1 10 From Eq. 4.26, 1, .. ~ .. 12.5 .. 0.8 (80%) S 48

_~x 100

From E. 4.23,

I,

From Eq. 4.24,

Ie .. W1ZW

Ip

..

0.2~~g.15 x X

100 .. 50%

100

.. O.250~IO.20 )( 100 .. 50% U1ustratlve Example 4.3. A cone penetrometer test was conducted on a sample 01 soil for the determination of the

42

liquid limit, and the following observations were recorded Cone. penetration

~~I~m~~~~~__4r__~____4-__-+_U __3___ '~O~----~20~~3~O~W~50~W ~--~m~Water cOli/em (w)%

51.5

53.2

55.2

58.1

595

NO of blows

Determine tbe liquid /imit. Fig. E-4.l. Solullon. Fig. E-4.3 shows the plot between !he cooe penetration and the water oontent. From the plot, the water cootent corresponding LO cone pcnetratioo of 25 mm is 58%. Thus

WI _

58%.

30

I" 'E

26

2s mm

..:: 21. --- - - - - - -

•

- - - ---

!:?22

~ 20

~ 18 ~ 16

,

u"

:W(. S8'O Of..

"~~--<51'---5~1---'~~--'5"'--~55~-'5~6-'-'~--~5;S~-o59'---6""O Water

cont1!llt_

Fig. E-4.3.

85

PLAsnCITY CHARACTF.¥ISfICS OF SOIL')

U1ustrative Example 4.4. A sample of clay has the liquid limit and the shrinlwge limit of, respectively, 60% and 25%. If the sample has a volume of 10 ml at the liquid limit, and a volume of 6.40 ml at the shrinXcge limit, determine the specific gravity of solids. Solution. Let Ms be the mass of solids, in gm. lbcrefore, mass of waler 81 the liquid limit = 0.6 Ms and mass of water at Ihe shrinkage limil = 0.25 M, Mass of waler losl belween the liquid lirnil = (0.6 - 0.25) Ms = 0.35 M, and the shrinkage limit RC
=

=

- 0.25 x 10.29 = 2.57 gm Volume of water al the shrinkage limit = 2.57 ml V, = 6.40 - 2.57 :::: 3.83 m Volume of solid particles, 29 . - 2.69 Therefore, specific gravity of solids, G.. M,V • '3°83 . p"" . Alternatively, directly from Eq. 4.10 (a), G -

(V2P..l~')-W' -

(6.40 x 1.0)10.29>-0.25 - 269

nlustralive Example 4.5. In an experiment for the dctenninalion of the shrinlclige limit, the following observations were taJcen. (0) Volume of saturated soil = 9.75 ml (b) Mass of saturated soil = 16.5 gm (c) W>lwne of dry soil after shrinlwge = 5.40 ml (d) Mass of dry soil afrer shrulkage = 10.9 gm Compute the shrinlwge limit and the specific gravity of solids. Solution. Given values arc VI = 9.75 ml,

V1 = 5.40 ml, Ml = 16.5 gm (M, -M.) - (V, - Vi) p. M,

and

From Eq. 45,

W, ..

Therefore,

w.' (l6.5-10.9)-\~~-5.40) x 1.0

From Eq. 4.11,

G..

•

M,:::: 10.9

gm.

5.6,O.~.35 .0.1147(11.47%) M,

V, p. - (M, -M.) • 9.75 x 1.0

'~·~'6.5

10.9)' 2.63

Illustrative Example 4.6. A soil has liquid limit and plastic limiJ of 47% and 33%, respectively. If the volumetric shrinkages at the liquid limi, and plastic limit are 44% and 29%, detLrmine the shrinkage limiL Solution. From Eq. 4.16, At liquid limit,

VS \IS ..

VI

~_

VI - V" ---v;;-x

V1- Vd

----v;- )(

100

100 .. 44

0.44 + 1.0. 1.44


86

or

Vd .. 0.694 VI

At plastic limit,

vs-

... (a)

Vp-VdxlOO_29

V,

V

~ .. O.29+1.0-1.29

.. .(b)

Vp .. 1.29 Vd

Let the volume at liquid limit, VI. be 1.0 ml. From Eq. (0), Vd:::: volume at shrinkage limit:: 0.694 rol From Eq. (b), Vp = volume at plastic limit = 0.895 mt

Volume

Water

conlent -

...

Fig. E4.6

From Fig. E 4.6 by proportion,

W,-W,

~

1.0 - 0.6,91 .. 0.895 - 0.694

0.47 - w,

033 -

W,

---0:306 - o:wt w, _ 0.06 (6.0%) Illustrative Example 4.7. The following index properties were determined for two soils A and B. Index property

Liquid limit Plastic limit Wateroonlcnt Sp. gr. of solids Degree of saturation

A

B

65 25 35 2.70

35 20 25

100%

2.65 100%

Which of he two soils (i) contains more clay particles, (ii) has a greater bulk density. (iii» has a grtXlter dry density. (iv) has a greater void ratio ?

PLASTICITY CIIARJ\CI'ERlSfICS OF SOILS

87

Solution. S. No. Plasticity index

PI .. 2.

w/-wp

Void ratio

e - wG 3.

Dry density

SOIL E 35-20 = 15%

0.35 x 2.7 .. 0.945

0.25 x 2.65 .. 0.663

2·~.;4~·0

Q.l!!!

P4" 1 +e 4.

saIL A 65-2.'1 = 40%

Bulk density p .. pd(l +w)

2.6i.e::31.0 _ 1.594 g/m l

.. l.388g/ml

1.388 x 1.35 .. 1.874 glml

1.594 x 1.25 .. 1.992 glml

As lhe plasticiLY index: of soil A is more Ih.m thm of soi l B, [I has more clay particles.

I'ROBLEMS A, Numericul 4.1. The consistency limits of a soil sample are: Liquid limit '" 52% Plastic limit '" 32% Shrinkage limit '" 17% If the specimen of this soil shrinks from a volume of 10 cm} at liquid limit to 6.01 an} at the shrinkage limit, calculate the specific gravity of solids. [Ans. 2.8OJ 4.2. A cone penetcmion test was carried out o n a sample of soil with the fol lowing results:

CO/Ie pellelratioll (mm)

I

Mois/ureCOIllelll (%)

16.1 50

I

17.6 52.1

I

19.3 54.1

I

I

213

22.6

57.0

I

58.2

Determine the liquid limit of the soil.

IA ns.6O%]

43. In a shri nlulge limit test, a dish with volume of 10.5 ml was filled with saturated Clay. The mass of the S!lCuraled clay wa<> 18.75 gm. Thc clay was dried gradually first in atmosphere and then in an oven. '[he clay was 12.15 gm and its oIolume 5.95 ml. Determine the shrinkage limit.

ma.~

of the dry [Ans. 16.9%]

4.4. A sample o f day has a void mlio of 0.70 in the undisturbed state and of 0 .50 in a rcmoulded Slate. If the specific gravity o f solids is 2.65, determine the shrinkage limit in each case. [Ans. 26.4%, 18.9%J

4.5. A fully saturated clay has a willer content of 40% and a mass specific gravity of 1.85. After oven-drying, the mass speci fic gravity reduces to 1.75. Determine the specific gravity of solids and the shrinkage limit.

[Ans. 2.80, 21.4%1 4,6. The Allerberg limits of a clay are : U "" 60%, PL = 45%, and SL = 2.'1%. The specific gravity of soi l solids is 2.70 and the naturai moisture content is 50%. (I) Wh:1I is ilS SiDle of consiSlency in naEUre ? (ii) Calculate the volume to be expet;led in the sample when moisture content is reduced by evaporation to 20%. Its volume al liquid limit is 10 cm). (Ans. consiSlency index "" 66.7%, 6.40 an)[

B, Descriptive and Objective Type 4,7. Discuss the imponance of Atlerberg's limiis in soil engineering. 4.8. What ate the main index properties o f a fine-grained soi l ? How are these determined in a laboratory ? 4.9. What do you understand by consistency of soil ? How is it determined ? 4.10. What are the different methcx1s for determination of the liquid limit of a soil? What are their relative merilS and demerits? 4.11. Describe the mcthcx1 for determination of shrinkage limit o f a soil. 4.12, Discuss various shrinkage parameters. How would you determine linear shrinkage ?

SOIL MECHANICS AND FOUNDATION ENGINEERJNG

88

4.13. What arc uses of (.:nnsislcncy lirnits'! Wh.ll nrc their limitations '! 4.14. Differcntime belwt,.'Cn: (a) Liquidity index and cunsistency index. (b) Flow index and toughness index. (el Plasticity and consistency. (d) Activily and sensitivil),. 4.15. State whether the following S(alernCnl~ nre true of false. (a) All the consistency limils Me determined fur the soil in distu rbed condition. (b) The liq uidity index cannot be more th:rn 100%. (e) The consistency index C:lll be neg'lIive. (d) Plastic limit is the water content of soil which represents the boundary between the plastic state and the semi·S(llid slate (e) Al shrinkage limit, the soil is fully saturated. (fJ The activity of a day minenll is a con~tanl. (g) The soils with son consist!!ncy hav
C. Multiple-Choice Questions 1. At shrinknge limit, the soil is (u) Dry (b) Partially ~aturiltcd (c) Satur;\ted (d) None of ahove 2. The shrinkage index is equal to (al Liquid limit minus plastic li mit. (b) Liquid limit minus shrinkage limit. (e) Plastit limit minus shrinkage limit. (d) None of ilbovc. 3. Toughness index of a soil is the nltio of tIl) Plasticity index to the !low index. (b) Liquidity index to the now index. (e) Co nsistency index 10 the now inUex. (d) Shrinkage index to the !low index. 4. A stiffelay has a consistency inde x of (a) 50--75 (b) 75- 100 (el Greater than 100 (d) Less than 50 5. The plasticity index of a highly plastic soil is about (al 10-20 (b) 20-40 (el Grater th~ln 40 (d) Less than 10 6. The activity of the mineral mon tmorillonite is (n) Less than 0.75 (h) Between 0.75 and 1.25 (e) Bctwcl:n 1.25 and 4 (d) Greater than 4 7. A soil sample has LL = 45%, PL'" 25% and SL "" 15%. For a natural water conten1 of 30%, th e consistency index will be (/1)75% (bl50% (c) 40% (ll) 25% H. For the soil wilh LL = 45%. PL :0 25% and ~h '" 15%, Ihe plasticity inu<:lx is (/I) 50% (b) 20% (c) 60% (if) 40%

IG

5 Soil Classification

Ih,

5.1. INTRODUCTION (1)1

(b)1

Soil classification is the arrangement of soils into different groups such thai the soils in a panicular group have similar behaviour. It i.. a sort of labelling of soils with different labels. M there is a wide variety of soils covering earth, it is desirable 10 systematize or classify the soils into broad groups of similar behaviour. It is more convenient to study the behaviour of groups than Ibm of individual soils. Cla<;sification of various commodities and species is also oommon in many other disciplines. For example, a chemist classifies the chemicals into various groups, and a zoologiSt classifies the specic~ into a number of groups. likeWise. a geotechnical engineer classifies the soils into various groups. For a soil classification system to be useful to the geotechnical engineers, it should have lbe following basic requirements: (I) It should have a limited number of groups. (2) It shouk! be based on the engineering properties which are most relevanl for the purpose for which the classification has been made. (3) It should be simple and should use the tenns which are easily un
5.2. PARTICLE SIZE ClASSIFICATION The size of individual particles has an important influena: on the behaviour of soils. It is not surprising


90

that the first classification of soils was based on Ihe panicle size. It is a general practice to classify Ihe soils into four brood groups. namely, grnve~ sand, silt sizc and clay size. While classifying the fine grained soils on the ba<>is of particle size, it is a good prllctice to write Sill size and clay size and not just silt and Clay. In general usage, the terms silt and clay arc used to denote Ihe soils that exhibit plasticity and cohesion over a wide range of water content. The soi l with clay-si7.c particles may not exhibit the properties associated with clays. For example, rocId1our has the particles of the size of the clay particles bul docs not possess plasticity. H is classified as clay-size and not just clay in the particle size classification systems. Any system of classification based only OD particle size may be misleading for fine-grained soils. The behaviour of such soils depends on the plasticity characteristics and not on the particle size. However, classification based on panicle siz.e is of immense value in the case of coarse-graincd soils, since the behaviour of such soils depends mainly on the particle size. Some of the classifi~tion system based on particle size alone are discussed below. (1) MlT System-MIT system of cL1SSification of soils was developed by Prof. G. Gilboy at Mass.'lChuseltcs Institute of Technology in USA. In this system, the soil is divided into four groups (Fig. 5.1 a). (I) Gravel. particle size greater than 2 mm. (it) Sand, particle size between 0.06 mm 10 2.0 mm. (iii) Silt size, particle size between 0.002 mm to 0.06 mm. (iv) Clay size, panicle size smaller than 0.002 mm (2~).

Boundaries between different types of soils corres!X>nd to limits when im!X>rtant changes occur in the soil properties. 'The particles less than 2~ size arc generally colloidal fraction and behave as Clay. The soils with panicle size smaller than 2~ are classified as cL'ly size. The naked eye can see the the plIrticle size of about 0.06 mm and larger. The soils with particle size smaller than 0.06 mm but larger than 21! are classified as silt-size. Important changes in the behaviour of soil occur if particle size is larger than 0.06 mm when it behaves as cohesionlcss soiL The boundary between gravcl and sand is abritrnrily kept as 2 mm. This is about the me of lead in the pencil. The soils in sand and Sill-sizc-rangc are further subdivided into three categories: coarse (C). medium (M) and fine (F), as shown in the figure. It may be nOled that MIT system uses only two integcffi 2 and 6. and is ea<>y to remember. (2) international Classlficalion System-The International Classification System was proposed for general use at Ihe Intemational Soils Congress held as Washington in 1927. This cla
Clay (size)

Sill (size)

I

M

0.002 (2_)

C 0.02

0.006

S",,'

I

I 02

0.06

M

I

C 0 .• F:: Fine

(0)

a,y

Ultra

any

c 0.2

j.4

0.6

MIT System

Sill j.4

0.006

C:: Coarse g,,,,d

Mo

c 2_

M

C 0.02

0.05

0.1

(b) International Oassjfieation

Sm,' Fine

Gravel

2.0 mm legend M:: Medium

Medium 0.25

(e) U.S. Bureau of Soils Oassification

Fig. 5.1. OllSSifiCiltion Systems.

0.2

C

0.5

Gravel VC 1.0 2.0mm VC:: ·Verycoarse

SOIL CLASSIFICATION

91

known as the Swedish classification system before it was adopted as InlermltionaJ system. However, the system was not adopted by the United States. In tbis system [Fig. 5.1 (b)1, in addition to sand, sill, and clay, a tenn mo has been used for soil particles in the size range between sand and sill. (3) U.s. Bureau of Soils Classification-This is one of the earliest classification systems developed in 1895 by U.S. Bureau of Soils (Fig. 5.1 (e)J. In this system, the soils below the size 0.005 mm are classified as clay size in contrast to 0.002 mm size in other systems. 1be soils between 0.005 mm and 0.05 mm size 'a rt; classified as silt size. Sandy soils between the size 0.05 mm and 1.0 mm are subdivided into four categories as very fmc, fine, medium and coarse sands. Fine gravels are in tbe size range of 1.0 to 2.0 mm. 5.3. TEXTURAL ClASSIFICATION Texture means visual appearance of the surface of a material such as fabric or cloth. The visual appearance of a soil is called its texture. The texture depends upon the panicle size, shape of particles and gradation of particles. The textural classificaCton incorporates only the particle size, as il is dimwIt to incorporate the other two parameters. In fad, all the classification systems b~d on the particle size, as discussed in Sect. 5.2, are textural classification systems. However, in soil engineering, the term textural classification is used rather in a restricted sense. The triangular classification system suggested by U.S. Bureau of Public Roads in oommonly known as the textural classification system (Fig. 5.2). lbe term texture is used to express tbe percentage of the three constituents of soils, namely, 5.1nd, sill and clay. 0100

~>-CIQY

o

, \

60~

\

\~

\ p

/

)0-----------Cloy

Silt

"

// I

Silty

toom

1000t--".---*"-""""""--;"";;____,~-"60;;----,)"'0----,:,"0Silt ('/. ) (Size O.OOS to 0.05 mm) fiB . .5.2. Textural cill85iftcalion System.

"'90,----,;;)">00

9'

SOli.. MECHANICS AND FOUNDA110N ENGINEERING

According to the textural classification system, the percentages of sand (size 0.05 to 2.0 mm), silt (size 0.005 to 0.05 mm) and clay (size less than 0.005 mm) are plotted along the three sides of an equilateral triangle. The equilateral triangle is divided into to zones, e.1ch zone indicates a type of soil. 1lle soil can be classified by determining the zone in which it lics. A key is given that indicates the directions in which the lines are to be drawn to locate the point. For example if a soil contains 30% sand and 20% silt and 50% clay, it is shown by point (P) in the figure. The point falls in the zone labelled Clay. Therefore, the soil is classified as clay. 'Ille textural classification system is useful for classifying soils consisting of different constituents. 'Ille system assumes that the soil does not contain panicles larger than 2.0 mm size. However, if the soil contains a certain percentage of soil particles larger than 2.0 mm, a correction is required in which the sum of the percentages of sand, silt and clay is increased to 100%. For example, if a soil contains 20% particles of size lager than 2 mm size, the actual sum of the percentages of sand, sill and clay particles is 80%. Let these be respcaively 12, 24 and 44%. The corrected percentages would be obtained by multiplying with a factor of l00/SO. Therefore, the corrected percentages are 15,30 and 55%. 1he textural c1assificatioo of the soil would be done based on these corrected percentages. In this system, the term loam is used to describe a mixture of sand, silt and clay panicles in various proportions. The term loam originmed in agricultural engineering where the suitability of a soil is judged for crops. The term is not used in soil engineering. In order to eliminate the term loam, the Mississipi River (USA) propC6td a Commission modified triangular diagram (Fig. 5.3). 'The term loam is replaced by soil engineering tenns such as silty Clay. The principal oomponent of a soil is taken as a noun and the less prominent KEY component as an adjective. For example, silty clay contains mainly particles of a clay, but some silt particles are also present. It must be noted that the primary soil type with respect to behaviour is not necessarily the soil type that constitutes the largest part of the sample. For example, the general character of a mixed soil is determined by clay fraction ii it exceeds 30%0Right Triangle Chart. Since the 1000;---;;:~~-''''':'---';;'-=--c;;;---;:'---;;;"" sum of the percenta'ges of sand, sill and SILT clay size particles is 100%, there is no need to plot all the three percentage. The percentage of sand particles can be Fig. S.J. Modified Triangular Di~ram. found by deduction from 100% the sum of percentages of sill and clay particles. It is possible t9 determine the textural classification by locating the point of intersed.OO of lines representing silt and clay. as shown in right.triangle chart (Fig. 5.4). The right-triangle chart is more convenient than the conventional lriangular chart as it involves only orthogonal arrangement of grid lines. 5.4. AASHTO CLASSIFICATION SYSTEM American Association of State Highway and Transportation Official (AASlITO) Oassification system is

SOtL CLASSIFICATION

93

useful for cl~ifying soils for highways. The particle size analysis and the plasticity characteristics are required to classify a soil. The classification system is a complete system which classifies both coarse-grained and fine-graillCd soils. In this system, the soils are divided into 7 types, designated as A-I to A-7. The soils A-I and A-7 are further subdivided into two categories, and the soil A-2, into four categories, as shown in Table 5.1. To classify a soil, its particle size analysis is done, and the plasticity index and liquid limit are determined. With the values of these parameters known, one examines the first (extreme left) oolumn of Table 5.1 and ascertains whether the known parameters satisfy the limiting values in tbat column. If these satisfy the requirements, the soil is classified as A-J-a. If these do not satisfy, one enters the lJJ 50 60 70 80 90 100 second oolumn (fcom the lerl) and determines SILT whether these satisfy the limiting values in Ihat column. The procedure is repeated for the next Fig. 5.4. Righi Trillngle chllrt. column until the column is reached when the known parameters satisfy the requirements. The soil is elassified as per nomenclature given at tbe top of that oolumn. The soil with the lowl!!;t number, A-I. is the most suitable as a highway material or subgrade. In general, the lower is the number of soil. the more suitable is the soil. For example, the soil A-4 is better than the soil A·5. In Table 5.1, the column for soil A-3 is to the left of the rolumn for soil A-2. This arrangement is only to determine the classification of the soil. This docs not indicate that soil A-3 is more suitable for highways than A-2 soil. Fine-grained soils are further rated for their suitability for highways by the group index (GI), detennined as follows:

GI - .(F -35)[0.2 + 0.005 (w,-4O)J + O.OI(F -15)(/, - 10) where

F:::: percentage by mass passing American Sieve

00.

200 (size 0.075 mm). expressed

... (5.1) a
a

wbole number. liquid limit (%) expressed a
WI ::::

Eq. 5.1 is somelimes expressed as GI

= 0.2 (F-35) + 0.005 (F-35) (WI -

40) + om (F-I5) (/,-10)

While calculating' Gl from the above equation, jf any term in the parentheses becomes negative. it is dropped, and not given a negative value. The maximum values of (F-35) and (F-15) are taken as 40 and that of (wI- 40) and (Ip - 10) as 20. The group index is rounded off 10 the nearest whole number. If the computed value is negative, the group index is reported as zero. The group index is appended 10 the soil type delennined Crom Table 5.1 . For example A-6 (15) indicates the soil type A-6. having a group index of 15. The smaller the value oC the group index, the better is the soil in that category. A group index of zero indicates 8 good subgrade. whereas a group index of 20 or greater shows a very poor subgrade. The group index must be mentioned even when it is zero to indicate that the soil bas been claMified as per AASlnu system.

1!

Table 5.1. AASHTO Classificadon System

Group CltUSi{icarion

Silt-clayMaJeria/s

Granular materials

Getleral Classificalion

More than 35% passing No. 2()() Sieve

(35% or less passing No. 200 Sieve (0.075 mm)

~-;:i...-A-l~

(0.075 mm) A-7

A---Z A-3

A-l-l>

A---l-4] A----2-S[

A---.2--6j

A-5

A---4

A--
A---.?-7

~ A-7--6

(0) Sieve Analysis;

Percent Passing (I) 2.00 mm (No. 10)

(ii) 0.425 mm (No. 40)

(ii,) 0.075

rnm

(No. 200)

50 '"" 3{) rna> 15"",

I '0.,.,

Simin

25 rna>

10 max

~="";:";'

(b) ~-~~ (,) Liquid limit

6 max

N.P.

(e) Usual types of signific8n1

Stooe Fragmenrs Gravel and sand

Fine Sand

I

I 3S max I 35 max I 35 max I 35 max I 36 min I 36 min I 36 min I 36 min 40 max

I

10 max

I

41 min

10 max

I

40 max

I

41 min

40 mal[

11 min

10 max

11 min

I

41 min

40 max

10 max

11 min

subgrade.

l'lF 41 min

~

11 min·

g: 3

Silty or aayey Gravel Sand

Silty Soils

aayey Soils

ConsIituenl materials (d) General rating as

I

Excellent., Good

• If plasticity index is equal 10 or less thaD (liquid Limit-30), the sal is A-7-5 (i.e. PL> 30%) If plasticity index is greater than (I.iquid limil-30), the sojl is A-7---6 (i.e. PL < 30%)

Fair 10 Poor

!A ~

g~ ~ ~

0

z

~ ~

~

.,

SOIL ClASSIFICATION

5.5. UNIFIED SOIL CLASSIFICATION SYSTEM The Unified Soil Classification System (USC) was rlrst developed by Casagrande in 1948. and later, in 1952, was modified by the Bureau of Reclamation nnd the Corps of Engineers of the United States of America. The system has also been adopted by Americ.1n Society of Testing Materials (ASTM). 1ne system is the most popular system for usc in all types of enginccring problems involving soils. The various symbols used are given in Table 5.2. Tuble S.2. Symbols used in USC System Symbols Primary

G

S M

c

o

p,

Secondary

Description Gravel Sao" Silt (Symbolh M obtained from the Swcdis word 'mo')

ao,

OrganiC poo,

w

Well.graded

P M C

Poorly graded Non-plastiC fines Plastic fines

L

Low Plasticity High plasticity

The system uses both the panicle size analysis and plasticity charaderistics of soils, like AASHfO system. In this system, the soils are classified into 15 groups (Thble 5.3). The soils are first cmssiried into two categories. (I) Coarse-grained soils-If more than 50% of the soil is retainOO 01] No. 200 (0.075 mm) sieve, it is designated as coarse--graincd soil. There are 8 groups of coarse--grained soils. (2) Fine-grained soils-U more than 50% of the soil passes No. 200 sieve, it is called fine-grained soil. There are 6 groups of fmc*grained soils. 1. Coarsc_grnined Soils-The coarse-grained soil., are designated a'i gravel (G) if 50% or more of coarse fraction (Plus 0.075 mm) is retained on No.4 (4.75 mm) s ieve; otherwise it is termed sand (S). If the coarse--graincd soils contains less than 5% fines and are well-grnded (W), they are given the symbols GW and SW, and if poorly graded (P). symbols GP and SP_ The criteria for well·grading are given in Table 5.3. If the coarsc*grnined soils contain more than 12% fines. these are designated as GM, Ge, SM Of SC, as per aiteria given. If the percentage of fines is between 5 to 12% dual symbols such as GW-GM, SP-SM, are used. Z. Fine-grained Soils-Fine-grained soils are further divided into two types . (1) Soils of low compressibility (L) if the liquid limit is 50% or less. These are given the symbols ML, CL and OL. (2) Soils of high compressibility (ff) if the liquid limit is more than 50%. These are given the symbols MIl, CII and OIl. The exact type of the soil is determined from the plasticity chart (Fig. 55). The A·line has the equation " = 0.73 (w,- 20). II scparntes the days from silts. When the plasticity index and the liquid limit plot in tbe hatched paction of the plasticity chan, the soil is given double symbol CL- ML. The inorganiC soil ML and Mil and the organic soils OL, OH plot in the same zones of the plasticity chart. The distinction between the inorganic and organic soiis is made by oven-drying. If oven drying dccrcnscs the liquid limit by 30% or more, the soil is classified organic (OL or Off); otherwise, inorganic (ML or MIl) Highly Organic Soils-Highly organic soils are identified by visual inSpection. These soils are termoo p"',(P,). 5.6. COMPARISON OF AASDTO AND USC SYSTEMS AASlITO system is for finding out the suit.1bility or otherwise of soils as subgrade for highways only.

.


Tuble 5.3. Unified Soil Classification Syslem Group Symbols

Major Division Coarsc-Graincd

Gravel (50% or

Soils.

more of coarse fraction retained on No.4 sieve

IMorethan 50% retained on No. 200 sieve (0.075 mm))

a""

Typical

GP

Well graded grovels Poorly graded gravels

GM

Silty grovels

GC

Gayey gravels

GW

Gravels

(4.75 mm)] Gravels

with fines

$and [more thon

50% ofooarse faction passing

Clean

s.",,,,

SW

Well-graded

"','"

Poorly graded

SP

"''''''

SM

Si[IY sands

SC

Oayey sands

ML

Inorganic sillS of low plasticity

No.4 sieve (4.75 mm») So"", with

grained Silts and clays Liquid 0< limit 50% or less passing No. 200 sieve (G.I175 mm))

[50%

more

CL

OL

SillS and days Liquid Limit greater than 50%

~:i ~

e;~~ ;.-il11

::;g ~

C" > 4 C~-1Io3

Not meeting both criteria foc GW AlIcrbergumits below A-line or plasticity index less than 4

Ancrberg

Limits in hatched area GM-GC

Auclberg Limits above A.Jine and plasticity

~~~l

index greater

~~~

to 3

thon?

ell :> 6 O_~'5 C~ _ I

NOI meeting both criteria for SW z~ Anerberg Limits Anerberg below A-line or Limits in plasticity index hatched area ,g~!1 [ less tban 4 SM-5C Atlcrberg limilS above A-line and plasticily index greater ~§:€£ than 7

·~l~

~ [".g>

~Hi!l

fines

Fine soils

Classification criteria

Mil

CH

Inorganic dayso£low 10 medium Imaslicilv Orgonicsills or low plasticity InorganlcsillS of high plasticity Inorganic days of high plasticity

na~

See Plasticity Chart (Fig. 55)

Orgnnic clays

011

of medium of high plasticity Peat. muck

Highty organic Soils

"

"'" oil""

highly organic soils

Visual-manual identification

SOIL CLASSIFICA110N

91

..

)0

U

\~

______ ."".,.,,-.""'"" " " /

OH

" '

0,

<

l"'li ' /,

Fig. 5.5. Plilsticity chm1 (USC). USC system is for detcrmining the suitability of soils for gencrJI use. Both the systems, however, have the same basis. They classify the soils according to the particle si:.r.c analysis and the plasticity characteristics. Both the systems divide the soils into two major categories, namely, coarse-grained and finc--grained soils. 'Ihe following differences between the two systems arc worth mentioning. (1) According to AASIITO system, a soil is tcnncd fine-grained if more than 35% passes No. 200 (0.075 mm) sieve, whereas in the USC system, if more than 50% passes Ihat sieve. In this respect, the AASl-rIU system is somewhat bcuer because the soil behaves as fine-grained when the percentage of fines is 35%, and thc limit of 50% in USC system is somewhat higher. (2) In AASH1U system, sieve No. 10 (2.0 mm si7.c) is used to divide the soil into gravel and sand, whereas in USC system, sieve No. 4(4.75 mm size) is used. (3) In USC system, the gravelly and sandy soils are clearly separated, whereas in AASHlU system, clear demarcation is not done. The soil A-2 in the laller system contains a large variety of soils. (4) Symbols used in USC system are more descriptive and arc more easily remembered than those in AASfHO system. (5) Organic soils are also classified as 01.. and 011 and as peat (PI) if highly organic in USC system. In AASHTO, thc[c is no place for organic soils. (6) USC system is more convenient to use than the AASHlU system. In the Jailer, the process of elimination is required whid1 is lime-consuming.

Tilble 5.4. Approximate Equivillence Between AASnm and USC System MSHTO System

usc system (most probable)

A-l-a

GW,OP

A-l-b A-2-4 A-2-5 A-2-6

SW, SM, OM. SP

A-2-7 A-3 A-4 A-S A-6 A-7-5 A-7.{'J

aM, SM OM, SM

GC,SC aM, OC, SM, SC

SP ML, OL, Mil, OH MIl, OH, ML, on CL on, MH,CL,OL CH,CL,OH

SOIL MECHANICS AND FOUNDATION ENGINEEJUNG

98

Table 5-4 gives approximate equivalence in both the SystCffiS. If the soil has been classified according \0 onc system, its classification according \0 the other can be determined. However, the equivalence is only approximate. For exact classification, the corresponding procedure should be used.

5.7. INDIAN STANDAIID CLASSIFICATION SYSTEM Indian Standard Classification (Isq syslcm adopted by Bureau of Indian Standards is in many respects simiLar \0 tbe Unified Soil Qassification (Usq system. However, there is one basic difference in llle classification of fine-grnincd soils. The finc- grained soils in ISC system are subdivided into three categorics of low, medium and high compressibility instead of two categories of low and high compressibility in USC system. A brief oUlline of Qassif}cation and Identificalion of Soils for general enginccring purposes (1S: 1498- 1970) is given below. For romplete details, the reader should ronsult the code. ISC system classifies the soils into 18 groups as per Tables 5.6 and 5.7. Soils are divided into three brood divisions: (1) Coarsc-grained soils, when 50% or more of the total materiaL by weight is retained on 75 microlllS sieve. (2) Fine-grained soils, when more than 50% of the total material passes 75 micron IS sicve. (3) If the soil is highly organic and contains a Large percentage of organic matter and particles ct decomposed vegctrltion, it is kept in a separate category marked as peat (P,), In aU, there arc 18 groups of soils: 8 groups of coarse- grained, 9 groups of fine-grained and one of pea. Basic soil components are given in Table 5.5. Symbols used arc the same as in USC system (fable 5.2).

Table 5.s, Basic Soil Components in ISC System Soil

(i) Coarse·

brained components

Soil componenis

Panir:le sire ran~ and den;riplioll

Symbol

Boulder

None

Roundl'tl to angular, bulky, hard, rock, particle; average diameter more than 300 mm

Cobble

None

Rounded 10 angular, bulky. hard, rock panicle; average diameter smaller than 300 mm bul ralained on 80 mm IS sieve

Gravel

G

Rounded 10 angular, bulky, hard, rock: particles; passing 8l mm IS sieve but retained on 4.75 mm IS sieve Course: 80 mm to 20 mm IS sieve

.'Ine : 20 mm to 4.75 mm IS sieve Rounded to angular, bulky, hard, rock particle; passing 4.75 mm IS sieve, but retained on 75 micron sieve Course: 4.75 mm to 2.0 mm IS sieve Medium: 2.0 mm [0 425 micron IS sieve Fine: 42.') micron [0 65 micron IS sieve

"'' ' ' (iI) Fine-grained components

Silt

M

a.,

Pllnides smaller [han 75 miaon IS sieve; identified by behaviour, that is, slightly plastic or non-plastic reganlles of moiscure and exhibits lillie or no strength when air-dricd.

c

Pllnicles smaller man 75 micron IS sieve : identified by behaviour, that is, it can be mOOe to exhibit plastic: propcnies within a certain considerable scrength when ail dried.

OrganiC maTtl!r

0

Orgonic moner deromposition.

in

various

sizes

and

stages Ii

NG

SOIL CLASSIFICPinON

99

SO

: to Diy 40

CL ",Is

the

;05

"

CI

CH

30

,""-,1"

,.,,~e

•

.9-

SC

S,

():,C>

vi'

20 ~

'"

:~ IS

£

MH 10 7

0'

------

4 ---- __

00

10

OH

MI l-ML

0'

01

ML -.2(' Ol

20

of

J5

30

liquid Fig. 5.6.

40

50

limit

PI~slicily

I"I)

',.

60

70

00

Chart (ISC)

\. Coarse-grained &ils----Coarse-grained soils are subdivided inlo grovel and sand. lhc soil is termed gfllvel (G) when more than 50% of coarse fraction (plus 75~) is retained on 4.75 mm IS sieve, and termed sand (S) if morc than 50% of the coarse fraction is smaller Ihan 4.75 mm IS sieve. Coarse-grained soils are further subdivided as given in Table 5.6 into 8 groups.

2. Fine-grained Soils---lbe fine-grained soils are fun her divided into three subdivisions, depending upon the values of the liquid limit:

ge

(0) Sills and clays of low compressibility-These soils have a liquid limit less than 35 (represented by symbol L). (b) Sills and clays of medium comprcssibility-These soils have a liquid limit greater than 35 but less

g' IS

30

75

'Y

'" ~.

'Y

:ic

u,

of

than 50 (represented by symbol I).

(c) SUts and Clays of high compressibility-These soils have a liquid limit greater th
5.8. nOUNDARY CLASSIFICATIONS Sometimes, it is not possible to Classify a soil into anyone of 18 groups discussed above. A soil may possess characteristics of two groups, either in particle size distribution or in plasticity. For such C.1SCS, boundary classifications occur and dual symbolS arc used. (a) Boundary classification for coarse-grained soils

The following boundary classification can occur: (I) Boundary classifications within gravel group or sand group can occur. The following classification are common.

GW--{;P, GM-GC, GW-GM, GW--{;C, GP-GM SW-5P, SM-5C, SIV-5M, SW-5C; SP-5M While giving dual symbols. first writc a coarser soil then a finer soil. 1(2) Boundary classification can occur between the gravel and sand groups such as GW-Sw. GP-Sp, GM-SM, and GC-SC The rule for ~ classification is to favour the non-plastic classification. For c1C.3mple, a gravel with

10% fines, C" = 20 and Ce = 2.0 and lp = 6 will be classified as GW-GM, and not GW-Gc.


100

lil.ble 5.6. Classification or Coarse-grained Soils (lSC System) Dil;siml (1) Come-

""'' ' '

soils (More than half of material is

larger than 75-mieron

Subdi\lis;CNI Gravel (0) (more than hal[ofcoorse fmaion is larger than 4.75 mm IS sieve)

IS sieve size)

dean

G=p

lAboratory Criteria

(l)GW

Well graded grovels

Co. grnterthan 4

(2)GP

Poo
Not meeting all gradation

graded

requiremcflIsforGW

gravels

(Fines less than 5%)

Typical

s)""bol C~

between I and 3

gravels Gravels

(;\)GM

with

Silty gravels

appreciable

Ancrberg Limits below

:Jmount of

A-line or /p

fi nes (Fines

11!SSIhrin4

more [han

12%)

(4) GC

ChJYcy gravels

Alterberg limits above A-line and fp

grater

than 7 Sand (S) (More than half of coarse fraction is smaller than 4.75 mm IS sieve)

Clean sands (Fines less than 5%)

(5)SW

(6) SP

Aucrberg Limits plotting above A-line with lp bclwen 4 and 7 are

requiring dual

symbols such as GP-GM, SW-SC,

symool

GM-GC

e"

Poorly-

Not meeting all gradation requirements for SW

gr.>dod

border line .",.,.

border-line cases requiring use ofduul

Wellgraded ",ds

e~

Remark When lines are between 5% 10 12%,

greellter than 6 between I and 3

"",ds Sands with appreciable amout of fines (Fines more than 12%)

(7)SM

silty

""'' ' (8) SC

Claycy "",ds

Atterbcrg Limits below A-Jincor Ip Jess Ihan 4 Attcrberg limits above Aline with Ip grealer Ih3n7

Alleraerg's urnils plotting above A-line with Ip between 4 and 7 are borderline'l::ases requiring use of double symbols SM-SC

(b) Boundary classllication for fine-grained soils.

(1) Within the same compressibility subdivision, such as ML--CL, ML-OL, CL-OL; CI-MI, MI-OI, CI-OI; MfI-CfI, MH-OfI; CII-OII. First write a coarser soil when there is a choice and then a finer soil. (2) Between low and medium oompressibil ity. such as ML-Ml, CL-Cl, OL-Ol (3) Between medium and high compressibility ML-MH, Cl -CH, 0/-011 (c) Boundary Classilicatlon between coarse-grained a nd li ne-gruined soils. Boundary classificalion can ocx::ur between a coarse-grained soil and a fine-grnincd soil, such as SM-ML, SC-CL

SOIL ClASSIFICATION

101

Thble 5.7. Clas."mClltion or Fine-grained Soils (lSC System) Group

Laboratory Crileria (ue Fig 5.6)

DhisiOlJ

Subtiil'isiOll

Symbols

Typicaillames

2) Finegruined soils (more than

Low· compressibility (L) (Liquid Limit less tnan 35%)

{l)ML

Inorganic silts with nOne 10 low plasticity

Atterbcrg limits plol below A-line or /p less than 7

(2) CL

Inorganic clays of low plasticity

Altcrberg limits plot above A-line andJp greater than 7

(3) OL

OrganicsiUs of low plasticity

Atlcrbcrg limits plot below A·line

(4) MI

Inorganic sillS ofmcdium plasticity

Atterberg limits plot below A·line

(5)CI

Inorganic clays of medium plasticity

Auerbcrg limits plot above A-line

(6)01

Orgaic silts of medium plasticity

Alterbcrg limits plot below

(7)MH

Inorganic silts of high compressibility

Auerberg limits plot below A-line

(a)eB

loorganic clays of high plasticity

Auerberg limits plot above A-line

(9)OH

Organic clays of medium 10 high plasticity

Altcrbcrg limits plot below A-line

1'<

Peat and oth~r highly organic soils

Readily identified by colour, odour, spongy feel and fibrous .

SO% """ 75~ IS sieve)

Inteonediate compre,<;sibility (I) (Uquid limit greater than 35% but less than 50%

High compressibiliy (11) (Liquid limitgrealcr than 50%)

(3) Highly organlcsoil

Anerberg limits pioting above A-line with Jp

-='10 7

(hatched zone) ML-CL

Remarks (1) Organic "d inorganic soils plotted in the same zone in plasticity chan are distinguished by odour and colour 0' liquid ,,~ aftcr limit ovcn-clrying. A roduaion liquid limit after " ovcn- drying to a than value three- founh of the liquid limit before ove,· drying

,,,'

positive

"

identification of organic soils.

",,,,,

(2) amon soils 01 India lie along a band partly A-line 'bo,,, and panly below tho Aline

.h,

A-lillC

See. plasticity chnrt (Fig. 5.6)

5.9. FIELD IDENnFlCAll0N OF SOILS

~ye

The soils can be identified in the field by conducting the following simple tests. The sample is filSt. spread on a flat surface. If more than 50% of the particles are visible 10 the naked (unaided eye), the soil is coarse-grained; otherwise, it is fine grained. The One- grained particles are


102

smaller than 751l size and are not visible to unaided eye. lbc fraction of soil smaUer than 7511 size., that is, the clay and sill fradion. is referred to as fines. (1) Coarse-grained Soilr-If the soil is coarse-grained. it is further identified by estimating the percentage of (a) gravel size particles (4.75 mm to 80 mm), (b) sand size particles. (75J.L to 4.75 mm) and (e)

silts and clay size panicles (smaller than 7511 size). Gravel panicles are larger than 4.75 mm size and can be identified visually. If the percentage of gravel is greater Ihan that of sand, the soil is a grovel; otherwise, it is sand. Gravels and s.1nds are further classified as cle.-m if they contain fines less than 5% and as dirty if they contain fioes more than 12%. Gravels and sands containing 5 to 12% fines are given ooundary classification. The fine fraction of the coarse-grained soils is identified using the procedure given below for fine· grained soils to determine whether it is silty or clayey. To difJercntiate fine sand from silt, dispersion Icst is adopted. When a spoonful of soil is poured in a jar full of wa:er, fine sand settles in a minute or so. whereas silt t.'1kcs 15 minutes or more. (2) Fine·grained soils--U the soil is fine-grained, the following tests arc conducted for identification 00 the fmetion of the sOil finer than the 425-micron IS sieve to differentiate silt from clay. (a) Dilatancy (reaction to shaking) test-A smaU pat of moist soil of aboul 5 ml in volume is prepared. Waler is added to make the soil soft but not sticky. "be pal is placed in the open palm of one-hand and shaken horizontally, striking against the other hand several times during shaking. If the soil gives a positive reaction, the water appears on its surface which changes t("l a lively roosistcncy and appears glossy. When the pat is squeezcd between the fingers, Ihe watcr and gloss disappear from the surface, It becomes stiff and ultimately crumbles. 'fl1e rapidity with which water appc.'1rs on the surface during shaking and disappears during squeezing 1<; used in the identification of fine-grained soils (pJbles 5.8). The larger the S:7..e of tbe particles, the quicker is the reaction. The reaction is called quick if water appears and disappears quickly. The reaction is tcnned slow if water appears and disappc.'1rs slowly. For no retlction, Ihe water docs nol appear at the surface. (b) Toughness test-The pHI used in Ihe dil:lt:lncy test is dried by working and remoulding until it has tbe consistency of pUlly. 'Ibe lime required to dry the pal depends upon the plasticity of the soil. 'Ibe pat is rolled on a smooth surface or between the palms inlo a threads of aboul 3 mm in diameter, The thrc.'ld is folded and re- rolled to reduce tbe water is soil, due to cvaporation by heat of hand, until the 3 mm diameter thread just crumbles. The water content at that stage is equal to the plastic limit and the resistance to moulding at that stage is called the toughness. After the thread crumbles, Ihe picces of the sample are lumped together and subjected to kneading until the lump also crumbles. lbe tougher the thread at the plastic limit and the stiffer the kneaded lump just before it crumbles, the higher is the toughness of the soil. The toughness is low if the thread is weak and the soil mass cannot be lumped together when drier than plastic limit. TIle toughness is high when the lump can be moulded drier than plastic limit and high pressure is required to roll the thread. The toughness depends upon the polency of the colloidal clay.

i

Table 5.8. Field Identification Tests

T", (a) DiJDlancy (b) lbughness

(c) Dry strength

ML

CL

OL

MI

CI

01

Mil

CH

OH

Quick

None 10 very slow

Sl~

Quick 10 slow

Noo,

Slow

Stow 10

None

NonclO very slow

None

Medium

Low

None

Medium

Low

Low 10 medium

lIigh

None of low

Medium

Low

Low

Medium to high

Low to medium

Low to medium

High 10 very high

(oj Dry st"mgl" I..I-Th, 1"" of the

w;J

Low "

medium

i, completely dried by ak drying, ,un drying

0'

Medium 10 high

ovcn'd'l'' : J

SOIL CLASSIFICATION

103

The dry strength is determined by breaking the dried pat and crumbling it betwcc.n finger.;. The dry strength is a mea<>ure of plasticity of the soil. The dry strength depends upon the colloidal frndion of the soil. The strength is termed high if the dried pat cannot be powdered at all; medium, if considerable pressure is required; and low, if the dry pat can be easily powdered. Table 5..8 can be used for the field identification of different soils.

S.IO. GENERAL CIIARACfERISTICS OF SOILS OF DIFFERENT GROUPS General characteristics of the soils of various groups as classified by ISC system and USC system are given in Table 5.9. The information given in the table should be considered as a rough guidance about the engineering properties of soils. For complete information. the tests should be oonductcd and the engineering properties determined. Thble 5.9. General PropeUes or Soils Soi/Group

Permeabiliry

Compres,~ibility

SlIear Slrenglh

Workability

Negligible

Exccllent

Excellent

Negligible

Good

Good

Negligible

Good

Good

Very low

Good to fair

Good

Negligible Very low Low

Excellent Good Good

Excellenl Fair Fair

Low

Good to fair

Good

(a) Gravels

GW GP GM GC

Pervious Very pcrvioUo'l Semi-pervious impervious Impervious

'0

(b)Snnds

SW

Pervious

SP

Pervious Semi.pervious to impervious Impervious

SM

SC (c)l..ow&.medium Plasticity silt &. clays

ML,MI

Semi-pervious to impervious

Medium

Fair

Fair

CL,CI

Impervious

Medium

Fair

Good to fair

OL,OI

Scmi-pervious to impervious

Medium

Fair

Fair

Semi.pervious to impervious Impervious

High

Fair to poor

Poo<

CII

High

1'<""

Poo<

OH

Impervious

High

"""

Poo<

(d) HIgh. plasticity silts&clllys MH

Note. Highly organic SOils (PealS) are not used In englneermg works.

ILLUSTRATIVE EXAMPLES Dlll'ltrative Example 5.1. A sample of soil was tested in a laboratOf)', and the following observations were recorded: Liquid Limil ::: 45%,

Plastic Limit

= 16%

SOIL MECHANICS AND· FOUNDA110N ENGINEERING

l<" U.S. Sieve No.

I

No.4

Percentage Pa&<>ing

I

100

No. 10 (2.0mm)

I I

91.5

I

I

No. 40 (0.425 mm)

I

No. 200 (0.075 mm)

I

SO.O

60.0

Classify the soil according to MSlflV system. Solution. Plasticity Index = 45 - 16 = 29% Referring to Table 5.1, and proceeding from the extreme left column to right, the first column in whidl the

properties fit is A-7. 1b a<>certain whether the soil is A-7-5 or A-7-6. the value of (wl- 30) is required. In this case, wl- 30 = 45 -30 = 15% As Ip > (WI - 30), the soil A-7-6, From Eq. 5.1. u\king F = 60,

Group Index = (F - 35) [0.2 + 0.005 (w/ - 40)J + 0.01 (F - \5)(Jp -10) GI " (GO - 35)[0.2 + 0.005 x 5)] + 0.01 (40)(19) = 13.3, S,ly 13.

The soil is classified as A-7-6 (13).

Illustrative Example 5.2. Classify the soils A and /1, with Ihe properties as shown below, according ID USC system. Soil

I

w/(%)

I

I I

45

fp(%)

29

I

I

% passing No.4 sieve

I

% possing No.2(){)sieve 59

100

B5

" more than 5()% passes No. 200 sieve, the soil

100

Solution. (a) Soil A. As is fine·grdincd. As WI is Less than 50%. the soH is of low plasticity. 'mc Atlerberg limits plot above the A-line in Fig. 5.5. 'Ille soil is classified as CL. (b) Soil 8. The soil is fine-grained. As the liquid limit is greatcr than 50%. the soil is of high compressibility. The Allcrberg limits plot below A-line. It can be either MH or OH. If the soil is OH, ilS liquid limit will decrease considerably on oven·drying. lIIustrntive Example 5.3. Classify th~ soil with the following properties according to ISC system. Liquid Limit

I

PlassidlY index

I

40%

I I

10%

% passing

4.75 mm sie\'e 60%

I

% passing

75JAsieve

I

45%

Solution. As more than 50% is rctClincd on 75", IS sieve, the soil is marse-grained. Coarse frJction

= 55%;

Gravel fraction

= 40%;

Sand frdcUon

= 15%

As more lhan half the coarse-fraction is larger than 4.75 mm IS sieve, the soil is gravel. The soil has more than 12% fines. it can be either GM or GC. As the Anerberg limits plot below A·lioe (Fig. 5.6), the soil is GM.

Illustrative Example 5.4. Fig E 5.4 ,fIIOWS the grain size distribwion curves for two soils A and B. Tht plasticity characteristics of the sails are given below.

son

A Soil B

Liquid Limit = 40%; Liquid Lilll/'t = 28%;

Plasticity Index = 10%;

Plasticity Index = 12% Classify lhe soils according to IS classification and COII/menl on their sheor strenglh.

I

SOIL Cu\SSJFlCATION

t05

~.o~~--~~~-v?-rH~ z

'"ffi

2of--l-----1f-r-:;l-""--t--t-i-t-i

Q.

(mm)

Fig. E-S.4. Solution. (a) Soil A. As more than 50% pffiSCS 75~ sieve. the soil is line-grained. The Allcrberg limits plot below A-line (Fig. 5.6) in the zone of intermediate compressibility. It can be either MI or 0/. If the liquid limit reduces \0 thrcc-founh of the original value or more on oven drying, it is IS; oIherwise MI. (b) Soil 8. As more than 50% of Ihe Iolal material is larger than 75 I-' sieve. the soil is coarse- grained. Coarse fmction

Gravel fmetion Sand fmelion

= 87%, = 37%;

= 50%.

As more than half of coarse fraction is smaller than 4.75 mm sieve, the soil is sand. As fines are more than 12%, the soil can be SA{ or Sc. As the Atteroerg limits plot above A· line (Fig. 5.6), the soil is Sc.

PROBLEMS

A. Numerical 5.1 Allerbcrg ]imil ICstS were carried out on 11 soil sample, with the following rC5ults: Liquid limit'" 40%; Plastic limil '" 2S%. Oassify Ihe soil according to Unified Oassifjeltion system and the Indian Standard classification system. [Ans. CL; CI] 5.2. The follOWing results were obtained [rom Ihe classification tests of a soil. Percentage passing 7Sjl sieve = 40% Liquid limil = 35%; Plastic Limit = 15% calculate the group index of lhe soil and dassify il aocording 10 AASl-ITO system. [Ans.4; A-6(4)] 5.3. The sieve analysis of a soil gave the following results : % passing 75~ sieve:: 4; % ret:lined on 4.75 mm sieve'" 50 Coefficient of curvature = 2; UniformilY ooefficienl = 5 Classify Ihe soil according to ISC sySlem. [Ans.GWJ 5.4. The sieve analysis of a soil gave the following results: %passing 75~ sieve ", 8; % retained on 4.75 mm sieve", 35 Coefficient of curvature '" 2.5; Uniformily ooefficient ~ 7 The fine frnClion gave the folJowing results : Plasticity index = 3; Liquid Limit = 15. [Am;. SW-SM} Classify Ihe soil according 10 ISC system. 5.5. Ascii has Ibe following charnCieristics: % ~ng 75~ sieve = 58%; liquirl Limit = 40% Plasticity Index = 10%; liquid lim!1 of oven-dried sample", 25% Classify the soil according 10 ISC syslcm. [Ans.OI]

106


B. Descriptive and Objective Type 5.6. What is the use of classll'ication o f soils? Discuss Indian Standard Classific.::uion system. 5.7. What IS the difference betwC\:n Ihe classiHcation based on particle size and the textural classification? Oiscuss the limit:llions of the two systems. Comp,m~ the AASHTO clns~ilic;\tion system nnd Unifk'<.l soil c lassification system. Why the latter system is morc commonly used? S.? Give the slcp~by-~1Cp procc.dure for r.:lassification of a soil by India n st:lndard classification system. 5.10. Discuss rickl idcnlilil'),stem. the soils with particle sib! larger than 4.75 mm arc classified as gravel. (c) The texture of a soi l dcpend~ onl)' on the particle sIze. (d) In AASHTO sy~ tcrn. the soil A-2 is better than the soil A-3. (e) The group index of b!ro indicates :t soil of very poor qunlity. (j) The group index of a wil cnn be negmivc. (8) According to USC system. the fine.grai ned soils are of 9 types. (h) The !'>Oil with pnrtieles size less th:m 211 arc clayey soils. (I) A co.arsc·gramcd sand is wcll.grnded if C.· = I to J and C~ is greater than 6. Oven-drying reduces th e liquid limit of an inorganic soil considerably. (k) The fine.grnincd soils with a high percentage of colloidal froction ha ve high d ry strength. (Ans.True,(d),(i),{k») 5.8.

v)

C. Multiple·Choice Questions t. IS classification ()fsoil is in many respects simi lar to «(I) AAS HTO classificmion (b) Tcxlurn! classitkation (t') Unified soil elilssilication (d) MIT clnssificmion 2. The maximum Sill! of pMticks of silt is (a) 75 11 (b) 60 11 (e) 2 11 (d) 0.2 11 3. The maximum Si7.1! of parl iclc.~ or clay is (0) 0.2 mm (b) 0.02 mm fe) 0.002 mm (d) 0.0CI02 mm 4. Acconling to IS classifiC<11ion system. the soils can be cl.1ssilit"(,,1 into «(I) 15 groups (b) 18 groups (e) 3 groups (d) 7 groups 5. The soils which pl01 above the A line in the pl.1sticity chart flrc «(I) cl!Jys (b) silts (e) sands (tl) organic soil s 6. A silty soi l gives a positive reaction in (a) Toughness tes t (b) Dilmancy test (c) Dry strength test «(I) None of above 7. A soil has the liquid li mit of 30. TIle cQrresponding plasticity index given b)' the A·li ne is (tl ) 7.3 (b) 7.5 (e) 9.0 (d) 9.5 8. The max imum value of the term (F. I 5) in the group index is taken as (a) 20 (b) 30 (c) 40 (d) 60

6 Clay Mineralogy and Soil Structure 6.1. INTRODUCTION The coarse-grained soils generally contain the minerals quartz and feldspar. These minerals are strong and electrically inert. The behaviour of such soils docs not depend upon thc nature of the mineral present. The behavior of fine-grained soils, on Ihc other hand. depends to a large extent on the nature and characteristics of the minerals presenl. The most significant properties of clay depend upon the type of mineral. The crystalline minerals whose surface activity is high are clay minerals. These clay minerals imparl cohesion and plasticity. The study of clay miner.lls is essential for understanding the behaviour of clayey soi ls. Clay mineralogy is the the science dealing with the structure of c lay minerals on microscopic, molecular and atomic scale. II also includes the study of the mineralogical composition and electrical properties of the clay particles. The study of clay minerals is important for particles smaller than about 2 micron size. Soil struclIlre means the geometrical arrangement of soil particles in a soil mass. It is concerned with the shape. si7..e and orientation of particles. If the individual particles are packed very close to one another, the void ratio is low and the soil is dense and strong. If the particles are so arranged that there are more voids, the soil is loose and weak. Engineering properties and behaviour of both coarse-grained and fine-grained depend upon the structure. This chapter is mainly devoted 10 clay mineralogy. The soil struclure is considered in the last section. In fact. clay mineralogy also discusses the structure of clayey soils nOi as a whole mass but at a particle level.

6.2. GRAVITATIONAL AND SURFACE FORCES The gravitational force in a soil particles is proportional to its mass. As the specific gravity of particles is approximately constant, the gravi tational force is proportional to the volume of the particle. TIle volume depends upon the particle size. Thus. the gravitational force on a particle is related to the particle size. In ' other words. the larger the particle size, the greater would be the gravitational force. Bonding or surface forces betwecn particlcs depend upon lhe surface area of the particles and not upon the volume. The surface area also depends upon the particle size .. However. the surface forces become more important only when the paticle size is small. As the particle size decreases. the effect of surface forces on a particle becomes more predominant than the gravitational force. The re lative magnitude of volume and the surface area can be judged if we consider, say, a cube whose each side is 10 mm (volume = 103mm\ When the cube is subdivided into smaller cubes. the ratio of the surface area to the volume increases, as shown in Table 6.1. The ratio increases ten thousand times when the side of the smaller cube becomes I micron. The magnitude of the surface area per unit volume (or mass) is known as specific l·urface. The particles of coarse-grained soils are larger than 0.075 mm size. For such soils, the ralio of surface area to the volume is relatively small. These soils do not possess pla~ticity and cohesion which are predominant only when the surface forces .are large. In fine-grained soils, the gravity forces are relatively insignificant compared


108

Tuble 6.1. Ratio or SUrfllce area to Volume Surface area

Side

Number of

Length

"be>

1.

10mm

1

2-

1 rom

10'

3.

0.1 nun

10'

6xlolxl 6 x 106 x 0.01

4.

0.01 nun

10

6 x 109 x (0.01)2

5.

0.001 mm (I,)

12 10

6 x 10 12 x (0.001)2

S.N.

~rfQCe

Vofumll!!

area

""=e

J 2 (mm /mm )

Ht'mml

600 mm2

Q.60 6.0 60.0 600.0 6000.0

with the surface [orces. The fine~grained soils possess the plasticity characteristics depending upon the surface area, the type of minerals and the nature of environment present around thc soil particle. A material in which the surface forces arc predominant is known as a colloid. ll1c lenn colloid has been derived from Greek words kolla and Didos, meaning a glucy material and alike. For colloids. the ratio of the surface area to the volume is very large. It varies between 6(X) to mm2/mml:r1le dayey soils with particles smaller than 2 micron size arc generally colloidal in nature. The colloids have very large speciflc surface.

las

6.3. PRIMARY VALENCE BONDS Primary valence bonds hold togethcr the atoms of a molcrule. These are of two types: (1) Ionic bond, (2) Covalcnt bon(1. 1. Ionic bond-In an atom, the electrons carrying a negative charge revolve about Ihe nucleus. Sane elements have an excess or a deficiency of the electrons in the outer shell. One alom joins another alom by adding some of the electrons to its outer shell or by losing some of electr0n5 from ilS outer shell Fer example, an atom of sodium has an exress electron in its outer shell and an atom of dllorinc has one deficient elearon in its outer shell. A molecule of sodium chloride is fonned by ionic bond when an atom of sodium combines with an atom of chlorine. TIle atom which loses an ion becomes a JXl!>itive io!] (cation) and that which gains an ion becomes a negative ion (anion). In ionic bonds, the forces bind the positive ions and negative ions.

The number of electrons required to oomplete the first six shells individually are respectively. 2, 8, 8, i.8, 18 and 32. The total number of electrons required to oomplete are, IhereCorc , 2. 10, 18. 36, 54 and 86. The deficiency or excess of electrons in a particular shell of an element is determined from the number of

electrons available and that required to complete the outersheU. For example, aluminium has 13 electrons. It has an excess of 3 electrons over the second shell (total 10 ekx:trons). IJkewise, oxygen whiCh has 8 electrons, lack 2 elec:Irons in the second shell (total 10 electrons). An atom of hydrogen has equal excess and deficiency. It has only one electron which can be oonsidered either as one deficient in the first shell or one excess elearon. Likewise, the alom of silicon has 14 electrons which has equal excess and deficiency of 4 each. It has an excess of 4 over the second shell or a deficiency of 4 in the third shell (total 18 electrons). See lbble 6.2 for ionic structure of various elements. The atoms of two different elements combine to satisfy their individual deficiency or excess. For example.. when aluminium and oxygen combine two atoms of aluminium (excess 6) combine with 3 atoms of oxygen (deficiency 6) to form aluminium oxide (Fig. 6.1).

~ Ai)

+, A.I

~

1

0

~

~~ 61 ~

~ ~ Fig. 6.1. Aluminium oxide

fl

CLAY MINERALOGY AND SOIL STRUCllJRE

109

Table 6.2. Ionic Strudure of Various AtORlli S. No.

Element

Hydrogen

Symbol

Number of

Defficiency in

E/~/ron

outershcll

Excess in oulershell

Remork

-I

• I

Can either lose or goin one ion

II

2.

Oxygen

0

3.

Silicon

Si

14

Aluminium f<""", Calcium Sodium Potassium Magnesium Chlorine

AJ

!3

F,

26 20 II I.

.. 5.

6. 7.

8. 9. 10.

Ca

No

M, K

CI

..

-2 -4

• I

.1

12 17

Can either lose or gain 4 ions

+3 .8 .2

.2 -I

2. Covalent Bond-Covalcnl bond develops between two atoms by sharing of electrons in their outer sheU. lWo atoms, each lacking one electron, may combine by sharing of a pair of electrons. Likewise, two atoms, each lacking two electrons, may combine by sharing four electrons. For example. the bond between two atoms of oxygen in a oxygen molecule is a covalent bond. Each atom Lacks 2 electrons in the outer sheU. The two atoms bond by sharing 4 electrons in their outer sheUs. In other words, a covalent bond occurs when there is sharing of electrons by atoms of like valence. 'The covalent bond occurs generally in clements of negative valences or in non-electrolytes. such as carbon. (A non-electrolyte does not form ions). Primary valence bonds are very strong. These do not break in normal soil engineering applications. lbcrefore, primary valence bonds are not of much relevance in soil engineering. However, the study of ionic structure is useful in understanding the behaviours of various atoms. 6.4. IIYDROGEN DOND

The hydrogen alom has only one elcctron. The number of electrons required to fill the first sheU is 2. The atom can be oonsidered either as a Oltion (with one excess electron) or an anion (with one electron deficiency). The bond between the hydrogen :+ cation (HI and anions of two atoms of aoother element is caUed the hydrogen bond. The hydrogen atom is attracted by two atoms instead of only one atom as suggested by its ionic struaure. The hydrogen atom cannot decide to which of the two atoms it should

:~ =~;t:~;. ~~e=t i~~: ~~:;:

~H

r~~~OGEN

•

L

,O'=> H+

0-2

~

H;r----L-{

or the hydrogen bond is the bond between the Fig. 6.1. Hydrogen Botrd. hydrogen atoms and oxygen atoms in a water molecule. The hydrogen atom links one molecule or water to the other (Fig. 6.2). A hydrogen alom can bond only two atoms of oxygen, as it is small in size and can fit in only two anions which are of large size (Fig. 63). In other words. only two anions can approach the hydrogen cation close enough to fonn a hydrogen bond. Only the strong electro-negative atoms, such as oxygen and chlorine, can join with hydrogen to form a hydrogen bond. The hydrogen bond is oonsiderably weaker than primary valence bonds. However, it is fairly strong and C3MOl be broken during nonnal soil engineering problems.

no

SOIL MECllANfCS AND FOUNDATION ENGINEERING

6.5. SECONDARY VALENCE BONDS

. ~

CQti"H'

Secondary valence bonds are intermolecular bonds which develop between atoms in one molecule to atoms in another molecule. A molecule is eleariC311y neutral, i.e., it has no charge. However. the construction of the molecule may be such that the centres of Ihe negative and pooilive charges do not exactly coincide. 1be molecule may behave like a small bar magnet, with

o-2-

-2

0

-

AnionJ

two electrical poles. Consequenlly, an electrical moment is developed inside Fig. 6.3. A Cllatioo joining the molecule. A molecule with such a structure is called a dipole. In nature, two anions. two dipolar molecules orient themselves in such a way that net attraction oc:cun;. The attractive forces so developed are known as Vander Waul Forces, after Vander Wool who POOlulalcd the existence of a rommoo attractive forces between molecules of all matters in 1873. Vander Wanl forces develop due to anyone of the following three effects. (1) Orientation effect-This effect occurs between the oppositely charged ends of permanent dipoles, as shown in Fig. 6.4. (2) Induction effect- Even in a non~polar molecule, a pole can be + induced. When a non-polar molecule is ;=====~ placed in an electric field, it gets + polarised and slans behaving as a ' - -_ _-'CJ dipole. Induction effect occurs between (a) (b) an induced pole and another dipole. (3) Dispersion effect-As all electrons oscillate, the centre of negative charges goc:s on changing (e) periodically. This results in the Fig. 6.4. Orienlalion Effect. fonnation of a temporary, fluctuating pole. Dispersion e[fect occurs between a fluctuating pole and another dipole. As all moleaJles behave as permanent or induced or fluctuating dipoles, Vander Waal forces are always present in molecules. These exist in all matters. TIle relative magnitude of orientation. induction and dispersion effects in a water molecule are 77%, 4% and 19% respectively. Thus the orientation effect is the most predominant effect. A common example of secondary valence bond is the attractive force between molecules of water: 'Ibe water molecules . act as a bar magnet because the positive and negative charges are not centrally located. It may be noted that all liquids arc not dipoles. Some of the liquids. such as kerosene and carbon tetrachloride, are non-polar, as shown by @---..----C!) construction in Fig. 6.5. Vander Waa! forces also develop between the surfaces of two parallel particles of clay mineral. separated by water. The magnitude of the forces depend upon the distance between the clay particles, structure of the minerals and the characteristics of water. The secondary val"ence bonds are relalively weak and are easily broken. The Vander Wanl forces play an important part in the behaviour of clayey soils. Fig. 6.5. Non-polar System

I 1_

_I I

.-----+-"

~I-_ __+-,' ~I-_ __+-,I

G

G

~

CU\Y MINERALOOY AND SOIL SfRUcruRE

III

6.6. BASIC SfRUCI1JRAL UNITS OF CLAY MINERALS Clay minerals are composed of two basic structural units: (1) Tetr.lhedraJ unit. (Z) Octahedral unit. I. 'Thtrahedrul Unit-A tetrahedral unit consists of a silicon atom (Si~ surrounded by four oxygen atoms (02-). forming the shape of a tetrahedron. Oxygen atoms are at the tips of the letrnhcdron, whereas the silicon alom is at its centre (Fig. 6.6). There is a nel negative charge of 4. An individual terabedron unit cannot exist in nature.

""0

0

Silicon

0,,,'0 ---- :- ---

0",,0

Oxygen (a) Si lico tetrahedron

(b)

Simptifled

~ S.

Ie) Sili ca

r epre ~ entotion

4x(_21 '_ 8

4X(.4 ),. .16 Id) Sil ica

6X(-2).-12 Net g _4

~heet

Fig. 6.6. Tetrahedral Unit.

A number of tetrahedral unit combine 10 form a sheet, with oxygen atoms at Ihe base of aU tetrahedra in a common plane, and aU the lips pointing in the same direction. Each oxygen atom at the base is shared by two tetrahedra. A SHiCll sheet is formed by ternhedrnl units. The three oxygen atoms at the base being common to two tetrahedra get their negative marge shared and the lip oxygen atom has two negative charges. Thus, there are 5 negative charges and 4 positive charges, leaving a nci negative charge of one per tctrnhedron. Fig. 6.6 (c) shows 4 tetrahedra units combined having a net negative charge of 4. Fig. 6.6 (d) gives a simple representation of silica sheet, commonly used in clay minerals. 2. Octahedrul Unit-An octahedral unit consists of six hydroxyls (OIrl) forming a configuration of an

and

• HYDROXYL • ALUMIN ......

(alOCTN-IEDRAL

•

3XHl=_3 I X(H)=+3 3X(-1)=-3

~

(bl SIMPlIFIED REPRESENTATICN

lNT

~ "

)K

•

(cJ GIBBSITE SHEET

6X{-'I'-6

c=::J.

4X{+3J.-1-12

(d) GIBBSITE

6X(-I)=-6

~

Fig. 6.7.
112

SOIL MECIIANICS AND FOUNDATION ENGINEERING

octahedron and having one aluminium atom at the centre (Fig. 6.7). As the aluminium (Ar·~ has three positive charges, an octahedral unit has 3 negative eh.'lrges. Because of net negative charge. an octahedral unit eannOi exist in isolation. Several octahedral units combine to form a gibbsite sheet. Fig. 6.7 (c) shows a gibbsite sheet formed by four octahedral units. The sheet is electrically neulral. Fig. 6.7 (d) shows a simple representation.

6.7. ISOMORl'lIOUS SUDS'nTUTlON It is possible that one atom in a basic unit may be replaced by another atom. The process is known a<; isomorphOUS subStitution (isomorphous means same form). For example. one silicon atom in a tetrahedral unit may be substituted by aluminium atom. This would occur if aluminium atoms are more readily available in water. A.. an aluminium atom has 3 positive charges whereas a siliron alom has 4 positive charges. there would be a nct unil charge deficiency of one positive charge per substitution. Likewise. magnesium (M{1 atoms may replace aluminium atoms in an octahedral unit and cause a reduction of one positive charge. Isomorphous subsLitution generally incI'C.:'lSeS the negative charge on the particle, owing to reduction of positive charges. A slight distortion of the crystal lattice also usually occurs due to isomorphous substitution.

6.8. KAOLrNITE MINEUAL Kaolinite is the most rommon mineral of the kaolinite group of minerals. Its basic structural unit consists of an alumina shect (gibbsite) (G) combined with a silica sheet (S). TIps of the silica sheet and one base of the alumina sheet form <1 common interface. 'The total thickness of the structural unit is about 7 Angstorm (A 0), where one Angstonn A ° is equal 7 to 10- 10 m or 10- mm. 'Ibe kaolinite mineral is fonned by stacking, one over the other, several such basic structural units. Fig. 6.8 shows two such unils. KAOLINITE The structural units join together by hydrogen bond, which develops between the oxygen of Fig. 6.8. Kaolinite Mincl1I1. silica sheet and the hydroxyis of alumina sheet. As the bond is fairly strong. the mineral is stable. Moreover, water cannot easily cnter between the structural units and cause expansion. The kaolinite mineral is elcctricaUy neutral. However, in the presence of water, some hydroxyle iOO5 dissociate and lose bydrogen and leave the crysml with a small residua] negative charge. l11e nat surfaces of the mineral attract positive ions (cations) and water. A thick layer of adsorbed water is formed on the surface. 'The kaolinite minerals generally have a hexagonal shape in pIon. with the side of the hexagon between 0.5 to 1.0 micron. The thickness of the mineral is about 0.05 micron. lbe specific surface is about 15 ro 2/g. The most common example of the kaolinile mineral is China Clay. HalloysUe is a clay mineral which has the same basic structure as Kaolinite. but in which the successive structural units are more mndomly !XIcked, and are separated by a single molCQllar layer of water. The properties of haUoysite depend upon this water layer. If the water layer is removed by drying, the properties of the mineral drastically change. Halloysite particles are tubular in shape, in contrast to the platy shape of kaolinite particles. The soi~ containing haUoysite have a very low mass density. 6.9. MONTMORILLONITE MINERAL Montmorillonite is the most common mineral of the montmorillonite group of minerals. 1bc basic structural unit consists of an alumina sheet sandwiched between two silica shccts. Successive structural units are stacked one over another, like leaves of a book. Fig. 6.9 shows two such structural units. The thickness of each structural unit is about 10 AO. The two successive structural units arC joined together by a link betwccn oxygen Ions of the two silica

CLAY MINERALOGY AND SOIL STRUCIURE

1!3

sheets. 1bc link is due to natural altr:1clion for the cations in the intervening space and due to Vander W.nal forces. lhe negmivcly charged surfaces of the silica sheet attract water in the space between two structural units. This results in an' expansion of the mioeral.. It may also cause dissociation of the mineral into indivKlual structural units of thickness 10 A o • The soil containing a large amount of the mineral montmorillonite exhibits high shrinkage and high swelling characteristics. The water;n the inlerventing space can be removed by healing at Fig. 6.9. Mootmonl1onite minel1ll. 200° to 300°C. Montmorillonite minerals have lateral dimensions of 0.1 I1 to 05 11 and the thickness of O.OCI1 I.l to 0.005 ",. 2 1lte specific surface is about 800 m /gm. The gibbsite sheet in a montmorillonite mineral may contain iron or magnesium instead of aluminium. Some of the silicon atoms in the silica sheet may also have isomorphous substitution . This results in giving the mineral a residual negative charge. 11 atlracts the soil water to fonn an adsorbed layer, which gives plasticity characteristics to the soil.

6.10. (LUTE MINERAL lliite is the main mineral of ·the illile group. The basic structural unit is similar to that of the mineral montmorillonite. However, the mineral has propenies different from montmorillonite due to following

reasons. (I) There is always a substantial amount of isomorphous substitution of silicon by aluminium in silica sheeL Consequently. the mineral bas a larger negative charge than that in montmorillonite. (2) The link between different structural units is through non- exchangeable pota<>Sium (Kj and not through waler. This bonds the units more firm ly than in montmorillonite (Fig. 6.10). (3) The latticc of illite is stronger than that of montmorillonite, and is, therefore. less susceptible to Cleavage. (4) Illite swells less than montmorillonite. However, swelling is more than in "[ • IONS kaolinite.

(f) ~~ ~e ~~:~ s~~~~nt ~:ctu~

F Ag-POTA55I"" tA

montmorillonite, as the potassium ions ILLITE just fil in between the silica sheet Fig. 6.10. Illite MinernJ. surfaces. The properties of the mineral illite are somewhat intermediate betweeD that of kaolioite and montmorillonite. The bond between the r.on-exchangeable K'" ions, though stronger than that in montmorillonite, is considerably weaker than hydrogen bond of kaolinite. The swelling of illite is more than that of kaolinite, but less than that of montmorillonite. 'll1e lateral dimensions of the mineral illite are the same as that of the mineral montmorillonite, equal to 0.1 fA. to 05 fA.. However, the thickness is much greater than that of montmorillonite and is between 0.005 Il and 0.05 ",. 'The specific surface is about 80 m2/gm,

6.11. ELEcrRlCAL CHARGES ON ClAY MINERAlS As mentioned before, the particles of clay carry an electric charge. This fact can be proved by inserting two eioctrooes in a biaker containing clay m ixcd with water. When the electrodes are connoctcd to an

SOIL MECHANICS AND FOUNDATION

114

ENGlN~ERJNG

electrical <;ircuil containing a battery and an ammeter, there is a deflection of the needle of the ammeter. This proves Lhat there is a flow of current through the medium. 1beoretically, a soil particle can carry either a negative charge or positive charge. However, in aClu~ll tests. only negative charges have been measured. The net negative charge may be due to onc or more of the following reasons. (1) Isomorphous substitution of one alom by another of lower valency. (2) Dissociation of hydroxyle ion (OlI) into hydrogen ions. (3) Adsorption of anions (negative ions) on clay surface. (4) Absence of cations (positive ions) in the lattice of the crystal.

(5) Prcsencc of organic matter. The magnitude of Ihe electrical Charge depends on Ihe surface area of Ihe particle. It is very high in small particles. such as colloids, which have very large surface area. A soil particle attracts the cations in the environment to neu!!TIlise the negative charge. 'lbe phenomenon is known as adsorption. : 6.12. BASE EXCHANGE CAPACITY The cations attracted to the negatively charged surface of the soil particles are not strongly attached. These em ions can be replaced by Olher ions and are, therefore, known as exchangeable ions. TI1e soil particle and the exchangeable ions make the system neutral. 11m phenomenon of replacement of cmions is called cation eXChange or base exchange. The net negative charge on the mineral which c::m be 5.:1tisfied by eXChangeable cations is termed cation-exchange capacity or base-exchange capacity. In other words, base--cxchange capacity is the capacity of the clay particles to change the cation adsorbed on the surface, Basc-cxchcmgc capaCity is expressed in teons of the total number of !JOS:ilive charges adsorbed per 100 gm of soil. 11 is measured in milliequivalent (meq). which is equal to 6 x lOw electronic charges. Thus, one rneq per 100 gm means that 100 gm of material can exchange 6 x lOw electronic Charges if the exchangeable ions are univalent, such as Na+. However, if the exchflngc.1ble ions are divalent, such as ea 2+, 100 gm of m ea 2+ > Mg+2 :> NH; :> I-r > Na+ :> U+ For example, AI3+ calions are more strongly attracted than ea'2+ cations. '01OS Al3+ ions can replace Q,2. ions. Likewise. ea 2+ ions can replace Na+ ions. The base formula of the clay mineral is altered by base exchange. For example, if calcium chloride is added to a soil containing sodium chloride. there would be an exchange of o?+ ions for Na+ ions, and the: sodium clay would tum into the calcium clay. Thus Sodium clay + Cl 2 = Calcium clay + NaC!. The properties of the clay therefore (.tlange due to base eXChange. The base eXChange capacity of the montmorillonite mineral is about 70--100 meq per 100 gm. However, that of kaolinite and illite fire respectively 4.0 and 40.0 mcq per 100 g.

6.13. DIFFUSE DOUBLE LAYER The faces of clay minerals carry n net neg
CLAY MINERALOGY AND SOIL srnUcruRE

115

charges or negative chargcs. The chnrgcs in clay minerals are due to molecular grouping and arrangement of ions. The electrical charges in the minerals are responsible for their behaviour when they come in contact with other panicles and with water prescnt in the soil. Clay deposits, because of their sedimentary nature, always exist in the presence of water. ~ecause of the net negative charge on the surface, the clay particles attract cations, such as potassium, calcium and sodium, from moisture present in the soil to reach an electrically balanced C
rO'YGEN

HYDROGE~

0

H

H/~ ~

Ce) MODEL

0 +

(b) RELATIVE LDCATm eel DIPOLE 'AIo\TER

MOLECULE

Fig. 6.11. Structure of a water molecule. (ul Model, (b) Relatillc location, (e) Dipole water molecule

charge, there is aHraction between the negatively Charged faces and the positive ends of dipoles [Fig. 6,12 (a)] . 1be secood mode of attraction between the water dipoles and the clay surface is through cations [Fig.

6.12 (b)]. Cations are attracted to the soil surface and waler dipoles are attached to these cations through their

=000 (a)

o00

&±!J (b)

(c)

Gl CAnON OIPOLE

o

Fig. 6.12. AltrllClion of water molecules 10 soil

SwfflCe.

negative charged ends. The third possible mode by which the attraction between the water and the clay surface occurs is by sharing of the hydrogen atom in the water molecule by hydrogen bonding between the oxygen atoms in the clay particles aod the oxygen atoms in the waler molecules [Fig, 6.12 (e)]. The cations attracted to a clay mineral surface also try to move away from the surface because of their thenna! energy, The nel effect of the forces due 10 attraction and thnt due 10 repulsion is that the forces of attraction decrease exponentially with an increase in distance from the clay particles surface. The layer extending from the clay particle surface to the limit of atlroction is known as the diffuse dQuble layer (Fig. 6.13). It is believed that immediately surrounding Ihe panicle, there is a thin, very tightly held layer of water about 10 A 0 thick. Beyond Ihis thickness there is a seoond layer, in which water is more mobile. This second layer extends to the limit of attraction, and is known as diffuse-double layer (Fig. 6.13). The water held in lhe diffuse-double layer. is known as adsorbed water or oriented water. Outside the diffuse double layer the water is nonnal. non·oriented. The total thickness of the diffuse-double layer is about 400 A 0,

116


For ;) given soil panicle, the thickness of the cation layer depends maiill)' on the valency and concentration of cations. The mono-valent cations. ~~~r-?!~~se-f such as Na+, lead to a thicker layer compared to r that by divalent cations, such as Cu2+. The number e Ie 9 e of monovalent cations required is twice the number $ e (tI @Cation of divalent cations. Increasing the conccntratior. of s : . $9 e 9 Anion cations clooe to the surface, reduces the thickness • I., e _ of the cation layer required to neutralise the negative Charge. $ e e (J Repulsion occurs between the like charges of r the two double layers of two particles. 'Ibe forces of repulsion between the two particles depend upon the characteristics of the double layers. An increase ~.§ in cation valency or concentration rcsulls in a 'u;"U decrc.1se of repulsive forces. However, V
6)'

·I.

• I ••

.ot~--------

6.14. ADSORBED WATER ll1e water held by electro-chemical forces existing on the soil surface is adsorbed water. As the a water is under the influence of electrical forces. its properties arc different from that of nonnal water. It is much more viscous, and its surface tension is also greater. It is heavier !han nonnal water. The boiling poin is higher, but the freezing point is lower than that of the normal water. The thickness of the adsorbed water layer is about 10 to 15 AO for colloids but may be upto 200 AG ~ silts. The attractive forces between the adsorbed water and the soil surface decrease exponentially with the distance until the double layer merges into normal water. The adsorbed water exists in an almost solidifi state. The pressure required to pull away the adsorbed water layer from the soil surface is very high; it m be as high as lO,
6.15. SOIL SI'RUCIlJRES lbe geometric..11 arrangement of soil particles with respect to one another is known as soil structure. soils in nature have different structures depending upon the particle size and the mode of fonnation. follOWing types of structures arc usually found. lhe first two types are for coarse-grained soils and types (3 and (4) for Clays. Types (5) and (6) are for mixed soils. (1) Single-grained Structure-Cohesionlcss soils, such as gravel and sand, are romposcd of bulky grains in which the gravitational forces ate more predominant than surface forces. When deposition of !Me soils occurs, the particles settle under gravitational forces and take an equilibrium position as shown in F1g. 6.14 (0). Each particle is in contact with those surrounding it. The soil structure so formed is known as singlrgrained structure. The arrangement is somewhat similar to the stacking of omnges on a grocer's counter OCIO that of marble pieces at a toys' shop.

CL\Y MINERALOGY AND SOIL smUcruRE

111

(a) Single Grained Structure

(b) Honey - comb Structure

Fig. 6.14. Soil slruclUre ill sallds and silts. (a) Single Graillcd Structure, (b) HOlley-romb Slructure

@

Depending upon the relative position of the particles. the soil may have a loose structure or a dense structure. Fig. 6.15 shows spherical particles in the looscst and those in the densest condition. It can be proved that for the loosest condition, the void ratio is ~ 090 am:: that for the densest state, IS 035 In actual sand deposits, as the particles are not exactly sphencal, the vOId raho between the loosesl and densest conditIOns vanes between 0 90 and 0 35 As mentIOned m chapter 3, the engmeerlng properties of sands tmprovc considerably wIth a

~n~~ra~ ~~d s:~~e~r ~~ I~=~t;~, ~~~tl~~g~~~s:~e

(a) LOOSE • (b) DENSE shear strength, and the lower Ibe compressibility and Fig. 6.15. Sphere<; ill J~sl and densest states. permeability. Loose sands are inherently more unstable. When subjected to shocks and vibratiOns, the particles move into a more dense state. Dense sands are quite stable, as they arc not affected by shock and vibrations. (2) Honey-Comb Structure-It is possible for fine sands or silts to get deposited such that Ibe particles when settling develop a particle-to-particle contact that bridges over large voids in the soil mass [Fig. 6.l4(b)J. The particles wedge between one another into a stable condition and form a skeleton like an arch to carry the weight of the overlying material. The slructure so formed is known as honey
(3) Flocculated Structure-Flocculated struclure occurs in d.IYS. The clay particles have large surface area and, therefore, the electrical (orces' are important in such soils. The clay particles have a negative charge on the surface and a positive charge on the edges.

::i:~~cJ~ha~~ac~d::sve~~~ th~t;:;tiV~~; charged faces. This results in a flocculated structure [Fig. 6.16(0)]'

~

~ ~ .c::==:o

-===-

~o:::=::==s-==-

Ca)

Flocculat~d Structur~ Fig. 6.16.

(b)

~tS:U~:~

Soil structure in cI~y (a) Phx:culaled Structure, (b) Dispen;ed Structure

SOIL MECHANICS AND FOUNDATION ENGINEERINO

118

Flocculent structure is fonned when there is a net attractive force between particles. When clay panicles settle in water, deposits fanned have a flocculated structure. 'The degree of flocculation of a clay deposit depends upon the type and concentration of clay particles, and the presence of salts in water. Clays settling out in a sail waler solution have 3 more []occulent structure than those settling out in a fresh water solution. Salt water acts as an electrolyte and reduces the repulsive forces between the particles. Soils with a flocculent structure arc light in weight and have a high void ratio and water content

However, these soils arc quite strong and can resist external forces because of a strong bond due \0 attraction between p<,rtic1es. The soils are insensitive 10 vibrations. In general. the soils in a Oocculated structure have a low' compressibility, a high ,penncability and a high shear strength. (4) Dispersed Structure-Dispersed structure develops in clays tlmt have been reworked or remoulded. The particles develop more or less 8 parallel orientation {Fig. 6.16 (b)l. Clay deposits with a flocculent structure when transported 10 olher places by nature Of man get remoulded. Remoulding converts the edge·to-face orientation to face-to-face orientation. The dispersed structure is fonned in nature when there is a net repulSive force between particles. 'The soils in dispersed structure generally have a low she~r strength, high compressibility and low permeabilily. Remoulding causes a loss of strength in a cohesive soil. With the passage of time, however, the soil may regain some of its lost strength. Due to remoulding, the chemical equilibrium of the particles and associated adsorbed ions and water molecules within the double layer is disturbed. The soil regains strength as a result of re- estoolishing a degree of chemical equilibrium. This phenomenon of regain of strength with the passage of time, with no change in water content., is known as thixotropy, as already disaJssed in chapter 4. (5) Coarse-grained Skeleton-A coarse-grained skeleta'i'! 'is a composite structure which is formed when the soil contains particles of different types. When the amount of bulky, cohesionlcss particles is large compared with that of fine-grained clayey in particles. the bulky grains particle-to-particle contact. These pmticles fonn a framework or skeleton {Fig. 6.17 (a)]. The space between the bulky grains is occupied by clayey particles, known as binders. In nature, the bulky grains are deposited first during sedimentation and the binder is subsequently deposited. As long as the soil structure is not Fig. 6.17. Composite SlrUcture (a).coRJSe Grnind Skeltion, disturbed, a coorsc-gr
CLAY MI NERALOGY AND SO IL STRUCTURE

B. Descriptive and Obj l.'ctivc type 6.2. Diffcrenti:uc between gravit3tional propclties 01

forc~

119

and surface forccs. What is the et"ft'Cl of increased surface area on the

)i{ul.~

63. What arc primary valent"\: bonds'! What is their imponancc m soil engmccring '! 6.4. What do you undcrl>t,md by hydrogen bond? Give examples. 6.5. Wh:lI arc secondary valence bonds'! Wrile a shorl nOle on Vander W331 forces. 6.6. Describe the constitution of the two basic structuml units rcqulft'(l in Ihe formation of clay minerals. Are these ele<:trically nCUlr:Il?

6.7. Discuss the charactcri~l1cs and the construction of Kaolinite. Montmorillonite and Illite mineral groups. 6.8. Write ~hon n(lte~ nn: (I) Base exchlmge capacity. (ii) lsomorphollssubstitution. (il') Adsorbed water (iii) Electricnl double I.lyer 6.9. What arc ditfcrent types ot soil Slnlctures which can occur in mllure ·1 Describe is brief. 6.10. STate whet hcr the followlllg statements arc InIC Of fillse. (a) The l11 il1(:nl l qU:lrtz b electrically act ive. (b) T he clay minerab li re rcspt.ll1sib le for plaslicty chnrnclC rislics of ~oi l s. (e) T he hydrogcn hond is stronger than secondary v~tl c n ce bo nds. (d) I SI' l11orJlhou~ ~ubstillition docs not change the electrical ct13rg<:: (1') The soib containing. thc minerallmlloyshe have .1 high unit weight. if) The miner'll !l\ulllmurillu11I tC. cause.> excessive swclhng and shrinkage. l1:) The nd~urbeJ water imparts phlsticity to SOils. (II) Honey-comb ~tructure occur~ in clayey soils. (0 Remouldcd tine-grainoo soils have a tlocculat
fA ns. T rue. (b). (e). (j). (g)]

C. Multiple-Choice Questions. 1. The behaviour of clay h govemed by ((I) Mass energy (b) Surf:lCe energy (e) Both (a) and (b) ((/) Nei lher (a) and (b) 2. Honey-combed strut:turc 1~ found in (a) Gravels (b) Co.lfSC sands (e) Fi ne ~ands :U1d SIltS (fl) day 3. TIle weakest bond ill ~otl~ I~ (b) Covalent bond (11) Ionic bond (tf) SecondJry valance bond Ie) Hydrogen bond 4. All O~'lahedrJl unit ha~ (a) Pour neg: llIvc charges (b) Thrcc negative c!mrgc.~ (e) One Il<::galive (If) No negative charge 5. In illi t<:: mineral. Ihebond be twecnstructural u11itsis \a) Hyd ro;:cn bo nd (b) PQt ~l ssi um i011 bo nd (e) Water l11ok.-cu lcs bond (tI) COV:l1e11l bond 6. The plas ticity charJcteri~lics of clays arc due 10 (f/) Adsorbed water (b) Free watcr (r) CapI llary wmer (tI) None of above 7. In tine l>:tnds and ~ihs, the most common type structure is (II) Smg!c grained (b) Honey comb (c) Flucculated (II) Disperred H. The base cxc!mnj,lc l·apacity of lhe mineral montmorrillonite is .. buul (/1) 70 mt-qI1QO g (b) 700 mav l OO g . (c) 7 meql100 g {(/) 40 meqf l OO g ~_ J~1~1~~m~W~~7m8~

7 Capillary Water 7.1. TYPES OF SOIL WATER The soil water is broadly classified into two categories: (1) Free water, and (2) Held water. Free water moves in the pores of the soil under the influence of gravity. 'll1e held water is rcwinoo in the pores of the soil, and il cannot move under the influence of gravitational force. Free water flows from one point to the other wherever there is a difference of total head. The rate at which the head is lost along the flow passage is equal to the hydraulic gradient. The flow of free water in soils is just like laminar flow in pipes. Because of very smaU flow passages in the SOil, the velocity head is generaUy neglected, and the total head is lakcn equal 10 the sum of the elevation head and the pressure head.

Free water is discussed in chapter 8. Held water is further divided into three types: (1) Structural water, (2) Adsorbed water, and (3) ,Capillary water. The structural waler is chemically combined water in the crystal structure of the mineral of the soil. lbis water cannot be removed without breaking the structure of the mineral. A temperature of more than 300 °C is required for removing the structural water. In soil engineering. the structural water is considered as an integral part of the soil solid. lhe water held by electrochemical forces existing on the soil surface is known as adsorbed water, as discussed in chapter 6. 1be quantity of adsorbed water depends upon the colloidal fraction in the soil the chemical com~ition of the clay mineral and the environment surrounding the particle. The adsorbed waler is important only for clayey soils. For coarse.- grained soils, its amount is negligible or zero. The adsorbed water is also sometimes called hygroscopic water. 1be amounl.of water in an air-dried soil is defined as hygroscopic water. Since air drying removes capillary water and free water, the remaining water is approximately equal to the adsorbed water. Hygrosropic water depen~ upon the humidity and temperature of air. It is assumed !.hal oven drying removes all the hygrosoopic water. The amount of water in an air-dried sample, expressed as a percentage of the dry mass, is known !:IS hygroscopic water oonteo!. 'fhe water held in the interstices of soils due to capillary forces is called capillary water. This Chapter discusses mainly the capillary water and its effect on soils. 7.2. SURFACE TENSION To under.;taod surface tension, let us oonsider a molecule of water surrounded by other molerules in the body of water, as shown in Fig. 7.1 (0). The forces due to moltx:ular attraction act all 'around, and the molecule is in equilibrium. However, al the free surface, as shown in Fig. 7.1 (b), the pull from the air above is smaller than the pull from the water molecules below and the equilibrium is disturbed.

la)

(b)

(a) (b) Fig. 7.1. Effed of Surrace tCr15ion .

CAPlUARY WATER

7 r

121

The forces tend to reduce the surface area of the air-liquid surface to a minimum. The surface assumes a curved shape to maintain equilibrium. 'l11e intcrfHcc behaves like a stretched membrane or a skin. The surface tension exists at the interface. Surface tension is defined as the force per urut length of a line drawn on the surface. It acts in the direction normal to that line. The surface tension of water at normal temperature is about 0.073 N/m at 20°C. It decreases with an increase in temperature. It is because of surface tension that a smaU needle can float on water, and insects can walk on it. Capillary water exists in soils so long as there is an air-water interface. As soon as the soil is submerged under water, the interface is destroyed, and the capiUary water becomes norma~ free water. The capillary water is always under tension (negative pressure). However, the properties of the capillary water are the same as that of normal, free water. 7.3. CAPILLARY IUSE IN SMALL DIAMETER TUBES Water rises in small diameter, capillary tubes, beatuse of adhesion and cohesion. Adhesion occurs because water adheres or sticks to the solid walls of the tube. Cohesion is due to mutual attraction of water molecules. If the effect of cohesion is less significant than the effect of adhesion, tbe liquid wets the surface and the liquid rises 1lI the point of contne. However, if the effect of cohesion is more predominant than adhesion, the liquid level is depressed at the point of contact, as in the case of mercury. If a glass tube of small diameter. open at both ends, is lowered into water, the water level rises in the lube, as the water wets the tube. Let 8 be the angle of contact between the water and the wall of the tube [Fig. 7.2 (a)]. T,

r,

,01

'b) Fig. 7.2. Capillary Rise

F" = Upward pull due to surface tension = (1~ cos 8) 1td where T, = surface tension and d diameter of the tube. F" = Downward force due to mass of water in the tube _ y.(,/4 d') x h. where h~ = height of capillary rise. For equilibrium,

F" - Fd (T.cos9) xd _ y.(xl4d')h. 41~cos8

h. -

4 T,cos 9

----:;::;t" - KP.:d

... (7.1)

For a clean glass tube and pure water, the meniscus is approximately hemispherical, ie. 8 = O. 1berefore,


122

h .. 41~ ~ y",d

... (7.2)

Taking T, .. 0.073 N/m. y", .. 9810 N/m'\ h~..

4 )( 0.073 3)( 1D....~ 9810 d .. - - d - melres

If d is in cenlimctcrs.

ht

3 )( dlO-

If he and d both arc in em,

h~_¥cm

where d is in mctres. ..

l

metres ... (7.3)

Capillary rise in tubes of non-unifonn diamcter depends upon the direction of flow of watcr. If a tube with a largc bulb is dipped in water. the water is lined due to capillary action. but it may not rise past the bulb where the diameter is d 2 lFig. 7.2 (b)J. The capillary rise is limited to a height of (hell because water cannot maintain equilibrium at a large diameter d2• If the same tube, with a large bulb is fiUed by pouring water from above or by lowering the tube below the water level and then raising when filled. an equilibrium is maintained at a height (hfh [Fig. 7.2 (c)l. '[be water is able to maintain equilibrium at the diameter d j above the bulb. lllUs the capillary rise in lubes of non-uniform di(lmeler is more if the flow is downward than when it is upward. The capillary rise docs not depend upon the shape and the diameter of the tube below the meniscus when tbe flow is downward. In upward flow. the capillary rise is terminated if the diameter of the lube is greater than that required for equilibrium. The height of capillary rise docs not depend ulX>fl the inclination of the tube. Even if the capillary tube is inclined, the vertical rise of water remains the same, equal to hf' In a capillary tube of uniform diameter, no water can be retained when lined. The upward forces (F..) due 10 surface tension arc balanced by downward forces (Fd) at tbe lower end [Fig. 7.3 (a)l. However, if (he (ube is necked, witb smaller diameter at lOp, the upward force (F..) is greater than the downward force (Ftil. and some T2 12 water can be retained in the tube [Fig. , . FcJ 7.3 (b)]. 12 ~o,-i T2

'd"

T,.

7.4. CAPILlARY TENSION 12 The water in a capillary tube is (a) under a negative pressure, commonly Ag. 7.3. CapiltAI)' Tube when lined called tension. The pressure variation in the capillary tube of Fig. 7.4 (a) can be dctennined as follows. lhe pressure at point D at the free surface is 2tmospheric Le., equal to zero gauge pressure. (In soil engineering, generally gauge pressures are used). The pressure at point C. which is at lhe s.'lme level as point D, is also zero. From laws of bydrostatic, the pressure at point B. which is at a height of hf above the free surface. is given by Po" - 'f .. A ... (7.4) The pressure is negative because it is less thnn mmospheric pressure. In other words, tbe wmcr at point B is under tension. The capillary rise at any point E is II, nod therefore the pressure is given by

CAPILLARY WATER

!l F-

Ca)

Cb)

Fig. 7.4. Pressure Variat ion.

PE .. -Yw h The capillary tension, therefore, varies linearly with the height of point above the water surface, as shown in Fig. 7.4 (b). The pressure al point F below the waler surface is, of course, positive (hydrostatic). As the capillary tube is open to atmosphere, the pressure at point A above the meniscus is atmospheric, i.e. zero. Therefore, the pressure difference across the two sides of the meniscus is equal 10 "twhe. The pressure difference is also known as pressure deficiency (P"). Thus p" .. "tw h~ Substituting the value of he from Eq. 7.2,

" - y. (4T,) 4T, y.d - d

...(7.5)

P

If the meniscus is not herni·spherical and it has diameters d 1 and be shown that

2 2)

" T ( p.'d;+d;

~

in two orthogonal di.red.ions, it can ... (7.6)

Capillary water can be likened to hanging of a weight 10 the inside walls of a chimney. The walls of the chimney support the load and transfer it as reaction to the base. The weight causes compressive stresses in the walls of the chimney. In a similar manner, the capillary water causes compression in the walls of the gJa<;s tube. The compressive force (F) is equal to the weight of suspended column of water. F -

(~h,)

y.

.. :(7.7)

The compressive stress in the wall of the tube can be determined from the contact area and the compressive force. The compressive stress is constant in the entire height he of the tube.

7.5. CAPILlARY RISE IN SOILS The water which falls on the ground as rain flows under gravity and passes through the soil and reaches a surface known as ground water lablt:. The soil is saturated below the ground water table. 1be ieveito which underground water rises in an observation well is called ground water table or simply water table. TIle ground water table is also called as the ground phreatic surface, a tenn deriVed. from the Greek word phretJs, meaning a well. Ground water, which is a fonn of free water, is not static. Il is a moving stream which flows under gravitational force. The water table is not horizontal. II takes Ute shape according to the topography. The water is drawn above the water table (abbreviated at WT.) due to capillary action. A soil mass consists of 8 \lumber of intercoOllecled interstices which act. as capillary tubes of varying diameters. Although the channels fonned by interconnected interstices are not circular in cross-section, the

SOIL MECHANICS AND FOUNDATION ENGiNEERING

124

results of capillary rise in circular tubes arc useful for understanding the phenomenon of capiUary rise in soils. The channels formed in the soil arc a sort of capillary lubes of varying diameter but not necessarily vertical. These capillary tubes may be inclined in any direction. Capillary rise in soils depends upon the size and grading of the particles. The diameter (d) of the channels in pore passage depends upon the diameter of the particle. It is generally taken as one·fifth of tbe

effect:~diameler

(D10)d::;:

ooan;e.grained soils.

ffkSQI<.'.W/k.w;;:x:...'VX~~~V ZONE OF AERATION

As the capillary rise is inversely Pl'OJXlrtional to the diameter of the lube, the capillary rise is small in coarse-grained soils, bul it may be vcry large in

fine-grained soils. tn some vcry [me-grained soils, it

ZONE

OF CAPILLARY

SATURATION

may be even more than 30 m.

yW.T. The space above the water table am be divided into two regions: (1) Zone of capillary saturation, in rig. 7.5. CapillHI)' zone. which the soil is fully saturated. (2) Zone of aeration, in which the soil is not saturated (Fig. 7.5). The height to which capillary water rises in soils is known as capillary fringe. It includes the zone of capillary saturation and a part oC the zone of aeration in which the capillary water exists in interconnected channels. The soil above the capillary fringe may contain water ~MOISTURE in the Corm of contact water (Fig. 7.6). In this Conn l water Corms a meniscus around the poim of contact. Surfaoe tension holds the water in contact with soil. Because of the tension in the capillary water, there is an equal and opposite Corce induced at the points of contacts which presses the particles together. The contact pressure depends upon the water content, particle size. angle of conlaCt and density of packing. The contact pressure . dccrcascs as the water cootenl increases because of an Fig. 7.6. increase of radius of meniscus. EventuaUy, a stage is reached when the contact pressure becomes zero as sooo as the soil becomes fully saturated. Terzaghi and Peck (1948) gave a relationship between the maximum height of capillary fringe and the effective size, as

~

0~-~

.. ~

where C = constant, depending upon the shape of the grain and impurities. ~ = void mtio. DlO = effective diameter, the size corresponding to 10% percentage finer. If D IO is in mm, the value of C varies between 10 to 50 mm 2, and the height (h)max is also given in mm.

If D IO and «ht)mu are in centimeters, C = 0.1 to 0.5 an 2•

Table 7.1 gives representative heights of capillary rise in different soils. Thble 7.1. Representative Heights or Capillary RIse S.No . 1.

23. 4.

s.

6.

SoU Type

fine gravel Coo",,,,,,, Fine sand Silt

C.y Colloid

Capillary rise(m) 0.02 to 0.10 0.10 to 0.1S 0.30 to 1.00 1.0 to to.O 10.0 to 30.0 more than 30.0

125

CAPIUARY WATER

7.6. SOIL SUcrION .As in the case of a capillary tube. the water in the soil above the water table has a negative pressure. The soil is in a state of reduced pressure, known as soil suction. TIle soil suction is measured in teno of the height of water column suspended in soil. Generally. it is expressed in terms of the common logarithm of the height in centimeters, and is known as PF value. For example, a soil suction of 100 em of water column can be represented as PF equal to 2, because. 100 = 102cm

and logl~

=2

A PF value of zero corresponds to a soils suaioo of 1 an, as loglo = O. Although the soil suction represents a negative pressure, it is customary to omit the negative sign. The soil suction can also be represented in pressure unit... using the relaHon, 1 em of water column = 0.0098 N/an 2

7.7. CAPILlARY POTENTIAL The tenacity with which the soil holds the capillary waler is measured in terms of the capillary potential or matrix potential. The capillary potential ("') is defined as the work done to take away II unit mass of water from a unit mass of soil. II is numerically equal to the tension (negative pressure) in the soil water but it is of opposite sign. Therefore, ... (7.9) 'i' =-p where p is tensioo in soil. [Note. Some authors express capillary potential as energy per unit mass in kJ/kg_ For example, if p = 1 bar = 100 kN/m2, height of water column -

~~

_ 10.2 m and lJI - 10.2 )( 9.81 - 100 kJ/kg]

It is worth noting that the capillary potential is always negative. The maximum possible value of '" is equal to zero when the soil tension is zero, which occurs when the water is at atmospheric pressure. As the water content in the soil decreases, the tension increases. This causes a decrease in capillary potential. The capillary potential is minimum when the water rontent is minimum . Water in the capillary fringe is seldom under equilibrium. It moves from a region of high potential (more water content) to a region of low potential (less water COIllent). The water starts moving as soon as the suction equilibrium is disturbed either due to evaporatioo of water or due to an increase in water content. The velocity of the capillary water is given by v _ k" . is ... (7.10) where k" = coefficient of unsaturated permeability, i, = suction gradient, which is equal to the potential difference per unit length.

7.8. CAPILLARY TENSION DURING DRYING AND WETIlNG OF SOILS Capillary tension develops not only in the soils abOve the water table but also in a soil when its water CODlent is reduced. When the water content of a saturated soil is reduced by drying, the water recedes into the interstices of the soil and (onos menisci. As the water content is reduced further, the menisci recede. The radii of curvature decrease, and there is a rorresponding increase in soil suction. Fig. 7.7 shows the relationship between the soil suction and the water conlent of a soil. The suctioo at a particular water content is more when the soil is drying than when the soil is wetling, and a hysterisis loop is formed. The reason for the differenrx in soil suction is that during drying the release of water [rom the larger pores is controlled by the surrounding smaller pores, whereas during welting it is not controlled by the smaller pores. The phenomenon is somewhat sUn ilar to the flow of capillary water in tubes of non- uniform diameter discussed in Sect 73. The process of drying is analogous to the flow of water in the downward direction, in which the capillary rise does not depend upon the larger diameter of the bulb.


126

The increase in soil suction with decreasing water content is continuous over the entire range of water content. lIS value is zero when the soil is saturated and is very high when the soil is oven dry. When a dry soil is subme.."ged under water, the meniscus is destroyed and the soil suction is reduced to zero. The capillary water changes 10 free

water. 7.9. ~~~~ AFFECTING SOIL

;g

The suction in soils depends mainly on the following factors: (1) Particle size-In general the smaller the particle size, the greater is the soil suction. 'The soils with Fig. 7.7. Drying lind Weldng Cycle. fine particles have a large number of small pores with small mdjj of menisci. It results in a large

capillary rise and hence greater suction. (2) Water content-As the water content of a soil decreases, the soil suction increases and it attains the maximum value wben the soil is dry. As discussed above, with 8 decrease in water content, water recedes into smaller pores resulting in the decrease of the radius of curvature of the meniscus. (3) HIstory or drying or wetting cycle-As diSCl.J$ed in the prereding section, for the same water content. the soil suction is greater during the drying cycle than in the wetting cycle. (4) Soli Structure-The soil structure governs the size of interstices in the soils. As the soil suction

depends upon the size of interstices, a change in the soil structure affects the soil suction. (5) 'Thmperature-A rise in temperature causes a reduction in surface tension (T,) of the water. Consequently, the soil suction decreases as the temperature increases. (6) Denseness of soil-As the denseness of a soil increases, generally soil suction ina-eases. When the soil is loose, with a low density, the pores are of large radius and the soil suction is low. (7) Angle of contact-The angle of contact between water and SOil. particles depends upon the mineralogical composition of soils. As tbe angle of contact (8) increases, the soil suction deaeases. The soU suction is maximum when the angle of contaa is zero.

t

(S) Dissolved salts-The surface tension of water increases with an increc'lse in impurities, such as saIl. Therefore. the dissolved salts cause an increase in soil

suction. 7.10. MEASUREMENT OF SOIL SUCTION Suction in a soil mass can be measured using the following methods: (1) Tensiometer Melhod-A lensiometer consists of a porous pot filled with water. The top of the porous pot is connected 10 a U- tube containing mercury. The pol is placed in the soil

h

l

1

Fig. 7.8. Tcnmomc1er..

127

CAl'ILLARY WATER

whose suction is to be detennined (Fig. 7.8). llIe soil draws wmer from the porous pot. The process continues till an equilibrium is attained, when the su<.1ion inside and that outside the pot arc equal. 111e suction (P") inside the porous pot can be calculated. using the manometer equation (see any text on Fluid MechaniCS). as O- 13.6h x 9.81 + (II + y) x 9.81 _ p" pot - -(12.6/1 + y) x 9.81 ... (7.11) where II = denection of mercury in manometer in metres, y = vertical intercept betwccn the mercury level in the right limb of the manometer and the centre of the pot. The soil suction (P") can be calculated using Eq. 7.11 once the values of hand y have been determined. llIis method is suitable for delennination of soil suction upto 0.8 bar or 80 kN/m2 or 800 cm of water. (2) Suction Plate Method-In this method, the soil sample is placed over a porous plate known as suction plate. The suction plate is in SOIL SAMPLE contact with water in the reservoir (Fig. .-#,'---, POROUS PLATE 7.9). The water reservoir is connected TO VAC UM P MP to a pipe. A mercury manometer is
~ 2. (TT-T~

... (1.12)

128

SOIL MECHANICS AND FOUNDA110N ENGINEERING

where h = soil suction. expressed in terms of the height of water column (log h _ PI')' 0> = rotational speed (rndinns per serond) '1 = radial distance from the centre of rotation to the water table '2 = radial distance from the centre of rotation to the middle of the soil sample. The test is conducted at various speeds to obtain a relationship between the water content and the soil suction. The centrifuge method can be used for determination of very high suctions, of the order of several thousands of kN/m 2• For accurate results, thin samples shaD be used. If the sample is relalive!y thick, it is subjected to an additional overburden pressure due to its own weight and erroneous results are obtained.

' .lI. FROST IlEAVE The water which migrates upward from the water table to the capillary fringe may freeze if the atmospheric temperature falls to the freezing point, and the i~ is formed. This results in an increase in the volume of soil, because when water is ronverted into ice. 1here is about 9% increase in its volume. If the porosity of the soil is 45% and the soil is sUlumted. the expansion of the soil would be (0.09 x 45) = 4.05%. In other words, there would be a hc.'lve of about 4 cm in every one metre thickness of the soil deposit. Due to frost heave, the soil at the ground surface is JiCted. This may cause the lining of light structure... built on the ground. The frost heave observed in most of the soils is much more thun a hc..'lve of about 4 cm (ler metre. This is due to the foct that when the ice lenses are formed in the soil due to freezing of water, the water film from the adjacent soil panicles is also removed. This disturbs soil suction equilibrium and more water is drawn up from the water table by capillary action to replenish the water deprived by the ice lenses from the soil particles (Fig. 7.11). This process may cause a frost heave of 20 \030% of the soil depth. G.$.

-===--- ::::::;?NSES III

_____ L ______ L

___ J~Y'L

fig. 7.11. Jee Lenses.

The soils which are prone to frost action are mainly silts and fine sands. These soils have large capillary rise due to relatively fine panicles. Moreover, water can easily flow through these soils because of fairly good penneability. In coarse~gmined soils and clayey soiL<;, the frost heave is relatively small. In coarse-grained soils, the frost heave is limited to about 4%, as there is very little capillary risco Clayey soils, on the other hand, have very large capillary rise, but their permeability is very low. lhe water cannot move easily thrOUgh these soils and, therefore, the frost heave is lim itcd. However, if the clayey deposited have fissures and crack.s, water moves easily and a large frost heave may occur in such soils. If the temperature persists below the freezing point for a long period. frost penetrates the soil further, and the depth of the affected soil increases. The depth upto which the water may fl'CC'Le is known as the frosl line. The basic condition for the formation of the frost he.1ve may be summarised as under: (I) The temperature in the soil is below freezing point and pen;ists for a long period. (2) A reservoir of the ground water is available sufficiently dose to the frost line to feed the growing ice lenses by capillary action. (3) The soil is saturated at the beginning and during the freezing period. (4) The soil has sufficiently rugh capillary poIenlial to lin the water above the ground water table. (5) The soil has good penneabilily so that water moves quiclc.ly through it.

CAPILLARY WATER

The cracks and fISSure also pennit rapid movement of water.

(6) The soil particles of size about 0.02 mm arc rn05t prone to frost heave. If a unifOlm soil contains more than 10% particles of the size 0.02 mm or if a well~grndcd soil a:mtains more than 3% particles of this size, the soil is prone to frpst heave. The foundations of structures should be carried below the frost depth to avoid possible frost heave after the completion of the structure. However. highways and runways have limited depth below· the ground surface and cannot be constructed below the frust line. In soch cases. other meao;ures are taken to reduce frost heave. as discussed in Sect. 7.13.

7.\Z. FROST nOlL After the occurrence of frost heave. if the temperature rises. the frozen soil thaws and free water is liberated. Thawing process starts from the upper layer and moves downwards. The liberated water is trapped

in the upper layer while the lower layers are stiU frozen. The strength of the soil in the upper layer is reduced due to its softening caused by an increases in water coment. The process of softening of soil due to Iibemtion of water during thawing is known as frost boil. Frost boil affects the structures resting on the ground surface:. The effect is more pronounced on highway pavements. A hole is generally formed in the pavement due to extrusion of soft soil and water under the action of wheel loads. In extreme cases., the pavement breaks under tramc. and there is ejection of subgrndc soil in a soft and soapy condition. '! Coarse-grained soils arc not affected much by (rost boil. as the quantity of liberated water is sman. and lhal too is drained away quickJy. The soils most prone to the softening effect are s ilty soils. These soils have low plasticity index and beoome very soft with a small inaease in water roment. Oayey soils are not affected as much as silty soils since the quantity of liberated water is small and the plasticity index is high.

7.13. PREVENTION OF FROST ACI10N The frost heave and frost boil cause great difficulties in the maintenance of highways and runways, as discussed above. The following measures are usually taken to mitigate the ill effects of the frost action. (1) The most effective method of prevention of frost action is to replace the frost~susccptible soil by coarsegrained soils such as gravels or coarse sands. In most cases, the method is not economic..111y fC1l<;iblc GROUND SURFACE owing to large quantities of soils involved. (2) The frost action can be prevented by providing an insulating blanket between the water table and t.he ground surface. . The insulati"& blanket ronsists of gravel, and has a thickness of W.T l7 15 to 30 an. '[be blanket reduces the capillary action and hence the migration of water and the formation of ice tenses (Fig. 7.12). (3) A good drainage system prevents the frost Fig. 7.12. Insulating Blanket. action in two ways: (I) It lowers the water table and thus increases the dislance between tbe ground surface and the water table. (il) The water liberated during thawing is drained away quickly by the drainage system. (4) Sometimes additives are used to reduce frost action. Dispersion agents, such as sodium polyphosphate, when mixed with soil, decrease the penneability of the soil. (5) Water proofing materiels and other chemicals are also used to cQ.aoge the adsorbed cations on the clay minerals to reduce tbe tendency of soils to attract tbe water dipole.

7.14. SHRINKAGE AND SWELLING OF SOILS Shrinkage A clayey soil shrinks when water evaporates from it. If water is added to such soils. swelling takes place. Shrinkage a,nd swelling are characteristics of clayey soil. The coruse-grained soils ha'!: v~ry liule shrinkage and swelling.


130

Shrinkage is due to tension in soil water. When tension (negative pressure) develops in water, compressive forces act on the solid particle. The compressive forces induced in the solid particles are similar to those induced in the walls of the capillary lube discussed in Sect. 7.4. When the water content of a soil mass reduces due to eV8lX'ration, the meniscus retreats. This causes oompression of the solid particles and hence a reduction in the volume of the soil mass. The Strc5SCS in pore water during shrinkage can be studied from the capillary tube analogy (Sect. 7.4). Let us consider a soil

mass consisting of spherical. solid particles, shown in Fig. 7.13. When the capillary spaces . . bclween the particles are completely filled Fig. 7.13. RClreahng or Mcmscus. with water, the menisrus forms a plane surface, as indicated by 1-1. The tension in water is zero. As evaporatjoo takes place, water is removed from the free surface and the meniscus retreats to the position 2~2. This process causc.s tension in the water and corresponding oomprcssive fo~ces on the solid grains. The tension developed depends upon the radius of the menisc..'Us. With further evaporation. the meniscus retrc.'lts to position 3-3 and the rndius decreases. This increases t.he compressive forces acting on the solid particles. Eventually, when the meniscus attains the minimum radius. shown by position 4-4. it is fully developed and the compressive forces induced are maximum. Funher recession of the meniscus docs nOI incremre the compressive forces, as there are n6 pores of smaller radius. The lower limit of the volume occurs HI the shrinkage limit. At the shrinkage limit. the soil is still saturated, but there is no free water at the soil surface. Further drying docs not cause a reduction in its volume as the soil resistance exceeds the compressive forces. As soon as tbe shrinkage limit is reached, the surface becomes dry. It is indicated by a change in the oolour of the soil surface to a lighter shade. There may be a small addition.'li shrinkage after the shrinkage limit, but this is usually ignored. Swelling When water is added to clayey soil which had shrunk by evaporatioo of the pore water, the menisci arc destroyed. The tension in soil water becomes zero. lbe compressive forces between the solid particles reduce considerably. Hnd clastic expansion of the soil mass occurs and this causes some swelling. However swelling mainly occurs due to attraction of dipolar molecules of water to the negatively charged soil particles. The swelling also depend.. upon a number of other factors, such as mutual repulsion of clay particles and their adsorbed layers and the expansion of entrnpped air. The mechanism of swelling is much more complex than that of shrinkage.

,I

ElTects or Shrinkage ond Swelling or Soils Shrinkage and swelling crc.'lte many problems. as discussed below. (1) Shrinkage and swelling cause the deformations and stresses in the structures resting on or in the soil. (2) High swelling pressures develop if the soil has an aa.:ess to water, but is prevented from swelling. The light strud1!res may be lifted if the swelling pressure ·is excessive. (3) In semi·arid regions. the clay near the ground surface is subjected to shrinkage during dry periods and the cracks are formed. During wet periods, the clay swells and the cracks are closed. This process of the formation and closing of the crocks may cause the development of fissures in soils. (4) If silt particles drop into the shrinkage cracks formed behind the retaining wall, particles later swell and force the rctaining wall out of thc plumb. It may cause the failure of the wall if it had not been properly designed to resist the pre:ssure so developed. (5) If the soil below the pavements has high Shrinkage and swelling properties, it creates the problems in the maintenance of highways and runways.

7.15. SlAKING OF ClAY When a clay that had been dried well below the shrinkage limit is suddenly immersed in water, it

CAPILLARY WATER

disintegrates into a soft. wei mass. The process is known as slaking of clay. Slaking can be explained as below. When the soil dries to a water content lower than FILLED the shrinkage limit. some of the voids gel filled with air WITH AIR (Fig. 7.14). Water enters these air-filled voids when the 'sOil is immersed in water. causes an explosion of SATURATED VOIDS the voids, and therefore disintegration of soil occurs. Fig. 7.14. Slaking of Clay. According to another interpretation. when water cnler.; the pores. it forms menisci which react against the air in the void. 1be entrapped air is subjecled to very high pressure and the soil mass disintegrates.

~

Jhis

VOIDS

7.[6. nULKJNG OF SAND if ~I damp sand is loosely dejXISited. its volume is much more than that when the same sand is deposited in a loose. dry slate. TIle phenomenon of increase in volume of sand due to dampness is known as bulking of sand. In damped Slate, cohesion develops between the particles due to capillary water. The cohesion prevents lhe particles from taking a stable position. A SOrt of honey-comb structure is formed. The effect is predominant when the waler content is between 4 to 5%. The increase in volume due to bulking is between 20 10 30% for most s.1Ods. If the damp sand is smurated by adding more water, the effect of capillary action is eliminated and the volume of the sand mass is decreased.

7.l3. CAPILlARY SIPHONING In an eanh dam with an impervious core. capillary siphoning may occur (Fig. 7.15). 1be water rises in tile outer shell due to capillary action. If the crest
-~-.,.L.---?0.,-

SIPHONING SPILLS OVER COAt)

Fig. 7.15. Capillary Siphoning.

capillary rise, water flows from the· storage reservoir to the downstream over the core. Omsiderable quantity I of stored water may be lost due to capillary siphoning. To prevent this, the aest of the impervious core should be kepi sufficiemly high. In other words. the difference of top level of the oore and water level in the reservoir should be more than the capillary risc in soil of the shell. ILLUSTRATIVE EXAMPLES llIustrative Example 7.1. What is the negative pressure in the water just below the meniscus in a capillary tube of diwlleter 0.1111111 filled with watet, The surface tension is 0.075 Nlm and wetting angle is 10

degrees. . Solution. From Eq. 7.1.

4 T. cos

hr ..

e

-----gp:;J ..

Negative pressure .. y",h c

"

4 )( 0.075 )( 0.9848 .. 0.301 m 9.81 )( 1000 x 0.1 )( 10-3

9.81 )( teXXJ )( 0.301

.. 2952.81 N/ml .. 2.9531tN/m 1


132

lJIustmllve Example 7.2. Estimate the cnpillQry rise in a soil with a void ratio 0/0.60 and an effective size of 0.01 mn!. Take C = 15 mm2, Solution. From Eq. 7.8,

e;lo"

he" 0.6 !50 .01 .. 2500 mm .. 205m illustrative Example 7:3. The PF of a soil is 2.50. Determine the capillary potential of the soiL Solution.

Soli suction .. (10)2.5 .. 316.23 an .. 3.1623 m

Capillary potential .. - 3.1623 x 9.81

)C

tal

N/m'

.. - 31.02 kN/m2

Dlustratlve Example 7.4. The capillary rise in a soil A with an effective size of 0.02 mm was 6() em. Estimate the capillary rise in a similar soil B wilh an effective size of 0.04 mm. Solution. From Eq. 7.8,

(h,h (D",), (h,), • (D",h

(~2

..

~:~

.. 2

or

(hen" JOan

llIustrative Example 7.5. The capillary rise in sill;s 50 em and IMI in fine sand is 30 em. What is the difference in the pore size of the twO soils ?

Solution. From Eq. 7.3,

he '" 0;: em

For sill,

(he)l .. 50 ..

°d~

or d L

For fine sand,

(heh • 30 -

°d~

or d,. - 10,0

Difference in pore size

..

6.0 )( lcrJ an X

10..3 an

• (10.00 - 6.0) x 11r'

• 4.00 x 10-3 em PROBLEMS

A, Numeriall

..

7.1 ~~~~ ~!,~~~a2 r~l ~~, sandy

soil which has a void ralio of 0.65 and the effective Si~:::' ~~~~l~i

7.2. The effective size of a soil Is 0,015 mm. Estimate the height of capillary rise. Take surface tension as 0.074 N/m. [Aos. 10 m]

7.3. ;,~~f~~~~e maximum capillary tension for

a

capillary tube 'of 0.1 mm diameter, Take s[1~~ ;~o:;m~

7.4. The glass vessel shown in fig. P 7.4 is filled with water. It hns two holes of diameter 0.01 em and 0.03 ern as shown. If a fully- developed meniscus is formed in the upper hole, determine the height h of the wall of the vesseL [Aos. 20.27 em] 7.5. In Prob, 7.4, if both the holes ore of the some diametcr, equal to 0.Q1 em, determine the cont9Ct angle in the lower hole if that in the upper hole is zero and h ;; 20,27 an, [ARS. 70.54"1

n. Descriptive and O~edlve Type 7.6. Whnl are different CDtegories of soil Wtltcr ? Dc:saibc in brief, 7.7, Discuss the phenomenon of capillnry rise in soils. What are the factors that effect the height of capillary zone? 7.8. What is soil suaion ? How is it measured? What are the factors thaI affectloH sualon?

CAPILLARY WATER

133

T h

1 Fig. P.7.4 7.9. Differentiat!! bl!tween frost heave and frost boil. Whm is their tHect on soils? How frost actiun can be prevented ?

7.1D. Write a note on shrinkage and swelling of soils. 7.11. Discuss the phenomena of slaking and bulking.

C. Multiple Choice Questions 1. Capillary rise in 11 small tube i~ duc tl) ((I) Cohesion (b) Adhesion (e) Both cohe.~ion muJ adhesion (d) Neither (a) nor (b) 2. The surface tension of water at nonnal temperatures is about (a) 0.73 dynes/Ill (b) 0.73 N/m (d) 0.073 kNlm (el 0.073 N/m 3. The capi llary rise in clay is usulIlIy between (a) 0.10 [md 0.15 m (b) 0.3 and 1.0 III (e) 1.0 and 10.0 m (d) greater than 10 m 4. A pF v~llu e of zero corresponds to a soil section of (a) I m (b) zero metre {el I em (d) lOem S. The frost heave in the following type of soils is gcner~llly high (a) Coarse sands (b) clays (e) Fine sands :llld silts (dJ gravels 6. Bulking of s:lIlds is usually (a) Less than 10% (11) Between 20 to 30% (el Greater thun 30% (d) Between 10 to 20% 7. The frost heave depth as percentage of the soil deplh in fine sands and sills is about (0) 4 to 5% (b) 5 to 10% (el 1010 15% . (d) 20 10 30% 8. A tension of 1kN/m2 corresponds to a cnpillary potential of Ca) I kJ/kg (b) 10 kJ/kg (c) 100 kJ/kg (d) 1000 kJ/kg ~_1~~~1~~~~~.~~~ &~

8 Premeability of Soils 8.1. INTRODUCTION A material is porous if it contains inlenitices. The porous material is permeable if the interstices are interconnected or continuous. A liquid can flow through a permeable material. Electron photomicrographs of even very fine clays indicate that the interstices are interconnected. However. the size, cross· seaian, and orientation of the interstires in diITerent soils arc highly variable. In general. all the soils arc permeable. The property of a soil which permitS flow of water (or any other liquid) through it, is calkd the penneability._In other words, the permeability is the ease with which water can flow through it. A soil is highly pervious when water can now through il easily. In an impervious soil. the permeability is very low and water cannot easily now through it. A completely impervious soil does nOI pennit the water to flow through it. However. such completely impervious soils do not exist in nature. as all the soils arc pervious to some degree. A soil is termed impervious when the permeability is extremely low. Permeability is a very important engineering property of soils. A knowledge of permeability is essential in a number of soil engineering problems. suCh as settlement of buildings, yield of wells. seepage through and below the earth structures. It controls the hydraulic stability of soil masses. The permeability of soils is also rrquircd in the design of filters used 10 prevent piping in hydraulic structures. As mentioned in chapter 7, free water or gravitational water flows through soils under the influence of gravity. Flow of free water depends upon the permeability of the soil and the head causing flow. This chapter deals with Darcy's law for flow of water, the methods for the determination of permeability and the [adors affecting the permeability of soils. further details of flow o[ water and seepage problems are discussed in the next chap{er.

8.2. HYDRAULIC HEAD 'The total head at any point in a flowing fluid is equal to the sum of the elevation (or datum) head, the pressure head and the velocity head. The elevation head (l) is equal to the vertical distance of the point above the datum. The pressure head (ply..,) is equal to the head indicated by a piezometer with its tip at that point. The velocity head is equal to ';ng. However, [or now o[water through soils. as the velocity (v) is extremely small, the velocity head is neglected. Therefore, the total head o[ water in soil engineering problems is equal 10 the sum o[ the elevation head and the pressure head. for flow problems in soils, the downstream water level is generally taken a'i datum. The piezometric level is the water level shown by a piezometer inserted at that point. l'he line joining the piezometric levels at various points is called a piezometric surface. The piezometric surface also represents the hydraulic gradient Hnc (HGL). The sum o[ the pressure head and the elevation head is known as the piezometric head. Fig. 8.1 shows two vessels A and B containing water at different levels and connected by a small lube containing soil sample. Let the length of the tube be L. lbe flow takes place [rom the vessel A with a high head to the vessel B with a low head through the tube. With datum at the water level in the vessel B. the

PERMEABlLTfY OF SOILS

Point

e

Elevation heod Pressure heod

-IZ

,f

hZ

Totol heod

h'

d

,e is d ;h

al ,d

of

,. he ye ilt.

:ly ",I

tor al

'he he

,be .gIl

Ihe

Fig. 8.1. Variolti I·[eads.

elevation head, the pressure head and the total hcad at three points I, 2 and 3 are also shown in the figure. The total head at point 1 is h and that at point 3 is zero. llle head h is known as the hydraulic head. It is equal to the difference in the elevations of water levels at the entry and exit points in a soil mass. Obviously, it is equal to the loss of head through thc soil. Thc hydraulic head is also known as the effective head. The loss of head per unit length of flow throujllhc soil is equal to the hydraulic gradient (I), i _ hl L .. (8.1) whcre h hydraulic head. and L = lcngth of the soil specimen. The variation of head at various points is represented by the line CD, known as the hydraulic gradient linc (H.G.L.) or pressure gradient line. If a piezometcr is inserted at any intcnnediate point 2, the water will rise upto the level of the hydraulic gradient line at that point. The line CD. therefore, represents a piezometric surface. It is generally assumed thai the loss of head over the length of the soil sample is uniform and, therefore, the variation of head is linear.

83. DARCY'S lAW The flow of free water through soil is governed by Darcy's law. In 1856, Darcy demonstrated experimentally that for laminar flow in a homogcneous soil, the velocity of now (v) is given by v _ ki ... (8.2) where k

= coofficient of penneability, i = hydraulic gradient.

The velocity of flow is also known as the discharge velocity or the. superficial velocity. Eq. 8.2 is known as Darcy's law, which is one of the comcr stones of soil engineering. The discharge q is ootaioed by multiplying the velocity of flow (v) by the total cross· sectional area of soil (A) nonnal to the d.iredion of flow. Thus q _ vA - kiA ... (8.3) The area A includes both tbe solidS and the voids. The coetrJcient of permeability can be defined using Eq. 8.2. If the hydraulic gradient is unity, the coefficient of permeability is equal to the velocity of flow. In other words. the coefficient of penneability is defined as the velocity of flow which would occur under unit hydraulic gradient. The coefficient of permeability has the dimensions of velocity [Ln]. It is measured in mmtscc. cmlsee. m/sec, m/day or other velocity units. The coefficient of penneability depends upon the particle size and upon many other faaors as

136


explained later. Table 8.1 gives the typical values of the cocflkicnt of permeability of different soils. Thble 8.1. 1yplcal Values of the Coefficient of Permeability Coefficient of S. No.

(mmlsec)

Drainage properties

to+ 1 to 10+2

Very good

penneabilily

Soil Type

Cleangruvel

to 10+1

..

Coarse and medium sands

10-

Fine sonds, loose sill

10- 10

10- 2

Fair

4.

Dense silt, clayey sillS

1O-~ 10 10-4

p"",

5.

Silty day, day

10-3 to 10-5

2

2

Good

Very poor

According to USBR, the soils having the coefficient of permeability greater than 10-3 mmJsec are classified as pervious and those with a value less than 10-5 mm/sec as impervious. The soils with the coefficient of permeability between 10-5 10 10-3 mm/sec arc designated as semi-pervious. 8.4. VALIDITY OF

DAI~CY'S

lAW

Darcy's taw is valid if the flow through soils is laminar. 'Inc now of water through soils depends upon the dimension of interstic.::cs. which, in tum, depend upon the particle size. In fine-grained soils, the dimensions of the interstices are very small and the flow is necessarily laminar. In coarse-grained soils, the flow is also generally laminar. However, in very coarse-grained soils, such as coarse grdvels, the flow may be turbulent. For flow of water through pipes, the flow is laminar when the ReynOlds number is less than 2000. For flow through soils, it bas been found that the now is laminar if the Reynolds number is less than unity. For now through soits, the characteristic length in the Reynolds number is taken as the average particle diameter (D). Thus

where p :: is the mass density and " is the coefficient of viscosity. Using Allen Hazen's equation (Eq. 8.30) fOr velocity, it can be shown that the maximum diameter of the particle for the flow LO be laminar is about 0.50 mm. 1bc value of the critical Reynolds number of unity is, however, oonscrvalive. .It has been demonstrated that the flow remains laminar even upto the Reynolds number of 75. It has been observed thai Darcy's law is 'laUd for flow in clays, silts and fine sands. In coarse So'u)(IS, gravels and boulders, the flow may be turbulent and Darcy's law may not be applicable. It is difficult to predia the exact range of the validity of Darcy's law. The best method to ascertain the range is to conduct experiments and determine the actual relationship between the velocity v and the hydraulic gradient. For Darcy's law 10 be valid, this relationship should be approximately linear. For flow through coarse sands, gravels and ~lders, the actual relationship between the velocity and the hydrauliC gradient is non-linear. Hough gave the following equation for the velocity when the now is turbulent. ... (8.4)

where n = exponent. with a value of 0.65 In extremely fine-grained soils, such as a colloidal clay, the interstices are very small. The velocity is therefore very small In such soiis also, the Darcy law is not valid.

8.5. DETERMINATION OF COEFFICIENT OF PERMEABILITY The coefficient of permeability of a soil can be determined using the fol1owing methods.

137

PERMEABILITY OF SOILS

(0) Laboratory Methods. 'Ibc coefficient of permeability of a soil sample can be determined by the

following methods : (I) Constant-head penneability test (il) Variable-bead permeability test. 1lle instruments used are known as permeameters. The fonner lest is suitable for relalh1cly more pervious

soils, and the latter for less pervious soils. (b) Field Methods. l11e coefficient of permeability of a soil deposit in-situ conditions can be delCrmined by the following fields methods : (I) Pumping-OUl tests. (il) Pumping-in tesls.

The pumping-oul tests influence a large area around the pumping well and give an overall value of the coefficient of permeability of the soil deposit. The pumping-in Icst innucnces a small area around the hole and therefore gives n value of the coefficient of permeability of the soil surrounding the hole. (e) indirect Methods. The coefficient of permeability of the soil can also be determined indirectly from the soil parameters by (I) Computation from the particle size or its specific surface, (it) Computation from the consolidation test data. The first method is used if the partiCle size is known. The second method is used when the coefficient of volume change has been determined from the consolidation test on the soil. (d) Caplllurlty-Penneubility test. The coefficient of permeability of an unsaturated soil can be determined by the capillarity--permeability test (Sect.. 8.16).

8.6. CONSTANT IIEAD PERMEABILITY TEST The coefficient of permeability of a relatively more permeable soil can be dClcnnined in a laboratory by the conslant-head permeability test. The test is conducted in an instrument known as constant-he3d permeameter. It consists of a metallic mould, 100 mm internal diameter, 1273 mm effeaive height and 1000 ml capacity aocording 10 IS : 2720 (part XVII). The mould is provided with a detachable extension collar, 100 mm diameter and 60 mm high, required during compaction of soil. The mould is provided with a drainage base plate

:a~a:~r :roh~sV~;n:~ ~~etm:a~~~ ~:!":

I

h

1

::: : h air release valve. The drainage base and cap have fillings for clamping to the mould. Fig. 8.2 shows a schematic skeLCh. The soil sample is placed inside the mould between two porous discs. 1bc porous discs should be at least

t

[en times more permeable than the soil. The porous discs should be dcaired before these are placed in the mould. The water lubes should also be dcaired. The sample can also be prepared in the

pemleameter by pouring the soil into it and tamping it to obtain the required density. The base is provided with a dummy plate, 12 mm thick and 108 mm in diameter, which is used when the sample is compacted in the mould.

AI- 8.2. ConsUlnl Head Pemaeamel1cr

SOIL. M ECHANICS AND FOUNDATION ENGINEERING

138

It is essential thai the sample is fully saturated. This is done by one of the following three methods. (l) By pouring the soil in the pcrmeameter filled with water and thus depositing the soil under water. (il) By allowing water to flow upward from the base to the top after the soil has been plaoed in the mould. 1ltis is done by attaching the COrlStant-head reservoir to the drainage base. The upward flow is maintained for sufficicnt lime till aU the air has been expelled out. (iit) Dy applying a vacuum pressure of about 700 mm of mercury through the drainage cap for about 15 minutes after closing the drainage valve. Then the soil is saturated by allowing dCllired water to enler from the drainage base. 1be air-release valve is kept open during saturation process. After the soil sample has becn saturated. the oonstant-head reservoir is connected to the drainage cap. Water is allowed to flow out from the drainage base for some time till a steady-stalc is established. The water level in the constant-head chamber in which the mould is placed is kept constant. The chamber is filled to the brim at the stan of thc experiment. The water which enters the chamber aner flowing through the sample spills over the chamber and is collected in a graduated jar for a convenient period. The head causing now (h) is equal 10 the difference in water levels between the constant-head reservoir and the constant-bead chamber. If the cross-sectional area of the specimen is A, the discharge is given by (Eq. 8.3) q .. kiA

q-

k~A

k_

~

... (8~)

where L ::; length of specimen, h "" head causing flow. The discharge q is equal to the volume of waler collected divided by time. The finer particles of the soil specimen have a tendency to migrate towards the end faces when water flows through it. This results in the formation of a filler skin at the ends. The coefficient of permeability of these end portions is quite different from that of the middle portion. For more accurate resUlts, it would be preferable to measure the loss of head hi over a length L' in the middle to determine the hydraulic gradient (I). Thus i_hilL'. The temperature of the permeating water should be preferably somewhat higher than that of the soil sample. This will prevent relea<;e of the air during the test. It also helps in removing the entrapped air in the pores of the soil. As the water cools, it has a tendency to absorb air. To reduce the chana:s of formation of large voids al points where the particles of the soil touch tbe permeameter walls, the diameter of the perrneamcter is kept at team 15 to 20 times the particles size. To increase the ratc of flow for the soils of low permeability. a gas under pressure is appUed to tbe surface of water in the constant-head reservoir. The total head causing flow in that case increases to (h + ply",,), where p is pressure applied. The bulk density of the soil In the mould may be determined from the mass of the soil in the mould and its volume. The bulk density should be equal 10 that in the field. The undisturbed sample can also be used instead of the compacled sample. For accurate results, the specimen should have the same structure as in natural oonditions. (See Oiapter 30, Sect. 30.13 for the laboratory experiment). The const.ant had permeability test is suitable for clean sand and gravel with k > 10- 2 mmJsec. 8.7. VARIABLE-lIEAD PERMEABILfIY TEST For relatively .less permeable soils, the quantity of water collected in the groduated jar of the constant-head permeability test is very small and cannot be measured accurately. For such soils, the variable-head permeability test is used. The permeameter mould is the same as that used in the oonstant-head permeabiUty test A vertical, graduated stand pipe of known diameter is fitted to the lOp of penneameter:-The sample is placed between two porous discs. The whole assemblX is placed in a conslant head chamber filled with water to the brim at the start of Ihe test. F)g. 8.3 shows a schematic sketch. 1be porous discs and waia


139

tubes should be dc-aired before the sample is plaa:d. If in-situ, undisturbed sample is available, the same can be used; otherwise

~~ns:~

is taken in the mould and rompacted to the required

The valve at the drainage base (001 shown in figure) is closed aoel a vacuum pressure is applied slowly through the drainage cap to remove air from the soil. The vacuum pressure is increased (0 700 mm of mercury and maintained for about 15 minutes. 1be sample is saturoted by allowing cleaired water to flow upward from the drainage base when under vacuum. When the soil is saturated, both the top and bonom outlets are c100ed. The standpipe is filled with water to the required heighL The test is staned by allowing the water in the stand pipe to flow through the sample to the ronstant-head chamber from which it overflows and spills out. As the water flows through the soil, the water level in the standpipe falls. 1be lime required for the water level to fall from a known initial head (h t) 10 a koown final head (hi) is determined. The head is measured with reference to the level of water in the constant- head chamber. Let us ronsidcr the instant when the head is h. For the infinitesimal smalltime dt, the head falls by tIh. Let the discharge through the samplc be q. From continuity of flow,

~

I 1 h,

STANO PJPE

112 h

h

SAMPLI

adh _ -qdt whcre a is cr06S-scctional area of the standpipe. or adh--(Axkx,)xdt

Fig. 8.3. Variabte Head PermeamelC r.

adh-....4.kx~xdl A Icdt

-dh

-;L-h Jntegrating,

or

,uj' aL

*

dl_

IJ

-

!: ~ 10,

h

(/2 -It) - log., (hllhv Ie. -

~

log., (hi/hi)

...(8.6)

where t _ (12 -11), the time intcrval during which the bead reduces from hi to h'2'

Eq. 8.6 is sometimes writtcn as ... (8.7) The rate of fall of water level in the stand pipe and the rate of flow can be adjusted by changing the area of the cross-seajon of the standpipe. The smaller diameter pipes are required for less pervious soils. The coefficient of penneability is reported at 27°C as per IS : 2720 (Part XVII). The void ratio of the soil is also general1y detennined. The variable head penneameter is suitable for very fine sand and silt with k ::: 10- '2 to 10-5 mm/scC.

(See Cllapter 30, Sect. 30. 14 for the laboratory experiment). Somelimes, the permeability test is conducted using the ronsolidomeler instead of the permeametet


140

mould (see chapter 12). The fixed-ring consolidometer is used stand pipe to its base.

a<;

a variable-head permeametcr by attaching a

8.8. SEEPAGE VELOCITY The discharge velocity v in Eq. 8.2 is not the actua1 velocity through the interstices of the soil. It is a fictitious velocity obtained by dividing the total discharge (q) by the total cross-sectional area (A.). The total CfOSS- sectional area consists of not only the voids but also the solids. As the flow can take place only through voids, the actual velocity through the voids is much greater than the discharge velocity. TIle actual velocity on a macroscopic scale is known as the seepage velocity (vs).

r

1

~T~r ~l~l

i----A--1

(b)

(.)

Fig. 8.4. Seepage Velocity

Fig. 8.4 (a) shows the longitudinal seaion through a soil sample in which the voids and the solid particles are segregated. However, it must be clearly understood that the voids and solids in actual soils fonn a complex system and it is not possible to segregate them . From the oontinuity of flow.

... (.)

q - vA _ v,A .. where A., is the area of flow through voids and v, is the actual seepage velocity. v, - v x (A/A.~) From Eq. (a). Multiplying the numerator and denominator by the length (L) of the specimen,

v, - v x (:. :

i)

...

(b)

The product (A x L) is equal 10 the lota1 volume V and the prodLKi (A .. x L), equal to Ihe volume of voids (V.) [Fig. 8.4 (0)]. There[ore,

v, - v x

V

'Y:'

... (0)

As the ratio V,IV is equal to the porosity,

v v _, n

... (8.8)

In other words, tlie seepage velocity is equal to the discharge velocity divided by porosity.

~

Using Eq. 8.2,

v, -

or

v, _ Ap xi

... (8.9)

wbere ~ - k/n ... (8.10) The ooefficient ~ is known as the coeDkienl of percolation.. Its value is always greater than the coefficient of penneability (Ie).


141

Strictly speaking, the seepage velocity is not be absolute velocity through the interstices. The interstices are tortuous and irregular in cross·section and cannot be represented as shown in Fig. 8.4 (a). The absolute velocity varies from point to point. Its direttion may also change and, at times, i! may be directly opposite to the general direction of flow. In fact, the problem is so complex that an analysis based on the absolute velocity is not possible. Although on the microscopic scale, the flow path is tortuous, on a macroscopic scale, it can be considered as a straight line. The seepage velocity can be taken a'> the maC'OSa>pic velocity at which the line of wetting progresses in the direction of flow. ObviOUSly, it is not equal to the absolute velocity as the water flows not in a straight line but it detours around solid particles. Fortunately, the absolute velocity is not of much practical use in soil engineering. lbe geotechnical engineer is interested in the macroscopic behaviour of the soil aDd not in its microscopic behaviour. The total discharge can be computed using either the discharge velocity (v) or the seepage velocity (VI). The discharge velocity is more convenient and is commonly used in soil engineering. In this text, when the tcnn velocity is used without any qualification, it means discharge velocity.

8.9. GENERAL EXPRESSION FOR LAMIHAR FWW For understanding the flow of water through soils, let us first consider the laminar flow through pipes. Fig. 8.5 shows a horizontal pipe of circular cross·section of radius R. Let us take a small cylindrical fluid element of radius r and length I, as shown in figure. The shear srress "'C on the surface of the fluid element is given by Newton's law of viscositya'>

Fig. 8.5. Laminar

now in a pipe.

, --~ (~ ) where 1.1. = coefficient of viscosity and

... (0)

~ = velocity gradient.

For steady flow, the net force acting on the element in the horizontal direction is zero. Therefore,

(Pl-pvrci - (2ttrl)"'C _ 0 Substituting the value of"'C from Eq. (a), and simplifying,

dv

- dPl-J>i)

d;-~

The pressure PI and P2 can be expressed in terms of piezometric heads hI and ~ as

PI- lwhl

Thus

dv

d; -

P2 - lw h2

and

-ryw(hl-hV --2-~-1-

Representing (hi - hiJ/1 by the hydraulic gradient (l).

dv

-r'(wi

d;-~

Integrating,

v _

-;f~ i ( ~ )

+ C

The constant of integration C can be obtained from the condition of no slip (Le.. v

=0) at the boundary


142

(i.e., r = R). Thus

'tw iR2 C-~

v .. -'1",1

Therefore

?

+ y",ifil

4J.l v_

Eq. 8.11 indicates that

4"

'h!.(R2

4"

_ ?)

... (8.11)

pipe

is of the shape of paraboloid of reVOlution, with the maximum velocity at its centre. The equation is known a.. Hagen-Poisseuille equation for laminar flow through pipes. The equation can be used to detcnnine the discharge q in the pipe a.. under. Discharge through a small ciraliar ring of radius r and thickness dr is given by the velocity distribution in a

y.i ..,

4; (,,-,.AI

dq - (2x,dr) v - 21Udr

q _

Integrating.

I'

Jo

2w

(~)

(R'--?)dr _ ,,!.ilt

41'

8"

Writing the radius R in terms of the hydraulic radius RH (i.e RH .. D/ 4 - RI2) and the area A

for n: R2,

q -

! ~ R~I

)C

A

... (8.12)

2 "

Ukewise, it can be shown thai the discharge through two parallel plates of width B and placed at distance d apart is given by (see any text on Auid Mechanics),

q_

l~(2Bd~ 3

Substituting

A=2Bd

nnd

"

RH - ;n: - d.

q .. 1. , (... i Rl, )( A .. . (8.13) 3 " Comparing Eqs. 8.12 and 8.13, it is observed that the general fonn of the equation for laminar flow through passages of different shapes is the same. The difference is only in the numerical value of the constants. The general equation for discharge in a conduit of any shape can be written as q - C,

C~i) RI, A

... (8.14)

where C, is a constant which depends on the shape of the conduit. Eq. 8.14 is sometimes called the generalized Hagen- Pouseuille eqULltion. This equation can be used in a modified form for the flOW" through soils ao; explained in the next section. 8.10. LAMINAR FLOW THROUGH POROUS MEDIA

Since the flow through porous media is laminar. Eq. 8.14 can be used. However, the area of flow passage in the cac;e of porous media is equal to the porosity times the total cross-sectional area and, therefore., Eq. 8.14 becomes q - C,

(¥) RI,

(NI)

when n is the porosity of soil. represented m. ratio. "(be hydraulic radius RJI for a p:trous medium can be written as

...(0)

PERMEABIUfY OF SOILS

143

R .. H

areaofflow A~ wetted perimeter ..

p:,

Multiplying the numerator and the denominator by the length of the passage (L).

R .. A~ xL.. volume of flow channel 1/ p~ x L surface area of Dow channel 100 volume of Dow channel may be taken as the volume of voids (V~), which is equal to e V, , where e is the void ratio aod V, is the volume of solids. The surface area of the Dow channel may be worked out 00 the basis of a hypothetical spherical grain of diameter D and having lhe same volume/area ratio as the entire

mass. Thus RII ..

V~

eV.

'JtrY/ 6

eD

A, .. T, .. e --;;[j2 .. 6

Substituting the above value of RII in Eq. (a) and taking n .. el l + e,

q . c,(~)(eN(~)A q · ~(~)(I: . )D'iA Replacing C,I36 by another cocfHc ient C,

q. c (~) (I: .) D'iA *.C(~)(I: e )D'i Using Eqs .. (8.2) and (8.3). the above equation can be written as

v. C(~) (I: .) D'i .ki where

k.C(~)(I:e)D'

... (8.15)

Eq. 8 .15 gives a general expression for the coefficient of penneabWty of soil. 8.11. FACffiRS AFFEcnNG PERMEABILITY OF SOILS The following factors affect the permeability of soils. (1) Particle size. As it is evident from Eq. 8.15, the coefficient of permeability of a soil is proportional to the square of the particle size (D). l11e permeability of coarse-grained soils is very large as compared to [hal of fine- grained soils. The permeability of coarse sand may be more than one million times as much that ofcJay. (2) Structure or soli mass. The coefficient C in Eq. 8.15 takes into 3CCOlUlt the shape of the flow passage. The size of the flow passage depends upon the structural arrangement. Hx the same void ralio, the permeability is more in the case of floca.J1ated structure as compared to that in the diSpersed structure. Stratified soil deposits have greatcr permeability parallel to the plane of stratification than that perpendicular to Ihis plane. Pcnncability of a soil deposit also depends upon shrinkage cracks. joints, fissures and shear wncs. Loess deposits have grealer permeability in the vertical direction than in the horizontal direction. The permeability of a natural soil deposit should be detcnnined in undisturbed condition. 1be distwbance caused duriog sampling may destroy the original structure and affect the penneability. The effect of disturbance is more pronounced in the case of fmc· grained soils than in the case of coarse-grained soils.


144

t

(3) Shape or Particles. The penneability of a soil depends upon the shape of particles. Angular particles have greater specific surface area as compared with the rounded particles. For the same void ratio, the soils with angular (~2• ..!2) YS 1+< particles are less penneable than those with I+eo He rounded particles, as the permeabiUty is inversely proportional 10 the specific surface. However, in a natural deposit. the void ratio for a soil with angular particles may be greater than that for rounded particles, and the soil witb angular particles may be actually more penneable. PERMEABILITY (k)----.(4) Void Ratio. Eq. 8.15 indicates that the coefficient of penneability varies as ;/(1 + e). Fig. 8.6. Variation of k with;, 1~: e and 1~: e For a given soiL, the greater the void ratio, the higher is the value of the coefficient of permeability. Based on other concepts. it has been established that tbe permeability or a soil varies as t? or e 2/(1 + e) (Fig. 8.6). Whatever may be the exact relationship, all soils have e versus log k plot as a straight line (Fig. 8.1). It must be noted tbat eacb plot in Fig. 8.7 is for a given soil. The permeability of a soil at a given void ratio may not have any relationship witb that of another soil at the same void ratio. Paradoxically, the soils with the largest void ratio (i.e. clays) are the least pervious. This is due to the fact that the individual void 0.9 passages in clays arc extremely small =~~ which water cannot flow 0.8

..i!.

1

1.0,--------------,

If the permeability of a soil at a 0·7 void ratio of 0.85 is known, its value at another void ratio of e can be determined using the following 0·6 equation given by Casagrande: 4;'

t

k - 1.4 ko,,, e'

... (8.16)

where ~.85 ::: permeability at a void ratio of 0.85, k ::: permeability at a void ratio of e. (5) Properties of water. As indicated in Eq. 8.15, the roefficient of permeability is directly proportional to the unit weight of water <"t ...) and is inversely proportional to its visoosity (~). The unit weight of water does not vary much over tbe range of temperature ordioarily encountered in soU eogtneering problems. However, there is a large variation in tbe value of the ooefficient of visalsity (~). The

0·5

S ~ § ::>

0·1., 0·3

0-2 ().1

O'~04

1COEFFICIENT OF PERMEAB1Ut-V (k) Fig. 8.7. Varilltlon cllOi k with e.

t'rJ mm/~c:

_____

k

PBRMEABILITY OF SOILS

145

coeffkient of penneability inaeases with an increase in temperature due to reductiO'I in the visrosity. It is usual practice (IS : 27111 Part XVU) to report the coefficient of permeabililJ at 27 D C. The following equation can be used for conversion of the penneability to 27D C.

k", - k,

-!;

...(8.17)

where kz1 = coefficient of penneabUity at 27 D C when viscosity is !In. and k, = coefficient of penneability of t DC when visa:Jsity is ~,.

Eq. 8.17 can be written as where

kn - e,k, e, is tbe corted.ion fadar. equal to

... (8.18) (jA.,/!lv).

The correction factor e, can be determined from the values of the coefficient of visoosity given in Thble 3.2. (6) Degree of Saturation. If the soil is not fully saturated, it contains air pockets fanned due to entrapped air or due to air liberated from percolating water. Whatever may be the cause of the presence of air in soils, tbe permeability is reduced due to presence of air which causes blockage of passage. Comequently, the permeability of a partially saturated soil is considerably smaller than that of a fully saturated soil. In fact, Darcy's law is not striclly applicable to such soils. The penneability of a partially saturated soil is measured in the laboratory by the capillarityI'.",,,ability test (Sect. 8.16). Adsorbed water. 1be fine-grained soils have a layer of ad
m

8.1l. COEFFlCIENT OF ABSOLUTE PERMEABILITY As discussed above. the coefficient of permeability of a soil depends not only on the properties of soil but also on the properties oC penncant (water). Attempts have been made to separate tbe effect of properties of permeant from the effect of the properties of soil. Another coeffident, known as the coefficient of absolute permeability (K), has been introduced. It is related to the coefficient of permeability (k) $ under: K - k(~/y.)

Using Eq. 8.15.

... (8.19)

K-C(~)Ii' 1+ •

Therefa-e, the coefficient of absolute permeability (X) is independent of the properties of water_ It

depends only on the characteristics of soils. The dimensions of the coeffident of absolute permeability can be determined from Eq. 8.19 as [K] - [

¥1[p,:] [~]- [L'j

It bas the dimension of area. The units of K are mm 2, cm 2, m2 or darcy. 1 darcy = 0.987 )( 10-' cm 2

146


The coefficient of absolute permeability for a soil with a given void rntio and structure is constant. It has tbe same value whatever may be the fluid.

8.13. PUMPING-OUT TESTS The laboratory methods for the determination of the coetlicient of permeability, as discussed before, do not give correct results. The samples used are generally disturbed and do nol represents the true in-sltu structure. For more accurate, representative values, the field tests are conducted. The field tests may be in the form of pumping oul test wherein the water is pumped out from the wells drilled for this purpose. The other type of Ibe field tests are pumping~in tests, wherein the water is pumped into tbe drilled holes, as diso.Jsscd in the, following article. For large engineering projeas, it is the usual practio: to measure the permeability of soils by pumping-out tests. The method is extremely useful for a homogeneous. coarse grained deposits COl" which it is diffirult to obtain undisturbed samples. In a pumping out test, the soil deposit over a large area is influenced, and therefore the results represent an overall coefficient of permeability of a large mass of soil. However, the tests are very costly and can be justified only for large projects. Ground water occurs in pervious soil deposits known us aquifers. The aquifers are reservoirs of ground water that can be easily drained or pumped out. An aquiclude is a soil deposit which is impervious. If an aquifer docs not have an aquiclude at its top and the water table is in the acquifer itself, it is called an unconfined aquifer. If the acquifer is confined between two aquicludes, one at its top and the other at its bottom, it is known as confined aquifer. The coefficient of permeability of the soil can be found using the equations developed below separately for unconfined aquifer and confmcd aquifer. (a) Unconfined Aquifer In an unconfined aquifer. a tube well is drilled as shown in Fig. 8.8. The well reaches the underlying impervious stratum. TIle tube used for the well is perforated so that water can enter the well The tube is surrounded by a saeen called strainer to check the flow of soil particles into tbe well. Waler is pumped oul of the tube well till a steady state is fC..1Chcd . AI that stage. the discharge becomes

Fig. 8.8. Unconfined Aquirer

constant and the water level in the well does not change. The water table, which was originally horizontal before the pumping was started, is depreSsed near the well. The water table near the well fonns an inverted cone, known as the cone of depression. The maximum depression of the water table is known as the drawdown (d). The expression for the coefficient of permeability can be derived making the following assumptions, known as Dupuit'S assumption.

PERMFABIUfY OF SOILS

147

(1) The now is laminar and Darcy's law is valid. (2) The soil mll'iS is isotropic and homogeneous. (3) The well penetrates the entire thidcness of aquifer. (4) The flow is steady. (5) The coeffICient of penneability remains constant throughout. (6) The flow towards the well is radial and horizontal. (7) Natural ground water regime remains constant. (8) lbe slope of the hydraulic gradient line is small, and can be taken as the tangent of the angle in place of the sine of angle, i.e. . dz ... (8.20) '-';t; ~

J.

Let

U5 consider the flow through a cylindrical surface of height

z at a radial distance of r from the centre

of the well. From Darcy's law,

q .. kiA Substituting the value of i from Eq. (8.20) and taking A equal to 2nI' z,

q. k

IS

,e

,11

"u.

..I ed

or Integrating,

!!!.. ..

(~)

2nkzdz

•

/
q 2d /

Irq

Iog.('';',)' k =

(2xn)

1

uJz

~ (zl- zl) q

2

--fL-----,;-1t(Zi - .:1)

or

log.. (r1ir )

... (8.21)

~ k • • (zl _ zl) log" ('';'')

... (8.22)

Near the test well, there is a rapid drop in head and the slope of the hydraulic gradient is steep, and asswnption (8) is not satisfied. The observation wells 1 and 2 should be drilled at considerable distance from the well for acx:urate measurements: The radial distance of the well should be at least equal to the thickness of aquifer (D). The observation wells are usually arranged in two orthogonal lines, one along the general direction of flow of the ground water and the other at right angle to this direction. An approximate value of the coefficient of pennenbility can be detennined if the radius of influence (R) is known or is estimated. The circle of influence, over whicb the effect of pumping is observed, extend) to a very large area. In fact, it gradually merges asymptotically 10 Ihe water table. The radiw; of influence varies between 150 to 300 m. According to Sichardt, it can be found using the relation

'1

R • JrnJdVli where R = radius of influence (m), d = drawdown (m) and J.: = Coefficient of penneability (m/SeC) According 10 Kozeny (1933), the radius of innuence;

R • [(12 rln)(qk/nJ""]'" where I is the time required to establish steady oooditions, and Eq. 8.21 can be written as

II

is the porosity.


148

k • ~ 109. (RI,.) where r... :: radius of test well, R = radi~ of influence, D :: depth of aquifer measured below tbe water table.

Eq. 8.23 gives an approximate

r

near the well is steep and Dupuit's assumption is not justified. Further. the value 'of the radius of influence (R) is

also approximate.

The piezometric surface is above the

top of the aquifer. In mnfined aquifer, the water pressure is indicated by the piezometric surface (PS). Thus the piezometriC surface is the water table equivalent for a confined

h:: depth of water in the well

P.S. = PIEZOMETRIC SURFACE

value of the mcmcienl of penneability. because the slope of the water surface

(b) Confined Aquifer. Fig. 8.9 shows a oonfined aquifer of thickness b and lying between the two aquicludcs.

... (8.23)

11 D

G.S.

--0;.:;--.. .

CONE

DEPRESSION

T ~.LL<====~ CONFINED

b

AQUIFER

;r,==='777n'7f.r.=c:'::l+.~~:>hJr",77777

aquifer. Initially, the piezometric surface is . horizontal. When the pumping is Fig. 8.9. Confined AqUIfer. started from the weU, it is depressed and a cone' of depression is fonned. The expression for the coefficient of penncability can be derived making the same $Sumptions as in the cze of unconfined aquifer. Let us consider the discharge through a cylindrical surface at a radial distance , from the centre and of height z. From Darcy's law,

q - kiA q ok Integrating,

(;l;-) (2ru-b)

[log.('ll. hqkb Izll k • q log. ('';'1) 2nb(z2- Zt)

or or

...(a)

k·

2.30q loglO (r2"rt) 2nb(z,-zl)

.(8.24) ... (8.25)

= height of water level in observation well (1) at a radial distance of '1 and Z2 = height of water level in observation well (2) at a radial distance of '2' As in the case of an unconfmed aquifer, an approximate value of k can be detennined if the radius of

where

%1

influence R is known or estimated. In this

C&'ie,

k • q log. (RI,.) 2nb(D-h)

... (8.26)

8.14. PUMPING-IN TESTS Pumping~in tests are conducted to determine the ooefficient of permeability of an individual stratum through which a hole is drilled. These tests are more economical than the pumping-out test. However, the

PERMEABILITY OF SOIlS

149

pumping-oul tests give more reliable values than that given by pumping-in tests. The pumping-in lests give the value of the coefficient of permeability of stratum just close to the hole, whereas the pumping-out lests give the value for a large-area around the hole. There are b~lcally two types of pumping-in tests: (1) Open-end tests, (2) Packer tests. In an open-end tests. the water flows oul of the test hole Ihrough its bottom end, whereas in packer tests, the water flows out through the sides of the section of a hole enclosed between packers. 1be value of the coefficient of pcnneability is obtained from the quantity of water accepted by the hole. The water pumped-in should be clean, as tbe impurities, such as sill, clay or any other foreign matter, may cause plugging of the flow passages. If the water available is tUrbid, it should be clarified in a settling tank or by using a filter. The temperature of the water pumped in should be slightly higher than the temperature of the ground waler 10 preclude the formation of air bubbles in stratum. (1) Open~nd 'Jests. A pipe casing is insencd into tbe bore bole to the desired depth and it is cleaned out. The hole is kept filled with water during cleaning if it extends below the water table. This is necessary to avoid squeezing of the soil into the bottom of the pipe casing when the driving 1001 is withdrawn.

T H

", ~,

lOT

t-'-'--I

",

(o)

~

~ (b)

1

S;ZW.T.

(
Fig. 8.10. Open-end /e!i.ts.

After the hole has been cleaned out, water is added to the hole through a metring system. The constant

rate of flow (q) is determined at which the steady conditions are established. The coefficient of penneabilily is detennined by the fOllowing equation (USBR, 19til). k -

s1rH

...(8.27)

= inside radius of the casing, H = difference of levels between the inlet to the casing and the water tabJe [Fig. 8.10 (a)l, q,. discharge If required, the discharge can be inaeac;ed by pumping-in water under a pressure p [Fig. 8.10 (b)]. In this ~, the value of H becomes equal 10 (H + ply",,). Foe aocmate results. tbe lower end of the pipe should be al a distance of not less than 10 r from the top where

r

as well as from the bottom of tbe stratum. The open-end test can also be conducted above the water table [Fig. 8.10 (e)]. In this case, however,

iI


ISO

is difficult 10 maintain a constant water level in the casing and some surging of this level has to be tolemted. Eq. 8.27 can also be used in this case. However, in this case H is equal to the difference of inlet level and the bottom end of the pipe. If required, the rale of now (q) can be increased by pumping-in water under a pressure A with a total head of (H + ply..,). (2) Packer Tests. The packer tests are perfooned in an uncased portion of the pipe casing. The packer tests are more commonly used for testing of rocks. The tests are occasionally used for testing of soils if the bore hole can stay open without any casing. (a) Single packer tests. If the hole cannot stand without a casing, single-packer lest is used. The packer Is p~ as shown in F4,g. B.l1 (a). Water is pumped into the hole. It comes out of the sides of uncased portion of the hole below the packer. If the casing is used for the full depth, it should have perforations in the portion of the stratum being tested. The lower end of the casing is plugged.

(a)

(b )

Fig. 8.U. Packcrtesl5.

When the steady ooooitions are attained. the constant rate of flow (q) is dctmnined. lbe value of the coefficient of penneability is found by the following equation (USBR, 19(1).

k -

k where

z!ui log, (LI,) z!ui sinh- (L I 2r) 1

if L

~

tOr

if lOr> L

••. (8.28) :t

r

.. .(8.29)

= inside radius of hole, L = length of the hole tested, H = difference of water levels al the entry and the ground water table for the hole tested below l'

the water table. sinh- 1 = arc hyperbolic sine. For the holes tested above the water table, H is equal to the difference of levels of water al the entry and the middle of the test section [Fig. 8.11 (b)].

If the water is applied under pressure (P). the value of II beoomc:s (H + pI., ...). ao; in the case of open-end tests.

After the test is oomplete. the packer is removed. If required. the hole is made deeper and again a packer is placed and the test repeated for that portion.

PERMFABILITY, OF SOILS

'"

stanJb~i~~~~~-:~: d:~le~~:erh~:t ~~

~i~::i:I:::'ris~~:; ~~ :~:edde~~:

T

1

TWo packers are fitted to a small diameter pipe, as shown in Fig. 8.12. The bottom of the pipe fitted with packers is plugged. Fig. 8.12 (a) shows the oonditions when the test section is below the groWKI water table and Fig. 8.12 (b), when above the ground water lable. The value of the coefficient of penneability is determine4 using Eq. 8.28 or Eq. 8.29. depending upon the value of L and r as

szW.T.

specified. (.)

The double-packer test is oonducted first

(0)

in the lowest portion ncar the bottom of the

Fig. 8.12. Double-prtcker test. hole and later repealed for the upper layCf5. The packer tests give better results when conducted below the water table than when above Ibe waler table. For reliable results, the thickness of the stratum should be at least five times the length (L) of the hole tested.

8.15. COEFFICIENT OF PERMEABILITY BY INDIRECT ME'llIODS The value of the coefficient of penneability of a soil can be estimated by using iodired methods, withoul conducting laboratory tests or field tests. The following methods are commonly used. (1) AUen Hazen's fonnuls. Allen Hazen conduded a large number of tests on mter sands of particle size between 0.1 mm and 3.0 mm, having a coefficient of unifonnily of less than 5, and gave the following relation:

k - C Dl.

...(8.30)

where k = coefficient of penneability (em/sec). 0 10 :::: effective size (em), C = constant, with a value between 100 and ISO If Ie. and DIO are taken in mm/sec and mm, respectively, the value of the constant C lies between 10 and 15. Although Allen Hazen formula was derived for unifoon sands in a loose state of compaction, it is frequently used used for other soils. However, the computed values may be in ern>r by 1: 100%. (2) Kozeny-Cannan equatloh. The coefficient of the permeability of a soil can be estimated using the Kozeny-Carman equal ion:

k_~'..L. (C,J.lS)7:Z

where

1 + e

... (831)

k = coefficient of permeability (croisee).

P ... := mass density of water (gm/ml), C, :::: shape factor, which can be taken as 2.5 for granular soils, J.l = coefficient of viscosity (poise), e :::: void ralio, g :::: 981 antsec2• T:= tortuooity, with a value of V2 for granular soil<; and S ::::;. surface area per unit volume of soil solids, known as specific surface (cm2/an~.

The Kozeny-Carman equation gives good results for coarse-grained soils such as sands. and some silts. However, when the equation is u.<;ed for clayey soils, serious discrepancies are observed. The acruracy for coarse-grained soils is about 20%. For computation of k from Eq. 831, the value of specirlC surface S is required. The specific surface (S) of a particle is equal to. the surface area of the particle per unit volume of the particle. It depends upon the shape and size of the particle. For a Spherical panicle of diameter D. specific surface (S) is given by


s_

(.10') _ ~ (.0'16) D

The specific swface of spheres unifonnly distributed in size between the mesh size

... (8.32) Q

and b, is given by

S _ 61.fiifi ... (8.33) For accurate results, the ratio alb should not be greater than 2. Ir the particles arc of irregular shape. the specific surCa<.:e can be determined indirectly from 8 comparison with the specific surface oC unifonn sphere of the same size, and using a factor known as angularity factor (J).

I ..

;:~~s~::=::s~e:u;!:;~~~~~

The value of f depends upon the angularity of the particles. Its value is usually taken as 1.1 for rounded sands, 1.25 for sands of medium angularity and 1.4 for angular sands. If Mh M2 ... Mil are the percentage of the total soil sample retained on different sieves. the overall specific surface oC the lotal sampk: is given by S .. j{M1S 1 + M2~ + ...... M"S,,) .•. (8.34) where S .. S2 ...... 5" are the specific surface of spheres uniformly distribute:! wilhin the corresponding

sieves. (3) Loudon's Fonnula. Loudon gave the following empirical formula.

IOg10 (k s')

-

a + bn

= coefficient of permeability (an/sec). S = specific surface (an2/cm\ n = porosity, expressed as a ratio. a = constant, with an average value of 1.365 at

... (8.35)

where k

lO"e,

b = constant, with an average value of 5.15 at lO"e. The Loudon fonnula is much more convenient to use than the Kozcny-Carman equation and gives approximately the same accuracy. (4) Consolidation test data. 1bc coefflcicnt of permeability of fine-grained soils can be determined .:ndirectly from the data ootained from a consolidation test conducted on the sample (see chapter 12). It is given by Ie - C~'t", m~ - C~p",gm.. ...(8.36) where Ie :::: coefficient of permeability (m/Sec). C.. = coefficient of consolidation (m 2 /soc), pw = density of water (kg/m~, g = 9.81 m/sec?, 2 m.. = coefficient of volume compressibility (em /N). y.., = unit weight or water (N/m\ This method is suitable for very fine-grained soils (Ie < Itr mm/sec) for which permeability test cannot be easily conducted in the laboratory.

8.16. CAPILLARITY-PERMEABILITY TEST The coefficient of permeability of soil in unsaturated condition can be determined from the capillarity-penncability test. The apparatus consists of a transparent tube made of lucite or glass, about 35 an long and 4 em diameter (Fig. 8.13). 1be sampie of the dry soil in powdered form is placed in the tube and screens are fixed at both ends. One end of the transparent tube is connected to high level waler reservoirs and the other end is open to atmosphere through an air-vent pipe. The air-vent pipe is connected to the screen at that end with a spring. The valve D connecting to the higher reservoir is initially closed. When the valve C connecting to lhe lower reservoir is opened, capillary action in soil occurs and it draws water into it. The wetled surface starts advancing towards the open end. Lei us oonsider the stage when the welted surfaoe has advanced by a distanoe of x. Let the negative capillary head be hrt as shown by an imaginary manometer in figw'C. (The manometer is imaginary and in actual tests, no manometer is used. It has been shown in the figure just to

1"

PERMEABILITY OP SOILS

IndiaJle the negative bead). The total bead causing flow is increased because of the negative bead (he) and is given by

Assuming a uniform hydraulic gradient over the entire length % • the velocity is given by Darcy's law. v .. 1 i .. k (hi; he)

...(a)

The welled surface moves (on a macrosoopic scale) with a seepage velocity (v,). given by Eq. 8.8 as V, ..

Fig. 8.13. Capillarity-Penneability test.

vln

Therefore, the seepage velocity is given by v .. !. . (hi + he) 'n X For partially saturated soils, the above equation is modified taking actual saturated porosity as Sxn. Thus v _ ~ (hi + h,) I Sn x where 1. "" roefficient of permeability in unsaturated condition S = degree of saturation, expressed as a ratio. v, _ dxI dl. we have Substituting

dx

k. (hi + he) --x-

xdx ..

k..(h~: he) dl

di - S;;

ht)! dl

Integrating,

!

or

- 2 - - --S-n-

1

Xdx .. i.(h l + Sn

..s-il

..s - .G

(12-11) ..

1

k.(hl+h.) (

21. (hi + h,) --S-n--

t2 - tl

)

...(8.37)

Eq. 8.37 can be used to detennine the coefficient of pcnneability (l..) if all other variables are given. As the capillary head (he) is also not known, there are two unknowns (ktt and ht) on the right-hand side of the equation. Therefore, one more equation is required. The SCCX)nd equation can be derived if the head is changed from hi to ~ when the water surface has advanced 10 about half the length of the transparent tube by closing the valve C and opening the valve D. Let %2 and x,l by the distances measured from the left end at the time t;2 and I). Eq. 8.37 becomes. for this case, as

.oi-..s 21. (h,+ h.) (I)-tV .. --S-n--

... (8.38)

The values of the unknown k,. and ht can be obtained analytically from Eqs. 8.37 and 8.38. A plol. is


154

made between;Z and I, as shown in Fig, 8,14. The slopes ml and mz of the lines give the

;~~~~!~~~~fl.hand sides of Eqs, 8,37 and ml _ 2 k.. (~ln + he)

... (8.39)

m, • 2k. (~': h,)

... (8.40)

;Z

4 x?

The values of k.. and he are detennined x,2 from Eqs. 8.39 and 8.40 after substituting the values of ml and mz obtained from the plot. The porosity n of the soil sample is determined from its dry density, as discussed in chapter 2, M, Gp,.. Pd -

fig. 8.14. Plot or t and x'l..

V - T-:;-;

e_Gp"'_l Pd and n .. _e_ I + e The degree of saturation (5") ' is obtained from the water rootent of lhe soil delennined after the test, using the equations developed in chapter 2. S .. wGle For accurate results, the capillary head (he) should be maintained constant almg the vertical wetting surface. It is done by slowly revolving the tube about its axis.

8.17. PERMEABILITY OF ITRATIFIED SOIL DEPOSITS A stratified soil deposit consists of a number of soil layers having different penneabilities. The average permeability of the deposil as a whole parallel 10 the planes of stratification and thai nonnal of the planes of stratification can be detennined as explained below. (Q) Flow ParaUel to Planes of Stratmcatlon. Let us consider a deposit consisting of two horizontal layers of soil of thickness H J and Hz as shown in Fig. 8.15.

I

If

1

"1 LAVER 0) For flow parallel to the planes of -q stratification, the loss of head (h) over a length L is the same [Of both the layers. Therefore, the L-_ _ - __ _ _" hydraulic grndient (I) for each layer is equal to the hydraulic gradient of the entire deposit. The system is analogous to the two resistances in ----i.~1 parallel in an elearical cira.1it, wherein the potential drop is the same in both the resistances. Fig. 8.15 From the continuity equation, the total discharge (q) per unit width is equal to the sum of the discharges in the iodividual layers. i.e., ... (a)

11Hz

~_LA_~E_R_(2)

I-I.---L.

1SS

PBRMEABIUrY OF SOILS

Let (kllh snd (kllh be the permeability of the layers 1 and 2 rc5ped.ively, parallel to the plane of stratification and (kh) be the overall penneability in that direction. From Eq. (a), using Darcy's law, ~

)( i )( (HI + Hv - (k")1 )( i '/(, HI + (k"h )( i )( H2

>.)(

k (kil HI +
. . .(8.41)

(b) Flow normal to the plane or stratlncatlon. Let us consider 8 soil deposit consisting of two layers of thickness HI and 112 in wbich the occurs normal to the plane of stratification (Fig. 8.16).

now

I'] '" 1T

i. Loyer 0)

to

Loyer III

I.

T

h

1

Fig. 8.16. Fl()'N oonnal to plane of stratification.

let (k..)1 and (k..h be the ooeffic.ient of permeability of the layers 1 and 2 in the direction perpendiruJar to the plane of stratification, and Ie., be the average coefficient of permeability of the entire deposit in that direction. In this case, the discharge per unit width is the same for each layer and is equal to the discharge in the entire deposit. The case is analogous to the resistances in series in an electrical circuit, wherein the current is the same for all resistances. Therefore, •.. (a) Using Darcy's law, considering unit area perpendk:ular to now, ... (b) h" )( ill )( I - (k..)1 )( (i..h )( I - (k"n )( (i..h )( 1 where i .. = overall hydraulic gradient, (i,,)1 = hydraulic gradient in layer I, (i~h = hydraulic gradient in lay~ 12

x

~

... (c)

~

...(d)

(i,), • [(k,)/(k,), (~h

•

I

[(k,)/(k,), I x

From Eq. (b), and

As the 100ai loss of head (h) over the enlire deposit is equal to the sum of the loss of beads in the individual layers,

WritiDg in teoos of hydraulic grandient (I) and the distance of flow, remembering h .. i )( L,

i" )( H -

(i~)1 )( HI

+ (i..h

)(

Hl

Us;ng Bqs. (c) and (d), . I" )(

H

(k,)

-

•

(k~)1 x 1,,)(

H I

+

(k,)

.

(k~h x ' .. )(

H 2


'" k,

[(Z;,

+

j-

(Z~

H - H, + H,

k..-~ HI H2 (k,), + (k,h

In general, when there are n such layers,

.t.. ..

HI + H2 + . .. + HII HI . H2 RIO (k,)' + (k,h + ... + (k,)"

... (8.42)

Evan (1962) proved that for isotropic (A;. .. kll) and homogeneous layers. the average permeability of the entire depooit parallel to the plane of stratification is always greater than that normal to this plane. For illustration, let us consider a deposit oonsisling of two layers of thickness 1 m and 2 m, having the coefficient of permeabJljty of 1 )( 10-2 em/sec and 1 x 10-4 an/sec, respectively. From

):I". """'I

8.41.

2

Ie

1)( 10-

... ..

X

100 + 1 )( 10.... )( 200 100 + 200

.. 0.34 x 10-2 ern/ sec From E
Hi OO + 2'x;OO

Ie" ..

~+~ .. 1.49 )( 10..... em/ sec

k, > Ie.. It may be noted that the average permeability parallel to the plane of stratification depends mainly on the penneability of the most permeable layer and its value is close to the permeability of that layer. On the other hand, the average permeability normal to the plane of stnllulCation is close to that for the most impermeable layer. In other words, the avemge flow parallel to the plane of stratifICation is governed by the most penneable layer and that perpendirular to the plane of stratification-by the lc$t permeable layer. Thus

ILLUSTRATIVE EXAMPLES Dlustratlve Example 8.1. In a oonstant head penneameter test, the following observations were taken. ))istaIIU between piezometer lappings c:: 100 nun Difference of waler levels in pinmne~rs ;:: 60 mm D~ter of thI! tesl sample "" 100 mm Quo.nlity of water colJectt!d = 350 ml [)uraliOf'l of the test = 170 seconds Detl:rmine the coelflC~t of permeability of the soiL

~


k =

In this case.,

q - 3501270 - 1.296 mVsec

" •

1.296 x 10.0 _ 0.0275 aD/ sec. (n/4) x (10)' x 6.0 Dlustratlve Example 8.2. ~ failing·ht!ad permeability test was conducted on a soil samplt! of 4 em diameter and 18 cm len~h. The hMd fell from 1.0 m 10 0.4IJ m in 20 minuta. If 1M cross· s«tiOMl aIU of the stand pipe was 1 em • detl:rmine tilt c~fficit!1U of permeability. 1berefore,


157


k •

~

log" (hi / hi)

_

1.0 x 18.0

Jog. (1.0/0.40)

(It/4) x (4.0)2 x 20 x 60 • 1.09 x 10-3 em/sec. lIlustratiye Example 8.3. A soil has the cOl'jficient of penlleahiUt)' of 4.75 x irrl mm/uc at Jife. Determille iT.I· vallie m 2r'C. 'fake the coefficient of viscosity at lO"e and 27'C as 8.0 milii poise and 8.5 mill; pnise. respecr;vt'I.I:

trt - Ie. ; ;


• 4.75 )( 10-2 )( 8.0/8.5 _ 4.48 )( 10 mm/sec.

'

n1ustratlve Example 8.4. Estimale the value of the coelflCienl of permeability of a soil with an effective ditvneter of 0.2 mm. . Solullon. From Eq. 8.30. k • C~o k • 125 x (0.02)2 • 0,05 ~1t1/sec.

Thking C • 125,

illustrative Example 8.5. The coefficient of J>Crmeability of a soil al a void raw of 0.7 is 4 x 104 em/sec. Estimate its va/~ at a void ratio of 0.50. Solution. From Eq. 8.15, As all the parameters remain constant, except e, k,n (0.70)' (1 + 0.50) k;:;- - (1 + 0.70) x (050)'

4 : _10--4 • 2.421

or

...s

ko., -

1.65 x 10...... em/sec.

Altematlve Method k.1.4ko.a.~e2

From Eq. 8.16,

4 x 10-' _ 1.4

ko.as -

or For e _ 0.50,

ko"

x (0.7)'

5.83 x 10-'

k _ 1.40 )( 5.83 )( 10--4 (0.5)2

• 2.04 )( 10...... em/sec Dlustratlve Example 8.6. A sandy la~r 10 m thick overlies an impervious stratum. The WQter table is in the sandy la~r at a depth of 1.5 m below the ground surface. Water is pumped out from a well at the raze of 100 litres per second and the drawdown of the water table at radial distances of 3.0 m and 25.0 m is 3.0 and 0.50 m, TeSJH!c/ively. De/ermine /he coeffICient of penneability. Solution. From Eq. 8.21 , In this case, Z2" 8.50 - 0.50 - 8.0 m and

Th

'

ere.ore,

k _

Zl"

8.50 - 3.0 .. S.sO m

100 x 10-' 1 (25/3) • [(8)' _ (5.50)') og,

- 0.002 m/sec

- 2 mm/sec.

'"

SQIL MECHAN.ICS AND FOUNDATION ENGINEERING

Illustrative Example 8.7. Dnennille the coejficielll of permellbWly of a confined aquifer 5 m thick which Rilles a .frcelli), tIi.~c/llIrge of 20 /itreslsec through (/ well of 0.3 til radills. The height of water in 'h e well which 1\'(1.\' 10 '" aIJo\'(! the base lJeJ()I1.~ pumping dropped to 8 m. Take the I'Mius of influence as 300 m. k = q /ogr (Rlr) 2xb(D-h)


, =

~~2~ l~~ (~~/O~

= 0.0022 m/"c.

lIIustrntive Example 8.8. De/ermine the average coefficient of pemllmbiliry ill 'he horh,ollral and wmical diret',j(JI1.~ for (I deposit ("(msi~'ril1g of llime layers of thickness 5 m. J til a/1d 2.5 m and having the cm'fficit:III.\' of perml'a{,ility of 3 x /0-1 /11I11/.H~C. 3 x JO-.~ IIIIIi/sec. and 4 x J(r 2 mmhec. respectively. Assume tile layer.\· an: i.Wllrvpic, Solution. From Eq. R.4J. taking /I = 3,

11, + (khh X 112 + (k"h x H~ H, + H2 + H.l 3 X 10- 2 X 5 x 10~ + 3 x IO- s x I X 101 + 4 (S + I + 2.S) X 10'\ 0.15 + 0.00003 + 0.10 0.0294 mm/sec.

_ (1.,,),

I.

!J -

X

X

10-2 x 2.S x 101

8.50

k"

From Eq. 8.42.

=

Z2

+ + H~ --.!.... + ---.!... +---.-2.. Hfl i

(',.),

k"

=

(1/1

aquifl'l' ille/illl'li 01

horiZOn/al. Tltl! dijJl'l'I'IICt! vf

(',J.,

~

2~5

3 x 10-2

3 x IO- s

--';-m

~

levels in

~7 6~b.f,~~)~\~it:~! ;:;~"~:!:e~:,,;;;;;~~:~(l~~.~;:~:

.;',::'~

.', :, : -:" :."."

~' ••. •.::.:.: -::":" . ',' .

depll!

Ihrrmgh Ihe al{III/CI'P.l'l' IlIIil widTh ifk = 0.7 nUll/sec. Thl! of aqmler I/ormal In Ihe

• '. : ' :.; :

direClirJII afpmv is 2.95/ III . Solution, Lenglh of aquifer between two '

= 601co)l

I.

From Dilrcy's law,

= /ilL dis~ha!ge

=

10°

= 60.926

~9026

:,:'.->','::

1

60 m Pig, B-8.9

observation wcllt;

Hydfllulir.: gmdienl

x 10.... mm/s

4 X 10-2

8'9~

100 10 Ihe

1I'(/fl!Y

~ = 2.S

5 + I + 103 ~+~+2.SX 10'

IIIustr-dtive Exumple 8.9. Fig, E show.\·

(',.),

".

-,

m

= 0.082

!;ler unit width.

q

=k iA = 0.7 x

10- 3 x 0.082 x (2.95 x I) = 0.169 X 10- 3 m3/sec

= 0.169

lit/sCi:,

15.


Illustrative Example 8.10. Fig. £-8.10 shows an upt:rimenlal set- up. If 30% of the effective head is lost in the soil A, determine the total head and the piezometric head at points 1 and Z. Determine the coefficient of permeobiliry of the soil B if thai of the soil A is 0.5 mllllsec. Solution.

=

T

0·4m

..L

~

=

Piezometric bead at (1) 0.3 + 0.3 + 0.4 1.0 m Datum head at (1) = - 0.60 m Total head at (1) = 1.0 - 0.6 = 0.4 m Head lost in Soil A = 0.3 )( 0.4 = 0.12 m Thtal bead at (2) = 0.40 - 0.12 = 0.28 m Datum bead at (2) = -0.30 m Piezometric head at (2) = 0.28 - (-0.30) = 0.58 m Loss of head in soil B = 0.7 )( 0.4 = 0.28 m Let kB be the roefficient of penneability of soil B. Since the discharge and area are the same in both the soils,

or

SOIL B

0 ·3m 0·3m

Fig. E-8.10.

kg )( 0.28/0.30 • 0.5 )( 0.1210.30

or

ks - 0.214 mm/sec.

llIustraUve Example 8.11. Determine the discharge per unit width of the slot in Fig. £-8 11 if the drawdown is 2.3 m. 11le coefJicient of permeability of the soil is 1 )( j(T2 mmlsec. Also determine the elevatjon of the water surface at a distMce of 30 m from the centre of the slot.

2·3m

------ ::==:-

T

Solution. Let us coosider flow at a distance of x from the centre of the slot. Using Darcy's law, discharge per unit width _ _.LJ._ _ _----Jl.L_ _ _.LJ.-1._

q _ k

(

dz ) ;& ,(z, I)

t. • ..-I '

_

_

----t160 m - -I Fig. E-8.n .

qdz_kzdz Integrating,

... (0)

Substituting the values given,

q The water surface z at x 01775

,

_ 1 , 10-' , 10--' (13.S' - 11 .2') '/ _ 000177" 1/1/ 2 (160 _ 0) m sec • , ""

= 30 m can be determined using Eq. (a). 'O 's _ 1 )( 10-2

, •

2

)( 10-3 (; - 11.2 2(30-0)

)

SOIL MECHANICS AND FOUNDIUION ENGINEERING

160

;. - 125.44 .. 2 x 30 x 0.1775 .. 10.65

or

or z .. 11.666 m. Illustrative Example 8.12. A capillarity--permeabilily test was conducted in two stages under a head of 50 em and 200 em al th~ end of entry of water. In the first stage, the wetted surface advanced from its initial position of 2 em to 8 em in 6 minutes. In the second stage it advanced from 8 em to 20 em in 20 minutes. If the, degree of saturation at the end of the test was found to be 90% and the porosity was 30%, determine the,

capillarity head and the coefficient of permeability. Solullon. From Eq. 8.37.

~ - >1 tz -

21<. (h, + h,) 'I .. ----s;;-;;-

~

6)(60

..

2I<.(50+h,) 0.9)(0.30

k. (SO + h,) - 0.0225 From

----s;;n--

(20)' _ (8)' 2k. (200 + h,) 20)(60" 0.9)(0.3 k,. (200 + h,) - 0.0378

or

From Eqs. (1)

... (1)

~ - xi 2!,,(h1 + he) I, _ I, -

Eq. 8.38.

.. .(2)

and (2).

200 + he .. 0.0378 .. 1 6ft 50 + he O.022S .

From

or

Eq. (1).

he .. 170.59 em k,. (50 + 170.59) - 0.0225

t .... 1.01 )( 10.... cm/sec. PROBLEMS

A Numerkals 8.1. (0) A CODStmt·head permeability test was run on a sand sample 30 em in length and 20 cml in area. When a loss of bead was 60 em, the quantity of waler ooIlecled in 2 minutcs was 250 mi . Dclennine Ihe mefficient ~ of permeability of the soil. (b) If the specific gravity of grains was 2.65, and dry mass of the sample, 1.1 kg, find the void ratio of the sample. [Ans. 0.052 an/Sec; 0.445]

8.1.

torr:~~~~:~n:~~IZnt:t f:~~::n~: :m~m~l~ ~ :a~r=~_:i:i. ~~ was 4 an in diameter and 30 em long. calculate the OJefficient of permeability of the sand.

[Ans. 0.0275 an/sec] of dense sand 10 III deep overlying iln impervious stratum. Observation holes were drilled al 15 m and 6.75 m from the well. Initially. the waler 1~1 in the well was 2.50 m below the ground surfooe. After pumping until steady conditions had been achieved. the waler level. In the cbservalion wells had dropped 1.95 m and 050m, respeaively. If the steady dischnrge was 5 litreslsec, determine the a:.efficienl of permeability. [Au. 0.698 x 10-1 cmIscc] 8.4. A penneameter of diameter 82.5 mm contains a column of fine sand 460 mm long. When water flows through il under a oonSlant head 81 a rate of 191 ml/minute, the loss of heod berween two points 250 mm apart is 3m nun, calculate tho coefficient of permeability. If falling head leSI is made on the same sample using 8 stand pipe of diameter 30 mm, in what time will the water level in stand plpe fall from 1560 mm 10 1066 mm above outflow level. {Aas. 3.92 x 10- 1 mmlsec; 59.1 sec]

8.3. During a pumping test, 8 well was sunk through 8 stratum

PERMEAB ILITY OF SOILS

161

8.5. Calculate Ihe coeflklent of pemleability of a soil sOlmp[e 8 em in height and cross-sectional area 60 cm 2. It is observed thnt in [2 minutes. 600 ml of water passed down under an effective constant hc~d of 50 em. On oven drying, Ihe test specimen weighs 750 gm. Taking 2.70 as speeific gravit~ of soil, calculate the seep:lge velocity of water during the test. lAos. 2.22 x 10-' em/sec; 0.33 emlsec.] 8.6. Fig. P-8.6 shows :J. eros.q·se<:tion through the simla underlying a site. Calculate the equivalent permeability of the layered system in the venical and horizontal din'Clioll. .. Assume thaI ench layer i~ isotropic. [Ans. 1.41 x 10-6 cm/sec: 0.081 emlsec1

Fig. P-S.6.

8.7. A glucial cl;lY deposit eontnins a series of sill partings in il at un average venical spacing of 2 m. If the silt layel'll are about 5 mm in thiekne.qs and have a permeability of one hundred limes thlll of the clay. determine Ihe ralio of the- horizontal and vertical penlle.1bi litics. [Aos. 1.244] ,8.8. In l\ flllling-head permeameler ir Ihe time intervals for drop in levels from II( to "2 and 1z2 to 11:1 are equal. prove thai

8.9. If the eITcrlivc gmin sile of the soil is 0.3 mm, estimate the cocfficielll of permeability. Take Hazen's C = 10. [Ans. 0.9 mm/sec[ 8.10. A soil ha~ a eodlicient of pcrme.1bilily of 0.5 x 10-4 emlsce at 20°C. Determine its vulue when the temperature rises 10 35°C. (~11O" '" 10.09 x 10-~ paiM: and ~IW'" 7.21 x 10-3 poise). [Ans. 0.7 x 10-4 emlsecJ 8.11. A dminage pipe beneath :I dam h;\s m..-come clogged with sand whose cocflicient of permeability is 10 m/day. It

~=tr~~~ ~~C;~n~~.:~s ~~wmt.h~':~l :~SSP~~lii~n~i I~~~~~~~ew;~ ~~es ~~e~,;~~ ~~a:~~~~ll:~l~~np:~ was filled with sand?

[Ans. 26.67 mJ

8.12. A soi l has the coefficient of permcnbility of 0.4 x 10-4 em/sec 1lt :I void ralio of 0.65 llOd a temper,lIu re of 30"C. Detemline the coefficient of permeability al Ihe same void ralio and a temperatu re of 20°C. At 20G C. p,..= 0 .998 glll/mi and ~ = 0.010 1 l>Oisc lind al 30~C. p .. '" 0.996 gmlml and ~ '" 0.008 poise. What would be the eoeflident ot' penneability al a void ratio of 0.75 and a temperature of 20°C? (Ami. 0.317 x 10-4 emlsee; 0.422 x 10-4 cmlsee]

B. Descriptive and Objedi\'e Iype 8.13. Whal is Darcy's law'! What arc its limil,ltions '! 8.14. WIMt afC differcnt methods for determination of the coellident of permeabilily in a laborutory ? Discuss their limitations. 8.15. Describe pumping-om methods for the determmation of tlte coefficient of pemlenbility m the field. What nre their adnntagcs and disadvantages? What arc Dupuit's assumptions " 8.16. Discuss open-end and packers methods for the determination of the coefficient of permeability. Compare Ihese methods with the pumping.out methods. 8.17. What is Alkn Ha;r.en·s funnula for th\'! eocflicient uf permeabilily '! What is ils usc'? Compare this with Kozcny-Carman eqUl.ltiun and Loudun's formu la. 8.18, 'De~ribe in brid the capillarity-pcrme,lbility t ~t ? Why the values o f Ihe cOfmciem of pcnneability obtained from this te~t diller frollllhose obUlined rrom other tests?


162

8.19. How would you (\ctermine the average permeability of a soi l deposit consisting of a number of layers ? What is its use in soil enboinecring? 8.Z0. Write whether the following statements are true or fnls<:. (a) The coellicienl of pcrnlcability of II soil increa$Cs with an increase in temperature. lb) The soils with [\ higher void ralio have alw3Ys greater pt!mll~ability than soils with a smaller void ratio. (el The coctlic.:icnl of pcnncability decreases with un increase in the specific surfncc. (d) For a given soil, the coefficient of permeability incrctlscs with an increase in void mtio. tel For a soil deposit co nsisting of isotropic layers, the cocftident of permeability parallel \0 the plane of st[;).lificalion is always greater than that normal 10 [his plane. if> The variable-head permeability tcst is used for fine- grain¢.! soils_ (8) The line joining the piezometric.: surra(:"cs i~ also known us the hydraulic grac.lient line.

IAns. True

C.

(a). (e), (tI). (to).

00, (g»)

Multiple-Choice Questions I. The pcrmellbility of sOil varies (a) inversely as square of grain size (b) liS SqUllrc of grain sizt/: (e) as grain size (tl) invt/:rsely as void ratio. 2. The maximum particle size for which Darcy's IllW is applicnble is (a) 0.2 mm (b) 0.5 mm (e) 1.0 mm (J) 2.0 mm 3. According to U.S.B.R .. n soil with n coemdent of pcrmeubiHty of 10-4 mmlsec will be classified as (a) Pervious (b) ImperviOUS le) Semi-pervious (e) Highly pccvious 4. The coefficient of permeability of clay is generally. (a) Between 10-1 lind 10-1 mmls (b) Between IO-~ and 10-4 millis (e) Between 10-:'1 and 10-11 mmls (JJ Less then ro-ll mm/s 5. A constant-head permeamcter is used for (a) Conrse-grained soils (b) Silty soils (e) Clayey soils {d)Organic soils 6 , The coemcient or permeability of a soil (a) increa.~es with a increase in temperature. (b) increases with II decrca.~e in temperature. (e) incrcase~ with II dt.'Crea.~e in unit weight of water. (tI) decreases with an increase in void rJtio.

1. A soil has a discharge velocity of 6 x 10-1 mls and a void r.llio of 0.50. Its seepage velocity is (a) 18 x 10-1 mls (h) 12 x 10-7 mls (C') 24 x 10-1 m/s (tl) 36 x 10-7 IIlls 8. In a pumping.out lest. tlte druwdown i.~ 5m. If the coefficient of permeability of the soil is IO-lmls, the radius of inlluence will be about (a) 250 m

(b) 300 m (rl) 200 m 9. For II sphere of 0.5 111111 diameter. the specific surface is I (a) 12 mm(b) 6 mm- t (c) 8 mm- I (rl) 9 mm- t (e) 150 m

~_I~~~~~~~~~~~~~~~a~

9 Seepage Analysis I. INTRODVcnON Seepage is the flow of water under gravitational forces in a pcnneablc medium. Flow of waler lakes place from a point of high head to a point of low head. The flow is generally hlminnr. , The path taken by a water particle is represented by a flow line. Although an infinite number of now lines can be drawn, for convenience, only a few arc drawn. At certain points on different flow lines, the total head will be the same. '111e lines connecting points of equal total head can be drawn. These lines arc known as equipotential lines. As flow always takes place along the steepest hydraulic gradient, the equipotential lines cross flow lines at right angles. TIle flow Unes and equipotential lines together form a flow net. The flow net gives a pictorial rcpresentalion of Ihe path taken by water particles and the head variation along Ihat path. Fig. 9.1 (a) shows a glass cylinder containing a soil sample of length L. A steady now occurs vertically downward through the soil sample under a head of II. The elevation head, the pressure head and the total head

(0)

Point

Elevation (he)

"eo:!

Pres~ure

Totol heod (H)

head(t-p)

Equipotentla t

Flow line~

0·151'1

I) O·5L

O.5L"'H,-O.5h ,,0.5"-O.5l

0 ·251'1

L+Ht-O.51"1 :0·51'1 Flow nel

(0)

Fig. 9.1. Vertic.11 flow through .$(Iii.

li n e~

SOIL MECHANICS AND FOUNDA:nON ENGINEERING

164

at points. A, Band C can be worked oul as shown in Fig. 9.1 (b) and 9.1 (e). The point B is at a height of 0.5 L above the datum. As the rate of loss of head is linear, the loss of he.'ld upto point B is hfl. Therefore, the total head at point B is IIfl. Fig. 9.1 (d) shO\Vs 0 simple flow net, in which five flow lines and an equal number of equipotcntinllincs are drawn. TIle equipotential lines are horizontal and the now lines arc vertical in this case. If a dye is inserted al a few points on the top of the soil sample, the paths taken by the dye represent the flow lines. 11lc flow nets in aclua! soil engineering problems are not as simple as shown in the

figure. In Ihis chapter, the methods for construction of flow nct and their uses arc discussed. 1be forces associated with seepage and their effect on the stresses are dealt in the following chapter.

9.2. l:APlACE'S EQUATION The simple method of construction of flow net as explained above cannot be used for soil engineering problems in which the flow is generally two-dimensional The Laplace equarion is used in the construction of the flow net in such cases. The follOWing assumptions arc made in the derivation of the Laplace equation: (1) The flow is two-dimensional. (2) Water and soil are incompressible. (3) Soil is isotropiC arfd homogeneous. (4) The soil is fully saturated. (5) The flow is steady, Le., flow conditions do not change with time. (6) Darcy's law is valid. Let us consider an element of soil of size dx by dz through which Dow is taking place (Fig. 9.2). The third dimension along y-a:ds is large. For convenience, it is taken as unity. Let the velocity at the inlet and outlet faces be v" and ( v.. +

~: . dx)

in x-direction and Vz and ( ".. +

"D -...

----. Yx +

~ . dz)

in z-direction.

~V, 1

~

d_

1

"" Fig. 9.2. Two-dImensional Row.

As the flow is steady and the soil is incompressible, the discharge entering tbe element is equal to thal leaving the element. Thus

v"dz + Vz dr

-

(~ + ~) dxdz •

(v" 0

+

~ . aX) dz + (Vz + ~

.

tit) dx

SEEPAGE ANALYSIS

165

(~ +

or

~)

.. 0

...(9.1)

Eq. 9.1 is the continuity equation for two-dimensional Dow. Let h be the total head at any point. Tbe horizontal and vertical componenlS of the hydraulic gradient are, respectively, i.o; ..

-~,

and

~

..

-~

The minus indicates that the head decreases in tbe direction of flow.

V.o; .. -k:.: ~,

From Darcy's law,

Substituting in Eq. 9.1, or

_~ if h

ac?-

k iPh

if h 'ac

_ k

/,. [Ph

.o;axZ+""a;--

As the soil is isotropic.,

kx .. k:.. {Ph

Vz ..

-kz ~

.. 0

0

Therefore,

a2 h

... (9.2)

ac?-+ai'-O Eq. 9.2 is the Laplace equation in terms of head h. ~

Sometimes. the Laplace equation is represented in termS of velocity potential

__ kh

~-

Therefore, and

*-

v, z

V

Substituting the values of

V;r

-

t, given by

-k~ -k

~

and v, in Eq. 9.1,

t.t t.t 0 ar'+ai'-

... (9.3)

Eq. 9.3 is the Laplace equation in terms of velocity potential. Laplace's equation can be solved if the boundary conditions at the inlet and exit are known. The equation represents two families of curves which are orthogonal to each otber. One family represents the flow lines along which the flow takes place. The other family represents the equipotenliallincs along which the potential @ or total head (h) is constant. The graphical representation of the laplace equation'> is, therefore, a flow

,ct.

9J. STIlEAM AND POTENTIAL FUNCTIONS Stream function N) is a scalar function of me coordinates 1; Z such that its partial derivatives satisfy the follOWing equations:

..[9.4(a)] ... [9.4(b)]

and As a stream function is a continuous function, its total differential is given by

ihI>-!':'I>.·u+!':'I>.·tt. ax az


166

SubSlituting the values of

~ and ~

from Eq. 9.4,

dtp .. -

v~

dx + v... dz.

If the stream function is constant along a curve, dip .. O. TIlerefore, -

II:

dr +

V.o:

dz. .. 0

(1!) _"-

or

dx...

...(9.5)

V.o:

111e tangent nt any point on the 'P-curve gives the directions of the rcsultnnt velocity (v) (Fig. 9.3). Hence, the 'P-curve repreSents the now line. The curves with constaO! values 'PL, 'P2 .... 'PII are the flow lines. Velocity potential ($) is a scainr function of x and z such that it derivatives sntisfy the following equations (Refer Sect. 9.2). /

*. ~

and

- v, -

-k* . .

[96(o)J

v: .. - k

y£ ... [9.6(b)J

Integrating, Eqs. (a) and (b), $ (X, z) .. - kh (x, z) + fez)
,oj

~:.,

.~~~ lb)

Pig. 9.3. flow CUNes.

Sil)Ce x and z can be varied independently,

f(x) .. g (x) - constant, say C Therefore, ¢I (x, x) .. - klJ(x. x) + C If the total head h is taken as a constant, it represents a curve for which 4' has a conslant value. This is an equipotential line. Assigning different values to h such as hi ,h2 • ... hll • we get a number of equipotential lines ¢II, 4>2, ,.. $". The slope along an equipotential line cp can be determined as under. The total differential is given by

d'''~'dx+~.dz If $ is a constant along a curve,

dcp .. 0

O-~'dx+~dz

Hence,

(dz) dX ... - ~

or

v, acp/oz .. - ~

From Eqs. (9.5) and (9.7),

(1!) (1!) __ "x

dx ...

dr.

V.o:

x "- --1 v,;-

Thus, the stream function and the potential function are orthogonal to each other. From Eqs. (9.4) and (9.6),

~ . ~ Ox .. az

or

n~

and

-~-~ ., ax

ariJz" iPz

... (9.7)

161

SEEPAGE ANALYSIS

_i'.!.. ~ ilxiJz

Therefore,

0:2

~ + ~ __ .i.t +

ar-

;v?

axaz

l i _0 araz

Thus., the stream function ("') also satisfies the Laplace equation. Determination of Discharge The discharge 6q between two adjacent flow lines 'tjI and (til + .6.",) can be determined as follows [Fig. 9.3 (a)]. The discharge is equal to the resultant velocity v multiplied by the nonnal distance (An) between 'tjI and ('I' + d\j». Obviously. discharge _ - v~ de + v", dz Therefore,

6.q -

J",:t+

A

'"

(- v" dr + v",dz)

Substituting the value of v", and v", from Eq. 9.4,

Aq •

f..... (£'ax!>. . d>: + £'!>.) ar . dz • f.•••• d\j> •

A 'I'

... (9.8)

In other words, the flow between two adjacent flOW" lines is ronstant and is equal to the difference of stream functions of the two lines.

9.4. CHARACfEIUSTICS OF FLOW NET Fig. 9.4 shows a flow field fanned between two adjacent flow lines and equipotential lincs. If velocity along the stream line represented by 'tjI,

V6

is the

Fig. 9.4. Flow field.

v",_v,cosa Vz

. ..[9.9(a)] ...[9.9(b)]

- -v,sina

The potential function !p can be written as

~.~~+~ . ~ Using Eqs. 9.6,

~

_ v", cos a + v", (-sin a)

Substituting the values of v", and v" from Eqs. 9.9,

a,

2

. 2

as-v.cos a+v,sm a-v6

... (9.10)


168 Likewise,

~

.. -Vz sinn + v'" coso.

~_v.. From Eqs.. 9.10 and 9.11.

Sin2a+vICXJS2a_ v,

~

~

~

~

... (9. 11 )

as - an

... (9.12)

as .. !J.n

The flow nct must satisfy Eq. 9.12. It is convenient to construd the flow net such that the change in stream fundion (A,\,) between two adjacent flow lines and the change in potential function (IJ.¢I) between two adjacent equipotential lines nrc constant. Therefore

6$/!J. ~J

..

con.'.tant

From Eq. 9.12,

!J. s /11 n .. constant Allhough any fixed ratio of flsll:ln can be used, for convenience, c.sl l::.n is kept unity. 1ltercfore, in actual pl1lctke, the [low nct consists of approximate squares, Os _ On ... (9.13) Thus the distance between two adjacent flow lines is equal to the distance between two adjacent equipotential lines. lbc ch..1rnctcristics of now nct can be summarised as under: (1) The fundamental condition that is 10 be satisfied is that every intersection between a now line and an equipotential line should be at right angles. (2) The second oondition to be satisfied is that the discharge (hLJ) between any two adjacent now lines is constant and the drop of head (M) between the two adjacent equipotential lines is constanl. (3) "be rnlio of the length and width of each field (lls/ tJ. n) is conslnnl. The ratio is gcnernlly taken as unity for convenience. In other words, the flow net consists of approximate squares. The flow net can be obtained by anyone of the following methods. as discussed in the following

sections. (1) Graphical method, (3) Soil Models, (5) Solution of L1place's equation.

(2) Electrical analogy method. (4) Plastic models,

It will be assumed that the flow is two-dimensional. In many of soil engineering problems, such as flow through a long carth dam, seepage under a long sheet pile and seepage below long gravity dams, Ihe flow is actually two-dimensional. In all such cases, vertical sections at different points along the length are identical. The velocity has components only in two orthogonal directions (x, z), the component in the third direction (y-dircction) is zero. However. if the length of the soil mass in the third direction (y-dircction) is small, the end effects are important and the flow is not truly two-dimensional and Laplace's equation. as derived above, docs not apply.

9.5. GUAPIllCAL METHOD The graphical method of flow net construction is the most commonly used method. lbe hydraulic boundary conditions which define the limiting flow lines and equipotential lines should be first identified and established. A reasonably gcxxl now net can be drawn by the gmphical method even by a novice with some practice. However. for getting a good flow net. a lot of prnctice and patience is required. Fortunately, the accuracy of the oomputation of hydraulic quantities, such as disch:Jrge and pore water pressure, docs not

SEEPAGE ANALYSIS

'69

depend much on the exaciness of the now net. A reasonably good estimate of hydraulic quantities can be made even from a rough flow net. 'fl1e following points should be kept in mind while sketching the now net. (1) Too many flow channels distrad the attention from the essential features. Nonnally, three to five flow d:13nnels are sufficient. (The space between two flow lines is called a flow channel). (2) The appearance of the entire flow net should be watched and not th:lt of a part of it. Small details can be adjusted after tbe entire Dow net bas been roughly drawn. (3) The curves should be roughly elliptical or parabolic in shape. (4) All transitions should be smooth. (5) The flow lines and equipotential lines should be orthogonal and form approximate squares. (6) The size of the square in a flow channel should change gradually from the upstream to Ihe downstream. The procedure for drawing the flow net can be divided into (he following steps: (1) First identify the hydraulic boundary conditions. In Fig. 9.5, the upstream bed lcvel GDAK represents 100% potential line and the downstream bed level CFJ, 0% potential linc. The first flow line KLM hugs the

fig. 9.5. Flow Net.

hydraulic structure and is formed by the flow of water on the upstream of the sheet pile. the downstream of lhe sbeet pile and at the interface of the base of Ihe dam and the soil surface. "lbe last - now line is indicated by the impervious stratum NP. (2) Draw a trial flow line ABC adjacent to the boundary line. The line must be at right angles to Ihe upstream and downstream beds. The location of the first trial line is determined from experience. An experienced person will make a good estimate of the first trial line and subsequent work would be reduced. (3) Starting from the upstream end, divide the first flow channel inlo approximate squares by equipotential lines. The size of the square should change gradually. Some of the squares may, however, be quite irregular. Such squares are called singular squares. (4) Extend downward the equipotential lines forming the sides of the squares. These extensions point out approximate width of the squares, such as squares marked (1) and (2). Other sides of the squares are set equal to the widths as determined above. Irregularities are smoothened Qui, and tile next flow line DF is drawn joining these bases. While sketching the flow line, care should be taken to make flow fields as approximate squares throughout. (5) The equipotenlial lines are further extended downward, and one more now line GlD is drawn, repeating the step (4). (6) If the flow fields in the last now channel are inconsistent with the actual boundary conditions, the whole procedure is repeated after taking a new trial now line.

SOIL MECHANICS AND FOUNDAll0N ENGINEERiNG

170

II is nOl necessary tbat the last flow channel should make oomplete squares. The flow fields in the last channel may be approximate rectangles with the same length to width ratio. In this ca<>e, the number of flow channels would not be full integer. In facl, the flow channels will be an integer only by chance.

9.6. ELECTRICAL ANALOGY METHOD According to Darcy's law, the discharge in a soil mass is proportional 10 the hydraulic head (h). According to Ohm's law, the current in an electrical conductor is proportional to the voltage (E). An analogy

exists between the two types of now. The analogous quantities in the two systems are given in Table 9.1. Thble 9.1 Analogous Quantities S.Na

Flow of water

kfA

1-

....w : q -

2. 3.

Disclurge,q Hend,h Length, L Nen,A

4.

5. 6.

Permeability. k

Flow o/CurrenJ Law:I_K'

~

' ,4

Current, 1 VoImge,E Length. L A1ea,A Conductivity, K

An electrical model is , made whose boundary conditions are similar to those of the soil modeL 1be equipotential lines are drawn by joining the points of equal voltage. The now pauem obtained from the elcctrical model are used in the construction of Oow net in the model. The following three types of electrical. analogy modelS are used. (I) Electrical Analogy Tray. A shallow tray, with a flat bottom , made of an insulating material is taken. The tray is filled with water. A small quantity of salt or hydrochloric acid or copper sulphate solution is added to water to make it a good conductor of electricity. The hyd~ulic boundaries are simulated on the tray. For the flow below a sheet pile shown in Fig. 9.6 (a), the boundary flow lines :)rt ABC and FG. An insulating material, such as ebonite or pcrspcx, is used to simulate the boundary flow lines. The insulating material is fixed to the tray by means of some non-conducting adhesive, such as plasticene or bee wax. The boundary equipotential lines DA and CE are simulated by some good conductor of electricity such as copper bars. For obtaining the flow pauern, an electrical potential difference of 20 V is applied to the two electrodes DA and CEo A VOltage dividing variable resistor, known as potential divider, is connected in parallel to the alternating current source to vary the voltage in the range of 0 to 20 V. A galvanometer (or any other null indicator) and a probe are connected to the variable potential ann [Fig. 9.6 (b)]. The position of the equipolentiallines is determined by locating the points of oonstant potential (VOltage). To trace the equipotential line corresponding to a given percentage of total potential (say }O%), the VOltage divider is set at that potential (2V). 1be 'Probe is moved in the tray till the galvanometer shows no ament flow. That position of the probe gives tbe point corresponding to 2V potential. By moving the probe, other points corresponding to that potential are obtained. A graph sheet is generally placed below the transparent plate to detennine the roordinates of the poinlS. A line joining all these points gives the equipotential line corresponding to 10% of the total head. likewise, the c:;quipotential line oorrespooding to 20% of the total head is3lbtained by changing the selling on the voltage, divider to 4V and repeating the procedure. Other equipotential lines can be drawn in the same manner. After the equipotential Ii"es have been draWl), flqw lines can be sketched manually. The flow lines should be orthogonal to the Cfluipotential lines and must. satisfy the actual hydraulic boundary conditions. Alternatively, the flow lines can be drown electrically by interchanging the boundaries. The copper strips are used for impenneable boundaries ABC and FG and insulating strips for VA and CEo The VOltage difference

SllEPAGE ANALYsrS

171

is applied acrlliS the new positions of

copper strips. The new equipotential lines, which are actually flow lines, are traced by locating the points with the help of probe. (2) Conducting Paper Method. A conducting paper is made by introducing graphite during its manufacture. One side of the graphite paper is coated with a non-conducting material and the other side wilh Ii positive aluminum coating. The paper is CUI to the shape of the hydraulic structure for which the flow net is F G 7 / / ) / / / / ; ; ?/Ta( ) ) J / ) ) ))??? required. The boundary equipotential lines, such as DA and CE in Fig. 9.6 (a), are given a coating of silver paint. When the paint has dried, the connecting wires are spaced out along GAlVA,NOMET6I the boundary strips in individual strands and are stapled in position. Direct current (D.C.) supply can be used as there are no polarization TRAYFIUEO WlrH WATER effects. A 2- V accumulator is used for fceding the circuit. The lines of equal potential arc traced, as in the electrical analogy tray. POTENTIALOIYIOER The conducting paper method is quicker and more convenient than the tray method. However, the accuracy is (b) low. As the transverse resistance of the paper is generally greater than the Fig. 9.6. Elcdric.,l Analogy Tray. longitudinal resistance, it causes error. The scales of the model in the longitudinal and transverse directions are sometimes kept different to account for difference in resistances. This makes the method more complex . (3) Potential Analyser Method. A potential analyser is made in the form of a mesh of resistances (usually. of 100 ohms). separated at each node by pins of negligible resistance. The mesh is CUI to the required shape. It is well insulated against temperature and humidity. A direct current with a VOltage difference of 1 V is applied to the appropriate boundaries o( the modeL The potential at any nodal point can be read with a high degree o( accuracy. '£be equipotential lines are then drawn through the points of equal potentiaL lbe method is quile convenient and gives fairly accurate results.

~:

•<

9.7. SOIL MODELS Row nel can be obtained from a small scale soil model of the hydra~lic structure. The soil model is placed between two transparent plates, about 100 rom apart. Fig 9.7 shows a soil model of an earth dam, with a horizontal filter at its toe. The flow lines are tmced direcUy by introducing a dye at suitable points on the upstream face of the dam. The equipotential lines can be drawn by connecting the points with the same piezometric levels. For this purpose, tiny piezometers are inslalled in the model at suitable points (not shown in figure). However, it is

SOIL MECHANICS AND FOUNDATION ENGINErrRING

more convenient to draw equipotential lines manually after tbe flow lines have been drawn. TIle accuracy of (he now net obtained from soil models is not good because of scale effects and capillary efIeas. Sometimes, viscous fluids are used in place of water to reduce

capillary effects. The main use of soil models is to demonstrate the fundamentals of flow nct and seepage in a laboratory. In practical problems. their use is rather limited, because of the time and effort required in the construction of these models.

Fig. 9.7. Soll model.

9.8. PLASTIC MODELS

'rt

A seepage flume of width of a few centimeters is used in thi... mcthoo. A model made of plastic is fastened to one side wall of Ihe flume, leaving II small space of 2.5 mm or less between ,Ihe model and the

olher side wall (Fig. 9.8).

SIDE Fig. 9.8. Plastic model.

GLYCERINE

')

PLASTIC

"

MODEL

VIEW

A highly viscous fluid. such as glycerine, is made to seep through the small space between the model and the side wall. The flow is laminar. As the fluid flows, it gives an accurate representation of seepage through soil. The flow lines can be observed directly by injecting II dye at suitable points. Plastic models can be constructed more quickly than soil models. The flow lines in such models are also better defined. Consequently, the flow net obtained is more acaJrate than that obtained from soil models. Different penneabilitics of the soil can be accounted for by varying the space between the model and the wall. Anisotropic soils can be represented by a zig-zag face.

9.9. FLOW NET BY SOLUTION OF LAPLACE'S EQUATION Laplace's equatiqn can be solved by numerical techniques, such as finite difference method. Relaxation method is generally used to find the potentials at various points. Once the potentials have been determined at different oodal points, the equipotential lines are drawn by joining the points of equal potentials. Potentials can be obtained very quickly if a high-speed digital computer is available. The Laplace equation (Eq. 93) can be written in fmite difference form, as ~+.+~+~-~.O

'10

.. ~~

where b ch and , .. are the potentials at the four adjoining points around the central point 0 with the potenlial " (Ag. 9.9). The aos,s-section of the earth structure, for which the flow net is required, is covered with a square grid with a number of nodes. The values of the potential (,) at various nodal points 2rc assumed, satisfying the

SEEPAGE ANALYSIS

173

hydraulic boundary conditions. As the assumed values are not correct. there would be a residual Ro at point 0, given by the equation, ... (9.15) " + ~ + +.J + ,. -4$0 • Ro Each node is oonsidered as a central node in tum and the residual determined. The ooject of the rela'tatioo method is to reduce these rcsidu.'lls !o uro. It must be borne in mind that the potentials at different nodes are inter-related and any change in potential at one node has an effect on the residuals at the adjacent nodes. The process is, therefore, quile tedius and timeconsuming. Howevcr, special relaxation techniques have been devised to reduce the effort. The final com~ct value of $ give the true picture of the variation of potential. The equipotential lines are drawn through the points o( equal potentials. 'Ibc flow lines are then drawn orthogonal to equipotcntial lincs. 9.10. FLOW NET IN EARTU DAMS

r--- --- 2 ------,

I

i

2

1

I

:I

1I

3

., 1

!

i

I

I

I L. ______ .~ _____ J

I

•

GRID AROUMl ·0· Ag. 9.9. Fillito Diffcl'<:llocGrid

wnn A HORIZONTAL FILTER

"ll1e methods of drawing a flow net discussed in the preceding sections are used when the boundary flow lines and equipotential lines are given. Seepage through an earth dam is a case of unconfined seepage in which the upper boundary of flow net is not known. In such cases, it becomes necessary to first locate the upper boundary before a now net can be drawn. Let US consider the case of a homogeneous eanh dam on an imperviFILTER ous foundation and having a hOrizontal filter at the downstream end (Fig. 9.10). The horizontal filter starts at point C. Fig. 9.10. EArth Dam with a horizontal filter The impermeable boundary CD is a flow line wh:ich forms Ihe,lower boundary of the flow oct. The upstream face AD is an equipotential line as the total head at every point on this face is equal to h. The discharge face cn is the equipotential line of zero potential. Thus, Ihn:c hyclnlulLc houndarv c(>nditicms :Ire known. The fourth boundary of the flow net is the lap flow line AB, which is not known in A the beginning. Below the line AB, the soil is saturated and the pressure every where on the AB is atmospheric. The line AB is known a<> phreatic line or seepage line. As the pressure PERVI().J5

~:~ isisU~u: t~: P::ati~I!~i~he ht~~~ +.Difnm~IM:ip~ERv~'OUr;srn7Tn\l""~5i::...~*·~~;'~;;" 'TE: ~=n~;w~:~~e

~~a~te;:::

;::nlS of soocessive equipotential lines and the phreatic line. Once the phreatic line has been located, the flow nct can be drawn by the usual methods.

-=1-(t -~Z' KOZENV's BASIC FMABOLA Ag. 9.11. Kozcny" Solution.

Kozcny studies the problem using the method of conformal !nmsformation. The boundary conditio[]S fa the now region ABeD are as under (Fig. 9.11).


174

Equipotential line, AD, has 4' _ - "" Equipotential line, BC, has , _ 0 Row line. DC, has 'P - 0 Flow line, AB, has 1p • q

Kozcny's solution represents a family of confocal parnbolas of flow lines and equipotential lines. The equation of Kozcny's basic parabola AD, with C as focus as well as origin, is x _

('Ik. _!q r')

1. 2

...

(9.16)

Kozcny's conditions arc not entirely fulfIlled by any practical earth dam. However, an earth dam with a horizontal drainage approximates the conditions at exit. An inconsistency occurs due to the fact that tbe upstream equipotential tine in an actual earth dam is a plane surface and not a parabola as assumed by Kozeny. OIs3grnnde (1940) recommended ilial the seepage line in actual dams can also be taken as ba<>ic parabola. provided the starting point for the parabola is taken al point E, sucb that AE '" 0.3 AF (Fig. 9.10). The distance AF is the projection of the upstream slope Oil the water surface. lbe coordinates of the phreatic line can be determined using Eg. 9.16. The origin is at C, which is also the focus. Substituting z = 0 in Eq. 9.16, the value of x is given by

xo ..

i (;) - ik

or q - 2kXo

2xo between the

focus and the diredrix is known as focal distance (s). Thus q - b Substituting the value of q from Eq. 9.17 in Eq. 9.16,

The distance

... (9.17)

x-~(¥-tr')-f-t or

i-2xs-? ..

0 ... (9.18) Eg. 9.18 can also be derived directly using the property of the parabola that the distance o( any point P on the parabola (rom the focus is equal to the distance from the directrix. (Fig. 9.12). lbus

FP - PO

~ .. s-x By squaring,

or

Xl + ? .. i i-2rs-?-O

+ x2_2sx

If x is taken positive towards left of F. the above

equation becomes

s' 2xs-r' _

+ 0 ... (9.19) The value of s can be determined using the coordinates of the starting point E (Fig. 9.10). Substituting x .. d and z _ " in Eq. 9.19. s2 + 2ds_h 2 .. 0

PARABOLA~

-2d=~ 2 Taking positive sign, s _ ..; (Jl + h 2 ) -d

... (9.20)

Fig. 9.12. Properties of PlIl
Once the value of $ has been determined, Eq. 9.19 can be used to determine the coordinates of the various points on the phreatic line. For diITerenl value of X, the corresponding z coordinates are computed and ploUed.

SEEPAGE ANALYSIS

175

An entrance correcLion is required for the phreatic line oblained by the above procedure. lllc actual flow line must start at point A and nOl point E. Further, the flow line must be nonnal to the upstream face which is an equipotential line. The entry correction is made by eye judgment as shown in Fig. 9.10. The actual phreatic line is sbown in solid line. Fig. 9.13 shows tbe entry correction when there is pervious grnvel 00 the upstream. The phreatic line in this case is horizontal at the entry, as it cannot rise above for being normal to thc inclined dis face of gravel. Once the phreatic line has been drawn, lhe flow

PHREATe

LINE

Fig. 9.13. Entry Com:ction for an 1.1.1$ face with gmvc:l.

net can be completed using the methods already discussed. Fig. 9.14 shows a typical [Jow net.

15

10

l~"m~~ Fig. 9.14. Flow Net in lin Ellrth dam.

Discharge through the body of the dam To determine the discharge through the body of the earth dam, let us consider the flow passing through the section PQ (Fig. 9.10). From Darey's law, discharge per unit length is given by q'"' kiA

q '"' k· ~ . (z x 1) From Eq. 9.19,

...(a)

s1Y:t

z'"' (2xs+ dz S dX - (2xs + i)Y1.

Therefore, Eq. (a) gives,

q '"' k (2.0 : s2)Y1. (2.rs +i)1'.!

or

q '"' k s

.. (9.21)

Eq. 9.21 is a simple «Iuation which gives approximate discharge through the body of Ihe dam. The discharge can also be obtained from the flow net, as explained Inter (Sett. 9.14). 9.U. SEEPAGE THROUGH EARTH DAM WITH SLOPING DISCllARGE FACE Fig. 9.15 shows an eanh dam without any filter on thc downstream ~ side. The downstrcam facc through which water escapes is inclined to the horizontal. In this case. the phreatic line cuts the downstream face. It i s r t h . . . nonnally not pennilled in earth dams as it may cause the failure of downslream Fig. 9.U. Flow Nel for c:anh dam without filler. slope due 10 sloughing action. The

SOIL MECHANICS AND fOUNI)A11QN ENGINEERING

176

down stream face of the dam acts as the discharge face. Fig. 9.16 (0) shows the downstream face when the phreatic line cuts the downstream fnce. 'Ihe downstream face makes an angle p with the horizontaL 'me angle is measured clockwise from the horizontal. In this case, the phreatic line can be drawn as in the case of the dam with a horizontal filter (Sect. 9.10),

Cd) Fig. 9.16.

laking the point C as the focus and also the origin. 'lbc phreatic line is given the entry correction as before. An additional correction at exit is required in Ihis casc, as the basic parabola goes outside Ihe downstream face, which is impossible. lbe actual seepage line meets the discharge face langcntially for p < 90°, ll1c seepage line has been shown by full line, whereas the theoretical basic parabola is shown by dotted line. In the case of borizontal filter, the angle p is 180<> [Fig. 9.16 (b)]. For a rock toe [Fig. 9.16 (c)J, the angle ~ is greater than 90<>. The phreatic line drops vertically in this case. Casagrande gave the charts for the exit O. correction. The basic parabola is shifted by distance 6.a 10 locale the point where the actual seepage line cuts the discharge face. The value o·3 of All is obtained [rom the value or Aa/(a + 6.a) after the distance (a + Aa) is obtained from the basic parabola. lbe value o[ .+ o· 2 Aa/(a + An) depends upon the angle p, given in Fig. 9.16 (d). The value is also available in the ·1 form of a curve (Fig. 9.17). It is wonh noting that the correction is zero when the angle fl is ISO. That is the reason why exit correction was o·0 30 90 126 156 not applied in the case of horizontal filler. The chart is applicable [or p :t 30<>. /l--

,

'-----....

I

"-

~

"'. 6

'" "'"

~

Fig. 9.17. Casagrnnd's Chart.

Obviously, An .. C (a + ALl) where C is the correction [actor obtained from the chart (Fig. 9.17)

9.12. SEEPAGE THROUGll EARTH DAM WITH DISCHARGE ANGLE LESS TllAN 30° If the angle p is less than 30<> (Fig. 9.1 8). point S at where the seepage line becomes tangential to downstream face can be obtained using Schaffemack's method. It is assumed that part CS of the seepage line is a straight line. A tangent at point S coincides over the length CS with tbe seepage line.

SEEPAGE ANALYSIS

177

Fig. 9.18

jz.;j;

q -

The discharge is given by,

~

But

... (9.22)

.. i .. tanp

z '" distance SP = 0 sin q = k (a sin ~) tan ~

and

Therefore,

~,

where SC '" a

... (9.23)

From Eqs. 9.22 and 9.23, kz

Integrating between x ..

..

0

j ,' or or

~

kosin~tan~

..

zdt - asinptan~dr cos P to x .. d., and between z ..

zdz = o~tanpj

.~,

0

sin P to h,

dx

t(h2 ....~lsin2p) .. asinptanp(d-ocosP)

h2 _d'- sin2

h'';''~ sin~

0' coo

_

p .. 20 ~ (d-o cos p) cos~

a'cos~ _ 2ad _ 2a'oos'~

p_

2 ad +

h2~ P ..

0

sm ~

+ 2d. 0"

V4d'-4(h'COSP/Sin'P)COSP 2cosp ... (9.24)

Once the value of 0 has been detennined from Eq. 9.24, the discharge can be found using Eq. 9.23. 9.13. SEEPAGE THROUGH EARm DAM WITH DISCHARGE ANGLE GREATER mAN 30° BUT LESS THAN 60°.

Eq. 9.24 was obtained on the basis of Dupuit's assumplioo that the hydraulic gradient is equal to dz/dr, Casagrande suggested that the actual hydraulic gradient for discharge angle greater Ihan 3Q°is given by


178

.

dz

I ..

(is

where distance s is measured along the curve. Based on tbis assumption, the discharge expression can be written as

q_k(~)Z

... (9.25)

Referring to Fig. 9.19, z = distance SP .. a sin j3

T h

1

I.

Fi.g 9.19. Earth Dam with dischJirge ~ng!e greater than 30".

~ - sin~

aod

q .. kasin'lj3

Therefore. Eq. (9.25) becomes

Ie

From Eqs. 9.25 and 9.26,

... (9.26)

~z .. kasin2~ zdz. .. aSin'lf3ds

! zdz .. aj

Integrating, or

culnp

i(h 2

-

h2 _ 02

or

02

(sinzj3)ds

(J

a2sin2~) .. asm'lf3 (S-o) sinz f3 ..

2tJS

sin2p _

2az smz 13

Z

h -2aS+ SinZj3.0

+2S:t" 4sl- 4,,2/sin2~ 2

or

a ..

S_Vsl_hz/sinzp

... (9.27)

The approximate length S of the straight line CE can be determined as

Therefore,

S-Vd'+h'

... (9.28)

a _ ~ - ~

... (9.29)

Once tbe value of a bas been determined, the discharge can be obtained from Eq. 9.26. For angle j3 > 60°, the error introduced due to approximation in Eq. 9.28 becomes large and this method is nOI normally used.

9.14. USES OF FLOW NET The flow net can be used for a number of purposes as explained below : (1) Discharge. The space between two adjacent flow lines is cal1ed a flow channel. Let Nt be the number

SEEPAGE ANALYSIS

179

of flow channels. The difference between two adjacent equipotential lines is called ~ quipotcntial drop. l...ct Nd be the number of equipotential drops. In Fig. 9.20, there are 5 flow channels and 10 equipotential drops.

n:l///

/

2/

I

I

31

!

!

/

lI \

"

\

\8\

\\

4! 5 6\ '1 : ~ 777;;;);;;}) 777/;; 777 77l!..17;)) J IIo1PERVlOOS

Fig. 9.20. Uses of Row Net.

Lei US consider the flow through the flow field shown hatched. From Darcy's law, the discharge through the flow field per unit length. dq - k . where and

(* 1

(dn x 1)

.. (a)

llh is equipotential drop in the flow field, /!..s and t:.n are dimensions of the flow field.

Substituting

Ah -

-k

in

Eq. (a),

dq - k·

~.

Nd

("-!!.) As

!fi .(~ )

Total discharge,

q - NI

Thking /!..s/6n = unity,

q - k.h."if;.

In Fig. 9.20,

Q - kxhx1o-0.5kh

6. q - kh.

N,

...(930)

The rotio (NINd ) is a characteristic of the flow net. It is known as shape [actor (p). It is independent of the penneability (k) of the soil. It depends only on the configuration or the shape of the soil mass. It is not necessary that NI and Nd be always full integer. The last flow channel may consist of rectangles, However, in the last flow channel, the \englhtbreadlh (/!..s/6n) ratio should be approximately the same for all flow fields. (2) Thtnl head. The loss of head (Ah) from one equipotential line to the next is hINd' The total head at aoy point (P) can be delennined as under. h, _ h - n x (hiNd) ... (9.31) where n is the number of the equipotential drops upto point P. In Fig. 9.20, n = 8 for point P. Therefore, total head at P is

I'"


hp - h-8 x (h/IO) - 0.2h It may be noted than if piezometers were placed at different points on the same equipotential line, water

would rise in these piezometers

(0

Ihe same elevatioo.

(3) Pressure head. The pressure at any point is equal to the total head minus the elevation head. As

mentioned above, the downstream water level is generally taken as datum. For example. for point P, the pressure head is given by where (hp)p

= pressure head

(hp)p _ hp - ( - (he)p) _ hp + (he)p at P and (he)p = elevation bead at P and hp is the total bead.

... (9.32)

Obviously, the pressure head at P is equal to the height of water colwnn in the piezometers at P, as shown in the figure. (4) Hydraulic gradient. The average value of hydraulic gradient for any flow field is given by i_MiAs ... (9.33) where tJ.s is the length of the flow field and Ah is the loss of head. The hydraulic gradient is generally maximum at the exit near point B where the length !u is a minimum. As the velocity depends upon the hydroulic gradient, it is also maximum at the exit.

9.15. FLOW NET FOn ANIS011~OrlC SOIL The coefficient of penncability of stratified soil deposits parallel to the plane of stratification is generally greater than that nonnal 10 Ihis plane. Such soils are anisotropic in permeability. Let us take the axes x - x and z - z parallel and perpendicular to the plane of stratification, respectively. Therefore kit > Ier From Darcy's law,

and

v" - k"i" -

-k,,~

...(a)

k:i, -

-~~

... (b)

v, -

Substituting the values of v.. and

Vz

in the continuity equation (Eq. 9.1).

_. a'h _ •. a'h

"'ax'

<'al-

0

k,a'h+k,a'h_ O

ax'

... (9.34)

al

As Eq. 9.34 is not Laplace's equation, the principles of flow net cOllStructiOn, as described in the preceding sedions, are nol applicable to anisotropic soils. Eq. 9.34 can however be converted to Laplare's equation by transformation. lei the x coordinate be transformed to the new coordinate XI by the transformation (Fig. 9.21).

x, - x,r,;;k.

... (9.35)

TD"-

'" 1.

c.)

C" Fig. 9.21. Trnnsfonnlllion of Coordinates.

SEEPAGE ANALYSIS

181

Eq. 9.34 can be written

a<;

( ~)t!!+i'h.O k. a2 a? 2

or

a h+i'h.o

a;;

... (9.36)

a?

Eq. 9.36 is the Lnplace equation in X, and z. Therefore, the principles of flow net construction can be used for anisotropic soils after transfocmmion. The cross-section of the soil mass whose flow nel is required is redrawn keeping the z·scale unchanged but reducing the x- scale by the ratio ~. The flow net is constructed for the transformed section by usual methods [Fig. 9.22 (b)]. The flow nCI for the actual section is obtained by transferring back the flow nct to the natural section by increasing the x-scale in the ratio ..ff;7iZ;. Obviously. the flow nct for the natural section docs not have the flow lines and the equipotcntial lines orthogonal to each other [Fig. 9.22 (a)J.

~ (.) NATURAL SECTION

(b) TRANSFORMED

FILTER

SECTfDN

Fig. 9.22. F\ownet ror anisotropic lOils.

The discharge through an anisOtropic soil mass can be obtained from an equation similar to Eq. 930, q • K h· (NINd ) •.• (9.37) where k' is the modified coefficient of permeability

as determined below.

Discharge through a flow channel on the transformed scale per unit width is given by Aq • K (M/Ax,) 6z Discharge through the same flow channel on the natural scale per unit width is given by Aq. k.(M/Ax)6z Since the discharge is the same in both the channels, K (M! Ax,) . 6z • k,' (MI Ax) . 6z

or

K • k, ' (Axl Ax)

Using Eq. 9.35,

K • k.' vr;:;7fJ

or

vr;r;

K • The discharge q is determined using Eq. 937 with a value of It obtained from Eq. 9.38.

...(0)

... (b)

... (9.38)


182

9.16. COEmCIENT OF PERMEABILITY IN AN INCLINED DllllicnON Let k,. and kt be the coefficients of permeability along x and z direction'> respectively and k~ be the coeIficient of permeability in inclined s-direction (Fig. 9.23). By partial difIercnUation.

¥s - ~ . ~ + Using the relations,

and

~

v% -

-kx~. v~ _ -~

VI _

- kl

-~ Now

*. .

_

V% _

~

•

¥Z

cos a

~,

Eq. (a) becomes

-~ . ~ - ~ . ~ VI

i .._'

(0)

and vl

... (b) _

~-cooa

VI

Fig. 9.23. PenneAbility in an indined direction

sin a

and~_Sina

Eq. (b) can be written as

.. .(9.39) ... (9.40)

vr;

Eq. 9.40 is equation of an ellipse with Vf; aod as semi-major and semi-minor axes, respectively. 1be directional variation of permeability am be determined from the ellipse (Fig. 9.24). A line moking an anglc a with x-axis gives the intercept ...nc; as shown in figure. 1bus k, can be found.

Fig. 9.24. Dirtdion:ll

Vllri:\lion

of

permeability.

9.17. FLOW NET IN A NON·HOMOGENEOUS SOIL MASS Sometimes, two different soils are used in a soil mass, thus making it non-hoqlOgencous. The Dow lines and equipotential lines get deflected at the interface. TIle flow net thus gels modified. Let the coefficients of the permeability of the Iwo soils be k1 and ~. We Sh.1U consider scparalcIy the two cases when (1) k] > k,. and (2) *1 < ~. Case 1. kl > k l • Fig. 9.25 (a) shows the case when the soil-l has permeability more than the soil-2. The flow lines get deflected towards the normal aner crossing the interrace. The phenomenon of deflection of the flow lines is somew.hat similar to refraction of light rays from a sparse medium to a dense medium. Lei aJ be the angle which the flow line makes with the normal in soil- l and a2 be the nogle, in soil-2. Let and +z be the two equipotential lines. 1be discharge through the flow channel between the two flow lines in two soils is given by

+J

&/1

= kl (~hl6sl) &II

= k2 • (illll &·v . &12 For continuity of flow across the interface, the discharge through the flow channel remains the same. Therefore. sod

or

&f2

kl •

(AhlAsl )

•

AqJ - Aq2 Ani - kz· (Ahlsv . An2

SEEPAGE ANALYSIS

183

--.......

tl tz J' , : l ,, kd, .f

~,'

(b)

(0)

fig. 9.25. Flownct in

1\

Ron-homogeneous soil.

... (9.4)) •

k l ' (IInI/"'I) - k, ' (lin';"',)

or

kl k, tan al - tan a2

lei

tan al .. ,(9.42) tana2 Eq. 9.41 must be satisfied at the interface by every flow line aossing it. Case (2) kJ < k1. Fig. 9.is (b) shows the case when the flow takes place (rom a soil of low pcnneability to that of high pennc.'lbility. At the interface, the flow line is deflected away from the nonnaL Using a procedure si!Dilar to that for the first case, it can be shown U).'l\

k; -

kl k, lanai· tan~ or As ~ > kl> the angle

kl

k; -

lanaI Ian a2

(same as Sq. 9.42)

az is greater than angle al and the flow

line deflects away from the oormaJ.

How Net for Non-Homogeneous Scdion Fig. 9.26 shows the flow net (or an earth dam consisting of two soils of different penneability. The now net is drawn using the following concepts. (1) The flow net consists of squares in soil-I. (2) The flow lines deflect at the interface, according ot Eq. 9.42.

k, <"'" Fig. 9.26. Non-homogeneous

llCCli~.

son. MECHANICS AND

)84

FOUNDATION ENGINEERING

(3) The now net in soiJs-2 consists of rectangles. The ratio of the sides of the rcaangIe can be determined as under: From Eq. 9.41,

k)

(~::) -

k,

(~~)

!;: - ~ Ani

6$-1,

(~::) .6.n:z

k1

652-1;

or 1n Fig. 9.26, as ~ > k J •

/>.'2

!J. n2 > 1.0

If the ratio of permeability is greater than 10,

now

net in the soil of higher permeability nced not be

drawn. Tbe 1005 of head in the soil of higher permeability is neglected. For example, in Fig. 9.26, if k t > 104 the flow net in soil-l is neglected and it is assumed thnt the now lines in soil-l are horizontal. The flow net will be constructed only (or soil-2, taking the interface as the uj:l>tream face. On the other hand, if kz > 10 *10 the flow net will be drawn only for soil-I. In Ibis latter case, the interface will act as D.

discharge face.

ILLUSTRATIVE EXAMPLES Illustrative Example 9.1. Determine the coordinates of the phreatic line for the earth dam shown in Fig. 9.14, Find the discharge through the earth dmn from the flow net and also analytically. Taire k 4.5 X 1fT' em/sec. Solution. From Eq. 9.20, taking d = 72.5 m and h = 30 m,

=

, _ >I(d'

+ h') - d

-~-72.5-5.96m The coordinates of the phreatic line are determined from Eq. 9.19.

i+7xs-il-o or

(5.96)' + 2x(5.96)-; - 0 35.52 + 11.92x-il - 0

1be I-coordinates are determined for different values of x as under. +72.5 m

30m

From Eq. 9.30,

q _ k h (N,INd ) ... 4.5 )( 10-6 )( 30 )( (4/19) q ... 2.84 )( 10-5 cwnecs/m q - f,s

Analytically, from Eq. 9.21, or q ... (~.5 )( 1O·~)(5.96) _ 2.68 )( 10-5 cumecs/m lIIustmtive Example 9.2. Determine the uplift pressure on the impervious concrete floor 0/ the weir slwwn iIJ Fig. 9.2. Also determine the exit gradient. Solution. How net in construaed as shown in Fig. E 9.2. Each equipotential drop 6h is 7.5/15 = 0.5 ffi, as nJ = 15.

185

SEEPAGE ANALYSIS

Fig. E-9.2

The total head at the two extremities or the floor are 7.0 m and 0.5 m. These are also equal to the pressure heads, as the underface or the floor is al the datum (dis level). Total uplift roree

U -

t

(hl +

~(7.0

flu 1... x area

+ 0.50) x 9.81 x (28.5 x 1)

U - 1048.4 kN The length (As) or the last C10w field ncar toe is 1.0 m. Thererore, exit gradient (I) _ 6.h/lls _ 0.5/1.00 _ 0.50 or

PROBLEMS 9.1. Determine the seepage discharge through the foundlltion of an earth dDm if the flow net has 10 cquipolcnlial drops and 3.5 flow channels. The length of the dam is 300 m Dnd the coefficient of permeability of the soil is 2.5)( 10'" cm/see. The

level

of water

above

the base of

the dam is

12

m on upstream and 4 m on downstream. lAns.66.23 )( 103 rnll yearJ

9.Z. In the experimental set up shown in Fig. P 9.2, now lakes place undcr a constant head through the soils A andD.

Fia. P9.2.

SOIL MECHANICS AND rOUNDA110N ENGINEERING

186 (I) Determine the piezometric head at point C.

(u) If 40% of the excess hydrostatic pressure is lost in (Jawing through soil B, whnt are the hydraulic bead and piezometric head Dt paim D. (iii) 1£ the coefficient of permeability of soil B is 0.05 cm/sec, determine IDe same for soil A. (iv) What is the dischnrge per unil area ? (Ans. 120 em, (li) 24 em, 64 an, (;i/) 0.033 cm/sec (iv) 0.02 m11scc. 9.3. A homogeneous canh dam is provided with a horizoolaJ filter drain 30 m long III ilS loe, as shown in Fig. P 93.

<,)

Determine the fcx::al length.

Fig. P9.3.

Also determine the seepage discharge per unit length jf the coefficient of permeability is 40 m/dOlY. IAns. s .. 3.99 m, q = 159.6 m1/dayj 9.4. A Stlndy stratum 5 m thick has II slope of 1 in 10 and lies between two impervious simta (Fig. P 9.4). If the piezometers inserted at two points 20 m apart indicate a pressure difference of 3.5m nnd the coefficient of permeability is 1.91 )( 10""" cm/sec, determine the seepage dischnrge. [Ans. 5.96 litccSolbour]

Fig. P9.4.

9.5. Water percolntes across a rcclilngulnr silly earth fill 30 rn long and 15 m wide. The fiJI is founded on an impervious strotum and the depth of watcr on one side is 5.0. Compute the seepage dischllrge. Ie = 0.15 crn/minute.. [Ans, 108 m3/dny] 9.6. A homogcneous dam is 21.5 m high and has a free board of 1.5 m. A flow net was constructed and the following results were observed. '" 12 No. of polcntinl drops No. of now chnnels =3 The dam has n horizonUlI fillcr of 15 m length Cnlculate the discharge/m length of the darn if the coefficient or permeability or the dam mnterinl is 2.7 )( 10~ rnlsec. . [Ans. 1.35 )( 10-5 culllCCS/m] 9.7.

~~i:::': ~:~It~ ~~:e~l:f~=b~ :y~~ ~~~:v::own in FifA~,:'~I~U!~~3 :~/:a;:~~

D. Descriptive and ObJedlv~ 'l)'pe 9.8. What is a flow net ? Describe its flow nel. 9.9. Explain the ~ or a now net.

pro~tties

and applications. Describe different methods used to construct the

SEEI'AGE ANALYSIS

Uri

Fig. P9.7. 9.10. Describe the electrical analogy method of flow net construction. 9.11. Prove that the discharge per unit width of .:m earth dam with Il horizOI1Ull filter Ilt its toe is equal to the coofficient of permeability times the focal length. 9.12. Prove that the discharge through on earth mass iii given by

q ... where

k..-t;'Nf

1r .. coefficient of permeability, Ii :: head, Nt = number of flow ch:mneis, Nd "' number of equipotcntial drops.

9.13. How would you draw the flow nct for a homogcneom earth dam without any filter 1 9.14. Whlll is entry correction of the flow nct 1 How is it donc 1

9.15. How would you conslructthe flow net when lhe soil is anisotropic 1 9.16. Explain the method of constructing the flow net in an earth dam consisting of two different zones. 9.17. Memion whether the fallowing sUitemems are true or false. (a) The flow lines and equipotential lines are orthogonal for an isotropic soil. (b) The number of equipotential lines and flow lines is always a full integer. (c) In two-dimensional flow, the velocity in the thi rd direction is zero. Cd) The velocity potential is equal to the totnI head. (e) The flow net for anisotropic soil can be obtained from Loplacc's equation. (/) The electrical analogy method can be used to obUlin directly flow lines. (g) Relaxation method is used 10 determine the potentiDls at various poinlS. (Ii) The upstream fDoe of an earth dam is an equipotential line. (I) The shape factor depends upon the type of soil. (J) When the flow pl\'iSCS from a soil of high permeability to that of low penneability, Ihe flow lines are deflected aWllY from the normal. ~ (1) The equipD(ential lines make equal vertical intercepts on the phreatic line. (I) The phreatic line of a homogeneous seccion always cuts the downstream face. (m) The phreatic line at the entrance may rise upward. (n) For an earth dam with a horizonUlI filter DC its downstream loe, lhe casagrande exit correction is zero. IA..... Tru', (Q~ (,~ (j), (g), (h), ('), (I), (n)]

C. Multiple-Choice Questions 1. The phreatic line in a homogeneous dam is (a) Circular (b) Ellipliad (c) Hyperbolic (II) Parabolic 2. If there is flow from a soil of permeability «1 to-that or k2, the angles Ih and 02 which the flow line makes witb the normal to the interface are related as

",

SOIL MECHANICS AND FOUNDATION ENGINEERING sin9t

kl

(tI) sina2 =

k2

c~se l

~

(c)

COSe2

=

k2

3. The pressure on :l phreatic line is «(I) cqunllo atmospheric pressure. (b) greater than Iltmospheric pressure. (e) less than atmospheric pressure. (d) nOI related to Ihe atmospheric prcssure. 4. A !low net ha.' 4 !low channels and 20 eq uipotential drops. the shape factor is (a) 1/5 (b) 5 (a) 80 (,I) None of above 5. For an isotopic soil, k,,/kz '" 9. For the transposed section. th e horizontal scale should be (Q) 1/9 (b) J/3 (e) nl ree limes (d) N ine times

6. The slarting point of the horizontal dminage is usually taken as .... of parabola (a)

Focus

(b)

(c)

Vertex

(d) Both (a) and (h)

7. If the flow net of a cofTcrd:lnl foundation

Origin

ha.~

m3/d) per m lenglh is (b) 0.1152 (el 1.0368 (d) 2.304 8. A fl ow net can be used 10 determine (a) Seepage. cocflicicnt of permeability und uplift pre.qsure (b) Seepage. coell1cient of permeability and exil gradient (c ) Seepage, exit grndient nnd uplift pressure (d) Seepage and ex it gmdient o nl y 9. For an an isOlropic soil with kx = 4kz. the value of the modified coefficient of permeability k' is ~)2kx W4kx (d 0.5 kx Cd) 0.25 kx 10. For a now net wilh Nt'" 5 und N,I = 20, the shape factor is (a) 0.25 (b) 4.0 (el lOO (d) 1.0 (An.<;. I. (d). 2. (b). 3. (a). 4. (a) 5. (b), 6. Cd), 7. (b), 8. (e), 9. (a) . 10. (a)] II = 6m. N.I = 6 and N,I = 18, k = 4 x IO-~ m/inin. then the sccp;tgc discharge (in

(0) 0.2304

10 Effective Stress Principle 10.1. INTRODUcnON The effective SlreSS principle enunciated by Karl Thrzaghi in 1936 fonns an extremely useful basis of the most importanf theories in soil engineering. 1be effective stress principle consists of two parts : 1. Oefmitioo of the effective stress. 2. Importance of the effective stress in engineering behaviour of soil This dlapter is devoted mainly to the fin! part. 1be socond part dealing with the importance of effective stress is discussed briefly in the follOWing article. The role of effedive stress on compression rflaraderistics and shear strength is dealt in detail in chapters 12 and 13, respectively. The methods for determination of effective stress in soils for hydrostatic conditions and for steady seepage conditions are discussed separately. The effect of seepage pressure on the stability of the soil masses in described. Piping failures and the methods for its prevention are also disrussed. 10.2. EFFECTIVE STRESS PRINCIPLE (1) DeOnilion of Effective Siress Fig. 10.1 shows a soil mass which is fully saturated. Let us oonsider a prism of soil with a O"OSS-sectional area A. The weight P of the soil in lhe prism is given by P _ Y,tII hA ...(a) where YUIl is the saturated weight of the soil. aod h is the height of lhe prism. Total stress (a) on the base of the prism is equal to the force per unit area. Thus

a-~-y,.h

... (10.1)

While dealing with stresses, it is more convenient to work in teons of unit weights rather than density. As discussed in chapter 2, y -.P . g 3 3 where Y is in N/m and p is in kgfm , g = 9.81 m/sef?Thus, Y,t/I - P,al X g - 9.81 p,., Generally, the unit weigblS are expressed in kN/m 3 and the mass density in kgfm 3• In that case,

Y,,,, - P7~ g -

3

9.81 )( to-

P,.

For example, if P,,,, - 2000 kg/m 3 , Y,tII - 9.81 x 10-3 )( 2000 _ 19.62 kN/m 3 Sometimes, Eq. (a) is approximated as

..(a)


190 YSN - 0.01

U ·:-:·· W<: ·.·

...(b)

Ps....

Y,Of .. 0.01 )( 2000 .. 20.00 kN/m 3 In that case For convenience, "Sq. (b) is sometimes used. Pore water pressure (u)

15

the pressure due to pore water

rilling(he VOldSO[(he,~:' ::

. (102)

I

·1

Pore wa ter pressure IS also known as. neutral pressure or h '. 5011 .... :. '~:. neutral stress, because It cannot resISt shear stresses Pore water pressure IS taken as zero when It IS equal to ==-~-'-~:~ . :~:' :.0 ' ._ • •:... ••_.:... • •_ . _. ' ' , atmospheric pressure, because in soil enginccring the pressures used are generally gauge pressure and not absolute pressures. Fig. 10.1. S.1turated soil mRSIJ. 1be effective stress (0) at a point in the soil mass is equal to the total stress minus the pore water

1

>'.. '. '.: .....

pressure. Thus

0 - 0 - 11

. . .(10.3)

For saturated soils, it Is oblaincd as

0" Y... h - '1... /1 o-('1'
J!r

m

a.y'h

whcre "t' is the SUbmerged unit weight. The effeaive stress is also represented by cr' in some texts. It may be noted that the effective stress is an abstrad quantity, as it cannot be measured directly in tbe laboratory. It is deduced from two physical, measurable quantities a and lL Thus the effective stress is a mathematical concept and not a physical quantity. (2) Importance of Effective Stress 'The effective stress controls the engineering properties of soils. Compression and shear strength of a soil are dependent on the effective stress. Thus

compression

• f( a)

and shear strength .. q> ( 0) where f and

As lhe effea.ive stress in a soil inaemes., the compression of lhe soil occurs. This causes seUlemenl of structures built on soils. The shear strength of 8 soil depends on its elTective stress. As the effective stress is changed, the shear strength changes. The stability of Slopes, the earth pre&SW'CS against retaining structure and the bearing capacity of soils depend upon the shear strength of the soil and hence. the effective stress. The importance of shear strength in soil engineering problem cannQ( be ovcr~mphasised. It is one of the most important properties of soils. As discussed in chapter 8, the pcnneability of soil depends upon the void ratio. With .tt change in effective stress, the void rdtio of the soil changes. Therefon-.. to some extent, the penneability of a soil is also g?vemed by the effective stress.

a

10.3. NATURE OF EFFECTIVE STRESS

Let us oonsider a physical model of a soil mass, fully saturated. as shown in Fig. lO.2(a). Let us lake a wavy plane X- X passing through the points of contact of solid particles. On the macroscopic scale, the wavy plane cannot be distinguished from a true horizontal plane as the individual particles are of relatively small siZe. 1berefort, for all practical purposes. the plane X-X can be assumed as horizontal.

191

EA'ECIlVE STRESS PRINCIPLE

The lOtal normal force P acting on the soil model is resisted partly by the interparticle forces at the points of contact (P"') and panly by the pore water pressure force (P..,) [Fig. 10.2 (b)]. 'Thus P - p. + p. ",(105) At every point of contact, the interparticle force F can be resolved into the normal component (N) and the tangential component (T) to the plane X-X [Fig. ID.2 (e)]. The interparticle forces are random in both

o Pm

~

(b)

,--~--,

I

Am

I

j.---A (e)

.... / ---..j

(d)

Fig. 10.2. Physical model of

1\

soil mass.

magnitude and direction throughout the soil mass. The tangential components, however, neutralise one another and the resultant of all the normal components is downward. The effective stress is the nominal stress transmitted through the soUd particles, and is given by (; _

sumar::~~~mpk:n~

0_ I:

",(lO,6)

Let the area of qoss-section occupied by the solid particles (minerals) be Am and that occupied by wale< bl: A_ [Fig, 10,2 (d)J

A - A", + A...

Therefore,

A ... - A - A", Let u be the pore water pressure. From Eq. ID.5, P _ Pm + P", .. l:N + IV

or

oA -

a

,.(10.7)

",(108)

Am + uA ...

where 0- is tbe actual normal stress transmitted at the points of contact of the solid particles, and a is the total stress (Eq. 10.1).

Eq. 10.8 .can be written as

a _ " (Am/A) +

U

(A.lA)

Using Eq. to.7,

o .. 0- (A",/A) + u (1 - A".IA)

'"where

0"0 Q",

_

Am/A.

Q ...

+ u (I-a",)

".(10,9)


1<>2

'Ille geolcdmical engineer is interested in the effective stress (0) not in the actual contact stress (0). Let us again consider the equilibrium in the vmical direction [Fig. 10.2 (d)}. We have P _ TN + uA ... aA .. 'EN + uA ... 0 _ IN/A + u (A,./A)

or

.. .(10.10)

In Eq. 10.7, as the area occupied by the interparticle contact (mineral to mineral) A. is very small (about 3% for granular soils). the area A ... be taken approximately equal to the lotal area A. In other words, A ... _A . Therefc::re, Eq. 10.10 becomes 0 " IN/A + u

Designating IN/A by the nominal effective stress, 0, 0

..

cr ..

or

a

+ u

(same as Eq. 10.3)

0- u

It must be nOled that the effective stress (0) depends upon the normal force (IN) transmitted at the points of contact, but it is not equal to the contact stress (fJ). It is equal to the total normal (orce N transmitted at the points of contad divided by the total area A, including that occupied by water. It has no physical meaning and, therefore. cannot be directly measured. It is much smaller (han the actual contact stress '&. The pore water pressure due to water in voids acts equaUy in aU directions (pascal's law). It docs not resist any shear stress. and, therefore, is also called the neutral stress. However. it is very important as tbe effective stress depends upon the pore water pressure. In clayey soils, there may not be direct contact between the minerals due to the surrounding adsorbed water layers. However, it has been established by actual experiments that the interparticle contact forces can be transmitted even through tbe highly viscous adsorbed water. The above equations whK;b have been developed assuming '-he soil as coarse-grained may be used for clayey soils as well. For surface active minerals, Eq. 103 is modified as 0 - u + (A' - R') ... [10.3
cr •

I DA. EFFECT OF WATER TABLE FLUcruATIONS ON EFFECTIVE STRESS Let us consider a soil mass shown in Fig. 10.3. The depth of the water table (W.T.) is HI below the ground surface. The soil above the water table is assumed to be wet, with a bulk unit weight of 'f. The soil below the water table is saturated, with a saturated weight of 'f_ Thtal down ward force (P) at section X-X is equal to the weight (W) or the soil. Thus, p ..

w ..

1 HI A + 'f,at 112A

where A is the area of ttOSS-section of the soil mass. Dividing by A througoout,

i.

1H} + 1,,# H2

The left-hand side is equal to the lotal stress CEq. 10.1). Therefore, The pcxe water

0

..

Y HI + y,atH2

pressure (u) is given by Eq. 102 as

! ! ! ! !U w = 1."t+ f sa• Hz' Fig. to.3

u .. y,.,H2 From Eq. 10.3,

0_ 0 - u .. (yHI + y_Hi) - y... H 2

0'

(j .. yHI

+ (y_ - Y...) H2

ErFECTIVE stRESS PRINCIPLE

J93 ... (10.11)

Eq. 10.11 gives the. effcaive stress at section X-X. Fig. 10.3 also shows ti
X-X.

(a) If the water table rises to the ground surface, the whole of the soil is S.'ltUf'dted, and a.y'(H\+H,).y'H As y' < y, the effective stress is reduced due to rise of water table. (b) If the water table is depressed below the section X-x,

... (10.12)

'0 - 1 H ... (10.13) In this case, the effective stress is increased. Thus, it is observed that the fluctuations in water table level cause changes in the pore water pressure and the corresponding chnnges in the effective stress. 10.5. EFFEC'IWE STRESS IN A SOIL MASS UNDER nYDROSTA"nC CONllrnONS Fig. 10.4 (a) shows n soil mass under hydrostatic conditions, wherein the wmer level remains constant. As the interstices in the soil mass nre interconnected, water rises to [he same clevmion in different

piezometers fixed to the soil mass.

r

'nlC

effective stress al various sections can be determined using Eq. 10.3.

'tw

WATER

SOIL.ltl'5.;lIt>'

(Yut~

SOIL.2!)'U\),

(lui?

C

Fig. 10.4.

(I) Water Thble above the soil surface A-A: Section A-A 0-1... 1-1, Therefore, Section n-B Therefore,

W~ler

Tabte aboYc soil

u - 1... H ... (a)

o - 'I,., 11 + ("f".)l HI.

a-

and u - 1... (N + HI)

[('1~",h - 'I ...} 11, - 'It' HI

... (b)

where 'I,' is the submerged unit of soil-I. Section C-C

0 -

'1 ... H + ('1~ ...)l H J + ('1~i'ihH2

u - 'I ... (H + HI + Hi> Therefore

... (e)

where '12' is the submerged unit weight of soil·2. (2) Water Table at the soli surface A-A Fig. 10.4 (b) shows the condition when lhe deplh H of water abq.ve Ihe section A-A is reduced lo zero. In this case, Ihe effective stresses at various sections are detcmlined as under: I Section A-A CJ_u_a_O ... (d) Section B-B 0 _ ('I....») H) , aod u _ 'I ... HI aDd

... (e)

194


St.!ction C-C

cr - 'fl'

Therefore,

... (j)

HI + '/2' 1-/2

Comparing Eqs. (a), (b), (c) with (d). (e) and (f). it is observed that the depth H of water above the soil surface does not contribute to the effective stress al all. In other words, the effective stress in a soil mass is independent of the depth of water above the soil-surface. It should therefore not be surprising that the marine soil dCpc6its, which are under a very large depth of water, have a low effective stress and correspondingly low shear strength. (3) Water Table In Soil-l fig. 10.5 (a) shows the case when the water table is at D-D in the Soil-l al depth H/. The effective stresses at various seclions arc determined as follows. .

(.)

(b)

Fig. 10.S. Wafe.r Table in (a) Soil-l and (b) Soit-2.

a-u-Ci_O

Section A-A Section D-D where

'(J

0""

Therefore, Section n-n Therefore,

Section C-C Therefore,

Ht ..

0,

u

=0

(j - '(I Hl' 0" • U ..

"\VhCIl

yllIl',

is the unit weight of soil above D.-D.

Yl HI' + (y,...h Ht" y",ll(

(Note. HI' + Ht .. HI)

a .. YI HI' + [(y,...)1 - y",l Ht .. YI HI' + YI' Ht" yllIl' + 'r,a)1 Ht" + (Y/ahHz U .. y..,(Hz + HI")

0" -

a .. ylH

t ' + YI' fIt" + yz' Hz a - ylH t ' + yz' Hz

(4) Water TobIe in Soil-2

Fig. 10.5 (b) shows the condition when the water table is at EE in Soil-2 at depth Hz'. The effective stresses at various sections are as under: Section A-A a-u-a-O Section 8-8 a .. yllIl, U" 0, Ci .. yllIl Section E--E 0" .. 11 II t + YiJI'";.' , U .. 0 (Note. lIz' + Hz" .., HV Section C-C

a .. yl IJI + yzflz' + (YfOlhHz/l

u .. Y... Hz" Ci .. "'1'1 HI + YzHz' + Yz' Hz"

EF1~nVE

srnESS PRINCIPLE

195

where Ya' is submerged unit weight of Soil·2. (5) Water Thble below C-C

Fig. 10.6 shows the condition when the water table is below C-C. As the pore water pressure is zero everywhere, the effective stresses are also equal to the total stresses. SectIon B-B

a •

a ...

y,H,

Section C-C 0 - a . . y,H, +YaHl The following pointS arc worth nOling in lhe five cases studied above. (1) The effective stress at any section goes on increasing as the Fig. 10.6. Water Table below soil. water table goes down. (2) The effective stress depends upon the bulk unit weight above the water table and the submergcd unit weight below the watcr level. (3) The effective stresses in a soil mass can be determined from the basic definitions, without memorising any Connula.

-----¥o-----

10.6. INCREASE IN EFFECTIVE STRESSES DUE TO SURCHARGE Let us consider the case when the soil surface is subjected to a surcharge load of intensity q per unit area. Let us,assume that the water table is at level B-B (Fig. 10.7). The stresses at various sections are detennined as under. Section A-A o_q. u_O, a_q i.e., all the points on the soil surface arc subjected to an effective stress equal 10 q. Section B-B a .. q + y, HI' U .. 0 1bcrcfore, a ... q + Yl H, Section C-C

q + y,H, + (YsahHl u ... y..,Hl

I I I I I /A

A

1".

B

"

0-

B

(l"satlz

1bcrcfore, a ... q + YI H, + Yl'1I1 C C From the above illustrations, it is clear that the effective stress throughout the depth is greater tban the case with no surcharge Fig. 10,7. Effect of SucchHrge. discussed in the preceding section. The difference is equal to the intensity q. Tn other words, the effective stress is increased by q throughout. 10.7. E.'FECTIVE STRESSES IN SOILS SATURATED BY CAPILLARY ACITON If the soil above the water table is saturated by capillary action. the effective stresses em be determined using Eq. 10.3. However, in this case the pore water pressure above the water twle is negative [Fig. 10.8 (a)]. The water table is at level B-B. Let us consider two cases: (1) Soil saturated uplO surface level A-A [Fig. 10.8 (a)] (2) Soil saturated upto level D-D [Fig. 10.8 (b)] (1) Soil saturated upto surface level A-A [Fig. 10.8 (a)] The pore pressure diagram is drawn on the right side. ~e

j

stresses at various sections are determined as under. o ,. 0, U,. -y..,HI Therefore, '0_ 0 -(-y..,HI) - y..,H I

Section A-A


196

Hi~

uSaturated Wet

}"'

"2 C (b)

Fig.tO.S. F1Teet ofcapil\ary rise.

If the soil was not saturated with capiUary action. the effective stress at section A-A would have been zero. "hus, the capillary action has increased the effective stress by ,(""HI" In other words, the negative pressure acts like a surcharge (q'Section D-D rJ (0" ..») HI' (Note. HI' + Ht" = HI)

=

I.f

Therefore,

=-

y",H(

= -y.-(lil

- HI')

a = [(Y,,,,)1 HI' - y" HI'] + Y", H I cr _ 'fl' HI' + y..,H1

If the soil had bccn saturated due to rise in water-table 10 A-A, the effective stress al section D-D would have been Yl' Il( Thus, tbe effective sIre&'> is increased by y..,HI due 10 capillary action.

&!clion 8--8

<1 -

(Ys«)' H L ,

u _ 0

1l1ererore. jj - (Ysa)\H\ - 'f1'1/\ + '1 ... 111 If the soil above B-B had been saturated duc to rise in water table to A-A, the effective stress would have been y/ H, " Thus the effective stress is increased by 't ... 1l1 by capiUary action

Section C--C

a - (Ysa), HI +

b...h Hz .

u - y... 1i2

Therefore, At this section also, Ole effective stress has also increased by ,(""HI. lt may be noted that the C[fectivc stress at all levels below the plane of saturation A-A, due to capillary water, is increased by y",H I • The capillary water pressure y... H I acts as if a surcharge. The effect is somewhat similar to the ronstant rompressive Stresses induced in the walls of the capillary tubes discussed in chapter 7

(2) Soil saturated upto level D-D [FIg. 10.8 (b)] . Let us now consider the case when the soil above the water table B-B is saturated only uplO level D-D upto a height Ht. The soil above level D-D is wet and bas a unit weight of y. The capillary rise in this case is HI". The stresses at various sections can be determined Il ..

Section D-D

ylH l',

(J ..

as unclcr.

0_0

(J _ Section A-A There is no effect of capillary water.

Il "

- y... flt

(j .. yIH I' + ,(... lit The effective stress due to capillary pressure is increased by y", H L".

Therefore,

Section

n-D

(J ..

yiN + (Yz..h

" • 0 Therefore,

lit ..

'(1'11/ + Yt'ITt + y... JJt"

EFFECTIVE SfRESS PRINCIPLE

197

The effective stress is increased by 'I ... Ht" due to capillary action. Likewise, it can be shown that the effective stress is increased by 'Iwllt at section C-C also. The following points may be noted from the study of both cases : (1) The capillary water above the water table causes a negative pressure '1 ... 11, where 11 is the capillary rise. This negative pressure causes an increase in the effective stresses at all levels below the saturation level. The increase is equal to '1 ... 11. The capillary action is equivalent to a surcharge q - y.H. (2) If the soil is saturated due to rise in water table, the e[fcx:tive stress depends upon the submerged unit weight; whereas for the soil saturated with capillary water, the e[fedive stress depends upon the saturated unit weight. In the latter case, the water does not contribute to hydrostatic pressure. (3) If the water table rises to the top soil surface, the meniscus is destroyed and the capillary water Changes to the free water, and the effective stress is reduced throughout. (4) Eq. 103 is applicable in all cases. However, it should be remembered that the pore water pressure in the capillary zone is negative. 10.8. SEEPAGE PRESSURE As the water flows through a soil, il exerts a force on the soil. The force acts in the direction of flow in the case of isotropic soils. The force is known as the drag force or seepage force. The pressure induced in the soil is lenned seepage pressure. Let us consider the upward flow of water in a soil sample of length L and cross-sectional area A under a hydraulic head of II [Fig. 10.9 (a)l. The expression for seepage force and seepage pressure can be derived considering the boundary \/ater pressures III and u2 aC1ing on the lop and hoIlom of the soil sample, as shown in Fig. 10.9 (bXI). The boundary water pressure can be resolved into two components, namely, the hydrostatic pressure and the hydrodynamic pressure as shown in Fig. 1O.9{bXil) and 1O.9(b)(iit). (I) The hydrostatic pressures III(S) and u2(s) are the components which would occur if there were no flow. If the samples were submerged under water 10 a depth of HI, lhese pressures would have

occurred.

T

I

(b)

Fig. 10.9. Verticall y UPWHrd now.

".

SOIL MECHANICS AND FOUNDXnON ENG INEERING

(2) The hydrodynamic pressures Ul(d) and u2(d) arc the components which arc responsible for flow of waler. This pressure is spent as the water flows through the soil. 'nlcse components cause the

seepage pressure. At the lop of Ihe sample,

III .. "1 (5) + III (d) 1... H , .. 1... /11 + 0

At the bottom of the sample,

1.12" tl2 (s)

+

U2

(d)

'(..,{H I + L + h) .. ,(... (H I + L) + l ... h The hydrodynamic pressure is due to hydraulic hc..1d h. The seepage force (1) ads on the soH skeleton due to 'flowing water through frictional drag. It is given by J - y.hA ... (10.14) The seepage pressure (Ps) is the seepage force per unit area,

p, .. l / A .. '1 ... h '(he seepage pressure (Ps) can be expressed in terms of the hydraulic gradient. From 10.15,

.. (10.15)

p, .. 'Yw h .. 'I ... . (hIL) . L

P." i1... L The seepage force (1) can be expressed as the force per unil volume .

J

) .. A"";L ..

YwhA

-:;u:- ..

Y..,

... (10.16)

V). as

h

L

or j .. h... ... (10.17) lbus, the seepage force per unit volume is equal to the product of the hydraulic gradient (I) and the unit weight of water. As the hydraulic gradient is dimensionless, the seepage force per unit volume has the dimensions of the unit weight (i.e.)!F/L)). It bas the units of N/mJ. For isotropic soils, the seepage force acts in the direct ion of now. 10.9. FORCE EQUILIBRIUM IN SEEIJAGE PROBLEMS Force equilibrium in seepage problems can be considered adopting either of the following approaches. (1) Considering the equilibrium of the entire mass and using the boundary pressures. (2) Considering the equilibrium of the solid particle or the mineral skeleton, and using the hydrodynamic pressures. (a) Vertical F10w

(I) Upwards . Fig. 1O.10(a) shows the forces acting on the soil mass shown in Fig. 10.9 (a). The unit weight of the soil used is the s..1turntcd unit weight. 'rne resultant force (If) on the soil mass considering the equilibrium of the entire mass, adopting the first approoch,

)!T I l4J lj] 1'.;1 ~I ,~H+h)' U,' ....

" 5

BOI..IiD4RY R : u,+

FO~CES ll2

w-

: LA,'_ Y."A h, )

u;' .... "

FORCES ACllNG ON SOLID PAATrClES

R:W"'-ui~w"_J :tA'f'-Y... A fb)

Fig. 10.10. Force 1,.'quUibirum in vcrtiCllI

now.

EFFECTIVE STRESS PRINCIPLE

199 R = (W + Ud - U2

R .. (LA

'(sat

+ '(wAHl) - '(... A (H J + L + h)

R .. LA l' - 1whA ... (10.18) or The figure on the left-hand side shows the force diagram. The resultant force R acts downwards. For stability of the mass, R must act downwards. Fig. to.IO(b) shows the forces acting on the solid particles, adopting the second approach. 1be unit weight of the soil used in this approach is the submerged unit weight. 1be resultant force (10 on the soil skeleton is given by R

.. W' - U2' R .. LA'(' - l..,hA

(same as Eq. 10.18)

In Eq. 10.18, the first term gives the submerged unit weight and the second term, the seepage force (Eq. 10.14). It must be noted that in the first approach, the seepage force (J) is not considered separately. It is automatically accounted for in the boundary forces. (iI) Downwards. Adopting a similar procedure, it can be shown that the resultant force when the flow is downward is given by ... (10.19) R '" 1..A y' + 1.., IIA (b) Inclined Flow

Fig. 10.11 (a) shown the flow through an inclined soil specimen. In this case also, the resultant force R can also be determined by adopting either of the two approaches discussed above for the vertical flow. As for the vertical flow, in tbe first approach, the resultant force R is the vectorial sum of the saturated weight

1b)

Fig. 10.11. Foroc equilibrium in Inclined flow.

(WSIl') and the boundary forces U. In the second approach, the resultant force R is the vectorial sum of the submerged weight (W. ub ) and the seepage force (J). 1be force triangles are shown in Fig. 1O.1I(b). Thus (1)

(2)

ii _ if,•• U ii - if,", • ]

... (10.20) .. (10.21)

Thus Ihe IW,o approaches give the same resultant force R. In Fig. 10.11 (b), the buoyant force is represented by D. Obviously,

B .. W,at - WslIb .. . (1 0.22) The two approaches give identical results. However, the first approach is more IX>pular. It is more convenient to determine the boundary forces than 10 determine the seepage forces. 1be seepage forces depend upon the direction of flow and change from point 10 point. It becomes diffio.lit to determine the seepage forces, especially in two-dimensional flow.


200

10.10. EFFECTIVE STRESSES UNDER STEA..DY SEEPAGE CONDTrIONS As the water flows through the soil, it exerts a seepage force on the soil particles. The seepage force affects the interparticle forces and hence the effective stresses. 'The effective stress is increased when the flow is downward, as the seepage force increases the interparticle forces. On the other band, when the flow is upward, the effective stress is decreased as the scepage force decreases the imerparticle fOrces. The two cases nre discussed separately below. (a) I)ownword Flow. Let US consider the case when the flow is downward (Fig. lO.12). The head causing flow in h. T1le pore water pressure at sections A-A and B-B are Fig. 10.12. DownwHrd Aow. indicated by the piezometers. The effective stresses at various sections are dClennincd using Eq. 10.3. Section A-A 0 .. 1... II"" u .. 1... 11.., Therefore, Section Il-B

1bercfore,

a-O a - (Yz,.)' H , + ..,••,11"", u - y... II",,!

a .. <'r1l..),I1\

+ 1..,H.., - l..,H...1

(j - (YI' Hl + Ywlll) + l..,H.., - l..,Il..,1 (j - Yl' HI +

y",(I1... + HI - H ...1)

For hydrootatic oomJitions, the effective stress is ll'IlI• as discussed in Sect. to.5, The second tenn indicates the effect due to [low. As (H"" + HI) > H",b the effective stress is incrtascd due to downward now:

Section C-C Therefore,

(J ..

1M.H.., + (Yu,)IH I + (1rtlhH2 ,

a - y... H..,

u ..

a

+ (ll' + l ...) /-II + ('12' + l ...)"2

a _ YI' HI + Y2' H2 + y", (HI + H M• + Hi)

or 0 - y,' III + Y2' H2 + y",h A oomparison with the effective Slres...es corresponding 10 hydrostatic conditions shows that the effective stress is increased by y... h. The conclusion that the effective stress is increased due to downward now c.1n also be drawn from inlitutive feeling that as the waler flows downward, it exerts a drag force in the downward direction and causes an increase in the interparticle forccs. "(b) Upward Flow. Fig. 10.13 shows the case when the now is upward. The piezometers at various elevations indicate the pore water pressure. Section A-A TIlercfore,

Section B-n 1bcrefore, 0'

(J -

1.., H"" , u" l ... II",

'0_0 0- 1", H..,' + (y'....)l III,

cr ..

U - y..,H",1

(Y.....)1 HI + l",II",-l... 1I""

.. (Yl' + y...) HI + 1",H... - y",H..,1 '0_ ll' HI + l ... (HI + fJ.., - 11",1)

As 11"'1 > (H) + H..,), the tenn 'r".(H, + H", - /-1,,'1) is negative, and the effective stress is less than that ~rom the corresponding hydrostatic conditions

201

EFFECllVE STRESS PRINCIPLE

Fig. 10. 13. UpWllrd Flow.

Section C-C

a '"' y... H ... + (Y~n)1 HI + (YttrhH2 • u" y... J/...2 o .. y,.,H.., + (YI' + y...) HI + (Y2' + y...) J/2 - y... H ..'l

cr ..

""tl' HI +

yi

Ih - (H,.rz - H ... - HI - Hi) Y",

a ..

11' H I + 12' H2 - 1..,h Thus, the effective stress is reduced by 1... h from the corresponding hydroo.talic conditions.

10.11. QUICK SAND CONDITIONS As discussed above, the effective stress is reduced due to upward now of water. When the head causing upward flow is increased, a stage is eventually reached when the effective stress is reduced 10 zero. The condition so developed is known as quick sand condition. Fig. 10.14 shows ..I soil specimen of length L SUbjected to an upward pressure. Let us consider the strcsscs developed a1 section C-C.

Fig. 10.14. Quick SIIoo condilioos.

a .. Y,(>I L .. (-y' + y..,)L, 1berefore,

U ..

y",H.. 1

..

(j .. (y' + ..,..,) L - y..,. (L + h)

(j .. ..,' L - ..,... h 111e second term can be written in terms of the hydraulic gradient a'i under. y..,11 .. Yw . (h I L) . L .. Y.., • i . L

Therefore,

(j .. y'L - "'wiL

..,..,(L + h)

202

SOil MECHANICS AND FOUNDATION ENGINEERING

The effective stress becomes zero if

1 'L - '1",; L

or i - 1 '11. . .. [IO.23(a)J The hydraulic gradient at which the effective stress becomes zero is known as the critical gradient (ic)' Thus i, - 1 '11. . .. (10.23) Substituting the value of the submerged unit weight in terms of void ratio from Eq. 2.24 (a),

ic ..

(~ : ~)

... (10.24)

Taking the specific gravity of solids (G) as 2.67, and the void ratio (e) as 0.67,

ir ..

~.6: ;6~

.. 1.0

Thus the effective stress becomes zero for the soil with above values of G and e when the hydraulic gradient is unity i.e. the head causing now is equal to the length of the specimen.

Alternative method The above expression for the critical gmdient can also be obtained from the equilibrium of forces. When the quick condition develops, the upward force is equal to the downward weight. Thus

1_ (L )( A) .. (h .. L)_Alw

or

(1,... - lw)LA .. A hlw

or

hlL .. l'/y..,

Ll' - hl.

i. .. y'/y..,

.. [10.23(b)]

... [same as Eq. 10.23 (a)]

111e shear strength of a cohesionless soil depends upon the effective stress (see chapter 13). The shear strength is given by s .. (j Ian .. where , is the angle of shearing resistance. 1be shear strength becomes zero when the effective st~ (j is zero. The soil is then said to be in quick (alive) condition. If the critical gradient is exceeded, the soil particles move upward, and the soil surface appears to be boiling. The quick rondition is also known as boilillg condilion. During this stage, a violent and visible agitation of particles occurs. The discharge suddcn1y increases due to an inaease.~ in the coefficient of penneabjJjt~ occurred in the process. If a weight is placed on the surface of soil, it sinks down. The soil behaves as a IJquld having no shear strength. When a natural soil deposit becomes quick, it cannot support the weight of man or animal But contrary to common belief, tbe soil does not suck the victims beneath its surface. As a mailer of fact, quick sand behaves like a liquid with a unit weight about twice that of water. A person can easily float in it with about one-third of his body out of quick sand. However, qUick sand is highly viscous and movement In it would require a great effort and energy. A person may die by drowning (suffocation) if he gets tired and let his head fall into the quick sand in panic. If a person is caught in quick sand conditions, he should keep his head high above the soil surface aod move slowly towards the bank. He should try to calch some tree on the bank and try to pull himself out of lhe quick sand. I! is to be emphasized that quiCk sand is not a special type of sand. It is a condition which occurs in a soil when the effective stress is zero. Any cohesionless soil can become quick when the upward seepage force is large enough to carry the soil particles upward. The quick sand condition may also develop in gravel when the hydraulic gradient exceeds the critical gradient. However, the discharge required to maintain quick condition in gravels is very large, which may nol be available. The required discharge depends upon tbe permeability of the soil

,m


The shear strength of cohesive soils is given by (sec chapter 13). s - c+otan$ The soil hIlS shear strength equal to the oohesion intercept (c) even when the effective stress is reduced to zero. The cohesive soils, therefore. do not become quick as soon a<; the effective stress is redua:d to zero. The quick sand conditions may be summarised as under: (1) Quick sand is not a special type of soil. It is a hydraulic condition. (2) A oohesionless soil beromes quiCk when the effective stress is equal to zero. (3) 111c critical grc!dient at which a cohesionlcss soil becomes quick. is about unity. (4) The discharge required to maintain a quick condition in a soil increases as the permeability of the soil increases. (5) The cohesive soil does not become quick when the effective stress is equal to zero, as it still possesses some strength equal to the cohesion intercept. (6) A quick condition is ma;;t likely to occur in silt and finc sand.

10.12. SEEPAGE ]'RESSURE AI'PROACH FOR QUICK CONDn'IONS The expression for critical gradient in the proceeding section has been developed using the first approach mentioned in Sect. 10.9, i.e., considering the boundary pressures. The same expression can . be developed using the serond approach, i.e., considering the h seepage pressure. Let us consider the equilibrium or the soil skeleton in Fig. 10.15. The downward force is given

-r

L,--------"h 1

by

,\

W...b-(AxL)y' The upward force is equal to the seepage force (&J. lO.14). J .. iyw (A xL) When the soil becomes quick, the resultant force is zero. 1lterefore, W.ub - J

lWSlJb

L

j I'ig.10.15. Seepltgc Pressure Approach

ALy' - iy .. (A xL)

L

i _ ... (some ... &J. 10.23) Y. Altematively. one can work with pressures instead of forces. The downward pressure due to the submerged weight of soil is given by (j .. y' L The seepage pressure is given by Eq. 10.16 as P. - iy",L The net effective stress would be zero when Thus

a - Ps'

iLy", - y' L i .. y'/y...,

... (same as Eq. 10.13)

10.13. EFFECT OF SURCHARGE ANIl SUBMERGENCE ON QUICK CONDmONS Fig. 10.16 shows a soil specimen submerged under water and subjected to surcharge load of intensity q.

Let us consider the stresses at section C-C.


l~c Fig. 10.16. Effect

T

I

or Surcharge.

o .. l",H... + q + 110lL u .. 1... H...1 or u .. l",(h + H", + L) (j .. ("f...,H... + q + 1,,,, L) - 1... (h + H..., + L)

aDd Therefore,

a•

The soil will become quick when O. Thus y...,(h + H..., + L) .. y..., lJ", + q + llal L

y...,h + y...,H", + Y... L .. y",H..., + q + Y,OI L y... h . y ' L+q h.y ' L+q ... ~ 1O.25) Y. Comparing Ihis equation with Eq. 10.23 (b) for the case when there is no surcharge as discussed in the preceding sections, it is observed that the head required to cause quick condition is increased by ql y.... In other words, the downward weight increases the stability against quick conditions. The critical gradient is, however, not affected by the depth of water (H...,) over the soil surface. Substituting q =:- 0 in Eq. 10.25.

or

h • L.!,. Y. hlL .. y'/y", i .. y'/y...

. .. (same as Eq. 10.23)

10.14. FAILURES OF HYDRAULIC STRUCfURES BY PIPING Hydraulic strualU'eS, such as weirs and dams. built on pervious foundations sometimes fail by formatioo of a pipe-shapcd channel in its foundation, known as piping failure. The failure occurs when water flowing through the foundation has a very high hydraulic gradient and it carries soil particles with it. There are two types of such failures : (1) Backward-erosion piping failure. (2) Heave..piping failure. (l) .Backwanl Erosion Piping. This type of piping begins with the removal of soil particles by flowing water near the exit points. A scour hole forms near the exit when the hydraulic gradient is high. 1l1e hole extends upstream along the foundati on. The failure occurs as soon as the scour hole approaches the upstream reservoir. The failure mechanism of backward erosion piping is not amenable to theoretical solulion.

EFFEcnVE STRESS PRINCIPLE

Backward erosion of soil is caused by the percolating water, and the piping begiru; when the hydraulic gradient at exit, known as exir gradient, exceeds the critical gradient (if), given by Eq. 10.24. The soil at the exit is removed by the percolating water. When the soil near the exit has been removed, the flow nct gets modified. There is more ooncentralion of the flow lines in the remaining soil mass, resulting in an increase

Ag. 10.11. Bnckward erosion Piping.

[

of the exit gradient. This causes further removal of the soil. This process of backward erosion continues towards the upstream reservoir and a sort of pipe is formed (Fig. 10.17). As soon as the channel approaches the reservoir, a large amount of water rushes through the channel so formed and the hydraulic sttuaure fails. Backward erosion piping may also occur in the body of earth structure. such as an earth dam. This lakes place when the phreatic line cuts the downstream face of the dam and the seepage pressure is high. It is indicat«l by a progressive sloughing of the downstream face. Such failures can occur even when the exit gradient is low. If the dowru;tream face has the slope angle equal to the angle of internal friction of the cobesionless soil, the critical gradient at which failure occurs is approximately equal to zero. In other words, the failure may occur even when the scclX"lge is almost horizontal towards the downstream face. Backward erosion piping may also occur along any weak bedding plane in the fOlJndation, or along the periphery of a conduit embedded in the e.1rth dam when the seepage pressure is high. Generally. backward erosion piping failure occurs when the exit gradient is greater than the aitical gradient. But, in exceptional casc..<;. it may occur even when the overall downward submerged weight of the soil is greater than the upward focce due to seepage. In such a case, some of the fine particles of the soil are carried by the percolating waler even though the most of soil particles are restrained. Thereafter, the seepage concentrates in the loosened soil and results in piping failure. (2) Heave Piping. Failure by heave piping occurs in the form of a rise or a heave of a large mass of soil due 10 seepage pressure. When the seepage force due to upward flow of water at any level is greater than the submerged weight of the soil above that level. the entire soil mao;s in that zone heaves up and is blown out by the percolating water. This type of failure is known as heave piping failure. Heave piping may occur on the downstream of a sheet pile cutoff wall of a hydraulic structure (Fig. 10.18). According to Terzaghi, heave piping occurs within a distance of D/2 on the dowmtream of the sheet pile, where D is the depth of pile below the ground surface. It occurs in the zone marked abc d when the upward seepage force is greater than the submerged weight of the soil in this ·zone. 1be seepage force can be determined from the flow n~. In Fig. 10.18, the equipotential line or potential 0.4 h passes through d and that of 0.3 h through c. The average excess hydrostatic pressure on the base c d of the prism abc d is 0.35 h. Therefore, upward seepage force, U _ Y... (0.35 h) (D/2 )( 1) per unit length and downward force due to submerged weight,

SOIL MECHANICS AND FOUNDATION E.NGINEERING

206

Fig. 10.18. Belly(' Piping.

W .. y'x(D/ 2xD) Heave piping would occur when U it W'. The failure is associated with an expansion of the soil which results in an increase in the permeability of the soil. The flow suddenly increases and ultimately leads to failure. The factor of safety with respect to heave piping can be obtained from the following equation.

F.~.~ U

Y. (h,) (D 12)

or where ha is Ihc average excess

... (10.26) hyd~latic

pressure at the base of the prism abc d

10.15. PREVENTION OF PIPING FAILURES

The ocamence of piping in and below a hydraulic structure such as an earth dam is disastrous. This may cause catastrophe. The following measures are generally adopted to prevent piping failures. (1) Increasing the palh of percolation. The hydraulic gradient (I) depends upon the path of per(:olatioo (L). If the length of the path is increased, the exit gradient will dcaease to a safe value. The length of the path of percolation can be intteased by adopting the following methods. (a) Increasing the base width of the hydraulic structure. (b) Providing vertical cut off wulls below the hydroulic structure. (c) Providing an upstream impervious blanket, as shown in Fig. 10.19. (2) Reducing Seepage. With a reduction of seepage through the dam, the chances of piping failure through the body of the dam are considerably reduced. The quantity of seepage discharge is reduced by providing an impervious core, as shown in Fig. 10.19. (3) Providing drainage niter. A drainage filter changes the direction of away from the downstream face. It prevents the movement of soil particles along with water. The drainage filter is properly graded, as discussed in the following section. The drainage filter may be horizontal or in the form of a rock toe (Sect. 9.11). It may also be in the fonn

now


""

Fig. 10.19. Prevenlion of Piping.

of a chimney drain. as shown in Fig. 10.19. A chimney drain is effective for stratified soil dcpooits in which the horizontal permeability is greater than the vertical penneability. (4) Loaded Filter. A loaded filter consists of graded sand and gravels. The function of the loaded filter is to incrc~ the downward force without inaeasing the upward seepage force. . The loaded filter is placed at the exit point where the water emerges from the foundation. For the sheet pile wal~ the filter is placed over the affected zone abc d in Fig. 10.18. 1be loaded filter increases the factor of safety against heave piping. The foclor of safety (F) is given by

F _ W'~ W

... (10.27)

where W is the weight of filter. The loaded filter should be of pervious material sum thai it does not increase the hydrostatic pressure. It should only increase the downward force.

10.16. DESIGN OF A GRADED F1LTER A graded filter consislS of layers of pervious material which permit now of water but prevent the movement of soil panicles. lbe soil particles in a panicular layer are coarser than that in the preceding layer. However. the difference of sizes of the panicles in the two layers should not be excessive otherwise the particles of the preceding layer will be carried into the next layer. The particle sizes of different layers are fixed according to the design criteria given below : (1) The filter material should be coarse enough so that !.he percolating water moves easily without any build up of water pressure in the filter. For the filter to provide free drainage. ilS ooefficient of permeability should be 25 times or more than the coefficient of permeability of the soil to be protected. known $ a base mateTial. As the ooefficient of permeability varies as the square of the particles Si7..c, the ratio of the particle diameters should be at least 5. Therefore. the Hrst criterion of design of graded filter Is that Du of filter material is greater Ihan 5DLS of base material. ... (10.28) or DI~ (f) > 5D Is (b)

where (f) stands for filter and (b) for base material. (2) The filter material should be fine enough that the soil particles of the base material are not washed through the filler. It would not be necessary to saeen out all the particles in the base materia).. If the mler openings restrain the coarsest 15% i.e. Dss size of base material. the soil particles are checked due to fonn~tion of a skeleton. The coarser 15% particles rollect over the openings in the filter material and form smaller openings to trap the smaller particles. ~ shown in Fig. 10,20, Therefore, the size of the openings formed in the filler must be less than Dss of the soil. It has been established that the diameter of the openings is about 115 of the DIS size of the filter. "lberefore. the seoond criterion is

~ DIS of filter

material <

DIl.~ of base of material

DiS(f) < S Das (b)

Combining both the criteria (I) and (2).

... (10.29)


208

/

"'-------,,

~~~L--',f---":::';Q~7 \ \

\

" .... , Fig. 10.20. Openings in filler

Dls(J) Dls(J) D85(b) < 5 < Dls(b) The U.S. Corps of Engineers have recommended that D",IJ) D,,(b) < 25

... (10.30)

... (10.30 (a)]

In a graded filter, each layer is designed considering it us a filler and the preceding layer as a base material. 111C p..1rticJe sizes of the layers increase in the dircction of flow. (Fig. 10.21).

",'. . . . r~~~. ".' c.) Fig. 10.21. Graded filler.

C"

(4) The material of the last layer should be coarse enough not to be carried away through the openings of the perforated drainage pipes, jf provided. If the Dss size of the last layer satisfjes the following criterion, lhe chances of washing of the filter material into the pipes are reduced. '

For circular holes in the pipe,

D8S of filler malerial Diameter of the hole > 1.2

... (10.31)

For starred openings,

Dasoffiltermateriai Width ofslol > 1.4 Generally,

Das of filter material size of opening

... (10.32)

for both types of the openings is kept equal 10 or grater than 2.0.

(5) The grain size curve of the filter material should be roughly parallel to Ihat of Ihe base material.

flFFEClWE STRESS PRINCIPLE

209

(6) To avoid segregation, filter should not contain the particles of size larger Ihan 75 mm. (7) For proper working, the filter material should not contain more than 5% of the fines passing 75~ IS sieve. (8) The thickness and area of Ihe filter should be sufficient 10 c..1rry tbe seepage discharge safely. If the filter has to work as a londed filter, the lotalthickncss should be large enough to provide adequate weight. 10.17. EFFECTIVE STRESS IN PARTIALLY SATURATED SOlLS In partially saturated soils. air is also prescnt along with water. Tn the discussion given below, it is a<>Sumed thnt air is in sufficient quantity such that there is continuity in both the air phase nnd the water phose. Because of meniscus formation, the air pressure is greater than the water pressure. It is assumed thnt the air pressure and water pressure are constant throughout Ihe void spaces. Thus, Ihere are three measurable stresses in a partially saturated, namely. total stress a, pore water pressure u'" and air pressure u". Let us consider the forces acting on the wavy plane X~, shown in Fig. 10.22 (a). The wavy plane passes through the points of contact of solid particles. The wavy plane can be taken as a plane as already mentioned. Fig. 10.22 (b) shows Ihe forces acting on the plnne. From equilibrium in the vertical direction.

c.c

(b)

Fig. 10.22. Pilrtially saturnled soil.

".(a)

aA - IN + u",A .., + u"A" where a = total stress, A total area of the plane, A", = area of the plane passing through water, A" = area of the plane passing througb air, and IN = summation of normal forces acting at the particle to particle contact points.

IN

From Eq. (a),

a -

0'

0 - (j

where

a = effective stress (-

INIA),

A

+

A", U'"

A

+

A" U"

A

+ u""a"" + u,,(1 - aw)

".(b)

a.,. .. A,./A,

and a" - A/A - 1 - a", The area at the points of contact is neglected as it is very small as compared to the area through the water and lhat through the air. Eq. (b) can be written as

a-

0- u,.,a", - u" (1 - Dw) ... (10.32) Eq. 10.32 cannot be verified experimentally as it is difficult 10 measure the area a... Bishop el al (1960) conducted a large number of tests and gave the following equation for the effective ~Ircss in partially saturated soils.

".(1033) where X (pronounced as Ch,) represents the fraction of Ihe area of the soil occupied by water. It pepends mainly on the degree of saturation S (Fig. 10.23). Its value is zero for dry soil and is unity for fully saturated soil. The value of X also depends upon the soil structure, the cycle of welting and drying, and stress changes.

SOIL MECHANICS AND FOUNDATION ENGrNEERING

210

'·0

/'

o· e

/

o·6

/

/

o· 2 o· 0

V

"-

20

40

Of

DEGREE Fig. 10.23.

60

SO

100

SATURATION (5) ~

Vllri~lion

of X with S.

ILLUSTRATIVE EXAMPLES

Illustrative Example 10.1. A sand dC/lOsit is 10 /II thick and overlies a bed of soft clay. The ground wotel' table is 3 m below the ground surface. If d,e sand above the grollnd water table has a degree of saturation of 45%, plot the diagram showing file variation of the total stress, porI! water pressure and tire effective stfJ!SS. The void ratio 0/ the sand is 0.70. Take G = 2.65. Solution

Bulk density,

p ..

(~

:

!e)

p _ (2.65;

p...

~'~7~ 0.70)

x HID _ 1744.12kglm]

'f .. 1744.12 )( 9.81 )( 10-3 .. 17.11 kN/ml

or

For saturated soils. S .: 1.0. and p ..

(2.~')+ +o.~·~O)

)( 1000 .. 1970.59 kg/m2

y • 1970.59 )( 9.81 )( 10- 3 .. 19.33 kN/ml Fig. E 10.1 shows the soil profile. The stresses at section D·B Dnd C·C are as under:

Section ,8-B

a .. 17.11 x 3 .. 51.33 kN/m 2,

a _ 51.33 Section C-C

u _ 0

kN/m2

a - 17.11 x 3 + 19.33 )( 7 _ 186.64 kN/m2

Fig. E to.l.


211

" =7

x 9.81 Fig. E 10. 1 shows the variation of stresses.

= 68.67

kN/m 2,

(J "" 186.64 - 68.67 = 1l7.97kN/m2

Dluslratiye Example 10.2. Delennine the increrue in the effective stress at section C-C ill Illustrative Example 10.1 whe/l the water table is lowered by 5 m. Assume that the soil above the water table has lhe degrte of soturation of 45% thmughout.

° =8 x

Solution.

= 175.54 kN/ml

17.1 I + 2 x 19.33

If = 2 x 9.81 = 19.62 kN/ml, (J = 175.54 - 19.62 = 155.92 kN/ml Increase in erfective stress = 155.92 - 117.97 = 37.95 kN/ml Illustrative Example 10.3. A soil profile consists of a swface layer of clay 4 1/1 'hick (y = 19.5 kNlnl) and a salld layer 2 111 Ihick ('( = 18.5 WI,,!,,) overlying all i/ll/,ermeaMe mck. The water table is at Ihe ground suiface. If the.. wate,. Ie.ve~ il.' a .\·tal/dpipe dr~e/l into the salld layer rise!!.' 2 //I above the ground suiface, draw the plm showmg lhe vanallOlI of 0, /j a"d o. Take Yw = 10 Will". (b) Determille the increase ill effective stress at the top of the mck whe" the arle.fiall head i/l the salld is reduced hy J m. Solution. Fig. E 10.3 shows the prol1lc.

Section 8-8 (Clay),

° = 19.5 x 4 = 78.0 kN/m

2

,

II

= 4 x 10.0 = 40.0 kN/m 3

a = 78.0

Section B-B (sand) Section C--C

- 40.0 = 38.0 kN/m 2 19.5 x 4 "" 78.0 kN/ml

a

=

/I

= 6 x ID.O

= 60.0 kN/ml. a =

78.0 - 60.0 "" 18.0 kN/ml

0= 4 x 19.5 + 2 x 18.5"" 115.0kN/m2

cr

u "" 8 x 10.0 = 80.0 kN/ml, :z: 115.0 - 80.0 "" 35.0 kN/ml Fig. E 10.3 shows the variation of slresscs. It is to be noted that there is a sudden change in u and

a at

leveIB-~

Fig. E 10.3.

(6) When artesian pressure is reduced by 1 m S«tion C-C

0==4 x 19.5 + 2 x 18.5 = 115.0 kN/m2 2 11=7 x 10.0 = 70.0kN/m2 , 115.0 - 70.00 "" 45.0kNlm

a=

Incrc3$e in stress"" 45.0 - 35.0 = 10.0 kN/ml.

MgI!:l~:I~:,ir:;n~:;::';a)~e~!f;Ia~~~ ~~~~k ~~n:iS;~9~f~g~::t;:~/~~:~):fro::'/I~)';/O;I~::~:~ Ii : r~'i~r

(p = 1.925 MglmJ ). The w{Uer table;s at tlr~ IIpper sllr/ace of the clay layer. Determille the effec'ive pressure at variOlfS levels immedialely after placemellt of a surclrarge load of 58.86 tNlml 10 the ground surface.

Solution. Fig. E 10.4 shows me soil profile. Sedion A-A

° = 58.86 kN/m

2

, II

=

o. a = 58.86 kN/m2

SOIL MOCIIANICS AND FOUNDATION ENGINEERING

B8

5S.S6kNlml

H

+ Jm

t

lm

!L B .CLV_ _ _ _"----"---------"-500d ClOy "-C- - - - - - " - - - -

3·5m

1

~

Grovel

"-0_ _ _ _ _-"--- _,1"$

",,).770'.175.11

kN/m2

kNlm2

kNIm2

'·ig. F. 10.4

St.'clio"

l~n

~tioll

C-C

u_ 58.86 -+- :t5 (1.65 x 9.81) .. 115.51 u _ 0, (i .. 115.51 kN/m2 u .. 115.51 -+- 3 )( (1.95 x 9.81) .. 172.90

~ction

1)-0

u .. 3 x 9.81 .. 29.43, 0' .. 172.90 - 29.43 .. 143.47 kN/m 1 a_ 172.00 -+- 3.5 (1.925 x 9.81) .. 238.99

a ..

Il .. 6.5 )( 9.81 .. 63.77, 238.99-63.77 .. 175.22 kN/rnz. lIItl'drulive Example 10.5, A sand deposit consists of two layers. The top layer is 2.5 m thick (p ,. 17()9.67 kg/l1,.f) and the bOltom layer is 35 m thick (PSN = 2064.52 kg/m)). The. waler table is at a depth of 3.5 11/ from 1111.' surface and the zone of capillary saturation is 1 m above the water table. Draw the diagrams, .~/ulII·iI/K 1/1(' vllriation of total Slre.fS, nemral stress and effective stress. ~lll1t1on, Fig. E 10.5 shows the soil profile.

. A

'"rI c

2·5m 2·5m

o

A

'IOPlJII'[A

,

X'"

~'''~~

[';;PIL bMii faAllPATQN '2 9OTTCJ04 LAVER 0

~:"·n .Nlm'

Fig. E 10.5

y .. 1709.67 x 9.81 )( 10-) .. 16.77 kN / m) .. 20.25 kN / m)

Y.", _ 206452 )( 9.81 )( 10-1 0 " 14 ..

uvd II-II (top luyer)

(j ... ()

o .. 16.77 )( 2.5 .. 41.93 kN / m' u .. 0

n ..

41.93 kN / ml

Level 8-11 (Ihlflum layer) 0 .. 16.77 )( 2.5 .. 4].93 kN/ m' u .. - \- )( 9.81 .. _ 9.81 kN/m2

a .. 4}.93 -( -9.81) Lewl ('--C

_ 51.74 kN/m%

o .. 16.77 )( 2.5 + 1 )( 20.25 .. 62.18 kN 1m'

u .0; (j _ 62.18 - 0 ... 62 ,I8kN / m i

EfFECTIVE STRESS PRINCIPLE

213 2

cr = 16.77 x 2.5 + 20.25 x 3.5 = 112.8 kN/m u = 2.5 x 9.81 = 24.53 kN/m 2, a = 112.8 - 24.53 = 88.27 kN/ml Illustrative Example 10.6. A 8 III thick layer of stiff saturated clay (y = /9.0 kN1mJ) is underlain by a layu of sand. The sand is IInder an artesian preSSllre of 5 Ill. Calculate lilt! IIIluimlim depth of the cui that ctur be made wilho,1I causing a heave. Solution. Fig. E 10.6 shows the cut Let H be the required depth or the cuI. Heave will occur when the effective stress a becomes z.cro at level A-A. Level D-D

lJ~L . '. Sond

Rg. E 10.6.

= 19.0 x (8 - 11) = 152.0 - 19H, 1/ = 5 x 9.8 1 = 49.05 cr = 152.0 - 19H-49.05 = 102.95 - 19H = 0 H = 5.42 m. 10.7. A 10 III thick layu of silty clay (p = /864.64 kg;"I"') ove,.lie~· a gravel layer. a

or Illustrative Exam ple The gravel is wuler WI w'lesiclII pl?SSlIre of 12.5 m. II i~' propOl"ed 10 excavate a founda tioll trellch 2 m deep. Detent,ine the facto/" of safety against heaving. (b) What wOl/ld be llle faclOr of safel), against heaving whe/l (/ IIl/iform IJressure of 98./ kNIII/ 2 is applied t() the footillg cQlu·tmcte,/ ill the above trench?

Solu tion. Factor of safcty

=

F =

D~~:~:r~::er:~ ~~~:pt:~ ;~~~:~~:~~~er ]864.64(9:.~lxxl~~\ x

(10-2)

= 1.19

(b) After Construction. Thc downward force is increased due to the uniform pressure of 98.1 kN/m2 F

=

1864.64 (9.8]

;8~O:\;.fO-2) +

98. 1

= 1.99

Illustrative Example 10.8. De/ermine the factor of safety againsl heave f(lilure in the hydraulic structure showlI ill FiR. Ex. 10.8. P = 1850 kgll// Solution. Average pressu re on the base of soi l prism, h" = 0.42h

= 0.42

x

~

= 3.36m

y' = 1850 x 9.81 x IO-J - 9.81

From Eq. 10.26, factor of safety, F = F

= 8.34

kN/m 2

~~ ~,

= 9.~~4x X3.;6 = 1.27

IDustrative Example 10.9. Determine the apl'fflximate Um its oftheft/ter material requiredfor Ihe soil of the bqse material which IUl.\· D,!o = 0.0/ mm alld DS5 = O. /0 mm. and the grading curve as shown ill Fig. E 10.9. From Eq. 10.28, Dl~(/) > 5DI!I(b)

Solution.

DIHI) > 5 x 0.0 1 > 0.05 mm

SOIL MECHANICS AND FOUNDATION ENGI NEERING

214

From Eq. 10.29,

00

II

90

V //

V /

O.10mm 80 70

/

0

Bose material--! .0

/

0

/

/

10 0 0.001

7

V /

/

//

Filter+/ j" /

Y

/

J

J

/ //

V'!' // II

/ /,6

kim mm )I"6.05mm / ~/o.5mm / / 0,1 / 7 0,01 Particle

10,0

size (mm) _ _ fig. E 10.9

Therefore, DIS of filter material should lie between 0.05 mm and 0.5 mm. As the gradation curve of the filter matcrial should be roughly parallel to that of the base material, the hatched portion indicates the limi~ of the material suitable as filler. IIIustralive Example 10.10.1[ excavation is carried out in a soil Wilh a porosity of 0.40 and the specific gravity of solids 0/2.65, determine the critical gradient. A 1.50111 layer of the soil is subjected to an upwlJld seepage head of 1.95 m. What depth of coarse sand would be required above the soil to provide a foeror tJ safety of 2.50 ? Assume that sand has the same porosity and sp. gr. of solids as the soil. Solution.

e -

6 - ~:: - 0.667

'"


eritreal gradient Sntumtcd density

~ Q.::.l ~ ~ ~ 099 I + e

I + 0.667

.

t~ : ;)

p,,,,:-

p"

265+0667) x 10= 1999/ml(1952kN/m) ,

=

~

Let x be the depth of sand layer Effective pressure at bottom Upward pressure at bottom Now factor of safety, FS.

=

(1.50+ x)(19.52 - 9.81)

= (1.50 + x) 9.71 = 1.95 x 9.81

_ (1.50 + x) x 9.71 = 2.50 1.95 x 9.81

. . . = 3.42m PROBLEMS A. Numerical 10.1. Determine Ihe 101al, neulml and ellcctive stresses at the bottom of the deposit shown in Fig. P 10.1. IAns. 199.14.83.39,115.75 kN/m2) G5.

l in

j':1·6g/ml

W.T.

io

3m ____'_._2_.0~g_/m_'________ 205m

,_1·70 g/ml

1m ==f;P=;:IB51i,,jqjzlmiii[:,===: 2m __~~~-~I.~85~g~l~m~I_______ Fig. P 10.1

10.2. The water table in a deposit of uniform sand is located at 2 m below the ground surface. Assuming the soil above the water table is dry. determine the effective stre~s at a depth of' 5 m below the ground surface. The void ratio is 0.75 and the specilic grnvity of solids is 2.65. {b) If the soil above the water table is saturated by capillary action, what is the effective slress at the thai depth? (Ans. 57.43 kN/m2; 65.83 kN/m 2j

IS

10.3. A deposit of fine sand ha.~ a void ratio of 0.54 and the specific gravity of wlid particles is 2.67. Compute the safe exit gradient, with u factor of safety of4. · [ADS. 0.271) 10.4. A deposit of silly clay lies between two layers of sand, as shown in Fig. P lOA. The lower sand layer is under anesian pressure of 4 m and the water level in the upper sand layer is 2.0 m below the ground surface. Detemline the effective stress at the bottom. "' .t6 kNlml t7 ~ 2,; (h) Also. determine the head above G.S. that would .... 20kN/ml cause heaving at the base of Ihe clay. [Aos. 32 kN/m 2,7.2 m] 10.5. The porosity of a sample of ~al1d in the lOOSe stale was --lLL___ 54% and in dense stale, 38%.· Find out the critical hydraulic gradient in hath the slates if the specilic gravity of the wil grJin was 2.60. Also lind out the Fig. P 10.4.

-------l~1

___y_"_"_'_N'_""______

'--fr

216

SOIL MECIIANICS AND FOUNDA110N ENGINEERING 3 3 3 S!lturated densili(..'S in kg/m . [Ans. 0.736; 0.992; 1737.33 kg/m ; 1992.56 kg/m 1 10.6. If 40% of the excess hydrost!ltic pressure is lost in flowing through soil II which has a coefficient of permeability of 0.05 cm/sec (Fig. P 10,6). determine the discharge veloci ty and seepage velocity through each soil and the hydraulic head at which instability occurs. IAns. Soil A : 0.02 cm/scc, 0,06 cm/scc; Soil B : 0.02 cm/scc. 0.053 r.m/SCC; 73.33 em]

Fig. P 10.6.

B. Descriptive Dnd Objective 'I)'pe 10.7. Define lolal slress, neutral stress and effective stress. What is the importllnce of the effective stress? 10.S. Prove that the effective stress (0) for a saturated soil can be expressed as (j _

o-u

where 0 ~ 10lal stress, II = pore water pressure. 10.9. Whal is the effect of surcharge and the capillary oclion on the efTective stress ? 10.10. Prove that rhe seepage force per unit volume is equal 10 the product of the hydraulic gradienl and the unit weight ofwalcr. 10.11. Discuss two different approaches of ronsidering the equilibrium in seepage problems. Which approach is more convenient nnd why?

10.12. What is the effect of the seepage pressure on the effective stress? Give examples. 10.13. Whot is quick SlInd ? How would you calculate lhe hydraulic grodienl required to create quick sand conditions in iI sample of sand ? 10.14. Explain the mechanics of piping in hydraulic structures. Wlult methods ore used 10 incrCllse the foCtor of safety against piping?

10.15. Why a filter is used on the downSircam of an carlh dilm ? How would you design a filter? 10.16. What are two different types of piping failures? Explain with the help of sketches. 10.17. What is effeCtive stress principle? 19.IS. Wrile whether Ihe following statements are correct. (a) The effective stress is the stress at the points of contact of the soil partidcs.

(b) 1lte effective stress stress CIIn be measured directly in the field. (c) The effective stress is equal to the lotal stress minus the pore water pressure. (d) The rise of wnter due \0 Cilpillnry action reduces the efTeclive stress. (e) The shear strength of n soil depends upon ilS effective stress. (J) In partially s:llurnted soils, the pore nir pressure is more thnn the pore waler pressure. (g) Quick sand is a type of sand.


217

(II) Quick sand cunditlons can develop even in gravel. (i) Backward-eroslo n piping is amenable to the thl.'Oretical solution. (j) Heave piping oceurs on the downst ream of the pile foc a distance of half the depth of the pile. (.1:) The piping cfln be checked by provid ing fI 100ldcd filter. (I) 'The gmded tiller consists of a layer of pervious materials which is well-graded. {m) The diumcler of the openings between particles is equnl to ubou t Dl.~ size of the panicle. (II) The graded tilter checks th e now of water. (0) Piping is the same ru; quick sand. (p) The effecti,,!! stress at various poinlS increases due to rise in the water table. (q) The effective stress in a fully submerged ,oil depends upon its submerged unit weight. (AliS. True, (c), (el. fJ) , (11). fJ). (.1:), (q)]

C. Multiple Choice Questions 1. The c ritkal grudicnt of a soil increases (tI ) with increase in void r.. tio (b) decrease in void ralio (e) decrease in spcdtie gravity (d) None of above 2. The ellit gradient is equal 10 the ratio of (a) total head to IOtal length (b) slope of now line (e) slope of equipotential line (lI) head loss to length of flow field at exit. 3. The effective st ress i."> (tI) actual contact stress (b) an abstract quantity (e) eljual 10 total stress (el) None of above 4. The em.:Clive stress "ontmls Ihe following properties of soils; «(J) Shear strength {b) Compressibility (el Permeability (d) All Ihe above S. Quick sand is (a) II type of sand (b) II condition in which II cohcsiunles~ soil lose~ ils strength because of upward now of water. (c ) a condition in which i.I cohesive );Qi] loscs its strength. (el) none of above. 6. The seepage pre5Sure is proportional to (a) hydraulic gradient (b) unit weight or wllter (e) length oflhe specimen (d) all the above. 7. A deposit of tine s,md has a porosity /I and spccilie gravity of solid is G. The crilical gradienl is equal to (tl) (a- I ) {I +n) (b) (0- 1) (I-n) (c) (G- I)/(I +n) (t!) (G- I )I(I-n) 8. For a void ratio of 0.60. the relntionship between lhe specific grnvi ty o f solids (a) and the hydraulic gradient (i) for the quick snnd condition is (a) a = 0.6 i + I (b) a = i + 0.6 {e) a = 1.6 i + I (d) a:: 1.6 i - I 9. For a soil deposit having /I :: 33% find G = 2.60. the critical gradient is

wIn

W lill

(e) 1.07

(lI) 1.10 ~_lm2~ 1m ~~~m~~~ ma~~~

11 Stressees Due to Applied Loads 11.1. INTRODUCTION Stresses are induced in a soil mru;s duc to weight of overlying soil and duc to the applied loads. These stresses are required for the stability analysis of the soil mass, the settlement analysis of foun<1.1tions and the determination of the earth pressures. The stresses due to self weight of soil have been discussed in chapter 10. These stresses are summarised in Section 11.3. lbe rest of the chapler is devoted to the determination of stresses due to applied loads. The slreSSeS induced in soil due to applied loads depend upon its stress-strain characteristic. The stress-strain behaviour of soils is extremely complex and it depends upon a large number of factors, such as drainage oonditiollS, water content, void ratio, rate of loading, the load level. and the stress path. However, simplifying assumptions are generally made in the analysis 10 obtain stresses. It is generally ru;sumed that the soil mass is homogeneous and isotropic. The stress-strain relationship is as.
(£) and Poisson's ratio (v). The modulus of elasticity can be determined in the laboratory by conducting a triaxial compression test (see Chapter 13). The stress-strain curve is plotted between the deviator stress (0) - oJ)' and the axial strain (£1). An unoonsolidated-undrained (UU) or an unconfined compression test can be performed for saturated, cohesive soils. A consolidated drained (CD) is usually conducted for cohcsionless ('1-.,.) soils. The value of modulus is generally laken as the

~).t ~::~tli~e:,t ~e: ~ ~~ ~m~~~~~

(<1j-Gj)1

- - - - - - - - - - - - - -- - -

initial tangent modulus or the tangent modulus at 1/2 to 1/3 of Ihe peak stress is also used. The value of Poissoo,'s ratio (v) for an elastic material varies from 0.0 to O.SO. For undrained conditions, the value of Poisson's ratio is 050. For

=::

::~it:~ ~:C~;:~~ ~~!~ri~l.l: ~~eO~~O~ +(~-~)f

outside the elastic range of 0.0 to 0.50 is also occasionally enoountered. It is difficult to ascertain the exact of Poisson's ratio. Fortunately, the effect of

value

SKont mod.Ilus

srRESSES DUE 'I'O APPLIED LOADS

219

Poisson's raUo on the oomputed stresses is not significant and an approximate value can be used without much error. Tubles 11.1 and 11.2 give typical mnge of values of modulus of elasticity and Poisson's mtio, respectively, for some soils. Tllble 11.1. 1)rplcal Values of E

E

Type of SQil

S. No. 1.

Sorraay

2

Hard day

3. 4.

Silty Sand

Loose Sand Dense Sand Dense grovel

S.

6.

MN/m

kN/m2

1.5-4.0 6.0-15.0 6.0-20.0 10.0-25.0 40.0--80.0 100-200

1500-4000 6000-15000 6000-20000 10000-25000 40000--80000 to 2 x lOS I x

105

Thble 11.2. 1)rplcal Values of Poisson's Ratio (v) Type 0{ Soil SatUnlted clay

1. 2. 3. 4. S.

0.4-0.5 0.1-0.3 0.3--0.35 0.30--0.50 0.20-0..30

Uns.oturuted clay Silt

Loose ",nd Dense sand

11.3. GEOSTATIC STRESSES

The method for the determination of total vertical stresses duc 10 self weight of the soil have been discussed in. chapter 10. The stresses due to self weight of soils are generally large in comparison with those induced due to imposed loads. This is unlike many other civil engineering structures, such as steel bridges, wherein the stresses due to self weight are relatively small. In soil enginccring problems, the stresses due to self weight are Significant. In many cases the stresses due to self weight are a large proportion of the lotal stresses and may govern lhe design. When the ground surface is horizontal and the properties of the soil do Dot change along a horizontal plane, the stresses due to self weight are known as geostalic stresses. Such a oondjtion generally exists in sedimentary soil deposits. In such a case. the stresseS are normal to the horizontal and vertical planes, and there are no sheariog stresses 00 these planes. In other words, these planes are principal planes. The vertical and horizontal stresses can be detennined as under. (a) Vertical stresses. 1be vertical stresses are detennined using the methods described in chapter 10. Let us oonsider the horizontal planeA---A at a depth l below the ground surface {Fig. 11.2 (a)]. Let the area of croS.IH;ection of the prism be A. If the unit weight of soil (y) is oonstant, the vertical stress (oJ is equal to the weight of soil in prism divided by the area of base. Thus Oz _

or

weiS!!.'! :;~prism

_

I (z AX A)

...(B.l)

IT the soil is strotified, having n layers of thickness l1o~ .... lll' with unil"weight 't1l '12) ... "t", the vertical stress is given by

-E,

... (11.2)


':>il

increI:se;a~~~1 n~e:~~~~e gi:n~:~~~ ~~: t~e~I:~tYwe~~h:~f :~:: -""''''''''''''''1''''''I_,.""",""",,,,G~.S~... above. In such a Cilse. the unit weigh! of soil cannot be taken as constant. In this case, the weight of soil in the prism is given by

w=j

yA dl

o

where dl is the thickness of a small strip of soi l al dcpth z. Therefore, the vertical stress is given by

f

lV

0

a ;: =A"== -

j ydz

a t ==

yA d,

Afbi

... ( 11.3)

o If the soil is stratified and also has a variable ani! weight, the vertical stress is given by ;1

'-1

Fig. 11.2. Gcostatic Slress.

...

J11 til + JY2 til + ... + JYndl

a;: =

... (11.4)

o 0 0 (b) Horizontal stresses. The horizontal stresses (ax and a,.) act on vertical planes, as shown in Fig. 11.2 (b). The horizontal stresses at a point in :I soi l mass are hig'hly variable. These depend not only upon the vertical stresses, but also on the type of the soil and on the conditions whether the soil is stretched or compressed laterally. In the treatment that rollows it wou ld be assumed that a.r = 0,. The ratio of the horizontal stress (0 ..) to the vertical stress (oz) is known as the coefficient of lateral stress or lateral stress ratio (K ). Thus

0,

K==~

=

or 0 ... Ko:. ... (11.5) In natural deposits. generally there is no lateral strain. The laleral slress coefficient ror this case is known as the coefficient of lateral presS/tl"e al rest (Ko). The value of its coefficient can be obtained from the theory of elasticity, as explained helow. In retaining structures (chapter 19). there is either stretching or contraction of soils and the value of K is different. The strain in x-direction is given by (see any text on theory or elastici ty or mechanics of materials) £.. ==

For conditions of no lateral strain,

E~,

i

[o.o:-v(o ,.

+ O~)]

= O.

Thus o.r = Taking u... = 0 ,. and simplifying , O.,(l-v) =

V

It

(0)" + OJ 0t

0,=(,:,)0, Os

where

= Koot

Ko =~

... (11.6) ... (11.7)

srRES.'iI!S DUE TO AI'I'LlE!) WAilS

221

The vulue of Ko can be obtained if the Poisson's ratio v is known or eslimated. Eq. 11.7 is not of much prnctiCl I ~ as the soil L'i not a purely elastk: material and it is dHncult to estimate the Il()i.'~,s()n r.uio. The value of ~ is detcnnined from actual measurement or soil prcs.,>ure or from experience. For :J sedimentary sand deposit, its value varies from 0.30 10 0.6, and for a nonnally consolidatctl d:lY, iL.. value generally lies betwccn 0.5 and 1.10. Table 11.3 gives the average values of Ko for diITcrcnl types of soil... Juky's fonnulu is oommonly used, according to which Ko • I - sin
lhhle 11.3. Values of Laterul S. No.

.·n.... . . u!"e Coemclent ut nest (Ko)

7'ypt a/Soil

K,

Loose sallil $llnd

0 5 -0.(0()

J)ellS~

11.:\~ )5U

3.

Cluy(tlrai/~)

11.5 - 0 .(,0

4.

Clay (UIIl/rail/cd)

(j.HO- I . 1

5.

Ovt>r< OIlSoIitltJlcd day

I.

2.

I .U_ :\.U

11.4. VElrnCAL STnl!:SSES nUE TO A CONCENTltATEU U)A.I) Boussincsq (pronounced as l]oo..si. ncsk) gllvc the theoretical solutiulls fur the s tres.... dis trihu tion in an elustic medium subjected 10 a !;onccnlmtcd I()tll.! on its surfllcc. 'I1m solutiun... :lrc !;ommonly used to obtain the st~ in a soil mass due to cxl!;mlJly applied loaUs. '1l1C folllW{ing :l':.... umplio{ls arc m:Klc. (/~), i.e., the ' ratio between the stres.... and stmin is constant. (2) The soil is homogeneous, i.e., it has identical propcrtk... at clilTercnl points. (3) The soil is isotropic, i.e., il hu... identical properties in all direction.... (4) The soil mass is semi·infinite, i.e., it extend" to infinity in the downward dirCl..1ion ;uw lateral directions. In other words, it is limited on ils top by .1 horizont:ll plane and eXlcnds 10 infinity in all other directions. (5) 111e soil is weightless and is free from rcsiclulll strc.-.scs before the 'I[lplicltiun (If the kXltl. [Note. The stresses due \0 self weight are oomputed scp
(I) 1be soil mass is :In clastic continuum, having a amslant valuc of modulus of clastldty

... (11.8) where R ::: polar distance betwccn the origin 0 and point I'. p '" angle which the line 01' makes with the vcrticul.

R • V~

Obviously,

+ 1 +?

R_~ and

sinp ... ,IR

The vertical stress

(o~)

and

,'l.2.i cas.p .. zl/l

where

at point P i.. given by

I·i~.

11.:\. SIn:S"CS ,Juc

In

a l1lflCCl"lllllted

1(\111


222

az-~~ .~

or

oJ ..

3Q

Zit .

(Z/R)' 3Q i' ----nr .. 23t . If

a,-~ . ?~

~ ·l· [(? :

a, -

a,_~.!,. ~.

or

az - 18 .

where

f8 ..

r

i')"

1

I

.•. (11.9)

[I + (d')'J>'

~

... (11.10)

3 ... (lLl1) 2lt[1 + (rlZ)2]~ The coefficient 18 is known as the Boussinesq influence coefficient for the vertical stress. The value of 18 ron be delennincd (or the given value of rtz (rom Eq. 11.11. The oomputed values are tabuJ31ed in Table 11.4.

Table 11.4. Values or Boussinesq's Coefficient (In)

,I, 0.00 0.05 0.10

18

,I,

0.4775

1.05

0.4745

0.15

0.4657 0.4516

o.w

0.4329

1.10 1.15 1.20 1.25

0.25 0.30 0.35 0.40 0.45

0.4103 0.3849

1.30 1.35

0.50 0.55 0.60 0.65 0.10

0.75 0.80 0.85 0.90 0.95 1.00

0.3577 0.3295 0.3011 0.2733

0.2466 0.2214 0.1978

0.1762 0.1565 0.1386 0.1226 0.1083 0.0956

lAO

18

,I,

0.0745 0.0658 0.0581

2.05 21.0 2.15

0.0513 0.0454 0.0402 0.0357 0.0317

l.45 150 1.55 1.60

0.0282

1.65

0.0179 0.0160 0.0144 0.0129 0.OU6 O.otOS

1.70 1.75 1.80 1.85 1.90

1.95 2.00

0.0251

0.0224 0.01Il0

0.0095 0.0085

18

,I,

18

2.W

0.0071 Q.(XJ70 0.0064 0.0058

2.25

0.0053

2.~

0.0048 0.0044 0.0040

4.50

0.0002

4.75 5.00

0.0002

0.0037

10.00

0.0000

2.35 2.40 245 250 255 260 265 210 275 280 2.85 290 295 3.00

3.25 350 3.75 4.00 4.25

0.0011 0.0008

0.0005 0.0004 0.0003

OJJOOI

0.0034 0.0031 0.0029

0.0026 0.0024

0.0022 0.0021 0.0019 0.0018 0.0016 0.0015

0.0844

The following points are worth noting when using Eq. 11.10. (1) The vertical stress does not depend upon the mcx1ulus of elasticity (E) and the Poisson ratio (v). But the solution has been derived assuming that the soil is linearly elastic. The stress distribution will be the same in all linearly elastic materials.

STRESSES DUE TO APPLIED WADS

223

(2) The intensity of vertical stress just below (he load pc>int is given by

0, • 0.4775 ~

... (l1.12)

(3) At the surface (z = 0), the vertical stress just below the load is theoretically infinite. However, in an actual case, the soil under the lond yields due to very high stresses. The load point spreads over a small but finite aTCa illld, therefore, only finite stresses develop. (4) The vertical stress (oJ decreases rapidly with an increase in r/z ratio. Theoretically, the vertical stress would be zero only at an infinite distance from the load point. Actually, at r/z = 5.0 or more. the vertical stress beoomes extremely small and is neglected. (5) 10 actual practice, foundation loads are not applied dirccUy on the ground surface. However, it bas been established that the Boussinesq solulioo can be applied conservatively 10 field problems concerning loads at shallow depths, provided the distance z is measured from the point of application of the load. (6) Boussinesq's solution can even by used for negative (upward) loads. For example, if the vertical stress decrease due to an excavation is required, the negative load is equal to the weight of the soil removed. However, as the soil is not fully elastic, the stresses determined are necessarily approximate. (7) The field measurements indicate that the actual stresses are generally smaller than the theoretical values given by Doussincsq's solution, especially at shallow depths. Thus, the Boussincsq solution gives conservative values and is commonly used in soil engineering problems. Limitations of Bousslnesq's Solution. The solution was initially obwined for determination of stresses in elastic solids. Its application to soils may be questioned, as the soils arc far from purely elastic solids. However. experience indicates that the results obtained are satisfactory. The application of Boussinesq's solution can be justified when the stress changes are such that oruy a stress increase occurs in the soil. The fC.'11 requirement for use of the solution is not that the soil be elastie (Le., fully recoverable), but it should have a constant ratio between stress and strain. When the stress decrease occurs, the relation between stress and strain is not linear and. therefore, the solution is not strictly applicable. If the stresses induced in the soil arc small in comparison with the shear strength of the soil, the soil behaves somewhat elastically and the Boussinesq solution can be used. For practical cases, the Boussincsq solution can be safely used for homogeneous deposits of clay, man-made fills and for limited thickness of uniform sand deposits. In deep sand deposits, the modulus of elasticity inaeases with illl increase in depth and, therefore, the Boussinesq solution will nOI give satisfactory results. In Ihis case, the assumption of proportionality between stress and strain cannot be justified. For such a case, non-linear elastic solutions or elastie-plaslie solutions arc required. The point loads applied below ground surface cause somewhat smaller stresses than are eaused by surface loads, and, therefore, the Boussinesq solution is nol strictly applicable. However, the solution is frequently used (or shallow footings, in which z is measured below the base of the footing.

11.5. nORIZONTAL AND SnEAR STRESSES DUE TO A CONCENTRATED LOAD The method for detenninmion of the vertical slress (0.:) has been discussed in the preceding section. In most soil engineering problems, only the vertical stresses are required. Occasionally, other stress components (on 0.1' .. and ""'.J'l:) are also required. 1bcsc compooenlS can be determined as follows: Fig. 11.4 shows an elementary stress block, indicating all the stress components. In all there are 9 stress components. namely. 0 .... 0,.. 0D "t..,.. ""'.co ~... ~"t"u and "t"Z)" However, the moment equation gives the following relatiOf\$.

t.,.

""'..,. -"Ssi

~ - "t.l)'i

""tu -"t".r:t.

".)01."=

and. therefore independent unknown components are only six 0 .... 0,. 0:. "..,., The equations for determination, of 0.: have already been given. The corresponding equations for other components are :

o .1Q. [~. •

2x

R'

('-2Vl,{ __' __ (2R ,)J' _.£}] 3 ' R(R.,) R'(JI.')' II' +


224

[D.

~ If

°Y·2n

u_

,

~~R+

+

i!..=M { R(R+z) 1 2

:)4: -~}]

_~[:!..l!_(l-2V)~ 1 2"

;r:y

'" _

If

:.

Jf(R+z)2

~.~

... (11.13) ,

Fig. 11.4. DlfJeRlnt SIRl" Components.

It may be noted that

o)(OOy

and

"t..,.

depend upon Poisson's ratio.

Cylindrical Cooroinates. Sometimes, it is more convenient to use cylindrical coordinates (r. e, z) instead of cartesian coordinates (r, y. z). The Boussinesq solution in lerms of cylindrical coordinates IS as under (F;g.11.5).

Fig. 11 .S. Cylindrical Coordin.1tes.

3Q

Vertical stress,

a, - OX·

Radial stress,

a

'

I

~

_.£[ 3%..' - ~l R'

2l<

Tangential stress,

ao - f:;(1 -2V) [R(R\ z) -

Shear stress,

"-ox'/1

3Q

Shear streSses where R ... ,,; +

"C..e -

l-.

as before.

..i'

'tze ... 0

.•. (11.14)

R(R + z)

~1

Sl1U!SSES DUE TO APPlIED LOADS

225

11.6. ISOBAR DIAGRAM An isobar is n curve joining the points of equal stress intensity. In other words, 3n isobar is a a)ntour of equal stress. An isobar is a sp::!tial curved surface of the shape of an electrical bulb or an onion. The curved . surface is symmetrical about the vertical axis passing through the load point. The isobar of a particular intensity can be obtained using Eq. 11.10. 'fhc calculations are shown in a tabular form. Table 11.5 gives calculations for an isobar of intensity 0.1 Q per unit area. Q

From Eq. 11.10,

0, -

18

~

Taking oJ - 0.1 Q.

~

0.1 Q - 18 . 18

=

0.1 z'

... (a)

For different depths z, the value of fs is computed from Eq. (a), as shown in the second row of 'Thblc 11.5. The values of rlz for computed values of IIJ are obtained from Eq. 11.11 or Table 11.4. Once the values of rlz have been determined, the radial distance r can be obtained os shown in table. It may be observed that r is zero at the load point. and it attains a maximum value at rlz = 0,75 and again decreases. As the isobar is symmctrical about the load axis, the other half can be drawn from symmetry. The shape of an isobar approaches a lemil1SQlte curve (not

Tnble 11,5. CalculaUons for Isobar of 0,1 Q 0.75

2.00

18

0.25 0.00625

0.50 0.25

1.25 0.1562

150 0.225

1.75 0.3062

,I'

1.50

0.05625 1.16

1.0 0.10

2.16

0.93

0.75

0.59

0.44

0.400 0.27

0~'i4

0.75

0.87

0.93

0.938

0.885

0.770

0.540

lkplh

%

2185 0.4775 0.000 0.000

Isobars are useful for determining the cITccI of the load on the vertical stresses al various points. The zone within which lhe stresses holve a significllnt eITect on the

sctt1cm~nt

pressure bulb. It is generally assumed that an isobar of 0.1 Q forms

of structures is known

(IS

the

a pressure bulb. The area outside the

pressure bulb is assumed to have negligible stresses. 11.7. VERTICAL STRESS DISTRIBUTION ON A 1l0RlZONTAL I'LANE The vertical stresses at various points on a horizontal plane at a particular depth z can be obtained using Eq. 11.10. Let us determine the stresses at a depth of 2 m. Therefore, 0, - I, . (Q/z') - 18 . (Q/4) - 0.2518 Q. The value of 0: are computed (see Table 11.6) for different values of rlz, after obtaining 18 from Sq. 11.11 or Table 11.4.

'Jable 11.6. Calculutlon for vertical stress at l = 2m

'I' I,

'"

050 0.25 0.4775 0.1194 Q

0.4103 0.1026 Q

1.00 050 0.2733 0.0683 Q

1.50 0.75

2.00

250

3.00

0.1565

1.00 0.0844

1.25 0:0454

1.50 0.0251

4.00 2.00 0.0085

0.0390 Q

0.0211: 0

0.01130

0.0063 Q

0.0021 Q


I

'

0· 00210

Fig. 11.7. Venical Stress on a horizontal p1:me.

Fig. 11.7 shows tbe vertical stress distribution diagram. The diagram is symmetrical about the vertical axis. The maximum stress occurs just below the load (r = 0), and it decreases rapidly as the distance r increases. The vertical slress distribution diagram on a horizontal plane can also be obtained graphically if tbe isobars of different intensity are avai11ble. The horizontal plane is drawn on the isobars diagram. The points of intersection of the horizontal plane with the isobar of a particular intensity give that vertical stress. 11.8. INFLUENCE DIAGRAMS An influence diagram is the vertical stress distribution diagram on a horizontal plane at a given depth, due \0 a unil concentrated load. In Fig. 11.7, if the concentrated load Q is taken as unity, the diagram becomes an influence diagram. The influence diagrams are useful for dctennination of the vertical stress at any point on that hOriz?"tal plane due to a number of ooncentrated loads applied at the ground surface.

'"

x

"

dO'

"'B-'e".

GeC~bce

~

I~: "t ~,1 " xox~;'x i I

' I

~

_~~~--~~~--_.. .........

ow

'

___ m

__

Fig. 11.8. Influence DllIgnlms.

Fig. 11.8 shows three influence diagrams, mark.ed II> 12 and T). due to unit loads applied at three points A', C' and 8' on the ground surface. The stress at any point A on the horizontal plane at depth z due to three loads Qb Q2 and Q3 is given by (O:).~ - QIOM + Q20AB + QJ 0AC

where and

... (11.15)

vertical stress at A due to unit load at A I 0AB = venical stress at A due to unit load at B ' 0AC = vertical stress at A due to unit load at C I OM ""

The values of

OM , 0AB

The computation,

work

and 0AC can be abtained from the influence diagram IL' 13 and h. is coosklembly _simplified using the reciprocal theorem, according to which

0AB - OBI!;> OAC - 0CA and 0BC - 0CB wbere the first suffix denotes the point where the stress is required and the second suffix gives the point above which the load is applied. Acoordingly, Eq. 1l.15 can be written as

...(11.16)

SfRESSES DUE TO APPLIED WADS

Zl.7

where OM = yertical stress at A due to uoit load at A oll!\ = vertical stress at D duc to unit load at A = vertical stress at due to unit load at

and oCA,

C

A

Therefore; there.is no need of drawing three influence diagrams in Ihis case. Only one influence diagram (11) with unit load at A' is sufficient. The values of of« and 0CA are determined from II diagram below the load points B' and C'. If the stresses al any other point, say point D, are required, then the influence line for load above that point (B' in this case) would be drawn. Alternatively, the influence line diagram II can be lraced on a paper and placed in such a way that its axis of symmetry passes through the point B'.

11.9. VERTICAL STRESS DlSTRIDunON ON A VERTICAL PLANE The vertical stress distribution on a vertical plane at a radial distance of r can be obtained using Eq. 11.10. In this case, the radial distance r is ronstanl and the depth changes. The values of r/Z are obtained for different depths l. The values of IB are obtained from Eq. 11.ll or Table 11.4 and the stresses computed as

Oz .. (IBIz?) Q. Table 11.7 shows the calculations for vertical stresses on a vertical plane al r ::: 1 m. Table 11.7. Calculutions of Vertical stresses ut r = 1m

,I, 18

0.25

0.50

1.00

1.50

2.00

2.50

4.0

2.0

1.00

0.667

0.50

0.40

0.20

0.0004

0.0085

0.0844

0.1904

0.2733

0.3294

0.4329

0,0064 Q

0.0340 Q

0.0844 Q

0.0845 Q

0.0683 Q

0 .0527 Q

0.017 Q

5.00

Fig. 11.9 shows the variation of vertical stresses on a vertical

~~,a~d~h~ ~~p~,e ;:~': t~~hS:;.:~.P~~t~~/~n:~:ll~::~~! --."...,,""'....,..t'"~'m"'I-"""_.k_ vertical stress first increases and then decreases. The maximum vertical stress occurs at r/z = 0.817. This rorrcsponds to the point of intersection of tbe vertical plane with the line drawn at 39°15' to the of load. vertical

axis the

11.10. VERTICAL STRESSES DUE TO A LINE LOAD The vertical stresses in a soil mass due 10 a vertical line load can be obtained using Boussinesq's solution. Let the vertical line load be of intensity q' per unit length, along the y-axis., acting on the surface of a semi-infinite soil mass, as shown in Fig. 11.10. Let us consider the load acting on a small length Oy. The load can be taken as a point load of q' Oy and Boussincsq's solution can be applied to determine the vertical stress at P (ol; y, z). From Eq.

,

Fig. 11.9. Stress on a venica1 plane

11.9, 60 _ 3(!~y) z

~. (r' + The vertical stress at P due to the line load extending from

... (0)

i')" .....QO

to +

00

is obtained by integration,

3t/i"-~

a, -

or

zn L

(r'

+

3q'i"a, - z;;-

f_

i')" dy

(x'

+

y' + 1)"

... (b)

SOIL MECHAN ICS AND FOUNDA11ON ENGINEERING

228

Fig.ILlO

Substituting ~

+ z?

= ,,2 in Eq. (b),

O"~

Let )'

= II Inn O.

Therefore. d)' =

3q'Z3

II

•

2n

cr.=~

Lei sin 9

= I.

<1\, + I)¥.

(I"

r 0

u.s sec's O

1

cos~9d9

.•. (c)

sect 9 dO.

a. = ~

Eq. (c) can be wrillen as

J-

2n _

."

2 1tu~ 1l1crcfore. cos 9 dB (It

0

us«l a dO

... (d)

=

Eq . (d) becomes

~ 4

cr~

=

cr1.

=~ 1t1l"

'u

J:

[1- -11"]' 3

a , = 3q'z? x!. 1tli~

0,

=

2.~'

( I - ,2) dt

=

o

~ 1t (Xl

+

[I ;xIA +

(2)1

... (11.17)

Eq. 11 .17 can be used to determine the vertical stress al point P. When the point P lies vertically below the line load. .r o.

=

Therefore. lhe expressions for lhe slresses

a~ 0'..

= '-'L 1tZ

. .. (11.18)

and t.o;: can be obr.aincd in II similar manner, starting from Eq. 11.13.

STRESSES DUE TO APPLIED LOADS

.l~

2q'

Ox ""

and

""""it (.i + zh 2 2q' xi'

'n-,,·~

...(11.19) ... (11.20)

11.11. VERTICAL STIIESS UNDER A STRIP LOAD The expression for vertiClil stress nt any point P under a strip load can be developed from Eq. 11.17 of tho tine load. The rxpression will depend upon whether the pint P ties below the centre of the strip load or nOl.. Note. The length of the strip is very long. For convenience. unit lcngth is considered. (1) Point P below the centre of the strip Fig. 11.11 shows a strip load of width B (= 2b) and intensity q. leI us consider the 10.1<1 acting on a small elementary width dx at a distnnce x from the centre of the load. This small load of q dx can be con-,idcrcd as a line toad of intensity q'. From Eq. 11.17,

.fIa- 11.11. Strip I...oIId, point P below ocnlre.

-~[--I-l' 1 + (xl::)2

6u, nz The stress due to entire strip lood is ootained as a _!9. ,

LetX/Z-tanlL

1U

f" __1 _ _ d< .... [1 + (xl:)2)2

1'hcref«e,dx=zsec'ludu

~

~

Substituting in Eq. (a), a .. )( 2 ' ':Ill Jo (1 + lan2 u}2 tb.4 where a = tan-I (bIz) = nngle made by extemilies of the strip at P.

or

a, ..

~;cos2udu

u, -

~

n

J.

f: C ~2u) '" +

···Ca)


or

0:

-! (29

+ sin29)

...(11.21)

(2) Point P not below the centre or the strlp Fig. 11.12 shows the case when the point P is nOi below the centre of the strip. The extremities of the strip make angles of ~l and ~2 at P. As in the previous case, the load q dx acting on a small length dx can be coMidered as a line load. The vertical stress al P is given by Eq. 11.17 as

I:===':~

Fig. 11.12 Strip load, point P noI al the oentre.

2qdx [

/l

0: - ---;u1 + Eq. (a) is simplified making the following substitution,

dx_zscc 2

x_ztanp

!!J. o. _ 2 q (2 sec2~) dfJ • ox

or

/lO: -

Integrating,

0, •

]'

... (a)

pllP

Therefore

¥

1

(x/Z)2

2 cos

[ 1 +

~dJJ

~n'~ r

" • ,1 (I + =2~) all " p + ~ sin 2p ]PI

'1. f

-*I 0:

Substituting ~2

-

-! [(~-~J

+

(sin~2cosf12

-

sin~ICOSPI)]

PI - 2 e,

...(b)

STRESSES DUE TO APPUED LOADS

231

If (131 + ~ ., 2
sinP2cosj32 - sin PI oos PI .. sin28cos2q>

Therefore, Eq. (b) becomes 0, .. ;-

The expressions for

0z

and

't;<;

0, .. ';

and

't;<; ..

[29 + sin 29 cos2'P]

... (11.22)

can be likewise derived.

[29 - sin 29 cos2'Pl

i [sin 28 sin2ql]

... (11.23) ... (11.24)

It may be mentioned that Eqs. 11.22 to 11.24 are geneml equations which can be used even for the case when the point P is below the centre of the load.

In this case,

P1 .. -131 .. 8

Fig. 1l.13.lsobara oIstrlp toad.

232

SOIL MECHANICS AND FOUNDATION IlNGINEERING

and

~J+~2-0

Therefore, Eq. 11.22 gives 0: _

.;

(28 + sin 29)

... (s.1me as Eq. 11.21)

Eq. 11.22 can be used to determine isobars of different intensity due to strip load. Fig. 11.13 shows the isobars. The isobar of load intensity 0.1 q is at a depth of about 6 B below the load. Fig. 11.13 can be used for determination of vertical stresses fit various points. 11.12. MAXIMUM SHEAR STRESSES AT l'OINTS UNDER A STRIP LOAD The shear stress at any point P below a strip load is given by Eq. 11.24 as

't,Q ,.

~ sin

29 sin 2
planes on which the shear stresses arc zero nrc known as principal plancs. TIlcrefore [or principal planes, "t.a:; = O.

~sin29sin2
0: -

02 - 0., -

~

o.

(29 + sin29)

~(29

- sin 29)

... (11.25) ... (11.26)

The maximum shear Slress is equal to half the difference of the principal stresscs. TIlUS "tmax -

~

(OL -

(3) -

~

sin 28

...(11.27)

Eq. 11.27 indicatcs that the maximum shear stress at P depends upon thc anglc 29 subtcnded by thc strip load at the point J~ Obviously, the maximum shear stress will remain constant if the anglc 28 docs not Change. Let us draw a circle with the centre 0 obt~Lined by the intersc('1ion of lines OA and on mnking angles

------.........

Fig. 11.14. Maximum shcarSLresses

SI'RESSES DUE TO APPUED LOADS

(9O-2B) with the ends of the load. as shown in fig. 11. I 4. As the angle subtended at the centre of circle is twice that at the circumference, the point P makes nn angle 2B. All the points on this circle will subtend an angle 28. From Eq. 11.27, US the maximum shear stress depends on the angle 2B, the circle is the locus of all points with sheur stress equal to 'tmax. The absolute maximum value of shur stress, ("t".....Jmax will occur when sin 2B = 1 in Eq. 11.27. Thus

«~)-

-;

'The locus of (tmax)mru: is a semi circle, which has the width of the loaded strip,

sin 28 • 1

oc

D,

as its diameter, In this

28 _ 90°.

11.13. VElmCAL STRESSES UNDER A CIRCUlAR AREA The loods applied to soil surface by footings are not conccntrntcd loods. These are usually sprend over a finite area of the foodng . It is generally assumed that the footing is flexible and the contact pres.o;;ure is unifonn. In other words, the io.'ld is ru;sumed to by I----R---I unlfonnly dlslnbuted over the area of the base of fOOlings,1 @ l n t e n S I I Y 2 Q Let us detcrmme the vertlCai stress at the pomt P at depth z below the centre of a umformly loaded Circular area (Fig 11 15) Let the mtenslty of Ihe lood be q per uml area dr r !o and R be the radiUS of the loaded area Boussmcsq's Q solullon can be used to determme Oz The load on the '. , ,,' I elementary nng of rndlUS , and Width dr L'> equal to q '\" : ':' ,,I (2M) dr. The load acts at a constant ra
I

,\<'\, 1 \\,

P;rom ;(:lx9'z",,u) I 0, - ~ . %' . [1 + (,I')'J" The vertical stress due to entire 10mJ is given by ,d,

the poUlt

3

0, -

Let

?

3qz

R

I. (,.' + ?-)"

'<~2 e";~' \).,tF

...(0)

___

+ ; - u. Therefore, 2r dr _ du

Fia:.

(Ii + l)

Eq. (0) becomes

0, -

Ie

3q?

/,1/

I ,:/

Y

JI.JS. Circular Lo.1d.

du

'}""

-i lit' (- ~) [u-"'L~' ". - -q? - q? 0, -

or

q

[(II' :?-)'" - (~'" 1

[~

-

[1 -{I

(R' :

+

1-)'" 1

;RI,)'

r

.. .(11.29) .. .(11.30)


234

r

wh.ere Ie is the influence coefficient for the drcular area, and is given by

E, - [

1-11

+

;RIz)'

1

... (11.31)

Table 11.8 gives the value of the influence coefficient lc for different values of Rlz.

Table 11.8. Innueuce Coefficients 1, for the Cin:ular Area

RJ,

/,

RJ,

I,

RJ,

/,

RJ,

/,

0.00 0.05 0.10

0.0000

0.65 0.70 0.75 0.80 0.85 0.90

0.4109

130 135

0.7734

1.9S

1.45 1.50

0.7891 0.8036 0.8170 0.8293

1.55

0.8407

2.00 2.50 3.00 3.50 4.00 5.00 6.00 7.00 8.00 9.00 10.00

0.9050 0.9106 0.9488 0.9684 0.9793 0,9857 0.9925

0.15 0.20 0.25 0.30

0.35 0.40 0.45 0.50 0.55 0.60

0.0037 0.0148 0.0328 0.0571 0.0869 0.1286 0.1592

0.4502 0.4880 0.5239

0.5577

1.00

0.5893 0.6189 0.6465

1.05

0.6720

1.10 1.15 1.20 1.25

0.6956 0.7175 0.7376

0.95

0.2079 0.2416

0.2845 0.3273 0.3695

0.7562

lAO

1.60

0.8511

1.65 1.70

0.8608 0.8697

1.75 1.80 1.85 1.90

0.8779 0.8855 0.8925

0.8990

0.9956 0.9972 0.9981 0.9987 0.9990 1.0000

Eq. 11.31 for the influence coefficient./c can be written in tenns of the angle 29 subteoded at point P by tbe load. Let tan

a = R/Z. Therefore, [

E, -

1-11

+

~n'e }YO .. (11.32)

Eq. 11.32 indicates that as e tends to 90°, the value of /( approaches unity. In other wor~, when a uniformly loaded area tends to by very large in comparison with the depth Z, the vertical slress al the point P is approximately equal to q. When the point P is not below the centre of the load, analysis becomes (;()mplicated and is outside the scope of this text. In that case, the isobars shown in Fig. 11.16 can be used to determine the vertical stress at any point. It may be noted that the isobar of O.lq cuts the axis of the load at a depth of about 4R (= W) below the loaded area. The zone within which tbe stresses is indicated by tbis isobar. as mentioned above, is known as the bulb of pressure. The reader should compare this pressure bulb with that below the strip load, which is much deeper. 11.14. VERTICAL STRESS UNDER A CORNER OF A RECTANGULAR AREA The vertical stress under a comer of a rectangular area (Fig. 11.17) with a uniformly distributed load of intensity q can be obtained from Boussinesq's solution. From Eg. 11.9, the stress at depth z is given by, taking dQ - q M- qd< dy,

STRESS DUE TO APPLIED LOADS

"

Fig. 11.16. Isobars for cin::ular loaded area.

_ 3ql

f rB

°t ~ """"27t Jo Jo

By integTlllion.

qdxd}' l + Z2)~/2

(x2 +

Although the integral is quite t.:omplicated, Newmark was able to perform it The results we re presented as rollows:

0t where m

=.!L[ 21t ~ 1//11

= BIz

The values of expressed as

and III

II

m: +1I2;;

2

•

III

+

/I

+

III II

+ I

+ sin- I (

"'m2 +

112

mn ] ... ( 11.33) + m2 n2 + I

=Uz

and n can be interchanged without any effect on the vaJues or

Fig. n .17. Vertical slTe!i3 under comer.

O"l.

Eq. 11.33 can be


236 (J~

= iNq

... (11.34)

where IN is Newmark's inllucoce coct1icicnl, given by

I

I N=21t

[11/1/

",1+,""+ 2 . _I ~. 11/2+,,2+ 1112,,2+ ,+sm

Table 11.9 gives the value.s of IN for different values of m and

mil

(

]

",,1112+112+11121/2

+ I

II.

Tahle 11.9. InDuence Coefficient;; IN ror Rectangular Area

0.2

0.4

0.6

O.B

1.0

2.0

3.0

5.0

10,0

0.2

0.0179

O.032g

0.0504

0.0610

0.0619

0.0620

0.0620 0.1154 0.1562 0.1850 0.2046 0.2399 0.2465 0.2491 0.2498

0.4

0.0328

O'(}('O2

0.0435 0.0801

0.0931

0.0547 0.1013

O.JJ34

0.1150

0.1154

0.0 0.'

0.0435

{I.GSOI

0.1069

0.1247

0.1361

0.1533

0.0504

0.0931

0.1247

0.1461

0.1598

1.0

0.0547

0.1013

0.1361

0.0610

0.1134

0.1561 0.1849 0.2044 0.2395

0.1150

0.2034

0.2378

0.2439

0.2461

0.0620 0.0620

0.1154 0.1154

0.1533 0.1555 0.1561 0.1562

0.1555 0.1841 0.2034 0.2378

0.0618

0.1598 0.1812 O.I!Wl

0.1752

2.0 3.0 '.0 10.0

0.1812 0.1999 0.2325

O.I!W9

0.2044 0.2046

0.2395 0.2399

0.2461 0.2465

0.2486 0.2491

0.1850

0. 1999

Fadum gave charts for detenmnal\on of Lhc inlluence factor IN (Fig. 11.18). These charts can be used in a design office. The t:harls t:an also be used for dClenninalion of the vertical stress under a strip load. in which case the length tends to infinity and the curvc for II 00 can be used .

=

o.26

0""

~

.l

V"

20

-

.1

..

./"' //

---------

1/ /,1%: /-~

0.1

n," ncO.6 nsO.S

n",0.4 n", n~o.2

V

"""'~V

°1~P

:I~F-

I n~ ~o

"mO••

0·'

I ID

Fig. 11.18. Flldum's chart.

1\.15. VERTICAL STRESS AT ANY POINT UNDER A RECTANGULAR AREA The equations developed in the preceding seclion can also be used for finding the vertical stress at a int which is not located below the comer. 11le rectan ular area is subdivided into rectangles such Ihat each

SI'RESS DUE TV APPUED WADS

rectangles has a comer at the point where the vertical stress is required. The vertical stress is delennined WJing the principle of superposition. 1bc following three cases can occur. (1) Point anywhe~ below the rectangular area. Fig. 11.19 (a) shows lhe location of the point P below tile rectangular area ABeD. The given rectangle is subdivided into 4 small rectangles AEPH, EBFP, IlPOD and PFCG, each having one comer at P. The vertical stress at P due to the given n:ctangular load is equal to that [rom the four small rectangles. 1bcrefore, using Sq. 11.34, E

(J)!

(2)

I

Hr----t-------- F \to)

I

())

I

(a'

(b, E

(e)

Fig.l1.l9. Vertical stress under a rccl3ngulllrarea.

0z - q [(IN)' + (/Nh + (IN>. + (IN)'] ...(11.35) where (iNh, (/Nb. (lNh and (/N)4 arc Newmark's influence fadors obtained from Thble 11.9 for the four "",
Now rea.angle ABCD = rectangle AEPF - rectangle BEPH - rectangle /X1PF + rectangle CGPH The last rectangle CGPH is given plus sign becal..lSC this area has been ' + (IN)'] ... (11.37) where (IN)!> (INn. (IN») and (IN). are the influence coefIkients for the rectangles AEPF, BEPH, /X1PF and CGPlf, n::spocti.vely. (3) Polnt below the edge or the loaded area. If tile point P is below the edge of the loaded arca AlJCD (Fig. 11.19 c). the given rectangle is divided inlO two small rectangles APED and PBCE. In this cme, 0, - q [(IN') + (IN),] where (IN)J and
•.. (11.38)

11.16. NEWMARK'S INfLUENCE CUARTS The methods for the delennination of vertical stresses under a strip., a circular and a rectangular area have been diScussed in the preceding sections. In practice, sometimes one luti to find the vertical SI~ uDder 8 uniformly loaded areas of other shapes. For such cases, Newmark's influence chans are extremely useful


238

Newmark's chart is based on the concept of the vertical stress below the centre of the circular area, discussed in Sect. 11.13. lei us oonsider a uniformly loaded circular area of radius R I • divided into 20 equal sectors (Fig. 11.20). The vertical stress at point Pat depth z just below the centre of the loaded area .due to load on one sector (hatched area (t» will be (1/20) of that due 10 load 00 fuU circle. From Eq. 11.29,

~ q [ 1 - { 1 + (~,/Z)'

0. -

r]

(a)

If the vertical stress (oJ is given an arbitrary fixed value, say O.OOSq, Eq. (a) b=me.s 0005

.

q

'" ]

20 [· {I

.. !L 1- _ _1_ _ + (R,Iz)' }

... (b)

Fig. 11.20. Concentric circles for Rl and Rz.

... (1139) Rl/z - 0.270 Solving Eq. (b). Thus every one-twenLieth sector of the circle, with a radius R] equal to 0.270 z, would give a vertical stress of 0.005 q al its centre. Let us now consider aoother concentric circle of radius R2 and divide it again into 20 equal sectors. Each larger sector is divided into two sub-areas. H the small area (marked 2) exerts a stress of 0.005 q at P, the vertical slress due to both area (1) and (2) would be equal to 2 )( O.cX)5 q. Thus.

2 x 0.005 q - !L [ I - { _ _1_ _ , }'" 20 1 + (R,h)

. .. (c)

]

Solving. R';z _ 0.40. .In other words, the radius of the second circle should be equal to 0.40 L Likewise, the radii of the third to the ninth circles can be determined. The values obtained are 0.52 Z, Z, 0.92 Z, 1.11 1, 1.39 z and 1.91 z. The radius of circle is 2.54 z. The radius for the tenth

94

0.64 Z, 0.77

circle RIO is given by

toxO.OO5q-!L20[I-{_I_,}'" ] 1 + (R,oIz)

or

RIO -

QC

1berefore, the tenth circle canoot be drawn.

Fig. 11.21 shows the complete NeWmarlc's influence chan, in which only ~ circles hove been drawn roc

z equal to the distance AB marked

OIl

the chart.

Use of Newmark's Chart. The chart can be used to determine the vertical stress at point P below the loaded area. A plan of the loaded area is drawn on a tracing paper to a scale such that the length AB( = 2 em in this case) is equal to the depth (z) of the point P below the surface. For example, if the pressure is required

=

=

at a depth of 1 m, the plan should be drawn to a scale of 2 an 1 m or R.F. 1150. The traced plan of the loaded area is placed over the Newmark chan such that the point P at which the .'pressure is required coincides with the cenlle of the chan. The vertical stress at point P is given by Oz"/)( n )( q ... (11.40) where I.: influence coefficient ( = 0.005 in this case),

239

srRE.SSES DUE TO APPUED LOADS r-~----'----'-- .--'-----'-~

Fig. 11.21. Newmark', Chart$.

n = number of small area units covered by the plan. Each area between two successive radial lines and two successive concentric circle is taken as one unit. q = intensity of load. The following points are wO£1h IlOting: (I) The fractions of the unit areas should also be rounted and properly accounted for. (2) If the plan of the loaded area extends beyonds the 9 ¥h circle. it may be assumed to approach the 10th circle for the purpose of counting the unit areas. (3) The point P at which the vertical stress is required may be anywhere within or outside the loaded an:8.

(4) If the depth at which the stress is required is changed, a fresh plan is required such that the new depth is equal to the distance AB on the chllrt.

11.17. COMPARISON OF STRESSES DUE TO LOADS ON AREAS OF DIFFERENT SHAPES The variation of venical stress with depth depends upon the shape and size of the loacted area. Fig. 11.22 shows the variation of the vertical stress with depth below the centre of circular, square strip loads. The vertical stresses are equal to the load intensity at Ihc surface and deaease rapidly with an increase in depth (z). In the case of circular and square Ioods, the vertical stress is about 10% of the surface load (q) at a depth of about 2B. However, in the case of strip loads, the stresses are much greater. Even at z = 38, the vertical stress is about 20% of the surface load (q). In other won1s. the pressure bulb in this case is much deeper.

and


l!T- -

o.oOJJ~o,-'O'i""'O~-""ro'-r-'OT.60"--r--'1o."''''-r--7I.00'1:~~~~ii~

I

TT

~

..

LA

~

rcu~~V/QUor

0." 0.6'3

0.,1-l-+I(C~,,-'<"-,.-.J,,-..,'-+-~-j,_"'_4"-'l/""':H ::~ g:~: '.0

I-+--sq~"a_"...,-X q:1-?-+AI~<'''''''1'Pf--H

2.00.08 2.5 0.06 3.0 0.010

triP

0.70

0.8'3

0.14 0.01 0.0"

0.31 0.25 0.20

~'.~: ~:~;

1-+---j,l1/'4'---++I--+-++-I-I •• Wid'h a, .'am"..

,., 1-1-/",t:---I-j-!.f-hHHHH 1/ 2.01-/!-+-+-+-+--+-I-4-H

,

2.5

3.0

I

-It

L-.L.L.L...L...L..L..L...L-l

u 'IlL!'

FIg. U.22. Companson

or cm:ular, lIqUllre and stop loads.

11.18. VER11CAL STRESSES UNDER TRIANGULAR LOADS Fig. t 1.23 shows a lriangutar load with a width of 2b and the intensity varying from 0 to q. The vertical (o~) at a point P(x,z) is given by

stI'CSS

Fig. 11.23. Triangular load.

°t -

2i [~ a -

sin 26 ]

where & Js the angle which the line PB marks with vertical, and a is tbe a ngle subtended by PA and PB at p. IC the point P is CXadly beknv the end B, x .. 2b and & = O. Therefore, <1

~

_!L (~(l) 2x

b

_~ Jl

Sl'RESSES DUE TO AP.PUED WADS

241

The above equations can also be applied to the case whcn the intensity of the load increases linearly from zero 81 one end to a maximum q and then decreases to zero (Fig. 11.24). For tho load shown in Fig. 11.24 (a),

I

1

2b

2b.1

I'

1

2b

.1

2b

~~. z

z

(b)

(a)

Fig. 11.24. Triangular load willi maximum intensity

~t

centre.

. .. (11.43)

When thc point P is exactly below the point B, 01 - az • a and x :: 2b. [Fig. 11.24 (b)]. 1ncrcforc, 0, - ~ [2b )( 20 + 2b(a - a)] ... (11.44)

11.19. VERTICAL STRESSES UNDER TRAPEZOIDAL WADS Fig. 11.25 (a) shows a trapczoidalload due to an embankment. which .consists of a triangular load over width a and a uniform load of intensity q over width b. The vertical stress at point P is given by

.~rI~ p

p (b)

(a) Fig.l!.25. Trapc:zoidalload.

b) (al + a,) 0, _; [(a\ + a0 + ~ aL] 0, • ; [ ( a :

0<

0, - ;;ra(al

+

au

~ '" 1

+ bad

Obviously. for the lrapezoidal load shown in Fig. 11.25 (b), the vertical stress at P,

... (11.45)

SOIL MECHANICS AND FOUNDA:nON ENGINEERING

242 0, _

... (11.46)

11.20. STRESSES DUE TO nOlUZONTAL LOAD (a) Line Loud QJ Fig. 11.26 shows a horizontal load QI per unit run acting on the soil surface shown by solid lines. In this case, il is more convenient \0 determine the rndial stress (0,) and the tangential stress (00)' 111C radial stress is given by Or "

2QIsin8 r(2u-sin2a)

0,

... (11.47)

where e = angle made by radial line r with the venical u = anglc made by soil surface with the vertical For horizontal ground surfucc.. a - ;"t/2. 'lbercfore. 0"

,

_

2QJ sin e ,(.-0)

Fig. 11.26. HOri7.ont:d lo.1d.

2Q\sinO 0, _ - . , -(b) Concentl'1llcd Load

... (11.48)

Q. In cartesian coordinates, the stresses duc to a horizontal load Q can be

written as

cr,. -

,,.,.· ~ iL 2

,. R'

x

[~_ R'

(1-2\1) + (l-2V)R1.13 -

(R .,)'

~}] n' (R

+ ,)

'" _f . .LIt x [~ + (I - 2v) n' 11 _x' R + ')}] Rl (R+Z)2 n2(R + 2:) "'~

(3

~~-~.¥ ~y:=~.*

... (11.49)

where x,y,z arc coordinates of point P and R is the dist::lnce Op, as shown In Fig. 11.4. This case is genemlly referred to as Ceruttj's problem.

U.21. STRESSES DUE TO INCLINED LOAD Fig. 11.27 shows an inclined load Q2 per unit run acting on the soU surface. The radial stress at a point at an angle 8 is given by

a • 2Q2 (COSPOOS8 + Sin pSin8) ... (11.50) , r 2a+sin2a 2ct.-sin2a where j3 = angle which the load Q2 makes with vertical, C1 ::: angle the soil surface makes with the vertical.

Fig. 11 .27. Inclined load.

'"

mESSES DUE TO APPUED LOADS

For the horizontal ground surface, a .. 1t/2. Thus 2Q, [ COS~COS8

(Jr "-;cr, _ When the lood is vertical,

~

:It

+

sin pSin9j !It

2~'ros(e_~)

. ... (11.51)

.. O.

or ..

2~2

COS e

... [(1 1.51 (a)]

11.22. WESTERGAARD'S SOLUTION Boussinesq'SOlution assumes that the soil depooit is isotropic. Actual sedimentary deposits are generally anisotropic. There are generally thin layers of sand embedded in homogeneous clay strata . Westergaard's solution assumes that there are thin sheets of rigid materials sand·wiched in a homogeneous soil mnss. These thin sheets are closely spaced and are of infmilc rigidity and are, therefore, incompressible. These permit only downward displacement of the soil mass as a whole without any lateral displacement. Therefore, Westergaard's solution represents more closely the actual Sedimentary dCJX>SilS. According 10 Westergaard, tile vertical stress at a point P at a depth z below the concentrated load Q is given by clbt Q ...(11.52) cr, +

(q,,)'l'" . ;:t

[C'

where c depends upon the Poisson ratio (v) and is given by c - V(1-2')I(2-2v) For an elastic material, the value of \I varies between 0.0 to 0.50. For the mse when v is zero, Eq. 11.52 is simplified considerably, taking c _ 1IV2. 1 Q .(11.53) o z " :re{l + 2(rlz)2JM'

?

0, -

1.

~

... (11.54)

where Iw is known as Westergaard influence coefficient.

1

1

... (11.55)

..... :re[1 + 2(r/z)2f2

The values. of I ... arc considerably smaller than the BOllSSinesq influence factor (10). Table 11.10 gives the values of I .... 'The values of 18 are also given for comparison. 'Thble 11.10. CompllrLson of lw and 18

riB

I

I,

lw '/l I. I.

I I I

0.0

0.1

0.2

03

O.4nS

0.4657

0.4329

0.3849

0.3183

O.lO9O

0.2836

0.2483

0.7

0.8

0.9

1.0

0.1762

0.1386

0.1083

0.0844

0.1142

0.0925

0.07S1

0.0613

I

0.4

I

0.3295 0.2099

I I I

2.0 0.0085 0.0118

0.5

I

0.2733 0.1733

I I I

3.0 0.0015 0.0038

0.6 0.2214 0.1411

I I I

6.0 0.0001

0.0005

Fig. 11.28 shows the variation of 18 and Iw with rh. The Westergaard influence factor is about 2/3


.-

Ag. 11.28. Comparison of IB and Iw.

of the Boussinesq values fo r small values of rlz. But for rlz more than 2.0, the Westergaard values are slighlly gre
11.23. FENSKE'S CIIARTS Just like Newmark's Charts which are based on Boussincsq's solution, Fenske's Omrts are based on Westergaard 's solution. The Fenske chart can be prepared using Eg. 1152.

1

Q 0: -

21t .

(cz)2p + (rlcz)2)3/l

The above equation can be integrated to obtain the vertical stress (oz) below the centre of a uniform circular load of intensity q and radius R as was done for the Boussincsq solution for derivation of Eq. 11.29. In this case,

o, _ q

[1_{_1}"2 1 + (Rlcz)2

1

···(1l.50

IT instead of the full circle, only I/Bth sector of the circle is considered, the stress is given by

0 . 9.. 1'8

[

__ 1_ _ ,n { 1 + (Riel)' }

1

... (1l.5~

Eq. 11.57 is similar 10 the equation used for Newmark's chart, with one difference that the depth used here is the modiLed depth cz. The radius Rl of the first circle can be determined for a constant value of 0%(say, 0.001 q). Thus 0001 . q

~

_ ~

8

[

1 ___ 1__ { 1 + (R,/czl' }

In

.. 0.127

R\ ,.. 0.127 (cz)

The modified depth cz is marked as the distance AB in Fig. 11 .29.

245


ou t A

B

t-- CZ ----l Fig. 11.29. I'enskc'$ Chart ( I _ 0.001).

Ukewise, the radii of other circles are determined. Unlike the Newmark chart, the radial divisions are also changed in Fenske's chan. There are 8 radial divisions [or the first circle and 48 radial divisions for the 18th circle, The radii of the circular arOi and the number of radial divisions are so chosen thai each influence

!9.

;6)

area unit is approximately a square. Thblc 11.11 gives the values of R/({:z) for different circles and their corresponding number of division. The method of using the Fenske chart is similar to thai for the Newmark: chart. However, in this case the d~tance AD represents the modified depth cz. 'The plan of the loaded area is drawn on a tracing paper to a scale such that the distance AD is equal 10 c times the depth z of the point P at which the stress is required. For Poisson's ratio of zero, the value of c is equal to 0.707. Table H.H. Values of Riel. ror Fenske's Chart

'7) ed

1 1 2 1 3 1 1 5 1 6 1 7 1 I 0.127 I 0.204 0.292 0.376 0.472 0.560 I 0.664 I 0.772 Di"~io,. I' 8 1 12 I 2D 1 24 1 321 32 I 40 40 C;",t, No·1 to 1 It 1 12 1 t3 1 14 1 IS 1 16 1 t7 1 18 Rl(cz) 1 1.032 1 1.176 1 1.332 1 1.512 1 1.712 1 1.952 1 2.236 1 2.592 1 3.044 o;"'~... I 48 I 48 I 48 I 48 I 48 I 48 I 48 I 48 I 48

Ci"kNo RJ (ez)

0.900

48 19

4.420 48

11.24, APPROXIMATE ME11JODS The methods discussed in the preceding seaions are relatively more accurate, but arc time· oonsuming. Sometimes, the engineer is interested to estimate the vertical stresses approximately for preliminary designs. The following methods can be used. (l) Equivalent Polnt·Lood Method. The vertical stress at a point under a loaded area of any shape can be detennined by dividing the loaded area into small areas and replacing the distributed load on each small


246

area by an equivalent point load acting at the centroid of the area (Fig. 11.30); e.g. in Fig. 11.30, Q .. qa2 for each area. The total load is thus converted into a number of point loadS. 1he vertiOl.l stress at any point below or outside the loaded area is equal to the sum of the vertical stresses due to these equivalent point loads. Using Eq. 11.10,

[Qd'D), <- Q, 0,' 1

or

0, •

;:

(',n? <-

••. <- Q. (',). 1

•

,:?,

... (11.58)

Q, (',),

T

+ . + . ..1

..

>Equival~nl

r--.-+-.-+-.--+-.-::!,,/ 5

6

7

8

-9

-10

"II

-12

point loads

&j. 11.58 gives fairly accurate results if the side a of the small area unit is equal to or less than one-third of the depth z of point P at which the Fig. 11.30. Equivalellt Point loads. venicai stress is required. (2) Two-to-one Load lli<;tribution Method. The actual distribution of load wilh the depth is complex. However, it can be assumed to spread approximately at a slope of two (vertical) to one (horizontal). Thus the vertical pressure at any depth z below the soil surface can be determined approximately by constructing a frustum of pyramid (or cone) of depth z and side slopes (2:1). The pressure distribution is assumed to be uniform on a horizontal plane at that depth.

i-ol·----B+Z-----.j., Fig. 11.31 . Two""o..()ne Distribution.

The average vertical stress crz depends upon the shape of the loaded area, as given below (see Fig. 11.31) (1) Square Aren (B x D),

0"%'"

qIi' zf

... (11.59)

(8 +

(2) Rectangular Area (D xL),

(3) Strip Area (width

n, unit length),

(4) Circular Area (diameter D),

... (11.6
q.

-18\(~)

Xx

1;

... (11.61) ... (11.62)

The above method gives fairly accurate values of the average vertical stress if the depth z is less than 2J times the width of the loaded area. The maximum stress is generally taken as 1.5 times the average stress determined above. (3) Sixty Degree Distribution. This method is similar to the preceding method. In tttis case, tbe pressurt

241

STRESSES DUE TO AI'PLIED WADS

distribution is assumed along lines making an angle of 600 with the horizontal instead of 63~0 (2 : 1). The method gives approximately the same results. 11.25. CONTACr PRESSURE DISTIUBUTION The upward pressure due to soil on the underside o f the footing is tenned cont;:la pressure. In the derivations of Ihe prcceding scctions, it has been assumed that the fooling is flexible and the cont."lct pressure distribution is unifonn and equal to q. Actual footings are not flexible as assumed. The aaual distribution of the rontact pressure depends upon a number of factors such as the clastic properties of the COOling material and soil. the thickness of footings. In faa, it is a soil- structure interaction problem. Borowicka (1936, 1938) studied the rontaa pressure distribution of uniformly loaded strips and circular footings resting on a semi-infinite elastic mass. assuming the base of the footing as frictionless. The analysis showed that the contact pressure dislribution depends upon thc relative rigidity (K,) of the footing-soil system. The relative rigidity is defined as K

,

_.!.6~ (I - >?) (~). E,

(1.)'

...

b

(11.63)

v" vf = Poisson's mtios for soil and footing material respectively, E., Ef = Mcxluli of elasticity [or soil and footing material, respectively. 2h = width (or diameter) of footing. I = thickness of fOOling. Fig. 11.32 shows the contact pressure distribUlion of circular and strip footings for different values of relative st iffness. For a perfectly rigid fOOling (K, _ 00), the oontact pressure is minimum .11 the centre. with

where

Ie

1

f.o-----Wldth ~ 2 b - - l

tl

I

j

o.oq '77"70'777777'?1

!a}Circulor

'001in9

' o)O t"''1

Fig. 11.32. ConIRC[ Pressure (Borowick..1).

9) 0)

I) 2)

..5

a value of about 0.5 q for the circular fOOling and 0.67 q for the strip footing. The contact pressure is very large at the edges. In fact, it tends to infinity. For purely flexible footings (K, - 0), the rontact pressure is uniform and equal to q. Borowick.a's rcsullS can be used to determine the contaa pressure on a cohesive soil which behaves like an elastic soil mass. In a cohesionlcss soil. modulus of elasticity increases with depth due 10 an increase in confining pressure. Such soils are non-homogeneous. Contact pn!SSure on saturated clay. Fig. 11.33 shows the qualitative cont."lct pressure distribuLion under flexible and rigid footings resting on a satumled CIHY and subjected to a unirormly distributed load q. When the footing is flexible, it deforms inlo the Shape of a bowel, with the maximum deflection at the centre. The contact pressure distribution is uniform. IT the footing is rigid, the settlement is uniform. The contact pressure distribution is minimum at the centre and the maximum at the edges. The stresses at the edges in real soils cannot be infinite as lhcoretlcaUy determined for an eltlstic mass. In real soils, beyond a certain limiting v;3!ue of stress, the plastic Dow occurs and the pressure becomes finite.

SOIL MECHANICS ANI) FOUNDATION BNG INEERING

248

( a ) FLEXIBLE

fOOTI NG

(tl)

RIGID

FODTlNG

Fig. 11.33. Conllld pressure on SIIturated clay.

Contact Pressure on sand. Fig. 11.34 shows the qualitative contact pressure distribution under flexible and rigid footings resting on a s
( ;a ) FLEXIBLE

FOOTING

(tl)

RIGID

f OOTING

Fig. 11.34. ConlllCt Pressure on Sand.

If the footing is rigid, the settlement is unifonn. 'The contact pressure increases from zero at the edges to a maximum at the centre. The soil, being unconfined at edges, has low modulus of elasticity. However, if the footing is embedded, there would be finite contact pressure al edges. Usual Assumption. As discussed above, the oonlact pressure distribution [or Oexible footings is unifonn for both clay and sand. The contact pressure for rigid fOOling is maximum at the edges for footings on clay, but for the rigid footings on sand, it is minimum al the edges. For convenience, the oont.act pressure is assumed to be unifonn for aU types of footings and all types of soils (Fig. 11.35) if load is

symmetric. 'I11e above assumption of uniform pressure distribution will result in a slightly unsafe design for rigid footing on clays, as the maximum bending moment al the centre is underestimated. It will give a . conservative design for rigid footings on s

249

(2) TIle soil mass is not clastic as the particles do not return to lhe origi nal position when the load is removed. (3) The , stress-strnin ratio for most soils is not constant. However. for m Ost soils the stress-strain ratio is approximately constant provided the stresses are well below the failure stresses, ilnd no unloading occurs. Although the applicability of elastic theories to soil problems is questionable, yet the results arc generally not much different from the observed values. A difference of 20 to 30% between the theoretical and the measured values may OC!;UT. This diffcren!;e is generally ignored considering many complexities of the problem. The eluslic theories are used as better theories are not yet available which can be used in a design

office. ILLUSTRATIVE EXAMPLES U1ustrative Example 11.1. A concelltrated load of 2000 leN i.f applied (j( the ground slIIface. Determille the vertical m 't!5S at a poi"l P which is 6 //I directly below tlte load. Aim calculate tile venical stress at a poillt R which i~' (If u depth of 6 11/ bllf til (I horiwllfal distance of 5 m form Ihe axis of the load. Solution. From Eq. 11 .9, (11

rlz

Point R,

rl z = 5/6.

I

[I

+ {rldl'li

3x2000

= O.

Point P.

3Q = 2nz?

0": =

21[(6)2

.

rI +

I

= 26.53 kN/m

015/2

2

3 x 2000 I 2 o~ = ~. (I + (516)2]~h - 7.1 kN/m

mm;trative Example 11.2. A IOllg strip footillg of width 2 III carrie~' a loatl of 400 kNlm, Calwlute the maximum stress at a depth of 5 /II below the cemre line of the fnotillg. Compare Ihe resllirs with 2 : I distributioll method. Solution. From Eq. 11.21, In this

ca~c.

/, = I m

=

and

o~

=

*

(29

+

sin29)

z = 5 m.

=

tan 9 115 0.2 and Taking q = 40012 = 200 kNfm2. (1:

2 9

=

2:

= 0.395

(0.395

T

radians

0.385) = 49.6 IcN/m 2

2 : 1 Distribution method. From Eq. 11 .6 1, 0 ;:

=

~:!

=

2~+\2

= 57. 14;.649 .6

Percentage error

= 57.1 kN/m

2

x 100 = 15.2%

Illustrative Example 11.3. There 6' {/ line load of 120 kNlm aCfing 011 the ground surface along y-axis. Determine the vertical ~·tress (If a poilll P which Iws x and z coordinates as 2 1/1 alld 3.5 m. respectively. Solution. From Eq. 11.1 7.

-EL[ _ _I_ ]' m:

a

I T (xld

t -

AI point P.

o

~

=2 x 1t

J' = 12.40kN/m2

120 [ _ _I _ _

x 3. 5

I

+ (2/3.5)2


250

llIustrative Example 11.4. The unit weight of the soil in a uniform deposit of loose sand (Ko = 0.50) is 16.5 kN/ml, Determine the geostatic stresses at a depth of 2 m. Solution. From Eg. 11.1, 0 ... yz .. 16.5 )( 2.0 .. 33.0 kN/rnl a., ..

From Eq. 11.6,

Ko o...

0.5 x 33.0 .. 16.5 kN/rnz

illustrative Example U.5. Determine the vertical stress at a point P which is 3 m below and at a radial distatlCC of 3 m [rom the vertical load of 100 kN. Use Westergaard's solution (v = 0.0).

1

oz" 1[[1 + 2 (r/z)2yn .

Solution. From Eq. 1153,

(J

Alternatively Using Eq. 11.54,

1

..

, Oz

. 100 .. 0.681 kN/m2

• [! + 2(3/3)'1'" ..

I"" .

Q

l

(3)'

~

Taking I.., from Thble IUD as 0.0613, Oz .. 0.0613 x ~ .. 0.681 kN/ml (3) Illustrative Example 11.6. Calculate the vertical stress at a point P at a depth of 2.5 1Il directly under the centre of the circular area of radius 2 m and subjected to a load 100 kArIn? Also calculate the vertical stress at a point Q which is at the same depth of 2.5 111 but 2.5 111 away from the centre of the loaded area.

0 .. q [ 1- { _ _1_ _ }'" ' 1 + (R/x)'

Solution. From Eq. 11.29

1

cr at P - 100 [ 1 - { 1 }"']_ 52.39 kN/m 2 , 1 + (2.012.5)'

From Fig. 11.16, the verLical stress at z Therefore,

0:

= 1.25 Rand r = 1.25 R is about 0.2 q.

at Q _ 0.2 x 100 _ 20 kN/m 2

Illustrative Example 11.7. An L--l>/lQped buil4Nlg in plan (Fig. E 11.7) exerts a pressure of 75 kNlm] tile soil. Determine the vertical stress increment at a depth of 5 m below the interior cartier P. Solution. The loaded area is subdivided into three small areas such that each small area has one comer at P. From Eq. 11.35,

For Ilre8 Al From Table 11.9, For area A z From Table 11.9.

0, - q [(INh + (INn + (INh] m-n-· 10/5_2.0

(IN), = 0.2325 111 _

15/5 .. 3,

For area AJ

(lNh = 0.2378 m .. 20/5 .. 4,

From Table 11.9,

(INh .. 0.2450

Therefore, 0, .. 75 {0.2325

12"

12 ..

1015 _ 2.0 1515 _ 3

+ 0.2378 + 0.2450] = 53.65 kN/mz.

0/1

:I~ : 10m

I

20m

~

Fig. E 11.7.

Hlustrative Example 11.8. A rectangular foundation 4 m by 5 111 carries a uniformly distributed foad af

STRESSES DUE TO APPLIED WADS

zoo kN/m 2. Determine the vertical stress at a point P located as shown in Fig.

'" E 11.8 OIId at a depth of

2.5 m. Solution. From Eq. 11.35,

0'%"

In this case For Al and Al,

In

q[(lN). + (/Nh + (INh + (IN)41

= 212.5 = 0.80, n = 2/2.5 = 0.80

(IN) • .. 0.1461, For A3 and

A.,

III

= 3(2.5 = 1.20, n = 2a.5

:: 0.80

(IN), - 0.1684 Therefore,

0%"

200l0.1461 + O.l46J + 0.1684 + 0.1684]

.. 12S.8kN/ m l llIuslrative Example 11.9. A T-shaped foundation (Fig. E 11.9) is loaded with a uniform load of 120 leN/n? Determine the vertical stress at point P at a depth of 5.0 m. Use NL'Wlllark's inflllence chart. Compare the OIISW'er by exact method

I,'m I 3m "I Fig. Ex. 11.8

Solution. The foundation plan is drawn on a tracing

paper with a scale such that the distance An in Fig. 11.21

represents 5.0 m. The plan is placed on the Newmark chan such that poinl P is al the centre of the chari. Number of area units occupied by plan = 63 From Eq. 11.40,

o~

.. { x n x q

o~ .. 0.005 x 63 x 120 .. 37.8 kN/m 2

Fig. E 11.9

Exact method. The loaded area is divided into 3 areas. such that they have onc corner at P. Area AI 11/ = 3/5.00 = 0.60 : /I = 1.5/5.00 = 0.30. (IN). = 0.0629 Area Al II! = 3/5.00 = 0.60: /I = 615.00 = 1.20, (fNh = 0.1431 Area A3 11/ = 3/5.00 = 0.60 ; 11 = 3/5.00 = 0.60. (fNh = 0.1069 From Eq. 11.35, o~ :: q I(lN)] + (fNh + (lNh J or o. = 120 [0.0629 + 0./431 + 0.1069J :: 37.55 kN/m2 2 lUustralive Example 11.10. A r~tonglilar loaded area '2 m x 2.5 m carries a load of 80 lcN/m (Fig. E 1/.10). Determine the vertical stress at point P located outside the loaded area at a depth of 2.5 In. Solution. From Eq. 11.37,

C1z .. [(IN). - (INn - (/Nh + (IN)4)]

Fig. E 11.10.

~IL

252

MECHANICS AND FOUNDATION ENGINEERING

For large rectangle AEPG, m .. 3.5012.50 .. 1040. n .. 2.5012.50 - 1.00 and (INh .. 0.1914 Area A z m =. 3.5/25 :: 1.40, n :: 0.5(2.5 ::: 0.20, (lNh :::: 0.0589 Area AJ m = 2.5(2.5 = 1.0, n = l.OnSO 0.40, (INh = 0.1013 Area A. m = 0.512.5 = 0.2, n = 1.0(2.5 = 0.40, (/N)4 = 0.0328 Thererore, Oz .. 80 [0.1914 - 0.059 - 0.1013 + 0.0328] _ S.12 kN/m l

=

Dlustrutive Example U.l1. A recllJngular foundation 3.0 )( ISO m carries a uniform load of 40 *Nlnt Determine the vertical stress at P which is 3 m below the ground surface (Fig. E 11.11). Use equivalent point load method.

~~~-3m----'1

1-01

r--- 1m ----l-- 1m ----1--1 m--l

l' ~m~

__ (1_I__-}__-(-21__-t.,--(-31----1

I IL_(7_)_---1__(8_)_~ 15

(£1

(51

(61

__ (9_)____'

Fig.E.ll.H.

Solution. Let us divide the loaded area into 9 small areas of size 0.5 m x 1.0 m. load on each area .. 40 )( (1.0 )( 05) .. 20kN The stresses at point P are determined due to 9 point loads, using Boussinesq's solution For loads (1) and (4),

r .. ,..; 1.5 2 + (0.25)2 .. 1.521

For loads (2), (3), (5), (6),

, •

For loods (8) and (9),

r _ ...; (0.75)2 + (0.5)2 _ 0.901 ;

For load (7),

r ..

~

V(1.5)2

I

rlz - 0.300 rfz _ 0.559

Therefore,

0-1:

In lhis case,

3x20· [ 2 a, • 2x (3)' x [1 + (0.507)']'" + (1

,

~

.

11.9).

,/z • 0.186

• 0.559,

+ (0.75)2 .. 1.677;

3Q

CEq·

rlz - 0.507

)( ,.,-----,-'-:-,"'"

[1+(,1%),),/'

2

4

+-(O-'.I'--86~)''']''"''n. I

+ [1 + (0.30)2fn + [1 + (0.559)2]5/2

0 , = 1.061 11.129

+

3.674

+

1.612 + 0.507]

1

= 7.34 kN/ro 2

Dlustratlve Example 11.12. Determine the vertical stress at a point P which is 3 m below the ground surface and is on the centre liM of the e~nt shown in Fig. E 11.12. T~ 1 .. 18 IcNlmJ•

SfRE$SES DUE TO APPUED LOADS

2-'3

Fig. E 11.12.

Solution. From Eq, 11.46, 0: ..

~

[a (a, +

aV

+ b

a,l

In this case, a = 12111 and b = 2 111, q = 6 x 18 = 108 kN/m2

a:z -

Ian al .. 2/3.0 .. 0.667;

tan (a, + all .. Therefore,

a: ..

3~~

.. 4.667;

~ Xx 1~

0,588 radians

(at + all '" 1.359 radians

[12 (1.359) + 2 x (1.359 - 0.588)} .. 102.3 kN/m!

PROBLEMS A. Numerical 11.1. A monument weighing 15 MN is erected on the ground surfoce. Considering the load ns determine the vertical pressure directly under the monument at

8

concentrated one,

a depth of 8 m below the ground surface.

(Ans. 111.9 kNlmlJ 11.2. A concentrated 1000 of SO kN acts on the surfoce of a homogeneous soil IllllSS of large extent. Determine the stress intensily at 8 depth of 5 m, directly under the load, and at a horizontal distance of 2.5 m. (Ans. 0.955 kN/m2; 05 5 kN/m2, 1l.3. Two columns A and B arc situated 6 m aport. Column A transfers a load of 5pO kN Dnd column B, a load of 250 kN. Determine the resultant vertical stress on a horizontal plane 20 m below the ground surface nI points 2 vertically below the points A and B, [Ans, 59.8 kN/m ; 29.9 kN/m2] 11.4. An excavation 3 m x 6 m for foundation is to be m3dc 10 8 depth of 2.5 m below ground level in a soil of bulk 3 unit weight 20 kN/m . What effect this excavation will have on the vertical pressure at a depth of 6 m measured from the ground surface vertically below the centre of foundation? IN for m '" 0.43 and II '" 0.86 is 0.10. 2

[Ans. decrease 20 kN/m J 11.5. A square foundlltion (S m x S m) Is to earry Il load of 4000 leN. Colculllle the vertieal stress at a depth of 5 m below the centre of the founootion. IN '" 0.084 fot 1/1 .. n '" 0.50. (ii) Also determine the vertical stress using 1 ; 2 distribution. [Ans. 53.76 kN/m2 ; 40 leN/ml ] 11.6. A wllter tower has 8 drcular foundlltion of to m dillOlCter. If the totn] weight of the tower, including the 4 foundation, is 2 x 10 kN, calculDte the vertical stress at 8 depth of 25 m below the foundation level. 2 [Ans. 231.9 kN/m J 2 11.7. A reclllngulllr foundlltion, 3 m x 2.1 m, is perfecUy flexible and carries II lood of 300 kN/m . Determine the vertical pressure at 8 depth of 5 m below a.point P shown in Fig. P 11.7. (Ans. 31.8 kN/m1 11.8. The contact pressure for a square footing 2 m )( 2 m is 400 leN/ml. Using 1 ;'2 distribution, determine the depth 2 81 which the contnct pressure is 100 kN/m . [Ans. 2 m] 11.9. A rerungular foundation 20 m x to m subjects the subgrade to a contaCl pressure of 2SO kN/m2. Detennine the vertical stress at a point P loalled 8t a depth of 5 m (Fig. P 11.9). (Use Thble for IN values) (Ans. 3.375 kN/m 21


l§±J 1)·7m

L

1--1 m

---+-- 1m ---1-1 m --l Fig. P 1l.7.

Fig. P 11.9.

11.10. A 1000 kN lood is uniformly distributed on II surface area of 3 m x 2.5 m. Find the npproximmc value of vcrlicrll stress al a depth of2 m. using. (i) 2: 1 distribution 2 (il) 60 0 distribution. (AIlS. 44.4 kN/m ; 392 kN/rn1 U.U . A concentrated lood of 1000 leN ncts vertically at the ground surface. Determine the vertical st TCS$ at D point which isot (I) a depth of 2.5 m and a horizonllli distance of 4.0 m 2 (il) DI a depth of ~.O and D radial distance of 2.5 m. [Ans. 3.2 kN/m ; 10.93 kNlm1

n. Descriptive and Objective lype ll.l2. State the assumptions made in computing Slte$eS belOW' the ground surface due to II point lood acting on it. Discuss mcirvalidity in practice. H.l3. Derive an expression for the vertical stress al a point 'due 10 3 poinl load, using Boussinesq's theory. n.H. What do you undersl.Ilnd by gcosttllic stresses? How nrc these determined ? 11.15. What is an influence diagram? What is its use in practice ? 11.16. Derive an expression for the vertical stress at a point due to a line load. Give examples of a line load. 11.17. How would you determine the stresses at a point due to a (a) Strip 1000 (b) circular lood. Compare the zones of influence due to the tlllO types of loads. 11.18. Describe the method of allculating the stress 81 a poinl below the oomer of a rectangular lcod. How is this method used for finding lIIe Slresses at points other than that below the comer 1 11.19. Discuss the basis of lIIe construction of Newmark's influence chan. How is il used ? 11.20. Explain Westergaard's theory fot the determination of the venicnJ stress at a point. How is it different from Boussinesq's solution 1 11.21. What is Fenske's chan? Explain its construction and use. 11.22. Oiscuss various approximate methods for the determination of the venical stress at a point What nre their limitations ? 11.23. What do you understand by contact pressure? What are lIIe factors that affect the contact pressure distributions? Drow the contact pressure distribution diagram for flexible and rigid footings on sand and d ayey soils. 11.24. Mention whether the following statements are true or false. (a) The vertical Slress due 10 a point 1000 depends upon modulus of elnsticity. (b) FOl' determination of the deformation, the secant modulus at Ute peak Sfress is used. (e) The Poisson rotio for most of the soils is zero. (d) The horiwntai stress can be more than the vertical stress. (e) The Boussinesq Influence coefficient just belOW" the pain! load is zero. (/J The maximum shear stress due to a strip load is constant at all points. (g) The wne of influence due 10 a circular load is deeper than thai due to 8 strip load. (h) While determining Newmark's influence ooemcient, the constant m and n can be interchanged.

255


(1) 'The Boussinesq solution always gives stresses greater than the Wes tergaard solutioll.
C, Multiple-Choice Questions 1. The stress developed at !I point in the soil exactly below a point load a\ the surface is (1I) proportional to the depth of point. (b) proponiooal to the S(lu!lre of the depth of point. (e) inVersely proporlionnl to the depth of point. (d) inversely proponiolllil to the square of the depth of point. 2. An isobar is a curve which (a) joins poims of equal horizontal stre.~s. (b) joins points of equal verlkal mess. (el Joins points of zero vertical ~tress. (d) joins points of maximum vertical stre.~s. 3. If the entire semi-infinite soi l mass is loaded wilh a load intensity of q at the surface. the vcnicnl stress at any depth is equal to (a) q (b) 0.5 q (e) zero (d) infinity 4. For a strip of width 8 subjected to a load intensity of q at the surface, the pressure bulb of intensity 0.2 q extends to a depth of (u) 38 (b) 6 B (e) 1.5 B (d) B S. Newmark's intluence ch'lrI can be used for the determina.tion or verlical slres.s under (II) circular load area only ( b) rcctangul:!r 100000 area only (el strip loud unly (tI) Any shape of loaded area 6. The WC~lergaard analysis is used for (a) homogeneous soiL~ (b) cohe.~iYe soils' (e) sandy soils (d) slratilicd soils 7. A concentrated load of JOOO kN acts vertically al a point on the soil surface. According 10 Boussinesq's equation the ratio of the venical stresses at depths of 3m and 5m is tal 0,35 (b) 0.70 (c) 1.75 (d) 2.78 8. A load of 2000 kN is uniformly distributed over an area of 3 m )( 2m. TIle average venical stress at a depth of

~a~ ~~~n~~/~J (e) 48 kN/ml

distrihution is

(h) 100 kNhnl (tl) 37 kN/m2 ~_l~~~~~~~~~~~~~&~

12 Consolidation of Soils 12.1. INTRODUCTION When a soil mass is subjected to a compressive force. like all other mmerlals. its volume decreases. The property of the soi l due to which a decrease in volume occurs under compressive forces is known as the compressibility of soil. 111e compression of soils can occur due to one or more of the following causes. (I) Compression of solid particles and wmer in the voids.

(2) Compression and expulsion of air in the voids. (3) Expulsion of water in the voids. Compression of solid particles is negligibly small. Compression of water in the voids is also extremely smail, as the walcr is almost incompressible in the fange of stresses involved in soil engineering. Therefore, the compression due to the first cause is not much significant. Air exists only in partially saturated soils and dry soi ls. The compression of the air is rapid as it is highly compressible. Funher. air is expelled quickly a" soon as the load is applied. However. the compression due to the second cause is not relevant for saturated soils. When the soil is fully satumted. compression of soi l occurs mainly due to the third cause. namely. expulsion of water. As this chapter is mainly concerned with saturated soils. only this cause is relevant. The compression of (l saturated soil under a steady static pressure is known as cOllsolidation. It is entirely due to expuh';oll of W(jfe/' Jm/ll the voids. It is similar to the action of squeezing of water from a saturared sponge under pressure. The soil behaves as a saturated sponge. As the consolidation of soils occurs. the water escapes. The solid particles shift from one position 10 the other by rolling and sliding and thus attain a closer packi·ng. It is worth nOling that the decrease in volume of soil occurs not due to compression of solids IX' water but due 10 the shifting of positions of the particles as the wfller escapes. Small volume changes may occur due to bending, distortion and fracture of the soild porticles. but such chooges are insignificant in the ordinary range of stresses involved in soil engineering problems. However. bending. distortion and fracture are indirectly responsible for a further decrease in volume due to shifting of particles. Settlement of a structure is its vertical. downward movemenl due to a volume decrease of the soil 01'1 which it is buill. In other words. the settlement is the gradual sinking of a structure due to compression of the soil below. A study of consolidation characteristics is extremely useful for forecasting the magnitude and time of the settlement of the structure. 111e compression of soils due to expulsion of air due to dynamic methods. such as rolling and tarnpin.. is known as compac tion (see chapter 14). This chapter deals mainly with consolidation of fully saturated soils. (Note. In geology, consolidation means hardening of soils due to solidification. It should not be confused with the word consolidation used in soil engineering).

CONSOUDATIOH OF SOILS

,. "

'. y. Y id

"

"'y

g.

12.2. INITIAL. PRIMARY AND SECONDARY CONSOLIDATION The coosolidation of a soil deposit can be divided into 3 stages : (1) Initial Consolidation. When a lo.1d is applied to a partially saturated soil, II decrease in volume occurs due to expulsion and compression of air in the voids. A smaU deaease in volume also occurs due to compression of solid particles. 'Ine reduction in volume of the soil just after the application of the load is known as initial consolidation or initial compression. For saturated soils, the initial oonsolidation is mainly due to compression of solid panicles. (2) Primary Consolidation. After initial consolidation. further reduction in volume occurs due to expulsion of water from voids. When a saturated soil is subjected to a pressure, initially aU the applied pressure is taken up by water as an excess pore water pressure. as water is almost incompressible as compared with solid panicles. A hydraulic gradient develops and the water starts flowing out and a decrease in volume occurs. The decrease depends upon the penneability of the soil and is. therefore. time dependent. TIlls reduction in volume is called primary consolidation. In fine-grained soils, the primary consolidation occurs over a long time. On the other hand, in coarse· grained soils, the primary consolidation OCQlrs rnther quickly due to high permeability. A.. w3lcr escapes from the soil, the applied pressure is grndually transferred from the water in the voids to the soild particles. Thus, tbe effective stress is increased. (3) Secondary Consolidation. The reduction in volume continues at a vcry slow rate even after the excess hydrostatic pressure developed by the applied pressure is fully dissipated and the primary oonsolidation is romplefe. This additional reduction in the volume is called secondary consolidation. lbe causes for seoondary consolidation are nOi fully established. It is attributed to the plastic readjustumem of the solid particles and the adsorbed water to the new stress system. In most inorganic soils. it is generally small In the discussions that follow the word consolidation means primary consolidations unless otherwise stated. lbe primary consolidation is the most important component of the total consolidation and forms the main Subject matter in Ihis chapter. 12.3. SPRING ANALOGY FOR PRIMARY CONSOLIDATION The process of primary consolidation can be explained with the help of the spring analogy given by Temlghi. Fig. 12.1 (0) shows a cylinder filled with a tight·fiuing piston having a valve. The cylinder is filled with water and contains a spring of specified stiffness. Let the initial length of the spring be 100 mm and tbe stiffness of spring be 10 mm/N. Let us assume that the piston is weightless and the spring and water are iniLiaIly free of stress. When a load P (say, I N) is applied to the piston, with its valve closed, the entire load is taken by water (Fig. 12.1 (b)]. The stiffness of the spring is negligible compared with that of water, and consequently, no load is taken by spring. From equilibrium, vot~.

Valve open

closed.

T

100mm

lL--L----.J (0)

.. 0

btl

Ib)

10)

Fig. 12.1. Spring Analogy.


258

Po.' + P¥ ... P where P w

= load

taken by water, p.

= load

... (12.1)

taken by spring, and P

= total

load.

ForP= 1 N. Eq. 12.1 becomes

P",+Pr-1.0

... (12.2)

Initially (t = 0) when valve is closed, Ps - 0.00. Therefore, P", - 1.0 If the valve is now gradually opened, water starts escaping from the cylinder. The spring starts sharing some load and a decrease in its length occurs. When a ponion (6P) of the load is transferred from the water to the spring, Eq. 12.2 becomes 6P + (1.0 - 6P) - 1.0 10~-,------------~ As more and more water escapes. the load carried by " IRe spring increases. Fig, 12.2 shows the transfer of the load from tbe water to the spring. Eventually, when the

steady conditions are established, the water stopS

t

c

escaping. Finally. at time t '" tr. the entire load is laken by spring. Thus, P w = 0 and P, = 1.00. This load caUS!!S a decreases in length of the spring %~~~----------~tf by 10 mm. The final length is 90 mm, as shown in Fig. TIME ( t ) - 12.1 (c). As the load carried by water is zero, it is again free of excess pressure. Now if the valve is closed and the Fig. 12.2. Load Sharing between' spring and waler. load P is increased [0 2N, the process of transfer of load repeats and finally the spring takes the complete load and its length becomes 80 mm. Likewise, the process is repeated till the final increment of the load has been applied.

9

12.4. BEDAVIOUR OF SATURATED SOILS UNDER PRESSURE The behaviour of saturated soils when subjected to a steady, static pressure is similar to that of the spring analogy model. The solid particles in the saturated soil behave like springs, while the water in the voids behaves like water in the cylinder. The perm~bility of the soil controls the flow of water and it can be likened to the valve in the pistOn. The pore water pressure (il) in the soil is analogous to the pressure carried by water in the cylinder. Finally the stress developed in the spring is analogous to the effective slress @) developed in the soil. When a pressure . .6.01 is applied to a saturated soil sample of unit cross-sectional area', the pressure is shared by the solid particles and water as

.6.0;' U - .6.01 ... (12.3) Initially, just after the
G

L)

t)

25.

CONSOUDATION OF SOILS '0

the void ratio is (e/)I' It must be noted that the curve shOW"n in Fig. 12.3 is drawn for applicahon of onc pressure increment .101' If the applied pressure is now increa<>ed to Aaz, the process of load transfer repeats and the soil attains eventually a different final void ratio (elh when the entire load is transferred to the solid particles. A curve can be dawn bciwccn the final void ratios and me coITesponding effective Stresses for different lood increments. (Fig. 12.4). It may be noted that as the effective stress increases, the final void ratio decreases, and. therefore, the .. 0 volume of the soil decreases. The reduction in Fig. 12.3. Variation of void nuio with lime volume is due to expulsion of water from the voids under excess hydrostatic pressure and is, therefore, primary consolidation.

[2.5. CONSOliDATION TEST

t (~t1 ---~ ---L--

The consolidation test is conducted in a JabOT:J I coosists of a loading device and a cylindrical container (t'f ---,---r--,---called consolidation cell. The soil specimen is placed in (pf) the cell between lOp and bottom porous stones. The ,I I I consolidation cells are of two types: (1) Ao.'lling or free K , 6i'2 ~3 6"4 lIi"S ring cell [Fig. 125 (b)J in which both the top and bottom porous stones are free 10 move. The top porous stone can Fig. 12.4. Plot between "/ and o. move downward and the bottom stone can move upward as tbe sample consolidates. (2) Fixed ring ccll [Fig. 12.5 (c)] in which the bottom porous stone cannot move. Only the top porous stone can move downward as the specimen consolidates. The fixed ring cell can also be used as a variable--head permeability test apparatus. For this purpa:;e, a piezometer is attached to the base of the cell. The inside surface of the ring should be smooth and polished to reduce friction. The ring imposes a condition of zero lateral strain on the soil sample. The internal diameter of the cell is usually 60 mm, but tbe

----:---T-

---:---r-l----r--

r

0,

-.l

Fig.

12~"i.

Comolidatioll T($I.


cells with a diameter upto 100 mm are also available. 'The thickness of the sample is fixed from the following considerations: (1) The thickness of the sample should be as small as possibJe to reduce side friction, but a minimum thickness of 20 mm is usually required to get uniform distribution of pressure on the Sc'lmple. (2) The diameter to the thickness ratio should be a minimum of 3. (3) The thickness of the specimen should not be less than 10 times the maximum sile of the particle. The thickness of the sample for a 60 mm diameter cell is usually taken as 20 mm. The specimens of diameter 50, 70 and 100 mm may be used in special cases. The consolidometer has arrangements for the application of the desired load increment, saturation of sampie and measurement of change in thickness of the s.'lmple at every stage of consolidation process. The consolidation cell is placed in a water jacket or water trough SO that water ha<> free access into and out of the sample. The cell is provided with a perforated pressure pad at its top for the application of load. The load is applied either by suspending weights from a hanger rcsting at the centre of the pressure pad or by a lever arrangement. The arrangement for saturation of the sample consists of a small water reservoir connected to the c.:ell with a plastic tube (not shown in figure). A dia1 gauge is used to measure the change in thickness as the consolidation takes place. The sample is kept submerged under water to prevent evaporation from its surface. Before conducting the test, the ·porous stones are saturated either by boiling them in distilled water for about 15 minutes or by keeping them submerged under water for 4 to 8 hours. 'The bottom porous Slone is fi rst placed in the consolidation cell and a filter paper is fixed on the porous stOlle. The ring containing the sample is then placed on the bottom porous Slone. Another filter paper is kept on the top of the sample and then top porous stone is placed. The loading pad is placed on the top porous SiOOC. The bolts are lightenoo SO as to hold tbe entire assembly, and then the consolid.'ltion cell is kept under the loading unit. It should be centred carefully so that the load is applied axially. The dial gauge is mounted and adjUSted. The mould assembly is connected to the water reservoir to saturate the Sc'lmple. The level of water in the reservoir should be approximately same as that of the sample. An iniual setting pressure of about 5.0 kN/m2 (for very soft soils, 2.5 kN/m2) is applied to the sample. The initial setting pressure is chosen such thm there is no swelling. The load is allowed 10 stand till there is no change in the dial gauge reading or 24 hours whichever is less. The rmal dial gauge reading under the initial setting pressure is noted. The first increment of load to give a pressure of 10 kN/m2 is then applied to, the specimen. The dial gauge readings are taken aft~ r 0.25, 1.0,2.25,4.0,6.25,9.0, 12.25, 16.0, 20.25,25,36,49, 64,81, 100, 121, 144, 169, 196, 225, 289, 324, 400, 500, 600 and 1440 minutes (24 hours). Sometimes, after 49 minutes, readings are taken at 1. 2, 4, 8, 10 and 24 hours. The primary consolidation in the sample is usually complde within 24 hours. The second increment of the load is then applied. It is usual practice to double the previous load in eacll increment. The successive pressures usually applied are 20, 40, BO, 160, 320 and 640 kN/m 2, etc., till the desired maximum required load intensity is renchcd. lbe maximum load intensity is governed by the actual loading on the soil in the field after the construction of the structure. After the consolidation under the final load increment is complete, the load is reduced to one-fourth
(See Chapter 30, Sect. 30.15 for the laboratory experiment).

,.

,I

,[

>d

,.

"

is

CONSOUDATION OF

spn..s

261

12.6. DETERMINATION OF VOID RATIO AT VARIOUS LOAD INCREMENTS The results of a consolidation lest are plaited in the fonn of a plot between the void ratio and the effective stress. It is, therefore, required to detennine the void ratio at various load increments. There are two methods : (1) Height of solids method. (2) Change in void ratio method. The first . method is a general method applicable to both s.'lturated and unsaturated soils. The second method is applicable only to saturated soils. (1) Height of Solids method. In this method. equivalent height of solids is determined from the dry mass of the soil. The height of solids is given by

Hs-t-(~w)'± where

... (12.6)

= height of solids, Vs = volume of solids, M, = dry mass of sample, G =specific gravity of solids, A =cross*scctional area

II~

of specimen,

From definition of void ratio,

e _ volume of voids _ V - V, volume of solids V,

... (a)

Eq. (n) can be written as, (AxH)-(AxH,) H-Il, (AxH,) -~

...(12.7)

where H is the total height (total thickness). Thus, the void ratio is determined from the total height (H) and the height of solids. The total thickness of tbe sample is measured at least once during the test, usually either before the start of the test or at the end of tbe test. At other stages of loading, the thickness H is worked out from the measured thickness find the difference in dial gauge readings. . .. (12.8) Therefore, 11- flo:t 1:6.H where flo = initial height and t:Jf = ch3nge in height. See Thble 12.1 for the illustration of tbe method. Table 12.1, Computation of Void ratios by Height or Solids Method 2 A = 50 cm , Volume = 125 ml, Given data 110 - 25 mm,

G = 2.67, Ms - 100.24 gm. Least counl of dial gauge = om mm

wf= 24.94%

CalculaJions

Observalions Applied pressure (kNlm 2)

Dial gauge reading

0.0 10.0

490 482 470 431 390 343

20.0 40.0

".0 160.0 320.0

640.0 0.0

295 249 364

Change in IhicJUless tJl (mm) - 0.08 -0.12

- 0.39 - 0.41

- 0.47 - 0.48 - 0.46 + 1.15

H=Ho±T.l:JI

H-Hs

25.00 24.92

10.75

24." 24.41 24.00

23.53 23.05 22.59 23.74

10.67 10.55 10.16 9.75 9.28

8." 8.34 9.49

e 10m Eq. (a) -(H-H,)!H, 0.754 0.748 0.740 0.713 0.684 0.651 0.617 0.585 0.666


From Eq. 12.6,

fl• ..

2.~~~2~.O xi>"

1.425 an

.. 14.25 mm

e .. /1 0- 11• .. 11 - 14.25 HI 14.25

From Eq. 12.7,

... (a)

Obviously, the iniUnl void ratio (eo) at the start of the test is given by

110 - fI,

e ." FOr an intermediate stage,

-----n;--

... (12.9)

H- H,

e .. ~

... (12.10)

After dctennination of the void ratio and the water content at the beginning and at the cnd of the tesl, the corresponding degree of satumtion can be found from the relation, S .. wGle. From the calculnted void ratiOS, a plot of "e' versus log 0 can be made, as shown in Fig. 12.8.

(2) Change in Void milo method. In this method, the final void ratio (e/). corresponding to oompletc

swelling conditions after the load has been removed, is determined from its water content, using the equation,

e, .. wG

assuming that the soil is fully saturated.

The void ratio corresponding to intermediate loading stages is determined as explained below. From the definition of void ratio,

e_v-v·_~_1 V, V, where V = total volume and V. is the volume of solids. Eq. (a) can be written as V - V, (l + e) A )( 11 - V, (1 + e) where A is the cross-sectional area of the specimen and /-I is its total height. 3y partial di[Jerentiation of (b), A dH _ V, de

From Eqs. (b) and

(e~

7f -

... (a)

... (b) ... (e)

1':'e

Ae_O;je)AH

... (12.11)

Eq. 12.11 can also be derived directly, taking the volume of solids as unity and the cross-sectional area also
6e

6H

""I"+e-H l!e -

As the void ratio

!!.jf1 AN

... (same as 12.11)

e, and the total height H of the sample arc known at Ihe end of the test, the void r"ollio

at any other stage can be determined from the change in thickness AH measured by the dial ga~ge. Thus, tbe change in void ratio (Ae) under each pressure increment is calculated from Eq. 12.11 bY ,working backwards from the known value of void ratio at the end of the lest afier swelling.

e,

Thus

(1 +e,)

Ae_~AH

... [12.11(')1

263

CONSOUDATION OF SOILS

where Ml is the change in thickness, as

me~urcd

with respect to the thickness Il, at the end of the test.

See Table 12.2 for the illustration of the method.

Table 12.2. Computation or Void Ratios by Change in Void Ratio Method Given Data flo .. 25 mm, G = 2.67,

e, From

WI )(

A = 50 cm 2•

Volume = 125 ml,

W, _ 24.94%.

H/ .. 23.74

~~7~)

6 e - (I

Least count of dial gauge =

om

)( 6. fl -

0.0702 6. H

... (4)

mm.

Obsef1l(1tions

CaJcu/miolls Challge ;11 thickness

Dial gauge reat/ing

H.Hoj:l:6.ff

6.J1(mm)

490 482 470 431 390 343 295

0.0 10.0 20.0 40.0 BO.O 160.0 320,0 640.0 0.0

= 190.24 gmt

G - 0.2494 )( 2.67 - 0.666.

Eq. 12.11 (0),

Applied pre.ssuree (kNlm,>

M~

....08 - 0.12 - 0.39 - 0.41 -0.47 - 0.48 - 0.46 +1.15

24.

364

25;.00 24.92 24.80 24.41 24.00 2.1.53 23.05 22.59 23.74

A, (from Eq. (d) +

0.006 0.008 0.027 0.029 0.033 0.034 0.032 0.081

0.754 0.748 0.740 0 .7 13 0.684 0.651 0.617 0585 0.666

f

12.7. CONSOLIDATION TEST RESULTS (1) Dial gauge reading-lime piOI. Fig. 12.6 (0) shows the plot between the dial gauge reading and lime for a typical load increment for clay and sand samples. The thicknes.
Cloy

~

~

g

"-.......... ..,.,-SOnd

--------Ti m~

(0-

Co)

Cb)

Fig. 12.6. (a) Dial gIluge reading- li me plot. (b) Final void f3tio-a plot

slows down as the time passes. There is practically no change in thickness after 24 hours. The oonsolidation al Ihal load increment is considered to the. complete at 24 hours. for sand, tbe change in thickness occurs very quickly and stops after a few minutcs. This is due 10 high penncability of lhe sand which pennits easy Dow of water.


264

The plot betwccn the dial gauge reading and time is required for determining the coefficient of consolidation. which is useful for obtaining the rate of consolidation in the field. (2) Final void ratio--efTectlve slres.q plot. The thickness of the specimen after 24 hours of application o[ the load increment is taken os the final thickness [or that increment. The final void ratio (el) corresponding to the final thickness for each increment is determined using the methods discussed in the preceding section. Fig. 12.6 (b) shows the plol between the fin.'li void ratios, (elb (efh, (e,») ... etc. and the corresponding effective stresses 0\.(20), ... for load increments 1. 2. 3, ... elc. As the sand is relatively less compressible, the change in void ratio is small. TIle plot between the final void ratio and the effective stress is required for detcrminaUon of the magnitude of the consolidation senlement in the field. The reader must carefuUy note the difference between Fig. 12.6 (a) and Fig. 12.6 (b). The former shows the process of consolidation under a particular increment. For each load increment, 8 plot like Fig. 12.6 (a) can be plolted. The latter shows the plot between the final void ratios reached under diffcrelll load increments and the corresponding effective stresses under those increments. (3) Final 'Void rutio---Iog CJ plot. Fig. 12.7 (0) shows a plot between the final void ratio and the effective Stress, which is similar to one in Fig. 12.6 (b). For convenience the suffix f has been dropped. The curve bas

Log
Ib)

Fig.12.7.(Q)~-aplor.

(b)~-logaplOl

concavity upward. The slope of the curve at different points is different. The slope decreases with an increase in effective stre..<;s. It is more common to plot the results on a semi-log graph. in which the final void ratio is plotted on the natural scale and the effective stress as abscissa on the log scale [Fig. 12.7(b). A The plot is practically a straight line for a nonnally consolidated clay (defined laler) within the range of pressure usually encountered in practice. (4) Unloading and Reloading plot. In Fig. 12.8, the curve An indicates the decrease in void ratio with an increase in the effective stress. 1l1e curve is similar to ,0 UNLOAD'NG E one shown in Fig. 12.7 (b). It is the '>i:: G loading curve. After the sample has reached I equilibrium at the effective stress of as
,,,

oz.

and swell lbe curve BEe is obtained in

"2

Fig. 12.8. Loading, untoa
,.,

CONSOUDA110N OF SOILS

unloading. This is known as the expansion curve or swelling curve. It may be noled that the soil cannot attain the void ratio existing before the start of the test, and there is always some permanent sct or residual deformation. If the specimen which has swelled to the point C is reloaded, the recompression curve CFD is obtained. As the load approaches the maximum value of the lo."ld previously applied corresponding to point B, there is reversal of curvature of the curve and then the plot DG continues as an extension of the first loading rurve AB. However, the reloaded specimen remains at a slightly lower void ratio at point D than that attained at B during the initial compression for the samc load.

12.8. BASIC DEFlNmONS The following basic definitions related to consolidation are of paramount importance. (1) Coefficient of ComJKtiSibility. The ooefficieot of compressibiUly (a~) is defined as deuea<;e in void ratio per unit increase in effective Stress. It is equal to the slope of the e - 0 curve at the point under coosidcration [Fig. 12.7 (a»).

-de

Thus

a,. - do -

-6e .6 0

... (12.12)

As the effective stress increases, the void ratio decreases. and therefore, the ratio del dO is negative. The minus sign makes a~ pooitive. For convenience, the coefficient of compressibility a~ is rcported as positive.

As the value of a~ is different at various effective stresses, while reporting its value, the effective stress to which that value corresponds must be mentioned. The coefficient of oompressibility decreases with an increase in the effective stress. In ocher words. Ihe soil becomes stiffer (less compressible) as the effective stress is increased and the curve becomes flaller. The coefficient of compressibility (a,,) has the dimensions of [L 2/F). The units are m2/kN. It may be noted that the units are inverse of that for pressure. (2) Coefficient of Volume change. The coefficient of volume change (or volume oompressibility) is defined as the volumetric strain per unit increase in effective stress. Thus m~

where

..

- A VIVo

~

... (12.13)

= coefficient of volume change, Vo = initial volume, AV "" change in volume, and A 0' "" change in effective stress.

m~

The reader should note that the coefficient of volume change is inverse of the bulk modulus used in solid medianics and fluid mechanics, For most clays, m~ _ 1 )( 10-3 to 1 )( lO-4m2JkN. The volumetric strain (.6VIVo) can be expressed in tenns of either void ratio or the truckness of the specimen as explained under: (a) Let eo be the initial void ratio. Let the volume of solids be unity. Therefore, the initial volume Vo is equal to (1 + eo). If Ae is the change in void ratio due to change in volume AV, we have AV - Ae. Thus AV

Ae

Yo - r:;e;; Therefore, Eq. 12.13 becomes

m~ .. - A e~(~ + eo)

...(12.14)

(b) As the area of cross-section of the sample in the consolidometer remains constant, the change in volume is also proportional to the change in height. Thus .6 V .. tJI AV .6H Therefore, where Ho

V;" Ho

=

initial height.


266

Therefore, Eq. 12.13. becomes

or

m~

..

- Mllllo

... (12.15)

~

... [12.15(a)J

!J.H .. -n1" HoAo

The relationship between

Q ..

and

In ..

can be obtained from Eqs. 12.12 and 12.14 as

.. 1 :veo

m~

...(12.16)

like Q .. • the coefficient of volume change m .. depends upon the effective stress at which it is determined. Its value also decreases with an increase in the effective stress. The unit of m" is the same as that of a,~ However, the coefficient of volume change m.. is more commonly used in practice than the coefficient of compressibility Q", (3) Compression Index. The compression index (C() is equal to the slope of the linear portion of the void ratio versuS log plOI [Fig. 12.7 (b)]. Thus

a

c( ..

JOg:

~(j~o)

... (12.17)

where 00 = initial effective stress, '0 = final effective stress, de "" change in void ratio

Sometimes, Eq. 12.17 is expressed as Cc

..

-a!:

6.

(J

10glO(~)

...

(12.18)

where 60 is the change in effective stress. The numerical value of C c can be easily determined from the difference in void ratio corresponding to one log cycle. Thus ... (12.19) The compression index is extremely useful for determination of the settlement in t!le fiek!. The compression index of a clay is related to its index properties. especially the liquid limit. Terzaghi and Peck gave the following empirical relationship for Clays of low to medium sensitivity «Sf $ 4). (a) For undisturbed soils,

Cc

,.

0.009 (wL - 10)

(b) For rcmoulded soils,

where wL = liquit limit (%). The value of Cc normally varies between 0.30 for highly plastic clays cmd 0.075 for low plastic clays. The compression index is also related 10 the insilu void ratio eo or water content (wo) as under C, _ 0.54 (co - 0.35) C, _ 0.0054 (2.6wo - 35)

... (12.21) ... (12.22)

The coeITicient of compressibility av may be calculated' from the compression index as under:

a,. - 0.435

50.

where 0", is tile average pressure for the increment. (4) Expansion Index. 'The expansion index or swelling index (CO') is the slope of the e-Iog obtained during unloading (BEC in Fig. 12.8). C _ de

• 10 (0+40) g lO~

a

plot

... 12.23)

1b7


As it is evident. the expansion index is much smaller than the compression index.

(S) Recompression Index. Recompression is the compression of a soil which had already been loaded and unloaded. The load during rerompression is less than the load to which the soil has been subjeaed previously. The slope of the recompression curve obtained during reloading ( CFD in Fig. 12.8) when plotted as e-log 0, is equal to the recompression index (C.). Thus

C _

~

-lie

10g(0+:0)

... (12.24) .

lbc reoompression index is appreciably smaller than the compression index C~. It is usually in the range of 1/10 10 1/5 of lhe compression index.

(6) NonnaUy consolldnted and Over-<:onsolidated clays. A normally consolidated soil is one which had not been subjeacd to a pressure greater than the presenl existing pressure. A soil is said to be over-consolidalcd if it had been subjected in the past to a pressure in excess of the present pressure. The {Xlrtion AD of the curve in Fig. 12.8 represents the soil in normally oonsolidaled condition. The curve in this ronge is also called the virgin compression curve. '!lIe soil in the range CD when it is recompressed represents overconsolidatcd condition. as the soil had bccn previously subjected to a pressure 02> which is greater than the pressure in the range CD. lbe maximum pressure to which an over-consolidated soil had been SUbjected in the past divided by the present pressure is known as the overconsolidation raLio (O.C.R.). For example, the soil indicated by the coodition at point C bas an over-consolidatioa ratio of o:0'J' It may be emphasized that normally consolidated soils and O\lerconsolidated soils are not different type<> of soils but thcsc are conditions in which a soil exists. The same type of soil can behave as nonnally consoLid..lted in a certain pressure range and an over-consolidatcd in somc other pressure rangc. For example, in Fig. 12.8, the soil which behaves as overc:onso1idat~ in the range CD would again behave as normally consolidated in the range DG. The liquidity index of a normally consolidated clay is gen,erally between 0.6 and 1.00, whereas that for an over-consolldated clay between 0.0 and 0.60. As the recompression index (C~) is very small as compared with Ihe compression index (C~), the soils in the overconsolidated Slate have smaller compressibility. The settiemenlS of the strudures built an over-consolidated clays are small. (7) Vnderconsolldated clays. If the Clay deposit has not reached equilibrium under the applied overburden loads, it is said to be underconsolidated This normally ocrurs in areas of recent land fill. 12.9. TERZAGill'S TDEORY OF CONSOLIDATION (1) Assumptions. Terzaghi (1925) gave the theory for the determination of the rate of consolidation of a saturated soil ma

268

remains constant during the load increment. In other words, the coefficient of compressibility and the coefflcienl of volume change are oonslant.

Comments on the assumplions '!be assumptions made by Tcrzaghi are not fully satisfied in actual field problems. The results obtained from the use of the theory to practical problems are approximate. However, considering romplcxity of the problem. the theory gives reasonably acrurnle estimate of the lime rate of settlement of a structure built on the soil. A brief oomment on the various assumptions and their effect is given below. Assumptions 1 to 3 are generaUy satisfied for fully saturated, clay deposits. However, the presence of air may affect the accuracy. Assumptions 4 and 5 are not fully satisfied. 1n fact. the coefficient of penneabilily varies at different points in the deposit. Its value dccreascs as the consolidation progresses due 10 an increase in the effective stress. Further, at vcry low hydraulic gradicnts. the Darcy law in not strictly applicable. Fortunately, the errors introduced due to these assumptions ore smalL The largcst error is probably due to the assumption 6. In the field. the consolidation is usually 3·dimensional and not one· dimensiomli. Howevcr. in the c.'lSe of deposits having large areas compared with their thicknesses have essentially one·dimensional consolidation and thc error is not mud). Assumption 7 is oot fully justified, as some secondary consolidation does occur along with the primary consolidation. However, for most inorganic soils, the secondary consolidation is small and does oot introduce much error. The actual relationship between the void ratio ood the effective stress is not linear. oontrary to the assumption 8. However, if 0 largc number of samples are taken from the same stratum and an average value of the coefficient of volume change (111,,) is taken for the appropriate range of the effective stress, the error introduced due to this assumption is not unduly high. The only justification for making this assumption is to get a relatively simple expression. The thoory becomes more complex when actual relationship is used. (2) Derivation or DllTerential Equation. The basic differential equation of one-dimensional consolidation can be derived as under: Let us consider a S
T

r~-'-''-'--T-'-'--'--'-'-1S!l-;I}--\f-*,.t¥--

T

H:2d

Fig. 129. Exoess HydroMnlic Pressllre.

Ilo:rl~"''''h

1 Wa..::r

tabl.

,..

CONSlOUDATION OF SOILS

excess hydrostatic pressure is independent of the position of the water table. Foe convenience. the water table is assumed at the level of the surface of the clay layer. Water starts escaping towards the upper and the lower sand layers due to excess hydrostatic pressure developed. The hydrostatic exress pressure at the top and the bottom of the clay layer, indicated by points C and E in the pressure diagram. drops to zero, However. the excess hydrootatic pressure in the middle portion of the cIay layer al D remains high. The curves indicating the distribution of excess hydrostatic pressure are known as isochrones. The isochrone CDE indicates the distribution of excess hydrostrltic pressure at time t. As the consolidation progresses, the excess hydrostatic pressure in the middle of the clay layer decreases. FiroUy al time t _ 'I. the whole of the exces.... hydrostatic pressure has been dissipated, and the pressure distribution is indicated by the horizontal isochrooe CFE. Let us consider the equilibrium of an element of the clay al a depth of z from its lOp al lime I. The consolidation pressure 6.0 is partly carried by water and partly by solid particles as 6.0" flo + (12.25) where 6. is the pressure c.1rried by solid panicles. and Ii is the excess hydro-static pressure (pressure units)

u

a

...

The hydraulic gradient (I) althal depth is equal to the slope of the i'>OChrone CDE at a horizontal distance the point C in the pressure diagram.

z from

i ..

Thus,

~

_ 0 (iily...,) ..

oz

oz

..!..

1...,

(El.)

...

a z

(a)

where Ii is the excess hydrosl3lic pressure at depth z. The expression for the hydraulic gradienl i can also be obtained as under. Let us consider a thin slice of clay laycr6.z al depth z (Fig. 12.10). The pressure difference (Ail) aa-oss this thickness is given by

flU_(U +¥z til)-U _*.til The unbalanced head across the thickness is given by

Ah-~-.!..(§!.. y.., yOO' az

dz)

The hydraulic gradient becomes

.!..(i§.)

i _ ~_ dz y..,

iJz

... [same as (a)]

T

1

From Darcy's law, lhe velocity of flow at depth z is bZ given by v_ki _k .

...!..(£'!.) az

1...

1be velocity of flow at the bottom of the element of UUctness Az am be written ~ v+

lberefore,

~'dz oz

-

v+

~['!'(i§.)l ilz y... ilz

Fig. 12.10. PrC:5Sure Difference on

/I

Thin Sliec.

dz . ..(1226)

OOIL MECHANICS AND FOUNDATION ENGINEERING

270

The discbarge entering the elemenl Qitt is Q", .. v{Ax )( by) where A x and Ay are the dimensions of the clemenl is plan.

The discharge leaving the element QOUI is Q"", .. ( v +

1; . dz ) (Ax )( Ay)

Therefore, the oct discharge squeezed out of the element is given by

AQ-Q_-Q" AQ - [

AQ ..

(v + Tz . dz) - v1Ax x Ay

¥z (ar x Ay )( Al)

... (b)

As the water is squeezed out, the effective stress increases and the volume From Eq. 12.13,

o~

the soil mass decreases.

where Vo '" initial volume of soil mass ( .. 6x Ay az) and li.o '" increase in effective stress. The decrease in volume of soil per unit lime

¥ . -m~(Ax

fly Al)

~

... (c)

As the decrease in volume of soil mass per unit time is equal to the volume of wntcr squeezed oul per unit lime, Eqs. (b) and (e) give

~ (Ax

Ay 6.z) ..

-m~

(6x !iy At)

~

i;--mv~ From Eq. 12.25,

Au .. A

aAa

---at ..

••• (<1)

0' - ; ;

ll§ iJt

-

8u

at

For a given pressure increment, at.. a .. O. Therefore. ~

a,

au

.. -at

i!!'. _ -m. (_~) az in

Therefore, Eq. (d) becomes

_ m.

(£¥.) at

... (12.27)

Equaling two values of iJvlaz from Eqs. 12.26 and 1267,

~ 1.

a'u _ m. (£¥.) az'

at

,a2 u au

c" a1- -

at

wbere c" is the coefficient of consolidation and is given by

... (1228)


271

k k .(12.29) c.. - - .. - 't",m.. gp ... m .. Eq. 12.28 is the basic differential equation of one-dimensional coosolidation. It gives the distribution of hydrostatic excess pressure Ii with depth z and lime t. 12.10. SOLUTION OF BASIC DIFFERENTIAL EQUATION OF CONSOLIDATION The solution of the basic differential equation of one- dimensional oonsolidation (Eq. 12.28) can be obtained using Fourier series. Let us express hydrotatic excess pressure U as

u - fdz)

. "(,)

... (12.30)

where II (z) and f2 (t) indicate some function of z and t, respectively.

uin Eq. 12.28, a' [fdz) I 1-" (z) -a,[/,(,)] 8z' - ,-

Substituting the above value of

c.

["(,)

a' 8z' [fdz)] ata [(, (/) ]

oe

r,w-

c'/,(/)

The lefl-hand side of the above equation is a function of x only and the right.hand side is a function of only. In other words, if the left-hand side is e.quallO some constant (say, - A2) when t is taken as a variable and the right-hand side is equ.1l to the same constant when z is considered as a variable. t

Thus,

a' [fdz) I 8z'

.cd

a' 8z'

_A' fdz)

... (a)

[f, (e)] - -A'c."(,)

...@

Eq. (a) has the solution given by

II (x)

• C 1 cos Ax + C2 sin Ax

... (c)

where C I and C2 are constants of integration.

Eq. (b) has the solution given by h(r)oo C)[A1C~1

...(4)

where C) is a constant of integration and e is the base of the hyperbolic or Napierian logarithm. Substituting the above solutions in Eq. 12.30,

uoo [CI cos Az + C2 sin Az] C3

u _ [Col

e-A.'lc.I

casAz + C$ sin Az] e_ A1. <"I

where C4 and Cs are other constants, sucb that

C,oo C1 Cl and C s - C2 C) The oonstants C 4 and Cs can be determined from tile boundary conditiom : (.) I =0 iii for any value of z

u-

where Uj is inilial hydrostatic pressure. I . 00 ii • 0 (iU) z =0 0

(il)

(i~)

,u _

z:H(=2ti)

,

uooO

for any value of z for any value of t foranyvaJueoft

For the boundary condition (iii), Eq. 12.31 gives C. "" O.

... (1231)

&lIL MECHANICS AND FOUNDATION ENGI NEERING

Therefore, Eq. 12.31 beoomes

... (e)

'ii .. 0 at z .. H.

For tbe boundary rondilion (iv),

Cs sin (AH)

Therefore,

e-

A1c

•1

..

0

The above equation is salisfied if AH .. mt, where n is any "integer. The equation can be written in the following form: 'ii .. 8 1 sin (nz/H) e-(,,2/ff)c~ 1 + 8 2 sin ('br::z11l) e_ (4,,21112)c. t + ... + B. sin (malH)

e_{,,2 ,,1/n1c~ t

+ ... . .• (1232)

where Db 8 2 ... B" arc ronstants. From boundary condition (I), when

or

t '"

0, 'ii .. 'iii' Therefore,

iii .. }: D" sin

.• .[12.32 (0)1

(n 1t zll1)

.-1

If m and

11

are two unequal integers. the foUowing identities hold good.

Ia sin mxsinnxd:c .. 0 Ia sin nx tb: .. xl2 2

and

In the above identities if (1tl IfI) is substituled for x. the differential dx changes to ('1[/11) dz and the limits of integration change to 0 to H. Therefore,

{

o

sin (m1tzlJI) sin (nxzlH)dz - 0

II

f o sin

sod

2

(1I1f.zI J1) dz - HI 2

Mulliplying both sides of Eq. 12.32 (o) by sin (n It z/H) and integrating between the IimilS o to H,

{iii sin (n 1tvJJ) dz .. o

i [8. . /'

III_I

sin(m1tzlH)sin (n l'tzlH) dJ:]

0

(m ",,)

+

.i, [B.. f,

sin' (n n z/1I) dz

1

The right-bald side of the 000ve equation ha<; been split into two parts: (/) when m .. n (ii) when m = .. Using identities (j) and (g), the above equation beoomcs

{ Therefore,

o

'iiI sin (n n z/1I) dz .. 8. . (HI2)

8""

2 I' JiJ Uj o

Substituting the above value of B" in Eq. 12.32,

sin (nxZ/11) dz

{RING

m


... (e)

U

-~~.. [~fo~in(n1tZlll)dz 1sin(n1tz/H)e-(,,1 J11/;l)C'~ , "-I

u .. ~-:L.. 2u· rut'

0'

1

l

(l-cosn:rc)(sinn:rczlH)e- (~ . 1/

"-I

in the

Only odd integers n are relevant, because for even intcger l-cos rut ..

n

~. f

... (x)

a

and for odd integers 1 - cos n:rc - 2 Substituting n .. 2N + I, where NN is _ _an integcr, Eq. (k) becomes

ii.

!u. ~

_ 1 _ [s'n (2N + 1)ru I 1/

'NL:o (2N+1)

l't

12.32)

1e

-«2N+ll·,llJllC' y l

... (12.33)

Eq. 12.33 is the required solution of the basic diffecential equation of one-dimensional consolidation. It gives the variation of hydrostatic excess pressure ii with depth z at any lime t in terms o f the initial hydrootatic excess pressure Ui ( equal to & o/y...). Substituting H • 2d, where d is the length of the drainage path, in Eq. 12.33,

_.§; n: N~O

32 (a)l

u Let

_ 1 _ [ . (2N+Oruj _« 2N+l )211.1/4;)C'~1 (2N + I) sm 2d e

T>-7 Ind the

..• (fJ •. .(g)

.. (12.34)

us introduce a non-dimensional p;.lramcter. known as time /at:tor (Tv). given by

Eq. 12.34 bca>mes,

U _

... (1235)

§ ~ l't

_1_ [ . (2N + 1)m: NL:o (2N+l»)( sm 2d

1 _(2N+l)l,.?r./4 e

... (1236)

A series of isochroncs indicating the variation of U with z Co1n be plotted for diffcrent values' of T~ . The shape of the isocbrones dcpends upon the initial distribution of excess pore water pressure ii; and the drainagc C(l{lditions at the boundaries of the cL'l)' layer. If both the upper and lower boundaries are free draining. the clay layer is known as open layer. If only one boundary of the clay layec is free draining, the laycr is called

half·dosed layer. limits

Fig. 12.11 (a) shows the isochrones for an open laycr of thickness 2d when the initial distribution of pressure is uniform. TIle upper balf of the diagram is also applicable for a half-closed laycr of thickness d (Fig. 12.11 (b)]. The hydraulic gradient at any pain! is equal to the slope of tbe isochrone at thai point. The progress of consolidation nt any point depends upon the porc water pressure at that point. The dcgree of consolidation (UJ at any point at depth z L.. equal to the ratio of the dissipated excess pore water pressure to the initial excess pore water pressure, i.e.,

u

... (h) lal

Fig. 12.11. \sochrones.


274

... (12.37) Substituting the value of Uliii from

Eq. 1236,

Uz·I- "

;

where

M.

2 . !ism

(Md') .-AlT. ...

...(12.38)

.0

i (2N

+ 1)

Eq. 1238 gives the degree of consolidation at a point. In practical problems, the main interest is to know the average degree of consolidation of the whole layer. The average degree of consolidation (U) is defined as U _ U j - V, U; whecc Vi is initial excess hydrostatic pressure over the entire depth,

Ui

1 l"-'2d JO U;

.•.(12.39)

dz

aod V, is the average excess hydrostatic pressure after time t over the entire depth, 1 1'V, - 2d 10 u

1

Eq. 12.39 can be written as,

dz

l"-

udz U_l _'2dJ __ _ o _ 1 l"'2dJ o Uj

r" U-l-;_ 10

Uj

dz

dz

... (12.40)

dz

For constant initial excess pore waler pressure Uj, Eq. 12.40 becomes

u-

1 1 - 2dUj

1'_ 10 u dz

... (12.41)

It may be noted that the average degree of oonsolidalion U is equal to the area of the hatched portion of the rectangle shown in Fig. 12.12. Substituting the value of

Uwhere M ..

u from Eq. 12.36 in Eq.

l"~ 2";. ~ Sin 1 - _1_ j" 2dui.0N_oM

~ (2N

12.41,

(AI') e-AIT• d -

+ 1) as before.

. .. (12.42) U - [(T.)

... (12.43)

Therefore the average degree of consolidation (U) depends

fig. 12.12. Average Degrece of Consolidation

CONSOUDAll0N OF SOILS

upon the non-dimensionaJ time fadar T~ The curves can be obtained between U and T, for different drainage conditions and the pressure distribution. Table 12.3 gives the values of T~ for different values of the average degree of consolidation (U). Case (1) is applicable for unifonn pressure distribution for ooth open layer and half-closcd layer. This is also

!)

applicable for triangular (linear) distribution of pressure in the case of open layers. Case (2) is applicable for triangular distribution of pressure for half·c1oscd layer with maximum pressure near the drainage face. Case (3) is also applicable for triangulru- distribution of pressure for half·closed layer but with the drainage face. Fig. 12.13 gives the rurve for the three cases considered. Thble 12.3. Variation or U with

~)

Presrure dislribuljOll

"[JOr ;;~'J,'

,'f,r:,

.

Curve (1) Ca.{e (1)

• • ,' j",-

~J Cun>e(3) Case (3)

T.

T.

T.

0.0

0.0

0.0

O.!

0.008

0.003

0.047

0.2

0.031

0.009

0.100

03

0.071

0.024

0.158

0.4

0.126

0.048

0.221

05

0.196

0.092

0.294

0.6

0.287

0.160

0.383

0.7

0.403

0.270

0.500

O.B

0567

0.440

0.665

0.•

0.848

0.720

0.940

U

0)

pressure ncar

T~.

r-~ Curve (2) Casc (2)

zelO

1.0

.1)

The time factor T" depends upon tbe coefficient of consolidation (e,,). time I and the drainage patb d (Eq. 12.35). lbe coefficient of consolidation represents tbe combined erred of the coefficient of permeability (k) and the coefficient of volume change (m~) as indicated by Eq. 12.29.

Thus

...(12.44)

As the codsoliclation progresses, both the coefficient of permeability (k) and the coefficient of volume change (m,,) decrease, but the ratio (kim,,) remains almost constant over a considerable mnge of pressure. Consequently, the coefficient of consolidation remains almost constant. The drainage path (d) represents the maximum distance that the water has to lravel before reaching the free-drainage boundary. For an o(X!n layer, the value of d is equal to half the thickness of the layer, whereas for a half-closed layer, it is equal to the thickness of the layer. The time (t) is measured from the instant tbe load is applied to tbe layer. , As indicated by 'Eq. 12.44, the time factor T" and hence the degree of consolidation depends upon t, m,. d and t. It also depends upon the distribution of pressure across the thickness. The rurve (1) in Fig. 12.13 is parabolic. 1llc relationship can be represented by tbe following empirical

equations.

27'

O~

o.


r-

'~

o.

~

o. ,

"'" ""'-",

'~ o. ,

~

---- I'---

'\

"-

"- \

"\ ~)

,

O.

L.

~

6

o.

,

\\

""

o. 8

\ [\ '\1\

o.9 0 0.01

O.Q)

().()
0.070. I T I / _ a . )

O.S

OJ'

~

LO

Fig. 12.13. V-To Curves.

T~ _ ~ U2

... (12.45)

(for U < 0.(0)

T" - - 0.933 \0£10 (1 - U) - 0.085

and

(for U > 0.60)

... (12.46)

where U is expressed as a ratio (not percentage).

Eq. 12.46 can also be written as, ... [12.46 (a)J T~ - 1.781 - 0.933 10g IO (100 - U%) These approximate relations are extremely useful when the curves between U and T~ arc not available. Limitation of the Consolidation Theory. One-dimensional consolidation theory is bused on a numbcr of assumptions which arc not realised in practice. The equation has the following limitntions : (1) The value of the coefficient of consolidation (c..) has been assumed to be constant. In reality, it changes with a change in the consolidation pressure. For accurate predictions of the time-rate of consolidation in the field, its value should be detennincd for thc cxpected pressure rangc. (2) The distance d of the drainage path cannot be measured accurately in the field. "n1C thickness of the deposit is generally variable. and an avemge value has to be estimated. (3) There is sometimes difficulty in locating the drainage face. Sometimes thin pervious scams which

CONOOUDATION OF SOILS

277

can act as good drairiugc faces arc missed in the boring operations. On the other hand. sometimes isolated sand pockets arc wrongly laken as drainage faces. (4) The equation is based on the assumption that the consolidation is one-dimensional. In field. Ihe consolidation is generally 3- dimensional. The lateral drainage may have a significant effect on the lime rate of consolidation. (5) 'llie initial consolidation $lnd the secondary consolid.ation have been neglected. Sometimes, these fonn an important part of the total consolidation. (6) In the field, the load is seldom applied instantaneously. The effect of the loading period has to be considered, as explained in Sect. 12.15. (7) In actual pradice, the pressure distribution may be far from linear or uniform. The theory becomes complicated when correct disJ:ribution is considered. NotwiUlstanding the above limitations. the consolidation theory is used to predict the time rate of settlement of the structures built on a soil. The results arc fairly occurate if the theory is applied with caution, keeping above limitations in mind.

12.11. DETERMINATION

m"

COEFFICIENT OF CONSOLIDATION

The curve between dial gauge reading and time I obtnincd in the laboratory by testing Ihe soil sample is similnr in shape to the theoretical curve between U and T~ obtained from the consolidation theory. This similnrity between Ihe laboratory curve and the theoretical curve is used for the detennination of the coefficient of consolidation (c~) of the soil. The methods are known as the filling mer}wds. '!be fallowing two methods are commonly used. (1) Square-root of time method. The method. devised by Taylor, utilizes the theoretic.'ll relationship betwccn U and ff.,. The relationship is linear up to the value of U equal to about 60% (Eq. 12.45). It has

been rurtbcr established that at U = 90%, the value of ff., is 1.15 times the value obtained by the extension of the initial straight line portion [Fig. 12.14 (a)]. The sample of the soil whose coefficient of consolidation is required is tested as explained ;n Sect. 12.5 For a given load increment, the dial gauge readings are taken for different time intervals. A curve is plotted between the dial gauge reading (R), as ordinate. and the.fi • as abscissa [Fig. 12.14 (b)], The curve AlleDE shows the plot. The curve begins at the dinJ gauge reading Ro III time to. indicated by point A. I\s the load increment is applied, there is an initial compression. 11 is obtained by producing back the Ro A

I R~ €

~1

o

~_\oI'; bo

u·/.

Theoreticat

'" ,:rimo r y consolidolion

,curve

b"~

901Jl!-Q===::::=~::,.. iT,

)]

IOOL

~)

RgO - - - - - - - - - - - -

~ E

(b)

.[T -

Rg. 12.14. Square.root d Time plot.

~~~~~i%~ion


278

initial linear part of the curve to intersect the dial-gauge reading axis at point A '. This corresponds 10 the corrected zero reading (N,,). The conso lidation between the dial glluge reading Ro and RI . is Ihe illilial compression. The TCf:l.aghi theory of consolidation is not applicable in this range. From the corrected zero reading point A', a line A' C is drawn such that ils abscissa is 1.15 times Ihat of the initial linear portion A ' 8 of Ihe curve. "Inc intersection of this linc. with the curve at point C indicates 90% of U. The dial gauge reading corresponding to C is shown as Roo and the corresponding absciss..1 as "rr;. The point D for 100% primary consolidation can be obtained from Roo as, R,. - RlOo

=

"*

(He - H<)())

The consolidation
The distance of the drainage path d is half the total thickness. The total thickncss may be taken as the average or the initial thickncss (II,) and final thickness (NJ~ of the sample. •• Th,,·

For single dminage,

d -- li2 -d=H

=

-2 '

[Hi

(¥)

+2

Ht ]

. .. [12.48(a)[

... [12.48(b}[

The test is repealed for different load increments -and an average value of c" obtained, as shown in Fig. 12.15. (2) Logarithm of time method. llle method given by Casagrande uses the theoretical curve between U and Log T,., as shown in Fig. 12.16 (a). The curve consists of Ihree parts : (i) an initial portion which is paraholic in shape, (ii) a middle portion which is almost linear, and (iii) the last portion to which the horizontal axis is an asymptote. c.., It is observed that the point of intersecti on of the tangent drawn at the point of inllexion on the curve and the asymptote of the lower portion gives the value of 100% consolidation. Logr __ The sample of the soil is tested as explained in Section 12.5. For a given load increment, 11 curve is , ploned betwccn the dial gauge reading R nnd log I Fig. 12. 15. Vnriation of C•. [Fig. 12.16 (b)}. Let Ro be the in.i[ial dial gauge before the application of the load increment. The corrected zero reading (R..) is ob[ained using the fac[ that the initial portion of the curvc is parabolic. Two points Band C arc selectcd cOITcsponding to some arbitrary time II and 4th respectively, illld having the vertical intercept a, as shown. Point A' is located such that the vertical intercept hetween B and A' is also equal to u. It represenls the corrected dial gauge reading Rr corresponding to zero primary consolidation. As a check, the procedure can be repeated hy selecting two other points (not shown) with the time ratio I : 4. It should also give approximately [he same location of point A ' . Obviously, the consolidation ' between the dial gauge reading Ro to RI " represented by A and A " is initial compression.

t


279

[s:

h.«"
c:.urv~

100

'

... Tv '0)

~

1

RIOO R.

----------- -!------~~¥---_

E

- - _____ ____ 1_ - - - - __ . :_ _ _ _ _ _ _

,

,

''30

'00

logt - _

'b) Fig. 12.16. Logarithm or tim e: Plot.

The fmal portion of the experimental curve is linear. The point F corresponding to 100% consolidation is obtained from the intersection of the two linear parts, as shown. The values of Rloo and 1100 are obtained corresponding to point F. TIle compression between the dial gauge readings Re and RiOO is the primary consolidation, and thai between R 100 and R, is the secondary consolidation. The point M corresponding to 50% primary consolidation (Rso) is located midway between Re and R 100 800 the value of time 1$0 is obtained. Thus

Re - Rso -

t (Re -

R 1(0)

From Table 12.3, for U = 50%, the value of Til is equal to 0.196. From Eq. 12.35, C. _ __

0.196'" I",

... (12.49)

1be distance d of the drainage path is detennined using Eq. 12.48, as in the first method. The test is repeated for different load inqements and an average value of CO' for the desired load range is determined, as shown in Fig. 12.15 .. Comparison of the two method... The two methods for determination of the coefficient of consolidation give comparable results for most of the soils. However, the following points must be carefully noted. (1) For some SOils, the square-root of time plot does not give a straight line for the inilial portion and, therefore, to 10000te the corrected zero Rc becomes difficult. For such soils, the log-of-time method is

better.

>d

00 p'

"

he

ge

(2) The square-root of time method is more suitable for soils exhibiting high secondary consolidation. In such soils the log I-plot does not show the characteristic shape required to locate the point COITtSponding LO 100% consolidation. (3) The square-root of time method is more convenient for a general case, as it requires dial gauge readings covering a much shorter period of time compared with the log-time method. The )atter method requires accurate plotting of the secondary coosolidation curve in order to locate tbe asymptote. Compression Ratios. The following definitions for different compression ratios are used. (I) Inldlll compnsslonJ8t1o (T,). It is the ratio of the initial oompression to the total compession. In tcnn.s of dial gauge readings. it is expressed m;

SOIL MECHANICS AND FOUND/maN ENGINEERING

... (12.50) where Ro = zero dial gDuge reading, Rt = corrected zero reading, and RI = final dial gauge reading. (il) PrImary compression mtlo (r,.). II is ralio of Ihe primary oomlxession 10 the 10lal oompression. In terms of dial gauge readings. it is given by

... 12.51) RlOo = dial gauge reading corresponding 10 100% primary consolidation. (ii.) Secondary comprtSSlon mtlo (rr)' It is the ralio of Ihe secondary compression 10 the total compression. In terms of dial gauge readings, RlOo - Rf ... (12.52) r, - ~ _ H w~cre

f

It can also be written as bcc.'1use

r, _ 1 - (ri + r;)

... (12.53)

r, .+ rj + rp - 1

12.12. PRECONSOLIDA110N PRESSURE The maximum pressure to which an ovcrronsolidatcd soil had been subjected in the past is known as the prcoonsolidation pressure or over-oonsolidation pressure (0..). When a soil specimen is taken from a natural deposit, the weight of the overlying material (over-burden) is removed. This causes an expansion of the soil due to a reduction in pressure. Thus the specimen A----..,B~E _p-::~R1Z.Qt!.!..A!:.G is generally preconsolidated or over"'" ......-- - ___ ..,- BISECTOR consolidated. When the specimen is loaded C ............... --H in the consolidation lest, the initial portion G<:!;r AD of the compression curve AIJCD (Fig. 12.17) is actually a recompression curve. ComequenUy, the initial portion AD is relatively flat. Il is followed by a straight line CD with a sleep slope which indicates the compression of a virgin (normally consolidated) soil. In the transition range BC. the slope gradually d1anges. The preoonsolic!ation Fig. 12.17. OelenniMtion of Oe. pressure (O() falls in this range. It can be obtained u<>ing the method given by Casagrande. The procedure consists of the following steps : (1) Detennine the point E 00 the curve where the curvature is maximum, i.e., the mdius of curvature is minimum. (2) Draw the tangent EF 10 the curve at E. (3) Draw a horizontal line EG at E. (4) Bisect the angle between the tangent EF and the horizontal EG, and draw the bisector Ell. (5) Produce back the straight line portion CD of the curve and determine the IX'int of intersection P of the bisector EN and the backward extension of CD. (6) Draw the vertical PJ Ihrough P which cuts the log a-axis at J. The IX'int J indicates the preconso!idation pressure a~.

"<:';-4Jv

NG

50)

In

51)

)tal

52) 53)

,.n


(7)

The vertical PJ cuts the curve at point K. lhe portion ARK of the curve represents the recompression curve and the portion KeD as the virgin compression curve.

12.13. CAUSES OF PRECONSOLIDATION IN SOILS Preconsolidation in a soil deposit may be due to one or more of the causes mentioned below: (t) Prcconsolidation may be due to the overourden which had been taler removed by erosion. (2) It may be due to loads of buildings and other structures which had been demolished. (3) It may be due to melting of glaciers which oovercd the soil deposit in the past. (4) Preconsolidation may be due to capillary pressure which acted on the soil in the past but was later destroyed due to a rise in water !.able. (5) Preconsolidation may be due to dessication of the clay deposit. During drying due to rise in tcmperoture, tension develops in pore water and the effective stress increase<>. After dessication, the soil remained prcconsolidmcd. (6) SustaineJ downward seepage forces cause an increase in the effective stress. When the seepage later stops, the soil becomes prcconsotidatcd due to reduction in the effective stress. (7) Pre--consolidmion may be due to tectonic forces caused by the movement of earth's crust which later became less severe.

12.14. FINAL SE'ITLEMENT OF A SOIL DEI'QSIT IN TIlE FIELD Computation of settlement of a soil deposit in the field ~islS of two parts : (I) Computation of magnitude of final settlement. (il) Detennination of the time-rotc of settlement. For .the computation of final settlement, the coefficient of volume change or the compression index is required, which is based on the plot between void mtio and the effective stress. For the time-rote of canpulation, the Tcrzaghi theory is used. It requires the coefficient of consolidation and is discussed in Sect. 12.10. (1) Flnal Seulement Using Coemclent or Volume Change Let US consider a small element of thickness Az at a depth z in the clay deposit of total thickness Ho (Fig. 12.18). Let the effective pressure increment causing the settlement be 60. From Eq. 12.15,

l>i'

I

I I I I I I i + I I I

w+///lI//l1lllZTP7/1M

e;s

j

Fig. 12.18. l./Iycr Subjected 10

>

of

tho

A}f .. mJlo(A 0) Representing the final seUlcmenl as As, and laking}fo .. AI..

As,_ mv 62 (AD) Total settlement of the oomplete layer,

Ao.


If both m" and 6,0 are constant, Sf - mIl t.. (j Ho ...(1254) In an actual case, as the pressure intensity 6,'0 and the coemcicnt of volume change my deaease with depth z, it is convenient to divide the stratum into a number (n) of small layers and to assume dO and mIl as constarll in each layer. The fmal setllement is given by

f"''

r

... (1255)

" - }: (m.), (d 0), (IU),

.~;

The settlement ',may also be obtained m, , - " ,.. ,m the method of grapbical integration. The variation of 6,0 and mil with the depth z is shown in Fig. 12.19 (a), and Fig. 12.19 (b) . respectively. Fig. 12.19 (e) shows the variat ion of the product (60 x mv) with the depth z. The final seltiement is equal to the area of the diagmm in Fig. 12.19 (e). :z. z z However, if the thickness of the clay layer Ca) (b) (e) is relatively small, the variations of 6. (j and m" with depth z may be asswned as linear (Fig. Fig. 1219. Graphical Integrnlion Method. 12.20). The values at the mid-depth may be taken as representative values, and Eq. 12.15 may be used as " - (m.). (dO).

H.

...(1256)

where (m ..)", and (6. 0)", are the values at mid-depth. (2) Flnal settlement U'>ing Void Ratio If e - (j plot for the soil is available, it can be used to determine the final settlement. The value of Ae oon-esponding to the given load increment is read off from the plot and substituted in Eq. 12.11-

dH-H.(~) 1 + eo

Y

)

r:r-.. [J:"...

~ '.

~

l

l (a)

(b)

Fig. 12.20

... (12.57)

where eo is the initial void ratio. TIle usual practice is not to use At! but to use the ooefficiem of compression index (Cf ) for nonnaUy consolidated soils and the coefficient of recompressklll index (C.) for preconsolidatcd soils as explained below. (a) Normally consolidated solis. As mentioned earlier, the compression index of a nonnally consolidated soil is constant. For such soils, it is more convenient to use compression index. From Eq. 12.18,

C f

_

-6.e 10810(00 + Acr}loo

or where tJ.e is change in void ratio when the effective pressure is increased from this value of Ae in Eq. 1257,

00 to @o

+ 40). Substituting

,.3

OONSQUDATION OF SOILS

Sf -

1

~~ eo

• Ho . 10gIO

(a ~oa a) o

..

(12.58)

(b) Preconsoliduted soils. The final settlements are small in the case of preconsolidated soils as the recompression index C, is considerably smaller thun the compression index. From Eq. 12.24,

-ae Therefore, Eq. 12 57 gives

- C, log

" - -

I

c,+

eo

ao 6a) (----0;;-+

. Ho . log

(a-o -+ ,;a) -

... (1259)

00

The above equation is applicable when (cio + a 0) is smaller than the perconsoli computed in two parts: (l) Settlement for pressure ~ to 'O~. (iJ) Settlement for pressure Or to (00 + II 0) For the first, part, the recompression index is applicable, whereas for the second part, the compression index is used. Thus

ao)

C, _ Crflo (00 + ', - - 1 - ' Holog\crc / oo) + -Ilog - -- -

+-

+-

In this case, the first part is relatively small and is sometimes neglected. ~

... (12.60)

12.15. TlMESE1TLEMENT CURVE The load is not applied to a soil deposit in the field instantaneously as assumed in the consolidation lhcory. First excavation is done for the footing. It causes a decrease in the stresses. After the excavation, the fOOling is constructed and the load of the super-structure is applied gradually, as shown in Fig. 12.21 (a) . 'lhe net load on the soil becomes zero, say at time t - to' It then gradually increases to full load P at t - tr Thus, there is expansion of soil due to removal of load in the beginning and it is followed by recompression. It is generally assumed that the nct compression of the soil during the period t = 0 to t _ to is negligible. The actual loading pcriod is from lime t - to to J - tpo During this period, the loading is assumed to vary linearly from zero to full load P.

The time-scUlemeDt curve in the field is obtained based on the a
SOIL MECHANICS AND rOUNDATION ENGINEERING

284

________________ lptP~-~,~F~u~"~LOO~d~'~p~--

-t"]

i L

r----7'lP-"--'---''-'--'c------ 11meo

'0' ,..,~ic__4c_+-~'-----_tlme

Ukcwisc, the settlemem at any other time can be obtained. Fig. 12.21 (b) also indicates the settlement at time 1/4 of the in')taneous curve. A smooth curve is drawn through all points so obtained. Thus the corrected curve OC is drawn. 1hc curve eM beyond the loading period is extended by making offset 8M equal to AC. The load after time tp is equal to the full load P. The offset, therefore, remains equal to the one·half of tbe looding period (/,/2). In other words, the horizontal offset between the oorreclcd curve and the instantaneous curve after the k>ading period is constant, and equal to 1/2.

Alternative method. The corrected curve can be obtained ea<>ily if the mle of settlement is not of interest during the construction period (tp ). In this case, the correde(l curve can be taken as the curve for instaneous loading in which the origin is t:lken at t _ t"l2. In other words, the whole load P is assumed to be applied instantaneously at half the loading period (t/2). 12.16. FlELD CONSOLIDATION CURVE lbe oomprcssion characteristics of in-situ soils are di(ferent from those obtnincd from the tests conducted on the soil samples. Even the so-called undisturbed S:JmpJes are also slightly disturbed when these are taken out. "Ille disturbance causes a slight decrease in the slope of the compression curve obtained in the laboratory after conducting' a consolidation test. Consequently, the slope·of the curve for in-situ soils is, expected 10 be greatcr than ,thnt obtained from the tcst. The mcthods for obtaining the field consolidation curve from the laboratory consolidation curve are discussed separately for normally consolidated, ~ over-consolidated and undcr-consolidated soils. Cc (a) Normally consolidated soli. Schmenmann ~ established thai the lab<),ralory virgin curve intersects the field oonsolidalioo line at n void ratio of 0.40 eO. where eo is the initial void ratio. The initial void ratio (eo) can be taken as the void ratio at the start of the laboratory test. Thus the field consoUdation lint mum pass through point Ag. 12.2Z. ficld Consolidation Curve of H.C. Clays.

CONSOUDATlON OF SOILS

C C04l'eSponding to a void ratio of 0.40 eo (Fig. 12.22). [Note. In some texts. il is taken as 0.42 eo]. The coordinates of point D represent the natural void ratio (eo) and the effective overburden pressure (00) before the sample was extracted. When the sample has been taken out. the overburden pressure reduces to zero, but

the water content (w) and hence void ratio remain essentially the same. The process is represented by line De_

at ed C. he

he in

ed

ed

en

:uRV(

When the undisturbed sample is tested in the laboratory. the compression curve (k.. ). represented by the curve ABC is obtained. The portion A 8 of thL.. curve represents the recompression ana the portion BC as the virgin compression. The upward extension of the linear portion BC intersects the horizontal line through eo at point £. For a nonnally oonsolidatcd soil, the point E always lies towards the len of the point D. Once it has been established that the soil is normally consolidated, the field consolidation Line or in-situ consolidation line (kt) is drawn joining the points D and C. 'rbe slope of this line DC is equal 10 the compression index of the in- situ soil. If the sample is remoulded and again tested, the compression curve (k,) is obtained. The slope of the k,· curve is somewhat smaller than that of the k..·curve. However. the downward extension of the k,.line also intersects the horizontal line drown from 0.4 eo at point C. (b) Preconsolidnted soli. In Fig. 12.23, the curve ABC is the laborotory compression curve (kw) for the undisturbed sample. The poinl C corresponds to a void ratio of 0.4 eo- The backward extension of the linear portion of the curve ABC meets the horizontal line eo D at point E. In case of preconsolidated soils, the point E lies towards the right of point D representing the in-situ condition ('00. eo). It is obvious that there is recompression of the soil from a pressure of ]: 00 to the prcconsolidation pressure (O~). A smooth curve !Xi is drawn from point D

!

f~~~i~~ ~~~ 1~::S~~i~7:;ou~~ ~

o~ is denoted by the point G. The point G is joined to the point C by a straight line. The curve DOC represents the field compression Log iF _ _ curve (kf ). For pressure between 00 to 0.., the Fig. 12.23. Fietd Consolidation Curve for D.C. CIIIYs. relevant index is the recompression index C, and for lhe pressure more than a... it is the compression index (Ct ) given by the krline. (c) Underconsolidated soils. In areas of rerent landfill, sometimes the soil is underconsolidated. It has not reached equilibrium under the effectivc overburden pressure (00)' For such soils, the prcconsolidation pressure @t) is less than thc overburden pressure.

Thus o~ < 00 or O.C.R. < 1 The total compression ror such soils when subjected to extemal loads is equal to the sum or the compression due to overburden pressure till equilibrium is reached and the additional compression due to external loads. Thus 6e- 6e L + lie2 where lieL = decrease in void ratio due to 00 tt.e:z = decrease in void ratio due to applied pressure. as found in nonnally consolidated soils. 12.17. SECONDARY CONSOLIDATION According to Terzaghi's theory of consolidation, the primary consolidation stops when the excess pore water pressure becomes zero. In actual practice, it has been observed that some consolidation continues even

SOIL MECHANICS AND FOUNDA'10N I!NGINEERtNG

286

after the full dissipation of the excess pore water pressure. This additional consolidation is known a<> secondary consolidation. In other words, the secondary consolidation is the consolidation which occurs after the completion of the primary consolidation. The causes of secondary consolidation are not well underslood. This is probably due to tbe gradual readjustment of the soil skeleton which oc(.:urs after the stresses caused during primary consolidation. There is plastic readjustment of the soil pnrticlcs 10 the new stress. (In Ihis respect. secondary con.
C, ..

10g~O~t~tl)

. .. (12.61)

where Ae is change in void roUo between time tl and 1'1 It is equal to the slope of the final linear part of the curve drawn between the void ratio and log time

(Fig. 12.24).

Primary consolidotion

100·/.

------------------------..:.~t_,

______==s_._'o_ndCIr y conSOlidation C -:!; '2 It , t:

Logt-_

FiS. 1224. Seconrbry CoosolidatiOfl.

Another pnramcl.cr, known as the coefficient of secondary consolidation (Cu ), is more commonly used. It is given by

C _~ .. ~. _ _1_ _ " I + ep 1 + t:p loglo (t2Itl) where ep = void ratio at the end of primary consolidation and At: = change in void mtio between time tl and t2' The magnitude of the secondary cotL'\Olidation is given by lime II

Ss .. C u x 1IIOglO(t2111) COfTcsponds to the end of primary consolidation.

... (12.62)

... [12.62(a)J

The rate of secondary consolidation depends upon the plastic characteristics of the soil. It is controlled by higbly·viscous, adsorbed water layer surrounding the clay minerals. As the secondary conso'ldation is highly erratic, its estimate l$ing the above equations is seldom accurate. For a p.'lrticular soil, the rate of secondary consolidation increases as the ratio of the pressure increment to the existing pressure is decreased. For standard consolidaLion lest, the ratio is kcp,t one. The rate also increases with a decrease in the thickness of the specimen used in the test. There are a number of other factors which control the rate of secondary consolidation. In general, the value of Co. for normally consolidated soils varies with the compressibility and hence with the natural water content. The rate depends upon the length of time the preload may have acted on the soil. It also depends upon lbe shear stresses and on the degree of disturbance of the sample. Serondary consolidation is important only for highly plastic Clays and organiC soils. In some organic


soils, the secondary consolidation may even be more than the primary consolidation. [0 over-consolidaled, inorganic clays, the secondary consolidation is usually small and hence it is neglected. 12.18. 3-D CONSOLIDATION EQUATION IN CARTESIAN COORDINATES Teralghi's tbeory of one-dimensional consOlidation discussed above is based on the assumption thai the soil is laterally confined and the consolidation takes place only in the vertical direction. In field, as the layers are not Laterally confined, the consolidation takes place in all the three-dimensions. In general, the consolidation in the horizontal direction is small and. therefore, neglected. However, in some special cases, such as in sand drains, there is significant radial drainage. in addition to the vertical drainage. For such cases, three-dimensional consolidation equation is required to determine the rate of consolidation. The equation for 3·D consolidation' is derived below, making the following assumptions : (1) The soil mass is homogenous. (2) The soil is completely saturated. (3) The soil particles as weD as the water in the voids are incompressible. The consolidation takes place due to reduction in voids caused by flow water. (4) Darcy's law, in generalised form, is applicable to anisotropic soils. (5) Pressure increment dO is applied instantaneously to I I develop an initial excess pore I Ptx,y;!:): water pressure Uj. Fig. 12.25 shows a parallelepiped \Vy dy ....Ly-""""iY.~............ of soil mass with sides li:c, dy, liz with its centre at point P (x, y, z). Let the ............. velocity components at the point P by /' v"' v,. and v:. 100 velocities on the s ix faces are obtained using the partial y Y:r._~.!tt.. derivatives and are shown in the figure. " 2 Fig. 12.25 The volume of water entering the parallelepiped per unit time (QJ is obtained from the products of the relevant velocities and areas. Thus

.,

Q; •

(v. - ~ .~ )dY

<4 +

(v,-~ ~) dr<4 +

(v,- ~

~)

x

dx dy

Likewise, the volume of water going out per unit time (Qo) is given by

Qo · (v"+~. ~)dy<4+ (vYaz2 +~.
Qo-Q; iJv" + ~ iJ"~) iJy + az dxdydz ( ax

... (a)

The volume of the parallelepiped V is equal to dx I dy, dz. It is also equal to V, (1 + e), where V, is the volume of the solids and e is the void ratio. Thus


v _ _ v __ dxdydz ~ 1 +e 1 +e V .. V1 (1 + e)

1* -fr [V. av

(1 +

e) ] - V. ~

dxdydziJe

...(b)

at-I+e'ai

Obviously the volume of waler squeezed oul per unit time is equal 10 the mange in volume of parallelepiped per unit time, From Eqs, (a) and (b),

dxdydz , I+e

~ at

_

(~

+

ax

~

+

ay

~)

az

dxdydz

E!=(l+e)(~+5+~) at ax ay az If ii is the

excess

hydrostatic pressure. the velocities in x,

~

... (c)

and 1 directions are obtained from Darcy's

law as

*.

vx-k~i~-~ %'i-*x:!:'~

ky

ky:!: . ~

vy "

ky iy

v: ..

k~i~-Is ~-k:t·~

"

Subslituting the above velocities in Eq. (c) , 2 (k"ax! a ii + Is- iPIi + k a ii) al :0;

~ .. .!....±...!

at

2

'Yw

...(d)

As soon as the pressure increment (~ 0) is applied, the pore water pressure develops, Initially, the load is entirely taken by pore water, but as the lime passes, water is squeezed out. The excess pore water pressure gradually decreases and the effective stress increases, as in the one-dimensional consolidation. Thus

~o-~o

+&

where ~ (; = effective stress, and ~ u: = pore water pressure. As any increase in effective stress (O) is equal to a decrease in excess hydrostatic pressure ii,

A'O .. iJe

Therefore, But

~

-

a5'"

- ~u.

ae

-~

coefficient of compressibility

Therefore,

0v

(Eq. 12.12),

iJe

au ..

-0"

From the rule of partial differentiation.

ae

ae

ai-~'

From Eqs. (d) and (e),

au

§.

iii - avo at

... (e)

CONSJUDATION OF SOILS

289

The above equation can be written in tenns of lhe coefficient of volume change mit'

a.

From Eq. 12.16,

nr~_~

Thus The equation can be written in tenns of coeITicienls of consolidation directions using Eq. 12.29.

au at -

Therefore,

il~ c~ ~ + e,y ~ + e,~

a2;i

c~.

e,,. and e"" in the three

a1;

al

... (12.63)

Eq. 12.63 is lhe general equation for lhrce·dimensio:1ai consolidation (3-D consolidation). 12.19. 3-D CONSOLIDATION EQUATION IN CYUf\JI RICAL COOJt Hl NATES Three-dimensional consolidation equation, obUlined in :11,' proceixling !\l',l::){m. can be twnsfonned into cylindrical coordinates (r, a, z) by making the followillg ELlrn,.,\Ulion: x-reasa, y_rsinO and Z ': 2: where r = radial distance (polar distance). and = angle made by the radius with the pole. Thus, lan yl x or arc Ian (y/x) ... (a)

a

a_

Differentiating Eq. Ukewisc., From Eq.

or

a-

r2_x'+l

and

(a~

(b~

2r.£!:..2x

ax

.£!:. ay .. r, .. as L a; - - 2

... (b)

or~- ~", cose

ax

sin S

1

='r

s""S~--~ aa

~

a; - 2(1 + I I.?-) Likewise.,

Therefore,

,.

aa

x

=.l.

sinS

,> ---;-

cosO

ay";J-~

au au.2!..+§...~ a;-a; ax as ax au aU.cose _.@..(Sin9) -ax"-ar as r

and

'a -iii ax2 -1\iJr

a ) (a;; rosa - -I . s109- rosS - 1..@ , as r ar ar

sine)


290

Likewise,

From Eqs. (c) and (d), In the case of axial symmetry,

C,,,,

at = iJu

at ~ Using Eq. (e),

ali

at ..

In the case of axial symmetry,

Thus,

(say)

= c\)' = c,,,

au

From Eq. 12.63,

(aljj ax2

C ,.,

+

ai

a21i) ai

(alii 1 c,., a,'l + -;

a2~ "" ae

aii .. at

al/i.

iJ'2jj

a:;z + c,"

c""

alii + +

C""

ail

alii

c,.: ail

a;: + ? iPU) a 82 au

1

a2/i.

+ c"" a?

0

c ,.,

(alii.!.. iJ1J. + r

£Ii) alii ar + c", ai

...(12.64)

Eq. 12.64 is the governing equation for consolidation in three- dimensions in Icrms of cylindrical coordinates for the case of axial symmetry. The equation can be split into two pans : 2

(I) Radial Row:

c11'

(aar21i

(il) Vertical flow :

c,~

(

+

.!.r £.E.) at

'" Jj{Jt

iJaJ) .. ¥!-

... (12.65) .. (12.66)

where c,r ,. k, I(m~ YM') and c,,,, - k~/(m" y..,). If U r and U~ are the avemge degrees of consolidation in radial and vertical directions, respectively, using

&j.12.43, ... (12.67) U, - fl.',) ... (12.68) and U" - [(T,) where T, and T" are time factors in radial and vertical directions. respectively, given by (using Eq. 12.44),

7; and

= c,.,I/(4R')

T" = c,.~(/Jl

... (12.69)

... (12.70)

where R is the radius of the drainage area, d is the drainage length in the vertical direction. It has been shown that the ovcraU avcmge degree of consolidation U under combined radial and vertical

directions can be expressed as (1 - U) •. (1 - U,.l (1 - U,) ·

... (12.71)

The value of U v can be obtained using the theory of one-dimensional consolidation as already discussed.

lbe value of U, js obtained as explained in the following section for the sand drains.

CONSOUDATJON OF SOILS

4)

,,1

56)

lng

291

1210. SAND DRAINS The main application of the mdial coru;olidation is in the design of sand drains used 10 ina-case the rale of drainage in the embankment. Sand drains are construaed by driving a casing (or a hollow mandrel) into the embankment and making vertical bore holes. The holes are backfilled with a suitably graded sand. The casing is withdrawn after !he sand has been fined. The drains are generally laid either in a square pattern (Fig. 12.26 (a)] or a triangular pattern (Fig. 12.26 (b)1. The spacing (5) of the drains is kept smaller than the thickness of the embankment (211) in order to reduce the length of the radial draimlge path. The 7.one of influence of each drain in a triangular pattern is hexagonal in plan, which can be approximated by an equivalent circle of radius R. where R = 0.525 S. In case of a square pattem, the radius of circle of innuence R is C
R"O·S25S

'01

ffiI Wi ~

=

(1) Free strain case. (2) Equal strain case.

blan~!: !;:~~I~re~as7~i~f ~ ::~~;n

Ag. 12.26. PatterM of s.,~ Drains. distribution of surface toads, but the settlements at the surface are uneven. The basic differential equation for radial drainage, given by Eq. 12.65, is

~_ at

~~lh:~e~ I~~i=

(aar2u+ !r 2

c ...

§.) ar

67)

68) 14), SAND

69)

BLANKET

70) SAND DRAIN

ical

.7 1)

sed.

'".:.": Fig. 12.21. Sand Dralns IrwaJladon.


292

In Ihe case of free strain case, the boundary conditions are as under

AI time I .. 0,

ii., 'ii;. At time t > 0, Ii .. 0

at T .. r.... and al r .. R,

W,:

=

0

The solution for excess JXlrc waler pressure Ii at any time t and at a r'Jdiai distance, is obtained by the solution of the differential equ.'1tion as

Ii.o±«> UI.
w~crc 12 ..

-2 Uda)Uo (a,. Ir ...) a [n

2

eH{J.~,,7,)

ifo (an) - ui (0.)1

Rlr", and Udal • Jdo) Y, (a) - Y, (0) J, (a) Uo(an) .. 10 (0 n) Yo (a) - Yo (an) 10 (a)

,~

Uo(arlr .. .. 10 (arlr...) Yo (a) - Yo

:I!'!:.I wt:cr~

(-Z)

10 (a)

./0 = Bessel function of [lISt kind of zero ordcr. .! I = n~"sel function of first kind of first order. 10 = Bessel function of second kind of 7.ero order. Y; = llcssci function of second kind of first order. n ! . ((2 .. arc roots of Bessel function which satisfy the equation

11 (an) Yo (a) - YI (an) 10 (a) .. 0

T, .. C", //(2 R)2 k,

c,,. .. .~

l,o. i~ic:1

[';1l'.

kJ,

nr~l.... 111"1...

k. i.. coefficient of permeability in horiZontal direction. average pore water pressure u"" throughout the soil mass may be written as

,u"" -

u~,.~ ... a2

4vf (n2 _ 1)

(a)

rn2V~(OJJ) _

UT(a)J

'The average degree of radial consolidation Vr can be detennincd from the equation

... (12.72) Fig. 12.28 shows the variation of V, with the time factor T, by dolled lines for different values of wheren _ Rlr....

/I,

(2) Equal Strain Case. This case occuno when the surcharge applied is rigid, such as heavy steel plates. !n this case. the settlements are uniform. but the distribution of pressure is non·unifonn. The problem was solved by Barron, who gave the expression for excess pore water pressure as

u

4u_ [, (,'-,:.)] u - (2R)2 F(n) R log. (rlr...) - - - 2 where F(n) _

~ )~) '1 - 1

u....

_ (),l

4n

= average value of pore

II"" _

iii

e;"

~ wmer pressure throughout out the embankment

in which A _

;~~)

.. .(12.73)

CONSQUDATION OF SOILS

293

The average degree of consolidation in radial direction is given by U, _ 1 _

tf.~T,IF(/f»

. .. (12.74)

Fig. 12.28 shows the variation of U, with T, by firm lines for 3 values of n. (EqU(lI

Tr

strain

can)

Url·1o} n .. o;

n~

!O

0.012

0.021

0.039

20

()'OZ6

0.046

o.oaZ

30

0.01..2 0·060

0.070

0.1]1 0.188 0.2'1'1

10

n ~ l"O

0.:331 0-431 0.'192 0.81..7 1.102 1.69)

Fig. 12.28.

V~rill1ion

of U, which T, .

It may be observed thai the curves for free strain and equal strain are nOI much different and they give

approximately the same results. Equal strains case is generally preferred as it is more convenient. Fig. 12.28 also gives the value of T, in a labular form for lhe equal slWin case.

Effed of Smeal'" Zone. A smear zone is formed around a sand drain due to the remouldining of clay caused during its construction. A decrease in the ooefficienl of permeability in the radial direction occurs due to remoulding. Burron extended the analysis of the equal-strain case taking into account the effect of smear zone. The analysis is based on the assumption thaI the clay in the smear zone bas zero excess pore water pressure on the inner boundary and the time-dependent excess pore water pressure on the outer boundary. Fig. 12.29 shows a cross-section through a sand drain having a smear zone. The radiaJ distance from the centre line of the drain weU to !he farthest point on the smear zone is equal to the radius of smear zone r~. 3)

The excess pore water pressure Ii in this cnsc in given by 2 2 -2fi2))IO&e ii - ~ u(J~ [lo&e(rI R) - (r _ ;;) + ~k" m 2R s It

«n

where

n]

... (12.75)

*" = coefficient of permeability of the soil in horizontal direction k. = coefficient of pcnneabilily of smear zone, D = r,/r". and n _ Rlr". where R = radius of influence

III •

.'

ttl _

82 log.. (niB) -

0' 43 + 4n2

+

k,(.,-o')

k; -.-,-

log.. B Fig. 12.29. Smear Zone

SOIL

294

MEC~IANICS

AND FOUNDATION ENGINEERING

and

The average degree of consolidation is given by

V, _ I _ ~ _ 1 _

e- ( 8T./mj

... (12.7~

The solutions for values of III are also lMlilable in the fonn of curvcs for different values of klolk, and B and n. It may be noted that for no smear zone. B = 1.0 and the results of equal-strain casc apply. "rne net effcct of the smear zone on the consolidation is thaI the influence area of the drain is reduced. As an upproximation. the effect of smear zone is sometimes taken indirectly into acrounl by reducing tht radius 'of influence R to 0.5 R. 'Inc following points regarding sand dmins may be noted. (1) Secondary consolidation is nol taken into acrounl in [he design of sand drains. In fact , the sand drains arc ineffective in controlling the sc(."(mdary consolidation for highly plastic and organk soils. (2) Sand drains tend to act os weak piles ond reduce the stresses in the clay. Consequently, the excess pore water pressure developed is gcncrnlly less os compured with that in the case when there arc !XI sand drnins. This factor is nO!. taken into nccount in equation given above. (3) The typical design parameteffi for the sand dl1lins arc os under: (0) (b) (c) (d)

Radius of sand drains well, r" = 0.2 to 0.3 m Spacing of sand drnins, S := 2 to 5 m Depth of sand drains. 2H = J \0 35 m Thickness of sand bl.'lOket = 0.6 to I m.

12.21. EFFECT OF LATERAL STRAINS ON CONSOLIDATION In field, the condition of zero lateral strain as assumed in Tcrzaghi's consolidation theory is not satisfied. The initiol excess pore water pressure (Uj) is, therefore, not equal to the increase in the vertical stress II! assumed in one- dimensiooal consolidation. Skemplon and Iljemun developed a semi-empirical method of calculating settlement, taking into the effect of lateral strain. The final settlement is expressed by them as Sf- fls_1

... (12.7

where SO«I is the final settlement based on the assumption of no lateral strain using one-dimensi consolidation theory. as explained in Sec. 12.14. The correction factor f.1 is given by

f'

0 111 ,•

"-

.1.0 1

{AO, +~(l A

- A)

jfo m" AOjdz

l

dz

... (12.7

where A is the pore pressure coefficient as discussed in ..shaptcr 13. If m" and A arc assumed 10 be constant. Eq. 12.78 rcducc.'I to fl .. A + (1 - A) a

... (12.7

/1603 tiz

where a--o- -

i'o

Ao! dz

lbe value of a dcpcnd<; upnn the sh:lpc or the lo.adcd area and the thickness D of the clay stmtum . relation to tile dimension of the luaued :Irc
-

r--

""

CONSOUOATION OF SOILS

Table 12.4. Values of coefficient

:ed. the

.and

•its.

0.

DIB

0.0

0.2."i

0.50

1.0

2.0

4.0

10.0

n (circular fooling)

1.0

0.67

0.50

0.38

0.30

0.28

0.26

0.25

n (Slrip/OOIing)

1.0

0.80

0.63

0.53

0.45

0.40

0.36

0.25

Note: (1) B"is e
The values of the pore pressure coefficient A depend upon the type of the clay. 'fhe typical values of the correclion factor Il are given in TaDJc 12.5. Fig. 12.30 gives the values of different values of A and DJB ratio, where D is the thickness of soil layer.

cess

Thble 12.5. Values of

j.l

for

~l.

Type of Soil

Heavily over-eonsolidaled day

0.25 to 0.4

Moderately overconsolidatcd cllly

0.4 to 0.7

Normally consolidated clay

0.6

Sofl, sensitive clay

LO to 1.20

to

1.00

1.2'

:ounl 2.77) ~onal

2.78)

0.6

,.0

,.'

.2.79) LH"ovily OVff tonsolidoted Fig. 12.30. Plot for vlllueof 11-.

ILLUSTRATIVE EXAMPLES lim in

tue of

Illustrative Example 12.1. Calculate the final sculemen! of the clay layer slwwn in Fig. E 12.1 due to an increase of pressure of 30 kN/nl at mid·height of the layer. Take y", .. 10kNlnl

SOIL MECHANICS AND FOUNDATION ENGINEERINO G,o;.

T 40m

cc"o.zz

Cloy

-------~--------1,,18 kN/m 3 '0;\.30

t

2.Sm

1

Fig. E 12.1

Also calculate the settlement when the water table rises to the ground surface. Solution. Initial pressure at the centre of the clay layer.

ao .. From Eq. 12.58,

4 )( 20 + 1.25 )( 18 .. 102.5 kN/m 2. c

Ilo

Sf" C, •

+

eo

_ ~

(ao ~ 6,0)

log lo

00

250 I

1 + 1.30)('

oglO

(102.5 + 30.0) 102.5

.. 0.0263 m _ 2.63 em When the water table rises to the ground surface,

00 .. Therefore,

4 )( (20 - 10) + 1.25 x (18 - 10) .. 50 kN/ml

0.22

sf" I + 1.30 )( 2.50 10gIO

(50+30)

-----so-

.. 0.0488 m .. 4.88 an As expected, the seulement inm:ascs due to the rise of Ihe water table.

Illustrative Example 12.2. A footing IuJs a size of 3.0 m by 1.50 m and it causes a pressure increment of 200 kN/m2 ai ilS base (Fig. E 12.2). Determine the consolidation settlement at me middle of me clay layer. Assume 2 : 1 fresswe distribution and consider Ihe variation of pressure across the depth of the clay layer. "Y..... 10 leN/III •

Il

l...

q"';;x:<;j}:x::.<:
~'V.I<.'W?

!--,.'m-l

¥

':16·0 kNlml

l5~t:'!·OkNlmJ

1m

T

L

0f·,m

CLAY

_ . _ _ _ __

L

. _ . _ _ _ ._3m

r: IS~N/m3

Ib;O·80 Cc =0-30

Fig. E J2.2.

Solution. Initial pressure at the centre of the clay layer,

00 .. 2.5 x 16 + 0.5 (18 - 10) + 1.5 )( (15 - 10.0) _ 51.5 kN/ml

'0


297

The pressure increase at the top, middle and the bottom of the clay layer are found as follows (Eq. 11.60).

(d 0), - (3.:x:. ;.O\\~51.:0i,O) .. 51.4 kN/ m2 (60)111 .. (3.~~ ;5\\~}·:O;.5) .. 27.7kN/m2 (&0)" - (3.:x:.

. 17.3

;.O\\~}:O;.O)

kN/m2

The average pressure can be found from the following equation (Simpson's rule).

or

d a ..

~

A (1

~. (5 1.4 +

..

[(dO), + 4 (.6.0)... + (60),,] 4)C 27.7 + 17.3J .. 29.9 kN/m 2

0.30 (51.5 + 29.9) sf" 1 + 0.80 x 3.010g 10 - 5-1-.5-

Therefore, from Eq. 12.58,

sf .. 0.09941 m .. 99.41 mm

hV/,,!p:l:t:~~:u~:~n!;n~' ~/;::Z';e~£e~,!':n/~e '~ t~~C~n:;:ru'::,a;r~::::! :;~;u:1:i;/:~~;emt:n~ of the ckzy layer. Tile clay is over-consolidated, with a preconsolidruion pressure of 75 kNlIIl. The values of the coefficients o/recompression and compression index are 0.05 and 0.25, respectively. Take initial void ratio as 1.40. Solution. From Eq. 12.60, ~J

= -, +-c,.eo

Ho 10gIO

(a,. =-)

~ = 0.05 x

0'

00

+ -, -c,.

(ll 1

./

2.0 1 <' 1 + 1.40 0.,10

.f/

= 7.34 x 10-.1 + 16.50

50

X

+eo

Ho 10glO

(a "a) o

- -+- 0"<,

+ 0.25 )( 2 1 ( 50 + 40 I + 1.40 oglo 75

1

10-) m

or sf = 23.84 x 10-'\ m = 23.84 mm lII11.dralive Example 12.4. A consolidation test was conducted on a sample of a normally cOlZSolidated clajl with an initial void ratio of 1.55, and the following results were obtained. 80

I",

320

640

1280

1.35

1.28

1.14

0.%

0.78

Plot e-loga curve. If the initial overburden pressure hence determine the coefficient of compression.

is 150 leN/m2, draw thcficld consolidntion line and

(b) If the thickness of the clay layer in the field is 4 m and the increase in the pressure due to loading is 50 kN/m 2, compute the settlement. (00.

Solution. Fig. E 12.4 shows the requjred plot The field consolidation line is drawn between the point eo) and the point D where the plot cuts the horizontal line through e = 0.40 eo = 0.4 x 1.55 = 0.62.

From the figure, From

Eq. 1258,

C( -

log1~ (!:/(JI) - IO~::~l~iS:oo)

-

0.844

SOIL MECHANICS AND FOUNDATION ENGINEERI NG 298

''' ,-~---,-,---.------,--,---r-~

---"

'."'r--,---t---t-+":tt---t----l---+--~

""'I---+- - -t---+-+--pc---L--L --1---1 \r Fi"ldCO~SOl ldOliO" " nt'

....... ,

Il0r--+-----..jf--+::".,'~t-~-~--~------J

1', \ : \. \ , 1.1Or-- +--+--+-i--jh-i,j--1----+-----l

\ \

\ \\

l.o0r--+---f---+-t-L--~~.c-l___-__+--l 1

\\

Jl.obc>rgbycUl've~\ ~,

"+-+---1

O.80t---t---I--+--}---+------l---~1

o.ror-----t--t--+-+------++-----+-----+~ \----J : '\

o. lt ~ - - - - ----0.60 10 20

-! - -

-100

--

0

1';0200

;r kNltnlJ Fig. E 12.4

0.844

_ ~)(4.0JoglO

( ~ 150 + 50 )

_ 0.1654 m .. 165.4 mm . Illustrative Example 12.S. A clay straalm 5 m thicJc has the initial void ralio of 1.50 and the effective 2 overburden pressure of 120 kNlm 2, When the sample is subjected to an increase of pressure of 120 kNlm , the void ratio reduces UJ 1.44. Determine the coefficient of the volume compressibility and the final settlement of

the stratum. Solution From Eq. 12.14,

-6.e/(1 + eo) mv - ~

or From Eq. 12.54,

s,-

"'v Il a Ho

sf ,. 2 )( 10..,( )( 120 )( 5 )( let mm .. 120 rom

299


llIuslrative Example 12.6. Determille the coefficiem of eonsolidntion of soil whose test data is plotted in Fig. E 12.6. The sample was 2.0 em thick alld had double draillilge.

Fig. E 12.6.

tin minvtu

Solution. The construction lines are shown doucc.l. From- the figure, ..rt;; - 6.55 or too - 42.90 minutes From

I':n.

~

0.848 cf

12.47,

c~.----;;;--

x (1.0)2 - 0.848 . 42.90

-

00198 2/. . em mm.

Illustrative Example 12.7. Determine the coefficient of consolidation of a soil whose data is plotted in Fig. E 12.7 The sample was 20 mm thick and had double drainage 100

191R£..:..191_~

18-1:0.20

~ ~: ~~ I-J.--.l--.I..+-J.I.U---l--"--.LLL IlllilL 111111-f---+-H-ttI

"~ 160'~----l-rl-++-,-H+--"I,,,++l-+tl Rc -R 50 : -t(R c-R100 1 "7(191-79):56 R'jQ" 191_56:135

0.,

2

)

4567891.0

2

3 4 109~10 t in

fig. E 12.7.

2

mil'\Ut~5

'og'--

1091100 678 100

----I----IH-+ttl

2 ) 4 56 800


300

Solution. The construction lines ore shown in Fig. E 12.7. R100 is obtained from the intersection of the two Slruight lines drawn. 1be poinl 50% consolidation (R!IO) and (he corresponding 'so are detennined.

R~

- Rso -

t

(Rc - R iI»)

Rc - 191, RiOO - 79

In this case,

R", - R, - t (R, - RHO) - 191 - t(l91 - 79) - 135

Therefore,

10g10 ISO - 0.58

From the plot,

t5() -

From Eq. 12,49,

c

3.802 minutes 0.196<1'

= 288.12 s

0.196 x (10)'

~ - ~ -

228.12

0086 mm'/scc

-.

muslratlve Example 12.8. A 3 III thick clay layer beneath Q building is overlain by a permeable stratum and is underlain by an impervious rock. The coeffICient of consolidation of the clay was found to be 0.025 cm 2/minute. The [mal expected settlement for the layer is 8 cm. (a) How much time will it lake for 80% of the total settlement 10 take place? (c) Determine the time required for a settlement of 2.5 em to occur. (d) Compute the settlement that would occur in ()nI! year. Solullon. (a) Length of drainage path, d Substituting the values,

U ., 80%,

From Table 12.3, when . Therefore, from Eq. (a),

or t _ 3.6 x 106 T"

U -

Po )(

U = 31.25%,

From Table 12.3, when Therefore, from Eq. (a),

From Table 12.3, when T"

100 - 31.25%.

T" _ 0.078

3.6 x 106

t ..

)(

0.078 .. 0.2808 )( 106 minutes .. 195 days 6

From Eq. (a),

... (a)

T" _ 0.567

3.6 )( Hf )( 0.567 _ 2.041 x 106 minutes - 3.883 years.

I •

(b) When the settlement is 2.5 an,

(c)

=3 m =300 em.

C~ - T~tfll 0.025 _ T~(300)2It

From Eq. 1235,

1 x 36S x 24 x 60 _ 3.6 X 10 T"

=0.146,

U

or

T" .. 0.146

=0.429.

Therefore, settlement = 0.429 )( 8 ::;: 3.432 em nlustrat lve Example 12.9. A clay kzyer 4 m thjck has a fiMI settlement of 6.0 em. The layer has double drainage. If the coefficient of consolidation is 0.02 cm1/m;'wte. determine the time required for different percentagf!S of consolidation from 10% upta 90% and hence plot the time-settlement curve. Solution. From Eq. 12.35.

If time is taken in years.,

T" _ C"tltf T, - 0.02 x 11(200)' or r - 2 x 10' T, 2 x 106 T.. 60 x 24 x 365 - 3.805 T"

The calculations are given in the tabular from below for U of 10, 20, 30, 40, 50, 60, 70, 80 and 90% and the corresponding T.. obtained from Thble 12.3. The settlements are ca}rulnted from the values of U and the final setllemcnt (Sf) i.e. s .. U )( Sf or s .. 6U ' U (%) l(ye4r)

10 0.008 0.030

20 0.031 0.118

0.270

30 0.071

40 0. 126 0.479

50 0.196 0.746

seem)

0.6

1.2

1.8

2.4

3.0

60 0.287 0.092 3.6

70 0.403 1.533 4.2

80 0.567 2157 4.8

90 0.848 3.227 5.4

301 CONSOUDAll0N OF SOIL.'i

2

o.0

2

3 456789

3456789

3 456 "ISo

t--t-

r-.

.0 .0

r-.... I

.0

~

.0

1\

C 5.0

f.-

~

~ 6.0

jj

7.0

!

8.0

9.0

0

1.0

0.10

o.01

Time (years)--rig. E-12.9.

Fig. E. 12.9 shows the settlemem-log t curve. lIIuslr~tive Example 12.10. An Qrea is /lnderlain by (1 stratum of clay layer 6 III thick. rhe laycr j',I' 2 doubly c/ra;'lCd and has the coefficient oJ consolidation of O.3m /month. (0) How 10llg would ;t take for a surcharge load to calise a settlement of 40 ,m if the same load causes

'III

a final sellfell/em of 60 ? ' (b) If the sand drains (8 ;:: 3 m and r..,

consolidation. Take

C,'r ""

= 0.30) are /lsed, determine the time required for 90% 2.0 ",l/momll. Assume the triangular layout of drains. Neg/ecl vertical consolidation

in Ihis case. Solution. (a)

U

_ 40/60 - 0.6667

i.e. 66.67%

From Table 12.3, for U ,. 66.67%, Tv - 0.364 From Eq. 12.35,

Ct. t _

1~, i'lt 0.364 )( (3)2/0.30 - 10.92 months.

(b) For triangular layout of drains, R - 0.525 x S

or From Fig. 12.28, for From Eq. 12.69,

R .. 0.525x3-1.575m n .. Rlrw .. 1.575/0.30 .. 5.25 U .. 90% and n .. 5.25, we have r

Tr';' 0.270

T, .. c, .. 1/ 4,r

0.270-~ 4 x (1.575)2

or 1 .. 1.34 months.


nIustrative Example 12.11. The laboratory consolidation data for an undisturbed clay sample are as follows. el .. 1.00, a) .. 85 kNlm 2, and e2 .. 0.80, '02 .. 465 kNlm 2. Determine the void ralio for a pressure 'OJ of 600 WI,,?

Solution. C .. ~ .. ~ .. loglo (O:YOI) logw (600/85)

Now

0.271 ..

1.~~9 e)

or

c} ..

0.77

illustrative Example 12.12. A clay layer 4 m thick is subjected to a pressure of 55 kN/m1. If the layer has a double drainage and wldergoes 50% consolido.tion in one year, tietermine the coefficient of consolidation. Take T~ .. 0.196. If the coeffICient of permeability is 0.020 IIIlyr, determine the sel/lemen! in one year and rate of flow of water per unit area in one year.

c, _ T,d'/I _ 0.196 x (2.0)'/(1)

Solution.

Cv ..

0.784 m2 /yr

k 0.020 )( 1000 m~ .. c"Y... '" 0.784 x llX)() )( 9.81

From Eq. 12.29,

.. 2.60 )( 10-3 m2/kN

s,- m" l-fo 60" 2.60

X

Settlement after one year

lO~J)( 4)( 55 .. 0.512m .. 0.5 x 0.572 .. O.286m.

Settlement rate Since U is proportional 10 ..fi for U < 0.60, the settlement (s) is also proportional to

¥C. Thus S2

\\Then t

= 1 year,

S :::

t

or

t ..

Cs2

0.286 m. Therefore,

1 C - (0.286)' - 12.226 Thus

I _

~ ..

12.226,' 2 )(

1~.226 S

..

24.:52 $

..

0.143 m/yr

Discharge per unit area per surface .. 0.143/2 .. 0.072 m3/yr/m 2 PROBLEMS

A. Numerical 12.1. A sntumted soil stmlum 4 m Ihick lies above an impervious stratum ond below II pervious stratum. It has a void ratio of 1.50 at lin initial pressure of 150 kN/m 2. (I) Compute the change in void ratio due to to an increase or stress or 50 kN/m2. Take Ct = 0.20. (il) Also compute the final settlement of the soil stratum due to above increase in stress. (iiI) Who! would be the time required for 50 percent consolidation? Thke Tv "" 0.20 and k ,. 3.0 X 10...4 cm/sec. 12.2. In a laboratory, Ihe consoIidntioo test

WIlS

[Ans. 0.025; 4 em; 34.18 minutes] performed on II specimen of Clay 3 an thick. The sample WIlS drained


303

lit top and bottom. The time required for 50% consolid:Jtion of the sample was observed to be 15 minutes. Determine the coefficien\ of consolidation of clay. QJlculate time required for 50% lind 90% consolidation for this clay deP9Sit in the field 3 m thick and drained at both ends. [Ans. 4.93)( 10:..4 crr?/sec; 104.17 days; 450.69 dllYs] 123. There is a layer of soft cloy 4 m thick under a newly constructed building. The overburden pressure over the 2 centre of the clay layer is 300 kN/m . Compute the settlement if there is an increase in pressure due to 2 construction of 100 kN/m . Take C~ '" 0.50, G :: 2.70. The water content of the deposit wa.<; found to be 50%. [Ans. 10.63 em] 2 Il consolidation test, on increase of 100 kN/m in the veniall pressure was applied to a saturated clay sample initially 2.5 em thick. The thickness of the sample reduced to 246 em after 24 hours. The sample was then relieved of jlressure and allowed 10 take up water. The final thickness was 2.465 cm and the moisture content was 30%. Assuming that the sample was saturated throughout the tcst, calculote the following: (l) the initial void rario. (fake G = 2.68) (il) the void ratio afler consolidation. (iiI) the void mtio after expansion [Ans. 0.83, 0.80, 0.804, 1.6 )( W ...4 m 2/kN] (iv) the coefficient of compressibility. In II consolidation test on a soil, the void ratio of the sample decreased from 1.2..'1 to 1.10 when the pressure is increased from 200 kN/m2 to 400 kN/m 2. Cnlculate the oocfficient of consolidation if the coefficient of 2 permeability is 8 )( 10-8 cm/sec. [Ans. 755 m /yearj The time required to reoch 60% consolidotion for a sample 1 em thi ck tested in consolidometer under conditions of double drainage wos found to be 35 seconds. Determine Ihe time required for a loyer 10 m thick to reoch the. 8 same degree of consolidation, if it has drainoge only on one side. [Ans. 1.4 )( 10 secondsJ A clay strotum 2.5 m thick lies over a sandy stmlum and hilS drainage on both sides. Colculate the values of the pore. . wllter pressure and the effective stress at the middle of the clay stratum when 60% of consolidation has 2 2 taken plocc under an increment oflood from 500 to 2500 kN/m . [Ans. 800 kN/m ; 1700 kN/m2} In a consolidation test, 0 fully smurnted clay sample was subjected to a load of 500 kN/m2. After 12 hours, the 2 average pore pressure was found to be 200 kN/m . Fina out the time required for 50% consolidotioo to take

12.4. In

125.

12.6.

12.7.

12.8.

cr.

place. Assume T~ _ (Jt/4) 12.9. There is clay layer 8 m thick with a layer of sand on either side. An undisturbed sample 2.5 cm thick of the clay when tested in the laboratory required 25 minutes to reach 50% consolidmion (T~ = 0.20). It is

[Ans. 8.33 hours] S7

proposed to OOnslroct a building at the above site. Estimote the time required for 00% consolidation 10 take place. (T~ = 0.85). (Ans. 7555.6 days]

12.10. Fig P·l2IO shows a bore hole log obtalfled dunng sub-surfuce explorntion. There are two Inyer.; of cloy of thickness 2 m each, loc.ated 6 m ond 15 m below fig. P 12.10. the ground surface. 100 bulk densities of sand and clay were found to be 2.0 gm/ml and 1.82 gm/ml, respectively. Compute the total settlement of each clay layer under II uniformly distributed load of 400 kN/m2 spread over a large area on ground surface. C~ "" 0.40, eo "" 1.08. [Ans. 32.45 an, 21.57 em, lotal ;: 54.02 an]

.

'~.': " .... , • . ' CQ/IRSE: SAND'.... ~ .. ,:.

lo°·15m ---========>:< :Em SILT

~~

1. m

Fig. P-12.11. 3 12.11. A 6 m thick fine silt strntum [C~ _ 7.2 )( 10- cm2/sec] underlies a coarse sand deposit and overlies a 0.15 an


304

2 5 thick fine sand 1l1fer under which there is a cloy stratum 3 m thick (Fig. P.12.1l) (cv _ 9 )( 10- cm /sec).

A pervious SITll!Um lies below the day layer. When the cloy layer reoclles a degree of consolidation of 20%, wh:lt would be the degree of consoIidlllion of the sill layer" (T.. = 0.03 for 20% U and U '" 82% for T" = 0.6). [Ans. 82%1 12.12. ,\ clay lOiyer of 10 m thickness underlies a sand stratum of 10 m arid overlies a pervious layer. The sand layer CJrri ~:l point load of 10 MN. Assume e = 0 .70, and G = 2.72, L.L. '" 60% and c" _ 2S X 10-4 cm2{sec, The waler mblc is located 5 m above the top of the day layer. Find how long would the clay take to settle 4.7 em. [AM. 36_'iS hours] 12.13. A 15 m thick hydraulically' isotropic clay stratum overlies nn impervious liilrulum. If [he coefficient of consolidation is 5 )( 10--4 cm 2/sec, find the time required for 50% and 'Xl% oonsolidolion (Tv., 0.20 and 0.85, respectively). (ii) In order to accelerate the rate of scnlemenl, vcnical sand drains of J m diameter were installed at 5 m centre to centre. Calculote how much the settlement is accelerated due to provision of sand drains. For 11 ::: 5, T, ::: 0.081 for U, '" 50% and Tr ::: 0.27 for UT . , 90%. [Ans. 9.0 )( 108 seconds ; 38.25 )( 108 seconds. 8 8 (ii) 0.405 )( 10 seconds; 1.35 )( 10 seconds]

n. Descriptive and

Objective type

12.14, Describe the consolidomcter test. Show how the results of this test are used to predict the rate of settlement and the magnitude ofscttlement. 12.15. Define the following terms : (i) Coefficient of oomprcssibilily, (il) Coefficient of volume change. (iii) Compression index, (il') Expansion index (1') Recompression index, L2.16. Di~ss the spring analogy for primary consolidation. What Dre its uses " 12.17. Differentilltl! between primary consolidation and secondary consolidation. 12.18. Differentiate between normally consolidated and the overconsolidmed soils. How would you detennine the Q\'crconsolidation pressure , 12.19. Discuss Tcrz.aghi's thcory of consolidation, stating the various assumptions and their validity. 12.20. What is the coefficient of consolidation" Whlll is its usc in the scllicment analysis" How is it determined ? 12.21. What is the time factor" lIow is it relaled 10 the average degree of ronsolidalion , 12.22. Discuss the limitations of Terlllghi's theory of consolidation. Why is theory used despife ilS limillltions , 12.23. What are different c;:Juses of preconsolidarion in soils? What is the effect of preconsolidation on the seulement7 12.24. How would you detennine the time-selllement curve in the field 7 12.25. Whar is field oonsolidation curve" !low is it obtained? 12.26. explain the phenomenon of secondary consolidation. Differentiate between the secondary consolidation indel( and the ooefficient of seu>ndary consolidation. 12.27. Explain the thoory of 3-dimensional consolidation. What is its practical use? 12.28. Describe sand drains. How are these designed? Discuss their uses. What is the effect of smear , 12.29. Discuss Skemptoo·Bjerrum's theory fOf calculating settlement, taking inlo &COOunt the effect of lmeml sltnins. 12.30. Write whether the following stlltementS are correct. When the soil is fully saturated., the compression occurs mainly due 10 compression of water. The initint consolidmion of a fully saturated soil is zero. The secondary consolidation is ncgtitlle in orgnnic soils. The friction in Ihe fil(ed ring cell is more Ihan thaT in Ihe fl
"'5


(J) The field consolidation curve is genemlty steeper th::ln the labor-llory C\Jrve. (A:) The actual settlement is always greater than that given by Terzaghi's theor: if lateral strains occur.

(A .... Truo.(b). (d). ifj. (,). (h).

IJ1J

C. Multiple Choice Questions 1. The coefficient of compressibility is the ratio of (a) Change in vuid ratio to change in effective stress (b) Volumetric strain to change in ell"ective stress.

(c) Change in thickness to change in effective stress. (d) Stress to strain. 2. With an incrense in the liquid limit, compression index (a) decreases {h) increases (e) remain constant cd) may increase or decrease 3. 1be recompression index is ahout . . of the compres.~lOn index (a) S limes (b) liS (c) 1/2 (d) 1120 4. When consolidation of a saturated soil sample occurs. the d'egrce of saturation (a) increases (b) decreases (el remains constant (d) may increase or decrease S. Consolidmion time of a soil sample (a) increase.~ with an increase permeabilIty. (b) increases with a dccrea.-«:s i'l compressibility. {d increases with a decrease in unit weight of water. (d) increases with a decrell~
008~

9. A fully saturated clay specimen is subjected to a pre!'oSure of 200 kN/m2 in the consolidation Icst. After a period of time when the average pore pressure is 60 kN/m2, Ihe degree of consolidation is (a) 60 (b) 70 (e) 30 (d) 50 ~_1~2~~~~~~~~~~~L~aM

13 Shear Strength 13.1. INTRODUcnON The shear strength of a soil is its maximum resistance to shear stresses just before the failure. Soils are seldom subjected to direct shear. However, the shear stresses develop when the soil is subjected to direct compression. Although shear stresses may also develop when the soil is SUbjected 10 direct tension, but these shear stresses are nol relevant, as the soil in this c..'1SC fnils in tcnsion and does not fail in shear. In field, soils are seldom subjected to tension, as it causes opening of the cracks and nssurcs. These cracks arc not only undesirable, but are also detrimental to the stability of the soil masses. Thus, the she.1r failure of a soil mass occurs when the shear stresseS induced due to lhe applied compressive loads exceed tbe shear strength of the soil. It may be noted that the failure in soil oocurs by re1.1tive movements of the particles lind nOI by breaking of the particles. Shear strength is the principal engineering property which controls the stability of a soil mass under loads. It governs the bearing capacity of soils. the stability of slopes in soils. the earth prc.ssure against retaining structures and many other problems, as explained in later chapters, All the problems of soil engineering are related in one way or the other with the shear strength of the soil. Unfortun.1lely, the shear strength is one of the most complex engineering propenies of the soil, llle current research is giving new concepts and theories. This chapter presents the bnsic concepts nod the accepted theories of the shear strength.

13.%. STRFSS·SYSTEM WITH PRINCIPAL PlANES PARALLEL TO TIlE COORDINATE AXES In general, a soil mass is Subjected to a three--dimensional stress system. However, in many soil engineering problems, the stresses in the third direction are not relevant and the stress system is simplified as two-dimensional. The plane suain conditions are generally assumed, in which the strain in the third (longitudinal) direction is zero. Such oonditions exist. for example. under a strip footing of a loog retaining wall. At every point in a stressed body. there are three planes on which the shear stresses are zero. These planes are known as principal planes. The plane with the maximum compressive stress (01) is called the major principal plane, and that with the minimum oompressive (03) as the minor principal plane. The t.hird principal plane is subjected to a stress which has the value intennediate between 01 and 03, and is known as the intermediate prinCipal plane. Generally, the stresses on n plane perpendicular 10 the imennediate principal plane are required in the analysis. Therefore. the stresses on the intennooime principal plane are nOl much relevant. Only the major principal stress (01) and the minor principal stress (0) are generally imponant. In solid mechanics, the tensile stresses are taken as positive. In soil engineering problems. tensile stresses rarely occur. To avoid many negaHve signs, compressive stresses are taken as positive and the tensile stresses as negative in soil engineering. Fig. 13.1 shows a plane which is perpendicular 10 the inlcnnediale principal plane. The major and minor principal stresses act on this plane. The major principal plane is hori7.ontu) and the minor principal plane is

SHEAR STRENGUI

vertical. Let us consider plane AB which is inclined al an angle counterclockwise) 10 the major principal plane AC.

8 (measured

Resolving the forces acting on the wedge ABC in the horizontal (x-d.ircctioo), aJBC .. aABsin8-'tABc.os8

°

where = nonnai stress on AB, 't = shear stress on AB. In above expression, the length perpendicular to the plane of the paper has been laken as unity, which is the general prnctice for two dimensional stress system. The above equation can be simplified as OJ

or

~-

"l

asin8 -"[cosEt

e-

oJ sin 8 - asin

'tcosO Ukewise, resolving the forces in the vertical (y- direction). OJ

"j

Fig. 13.1.

.. .(0)

AC- oABc:osO + 'tABsine ...(b)

or OJ cosO - ocosO + 'tsinO Multiplying Eq. (a) by cos 8 and Eq. (b) by sin 8. and subtraaing, (01 (01 -

OJ) sin

0 cos 0 .. q(ros8sinO - ros8sinO) + 't(sin2 e +

oJ) sin 8 cos 8 .. 't ..

2

oo(

0)

"t

~

... (13.1)

(0, - oJ) sin28

Substituting the above value of't in Eq. (a). olsine - asine _

(OJ;Oll

sin28cosO

oJ" 0- (01 - 0)cos2 e 0 - 0)

+

(01 _ 0)(1

o .. (01 ;

0)

+ C;S28)

+ (01 ; oJ)

c:os 2 0

Eqs. 13.1 and 13.2 give the stresses on the inclined plane AB, making an angle

... (13.2)

e

(measured

counterclockwise) with the major .principal plane AC.

13.3. MOJIR'S CIRCLE Otto Mohr, a Gennan scientist, devised a graphical method for the detennination of stresses on a plane ioclined to the principal planes. The graphical construction is known as Mohr's circle and is extremely useful. In this method, an origin 0 is selected and the normal stresses are plotted along horizontal axis aDd the shear stresses on the vertical axis. As the compressive s~ are taken positive in soU engineering, these are plolted towards the r:ight of tbe origin. i.e. along positive x-axis. The shear stress is generally taken as positive if it causes a counterclockwise couple at a point inside the wedge ABC in Fig. 13.1. Thus the shear stress marked on the plane AB is pa;.itive. The positive shear stresses are plotted upward from the origin, i.e., along positive y-axis. Let us draw the Mohr circle for the stresses shown in Fig. 13.1. In Fig. 13.2, tbe JXlint E represents the minor principal stress oJ ani;! the JXlint F. the major principal stress al' The point C is the middle point with the normal stress coordinate equal to (01 + oJ)I2. The circle is drawn with C as the centre and EF as diameter. 1be circle is known as Mohr's circle. Each point on the circle gives the stresses a and 't on a particular plane. It can be shown that the point D on the circle gives the stresses on the plane AD inclined al an angle to the major principal plane. The line DE makes an angle 9 witli a-nxis. 1be angle DCF subtended al the centre is obviously twice the angle DEC.

e

308

SOIL MECHANICS AND FOUNDATION ENGINEERIN.G

From tbe figure,

Oh_OC+CH

or

OH ...

°1 ;

°2

+

(0) ;

DH ... CD sin2e ..

ond

0

1

(3)

;

cos 2 9 ...

"3

sin29 ..

(1

't'

Comparing tbe above equations with Eqs. 13.2 and 13.1. it is obvious that the coordinates of the point D

represent the stresses on the inclined plane AD in Fig. 13.l. The point E is a unique point, which is known as the pole (P) or the origin of planes (OP). If a line is drawn. from any point (say D) on the Mohr circle parallel to the plane ( say. Ali) whose stresses arc represented by that point, it will intersed the circle at the T pole P. When the major principal plane is honzonta! (parallel to x-axts), the mmOf prinCIpal planl! IS vertical I ~Mlnor pror>c.pal plone and the pole lies at the pomt E which mdlcates tne minor V... pnnclpal stress Once the pole has bccn located, the : ... --~ stresses any other plane making an angle 0. With lnt: [._--i I major pnnclp.'ll plane can be determmed gl11plm:,,"\lly by __ ~I draWing a line through the pole and maklOg an ano-Ie ex 0 .-P e 2£ I with u-axis. rae coordinates of the point obtained by (he intersection of this line with the circle give the stresses on that plane. The line OD represents the magnitude of the resultant stress on the inclined plane AB. '1l1e anglc of the obliquity of the resultant with the nOlmai of the plane An is equal to the angle (3. R~. D.2. It may be mentioned that the cqumions developed in Sect. 13.2 and the Mohr circle discussed in this section are based on the principles of mechanics.These arc valid for all materialS irrcspective of thci~ sfress-strain characteristics. The same relations are also applicable in solid mechanics.

ill

13.4. STRESS·SYSTEM wml PIUNCIPAL PLANES INCLINED TO TilE COORDINATE AXIS

Fig. 13.3 (b) shows a stressed element in which the principal planes are inclined to the coordinate axcs. In other words, the principal planes are not vel1ical a~ horizontal. The stresses on a plane inclined at an angle to the major principal plane can be determined using Eqs. 13.1 and 13.2 or using the Mohr circle, as

e

'

~

"3 T t.4inor

"'0"'01 " 0 0 ' ·

r.1c)or

Fig. 13.3. Inclined

':i

"i

Prindpat Planes.

principel

plont

SHEAR STRENGTI-I'

309

these relalions are indcpt'ndent of the inclination of the principal planes. However, the orientation of the principal planes is different from that in Fig. 13.2. In Fig. 13.3 (a), the poinLs E and F are loc.1tcd and the Mohr circle is drawn, as in Fig. 13.2. The JXlle P is located by drawing a line EP from E parnllel 10 minor principal p13ne Be or by drawing a line FP from F parallel 10 the major principal plane AC. Tbe stresses on the plane AB inclined at an angle B to the major principal plane can be determined by drawing a line PD through the pole P and parallel to AB. The coordinates of poiQt D give the stresses 0 and "t on the plane AB. It may be noted that the line ED also makes an angle e with a-axis. 13.5. SI'RESS-SYSTEM wrm VERTICAL AND HORIZONTAL PLANES NOT PRINCD'AL PLANES Fig. 13.4 (0) shows an element ABC in which the vertical and horizontal planes BC and AC are subjected to shear stresses in addition to the normal stresses. Hence, these planes are not the principal planes. The stresses on the plane AD inclined al an angle B 10 plane AC can be detennined from the equilibrium of forces. Resolving the forces in x·direction [Fig: 13.4 (b)~

OJ incipc:;i

-, ~

..

t'

(.,

(b)

Co, Fig. 13.4. Vertical Planes not Principat Plarw!S.

aABsin8 ..

OIlIJC

+ "t.ryAC + "tABoose

... (a)

asin9 .. all sin 6 +""C.ry oosB+ ""Coos8 . Resolving the forces in y-dircclion, aABros8 +"t" ARsinB .. "t"..,.DC + a,.AC arosB + "[sinS _ "t"xysinB + o,cosB

or

...(b)

ocose .. "t"..,.sinS + o,ros8 - ""[sinB

Multiplying Eq. (a) by sin e and Eq. (b) by cos B and adding a .. a ll sio2 S + "[...,sin8cosB + "[sin8cosB + "tzy sin S cos 8 + O,OOS2 S _ "tsinBc:osB 0

..

a x sin 2 B + o,cos2·9 + 2 "t"xy sin 9 cos 9

o .. all (

0"

a,,;

1- ~2

8) +

a,

(1

+

C;S 29)

or + (Or;a,,) cos26 + ""C.rysin26

Eq. 13.3 gives the nonna! slress 0 on the plane AB.

+ "t"..,.sin

2e ... (13.3)


310

Substiluting the above value of

0

in Eq. (0) and simplifying,

¥

1:"

Eq. 13.4 gives the stress 1: on Eq. 133. can be written as.

the

.. (13.4)

sin 26 - "'"""coo2e

plane AB.

0- (0",; or) _ (Or; oJ coo29 + 1:..,.sin28

... (13.3(a)J

Mohr's clrde Squaring Eqs. 133 (0) and 13.4 and adding we get

[of'; 0,) r+.'. [( 0,; 0.) ,

+ [(

[ ( ~)l' 2

0-

Eq. 13.5 is the

0, ; 0.)

",,'29 n;,sin'29 + 2 ( ¥ )""29 '., sin29 1

sin'29 + .;, ",,' 2 9 _ 2 (0,

;

0.) sin 2 9 ('., OOS 2 9)

,(~), 2 +-.:q

... (135)

l

+'t_

1

~uation of a c ircle whose centre has the coordinates ( ( o. ;

0, )

.0

1and whose radius

l(~) ~)

is + . lbe coordinates of any PJint 00 the circle represent the nonnal and shear stresses (0,1:) on a particular plane. The circle is Mohr's circle.

Conslruction of Mohr Circle To draw the Mohr circle in this case, the nannal stresses 0:r and 0, are marked as PJints Q and R respectively on the o-axis [Fig. 13.4 (c)]. At PJinl Q. a perpendicular QS is drawn, such that QS - 1:JI:}" The shC"ar stress 1:", is positive on the plane BC as it causes a counterclockwise moment at a point inside the wedge. likewise, the perpendicular RT is equal to 1:..,.. However, this is negative because the shear stress on the plane AC causes a clockwise moment. Point C is at the middle point of QR and ha.. the coordinates [ ( 0%

;

0, )

.0 ]. It also lies on the lineVjdrawn

~) 'f~]

.

through point SandT. A circle is drnwn with its amtre at C and its mdius equal to f ( + The circle passes through points Sand T. It may be noted that the point T in the :Johr circle represents 'the stresses on the plane AC and the point S. on the plane The line CD drawn at an angle of 29 10 the line CT interseas tbe Mohr circle at D. Thererore, the point D gives the stresses on the inclined plane AB.

nco

Principal Planes. As the principal plnnes are the planes with zero shear stresses, from Eq. 13.4,

o. tan 2

ep

-

cry; (0,.

0",

sin29

~~:r)12

p

_

'txy

coo29p ... (13.6)

where 8p is the angle whim the princip.11 plane makes with the plane AC in Fig. 13.4 (b). As the angle subtellded at the centre is twke this angle, the line CU indk:alcs one of the principal planes and the point U represents the major principal stress. This can also be ~roved from the above value of Ian 20r We have

SHEAR

SfRE~Gll'l

311

ros29p

_

±

r=(O=,-=O:::J:;::/=2=

"(¥)'+.:. Substituting these values of sin 201' and the ros 29p in Eq. 13.3,

o-~±(¥)x

(oy-o,,)/2

"(¥)'+.:. or

a" + a,.

o--Z-

:t

"(¥)'+.:.

Therefore, the two principal stresses are as under.

"(¥)'+.:.

Major principal stress.

".(13.7)

The point U gives the major principal stress (01)' Minor principal stress.,

" .(13.8)

The point V gives the minor principal stress (0) AtsO. because tan 2 61' = tan, (281' + 180°), the second principal plane is indicated by the line

cv.

13.6. IMPORTANT CHARACTERISTICS OF MOHR'S CIRCLE The following important characteristi~ of Mohr's circle should be carefully noted, as these are required for further study. (1) The maximum shear stress '' t...... is numerically equal to (01 - 0)12 and it occurs on a plane inclined at 45° to the principal planes (Fig. 13.5). (2) Point D on the Mohr circle represents the stresses (0, 't) on a plane make an angle 0 with the maj,or principal plane.

6)

:le U Fla. 13.5. Characteristics cI Mohr's Otde.

9JIL MECHANICS AND FOUNDATION ENGINEERlNO

1I2

The rcsullanl stress on that plane is equal to ~ and its angle of obliquity with the normal of the plane is equal to angle fl. given by ~. tan-I ('/ 0) ... (13.9) (3) The maximum angle of obliquity Pmax is obtained by drawing a tangent to the circle from the origin O.

~

_ 5io-1 (01 - OJ )12 _ sin-1 (01 + oiJlZ

(01 (Jl

+

... (13.9)

OJ) OJ

(4) The shear stress T.! on the plane of the maximum obliquity is less than the maximum shear stress "tmu.

(5) Shear stresses on planes at right angles to each other are numerically equal but are of opposite sigm. as shown in Fig. 13.4 (c). (6) As the Mohr circle is symmetrical about o-axis, it is usual practice to draw only the top half circle

for convenience. (7) There is no need to be rigid about sign convention for plolting the shear stresses in Mohr's circle, These can be plotted either upward or downward. Although the sign convention is ~uired for locating the orientation of the planes, the numerical results arc not affected. 13.7. MOHR·COULOMB TIlEORY The soil is a particulate material. The shear failure occurs in soils by sUppage of particles due to shear stresses. The failure is essentially by shear, but shear stresses at failure depend upon the nonnal stresst.s on the potential failure plane. Acrording 10 Mohr, the failure is caused by a critical combination of the normal and shear stresses. The soil fails when the shear suess ('tj') on the failure plane at failure is a unique functioo of the normal stress (0) acting on thai plane. ' , . /(0) Since the shear stress on tbe failure plane al failure is defined as the shear strength (s), the above equation can be written as

, . /(0) ... (13.11) The Mohr thoory is roncemed with the shear stress at failure plane al failure. A plot can be made between the shear stress "t nnd tbe nonnal stress a nt failure. 1be curve defined by Eq. 13.11 is known as the Mohr envelope [Fig. 13.6 (a) 1. There is a upique failure envelope for each malerial. Failure of the material occurs when the Mohr circle of the stresses touches tbe Mohr envelope. As disrussed in the preceding sections, the Mohr circle represents all possible combinations of shear and normal stresses at the stressed point. AI the point of conlal;i (D) of the fairure envelope and tbe Mobr circle, the critical combination of shear and normal stresses is reached and the failure occurs. The plane indicated by the line PD is, lherefore, the failure plane. Any Mohr's circle which does not cross the failure envelope and.

T

c.) T

Cb)

T

Cd

Pia.

13.6.

F.itu~

Envdopea.

SHEAR STRENG1l(

313

lies below the envelope represents a (non-failure) stable oonditioD. The Mohr circle cannot aoss the Mohr envelope. as the failure would have already occurred as soon as the Mohr circle touched the envelope. The .shear strength (3) of a soil al a point 00 a particular plane was expressed by Coulomb ru; a linear function of the oormal stress on that plane, m s • c + a (13.12) In other words. the Mohr envelope is repluced by a straighl line by Coulomb as shown 10 Fig. 13.6 (b). In Eq. 13.12, c is equal to the intercept 00 't-axis and , is the angle which the envelope makes with a·axis [Fig. 13.6 (c)]. The oomponent c of the shear strength is known as cohesion. Cohesion holds tbe particles of the soil together in a soil mass, and is independent of the normal stress. The angle til is called the angle of in~rnoJ friction. It represents tbe frictional resistance between the particles, which is directly proportional to the normal stress. As mentioned before, the failure occurs wben the stresses are such that the Mohr circle just touches the failure envelope, as shown by point B in Fig. 13.6 (c). In other words, shear failure occurs if the stresses a and 't 00 the failu~ plane~_as point B. If the stresses plot as point A below the failure envelope. it represents a stable. noo-fai1fi.e oondlticm. On the other hand. a state of stress represented by point C above the failure envelope is not possible. It may be noted that a material fails along a plane when the aitlcal combination of the stresses a and 't gives the resultant witb a maximum obliquity (f\uJ. in wbich case tbe resultant just touches the Mohr circle.

tan,

...

13.8. REVISED MOBR·COUWMB EQUATION Uiter research showed that the parameters c and 4' in Eq. 13.12 are not necessarily fundamental properties of the_SOil as was originally assumed by Coulomb. These parameters depend upon a number of factors, sum as the water content, drainage conditions, ronditiollS of testing. The current practice is to cmsider c and ., as mathematical parameters which represent the failure COllditions for a particular soil under given conditions. That is the reason why c and • are now called cohesion i~ercept and the angle of shearing tWstau!;C. These indicate the intercept and the slope of the failure envelope, respeaively. Terz.aghi cstablished that the Donnal stresses which control the shear strength of a soil are the effective stresses and not the total stresses. In terms of effed.ive stresses., Eq. 13.12 is written as s _ c' + tan.' ... (13.13) where c' and .' are the ootlesion intercept and the angle of shearing resistance in tenus of the effective

a

Mresses.

Eq. 13.13 is known as the Revised Mohr-Coulomb equation for the shear strength of the soil. The equation has replaced the original equation ceq. 13.12). It is one of the most important equations of soil en~neering.

The Mohr-Coulomb theory shows a reasonably good agreement with the observed failures in the field and in the laborntory. 1be theory is ideally suited for studying the behaviour of soilo; at failure. The theory is used for eslimation of the shear strength of soils. However. even this theory is not perfect. H has Ihe following main limitations : (1) It neglects the effect of the inlennediale principal stress (o2), (2) It approximates the curved failure envelope by a straight line, which may not give COlTCC't resUlts. (3) When the Mohr envelope is curved, the aQuai obliquity of lhe failure plane is slightly smaller than tbe maximum obliquity. Therefore, the angle of the failure plane.. as found, is nO! correci.. (4) For some clayey soils, there is 00 fixed relationship between the nonnal and shear stresses on the plane of failure. The theory cannot be used for such soils.

13.9. DIFFERENT TYPES OF TESTS AND DRAINAGE CONDnl0NS The following tests are used 10 measure the shear strength of a soil. (1) Direct shear test (3) Unconfined Comprcssioo test

(2) Triaxial Compression test (4) Shear V
SOIL MECI-IANICS AND FOUNDATION ENGINEERING

314

The shear

problem. In

must be conducted under appropri
tcs(

sh~lr

(1) Consolidation stage in which the normal stress (or confining pressure) is applied 10 1he specimen and

it is allowed to consolidate. (2) Shear stage in which the shear stre.'>S (or deviator stress) is applied to lhe specimen to shear it. Depending upon the drainage conditions, there are three types of tests as explained t>clow :

(1) Unconsolidated-Undndned Condition. In this type of test, no drainage is permitted during the consolidation siage. The drainage is also not pennitted in the shear stage. As no time 1.<; allowed for consolidation or dissipation of excess pore water pressure, the tCSt can be conducted quickly in a few minutes. 1'he lest is known as unoonsolidatcd--undraincd test (UU test) or quick
13.10. ManE OF API)LlCA1l0N OF SIIEAR FORCE The shear force in a shear lest is applied either by increa'>ing the shear displacement at a given rate (X" by increasing the shearing force at a given rate. Aaxxdingly. the shear tests are either strain-controUcd or stress· controlled. (l) Sirain controlled tests. In a stmin-conlrolled lest, the test is conducted in such a way thai the shc.aring slroin increases at a given ralc. Generally. the rale of increase of the shearing strain is kept oonstant, and the specimen is sheared al a uniform strain rale. 'Ibe shear force acting on the specimen is measured indirccUy using a proving ring. The rate of shearing slJ"din is controlled manually or by a gear system attadted to an electric motor. Most of the shear tests are conducted as strain-controlled. The slress--strain dtaracterislic are easily obtained in !hese teslS, as the shape of the siress-strain curve beyond the peak point can be observed only in a strain- contro lled lest. A strain--conlrolled test is easier to perform than a stress- controlled lest. (2) S(~ss-Controlled tests. In a stress--controlled test, the shear force is increased at a given rate. Usually, the Me of increase of the shear force is maintained constant. The shear load is increased such thaI the shear stresses iT1Ct'C3SC at a uniform rate. lbe resulting she.'lI' displaa:.ments are obtained by means of a dIal gauge. Stress--conlrolled tests are preferred for conducting shear tests at a very low rale, because an applie..J load can easily be kept constant for any given period of time. Funher. Ihe loads am be conveniently applied and removed. TIle stress-controlled tcst represenls the field conditions more closely.

13.11 DIilECT SHEAR TEST (a) Appal"ullL'i. A direcl shear Ie..<;t is oonductcd on II soil specimen in a s hear box which is split into two b.1lves along a hori7.ontal plane al ilS middle (Fig. 13.7). The shear box is made of brass or gunmetal. It is

SHEAR STRENGTH

315

Ag. 13.7. Din:C( Shear Test.

tither square or circular in plan. A square box of size 60 x 60 x 50mm is commonly used. The box is divided horizontally such that the dividing plane passes through the .cenlre. The two halves of the box are held together by locking pins. Suitable spacing screws to separate the two halves are also provided. The spacing screws are fixed to the upper h..1lf and they bult against the top of the . lower half. The box is provided with the gripper or the grid plates which are toothed and fitted inside it. lbe gripper plates are plain (without perforations) for undrained tCSlS and perforaled for drained tests. Porous stones are placed at the top and Ihe bottom of the specimen in drained tests. A pressure pad of ~ or gun metal is fitted into the box at its top to transmit the nonnal load to the sample. The normal load from the loading yoke is applied on the top of the specimen through a steel ball bearing upon Ihe pressure pad. The lower half of lhe box is fIXed to the base plate which is rigidly held in position in a large container. The large container is supported on rollers (roners not shown). 1be oontainer can be pushed forward at a constant rate by a geared jack wbich works as a strain-controlled device. The jack may be operated manually or by an electric motor. A loading frame is used to support the large container. It has the arrangement of a loading yoke and a lever syscem for applying the normzl load. A proving ring is fitted to the upper half of the box to measure the shear force:lbe proving ring butlS against a fixed sup[K>M. As the box moves. the proving ring records the shear force. Ihe shear displacement is measured wiLh ~ dial gauge fitted 10 the oontainer. Another dial gauge is fitted 10 the top of the prt.."i.')ure pad to measure the change in the thickness of the. specimen. (b) Test. A soil specimen of sizc 60 x 60 )( 25 mm is taken. It may be either an undisturbed sample or made from oompactcd and remouldcd soil. 'lbe specimen may be prepared directly in the box and comp:IClcd. The base plale is attached to the lower tullf of the box. A porous stone is placed in the nox. For undrained tests, a plain grid is kept on the ~rous stone. kccping its segrcgmions ~t right angles to the direction of shear. For drained tests, perforated grids arc used insle.'ld of plain grids. The mass of the base plate. porous stonl! arxl grid is taken. The specimen if made scpar;\lely is trall'iferred to the box and its m:lSS taken. The upper grid, porous slone and the pressure pad are placed on the specimen. '100 box is placed inside the large container and mounted on the loading frame. The upper half of the box is brought in CQlltaci with the proving ring. 1be loading yoke is mounted on the steel ball placed on the pre.<;surc pad. The dill! gauge is filled to Ihe container to give the shear displacement. 'The other dial gauge is mounted on the loading yoke to record the vertical movement. The locking pins are removed and the upper half box is slightly raised with the help of spacing screws. The space between the two halves is adjusted, depending upon the maximum particle size. TIle space should be such that the top half of the 00" does not ride on soH grains which oome between the edges. The normal load is applied to give a normal slress of 25 kN/m2. Shear load is then applied at a oonstanl rate of slrain. For undrained tests, the rate is generally between 1.0 mm to 2.00 mm per minute. For drained

316

SOIL MECHA.~ICS AND FOUNDATION ENGtNEFRING

tests, the strain rale depends upon the type of soil. For sandy soils, it may he taken flS 0.2 mm/minute; whereas for clayey soils, it is generally between 0.005 to 0.02 mm/min. Inc &'lmplc sllears along Ihe horizontal p.k'lne between the two halves. The readings of the proving-ring and the dial gauges arc taken every 30 seconds. The lest i~ continued till the specimen fails. The failure is indicated when !.he proving ring dial gauge begins to recede afler having re.'lchcd the maximum. For the soils which do not give a peak. point, the failure is assumed to h:lVe occurred when a shearing slmin of 20% is readIed. AI the end of the test, the specimen is removed from the box and its waler conlent found. The lest is repeated under the nonnal stress of 50. 100, 200 and 400 kN/m 2 . lne range of the normal stress should cover the range of loading in the field problem for which the shear parameters are required. The shear stress at any stage during shear is equal to the shear force indicated by the proving ring divided by the area of the specimen. A plot can be made between the shear stress and the shear strain. 1bc shear strain is equal to the shear displacement (till) divided by the length of the specimen (L). 'The shear stress is obtained from the shear load indicated by the proving ring and the cross-sectional area. Direct shear tests can be COnduCled [or anyone of the three drainage conditions. For U-U test. plain grids are used and the sample is sheared rapidly. For CU test, perforated grids are used. The sample is consolidated under the normaiload and afler tbe completion of consolidation, it is sheared rapidly in about 5-10 minutes. In a CD test, the sample is consolidated under tbe normal load and then sheared slowly so tbat excess pore water pressure is dissipated. A CD t~t may take a few hours for oohesionlcss soils. For cohesive soils, it m~y take 2 to 5 days. 'Il$ direct shear test is generally conducted on co~ionless soils as CD lest. It is convenient to perform and it gives good results for the strength parameters. It is occasiooally used to determine the strengt!! parameters of silt and clay under unconsolidated-undrained T and consolidated drained conditions, but i: does not offer the flexibjijty of a triaxial compression test, as explained later. 13.12. PRESENTATION OF RESULTS OF DIRECT SHEAR TEST Loosp 50nd (0) Stress-Strain Curve. A stress-strain curve is a plot between the sbear stress l: and the shear displacement (MIlL) {Fig. 13.8 (o)J. In case of dense sand (and also ~-------------------'1. over-consolidated clays), the shear stress attains a peale. '0) vallie at a small strain. With further inc:rease in strain, the shear stress decreases slighlly and becomes more or less IlII constant, known as ultimate stress. In case of loose sands (:md nonnally consolidated clays). the shear stress increases gradually and flOally allains a constant value, known as the ultimate stress or residual strength_ It has been mservcd that the ultimate shear stress attained by both dense and loose -Av s•.-mds tested under similar conditiOll!l is approximately the same. The figure also shows the stress-strain aJNe of a medium dense s.'lnd. Loost'sond Generally, the failure strain is 2 to 4% for dense sand and 12 to 16% for loose sand. Fig. 13.8 (b) shows l?e volume changes with an increase in shear strain for CD teslS. Since Ihe cross-sectional area of the specimen remains unchanged, the volume change is proportional to the change in thickness measured by the dial gauge. In case of dense sands (and Shoi~ over-oonsolidaled clays), the volume first d~ slightly,

'"

Ie)

Fig. 13.8. SIre.ss-Slmin ClIr.u

SHEAR STRENGTH

317

but it increases with further increase in strain. In the case of loose stmds (and nonnally consolidated clays), the volume decreases with an increase in'. shear strain. The figure also shows the curve for medium dense

sand. It may be observed that the void ratIO of an initial loose sand decreases witll an ini=rease in shear strain, whereas that for the initially dense sand increases with an increase in strain [Fig. 13.8 (C}}. lbe .v9id ratio,at which there is no change in it with an increase in strain is known as the critical void ratio. If ' the sand initially is at the critical void ratio. there would be practically no change in volume with an increase in shear main. (b) Failure Envelope. For obtaining a failure envelope, a number of identical specimens are tested under different nonnal stresses. The shear stress required to cause failure is determined for each nonnal stress. The fmlme envelope is obtained by plotting the points corresponding to shear strength at diffcreni normal StI'eSSC1> and jOining them by a straight line rpig. 13.9 (a)]. The inclination of the failure envelope to Is

0'<

d

.:;,9-i- ¢o.-.t.v.

T

~,..\\"

n h

oi,..

(a·)

c.::

d'lyy-"

1

~~ (e)

Fig. 13.9. Fa.ilure Envelope.

the horizontal gives the angle of shearing resistance ¢' and its intercept on the vertical axis is equal to the cohesion intercept c. For dense sands. the failu re envelope can be drawn either for peak stress or for ultimate stress. The values of the parameters q. and c for the two envelopes will be different. For loose sands. the failure envelope is drawn for ultimate stress, whieh is usually taken as the shear stress at 20% shear strain. (c) Mobr-Cirde. In a direct shear test, the stresses on planes other than the horizontal plane are not known. II is, therefore. not possible to draw Mohr stress circle al different shear loads. However. the Mohr circle can be drawn at me failure condition assuming that the failure plane is horizontal. In Fig. 13.9 (b) . the point B represents the failure condition for a particular nonnal stress. The Mohr circle at failure is drawn such that it is tangential to the failure envelope at B. The horizontal line BP gives the direction of the failure plane. ·The point P is the pole. The lines PD and PA gi~e the directions of the major and minor principal planes, respectively. The principal planes are also shown in Fig. 13.9 (c).

Merits and Demerits ot Dired Shear Test 1be direct shear test has the following merits and demerits as compared to the triaxial compression test (dacribed in the foUowin, section).

SOIl. MECHANICS AND FOUNDATION ENGINEERING

318

Merits. (I) lbe sample preparation is easy. The test is simple and convenient. .(2) As the thickness of the sample is relatively small, the dminage is quick and the pore pres.sure dis.sipate5 very rapidly. Consequently, the consolidaled-drainoo and the consolidated.. undrained tests lake rclntivcly small period. (3) It is ideally suiled for conducting drained tests 00 rohcsionless soils. (4) TIle apparatus is relatively cheap.

Demerits. (1) lbc stress conditions are known only al failure. The conditions prior to failure arc indeterminate and, thereiore, the Mohr circle c.'moot be drawn. (2) The stress distribution on the failure plane (horizontal plane) is nOl uniform. The stresses are more at the edges and lead to the progressive failure, like tearing of a paper. Consequently. the full strength of the soil is not mobilised simultaneously on the entire failure plane. (3) 'lllC a~ under shear gradually decreases as the test progresses. Out the corrected area cannot be determined and, therefore, the origin:li area is taken for the computation of stresses. (4) 11lt: orientation of the failure plane is fixed. This plane may not be the wcakest plane. (5) Control on the drninage conditions is very diffICUlt. Consequently, only drained tests can be conducted on highly permeable soils. (6) 1be mea'>uremenl of pore water pressure is nol poosible. (7) '111e side walls of the shear box all1')C lateral rcstrnint on the specimen and do nO( allow it to deform l:lIcl"'.lUy.

13.13. DIFFERENT TYPES OF SOILS On the basis of shC:lr strength, soils can be divided into three types. (I) Cohcsionless soils. (2) Purely cohesive soils and (3) Cohesive-frictional soils. 1. Cohesionless soils. ll1csc are the soils which do not have cohesion i.e., c' - O. 'fbcsc soils derive the shear Slrcngth from the inlergranular friction. 'Ibcse soils are also called frictional SQils. For example, sands and gravels. 2. I'urely cohesive soils. 1bcse are the soils which exhibit oohcsion but the angle of shearing resistance , :: O. For cxample, satumted clays aDd silts under undrained conditions. lhese soils are also called ' .. '. 0 soils. 3. Cohesive-frictional soils. These arc composite soils having both c' and ". These are also called soils. For example, clayey sand, silly sand, sandy clay, etc. (Note. Sometimes, cohesive-frictional soils are also called cohesive soils. Thus any soil having a value of c' is caned a cohesive soil.l

c-,

13.14. TRIAX1AL COMPRESSION TEST APPARATUS The triaxial romprcssion lest, or simply triaxial test, is used for Ihe dctcnnination of shear characteristics of all types of soils undcr different drainage conditions. In this test, a cylindrical specimen is stressed under conditions of axial symmetry, as shown in Fg. 13.10. In the flrsl stagc of lhe test, the specimen is SUbjected to an an round confining pl'C&Sure (o..) on the sides and at the top and the bottom. This stage is known as the oonsolidation stage. In the second stage of the lest, called the shearing stage, an additional axial stress, known as tl)e deviator stress (001)' is applied on the lOp of the specimen through ~ rom: ·Thus; the lolal stress in the axial direction at the

-0 "C+'ij

111

-

-

--

"C-

--

't

-

t t t
(.) ELEVATION

<1.

'0'

1/..,.<1.

'" I

-«

I , "-

(b) PLAN Fig. 13.10.

SHEAR STRENGTII

319

time of shearing is equal to (or + 0d)' II may be noted that when the axial Slress is increased, the shear stre.sses develop on incHncd planes due to compressive stresses on the top. The vertical sides of the specimen are principal planes, as there arc no shear stresses on the sides. The coofining pressure Oc is equal to the minor princiP.11 stress (oJ). The top and bottom planes are the major principal planes. The IOlal axial stress which is equal 10 the sum of the confining pressure and the deviator stress, is the major principal stress (01)' l3ccause of axial symmetry, the intcnncdiatc principal stress (o:z) is also equal to the confining pressure (or)' [Note. The above ~terpretalion of the stress conditions in the triaxial test is 1I0t strictly correcl according 10 the theory of elasticity. In the case of cylindrical specimens. the three princ ipal ' stresses arc the axial, radial and the circumferential stresses. lhe state of stress is statically indclcrrninute throughout the specimen. For convenience. in the triaxial test. the circumferential stress is taken equing the thcory of manometers. As the mercury surface in the upper pot is open to atmosphere. the (gauge) pressure there is zero. Prom the manometer equatjon,


31Jl

r

1

Fig. 13.12. Mercury Control System.

where cr( '1w

the cell pressure at Ihe centre of the specimen. the unit weight of water. and Y~I the unit weight of mercury. The above equation ctJn be simplified lIS

Ilv

Oc ,., ,/", (hI + (112 - hlh... .,:(13.14) Tbe upper pot is sUP1>ortcd by a spring. When the volume of the s{XX-imen decreases due to oonsolidation or when the 'XatF lC
... (13.15) where A = cross.sectional area of the mercury pot, and W = weight of unit length of the lube filled with mercury which is also lifted above the floor. 2. Pore water Pressure Measurement Device. The pore water pressure in the triaxial specimen is measured by attaChing it to the device shown in Fig. 13.13. It consists .of a null indicator in whict]. no· flow condition is maintained. For accurate mca,;urcment. no· flow condition is essential because the flow of water from tbe sample to the gauge would modify the actunl magnitude of the pore water pressure. Further, the flow of water leads to a time lag in the attainment of a steady state in samples of cohesive soils because of low pcnneability. The null indicator is essentially cs connecting the specimen anJHhe null·indicator should be such that these undergo negligible volume changes iIDdQ' pressure and are free from leakage. . Any, change in the pore·water pressure in the specimen tends to cause a movcmcnt of the mercury level in the null-indicator. However, the no-flow conditioo is maintained by making a corresponding changc IUl tile

SHEAR SfRENG1lJ

321

Fig. 13.13. Pore Water Pressure MeIISurcnlell[ Device.

4) DO

he >01

lIS

Ith of

IS)

is ow rter OW

low .

other limbs by means of the oontroi cylinder. Thus Ihe mercury levels in the two limbs remain oonstant. The pressure applied by the control cylinder is recorded by pressure gauge or the manometer. rr the specimen is partially saturated, a special fine, porous ceramic disc is placed below the sample in the triaxial cell. The ceramic disc pennits only pore wMcr to flow, provided the difference bctweCl't the pore air pressure and pore water pressure is below a certain value, known as rhe air-entry value of the ceramic disc. Under undrained oonditions, the ceramic disc will remain fully saturated, provided the air-entry value is high. It may ·be mentioned that if the required cerdlTlic disc is not used and instead the usual coorst; porous disc is used, the device would measure air pressure and not water pressure in a partially saturated soil. In modem equipment, sometimes the pore water pressure is measured by means of a lransduccr and not by conventional null indicator. 3. Volu'm e Changes Measurement. 'k>iwne changes in a drained test and during consolidation stage of a TRIAXIAL CEll consolidated undrained test are measured by means of a burette connected to the specimen in the triaxial cell. For PEDESTAL accurate measurements, the water level in the burette BURETTE should be approximatcly at the level of the centre of the specimen (Fig. 13.14). During consolidation slage, 'the volume of the specimen decreases and the water level in the burette Fig. 13.14. Volume aumge Measurement. rises. 'The change in the volume of the specimen is equal to the volume of tbe water increased in the burette. During shearing of specimens of dense sand wben the VQlume of the sample increases, the water flows from the burette La the specimen. The increase in volume of tlJe specimen is equal to the volume of water decreased in the burette.

U.lS. TRIAXIAL TESTS ON COHESIVE SOIlS ,is lrol

ewling Ider

evel the

The following procedure is used for oonducting the triaxial tests an cohesive soils. (a) Consolldated·undralned test. A dcmred, coarse porous disc or slone is placed on the top of the pedestal in the triaxial test apparatus. A filler paper disc is kept over the JX>rous stone. The specimen of the cdlcsive soil is then placed over the filter paper disc. The usual size of the specimen is about 37.5 mm diameter and 75.0 mm height. A porous stone is also placed on the top of the specimen. Deaircd vertical filter Slrips are placed at regular spacing around the entire periphery such that these touch both tbe porous stones. The sample in then enclosed in a rubber membrane. which is slid over the specimen with the help of 11 membrane stretcher. The membrane is sealed to the specimen with Q-rings.

322

SOIL MECHANICS AND FDUNDA"110N ENGINEERING

'The triaxial cell is placed over the base and fIXed to it by lightening the nuts. The cell is then filled with water by connecting it to the pressure supply. Some space in the lOp portion of the cell is filled by injecting

oil through the oil valve. When excess oil begins to spiU out through the air-vent valve, both the valves (oil valve and air-vent valve) are closed. Pressure is applied to the water HUed in the cell by connccting it to the mercury-pot system. As soon as the pressure acts on the spcx:imcn. it starts conSOlidating. The specimen is connccted to Ibe burette through pressure connections for measurement of volume changes. The consolidation is complete when there is no more volume change. When the consolidation is complete. the specimen is ready for being she.1red. The drainage valve is closed. The pore water pressure measurement device is attached to the specimen through the pressure connCctions. The proving ring dial gauge is set to zero. Using the manual control provided in the loading frame, the ram is pushed into the ceU but not allowed 10 touch the loading cap. The loading machine is then

run at the seleded speed. 1be proving ring records the force due to friction and the upward thrust ading 00 the ram. The machine is stopped, and with the manual control, the ram is pushed further into the cell bringing it in COnt.1ct with the loading cap. The dial gauge for the measuring axial defannation of the specimen is sct to zero. The sample is sheared by applying the deviator stress by the loading machine. The proving ring readings are genemlly taken corresponding to axial strains of 1/3%, 213%, 1%, 2%. 3%. 4%. 5%, ... until failure or 20% axial strain. Upon completion of the test, the loading is shut ofT. Using the manual control, all additional axial Slrc$ is removed. The ccll pressure is then reduced to zero, and the cell is emptied. The triaxial ccll is unscrewed and removed from the base. O-rings arc taken out, and the membrane is removed. l'he specimen is thCII recovered after removing the loading cap and the top porous stone. 1be filler paper strips arc peeled ofT. The post- shear mass and length are determined. The water content of the specimen is also found. (b) Unconsolidated Undrulned tcst. The procedure is similar to that for a consolidated-undrained tCSl, with one basic difference that the specimen is nO( allowed to consolidate in the first stage. llte drainage valve during the test is kept closed. However, the specimen cm be connected to the pore-water pressure measurement device if required. Shearing of the specimen is started just after the application of the ceU pressure. 'nle second stage is ex.1C11y the same as in the consolidated-undrained test dc.c;cribcd above. (c) Consolidated Drained test The procedure is similar to thai for a consolidated-undrained test, with one basic difference that the specimen is sheared slowly in the second stage. After the consolidation of tbe specimen in the first stage., the drainage valve is not closed. !t remains connected to the burette throughout the test. The volume changes during the shearing stage are measured with the help of the burette. As the permeability of cohesive soils is very low, it takes 4·5 days for the consolidated dmined test.

13.16. TRIAXIAL TESTS ON COlIESIQNLESS SOILS Triaxial tests on specimens of cohesionlcss soils can be conducted using the procedure as described fa cohesive soils. As the samples of cohesionlcss soils cannot stand of their own. a special procedure is llSC{I fa preparatioo of the sample $ described below. A metal fonner, which is a split mould of about 38.5 mm internal diameter, is used for the preparntim of the sample (Fig. 13.15). A coarse porous stone is pJaccd on the top of the pedestal of the triaxial base, and the pressure conneaion is attached to a burette (not shown). One end of a membrane is scaled to the pedestal by O-rings. The metal fonner is clamped to the ba<;e. lbe upper metal ring of the former is kept inside the lop end of the rubber membrane and is held with the help of a clamp before placing the funnel and the rubber bung in position as shown in figure. The membrane and the funnel are HUed with deaired water. "fbe cohcsionlcss soil which is to be tested is saturated by mixing it with enough water in a beaker. The mixture is boiled to remove the entrapped air. The saturated soil is deposited In the funnel, with a stopper in posit jon, in the required quantity. The glms rod is then removed and the sample builds up by a continuous rapid flow of saturated soil in the former. The

SHEAR SfRENGTI-I

32J

funnel is lherr removed. The sample may be compaded if required. The surface of the sample is leveled and a porous stone is placed on its lap. The loading cap is placed

genUy on the lop porous stone. Q·rings are flXed over the top of the rubber membrane. A small negative pressure is applied 10 the sample by lowering the burctle. The ncgauve pressure gives rigidity to the S3mple and it can stand without any lateml support. For sample of 37.5 mm diameter, a negntive pressure of 20 an of water (or 2 kN/m1 is sufncicnt. As soon a> the negative pressure is applied, the consolidation of the- sample occurs and it slightly shortens. 1be diameter of the upper porous stone should be slightly smaller than thut of the specimen so that it can go inside when the sample shortens; Olherwise, a neck is formed. 1be split mould is Ihen removed, and the diameter and the beight of the sample are measured. The thickness of the membrane is deducted from the total diameter to get the net diameter of the sample. The cell is then placed over the base and clamped to the base. II is then filled with water. The rest of the procedure is the same a<; for cohesive soils.

fig. 13.IS. Preparation eX Sample of Cohesionlcss Soil.

1.1.17. MERITS AND DEMERITS OF TRIAXIAL TEST The triaxial test has tbe following merits and demerits. ,I

Merits. (1) There is complete conlroLover lhe drainage conditions. Tests can be easily conducted for all three

types of drainage renditions. (2) Pore pressure changes and the volumetric changes can be measured directly. (3) The stress distribution on the failure plane is uniform. (4) The specimen is free to fail on the weakest plane.

(5) The state of stress at all intermediate stages uplO failure is known. lbe Mohr circle can be drawn at >II

'"

;al

he

aoy stage of shear.

(6) The test is suitable for accurate research work. The such

Wi

extension test and tests for different

stres.~

apparnt~

is adaptable to special requirements

paths.

Demerits. cd

\ir. ISS

be

(1) The apparatus is elaborate, costly and bulky. (2) The drained lest takes a longer period as compared with that in a direct shear lest. (3) The strain condition in the specimen are not uniform due to Crictional restraint produced by the loading cap and the pedcs181 disc. This leads to the formation of the dead zones at each end of tbe specimen.


32A

The non-uniform distribution of stresses can be largely eliminated by lubrication of end surfaces. However, non-uniform distribution of stresses has practicaUy no effect on the mealOured strength if length/diameter ratio is equal to or morc than 2.0. (4) II is not possible \0 determine the cross-sectional area of the specimen Ilccuratcly al large strains, as lhe assumption thai the specimen remains cylindrical docs not hold good. (5) The Icst simul..'1ICS only axis-symmetrical problems. In the field, the problem is generally 3-dimcnsional. A general test in which all the three stresses are varied would be more useful. (6) 'Ibc consolidation of the specimen in the tcst is isotropic; whereas in the field. the consolidation is

. generally anisotropic. Despite the above-mentioned demerits. lhe lriaxia) lest is extremely uscrul. It is the only rcli.'lble test ror accurate detcrmin:"ttion or the shear char
13.18. COMPUTATION OF VARIOUS PARAM ....IERS (a) Post-Consolidation Dimensions. In consolidated-drained and consolidated-undrained tests, tbe consolidation or the specimen takes place during the first stage. As the volume or the specimen decreases, its post-conSOlidation dimensions arc different from the initial dimensions. The post coru;olidation dimensions can be determined approximately assuming thllt the snmple remains cylindric.'ll mld it behaves isolropically. Let Li. D" and Vi be the length. diameter and the volume or the specimen berore consolidation. Let Lo- Do and Vo be the corresponding quantities aftcr consolidation. Thcrerore, volumetric Change, The volumetric change (A V,) is mc.'lsurcd with the help or burette.

a Vi

£" " -v,:-

Volumetric strain,

For isotropic consolidation. the volumetric strain is three times the linear strain (£/), Thus £/ .. £,,/3 Lo-Li-A~-Li-L,x~

Thus

Lo • L, (1 - EI) .. L, (1 - E,/ 3)

... (l3.16)

Likewise, Do - Dj (I - E,/ 3) Ihe post consolidation diameter Do can also be computed after La has been detennincd rrom the relation, (j[/4 . D~) x La - Vo

·~

Do - V~-

... (!3.I~

(x / 4) )( [,0

(b) Cross-S(,.'Clional Area During Shl!ar Stage. A<; the sample is sheared. its length decreases and the diameter increases. The cross-sectional arc.'l A at any stage during' $hem can be determined assuming that the sample remains cylindrical in shape. lei tJ.L o be the change in length and avo be the change in volume. The volume of the specimen at any stage is given by Va :!: avo. Therefore,

A (Lo -

aLo) ..

Vo:!: avo

A .. Vo:!: AVo _

Lo-aLo

VO(I:!: a~o) Lo

(Ho) To 1 -

Eq. 13.18 is the general equation which gives the cl'06S-scctiomtl area of the specimen. The above equation can be written
SHEAR STRENGTH

325

... (13.19) where £\ is the axial strain in the sample. For 3D undrained tcst, as the volumetric change (ilVo) is zero, Eq. 13.19 becomes A - 1

le IS

~o £1

... (13.20)

The strcsscs in the specimen at various stages of shear should be calculated using the craiS-sectional area A as found above. The correction so applied is known as area CQrrection. (e) Stresses (I) DeviOlor stress. 'The deviator stress (od) acting on the specimen wht::n the axial load applied by the loading maChine is P can be obtained as Od -

~

••• (13.21)

The deviator stress (oJ) is equal (CI\ - (13) It may be 00100 thai the lood indicated by the proving ring is slightly more than P because of friction on the ram and the upward thrust on the ram due to pressure of the water in the cell. The correa ion can be

determined separately. A more convenient procedure is to lift the ram above the specimen when the cell pressure has been

16)

on,

applied. The machine is started keeping the strain ralc the same as to be used in the actual test. ')be proving ring records the load. To account for correction, the dial gauge on the proving ring is set to zero to indicate zero load. This automatically compensates the rom friction and the upward thrust on the rom due to ceU pressure. Thus the lond indicated by the proving ring during shear would be equ.11 to the load P applied to the specimen. (iI) Principal stresses. The minor principal stress (oJ) is equal to the cell pressure (o~). The major principal stress (01) is equal to tbe sum of the cell pressure aocl the deviator stress. Th~

~-~+~-~

or

0l-O)+OJ

(iii) Compressive strength. The deviator stress at failure, (0',- aiJ" is knovorn as the compressive strength of the soil. 17)

~n5e

sand

(O.e.e.)

13.19. PRESENTATION OF RESULTS OF TRIAXIAL

TESTS

the Ihe

(a) Stress-Struin CUn'es. Drained Test. Fig. 13.16 (a)

[be

shows the stress-strain curve for a drained test. The y·axis Shows the deviator stress (01 - (3) and the x-axis, the axial

.18)

... (13.22) cr1-cr3

struin (£\). For dense sand (and over-COl'L'iOlid
,,.

SOIL MECHANICS AND FOUNDA'nON ENGINEERING ewEll Cot6OL lDATED U AV (OCC)

{ .. , STRESS. STRAIN

(LOOSE SAHD J

C1/ERC\lII SO!. IDATED CL AY (DENsE SAND )

(b) PORE

PRESSI..JIE

Fig. 13.17. Stcl"$$-5lrnin Curve, CU Test.

the volume deccca
atmospheric pressure). (b) Mohr Envelopes. For drawing the failure envelopes, it is necessary to test at least three samples at three different cell pressures in the stress range of interest. For dense sands and o vcr-coosolidatcd clays, the failure envelope can be drawn eilher for the peak stress or for the ult imate stress. For loose sands and nonnally oonsolidatcd clays, the failure envelope is drawn for lhe ultimate stress, wh ich is usually taken at 20% strain. Further, the failure envelope can be drawn either in terms .o f effective stresses or in tenns of total stress. Of course, the two envelopes will give diffeNnt values of strength pArameters (c and cp) .

Brillleness JB is expressed

I 8

as

_ Peak strength - Ul!imatc strength Ultimate strength

(I) Effective Stresses. Fig. 13.18 (0) shows the failure envelope for a nonnally oonsolidatcd clay in terms of effeaive stresses obtained from a consolidated drained test. The failure envelope has an angle of shearing resistance of and pmses through origin. First the Mohr circles for the three tests are drawn in IConS of effective stresses corresponding to failure conditions. 1ben the best common tangent is drawn to the three circles. The common tangent is the

+

(0) FO 'l ur~ envtlOpt' for ncrmoll~ (ef trc\;v" s!ressn)

con..al,d
~kwsss

T

(b)Foiluff "rwt'lop" (r!lf~livt

tor O\'O!,_conSOlidated

d oy

sl,,,sSt'S)

Fig. 13.18. Failure Envelopes (effective slreS!i).

'0

SHEAR SfRENGTH

327

failure envelope. As each circle represents a failure. there must be at least one point on it which gives the stresses satisfying the failure criterion. Obviously. the common tangent joins all such points of the three

circles. 'Tbus for nonnally consolidated clays, shear strength is $ -

otan~'.

Fig. 13.18 (b) shows the failure envelope for ovcrconsolidatcd clay in terms of effective stresses. The failure envelope is slightly curved in the initial portion, but, for convenience, it is approximated as a straighl Une. lhe failure envelope has an intercept c' on the 't- axis. The angle of Shearing resistance is ". In the case of ovcr-mosolidated clays. shear strength is

iy

'"

Iv

"

s_c'+otancp' The failure envelopes il!.. terms of effective stresses can also be drawn from the results of a consolidated-undrained lest (CU test) when the porc water pressure measurements are also laken. The shear strength parameters c' and 'P' obtained Crom the oonsolidatcd·undrained tests and that from consolidated drained tests are approximately equal. Drained tests on dense Sc1nds and ovcr-consolidated clays give slightly higher values oC ~' due to extra work required during dilation (increase in VOlume), but the difference is small. and, therefore, usually neglccted. (iJ) TOUll stresses. 11lC failure envelope in terms of total stresses can be dmwn from the test resulls oC a consolidated·undrained test. [Note. In consolidated·drained lests, the total strc.sscs are also equal to the efJeaive stresses, as the pore water pressure throughout is zero]. The failure envelopes are similar in shape to that in terms of effective stresses but the values oC the strength parameters are quite different. Fig. 13.19 shows the Cailure envelopes for effective stresses and aLw for total stresses for a normally consolidated clay. The angle of shearing resistance in terms of total Stresses ('e.,) is much smaller than that for the effective stresses

(+'). In the case oC normally consolidated clays, shear strength is s" otan4l.... Fig. 13.20 shows the failure envelope for an overconsolidated clay in terms oC total stresses. The ~ngle oC Shearing resistance (41 ..,') is much smaller than the angle 41' obtained in terms of etIcctive stresses. In the case oC overoonsolidated clays. shear strength is

s .. e.... +

otan~r..

The angle of shearing resistance angle of shearing resistance.

q..... obtainw

~nv~topr

(Tolill

f
Crom the lotal stress envelopes is also known as apparent

~bnsss. Failur~

TOTAl STRESS

for

0'Irf - consolidat~d

strrss~s)

Fig. 13.20. Strength Envelope.

clay


328

Fig. 13.21.

Fig. 13.21 shows (be failure envelope in terms of total stress obtained from an unconsolidated-undrained on a normally coosolidated clay. The failure envelope is horizontal (~ .. 0), and has a cohesion intercept of c", In Ihis case, shear strength is s - cu' The failure envelope for an over-consolidated Clay is also horizontaL but (he value of e" wiU be more, depending upon the degree of overronsolidation. lest

For an unconsolidmcd-undl"Jincd test, the [ailure envelope cannot be drawn in terms of effective stresses.

In aU the tests conducted at different oonfining pressures, the effective stress remains the same. This is due to the fact that an increase in confining pressure results in an equal increase in pore water pressure for a saturated soil under undrained conditions. Thus only one Mohr circle (shown dotted) in terms of cfJectivc stresses is obtained from aU the three tests. It may be noted that the deviator stress at failure is the same for aU specimens.

13.20 EFFECT OF CONSOLlIlATION PRESSURE ON UNDRAINED STRENGm As di~ in the preceding section. the shear strength of a cohesive soil under undrained conditions depends upon the consolidation pressure. If a rcmoulded specimen of saturated Clay is first consolidated under a confining pressure. say (o()J, and then sheared under undrained oonditions. with different confining pressures, the deviator stress is independent of tile cell pressure. lhe failure envelope is horil..ontal and the cohesion intercept is C"l [Fig. 13.22 (a)].

Fig. 13.22. Effect of

Consotidmion Pressure on eM

The reader should c."lrcfully note the difference between the conventional consolidated-undrained test discussed earlier and the lest describe
329

SHEAR srRENGTIf

Fig. 13.23. CII

-

0 Plot

roc NormAlly consolidated Clay.

For over-consolidated clays, there is a disoonlinuily in the plot between c.. and the preconsolidalion pressure (0..), m; shown in Fig. 13.24.

"""""-"

CONSOLIDATED

0

at a pressure equal to

CLAY

Fig. 13.24. COt -0 Plot for Ovcr-oonsolidaloo ClAY

13.21. RElATIONSJIIP nEJWEEN UNDRAINED SnEAR STRENGlll AND EFFECflVE OVERBURDEN ]·RE..~URE A relationship can be obtained between the undrained s strength (c .. ) and the effective overburden pressure oJ for II normally consolidated natural deposit. Fig. 13.25 shows the Mohr circle in tenos of total stresses obtained from a conventional CU test conducted at a confining pressure of oJ on a nonnaUy consolidated Clay. By geometry,

OJ + c..

-c-.- -

1 sin 4'....

1 _ 1 -I-sin, .... -OJ _ _ -c.. +.... sin to...

Fig. 13.25. Re.IMionship between CII,O) Hnd

~" ,

1

0<

OJ

I-sin ....

The ratio c.. /OJ is a constant for a given Clay. Skempton suggested that a similar constant ratio exists betwccn the undrained shear strength of normally oonsolidated natural deposits and the effective overburden pressure. It has been established that the ratio (c.. / 0) is oonstant provided the plasticity index

elj

0·3

5f 0·2 0·'

-1t=O.1I+0.0031PI

e:-:,,"""'..::--=,,,~..:-:::,,-',::::oo:-;,,""o~"o

0·0o~"~ "c-+.",--..

Fig. 13.26. Plot Iktw~n CJo3 Rnd PI.


330

(PI) of the soil remains constant (Fig. 13.26). An approximate value of the undrained shear strength of a normally consolidated deposit can be obtained from Fig. 13.26, if the plasticity index has been determined. The relationship is expressed as (Skempton, 1957).

~ _ 0.11 + 0.0037 PI a where c.. = undrained cohesion intercept, (j = effective over-burden pressure PI = plasticity index (%) The value of the ralio (c .. / 0) determined in a consolidated-undrained test on undisturbed samples is generally greater than actual value because of anisotropiC consolidation in tbe field. The actual value is best determined by in-situ shear vane test, as expLained later.

13.22. UNCONFINED COMPRESSION TEST The unconfined compression lest is a special form of a triaxial lest in which the confining pressure is zero. 'Tbe test can be conducted only on clayey soils which can sland without confinement. ··The test is generally performed on intact (noo- fISSured), saturated clay specimens. Although the test can be conducted in a triaxial test apparatus as a U~ test, it is more convenient to perform it in an unoonfined compression testing machine. There are two types of machines. as described below. (1) Machine with a spring. Fig. 13.27 shows the unconfined oompression testing machine in which a loaded spring is used. It consists of two metal cones which are fixed on horizontal loading plates B and C supported on the vertical posts D. The upper loading plate B is fixed in position, whereas the lower plate C can slide on the vertical posts The soil specimen is placed between tbe two metal cones. is turned, the plate A is lined When the handle upward. As the plate A is attached to the plate C, the Latter plate is also lifted. When the handle is turned slowly, at a speed of about half a turn per second, a oompressive force acts on the specimen. Eventually, the specimen fails in shear. 1be compressive load is proportional to the extension of the spring. Spring _-++_~ The strain in the specimen is indicated on a chart 0 fixed to the machine. As the lower plate C moves Fix~ plate upward, the pen attached to this plate swings sideways. Sa I Pivot The lateral movement of the pen (in arc) is proportional Mewing p~te~ to the stmin in the specimen. ~rr==:J::JIT:~~ The chart plate is attached to the yoke Y. As the r yoke moves upward when the handle is rotated, the chart plate moves upward. The pivot of the ann of the pen also D)oves upward with the lower plate. 'The vertical Fig. 13.27. U~nfined ~pression Testillg movement of the pen relative to the chart. is equal to the MRChllle (Spnllg Type). extension of the spring and bence the compressive force. Thus the chart gives a plot between the defonnatioo and the oompressive force. Springs of different stiffncsscs can be used depending upon the expected compressive strength of the specimen. (2) Machine with a Proving Ring. In this type of the unoonfined compression testing machine, a proving ring is used to measure the compressive force (Fig. 13.28). There nre two plntes, having cone seatings for the specimen. The specimen is placed on the bottom plate so that it makes contact with the upper plate. The dial gau~e and proving ring are set to zero. The compressive load is applied to the specimen by turning the handle. As the handle is turned, the upper

SHEAR SfRENGTH

331

plate moves downward and causes compi'CSSion. (In some machines, the upper plate is fixed and the compressive lood is opplied by raising the lower plate). lhe handle is turned gradually so as to produce an axial strain of 1(2% to 2% per minute. The shearing is continued till the specimen fails or 1i11 20% of the axial strnin. occurs, whichever is earlier. The compressive for~ is determined from the proving ring rcuding, and the axial strain is found from the dial gauge reading. Presentation of Results. In an unconfined rompression test, the minor principal stress (03) is zero. The major principal stress (0\) is equal to the deviator stress, and is found from Eq. 13.21. 0\ -

Proving ring

.

~~Ig~

PIA

Moving ptate

where P = axial load, A = area of cross-section. and

Conical seatings The axial stress al which the specimen fails is known as the unconfined oompressive strength (q~). The stress- strain curve can be ploued between the axial Stress and the axial strain al different stages before failure. While calculating the axial stress, the area of cr(liS-section Fig. 1328. Unconfined Compression Tesling of the specimen at that axial stmin should be used. The Mllchine (Proving Ring Type). corrected area can be obtained from Eq. 13.20 as . A - AoI(I-;) The Mohr circle am be drawn for stress conditions at failure. As the minor principal stress is zero, the Mohr circle pas.'iCS through the origin (Fig. 13.29). The failure envelope is horizontal (+.. - 0). The cohesion intercept is equal to the radius of the circle, i.e. ,to

°1

q..

. .. (13.25)

s-C"-T-T nit ~m

'" Fig. 13.29. Mohr Circle (or Unconfined Compression Test.

Merits and Demerits of the test Merits (1) The test is convenient, simple and quick. (2) It is ideally suited for measuring the unconsolidated-undrained shear strength of intact, saturated clays. (3) The sensitivity of the soil may be easily determined by conducting the test on an undisturbed sample and then on the remouldcd sample.


332

Demerits (1) The test cannot be conducted on fissured clays. (2) The lest may be misleading for soils for which the angle of shearing resistance is not zero. F()( such soils, the shear strength is not equal to half the compressive SlrCngth. (See Chapler 30, Sect. 30.17 for the laboratory experiment). 13.23. VANE SIIEAR TEST 'llie undrained shear strength of soft clays can be dctcnnined in a labnratory by a vane shear test. lbe test can also be conducted in the field on the soil at the ooHom of a bore hole. The field test can be performed even without drilling a bore hole by direa. penetration of the vane from the ground surface if il is provided with a strong shoe to protect ;1. Th" apparatus coosists of a vertical steel rod having four thin stainless stccl blades (vanes) fixed at its bottom end. IS : 2720--XXX-1980 recommends thai the height H of the vane should be equal to twice the overall diameter D. lhe diameter and the length of the rod are recommended as 2.5 mm and 60 mm respectively. Fig. 13.30 (a) shows a "mm vane shear test appal"'dlus. For conducting the test in the laboratory, a specimen of the size 38 mm diameter and 75 mm height is taken in a cootaincr which is fixed securely to the base. TIle vane is gradu.'llly lowered into the specimen till the top of the vane is at a depth of to to 20 mm below the lOp of the specimen. 'Ibe readings of the strain indicator and torque indicator are taken. Torque is applied gradually 10 the upper end of the rod at the rate of about 6 0 per minute (i.e. 0.10 per second). The torque ading on the specimen is indicated by a pointer fixed to the spring. The torque is ElMTON continued till the soil fails in shear. The shear strength of the soil is determined using the formula derived below. Derivation of Formula. In the deviation of the fonnula. it is assumed 05mm that the shear strength (5) of the soil is oonstant on the cylindrical sheared PLA;-U-surface and at the top and bottom faces of the sheared cylinder. The torque clpplied (7) musl be equal 10 the sum of the resisting torque at the sides (TI ) and that at the top and bottom (Tv. Thus,

T

f

r

-----L-l f

'.>l

... (a) The resisting torque on the sides is equal to the resisting force developed on the cylindrical surface multiplied by the radial distance. Thus,

TI - (snDH) x DI2

... (b)

(OJ

Fig. 13.30. Vane Shear Te5I.

The resisting torque T2 due to the resisting forces at the top and bottom of the sheared cylinder can be determined by the integration of the torque developed on a circular ring of radius r and width dr IFig. 13.30 (b)J. Thus.

T, - 2

on

fo ['Cw)drJ' 0'

T2 - ru ~ From Eqs. (a), (b) and (c),

T - . , [0' HI2 +

4",

[If a "3 Jo

... (c)

0'16J

SHEAR STRENOTII

333

or

, -

T .(0' H/2 + d16)

=

... (13.27)

=

For ex.1mple, if D 1.2 cm, and /J = 2.4 an, S 0.158 T where T is in N-cm and S in N/an 2. Eq. 13.27 is modified if the top of the vane is above the soil surface and the depth of the vane inside the sample is N I • In such a case. ... (13.28) The shear strength of the soil under undrained conditions is equal to the apparent cohesion cu' The vane shear test can be used to detennine the sensitivity of the soiL After the initial test, the vane is rotated rapidly through several revolutions such that the soil becomes remoolded. The test is repealed on the remooided soils and the Shear strength in remoulde
(J)~:~~~~:

Merits and Demerits of Shear Vane Test Merits. (1) The test is simple and quick. (2) It is ideally suited for the determination of the in-situ undrained shear strength of non-fissured, fully saturated clay. (3) The test can be conveniently used to detennine the sensitivity of the soil. Demerits. (1) The test cannot be conducted on the fissured clay o r the clay containing sand or silt laminations. (2) The test does nol give accurate results when the failure envelope is nol horizontal. 13.24. PORE PRESSURE PARAMETERS A knowledge of the pore water pressure is essentiaJ for the determination of effective stresses from the total stresses. The pore water pressure is usually measured in the field by installing piezometers. However, in some cases, it becomes difficult and impractical to install the piezometers and measure the pore water pressure directly in the field. For such cases, a theoretical method for the determination of the pore water pressure is useful Skcmpton gave the pore pressure parameters which express the response of pore pressure due to changes in tbe total stresses under undrained conditions. These parameters are used to predict pore water pressure in the field under similar conditions. The expressions for pore pressure parametelS are derived separately for isotropic consolidation, for deviatoric stress and for the combined effect.

FORE

PRE~E ~

(.)

u"

~

PORE

PRESSURE

PORE

PREi~RE

u".... c ul

!NCREASE =..o.u,

Fig. 1331. Pore Pressure Under Isotropic ConsolidAtion.


334

( I) Ilore Pressure Under Isotropic Consolidations. Let us consider a small clement of a saturated soil mass which is in equilibrium Wldet three principal stresses 01' Oz and 0) [Fig. 13.3. (a»). leI the initial pore pressure be 110- When the element is subjected 10 an equal increase 60) in all the three directions, lei the increase in the (Xlre pressure be Au) [Fig. 13.31 (b)]. Consequently, the inaease in the effective stress in each direction would be equal 10 (.1.0)-6u). Let the initial volume be Vo and the porosity be n. As the soil is saturated, the volume of water would be equal to nVo. If C~ is the roefficicnl of compressibility of the soil skeleton, the reduction in the volume the soil mass due to an increase in the effective stress (110) - Au) is given by

or

6Vo .. C.Vo(.6.a)-A u) ... (n) If C" is the coefficient of volume compressibility of pore fluid under isotropic condiLiollS. the reduction in the volume of voids is given by !J,. V" - C" (n Vol 611) . • • (b) Assuming thnt the solid particles are incompressible, the reduction in the volume of the soil mass is equal to the reduction in the volume of voids. Therefore, from Eqs. (0) and (b),

C.

VO

(6

0) -

Au) - C" (n Vol 6

Il)

6u)[nC" + CAVo - C. Vo 6o) or

C,Ao) 6u) - nC" + C,

... (13.30)

6u) .. B60)

where B is known as the pore pressure parameter for the isotmpic consolidation, and is given by B - 1 + n tc,.IC,)

... (13.31)

In a fully saturated soil, the compressibility of the pore water (Cl') is negligible oompared with the

8

compressibility of the soil milSS (C.). 'llierefore, the ratio (Cl' IV,) tends to zero and the coefficient B becomes equal to unity. In other words, the change in pore water pressure is equal to the change in the total stress, i.e.

o· 8

In a partially saturated soil, the compressibility of the air in the voids is high. The mtio (Cl' I V. ) has a value greater than unity, and, therefore, the pore pressure coefficient B has a value of less than unity. The pore pressure coefficient 8 increases with the degree of saturation, but the variation is not linear (Fig. 13.32). There is a steep rise in the value of B al S = 80% or so.

o· 4

(

/ I I

6

o· 2 0·0

o

20

-V

./ 100

(2) Pore Pressure Under Deviator Stress. Let us consider the element of a saturated soil which is in equilibrium under three principal stresses 01. °2 and 0) (Fig. 13.33). When the element is subjected to a deviator FIg. 13.32 Variation ot B with S. stress Od(- 01- 0), I ~ the increase in pore water pressure be 6 11d. 1be changes in the effective stresses in the three directions are given by

60\ -

60d - 6ud

6~ - -6Ud

...(0) ...(b)

~IfEAR

SfRENGTI1

(b)

(.)

Fig. 13.33. Port: Pressure Under Deviator Stress.

and lla3" - 6ud ... (c) In an elastic material. the volumetric strain 6 VolVo is equal to the sum of the linear strains in three directions., and is given by d Vo "'"'YO " 6£1 + 6£2 + 6£3

°t _

where £1 '" strain in the direction-I .. 6. v (6 °2 + 6(3 ) E E E 0 E2 = strain in thedirection-2 .. 6. 2 -v (6.0 1 + /l03)

E

and

E

E

E3 '" strain in the direction-3 .. /l 03 _ V (/l at +

E

Therefore,

6

o v: ..

1

~ 2v

E

.6~) E

(6. oJ + 602 + 6.(3)

6Vo .. ~ (601 + A~ + 60(3) Vo E 3

...(d)

Because the soil is not a purely elastic material. Eq. (d) for soils is modified as d VO

Va"

C,

(60 J + 6.02 + 60(3) 3

... (e)

where C, is the coefficient of volume compressibility of the soiL Substituting the values of 6.'01, A~ and 6003 from Eqs. (0), (b) and (c) in Eq. (e), 6 Vo.. (6 VO C,

0d -

A Ud - /l 3

ud - /l Ud) I

d Vo _ S.·(dOd _ 3du,,) Vo 3

... if)

As in the case of isotropic consolidation, the reduction in the volume of OUi.d in voids is given by d~-~0~d~

As the change in the volume of the soil mass is equal to the reduction in the volume of voids,

~

(6 ad - 36i4d)Vo" C,,(nVo)Aud AUd(nC~ + C,)Vo" C,Vo(6a,r'3)

.~

,,.


6 ". '

(, +

ln~. )

x (

t) .

(60.)

Because a soil is not perfccUy elastic, the constant 1/3 is replaced by A in the above expression. Thus

ne.)

A

6.11<1" - ( x (L\Od) 1+-

e,

Using Eq. 1331, the above expression is written as Au" .. ABl!.oJ Eq. 13.32 can also be wrinen as lJ.IIJ"

... (1332)

A !J.Od

A .. A

... (1333)

x B

... (13.34) La A, as B is unity. The value of the pore pressure ~rameler A can be determined experimentally in a triaxial test. A ~turatcd soil specimen is set up under a certain confining pressure and the pore water pressure is measured. A deviator stress, lloJ, is then applied to the specimen and the change in pore water pressure tJu d is measured under undrained conditions. lJle value of the coefficient A, can be determined using Eq. 13.33. The coefficient A is also equal to A, as B is equal to unity. '·0 Although the value of the parameter A can be delennioed at nny slage of the loading, its value at failure (A f ) is of main interesl. The parameter Af is required for the o. determination of the shear strength which depends upon the effective stress al failure. The value of Af depends upon such factors as the degree of saturat ion and over- At 0 ----- - -consolidation ratio. For a heavily overconsolidaled clay, there is a L~;--...L--:----7;c--':;tendency for Ihe soil to increase in volume (dilate) when -O·5~ to -..... 8 16 32 where

For II fully salurntcd soils,

A is aLo;o equal

t

:~odt~~a:~l ~~!a~~lii:·~~ :~I:m':,"a~e~:~~

OVER - CONSOLIDATION RATIO pore pressure develops. This gives a negative value of Fig. 13.34. AI. Fig. 13.34 shows the variation of AI with over-oonsolidation ratio [or a particular soil. Hr overconsolidation ratio of greater than 4.0, AI is negative. For a highly sensitive clay, the deviator stress may cause collapse of the soil structure, and the value At may be even greater than unity. Table 13.1 gi·ves the values of AI for different soils. Table 13.1. Values of AI S.No.

7'y~ofsoil

4. 5.

Sensitive day Normally consolidated clay Lightly overconsolidcllecd clay Highly overconsolidated clay Very looocfincsnnd

6.

Medium fine sand

7.

Dense fine sand

1.

2.

,.

A, 1.5-25 0.5- 1.3 0..3-0.7 -0.5-(1.0 2.0-3.0 0.0--1.00 -0.3---0.0

. . ..

SHEAR SfRENGTII

337

(3) Pore pressure due to both isotropic Consolidation and Deviator stress. In 8 conventional triaxial tcst, the specimen is fust isotropically ronsolidnted under a pressure of 603' and then it is sheared under a deviator st~ of (601 - dO)). The pore pressure developed can be obtained by combining Eqs. 13.30 and 1332. as dU- dU) + dUd

du .. BdO) +AB (dOl-dO)) dU" B[6.o) + A (dOl-dO)] dU" B6.03 +

A (dOl-d03)

... (13.35) ... (13.36)

A_AxB

where

For a fully Sllturnted soil. as 8 - 1.0, ... (13.37) du .. do) +A (A01-dO)) The above equations can be used for determination of the pore water pressure in the field when the values of the parameters A and B have been determined. 13.25: MOIIR.COULOlim FAILURE CRn'EIUON lbe Mohr-Coulomb equation was discussed in Sect. 13.6. The Mohr- Coulomb failure criterion can be written in terms of principal stresses. Fig. 1335 shows the failure envelope of a soil. It is tangent to the

T

A~1:~~~L---
Mohr-circle at failure. The slress on Jhe failure plane al failure is indicated by the poim 8 on the Mohr circle. The failure plane is indicated by the line BE. The failure plane makes an angle 6, with the major principal plane. The failure envelope has an intercept c on the .-axis and it cuts the o-axis (when exlended back) at point A. From the triangle ABC,

sin

t -

~

sin 'P - __-,1_/2-'.(,-U(",:=-,U-,:"-)U=-") crotcp + - - 2 (01 - 0))" 2ccosq>

In terms

+

(0 1

+ O:3)Sinq>

or effective stresses, the above equation becomes (0'1 - a) .. 2c'oosq>' + (01 + (3)Sinq>'

where c' and .' are shear strength parameters in terms of effective stresses. Eq. 13.39 may be written in a slightly modified fann as

a; - al

sinq> ..

a3

sinq> + 0) + 2c' oosq>'

... [13.38(0)]

... [13.38(b)] ... (13.39)

SOIL MF£HANICS AND FOUNDATION ENGINEERING

338

_ _ ~ 2c'cos
.. (13.40)

Tan2 (45" + 4112) is caUed flow ratio (N.). 1bus For cohesionlcss soil,

c' ...

o.

-°1

-(~) 0) 1 _ sin $'

. .. (13.41)

Therefore, -

... (13.42)

When the stresses in a soil mass satisfy the above failure criterion, the failure is imminent. The soil mass is said to have reached a state of plastic equilibrium. The failure criterion is extensively used for the detcnnination of earth pressure on retaining walls and in many other applications. The angle S, can be determined from the angle of shearing resistance ¢I'. As the failure depends upon the effective stresses and not thc total stresses, thc angle of shearing resistance affecting the orientation of the failure plane is " and nol the apparent angle ~. FrCllTl the triangle ABC. 2S,_ 90 + 41' 8f ... (45" + 41'/2)

or

... (13.43)

13.26. MODIFIED FAILURE ENVELOl'E Mohr's failure envelope is gcneraUy obtained by dr-dwing a common tangent \0 Mohr circles at failure [Fig. 13.36 (a»). This method of obtaining the failure envelope is nOI convenient, as il becomes difficult to draw the required tangent touching aU the circles. A modified failure envelope ~ explained below is more convenient. A modified failure envelope is a plot between p and q values at failure [Fig. 13.36 (b)], where

•

Fig. 13.36. Conventional lind Modified Flliluree Envelopes.

SHEAR SfRENG1lI

339

... [13.44 (a)J 01 -

03

... [13.44 (b)J

q--2-

As discussed in Sect. t 3.6, the coordinates of the lop point of the Mohr circle corresponding 10 the maximum shear stress (0\ ... oV/ 2 and (0) - (3)/2 and are. therefore, equal to p and q. In Fig. 13.36 (a), the points 1, 2 and 3 give the maximum shear stresses reached in the three tests at the time of failure. These points are transferred 10 p-q plot in Fig. 3.36 (b), and a line is drawn through these points. The line makes an angle a' with the p-axis and has an intercept a' on the q-mds. 1'his line is known as the modi[led failure envelope, and has the following equation in tcnns of effective stresses. q _plana' ... a'

are

Cil;O)) _CJI; (

3

lana' ... a'

... (13.45)

(0\ - (3) - (01 ... (3) lana' + 20'

... (13.46)

)

A rci1ltionship between the shear strength pammcters c' and 41' and the p.:1ramelcrs of the modified envelope oj and a' (:Un be obtained by comparing Eqs. 13.39 and 13.46.

sin 'P' - lana' c' coscp' _ a' or

and

The values of parnmclcrs c' aoo eqll3ti~

q.' are

c' _ a'

... (13.47) ... (13.48)

secq,'

obtained from the intercept

a' and the slope a ', using above

_

It may be noted thai points 1, 2,3, elc. represent the maximum she..'lr stresses which arc greater than the shear stresses on the failure plane. 111e Mohr envelope gives the shear stresses on the failure planes, which are represented by the points of tangency; wherea-; the modified failure envelope joins the points of maximum shear stresses at the time of failure. The main advantage of the modified failure envelope is thaI the stress conditions al failure are rep~ nted by one poinl instead of a Mohr ciffle. As the averaging of scattered points is easier tban drawing a common tangent to a number of eircl~ it is more ~ convenient than the Mohr-Coulomb failure envelOpe. / Stress poth Despite this, Mohr-Coulomb failure envelope is used more commonly than the modified failure envelope. A Mohr envelope is more popular because it not only gives the shear strength parameters, but also' the principal stresses at the time of failure and directions of the failure plane. It \\I gives a beller insight into the whole phenomenon. The modified failure envelope is also known as the Kf (0) -Une. It can be drawn either in terms of effective stresses or in terms of total stresses. q

13.27. STRESS PATI1 A stress path is a OJNe which shows the changes in stresses as the load acting on the soil specimen changes. Lambe's stress path is a oommonly used stress path. It is a line drawn through the points representing the maximum shear stresses acting on the specimen ao; the load is changed. In other words, it is the locus of points of maximum shear stresses experienced by a specimen ao; the \ood changes take placi.

Stres~

(0)

Fig. 13.37. Stress Path.

path

SOIL MECIIANICS AND FOU NDA'10N ENGINEERING

340

Fig. 13.37 (a) shows the Mohr circle I. II. III which represent the stress conditions of a specimen as the load on the specimen is increased such that the minor principal stress remains constant. The line joining points I, 2, 3 at the tops of the Mohr circles is the stress palh. It is more convenient 10 draw the stress path on a p-q plot. as shown in FIg. 13.37 (b). There is no nOed to draw the complete stress circles. Only the stress points 1.2, 3, etc. corresponding to maximum Shear stresses are plotted. The direction of arrow on the stress path indicates the direction of the stress changes. There are basically three types of stress paths.

(1) EfTecti...e Stress Pilth (ESP). It is plotted between effective SlrCSSCS (OJ +0)12 and (01 ~(3)/2 (2) Total Stress Path n'Sp). It is plotted between total stresses (oJ + 03)/2 and (oJ - 0))12. (3) Total Sh"tSS minus static pore pressure path rfSS)'). It is a plot between [(oJ +(3)12 - u~J and (OJ - (3)12, where 148 is the sUltie pore waler pressure. It may be noted thaI the vertical axis for aU the stress paths is the S
03

0\ - 03

-2---2-- q

If the principal planes arc horizontal Clnd vertical, the vertical stress (0,,) and horizontal, stress (a,,) are the principal stresscs. In that cilSe, the above three basic stress pathS can be drawn between (1)

(2) 0,. ;

(3)

°h ,

respectively.

In the norm al triaxial test, as the static pore waler pressure (u8 ) is zero, the stress paths (2) and (3) coincide. However, if a back pressure is used in testing, the two stress pathS wiU be different. In fK!kl. the static pore water pressure depends upon the paoition of the waler table with rcspea \0 the point.

"'---!:.---;t-P,P I--us----t '0)

Fig. 1338. Different

Slres~

Pl1tlis.

Fig. 13.38 (0) shows the stress paths for a drained lest. The point A corresponds to the stress conditioo when only the confining pressure acts on the specimen. ( A<; 01 _ 0l,P - 0). The point F represents the failure. Fig. 13.38 (b) shows the stress paths for a consolidated undrained test on a normaUy consolidated clay and Fig. 13.3S (c), on an over-consolidated clay. Fig. 13.39 shows the effective stress pathS A-I for conventional triaxial compression test (0/0 = constant) i.t!. 6 o~ = positive and 60" = O. The typical example is the case of a footing subjected to tbe vertical lo..1c1s. The stress path A - 2 is for the unloading case, when 60/0 = negative and 60 .. = O. The typical example is the case Qf active earth pressure on the retaining waUs. TIle stress path A - 4 is for the loading case in extension test when 60" = 0 and 60h = positive. It represents the pa
ING

SHEAR SfRP..NG11I

341

'he ling

">th

Yertij:ol compression

1"'::::---7f:.----------- P.~ and

Yerticol eJlension lo;rv<~l

-Q

If'~flf!"

"h

Rr, -~

Fig. 1339. Loading aoo Unloading Stress Paths.

The stress path A - 3 is for the unloading case in the extension test, when 60" II represents the case of the foundation excavation.

(3) the

= 0 and 6o~ = negative.

The follow-ing points about the stress path are worth noUng. (1) A n..egative value of q indicates Ihat the horizontal stress is greater than the vertical stress; and a positive valuc, that the vertical stress is greater than the horizontal. (2) TSP to the right of ESP indicates a positive pore water pressure and TS? to the left of FSP, a negative pore water pressure. (3) TSSP to tile right of ESP indicates a positive cxcx:ss pore pressure and TSSP to the left of ESP, a negative excess pressure. When both coincide, excess pore pressure is zero. (4) Lines from any location on a SlrtSS path drawn at 45" 10 the axis CUI the horizont.'li axis at a stress equal to the horizontal stress. Note. The slreSS p;!lh can also be drawn using the space diagonal (Rendulic stress path). which is ourside the scope of the text.

13.28. SUEAR STRENGTD OF PARTIALLY SATURATED SOn.s The eITedive stress in a partially s.'1turnted soil is determined using Bishop's equation (see chapter 10).

a-

tion Ibe clay

nplc

e. It

0- u" - X(u ... - /I.,) ... (13.45) The parameter X depends upon the degree of saturmion. As it is difficult to determine accurately Ihe effective stresses in partiaUy saturated soils. it is common practice to draw the failure envelope in terms of total stresses. Unoonsolidated-undraincd triaxial tests are conduaed on a number of samples with a given degree of initial saturation (S). 1l1c Mohr circles in terms l' of IOtal stresses are drawn at failure and the FAILURE ENVELOPE failure envelope obtained (fig. 13.40). As an increase in ooilfining pressure causes an increase in the solubility of air in voids. the degree of saluration increases. and the failure envelope (ends to the horizontal. For design purposes, the curved envelope is approximated as a straight line.

Fig. 13.40. Sm:ngth Envelope.


342

s_c...

where

+ot.·m~ ....

... (13.50)

$ = shear strength. Cus = oohcsion intercept. IP.. = anglc of shearing resistance.

As it is difficult to draw the correct failure envelope. the values obtained are necessarily approximate. As the initial degree of So1turntion is -increased, the shear stress to cause failure is decrcascd (Fig. 13.41). Hx a fully saiW1lted sou. the total stress failure envelope is hori7..ontaI. as already disrussed for fully saturated soils. ollIe tests arc conducted simulating the fl.c ld

conditions. However, an actual soil deposit in field may

r

5=80"1,

5=90'/,

;::::::----

not remain unsaturated if it has an accrss to water. For such a case, the soil sample is s
13.29. HVORSLEV'S STRENGTfl THEORY Acrording to Hvorslev's hypothcsis, the shear strength of remouldcd, saturated clays is given by

s • c~ +

a tan,~

... (1351)

where c~ = true cohesion, ,~ true angle of intemal = e[feaive stress on the failure plane at friction, failure. The constants c~ and .p~ arc known as the

a

I1vorslev shear strength parameters. The true cohesion c~ depends upon the water content (and hence void ratio) of the soil at failure, whereas the true angle of internal friction ,~ is approximately constant for a soil. It depends upon the plasticity index of the soil. The angle ,~ decreases with an increasc of plasticity index of the soil. At a givcn water content, the true cohesion c~ is diredly proportional to the equivalent consolidation pressure a~. and may be written as

c, _

K(j~

fig. 13.41. Fllilure Envelopes for DilTcrent Degree oCS.1turntion.

1)

SHEAR SfRENonl

343

where K is known lIS Hvorslcv coefficient of cohesion. Accordingly, thc shear st·cngth can be expressed

o

s _ Ko.,+ lan q>., ... (1352) Dishop and Henkel (1962) suggested a method for determination of c~ and '., from a series of coosolidatcd-undraiocd triaxial tests on normally coosolidated and over-ronsoHdated specimens. 1be two failure cnvelopes ar:e obtained as usual and are shown in Fig. 13.42 (b). The watcr oontent at failure for the two types of specimens is plotted against the maximum principal stress as shown in Fig. 13.42 (a). For determination of the true failure envelope, any circle (say left circle I) for the over-consolidated clay in Fig. 13.42 (b) is chosen. The point corresponding to its maximum stress (0,), is projected upward to the curve in Fig. 13.42 (D) to get the point 1 on thc ruNe for ovcr-coru;olidated clay. The point 1 is projected horizontally across at constant water content to obtain point 2 on the ruNe for the normally coosolidatcd clay. The point 2 is projected downward to obtain the point (01)/1 in Fig. 13.42 (c). Through this point, a Mohr circle II is drawn to IOuch the failure envelope for normally' consolidated Clay. In Fig. 13.42 (c), the left circle I is the same as the circle I in Fig. 13.42 (b). 1bc common tangent to the circle I and n in Fig. 13.42 (c) is the true failure envelope. The pammeters c~ and ,~ are obtained from this envelOpe. The true failure envelope has been obtained using the o:mccpt that two samples can exist at the same water content, one as normally consolidated and one as over-consolidaled. As the watcr contents at points 1 and 2 are equal. the true cohesion is the same and the difference between the shear strength of the two s:lrnples is due to the internal friction only. The fundamental properties o[ soiLo;; can be studied in termS of Hvorslev shear strength parameter. However, the theory is generally used only for research purposes. For practical use in engineering problems, the Mohr-Coulomb theory is commonly used.

a, - w,

13.30. LIQUEFAcnON OF SANDS As dlscussed earlier. the shear strength of sandy soils is given by the Mohr-Coulomb equation (Eq. 13.13). taking the cohesion intercept as zero. Thus s .. 0 tan,' ... (13.53) If the sand deposit is at a depth of z below the ground and the water table is at the ground surface. the effective stress is given by (sec Olapter 10).

0- "t_ Z - "t",z- y' Z Therefore. s _ y' z lan " If the sand deposit is shaken due to an earth-quake or any other oscillatory load, extra pore water pressure (u') develops, and the strength equation becomes s _ (y'z-u') tan It can also be expressed in the term of extra pore pressure head II, where u' .. y". h. Thus

+'

s.

(y'z-y-")"'n~'

.. . (13.54)

As indicated by Eq. 13.54, the shear strength of sand decreases as the pore water increases. Ultimately, a stage is reached when the soil loses all its strength. In which case,

y'z-y",h .. 0

or

~ z

.i. y.

Expressing h/z as critical gradient.

.

(G-lh.

1

Ie, -

--r-+e . 1:

.

G - 1

le,"~

... (13.55)

The phenomenon when the sand loscs its shear strength due to oscillatory motion is known as

SOIL MECI-IANICS AND FOUNDATION ENGINEERING

344

liquefaction of sand. The structures resting on such soils sink. In the C&'>e of partial liquefaction. the structure may undergo excessive settlement and the complete failure may not occur. The soils most susceptible to liquefaction arc the saturated, fine aod medium sands of uniform particle size. When such deposits have a void ratio greater than the critical void ratio and are subjected to a sudden shearing stresses, these decrease in volume and the jXl!'c pressure u' increases. The soil momentarily liquefies' aDd bchaves as a dense fluid. Extreme care shall be taken while constructing structures on such soils. If the deposits are compacted to a void ratio smaller than the critical void ratio, the chances of liquefaction are reduced.

13.31. SHEAR CHARACTERISTICS OF COIIESIONLESS SOII.S The shear charnclcrislics of cohcsionlcss soils ~ be summarized as given below. The shear strength of cohcsionless soils. such as sand.. and 000- plastic SillS, is mainly due to friction between particles. In dense s.1nds, interlocking between particles .also contributes significantly to the strength. lbe stress-stroin curve for dense sands e::hibits a relati ....ely high initial tangent modulus. The stress reaches a maximum value at its peak at 8 comparatively low strain and then decreases rnpidly with an incrct1sing strain and eventually becomes more or lcs.~ constant, as discussed earlier. '[he stress-strain curve for loose sands exhibits a relati ....ely low initial tangent modulus. At large stmins, the stress becomes more or lcss constant. The dense sand shows initially a volume decre..1SC in a drained test, but as the strain increrlScs, the ....olume starts increasing. The loose sand shows a volume decrc..1sc throughout. and

1~It~ea:e~~~~~:~~~i=~:~;lg:.,

;;~~:O~:,I~~3~~/~~~~~~r:7~e~:~I~~~~~ specimen shows

~

clear fadure plane and the failure

~;en :ilt~;eb::~:~~ur:;I:~~'~!)~1n

'"3

:~.:~..:';' ":::~ .::..

! • ' .... '~"

':~.

•

\~ :'.~: · ·:~:i ' ; ' ' .•. ::;

~ /~'.\\'>'~'.':

Cj

is be :....:.:/\\:.: Cj drawn eithcr for lhe peak stresses or for the 9 '"i ultimate stresses. 111e value of thc angle of shearing rcsistance (If) for lhe ft1i1ure en....e lope for peak stresses is considerably greater than Ihat for (a) LOOSE SAND (b) DENSE SAND the ultimate stresses. In lhe case of loose 5.1nds, as Fig. 13.43. Types of Failure. the peak stress and the ultimme stress are identica l, there is only one failure envelQpe. The. angle of shearing resistance in very loose state is approximmely t:qual to lhe angle of repcse. The angle of repose is the angle at which a heap of dry 5.1nd stands without any support. It has been established tbal air-dry sand gi ....es approximately lhe same value of " as the. saturated s..1nd. As it is easier to perform tests on dry sand, tCSts can be performed on dry sand instead of s..1turated sand. I[ the failure envelope is slightly non-linear, a straight line may be drawn for the given pressure range and the angle of shearing resistance is taken as the slope of this line. The cohcsion intercept, if any, is usually neglected. The angle of shearing resistance of smlds in lhe field can be determined indirectly by conducting in-situ tests, such as tbe standard penetration test (SPT) as eKpl.:1ined in Chapter 17. The factors that affect the shear strenglh of oohesioolcss soils arc summarized below: (1) Shupe of particles. The shearing strength of s.1nds with angular particles ha .... ing sharp edges is greater lhan that with rounded particles, oLhcr p.1l'3meters being identical. (2) Gradation. A well-graded sand exhibits greater shear strenglh than a uniform sand. (3) Denseness. The degree of interlocking increases with an increase in density. ConscquenUy, lhe greater the denseness, the greater lhe strength. 'The value of .' is related 10 the relative density (D,) as " - 26'" + 0.2 D,. However, the ultimate ....alue of " is not affected by denseness.

~EAR

"5

STRENGTII

(4) Confining pressure. The shear strength ina-cases with an increase in oonfining pressure. However, for the range of pressures in the common field problems, the effcct of confining pressure on the angle of shearing resistance is nOI Significant. (5) Deviator stress. 'Ole angle 4" decreases under very high stresses. As the maximum deviator stress is 1 increased from 500 10 5CXX) kN/m , the value of " decreases by about 10%. This is due to the crushing of particles. (6) Intennediate principal stress. The intermediate principal stress affects the Shear strength to a small cxtent. The friction angle for dense sands in the plane strain case is about 2" to 4" greater tban that obtained from a standard triaxial test. However, for loose sand, there is practically no difference in the two valucs. (1) ~dJng. The angle of shenring resistance of sand is independent of the rdte of loading. 1be increase in the value of from the slowest \0 the fa<>test possible rate of loading is only about 1 to 2%.

+'

The angle of shearing resistance in loading is approximately equal 'to that in unloading. (8) Vibmtions and Repeated Iouding. Repe.:lIed loading can cause significant changes. A stress much smaller than the stalic failure stress if repeated a large number of times can cause a very large strain and hence the failure. (9) Type of mlncmls. If Ihe S<1nd contains mica, il will have a large void ratio and a lower value oC 4". However, it makes no difference whether the sand is composed of quartz or feldspar minerals. (10) Capillary moisture. lhe sand may have apparent cohesion due to capillary moisture. The apparent cohesion is destroyed as soon as the sand becomes saturated. A person can ca<>ily walk on damp s.1nd near the sea beach because it possesses strength duc to capillary moisture. On the same sand in sacurated conditions, it becomes difficult to walk: a<> the capillary action is destroyed. Table 13.2 gives the representative values oC,' for different lypes of oohesionless soils. 'Thble 13.2. Representative Values of " for Sands and Silts S. No, l.

2.>

,.

.. 5.

&>1 Sand, round grains. uniform Sand, !lngulnr, well.gmdcd Sandy grlIVC\S Silly sand Inorganic silt

.'

zr to 34° 33° 1045° 35" 10 50° 27" 10 34° zr 1035"

Note. Smaller values are for loose oonditions and larger values are for dense conditions.

13.32, SHEAR CIIARAcrERlSTlCS OF COHESIVE SOILS 'llie shear characteristics of cohesive soils are summarized below: The shear characteristics of a cohesive soil depend upon whether a soil is normally consolidated or over·consolidaled. The stress· strain curve of an ovcr-consolidated clay is similar to thai of a dense sand and that of a normally consolidaled clay is identical to thai of a loose sand. However, the strain required \0 reach peak. stress are generally greater in clay than in sand. lbc high strength at the peak point in an over·consolidated clay is due to structural strength; whereas in the dense sand, it is mainly due to interlocking. In over·consolidated Clay, strong structural bonds develop between the particles. Loose sands lend to increasc in volume at large strains whereas normally consolidated clays show no tendency to expand after a decrease in volume. The effective stress parameters (c',4") for an ovcrconsolidated clay are determined from the failure envelope. s-c'+otanf' However, for a normally consolidated clay, the failure envelope passes through the origin and hence c' = O.

SOIL MECHANICS AND FOUNDATION ENGlNEEIlING

For ·c1ay, when tcsted in unconsolidated-underained test, the failure envelope in terms of tOl.al stresses is horizontal (f.... 0), and the shear strength is given by s .. c.. For heavily over-COll.'iOlidaled clays, negative pore pressure develops during shear, and the undrained strength is more than the drained strength. The $.. = 0 concept, therefore, leads to the results on the unsafe side. Funher, tbe continuity of heavily over-consolidated clays is commonly disrupted by a network of hair cracks and the average pressure is reduced. lbis results in a substantial decrease in its shear strength. The general behaviour of compacted (partially saturated) clays is similar to a modcrately over-consolidated clay. A total stress analysis is used for sueh soils. The factors affecting the shear strength of cohesive soils may be summarised as under: (1) Structure of clay. The clay exhibits a definite structure. Even a nonnally-consolidated clny exhibits a small peak due to structural strength. In case of o\'cr-coosolidated clays, the structural strength is predominant. (2) Clay content. lbe ultimate friction anglc .' of the oohcsivc soil depends upon the clay content. As the clay rontent increases, the angle decreases. Further, the difference between the 4" at peak and that at ultimate condition increases with an increase in clay cootent, because the clay particles do not reach a fully oriented faa>to-face alignment at peak. (3) Drainage condilions. As the oohcsive soils have low penncabiUtYJ the shellr strength will depend whether it is in drained condition or in undrained oonditions. 1be cohesive soils have very low strength just after the application of the load when undrained conditions exist. (4) Rate of stmin. In the case of normally consolidated clays, the effect of rate of strain upon the anglc of shearing resistance is relatively small. The value of " may decrease by about 10% if the stmin rate is reduced by n factor of 10. However, in some cases, the angle " is found to. increase with a deaease in rote of suain. In the ~ of over-consolidated clays, some of the shear strength is always lost when the rate of strain is decreased. (5) Intermediate principal stress. 1be valucs of and .' are affected very little by the magnitude of the inlenncdiate principal stress. (6) Repeated loading. For Clays tested at constant water rontent, the shear strength is increased due to II Jorge number of repetitions of the stress. However, if the stress intensities are too high, the Ctlmulativc deformations may result in failure. (7) Confining pressure. The shear strength of clays increases with an ina-ease in the confining pressure, provided there is enough time available for the pore water pressure to dissipate. (8) Plasticity index, The value of " decreases with an ina-ease in plasticity index of the Clay. The following relation is commonly used.

c

sin f' - 0.814-0.234log,olp (9) Stress history. The values of strength parameters depend upon the stress history. (10) Disturbance. The shear strength of disturbed sample is lcs<> than that of the undisturbed samples. Table 13.3. gives' the representative values of c.. for different types of cohesive soils for undrained CODdilions. Thhle 13.3. Representative Values of c.. for day

s.

No.

Soil

l.

Very soft day

2 3.

Sofl to medium clay

4. 5.

Stiff clay Very stiff Hard

Cohesioll c.. (kNlm1 < 12

12 -25 50 - 100 100 -200 > 200

34'

IG

Si-IEAR srRENOTIi

is

13.33. CHOICE OF TFST CONDITIONS AND SHEAR PAllAMEfERS Test oooditions and shear parameters should be chosen to represent the field conditions as closely as possible.

cd

In case of coan;c grained soils, drainage takes place quickly during the application of the load. Tests on

,fe

these soils are gcnernlJy carried out under drained condjti~ and therefore consolidated drained tests are more common. However, in some special cases when large masses of saturated, fine sands arc subjected to

lir ly

at

Iy ,d s>

Ie js

te )[

e, ,e

quick loadings due to earthquakes or othec causes, the undrained conditions are more relevant. In case of fmc-grained soils, drainage takes plucc very slowly. Immediately aOer the application of the load, undrained conditions are suitable. However, for rmal stability problems, drained tests are relevant, as the water content of such soils adjusts itself to the new coodilions in due course of time. Consolidated-undrained tests are required for the cases when the soil gets consolidated under a certain loading and then additional T load is applied; for example, in the cnse of earth dams, ~o wherein the soil gets consolidated under self weight ~ before additional the reservoir is filled and thc watcr pressurc ,,". 'Pcd .pc'u CU causes stresses. Fig. 13.44 shows thc shear strength envelopes obtained for a consolidated drained (CD), consolidatC
D1ustratlve Example 13.1. A series of direct shear rests was conducted on a out till the sample failed The following results were obtained Sample No.

soi~

each test was carried

Nomwlsfress

ShMrSfressu

15 30 4S

32

2 (kNlm )

Determine the cohesion intercept and the angle of shearing resistance.

2 (kNlm )

IS 25

SOIL MECHANICS AND FOUNDATION ENGINEERING 40

]0

i

,.

Z

.." T

.0

1

". W

10

10

Fig. E-13.1.

Solution. Fig. Ex. 13.1 shows the plot between the shear stresses and nannnl strcs..<;es at failure. From the plot, c .. 11 kN/m 2, 1(1 .. 26°

IIlustrutive Example 13.2. The following results were obUlined from a series of consolidated undrained testJ on a soi~ in wldch the pore water pressure was not determined Determine the cohesion intercept and the angle of shearing resiSUlnce. Sample No.

COllfilling

Devialor Slress

preessun!

aI/ai/u"

2 (1cNIm )

(lcN/m2)

600

100 200

7S
300

Solution. The major principal stresses in the three test are 700, 950 and 1170 kN/m 2, respectively. Fig. Ex. 13.2 shows the Mohr circle at failure in tenns of total stresses.

600

1100

A common tangent is drawn to the Mohr circle.

from the plot,

c _ 150 kN/n?,

+= 24°.

mustrative Example 13..3. A series of consolidated· undrained over-consolidated clay and the fol.lowing results were obUlined.

(cti) triaxial tests was conducted on an

SHEAR S rRENG11-I

Sample

Cdl pressure

No.

(1cNIm )

Det'imorsl/'Css 2 (lcNlm )

2

125

-70 - 10 + 120

510 620 850

250 500

Plot the strength envelopes in terms of total stresses and effective stresses, and hence determine the strengtl, parameters. If (he soil was preconso/idaled to a pressure 0/ 1000 leN/ml, plot the variation 0/ the pore pressure parameter AI Wilh the over-consolidation ralio. Solution. The table below shows the required calculations. Sample

a,

No.

-(7)-u

125

635

-70

250 500

870 1350

- 10 + 120

195 C = 12.'i + 70) 260 380

IT

T

tkNlm2)

01

AI

705 ('" 635 + 70)

880

- 0.137 (= -70/510) - 0.016

12.'"

+ 0.141

- UIOd

(Tolol

slrrs~s)

O.CR. 8.0 ( '" 1000/125) 4.0 2.0

SJIL MECHANICS AND FOUNDATION ENGINEERING

350

Fig. Ex. 13.3 (a) sbows the su:n!I~9~~C~O~ l~I~~~~f total stress~. From the plot, Fig. Ex. 13.3 (b) shows the strength envelope in terms of effcctive stresses. From the plot,

,'_ 28.5°, c· _ 50 kN/m 2 Fig. Ex. 13.3 (c) shows the plot belween AI and O.CR.

lllustratin Example 13.4. The following results were obtained from a consolitkAted-undrained (CD) test on a nonntllly consolidaJed clay. Plot the strength envelope in terms of total stresses and effective stresses and determine the strength parameters. Sample No.

Gel/pressure (iN/m2)

Deviator stress (1eN/"l)

PQre7:;"'~)UfJ!

250 500 750

152 300 455

120 250 350

Solution. The table below shows the calculations for principal lotal stresses nnd effective stresses.

Sample

0,

No.

250 500 750

402 800 1205

cr,

"'

120 250 350

130 250 400

282 550 855

Fig. Ex. 13.4 (a) and Fig. Ex. 13.4 (b) show the plot in terms of cffeaive stresses and total stresses, respectively. 600

'1 ~ r

400

lOa

Etfect;.,.

Total

stresSots

From the plot in tcnns of total stresses, ¢I

stresses

Fig. Ex. 13.4 (b)

Fig. Ex. 13.4 (a)

= 14°

2r

From the plot in terms of effcaive stresses. .' = Dlmtrative Example 13.5. A sample at'a soil failed in a triaxioJ rest under a devitJtor stress of 200 kN/nr when the confming pressure war 100 kN/m . If for the same sample, the confming pressure had been 200 kN/m2, what would have been the deviator Stress at failure? Asswne the soil has (a) c = 0 and (b) • = Solution. 1be stress 01 in the first test = 200 + 100 = 300.

o.

(a) Fig. E 13.5 (a) shows the plot for c = O. The envelope has bccn drawn tangential to the first circle. ]be second circle has been drawn 'i:.ucb. that it pa-;ses through 0) = 200 and is also tangential to the envelope. From the plot, • _ 30°.

----'--1400

351

SHEAR SfRENGTIi

Alternative method sin ~ _ [ : :

From Eq, 13.38 (0),

In the seoond lest, • remains equal to 3O .

01 - 200 0.50 .. 01 + 200 01 -

(b) From the piC){ when

0) ..

_ 0,5

•

(0, - ol)/2

Sin ell .. (01 +

Therefore, from Eq. 13.38 (a),

Therefore,

~~

D

0)12

or 01 .. 600 kN/m

60J - 200 .. 400 kN/m

2

2

41' ... 0,

0, - oJ

.. 200, as shown in Fig. Ex. 13.5 (b)

100 100

'F---io.'

I

~~J.n------~

'00],.0 0

,(

,{s:')oo

cr(kNIm2)

(b)

(0)

Fig. Ex. 13.5.

Illustrative Exumple 13.6. A sample o{ dry cohesion less soil war tested in a triaxial machine. I{ the Qngle of shearing resistance was 3tJ> and die confining p~sure, /()() lcN/m 2, determine the deviator stress at which the sample failed Solution. For dry cohesionless, the cohesion intercept is zero. . (01 - 0))12 Sin ~ .. (01 + 0))12

From Eq. 13.38 (a),

"

sin 360

Deviator stress at failure,

..

0, -

100 01 + 100

(Jd .. 385.4 - 100 .. 285.4 kN/m 2

Illustrative Example 13.7. The stresses on a failure plane in a drained test on a cohesion less soil me as under: 2 Normal stress (0) = 100 kN/m . Shear st~s ('t) = 40 1eN/,,? (a) Determine the angle of shearing resistance and the angle which the failure plane makes with the IMjor principal plane. (b) Find die major and minor principal stresses. Solullon. Fig. Ex. 13.7 shows the Coulomb failure line passing through the origin and the point A with coordinates (100, 40).

lan.' .. ; : .. 0.4

or ... 21.80 0

From Eq. 13.43, the angle whicb the failure plane makes with the major principal plane, S .. 45 + f,'/2 .. 45 + 10.9 .. 55.9 0

SOIL MECHANICS AND FOUNDAll0N ENGINEEIUNG

352

kNlm2

Fig. lli:.13.7.

(b) The centre C of the Mohr circle 'is located by drawing a normal AC to line OA at A. Mohr's circle is drawn through point A, with its centre at C. The circle ruts the a·axis at points B and D. From the figure. The stresses 0) and

O'J

0) .. 73 kN/ m2 , 01 .. 159 kN/m 2 can also be obtained analytically as follows.

OA -

Length

-"ocr

+

40' _ 100.7

AC .. OA tan ,', ... 107.7

I( Ian 21.8" .. 43 kN/ m2 DC ... OA sec ¢I' .. 107.7 x sec 21.8 0 ... ll6 kN/m 2

DD ... DC + AC ... ll6 ... 43 ... 159 kN/ m2 ... (11.

on ... oc - AC ... 116 - 43 ... 73 kN/m2 ... 03' illustrative Example 13.8. An undrained trit1xial compression test was conducted on a sample of compacted clay. Pore-water pressure was measured alter 1M. npplication of ,he cell pTt!ssure and also 01 failure, as givcn below. Find the pore prcssure coefficients A and B. (a) Consolidation stage. Change in cell pressure Change in pore waJer pressure

= 0 to 100 leN/m l == - 60 to + 10 kNit,?

(b) Shearing stage.

'" 500 kNlm 2 '" - 70 kN/m 2•

Deviator stress at failure Pore pressure at failure

Solution. From Eq. 13.30, II U3 or

From Eq. 13.32, 0<

_

D II 03

[1O-(-60)J - B[I00-0J or B - 0.70 II Ud - AD II Od (-7fl-1O) - A x 0.7 (500) or A - -0.23.

JIIustratJve Example 13.9. A shear vane of 7.5 em diameter and 1J.0 cm length was used to measure the shear strength of a soft clay. If a torque of 600 N-m was required to shear the soil, calculate tile shw strength. The vane was then rotated rapidly to cause remoulding of the soil. The torque required in the remon/tied state was 200 N-m. Determine the sensitivity of the soil. Solution. From Eq. 13.27, S -

T 600 )( 10-3 :r.([)2flI2 + d16) - 1([(7.5)2)( 11.012 + (7.5)3/6] )(10~

s _ 503 kN/m 2

'G

35'

,SHEAR SI'RENGTIi

In the remoulded state, $,..., ..

From Eq. 13.29, sensitivity

..

l 200 X lcr .. 168 kN/ml Jt(7.5fz x 11.0/2 + (7.5)3 /61 )( to"'

~~

.. 3.0

Dlustrative Example 13.10. A series of triarial tests was condm:ted on samplcs of a cohesionless soil and the following readings were taken for the deviator stress (leN/m2) at different strains. Draw the srressSl7mn curves and hence obtain the values 'Of lhe seCllJ1t moduli at one-half the u.ltinuue stress. Samp!' NA

1

, 2

~

I

2

3

4

5

6

7

8

9

\0

15

27 37 52

36 4. 65

43 58 75

4. 65 89

54 72 88

57 76 96

61 .7.

63 81 98

64 82 99

(1tNIm)

50 100 150

22

32

.6

Solution. FIg. Ex. 13.10 shows the stress-stram curves for the soil at three different confirung pressures. From the plot, the values of secant moduli arc obtained.

of

]

4

S

6

7

8

9

S!rom('/.)_

Fig. B·I3.l0.

(E,)] for at .. 50 IcN/m 2 is 1250 kN/m2

(E,h for o( .. 100 IcN/m 2 is 1750 kN/m 2

(E,h for a c

-

15£1 kN/m2 is 2700 kN/m 2 PROBLEMS

A. Numerical 2 13.1. The principal stresses al a point in a m
SOIL MECHANICS AND FOUNDA'10N ENGINEERJ1\G

354

kN/m 2, the sample failed at a deviator stress of 500 kN/m 2. The pore water pressure was 150 kN/m 2. The failurt plane OCCUlTed at an onglc of «J0 to the horizontal. Detcnnine the norrrui and shear stresses on Ihe failure p12Dt at failure. Also calculate the the maximum shear stress. (il) Ir the same specimen were tested in a drained test with a confining pressure of 200 kN/m2, whm would have 2 been the deviator Sl/'CSS 8t failure? c = 44 kN/m . fAn.'!.226.5; 108; 125; 552.4 kNlm1 11.4. A cylindrical soil sample failed at an axial lood of 140 kN/m2 in an unconfined compression ICSl. The failln plane makes an angle of 54° with horizontal. Determine the soil propcnics. (Ans. 'P = 181 13.5. The following resull<; were obtained from an undrained shear box lest o n a soil.

Normal Load (N) Failure load (N)

2.10

500 460

320

750 610

Determine the strength parameters in terms of 10lal stresses. The cross-sectional :IreD of [he shear box was

36 cm 2,

lAns. , '" 29°, c '" 55 kN/m2j

13.6. The results of 0.

CU lest on a compacted soil

Sample No.

03

are given below: , (0"1 -('1)

(kN/m'l

(kN/m'i

70 350

2.1 0 550

- 10 >90

:=~ne the cohesion intercept and the angle of shearing resi stance in ;~: ~i.~(1~5to~~/~~~ ..~bioe~~~~ 13.7. A shear vane, 7 _
IAn.... 963 kNJmlJ

13.10. A direct shear test

WlIS performed on a 6 em x 6 em sample of dry sand. The normal load was 360 N. Tbe failure oocurred at a shear 1000 of ISO N. PlOI the Mohr strenglh envelope, and determine ,. Assume c = O. Also determine the principal stresses at failure. [AilS. 26.57°; 69.1 kN/m2; 181.1 kN/1D1 2 2 13.11. An embankment is constructed of soil, with c' '" SO kN/m , . ' = 20· and., '" 16 kN/m . Determine the pm water pressure, effective stress. shear Slrength of the soil al the base of.embankment jlL~t after the fill has been raised from 3 m to 6 m. Thke pore prfSSure coefficients A and B os 0.50 ond 0.80, respectively, ond the lotmJ pre..o:.sure as one·holf of the vertical pressure. [Ans.. 28.8; 67.'2:, 74.5 kN{m1 13.12. A series of shear tests was performed on 0 soil. Each test wa,c; corried out until the soil somple sheared and the principill stresses for each test are as follows.

Test

01

(kNlm1 875 1160 1460

Plor the Mohr circle of stress and determine the strength envelope and ongle of intemol friction of the soil.

[Ans.29'j

B. Descriptive and Objective 1»pe 13.13. Whot is Mohr's strength theory for soils? Sketch Iypical strength envelopes for

D clean

sand.

13.14. Describe direct shear test. What are its merilS alld demerits?

13.15. Describe the triaxial shear lest. What are the advantages of trioxial shear lest over the direct shear test? 13.16. What is unconfined comprcssioo test? Sketch the apparalus used. What is ils advantoge over 0 triaxiol test?

~I NG

SHEAR STRENGTH

have

13.17. 13,18. 13.19. 13.20.

llilurc • 18°1

13.21. 13.22.

'.Iilure plane

~/m2J

13.23. 13.24. 13.2S. 13.26. 13.27_ 13.28. 13.29. 1330.

35S

Define slow, qUick and conMllidated quick triaxial shear [cst, illuslrllting the ir u e by at least one field example. What is Mohr' s d Tde'! Discuss LtS Lmportant chamclcrislics. Write revised Mohr-Coulomb equation. How docs it dIffer from the original C
C. Multiple

Choice Questions I. The shear Strength of plasllc undrained clay depends upon «(I) Internal friction (b) Cohesio n (el Both (a) ,met (h) (d) Neither ((j) and (b) 2. Whcn drainage is permined throu ghout the triaxi al test, tlt e te.~t is known as {al Quick test (b) Drnined test (e) Consolidated undrnined test (tI) None of Cal, (hl and k)


356

3. The shear strength of a cohesionlcss soil

IS

(a) proportional to lhe angle of shearing resistance. (h) inversely propol1ional to the angle of shearing re.~istancc.

(el proportional Iu lhe tangent of lhe angle of shearing resisl:mcc. (d) None of ab(lVc. 4. The nngle of lhe failure plane with Ihe moljor principal plane is given by (a) 45° + ¢I' (b) 45° + ¢I'12 (c) 45 0 - ¢'/2 (tI) 45 - (I' where 41' is the angle of shearing rcsislam:e. S. For a heavily ovcH.'Onsolid.'lICd clay. the pore pressure coefficient AI is in the range of

{al 0.7 to 1.3 {el - 0.5 to 0.0

(b) 0.3 to 0.7 (ll)

-1.0 to -0.50

6. If lhe angle of the rnooifietl failure envelope is a', lhe :mglc of she:lring resistance ¢I' is given by (a) cos (j)' = Ian cr.' (h) sin qJ' = Ian ,,-' (el Ian $' = sin rl (tf) None of above 7, Coulomb's eqLlation for shear strength can be reprcscmcJ as (tI) c = S + (J tan Q (h) c = s - 0' toln ¢J (cls=(J+ctanl!l Vf):1 = c-(J!:Jn¢J 8, For saturated, nom1ally consolidated soi ls, Skempton's pore pressure coefticicnts can be represented as (a)AI,B<1

(b)A>I,B>1 (d)A< I.B >1

9, In an undrained triaxial t:o mpression tes!. the sample fa iled at a deviator stress of 200 kN/m 2 when the cell pressure was 100 kN/m 2. 111e cohe.~ion intercept IS (af200 kN/m2 (b) 100 kNfm2 (c) 300 kN/m2 (tf) 50 kNfm 2 10. A dry sand specimen was tested in a triuxial machine· with th e cell pressure of 50 kPa. If the deviator stress at failure was 100 kP.,- lhe angle of ~hearing re~istancc is (a) 30" (h) IS" (e) 45" (if) 60"

IAlis. I. (b), 2. (b), 3. (c). 4. (b), 5. (e). 6. (b). 7. (b), 8. (al, 9. (bl. 10. (a}1

NO

14 Compaction of Soils 14.1. INTRODUCTION

Compaction me3rui pressing the soil particles close to each other by mechanical methods. Air during COOlpSction is expelled from the void space in the soil mass and, therefore, the mass density is increased. Compaction of a soil mass is done to improve its engineering properties. Compaction generally increases the shear strength of the soil, and hence the stability and bearing capacity. It is also useful in reducing the COOlpressibility and permeability of the soil.

"'"

Compactioo is an entirely different process tban consolidation disaJssed in chapter 12. It is important to note the fOUowing basic differences between the two processes. even though both the processes cause a rOOucUon in Uie volume. .

(1) Coosolidation is a gradual process of reduction of volume under sustained, static loading; whe~ canp3d.ion is a rapid process of reduction of volume by mechanical means such as rolling, tamping and

:0)]

vibration. (2) ConsoUdation causes a reduction in volume of a saturated soil due to squeezing out of water from the soil; whereas in compaction, the volume of a partially saturated soil decreases because of expulsion of air from the voids at the unaltered water content (Fig. 14.1).

Ca) COMPACTION

Cb) CONSOLIDATION Fig. 14.l.

(3) Consolidation is a process wbich occurs in nature when the saturated soil deposits are subjected to stade loads caused by tbe weight of the buildings and other structures. In COOlrast, rompaction is an artifieial process whicb is done to increase the density (unit weight) of the soil to improve its properties before it is put to any use. Compaction of soil is required for the oonstrudion of earth dams, canal embankments, highways, runways aoo in many other engineering applications. This chapter deals with various methods of compaction aod their effccts on the engineering properties of the soil Various other methods of site improvement are also discussed. (Stabilisation of soilS is discussed in chapter 15).

SOIL MECHANICS AND FOUNDATION HNOlNEI!RING

358

14.2. STANDARD I'ROcrOR TEST To assess the <1mounl of comp.'lClion and the willer content required in the field, compaction tests arc done on the s.1mc soil in the I:lboralory. The IcslS provide a relationship between lhe waLer content and the dry density. The waLer content at which the maximum dry densiLy is attained is obtained from Ihe relationships provided by the lc~lS. i'r<)(.1or (1933) used a standard mould of 4 inches intemal diameter and an cITcclive height of 4.6 inches, with a alpncily of 1/30 cubic fool. The mould had a detachable base plate, and a rcmovublc collnr of 2 inches height al iL<; lnp. '111c soil was compacted in Ihe mould in 3 equal layers, coch layer was given 25 blows of 5.5 pounds rnmmcr r'llling through a height of 12 inches. A curve was oblained between the dry density and the water content. IS : 2720 (part VII) recommends esscntially the s.'lmc srccifications as in Slandflrd Proctor test, witt) some minor mooific:ltions and metrificmion. The mould recommended is of 100 mm diameter, 127.3 mm

"""'iT

~

Con"

~

~=====mJ-bra2td

Th,.. 1"9'

on

BeSl! plaIt (0)

Fig. 14.2. StanJ.,rd Proctot Test.

height and 1(XX) ml C3fX1city [Pig. 14.2 (0)]. The rammer recommended is of 2.6 kg mass with a frce drop of 310 mm ana a face diameter of 50 mm. '!"he soil is compacted in three layers. The mould is fixed to a detachable base plate. lbe eoll.'lr is of 60 mm heigh!. If the percentage of soil retained on 4.75 mm sieve is more than 20%, a larger mould of internal diameter 150 mm, effe{.'1ive height of 127.3 mm and capacity 2250 ml is recommended. Procedure. About 3 kg of air-dried, pulverised soil passing 4.75 mm sieve is taken. Water is added to the soil to bring its water content to about 4% if the soil is ooarsc-graincd aoo to about 8% if it is fmc-grained. The watcr content should be much less than the expected optimum water content (fable 14.1). The soil Is mixed thoroughly and covered with a wet cloth and left for maturing for about 15 to 30 minutes. Table 14.1. Range or Optimum Water Content Saud

61010% '1l1e mould is cleaned, dried and grctlSCd lightly. The mass of the empty mould with the base plate. but without collar, is mken. The collar is thcn fitted to the mould. The mould is placed on a solid base and filled with fully matured soil to about onc-third its hcight. The soil is compacted by 25 blows of the rammcr, with

COMPACfJON

Ih

or SOILS

359

a free fall of 310 mm. (Tlle number of blows required for lhe bigger mould of 2250 ml cupacily is 56 instead of 25). The blows arc evenly distributed over the surface. 'Ibc soil surface is scratched with a spatula before the second layer is placed. The mould is filled to about two-thirds height with the soil and compacted again by 25 blows. Likewise, the third layer is placed and compacted. The third layer should project above the top of the mould into the collar by not more than 6 mm. lbc collar is rotated to break the bond between the soil in the mould and thai in collar. 'The coUar is tben removed, and the soil is lrimmcd oIT flush with the lOp of tbe mould. The mass of the mould, base plate and the compacted soil is taken, and thus the mass of the compacted soil is delennined. The bulk density of the soil is computed from the mass of the oompacted soil and the volume of the mould. Representative soil samples are laken from the bottom, middle and lOp of the mould for dctennining the water content. The dry density is computed from the bulk dcruiity and the water content. Bulk mass density, wbere M

p _

~

gm/mI

... (14.1)

= mass of compacted soil (gm), V = volume of the Dry density, Pd"

mould (ml).

Tf-;

... (14.2)

where w is the water content. The soil removed from the mould is broken with hand. More water is added to the soil so as to increase the water content by 2 to 3%. It is thoroughly mixed and allowed 10 mature. The test is repeated and the dry density and the water content are detennined. Compaction Curve. A compaction curve is plolted between the water content as abscissa and the corresponding....dry density as ordinate (Fig. 14.3). It is observed that the dry density initially inaeases with an increase in water content till the maximum density 1.95 (p.qnux is attained. With further increase in water content. the dry density decreases. The water content 1-90 corresponding to the maximum dry density is known as

,,

:;:=te:a~~.M~~I:nl

(D.W.C) or the optimum

~

£

a water content lower than the optimum, the soil 1-85 is rather stiff and has lot of void spaces and, therefore, the dry density is low. As the water coment is increased, ~ 1-80 t~e soil particles get lubricated and slip over each other, : . aOO move into densely packed positions and the dry 0 AI

.i'

of )8

~tcr

:i~beiso~~=';h~I:~~~~~~t ;a:~t~c:=n~;:;

1.75

Ito

~.pied by solid

density, as it occupies the space that might have been particles, as further explained in Sect.

1.10 6

I is ·.1).

Ites.

bul lled wilh

-'!~t~! ____ _

\\

; I

:

'--Z~ro

\

-air VOId (100 ·'.Saturatlon lin~

,, ,, ,,

,\ \

1 8

10

12

I~O.M~.

18

20

22

WOler conler'll f'Or a given water conlenl. theoretical maximum fig. 143. Compaction Curve. density, (P4w- , is obtaillCd corresponding to the oondition when there are no air voids (i.e. degree of saturation is equal to 100%). The theoretical maximum dry density is alSo known as saturated dry density (p,JSaI' In this condition, the soil becomes saturated by rcduaion in air voids to zero but with no change in water content. The soil could also beoome saturated by Increasing the water content such that all air voids are filled. As we are interested in the dry density at a given water content, the laUer case is not relevant here. An expression for the theoretical maximum deruiity is developed below. From the equations developed in chapter 2, the dry density (Pd) is expressed as


Gpo

Pd-~

Gpo Pd - I + (wGIS)

As e .. wGIS,

... (14.4)

It may be mentioned that compaction methods cannot remove all the air voids, and, therefore, the soil never becomes fully saturated. Thus, the theoretical maximum density is only bypotheticaL It can be calculated from Eq. 14.4 for any value of w if the value of G is known. The line indicating the theorel.ical maximum density can be plolted along with the compaction curve, as shown in Fig. 14.3. It is also known as zero air void line or 100% saturation line. Likewise, the lines for otber degrees of saturation, say 80%, 90% etc. can be drawn. For example, for S = 90%, Eq. 14.3 becomes Gpo p,' 1 + (w G/O.90) .. (145) Instead of drawing lines corresponding to different degrees of saturation, it is sometimes more convenient to draw lines corresponding to different percentage air voids (n..). From equations developed in chapter 2,

p, • (\-:-)wGo"0

... (14.6)

For theoretiali maximum density, nil _ O. Therefore,

(P.du_ - 1 ~p;G

(same as Eq. 14.4)

Thus, the zero-air void line and 100% saturation line are identical. The lines for other percentages of air voids, such as 10%, 20% etc. can be drawn. For example, for 10% air voids, Eq. 14.6 gives 0.90 x G pw ... (14.7) Pd-~ Il may be noted that 10% air-void line and 90% saturation line are not identical. The water renteo! at which the soil is compacted in the field is controlled by tbe value of the optimum water oontent detennined by the laboratory compaction test. The amount of compaction in the field should be approximately equal to that in the laboratory. The standard Prodor test desaibed above is adequate 10 represent the compaction of fills behind retaining walls and in higbways and earth dams where light rollm are used. In such cases, the optimum water content obtained from the standard Proctor lest can be used as II control criterion. However, in situalions where heavier compaction is required, for example in modern highways and runways, the standard Proctor tcst does not represent the equivalent compaction in the laboratory. For such conditions, the modified Proctor test, as described in the following section, is used to represent the compaajon in the field. (See Chapter 30, Sect.

3O.1~

for the laboratory experiment).

14.3. MODIFIED PRocroR TEST The mooified Proctor test was developed to represent heavier compaction than that in the standard Proctor test. The test is used to simulate the field conditions where heavy rollers are used. The tcst was standardised by the American Association of Stale Highway Officials and is, therefore, also known as modified AASHO·tesl. The Indian Standard Code IS : 2720 (part VllI) gives the specifications for heavy compaaion based on this test. In the modified Proctor test, the mould used is the same as in the standard Prodor test. However, the rammer used is much heavier and has a greater drop than that is the standard Proctor tcst. Its mass is 4.89 kg and the free drop is 450 mm. The face diameter is 50 mm as in the standard Proctor test. The soil is

COMPAcnON OF SOILS

361

compacted in five equal layers, each layer is given 25 blows. The compaClivc effort in the m~ified Proctor lest, measured in kJ/m l of soil, is about 4.56 times that in the standard Proctor lest. Thus, a much heavier compaction is attained. (Compactive effort in modified Pactor test:: 2700 kJ/m 3; in standard Proctor test :: 592 kJ/m~. If the percentage of soil retained on a 4.75 mm sieve is more than 20%. the larger mould of 150 mm internal diameter, effective height of 127.3 mm and capacity 2250 ml is used. In this case, 56 blows are required for each layer. The rcst of the procedure is similar to that in the standard Proctor test. The dry densities arc obtained for different water contents and the oompaction Olrve is drawn. Fig. 14.4 shows the compaction curve for the modified Proctor test (curve No.2). The curve is higher than and to the left of that obr.aincd from a starxlard Proctor test (curve no. 1). The heavier compaction increases the maximum dry density but dca-eases the optimum water content. TIle percentage increase of th.e dry density is between 3 to 18% for most soils; the percentage increase is more for clayey soils than for th.e sandy soiIs. \ "

"

, \

~ ,

(2)

\,

,

\\

II)

5tondord

(2) Modlf oe d

,

proctor proctor

tt1~! tes l

"_Zt1fOC""",dline

'"

IS . IOO ·I.)

~"'"

fig. t4.4. Compaction Crves of 5randard Proctor Test and Malified Proctor Test. Fig. 14.4 also shows the zero air.void line. It may be noted that the ma~mum dry density attained even in the modified Proctor test is lower than the theoretical maximum dry density indicated by the zero air-void line. The line of oplimums Shown in the figure joins the points indicating the maximum dry density. It is roughly parallel to the zero air-void line.

14.4. COMPACJ10N OF SANDS . The oompaction curves shown in Fig. 14.3 and 14.4 are obtained for soils which contain at least some percentage of cohesive soils, In case of pure sandy soils, the effect of water conlcnl on the dry density is nOI well defined when the water rontent is below the optimum value. There is a large scattering of the points on tbe compaction curve. Generally, the dry density decreases with

Air dry

Maximum bulking

an increase in the watcc content in this range (Fig. 145). The dry density decrea<>es due to capillary tension in pore water. The capillary tension resislS the tendency of soil particles to lake a dense state aod hence the volume increases. The phenomenon is known as the bulking of sand The maximum bulking occurs at a watpr content of about 4 to 5%. With further increase in the waler content,

Fig. 14.5. CaOlpaction CUf1ICC (or Cobesionle3S Soit.

SOil MECHANiCS AND FOUNDATION ENGINEERING

the dry density increases aurcmcnt of compactness (or denseness). The dry density of the sand is measured in the embankmcnt and its relative density is determined if the dry densities in the loosest and densest stales are known.

14.5. JODHPUR MINI COMPACI'OR TEST 'Illc Jodhpur Mini Compactor test was developed by Prof. Alam Singh (1965). A small mould of internal diameter 79.8 mm (cross- sectional area = 5000 mm). effective height 60 mm and a capacity of 3OOm1 is used. The rammer used is of 25 kg ma'IS and is known as the dynamic ramming tool (DR]). 111e mass slides down a stem through a height of 250 mm and falls ovcr a fOOl. of 40 mm diameter and 75 mm heighl and compacts the soil. The test is suitable for both fine-grained soils and coarse-grained soils (minus 4.75 mm sieve). The procedure for conducting the test is similar 10 that in the standard Proctor test, but tbe soil is compacted only in 2 layers. Each layer is compacted by 15 blows of the dynamic ramming tool uniformly distributed over the soil surface. The compactive effort is approximately equal to thai obtained in the standard Proctor test. It is daimcd that the optimum water content and the dry density obtained in the tcst are almost equal to that in the standard Proctor test. It is recommended that, for fine- grained soils, a fresh soil sample shall be taken for each test after allowing a suitable maturing time. 14.6. DARVARD MINIATURE COMPACTION TFST In Harvard miniature compaction test. romp..1dion is done by the kneading action of a cylindrical tamping foot of 0.5 inch (12.7 mm) diameter. The tamping fOOl. operates through a pre-sct compression spring

to give the tamping force to a predetermined valuc. The mould used is of

116 ioch (33.34 mm) diameter aod

of effective hcigbt of 2.816 ioch (71.53 mm). The capacity of the mould is 1/456 cubic foot (= 62.4 ml). l11e number of layers, the tamping force and tbe number of tamps per layer are selected depending upon the ty~ of the soil and the amount of compaction required.

14.7. ABBOT COMPAC110N TEST In the Abbot compaction test, a metal cylinder (mould) of 5.2 em internal diameter and 40 em effective height is used. '£be cylinder is damped 10 the base. The soil is taken in the cylinder and compacted by a 2.5 kg rammer having a circular face of 5 cm diameter. l11e rammer is lifted up and dropped inside the cylinder through a height of 35 an above the base. 14.8. FAcroRS AFFECTING COMPACTION The dry dcnsity of the soil is increased by compaaion. The increase in the dry density depends upon the following (octors : (I) Wate.- ContenL At low water content, the soil is stiff and offers more resistance to compaction. As tbe water content is increased, the soil particles get lubricated. The soil mass becomes more workable and the particks have doser packing. The dry density of the soil inaea<;es with an increase in the water content till the optimum water content is rel\Ched. At that stage, the air voids attain approximately a constant volume. With further increase in water content, the air voids do not decrease, but the total voids (air plus water) increase and the dry density deae&;CS. Thus the higher dry density is achieved upto the optimum water content due to forcing air out from the soil voids. Aner the optimum water content is reached, it becomes more difficult to force air out and to further reduce the air voids. "be effect of water content on the dry density of the soil can also be explained with the help of electrical double layer theory (Chapler 6). At low water contenl, the forres of attraction in the adsorbed water layer arc

COMPACIlON

.,.3

or SOILS

large, and there is more resistanre to movements of the particles. As the water content is increased, the electrical double layer expands and the interpartkle repu~ive forres incJ'eaI;e. The panicles easily slide over one another and are closely packed. This results in higher dry density. (2) Amount of Compaction. As discussed earlier, the effect of increasing the amount of oompactive effort is to inacac;e tbe maximum dry density and to dcaease the optimum water content (Fig. 14.4). At a water content less than the optimum, the effect of inaeascd compaction is more predominant. At a water content more than the optimum, the volume of air voids becomes almost constant and the.: effed of increased compaction is not Significant. It may be mentioned that tbe maximum dry density does not go on increasing with an increase in the compactive effort. For a certain inacase in the compective effort, the increase in tbe dry density becomes smaller and smaller. Finally. a stage is reached beyood which there is no further increase in the dry density with an increase in the compactive effort. The line of optimums which joins the peaks of the compaction curves of different compactive efforts follows the general trend of the zero-air void line. This line correspoods to air voids of about 5 %. (3) Type of Soil. The dry density achieved depends upon the type of soil. The maximum dry density and the optimum water content for different soils are shown in Fig. 14.6. In general, coarse-grained soils can be compacted to higher dry density than fine- grained soils. With the addition of even a small quantity of fines to a coarse-grained soil, the soil attains a much higher dry density for the same campaClive effort. However, if the quantity of fines is increased to a value more than that required to fill tbe voids of the coarse-grained soils, the maximum dry density decreases. A well graded sand attains a much bigher dry density than a poortly graded soil. Cohesive soils have high air voids. These soils attain a relatively lower maximum dry density as compared with the oobesionless soils. Such soils require more water than oohesionless soils and, therefore, the optimum water oonlenl is high. Heavy Clays of very high plasticity have very low dry density and a very high optimum water content. (4) Method of Compaction. 1be dry density achieved depellds not only upon the amount of compactive effort but also on the method of oompaction. For the same amount of compactive effort, the dry density will depend upon whether the method of compaction utilizes kneading action, dynamic action or static action. For example, in Harvard Minimure compaction test, the soil is compacted by the kneading action, and, therefore, the 2.00

III

~

~

(l)W~'-,md.d

"'"'

(2)l.OW-PlasticitYSi\t O}l.ow-plos!itily

clay

(l.)High-plostic,ty ctoy

'.50 '~°lro---t----;;n;__-"::r'--~20;---~2:.'--~"';Weier content (.,.)-_ Fig. 14.6. Compaction Curves for Different Soits.

SOIL MECHANICS AND FOUNDATION ENGINEER)NG

364

compaction curve obtained is different from that obtained from the other conventional tests in which an equal compaetive effort is applied. Different methods of compaction give their own compaction curves. Consequently, the lines of optimums arc also different. (5) Admixture. TIle compaction characteristics of the soils are improved by adding other materials, knO\Vn as admixtures. The most commonly used admixtures are time, cemcot and bitumen, as diso.lssed in chapler 15. The dry density achieved depends upon the type and amount of Ddmixtures.

14.9. EFFECT OF COMPACTION ON PROPERTIES OF SOn..s The engineering properties of soils are improved by compaction. 'The desirnble properties are achieved by proper selection of Ihe soil lype, the mode of placement and the method of compaction. The effect of compaction on various soil properties is discussed

below. In the (ollowing discussions, the dry of optimum means when the water cooten! is less than

e

:~ :I~~t: :~:~ ~:;p~~~;:;= when

1

~ FlOCCLLATEO

'SPERSEO

HIGH COMPAC riVE EFFORT CURVE

(1) Soil structure. The water content at which 'SPERSEO the soU is compacted plays an important role in the ;;; . LOW COMPACT1YE engineering properties of the soil. Soils compaded , . EFFORT CURVE at a water content less than the optimum watcr CAJ content gcnerally have a flocculated structure, ~ FLOCCULATED regardless of the method of compaction. Soils compacted at a water content more than the optimum water content usually have a dispe~ WATER CONTENT ("1.) structure if the compaction induces large shear Fig. 14.7. Soil Struaure in CompllCled Soils. strains and a flocculated structure if the shear strains are relatively small. In Fig. 14.7, al point A on the dry side of the optimum, the water oontent is so low that the attractive forces are more predominant than the repulsive forces. This results in a flocculated structure. As the water content is increased beyond the optimum, the repulsive forces ina-ease and the particles gel oriented into a dispersed structure. If the compactive effort is increased, there is a com:spooding increase in the orientation of the particles and higber dry c!ensities are obtained, as shown by the upper OJrve. (2) Penneubility. The penneability of a soil depends upon the size of voids, as discussed in chapter 8. The penneability of a §Oil decreases with an increase in water conteot 00 tbe dry side of tbe optimum water content. There is an improved orieotatioo of the particles and a corresponding reduction in the size of voids which cause a decrease in penneability. The minimum permeability occurs at or slightly above the optimum water content. Afier that stage, the permeability slightly incrcmes, but it always remains much less than that 00 the dry side of the optimum. The slight increase in the dry density is more pronounced than the effect of improved orientation. If the compactive effort is increased, the penneability of the soil decreases due to inaeased dry density and better orientation of particles. (3) Swelling. A soil compacted dry of the optimum water content has high water deficiency and more random orientation of particles. Consequently, it imbibes more water than the sample compacted wet of the optimum, and bas, therefore, more swelling. (4) Pore water pressure. A sample compacted dry of the optimum bas low water content. The pore water pressure developed for the soil compacted dry of the optimum is therefore less than that for the same soil compacted wet of the optimum. (5) Shrinkage. Soils compacted dry of the optimum shrink less on drying compared with those compacted wet of the optimum. TIle soils compacted wet of the optimum shrink more because the soil particles in the dispersed structure have nearly parallel orientation of particles and can pack more efficiently.

:~ct

~

-

COMPAcnON OF

SOI~

365

(6) Compressibility. The flocculated structure developed on the dry side of the optimum offers greater resistance to compression than the dispersed structure on the wet side.

Consequently.

the soils on (he dry side

are less compressible. However. the compressibility of the soil depends upon a number of other factors. It increases with an increase in the degree of saturation. The compressibility of a soil compacted on the wet side of the optimum is also influenced by the method of compaction. If the compaction is of kneading or impact Iype, it creates a more dispersed structure with a corresponding iocrease in the oomprcssibility. If the compaction causes very large stresses, the compressibility increases due to breakdown of ~ the structure and greater orientation lor the ~ DRy SlOE

t

particles.

tn

com~ct:7r;'~~r:~ ;tl:i;:~::e ~I:~~

a:

stress-strain curve than those on the wet side

(Fig. 14.8). The modulus of elasticity for the

~

~

WET SIDE

soils oompaded dry of the optimum is therefore high. Such soils have briUle failure

:~

:c~c~-~o~~~~ c~~s~

: 00:=1: AXIAL STRAIN. ______ relatively flatter stress-strain curvc and a Fig. 14.8. Stress·Slmm Curves. corresponding lower value of the modulU'i of elasticity. 'lbe failure in this case occurs at a large strain and is of plastic type. (8) Shear Strength. 10 general. at a given water oontent, the shear strength of the soil increases with an increase in the compact.ive effort till a aitiQlI degree of saturation is reached. With further iocrease in the compactive effort, the shear strength decreases. 1be shear strength of the oompacted soils depends upon the soil type, the moulded water content. drainage oonditions, the method of compaction, etc. The she.'lr strength of the oompaded silts and clays at the moulded water rontent and at a water content when fully saturated are quite different, as discussed below. (a) Shear strength at . moulded water content. Two samples are compacted to the same dry density, one dry of the optimum and the other weI of the optimum, and tested for shear strength. Fig. 14.9 shows the Mohr-C'...oulomb failure envelopes. The soils compacted dry of thc optimum have a higher shear strength at low strains. However, at large strains the

~:t~ro~~~: ; : ~~m~:! ~~e~~hd~

tE

approximately equal for both the samples. ffi~ On the wet wide, the shear strength is oc further reduced if the compaction is. by 11'1 DRY SlOE kneading action. It causes a greater orientation ~ towards a dispersed structure than that by static compaction methods. ~ (b) Shear strength after saturation. Two (/) samples are oompaaed to the same dry density, one dry of the optimum and the other wet of the optimum, and then soaked in water, NORMAL' STRESS (c;""") ______ without any VQlum~ change, to have full . saturation. The samples are then tested for Fig. 14.9. Failure EnvelClpC$. shear strength. The samples compacted dry of the optimum show greater strength. However, the difference in the strength of the two samples is much smaller than that prior to saturation. The difference in the water deficiency of the two samples and the consequent pore water tension is greatly reduced after saturation.


366

If swelling is pennittcd during saturation, the difference in strength of the two 5.1mplcs is further reduced. In some cases, the sample compacted on the wet side may exhibit even more strength. The drained shear strength of the two samples is almost equal.

14.10. METDODS

m' COMPACTION

USED IN FIELD

Several methods arc used for compaction of soil in field. The choice of the method will de~nd upon the soil type, the maximum dry density required. and economic consideration. Some of the more commonly used cooventiooai methods are disctf;;sed below. Other methods of compaction,. such as vibrofiotation, pounding, are discussed later. (1) Tampers. A hand-opcrated tamper (or rammer) consists of a block of iroo (or stone), about 3 to 5 kg in mass, attached to a wooden rod. The tamper is lifted for about 0.30 m and dropped on the soil to be compacted. A mechanical rammer is operated by compressed air or gasoline power. It .is much heavier, about 30 to 150 kg. Mechanical rammers have been used upto a ma<;:> of 1000 kg in some special cases. Tampers are used to compact soils adjacent to exisling structures or confined areas, such as trenches and behiod the bridge abutments, where other methods of compaction cannoc be used. Owing 10 very low output, tampers are IlOt economical where large quantities of soils are involved. Tampers can be used for aU types of

soils. (2) Rollers. Rollers of different types are used for compaction of soils. The compaction depends upon the following factors. (I) Contact Pressure. In general, the compaction ina-eases with an increase in the contact pressure. For a smooth-wheel roUer, the contact pressure depends upon the load per unit width and the diameter of the roller. . (il) Number of passes. 'The compaction of a soil inaeases with an increase in the number of passes made. However, beyond a certain limit, the increase in the density with an inaease in the number of passes in not appreciable. From economy consideration, the number of passes is generally restricted to a rcasooable limit between 5 to 15. (UI) Layer thickness. The compaction of a soil increases with a decrease in the thickness of the layer. However, for economy considc~tion, the thickness is rarely kept less than 15 em. (i.,) Speed of roller. The oompaction depends upon the speed of the roUer. The speed should be SO adjusted that the maximum effect is achieved. Types of Rollers (a) Smooth-Wheel Rollers. A smooth---whecl roller generally consists of three wheels; two large wheels in the rear and one small wheel in the front. A tandem type. smooth--wheel roller consists of only !wo dnuns; one in the rear and one in the front. The mass of a smooth--wheel roller generally varies between 2 to 15 Mg. These rollers are operated by internal combustion engines. [Note. Some authors express 1000 kg mass as one tonne (11). As lonne is not a standard SI unit, it is better to express 1000 kg us 1 Mg]. Smooth-wheel rollers are useful for finishing operations after compaction of fills aOO.. (or compacting granular base oourses of highways. These are Dot effective for compactioo of deep layers of soils, as the resulting compaction pressures induced are low. Further, these rollers also cause stratifical;~m in deep layCl$ due to non-uniform compaction. These rollers are generally used 10 "seal" the surface 01 tbe fill at the end of day's work to provide a smooth surface to quickly drain oIT any rain water. (b) Pneumatlc·tyred rollers. Pneumatic-tyred roUers use compressed air to develop the required inllatioo pressure. 1bc contact pressure depends upon the area of conlact and the inDation pressure. The roller generally consistS of 9 to 11 wheels fixed on two axles, with the pneumatic tyres so spaced that a complete coverage is obtained with each pass of the roller. The rollers arc available in a wide range of load sizes. The gross mass of the roller varies between 5 to 200 Mg. However, the rollers with mass more than 50 Mg are rarely used. The inflation pressure varies between 200 to 1000 kN/m 2 . The rollers are available as a self-propelled unit a<> well as a towed unit.

COMPACIlON OF SOILS

367

The roller compaclS the soil primarily by 1cne.1ding action. 1bcse roUers arc effcaive for compacting cohesive as weU as cohesionlcss soils. These rollers are the best type of equipment for general use. Light rollers (mass upto 20 Mg) are effective for compacting soil layers of small thickness upto 15 cm, whereas heavy rollers are useful for layers of thickness uplO 30 em. Sometimes, the rollers arc dcsigned to produce a wobble eITect, due to which a slightly weaving patb is lraeleed. This improves Ihe compaction of the soil. Pneumatic-tyred rollers are generally provided with a weight box or ballast box. TIle box can be filled with ballast 10 increase the weight of tbe roller. (c) Sheep-foot rollers. In ancient time before tbe advent of tbe rollers. it was usual practice to pass a flock of sheep on the newly formed soil fill to cause its compaction. The same principle is used in the design of sbccp-fool rollers. The shcc~fOQ( roller consists of a bollow drum wilh a large number of small projeaions (known as fect) on its surface. These projections penetrate tbe soil layers during the rolling operations and cause compaction. The drums are mounted on a steel frame. The drum can be filled with water or ballast to increase the mass. Sheep-fool rollers are available both as a self- propelled unit and a lowed unit. As rolling is done, most o( the weight of the roller is imposed through the projections on the soil. The contact pressure is generally betwccn 700 to 4200 kN/m 2. The roller may sink inLo the soil if the contact pressure is marc than the bearing capacity of the soiL ·Ibe shcc~foot rollers are ideally suited for compaction of cohesive soils. The rollers compact the soil by a combination of tamping and kneadin;J action. When the roller is passed for the first time. the projections penetrate the soil layer and the lower portion of the lnyer is compacted. In successive passes, compaction is obtained in the middle and the top jXXtioo of the layer. This continually rising effect of the compaction is caUed wa/Jdng-OUl of the roller. The depth of layer that can be compacted depends uJXIn the length of the projections and the weight of the roUer. Small rollers can compact layers o( 15 em thickness, wherta'> heavy rollers can compact layers of 30 em thickness. In general, the thickness of the layer compacted is kept not more than 5 cm greater than the length of tbe projection. (3) Vibratory compaclors. In vibratory compactors, vibrations are induced in the soil during compaction. The ~mpactors are avaiJable in a variety of forms. When the vibrator is mounted on a drum, it is called a vibratory roller. These roUers nrc available both as pneumatic type and the smooth-wheel type. In a smooth-wheel type, a separate mOlor drives an arrangement of eccentric weighlS to create high frequency, low amplitude, up- and-down osciUatioos of the drum. These roUers are suitable for compacting grtJIW/ar soils, with no fines. io layers upto I m thideness. However. if there is appreciable percentage of fines, the thickness has to be reduced. In a pneumatic-tyred vibratory compactor, a separate vibrating unit is attached to the wheel axle. The ballast box is suspended separately from the axle so that it does nO( vibrate. 1bese compactors are suitable for compacting granular soils witb thickness of layer of about 30 em. Another form o( a vibratory compactor is a vibrating-plate compactor. In this system, there are a number of small plates, each plate is operated by a separate, vibrating unit. Hand- operated vibrating piHtes are also available. The effect of the vibrating plates is limited to small depths. Their main use is 10 compact granular base courses for highways and runways where the thickness of layecr; is small. Vibratory compactors can compact the granular soils to a very high maximum dry density. 14.11. PLACEMENT WATER CONTENT As the methods used for compaction in the field are differenl (rom that for compaction in the laboratory, the optimum water content in the field may oot be necessarily be the same as in the laboratory. The laboratory value may be laken as a rough guide for placement water content in the field. The ideal placement water content when the pneumatic-tyred rollers are used is approximately equal to the optimum water content as obtained from a standard Proctor test. The placement water content when the sheep-foot rollers, smoothwhccl rollers and vibratory rollers arc used. is of the order of the optimum water content obtained in the modified Proctor test. For important works, a full-scale test is conducted in the field to determine the placement water content,

368


the thickness of layer. ma
... (14.8)

For cohesive soils. the dry density of the order of 95% of tbe maximum dry density of the standard Proctor test (i.e. 95% relative compaction of the stondard Proctor test) can be achieved LLSing a Sheep-foot roller or a pneumatic-tyred roller. However, if the soil is very heavy clay, only sheep-foot rollers are effective. For moderately cohesive soils, the dry density of the order of 95% of that in the modified Proctor test am be achieved using pneumatic lyre
OOMPAcnON OF roiLS

generally determined using the sand·bath method, alrohol method or the calch..m carbide method, as discussed in chapter 2. 1be nuclear methods are also beiog used increasingly. The water content can also be detennined indirectly using a Proctor needle (also known as plasticity needle). The Proctor needle consists of a rod aUacbed to a spring-loaded plunger (Fig. 14.10). The stem of the plunger is marked to read lbe resistance in newton. A sliding ring on the stem iodicates the maximum resistance recorded during the t(St. The needle-shank: bas graduations to indicate the depth of penetration. The equipment is provided with a series of needle points of different cross·sectional areas (0.25, 0.50, 1.0 and 2.5 to obtain wide range of the penetration resistance. For cohesive SOils, the needle points of larger aoss-sectional area<> are required and for cohesionless soil, those of smaller aoss· sectional areas are used. The needle point used should be such that it is neither too small for accurate measurement nor 100 large. A suitable needle point is selected and screwed to the needle shank:. After the soil has been compacted at a given water content in the compaction test in the laboratory, the Proctor needle is forced 7.5 em into it at the rate of about 1.25 an/sec. The maximum force used is found from the rompression of the spring. From the known area of the needle point, the penetratioo resistance per unit area is mmputed. A number of such measuremcnts are made in the laboratory during the mmpaction test, and a ailibration curve is obtained between the penetration resistance (R) and the water content. as shown in Fig. 14.11. It is found that for a given degree of compaction, the penetration resistance decreases with an increase in watcr content.

an1

STEM SLIDING RING

a

Fig, 14.10. Proc:IOr Needle.

10 14 12 WATER CONTENT (0/.) _ _ _ Fig, 14.11. Calibration CuM) ror Penetration n;sistanoc R.

"

Th determine the watcr content of the compacted soil in the field, lhe soil is compacted in the standard compaction mould in thc field in the same manner as was used during tbe calibration of the needle. The

penetration resistance of the rompacted soil is measured. The moisture rontent is then obtained form the calibration curve. This method of the detenninalion of the water rontcnt is quite rapid and reliable for fine-grained soils. However. it does not give accurate results for cohesionless soils and for soils having a large percentage of gravels and stone pieces. 14.14. VIBROFWTATION METHOD VibroflOl:alion is used for compacting thick dcposits of loose, sandy soils upto 30 m depth. A vibroflOl consists of a cylindrical 'tube, about 2 m diameter, fitted with water jets at the top and the bottom. It oontains

370


a rOlaling eccentric mass which develops a horizontal vibratory motion. The vibroOot is sunk into the loose soil upto the desired depth using the lower waler jet [Fig. 14.12 (a»). As water comes out of the jet, it creates a momentary quick condition ahead of tbe vibroOot due to which the shear strength of the soil is reduced. 'The vibroOot settles due to its own mass. Whcn the desired depth has been reached, the vibrator is activated. 1be vibroOot then vibrates laterally and causes the oompadion of the soil in the (a) (b) (e) horizontal direction to a radius of about Fig. 14.12. Vlbroflotalion. 15.m. The water from the lower jet is transferred 10 the lop jet and the pressure is reduced so that it is just enough to carry the sand poured at the top to the bottom of the hole [Fig. 14.12 (b)]. Vibration continues as the vibroDot is slowly raised to the surfOlce. Additional sand is continually dropped into the space (cralor) around the vibroOol. By raising the vibroOOl in stages and simultaneously backfilling, the entire depth of the soil is compacted [Fig. 14.12 (e)l. The spacing of the holes is usually kept between 2 to 3 m on a grid pattern. 'fh(f relative density (density index) achievcd for the sandy soils is 70% or more. [n soO, cohesivc soils, vibrofiotation is not effective. For cohesivc SOils. it can be used to (onn a sand pile to reinforce the deposit and to accelerate consolidation and thus improve its engineering properties. 14.15. TERRA PROBE METIIOD Terra probe method in many respects is similDr to the vibroflotation method. The terra probe consists of an open-ended pipe, aooul 75 cm diameter. 11 is provided with !'I vibratory pile drive. The vibratory pile driver when adivated gives vertical vibrations to the terra probe and it goes down. After reaching the desired depth, the terra probe is graduaUy raised upward while the vibrodriver continues LO operate. Thus, the soil within and around the terra probe is densified. The terra probe method has been successfully used upto depth of 20 m. The spacing of the holes is usually kept about 1.5 m. Saturated soil conditions are ideal for the success of the method. For the sites where the water tDble is deep. water jets are fitted to the tcrra probe to a<;&ist the pcoctration and dcnsillcation of tbe soil. The terra probe mcthod is considerably faster than the vibrofiorntion method. As it does not require backfilling of sand, it can even be used at offshore ' locations. However, the method is less cffective than vibrollotation method. In the terra probe, the zone of influence is considerably smallcr and the relative density achieved is also lower. 14.16. COMPACTION BY POUNDING To dcnsify large deposits of loose, sandy soils, the pounding method bas also been recently used. The method is also known as heavy tamping, dynamiC COmpal.1ion or high-energy compaction. Pounding is done by dropping a heavy mass (2 to 50 Mg) from a large height (7 to 35 m) on the ground surface. The actual mass and the hcight are selected depending upon the crane available and the depth of the soil deposit. A closely spaced grid pattern is selccted for the pounding locations. At each location, 5 to 10 poundings are given. The pounding method is used to compact the soil deposits to a great depth. It is very effective for densifying loose sandy deposits. Recently, the method has been successfully used 10 compact fine-grained soil

371

COMPAC110f'l OF SOILS

deposits as well. 1be deplh (D) in metres upto which the method is effective ClIl be determined from the following relation:

D - C

=

VMH

=

.. .(14.9)

=

where C coefficient (0.5 to 0.75). M mass (Mg), H height of drop (m). While using the pounding method, care shall be taken that harmful vibrations arc not transferred to the adjacent buildings. The radius of influence (R) in metres beyond which no hannfu l vtbrntioos are transmitted can be determined from the relation.

R _

!30VMH

.. . (14.10)

where M = mass (Mg), and H = height of drop (m). 14.17. COMPACTION BY EXPLOSIVES . Buried explosives are sometimes used to density cohesionless soils. The shock wave and vibralions produced by explosives are somewhat similar to thaI produced by vibratory, compaction equipment. The methoo is quile effective when the cohesionless soil is fully saturated. The shock waves cause liquefnction of sand, which is followed by densification. 10 partially saturated cohcsionlcss soils, compressive stresses develop due to capillary action and prevent the soil particles from laking closer positions. The method is nOI effective for partially saturated soils. 1he deplh upto which the blast is effective is limited to about 25 m. The uppennOSI zone of the soil upto a depth of about 1 m gets displaced in a random manner and is, therefore, not properly densificd. ThL.. zone should be compacted using the conventional methods by rollers. Explosive charges usually consist of about 60% dynamite and 30% spedaJ gelatin dynamite and ammonite. The charges are placed at two-thirds the thickness of the stratum to be densified. The spacing of the explosive points is kept between 3 to 8 m. Three to five blasts are generally required at each location. The radius of influence (R) of compaction can be determined using the relation

R - (M/C)~ where R

...(14.11) .

= radius of influence (m), M = mass of charge (kg), C constant ( =0.04 for 60% dynamite)

14.18. PRECOMPRESSION As discussed in chapter 12, prccompression improves the properties of the cohesive soils. In this method, the soil is preloaded before the application of the design loads. Preloadiog causes settlement before actual construction begins. The preload is generally is the form of an earth fiU which .is left in place for a long time so a'i to induce the required settle!llent. After the required compression hml been achieved, the preload is removed prior to the construction. A monitoring system consisting of settlement plates and piezometers may be used to check the progress of settlement. The prcrompression method is effective for compaction of silts, clays, organic soils and sanitary land fills. The preload must be carefully selected so as not to cause shear failures in the soil. The stability of the soil deposit under preload should be checked. Sufficient soil data should be collected to predict the rate arxl magnitude of the settlement. Sometimes, venical sand drains are used to decrease the lime of selllemcnt.

14.19. COMPACTION PILES Cohcsionless soils can be den..ified by constructing compaction piles. A capped, pipe pile is driven into the soil. 1be soil surrounding the pile is compaaed due to vibrations caused during driving. 1be pile is then extracted and the hole formed is backfilled with sand. Thus the compaction pile is formed. (For more details,

see chapter 25.) 14.20. SUITABILITY OF VARIOUS METIIODS OF COMPAcnON The suitability criteria of various methods of compaction can be summarised ~ under: (1) Cohesionless Soils only. Smooth-wheel rollers are suitable for compacting layers of small thickness


(2)

(3) (I) (ii) (ii!)

in base courses. Vibratory rollers, vibroflotation, terra probe, blasting, compaction piles and explosives are effective for compacting dejXlSilS of large thickness. Cohesive Soils only. Sheep-foot rollers are suitable for compaction of cohesive soils. Precompression is also quite effective. Both cohesion less and cohesive soils. The following methods are universal. 'Illcsc can be used for both cohcsionicss soils and cohesive soils. Tampers are effective for rompacting soils in a conrmed space of al1 types. Pneumatic-tyrcd roUers are extremely useful for compacting aU types of soils. Pounding method has a great promise for compacting all types of soils.

ilLUSTRATIVE EXAMPLES lIIustrati.-e Example 14.1, A sample of soil was prepared by mixing a quantity of dry soil with 10% by mass of water. Find the mass of this wet mixlure required /i) produce a cylindrica~ compacted specimen of 15 cm diameter and 12.5 cm deep and having 6% air content. Filld also the void ratio and the dry density of the specimen if G = 2.68.

Solution.

Air content, a c - V.,.IVv - 0,06 V.. _ 0,06 V y , Hence V.., _ O,94Vy

Thus

Va - 0,06 (

o~~)

- 0.0638 V..,

Volume of specimen (l') - nl4 x (15)2 x (12.5) - 2208,9 ml v _ Vs + V.., + V.. Now, with usual notations, 2208.9 _ V, + V.., + 0,0638 V.., _ Vs + 1.0638 V ... Writing volumes in tenns of mass,

~ 1.0)

2208.9 - (2.68

+ 1.0638 (

~)

Substituting M ... - 0,1OM" 2208.9 -

(2~C:S)

+ 1.0638 x OJ All

M, - 4606.54 gm. Mass of wet

soi~

Bulk density, Dry den<>ity,

M ... - 460.65 gm

M .. Ms + M... - 4606.54 + 460,65 - 5067.19

P- %- - 5:;8~ . Pd -

~

- I

2294gm/ml

2~2~1O

.. 2.085 gm/ml

e _ G p... _ 1 _ 2.68 x 1,0 _ 1 _ 0.285

Therefore,

Pd 2.085 llIustnative Example 14.2. The following results were obtained from a standard compaction test on a sampk of soil. WaterronJetJt(%)

0.12

0.14

0.16

0.18

0.22

Mass of wet soil (leg)

/.68

1.85

1.91

1.87

1.85

The volume of the mould used was 950 mi. Make necessary calculations and plot the compaction curve

373

COMPACTION OF SOILS

and obtain the maximlUlI dry density and the optimum Waler content. Also calculate the void ratio, the degree of saturation and the thecretical maximum dry density (G = 2.70). Solution. Calrulations are shown in tabular form. Water oontent(w)

0.12

0.14

0.16

0.18

0.20

0.22

Mass of wet soil (M) (kg)

1.68

1.85

1.91

1.87

1.87

1.85

1.77

1.95

201

1.97

1.97

1.95

158

1.71

1.73

1.67

1.64

1.60

0.71

0.58

0.56

0.62

0.65

0.69

0.46

0.65

0.77

0.78

0.83

0.86

2.04

1.96

1.89

1.82

1.75

1.69

Bulk density

p-MIV -

o.~o

Dry density fJd-e{(I.+w) \bid ratio

e-~-l PJ

Degree of saturation

S _ weG 1beoretiall mmaximum dry density (pd)lheomax

0=

~

Fig. Ex. 14.2 shows the compaction curve. 2.O° rrTT-n-,-,-,-rrTT-n-,-,rrrr"-,-,rrrrrr,,

I ::: H-t+t+-t1+H-t+t+-t1+l-+++-t1H-H-++++ H-t+t+-t1+H-t+t+u,.-t1+l-+++-t1H-H-++++ 1.90

E

~ 1.60 H-+++t-f-j+H-H'H-f1+H-.pf~H-H-+++l

I.'" 'ii

~

0

1.40

qp 1.20 1.10 1.00

0

CUO

..,.

W:atfer content _ _ _ Fig. E-14.2-

From the plol. (Pd)_ .. 1.74 gm/ml, O.We. = 15.2%, Illustrative Example 14.3. The ma-cimum dry density of a sample by the light compaction test is 1.78 glml at an optimum waler content of 15%. Find dIe air VlJid\' and the degree of saturaiion. G = 2.67. What would be the corresponding value of dry density on the

SoIUtl~

zero air void line al O. We. ?

G P... G p ... Pd .. ~ .. 1 + (M-G/S)


374

1.78 ..

0'

1.78 +

2.67 x 1.0 1 + (0.15 )( 2.6?/S)

O.~13 .. 2.67

or S .. 0.801

Pd. (l~:,,}~

Now

n.. .. 0.006

••

or 80.1%

(l~:,,}O~l;:\~;O

. 1.78

or 6.6%

(y.i)lhec>nurc " 1 ~ P:o .. 1

+2'~~15X :·~.67

.. 1.91 g/ml

PROHLEMS A. Numerical 14.1. A cylindrical specimen of a oohesivc soil of 10 em diameter and 20 an length was prepared by COIllp
a mould. If the wet mnss of the specimen was 3.25 kg and its water content was 15%. determine the dry density and the void ratio. If the specific gravity of the particles was 2.70, find the degree of saturation. (Aos. 1.80 gmlmJ; 0.50; 81%J 14.2. The following are the resulls of a swndard compaction lest performed on a sample of soil. Water Con/ellt (%)

21.2

Mass of wei soil (kg)

1.92

If the volume of the mould used was 950 c.c. and the specific gravity of soil grains was 2.65, make necessary calculations and plO! the water contcnt-dry density curve and obtain the optimum w.ller content and the maximum dry density. [Ans. 15%; 1.83 grw'm1J 6 14.3. An earthen embankment of 10 m) volume is 10 be constructed with a soil having n void ratio of 0.80 after compaction. There are three borrow pits TIlllrked A. D and C, having soils with void ratios of 0.90, 1.50 and 1.80, respectively. The COSt of excavation and trnnsportlng the soil is Rs. 0.2$, Rs. 0.23 and Rs. 0.18 per ro3, respectively. Calculate the volume of soil to be eXQlv8ted from each pit. Which borrow bil is the I'IXGt economical? (0 = 2.65). [Ans. 1.055 x 106; 1.389 x 106; 1.555 x 106 m3; A]

B. Descriptive and Objective Type 14.4. Differentiate between consolidation and compaction. Give examples. 14.5. Describe Standard Proctor lest and the modified Proctor lest. How would you dwde the type of the test to be conducted in the laboratory ? 14.6. What is il compaction curve? Give its salient features. What is a zero-air void line ? 14.7. What are the factors that affect compaction? DiS\-=uss in brief. 14.8. What is the effect of compaction on the engineering propenies 9£ the soil? How would you decide whether the soil should be compacted the dry of optimum or the wet of optimum ? 14.9. What are the different melbods of compaction adopted in the field? How would you select the type of roller 10 be used? 14.10. Write shan notes on (a) Placement water content (b) Rclolivc compaction (e) Compaction control (d) Terra probe (e) Vibrofiotalion (j) Comp
m

COMPACTION OF SOILS (d) As the comp;l(.1101l is increased, the optimum waH..... conlenl increast!S. (e) n,e modern highways and runways have compaction equal to that attained in a standard Proctor tcst. (j) Vibrollotalion is effcctive for highly cohesive soi l. (8) The Proctor nccdle can be uSt.>(/ to determine the dry density nchievcd in the field. (II) The rcl:l!ive compaction is Ihe same as the relative density. (i) lllc pneumalic.lyn:<1 rollers can be used for bOlh cohe~ionle!>~ and cohesive soils. fJ) llle water content of the soil in lhe tieltl i~ lliways ke]>I equlll to the optimum water (;on tent. (k) The soils comp:lclcd dry of the optimum have hi gher modulus of elasticity than those on the wet side. (I) The core in an earth dam is generally compacted wet side of the optimum. (III) The perme..1bil ity of the ~0I1 dccrea\cs by eoml)action. (/I) TIle Jodhpur mini-compactor test gives lower dry density tl'mn the standard Proctor test. (0) In th e Jodhpur 11lini-compaelOr test. knea.ding m:lion !,lkes pilice. (p) The number of passes made by a roller is generally more than ten. (q) n,e shear strength 0 1 u soil i1lways mereasc~ with an Inere,L~e in compaction. (r) In pneumati .... tyres. the contact I)TCssure depends upon the mllation pressure. (s) The thickness of lhc hlyer during compllcuon IS kep t :lbout 10 cm. (I) The smooth-whed roliers !.:an be used for compaction of Jeep fills [A ns. True.

tel.

(I). (k), (I), (m), (r)]

C. Multiple Choice Questions I. Pneuln3tic-tyrcd rollers arc useful for compilcting (a) Coheslve soils (h) CohesionJess soils (e) BOlh (a) aoo (IJI (l/) For soils III conli ned space 2. Vibrollotation technique is best suited for compactlllg (a) Coarse sand and grnvds (b) Silts Cc) Clays (/) organ ic soils 3. Preeompres~ion method is useful for compacting (a) Silts (h) Clays ee) Organic soils (l/) Alilhe above 4. The line of opli mullls generally cOrTC~ponds to percentage air voids of tlbout (a) zero percent (b) 5 perx:ent {el 10 percent «(1) 20 pt."fcent S. The mnge of opiimum water contents for the st:lll
~1~~~~~~~~~~~~~

15 Soil Stabilisation IS.I. INTRODUCTION Soil stabilisation is the process of improving the engineering properties of the soil and thus making it more stable. It is required when the soil available for constructioo is 001 suitable for the intended purpose. In its broadest senses. stabilisation includes compadion, preconsolklalion, drainage and many other suet processes. However, the tenn stabilisation is generally restriaed to the processes which alter lhe soil nuueriaf inelf for improvement of its propenles. A cementing material or a chemical is added to a natural soil for the purpose of stabilisal1oo. Soil stabilisation is used to reduce the permeability and oompressibility of tbe soil mass in earth structures aod 10 increase its shear strength. Soil stabilisation is required to i.ncrea'ie the bearing capacity of foundation soils. However. the main use of stabilisation is to improve Ibe natural soils for the coostructioo of highways and airfields. The principles of soil stabilisation are used for OODlrolling the grading of soils and aggregates in tbe construction of bases and sub-bases of the highways and airfickts. Soil stabilisation is also used to make an area trafficable within a short period of time for military aod other emergency purposes. Sometimes, soil stabilisation is used for city and suburban streets to make tbern more noise-absorbing. This chapter deal.;; with the various methods of soil sr.abUisation and their effects on the engineering properties of the soils. !Sol. MECDANICAL !rrABILISATION Mecbanical stabilisation is the process of improving the properties of the soil by changing its gradation. 1Wo or more types of natural soils are mixed to obtain a composite material which is superior 10 any of its components. To achieve lbe desired grading. sometimes the soils with ooarse particles are added or the soils with fine particles a re removed. Mecbanical stabilisatioo is also known as granular stnbiJisation. For the purpose of mechanical stabilisation, the soils arc subdivided into two categories: (1) Aggregates : Tbesc are the soils which have a granular bearing skeleton and have particles of tbe size larger than 751'. (2) Binders : ThC'SC are the soils which have particles smaller than 75" size. They do not possess a bearing skeleton. The aggregates consist of strong, we~·graded, angular particles of sand and gravel which provide internal friction and incompressibility to a soil. The binders provide cohesion and imperviof.lSl)(:$ 10 a soil. These 3Jt composed of silt and clay. The quantity of binder should be sufficient to provide plasticity to the soil, but it should not cause swelling. Propcr blending of aggregates and binders is done in order to achieve required gradation "Of the mixed SOlI. 1be blended soil should possess both internal fridian and cohesion. 1be material should be workable

SOIL STABILISATION

377

during placement. When properly placed and compacted, the blended material becomes mechanically stable. The IOOd
'The mechanical stability of the mixed soil depeods upon the following factolS. (1) Mechanical strength or the aggregate. The mixed soil is stable if the aggregates used have high strength. However, if the mixture is properly designed and compacted, even tbe aggregates of relatively low strengtb can provide good mechanical stability. (2) Minerai composition. The mechanical stability of the mixed soil depends upoo the composition of the minerals in it. 'The minerals should be weatber.resisLant. $odium sulpbates and sodium carbonates cause large volume changes due to their bydration and dehydration. These are detrimental to the stability. (3) Gradation. The gradation of the mixed soil should be such that the voids of the coarser particles are filled with finer particles so that a high density is oblained. According to Fuller (1907), the max:imum density is achieved if the particle size distribution of the mixture satisfies lbe following criterion: p n

'" ,r or xl xl

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=

(dID)O',. x 100

... (15.1)

wbere p = perrentage of the soil mixture passing sieve of size d, D ::::: maximum particle size. The U.S. Bureau of Public Roads recommends that the value of the exponent in Eq. 15.1 should be taken as 0.45 instead of 0.50. It is found by experience that to obtain sufficient cohesion in the mixture, it is necessary to bave a greater proportion of the material passing 7511 sieve than that given by Eq. 15.1. Generally, an ideal mixture would include about 25% binder. The basic requirement is that the mixture shall contain a fair proponioo of different size particles. It should contain sufficient quantity of fines to provide cohesion, especially in the wearing

surface. (4) Ploslicity characteristics. Sailo; with high liquid limit and plasticity index are suitable as bindclS for soils used [or surfacing. Such soils possess greater cohesion and better moisture retention capacity. They provide a better seal against the downward movement of surface water. For soils used in base courses, tbe requirement of plasticity characteristics are quite different from those for surfacing. The soil in base course should have low plasticity to avoid excessive accumulation of water and the resulting loss of strength. As the soil .available at site may seldom meet hoth the gradation and the plasticity characteristiCS, it usually beromes necessary to mix soils from different sources to obtain the desired mixture. This is normally done by trial mixcs. As far a<> possible. the maximum use of locally available soils should be made for

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pc

(5) Compaction. The mechanical stability of the stabilised soil mass deperxls upon the degree of compaction attained in the field. NonnaUy, tbe oompaction is done at or near the optimum water content. Uses or Mechanical Stabilisation. It ' ~ the simplest method of soil stabilisation. It is generally U'ied to improve the subgrades of low bearing capacity. It is extensively used in the construction of bases, sub-bases aod surfacing of roads.

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15.3. CEMENT S1'ABn.ISATION Cement Stabilisation is done by mixing pulverised soil and portland cemeot with water and compacting the mix to attain a strong material. The material obtained by mixing soil and cement is known as soil-cemenl. The soil-cement becomes a hard and durable structural material as the cemen.t hydrates and develops strength.

are tit ,00 ble

1)pes or Soll-cement MitdleU and Freitag (1959) have divided the soil-cemenl into 3 categories. (1) Normal SoU·Cement_ It COOSislS of 5 to 14% of cement by volume. The quantity of cement mixed with soil is sufficient to produce a hard and durnble construction material. 1be quantity of water used should

318


be just sufficient to satisfy hydration requirements of the cement and 10 make the mixture workable. The normal soil-cement is quite weather-resistant and strong. It is commonly used for stabilising sandy and other low plasticity soils. (2) )Iastic Soil-Cement. This type of soil-cement also contains remcot 5 to 14% by volume, but it has

more quantity of water 10 have wet co~islency similar to thai of plastering mortar at the time of placement. The plastic soil-ccmcnt can be placed on sleep or irregular slopes where it is difficult to use nonna\ ro.1d-mak:ing equipment. It has also been successfully used for water-proof lining of canals and reservoirs. The plastic soil-ccment can be used for protection of sleep slopes against erosive action of water. (3) Cement-Modified soil. II is a type of soil-cement thai contains less than 5% of cement by volume. It is a semi-hardened product of soil and cement. It is quile inferior to the other two types. As the quanlity of cement used is small, it is nOl able to bind all the soil particles into a coherent mass. However, it interacts with the silt and clay fradions and reduces their affinity for water. It reduces the swelling characteristics of the soil. The use of cement-modified soils is limiled. The following discussion is ooofined to the fir.>t two types of soil-cement.

Factors affecting cement stabilisation The factors affecting cement stabilisation can be summarised as under: (1) Type of soil. Granular soils with sufficient fines are ideally suited for cement stabilisation. Such soils can be easily pulverized and mixed. Ihey require Ihe least amount of cement. Granular soilS with deficient fines, Such as beach sands and wind-blown sands. can also be stabilised but these soils require more cement. As it is difficull to move road-making equipment over such soils when dry, il is desirable to keep them wet for better traction. Silty and· clayey soils can produce satisfactory soil-cement but those with a high clay-content are diffiOJh to pulverise. Moreover, the quantity of cement increases with an increase in clay content. The quality of soil-cement in this case is not good, as it may have high shrinkage properties. Organic matter, if prescnt in colloidal fonn, interferes with the hydration of cement and causes a rcdudion in the strength of soil-cement. The trouble is more common in sandy soils than in clayey soils. The soil sbooJd be treated with calcium chloride to remedy the situation. Sodium hydroxide is also effective in correcting the ill cffects due to organic mailer. Sometimes, addition of a small quantity of silt or clay to a sandy soil may aid in the cement reaction. (2) Quantity or cement. A well-graded soil requires about 5% cement, whereas a (XlOrly graded, unifonn sand may require about 9% cement. Non-plastic silts require about 10% cement, whereas plastic clays may need about 13% cement. The actual quantity of cem~nt required for a particular soil is ascertained by laboratory tests. For base courses, samples arc subjected to durability tests for determination of the quantity of the cement required. It oonsists of 12 cycles of freezing and thawing or 12 cycles of wetting and drying. 'The maximum volume change (swelling plus shrinkage) of 2% is generally permitted. Sometimes, the quantity of cement is determined according to the minimum unoonfincd compre&ive strength. Generally, a minimum strength of about 1500 kN/m2 for clayey soils and of about 5500 kN/m2 for sandy soils is Specified. High strength is ootained by decreasing the water-<:emcnt ratio. This is done by increasing the cement content for tbe same water content. As a rough guide, the cement content can be taken as 6% for sandy soils and 15% for clayey soils. (3) Quantity or Water. The quantity of water used must be sufficient for hydration of cement and silt-clay cement and for making the mix workable. GeneraUy, the amount of water ascertained from compadion consideration is adequate for hydration a<; weU. Water used sbould be clean and free from harmful salts, alkalies, acids or organic mailer. In general, the water which is potable is also satisfadory for soil-cement. (4) Mixing, Compaction and Curing. The mixture of the soil, cement and water shOuld be thoroughly mixed, as the success of cement stabilisation depends mainly on thorough mixing. If it is nOi. properly mixed,

SOIL STABIUSATION

319

it may result in a non-homogeneous, weak product. However, the mixing should nol be continued after the cement has started hydrating, as il would resuU in a loss of strength. Soil-cement should be properly compacted. Compaction is generally done as for soil alone (Chapler 14). For good results, fine- grained soils should be compacted wet of optimum, and coarse- grained soils, dry of optimum . After compaction, the surface is finished by a rubber-Iyred roUer. Soil-cement should be protected agaiost loss of moisture by providing a thin bituminous coaling. Sometimes, other materials, such as water-proof paper, mow straw or diet, are also used. (5) Admixture. To increase the effect iveness of cement as stabiliser. admixtures arc sometimes added to soil cement. Admixtures may permit a reduction in the amount of cement required. These may also help stabilisation of soils which are not rcsporriive to cement alone. Lime and calcium chloride have been used as admixturc.s for clays aoo soils containing harmful organiC matter to make them more responsive to cement. Fly ash acts as a pozzolana and is effective for stabilisation of dune sand. Sodium carbonate and sodium sulphates have also been used as admixture.

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Construction Methods '[lJ.e construction of soil-cement bases and sub-bases is done using the following methods. (1) Mix-in-place method. In this method of construction. mixing of soil-cement is done at the place where it would be finally placed. 11 consists of the following steps: (I) The subgrade is cleared of all undesirable materials such ao; boulders, debris. stumps. It is then levelled to the required formation level. (ii) The levelled subgracle is scarified to a depth equal to the proposed thickness of the soil cement (ilt) The scarified soil is then pulverised till at least 80% of the soil passes 4.75 mm IS sieve. It can be done either manually or with the help of a machine. Pulvcrisatioo of highly plastic soil can be done easily if about 4% lime is added to it. (iv) The pulverised soil is properly shaped to the required grade and the required quantity of cement is spread unifonnly over the surface. It is then intimately mixed dry with rotary tillers or special soil mixers. (v) lbe required quantity of waler is sprinkled over the surface and wet mixing is done till the mixture has a uniform colour. 1be operation should not last longer than 3 hours. The surface is then properly graded using towed graders. (VI) Compaction is done using suitable methods. The thickness of the layer should not be more than 15 em. Compaction should not take more than 2 hours. After cornpaction, the surface is properly finished. (vii) The compacted soil-cement is moist cured for at least 7 days by providing a bituminous primary coot. Altematively. it is k.ept damp by frequent applicatioo of a light spray of water. The mix·in-piace method of construction is quite simple, cheap and easily adaptable to dj(ferent field conditions. The maio disadvaotage is that the mixing is not uniform and high strength cannot be achieved. (1) Plllnt-mix method There are two types of plants used in the plant-mix method of construction. (a) Stationary plant. In this method, the excavated soil is transported to a stationary plant located at a suitable place. The required quantity of cement is added to the soil in the plant. Mixing is done after adding water. The lime required to obtain a uniform mixture depends upon the type of soil. 1be mixed material is then discharged into dumper trucks and transported back to the subgradc. It is spread and properly compaaed. The Slationary plant is useful for obtaining a uniform mix. In this methOd, the depth of treatment can be belter controlled. However, the method is quite expensive ao; compared with mix-in· place mc·thod. The material bas to be compacted as delivered and oot as a complete section of the road. A further disadvantage is that the work may have to be stopped even after a minor breakdown in the plant. (b) Travelling Plant. A travelling plant can move along the road under construction. The soil, after


placement of cement o ....er- iL, is lifted up by an elevator and discharged into tbe hopper of the mixer of the travelling plant. Water is added and proper mixing is done. The mix is then discharged on the subgrade and spread by a grader. It is then properly compacted. 1be travelling plant method, like stationary plant, is useful for acOJrate proportioning aod unifonn mixing. The depth of treatment is also properly rontrollcd and a unifonn subgrade surface is attained. However, the initial cost is vcry high.

1504. LIME !mUIILISATION Ume stabilisation is donc by adding limc to a soil. It is useful for stabjlisation of clayey soils. Whcn lime reacts with soil, there is exchange of cations in the adsorbed water layer and a decrease in plasticity of the soil occurs. The resulting material is more friable than the original clay, and is. therefore, more suitable • subgrade. . Lime is produced by burning of lime stone in kilns. The quality of lime obtained depends upon the parent material and the production process. There are basically 5 types of limes. (I) High calcium, quick lime (CaO) (il) Hydrated, high calcium lime (OH)2] (iii) Dolomitic lime (Cao + MgO) (iv) Normal, hydrated dolomitic lime [Ca (OH)2 + MgO] (v) Pressure, hydrated dolomitic lime [Ca (OH)2 + MgCh]. The quick lime is more effective as s{.1biliser than the hydrated lime; but the latter is more safe and convenient to handle. Generally, the bydrated lime is used. It is also known as slaked Jime. lbe higher the magnesium content of the lime, the less is the affinity for water and the less is the heat generated during mixing. The amount of lime required for stabilisation varies between 2 to 10% of the soil. However, if lhc lime: is used ooly to mOOify some of the pbysico---dlemical characteristics of the SOil, tbe amount of lime is about I to 3%. The following amount may be used as a rough guide. (I) 2 to 5% for clay gravel material having less than 50% of silt· clay fractioo. (il) 5 to 10% for soj~ with more than 50% of sill
rea

Chemical and Physical Changes In lime stabilisation When lime reacts with wet soi~ it alters the nature of the adsorbed layer by base exchange. Calcium ions replace the sodium or hydrogen ions. The double layer is usually depressed due to an inaease in the catioo ooocentratioo. However, sometimes the double layer- may expand due to high PH value of lime. Ume reacts chemically with available silica and alumina in soils. A natural cement composed of calcium alumino silicate! complexes is fonned, which causes a cementing action. 1be reaction depends upon the effective .concentration of the reaaants and temperature. In lime stabilisation, the liquid limit of the soil generally decreases but the plastic limit increases. Thus, the plasticity iOdex of the soil deaeascs. The soil heoomes more friable and workable. The strength of the lime-stabiUsed sOil is generally improved. It is partly due to a decrease in the plastic properties of the soil aDd partly due to formation of cementing material Increase in the unconfIned compressive sLrCngth is sometimes as high asc60 times. The modulus of elasticity of thelsoil also inaea-;es substantially. Addition of lime- causes a high concentration of calcium ions in the double layec. It causes a deaease in the tendC09Y of attraction of water. Consequently, the resistance of the soil to water adsorption, capillary rise

SOil SfABIUSAllON

381

,e

and volume changes on wetting or drying is substantially increased. TIle lime-stabilised bases or sub-bases fonn a water resistant barrier which stops penetration of rain water. There is an increase in the optimum water content and a reduction in the maximum dry density. In swampy areas where the water content is above the optimum, application of lime to soil helps in drying of the soil. Cyclic freezing and thawing can cause a temporary loss of strength, but because of subsequent healing aaion. there is no Joss of strength in long run.

,.

Construction Methods Construction methods used in lime stabilisation are similar to those used in cement stabilisation. However, the following differences should be carefully noted. (I) As the reaction in the ClSC of lime is slow, there is no maximum time limit between the addition of lime to the soil and the completion of compaction. However. care should be taken to avoid carbooation of lime in the process. (iI) lime may be added in the form of slurry insIead of dry powder. (iii) A rest period of 1 to 4 days is generally required after spreading lime over a beavy clay before fmal mixing is done. l11is facilitates proper mixing of lime and soil. (iv) lbe soil-lime is compaaed to the required maximum dry denSity. After compaction, the surface is kept moist for 7 days and then covered with a suitable wearing coat. Sometimes, the wearing coat is applied soon after compaction to help hold the moisture.

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lS~. BITUMINOUS SD\IlILISAT10N Bitumens are non-aqueous systems of hydrocarbons thut are soluble in carbon di-sulphidc. Tars arc obtained by tbe destructive distillation of organic materials such as coal. Asphalts arc materials in which the primary compOnents are natural or refined petroleum bitumens. Bituminous stabilisation is generally done with asphalt as binder. As asphalls are normally too viscous 10 be I1'>ed direc".iy, these are used as OJt-back with some solvent, such as gasoline. These are also used as emuisions, but in this form they require a longer drying period. Any inorganic soil which can be mixed with asphall is suitable for bituminous stabilisation. In cOOcsionless soils, asphall binds the soU particles togcther and thus serves as a bonding or cx:menting agen!. In cohesive soils, asphalt protcas the soil by plugging its voids and water proofmg it. It helps the cohesive soil to maintain low moisture content and to iocrcasc the bearing capacity. The amoWlt of bitumen required generally varies between 4 to 7% by weight. The actual amoont is dctennined by trial

Types of soil-bitumen According to the Highway Research Board of USA, there are foor types of soil-bitumen. (I) Soli-bitumen (proper). This is a water-proo~ cohesive soil system. The best results are obtained if the soil satisfies the following criteria. (0) Passing No.4 (4.76 mm) Sieve 50%. (b) Passing No. 40 (0.425 mm) Sieve 35 to 100%. (e) Passing No. 200 (0.074 mm) Sieve 10 to 50%. (d) Plastic limit less than 18%. (e) Uquid limit less than 40%. (/) The max:imum size of thc particle should not be grcuter than one- third the romp..'\ctcd thickness of the soil-bitumen . The quantity of bitumen varies from 4 to 7% of the dry weight. (2) Sand·bltumen. This is a bitumen stabilised cohcsionlcss soil system. The sand should be free from vegetal mailer or lumps of clay. The sand may require filler for its mechanical stability. However, it should not contain more than 25% minus No. 200 sieve material (i.e the rpaterial finer than No. 200 Sieve) for dune sands and not more than 12% in case in other types of sand.

SOIL MECtlANICS AND FOUNDATION ENGINEERING

The amount of bitumen required varies from 4 to 10%. (3) Water-Proofed Cia)' Concrete. A soil possessing a good grndation is water proofed by a uniform distribuLion of 1 to 3% of bitumen in this system. Soils of three different gradations have been rcrommended. For the three gradations, the percentage paSSing No. 200 sieve varies between (I) 8 to 12; (il) 10 to 16 and (jil) 13 to 30. (4) Oiled earth. In this system, a soil surface consisLing of silt-clay material is made w~ter proof by spraying bitumen in two or three applications. Slow or medium curing bitumen or emulsions are used. The bitumen penetrates only a short depth into the soil The amount of bitumen required is about 5 litres per square metre of the soil surface. J.'actors affecting bituminous stabilisation (I) Type of soil. Bituminous stabilisation is very efTcctive in stabilising s.1ndy soils having little or no

fmes. If a cohesive soil has the plastic limit less than about 20% and the liquid limit less than 40%, it can be effectively stabilised. Ilowever, pl.::lstic clays cannot be properly treated because of the mixing problems and large quantity of asphall required. Fine-grained soils of the arid regions whicb h.wc high PI( value and contain dissolved salIS do not respond weU. (2) Amount of asphalt. lhe quality of the bitumen·stabilised soil improves with the amount of asphalt upto a certain limit. 1iowever. if the amount of the asphalt is excessive. it rcsullS in a highly fluid mixture that cannot be properly mmpactcd. (3) MjxJng. The quality of the product improves with more thorough mixing. (4) Compaction. The dry density of the bitumen-soil depends on the amount :md type of compaction. It also depends upon the volatile content. In modified AASHO test, the maximum dry density occurs at a volatile content of about 8%. For samples cured and then immersed in water. the maximum strength occurs al a moulding volatile content corresponding to the maximum compacted density. Construction Methods Construction methods for bituminous stabilisation are similar to those used for soil-cement stabilis."ltion. However, the following poinlS should be noted. (1) The optimum volatile rontent for oompaction is genernlly much greater than that for Slabilily. The volatile content required for thorough mixing may be even greater, especially for Clayey soils. II is, therefore, necessary to aemte the mix between mixing and compaClion and between compaction and application. (2) 1b obtain a high stability, the layer method of .construction is preferred. Each layer is kept about 5 cm thick. When the lower layer has dried up. the su~uent layer is laid. The total thickness for bases is kept between 10 to 20 an. (3) In the mix-in-place method. tbe bitumen Lt; sprclycd [n severnl passes. Each layer is partially mixed before the next pass. This method preVCl1L'i the satunuion of the surface of the subgradc. (4) Climatic conditions influence the amount of bitumen that can be applied, as the amount of fluid (moisture) already prescot is the soil depends upon the climatic conditions.

15.6. CIlEMICAL STABILISATION In chemical stabilisation. soils are stllbilised by adding different chemicals. TIle main advantage of chemical stabilisation is that setting time and curing time can be controlled. Chern ical stabilisation is however genernlly more expensive than other types of stabilisation. The follOVo'ing chemicals have been succx:ssfully used. (1) Calcium Chloride. When calcium chloride is added to soil. it causes colloidal reaction and alters the charaeteristics of soil water. As calcium chloride is deliquescent and hygrosropic. it reduces the loss oC moisture from the soiL It also redures the dlances of frost heave. as the freezing point of waler is lowered. Calcium chloride is very effective as dust palliative. As the soils treated with calcium chloride do not easily pick up water, the method is effective for stabilisation of silty and clayey soils which lose strength with an increase in water cootent.

383

SOIL STABIUSA110N

calcium chloride causes a slight inCrc.1sc in the maximum dry density. However. the optimum water content is slightly lower than that for the untreated soil. It C,1U$CS a small decrease in the strength of the soil. However, if the compacted soil is put to watcr imbibition, water pick up is reduced and the strength of the treated soil is greater than that of the untreated soi\. It may be noted that most of the benefits of stabilis.1tion require the presence of the chemical in the pore nuid. As soon as the chemical is leached out, the benefits arc lost. The performance of treated soils depends to a large extent on the ground-water movement. The construction methods are similar to those used for lime stabilisation. lbe quantity of calcium chloride required is about of the weight of the soil.

4%

(2) Sodium Chloride. TIle action of sodium chloride is similar 10 that of calcium chloridc in many respects. However, thc tendency for attraction of moisture is somcwhat lesser than that of calcium chloride. When sodium chloride is added to tbe soi~ crystallisation occurs in the pores of the soil and it forms a dense hard mat with the stabilised surface. '£l'le pores in the soil gel filled up and relaro further evaporation of water. Sodium dlloridc also cbecks the tendency for the formation of shrinkage cracks. Sodium chloride is mixed with the soil either by the mix-in-place method or by the plant-mix method. It should not be applied directly 10 lhe surface. The quantity of sodium chloride required is about 1% of the soil weight. (3) Sodium Silicate_ Sodium silicates, as well as other alkali silicates, have been successfully used for soil stabilisation. The chemical is used as solution in watcr, known as water glass. The chemical is injeded inlO the soil. Sodium silicate gives strength 10 soil when it reacts with it. It also makes the soil impervious. It also acts as a dispersing agent. The maximum COlilpadcd density is increased. The qunntity of the chemical required varies between 0.1 to 0.2% of the weight of the soil. This method of stabilisation is relatively inexpensive, but its long-term stability is doubtful. Thc treated soil may lose strength when exposed to air or to groWld water. (4) Polymen.. Polymers are Jong-dlaincd molecules formed by polymerising of ccnain organic chemicals called mOllf)mers. Polyrnm may be natural or synthetic. Resins arc natural polymers. C'..alcium acrylate is a commonly used syntheLic polymer. When a polymer is added to a soil, rcnction takes place. Sometimes, the monomers are added with a catalyst to the soil. In thaI case, polyrnerisation occurs along with the fC.1ction. (5) Chrome Lignin. The chemical lignin is obtained as a by- prooud during the manufacture of paper from wood. O1rome lignin is formed from black liquor obtained during sulphite paper manufacture. Sodium bicarbonate or potassium bicarbonate is added to sulphite liquor to fonn chrome lignin. It slowly polymeriscs illto a brown gel. Whcn the chemical is added 10 the soil, it slowly reads to cause bonding of particles. The quantity of lignin required varies from 5 to 20% by weight. As lignin is soluble in water. its stabilising eITect is not permanent. (6) Other Chemicals. (I) Some water proofers such as alkyl chloro silanes. siliconates amines and quaternary ammonium snIts, have been used for water proofing of soils. (it) Coagulating chemicals. such as calcium chloride and ferric chloride, have been used to increase the electrical attraction and to fonn flocculated structure in on:Ier to improve the permeability of the soil. (iiI) Dispersant, such as sodium hexa-metaphosphate, are used to increase electrical repulsion and to cause dispersed structure. The compacted density of the soil is increased. (iv) Phosphoric acid oombined with a welling agent can be used for slabilisalion of cohesive soils. It reacts with clay minerals and forms an insoluble aluminum phosphate: 15.7. TIIERMAL S'D\BILISATION Thermal change causes a marked improvement in the properties of the sOil. Thermal stabilisation is done tither by heating the soil or by cooling it. (a) Heating. As the soil is heated, its water content decreases. Electric repulsion between clay particles is decreased and the strength of the soil is increased. When the temperature is inaeased to more than 100°C, !he adsorbed water is driven off and the strength is further increased.

384


When the soil is beated to temperature of 400°C to 600°C. some irreversible changes occur which make the soil non-plastic and non-expansive. The clay clods are converted into aggregates. With further increase in temperature, there is some fusion and vitrification, and a brick-like material is obtained which can be used as an artificial aggregate for mecbanical slabilisatio~. This method of stabilisation is quile expensive because of large heat input. It is rarely used in pr3dice. (b) Freezing. Cooling causes a small l~ of strength of clayey soils due to an increase in interparticle repulsion. However, if the temperature is rcdured to the freezing point. the pore water freezes and the soil is stabilised. Ice so fonned ads as a cementing agent. Water in cobesiooless soils freezes at about O°C. However, in cohesive soils, water may freeze at a much lower temperature. 'Ibc strength of the soil in~ as more and more water freezes .. This method of stabilisation is very cosUy. This method is used only in some special cases. It has been successfully used to solidify soils beneath foundations. The method is commonly used when advancing tunnels or shafts through loose silt or fine sand. Freezing may cause serious trouble to adjacent structures if the freezing from penetrates these areas. It may cause excessive heaving. The method should be used after considering the arove aspects. 15.8. ELECfRICAL STABILISATION Electrical stabilisation of clayey soils is done by a process known as electro-osmosis. As a direct current (D.C.) is passed through a clayey soil, pore water migrates to the negative electrode (cathode). It occurs because of the attraction of positive ions (cations) that are present in water towards cathode. The strength or the soil is considerably increased due to. removal of water. (For further details of electro-osmosis see chaptet 16). Electro-osmosis is an expensive method, and is mainly used for drainage of oohesive soils. Incidentally, the properties of the soil are also improved. 15.9. STABILISATION BY GROUTING In this method of stabilisation, stabilisers are introduced by injection into the soil. As the grouting is always done under pressure, the stabilisers with high viscosity are suitable only for soils with high penneability. This method is not suitable for stabilising clays because of their very low permeability. The grouting method is casHier as compared with direct blending methods. TIle methlXl is suitable roc stabilising buried zones of relatively limited extent, such as a pervious stratum below a dam. The method is used to improve the soil that cannot be disturbed. An area close to an existing building can be stabilised by this method.

1ypes of Grouting Depending upon the stabiliser used, grouting techniques can be classified as under: (1) Cement Groutmg. A cement grout consists of a mixture of cement and water. U the hole drilled ill the soil is smooth, tbe water-cemenl ratio is kepI low. Sometimes, chemicals are added to grout to increase iI fluidity SO that it can be injected into the soil. Cement grouting is quite effective for stabilising rocks with fissures, gravel and coarse sand. (2) Clay Groulmg. In this metbod, the grout used is composed of a very fme-grained soil (bentonite clay) and water. The bentonite clay readily adsorbs wafer on its surface. The viscosity, strength and flow characteristics of the grout can be adjusted according to the site conditions. Clay grouting is suitable itt stabilising sandy soils. Sometimes, other chemicals are added to clay grout. Clay cement grout is a mixtwe of clay, bentonite and cement. Clay-cement grout is a mixture of clay, bentonite and cement. CIay-chemical grout is a mixture of clay and sodium silicate. It is effective for medium and fine sands. (3) Chemical Grouting. The grout used consists of a solution of sodium silicate in water, .Jqlown as water glass. ~e solution cootains both free sodium hydroxide and colloidal silicic acid. An insoluble silia gel is fonned. ¥ the reaction is slow, calcium chloride is generally added to accelerate the reaction.

SOIL STADILlSA1lON

385

The melhod is suitable for medium and [me sands, However. the effect of chemical grouting is not permanent. (4) Chrome-lignin grouting. 100 grout used is made of lignosulphatcs and a hexavalant chromium compound. When it is combined with an acid. lhe chromium ion changes valence and thereby oxidises the lignaiulphates inlo a gel. The method can be used 10 stabilise fine sand and coarse silt. (5) Polymer grouting. Various polymers have been successfully used in grouting of fine sands and silts, (6) Bituminous grouting. Sandy and silly soils have been grouted successfully using emulsified asphalt. Slow-setting emulsions arc generdlly preferred, as these can travel a large distance into the stratum.

15.10. SThllILISA110N BY GEOTEXTILE AND FABRICS The soil can be stabilised by introducing geotextilcs and fabrics which arc made of synthctic materials, such as polythclem; polyester, nylon. The gcotextile sheets are manufactured in diITcrent thicknesses ranging from 10 to 300 mils (1 mil 0.0254 mm 25.4 j,l), The width of the sheet am be up to 10 m. These arc available in roUs of length uplO about 600 m. Gcotextilcs are manufactured in different pattcrns, such as woven, nOn-woven, grid, and hybrid. '[he woven gcolextilcs arc made from continuous mono-filament or slitfilm fibers. The don-woven geotextiles are made by the use of thermal or chemica l bonding of continuous fibres and then pressed through roUers into relatively thin sheets. The grids of geotextilcs arc made from a sheet of polymer by punChing it and then elongating it in at least one direction. 'Ihc hybrid geotextile arc nothing but combinations of woven, non-woven and grid. The geolextilcs are quite permeable. Their permeability is comparable to that of fine sand to coarse sand. These ar quitc Sleong and durnblc. These arc nOI affected by even hostile soil environmem. The use of gcotextilcs in geotechnical and coostrudion engincering has increased considerably in the last 2 decades. Gcotextiles are being increasingly used for the site improvement, soil slabilisation and various other related works. While selecting geotextiles for a partirular job. due importance should be givcn to the major function thai the geotextile bas to perform, as explained below, 1. Geotextiles as sepurator.>. Gcolexliles arc commonly used as separators between two layers of soils having a large difference in particle sizes to prevent migration of small-size particles into the voids of large-size particles. The main usc as separators is in tbe construction of highways on clayey soils. As the particle size of granular base .course of the highway is SE BASE COUR much larger than that of the subgrade (clayey soil in this ~). it is the usu.al prac~~ to pr~vi~e an.interven.ing ~#jl;k§AInt;BSk' SOIl layer of a SOil contJ).lOmg gram-sIZeS mlermecliate .. ::. .' .' ' ..,' .. ' : '. '. '. '.'. ' .. between that of the subgrade and the base course to ' .. ;' : ~- : '.:'. SUSGRAOE ~,' ::-. ::-'- ,': . GEOTEXTllE prevent migration of clay particles' into the base course. . . . . Fig. I S.I. Instead of the intervening soil layer, gcotextile can be provided to serve the Sllme purpose. The size of perforations should be according to the requirement. Thus a gcolextile sheet is used between the subgrade and the base course (Flg. 15.1). 2. Gootextlle as FlUer. It is the usual practice to provide a properly graded filter to prevent the movement of soU particles due to seepage forces The filter is so designed that the particle size of the filter is small enough to hold the protected material in place, as discussed in chapter 10. 1f the filter material is not properly selected, the particles of the soil move into the pores of the filter and may prevent proper functioning of the drainage. It may also lead to piping. Geotcxlilcs. can be used as filters iDStead of conventional filter. When the silt·laden turbid water passes through the geotcxtile, the silt particles are prevented from movement by the gcotextile. 1be modification in the soil aod void of tbe geotexlile occurs, and Fig.IS.2.

=

=

r

11:


386

after some time an equilibrium stage is attached. For relaLivc1y thin gcotcxtilc sheet, most of the filtration occurs within the soil just upstream of the gcotextileJabric. Fig. 15.2 shows the usc of geotcxtilc as filter on the upstream and downstream of the core of a zoned earth dam. It prevents tbe migration of the particles of the core inla the sbells. 3. Geotextile as Drain. A drain is used to convey water safely from one place to the other. As the goolextiles are pervious, tbey themselves fundion as a drain. They bave a relatively higher water-carrying capacity as comP'lred to Ib.:ll of the surrounding soil. . Drainage occurs either perpendicular to the plane of the sheet or in-plane of the sheet. In the first case, it fuoctions primarily as a filter. In the latter case, it ads us -a water carrier, and a relatively bulky geotextilc or a composite system of gcolcxtile is required. Fig. 15.3 shows a Iypical application where gcotexti1c is used for drainage behind a retaining wall. Fig. 15.3. In all the above applications of the geotexLile, the following advantages are generally adlievcd. (1) The installation is generally easicr and (aster. (2) The system has greater stability. (3) The quantity of soil to be excavated ,md disposed of is less. (4) '[be load on the structure is less. 4. Geo1extile as Reinforcement ror slnmgthenlng soil. Gcotextilcs have a high tensile strength. '[besc can be use<.t to increase the load-carrying capaCity of the soil. Geotextiles are used as reinforcement in the SOil, which is poor in tension but good in oompression. The action is somewhat similar to that of steel bars in a reinforced concrete slab. Geotextiles when uscd a<> rcinforcc- ~EOTEXTllE ment for soils have solved many EMBANKMENT construction problems on soft and ~ compressible soils. Fig. 15.4 shows the reinforcement of an embankment with ""'------------------">. geotcxtiles. The geotextiles/ havc been Fig. 15.4. used in the construction of unpaved roads over soft soils. These are laid over the soil and the base courre of the road is placed directly over it. When the vehicles pass over the road, the gcotextile deforms and iLS strength is mobilised. 5. Geotextlles used a.... reinron:ement in retaining walls. Geotcxtile can be used as reinforcement in the construction of earth-retaining structures. Gcote:dilcs are used to fonn the facing of the retaining wall as well as reinforcement. Such retaining walls are also ClUed fabric reinforced retaining walls (Fig. 15.5).

..,

-

,rGEOTEXTllE /

/

L.'D~~-===_/

I C ·:·:·

FORM

(a)

rGRANULAR _MATERIAL

:r~~~~~.~.:·~V (b)

(0)

(d)

Fig. 15.5.

Fig. 15.6.

)rPE

of ,Is

SOIL STABILISATION

387

The following procedure is used for the construction of the fabric-reinforced wall. (I) First the ground surface is levelled and the first gcotextile sheet of th(·. required width is laid over the surface such that about 1.5 m to 2 m of the sheet at the wall surface is draped over temporary wooden form (Fig. 15.6 a). (ii) Grooular material is placed over the geotextile sheet aod compacted with a roller of suitable weight. (iii) After compaction, the sheet is folded as shown in Fig. 15.6 (b). (iv) The second gcotextile sheet is placed over the compaaed layer over the granular material and draped over the wooden fonn as shown in Fig. 15.6 (c), and the process is repeated. (v) The front face of the wall is protected by the use of shotcrel.e or gunite. Shotqete is the cement concrete with a low water content. It is sprayed over the soil surfaoo at a high pressure. Fig. 15.6 (d) shows the completed wall. The design of fabric reinforced retaining walls is similar in principle to that of reinforced earth discussed. in the following section.

15.11. REINFORCED EAR'm The soil can be stabilised by introducing thin strips in it. In reinforced earth. thin metal strips or strips of wire or geosyntbctics are used as reinforcement to reinforce the soil. The essential feature of the reinforced earth is that friction developi between the reinforcement and the soil. By mc.·lnS of friction, the soil transfers the forces built up in the earth mass to the reinforcement. Thus tension develops in the reinforcement when the soil mass is subjected FAONG to shear stresses under loads. UNIT The main application of the reinforced earth is in the reinforced earth wall. The wall consists of a facing element, reinforcement and the back fll.l (Fig. 15.7). At the exposed vertical surfare of the earth mass, facing elements are used to provide a soct of barrier so that the soil is contained. The facing units are generally prefabricated from units which are small and tight so thai they can be easily transported and placed in position. These are usually made of steel, aluminum, reinforced concrete or plastic. These should be strong enough 10 hold back the back fill. Moreover, these . FIg. 15.7. should be such that the reinforcement can be easily fastened to tbem. The facing units generally· require a small plain concrete fooling al the bottom so that they can be easily built. The reinforcement is connec:ted to the facing clement and extended back into the backfill zone. The friction developed in the reinforcement restrains Ihe facing element. First a layer of reinforcement strips is p1.1ccd al the level ground surface and the backfilling is done with a granular soil. The soil with less than 15 percenl passing No. 200 sieve is used. The entire process of laying strips and backfilling is continued till the required height of the reinforced earth wall is attained. Galvanised steel strips arc commonly used as reinforcement. EaCh strip is about 50·100 mm wide and several metres in length. Tbc thickness is upto 9 mm. Sometimes metal rods, wires and geotextiles are used as reinforcement. Design of relnforced earth wall lbc following assumpcions are made: 1. The backfill is horizontal, without any surcharge. 2. The earth pressure acting on the facing element is the same as that acts on a rigid vertical face retaining concrete wall.


388

3. Rankine's earth pressure theory for active pressure (discussed in chapter 19) holds good. 4. The failure plane makes an angle of (45" + ~) with the horizontal, where, is the angle of shearing resistance of the backfill material. B C Let u.
1"1 ' h

~a~:Ya~:g~e :r~~~' +3 ~~~ur:iihla~~eA~=~al~~:

h

active pressure acting on the wall at any depth Z below the

h

soil surface is given by

unit length of the waU is given by

... (15.2)

where K... is Rankine's active earth pressure ooefficienl, cquallo 1002 (45 0 + ~!2). as discussed in chapter 19.

Let us divide the height H of the wall into small equal

\

h

\b

h

h

P" _ ~YH2 KA

\

1" H

p,'yZK. . .. (15.1) The pressure variation is linear. The total pressure per

\

,

RANK1"E FAILURE SURFACE

'RANKINE

_

,,~·!t __'~ illr~

PRESSURE

A

!-.....----i

G 2 A:lan (45'- T)

I(

Fig.

l:'i.S.

heights h as shown in Fig. 15.8. Let ZI be the depth of ony reinforcing strip i. The total earth pressure acting on the strip is represented by the area abed of the pressure diagram. The average pressure P; on the strip is given by

Pi - '1 Z;KA

•••

(15.3)

LeI us assume that the reinforcing stri!l) arc placed at intervals of h in the vertical direction. Let the spacing of the strips in the direction perpendicular to the plane of paper be s. 1'11us one reinforcing strip is subjected to the earth pressure on the area of (h x s). Therefore, the ten<>ion in the strip i is given by Ti - PI A - ('1 Z;KAHh )( $) ... (15.4) Using the same procedure, the tcosion in other reinforcing strips can be determined. Of course, the tension increases as the depth incre3SCS. 'The sum of the tension in all the reinforcing strips is equal to the total earth pressure on a length of s. Thus

I

1; -

s p.

. .. (15.5)

i.l

Length of reinforcing strip. 1bc reinforcing strips should extend well beyond the active zone inlO tbe backfill to have proper grip length. The length of the strip lying between the wall AD and the failure plane AC is Dot effective for computing the grip length. The effective grip length lying 00 the right-hand side of the failure plane AC should be able to provide a suitable factor of safely against failure. lei us again consider the reinforcing strip at depth Zi . If Fi is the frictional resistance on the reinfordng strip of length L", then .. (15.6) where F, is the fattar of safcty (usually lakcn as 2) and T; is the tension in the strip. Jf 6 is the angle of surface friction, then F/ .. (y ZI) tan 6 x (surrace arca of strip) The surface area of the strip of width b Is taken equal to (2 bLc) as the resistance develOps on both faces of the strip. Thus F, _ (y Z,) tnn b (2 b L.) ... (15.7) Substituting the value of F; in Eq. 15.6 and simplifying L

..

..

~ 2'1Z;btan6

(15.8)

38.

SOIL STABILISATION

Eqs. 15.7 and 15.8 are applicable 10 rectangular strips. If round bars are used,

L.~

~ 1tdyZ;tanb where d is the diameter of the bar. If continuous reinforcing sheets are used,

L-~ ~ 2yZ tanb

... [15.8 (a)]

... [15.8 (b)]

j

The angle of surface fridion b depends upon the density and type of the backfill material. It also depends upon Ihe roughness of the reinforcing Sirip. 1be value of b usually varies between 0.5 ., and .. where 4' is the angle of shearing resistance of the backfill material

Cross-sedionnl area of the reinforcing strip The aoss*scctiona! area of lhe reinforcing strip is determined if the allowllble tensile stress (/,) is known. The cross*scctional area is obviously equal to the tensile force divided by the allowable tensile stress. Tbus

A,.I'!..F,Ti

t.

Eqs.

t.

A,. F,(yZiK.,h,

or

t.

... (15.9)

and 15.9 give different lengths L", and the aoss--sectional areas A, for different reinforcing strips. For convenience, it is the usual practice 10 adopt the same length and the cross-section for aU the reinforcing strips. Thus the value oorresponding to the maximum tension at the base of wall is used for all strips. Hence Zj - fI for all strips. 15.8

PROBLEMS A. Descriptive and Objective type Questions IS.1. What is soil smbilisntion ? Whnt are its uses ? IS.2. What is mechanical stabilisation ? What are the raaors that aerea the mechanical stability of a mixed soil? IS.3. Dc:;cribe in brief cement stabilisation. What arc the factors that affect the smbility or soil cement? Discuss construaion methods. J5.4. Discuss the use or lime in Slooilisation or soils. What arc the chemical and physical manges which take place in lime stabilisation? 15.5. Write a short note on bituminous stabilisnlion. What are different types or soil bitumen? Dcsaibe the raaors affecting bituminous stobilisation. IS.6. What are different types of chemicals used in stabilisation of soils? IS.7. Write short notes on: (I) Thennal stabilisntion (u) Electrical stabilisntioo. (iil) Grouting (iv) Geotextile. 15.8. Write ~hethcr the ronewing stalements art true or false: (a) Mechanical stabilisation requires addition of chemicals to soils. (b) Cement stabilisation is more suitable for fine grained soils than coarse-grained soils. (e) Lime Slabilisalion is suitable for coarse-grained soils. (d) Bituminous Stabilisation can be used for both coarse- grained and fine-grained soils. (e) The effect or lignin as a SUlbilisntion agent is permrment. (j) Thermnl stabilisation Is quile inexpensive. (g) Elcctro-osmosis is used ror stabilising highly cohesive soils. (h) A ~olextile is embedded in the soil to give it stobility. [Ans. True, (d), (g). (II)} (.) Cement groutIng can be used for clayey soils.


390

n. Multiple Choice Questions 1. Cement stabiliS:llinn is gCOCflllJy used for stabilising (a) Sands (Il) Sills ee) Clays (t1) All the above 2. Lime st(lbilismion is generally usC
4. The following methods of mix ing cement in the stabilisation of the soil is generally the best; (tI) Mix.in-placc method (/1) Sl:IIionary plant method (c) Trnvclling plan! method (d) NOllc of above. S. For stabilisation of heavy clays. the following method is generally QlO1II'eft"ective (a) Mechonicai stabilisation (h) Thermal sl
lAos. 1. (0). 2. (e ). 3. (d). 4. (e). 5. (d). 6. (b). 7. (d). 8. (e). 9. (cf). 10. (e)]

16 Drainage, Dewatering and Wells 16.1. INTRODUCTION Drainage is the process of removal of gravity water (free water) from a soil mass in order to keep it in a stable condition. Drainage may be classified into two categories: (I) Surface Drainage, (iI) Sub-surface Drainage. Surface drainage is the method of collection and divclSion of the surface run off. Subsurface drainage oonsists of collection and dis[XlS81 of the ground water. Subsurface drainage is also known as dewatering. 11 is process of removal of water (rom a foundation pit when it is situated below the ground water table or when it is surrounded by a colIer dam. The purpose of dewatering is to keep the excavation dry so that concreting can be done. Dewatering is temporary if it is done al the lime of construction. It is followed by restoration to its original water table after tbe structure has been compleled. Pennanenl dewatering is required for removing subsurface gmvilationai water throughout the life of structure. It may be necessary to keep the water away from the structure to ched: dampness or other ill effects. Subsurface drainage not only facilitates construction, but it also helps in improving the properties of the soil. This helps in the stabilisation of soils. In case of fine·grained soils, although the quantity of water removed is not much, improvement in the properties of the soil is significant. Subsurface drainage also helps in reducing the hydrostatic pressure acting on the base of the structure. The method of sub-surface drainage to be adopted at a particular site would depend upon the characteristics of thc soil, the pa;ition of the water table and the time period the system has to operatc. Various methods of sub-surface drainage are discussed in this chapter. Theory of wells is alSo dealt with. The methods of surface drainage are outside the scope of this te"t. 16.2. INTERCEPTOR DITCIIES Intcrceptor ditches arc used (or excavation of limited depth made in a coarse soil. These ditches are constructed around the area to be dewatered. 1be ditches must penetrate deeper than the level of the work area (Fig. 16.1). At suitable locations, sump pits are constructed along the ditch for installation of the pump to remove the water collected.

Fig. 16.1.llIleroepcocOitcb.

SOIL MECHANICS AND FOUNDATION ENGINEERINQ

392

If the soil is fine sand of low permeability, boiling may occur in sumps and ditches. This may be prevented by placing niter layers on the sides and at the bottom of thc ditches and sumps. Interceptor ditches are mail economical for carrying away the water which emerges 011 thc slopes and near the botlom of the foundation pit. The method can be effectively used for rock fonnalion, gravel and coarse sand. In fine sands and sillS, there may be sloughing, erosion or quick COIldiliorlS. For such soils, the melhod is confined 10 a depth of 1 102m.

16.3. SlNGLE-STAGE WELL POINTS A well point is a perforated pipe about I m long and 5 em in diameter. 1be perforations are covered with a screen to prevent clogging. A jetting nozzle is provided at its lower end. A conical steel drive point is flXed to the lower end of the well point to facilitate instaUntion (Fig. 16.2). A ball valve is also provided. near tbe lower end which permits 00'0'1 of water only in the downward direction during installation. The well point is wnnccted to the bottom of the riser pipe of the same diameter. Risers of different well points are connected to a horizontal pipe of 15 to 30 em diameter, known as header. The header is connected to a specially designed pumping unit. Tbc spacing of the well points depends upon the type of soil and the depth of water. Generally. it varies between 1 to J m. Well points can be installed in a drilled hole, but generally these are installed by jelling. Water is pumped through the riser pipe in the downward direction. As it discbarges through the nozzle, it displaces the soil below the tip. Jelting is continued till the required pcnetralion of the tip is achieved. The advantage of installation by jelling is that the water under pressure w~hes away soil fines near the. tip Ball volve and lca~es a relatively coarse material. It forms a natural filter around !be tip. The hole formed ncar the tip is filled with coarse sand. poiot After the well points have been installed around the area (a) Elevation. (b) Sec!ion to be dew3tered, pumping is started, Each well point lowers FIg. 16.2. Well POInts. the water table around it and fonns a small cone of depression (Fig. 16.3). Various cones of depression join, and a common drawdown curve is oblained. The water table is thus lowered (Fig. 16.4.) Well points are suitable for lowering !be water table by 5 to 6 m in soils wilh a coefficient of penncabilily between I x 1~ to 1 )( 10-6 m/sec. 1he saea1 normally provided wilh (be well points can prevent medium sand and roarse sand ·from entering the well point. If the stratum to be dewatered consists of

,.,

PIPE

Fig. 163.

Well·point Installation.

..

DRAINAGE, DEWATERING AND WElLS

--.-"~=

393

~--,. ~ sz.

wellpoinl. 1t is essential to oontinue pumping once it bas been started until the excavation is complete. If it is stopped in between, it may prove to be disastrous.

~ _ ___ ~.!.:_

_ _ _ __ __

Tr~l'ICh-z"

16.4. MULTI·STAGE WELL POINTS

grca:~nru!~ :~I:u:~~a:el:~ ~;:!~o~~~~

'-WtUpoinl

In this melhod, two or more rows of well points are Fig. 16.4. Effect installed at different elevation'>. Fig. 165 shows a two-stage well poiDt system.

of

~

Well Points.

Fig. 16.5. Multi-lIllge Well Points.

The installation of weU points is done in stages. The first stage well points (marked I) are located near tbe perimeter of the area, as in a single-stage well point system. These are put into operation and the water table is lowered by about 5 m and the area is excavated. The well painlS of the second stage (marked II) are then installed within the area already excavated. Water table is further lowered by about 5 m and tbe excavation of further 5 m depth is done. lbus the total depth of excavation becomes about lO . If required, tbe third stage of well points can also be installed to further lower the water table. The method is useful for excavations uplO 15 m depth. Excavations exceeding 15 m depth are generally dewatered by a deep well system (Sect. 16.7). 16.5. VACUUM WELL POINTS Well points cannot be used successfully for draining silty sands and other fmc sands with an effective size less than about 0.05 mm. The coefficient of permeability of such soils is generally between 1 x 10-5 to 1 x lO-7 m/sec. These soils can be effectively drained by using V3Q1um well points. For installalioo of a vacuum well points, a hole or about 25 em diameter is rormed around the well point and the riser pipe by ~IoI,!.M'I'Zl In:rk/".J.J.J,,M:,,J.,.. jetting water under pressure. When water is still flowing, medium to coarse sand is filled into the bole upla about 1 m from the top. Sto! The top 1 m portion of the bole is then filled by tamping clay into it. It fonns a sort of seal (Fig. 16.6). Any other impervious material can also be used instead of clay to form a seaL Well point Sond lilt~r spacing is generally closer than that in a conventional system. (voids \,Inder vO~\,Ium) When the header is connected to a vacuum pump, it creates a vacuum in the sand filter around the well point. As the pressure on lhe water table is equal to the atmospheric pressure, the head causing flow is inaeased by an amount equal to the vacuum pressure. The hydraulic gradient Increases and it overcomes the flow resistance in the soil pores. The ground water flows to the region of vacuum in the well points and drainage DCO.lf$. As the effective pressure on the soil is increased, Fig. 16.6. Vaccum Well Point


394

consolidation takes place. It makes the soil stiff. However, the process is slow and it may take severol weeks for the soil to become stiff enough for carrying Qui the excavation work.

16.6. SIIALLOW WELL SYSTEM In a shallow well system, a hole of about 30 an diameter is first bored into the ground, using a casing.

A filler lube of about 15 em diameter, covered with a special wire mesh, is then lowered into the casing. 1be casing is gradually withdrawn and suitable filler material is added in to the annular space between the casing and the Ulter lube. 'Ibis forms a filter well. A suction pipe is lowered into the filler well. A number of such wells may be installed. The sud ion pipes of all these wells are connected to a rommon header A pumping unit is attached [0 the header. As the pumping is stancd. the drainage OCOJI"S. TIle suction lift of ~hc well should not be more than 10 m for its proper working. Shallow well system is rarely used in practice. Well-point systems, as discussed earlier, are more economical uplO a depth of 10 m than a shallow well system.

16.7. DEEP WELL SYSTEM A deep well is about 30 10 60 an in diameter, bored to a depth of 15 to 30 m . It is provided with a casing which is perforated ooe in the pervious zones penetrated. Coarse filter mnterial is placed in the annular space between the casing and the walls of the hole. The spacing of deep wells varies betwccn 10 to 30 m, depending upon thee area to be dcwatered and the IQCalion of the water table. A submersibJe pump is placed inside the casing near the bottom. The pump is driven by a motor mounted on the top of the casing through a vertica"'shaft (Fig. 16.7). However, if the motor is submersible, it can be directly attached to the pump. As the pump is placed al the bottom, there is no restriction on the height to which the water can be lifted, unlike a shallow weU system. Deep wells are located on the outer periphery of tbe area to be excavated. A row of well point is also insLalled at the toe of side slopes of the deep excavation to intercept seepage between the deep wells and to prevent sloughing of the slopes near the toe. Fig. 16.7. Deep Well System. As in the case of well point systems, pumping once started must be continued until the entire excavation work is completed. If the pumping is stopped in between, tbere would be a rapid development of hydrostatic and sccp3ge pressure in the excavation, which may prove 10 be disastrous.

16.8. HORIZONTAL WELLS Horizontal wells of about 5 cm to 8 em diameter have been used for drainage of hill sides. These wells are drilled ioto the hill at a slightly upward slope. A perforated casing is installed in the well to collect and dischm-gc water. Horizontal wells have been successfully installed for horizontal length of 60 m. Large horizontal tunnels have also been successfully used to tap deep aquifers beneath hill sides. A combined system of vertical wells and horizontal wells can be used to drain stratified soil de~its. In tbis system, the vertkaJ. weUs intercept the aquifer and discharge the water into the borizontal well. 100 water collected by the horizontal well is discharged at a suitable point.

16.9. ELECTRO·OSMOSIS Elcctro-osmos~ is a method of drainage of cohesive soils in which a direct ament (D.C.) is used. When a direct current is pa<>sed through a saturaU:d soil between a positive electrode (anode) and a negative electrode (cathode), pore water migrales 10 the cathode. The cathode is a well point which collects the water drained from the soil. The water oollected is discharged. as in a conventional weU-point system.

39'

DRAINAGE, DEWATERlNG AND WEllS

The phenomenon of electro-osmosis can be explained witb the belp of the electrical double layer. Cations (positive ions) are formed in pore water when the dissolved minerals go into solution. 1besc cations move tow"ards the negatively charged surface of clay mineral<; to satisfy the elecuical charge. As the water molecules ad as dipoles, the cations also attract the negative end of dipoles. When the cations move to the cattxxlc, they take with them the attached water molecules. In fact, tbe entire outer part of the diffuse double layer which is loosely adsorbed to the

:i1ne.~~~estheg~rshm~pa~o:f u!

PUMP__(-)

layer about 10 A O thick remains attached to the particle. Fig. 16.8 shows an installation in wbich electro-osmosis is used. Anodes are in the fonn of steel rods located neaf the toe of the slope of the ROW OF WEllS RCIN OF WElLS excavation. Cathodes arc in the Conn of (CATHOOES) perforated pipes, resembling well ( CATHODES) points, installed in the soil mass about Fig. 16.8. 4 to 5 m away from the slope of the cut. The electrodes are so arranged that the natural direction of now of water is reversed and is directed away from the excavation. This arrangement is required 10 prevent sloughing of the slopes. In many cases, mere reversing of the direction of flow helps in inaeasing the stability of the slope even if there is no significant decrease in the water contcnt of the soil. The system requires about 20 to .30 amperes of electricity per well at a voltage of 40 to lBO. The consumption of energy is between OS to 10 kWhlm] of soil drained. Because of specialised equipment and high electricity consumption, drainage by electro-osmosis is expensive compared with other methods. 'The method should be used only in exceptional cases when other methods cannot be used. It is normally used 10 drain water in a cohesive soil of low perme.'lbility (k _ 1 x 10-5 to 1 )( 10-3 m/sec). Electro·a;.mosis alsQ helps in increasing the shear strength of the cohesive soil.

16.10. PERMANENT DRAINAGE AFfER CONSTRUCI10N When a usable part of a suuaure Lies below the ground water table, it should be made water proof. Suitable construction techniques should be used. As far as JXlSSible, basement walls and floor should be cast monolithically. If the joints are provided, these should be as few as possible. All joints shall be provided with water stops. Permanent drainage is done by providing foundation drains and blanket drains, as explained below. (0) FOUndatJOD Drains. When ground water flows towards the struaure, provisions are made to G. S. carry the water away from the foundation. Fig. 16.9 shows an arrangement of foundation drain in which perforated drains are used. The drains are BACKFIll surrounded by a filter. The drains should be placed 0B'!l!~"!. _W~T.... _ at an elevation higher than the bottom of the IIA:;DEHT footing to avoid the possibility of carrying away fine soil particles. The water collected in the drain is disposed of by gravity to a nearby storm drain, nullah or any other drainage facility located at a lower level. If Fig. 16.9. Foundalion Drain. no such drain is available, the water is collected in a specially COOSlructcd sump well and pumped out. The drainage system should be provided with some periodical cleaning arrangement.


396

Foundation drains arc eITective when the depth below the water table is not too much. When the water table is very high, suitable interceptor drains are installed at some distance away from the structure to lower the water table in stages. (b) 8lanket Drains. The blanket

::ns(F~. '76~i~ ~n~~~~e ~:::~ consists of coarse sand, gravel and

'~ --. \\

\~~~~~~~~~~~~~~F~LO;OR

crusbed It provides a bighly pervious stones. drainage path. The water coming out of the blanket drain is collected and drained away by gravity. Alternatively, it can be oolJeded in a COARS£ sump pit and then pumped out. Blanket drains are quite effective Fig. 16.10. Blankel Drain. in reducing the uplift pressure on the floor. 'llle possibility of upward seepage flow lhroogh the basement floor is also considerably reduced.

SAND

16.11. DESIGN OF DEWATERING SYSTEMS The design of a dewatering system consists of the determination of the number, size, spacing and penetration of the well points or wells. 'lllese parameters depend upon the expected rate of discharge, the type of the soil and the drawdown in the wells. Collectors and pumps should have sufficient capacity to serve the intended purpose. It is extremely important to establish a fundamental relationship between the discharge and the corresponding drawdown. 1be rale of discharge is computed using Darcy's law. It is assumed that the stratum is homogeneous and isotropic. In case the stratum is anisotropic, it is assumed to have been transformed into an equivalent isotropic stratum using the method di.scussed in Chapter 9. It is furthcr assumed that the flow is continuous and steady. Equations for the rate of discharge and COI1'CSpOOding drawdown for different types of wclls are developed in the following sections. The well may be either gravity well or artesian well. A gravity well penetrates a homogeneous. pervious stratum aquifcr in which the water table is located. An artesian well penetrates a homogeneous, pervious stratum which is bounded by impervious strata above and below and in which the piezometric surface is above the top of the pervious stratum. A combined artesian· gravity type of Dow occurs when the water table is an artesian well falls below the lOp of the pervious stratum. The equations developed may also be used for the detennination of discharge from wells for irrigation and other purposes. However such wells are CODStructed for supplying water and not drainage. In this text, the use of wells for dntinage is of main oono::m. i6.12. DISCHARGE FROM A FULLY PENETRATING

SLOT When the drainage wells are closely spaced in a straight

line, an approximate equation for the discharge can be obtained by considering the line of wells equivalent to a long slot. The flow in the slol may be gravity, artesian or combined. , (a) Gravity Flow. Fig. 16.11 shows a slot in which gravity flow occurs. 100 slot penetrates the pervious stratum in which the water table is located. ' The stratum is bounded by an impervious stratum at its base. The gravity flow is also known

----'S7J!,!':7.::~

-----

as UJtCQnfine..d flow. The equation for the discharge is derived based on Dupuit· Forcbheimcr assumptions (see Chapter 8). According to one ot

ilig.l6.11.GravilyAOW'lnaSiot.

DRAINAGE, DEW..xrERiNG AND WEu.5

397

these assumptions, the hydraulic gradient at any point is taken equal to the tangent of the angle with the boriwntal which the drawdown curve makes at that point. 1l is further assumed that the stratum is isotropic, homogeneous nod Darcy's low is applicable. Lei us oonsider the flow through a vertical section of height z located at a distance x from the slot. From Darcy's law, "'-

q .kiA

q-k·~·('·y)

... (a)

where y is the length of the slot perpendicular to the plane of paper. Trans(X)Sing Eq. (a),

z dz -

fy dx

Integrating and substituting the boundary conditions, x

["2I]"

h -

=0, z =" and x =L. z =fI.

fy [xt;

... (b).

[H'~"']_ ~ q_k(H2~h2).

y

... (16.1)

The drawdown level at a vertical section at a distance x can be ootained from Eq. (b).

I]" - fy ["2,

[xt

/1'-I_~(L_X)

... (16.2)

Substituting the value of q from Eq. (16.1),

II' -I _ 2(L-x) ky

. k(/1'-h') . y 2L

H'-l-¥.(H'-h') As the flow at the

... (16.3)

face of the slot is almost vertical, Dupuit Fordlheimcr assumption is not strictly valid.

The actual drawdown level is given by the following equation

/1'

-1_

¥

[/1'-(h + h,)']

.. .(16.4)

The value of the height hi depends upon the ratios hlH and LllI. (b) Arteslun Flow. Fig. 16.12 shows artesian flow in the siOl.. The artcsian flow is also known ~ confmed flow. In field. sucb a condition occurs when closely spaced wells are installed ncar aod paraUelto the bank of a river such Ihm the pervious stnllum is exposed. 1bc equation for discharge can be derived as in the caseofgmvity flow. However, in this ClSe, the area of Dow is equal 10 (I x y) and not (z x y). where t ~ (be thickness of the pervious smatum. Therefore, Eq. (a) becomes

A.g 16.12. Anesian Row in 8 Slot.


398

q_k(~)(IXY) dz _..!L klY

Integrating and substituting the boundary conditions.

[zt: -

k;y

[xfa

(H - h) _

ty-

(L - 0)

q -

4!-

(H - h)

... (c)

... (16.5)

The drawdown level at a distance x from lhe slot can be obtained from Eq. (c), as [zt -

k;y

[xt

(H - z) -

ty-

(L - x)

,Substituting the value of q from Sq. 16.5.

H _

z _ (H - hi(L -X)

... (16.6)

(c) Artesian-Gravity Flow. Fig. 16.13 shows an artesian gravity type of flow. The flow ncar the slot is gravity flow, whereas that away from the slot is artesian flow. Let Lv be the distance of the point al which Ml1!I'.o.~.

T t

1 Fig. 16.13. Artesian.Gravity Flow in a. Slot

the flow changes from artesian flow to gravity flow. The distance La can be determined by equating the discharge in the gravity flow portion to thaI in the artesian Dow portion. From Eqs. 16.1 and 16.5, k(1- h') y _ kly(H - I) 2LG L -La (L - LG)(i' - h1 - 2LG I(H - I) LG [21 (H - I) •

L

L (1

(i' - h1J _ L(1 _ h')

- h1

G-1.IH_?_1i'

... (16.7)

DRAINAGE, DEWATERING AND WEUS

q.

From Eq. 16.1.

399

k(t' _ h') .Y 2LG

Substituting the value of LG from Eq. 16.7,

q. k(t',-hl y ' (2/11- t'-h') 2L (r - h) kY(21ll _ t' _ h')

q.

... (16.8)

2L

Eq. 16.8 can be used for the detcnnination of the discharge. TIle drawdown levels can be obtained as under.

4) X' • (LG

(i) Gravity Dow portion (x s

From Eq. 16.3,

t' _

- x) (t' _ LG

Ii')

Lc;fl-X') • (LG - x)(t' - h')

0/

LGX'· x(t' - h') + LGh'

z(u) Artesian Dow portion (x

From Eq.• 16.6,

~

vi:;. (p - h

2

)

+ h2

... (16.9)

4)

H - z -

(lI-/)[L -LG-(x-LG)) (L _ Le)

or

... (16.10)

16.J3. DISCHARGE FROM A PARTIALLY PENETRATING SLOT When the thickness of the stratum is too large for fuli penetration of the slot, a partially penetrating slot is used. The flow in a partially penetrating slot can be gr-dVity now or artesian flow. (Il) Gravity Dow. (Fig. 16.14). The discharge qp can be computed ll'Jing the results of model studies conducted by Chapman (1956). SWT G.s·.

r...

~;;=1

1. 1 ~L~l ,,,"

;;;;""""";,, ,")}

l!L , ,

Fig. 16.14. Partially Penetrating Slot (Gravity Row).

q,'

The symbols paper.

are

[

- h) 0.73 + 0.27 (11 11

given in Fig. 16.14. As before,

y is

2L (11 - h)

... (16.11)

the length of the slot perpendicular 10 the plane of

400


The maximum residual head (hD) downstream of the slot is given by

hD - h [

I;! (H -

h) +

11

... (16.12)

Eqs. (16.11) and (16.12) arc valid for the ratio L/H equal to or greater than 3. It may be noted that hD is

greater than h. (b) Artesian now. Fig. 16.15 shows artesian flow in a partially penetrating slot of depth W. The

discha
~

~~=T

given by (IIanon,T

TT

ill

W

H

~L~l

; J,;::;)';;;;; ;:)

>; ;;)')'J;;n·;;> )));on:;;

Fig. 16.15. Partially Penetrating Slot (Atn::sian Flow.)

kry(H-h)

••. (16.13)

qp-~

where E... is the extra-length factor, which depe0d9 upon the rntio WIt and LII (Fig. 16.16)• 0

, .'

:JL t

.

t·o.,~

.

/'

/<-t

-

o -

0

/

"f--

,.

o.oo~

00\

0.,

..

.so

>0

E"'H~

"

Fig- 16.16. Variation or Extra-Length Factor.

The maximum residual head ho on the downstream of tbe slot is given by

hD _ E. (H - h) + h L + E... ~ in the case of gravity flow,the head hD is greater than that at the slot.

... (16.14)

16.14. DISCHARGE IN A SWT FROM DOTll SIDES tn most of the practical cao;es, the flow towards a slot is from both sfdes and not only from one side. The equations for discharge from a slot from both sides are given below.

DRAINAGE, DEWXJ'ERING AND wnll.S

401

(a) Fully Penetrllting Slot. The discharge from a fully penetrating slOl from l.oth sides is twice tile discharge from one side. Equations developed in Sect. 16.12 can be used. For gravity flow, Eq. 16.1 gives the discharge from one side. Therefore,. the disch.'1rgc from both sides is given by ... (J6.J5) For artesian flow, Eq. 16.5 can be used to give the discharge from both sides as

q ., 2ktvri - h)

.. .(16.11l)

(b) Partially Penetrntlng SloL Chapman (1956) gave the following equfltiOns for the discharge of a partially penetrating slot from both sides.

For gravity flow,

qp -

[

0.73 + 0.27

~1""-22 lf L (l-/ - II )

... (16.17)

In case of artesian [low, the discharge is given by the equation, qp -

2kly(N- h) L + }.I

... (16.J8)

where }. is a factor, which depends upon the ratio (WIt), a'> given in Fig. 16. 17, W = depth of slot in the aquifer, and t = thickness of aquifer. All other notations are the same as in Sect. 16.12. 0

,

----

16.15. WELL IIYDRAULICS //' Wells and commonly used for the dewatering of ground water. A well is a circular hole of a V 1 suitable diameter made in an aquifer. As the pumping is done from the well, a cone of O. depression is created aU around the well. The o. equations for discharge can be developed using (.Y!.) / Darcy's law. However. the main usc of wells is to O.6 supply water for drinking, irrigation and other o.1 / purposes. Ground water is an important source of water. It i.. exploited through open wells, tube 9 wells, springs and horizontnl galleries. Ground water is the water that falls as 0 I.S precipitation and then infiltrates the soil below the ).-water table. The ground water reservoir is formed Fig. t6.17. Variation of).. in the voids of the water-bearing strata, known as aquifers. These aquifers act as oonduits for transmission of ground water. As the discharge from wells depends mainly on the permeability of the soil, pumping out tests as discussed below arc frequently used to determine the coefficient of permeability or the soil, as explained in chapter 8. As already expl
,

,

I'

•

/

/

,I r

'"


16.16. TERMS USED IN WELL I1YDRAULlCS Various tcchnical tcnns used in well hydraulics are defined below. (I) Specific Yield. The specific yield (Sy) of an aquifer is defined as the mUo of the volume of water drained by gravity to the total volume of the aquifer. It is usually expressed as a percentage. Thus

S • Volumeofwaterdrainedby.grovity x 100 y Total volume of the aqUifer

Ywy

Sr· V

... (16.19)

x 100

lhe total volume of water in an aquifer depends upon porosity. However, a high value of pom;ily docs not necessarily indicate that the aquifer will yield large volume of water. '1l1e qu.'lntity of water which can be obtained from the aquifer is that which flows under gravity. Therefore, it depends upon the penncability and hence specific yield. The specific yield is always less than porosity. It is sometimes called effective porosity. Specific yield of most of the aquifers, such as sand and gravel, vllries between 15 to 30%. (2) Specific Uetentlon. The specific retention (SR) of an aquifer is the ratio of the water retained in the soil after drainage to the total volume of aquifer. It is also expressed as a percentage. Thus S R

..

Volumeofwatcrretain~

x 100

Total volume of the aqUIfer

Y,,,, SH." V

... (16.20)

x 100

When a saturatc
For a saturated soil, the total volume of waler V... is equal aquifer (Y). Thus V... .. n V BUI

Therefore

10

porosity (n) limes the total volume of the

V... • V...,.. + Vo.R V... V...y + Wo.R n.-y_--y--

n .. Sy + SR ... (16.21) Thus, the porosity of the stratum is equal to the sum of the specific yield and the specific retention. (3) Stornge Coefficient. The storage coefficient is defined as the volume of water released (or stored) by an aquifer per unit surface area per unit change in the com ponent of the head normal to the surface. In an unconfined aquifer, the storage coefficient corresponds to its specific yield. The storage coefficient for a unilthickncss of the aquifer is equal 10 the specific yield, provided gravity drainage is complete. In a confined aquifer, the storage coefficient alSo depends upon the compression of the aquifer and the expansion of the contained water when the pressure is dccre.'lsed during pumping. Typical values of storage coefficient fOf unconfined aquifers range from 0.02 to 0.03 and that for confined aquifers range from 0.00005 to 0.005. The actual values can be obtained from the pumping oul tests. (4) Transmissibility Coefficient. The coefficient of transmissibility (1) is defined as the rate of flow of water through a vertical strip of aquifer of unit width and extending to the fully satur.llion height under oo:t hydraulic gradient. Obviously, the cocITicient of transmissibility in a confined aquifer is equal to the product of the coefficient of permeability and the thickness I of aquifer. Thus T .. k x I ... (16.22) lIS units arc m1Jsec or cm1/scc. The coef.ficient of transmissibility of a welhn an unconfined aquifer is equal to the product of tbe coefficient of permeability and the average saturated thickness t". lous

DRAINAGE, DEWATERING AND WELLS

."

T _ k x

•. (16.23)

10

when t,," (/1 + h)12 where H is the height of the original water table abov..; the impervious stratum and h is the beight of water in the well after drawdown. 16.17. DISCHARGE FROM A FUlLY I'ENETRATING WELL The water table is initially horizontal. When the pumping is stane
? .. qlo~~ /r...)

+ h2

... (16.25)

The drawdown at a radial disl.'lnce r can be computed from the equation

1/' -

i' -

-!k log. (Rlr)

If there are two observation wells at radial distances are hi and h'b respectively, Eq. 16.24 can be written as

... (16.26) rl

and

r2 ,

and if the depths of water in these wells

Jtk(h~ - hi) ... (16.27) q.. log., (r2/rJl Eq. 16.24 can be written in tenns of the effective length of the strainer (L). Let d be the druwdown at thewen.Le.,H-h=d H+h_d+2h Thus ",(II + h)(l1- h) Therefore, qlog. (Rl r.)

q -

",d(d + Zh) log. (Rlr.)

equal to h, ",d(d + 2/.) ... (16.28) log. (Rl r.) (%) Artesian Flow. The artesian flow occurs in a confined aquifer (see Fig. 8.9). Eq. 8.26 developed in chapter 8 can be written as As the effective length L of the strainer in the well is

q -

q _

2~~[~/;.)h)

where, Ie = coefficient of pcnneability. t = thickness of aquifer, /J "" depth of the bme of the aquifer below the piezometric surface, h ::: depth of water in the well, R ::: radius of influence, r ... = radius of well. The elevation z of the draw down curve at a radial distance r from the well is given by

... (16.29)

SOIL MECHANICS AND FOUNDA'10N ENGINEERING

z _

&, Iog..(rlr...) + h

.(16.30)

The drawdown at that distance is given by Il - z _

&, log.. (IUr)

... (1631)

Sometimes, Eq. 16.29 is written in terms of the coefficient of transmissibility (7) and the drawdown (d). Using Eq. 16.22 and taking (Il - h) equal 10 d. 2;tTd ... (1632) q • log. (Rlr.) If h t and 1'2 are the depths of water in two observation wells situated at distances rl and r2' Eq. 16.29 becomos 2nkt(h 2 - hI) ... (16.33) q log., (r:yrt) (3) Artesian.Gra"ity Flow. Fig. 16.18 shows a wen in which artesi..1n·gravity flow occurs. As already explained, the flow is gravity type ncar the wen and artesian type away from the well. P. S.: PIE20METRIC SURFACE

Fig. 16.18. Artesian·Grnvity Flow.

The discharge can be obtained from the following equ.'1tion developed by Musk..'1t (1937), 1tk{2tll _

r _h2)

q. log. (Rl r.) The elevation z at a rndial distance r can be determined from the following equation.

~

z - 1og.,(Rlr...) log.,(rl r ...) +

V',.. - L.!!L 10g.,(Rl r...) log., (Rl r)

... (16.34)

... (1635)

The radial dist.1ncc Ro of the point at which the flow ehanges from artesian type 10 gravity type is given by 10. (R) (il - "')log. (R) + 21 (N - I) log. (r .) ... (16.36) ~ G 2IH_I _ h7 Eqs. 16.34 and 16.35 have been based on the assumption that the head al the well is at the same elevation as the water surface in the well. This is not true if the drawdown is relatively IClrge. For such cases, the head at and in the close vicinity of the well is greater than the water depth in the well.

16.18. mSCllARGE FROM A PARTIALLY I)ENETRATING WELL The discharge from a panial1y penetrating well depends upon the depth of penetration in the aquifer. Like a fully penetrating well, the flow can be gr.avity flow or artesian flow.

405

DRAINAGE, DEWATERING AND WElL')

(a) Gravity Flow. Fig. 16.19 shows a partjaUy penetrating well with gravity flow. The discharge is given

by the foUowing equation.

Fig. 16.19. Partilllly Penetnlting Well (Gl'lIvity Flow).

2."{ k (If']Jr - h)

q, • [ ,

W log.,

(It Wl2r...) +

... (16.37)

0 '] H

where fly/ is the height of the draw down curve at a distance of 211 from the well and W = depth of penetration of Ihe well in Ihe aquifer. The other notations are Ihe same as in the preocding sections. (b) Artesian Flow. Fig. 16.20 shows 3 partially penelrating well with artesian flow. The discharge is given by the following equation.

Fig. 16.20. P;u1i.ally Penetrating Wet! (Artesilln Flow).

2xkJ(H - h)C ... (16.38) qp log.. (Rlr... ) wbere C is the correction foetor for partial penetration. It is equal to the ratio of the discharge from the partially penetrating well to tbat for a fully penetrating well for the same drawdown.

The value of C can be obtained from the follow ing equation given by Kozeny (1933).

C •

.'!'(

(1 + 7

~ cos • W 2W 2t

x 180 ) It

... 16.39)

where W = depth of penetrations in the confined aquifer. The other notations are tbe same as in tbe preceding sections. 16.19. INTERFERENCE AMONG WELLS When two or more wells are located close to one another. their drnwdown curves intersea within their zones of influence. Consequenlly, the discharge in the individual well is deacascd. The phenomenon is known as interference among wells. The discharge from an individual well depends upon the spacing of tbe

wells and their oriemation.


(a) Artesian Wells. (.) Two wells. Fig. 16.21 shows two artesian wells having a spacing of B. The dOUed lines [curves (1) and (2)] show their individual cones of influences. 'Ilte combined cone of influence is shown by the curve-3. The discharge through each weU is given by

(1) Drawdown for we-II (2) Orawdown for wrll no. 2

(3)Composite drawdown

fig. 1621. Inrc:rfc:n::ncc llmong Wells.

2Jtk1(H - h)

q.

(

Fi' )

... (16.40)

lo~ Br""

where R is the radius of i~fluence (R > > B). All other notations are the same m before.

(i.) Three wells in the same line. If there are three identical wells in the same line having a spacing of B, tbe discharge through the two end wells is given by 2dl (11 - h) log. (Blr.) The discharge through the middle weU is given by 2dl(H - h) log. (BIZr.)

... (16.42)

(ii,) Three wells ronning a lriangle. If there are three identical wells located at tbe apexes equilatcml triangle, with side B, the discharge from each well is given by

q • 2. k1 (11 - h) log. (R'IB'r.)

~[

an

... (16.43)

(b) GravIty Wells. The discharge from gravity wells caD be obtained by using the equatiOflS given above [or the artesian wells and making the following substitution : (/-Pl2t) for 11 and (h 2/2J) for h. For example, the discharge from ench well when the two wells are spaced B apart is obtained from Eq. 16.40 as q. 2xkJ(H2/2J _ h2l2t) 10g.(Fi' /Dr.) xlc(H2 _ hZ)

q - log..(R2/r""B)

.. . (16.44)


''''

16.20. SI'IlERlCAL FLOW IN A WELL The equations developed in the preceding sections arc for radial flow. Fig. 16.22 shows a well which penetrates up(Q the top surface of the confmed aquifer. 1be flow in the well is spherical. In this case, the ,...,.,~_ _~G~.~

--=.:=.,:.::..=- --- ---

--- ~=::--':~:------ T

/-

I

"

,}

I

1

I};

Fig. 16.22. SphcriOit ROII/.

equations given in the preceding secLions do not apply a'i the Oow is not radial and the Dupuit assumption is not vaUd. The discharge in the case of spherical flow is given by ... (16.45) q - Z.b.(I1- h) where r", is the radius of the well. The discharge in case of spherical flow is mudl less as rompared 10 lhal in a fully penetrating well. As the weUs with spherical flow arc not very effective, thcsc arc rarely used in practice.

16.21. DISCllARGE FROM AN OPEN WELL An open well is a.vertical hole of a large diameter 2 10 10 m. It is of shallow depth, and il draws water only from one pervious stratum. 'fbe well penetrates to such a depth below water table that there is at least a water depth of 3 to 4 m even in a dry season. Generally, the depth of an open well is limited to 30 m. The sides of the well may be lined with bricks or stones. In case of an unlined well, the discharge is from the bottom and sides. However, in the case of a lined well, it is mainly from the bottom of the well. 1be discharge capacity of an open well is generally limited to 0.004 aJmCQi, because the well can be excavated to a limited depth. Moreover. the water can be withdrawn only at a small velocity. If the velocity is greater than th-:- critical velocity, the soil grains may be dislodged and subsidence may occur. Open wells may be dug, bored, drilled or driven. Large diameter open wells are generally dug manually or mechanically. Safe yield of an open well is the rate of flow 31 which the water percolates into the well under safe maximum worldng head. As the water is pumped out, the water level in the well falls. The difference between tbe water level in the well and the original water table is the depression head. The critical depression head occurs when the velocity is so high as to dislodge the

soil particles. The safe maximum

WOrk.

I

MV;C"" 1--=-":-":---

ing bead is usually ; taken as ooe- third of the critical depression head. The yield from an open well is determined by the H follOWing two methods: (1) Pumpmg test at constant level, (2) ReaJperalion test. 1. Constant level Pumping tesL In this method, the water level in the well is depressed by pumping. out water till the maximum safe depression head is reached. The mte of pumping is so adjusted that the water level in

WIO

............

....

-------i£J
______

Fig. 16.23. Pumping TCSI.

/'

SOIL MECUANICS AND FOUNDATION ENGINEERING

the well remains constant (Fig 16.23). AI that stage. the rate of withdrawal is equal to the pcroolalion rate. TI1C rate of pumping per hour gives the yield per hour of the well al a particular drawdown. The discharge is generally expressed as

q-Kxlf

q .. (KIA) x A x /I

... (16.46)

3

where q = discharge (m /hr). K = Constant, depending upon the soil, A = Cross-sectional area of the cavity of flow at thc well bottom (m 2). It is generally taken a<> 413 times the actual cross-sectional area of the well bOllom. fI depressioo head. l h c ratio KIA is the specific yicld of the open well (m)/hr) per m2 of the area through which water percolates under I m of depression head. {The value of KJA given by Mamol is equal to 0.25 for clay, 05 for fine sand and 1.0 for coarse sand. J

[Note. The specific yield of lube well is not the same as Ihat of an open well.} 2. Recuperution Test. In a recuperation test, the water level is depressed to a level below the normal level by pumping. As the pumping is stopped. water level in the well stnrts rising. The time required for the water levc\ to rccupcrnte is noted. lhe equation for tnc discharge is derived below. Let A-A be the water level before the pumping and 8-8 be Ihe water level when the pumping is Slopped (Fig. 16.24). LeI fit be the depression head when the pumping is stopped, and 11 be the depressioo head at a time t aOcr the pumping is stopped. lei 112 be the

.5 .

Pig. 16.24. RccupcrdtioJl Test.

depression head at a time T ufter the pumping is StOpped. Volume of water entering the welt when the head recuperates by dJ I is given by dV.A till

... (a) where A = Cross-sectional area of the well bottom. The volume of water entering the well in time tit during which head rccupcr.ltCS by dl/ is also given by

dV - qd.

...(b)

where q = rate of discharge in the well N; the mte of dischnrge depends upon the depression hend, it C.1n a1<>o be wrillen as

q • K· II dV. KHdt

Thus From Eqs. (a) and (c).

... (0)

Klldt.-AdH lhe minus sign indicates thnt the hend decreases us time Integrating Eq. (d) between the limits (t

~{ A

~ o.

dt __

0

= 0, II

/!!2 ~II

= 1-1\ and

... (d)

increases.

t t

= T, 1-1 - lJ.iJ ,

11-1\

l' - fiog.IJJ!% - log. (11,111,)

~

.t

log.. (H I I1l2)

... (16.47)

DRAINAGE, DEWATERING Arm WEUS

Knowing the values of Ill> 1i2 and 1~ the value of the specific yield (KIA) can be romputed. Once tbe value of K/A has been detenninoo, the discharge can be obtained (rom Eq. 16.46 foc any other depression head Ii as

q - (KIA) )( A )( H Although recupcmtion test is not as reliable as the oonstant level pumping method, it is used when it is difficult to regulate the ffite of pumping 10 attain a constant water level in the well required in the constant level pumping Icst. .

J6.22. ADVERSE EFFECTS OF DRAINAGE As the water is drained oul from the soil. the neutral slreSS decreases and the effective stress is increased. '£he increased effective stress may cause consolidation settlement. Therefore. tbe structures already existing in the zone of influence may experience undesirable settlement. 'Ibis undesirable effect can be: minimized by artificial recharge i.e. by pumping water into the ground near the existing structures through a system of well points. Thus a constant ground water level is maintained nc.'l.r the structures. Another adverse effect of the drainage is that cavities are forme
ILLUSTRATIVE EXAMPLES Illustrative Example J6.1. A slot is made in an uncon[uted aquifer to drain water: The flow to the slot occurs from both sides. If the water table is at a height of 12 m above the base and the drawdown is 4 m, find the dischflrge per metre length, assuming that die distance of the slot from both sides is 100 m. Take k _ 5 x lO"""mlscc.

k (IP~ h

2

Solution. From Eq. 16.15.

q _

)

•

Y

2 ... 5 )( 10-' (122 _ 8 )

----roo--)(

1 - 4 )( 10-< m', / sec

Illustrative Exumple 16.2. A welf flllly penetrates an unconfined aquifer hQl'ing a saturated thickness of 10 m. If die radius of the well is 10 cm, and the drawdown is 3 In, determine die dischflrge. Take the radius of influence as 300 m and the coefficient of permeability as 8 x 10""" mlsec.


q -

1tkVP_h2) log., (Rl r...)

or

U1ustnltive Example 16.3, Calculate the discharge from a fully- penetrating tube well, having the following particlIlar : (I) Thickness of confuted aquifer = 25 m (Ii) Tube well diameter = 30 em (iil) Drawdown =3m (iv) Radius of influence = 150 m (v) CoefJicient of permeability :: 30 m/day Solution. From Eq. 16.29.

q _

2~~[Z/~.)h) 21t )( 30 )( 25 )( (3) '"8< (ISO/O.I5)


410

.. 21)46.5 ml/day .. 0.0237 m J lsec.

llIustratlve Example 16.4. A tube well is driven in a confined aquifer of 25 m thickness located 20 m below the ground surface. The water table is 15 m below the ground surface. If /he discharge of the well is 0.05 nr/sec when the depression head is 10 m, find the diameter of the tube well. Take the radius of influence as 300 m and k _ 3 )( to-4 m/sec.

q.

2~: )Z/~.)h)

10g.(RI,.) •

2"kl~ - h)

Soludon. From Eq. 16.29.

1og,;(Rl r...) .. 2n x 3 )(

~~;

j(

25 x 10 .. 9.42

RI T.., .. 12332.58 or

T.., ..

300/ 12332.58 .. 0.0243 m

illustrative Example 16.5. A fully penetrating well draws water from a confined aquifer of thickness 12 m and o{ permeability 1.5 x 10-J mlsec. If the discharge is 0.03 mJ/sec, compute the drawdown at 30 m from the centre 0/ the well. Take RI T.., = 2000. SoIuUon. Eq. 16.31,

1/ - z .. ~ log., (Rl r ...)

2" x

1.5°~31O" x

12 log. (2000) • 2.017 m

illustrative Example J6.6. Fig. £ · 16.6 shows an artesian-gravity well with a rodi~ of influence of 300 m. If the diameter of dIe well is 30 em, compute the disc/large. Take k. = 5 x leT" nt/sec.

Fig. E-l6.6.

Solution. From Eq. 1634, or

q. "k(21H i' _ h2 ) log. (RI,.) _ K )( 5 )( 10'" (2 )( 12 )( 15 - 122 -

q

log. (300/0.15)

g2) _ 00279 •

3/

m sec.

Dlustratlve Example 16.7. A well 25 em diameter partially ~etrate.s a confined aquifer by 9 m. 1/ the totallhicJcness of the aquifer is 20 m aNI the disduJrge is 0.04 mJ/sec, compute drawdown at the well. Take = 200 m, and k =4 x 10-; m/see.

R


qp _ 2 X~i7;;T~) C

where C is given by Eq. 16.39 as


411

c • ~t (I

+

• .2..(1 + 20

Therefore;

7~ oos x W • 180) 2W 2t :rt

7~ cos:rt 9 x 180) .. 0.649. 2x9 2x20xx X

2x x 4)( to....3)( 20 (H h»)C 0.649 q, • log. (200/0.125)

H _ h .. 0.04 )( 7.378 .. 0.905 m 0.326 Dlustratlve Example 16.8. Two tube wells fully penetrate 12 m thick confuted aquifer and are 150 m apart. The wells have following particulars : (,) Diameter = JOcm (ii) Depression head = 4.5 m (ii,) Rtufjus of influence z 300 m (iv) Coefficient of permeability = 1.5 x mlsec. Detennine the discharge from Cllch well and compare it with for no interference. 2dt(H - h) SolutlM. From Eq. 16.40, q. log.(R'/Br.)

lrr

2:rt)( 1.5 )( 10-3)( 12)C 4.50 .. 0.0613m)/sec 1og.(3OO'/150 x 0.15) For no interference. Eq. 16.29 gives

q.

2xkt(H-h) log. (R/r.) 2" )C 1.5 )( 10-3 )( 12 )C 4.50 log. (300/0.15)

OM

00669 )/ . m sec

Percentage reduction is discharge

•

(0.~.~~0613) x

100 • 8.37%

lllustl'1llive Example 16.9. Determine the discharge from a well with spheriaJI flow and having the following partiaJlars : (i) ThicJou!3S of confmed ~uifer = 10 m

= 100 m

(il) Radius 0/ influe"ce (iii) Radius 0/ well

= 0.1 m '"' 1 )( 10-4 m/sec (iv) Coefficient of permeability (v) Drawdown =J m What would have been the discharge if the well bad penetrated the confined aquifer fully ? Solution, From Eq. 16.45,

q. 2 x kr. (H - h)

q .. 2n)( to.... )( (0.1 )( 3) .. 1.88 x 10.... m)/sec If the well had penetrated fully, the discharge would have been given by Eq. 16.29 as, 2nkt (H - h)

or

q • log. (100/0.1) q ..

2lt )( l~ )( to )( 3 ~(lOO/O.I)

(f"" ) .. 1:1.29 x 1 m /sec

The ratio of the discharge in spherical flow to that In a fully penetrating well is given by


412

r ..

i.e.

2;~7

,..

ll.s .

7%

the discharge in spherical flow is about 7% of that in radial flow.

llIustrative Example 16.10. lksiq: an open well in coarse sand (specific yield = 1.0 mJ' hrfml under unil drallf.'down) for a yield of 0.003 III /5, when operated under a depression head of 3 m. Solution. From Eq. 16.~ q .. (K/~) )C A x 11 0.003)( 60 )( 60 .. 1.0 )C A. x 3 or A .. 3.6m If d..., is the diameter of the well.

i ..

(11:/4) x (d.. 3.60 or d... .. 2.14 m illustrative Example 16.11. During a recuperation test, the water level in an open well was depressed by 2.4 m which recuperated by an amount of 1.5 m in 6() minutes. Determine the yield from the well of 3 III diameter under, a depression head of 3.0 m.

t ~ .. +


~

From Eq. 16.46,

q .. (KIA) x A x H

..

log,. (HI I 1/,)

log.. (2.40/ 0.90) .. 0.98\

.. 0.981 )( xI4 )(

(3f )(

3.0 .. 20.80 m"/ hour

PROBLEMS

A Numerical 16.1. A slOi is made in a confined aquifer 2 m thick to drain water. 111e flow to the SIOl ou:urs from both sides. If the Wllier U1ble is III II height of 10 m above the base and the dmwdown is 3 m, find the dischorge per metre length. The disUlnce of the slot from both sides is 120 m. 'HIke k = 4 )( 10-4 mJscc. fAns. 0.04 lit/Scc] 16.2. A 30 em di~ler well penetrates 20 m bcloYl the waler table. The draw down al 100 m is 0 .50 m and that at 30 m is 1.10 m. If the discharge is 100 litJsec, determine the transmissibility of the aquifer. 2 [An.... 0.032 m /sec.\ 16.3. A well penetrates an unoonfined aquifer having a saturated depth of 90 m. When the drawdown is 10 m, the discharge is 4 lit/sec. Determine the discharge when the drawdown is 15 m. Take the radius of influence the same in both the cases. [Ans.. 5.82 lit/sec] 16.4. A well is sunk through a layer of sand of thickness 12 m lying over an impervious stratum. When the water was 3 pumped at a constant rate of 5 m /minute, the Wllter levels in the observation wells situDled at a distance of 15 m and 30 m from the pump well wen: 2.9 m and 2.7 m below the ground Wllter table, respectively. Find the permeability of the sand. fAns.. 5 mmJscc] 16.5. Find the coefficient of permcubllily of an unoonfined aquifer of thickness 18 m when a well of diameter 20 ern discharges 50 lit/sec under a depression head of 45 rn. Take the radius of influence as 300 m. [Ails.. 9 x 10-4 mJsec] 16.6. calculate the discbDrge through II fully penetrating tube wcll of the following ptlrticulnfS : (I) Thickness of confined aquifer (il) Diameter of the well (iiI) Drnwdown

(iv) Coefficient of permeability (v) Radius of influence

= 25m

= 25 an =3 m = 0.29 x 10-3 m/sec. = 200 m

lAns. 18.51 lit!sec.j 16.1. A fully penetrating well of diameter 30 em draws water from a confined aquifer of permeability 2 )( to"3 m/sec nnd thickness 12 m. If the steady disc;:harge is 0.03 cumecs, compute the drowdown at 10 m from the centre of the well. Take radius of infl uence as 500 m. tAns. 0.178 m]

DRAINAGE!, DEWATERING AND WEUS

413

16.8. A fUlly penetrnting well of diameter 30 cm draws water from a 25 m Ihick mnfined aquifer. The steady state drawdown at 10 m and 50 m were observed 10 25 m and 0.50 rcspea ively. D:tennine the steady-stage 3 [Ans. 29.261il/sCC] discharge. Thke k = 1.5 x 10- m/SOC. 16.9. Design an open well in a fine sandy soil (specific yield = 05 m3/OOur/m1 under unit depression hr.ad) to yield 0.004 cumecs under a depression head of 3 m. [Ans. 3.5 m dial 16.10. calculate Ihe overage yield of an open well of 3 m diameter from the recupeltltion test wherein the water level is depressed to the extent of 2.0 m and recuperation rate is 1.0 m per hour. 1be allowable depression head in the well in 3 m. [Ans. 14.69 mlftlt]

8. Descriptive and Objective Type 16.11. What do you know about drainage of soils? What are its uses ? What are its ill effects? 16.1Z. Explain the working of a single-stage well point system. What are its limitation? lZ.13. Write shen notes on: (0) Muhi-SUIge well point system. . (b) v.x:uum well points. (e) Deep well system. 16.14. Wh31 is clectr0-081Jl(l';is? Whlu arc its advantages and disadvantages as compared with the oonventional drainage systems? 16_15. What is the function of pcrmnnenl drainage systems installed after construction? Describe in brief the foundation drains and blanket drains. 16.16. Differentiate between (a) Asi()(anda welJ (b) Omvity now and anesian now (e) Fully pcnctmting and panially pcnctrnting slots. (d) Coefficient of pcnneabilily and tmnsmissibility coefficient. (eo) I\n open well and a tube well (/) Specific yield and specific retention. 16.17. Derive the expression for discharge from a fully. penetrating slot when the now is (a) gravity flow (b) ancsian now (c) anesian-gravity now. l6_18. Derive the expression for discharge from 9 fully-pcnetmting well in (a) a ronfined aquifer (b) an unconfined aquifer. 16.19. What do you understand by spheri cal now? Why this type of flow is not used in practice? 16.Z0. What is mutWlI intcrfe·rence of wells? How can this be avoided? 16.21. How would you detennine the yield of an open well? 16.2Z. Which of the following statements are lrue? (a) FJectro-osmosis is gL'fleroJly used for coarse-grained soils. (b) The diSChnrge from a slOl is doubled if the now can occur from both sides. (e) Tn a confined aquifer, when the well fully penetrates the aquifer, the now is artesian. (d) A double-stage well point can be for lowering the water table by 10 m. (e) Foundation drains are more effective than blanket drains if the depth below the water table is not mucb. (/) An open well gives more discharge than a tube well. (Am. Tru', (b), (el. (d), (')J

C. Multiple Choice Question

414 (b) 10 m

(l)5 In

(tf) 20 m

(el 15 m

4.

~~eC~~~:C{'Jr

tronsmissibililY

COCmC~~~I:;~/SCC

(d cmJ/scc (d) None of above S. The mdius of influence of the gmvily wells is generally assumed as (0) 100 m (h) 300 m (e) 1000 m (Ii) 50 m 6. The specific yield of an open well in c{)
{al 1.00 II;'

(b) 0.5 11;1

(e) 0.2511;1 (d) 5.0 h;' 7. Select the incorrect statement (a) The confined aquilcr is bounded 8t lOp and bouum by impervious st ..lia. (b) The pressure of water in the confined ll
PART-II

EARTH-RETAINING STRUCTURES AND

FOUNDATION ENGINEERING

17 Site Investigations 17.1. INTRODUenON Site investigations or subsurface explorations are done for obtaining the infonnalion about subsurface conditions at the site of proposed construction. Site investigations in one form or the other is generally required for every big engineering project. Information about the surface and sub-surface features is essential for the design of structures and for planning construction tcchniques. Site investigations consist of determining the profile of the natural soil deposits at the site, laking the soil samples and detcnniniog the engineering propenies of the soils. It also includes in-silu tcsting of the soils. Site investigations arc general1y done to obtain the information that is useful for one or morc of the

following purposes. (I) To select the type and depth of foundation for 11 given structure.

(2) (3) (4) (5) (6) (7) (8) (9)

To determine the bearing capacity of the soil. To estimate the probable maximum and differential settlements. To establish the grouncl water level and to determine the properties of water. To predict the i..1tcral earth pressure against retaining walls and abutments. To select suitable construction techniques. To predict and to solve potential foundation problems. 1b ascertain the suitability of the soil as a construction material. To investigate the safety of the existing structures and to suggest the remedial measures.

The relevant information is obtained by drilling holcs, taking the soil samples and determining the index and engineering properties of the soil. In-situ tests are alw conducted to determine the properties of the soils in natural cooditiollS. This chapter discusses various methods of sub-surface explorations and in-situ testing.

416


17.2. PLANNING A SUBSURFACE EXI'LORATION PROGRAMME A sub-surface exploration programme depends upon the type of the structure to be built and upon the variability of the strata al the proposed site. lbe extent of sub-surface exploration is closely related to the relative cost of the investigations and thai of the entire project for which it is undcnakcn. In general. the more detailed the investigations arc done, the more is known about the sub-surface oonditions. As a result, the grc.1ler economy can be achicved in (he ronslruction of the project because the element of uncertainty is considerably reduced. However. a limit is rt.'lchcd when the rust of investigations ootweighs any saving in the cost of the projcd. lind it increases the overall cost. It would 001 be economical to have investigations beyond thallimil. The extent of invcstigations would also depend upon the location of the project. A small h'"IUse in an already built-up area would DOt n:quire muCh exploration. On the other hand, if the house is to be built in a newly developed area, a detailed investigation would be required to ascertain the location of different soil stmta and their physical characteristics. If a multi-storeyed building is to be constructed, extensive sub-surface explorations would be necessary. These buildings impose very hc.wy loads and the zone of inOuence is also very doep. It would, therefore, be morc desirable 10 invest some amount on sub-surface exploration than 10 overdesign the building and make it costlier. Planning of a sub-surface explorntion programme is a difficult task. I3esides a thorough knowledge of soil engineering, it requires experience and engineering judgment. Sometimes, the exploration programme bas to be changed as the investigations progress. As the variability of the soil strota is found to ina-case, the extent of investigations is also increased. On the other hand, if the site is found to be underlain by uniform deposits, the extent of investigations is decreased. In general, the aim of the investigations should be to get the maximum information that is useful in the design and construction of the project at a minimum C05I. The cost of site investigations geocr-dlly varies between 0.05 to 0.2% of the total cost of the entire structure. In some unusual conditions, the COSt may be evcn uplO 1%. 17.3. STAGES IN SUB-SUIU'ACE EXPLORATIONS Sub-surface explorations arc generally carried out in three stages: (I) Reconnaissance_ Site reconnaissance is the first step in a sub-surface explorntion programme. It includes a visit to the site and to study the maps and otber relevant records. Il helps in deciding- future programme of site investigations, scope of work, methods of exploration to be adopted, types of samples to be taken and the laboratory testing and in-s itu testing. (2) Preliminary Explorutlono;. The aim of Q preliminary exploration is to determine the depth, thickness, extent and composition of each soil slrnlum at the site. The depth of the bed rock and the ground water table is also delennined. The preliminary explorations are genernlly in the fonn of a few borings or test pits. Tests are conduacd with cone penetrometers and sounding rods to obtain infonnation about the strength and compressibility of soils. Gcophysical methods are also used in preliminary explorations for locating the boundaries of diITerent strata. (3) Detailed Explorations. The purpose of the detailed explorations is to detenninc the I!ngineering properties of the soils in different slnlta. It includes an extensive boring programme, sampling and testing of the samples in a laboratory. field lests, such as vane shear tests, plate load tests and permeability tests, are conducted to determine the properties of the soils in natural slate. The tests for the determination of dynamic propenics are also cmried out, if required, For complex projects inVOlving heavy structures, such as bridges, dams, multi-storey buildings, it is esscnt.ial to have dClaiJed explorations. Ilowcver. for small projects, especially at sites where tbe strata are unifonn, detailed investigations may not be require. The design of such projects is genemUy based on the data collected during reCQnn.'lissance and preliminary explorations.

SI"l"B INVESfIGATIONS

417

17.4. RECONNAISSANCE The geotechnical engineer makes a visit to the site for a careful visual inspection in reconnaissance. The infonnation about the following features is obtained in reconnaissance. (1) The general to(X>gr3phy of the the site, the existence of drainage ditches and dumps of debris and sanitary fiUs. (2) Existence of settlement aacks in the structure already buill near the site. (3) The evidence of land slides, creep of slopes and the shrinkage cracks. (4) The stratification of soils as observed from deep cuts ncar the sitc. (S) The location of high flocxl maries on the nearby building and bridges. (6) The depth of ground water table as observed in the wells. (l) Existence of springs. swamps, etc. al the site. (8) The drainage pattem existing at the site. (9) Type of vegetation existing at the site. Tbc type of vegetation gives a clue to the nature of the soil. (to) Existence of underground water mains, power conduit, etc. at the site. In addition to making site visits, the geotechnical engineer should study geological maps, aerial photographs, toposhcet. soil maps and the blue prints of the existing buildings. Maps and publications of various agenCies give a lot of infonnalion about the geologic character of the area. The gcotcdmical engineer should also get infoonation about the type of structure to be built and its proposed usc. In the case of a multi.storeyed building. the infonnation aboUt the column loads and their approximate locations should be obtained. In the case of bridges, the span length and the Joad carned by tbe piers and abutments should be a'iCenained. In the case of a dam, the gcotechnical engineer should gel informalion about the type of the dnm, its height, base width and other salient characteristics. The infonnation obtained during reconnaissance is helpful in evolving a suitable sub-surface investigation programme.

17.5. DEJYl11 OF EXl'LORAll0N The depth of exploration required at a partiadar site dependS upon the degree of variation of the subsurface data in the horizontal and vertical directions. It is not possible to fIX the number, disposition and depth of borings without making a few preliminary borings or soundings at the site. The geotcctmical engineer baving a long experience and good engineering judgment may give some guidelines. The depth of exploration is governed by the depth or the influence zone. The depth of the influence zone depends upon the type of the structure, intensity of loading, shape and disposition of the loaded area, the soil profile, and the J*lysica1 characteristics of the soil. 1be depth uplO which the stress inacmcnl due to

b GS

J~~

BOREHOLE~ Fig. 17.1. Oeplh or Exploration.

T

\·58

j

418


superimposed loads can produce signincnnt settlement and shear stresses is known as the significant depth. 'Inc depth of exploration should be at least equal to the significant depth. The significant depth can be dctcnnincd using the method discussed in chapter 11. The significant depth is generally taken as the depth al which the vertical stress is 20% of the load intensity. According to the above criterion, the depth of exploration should be about 1.5 times the width of the square footing (Fig. 17.1) and about

Fig. 17.2. Depth

or Exploration for Closely-spaced Footings.

3.0 times the with of the strip footing. However, jf the footings are closely spaced, the whole of the loaded area acts as: a raft foundation. In thai case. the depth of boring should be alleast 1.5 limes the width of the entire loaded area (Fig. 17.2). In the case of pile foundation, the depth of explornqon below the tip of bearing piles is kept al least 1.5 times the width of the pile group. However, in the case of friction pilcs. the depth of exploration is taken 1.5 times the width of the pile group measured from the lower third point (Fig. 17.3). It is more logical 10 relate the increase in stress to the in-situ stress. The depth of exploration is usually taken upto the level at which the increase in stress is 1!20th of the in-situ stress before the SORE appllcation of the load. HOLE 0 When the foundations are taken up to rock. it should be ensured that large boulder.; are DOt miS{akcn as bed rock. The minimum depth of core boring into the bed rock should be 3m to establish il as a rock. Ln ca<>e of multi-storeyed buildings, the depth of exploration can be taken from the following formula (Sowers and Sowers, 1970),

r-

71

,·59 t---9--1

TIl

D - qsi'"' ... (17.1) where D = depth of exploration (m). C = constant, Fig. 173. Depth of Explornlion for Friction Piles. equal to 3 for light steel buildings and narrow concrete buildings. It is equal to 6 for heavy stccl buildings and wide concrete buildings. S = number of storeys. If loose soil or reccnLly deJXlSited soil or a weak stratum is encountered, it should be explored thoroughly. Explorations should be carried to a depth at which the nct increase in the vertical stress is less than tbe aJowable bearing pressure of the soil. For two adjacent footings. each of size B )( L, spaced at a clear spacing A, IS: 1892-1972 suggests thai the minimum depth of boring should be 1.5 B wben A :t 4B; and it should be 1.5 L when A < 28. Faadjacent rows of such footings, the minimum recommended depth of exploration is 4.5B when A < 28; it is 3.5 8 when A> 2D :md it is 1.5 B when A ~ 48.

SITE INVESfIGAll0NS

419

For explorations of deep excavations, the depth of exploration below the proposed excavation level should be at Ic.1st 1.5 times the depth of excavation. In case of road cuts, it is taken at lcac;t equal to the width of the cut. In case of road fiUs, the minimum depth of boring is 2m below tbe ground surface or equal to the height of the fill, whichever is greater. . In case of gravity dams. the minimum depth of boring is twice the height of the dam . • 17.6. \ATERAL EXTENT OF EXPLORATIONS .The lateral extent of exploration and the spating of bore holes depend mainly on the variation of the strata in the horizontal direction. The cxplordtion should bc cxtensive so as to reveal major changes in the propertics of the sub~surface strata. For small and less important buildings, even onc bore hole or a trial pit in the centre may suffice. But for compact buildings, covering an area of about 0.4 hectares, there should be at least 5 bore holes, one at the centre and four near the comers (Fig. 17.4).

,..-----&-----,

i

! I

i i

o C'7-----f':!) Fig. 17.4.

~ I I

' Jl __ _ Fig. 17.5.

For large, multi-storeyed buildings, the bore holes should be drilled at all the comers and also al important loc.1tions. The spacing between the bore holes is generally kept between 10 to 30 m, depending upon thc variation in the subsurface conditions and loading (Fig. 17.5.). For highways, subsurface explorations are usually carried out along the proposed ceotre line or along the propose ditch line. The spacing of bore holes usually varies betwccn 150 and 300 m. If the sub~st ra[ a is erratic, the spacing may be reduced to even 30m. In case of concrete dams, the spacing of bore holes generally varies between 40 and 80 m. 17.7. OPEN EXCAVATION M~"TIIODS OF EXPLORATION In Ibis method of exploration, an open excavation is made 10 inspect the sub~strala. Tbe methods can be divided into two categories: (1) Pits and Trenches, (2) Drifts and Shafis. (1) Pits and Trenches. Pits and trenches are excavated at the site to inspect the strata. TIle size of the pit should be suCficient to provide necessary working space. IS : 4453-1967 recommends a clear working space of 1.2 m x 1.2 m at the boltom of the pit. The depth of the pit depends upon the requirement of the investigation as already discussed. Shallow pits up to a depth of 3 m can be made without providing any lateral support. For deeper pits, especially below the ground water table, the lateral support in the form of shccting and bracing system (Fig. 17.6) is required. I\s the depth of the pit increases, its cost increases rapidly. For depths greater than 6 m, bore holes arc more economical than open pits.

4'"


Deep pits should be properly ventilated to prl.."vCDt accumulation of dead air. If water is encountered in a pit, it should be suitably dcwatered. GS Trenches are long shallow pits. As a trench is continuous over a considerable length, it provides exposure a long a line. 1he trenches are more suitable STUD FRAME\ than pits for exploration on slopes. Test pits and trenches can be excavated manually or tical 2·0 ~~~inq SOmm mechanically. Adequate precautions should be taken thie against possible accidents due to caving of the ground. (2) Drif'ls lind Shufts. Drifts are horizontal tunnels made in the hill-side to detennine the nature and structure of tile geological formation. IS: 4453--1980 recommends that a drift should have the minimum clear dimensions of 1.5 m width and 2.0 m height in hard 2·0 rock. In soft rock, an arch roof is more advantageous Ihan a flat roof. If the ground is unable to stand of its own, supports have to be provided to carry the load of TOM OF r~OT PIT the roof and the sides of the drift. Drifts arc useful for establishing the minimum SECTION excavation limits to reach sound rock and for locating faullS and shear zones and buried channels in the river section. However, drills are generally expensive. These are used only wben other methods do not provide the required infonnation. Drifts are also known as adits. Shafts are large size vertical holes made in the geological formation. These may be rectangular or circular in section . The minimum width of a rectangular shaft is 2.4 m and for a circular shaft, the minimum Fig. 17.6. diameter is 2.4 m. In weak ground, the sides of the shaft should be properly supported. Deep shafis should be properly ventilated. Shafts are used to reach a particular strata at a depth of 4 m or more. Shafis are also used to extend the exploration below the river bed already done by means of tunnels.

rm

t

m!

!

1

D

17.8. BORINGS FOR EXPLORATION When the depth of exploration is large, borings are used for exploration. A vertical bore hole is drilled in the ground to get the information about the sub-soil strota. Samplcs are taken from the bore hole and tested in a laboratory. The bore hole may be used for conducting in-situ tests and for locating the water table. Extensometers or pressure meter may also be installed in the bore hole for the measurement of deformatioo in the sub-Slrata. Depending upon the type of soil and the purpose of boring, the following methods are used for drilling the holes. (2) Wash Boring (3) RotaI)' Drilling (1) Auger Boring (5) Core Boring. (4) Percussion Drilling A few holes are drilled during the preliminary investigation. In the detailed investigations, a large number of holes are drilled to thoroughly investigate the sub-soil strata. The rcsuHs of boring arc presented in the form of boring-log and sub-surface profiles (Sect. 17.22). 17.9. AUGER BORING An auger is a boring tool similar to one used by a carpentor for boring boles in wood. It consists of a

SITE INvEsnOATIONS

shank with a cross-wise handle for turning and having central tapered feed screw [Fig. 17.7 (a)1. TIle augers can be operated' manually or mcd:lanically. The hand augers used in boring are about 15 to 20 em in diameter. 1besc are suitable for advancing holes UplO a depth of 3 to 6 m in soft soils. The hand auger is attached to the lower end

T

421

iT

of a pipe of about IS' mm diameter. The pipe is provided with a eross-arm al its top. The hole is advanced by turning the crossarm manually and al the same time applying thrust in the downward direction. When the auger is filled with soil, it is taken out. If the hole is already driven, another type of auger, known &'> post-hole auger [Fig. 17.7 (b)} is used for taking soil samples. Mechanical augers are driven by power. 1bese are used for maklng holes in hard strata to a great depth. However, for depths greater than 12 m, even mechanical augers become inconventicnt and other methods of boring are used. Continuous rughl augers are special type of mechanical augers which arc provided with a central hollow tube. When the lol Ib) bole is advanced, the central tube is kept plugged. As the auger is turned into the ground, the cuttings rise to the surface through the Fig. 11.1. Augers. spiral. During sampling, the plug is removed and a sampler is inserted Cor taking lhe samples. The main disadvantage of using a (X)ntinuous flight auger is that it becomes difficult to 3SCe11ain the depth from which tbe cuttings coming on the ground have been removed. Auger boring is generally used in soils which can stay open without casing or drilling mud. Clays. silts and panially saturated sands can stand unsupported. For soils which cannot stand unsupported, especially for sandy soils below water table, a casing is nonnally require(!. For such SOils, the method of auger boring becomes slow and expensive. Auger boring cannot be used when there are large cobbles, boulders or other obstructions which prevent drilling of the hole. Auger borings are particularly useful for subsurface investigations of highways, railways and air fields, where the depth of exploration is small. 1he investigations arc done quite CABLErapidly and economically by auger boring. (Attache<1 to winch) The main disadvantage of the auger boring is that the soil samples are highly disturbed. Funher, it becomes difficult to locate the exact changes in the soil strata.

17.10. WASH BORING In wash boring, the hole is drilled by first driving a casing, about 2 to 3 m long, and then inserting into it a hollow drill rod with a chisel-Shaped chopping bit at its lower end. Water is pumped down the hollow drill rod, which is known as wash pipe. Water tmerges as a strong jet through a small opening of the chopping bit. The hole is advanced by a oombinalioo of chopping adion and the jetting action. &'> the drilling bit and the accompanying water jet diSintegrate the soil. The water and the chopped soil particles rise upward through the annular space l>etween the drill rod and the casing. The return water, also known as wash water, is laden with the soil cuttings. It is coUected in a tub through a T-shaped pipe fixed at the top of the casing (Fig. 17.S).

TUB

CASlNG

BlT

Fig. 17.S. Wash I3ocing.

422


The hole is further advanced by alternately raising and dropping the dlopping bit by a winch (not shown). The swivel joint provided al the tOp of the drill rod facilitates the turning and twisting of the rod. The process is continued even below the casing till the hole begins 10 cave in. AI that !'>tagc, the bouom of

the casing can be extended by providing addilional pieces at tbe top. However, in stable, cohesive soils, the casing is required only in the t()P ponion. Sometimes, instead of a casing, special drilling fluids made of suspension or emulsions of fal Clays or bentonite combinoo with some chemical additives are used for supporting the walls of the hole '[he wash samples collected in the tub do not represent the soil in its lrue condition. There is oomplete break down of particles by chopping action. 1here is also mixing of the particles and the loss of fine particles in transportation. '{be samples are of lillie practical use. However. some indication about the changes in strata is provided by the reaction of the chopping bit as the hole is advanced. It is also indicated by a change in colour of the wash water. The wash boring is mainly used for advancing a hole in thc ground. Once the hOle has been drilled, a sampler is inserted to obtain soil samples for testing in a laboratory. The equipment used in wash borings is relatively light and inexpensive. The main disadvantage of the method is that it is slow io sliff soilS and coarsc·grained soils. It cannot be used efficiently in hard soils. rocks and the soils containing boulders. "[he method is oot suitable for taking good quality undisturbed samples above ground water table, as the w,lsh water enters the sirota below the bottom of the hole and causes an increase in its water romenl. 17.11. ROTARY DRILLING In the rotary drilling method. the bore hole is advanced by rotating a bollow drill rod which has a cutling bit at its lower end. A drill head is provided at the top of the drill rod. It consists of a rotary mechanism aDd an arrangement for applying downward pressure. As the drilling rod is rot.1tcd, thc CUlling bit shears off chips of the material penetrated. A drilling fluid under pressure is introduced through the drilling rod to the bottom of th,e hole. The fluid carries the cuttings of the material penetrated from Ihe bottom of the hole to the ground surface through the annular space between the drilling rod and the walls of the hole. The drilling fluid also cools the drilling bit. In case of an uncasod hole, the drilling fluid also supports the walls of the hole. When the soil sample is required to be taken, the drilling rod is raised and the drilling bit is replaced by as.1mpler. Rotary drilling can be used in clay, sand and rocks. Bore holes of diameter 50 mm to 200 mm can be easily made by this method. The mcthod is not well adapted for use In materiaL<; containing a large percentage of particles of gravel size and larger. The particles of this size start rotating benealh the drill rod and it becomes difficult 10 advancc thc hole.

17.12. PERCUSSION DRILUNG The percussion drilling method is used for making holes in rocks, boulders and other hard slmta. In this methods, a heavy chisel is altemately lifted and dropped in a vertical bole. The material gets pulverised. If the point where the chisel strikes is above the water table, water is added to the bole. The water forms a slurry with thc pulverised m a leria ~ which is removed by a sand pump or a bailer at intervals. PcrOJSSioo drilling may require a casing. Percussion drilling is also used for drilling of tube wells. lhe main advantage of percussion drilling method is that it can be used for all types of materials. It is particularly useful for drilling holes in glacial tills containing boulders. One of the major disadvantages is that the material at the bottom of the hole is disturbed by heavy blows of the chisel. It is not possible to gel good quality undisturbed samples. Further, the metbod is generally more expensive than other methods. Moreover, it becomes dim~lt to detect minor changes in thc properties of the strata penetrated.

17.13. CORE DRILLING The oore drilling method is used for drilling holes and for obtaining rock roccs. In this method, a am barrel fitted with a drilling bit is fixed to a hollow drilling rod. As the drilling rod is rotated, the bit advances

SITE INVEsnGATIONS

423

and cuts nn rmnular holc mound un inluct core 1l1C core is ·then removed from its bottom and is retained by a core lifter and brought to Ihe ground surfllce. Wmer is pumped continuously inlo the drilling rod 10 kcep the drilling bit cool aTKJ to carry the disintegrated material 10 the ground surface. The core drilling may be done using cilher a diamond studded bit or a OJtting cdge consisting of chilled shot. The diamond drilling is superior to the other Iype of drilling. but is costlier. The core barrel may consist of a single tube or a double tube. A double-tube barrel gives a good quality sample of the rock.

17.14. TYPES

O.~

SOIL SAMI'LFS

Soil samples are obtained during sub-surface explor-
17.15. DESIGN FEATURES AFFECl'lNG TUE SAMPLE DISTURBANCE The disturbance of thc soil depends mainly upon the following design features : (1) Area ratio. The area rntio is defined as A .. Maximum cross-sectional area of the CUlling edge x 100 , Area of the soil sample Fig. 17.9 shows the lower portion of a sampler. Tbc area ratio can be expressed as

ai-ol

A, - ~ x 100

... (11.2)

where D J = inner diameter of lhe cuning edge, D z = outcr diameter of the CUlling edge. For obtaining good quality undisturbed samples, the area mtio should be 10 percent or less (Hvorslev, 1949).

(2) Inside c1cllnmcc. The inside clearance is defined

SAMPLING

TUBE

Cj

_

D};;1

DJ

x tOO

... (11.3)

where D J = inner dinmcter of Ihe S
CUTTING

EDGE

Fig. 17.9. Soil 5.,mpler.

424

OOIL MECHANICS AND FOUNDATION ENGINEERING

Co.

~;4D4)(

100

... (17.4)

where D4 = outer diameter of the sampling tube. For reducing the driving force, the outside clearance should be as small as possible. Nonnally. it lies betwccn zero and 2 percent. (4) Inside wall friction. The friction on the inside wall causes disturbance of the sample. Tbc inside surface of tbe sampler should be smooth. It is u<;ually smeared with oil before usc to reduce friclion. (5) Design of non-return valve. The non-return valve provided on the sampler should be of proper dcsign. It should have an orifice of large area to allow air, water or slurry to escape quickly when the sampler is driven. It should immediately close when the sampler is withdrawn. (6) Method of applying force . lbe degree of disturbance depends upon the method of applying force during sampling and upon the rate of penetration of the sampler. For obtaining undisturbed samples, the sampler should be pushed and not driven. 17.16. SPLIT-SPOON SAMI'LERS The most commonly used sampler [or obtaining a disturbed sumplc of the soil is the stundard split-spoon sampler (Fig. 17.10). If consists mainly of three parts (I) Driving shot; made of tool-steel, about 75 mm long,

~7?l'7'\\M \\ ,;j.(j~Spring

Fig. 17.10. Stand.1rd Split Spoon S,1mplcr.

Core Catcher.

Fig.I'.]\' Spring; Co~ Catdl<,lr.

(ii) steel tube about 450 mm long. split longitudinnlly in two hnlvcs, and (iii) coupling at the top of the tube about 150 mm long. 'Jbe inside diameter of the split tube is 38 mm and the outside diameter is 50.0 mm. The coupling head may be provided with a check valve and 4 venting ports of 10 mm dia to improve sample recovery. lbis sampler is also used in conducting standard penetration lest (Sect. 17.22). After the bore hole has been made, the sampler is :lItachcd to the drilling rod nnd lowered into the hole.

425

SITE INVEsnGA1l0NS

The sample is collected by jacking or forcing the sampler into the soil by repented blows of a drop hammer lnc sampler is then withdrawn. llle split lube is separated after removing the shoe and the coupling and the sample is .taken out. It is then pl..1ced in a container. scaled, and transported to the laboratory. If the soil encountered in the bore hole is [ine sand and il lies below the waler table, the sample recovery becomes difficult. For such soils, a spring-core catcher device is used 10 ;lid recovery. As the sampler is lifted, the springs close and fonn a dome and retain the sample (Fig. 17.1 I). While taking samples, care shall be wken to ensure that the water level in the hole is maintained slightly higher than the piezometric level at the boltom of the hole. It is necessary to prevent quick sand conditions. The split tube may be provided with a thin metal or plastic tube liner to protect the sample and to hold it together. After Ule sample has been collected, the liner and the sample it contains arc removed from the tube and the ends are sealed.

::--''''''''000< ~\rotottor1

0)

~\~ ~ IH

Sec ti cm 0_0 Pig. 17.12. Scraper Bucket Sampler.

17.17. SCRAPER BUCKET SAlVIPLER If a sandy deposit contains pebbles, il is not possible to obtain samples by standard split-spoon sampler or spLit-spoon s:!mpler fitted with a spring core catcher. The pebbles come in-between the springs and prevent their closure. For such deposits. a sa-aper bucket sampler can be used . A scraper bucket sampler consists of a driving point which is Dttached 10 its bottom end (Fig. 17.12). There i<; a vertical slit in the upper portion of the s.1mpler. As. Ihe ~ l>ampler is rotated. the 2Smm

~~~~eo~:;IC;o:~ T

enter the sampler through the slit. filled with the scrapings, it is lifted. Although the s.1mplc is quite disturbed, it is still representative. A scraper bucket sampler can also be used for obtaining ihc samples of cohesion less soils below the water table.

BOUchec\c

rn

17.18. SHELBY TUDES AND TIIIN-WALLED SAMPLERS Shelby tubes are thin wall tube samplers made of seamless steel. The outside diameter of Ihe tube may be

I I

::~~ :ts~~e l~m:~; ~~:it:~~~~I~:C: 7~~~:

I

The bottom of the tube is sharpened and bevelled, which acts as a cutting edge [Fig. 17.13 (a)]. The area ratio is less than 15% and the inside clearancx: is between 0.5 10 3%. Fig. 17.13 (b) shows a thin-walled sampler (1S :

~~~~~::;~~ ;;'~yl:fl~h a~~ ~~et~U~; t~~ ~~el~ial~~~r ~~~ clayey soils. The diameter generally varies between 40 and 125 mm, and the thickness varies from 1.25 to 3.15 mm.

I I

1m

\1

0

I (b)

Fig. 17.13.

(II)

(0)

Slielby

Tube, (b)

Thin-waited Sampler.


426

Rod

The sampler lube is attached to the drilling rod and Iow-ered to me boUam of the bore hole. It Lo; then pushed into care shall be taken 10 push the tube into the soil by a continuous rapid motion without impact or twisting. The tube should be pushed to the lcngth provided for the sample. At least 5 minutes after pushing the tube into its final position, the tube is turned 2 revolutions 10 shear the sample oIT lit thc bouom before it is withdrawn. 'lbe tube is taken out and its ends arc scaled before transportation. Shelby tubes are used for obtaining undisturbed samples of (,;Clay. lhe soil.

17.19. PISTON SAMPLER A piston s.1mpler consists of a thin-walled tube with a piston inside. The piston keeps the lower end of the s.1mpling tube closed whcn lbe sampler is lowered to the bottom of the hole [Fig. 17.14 (a)). After the sampler has been lowered to the desired depth, Ihe piston is prevented from moving downward by a suitable arrangement, which differs in different types of piston samplcrs. The lhin tube sampler is pushed past lhe

Casing

Somptt'f

177;17\Jl---Piston -Soltorn 01

"'" (0'

rnis~~~o :';;~ ~~lhsa~t~o~F~~\~. ~~~~. The piston remains 1be presence of the piston prevents rapid squeezing of the son soils into the tube and reduces the disturbance of the s.1mple. A vacuum is created on the top of the sample, which (b, helps in retaining the sample. During the withdrawal of the sampler, the piston provides protcction against the water Fig. 17.14. Piston Sampler. pressure which otherwise would have occurred on the top of the sample. Piston samplers are used for getting undisturbed soil samples from soft and sensitive clays.

17.20. DENISON SAl\{PLER The Denison sampler is a double-walled sampler. l11e outer barrel rotates and O.Its into the soil. The sample is obtained in the inner barrel. The inner barrel is provided with a liner. It may also be provided with a basket-type core retainer. The sampler is low-ered to the bottom of the drilled hole. A downward force is applied on tbe top of the sampler. A Quid under pressure is introduced through the inner barrel to cool the coring bit when the outer barrel rotates The fluid returns through the annular space between the two bamls. The rotation of the outer barrel is continued till the required length of the sample is obtained. The Denison sampler is mainly used for obtaining samples of sliff to hard cohesive soils and slightly cohesive sands. How-ever, it cannot be used for gravelly soils, loose oohesionless sands and silts below ground water table and very soft cohesive soils. The Denison sampler gives a sample 5 (140 mm) in diameter and 20 incbes (508 mm) long. Care is

f

needed in adjusting tbe speed of rotation, the pressure on drilling bit and the velocity of wash water ~hen drilling in soils and very friable rocks.

17.21. UAND·CARVED SAMPLES Hand-carved samples can be obtained if the soil is exposed, as in a test pit, shaft or tunnel. Hand-carved samples are also known as cJumk samples. 'llle soil should bave at least a truce of cohesion so that it can stand unsupported for some time. To

SITE INVESl1GATIONS

427

BOX

Fig. 17.15. Hand-Cal"led Sample.

obtain a sample, a column of soil is isolated in the pit. The soil is carefuUy removed from around the soil column and it is properly trimmed. An open-ended box is then placed over the soil column. The space between the box and the soil column is filled with parnfJin. A spooe or a plate with sharp edges is inserted below the box and the sample is cut at its base (Fig. 17.15). The box filled with the soil sample is removed. It is turned over and the soil surface in the box is trimmed and any depression is filled with p.1raffin. A chunk sample may be obtained without using the box if the soil is cohesive. A column of soil is isolated. lbe block of soil is carcrully removed from the soil column with a sh
,

~\

17.22. STANDAIU> I'ENETRATION TEST The standard penetration test is the most commonly used in-situ test, especially for cohcsionlcss soils which cannot be emily snmpled. The tcst is extremely useful for determining the relative density and Ihe angle of shearing resistance of cohcsionless soils. It can also be used to determine the unconfined compressive strength of cohesive soils. The stand.1rd penctration Lest is conducted in a bore hole using a standard split-spoon sampler, described in Sect. 17.16. When the bore hole nus been drilled 10 the desired depth, the drilling lools are removed and the sampler is lowered 10 the bottom of the hole. "Il1e sampkr is driven into the soil by a drop hammer of 63.5 kg mass falling through a height of 750 mm at the rate of 30 blows per minute (IS : 2131 - 1963). The number of hammer blows required 10 drive 150 mm of the sample is counted. The sampler is further driven by 150 mm llnd the number of blows recorded. Ukewise, the sampler is oncc again further driven by ISO mm and the number of blows recorded. 'Ibe number of blows recorded for the first 150 mm is disregarded. The number of blows recorded for the last two 150 mm intervals are added 10 give tbe slandard penetration number (N). In other words, the standard penetration number is equal to the nUmber of blows required for 300 mm of penetration beyond a scating drive of 150 mm. ]f the number of blows for 150 mm drive exceeds 50, it is lalten as refusal and the test is discontinued. The standard penetration number is corrected for dilatancy correction and overburden rorrection as explained below. (a) Dililtancy Correction. Silty fine sands and fine sands below the water table develop pore pressure which is not easily dL-.sipated. The pore pressurc increases the resistance of the soil ond hence the penetration number (N). Terzaghi and Peck (1%7) recommend the following correction in the case of silty fine sands when the observed value of N exceeds 15. lbe corrected pcnctrntion number,

N~

- 15 +

t

(NR - IS)

where NR is the recorded value, and Ne is the corrected value.

IfNR siS.

Ne - NR

.. (17.5)


428

(b) Overburden Pressure Correction. In granular soils, the overburden pressure affects the penetration rcsist.1DCe. If the two soils having same relative density but different confining pressures are tested, the ODe with a higher confllling pressure gives a higher penetration number. As the confining pressure in cohesionicss soils increases with the depth, the penetration number for soils at shallow depths is underestimated and that at greater deplhs is overestimated. rOOf unlfonnity, the N·valucs obtained from field tests under different

effective overburden pressures are COlTected to a standard effective overburden pressure. Gibbs and Holtz (1951) recommend the use of tbe following equation for dry or moist clean sand.

Nt - NR x 003s:!70 where

NR :: observed N·value, He '" corrected N-value,

... (17.6)

00 = effective overburden

pressure (kN/m~

Eq. 17.6 is applicable for 00 :s 280 kN/m2, The ratio (N/ NR) should lie between 0.45 and 2.0. If (N~/NR) ratio is greater than 2.0, Nt should be divided by 2.0 to obtain the design value used in finding the bearing capacity of the soil. The correction may be extended to saturated silty sand and fine sand after modifying theNR according to Eq. 17.6, i.e. Nt I. I. 0.' 00 obt.1ined from Eq. 17.6 would be taken asNR in Eq. 17.5. Thus the overburden correction is applied first and then the di1ntancy COITCCtion is applied. Peck, Hansen and Tbombum (1974) give the chart for 0 correction of N·valucs \0 an effective overburden pressure of 96 kN/m2, According to them. N .. 0.77 NR log

(1~5)

for

00

00

01:

24 kN/m 2 ... (17.7)

II

Fig. 17.16 shows the corrcclion diagram. AI 00 = 0.0. the value of NINR is 2.0. lbe correction given by Bazaraa (1967), and also by Peck and Sazama (1969), is one of the commonly used corrcctiO{lS. According to them, 4NR N .. 1 + 0.0418 Co and

and

00

< 71.8 kN/m

2

... (17.8)

4NR

N - 3.25 + 0.0104 00

V

0

"

ISO

0

if

/

/ I

Fig. 17.16. OverbUfden Correction Diagram.

. _ 2 if 00 > 71.8 kN/m

if '00 " 71.8 kN/ml

... (17.9) ... (17.10)

Correlation of N with Engineering Properties The value of the standard penetration number N depends upon the relative density of the CQhesionless soil aod the unconfined compressive strength of the cohesive soil. If the soil is compact or stiff. the penetration number is high. The angle of shearing resistance (4l) of the cohesion1ess soil depends upon the number N. In general, the greater the N-value. the greater is the angle of shearing resistance (Fig. 17.17). Table 17.1 gives the average values of .. for different ranges of N. The consistency and the unconfined shear strength of the cohesive soils can be approximately determined from the SPT Dumber N. As the rorrelatioo is not dependable, it is advisable 10 determine the shear strength of the cohesive soils by conducting shear tests on the undisturbed samples or by conducting in·situ vane shear test (Sect. 17.24). Table 17.2 gives the approximate value of the unconfiDed shear strength for different ranges of N. The unconfined compressive strength can also be determined from the following relation.

SITE INVESTIGATIONS

429 10

/

/

/

/

0

0

0

/

/

V / v

0

0 18

30

12 An gle 01

14

)6

Fig. J7 .17. Variation

"

"

"

resis tellc e (¥J

sne e r i ng

..

of.' with N. ... (17.11)

qu - 125N OII l

where qu is unconfined compressive strength (kN/m1.

Thble 17.J. Correlation between N and

~

N

Dens.ness

0-4

"''l'Loo;e Loo;e

27"-35"

Medium

30"-40"

De"se

35"--45" > 45"

4-10 ID-30 30-50 > 50

Very Dense

25"-32"

Thble 17.2. Correlation between N and qll N

Consistency

0-2 2--4

Very Sofl

4--8 8-15 15-30 >30

Soft Medium Stiff VerySliff

",,'

Q.

kNlm'l <25

25-50 5D-lOO 100-200 200-400 > 400

17.23. CONE PENETRATION TESTS Sounding methods are frequently used to determine the penetration resistance and the engineering propenics of the soil. The sounding methods mainly consist of the cone lest and the standard penetration tCSI (Sect. 17.22). The roDe lest was developed by the Dutch Government, Soil Mechanics Laboratory at Delft and is,


430

therefore, also known as Dutch rooe Test. The test is conducted either by the static method or by dyn.1mic method. as disrussed below. (a) Static Cone penetmtion tesl. 1be Dutch cone has an apex angle of 60° and an overall diameter of 2

mm, giving an end area of 10 00 (Fig. 17.18). For obtaining the cone resistance, the roDe is pushed downward at a steady rale of 10 mm/sec through a depth of 35 mm each lime. The cone is pushed by applying thrust and not by driving. Mer the cone resistance has bccn determined, the cone is withdl1lwn. The sleeve is pusbed on to the roDe and both are driven together into the soil and the combined resistance is also dctcnnined. 1be resistance of the sleeve alone is obtained by subtracting the cone resistance from the combined resi'itance. A modification of the Dutch cone penetrometers is the Refined Dutch cone. It has got a friction sleeve of limited length above the cone JX>int. It 1Smm dio is used for obtaining the point resistance of the cone and the frictional resistance of the soil above the cone point. For effective use of the cone penetration test, some reliable calibration is required. This consists of comparing the results with those obtained from conventional tests conducted on undisturbed samples in a laboratory. It is. also convenient to compare the cone test results with the standard . Fig. 17.18. Dutch Cone. penetration test results. As the sHlndard penetration tests have been more commonly conducted in the past, good correlation studies arc available between the SPT number (N) and the engineering properties of the soil (Sect. 17.22). If the cone penetration results are rel.1ted to the SPT number H, indirect rorrelations are obtained between the cone test results and the engineering properties of soil. The following relations hold approximately good between the point resistance of the cone (qJ and the standard penetration number (N). (I) Gravels qr = 800 H to 1000 N .[17.12(a)J (il) Sands q( = 500 N to 600 N ... [17.12(')J (iii) Silty sands qc = 300 N to 400 N ... [17.12(')J (iv) Silts alX! Clayey silts qr = 200 N ... [17.12(.)J 35.7

k7mmol

where qr is in kN/m 2 (b) Dynamic cone Test. The test is conducted by driving the cone by blows of a hammer. The number of blows for driving the cone through a measure of lhe dynamic cone resistance. Dynamic cooe tests are performed either by using a 50 mm cone without bentonite Slurry or by using a 65 mm cone with bentonite slurry (IS : 4968-part ! alX! 11-(976). The driving energy is given by a 65 kg-hammer falling lhrough a height of 75 an. The number of blows for every 10 an penetration is reoordcd. lbe number of blows required for 30 cm of penetration is taken as the dynamic cone resistance (N~",.). If the skin friction is to be eliminated, the test is ooooucted in 3 cased bore hole. When a 65 mm cone with bentonite slurry is used, the sct-up should have arrangements for circulating slurry so that the friction on the driving rod is elim inated (Fig. 17.19). The dynamic oone resistance (Nt"') is correlated wilh the SPT number N. The following approximate relations may be used when a 50 mm diameter cone is used.

specified distance is a

HoIe5

lmm d,o

60mm dio

Fig. 17.19. Dynamic Cone Test.

431

SITE INVESTIGATIONS N~",

_ 1.5 N

Nt/If - 1.75 N N~",

_ 2.0 N

for depths upto 3 m

... [17.13(a)]

for depths between 3 to 6 m

[17.13(b)]

for depths glcater than 6 m

.. [l7.13(c)J

The Central Duilding Research Institute, Roorkee, has developed the following corrclmion bctwccn the dynamic cone resistance (Nbc,) of 65 mm diameter cone without using bentonite slurry and the S£Yf number (N) N~1>r .. 1.5 N for depths UplO 4 m ... [17.14(a)J

J

N~/JI'

- 1.75N

for depths betwccn 4 to 9 m

... [17.14(b)J

N~",

- 2N

for depths grealer than 9 m

... [17.14(c)]

The above relations are applicable for medium to fine sand.

17.24. IN·SfllJ VANE SnEAR TEST

I

I

BLADES

In-situ vanc·shear test is condu(..1cd to deICrmine the shear strength of a cohesive soil in its nnlurol condition. The appar'Jtus used is similar 10 one used in a l:1boralory (Chapter 13). It consists of four blndes, 100 mm (or 150 mm or 200 mm long). attached at right angles to a stccl rod. The steel rod has a torque-measuring device at its top. The height-diameter rnlio (/liD) of tbe apparatus is generally equal to 2 (Fig. 17.20). for conducting the lest, the shear-v:me is pushed into the ground at the bottom of the bore hole. When a torque is applied through the handle lit the top of the rod, the soil is shc.'lrcd along a cylindrical surface. The torque required to shc.'lf the cylinder of the soil is measured by means of a spring balllncc. The undrained shear strength s" of the soil is determined from Ule cqlk1t ion developed ill chapter 13.

T '" - _(dU/2 + ri'16)

... (17.15)

,.-

~,

Ti

1~. H :

"

V

0/""1

where T::: torque applied, JJ ::: height of the vane, Fig. 17.20. In·situ Vane ShcarTcsl. D ::: diameter of the soil cylinder sheared. The vane-shear test is extremely useful for determining Ihe in-situ shear strength of very soft and sensitive clays, for which it is difficult to obtain undisturt>cd samples. The test can also be used even for determining the shear strength of stiff. fi'iSurcd clays. However. the method cannot be used for Sc'lndy soils.

[7.25. IN·SITU TESTS USING A I'RESSURE METEII Menard developed a pressure meter. or sub-soil deformeter, which can be used for determining tbe stress-deformation characteristics of the soils in the n.mural conditions. 'lbe pressure meter consists of an innatoole cylindrical probe which is COllncctcd to a water reservoir (Fig. 17.21). The probe is inserted into the bore hole and it is inOalcd by applying water pressure. 'lbe probe presses against the unlined walls of the bore hole. As the pressure is increased, the soil deforms. The volumetric deformation of the bore hole is obtained by noting the fall in water level in the water rcst::rvoir. Fig. 17.22 shows a typical prcssurc.volumetric strain curve. The soil is initially in elastic phase but it enters the plastic phase at high pressure. Aftcr the plastic swge, thcre is no choogc in the volume with further incre.1sc in pressure. 1be prcssurc.dcfonnotion dola obtaincc.J from the test may be used to ddcrmine modulus of defonnation, undrained shear strength, "ngle of shearing resistance and other engineering properties of the soil. IS : 1892·1979 describes the usc of pressure meier.


.432

Gas

pr~ssure

~TlCO

I_

rE

GU.'"

Fig. 17.21. Pressure Meter. onSEl~VAnON

'1

cell

Probe (lllE'asuring cell) Guard cell

17.26.

ELASTC

PRESSURE (p) ___

Fig. 17.22.

OF GROUND WATER TABLE

The ground water level affects the pore water pressure and hence tbe shear strength of the soil. As tbe ground water level changes seasonally, it becomes necessary to establish the highest and the lowest water level (or the proper design of structures. The depth of the ground water !.able is usually determined in an exploratory investigation. In soils of high permeability, the level of ground water in a bore hole stabilises in about 24 hours. The depth of the water table is mensured by lowering a chalk·coated tape in the bore hole. The depth can also be mea
Fig. 17.23 ClWIgl7lndc PictOmela.

SITE INVESrIGATIONS

433

Go5. Casing

_____

sz..~:!.:._

-I~---3 h,

_____ 2 ______ 1

h, ____ .0

1:1 0

Fig. 17.24.

Ho ::: height above level 0 - 0 H2 = beight above level 2 - 2

h; - hi - h2

~ - hi - h2

hl

... (17.16) - h2 - h] The depth of the water table is taken as the average of three levels given by heights 1100 112 and H) i.e. level of the water table is given by

aoo I/J :::: height above level 3 • 3

-t

[(level 0 - 0 + Ho) + (levcl2 - 2 + Hi) + (level 3 - 3 + /lJ)J

... (17.17)

[7.27. GEOPHYSICAL METHODS A number of geophysical methods are used in preliminary investigations of sub-soil strata. 1be methods can be used for the location of different strata and for a rapid evaluation of the subsoil charaaeristics. However, the methods are very approximate. For detailed and reliable investigations. the conventional methods of driving a hole, taking a sample and testing it in a laboratory, as already described, must be resorted 10. The geophysical methods can be broadly divided into the two categories : Seismic methods and Electrical resistivity methods, dcsaibed in the following sections. [7.28. SEISMIC METlIO))S The seismic methods are based on the principle that the elastic shock waves have different velocities in different materials. At the interface of two different materials., the waves get partly reflected and partly refracted. Seismic methods of subsurface explorations generally utilise the refraCted waves. The shock wave is created by a hammer blow or by a small explosive charge at a point P (Fig. 17.2S). The shock wave travels through the top layer of the soil (or rock) with a velocity VI> depending upon the type of materia} in layer-I. 1be observation of the first amval of the waves is recorded by geophooes located al various points, such as A, B, C. The geophones convert the ground vibration into electrical impulses and transmit them to a recording apparatus (not shown). The basic equations of the refraction survey are derived based on the assumption that the subsurface strata are such that the velocity of the shock waves increases as the depth increases. In other words, it is assumed that Vl :> V2 :> V t in Fig. 17.25. At geophones located close to the point of impact, such as point A, the direct waves with velocity V 1 reach first.

SOIL MECHANICS AND FOUNDJUlON ENGINEERING

434

Fig. 17.25. Seismic melhod.

n,

AI points which are located away from the point of impact, such as poiot the refraded waves reach earlier than the dircct waves. These waves start from point P, travel with velocity VI in the upper layer, get refracted at the interface. move with much higher velocity V2 in the second layer, emerge again at tbe interface and travel back to the ground surface al a lower velocity VI in the upper layer. AI points further away from the point of impact, such as point C, the waves which are refracted twice, once at the interface of the layers I and II , and once at the interlace of tbe layers II and 1lI. reach earlier. For the determination of the thickness of different layers, a distance-time graph is plotted (Fig. 17.26).

The time (t) of arrival of the first impulse at various geophones is taken as ordinate and the distance (X) of the geophones from the point of impact P is taken as abscissa. Obviously, the velocity in any layer is equal to the reciprocal of the slope of the corresponding line. TIle slopes of the various lines are detennincd and the corresponding velocities computed.

VJ::: 1anoC3

iii

~

;1

'2

,

~=ttna(2 I

~:

v,=tano<,:

«it

:

9;;t

X2 DISTA.NCE

FROM

" ___ SOURCE (X)

Fig. 17.26.

Upto a certain distance XI> the dircct waves in the layer I reach fir.;t. At this point, the fjr.;1 two lines in Fig. 17.26 intersect, which indicates that the direct wave traveling a distance Xl with a velocity VI and the refracted wave traveling with a velocity Vi in distance 2111 and with a velocity of V2 in distance XI reach simultaneously, where H I is the thickness of the layer I. Thus ~ _ 2HI

VI

VI

...

~ V2

SITE INVESTIGATIONS

'"

or

... (17.18)

Eq. 17.18 gives reliable results when t~ waves are produced by a sinusoidal force and not by impact. The following empirical equation gives more reliable results for impact shock. HI -

XI·~

T V~

... (17.19)

Likewise, the thickness of lbc second layer (H,) is obtained from the distance X 2 colTCSpooding to the point of intersection of the second and (he third line in Fig. 17,26. It is given by the relation

X

Ih - O.85H 1

rv;-:v;

+"'22 V b , •

... (17.20)

TIle pnx:edurc is oontinued if there are more 111.1n three layers. The type of material in various layers can be determined by comparing the velocities obtained with the standard velocities given in Thble 17.3. Thble 17.3. Velocities in Different Types of Strata 1)p' of

Rock/Scil ltlodl}' (m/sec)

CrOJlile

4000

Sand Stont:

Shale

1500

1300

Hard Clay

Loa«

Loos<

Loo"

gravel (Wet)

sOlid (11<1)

salld
600

500

500

250

10

10

10

10

10

10

10

6000

3000

3000

1500

1000

1500

600

Umltallon or the seismic methods (1) The methods cannot be used if a hard layer with a greater seismic velocity overlies a sofier layer with a smaller seismic velocity. (2) The methods cannot be used for the areas covered by concrete., asphalt pavements or any other artiflcial hard cm;t, having a high seismic velocity. (3) If the area contains some underground features, such .as buried conduits, irregularly dipping strata, and irregular water table, the interpretation of (he results becomes very difficult. (4) If the surface layer is frozen, the method cannot be successfully used, as it corresponds to a case of harder layer overlying a softer layer. (5) The methods require sophisticated and costly equipment. (6) For proper interpretations of the seismic survey n:5ulJs, the services of an expert are required. Despite above limilntions, the mcthod is extremcly useful for the determination of the thickness of various strata and their characteristics. 1bcse surveys are U'ieful for ool8ining preliminary information about the types and depths of various strata at a given site.

17.29. ELECTRICAL RESISTIVITY M~"TIIODS The electrical resistivily (P) of a conductor is expressed as p _ RAIL

... (17.21)

where R = electrical resistance (ohms), A = area of cross-sea ion of tnc conductor (cm1, L = length of conductor (cm). p = electrical resistivity (obm-centimeter). (II may be nOled lbal the electrical resistivity is the reciprocal oC conductance.) The resistivity of a material depends upon the Iype of material, ilS water content and the concentration of dissolved ions and many other factors. Rocks and dry soils have a greater resistivity than saturated clays. Table 17.4 gives approximate values of resistivity of different rocks and soils.

436

SOIL MECHANICS AND FOUNDATION ENG INEERING

Table 17.4. Resistivity of Different Rocks and Soils 1)"" of rock/soil

Resistivity (Ohm-m)

Soulld

mok

Weatllered Rock

1500 > 5000

Gravel

Sam'

Clayey sOlid

1500

500 to 1500

500

'0

10

2500

4500

Satura/eli ofay olldsill

200 to 100

10

The electrical resistivity methods arc of the following two types: (0) Electrical pronling Method. llle method is also known as the resistjvity mappillg method four

electrodes are used al a oonstanl sp..1cing a (Fig. 17.27). To conduct the lcst. fOUf electrodes, which arc usually in the form of meL11 spikes, are driven into the ground. The two outer electrodes arc

AM""ETER

known as current electrodes. The two inner

electrodes

arc

c.111cd

potential

electrodes. The meno resistivity of the strata is detennined by applying a D.C. current to the outer electrodes and by measuring the voltage drop between tile inner electrodes. A current of 50 to 100

".,j~""".,.f"""""""""""""p'

miUiamp is usually supplied. l11c mean resislivity (p) is given by

CURRENT ELECTRooe::

Fig. 1727. Electrical1lrofiling MelllOd.

the fonnula

,

2l'taV

p---

... (17.22)

where I ::: current supplied, a = spacing of electrodes, and V = voltage drop.

Eq. 17.22 gives the mean resistivity upto a depth of a below the ground surface, as the depth of current penetration bclow the ground surface is approximately equal to the spacing of electrodes. 'n1C elcctrodes arc moved as a group, and different profile lincs nrc run U(,.TOSS the urea. The test is repcate
!

,,

changes

The method is useful in studying the changes in the strata with incre.1sing depth at a point. The Fig. 17.28. Electrical Sounding melhod. method can indiQlte sub-surface variation when a hard L'lyer overlies a son layer or vice-versa. It can also be used to locate the water table.

SITE INVESI'IGA1l0NS

Limitation of the electrical resistivity methods. (1) The methods are capable of detecting only the str
accu~dteisr:~n~~n~t~O ~::~. = p : : ha~~ o~~ (m)

&JL T_"'..'.•

,"_r:-,-_-._LL '-,_P_',

should be identified by a code number. The loca1ion of each bore hole should be fixed by 2·0 measurement of its distance or anglcs from 3·0 some permanent feature. All relevant data for 4.0 the bore hole is rcoordcd in a boring log (Fig. 5.0 Vfi/lDE=SAND 17.29). A boring log gives the da>Cription or classification of various strata encountered at 6·0 different depths. Any additional information that '·0 is obtained in the field, such as soil consistency, 11·0 uoconfined compression strength. standard '1.0 penetration test, cone penetration test, is also 10.0 indicated on the boring log. H should also show rig. 17.29. Boring Log. ( 1 tlnl = to kN/ ml). the water tnblc. If the laboratory tests have been conducted, the information about index propenies, compressibility, shear strength, permeability, etc. should also be provided. The data obtained from a series of bore boles is presented in the form of a sub-surface profile (Fig. 17.30). A subsurface profLIe is a vertical section through the ground along the Line of exploration. It indicates the boundaries of different stmla, aloog with their classification. It is important to remember that conditions between bore holes are estimated by interpol,ltion, which may nOl. be correct. Obviously, the larger the number of holes, the more accurate is the sub-surface profile. The site investigation report should contain the discussion of the results. 'Inc discussion should be clear and concise The recommendations about the type and depth of foundation, allowable soil pressure and expected settlements should be specific. The main findings of the report are given in conclusions. A soil exploration report generally oonsists of the fonowing. (1) Introduction, which gives the scope of the investigation. (2) Description of the proposed structure, the location and the geologic-'lI conditions at the site. (3) Details of the field exploration programme, indicating Ihe number of borings. their location and depths. (4) Details of the methods of exploration. (5) General dcsaiption of the sub-soil conditions as obtained from in-situ tests, such as standard penetration tcst, cone lest. (6) Details of the laboratory test conducted on the soil samples obtained and the results obtained.

SOIL MP£HANICS AND FOUNDATION ENGINEERING

438

DEPTH (m)

o

LOOSE

SANe

: SAN9Y

- . SILT:

Fig. 1730. Subsurface Profile.

Depth of the ground water table and the changes in water levels. Discussion of the results. Recommendation about the allowable bearing pressure, tbe type of foundation or structure. Conclusions. The main findings of investigations should be clearly stated. It should be brief bul should mention the salient points. Umilations of the investigations should also be briefly stated.

(T) (8) (9) (10)

PROBLEMS A. Descriptive and Objective Type 17.1. What do you undcnownd by site invcstiglltion ? What are the different purposes for which sile investigations are

done? 17.1. What is reconnaissance '1 What type of information is obtained in reconnaissance? What is its use?

11.3. How would you decided the depth of exploration and the lateral extent of the investigations? 17A. Describe open excavation methods of exploration. What are their advantages and disadvantages ? 11.5. Describe various methods of drilling holes for subsurface investigations. 17.6. 17.7. 11.8. 17_9. 17.10. 17.11. 17.1Z. 17.13. 17.14. 17.15. 17.16. 17.17.

What do you understand about disturbed and undisturbed samples? How would you oowin undisturbed samples? What are the factors that IIffcct the sample disturbllnce? now are these effects minimised? Describe the split-spoon sampler. What is its usc? Discuss various Iypes of soil samplers for obtaining undisturbed samples.. How would you obtnin a hand-carved sample? Discuss stlndard penetration test What are the various oorrections? Wh81 is the importance of the test in geotechnical engineering? Describe cone penetration tests. How these tests differ from standard penetration test? How would you oonduct an in ..situ vane shear test? What is its use ? Discuss the various methods for dClennining the level of the ground water table. Describe, in brief, various geophysical methods. Discuss their limitations and uses. Describe the salient features of a good suh-soil investigation repon. Write whether the following statements are true or false. (a) A disturbed sample is nol a trul y representative sample. (b) An undisturbed sample is absolutely undisturbed. (e) The samples obtained from wash water of wash borings are disturbed samples. (d) The sample obwioed by a split spoon sampler is an undisturbed sample.

SITE IN VESTIGATIONS

439

Ie) The sHlndilrt! penetration Icst IS mure useful for cohcsionless soil than cohesive SQils. (j) The static cone tcst is more useful tor whcsive soils dmn cohesionlc.~s soils. (8) Dilatancy correctiun of SPT numbe r is clolle for clayey soils. (II) In-situ v;lnc shear test is cxtremely useful for oblaining the shear strength of very soft cohesive soils. (I) The Cas3gr:mcle piezometer is used lor deiennining the water level in l'tJhcsionless soils. (j) SeismIC methods can be used if the seismic velocity of diftcrcllt str-lla increases as the depth is increased. (k) Gcophyslcal methods are useful for preliminary H1vestigations. (/) A soil proliic gives an ;\Ccumle prome of different Mrata. (m) A bClrmg log mdicates differcl1l SlratH along the depth of II bore hole. (Ans. Tl1le, (a). Ie). (e).Ij). (11),

(j). (k). (1II}1

U. Multiple Choicc Qucstions 1. The

.~I :mdard

pcnctr.ltion lest is useful to

mea.~ure

(a) shear Mrength of soft days

she.lr Mrength of sands (c) consistency ofehlYs
(c)

sandy

(tl)

All

soil~

the above

4. The seismic refraction mel hods cannot be usell if the wave velO\: it y in Ihe lower layer is . . .. Ihat in the upper layer.

(M less than (ti) more lhan thice 5. Selcctthe incorrect statement: For u good quulity ~oil sample. (a) The are.1 ratIo shou ld be low (b) The CUlling edge shoulll be thick (e) 111C inside ckamnce should he ~mall (J) The outside clellrance should be small 6. If the ill1ual vlllue or thc standard pcnclnllion number (N) is greater than 15 for line snnds below water table. «j) grealer IhUI1 (e) more thun four limes

;:,; ::rr:T~ ';'~~ jf k) 15

+

(N ~ 15)

N b

(N;

(iJ)

15 _

15)

(dJ

I5+(¥)

7. The height-diameter mtio for the in-silu vane is ~)ID

(~I.~

(e) 2.00

(d) 3.0

[Ans. I. (/J), 2. (b). J. (a). 4. (b). 5. (h). 6. (e). 7. (e)1

18 Stability of Slopes 18.1. INTRODUCTION An earth slope is an unsupported, inclined surface of a soil mass. Earth slopes are fonned for raHway fonnations, highway embankments, earth dams, canal banks, levees, and at many other locations. Fig 18.1 shows some of the examples of eanh slopes.

SIOP.~SIOpe~ Y~;;k\,*ilWAYTRACK))(\\

(al

(bl

~SIOPe

~

~'-SIOP~~ (c) CANAL

(d) EARJH

DAM

Fig. 18.1 . Earth Slopes.

The casl of earth work would be minimum if the slopes are made steepest. However, very steep slopes may not be stable. A compromise has to be made between economy and safety, and the slopes provided are neither too steep nor too Oat. In other words, the steepest slopes which arc stable and safe should be provided. The failure of a slope may lead to loss of life and property. It is. therefore~ essential to check the stability of proposed slopes. With the development of modem methods of testing of soils and stability analysis. a safe and ccooomical design of a slope is possible. 1bc geotechnical engineer should have a thorough knowledge of tbe various methods for checking the stability of slopes and their limitations. The failure of a soil mass occurs along a plane or a curved surface when a large mass of soil slides with respect to the remaining mass. In general, there is a downward and outward movement of the soil mass. A slope failure occurs when the forces causing failure are greater than the shearing resistance (shear strength). developed along a critical surface of failure. lbe factors leading to the failure of slopes may be classified into two categories.

STABILITY OF SLOPES

441

(1) The factors which cause an increase in the shear stresses. The stresses may increase due to weight of water causing saturation of SOils, surcharge loads, seepage pressure or any other cause. The stresses are also ina-easrA due to steepening of slopes either by excavation or by natural er05ion. (2) The factors which cause a decrease in the shear strength of the soil. 1be loss of shear strength may occur due to an increase in water oontent, increase in pore water pressure, shock or cycliC loads, weathering or any other cause. Most of the naturnl slope failures occur during rainy seasons, as tbe presence of water causes both increased stresses and the loss of strength.

18.2. BASIS OF ANALYSIS The soil mass must be safe against slope failure on any conceivable surface across the slope. Although Hie metlxx1s using the theory of elasticity or plasticity are also being ioacasingly used, the most oommon methods are based on limiting equilibrium in which it is assumed that tbe soil is at tbe verge of failure. The methods of limiting equilibrium ar~ statically indetecminate. As the stress-strnin relationships along tbe

assumed surface are not known. it is necessary to make assumptions so that the system becomes statically determinate and it can be analysed easily using the equations of eqUilibrium. The following assumptioIl') are generally made. (1) The str<:ss system is assumed to be twSumptions are made regarding the magnitude and distribution of forces along various planes. In the analysis, the resultant of all the actuating forces lrying to cause the failure is detennined. An estimate is also made of the available shear strength. The factor of safety of the slope is determined from tbe available resisting forces and the actuating fora:s.

18.3. DIFFERENT DEFlNITIONS OF FAcroRS OF SAFIITY Three different definitions of the factor of safety are used.

(4) Factor of safety willi respect to shear st[t!nglli In common usage, the faaor of safety is defined as the ratio of tbe shear strength to the shear stress along the surface of failure. The factor of safety as defined above is know n as the faaar of safety with respect to shear strength.

...(18.1)

Thus where F, 'tnt

= faaor of safety with respea to shear strength, s =shear strength. = mobilised shear strength (equal

to applied shear stress).

Eq. 18.1 can be written in terms of the cohesion intercept and the angle of shear

F._~ em + alan,,,,

where ClIO :: mobilised cohesion. Rearranging Eq. 18.2,

'm =

res~

as ... (18.2)

mobilised angle of shear resistance, (j = effective pressure.


442

-F; + i.i~~4t_

ClIO

+

atan4t... . .. (IS.3)

em - elF,

Therefore,

... (18.4) and tan 4t". - tan +IF. Eqs. 18.3 and 18.4 indicate that the factor of safety wilh respect to the cohesion intercept and that with respect to the angle of shearing resistance arc equal to the factor of safety with respect to the shear strength. (b) Factor of safety with respect to cohesion The factor of safely with respect to cohesion

(F~)

is the ratio of the availablc cohesion intcrcept (c) and

the mobilised cohesion intercept. ... (IS.5)

Thus where

intcrcept, ClIO = mobilised cohesion intercept, Fe = factor of safcty with respect to cohesion.

C

= cohesion

(c) Factor of sufely with respect to friction TIle factor of safety with respect 10 fridion is the ratio of the available frictional strength to the mobilised

frictional strcngth. Thus

F ...

:t~n+~

F.- t~n:m

whcre F.

+. .

...(IS.6)

=factor of safety with respect to friction, " =angle of shearing resistance,

= angle of mobilised shc.1ling resistance. For small angles, Eq. 18.6 can be expressed as F. -

t

... ([IS.6(o)J

In the analysis of stability of slopes, generally the three factors of safety are taken equal, i.e. F. - Fe - F." However, sometimes when greater reliance is placed 00 the parameter than the parameter c, the factor of safety with respect to cohesion is taken greater than that with respect to friction. In such a case, the factor of safety with respect to friction is usually taken as unity Le. ......

+

+.

18.4, TY1'ES OF SLOPE FAILURES A slope may have anyone of the foUowing types of failures. (1) Rotational failure, 'Illis type of failure occurs by rotation along a slip surface by downward and outward movement of the soil mass (Fig. 18.2). lbe slip surface is generally circular for bomogeneous soil conditions and non-circular in case of non- homogeneous conditions. Rotational slips arc furtber divided into 3 types. (a) Toe failure, in wbich the failure occurs aloog the surface that passes tbrough the toe [Fig. 18.2 (a)J. (b) Slope failure, in which the failure occurs along a surface that intersects the slope above the toe [Fig. IS.2 (b)J. (c) Base failure, in which the failure surface passes below the toe [Fig. 18.2 (c)J. The slope failure occurs when a weak plane exists above the toe. The base failure occurs when a weak stratum lies beneath the toe. If a strong stratum exists below the toe, the slip surface of the base failure is tangcntial to that stratum. In all other cases, the failures are generally loe failures. Toe failures are most common. (2) 1hmslaUonai Failure. A OOf1S!ant slope of unlimited extent and baving unifonn soil properties at the

44'

srAIlILffY OF SLOPES

{c)8ose

!oi\ure

Fig. 18.2 Rotational Fllilure.

same depth below tbe Cree surCace is known as an infinite slope. In practice, the slopes which arc oC considerable extenl and in which the conditions on all verticals are adequately represented by avemge conditions arc designated as infinite slopes.

1 I

Fig. 183. Translational

Failure.

Translational failure occurs in an in(initc slope along a long Cailure surface parallel to lhe slope (Fig. 18.3). The shape of tbe failure surface is influenced by the presence of any hanI stratum at a shallow depth below the slope surface. Translational failures may also occur along slopes of layered materials. (3) Compound Failure. A compound failure is a combination of the rotational slips and the translational sUp (Fig. 18.4). A compound failure surface is curved at the two ends and plane in the middle portion. A compound failure generally occurs when a bard stratum exists at considerable depth below the toe.

1 J

Fig. 18.4. Compound Failure.

(4) Wedge Failure. A failure along ao incUn'C

444

PLANE OF WEAKNESS

::;;~ ~ Fig. 18.5. Wedge Failure.

in the form of spreads and flows may atso oocurs. 18.5. STABILITY OF AN INFINITE SLOPE OF COHESIONLESS SOn.s The stability mteria of an infinite slope of cohesionless soils will depend whether the soil is dry, or submerged or has steady seepage, as explained below. (1) Dry Soli. Fig. 18.6 (0) shows a section of an infinite slope having a slope angle of i. leI us mnsider the prism ABeD of the soil. with the inclined length An equal to b. The horizontal lengtb of the prism is b COS L The height of the prism is H [Fig. 18.6 (b)].

~ B

.

b

,-

H

0

i

,-

C

Ca)

IbC~t

r

1:tI

NGW

'~-----,V

~

o

0

0-

Cc)

Fig. 18.6. infinile Slope In Dry Sand. Volume of prism per wit length _ Hb cos i Weight of prism per unit length, W _ Y(llb cos r) TIle weight of the prism can be resolved into the nonnal compooent N and tangential components T to plane CD.

445

STABlLITY OF SLOPES

N .. Woos; .. 1 Hb ros 2 j

Thus

T .. Wsin i ""1 Nbcosisin i The normal and shear stresses are given by

""..

a ..

~

2 .. YHbbOOS i .. yHcxxli

"t ..

~

..

YHbcr;isini .. yHc.osisioi

... (18.7) .. (18.8)

tend to Qluse the shear fail ure along CD. This tendency is opposed by the shearing resislance developed along the plane CD. As the soil is dry, there is no pore water pressure. The shear stresses

Therefore,

"t

s .. alan ,' .. atan,'

s .. (yHcos 2 ,)tan+' The factor of safety against shear failure is given by

2 F .. ! .. C:tHcos I)tanf' '''t yHcos isin i F, ..

~

... (18.9)

Eq. IB.9 indicates that the slope is just stable wben .' .. i. The factor of safety is greater than unity when i

is less than ,'. For the slope angle i greater than cp', the slope is not stable. It is worth noting that the factor of safety of an infinite slope of a oohesioolcss soil is independent of the height H of the assumed failure prism. '£be angle .' in Eq. 18,9 should correspond to the actual relative density of the soiL As the soil in the surface layers is in a relatively loose slate, the angle .' oorrespooding to the loose state is generally taken. Eq. 18.9 can be represented graphiQllly [Fig. 1B.6 (c»). The ordinate PQ is equal to a tan .' and represents the shear strength, The ordinate PI Q represents the shear stress 't equal to a tan i. Obviously, tbe factor of safety is given by

F,.

~_~_~

(2) Submer-ged Slope. If the slope is Submerged under water, the norma! effective stress and the shear stress are calculated using the submerged urnt weight and oot the bulk unit weight as was used for dry soiL Thus, from Eqs. 18.7 and 18,8.

0 .. y'Hcos2 j and "t - l'Hsinicosi where y' is the submerged unit weight. Therefore, the fador of safely is given by

.. .(18.10) ... (18.11)

F _! .. (y'Hcos2 ,) lan" f 't y'HsinirosI F _~ .. .(18.12) , tan i Comparing Eqs, IB.9 and 18.12, it is observed that the factor of safety of a submerged slope is the same as that in dry condition. (3) Steady Seepage along the slope. Fig. IB.7 (a) shows an infinite slope with steady seepage parallel to the surface. Fig. 1B.7 (b) shows a free-body diagram of the prism ABCD of the soil. The forces acting on the venical sides of the prism due to water and soil are equal and opposite and, therefore, Qlncel. The weight of the prism W is taken corresponding to the saturated conditions.


Therefore,

W .. y,.Hbcosi N - Woos; _'jptHbro:;'l I'

T- lYsin; .. y_Hbsinicosi AI tbe base of the prism, there is an upward (oree due pressure (u), given by

10 waler

'oj

u '"' y",ffcos 2 ; Uplift force, U .. (y", H Thus, the nel nannal force

or

COi'l i)

b

If is given by

N -N - U .. YIQlHb~2; - (y",ffco!l?t)b N .. y' bH cos2 i, where i is submerged unit weight.

-IT}

The effective stress is given by

o

=

~ ..

(b)

y' ll cos'lj

Shear strength, s .. (; lan 4>' .. y' lJ cas2 i tan 4>' The shear stress is given by 't

~

f.

now

,~

li~

mffil

,oJ

Fig. 18.7. Steooy Seepage along the Slope.

y,,.Hsinicosi

Therefore, tbe factor of safety is given by

F .. ! .. y' llcaf?itan+' I

"t

YrQlHsinicosi

F=~ .<

1."" Ian

... (18.13)

i

As tbe submerged lUlit weight y' is about one-half of the saturated unit weight, the factor of safety of the slope is reduced 10 about one-half of that corres!Xlnding to the condition when there is no seepage. The angle ti' in the wet condition of a cohesionless soil is approximately the same $ in dry condition.

18.6. STABILITY ANALYSIS OF AN INFINITE SLOPE OF COHESIVE SOILS The stability analysis of an infinite slope of cohesive soils is similar to that, in the case of oobesionless soils, with one basic difference that the sbear strength of a cohesive (c - , soil) is given by s_c'+(1lan+'

(a) Dry Soli. Taking values of (1 and

't from Eqs. 18.7 and 18.8, we have s .. c' ... (yHoos2 i)tan+'

"t ..

yHcosisini

1bcrefore. the faaor of safety F, is given by F _ c' ... (yHcos 2 l)tan"

,

yHrosicosi

Thus, the fador' of safety of an infinite slope in cohesive soils depends not only on

y,H and c'.

... (18.14)

+' and i but also 00

Fig. 18.8 represents a graphical method for the determination of the factor of safety. 1be line RP2 represents the failure envelope. When the slope angle i is less than 41', the slope is always safe as shown by

lintrOp,. When the slope angle i'l is greater than +'t the slope line cuts the failure envelope. At !Xlint P. the slope is just stable. For normal stress greater than that indicated by point P, tbe shear stress is greater than

447

STABILITY OF SLDPES

the shear strength and the slope is not stable. A'i the nonnal stress 0 depends upon the height 11 of the slope, an expression for the height can be found when the slope is just stable. Therefore, equating the shear stress and the shear strength corresponding to point P,

lfloosisini .. c' + lHoos2 jU,lOCP' 11100s

2

j

(:~ II ..

-

tanCP') _ c'

c' 1 (tan j - tan q>') OOS2 j

The height at which the slope is just stable is known as the critical height (H~). Thus

Fig. 18.8. Infinite Slope in Cohesive Soils. c' ... (18.15) l(tan; - tanq:>')CfYii,2i For heights less than the critical beight, the fador of safety is given by Eq. 18.14. fir.

(b) Submerged Slope As in the ClSC of cohcsionless SOils, the normal and tangential components of the weight are taken for submerged unit weights and not for bulk unit weights. Thus. from Eq. 18.14, F .C'+l'Hcos2 itanp' ... (18.16) • l'Hcosisini The va lue of in Eq. 18.16 should be laken a:>rrespotiding 10 the submerged conditions, which may be quite different from that in dry condition in case of cohesive soils.

.1

(e) Steady Seepage along the Slope

100 case is similar

10

that for a oohesionless soil. In this cast; the factor of safety is given by c ' + J'Hcos2 jtanq>' F, " l'IIIHcosisini ... (18.17)

The critical height is obtained corresponding to a factor of safety of unity. Thus c' + '(' H~oos2jtanlP' - 11~H(oosisini

H~ cos2 i (1.111 tan i - 1 I tan !p') .. c'

IJ~

c'

... (18.18)

- ('(,lII tan ; -l'tan
Somctimes. Eq. 18.18 is written as

II, - 1,M

[tani - (t) tan,' j=' i

Eq. 18.19 indicates thai the cffect of the angle of shearing resistance 18,15 of the dry soil.

... (18.19)

.p' is reduced as compared with Eq.

18.7. WEDGE FAILURE A wedge failure occurs when a soil deposit has a specific plane of weakness. The stratified deposits generally fail along the interface. Fig. 18.9 shows a soil mass resting on an inclined layer of impermeable soil . There is a tendency of the upper mass to slide downward along the plane of contact AB.


Fig. 18.9.

The force trying to cause sliding is the langenlial component T of the weight (W) along the plane of contact.

... M

T-W~na

where a is the angle which the plane A/J makes with horizontal, W is the wcight of wedge per unit length perpendicular 10 the plane of paper. The force Icnding 10 resist the sliding depends upon the oohesion c and the frictiooal foroe and is given by ... (b) S - cL + (Woos a) tan where L is the length of lbe failure surface AB. Thc factor of safety against sliding is obtained from (a) and (b) as

+

F, _ cL + (::aa)tanq>

... (18.20)

18.8. CULMANN'S METHOD Culmann's mctlxxl is used for the approximatc stability analysis of homogeneous slopes. A plane failure surface passing through the toe is assumed. A plane failure surface is not a correct assumption for a homogeneous soil However. it is a simple failurc mechanism and is described for purpose of illustration and for determination of the approximatc valuc of the factor of safety. Let us consider the equilibrium of the triangular wedge ABD fonned by the assumed failure surface AB (Fig. 18.10) The wedge is in equilibrium under the three forces:

,~m ~c

(90-<)

R\~OJr-r{\

"

W (o(- ~)

Fig. 18.10.

(1) Weight of the wedge (W). (2) Cohesive force (C) along the surface AB. (3) Reaction R. The reaction R is inclined at angle tj:I", to the nonna!. The triangle of forces is also shown in figure. 1be magnitude and direction of Wand C are known. 1be direction of R is also Imown. The weight of the wedgc is given by

449

STABILITY OF Sl.OP£S

1 W--2'YL

(II). (. ) --;----: SlOl

SID

.. .(a)

I-I).

C .. e,.L

and

where H = height of slope. e". = mobilised cohesion. q.". = angle of mobilised friction, and L = length of failure surface AB. [Note. Area of a triangle AnD = ~ d b sin A where d and b are the length of sides DA and AD, respectively.] C

From the law of sines.

W"

sin(a-t",) sin(a-'".) sin(cjI". + 00°)" ~

... (c)

Substituting the values of Wand C from Eqs. (a) and (b) in Eq. (c),

c,.,L

sin (a - cjI",) a)" ~

lI2yL(H/sint) sin (i

or

(~)

.i

coseci sin(i - a)

...(d)

sin (a-cjIm) scc'm

... (18.21)

'£be left-hand side of Eq. 18.21 is known $ the stability number (S,,). The most dangerous plane is thai for which the angle a is such that the stabillly number becomes a maximum, i.e..

dd~)

t

_

0

[sin (i - a) sin (a - q:>",)] .. 0

sin(i - a) cos (a - '".) - sin (a - cjI",)cos (i - a) .. 0

or

tan (i - a) .. tan (a -t",)

i-a-a-+. (i + ~.)

"'--2where

... (e)

a.. is the critical slope angle.

From Eqs. 18.21, taking the value of

(~L

a.. from

Sq. (e),

-i =isec~_

+z")} 1[sm{(' +/-) - ~-) 1 x

.

. sec .,.sm . (i- -~_) 2 - sm (i-~_) -2-

.. '12 COSCC

I

.. '21 COSCC

I

C_) (Yii ........

[sm{ i - (i

.

sec t",

[ i-cos

l-cos(i-~.)

(4 sin;cos',.) 4 c'" sin i cos , ...

II· y[1 - cos(i - ~.)J

(i -

2

~-)

1 ... (18.22) . .. [18.22(a)J

where H is the safe height of slope. 1be Culmann metbod gives reasonably accw-ale results for homogeneous slopes which are vertical

0f


450

nearly vertical. The critical surface for general slopes is not a plane and, therefore, the aitical slope little practical use for such slopes.

a~

has

18.'. ~. - 0 ANALYSIS In case of fully saturated clay under Wldrained conditions, the stability of the slope can be cbeckcd in terms of (()(at stresses. Such a condition occurs in slopes immediately after construction. In Ibis case. =0 aooc=c... The failure surface is assumed to be a ciroJlar arc AD (Fig. 18.11). 1be figure shows a failure surface with centre 0 and radius r (i.e. DB '" OA = r).

+..

Fig_ 18.11.

The total weight W above the failure surface causes instability. For equilibrium, the shear strength to be mobilised along the failure surface can be e~ as

~m

-

f; - ~

where F is the factor of safety. Taking moments about 0, W)( d - (c./F)L.r

... (18.23)

where LQ is the length of arc AD and d is the lever arm of Wabout O.

Thus

F _ c"LQr

.. .[18.23(0)]

W x d If a tension aack develops and water enters the crack, the hydrostatic pressure force P w acts on the portion Be of tbe arc al a height of hl3 from C, where h is the depth of tension crack. equal to 2 c,l'y (see chapter 19). The arc length in tbat case sbould be taken equal toAC. Eq. 18.23 may be modified aocordingly.

18.10. FRlcnON CIRCLE METUOD The friction circle method is useful for the stability analysis of slopes made of homogeneous soils. In this metbod, the slip surface is assumed to be an arc of a circle. (0) D'l' Soil Fig. 18.11 shO\Vs a circular failure surface ABC of radius r with its centre at O. The three forces acting on tbe sliding wedge AEBD are given belO\V: (1) Weight (W) of the sliding wedge. (2) Cohesive force C developed along the sUp surface ABD. (3) Reaction R on the Slip surface.

451

STABILITY OF SLOPE;.S

The reacUon R is inclinCd at angle '''' 10 the nonnal to the slip surfacc. As the dircaioll of the nonnal chnnges, the direction of R also changes. With tbe centre as 0, 0 small circle, known as friclion circle, is drawn with a radius r sin ''''. All lines which arc tangent to the friction circle make an angle ,,.. with the nonnal of the slip surface. These lines represent the direction of the combined normal and mobilised frictiooal forces on the slip surface. The value of,,,. is obtained from Eq. 18.6, aOcr choosing a value of F•. Thus the reaction R is tangential to the friction cirdc. [Note. Acttml1y, the reaction R is lOngential 10 lhc friction circle of a slightly larger radius of K T sin f .... whcre K is a factor with a value g~ter than unity. as it is evident that the two reactions dH [Fig. 18:~2 (0)] inlersect slightly outside the friction circle of radius T sin !P.... lIowevd. this discrepancy is generally disregarded]. The cohesive force e". is equal to e", Ltl wbere e", is the mobilised cohesion and Lo is the length of the circular surface arc. It is oonvenient to replace this force acting along the arc by an equivalent force C acting aJoog a lioc. The Corce along arc AEB is also equal in mogniludc to the force e", )( Lc where Lt is Ihe length of the chord AD. The line of action of Ihis force can be determined by tnking moments of lhe aclual force and the equivalent force about O. (L~ )( c",) )( a _ (e". )( LJ )( r

a -

T

L. r;

... (18.24)

~ Obviously. the distance a is gremer than T, as L" > L~. The intersection of Ihe weight IV Dnd the cohesive force C". C5lablisbcs a point P through which the

,.)

Ib)

.' ~~ " o

,,'

F,

Fig. 18.12. Friction'"~Circle Method.

SOIL MECHANICS AND FOUNDATION ENGINEERING reaction R must m..', TIlC direction of R is ohlnincd by drawing a line tangential to the t-circle, The forces Cm and R can be determined from the force triangle. Fig. 18.12 (b) shows the Corce triangle. The wcight vector W is drawn first. 111c triangle is completed by drawing the vectors R and Cm along the directions already established. From the force triangle, the value oC the cohesive force C". is determined. 'll1e mobilised oohesion is equal \0 the cohesive force C", divided by the length of lhe chord L r • '£1ms

e","

c.

. .. (18.25)

Y;

1ne factor of safely with rcspcclto oohesion is given by Eq. 18.5. If the value of F~ obtained from Eq. 18.5 is not equal to the assumoo value of F., the analysis is rcpeated. 'me procedure is repeated after taking another trial surface. 111e slip circle which gives the minimum factor of safety (F,.) is the most critical circle. Generally. the analysis is repeated 3·4 times to oblain a curve between the assumed value of F+ and Ihe compuled value of F~ as shown in fig. 18.12 (c). 111e fac.:tor of safety with respect to shear strength F, is obtained by drawing a line at 45°, which gives

Fc - F . .. Ft. For a purely cohesive soil. cjI = 0 and the friction circle reduces to a poinL The factor of safcty is determined from the resisting moment due to C and actuating moment due to W (See Sect. 18.9). Sometimes, the factor of safclY with respect to friction (F'> is assumed to be unity aod the factor of safety with respect to only cohesion is obtained. . (b) Submerged slope The above discussion refers to inlergranular (effective) forces only. II the soU is dry and there is no submcrgence. the dry unit weights are oonsidered. [f the soil is submerged, the submerged unit weight is used when calculating the weight of lhe wedge (W). The neutral forces acting on the wedge are given below:

(1) Neutral part of the weight. W.. • V'i ... where V is the volume of the wedge. (2) Water pressure (lI) acting on the slope AD. (3) Resultant water pressure (Us) acting on the curved surface AEH. The three neutral forces are in equilibrium among themselves and do oot draw upon the shear strength of the soil. (e) Sudden dmwdown conditions When the water standing on the slope AD is suddenly aDd quickly removed, the water pressure force (ll) disappears. However, if there is 00 time for drainage 10 occur from the soil in the slope. the soil remains submerged as before and the neull'81 part of lbe weight (W..) is still acting. Thus, the equilibrium of tbe neutral forces is disturbed, although the equilibrium of lbe intergranular forces remains unaffected. '!llC equilibrium of the neutral forces can be maintained only if the soil can mobilise additional cohesion (co) such thai the additional oohesive force CilLo. the neutral part of the wcight (W..) and the new resultant water pressure (Us) are in equilibrium. 1bc additional cohesion can be determined by drawing another force triangle. Tbe total oohcsion mobilised (c",') in this case is equal to the cohesion mobilised for intergranuiar forces and that for neutrnl forces. ... (18.26) c,/: .. cM + c"

Thc factor of safety with respcd to cohesion is given by F~

_

~

_ _c_

... (18.27)

cm+c" Thus the factor of safety of the slope is considerably reduced during sudden drawdown conditions. Cm

STABILITY OP SLOPES

\

453

18.11. STABILITY CHARTS The stability number (5,,), as defined in Sect. 18.8, is given by ... (18.28)

The reciprocal of lhe stability number is known as stability factor. The stability number is a dimensionless quantity. Taylor dCICmlined the values of S,. for finite slopes using the friction circle method. Slopes that are of simple sections and of homqgeneous soils may be analysed using the slope stability charts given by Taylor. The charts are prcp..1red indicating the stability number, and slope angle i for various values of
+'"

0·35r----r-,..----,---r----,.---r-----,r----r---,

vi ~ 0·30'f----l--+-+-+-+--+--f--j--I " 0·25 U~I~ v'f

0·20f----!,=,,\--+--I--+--+:_'f'c7"'-f7"~

8 0.15 ~

:a al

0·10 0·0

SLOPE

ANGLE

i

Fig. 18.13.

parameter DI beoomes also important (Fig. 18.14). The parameter Dr depends upon the depth of the hard stmlum below the lop of the slope, and is given by Dr _ Depth of hard

st~~;h~~~O:O~ lOp of the slope

When the slope is steep, the failure surface passes through the toe, whereas for the flatter slope. the failure extends below the toe. The chart in Fig. 18.13 is based on the most aitical circle pa5Sing through the toe of the slope. For slope angle i greater then 53°, the toe failure occurs. For; :$ 53°, and small values of 'Pm , a more critical surface may pass helow the toc. The chart in Fig. 18.14 is appl icable for
=

=

U... (1) The stability number can be used to detcnnine the factor of safety of a given slope. For the known values of j and

'54

F, - -::. - So ~ /I

... (18.29)

If CJIm .. 0, the chart in Fig. 18.14 (or Table 18.2) is used to detennine the stability no. (S.. ) for the given values of j and Dt The chart can also be used to determine the distance nH from the toe where the slip circle culS the horizontal line. 0.20' , 0.19

- - , - -- , - - . , - - - - , - -- , - - , - - - ,

f--+-+--+--I---j--f----1

ul~:::~~"'~""~~ t' O.l6~~"'!"'''1.1.~ ;..~.x. '"

I

.

X~·

~y _ "n.l

::: 1/ X

/.- ----I', V

""11 /

'<;'

. / rx

~

'" /

:::~ I / '.00:'

-...Q.:J

.

/

,:'

1,iY'-+--+--+--+--

/

Us~

I--

.~.

0,--3

lull

lines

choin dolled

for

Sn

USe' doshed

lint'S lor n

Fig. 18.14. Stability Chart

lor

(+'" -

linn

Sn

0).

(2) 1be stability charts can also be used to determine the steepest slope for a given factor of safety. 10 this case, the stability number is comJXlled from the relation

S" ..

F~H

For the computed value of S,. , the value of i is read from tbe stability chart for the given value of'm. Tables 18.1 and 18.2 give the values of stability numbers. Thble 18.1. StablUty Numbers (see Ag. 18.13)

~

..,,'

90" 75 0

"

3D"

15"

0"

'"

10"

,,"

0 .261 0.219 0.191 (0.170) (0.156) (0.145)

0.239 0.195 0.162 0.136 (0.110) (0.068)

0.218 0.173 0.138 0.108 0.075 0.070

0.199 0.152 0.116 0.083 0046 (0.023)

20"

25"

0.182 0.134

0.166 0.117 0.079 0.044 0.009

0.097 0.062 0.025

[Note. figures In brackets are for ttie most dangerous Circles through the toe when a more dangerous arcle eXIsts below the toej.

.,5

STABIUTY OF SLOPES

ThbIe 18.2. StabUity Numbers for CohesIve soils (IJm .. 0) a nd i :so 53° (see Flg. 18.14)

~ .,.""

30" 225"

,,"

75"

1.0

/ .50

2.0

3.0

0.181 0.164 0.133 0.113 0.083 0.054

0.181 0.174 0.164 0.153 0.128 O.OSO

0.181 0.177 0.172 0.166 0.150 0.107

0.181 O.lSO 0.178 0.175 0.167 0.140

0.181 0.181 0.181 0.181 0.181 0.181

The following points should be carefully noted. (1) lfthe factor of safety with respect to friction Ftpisunity. til". - til. (2) If the factor of safety with respect to shear strength is required, a trial and error procedure is adopted. A value of F. is assumed, and the value of is used to determine Fe from the stability chart. If Fc is not equal to a'iSumed value of F .. another value of F If> is assumed and the procedure is repeated. At least 3-4 trials arc required to obtain a curve between Fc and F.. and to gel the correct value of Fe from the curve, as shown in Fig. 18.12 (e). (3) For a submerged slope, tbe stability number is computed using the submerged unit weight (y'). The angle of shearing resistance should also be for the submerged conditions. (4) For a sudden drawdown case. the stability number is oomputed using the saturated unit weight ('i~a)' The weighted angle of internal friction as obtained below is used for finding out the stability number.

+".

tan,. _ .

or

,

...

_

L

'f..

(-F1• tan
L, _ L 'fsDI'"

'isa

(i:.) F.

... (1830)

...

[18.3O(a)1

where . ' is the effective angle of internal friction. (5) For purely frictional soils, tile cohesion inlercept (e) is zero. As the stability Dumber reduces 10 zero, the stability charts cannot be used for such soils. (6) The values of shear strength parameter (c and .) should be obtained from tile teslS conducted in the laboratory simulating the drainage conditions in the field. For example, for checking the stability of a slope of oohesJve soils just after the constructions, the relevant drainage colXlltions are unconsolidated undrained conditions For long tenn stability, oonsolidated drained conditions are relevant.

18.\2. SWEDISH CmCLE METHOD The actual shape of a slip surface in the case of fmile slopes is curvilinear. For convenience, it is approximated as circular. The assumption of a circular slip surface and its application for stability analysis of slopes was developed in Sweden. The method is known as the Swedish circle method or the method of slices. Fig. 18.15 (a) shows a slOpe. Let AD be a circular surface with radius r and centre O. The trial failure wedge above the slip surface is divided into vertical slices by drawing vertical lines, as shown. '[be slices are usually of equal width, but not necessarily so. In case of non-homogeneous slopes where the s lip surface pa
456


(b,

{o,

Fig. 18.1S. Swedish Circ:te Melhods.

(1) Weight (W) acting vertically through its centre of gravity. (2) Cohesive force (C) acting along tbe curved surface in the direction opposite to the direction of

probable movement of the wedge. (3) Reaction (R) at lhe base inclined at angle 4' 10 the nonnat. assuming the slippage is imminent. {4} Reactions on the two vertical sides of the slice due to adjacent slices. However, in the Swedisb circle method, it is assumed thai the reactions on the two sides are equal and opposite and are, therefore, in equilibrium and do not affect the stability of the slice. Accordingly, only the first three forces are considered for the analysis. The weight W is resolved into its normal component (N) and tangential component (1). Let us take the

moments about the centre of rotation 0 of all the 3 forces. Actuating or overturning moment, Mo - T )( r

...(a)

The moment due to N-componenlS is zero, as those components always pass through O. Resisting moment. MR '"' (C AL) )( r + R (rsin +) where tJL is the length of the curved surface of the slice.

.•.(b)

Resolving the forces is radial direction.

From Eqs. (b) and (d).

N- Rca;, or R -Nlros; Rsin+ .. Ntan+ MR ,", (CAL)r + Nrtan+

... (c) ...(d) ... (e)

The factor of safety for the slice is equal to the ratio of the resisting moment (MR) and \.he ovenuming moment (Mo). Thus

F, '"' dcAL ;rNtan+1 .. cAL

+rNtan +

The factor of safety of the entire wedge is given by

F~ .. l:c AL

;i

Ntan ;

... (18.31)

,>7

srADILITY OF SLOPES

If c and

+are oonstant,

cL,,+tan,IN F, ~T

.. . 18.32)

where L" = length of the entire slip surface _ I AL lbe components N and Tare detennincd ' by drawing force triangles as shown in Fig. IS.15 (b). If tbe angle a wbich the noonal makes with the vertical ~ measured, tbe components can be oomputed as under. N-Wrosa, and T-Wsino. lbe length IlL of the arc ~ given by b sec a, where b ~ the width of the slice. Therefore, Eq. IS.31 can be written as F. _ 1: c b sec a

;;J::;'

0)

tan t

... (IS.33)

It may be noted that the tangential oomponent T may be negative Le. in the direction opposile 10 that of movement for some of the slices near the toe. The procedure can be summarised as under: (1) lake a trial wedge and divide it into 6 to 12 vertical slices. (2) Delennine the weight of each slice and its line of action. For convenience, the weight is generally taken proportional LO the middle ordinate of the slice and il is assumed to have line of its action through the middle of the slice. (3) The weight is resolved (analytically or graphically) into nonnal and tangential components. (4) The curved length AL of each slice is measured or oompuled. (5) The factor of safety is detennined from Eq. lS.33 or Eq. IS.31. The calculations are generally done in a tabular fann. 1be stability analysis is repeated for a number of lrial surfaces. The circle which gives the minimum factor of safety is the most crilical circle.

Location or Most Critical Circle In order to reduce tbe number of trials to locate tbe most aitical circle, the Fellenius line AB can be drawn (Fig. IS.16). Fellenius has shown that the centre of !.he most critical circle lies on this liDe. For drawing the Fellenius line AB, the point B is located at a depth H and at a distance 4.5 H from point P at tbe toe of the slope, where H ~ the height of the slope. The point A is located by drawing two lines PA and QA, where PA makes angle a with the slope line PQ and QA makes angle ~ with the horizootal at Q. The angles a and p are obtained from !.he table given in Fig. IS.16. The angles depend upon the slope. The ceDtre of the most criUml circle may lie anywhere on the line.AB or its extension. The centres of

Fig. 18.16. Fellcnius Unt.


.58

trial circles are taken or. this line shown as 0",

a and 0'. The factors of safety obtained when these were the

trial centres are genernlly plotted ao; nonnals to the line AB to obtain a curve of F•. The centre axrespooding to the minimum factor of safety indicates the most ailical circle. For a purely cohesive soil (, - 0), the point A itself represents the centre of most critical circle. The Swedish circle method is a general method of slope stability analysis. It can be used for non~ homogeneous soil masses, stratified deposits, fully submerged or partly submerged conditions. 'The method is

also applicable when seepage occurs and pore pressure develops in the soil ma<>ses, as explained taler. However, the method is necessarily an approximate one, as it neglects the effect of forces ac'Jng on the sides of the vertkal strips. Forunalely. the method errs on the safe side, i.e. the fador of safety ootained is generolly less than thai obtained from the more accurate methods, such as Bishop's method, which also consider the forces on the sides of the vertical strips. Use of N-Cune and T-Curve The values of l:.N and IT in Sq. 18.31 can also be detennined by drawing N and T curves. The sector AnD (fig. 18.17) has been divided into 6 slices having equal width. The weight W of each slice is proportional to its central ordinate (Le. height at mid-width), shown dotted. The weight of each slice is

Fig. 18.17. N and T CUlVe8.

resolved graphically into nonnal and tangential components by drawing a normal to the slip cirel.. I.brough the point where the central ordinate of the slice intersects the circle, and drawing the force triangles (shown dotted). A horizontal base line is selected and N components of the various slices are plotted directly below the vertical tines of the sector. The extremities are joined by a smooth curve, and thus N-OJrve is obtained. Ukewise, T-curve is obtained. It may be DOted that Tr·component in this case is negative and plotted below the base. Now "EN-ANxl where AN = area of the N-diagram (m\ .., = unit weight of soil.

. .. (18.34)

459

STABILJII'Y OF SLOPES

Ukewise., I.T .. AT )( 1 [Note. If AN and AT are measured in an Z, they are multiplied by

:l

... (18.35) where x is the scale (1 em = x

metres)]. The areas AN and AT can be measured by ' means of a planimeter or by using a graph paper.

Use or Rectangular Plot Prof. AIam Singh (1962) devised a simple method for determination of"i:.N and l:.T without using a planimeter. Let ZI, Z2o'" Zs be the end ordinates of the 6 slices shown in Fig. 18.17 (a). Let b the width of each slice. The total weigbt IW of the sliding wedge may be written as IW .. by

0)] - 2Z,) - • (Z,. - 2Zs) - • (ZS' - 2z.) - • (Z -•• 2zs) - • (Zs. -2[( -0.2Z,)- ' (Z,.

I W .. by [ZI -+ Z2 -+ Z3 -+ l4 -+ lsI If the last slice is of smaller width equal to m )( b, the above expression is modified as I W .. by [ ZI + Z2 -+ l) + Z4 + (

¥)

Zs]

... (18.36)

... (18.37)

The value of IW can be determined by drawing a rectangular plot of width b and having boundary ordinates 1 b l2 .. . etc. as abscissae. In the case of last slice being of widtb mb, the last ordinate is multiplied by (1 + m)/2 before plotting. The area of the diagram is proIX'rtional to IW. If the vertical ordinates are resolved along the normal and tangential directions, the N-components and T..c;omponents a(e obtained. The rectangular plot of N-components is then drawn, tak.ing the width of the plot equal to tbe width b of the slice [Fig. 18.18 (b)]. The normal oortlIX'nents N"N2o etc. are plotted as abscissae. Likewise., the rectangular plot of T-comIX'nents is drawn. In T-plot, as Tl oomIX'nent is negative, it is plotted in the opposite direction. The net area of T- plot is shown hatched [Fig. 18.18 (c)]. The values of


460

TN and IT are obtained using Eqs. 18.34 and 18.35, where "( is the unit weight and AN and AT are the areas of tbe N.plot and T- plot, respectively. The rectangular plot method greatly simplifies the calculations for determinations of the areas of N and T- diagrams. It may be noted tbat the weights in the rectangular plot are proportiona1 to cod ordinates., and not the mid-ordinatcs.

18.13. STABILny OF SLOPE UNDER STEADY SEEPAGE CONDmONS The stability of slopes of an earth dam is investigated for the steady seepage conditIons, sudden drawdown conditions and during construction conditions. The steady seepage condition is coosid\%ed in Ibis section. When the rescNOir on the up>lream of the dam is filled, water starts seeping through the dam. After sometime, the steady seepage oonditiol'l'> arc established and a well-defined phreatic line is formed. The soil below the phreatic line is saturated and subjected to pore water pressure. On the upstream slope, tbe seepage forttS are directed inwards and hence: tend to increase the stability. However, on the down-stream slope. the direction of the seepage forces is such that they decrease tbe stability. 1be steady seepage condition is, therefore, aitical for the downstream slope of an earth dam. Fig. 18.19 (0) shows the downslream slope of an earth dam provided with a horizontal fIlter al its loe. The flow net is drawn using the methods discussed in Chapler 9 . 'The boundary pore pressures acting on the slip surface arc oblained from the flow net. Fig. 18.19 (b) shows the enlarged view. The pore pressure", at point-l where Ihe first cquipotentialline (h/9) cuts the slip surface is equal to the vertical distance between the point-l and the point P where the equi(K)l:entiai line intersects the phreatic line As the pore pressure acts normal 10 the surface, a line equal 10 "I is drawn normal 10 the slip surface at point-!' likewise, the pore pressure "2 at point-2 is found. The pore pressure diagram is then drawn joining Ihe extremities of all these lines. The pore pressure diagram is shown hatched in Fig. 18.19 (b) . Eq. 18.32 can be used to detcnnine tbe factor of safety. However, in this case the IOtal weight of the slice is due to bulk\.unit weight above the phreatic line and the saturated unit weight below the phreatic line. The N-componcnt of the weight is reduced due to pore pressure below the phreatic line. 1lterefore,

o

.. 8h

(b)

EQUIPOTENTIAL LINES

Fig. 18.19. Stelldy Seepl'lge Condltions.

461

srABILITY OF SLOPES

cL. + laD' I(N - U) F -

... (19.38)

IT

where W is obtained from the area of the pore pressure diagram using a plaimcter or by rectangular plot method. It represents the lOtal force due to ?Jre pressure. Alternatively, Eq. 18.38 can be wrillen as

Fs _ Ie!J.L + Ii~

-

uD tan,

... (18.39)

where u is the average pore pressure on tbe slice and I is the curved length of the base of the slice. (Note. J _!J.L _ b sec a). J8.14. STABILITY OF SLOPE UNDER SUDDEN DRAWDOWN CONDmONS The critical cooditioo for the stability of the upstream slope of an eanb dam is when there is a sudden dmwdowo in the reservoir upstre.1m. If the soil is of low pcnncability, 00 appreciable ch..·mge in tbe saturation level inside the slope takes place when the reservoir level goes down. The weigbt of water which is still present in the soil tends to cause sliding of the wedge, as the water pressure wbich was acting on the upstream slope to balance this weight has been suddenly removed. Acrording to another interpretation, the shearing resistance of the soil is considerably reduced due to pore pressure existing in the soil, whereas the disturoing [oree due to saturated weight of the soil remains the same. The flow net for the sudden dmwdown condition enn be drown using the electrical analogy method (Clap. 9), and the pore pressure acting on tbe base of the various slices dctcmlined. "{lie stability of the slope is investigatc:d using Eq. 18.38 or Eq. 18.39. However, the method is not coovenienL

\

Fig. 18.20.

The pore water pressure (u) can also be estimllled using Skempton pore pressure coefficients (Otapter 13). Fig. 18.20 sbows the upstream slope of a dam . The pore.water pressure at ally point P before drawdoWll is given by

.•• (18.40)

"" - y.(h + h. - h') where h = height of soil above P, h"" :c: height of woter column above P, and h' = loss of bead due to seepage, indicated by the equipotential line passing through P. It is assumed that the total major principal stress (oJ at P is due to the weight above. When there in a drnwdown, the major principal stress decreases by /l01 - -"f""h..,

and the change in pore water pressure is givell by /lu - O!J.O l -

where

-By..,h..,

Ii is the overall pore pressure coefficient, related to coefficients A and B, as jj - 8 [ 1 - (1 - A) ( 1 -

!~) 1

... (18.41)


462

Therefore, tbe pore water pressure at P immedintely after drawdown is given by

.-.+6.-~0.~-~-i~~ • - 1. [h • ~ (1 -

or

The value of

i) -

h']

...(18.42)

B is sligbtly greater than unity. A con<;ervalive value of Bequal to unity is generally taken.

• - 1. [h - h'] Thus the pore water pressure u at various points on the slip surface is detcnnined. The factor of safety is then obtained from Eq. 1839. ' [bus

... (18.43)

Approximate method An approximate alternative method for investigating the stability of the upstream slope under sudden drawdown conditions is to oonskler tbe saturated unit weight of soil (or calculating the driving forces (i.c.l: 1) and the submerged unit weight for calQllsling the resisting forces (i.e. IN') in Eq. 18.32.

F, _ cL Q + ~a; ~IN

Thus

... (18.44)

In other words, it is assumed that the full pore pressure acts even after the drowdown and the soil is fully saturated. Below the drawdown level, as the water pressure on the slope is still aaing. the submerged unit weight is used for both the driving forces and the resisting forces. This is similar to the case of a submerged slope as discussed in Sect, 18.5. Ir the slope material has high coefficient of permeability, it drains as fast as the reservoir level goes down, tben both the driving and resisting forces are calculated using the bulk unit weight. If the slope material is of medium permeability, the saturation line moves downward at a rate depending on the permeability of the soil. A series of flow nets can 00 drawn for different posillons of saturation line and the correspooding pore pressures determined and the fador of safely is estimated.

18.15. STABlL1TY OF SLOPES DURING CONSTRUC-nON When an eartb dam is buill of the soil of low penneability, excess pore pressures develop in the air and water voids due to compaction carried out during construction or due to its own weight. The pore pressure developed depends upon the placement water rontent, method of compaction, weight of the overlying layers and the rate of dissipation of pore pressure during construction. When the placement water cootent is more than the optimum aod there is 00 proper drainage, injtiai pore pressure at any point may be as high as 100 percent of the weights of the overlying layers. An estimation of initial pore pressure may be made using Bishop's method of predicting pore pressure. According to which, the pore pressure (1.1) at nny point can be written as ... (a) where Ito :::: initial pore water pressure, and II 1.1 = change in pore water pressure. In terms of the change in total major principal stress Il 01' Eq. (0) can be written as 1.1 - Uo +B A 01 where is the overall pore pressure oocfficient. The increase in lotal major principal stress II 01 is approximately equal to the fill pressure (yh). Thus

B

• - •• i (1 h) ... (18.45) As the soil is partially saturated when compacted, the initial pore water pressure (1.10) is generally depend upon the placement water content. For high water content, negative. The actual values of Uo and 1.10 may be zero. Thus

B

• - i (1 h) ... (18.46) The value of jn Eq. 18.46 must be that corresponding to the stress condition in the dam. It can be determined from undrained triaxia l tests on rompactcd specimens with pore pressure measurements.

B

463

SfABILn'"Y OF SLOPES

Alternative method The pore pressure during construction can be delermined from Hill's equation:

p" V" u - h~ Vw

... (18.47)

where p" = air pressure in the voids of a soil mass after initial compaction (absolute pressure).

and

V" :: volume of air after initial compaction, in percent, Vw = volume of water after initial compaction, in percent, he "" Henry's constant of solubility of air in watcr by volumc ( := 0.02 at w aC) u :: pore water pressure wilen the soil mass hm; been consolidated to complete saturation.

When the soil is not consolidated 10 complete saturation, the pore pressure can be determined using tbe following
... (18.48) where d :: embankment compression, in percent, of the original total embankment volume. A graph is generaUy plotted between the effective stress 0 and the percent consolidation d, from tbe results of consolidation tests conducted on the samples. The values of u are obtained from Eq. 18.48 for different values of d. The total stress is given by

A plot is then made between the total stress 0 and lhe pore pressure IL lbis plQ( is used for the determination of pore pressures at various points in lhe dam during construction from the values of the tOUlI stress o. Obviously, lhe total stress at any point in the dam is
18.16. BISHOP'S SIMPLIFIED MEllIOD The conventional Swedish circle method (Sect. 18.11) satisfies only the overall moment equation of equilibrium . It neglects the moment
F& - sIt",

... (18.49)

The (orces acting on !...Je slices are given below [Fig. 18.21(a»). (1) Weight of slice, W .. 'f h b, where h is the average height. (2) Nonnal force 00 the base, N' - N - ul, where u is the pore pressure, and I is the length of the base (=b=a). (3) Shear force on lhe base, T _ 't.. I. (4) Normal forces on the sides EI and £2. (5) Shear forces on the sides Xl and X2. (6) Any external force adiog on the slice. The proolem is statically indeterminate. Th solve the problem, a number of simplifying assumptions are made regarding the interslice forcesX"Xz,E t and £2_ Fig. 18.21 (b) shows a slope. Tldcing moments about O.

ITT - IWrsina

SOIL MOCIIANICS AND FOUNDATION ENGINEERINO

1-'-1

"QI-',

~t

~

w

l::tlMe":

"

H::H-U[

Fc,N'::N-ut C.)

,b)

Fig. 18.21.

T ..

Du' 1bcrcfore,

't,..

f-

f.;

Ie

t

l:f;)(I)(T-~WXrXSina

F II.;:m1a .. ICCI +W~i~~/,)I

or

6

..

F .. Ic'1 + tan . ' 1:.N' wbercN' .. 01 ... (1850) , IWsina Sq. 18.50 is exact. However, approxim3tiOns are introduced in tbe determination o[ the normal component N. In the oonvenlional Swedish circle method, the resuUnnt of inLer-slice forces b Laken as zero. Therefore, N' .. !V cos a. - uJ

Thus,

F, _

f~

c L. + "'n~+~~~:= a- uQ

_ c' L. +

+/

"'"~

(same as Eq. 18.38)

(N - U)

In Bishop's simplified method, it is assumed thai the ra;uilant forces Oil the sides of the slice are horizontal, i.e,XI - X2 .. O.

Resolving the forccs in the vertical direction,

W .. N'cnsa + ulcosa + Tsina + Xl - X 2

Substituting

T ..

-f.; x I

w.

N'rosa + ullXlSa + (d I +

and

XI -

W .. Ncosa+ ulrosa + _ "or

X2

..

0, we have

~tan+') sina

%! sino. + !f;tan+' sina

.) c'tsina N ' ( coso.+ ~ F, sm a .. W-ufcosa-~

46'

STABIlITY OF SLDPES

W-uloosa- c'lsina N' _ _ _ _-,--,,-~F..!,_

oosa +

~

... (18.51)

sino.

Substituting the above value of N' in Eq. 18.50,

W- uloosu - C"SinUl Ic'l+tan,'I

[

coso. +

~ . F,n,.F' sma

~-----~~t~w~s~m-u----~

-

IW~ina

lan~' (IVSCCU c'bscca +

I

1

[

1 [{, - In'sina I c bSIXa -

ulF,

c',tanutan+,jF, } 1

+

T) 1

tan4l'tana +

lan"lana F,

[{ c'bSIXa(t + tan a tan ,'IF,) -

1

- IWsma x 1:

I

tan.'(IVSCCU-UI)]

+

1 +

tancflana F,

c'bsccatana~}'

tan4l'tana + F, + tan4l'(Wscca -

___l_t[ IWsina

c'bseca

+tan,'(Wscca

1+ tan.~:ana

-~ t[IC'b + (IV - ub)tan~'1 sma F, _

UbSCCa)]

1 + tan 41~~an a ubsecal]

1 + tan.;,~taDa

:::'' ON]

1+~

I...!.. [c'b + (lY - ub)tmt,']

ma

IWsioa

+'

where ma = (1 + tan tan o./F,) cos a Sometimes, the pore pressure u is expressed in tenns of pore pressure ratio r.,. In that case,

... {I8.52) ... (1853)

u _ r.,yh .. r.,(Wlb)

Eq. 18.52 gives the factor of safety of the assumed failure surface. As the factor of snfcty (F,) appears on both the sides, a process of successive approximation is required. A value of P, is assumed and the analysis is done. The value computed from Eq. 18.52 is compared with the assumed value. If the two values


466 ~iffer,

the process is repeated till convergence. As the convergence is rapid, only 34 trials arc required. A

c'omputer may also be used.

The effective stress analysis is generally done, but the total stress analysis is also paiSible. The factor of safety determined by Bishop's simplified method is an underestimate and, therefore. it errs on the safe side. The error is genemUy less than 2% and not more than 7% even in an extreme case. 18.17. OTHER METHODS OF ANALYSIS

A number of investigators have developed different methods for the analysis of slopes. The methods are similar in nature to tbe Bishop methods but diller in handling of interstice forces. (1) Spencer (1967) assumed that the forces on the sides of the slices are parallel [Fig. 18.22 (a)]. The

method satisfies Ihe requirement of equilibrium of forces and moments. (2) Morgenstern and Price (1965) developed a method in which all boundary and equilibrium conditions are satisfied. The analysis imposes normal and shear forces on tbe sides and also includes water pressure effects [Fig. 18.22 (b)]. The method satisfies the requirement of equilibrium of forces and moments which act on cl slice. The failure surface may be of any shape. .

' if "

,

'01-U' ," 1

_ _ E,

~~=::l

~ (,J

~ (OJ

Fig. 18.22 The soil mass above the failure surface is divided into sections by a number of vertical planes. The problem is made statically determinate by assuming a relationship between the forces E and X on the vertical sides, as .

X. ).[(x)E where I(x) is an arbitrary function describing the pattern in which the ratio XIE varies across the soil. The value of the scale factor ). is obtained as a part of the solution along with the factor of safely. The values of the forces E and X and the point of application of E can be determined at each vertical boundary. For any assumed value of I(x), it is necessary to check that the solution is physically reasonable and no shear failure or tension develop within the soil mass above tbe failure surface. As the choice of f(x) docs nOI influence the computed values of the factor of safety by more than S%, for oonvenience I(x) is taken as unity. The analysis involves a complex process of iteration for the values of ). and Fs. The usc of a computer is essential. However, slope stability charts which simplify the solution have been developed for the design office usc. (3) Bell (1968) developed a method in which alllhe conditions of equilibrium are satisfied. The soil mass is divided into vertical. slices. The failure surface may be of any shape. The statical determinacy is obtained by means of an assumed distribution of normal stresses aloog the failure surface. The soil mass is considered as a free body, as in the case of the friction circle method. (4) Cousins (1978) developed charts which have applicatioo to 8 wide range of field problems. The cbarts are quite convenient, as these deal in easily calculated parameters. Cousins charts are based upon the friction

SfABIut:Y OF SLOF£S

467

circle method of analysis discussed earlier. However, !he charts are for a general case. These can be used for the evaluation of long·term stability analysis in tenns of effeaive stresses. The charts can be used even in case of tension crncks. (5) Stability charts for tbe analysis of earth dams under sudden drawdown conditions have been developed by Morgenstern (1963). The charts are based upon an effective stress analysis using the method of slices.

18.18. IMPROVING S'D\BILITY OF SLOPES The slopes which are susceptible to failure by sliding can be improved and made usabJe and safe. Various me!hods are used to stabilise the slopes. The methods generally involve one or more of the following measures, which either reduce the mass which may cause sliding or improve the shear strength of the soil in the failw-e zooe. (1) Slope flattening reduces the weight of the mass tending to slide. It can be used wherever p
ILLUSTRATIVE EXAMPLES Dlustratlve Example 18.1. A. long natural slope in an OW!f'C(}nsolidated clay (c' _ 10kN/ m 2, ,' _ 25 0 , 2 1_ - 20 kNlm ) is inclined at J{r to the horizontal. The water table is at the surface and the seepage is para/lei to slope. If a plane slip had developed at a depth of 5 m below the surface. determine the factor of safety. Take '(... _ 10 kN/mJ.

w


F~ -

c' + y' llOO52itant' ls.Hc:osisini

_ 10 + (20 - 10) )( 5 )( 0052 10° tan 25° _ 1.90 20 )( 5 )( cos 10° sin 100 f DlustcaUve Example 18.2. A. vertical CUI is made in a clay depasif (c 30/cN/m2, , Fin the maximum height of the cut which can. be temporarily supported.

=

F~ ~ H


S" -

TakingF~

S" - 1.0)(

- 1.0,

=0, '( =16 lcN/mJ).

~6)( H

From stability chart (Fig. 18.13 or Thble 18.1), S" _ 0.261. Substituting the value of S" in Eq. (a),

.. .(a)


468

l~x

H .. 1.0 x

0.261 .. 7.18m

D1ustratlve Example UU. A cut of depth 10 m is made in a cohesive soil deposit (c = 30 leN/m2, " = o and y = 19 kNlm). There is a hnrd stratwn under the cohesive soil oJ a depth of 12 m below the original ground surface. If the required facror of safety is 1.50, determine the safe slope. Solution. In this

ca<;e,

DrH = 12 m and H

From Eq. 18.28,

Stl .. Fe

= 10 m.

~H

Therefore, D, .. 1.20.

.. 15

:Ie

~

)(

10 - 0.105

From Fig. 18.14, for Sit = 0.105 and D, = 1.20, we have j ::; 15°, D1ustratlve Example 18.4. Determine the foctor of safety with respect to cohesion for a submerged lO°and Y6'" = 18 kNlm 3). embankment 25 m high and having a slope of 41r. (c = 4() kN/m 2, Solution. From Eq. 18.28, using submerged unit weight,

+..

S _ IJ

For

40 0.195 Fe )( (18 9.81»( 25 .. ----p;-

... (e)

+= 10° and i = 40°, Table 18.1 gives S" = 0.097 by interpolation.

Therefore,

0.097

= O.;~5

or Fe :::: 2.01.

Illustrative Example IS.5. Determine the /acUJr of safety with respect to cohesion if an embankment 25 m high and having a slope of 40° is subjected to sudden drawdown. (c '" 40 kN/nl, 4' '" 10°, "(,at '" 18 kN/ml). Solution. From Eq. 18.30,

TakingF.u 1.0,

tan,. - L (F~+ tan,,) "(,//I tan ,. _ (18 ~t81) (tan WO)

From Table 18.1 (or Fig. 18.13), SI! = 0.13 for From Eq. 18.28, using saturated unit weight, 0.130 ... Fe

X

(JIm'"

~

)(

or t. _ 4.59"

4.59° and-; '" 40°.

25

Fe ... 0.684 < 1.00 ·50

The slope is not safe.

"-

DIustrative Example 18.6. Determine the factor of ·40 safety with respect to shear strength of a slope 10 m high and luwing an inclination of 4()0 of a soil with c '" 30 kN/ml, 4' = u1' and"( '" 19 kN/mJ. ·30 Solution. The factor of safety is obtained by trial and , error. Lei us assume F I¥ = 1.30 for the fU'St trial. TI From Eq. 18.4,

lao

4'......

~ ... ta~;:

'III . . 7.72°

From Thble 18.1, for

t ..... 7.72°

0.U5 ..

.to/,~P'

/

and i ... 40°,

S" ... 0.115

From Eq. 18.28,

·20

'C

X

Fe~H

100

1-10

I 20 Ft

1·30

-

Fig. E·18..

1.lo0

1·50

STABIurY OP SlDPES

469

0.115 • Fe )(

~ )(

10

or Fe - 1.37

Ukewise, for F+ =- 1.40, the value of Fc is . found 10 be 1.32 and for F+ ::: 1.20, Ihe value of Fe is computed as 1.44. Fig. E-18.6 shows the plot between Fe and F+. The line drawn at 45° cuts the plot al the

required point.

Fe - F'f • F, - 1.34.

From the plot,

E_18~~.~n;:::m~X=~=t!~;i ~ai:t~ ;;:::r ::;;:fo~: i:J~~~:"::;::!I~a~~i:~ ~ !~~~lg~ 3 = 20" and y.: 17 kNlm

•

Fig. £-18.7.

Solution. Weight of triangular wedge ABD,

w_ Sliding force T

i)(

2.86)( 2.5)( 17 .. 6O.8kN

=- oomponent of W parallel to the plane AD ... W sin 25° .. 25.7kN

Length of plane AB, Resistance to Sliding,

L .. 251sin 25° - 5.91 m s ... cL + (W cos 25°) tan 4' .. 6.0 )( 5.91 + 60.8 x 0.906 )( 0.364 ... 55.51 kN

Factor of Safety,

F, -

t . 5~~71

... 2.16

=

=

Dlustratlve Example 18.8. A vertical cut is made through a homogeneous soU mass (c 20 leN/m2" 20", Y = 16.5 kNlnf). Using Culmann!f method, deJermine the sale depth 01 the cUI, laJdng alactor o/sa/ety 0{2.0.

(b) Also determine the sale depth using stability charts. 4c. sin i cos q,,,, • y [1 - oos (i - • .lJ

Solution. From Eq. 18.22 (a),

H

Now

c,. _ clFe ... 20/2 ... 10 IcN/m2

tan,.. _ tan Therefore, (b) For Therefore,

H ... i = 90",

q, = 10.31°,

,12 . .

we have

H ..

13020°/2 ... 0.364.1 ... 0.182

~.~ [~o _s::t~ ~ 1~~3~!;]

.,c;,. ..

SOl = 0.217

2O:~.217

165

or ' .. '"

.. 2.91 m

(See Table 18.1) .. 2.79 m

10.31°

47<1


Dlustratlve Example 18.9. An unsupported slope is shown in Fig. E~18.9. Determine the factor of safety against sliding for the trial slip surface. Take c = 50 kNlm2, = O. The weiglu of the wedge ADD is 2518 leN and acts at a horizontal distance of 11 m from the vertical AO.

R

lJ"

and,

Solution. Length of arc AD. .. Actuating moment, Mo .. .. Resisting moment, M, .. .. Therefore, faaor of safety

Lo _ 2 ~~ 24 x 65

27.21 m 2518 x 11 27698kN-m c Lo r .. 50 x 27.21 x 24 32652 kN- m Fe (- F,)

•

~;~;

• 1.18

:65'

................. (~.?.:;

I

...........,.,.,

,,

I

T 14m

1

"

'

I

I I I I

~11

I I

:, 50' Fig. E-18.9

Alternative method The factor of safety can be determined from the mobilised cohesion (c"J. According to lhis method, e", L.. r .. 27698 c", ..

27.;~6~ 24

.. 42.41 kN

Fe .. c/c", .. 50/42.41 .. 1.18 Therefore, Ulustmtive Example 18.1(t. Fig. £-18.10 shows a slip surface with a radius of 22 m in a slope with a height of 14 m and an tllIgle of inclination of 45". If t_ :: 15", 1 :: 18 IeNlm3 and c = 4() leN/ml, detennifll! the factor of safety with respect III cohesion using the friction circle method.

Fig. E-18.10.

STABIUfY OF SL..QPES

471

The weight of the sou wedge is 15{)() /eN and it oc:ts at a horizorual distance 0/10.3 m from A. Solution. rsin 22 x sin 15 0 _ 5.1 m. The friction circle is drawn with a radius of 5.1 m.

+_

From Eq. 18.24,

a .. r x

~

t

- 22 x (2

XX

:~~~ ;2~

.. 23.20 m

A line is drawn pamllel 10 Ihe chord at a distance of 23.20 m. The 10 lal cohesive force passes through P which is at the intersection of the vertical line through W. The reaction R also passes through P and is tangential to the friction circle. The figure also shows the force triangle, From the force triangle, C = 600 kN

c". .. From Eq. 18.5,

F~

..

f . ~2

t.

.. 25.7 kN/ml

:'7 - 1.56

Ulllstrative Exam.r.le 18.11. Fig. £·18.11 shows a lrial slip surface through a soil mass (c "" 20 lcN/m 2, IjI ::: 30°, y = 20 /eN/m j. Determine the factor of safety using Swedish circle method.

T T_DI,,"m

~

Apl6m 2

7T:16X20

--=Flg-.:-C_,'-'S.1--"

:320kN

SOIL MECHANICS AND FOUNDATION BNGINEBRING

472

Solution. 'Ilte sliding wedge is divided into 6 vertical slices of equal width of 2 m. The calculations arc shown in tabular ronn below. Tuble Example 18.11. Weiglrt (W)

Sli~

No.

Average

ordillate

llidth

Volume

1.

O.9m 2.4 m

2m 2m 2m 2m 2m 2m

1.8 m.1 4.8 m) 7.2 m)

2 3. 4. 5. 6.

3.6 m 4.5 m 4.0 m l.80m

T .. Wsina (kN)

WeightW (kN) 36 96 144 180 160 72

9.0 ml 8.0 m3 3.6 ml

9" 19" 29" 42" 55"

15 46.9 87.3 107.1 59

1"f" 315.3 kN

Nil Wcosa (kN) 36 94.8 136.2 157.4

118.9 413 IN = 584.6 kN

[Altcmnlively, from diagrams, IT = 320 leN and "£N =: 580 leN] Length of nrc, L" .. (2 It )( 12/360) )( 71" _ 14.86 m. From Eq. 18.32.

F, .. cL .. + ; ; . EN

.. 20 )( 14.86 ;It;~ 30° )( 584.6 .. 2.01. D1ustrutlve Example 18.12. Determine the factor of safety of the slope shown in fig. £-18.12. The slip surface paucs tllTOugll two layers of soils. The pore water on the slip surface is shawn by the dotted lhll!. Take y""

=:

}O leN/mJ,

T

rI Fig. &18.12.

471

STABIUfY OF SLOPES

Solution. The sliding wedge is divic\cd into 8 slices of equal width 4 rn. One ver1icalline is drawn aI the interface of the layers between slices (7) and (8). The weight of each slice is determined from the area of the slice in the layer and the oorrCS{X>l1ding unit weight of the soiL The pore water pressure is detennined a.. the vertical ordinate of the dotted line above the curved surface. The calculations are shown in tabular ronn. Total length of arc (L.J .. (2 1C x 28/360) x 75 - 36.63 m Length of arc in I-layer - (36.63/75) x 65 .. 31.75 m Length of arc in II-layer .. (36.63/75) x 10 .. 4.88 m 1: (N - ul) in 1 layer (lSI to 7th slice) :::;: 2388 k:N

1: (N - ul) in II layer (8th slice) From Eq. 18.39,

= 198 kN

F, _ 1:cAL + I.I(~ - ul) tan , F& .. (30 x 31.75 + 25 x 4.88) 1;0~ x tan 20 + 198 tan 10 0

0 ..

1.81

illustrative Example 18.13. Find the facwr of safety against sliding for the slope shown in Fig. E-18.13, using Bishop's simplified method.

II Loy« c:a 40kNlm1

. ;ZOO

__ ;19 kN lm3

I Loyt r

,.0

+sJcf'

"t ,, 21kNI~

Fig. E-18.13.

Solution. The slip circle has been divided into 5 slices; the fIrSt four stices are in Layer I. The widtb of the first slice ~ 5.6 m and that of other slices is 6 rn. CaIrulations are done in tabular form. 1be value of mo in oolumn (11) ba.. been oomputoo using Eq. 18.53. Inu" (1 + tan4"tanalF,)oosa. For first trial, F, is assumed to be 1.20.

From Eq. 18.52,

~-"-[c' b m

+ (W - ub)tan

~'J

F, - -cc,,-"----,;W"'sm"" o -ca- - -

~

Table Example 18.12. ll
Average orriiNue

mdth

=bseca

Weight W(kNi

Th/u...

"

N.W

N-ul

(kNi

(kN)

(kN)

(9)

(10)

(11)

(12)

822

-33.1 -32.1

(bi

3 (m )

(5)

(6)

(7)

(8)

128

1m

_150

4.14

41.4

123.6

(1)

(2)

(3)

(4)

1.

1.60 m

4 m

6.40

'" WsillQ (kNi

2.

4.6 m

4m

18.4

368

3m

- 5'

4.02

120.6

366.6

246.0

3.

6.8 m

4m

27.2

544

4.2m

3'

4.01

168.4

543.2

374.4

+ 28.4

4.

7.'1 m

28.8 6.6

576

S.2m

10'

4.06

211.1

671.6

4605

+ 118.4

1.4m

4m 4m

6.0 3.6

4m 4m

24.0

480 274

5.0 m

19'

4.23

211.5

712.9

501.4

+2455

6.

4.2 4.8

4m 4m

16.8 19.2

336

4.2 Itt

28'

453

190.3

618.8

4285

... 329.1

~

7.

1.6 4.8

4m 4m

6.40 19.2

1.8m

38'

5.08

91.4

386.4

295

+ 303.5

> z

8.

2.6

4m

10.4

46'

5.75

198

198

~

5.

14.4

~ 682

r-----mI

:: 128

I

~

:~ 198

I T _ UOZ.l

~

"

f.l

"g z 11" 0

z ~

0

~

z

0

SfABILITY OF SLDPES

Ht-ighl (lI?

Slice No.

47>

"''''h "'''''''' (b)

(I)

(2)

(3)

(mJ) (4)

Weight (>H) (5)

(6)

1.

2.6

5.6

14,56

305.8

- 2"

2.

6.6

6.0

39.60

831.6

3.

6.2 3.6

6.0 6.0

37.20 21.60

4.

3.0 8.0

6.0 6.0

18.0 48.0

5.

6.4

6.0

38.4

A\.'erage

ordinate

c'

ub

(7)

(8)

Wdna tan .'

.-

,,"

781.2 410.4

"'"

(W - ubi tan (12).'

(9)

(10)

(1I)

0.577

-10.7

0'.984

176.4

05Tl

+ 130.0

1.063

479.8

OSTI

+ 465.6

1.107

687.6

05Tl

.776.3

1.088

7443

.597.6

0.820

2655

~ 318 912

3'-

729.6

55"

r-wo 4.0

0.0

0364

1: 1958.8

1 [ 1 1 1 - 1958.8 0.984)( 176.4 + 1.063 )( 479.8 + 1.107 )( 687.6 •

1.~

)( 744.3 +

O.~

(40 )( 6 + 265.5) ]

- 1.30 The assumed value

of P, is not correct. The

process

may be

repeated

after taking

P, "" 1.25.

PROBLEMS A. Numerical 18.1. What inclination is required where a lilling 12 m high is to be ronstructed having a factor of safety of 1.251 2 l IS°, '1 = 17.0 tN/m . The stability number for .... = 12" is equal lo 0.063 when The soil has c '" 20 kN/m • the slope is 30° and 0.098 when the slope is 45·. [Ans. 36°). l 18.2. A cutting of depth 10 m is to be made in soil which bas c = 30 tN/m2, T = 19 kN/m and • = O. There is a hard suatum below the original soil swfaa: at a depth of 12 m. fiod the safe slope of cutting jf the factor of o safely is 1.50. For D,s 1.20, S,. .. 0.143 for i = JO and S,. • 0.101 for i = 15°. [Ana. 17"]. l 18.3. A vertical OJt is to be made in clayey soil for which lests gave c '" 30 kN/m2, 1 '" 16 kN/m and = O. Fiod the maximum height for which the CUI may be temporarily unsupponcd. For 0, and i :: 90°, the value of the stability number is 0.261. [Ans. 7.18 ro] 2 l 18.4. When is Ihe factor of safety for a 45° slope 12 m rugh in a clay (c = SO kN/m , T .. 18 kN/m and = 0) having a rock: stratum al a depth of 12 m below the toe? For Df Z 2.0 and i '" 45·, the wlue of S" is equal 10 0.177. IAos.1.3J 11.5. Determine tbe factor of safety with respect 10 cohesion only for a submerged embankment 25 m high whose u~m face Iuls an inclination 0145°. 1be soil has the following properties; c :a 40 kN/m2, • = 10°,16. "" 18 l kN/m . The relevant stability number is equal to 0.108. (ADs. 1.85) 18.6'. What is the factor of safety if the embankment in Prob. 18.5 experiences the effect of sudden drawdown 1 FOI' ...... 4S and j ::: 45°, the value to the stability number is 0.136. [Aos. 0.65] 2 18.7. A CUI 10 m deep is to be mooe in a stratum of rohes.ive soil (c ::E 35 kN/ro , y = 18.5 kN/ml and 0). The bed rod:: is located IS m below the original grouod surface. Determine the factor of safety against failure if the slope is 30°. For Df::: 1.5 and i " 30°, the stability number is equal to 0.164. [Aos. 1.15] 18.8. An embankment 10 m high is inclined al 35° to the horizontal. A stability analysis by the method of sliocs gave the follOWing forces:

+'"

+""

+

+

+..


476

l:N _ 900 kN, l:T: 420 kN, 1:U a 200 kN. If the length of the failure arc is 23.0 m, find the factor of safel)'. The soil has c .. 20 kN/m2 and

+_15".

[Ans.l.54J 18.9. A dam of homogeneous section is 25 m high with upstream slope of 2S to 1.0 and downstream slope of 2 to 1. There is a 12 m long horizontal filler 31 the downstream end. TaJcing a free board of 3 m (i.e. water depth 22 m), dclennine the (a) faClor of safel)' of downstream slope under steady seepage conditions. (b) fador of safety of upstream slope under sudden drawndown ronditions. 18.10. For the dam in Prob. 18.9, determine the factor of safety of downstream slope under steady seepage condition using Bishop's simplified method. 18.11. A 40" slope is excavated to a depth of 8 m in a deep layer of saturated etay (c ::: 70 tNh1? and • ::I 0, Y = 19 kN/m\ Determine the factor of safety (or the trial failure surface shown in Fig. P. 18.11. (Ans.2.87]

T 8m

Fig. P. 18.11 18.12. Determine the faCIO!' of safcly of [he slope given in Prob. 18.11. If too soil has c = 20 kN/m2; 3 kN/m . Use the friction circle method.

+= 15°, Y ::: 20

n.

Descriptive and Objective TYpe 18.13. What are the assumptions that are generally made in the analysis of the slability of slopes?

Di~ briefly their validity. 18.14. What are different fsctM o( safety used in the stability of slopes? 18.1S. What nre differenl types of slope failures? 18.16. Derive an cxpressioo for the (aaor of safety of an infinite slope in a c:ohcsionless soil. What is the effect of steady seepage parallel to the slope on the stability? 18.17. Discuss the mcthod for dJealng the stability of an inlinire slope in a cohesive soil. What Is ulrical height?

18.18. Describe Culmann's method for the stability analysis or homogeous slopes. What are its limitations? 18.19. Discuss the friaion circle method for the SIlIbility analysis of slopes. can this method be used (or purely cohesive soil? 18.20. What is a stability number? What is its utility in the analysis of slability of slopes? Discuss the uses of stability

""o1S. 18.21. How a slope is analysed using Swedish circle method? Derive an expresion for the factor of safety. 18.12. Describe Bishop's simplified method. What are its advantages over conventional S..vedisb circle method? Derive an expression for the factor of safety. 18.23. Discuss the various methods for improving the stability of slopes. l8.24. Write whether the following statements are true or false. (a) The friction circle method can be used for 8 non· homogeneous soil mass. (b) The Stllbility numbels can be used for the analysis of purely cohesionless soil Slopes. (c) The factor of safety of an infinite slope of a cohesive soil depends upon Ihe height H of the slope. (d) Culrnann's method assumes that the failure surface is 8 plane. (I:') The upstream slope of an earth dam is critical during sudden dl"8wdowt1lX1nditions. (J) Bishop's simplified method oonsIders ali the forces acting on the sides of the slices. (g) In Bishop's simplified method, a process of successive approximations is required. (h) The total stress analysis can be used for the stability of slopes.

471

STABILITY OF SLOPES (!) The r.:onvcntional S..... I,.-dish circle method always cIT'S on the safe side.

IAns. True. (d. (i/), (e), (g),(/.). (,)).

C. Multiple Choice Questions I. The method of slices for the stability of slo pe (a) can be used for stratified soils. (b) can be u!!.oo when seepage occurs ,md the pore pressure eXISls within the soil. k) gives the factor 01 saf!;ty ba~cd on mOnlents and not the forces. (iI) All the aboVl:. 2. Taylor's stability charts liTe based on the total stresses using the toJ frictIOn cirde method (b) method of ~Iice.'> (e)~" .. 0 lln;l!ysls {il ) nonc ofille above J. In stability analysis. the term mobilised shear strength is referred to as (a) she.lr strength (b)muximum shcarstress (e) applioo shcOlf ~tress (d) none of the ;!bovc. 4. Bishop's simplified melhod of slit,:es sIallc~ I,.'quations. excepl the horiwntal forces t.:(luiltbrium. 5. The loll owing .. s~umption is nOi [Mde for the frktlOn circle method of slopt.: 5tability analysis; (a) FI'H.:rion i.~ fully mobilised (b) Total Stress analy~;s IS applkablc (d The resultant is r:mgelll;al 10 the fricrilln circle \d) The resultant pa.'>scs th rough the centre of friction circle 6. The fnelOr of ~ali;ty 01 lin intinite slope in a ~alld deposit i~ 1.732. If the angle of shcilring resistance is 30". the safe slopc i~ la) 11).45 0 (b) 75 ,4 0 Ie) 11(4 0 (d) 71.6 0 7. Identify the ;rn.;orrect statement The stl1bility of It ~Iupt.: is dcrrc .. ~ed by {a) Removal of 11 pan of slope hy excavution (b) Shock cau~d by an earthquake (c) Pun: wlltcr pres~urc in the soil (e!) Providing a berm ;Illhc tac II. FlJf the compolalion of N-component for sudden dmwdawn conditions by Hpproximnte method. the weight is (a) Saturntoo unit weight (b) Submerged unit weight (e) Bulk unit weight (d) Dry unit weight ~_1~1~1~~~~~~~~~&~

19 Earth Pressure Theories 19.1. INTRODUcnON As discussed in the preceding chapter, a soil mass is stable when the slope of the surface of the soU mass is flatter than tbe safe slope. At some locations where the space is limited, it is not possible to provide flat slope and the soil is to be retained at a slope steeper than the safe one. In such C$eS. a retaining strudure is required to provide lateral support to the soil mass. Generally, the soil masses are vertical or nearly vertical behind the retaining structure. Thus, a retaining wall maintains lhe soil at different elevations on its either side. In the absence of a retaining wall, the soil on the higher side would have a tendency to slide and may

not remain stable. The design of the retaining structure requires the determination of the magnitude and line of action of the lateral earth pressure. The magnitude of the lateral earth pressure depends up:)n a number of fadors, such as the mode of the movement of the wall, the flexibility of the wall, the propenics of the soil. the drainage conditions. It is a soU-structure interaction problem, as the earth pressure depends upon the flexibility of wall. 1be earth pressure theories whidt consider soil-structure interaction are complicated and require a computer. For coovenience, the retaining wall is assumed to be rigid aod the soil-structure interaction effect is neglected. The lateral earth pressure is usually computed using the classical theories proposed by Coulomb (1773) and Rankine (1857). The general wedge theory proposed by 'Ierzaghi (1941) is more general and is an improvement over the earlier theories. However, the !hooey is quite complicated. The design of rigid retaining walls and Oexible retaining walls, such as sheet pile walls and bulk heads, is discussed in tbe next chaptcr.

19.2. DIFFERENT TYPES OF LATERAL FARTH PRESSURE Lateral eartb pressures can be grouped into 3 categories, depending uJX>n the movement of the retaining wall with respcd. 10 lbe soil retained. The soil retained is also known as lbe backfill. (1) A~rest pressure. 1be latcral earth pressure is called at-rest pressure when the soil mass is not

"'SSN~ PA~RE \...~

ACTIVE

PRESSURE

MOVEMENT t7MAOS LEFT (b'-

A,. 19.1.

EARTII PRESSURE TItEORlES

479

subjected 10 any lateral yielding or movement. This case occurs when the retaining wall is firmly fixed al its top and L.. not allowed to rotate or move laterally. Fig. 19.1. (a) shows the basement retaining walls which are restrained against the movement by tbe basement slab provided al their tops. Another example of the ai-rest pressure is that of a bridge abutment wall which is restrained at its top by the bridge slab. The at-rest condition is aLso known as the elastic equilibrium, as no part of soil mass has failed aocI attained the plastic equilibriwn. (2) Active prasure_ A state of active pressure occurs when the soil maurc acting on an anchor block. Variation of ]'ressure Fig. 19.2 (a) shows the variation of eartb pressure with the wall movement. Point B represents the case wbcr. there is no movement of the waU. It indicates the at-rest pressure. Point A in Fig, 19.2 (a) indicates the active pressure. When the wall moves away from the backfill [Fig. 19.2 (b)]. some portion of the backfill located immediately behind the waU tries to break away from the rest of the soil mass. This wedge-Shaped portion, known as the failure wedge or the sliding wedge, moves downward and outwards. The lateral earth pressure exerted on the wall is a minirnwn in tbis case. The soil is at the verge of failure due to a decrease in the lateral stress. 1be borizontal strain required to reach the active state of pla5tic equilibrium is very small. Lambe and prfSsun~

Eorth

M~nt

o Mowment oway

II1XfIIiU .

I

rHo

+ Mo"'M\tm ta.\o:Il1Is

litl_

moYt'ment

(0)

Foiluft' wedge

lblA,INe

(e) Possiw

Fig. 192


480

Wbiunan (1969) have shown that in dense sand, the horizontal strain required is about 0.5%. For example, for a wall of 5 m height, a movement of 0.025 m would develop active earth pressure. Point C in Fig. 19.2 (a) indicates the passive pressure. 'When the wall moves towards the backfill [Fig. 19.2 (c», the earth pressure ina-eases. The failure wedge moves upward and inwards.1be maximum value of the eanh pressure is the passive earth pressure. The soil is at the verge of failure due to an increase in tbe lateral stress. Lambe and Whitman (1969) found lhal very lillie horizontal strain (about 0.5%) is required 10 reach one-half the maximum passive pressure in dense sand, bul much mo.-c borizontal strain (about 5% in dense sand, and 15% in loose sand) is required to reach the full passive pressure. However, their data are based on the triaxial shear lest results and Ihe magnitude of the horizontal strains required in the field may be somewhat different. It may be summarised that the state of shear failure corresponding to the minimum earth pressure is the active state and Ihat corresponding 10 the maximum earth pressure is the passive state. These are the two extreme conditions of pla.<;tic equilibrium. For all intermediate stales when the soil is not in plastic equilibrium. it is said to be an elaslic equilibrium. The ai-rest conditioo is a special case of an elastic equilibrium wilen tbe Slale of stress corresponds to the condition where there is no movement. IL indicates the in-silu condiljons.

.Y!iL. 19.3. EARTH PRESSURE AT REST The earth pressure at rest was discussed briefly in Chapter 11. However, the emphasiS there was on the detenninaHon of the horizontal Slresses in lbe soil mass. The expressions for earth pressure at~rest would be used in this chapter for the determination of the magnitude and line of aeLion of the total forces due to earth pressure 00 the retaining structures. 1be methods for estimation of the coefficient of earth pressure at rest ' - - - - - - ' (~) have been discussed in chapter 11. Fig. 19.3. Fig. 193 shows a retaining wall in which no movement takes place. l11e vertical effedive stress at point A at a depth Z ~ given by ... (19.1) The horizontal intergranular (effective) stress can be obtained using the coefficient of earth pressure at-rest (XO). which is equal to the ratio of the horizontal stress to the vertical stress, Thus ... (19.2) The stress

a.., is usuaUy represented ac; Po. indicating the lateral pressure at rest.

Thus Po - Ko Oz .. (19.3) II may be noted thaI the oocffident of lateral pressure at rest (XO) relates the effedive stresses. The total lateral pressure (PII) is equal to the sum of the intergranular pressure (PO) arxl the pore water pressure (u).

Thus

Ph .. Po + u

... (19.4)

In Fig. 193, the lateral pressure at depth Z is, therefore. p, - Ko(yZ - y.Z.) + y.Z.

•.• (195)

As Eq. 19.5 indicates, the pressure distribution is lriangular with zero pressw-e at the top (Z = 0). and the maximum pressure at the bottom of the wall. Fig. 19.4 (a) shows the pressure distribution when the soil is dry. The pressure at the bottom of the wall at depth J1 is given by

EARUi PRESSURE TIlEQRIES

481

(o)

(b)

For or;tive pressure. $l.bsmute Kc pressure ,substitute Kp lor 11.0.

for

Ko

cnd

lor

JXlU,",

Fig. 19.4.

Ph - Ko'YH The total pressure force per unit length of the wall is given by

p -

r.

o

K"yz dZ

P-~K"yH'

... (19.6)

In Fig. 19.4 (b). the depth of water table is at depth d below the surface. 'The pressure at depth Z > d is given by p, - K" [yZ - y.(Z - d)] + y.(Z - d) p, - K" yd + K"y'(Z - d) + y.(Z - d)

The pressure at the bottom (Z = 11) of the wall is given by

... (19.7) p, - K" y d + K" y' (H - d) + y. (H - d) The total pressure force (P) can be detennioed from the pressure distribution diagram. If the water table is at the ground surface [Fig. 19.4 (c)], the pressure at the bottom of tbe wall is given by. laking d = 0 in Eq. 19.7. Ph - Ko'Y' H + 'YwH ... (19.8) The resultant pressure (P) acting on the wall is delennined from the pressure distribution diagram. The point of application of the resultant pressure P is determined from the pressure distribution diagram. For triangular pressure distribution. it acts at height Hf3 from the base. 19.4. RANKINE'S EARTH PRESSURE THEORY Rankine (1857) tDnsidered the equilibrium of a soil element within a soil maos bounded by a plane

surface. The following assumptioos were made by Rankine for the derivation of eartb pressure.


(1) (2) (3) (4) (5)

The The The The The

soil mass is homogeneous and semi-infinite. soil is dry and cohesion less. ground surface is plane, which may be horizontal or inclined. back of tbe retaining wall is smooth and vertical. soU element is in a state of plastic equilibrium, i.e., at the verge of failure.

Expressions for tbe active eanh pressure and the passive earth pressure are developed as explained below: (a) Active Earth Pressure. Let us consider an clement of dry soil at a depth Z below a level soil surface [Fig. 19.5 (a»). Initially, the element is at-rest conditions, and the horizontal pressure is given by o" .. ~c .. wbere 0 .. is the vertical stress at C, and c~ is the horizontal stress at C. or course, c .... yZ. The stresses 0" and c .. are, respectively. the minor and major principal stresses. and are indicated by points A and B in the Mohr circle [Fig. 19.5 (b)]. Let us now ronsider the case when the vertical stress remains constant while the horimotal stress is deacased. The point A shifts to position A' and the diameter of the Mohr circle increases. In the limiting condition, the point A shifts to position A" when the Mohr circle [marked (3)] touches the failure envelope. The soil is at the verge of shear failure. II bas (b) attained the Rankine active state of plastic equilibrium. The horizontal stress at that state is Fig. 19.5. the active pressun.: (pQ)' Fig. 19.6 shows the Mohr circle when active oonditions are developed. Point E represents the active condition. From the figure,

T

Fig. 19.6.

As

p.-OE_DC_CE CE - CD _ DC,u,,', pQ" OC - OCsin,' .. OC(I-sin,')

Also

... (aJ

c .... OB .. OC + CB .. OC + OCsiD.' a~-OC(l+sin,')

... (bJ

IG

EARn)

PRESSU~E

48'

THEORiES p(J~

From Eqs. (a) and (b),

"0:""

0'

p" ..

1 + sin,' + sin,' ( ~) 1

Ov

... ([9.9) p,," K"yZ where KQ is a coefficient, known as the coefficient of active earth pressure. It is a function of the angtc of shearing resistance (,'), and is given by

Ka"

! : :: :: ..

l lan ( 45" -

i

c~mple, if .' = 30", K" .. Eq. 19.9 can be used to dClcnninc the active earth For

-f)

...

(19.10)

c;y

pressure on the retaining waU. The pressure distribution is similar to one shown in Fig. 19.4 (a) in which Ko is substituted for KoFig. 19.7 shows the failure planes. These are inclined at (45 0 + ,'/2) to the major principal plane whim is

~

hoqzootal.

- .-

When the wall moves away from the back fill, the failure wedge moves downward and the resisting forre due 10 the shearing strength of the soil is developed in Fig. 19.7 the upward direction along the failure plane (Fig. 19.8). The resisting force causes a decrease in the earth pressure acting on the wall. The decrease in earth pressure continues till the maximum resistance has been mobilised. The earth pressure does oot dec~ beyond this point and the active state is read1ed and the soil has attained plastic equilibrium.

-

Movement o f wall Fig. 19.8.

484


Fig. 19.9.

Now

Pp

"

OC+ CE .. OC. CD .DC+ OCsin,'

Pp" OC(l + sin,')

0<

Also

DB .. OC - Be .. DC - CD .. DC - OCsin a~ ..

OC(l-sin,')

+'

... (a) ... (b)

,.

Fig. 19.10,

From Eqs. (a) and (b),

I!R..~ 0v .. 1 - sin,'

pp-(!~::~::)o~ or Pp .. Kp"1 Z where Kp is the coefficient of passive earth pressure, given by

For example,

if" .

Kp ..

30".

~

2 0 : :::: .. too (45 + +'/2)

...(19.11)

.. .(19.12)

Kp .. 3

1be coefficient of passive pressure (Kp) depeDds upon ,'. '[be pressure distribution is similar to tbal shown in Fig. 19.4, in which Kp is substituted for Ko.

F.ARni PRESSUR.E THEOR.IES

The angle which the failure plane makes with the major principal plane is equal to (450) + ,'12). As the major principal plane is vertical, the failure plane make an angle of (45 - *'/2) witb the hori7..onlal (Fig. 19.11) which is the minor principal plane. When the wall moves towards the back fill, the lateral earth pressure' increases because the resistance builds up in the direction towards the wall. The pressure reaches a maximum value when the full shearing

Fig. 19.11.

Fig. 19.12.

resistance ha'l been mobilised (Fig. 19.12). Further movement of the wall does not increase the pressure, and the passive state is reached and the soil has attained plastic equilibrium.

19.5. RANKINE'S FARm PRFSSURE WHEN mE SURFACE IS INCLINED Two stresses are called conjugate stresses when the direction of one stress is parallel to the plane on which the other stress acts. Rankine assumed that the vertical stress on an clement of the soil within the inclined backfill and tbe lateral stre..'IS on the vertical plane of the element are conjugate stresses. In other words, he assumed that the lateral stress is parallel to 100 inclined backfill. Let us consider an element of soil at depth Z below the soil surface inclined al angIe j to horizontal (Fig. 19.13). The angle i is known as the angle of surcharge. The intensity of vertical stress (ov) on the element is given by

1

Ov.1(ZbbClJ5i)

or ov - 1 Z cos i

... (19.13)

The other conjugate stress is the lateral stress (oJ. Fig. 19.13. It may be mentioned that the vertical stress 0v is not tbe principal stress, as a shear stress also exists on the inclined plane at the top of the element. Likewise, the lateral stress 0 ... is also not a principal stress. A relationship bctween the latcral pressure and the vertical stress can be oblamed (or the active and passive cases as given below. (a) Active Earth Pressure. Fig. 19.14 shows the Mohr circle corresponding to the active limiting

486


r OA='¥

OB=Pa

Fig. 19.14.

conditions. lbe vcnical stress o~ is represented by the line OA making an angle i with the horizontal. At any depth, the value of 0'" is constant and equal to that given by Eq. 19.13. If the lateral expansion of the soil is suriicicnt to induce the state of active plastic equilibrium, the Mohr circle must paiS through A and it should be tangential 10 the failure envelope. The origin of planes P is obtained as the point of intersection of OA with the Mohr circle. As the reader would recollect (chapter 13). the origin of planes is located by drawing from the point representing a stress (vertical Slress, in this case) a line parallel to the plane 00 which it acts (plane inclined at i, in this ca<>e). A vertical line through P nus the circle at B below the o-axis. The conjugate stress, which is the active pressure (PrJ, is represented by DB. Numerically, tbe conjugate stress is also equal to OP.

From the figure,

p"

0:-

DB OP OD - DP OX-QA-OD+DA

... (a)

The ratio p,.lo" is known as the conjugate stress ratio. Now

OD - Oecosi

,nd B", ,nd

DA • DP • IX -

Therefore,

VA • DP • "r("O-=C-s;-n""~''')2-_-(''''OCC=-s'--in-=i)2

oe

_ DC P.

From Eq. (a),

sini

Ysin 2 ,' _

OC oos i

sinz;

... (0)

OC Vrs~in2~.-'-s-in~2i

0: - OCcosi + OCY Sin2 ,' _ sin'l; p"

cos;_YSinz"_sin2 ;

0: - cos; + Ysin2 " Y(1 + Y(1 -

cos; cos;

or

... (b)

vPC' - DC'

PC-FC-OCsin~jI'

Pit

cos;

···(dJ

_ sin 2 ;

cos2 ,')

Ya:Jt,,2i

cos2 ,')

(l _

cos 2 p'

0: - cosi + Ycos1j _ cos2 ,'

cos'}. z)

(1 _ COS 2 1)

EARTIi PRESSURB THEORIES

or

487

Z p._COSi_VCOSZi_COS Cl.)Sj

or

+

VCl.)S2j

+,

_

(lZCOSI)

cu2 t'

P. - K.yZ

... (19.14)

where K. is the coefficient of active pressure. given by

. K .... COSI

cas j X

_

"rCOS--.,7j ---COS--",,-: ..

cos; + VCOSZj _ fXYi>2+'

In must be noted that P. is parallel 10 the inclined surface. For the special case, when i = 0 K _ l....=...!!!!.£ • 1 + sin 41' The conjugate stress in thai case become principal stresses, as already discussed.

. .. (19.15)

(same as Eq. 19.10)

Direction of fliliuR planes The inclination of the major principal plane is indicated by the line PE in Fig. 19.15. It makes an angle 9 1 with the horizontal. The lines PF and PP indicate the failure planes. 1be failure plane PF makes an angle III with the horizontal.

Fig. 19.1S.

From the triangle PEG, the exterior angle,

8/1 .. III + 8 1 al- 8/1- 8 1 where 8/1 is the angle which the failure plane PF makes with the major principal plane, and is equal to (45' + .'/2). 0 Thus al - (45 + .'12) - 8 1 10 the triangle OPC, the exterior angle

'¥ is given by

'1'''' 28 1 + i Therefore,

or 8 1

0

UI ... (45 + +'12) -

The angle 'I' is obtained from tbe triangle PCD as

"

Y

(Y)

... (19.16)


488

CD OCsin; OCsin; sin; Stncp .. CP .. ~ .. OCsio,' .. sin,' .

1p ..

i)

. _I ( sin sm sin,'

... (19.1~

The other failure plane PF' makes an angle (X2 with the horizontal, which is given by (XI- Bf2+ BI (Xz .. (45 0 + ,'/2) + ('$ - 1)/2

... (19.1~

(b) Passive Earth Pressure. This case is similar to lhe one for the active case with one basic differeno: that the vertical stress is the smaller of tbe two conjugate strc..'ises. In Fig. 19.16, OA represents the vertical stress (a~). The point P shows the origin of planes, and DB represents the passive pressure, From the figure,

T

OA= by

08=,.

Fig. 19.16.

ee.

DB

OP

OD+DP

a~-OA-OA-OD-DA

Now

OD .. DCoosi

and

DP _ DA _ y'7(A:::C):;-j---;'(DC)==' _ ,; (FC)' _ (DC)'

.. V(DC sin ,,)2 _ (DC sin i)2 .. DC Vsin2 ., _ 5m2 j Therefore,

&.

OCcosi +

a~ .. OCoosi _

OC"rs,in:zr .::-·-_-S1'-·n''-;i

DCV 5in2 ,' _

8m 2 ;

Vsin2 q>' - sin2 i cosi - Vsin1 q>' _ sin1 i cosi + Vcos2 j _ OOS2rp' cos; - Voos2j - axi!p' cos; +

or

Pp" (yZCOSI)

cos;

+

cosi _

"roos.....",-c i -_-cos--.-:zp .....•

VCXY.i2 j

_ ca;2q>'

... (19.1~

HARTIi PRBSSURB THEORIES

489

where Kp is given by

cos; + Kp - cosi 'OOSi _

Ven??i

_ cos2 cp'

Vcns2i _ ~2cp'

... (19.20)

for the special case, when j - 0

K .. p

8) co ,,1 ~,

~

(same as Eq. 19.12)

1 - sin.'

Inclination or radhu'e planes. In Fig, 19.17. the line PB indicates the major principal plane, and the line PE shows the direction of the minor principal prime. The minor principal plane makes an angle 9:2 with the horizontal. From the triangle PCO,

282-'lV+i

~A..oR

P.

PLANE

1i'ig.19.17.

The inclination of the failure plane PF with the horizontal is equal to

~ oo!ained from

the relation

8/1-a..z+ e2 where ell is the angle which the failure plane PF makC!S wilb the minor principal plane

ell -

Thus

~

or

a, - (45

The angle

..

82

0 -

,'12) -

(~)

... (19.21)

'IV is obtained from the triangle PCD as

._,( sin,' Sini)

. ..(same

'$" 8m

~

Eq. 19.11)

The other failure plane PF' makes an angle al with the horizontal. given by at ..

81l + 81

a, _ (45

0

-

,'12) + (

~)

... (19.22)

(c) Pressure DlstribuUon. Fig. 19.18 shows the p~re distribution when the soil surface is inclined at

SOIL MECHANICS AND FOUNDATION ENOINEERJNO

'90

Fig.19.18.

an angle i. The lateral pressure is parallel to the soil surface. The maximum pressure occurs at the boUom of the wall and is given by p. where p .. Po in aa.ive case and p .. Pp in passive case. From Eq. 19.14,

Po .. Ko YII Po - cos;

~

oos; - " co~,2; _ C(62.' "rxlflj _ cos2 ,' yH

005; +

Total active pressure,

Po"

Likewise,

pp-Kpyll

Ko yIP

... (19.23)

2

Pp _ cosi cos;

-+-" cos2 j

2

-

cos

.'

yH

ensj - " ens ; - co}.' p, ..

Total passive pressure,

}Kp yIP

.. .(19.24)

For soils below the water lable. the submerged weight y' should by used instead of bulk unit weight y. (d) Inclined Back otwaU. Sometimes, we come across a retaining wall with an inclined back [Fig. 19.19 (a)]. Rankine's theory am be used to determine the tanh pressure with some modificalia,.. A venical plane AC is taken through A aod tbe active pressure Po is determined on this vertical plane from Rankine's theory. The total pressure Pion the wall is the resultant of the pressure Po and the weight W of the soil wedge ABC. Thus

'a)

(b)

Fig. 19.19.

,<)

EAR'll1 PRESSURE THEORIES

491

II acts at a height of 1//3 from the bottom. In case of the wall with an inclined back and the backfiU with a surcharge angle i [Fig. 19.19 (b)}, the aC1ive pressure P,. is inclined to the vertical plane AC at an angle i, and the total pressure PI is the resultant of the pressure PQ and W. Fig. 19.19 (c) shows the retaining wall with it back inclined towards the backfill. In this case, the vertical plane AC is on the other side of the back: face, and, therefore, W is imaginary.

19.6. RANKINE'S FARm )'RESSURE IN COHFSIVE SOILS Rankine's original theory was for cohcsionless soils. It was extended by Resal (1910) and Bell (1915) for cohesive soils. The treatment is similar to that for cobcsionlcss soils witb one basic dilference lhal the failure envelope has a cohesion inlercept c', whereas that for cohesionlcss soils is

,.

zero. The following treatment is limiled 10 the case when Ute backfill is horizontal. (a) Active Case. Fig. \9.20 shows the Mohr circle in which point B indicates the vertical stress le'cot+;'1 and point E shows the active pressure. The failure envelope is tangential to the circle. 1be relationship betwccn PQ and o~ can be obtained as under. From triangle FeD,

sin+' ., Sin

(°1 ;

1-

~

~-

,

c

L~~

FOe;> OC

. (01 - (3)12 - e' COI~' + (01 + (3)12

(3) _

(1 - sin+') ..

°3-

(al ;

sin~' + c' CC6~'

(3)

-f !: ::::

(1 + Sin,') + e'cos+'

°1 -

i~':::

... (a)

03- 011302(450 - ~'12) - 2e' tan (45 0 - ,'12)

or

As 03 is equal to the active pressure (PQ), and 01 is equal to the vertical stress a~ (- 't Z), Eq. (a) becomes

P,.

. (~) 1 + sin,'

y

z-

20'=<1>'

1 + sin 4'

p,.-K yZ-2c'..fK;,

... (19.25)

Q

K

where

It can be shown that 1

Q

~i:'

_

!: :::: _

tan 2 (45° _ +'12)

"-

tan (45

0 -

,'12) -

..fK;,

Eq. 19.25 indicates that at Z::: 0, the active pressure is given by ... (19.26) Po' -20' VK. The negative sign shows that tbe pressure is negative, i.e. it tries to cause a pull on tbe wall. This tensile stress decreases with an increase in depth. and it becomes zero at a depth Z~, given by

492


0- K.yZ, - 2e' VK. 2e'

Z, - yVK.

... (19.27)

The depth Zc is know n as the depth of tensile aack.. The tensile stress eventuaUy causes a ttack to from along the soil-wall interface. The pressure at the depth H is given by,

pQ - KQyH - 2c'..t'K;. Pressure DistrlbuOon Fig. 19.21 shows the pressure distribuHon against tbe retaining wall. The total pressure on the retaining wall of height H is given by integration of Eq. 19.25 as

fig. 1921.

P.-r. (K.yZ-2e'VK.)dZ P" - K"

I!f- -

2e'

VK;. H

... (19.28)

Eq. 19.28 is applicable before the fonnation of crack After the occurrence of the tensile crack, the fora: on the wall is caused only bY' lbe pressure from

Z - Ze to Z = H, i.e. tensile stresses are neglected.

Thus

p. - i(H - Z,)(K.yH - 2e'VK.)

Substituting the value of Z., from Eq. 19.27, p. -

t (H - y~) (yHK. - 2e' VK.)

P

!yJiK

Q

-

It acts at a height of (H - Zc)l3. For 0, Eq, 19.29 reduces to

+."

p. -

Q

t

-

2d H..fiC + 2 (c,)2/.,

yH'- 2e'H + 2(e')'/y

For soils below the water table, the submerged unit weight.,' and the corresponding value of should be used.

... (19.29)

.. .(1930)

+' and c'

Critical height of UIUUpported nrlical cuL As shown in Fig. 19.21, the pressure is negative in the lop region. It becomes zero at a depth ZC" If the

493

IlARTH PRESSURE THEORIES

wall has a height of 2Z". the total earth pressure is zero. This height is known as the critical height (He). H, _ 2Z, ... (1931) If the height of an unsupported vertical alt is smaller than He' it should be able to stand. However, the conditions in unsupported vertical cut are different from those near a retaining wall. In the vertical cut, the lateral stress is everywhere zero, whereas in the retaining wall, it varies from - 2c' ffa to + 2c VK;. Because of this difference in the stress condition, the safe height of the vertical cut is slightly smaller than Ilk'lt given by &j. 19.3l. Substituting the value of Ze from Eq. 19.27,

He - 2 x 2e'/(.,..rK;.) - 4e'/(.,VK;) For

t' .. 0,

.(19.32) ... (19.33)

He - 4e'l.,

(b) Passive Case An expression for the passive pressure in a cohesive soil can be dctennined by referring to Fig. 19.22. The Mohr circle bas bccn drawn such that the point JJ represents Ov and point E as pp. In triangle FCD,

Fig. 19.22.

.

,

CD

(01 - 0))/2

sm~ "Ci-c'cot~+(ol+o)12 (01 - 03) .. (01 + a)sin t' +

2c'cos~'

01(1 - sin.') .. 03(1 + sinq,') + 2c'cos " ~

01 .. 0'

0)

2c'ca;t' (1 _ sin t') + 1 _ sin "

Pp - ovtan1(45° + "/2) + 2c' tan (45 0 + .'12) Pp" yZKp + 2c'

VI<;

... (19.34)

where Kp = coefficient of passive pressure, given by

1 Kp" tan (45° + ,'12) ..

~

:

::~::

... (19.35)

The failure plane makes an angle of (45 0 + "12) with the vertical (major principal plane) and of (45 ,'12) with horizontal (minoc principal plane).

Pressure distribution Fig. 19.23 shows the pressure distribution obtained from Eq. 19.34.

0

SOIL MECHANICS AND FOUNDAll0N ENGINEERING

494

----j
I--

Fig. 19.23.

WhenZ=O.

pp-2c'vx;,

When Z = II, PI' - YII Kp + 2 c' YK.; ... (19.36) The pressure, unlike active case, is positive throughout the depth. Tbe total pressure on the retaining wall of height 11 is given by PI' - ifl(yI-lKp) + 2c'..rK; x /I PI' - Kp 1112/2 + 2e' HYK;

.. . (19.37)

19.7. COULOMB'S WEDGE THEORY Coulomb (1776) developed a m~hod for the determination of the earth pressure in which he considered (he equilibrium of the sliding wedge which is formed when tbe movement of the retaining wall rakes place. As discussed before, in the active case, the Sliding wedge moves downward and outward relative to the backfill, whereas in tbe passive case. the sliding wedge moves upwards. The lateral pressure on the wall is equal and opposite to the reactive force exerted by the wall in order to keep the sliding wedge in equilibrium. The analysis is a type of limiting equilibrium method: The follOWing assumptions are made: (1) The backfill is dry, cohesionless, homogeneous, isotropic and ideally plastic material. (2) The slip surface is a plane surface wbich passes through the heel of the ,,:,all. (3) The wall surface is rough. The resultant earth pressure on the wall is inclined at an angle l:J to the nonnal to the wall, where l:J is the angle of the friction betw.een the wall and the backfill. (4) The Sliding wedge itself acts as a rigid body. The magnitude of canh pressure is obtained by considering the equilibrium of the Sliding wedge as a whole. In Coulomb'£ theory, a plane failure surface is assumed and the lateral force required to maintain the equilibrium of the wedge is found using the principles of statics. The procedure is repeated for several trial surfaces. The trial surface which gives tbe largest force for the active case, and the smallest force for tbe passive case., is the actual failure surface. The method readily accommodates the friction between the wall and the backfill, irregular baclcftll, sloping wall, surcharge loads, etc. Although the initial Iheory was for dry, cobesionlcss soils, it has now been extended 10 wel soils and oohesive soils as well. Thus Coulomb's theory is more general than the Rankine theory. 19.8. COULOMB'S ACTIVE PRESSURE IN COUESIONLESS SOILS Fig. 19.24 (a) shows a retaining wall with an inclined back face and a sloping dry granular backfill In active case., the sliding wedge ADD moves downward, and the reaction R ads upward and inclined at an angle .' with the nonnal. 1be sliding wedge ABD is in equilibrium under three forces:

49'

BARlli PRESSURE THEORIES

(0)

lb)

Fig, 19.24.

(1) Weight of the wedge (lV). (2) Reaction R on the slip surface BD. (3) Reaction PI> from the wall. It may be noted that, at failure conditions, the shearing resistance on the failure surface is fully mobilised and the reaction R bas the maximum obliquity. It is therefore inclined at an angle ¢l'to the nonnal to the failure plane. Further, PI> acts upward, and it is inclined at an angle 6 with the normal as shown ill Fig. 19.24 (a). Fig. 19.24 (b) shows the force triangle. As the magnitude of one force (viz. weight W) and the directions of all three Caras are known, the force triangle can be completed. TIle magnitude of Ptl is detennined from the force triangle. The pressure acting on the wall is equal and opposite to PII' TIle procedure is repealed afier assuming an other failure surfacc. The surface that gives lbe malrimum vaJue of p. is the critical failure plane, and the C()(l'eSpooding force is the active foroe. Coulomb's method does not give the point of application of the resultant earth pressure (P~. 1bc point of application is found to be approximately at the point of intersection E of the back of the retaining wall with a line CE drawn from the centroid C of the failure wedge and parallel to tbe failure surface. As this procedure is cumbersome, for convenience, the pressure distribution is sometimes a is inclined at angle 6 below the normal as shown in Fig. 19.24 (a), However, if the wall is supported on a soft, compressible SOil, it may sctUe to such an extent lhat the movement of the wall will be downward relative to the back.ftll and the relative movement of Ihe wedge will be upward. In such a case, the force P. would be inclined at an angle 6 above the oonnal 10 the wall. (2) The angle 6 is the friction angle between the soil and the wall It may be dctennined by means of a direct shear test. For concrete walls, 6 is generally taken as 2/3,'. The value of 6 cannot exceed ,', because in that case the failure will ocrur in soil. IC the friction angle 6 is zero, and tbe wall is vertical and the ground surface is horizontal, the Coulomb method gives identical results with tbe Rankine method. (3) Coulomb's theory assumes the failure surface to be a plane. TIle actual failure surface is slightly curved. Fortumltely, for the active case, the error is' srnan, and therefore the failure surface may be assumed to be plane without any significant error.

001L MECHANICS AND FOUNDATION ENGINEERING

496

Expression for Active Pressure From Fig. 19.24 (b), using the law of sines,

P,

W P + 6 - a + ,') p _ Wsin(a-t') .. sin(180"-lJ+6-a++') where p .. _ total active pressure force. The weight Wof the wedge ABD can be detennined from Fig. 19.2.5 as siD (<< - +') - sin (180" -

... (a)

r H

J~ Fig. 19.25.

W_~xBExAGXY TalciogAG _

m.

and BE = L, we bave W -

W.

0'

~L

x m x 'f

I(m x y)(OO + GE)

Now

AD - H/sinp.

From triangle ABG,

~ _ sin (180" -

(Il

... (b)

+ a)] - sin(p + 0)

m _ AB sin(P + «) _ HSi~~p+ «)

Therefore, From triangle ABO,

... (c)

m sin(~ + a - 90')· sin(~ + a) BG

BG _ msin(p + a - 90 0 )/sin(j} + 0) From uiangle AGE,

...(4)

GE m sin (90° - a + i) - sin (<< - I)

GE' . m sin (90" - a + ,) sio(a - i)

...(e)

Substlluting the values of m, 00 and GE from Eqs. (c), (4) and (e) In Eq.

W= -2"~ sin p

.y

x

[Sin
(b~

Sin(9O"-a+,, ] III

sin (a _ ,)

EARTII PRESSURE TIIEORIES

or

497

W .. 1,PysinW sin2~

2

+oi [ sinW + a-90°) sinO~ + a)

+ sin(90~ - a + sin(a-/)

Q]

Substituting the above valuc of W in Eq. (a), P .. oil

1 IPysinH3 2 )( [

sin (a -,') sin(1800-~+6-a++')

+a)2 )(

sin2~

Sinm + a - 9( sin(p", a)

0 )

+

sin (90 0 - a + sin(a I)

I)]

The active pressure force PtJ will be a maximum when the failure plane makes an angle a with the horizontal such that

iJPfi ..

au

a

Mullcr-Drcslau gO\'c the solution, which is quite complicated. lbe maximum value of PtJ thus obtained is Coulomb's active force, given by ... (19.38) where K" = Coluomb's active earth pressure coefficient, given by 2 K" .. sin (p + ,')

. ' • . (0_

sm

p SID P

5)[ 1

+

2 ... (19.39)

./SinW+6)SinW-Oj

V

sin (P

_ 6)

sin (~ + I)

The lioc of action of P" wilJ be at a height of 11/3 above the base of the wall and it will be inclined at an angle (I to the normal drawn to the back of the wall. 19.9. REHBANN'S CONS11tucnoN FOR ACTIVE PRESSURE Rchbann (1871) gave a grnphicul method for the determination of the lotal active prtSSW'C according to Coulomb's theory. It is based on Poncclct's solution (1840), and is, therefore, also known as Poncclct's method.

Fig. 19.26.


498

Fig. 19.26 shows the graphical oonstrudion. The line BD is drawn at an angle q.' to the horizontal. The line BL, drawn at an angle" with the line BD. is known as the earth pressure line. The angle 'tV is equal to (~ - 5). A semi-circlc BMD is drawn on BD as diameter. The line AH is drawn parallel to BL, intersecting the line BD at J!. A perpendicular HM is drawn at H. intersecting the semi-circle at M. With B as centre and 8M as radius, an arc MF is drawn, intersecting DD at F. The line FE is drawn parallel to BL, intersecting the ground surface at E. With F as centre and FE as radius, an arc is drown to intersect BD at N. The line BE represents the critical failure plane. The total active pressure P" is given by P D - Y (areaoflriangleNEF) p. _ y (112 x NF x x) ... (19.40) where x is the perpendicular distance EG between E and BD. Proof. The proof of Eq. 19.40 is as under:

The triangle BEF and the force triangle in Fig. 19.24 (b) are similar. Therefore, p. W '" EF /iii

(EF) BF

or Po " W

... (a)

From triangle EFG,

iF .. sin '$.

Now

M . ffi-m . ffi-~-~ - ffi - ~+M

... (b)

or EF .. x cosec '$

BF ... BD - x cot ('f - i) of. x cot '$ BF _ L - x cot (4" - i) ... x cot '$

... {e}

where L = Length of BD. Now W .. ., (VOlume of triangle ABE) Taking unit length of the wall. W .. ., ( area of triangle ABE) .. y [area of triangle ABD - area of triangle BED)

W - l~xLxm-lxLx~-t.,L~-~

...(d)

where m is the perpendicular dislance from A to BD. From Eqs. (a), (b), (e) and (d),

Po .,

i

y

L (m - x) x [L _ x

In Eq. (e), substituting

cosec '" ..

rot~~/f + x

Po"

For maximum value of po.

t

1 L (m - x) (L

~xdx)

iJiJ:o .. 0

(L - dx) (me - 2cx) - (m - x) ex (-d) - 0 IX

0'

m L - xL .. xL -

~

(m - x) -

~

dr

(L - dx)

••• (e) •.. if)

cotW-·)-oot'l'-d,

and

cot 'tV]

C

mL - 2xL +

0

ie

EARTH PRESSURE THEORIES

499

Substituting the value of d from Eq. (f).

~ (m Using Eq. (c),

- x) -

~ rL -

x cot (+' - i) + x rot

~]

T-T-~XBF

Area of triangle ABD - Area of triangle BED::: Area of triangle BEF or Area of triangle ABE ::: Area of triangle BEF In other words, for the maximum value of P the failure plane BE should be such that the triangles ABE and BEF have equal areas. Let us examine whether Rehbann's ronstruruoo satisfies the above criterion. From the properties of the Q •

circle,

BH x HD _ (HM)'

Adding (BfJ)2 to both sides, (BH)' + BH x HD _ (HM)' + (BH)' Bff (Bff + liD) _ (BM)' Bff x BD _ (BF)'

or

BHIBF - BFIBD As Ali is parallel to EF. triangles BIU and BEE are similar. Therefore. BHIBF - Bl/BE

... (.) ... (u)

From Eqs. (,) and (il). BFIBD _ H//BE i.e. JF is pa~llel to AD. Thus. the figure AJFE becomes a parallelogram. Therefore, the perpendicular from point A OD the diagonal JE and the perpendicular from point F on the same diagonal would be equal in length. Consequently, the areas of triangles ABE and BEE which have the same base BE would be equal II proves that Rehbaf1l1's construction satisfies the required criterion. Expression for total pressl1l'l! from Rehbann's construction From Eq. (Q~ p. - W x (EFIBF) - y (areaoftriangleABE) x EFIBF or PtJ - 'f (areaoftriangleBFE)x EFIBF - 'f x

or

t x BF x x x EFIBF

PtJ-txYXEFXX PQ

-

t x Yx

NF )( x

... (19.41)

(same as Eq. 19.40)

Eq. 19.41 can be written in teoos of various angles.

Frob similar triangles £ED and AND, EF

AH

w-Ai)

,

(EF)' _ (AH~ED)

From triangle flAB,

AH

... (ii<)

AB

sin (180° - ~ - ") - sin,!,

500


All .. AD Also

Bl1

""Ali ..

sin ;fn ~ ~')

~

sin 'I'

BH_AB~ sin'!:'

From triangle ADD,

BD

Ali -

~

sin(ip' - i)

BD - AB~ sin ($' - i)

The ratio AD/ED can be written as

rv _AE;oED '" ~+ * _ ~+

From Rehbann's constructiOn,

I,.

ff'D+

1

1

m;02

BI-I)( BD ..

BF .. vBIJ x ED

M; - ~ Bfl/BD Therefore,

~

From Eq. (iii),

EFl ..

.. "BI-IIDD + 1

(Afrl )(

EP2=AFr x

(EDIAD)2

1 (~BIIIBD + 1)'

Substituting the value of All, 811 and DO,

EF2"AIf[~f x SIO'\j.t

1

[V~'~l]' sin(~ + ,) + SID1p

From Eq. 19.41, substituting AD .. 11 cosec

P" ..

~

andx .. EFsin'l',

}y (HOOSCCp)2sin'\j.t x[Sin!!;")f

[v,in""

1

+ 6)';0(<1>' Sln'l'sm(p + I)

P where

Q

'"

i K"

2

'I 1-/

.J

+

1]' ... (same as Eq. 19.38)

EAR'm PRESSU.RE THEORIES

As'l'.~-b,

SOl

2

K" _

sin

Sin2~ sin(~ _

6)

(~

2 ... (same

+ ,')

as

Eq. 1939)

[1 Vsi~(" 6)s~nW .0] sm(p - 6)sm(p +

+

-

+ I)

19.10. CULMANN'S CONSTRUCTION FOR ACI'IVE I'RESSURE Rehbann's construction becomes inconvenient when the slope angle i approaches the ~ng\c ~'. Culmann 11 meth
(1866) developed

Fig. 19.27

(1) From point B (Fig. 19.27), a line BD is drawn at an angle

t'

to the horizontal. As the weight of the

wedge is plolted along this line, it is also known as the weight line. (2) A line BL is drawn at an angle 'll' with the line BD, such that'll' _ 1.1 - fi, where P is the angle which the back fare makes with the horizontal and 6 is the angle of frictioo. (3) A failure surfare OF is assumed, and the weight (W) of the failure wedge ABF is computed. (4) The weight (W) of the wedge is plotted along BD such that BP = lV. (5) A line PQ is drawn from point P parallel to BL to intersect the failure surface BF at Q. (6) The length PQ represents the magnitude of P" required 10 maintain equilibrium for the assumed failure plane.

:)

(7) Several other failure planes llF", BF', BF" " etc. are assumed and the procedure rcpc.'lted. Thus the points Q", Q', Q"', etc. are located. (8) A smooth auve is drawn joining the points Q", Q, Q',Q"', etc. The curve is known as CUll1llllUl's line. (9) A line (shown dotted) is drawn tangential to the Culmann line and parallel to BD. Point T is the lXlint of tangency. (10) The magnitude of the largest value (Pawl) of P" is measured (rom the tangent point T to the line BD and paralielto BL. It is equal 10 Coulomb's active pressure (P.. ). (11) The actual failure plane passes through the point T (shown dotled). meet of Unlronn Surcharge or Une loud Cuimann's method can be easily extended to include the eITed of uniform surcharge or the line load applied to the backfill. Each such iood that falls within the assumed failure wedge is included by adding it to the weight of the failure wedge.

SOIL MECHANICS AND fOUNDATION ENGINEERING

502

o

Fig. 19.28.

Fig. 19.28 shows the modification when a line load q' acts on lhe baclUlil. The Culmann line upto point Q .is similar to the case when there is no line load. However, at point Q. it becomes discoo.tinuous, m suddenly the load of the wedge is inc:rea..ed due 10 line load q'. In this case, tangents are drawn to the two segments of the a.uve, and the value of Po is determined. The value of (Po + !J.PJ is measured from the line BD to the maximum point on the Culmann line that includes the line load.

l!Ul. COULOMB'S ACTIVE FARTH PRES:SURE FOR COHFSIVE SOILS The Coulomb wedge theory ~n be extended 10 oohesive soils. In addition to the three forres (W, p .. aDd R) oonsidered for the cohesionless soils in the preceding seclioos, there are two more forces, namely. (1) tbe cohesive force C acting on the failure plane BE (Fig. 19.29), (2) the adhesive force CD acting on the back. of

'q w

~

~

Co

,.

0)

o

Fig. 19.29.

the wall. In all there are 5 forces which keep the wedge in equilibrium. The resultant pressure Plan the bad of the wall acts at an angle 0 to the nonnal. The force polygon for the 5 forces is shown in Fig. 19.29 (b). The magnitude of W is determined from the weight of the wedge ABE. The magnitudes of C and Cn are respectively c x BE and c n x AB where c and c.. are unit cohesion and unit adhesion. Knowing the directions of all the five forces and the magnitudes of 3 forces, the force polygon is completed as shown. The total pressure p .. on the wall is the vector sum of PI and C... By analysing several trial wedges, the maxinium value of PQ can be determined.

503

iNO

The resultant pressure P fJ is assumed to ad on the back of the wall at the point of intersectioo of a line drawn parallel to the critical surface and passing through the centre of gravity of the wedge, with the back of the wall. For convenience, sometimes it is taken at a point at one-third the height of the wall from its base. Note. (1) Sometimes, the resultant pressure 00 tbe wall is taken as maximum value of Pl' without combining it with C... (2) In oobcsive soils, tension cracks develop to a depth of Z~. It is generally assumed thaI upto depth of lension crack, no cohesion, adhesion or friction acts. 19.12. TRIAL WEDGE METHOD The trial wedge method assumes that the rupture surface is plane. The trial wedge method is a general method wbicb can be used to oompute active pressure and passive pressure for both Rankine's condition and Coulomb's condition. The method can also be applied 10 the cases when the soils are cohesive. The trial wedge method for oomputing Coulomb's active pressure for cohcsionless soils is discussed below.

int

JT~a I: b

NO

w

ne

IX! he of

,

(Pa)max

'-PRESSURE (a)

"

lOCUS

(b)

Fig. 19.30. In Fig. 19.30, the line BF sbows an assumed failure plane. 1be wedge ABF is held in equilibrium by the reaction R ooing OIl tbe rupture plane BF and the force PfJ • The force P fJ acts at obliquity 6 below the oonnal to oppose the movement of the wedge. The reaction R acts with obliquity below the normal. The force triangle abc is shown in Fig. 19.30 (b). The valuc of PfJ is detennincd from the force triangle as the distance be. Another trial surface, such as BF' is assumed and the force triangle a b' c' is drawn. A common vertical line for the weights is taken for all the force triangles. The value of p ... is again determined. The procedure is repealed, taking scveral leial planes, and the oorresponding force triangles are drawn and the values of P fJ determined in each case.

+'

cit.

A curve, called pressure locus (shown dotted) is drawn through the points of intersections c, c', etc. of P fJ and the corresponding R. The maximum pressure vector (PJmax gives the magnitude of the Coulomb active JXC&"ure. The failure plane corresponding to this vector is the actual failure plane. However, it is difficult to localed the failure plane precisely. The above procedure is general. Various modifications can be made as under: (1) If the adive pressure corresponding to Rankine's condition is required, the plane AB on which the pressure is computed is vertical. lbe resultant pressure on tbis plane is parallel to the ground surface. The rest of tbe procedure remains unchanged. (2) If the ground surface is irregular, the trial wedge method can still be used. Of oourse, there would be some difficulty in calculating the weight Wof the wedge.

SOIL MECHANICS AND rOUNDA110N ENGINEERING

504

(3) The method can also be used to determine the active pressure against the back: of the wall if the back fill carries a surcharge load distributed over the ground surface or a line load 3Cting on tbe fill. The weight of the wedge would include all such loads. (4) If the soil is oohcsivc, the force polygon would also include tbe cohesive Carre C on tbe failure plane and the adhesive force CD on the wall. I! may be noted Ihnl the trial wedge method does not give the point of application of Ptr 1llc pressure variation is assumed to be triangular and it is assumed thot P" aas at onc~third height. However, in irregular ground surface, the pressure variation is not triangular and it would result in some error.

19.13. COUWMll'S PASSIVE EARTII PRESSURE FOR COllESlONLESS SOlLS 'rbc failure surfucc in Coulomb's passive

Slate is assumed to be a plane. Fig. 19.31 shows the case when the passive conditions develop. In this case, the failure wedge moves upwards. The directions of Rand Pp which oppose the movement are also shown. The reaction R acts at 41' to the normal in the downward direction aod the reaction prcs.sure Pp acts at an angle b to lhe normal in the downward dircaion. The procedure for computing Coulomb's passive pressure is similar to one for the active case (Sect. 19.8) However, there is one basic difference. In this case, the critical failure surface is that which gives the minimlllli value

'------'==¥t:::..._

I.,

10'

fo1g.19.31. of Pp • Fig. 19.31 (b) shows the force triangle. The value of Pp is determined from this triangle. The procedure is repeated aller assuming a new trial failure surface. The minimum value of Pp is the (".ou!omb passive pressure. Using the procedure similar to that for the active case, it can be shown that the passive prc.c;.sure is given by

... (19.42) where

Kp _

sir12(p - 41')

sinlp. sin(1l + b)

2

[I _Vsi~W + b)s~n(+' + t1] sm(1l b)sm(p I} +

... [19.42(a)1

+

The resultant passive pressure Pp octs at a height of U/3 measured from the bottom of the wall. It would be inclined at an angle b to the normal. as shown in Fig. 19.31. However, when the retaining wall moves up relative to the soil, the friclion angle b is measured below Ule oorma! and b is said to be negative. The negative waU friction produces a value of passive pressure lower than that for the usual positive wall

friction. It is worth noting Ihal the wall friction decreases the active pressure, but it inacases the passive pressure. Moreover, the wall friction has a greater influence on the pac;sive pressure than on the active pressure. When exceeds II.J .', Coulomb's assumption of plane failure surface is not justified in the passive case. It gives much greater value of Pp compared to that obtained for the actual curved surface. As the passive pressure is generally required to provide the stability 10 a rel.:lining wall subjected to the active pressure on the other side, the higher value of Pp obtained from the plane failure surface in unconscrvative or unsafe. For such cases, the failure surface should be I.:lken as a logarithmic spiral or a circular arc as explained in Sc.ct. 19.14. 1)

10

BARnI PRESSURI3 THEORIES

:X

Rchbo.no's construction. Rchbann's construction can be used for the dclcnninnlion of the p.'lSSive pressure. 1I0wever, in this case, Ihe +-line is drawn at an angle - +' (i.e. below horizontal) (0 intersect Ihe extension of ground surfoce at point D. Culmonn method. '[he Culmann method and trial wedge method may also be used. From a series of scvernl force triangles corresponding 10 the various trial surfaces, a pressure locus or a Culmann line for pussivc pressure is obtained. The Coulomb passive pressure is the minimum value of pressure so obltlincd.

"JS

19.14. PASSIVE 1'RESSURE UY 'I1tE FRICTION CJRCLE Ml!.lHOD /I.s discussed above, Ihe rupture surfuce cannOI be tlSSumcd 10 be a plane in Ihe passive case, especially when [, > t'/3. as il givcs uns.'lfc values of the passive pressun:. The actual rupture surface resembles more closely a log-spiml in such 0 case. Tcr13ghi's general wedge theory (1943) enn be used to determine the passive pccssurc. However, Ihe melhod is quite involved :md beyond the scope of Ihis lext. The friction circle method is somcwhot easier and is described below for both cohesionless aod oohcsive soils.

(0) Cohcslonlcss Soli TIle lower portion DF of the slip surface DFD is assumed to be an arc of a circle which joins, without brcnk, the plane slip surface FD [fig. 19.32(a)]. The plane slip surface portion FD is inclined at (45 0 - ,'/2) to the horizonlIlJ. It is assumed that the upper portion CFD of the sliding wedge AnD is in Rankine's passive

o

eirel~

T

1 I,)

Ib)

Ag. 19.32(0), (b) and (c)

'd

SOIL MECHANICS AND FOUNDATION ENOINEERlNO

stale of plastic equilibrium. Planes AF and DF are the boundary failure planes of the Rankine's passive zone MD. The plane FD is tangential to the cirOJiar arc BF at F. Let US consider the equilibrium of the free body ABFC [Fig. 19.32 (b)]. The body is in equilibrium under 4 forces. (1) Weight Wof the free body ABFC of soii (2) Resultant pressure P due to Rankine's passive zone CFD. (3) Reaction R on the curved surface. (4) Passive force PI"

The friction circle method uses the OODCCpt that the reaction farre R acting along the failure arc acts at an angle " with the normal, and is therefore tangential to a circle of radius r sin .' drawn at the centre of the failure arc. Thus the resullant reaction R is tangential to the friction circle. Actually, R is tangential to a circle of radius K r sin ,', where K is a factor Slightly greater than unity, but for oonvenieoce K is taken as unity. The Rankine passivc pressure P acts on plane CF at a hcight of h/3, where h is equal to the beight CF and is given by P

=tKpyh2.

The passive pressure Pp acting on AB is also asswncd to act at a height H/3, where H is the beight of the wall. The weight W of the wedge ABFC is determined from its area and the unit weight of the soil. Knowing the magnitudes and directions of Wand P, their resultant S can be determined as shown in Fig. 19.32(c). Now tbe system is reduced to 3 forces, namely, Pp'R and S. The three forces must meet at a point Q for the equilibrium [Fig. 19.32(b)]. Knowing the directions of Pp and R and the magnitude and direction of S, the force triangle can be drawn and the value of Pp computed. The procedure is repeated for various trial surfaces and the value of Pp determined. The minimum value of Pp thus obtained is the required passive pressure. Of oourse. the pressure acting on the wall would be equal and opposite to PI' so found.

(b) Cohesive Soli For a c---+ soil, the analysis is done in two parts :(1) FitSt Pp is detennined as in the case of rohesionless soils, as explained above. (2) Additional pressure Pp ' is detennined considering only the cohesive aoo adhesive forces, as describe(! below. The total passive force in tbis case is (PI' + Pp '). The minimum value of the force (Pp + PP') is the required passive pressure. It is determined after evaluation of Pp '. Fig. 19.33 (a) shows a retaining wall with a cobesive fill. The free body ABFC is in equilibdum under the additional forces Pp',C,R' and PcThe Rankine p&\Sive force Pc due to cohesion on the plane CF of height h is given by

Pc - 2c VK;h It acts at the mid-point of CF. The passive pressure Pp' acts at the mid-point of the face An at an angle 6 to the normal. The force C is equal to the swn of the cohesive forces parallel to the chord BF of length L e •

e _c

)(

L~

The fon::e C is located at a distance a from the centre, given by

a - rL,ILc where L(I is the length of the arc BF. 1be adhesive force C... on the wall is given by

lG

EARTH PRESSURE TIlEORIES

at

Frltlion

cjrcl~

,f

a

"

'-./2

F

)f

l· ,f

(0)

e 01

Fig. 19.33.

'b)

C... -c,..xAB where c'" is the adhesion along the wall. From the force diagram [Fig. 19.33(b)], tbe resultant C of the force C and the wall cohesion C w is first determined.

The resultant P/ of the forces C and Pc is then delennined ~ shown. Point (1) in Fig. 19.33 (a) shows the position through which P/ passes. Now the forces Pp ' , p~' and R' are coosidered. Point (2) indicates the position through which the three focccs act. It lies at the inlecscction of P/ and Pp'. TIle direction of R' is now determined as it ~ through point (2) and is tangential to the fridioo circle. Once the direction of R' is determined, the force polygon in Ag. 19.33 (b) is completed and the value of PI" is determined. The total passive pressure is equal to the sum of PI' and pp'. where PI' is the friaional part as calculated for cohesionless soils and PI" is the cohesion part. The above procedure is repeated laking different failure surfaces. The failure surface which gives the minimum value of (PI' + PI") is the required surface, Thus the passive pressure force is found. 19.15. DETERMINATION OF SUEAR STRENGm PARAMETERS Retaining wall problems represenl plain strain conditions in whiCh intennediate principal stress bas a

'06

SOIL MECHANICS AND FOUNDATION ENOINEERING

value in-between the major and minor principal stresses. The angle of internal Cridion (,') used in tbe

computation of latcra1 earth pressure should correspond to the pJ3n(f strain condition. As tbe determination of this angle in the laboratory is diffiaJll, tbe angle of friction is ' determined from the triaxial test and a

correction is applied. Bjenum et al (1961) found that the angle of internal friaion in plain strain (fp) is about 10% more than lIlat in the conventional triaxial leSt (,,). Meyerbof (1963) gave tbe following relation for

+,.

~,-

...

(1.1 - O.l~)~,

(19.43)

where B is the width and L is the length of retaining wall.

Shear strength p."lrn,tnetes c and , in terms of either total stresses or efIcctivc stresses may be used according to the actual drainage conditions. However, as in practice most back fills 8fe of granular pervious materials, consolidatcd-drained conditions eKisl and the cohesion intercept is generally zero. ILLUSTRATIVE EXAMPLES Illustrative Example 19.1. Determine the toterol earth pressure at rest per unit length. of the wall shown ill Fig. £·19.1(a).Also detennine the Ideation oflhe resulltv1tearthpressure. TakeKu- I-sin,' amiy ... _lO kN/m'. Solution.

At point B

Ko - 1 - sin f' .. 1 - sin 30° .. 0.50. 0, _ 2 x 17 .. 14 kN/m l ,

u• 0

Po" Ku 0," 0.5 x 34 .. 17kN/m2 At point C

a, _ 2

x 17 + {19 _ to} )( 2 _ 52 kN/ml

Po .. 0.5 x 52 .. 26 icN/m 2

u .. 2 )( 10 .. 20kN/m2

Fig. E·19.1 {b} shows tbe pressure distribution diagram. The diagram Let P"P2>P) and p. be the lotal pressure due to these parts. Thus

1'·30'

L -______

been divided into tour parts.

T

¥:!1kNlm 3

t,m

.':JO"

r

~'_~_t·_"_·N_'_~_

~

Fig.1!--19.1.

Pi ..

~

p) ..

~)(

)( 17 x 2 .. 17 leN. 9 x 2 .. 9kN.

P2

-

p • ..

2 )( 17 .. 34 kN

~

)( 20 )( 2 .. 20kN

Thtal p .. IDkN The line of action of P is delennined by taking moments about C.

PZ ..

17 x 2.667 + 34x 1.0 + 9 x 0.667 + 20 x 0.667

Z ..

453 + 34 ;; 6 + 13.3 _ S:06 _ 1.23 m from base.

EARTIi PRESSURE TIiEORrES TakeD~::;~v~:;,mple

509

19.2. Determine the active pressure on the retaining wall shown in Fig. £-19.2 (a),

f',,)S'" Y:I7kNlm)

,,)

T

F r

Fig. E-19.2.


~ .. ~

For the upper layer,

K.. ..

For the bottom layer. At point B

: : : ::

~

: :: ~~: .. 0.271

K.. -

~

: : : ~: .. 0.238

a, ..

2.5 )( 17 _ 42.5 leN/m 2•

u - 0

p.. .. 0.271 )( 42.5 .. 11.5 kN/m 2

Below the interface. Pel is given by p.. .. 0.238 )( 42.5 .. 10.1 kN/m 2 0, .. 2.5

At point C

Ie

17 + 25 )( 8.0 .. 625 kN/m 2

u .. 2.5 x 10 .. 25 leN/m 2 Pel .. 0.238 )( 62.5 .. 14.9 leN/m2

Fig. E-19.2 (b) shows the pressure distribution. The forces Ph Po p) and P4 are detennined from the pressure distribution diagram.

PI"} x 2.5 x 11.5 .. 14.4kN. p) ..

t x 2.5 x

P2 .. 25 x 10.1 '"' 253kN P4 ..

4.8 .. 6.0kN.

Talal

t )(

2.5 )( 25 .. 313 leN

P - 77.0kN

Taking moments about C,

Z ..

14.4 )( 3.33 + 25.3 )( 1.25;.~.o )( 0.833 + 31.3 )( 0.833 .. 1.44m

Illustrative Example 19.3. Determine the active pressure on the wall shown in Fig. £-19.3, using Rankine's theory. Solution. From Eq. 19.15, . CDS i _ .Joos2i _ cos2" K,," COSl x cos; + .Joos2, _ cx>s2,' K .. ros IS"

..

Ie

cos IS" - .Jcm/ISo cos 15° + vCDS l l5°

2

0

cos 30 cosl300

..

0.373.

SOIL MBCHAN1CS AND FOUNDATION ENGINEERING

510

,.

'. '13 m

...l

T 'm

.=30· '1:19 kNlm3

1

Fig. E-19.3.

P., ..

From Eq. 19.23,

kK., yH'l .. k)( 0.373 )( 19.0)( (4)'l -

56.7kN

lhc active pressure aas at a height of 4/3 m and inclined at 15° to normal Dlustrative Example 19.4. Determine the stresses at the top and the bottom of the cut shown in Fig. E-19.4. Also determine the maximum depth of potential aack: and . ' ~ 12° the maximum depth of unsupported excavation. c: . 20k Nlm2 ... . 18 kN/m3 Solution. From Eq. 19.25, p .. .. ~ "'t Z - 2 c' VK;,

~i~ l~~o

~ : ~~:

where

K.. - / :

Thus

P.. - 0.656 )( 18l - 2 )( 20 x v'()]36

or

p.. _ll.81Z _ 32,4

AttopZ=O,

_

.. 0.656

p., _ _ 32.4 kN/ m'l

I

At bottom Z = 4, Pa .. 14.8 1cN/ m'l From Eq. 19.27, depth of aack,

,

-1

14.8

r-

Fig. E-19.4.

Zt- ~~ _ 2_~ _ 2.74Sm y v K., 18 vO.656 From Eq. 19.32, maximum depth of unsupported excavation, Ht

4c'

-

yVK" ..

5.490 ...

Illustrative Example 19.5. A 5 m high retaining wall is shown in Fig. E-19.5. Determine the Rankine active pressure 011 the wall. (a) Before the formation of the·crack. (b) After the formation of the crack. Solution.

K .. I-sin.' .. l-sin30° -0.333 ., 1 + sin+' 1 _ sin 30°

From Eq. 19.25,

Po p., P.. p.,

At top, Z = 0, At point .D, where p., .. 0.0,

-l("lZ - 2c'

IK.

.. 0.333 )( 17.5Z - 2 x SYo.m .. 5.83 Z - 5.77 .. -5.77 kNI m'l

EAR'I1i PRESSURE THEORIES

S1l

LJl c.)

Fig. E·19.S.

5.83Z - 5.77 .. 0

or Z .. 0.99 m

PQ .. 5.83 x 5 - 5.77 .. 23.38 lcN/m2

At bottom Z "" 5 m, Before formation of crack Negative pressure,

PI ..

tx

Positive pressure,

P2

4-

..

0.99 x 5.77 .. 2.86 kN

x 4.01 x 2338 .. 46.88 leN

Net p.' 46.88 - 2.866 • 44.02 kN Uoe of action of Pa is determined as under, by taking moments about C.

Z ..

46.88 x 4.01/3 :w~O~ x (4.01 + 0.67) .. 1.12 m

The same result would be obtained using Eq. 19.28. After fonnation of Crack After the (ormation of the crack, the negative pressure is eliminated. The pressure distribution is given by tbe area bed.

PQ

..

4-

x 23.38 x 4.01 ......6.88 kN

It will act at a height of 4.01/3 m above base Alternatively, directly from Eq. 19.29,

t H2 K 2e' H VK;, 2 .. t x 175 x 52 x 0.333 - 2 x 5 x 5v'll333

Pa ..

1

Q

+

-

(C')2/1

+ 2 x (5)2/17.5

PQ . . 72.84 - 28.85 + 2.86 .. 46.85 kN. U1ustratlve ElI:ample 19.6. DetermilU! the Ran/ci1U! passive force per unit length of the wall shown in Fig. £·}9.6. The water table is at the level of B. Take 1", = 10 kN/m1•

or


K .. p

~ 1 - sin 4"

For top layer I,

(i<,.h ..

! : ::~: .

3.00

For bottom layer D,

(Kph"

!: ::;:: .

2.37


Sl2

T

I

.',,30°

~~

2m

c'"O

B 't,,\6kNlm 3

II

.'=24°

c',,(IkNlm 2 lel9kNlm 3

t

96

-{:

2m

1

~4~-J0~

!o,

'0' Ag. E-J9.6.

From Eq. 19.34. At point A.

At point B.

Pp .. Kp Y Z + 2e'

z-o. Z .. 2 m.

pp-o. a,," 2

VK; )( 16 _ 32 kN/ml

Top layer,

Pp .. 3 )( 32 _ 96 kN/ml

Bottom layer,

Pp .. 32 )( 237 + 2 )( 10 )(

AI point C,

a. - 2)(

V2.TI _

100.6 kN/ml

16 + 2)( (19 - 10) _ SOkN/ml

Pp - 50 )( 237 + 2 )( 10 u _ 2 )( 10 _ 2OkN/m1

V2.37 ..

149.3 kN/m 1

Fig. E-19.6 (b) shows the pressure distribution.

TQ(al pressure P - PI + P2 + P3 + P4

-t )(

2 )( 96 + 106.6 )( 2 +

t )(

42.7 )( 2 +

~

)( 2 )( 20 .. 371.9 kN

E_19~~.~tr:t~v;s ~,:~e 19.7. Delermine the Coulomb active force on the retaining wall shown in Fig.

i

a

15-

"'S-20· .30·

'¥.11-~ IcN/m3

~.75

Fig. &-19.7.

Solution. From Eq. 19.39.

sin' (p + .')

K. _

sm". t' .

'(.-')[1 sm t' U

10

BAR111 PRESSURE THEORIES

sinl (75 + 30°) K. _ _ _ _ _~~~~:::::::~~, 0

, '750 ' MSO _ 20°) [1

-.,

~ KII ..

(30° • 20°) ';0 (30° - 15°) 1 • V';0 ~w-~~w·~

0.933

0,933

)C

0.819 [ 1 +

'2 ..

V

°;:;9)(x

0.548

°i~9 J

iX"yH1

From Eq. 19.38,

p" ..

or

P" .. ~ x 0548 x 17.5

]I(

(5)2 .. 119.9kN

This will ad at a height of 5/3 m and inclined at 20° to normal in the direction shown. 1be reader should note that the direction of P a is equal and opposite to that on the wedge. as disoJssed in Sect. 19.9. Dlustntln Example 19.8. Determine the active thnut on the retaining wplJ shown in Fig. E-19.8. The backfill is cohesionless (qt' = 30", Y = 19 kNlmJ, 6 = 20°). Solullon. The q1ethod of trial wedges is used (Sect. 19.12). Several trial planes such as .8-1, B-2, B-3, B-4 are chosen. The sliding wedge is in equilibrium under the three forces W, R aoo Pp' TIle weights of the wedges are romputed as given below. WcdgeABl, Wt .. (6 x 312) x 19.0 .. 171.0kN WcdgeAB2, WI" (6 x 612) x 19.0 .. 342.0kN WedgeAB3,

Wl

WedgeAB4,

w" ..

..

tx tx I_

2 x 9 )( 19.0 + 342 .. 513 leN

2 x 9 x 19.0 + 513 -684kN

3m

3m

T

f 'm

1

Ca)

,. Fig. 8-19.8.


514

Fig. E-l98 (b) shows the force diagrams drawn along the common weight line. The pressure locus is shown dotted. The maximum thrust is given by P(I - 178.5 leN D1ustratlve Example 19.9. Determine the QCtive thrust on Ihe retaining wall shown in Fig. E-19.9, using

Cuimann ~ method.

t--2m---J.-lrn---+lm-+-2m---\-t---- 2m---t

T J 6m

4:350 Y,,20kN/m3 ~

",gOO

~_"O

Ai. E-19.9.

Solution. The line BD is drawn at an angle of 35 0 to the horizonLal. and the line & at an angle 1p = to the line BD. In this case, 'I>_~_6_90·_0_90· ~l. B-2. B-3, and B-4 are the four assumed trial failure surfaces. The weights are calculated as un

i t)( wl-t)(

WI -

x 2 x 6 )( 20 .. 120kN

W2 -

4)( 6)11 20 .. 240kN

W• ..

tx

6)( 6)( 20 .. 360kN 8

x 6 x 20 .. 4801cN

The weight (Wo) of wedge AlJO, upto tbe line lood, is 180 kN.

The weight including line load, Wo .. 280 leN The weights are plotted along the line BD. • From each of the points SO located on BD, lines are drawn parallel to BL to cut the oorrespond' assumed failure surface. On the failure plane BO, two points e and e' are obtained, one for wedge without line load and one for wedge with !he line load: Culmann's line is drawn joining the points a, e, e', b, C, d. It is a disoonLinuous curve because of the ' load, A tangent is drawn to the Culmann line such that it is parallel to BD. From the point or tangency, a ' EF is drawn parallel to Bt. The length of EF gives the required Pr

PII

-

156kN

lING

BAR11I PRE$URE THEORIES

>sis

515

PROBLEMS A. Numerical

:sing

19.1. ~~cr;on:!~ ~~V~~~b~::r:nc;~'::!lt~r:!t~n~v~n~c:~ning 19.2. For the retaining wall in Prob. 19.1, determine the active pressure per unit run.

willi

4mhjg~A:t~~ ~i IAns. 56.7 kN]

19.3. Determine the active pressure and passive pressure, using Coulomb's theory, on the wall shown in Fig. P. 19.3. (Ans. 73 kN, 1413 leN]

Fig. P-193,

19.4. .,:=~r :j~:~~S~:I:I.~~~=~hIlC~h~~ ~n ~~h~~pc~~n~~n:h~;;:l~ i: ~~~~y~e l'::,::t:: :Jer:

20°, Neglect tcnsion. (Ans. 181.3 kNJ 19.5. A retaining wall with II vertient smooth back is 8 m high. It Supports II cohcsionlcss soil (y :c 19 kN/m 3, 4t := 30°). The surface of the soil is horizontal. Detennine the thrust on the wan. [Ans. 202.7 kN] 19.6. Determine the resultant thrust on the wall of Prob. 19.4 if weep holes are blocked and water builds up behind the wall until the water table is 3 m above the base of the wall. [Aos. 2043 kN] 19.7. A retaining wall is 7m high, with its back face smooth and venical. It retains sand with its surface horizontal.

~nfct::e~~ih ~~~:::e~~~e:~!~~ ;t~~8~Jh:~ ~e =ba~~I~S !a~ld~~l. saturatoo 2 [Ans. 42 kN/m2, 49 kN/m1, 25.7 kNtm )

19.8. A vertical retaining wall to m high supports a rohesionless soil (y = 18 kN/mi. The upper surface of the back fiU rises from tile aest of the wall at an angle of 15° with the horizontal. Dctennine the total active pressure by Culmann's method and check the pressure obtainoo by Rebbann's construClion. Take t .. 30° and (\ :::: 20°. [Ans. 333.5 kN)

jng tbe dne ~ ne'

B. DescrIptive and Objective 1)'pes 19.9. What are different types of earth pressure? Give examples. 19.10. Define earth pressure at res!. Show the earth pressure distribution on a retaining wall, assuming the soil is dry. 19.11. What are the assump'ions of Rankine's theory? Derive the expreWoos for active pressure and passive pressure. U.12. Derive an expressioo fot aaive pressure when the ground surface in inclined. 19.13. What are the assumptions in Coulomb's theery? Compore Rankine's theory and Coulomb's theory. 19.14. Describe Rehbann's construction. What is its use? 19.15. Discuss Culmann's method ror the detennination of active earth pressure. d.lIi. explain the trial wedge method. 19.17. How would you determine passive pressure when ~ :> 1/3 'P'? IJ.II. Write whetber tbe following statemenls are true. (Q) The active pressure Is the minimum pressure which develops when the wall moves away from the fill.


51'

(b) The shear resistance dtvcloped along the railure surface is a minimum when the active conditions develop.

The Rankine pres~urc is lliways noonnl to the wan surface. (t!) Culmano's construction ilt more genera! than Rehbann's construction.

(e)

(t' ) Culmann's ConSlnll'l ion I:an be used even for passive case.

(/) Coulomb's Iht.'Ol')' always gives conservative results. (g) For~':: 30". the passive pressure is three limes the octive pressure, accordmg to Rank ine's theory. (II) The aClIye pressure Increases if a dry soil becomes submerged. (i) TIle basement walls are gencmlly dcsigned for ai-rest pressure.

(Aos. True, (a). (e). (d), (e), (I)]

C. Multiple Choice Questions 1. The inclinatIOn of the failure plane behind a veniea] wall in Ihe passive pressure case is inclined to the ho rizonlu l

;It

(b) 45° - ¢I «(/) 45° .. ¢I 2. The yield of a rctaini ng wall required to reach plaslic equilibrium in aClive case is (aJ more than tllat in the passive case (b) less than that in th e passive cn.~c. (e) equal 10 Ih.\1 in th e passive case. (d) No ne of abov;,:. 3. The active earth prt:ssure coeflic ient K" generall y refers to (a) effective slrcsse~ (b) lotal stresses (c) neutral stresses (d) All the above 4. The active pressure cau!>ed by a cohesionless backfill on a smooth vertical retaining wall may be reduced by (a) compacting (1)(: backl1ll. (b) providing a surcharge load on the backl1l1. IC) satura ti ng Ihe backfill wtth waler. «(I) None of the above. S. The IOIaI aclive p re~sure after the devdopment of tension cracks is equal to {lJ)1YHl Ka - 2,:·'H.fKc/

(a) (t')

45 0 - ¢112 4.5° .. ¢!12

(b)1YH 2 Ka .. 2c'H..JI(;/

(cJ~YH2KII-

2(:'Hffi - 2

«(1)1 yH 2 Ka _ 2c'H..JKa +

¥

~

2 Y 6. The rud ius orlhe friction circle is cqual to WR~,

WR=.

(c) R tan ill (ll) R ill 7. If u uniform surcharge of 120 kN/m2 is placed on the backl111 with ¢I' = 30~. Ihe increa.~e in pressure is (a) 12 kN/m 2 (b) 30 kN/m2 (el40 kN/m2 {(I) 120 kN/ml [Ans. I. (a), 2. (b) 3. (a), 4. (a), 5. (d). 6. (a), 7 . (eJI

UNG

::lop.

I. (i»)

11he

20 Design of Retaining Walls and Bulkheads ltI.l. INTRODUcnON (0) Des.lgn of Retaining WBlis. Retaining walls are relatively rigid walls used for supporting the soil mass laterally so that the soil can be retained at different levels on the two sides. The lateral earth pressures ad.ing on the retaining walls have been discussed in the preceding chapter. The types of retaining walls and their dcsiWl features are explained in this chapter. However, the design is limited 10 the determination of the sbear fooces and bending momenrs. Actual structural design is outside the scope of Ihis text. (b) Bulkheads. Sheet pile walls, or bulkheads, are special type of earth retaining structures in which a coolinuous wall is constructed by joining sheet piles. Sheet piles are made of timber, steel or reinforced

by

concrete and consist of special shapes which have interlocking arrangements. Sheet pile walls are used for

water front structures, canal locks., coffer dams, river protection. etc. Sheet pile walls are embedded 10 the ground to develop passive resistance in the front to k.eep the wall in equilibrium. \1Irious types of sheet pile walls and their analysis and design are discussed in this chapter.

• (c»)

ltIJ. lYPES OF RETAINING WALlS The moot common types of retaining waDs are classified as under: (1) Gravity RetaLnlng Walls. These walls depend upon their weight for stability (Fig. 20.1 (4»). The walls are usually constructed of plain ronaete or masonry. Such walls are not economical for large heights. (1) Seml·Gravlty Retaining Walls. The size of the section of 8 gravity retaining wall may be reduced if a small amount of reinforoement is provided near the back face [Fig. 10.1 (b)] . Such waDs are known as semi*gravity walls. (3) C.nlUever RetalnLna Walls. Cantilever retaining walls are made of reinforced cement ooncrete. The wall consists of a thin stem and a base slab cast monolithically (Fig. 2O.1(c)J. This type of wall is found to be economical upto a height of 6 to 8 m. (4) Counlerfort Retaining Walls. Counterfort retaining walls have thin vertical slabs, known M oounterforts, spaced across the vertical stem at regular intervals [Fig. 2O.1(d)]. The counterforts tie. the vertical stem with the base slab. Thus the vertical stem and the base slab span between the counterforts. The purpose of providing the counterforts is to reduce the shear force and bending moments in the vertical stem and the base slab. The counterfort retaining walls are coonomical for a height more than 6 to 8 m. [NOIe : Counterforts are on the side of the back fill).

%OJ. PRINCIPLES OF THE DESIGN OF RETAINING WALlS Before the actual design, the soil parameters thaI influence the tarth pressure and tbe bearing capacity of the soil must be evaluated. These include the unit weight of the sou. the angIe of shearing resistance. the


518

'0)

Id) Fig. 20.1. Dlffc:n:nt Types of Retaining WIllis.

cohesion intercept and the angle of wall friction. Knowing these parameters, the lateral earth pressure and the bearing capacity of the soil can be determined. Methods for the computation of earth pressure have been discussed in chapter 19. The bearing capacity theories are explained in chapter 23. With tbe earth pressure known, the retaining wall as a whole is checked for stability. Fig. 20.2 shows a retaining wall with a smooth

1 li

back face and no surcharge. The active pressure PQ acts horizontally, as shown. The front face of the wall is subjected to a ~ive pressure (Pp) below the soil surface. However, it is doubtful

whether the full passive resistance would develop.

H

Moreover, o[ten Pp is small and therefore it may be neglected. This gives more conservative design. The weight W of the wall and the active pressure P" have their resultant R which strikes the base at point D. There is an equal and opposite reaction R' at the base between the wall and the foundation. For oonvenience, R' is resolved into the vertical and horizontal components (R/ and RH'). From the equilibrium of the system, Fig. 20.2.

'INO

DESlGN OF RETAINING WALLS AND BULKHEADS

'19

and RJI' - p .. The third equation of equilibrium, namely the moment equation, is used to ~etermine the eo:entricity t! of the fOl"Q: R,t' relative to the centre C of the base of the wall. Obviously, by laking moments about the toe, R~' x i _ W x a - p .. (HI3) Rv' - W,

••• (20.1)

where x is the distance of the point D from the loe. Thus, ccccntricity, t! _ bl2 - i

... (20.2)

where b = width of the base. For a safe design, the following requirements must be satisfied. (I) N. Sliding

The wall must be safe against sliding. In other words, IA Ry > RJI where Ry and RII are vertical and horizontal componcots of R, respectively. The factor of safety against sliding is gfven by ~Rv

F, -

. •. (203)

Ii;;

where IA =coefficient of fridion between the base of the wall and the soil ( = tan 0). A minimum factor of safety of 1.5 aga~lDSt sliding is geoerally recommended. (2) No Onrturnlng The wall must be safe agaillSt ovenurning about toe. The factor of safety against overturning is given by F, _

~ MR

... (20.4)

~Mo

where I MR = sum of resisting moment about toe, aoo I Mo sum of overturning moment about toe.

=

the
.ure

In Fig. 20.2,

I

The factor of safety against overturning is usually k.ept between 1.5 to 2.0. (3) No bearing capadty rallure The pressure caused by Rv at the toe of the wall must not exceed the allowable bearing capacity of the

soil. The pressure distribution at the base is assumed to be linear. 100 maximum pressure is given by

p~ - ~ (I

H

1

... (20.5)

+ ,.,./b)

... (20.6)

The factor of safety against bearing failure is given by ... (20.1) where q... :: allowable bearing pressure. A factor or safety or 3 is usually specified, provided the settlement is also within the allowable limit.

(4) N. 1 _ 1bece should be no temion at the base of the wall When tbe

eccentricity (e) is greater than b!6, tension

52"


develops at the heel. Tension is nOI desirable. The tensile strength of the soil is very small and the tensile crack would develop. The effeaive base area is reduced. In such a case. the maximum stress is given by Pma ·

Rv ) 3"4 ( b-2e

... {20.8)

20.4. GRAVITY RETAINING WAILS As in design of all other structures, a trial sectJon is first chosen and analysed. If the stability cheW

yield unsatisfactory results. the section is changed, and rechecked. Fig. 20.3 shows the general proportion of a gravity retaining wall of overall height H. The top width of the stem should be at least 0.3 m for proper placement of concrete in the stem. The depth (D) of the foundation below the soil surface should be al least 0.6 m. The base width of the wall is generally between 0.5 H to 0.7 Hi with an

T

average of 211/3. The earth

1

Pa (R.lHtclNE)

30

1

pressure can be computed using either Rankine's theory or Coulomb's theory. For using Rankine's theory, a vertical line AB is drawn through the heel point A. II is assumed that the Rankine active

conditions exist along the vertical line

~~el~;;:~r, oteR=~:~iO~:;j~~

Fig. 203. Gravily wall-Rankjne P!usure.

along AD is theoretically justified only if the shear zone bounded by the line AC is not obstructed by the stem of the wall. where AC makes an angle 11 with the vertical, given by " _ (45 0 + i/ 2) _ ,'12 _ sin- 1

(

~in i )

... (20.9)

sm.'

where i is the angle of surcharge. The angie a which the line AC makes with the horizontal is given by, 0

a - ( 45 +

~) - ~

+ sin-

1

(s~~n~, )

... (20.10)

When i :: 0, the value of a is equal to (45 0 + ,'/2) (Fig. 20.4). While checking the stability, the weight of soil (WI) above the heel in the zone ABC should also be taken into consideration, in addition to tbe earth pressure (P.,) 00 tbe vertical plane AD and the weight of the wall (We)' Coulomb's theory can also be used for the determination of earth pressure (Fig. 20.5). As the Coulomb theory gives directly the lateral pressure on the bad:: face (P.,), the forces to be considered are only P., (Coulomb) and the weight of the wall (W(). In this case, the weight of soil (W,) is not to be considered separately. Once tbe forces acting on the wall have been

I

I

1 1

,

,

I

I

:-+-~

1/

L--::;",-'f1, 45'-

Fig. 211.4.

f

"

INO

DESIGN OF RETAINING WAl..l.S AND bULKHEADS

sile

detennined, the stability is checked using the procedure discussed in the preccaing section. For oonvenience, the section of the retaining wall is divided into rectangles and triangles for the computation of weigbt and the determination of the line of action oC the weights. Seml-anvlty Reta1n1na walls. The base width of the semi-gravity retaining walls is slightly smaller than that of a aHTCSpOnding gravity wall. The re;t of the design procedure is the same as that for gravity retaining

).8)

dt.s

of

walls.

10.5. CANTD..EVER RETAINING WALLS Fig. 20.6 shows a cantilever retaining wall. The

52.

T

P. (COULOMB)

.--

H

1

. gencml proportions for an overall height of H are also Fig. 2O.S. Grav,!}, wall--Coulomb Pressure. shown. The top width of the stem is at least 0.3 m. 1be width of the base slab is kept about 2H!3. The width of the stem at bottom, the thickness of tbe base slab and the length of the toe projection, each is kept about

O.1H.

T Pot (RANKINI

9)

0) Fig. 20.6. CAntilever Retaining wall.

The earth pressure is computed using Rankine's theory on the vertical plane AB, provided the shear zone bounded by the line AC is not obstruaed by the stem of the wall The line AC makes an angle" with the vertical given by Eq. 20.9. Fig. 20.7 shows the forces ading on the wall. The Rankine pressure Pa acts at an angle i with the horizontal. It is resolved into the vertical and horizontal 'components Pvand as shown. The passive pressure Pp is also shown, but generally it is neglected. For oonvenience, the weight of soil (W~) over the slab is divided into two pans (I) ,and (2). Likewise, the weight of stem is divided into two parts (3) and (4).

p" .

«(I) Factor of safety against sliding The factor of safety against sliding may be expressed as

IF, F~ - IFd where I FR = sum of the horizontal resisting forces.

... (20.11)


522

T o i

Fig. 20.7. Forces on a Cantilever wall.

1: F d = sum of lbe horizontal driving forces. Eq. 20.11 can be written as

and

... (10.12) where b = base width, I:V = sum of all the venical forces, Wn W, aDd PI" P" .. P"sin I and

Ph" P"OOSi . Pp = passive force in the front of the wall (.. 112 Kp2 Y2 Jj2 + 2c2 Vi(,,; D) where c2. 12 and 4Iz are parameters of the foundation soil. The factor of safety can also be detennined from Eq. 203 if 1.1. is given. IT the required factor of safety of 15 against sliding is nol obtained, a base key is generally provided' (Fig. 20.8). The key inaeases the passive resistance to Pp' where

Fig. 20.8. Key In a retaining wall.

DESIGN OF RETAINING WALLS AND BULKHEADS

Pp' .. iYl(D1)1 Kpl + 2c1 Dl

vx;;.

.(20.13)

where Dl is the depth of the boliGm of the key wall (rom soU surface. Generally, the base key is constructed just below the stem and some of the main steel of the stem is extended into the key. The fridion angle th and C2 are generally reduced to about one-half to two-thirds of the values for extra safety, as the full passive resistance is doubtful.

Factor or safety against Onrturnlng Eq. 20.4 can be used to obtain the fador of snfety against overturning, EM, Fo" T.M" where 1: MR = sum of the resisting moments about toe, IMo ::=: sum of the overturning moments about loe. The only overturning force is P", acting at a height of H/3.

Mo" p lo )( HI3 ... (20.14) The resisting moments (MR) are due to weights Wb W2:I W). W.. and Ws of the soil and the OOOQ'ete. The

vertical component of pressure P" also helps in resisting moment. lIS resisting moment is given by ... (20.15)

Mil" p,,)( b Therefore

Ml +M2 + M) + M.. + Ms + M~ F,," )( H/3 lo

r

... (20.16)

where Ml • M1 •... Ms are the moments due to Wh W2 .. Ws about toe. }o'actor or sarety against bearing capacity failure The sum of the vertical forces acting on the base is equal 10 IV. The horizontal force is Plo The resultant force (R) is given by

R • V(E V)' + (p.)' The net moment of these forces about toe B is given by

IM .. IMR-l:.Mo ')be distance i of the point E, ,from the toe, where R strikes the base is given by

_ x ..

IM

IV

... (20.17)

Hence, the eccentricity e of R is given by

e-bl2-i ... (20.18) If e > b/6. the section should be dlanged, as it indicates tension. The pressure distribution under the base slab is detennined as

and

Pmu.-T

IV (1 +b'60)

.•. [20.19(a)J

pnUn-¥(I-¥)

... [20.19(b)J

The factor of safety against bearing capacity failure is given by Eq. 20.7. 20.6. COUNTERFORT RETAINING WALLS For oounterfort retaining walls, the general proportions of the stem and the base slab are almost the

,,.


the

same as that in the cantilever walls. 1be oounterforts arc about 0.3 m thick: and bave centre-to-cenlre spacing of 0.3 H 10 0.7 JI. The analysis is also similar to that of a cantilever retaining wall. The pressure p_ and Pmln are determined, as in the case of cantilever walls. ]be basic differeoce between the COW1terfort retaining wall and the cantilever retaining wall is in the detcnnination of the bending moment and shear forces. (1) In cantilever retaining walls, the stem acts m; a vertical cantilever fIXed at base whereas in the CQUoterfort retaining walls, it acts. as a oontinuous slab supported between the counterfocts. 1be slab has positive momcots in the middle and the negative moments at the supp:>rts. The reinforcement is provided in the horizontal direction on the front side of the stem in the middle and on the rear side at the supports. In cantilever walls, the main reinforcement is in the vertical direction at the rear face. (2) In cantilever walls, the toe slab and the beel slab both act as cantilevers subjected to the upward pressure. The reinforcement is provided at the bottom face. In counterf~It retaining walls, although the toe slab acts as a cantilever, tbe heel slab acts as a continuous slab supported on the counterforts. The main reinforcement is al the lop face in the middle portion and at Ihe bottom face near the supports. (3) In couoterfort retaining walls, the rounterforts are designed as cantilever of varying section and ftxed at the base. The maln reinforcement is provided at the back face of the counterfort. In addition, the vertical and horizontal ties are provided in the counterforts to jOin the base aod tbe stem to the counterforts. The structural design of the counterfort and cantilever retaining walls is outside the scope of this text.

20.7. OrnER MODES OF FAILURE OF RETAINING WALLS In addition to the Ihree types of fallures., viz, slicting, overturning and bearing failures. a retaining wall may fail in the following two modes if the soil below is weak:. (1) ShaDow shear Failure. This type of failure occurs along a cylindrical surface ABC passing through the beel of the retaining wall (Fig. 20.9). The failwe tal.es place because of excessive shear stresses along the cylindrical surface within the soil mass. However. it has generally been found that the factor of safety against horizontal sliding discussed in Sed. 20.3 is lower than that for the shallow shear failure. Consequently. if the factor of safety against sUding (F,) is greater Ihan about 15, shallow shear failure is not likely to occur.

c

L-------Jc 8 fig. 20.9. Shallow Shear Failure.

Fig. 20.10. Deep Shear

FaillJ~

IG

re

Ie

ob is

Ie

DESIGN OF tmrAININO WALlS AND BULKHEADS

(2) Deep shear (aDure. This type of slope failure occurs along a cylindrical surface ABC (Fig. 20.10). when there t... a weak layer of soil underneath the wall a depth of about 1.5 times the height of the wall. The critical failure surface is determined by trial and error procedure. For the backfills having slope i less then 10°, it has been found that the aitical failure surface DEF passes through the edge of the heel slab. The minimum' factor of safety is found by trial and error, taking different circles, and determining the resisting forces and the driving forces along the failure surface (See reng, 1962), When a weak soil layer is located at a Shallow depth below the retaining wall, the possibility of deep shear failure should be investigated. The possibility of excessive settlement should also be looked into. Sometimes, piles are used to transmit the foundation load to a firm layer below the weak layer. However. care !hall be taken in the design of piles so that the thrust of the sliding wedge of soil does not cause bending of the piles. 20.8. DRAINAGE OF THE BACKFILL When the backfill becomes wet due to rainfall or any other its unit weight increases. H increases the pressure on the wall and may create unstable conditions. Further, if the water table also rises, the pore waler pressure (u) develops and it causes ex~ive hydrostatic pressure on the wall. To reduce the

reason,

development of excessive lateral pressures on the wall, adequate drainage must be provided.

BACKFIL L

Weep holes are generally provided in the

FILTER

walls. The weep holes are of about 0.1 m diameter.

MATERIAL

The spacings of the holes generally varies between WEEP t-I)LES 1.5 m to 3 m in the horizontal direction. As the bacldill material may be washed into weep holes

and may clog them, filter material is placed around the weep holes (Fig. 20.11).

Perforated pipes are aIro frequently used for the drainage of the backfill. 'These pipes are laid fig. 20.11. Weep Holes. near the base (Fig. 20.12). The water is collected from the backfill and discharged at a suitable place at the ends. The filler material is placed around the pipes. These days, a filter cloth or a geotextile fabric is also used to serve tbe purpose of a filter material. All drain pipes should be provided with clean-outs for cleaning when cloggoo. Fioe-grainoo soilS cause large earth pressure against retaining walls and are, therefore, rarely used as a backfill material. As far as possible, good draining, granular material should be used, at least in the sliding wedge portion of the wall. In case a fmc-grained material cannot be avoided, some form of filter of coarse permeable material is placed behind the retaining walls to prevent the development of excessive pore water pressure. Fig. 20.13 shows two types of drainage filters commonly used. The water percolating into the filter is discharged through the weep holes. The inclined filter is found to be more effective tban the vertical filter. .

FILTER MATERIAL

r''--------'=q-,.., ~~~FORATED Fig. 20.12. Perforated Pipca.

Fig. 20.13. Drainage Rlter.

SOIl. MECHANICS AND FOUNDATION ENGINEERING

''''

BULKHEADS 20.9. 1YPES OF SHEET PILE WALLS Sheet piles an:: generally made of steel or timber. However, sometimes reinforced cement concrete sheet piles are also used. The use of timber piles is generally limited to temporary strudures in which the depth of driving docs not ~xceed 3m. For permanent strudures and for depth of driving greater than 3 m, steel piles are more suitable. Moreover, steel sheet piles an:: relatively water tight and can be exlrn(.1cd if required and re- used. However, the coot of steel sheet piles is generally more than that of timber piles. Reinforced cement concrete piles are generally used when these are to be jetted into fine sand or driven in very soft soils, such as peat. Foe tougher soils, the roncrete piles generally break. off. ~D SOCKET JOINT

~IPtLE IIPILE~ Fig. 20.14. Plan cC • Sheet Pile Wall.

Fig. 20.14 shows the plan of a typical steel sheet pile wall, in which 2 sheet piles are shown with jOints. Based on its structural fonn and loading system, sheet pile walls can be classified into 2 types: (1) Cantilever Sheet piles and (2) Anchored Sheet Piles. (1) Cantilever sheet plies

Cantilever sbcct piles are further divided into 2 types : (a) Free canUlever sheet pile. [Fig. 20.15 (a)]. It is a sheet pile subjeded to a oonccntrnted horizontal lood at its top. 1bcre is no backfill above the dredge level. 1be free cantilever sheet pile dc.rlves its stability entirely from the lateral passive resistance of the soil below the dredge level into which it is driven.

~J-I Anchor

rored e

level

pI Free cantilever Shut pile

(b) ContHever Sheel pile

tt) Anchored

Shut pile

fig. 20.15. Type:.otSheet Pile WAn •.

(6) CantUever sheet pile. [Fig. 20.15 (b)]. A cantilever sheet pile retains backfill at a higher level on one side. The stability is entirely (rom the lateral passive resistance of the soil into which the sheet pile is driven, like that of a free cantilever sheet pile. (2) Anchored sheet pUe; Anchored sheet piles are held above the driven depth by anchors pro1fided at a suitable level [Fig. 20.15 (e)]. 'The anchors provide (orces for the stability o( the sheet pile, in addition to the lateral passive resistance

INO

lee! a of ilcs ~rxJ

en.

x:b

DESION OF RETAINING WAllS AND BULKHEADS

of I.bc soli into wbich the sheet piles are driven. The anchored sheet piles are also of two types: (II) Free-e:arth support plies. An anchored sheet pile is said to have free.earth support when the depth of embedment is small and the pile rotates at its bottom tip. Thus there is no point o[ oontrafiexure (or inflexion point) in the pile. (b) Flxed.earth support piles. An anchored sheet pile has fixed earth support wben the depth of embedment is large. The bottom lip of the pile is fixed against rotations. There is 8 change in the curvature of the pile, and hence., an inflexion point occurs. lUl FREE CANTILEVER SIllili"T PILE The free cantilever sbeet pUe rotates about a point 0 below the dredge level. The actual pressure dhtribution is shown in Fig. 20.16 (a). Blum (1931) gave a Simple solution. The passive resistance of the soil

IS.

1)

21 8 IS

(.I) ACTUAL PR£SSURE

DISTRIBUTION

~

(b) ASSUMED

PRESSURE

_ _DlSTRfBUTKW ~_~_

Cd MOVEMENT

DIAGRAM

CIl the kR side is idealized as a right angled triangle AOE [Fig. 20.16 (b)1. The distributed pressure acting 00 !he right side below the pivot 0 is replacx:d by an equivalent concentrated load PI acting at lX'iot O. In calOJlations that follow, however, the magnitude of the force PI is not required, For equilibrium, the moment of all the forces about 0 must be zero, i.e.

M, - F(h + d) -

[t

yd(K, - K.)d

1x ~ -

0

.. .(20.20)

where F is the horizontal force, h is the height of wall above the dredge level, d is the depth of embedment. Eq, 20.20 can be solved for d. The actual depth to be provided is generally taken as 1.2 d. The point of the maximum bending in the sheet pile can be detennined as under. The bending moment at depth x below the dredge level is given by

M, - F(h + x) For maximum bending (MIIW\X).

d~z)

f

(K,. -

K,,)

••. (20.21)

_ 0

F _ Y(Ke; KJ (3.<') _ 0

'"

.x - VY(K,u:.

K,,)

... (20.22)


The maximum 8.M. (M1IlP) is obtained by substituting the value of x from Eq. 20.22 iota Eq. 20.21. The section modulus of tbe sheet pile can then be determined as

S _ MIIlP ... (20.23) a. when:: all = aUowable bending stress in the pile. Fig. 20.16 (c) shows the bending moment diagram .

20.11. CANTILEVER SHEET PILE IN COfiESlONLESS SOILS Fig. 20.17 (a) shows a cantilever sheet pile in a oohcsionless soil deposit The pile rotates about the point

0'. TIle pressure above 0' is passive in the front and aaJve on the back side. However, the pressures below B_ _ __

T h

+~ d

ACTIVE

T

1

\ PASSIVE (.)

(b) ACTUAl Fil. 20.17. Cantilever

~

PRESSURE

DISTRIBUTION Pile.

the point 0' are reversed i.~. there .$ aaive pressure in the front and passive on the back side. Fig. 20.17 (b) shows the actual pressure distribution. As the analysis taking actual pressure distribution is quite complicated,

the pressure distribution is generally simplified as shown in Fig. 20.18. In Fig. 20.18, the pressure is zero at point 0 1 at a deptb a below the dredge level. The pressure diagram BCOl shows tbe active pressure. The pressure at the dredge level is given by PI -yhKII

~

~.

1

l~,,"~:~__ -- - --A

ij F

Il(kp-")(d-·).lr(h+d)ko-1d~ (~) ASSUMED PRESSURE

Pig. 20.18. Simpllncd

I

DISTRIBUTION

Press~

DI.trlbution.

DESIGN OF RETAINING WALLS AND BULKHEADS The depth

a of point 0 1 of zero

529

pressure is given by PI - ya{Kp - K,.} _ 0

a _ -P -'... (20.24) 1 (Kp-K.) lei the lotaJ active pressure above point 0 1 be PI acting at a height of ZI above 0 1, The passive pressure is given by the diagram 0 1 £0. The passive pressure intensity at the bottom tip A QrI be elCpressed as P2 - nK, - K.)(d - a) - 1(K, - K.)b where b .. d - a, in which d is the depth of point A below the dredge level. The passive pressure is indicated by the diagram OAF on the back side. The intensity of pressure at the lip A is given by Pl - y{h +d)Kp - ydK" Pl - Y(II + b + a) Kp - y (b + a) K,.

or

From the equation of equilibrium in the horizontal direction, PI + p) - P 2 - 0 The total pressure p ) and P2 can be expressed in tenns of P3 and P2 as follows:

PI + tm(P2 +

P~-~P2b

... (20.25)

- 0

In Eq. 20.25, the equivalence of areas has been taken as shown in Fig. 20.19. The height of the point £ aboYcthetipA is taken as m.

b) d, at Fig. 20.19.

From Eq. 20.25,

1I2p.p - PI P2b - 2P1 m-1I2
... (20.26)

Taking moments of all the forces about A , Pdb + ZI) - iP2 b (b/3) +

t

m
~

- 0

Substituting the value of m from Eq. 20.26, ,

_~+
Pdb + ZI)

.-


6

b4 + C 1 b l

-

2

+Pl) [Plb -2P 1 6 P1+p)

C2 b 2 - Clb - C 4

_

j_ 0

0

C • -

... (20.27) .(20.28)

PI (6Z I P4 + 4Pd

11(K, - K.)]'

SOIL MECHANICS AND FOUNDATION

530

ENGJNEER!N~

... (20.29.

in which

Eq. 20.28 is solved by trial and error to determine b. The value of d is equal to (b + a). 'Ibe depth dy for a factor of safely of unity. The required depth (D) is usually laken as 1.2 d to 1.4 d. Thus . D _ 1.2d to l.4d ... (20.3I! This gives a factor of safety of about 1.50 to 2.0. AHcmativcly, a (actor of safety can be applied to the passive resistance. In that case, the value of Kp J

JS

usually taken ~ to ~ of the normal value while oomputing b from Eq. 20.28, and the required depth D is taku equal to d. In the above discussions. the depth of water table is not B considered. If the water table on the front side is at the same level as on the rCo:"lf side, the analysis remains unaltered cxu:pt that the submerged unit weight ('I ') should be used fOf the soil below the h water table (see Illustrative Example 20.6). However, if the p difference in the two levels is greater than 1 m, the pressure due to 1 water on the sheet pile should be found from the flow nct and A properly accounted (or in the analysis. T 8: Approximate Analysis. The exact analysis of the cantilever? 2 sheet pile as discussed above> is quite involved. An approximate --.L A value of d can be obtained using a simplified pressure diagram as I KpJ'd KaJ'(h+d) shown in Fig. 20.20. In this analysis, the resistance of the pile below the point 0 is replaced by a concentrated force P3. (Note that the pressure distribution extends upto tip A). Fig. 20.20. Approllimate Analysis. From the equilibrium in horizontal direction.

T 1

I

---1

PI - P2 + P 3 .. 0

Taking moments about point A,

P 1 (h ; d) - P2 x

~

.. 0

t

Kp

Substituting the values of PI and P'b

iK. y(h + d)'

~-

yd'

x

~.

0

(Kp - KJd'- 3h K.d' - 3h'K.d - K.h'. 0 rq. 20.31 is is solved by trial rmd error for d TIle value of d so obtained is usually increased by 20 to 40%. Thus

... (2031

D., 1.2d to 1.4d. 20.12. CANTILEVER SHEET ]'ILE PENETRATING CLAY Fig. 20.21 shows a cantilever sheet pile penetrating clay ($ = 0) below the dredge level. The backfill of cohesionless soil (c = 0). Lei tbe bulk unit weight of the backfill material and clay be, respectively, YI aJ y. The rohesion intercept of clay is c. The pressure PI at the dredge line on the back side is given by PI" y1hK", Below the dredge level but above the point of rotation 0, the passive pressure acts from left 10 right and II active ~ure acts from right to kft. Therefore, tbe prc$ure at depth Z below tbe dredge level is given by

P2 ., Pp - p" 1'2 - (KpYZ +2c«,) - [K.y(Z + h)- 2cVK:;]

~ING

DESIGN OF RETAlNING WALLS AND BULKHEADS

20.29) hdis

!D.30)

C()-1ESJONLE$ SOIll cpO

4\

r'1

Jh

'?i~~j----'L----'C L

-"J Fig. 20.21. Cantilever Sheet Pile in Clay.

For, - 0, we have

Kp_K.. _l.O.

Therefore, P2- 4c -yh Ukewise, the pressure Pl from righl 10 leO is given by P3 - Kp(h + d)y +2cYK; - (K.. yd - 2cv'K';;]

or

For, = 0, we have P3 - 4c+ yh From equilibrium in the horizontal dircaion, considering equivalent areas as in Fig. 20.19. PI - [P2 )(

10.31)

:fill is

II and

"'!he

.,

dJ

+

iP2

+ Pl] )( ml2 - 0

P I- (4c-yh)d+ (Bc])( ml2 - 0 (4c - yh)d - PI

... (2032)

4c

Thking moments of all the forces about A, PI

(2\

+ d) - [4c - yh] )( d )(

~

+

~

)( (Sc) )( m )(

~

_ 0

. . . (20.23)

Substituting the value of m from Eq. 20.32,

-

,p

PI (ll + d) - (4c-yh)-z +

34 c [(4C-

Y h)d - P,]' 4c - 0 ... (20.34)

The above equation can be written as

Jl

(4c-yh)-2P J d-

PI (12cZI + PI) 2c+yh .0

... (2035)

Eq. 20.35 can be solved for d. The actual depth D is kept 40% to 60% more. Thus D _ l.4d to 1.6d

cn

Alternatively, the depth d caD be computed using a reduced value of or 2c/3 in Eq. 20.35. In tbis ca!it, the actual depth D would be equal to the computed value of d, as the ractor of safety has already been applied to c.

roIL MECHANICS AND FOUNDATION ENGINEERING

If the water table exists on both the sides, modification can be done as in the case of cohesioolcss deposits. The submerged weights are used below the water table (see Illustrative Example 20.8). 20.13. ANCHORED SHEET PILE wnn FREE·EARTH SUPPORT The stability of anchored sheet pile depends- upon the anchor force in addition to that upon the passive earth pressure. The embedment depth is considerably smaller than that in a cantilever sheet pile. 1berefore, by this metbod, the lotal length of the sheet pile is reduced. Of course, the additional (;:(lSt of anchors is also to be considered while judging the eronomy of the two types of construction.

Fig. 20.22. Free Eanh Support.

Fig. 20.22 (0) shows an anchored sheet pile with free earth support. The deflected .shape is also shown. As already mentioned, there is no point of contraflexure below the dredge level. Thus, below the dredge level, no pivot point exists for the statical system. The statical an.1lysis is based on the assumption that tbe soil into which the pile is driven does not produce effective reslraint to induce negative bending moment at its SUppo!l The equations for the depth d are derived separately for the oohesioniess and cohesive soils. (2) Coheslooless SoDs. Fig. 20.22 (a) shows the forces acting on the pile, assuming that the material above and below the dredge level is cohesionless. ••. (1D.36) From equilibrium,

where T is the tensile force in anchor. The depth a to the point of zero pressure can be delennined as under.

yK,,(h + a) - yKpa .. 0 "1 (K, - K.) - 1K.h

a ..

~

... (20j~

(K, - K.)

~ Plb

~

1berefore,

P2"

where

P2" y(K, - K,,)b

..

y(K, - Kfl)b

2

...(~

Taking moments of all the forces about anchor point M,

Pda + h - e -

Zd -

P2 (h - e + a + 2b13) .. 0

... [20.38(a)J

The distance Zl is determined as in the case of cantilever piles. Substiluting the value of P2 from Eq. (a) in Eq. 20.38 (a),

PI

("

+ h-e -ZI) -1(K" - K.)b x bl2 (h - e +" +2b/3) - 0

... (20.38(b)(

DESIGN OF RETAINING WAllS liNG

mess ssive Ifore, also

~D

BULKHEADS

533

The above equation can be written as

.,

b'(K" - K,,)y/3+ b2 (K" - K,,)y/2(g + a) - Pd - 0 l

3Pd

2

where f - a + h - e - II and Eq. 20.40 can be solved for b. 1ben d is detennined as d _ b + a

...(20.39) -

... (20040)

b + 15b (g + a) - y(Kp _ K,J - 0

g _h - e

The actual depth D is taken equal to 1.2 to 1.4 times d. The force in andlor rod can be obtained from Eq. 20..36 as, T - PI - P2

The values of PI and P2 are obtained from pressure diagrams. (b) Cohesive Soils Let US now consider the case when the anchored sheet pile is driven in clay <+ :: 0), but has the backfill of cohesionless, granular material (Fig. 20.23). The pressure dislribution above the dredge line is the same as lhal in the case of cohesionless soils. However. below the dredge tine, the pressure is given by

own. eve!. into

+1

!'P""

"<1

<:=0

:edge 0.36) Fig. 20.23. Anchcred Sheet Pile driven In Clay.

P2 - (KpYZ + 2c

K" -

..nr;) - [K.(Z + h)y -

2c ~)

Ko - \.0. Therefore

P2 - 2c + 2c - yh - 4c - yh

0.37) .•.(aJ

8(a»

.8(b»

From equilibrium of forces,

or PI - 1'2 x d - T Thking moments of all forces about M. PI Xf-P2d(g + d/2) _ 0 Substituting P2 - 4c - yh. PI'! - (4c - yh)d(g +dl2) _ 0

d'

2PI!

+ 2gd - 4c _ yH

.. .{20.41)

•.. (20.42)


534

Eq. 20.42 can be solved for d. The actual depth (D) provided is 20 to 40% more than d. It may be noted that the wall beromes unstable when P2 - 0, i,e., 4c - YH .. 0

~

-*-

0.25

... (20.43)

The left band side is equal to the stability number (Sft) defined in chapter 18. In other words., the walls becomes unstable when SIr is equal to or less then 0.25. IT the adhesion of clay with the sheet pile (ca) is considered, Eq. 20.43 is modified as Sn"

Taking,

Vl

-:{H

+

S;

... (20.44)

V I + calc - 1.25,

S" - 0.25 )( 1.25 .. 0.31 'illerefore, the minimum stability number (S,,) required is 0.31. If the factor of safety required is F, the stability number (SII) should be equal to 0.31 F or more. 20.14. ROWE'S MOMENT REDUcnON CURVES As sheet piles are relatively flexib le, these deflect considerably. Their flexibility causes a redistribution of lateral earth pressure. The net effed is that the maximum bending moment is considerably reduced below tbe value obtained for the free-carth supports discussed in the preceding sCction. Rowe (1952) developed a theoretical relation between the maximum bending moment and the flexibility of the sheet pile and gave moment reduction curves. The relative flexibility (p) is defi~ as p _ (h

~/D)4

_ 1.1 )( 10--6

~

... (20.45)

height (m), D = actual driving depth (m), E = Young's modulus of the pile material (MN/m1 and I = moment of inertia of the pile (m4/m~ H = total length of the pile. For anchortxl sheet piles in cohesionless soils, the relative density is important. The relative depth of anchor factor, f} = e/H is also relevant. For anchored sheet piles in cohesive soils, the stability number (S~), as given belOW, is also required. where h

= retained

So - 1.25 cI(y h) The relative height of piling factor a

... (20.46)

= h/H is also important for cohesive soils.

,., ...-----..---,--------,----,

,.,f----+---+------+----I O.~,L.,=---:l:----_".f.,=---:!;------!. l09

I

Fig. 20.24.

INO

DesiGN OF REI'AINING WAllS AND BULKHEADS

.43)

Fig. 20.24 shows a Iypical moment reduction curves Cor rohesionless soils. The ratio M/MOWI. is dclermined directly for the given value of p. 1be curve (a) is for loose sand (relative density = 0) and the ClIVe (b) for dense sand (relative dCrL'iity = 100%). The value of Mmu being known from the free-earth sup(X)lt analysis, the design moment Mil can be computed.

'al.Is

•) ;,;

5"

(For more details, the original paper may be consulted) .

IUS. ANCHORED SUEET PILE wml FlXED-EARm SUPPORT Fig. 20.25 (a) shows the deflected shape of an anchored sheet pile with fixed-earth support. The elastic : line manges its OJrvature at the inflexion point I. The soil inla whidl the sheet is driven exerts a large

_44)

.

..

(he

lof .he

ility A (0)

.45)

Fig. 20.25.

Iof

-46)

restraint on the lower pan of the pile and causes a change in curvature. Fig. 20.25 (b) shows the pressure distribution, Blum (1931) gave a mathematical relationship between (i/h) and, (Fig. 20.26), where j is the dcplb of the point of inflexion I below the dredge level and h is lhe beight of sheet pile above the dredge leveL Thus inflection point I is located. Foe simplicity, the lower portion of the pressure diagram on the right hand side in Fig. 20.25 (b) is repJaced by a concentrated force Rk at point K and the diagram shown in Fig. 20.27 (0) is used in the analysis. The magnitude of Rk is initially unknown, but it is 0.3 automatically excluded from calOJwtions when the moments art taken about K. Once the depth has been found, Rt can be dclennincd from the equilibrium equation in the horizontal direction. 02 . As the exact analysis of the anchored sheet pile with 0 ·( rlXed-earth support is complicated, an approximate method, known as U/uivatem-beam method is generally used. It is aumcd !hat the sheet pile is a beam which is simply 0·0 2040supported at the anchor point M and fixed at the lower end 253035K. Fig. (b) shows tbe bending moment diagram. The bending moment is zero al the inflexion point I. Fig. 20.26. 1bettetically, the lower pan lK of lhe pile can be removed BOO the shear force can be replaced by a reaction RI . Thus, a simply-supported beam .B1 is obtained [Fig.

~

w.n

m

20.27 (c)]. The following procedure is used for the analysis of the sheet· pile with eIJlivalent beam method.

fixcd~carth support, using


""

t !

K 8 .M . Oio2 ro m

,.,

",

~ t

T ,III

'_R, I - RI

S f!Om(2)

K

-RI(

",

Fig. 20.27. S implified Pressure Diagram.

(a) Upper Beum BI

(1) (2) (3) (4)

Dctcnninc the pressure PI at the dredge level. Estimate the angle of she.1ring resistance ijI' of the soil. Determine the distance i of the point of inflexion from Fig. 20.26. Determine the distance a of the point of zero pressure from the equation, a ___ P_Iy(Kp - KJ (5) Determine the pressure po at the point of inflexion from the relation,

Po -

~ (a

- i)

... (20.4~

... (20.4!

(6) Determine the reaction RI for the beam IB by laking moments about the point M of anchor of all III forces acting on IB [Fig. 20.28 (a»). (b) Lower Beam IK (7) Determine the pressure P2 (rom the relation P2 - y(Kp Alternatively,

-

K.)(d - a)

... (20.4~

pz - (D p~ i) x (d - a)

(8) Determine the distance (d - a) by taking moments of the fotUS on the beam IK about K [Fig. 20.28 (blj The reaction RI on the lower beam is equal and opposite to thot on the upper beam. (9) Calculate d from Eq. 20.49 and hence find D _ 1.2 d.

(10) Determine the tension T in anchor by considering the equilibrium of beam Ill. Thus T - PI - R, whcre PI = total force due to pres.<;ure on lB. 20.16. DESIGN OF ANCnORS 'l'hc anchors used in sheet pilc walls arc of the following types: (I) Anchor P!.1tCS and Beams (also, known as deadman) (Fig. 20.29). (2) 'Tie backs. (3) Vertical Anchor piles. (4) Anchor beams supported by baUer piles (Fig. 20.30).

... (20.511

E.ERINO

S37

DFSIGN OF RETAINING WAllS AND BULKHEADS

1

trs B

T

p,

(oj Top

beam

TX~I'O d-o

1

"

(b)BOClom

K

",

""

brOm

Fig. 20.28.

20.47)

20.48)

The design of anchor plates and beams is diSCUSSC
illlhe

ANCHOR BEAM

10.49) TENSION PILE

8 (b)] Fig. 20.29. Anchor Plal(s.

1).50)

Fig. 20.30. Batter Piles.

The resistance offered by an anchor plate or a beam is derived from the passive resistance of the soil in (root of the plate. For full JXlSSive resistance to develop, tbe anchor plate must be located in zone CDE. Teng (1962) gave the (ollowing equations for the ultimate resistance of anchor plates in granular soils located at or

near the ground surface. Let D be the length of the anchor perpendicular to the aoss section and lei h be the height of the anchor. (Q) For continuous plales or beams with Blk 'Z: 5, the ultimate resistance is given by

p • • IJ(P,-P.)

Po. - B( iyH2K

p -

tYff

K,,)

538

SOIL MECHANICS AND FOUNDATION ENOINEERING

or

P_I!!.P..(K_K' II 2 p tl/

... (20.51)

where 1/ is the depth of the lower face of the anchor beam from the ground surface.

(b) For plates or beams with Blh < 5, the ultimate resistance is given by P,. - B(Pp where Ko

Thus

= coefficient of earth at p. -

-

PJ

+

iKoY(v'i(,;

+

VK;)Jilum+

rest ( :: 0.40).

I.!f!!. (K,- K.l

+

tKoY(v'K,

+ >!K;)fflJ1n.

••• (20.52)

The allow-able resistance is taken as Po -

~

... (20.53)

where FS = factor of safety (generally taken equal to 2.0),

The centrc-to-CClltre spacing of anchors is obtained from the relation, s _ PalT ... (20.54) where T '" tension in sheet pile per unit length as obtained from the analYJja of anchored sheet pile. ll..LUSTRATIVE EXAMPLFS

Fig. E-20.I.

where

KII -

Sin'(jl + .')

.'~ . (~- 6) [I sm sm

+

V'in(.'

+ 6)sin(i' -I) sin(p-6)sin(~+')

2

1

539

DESIGN OF RETAINING WAllS AND BULKHEADS

or

K"

-

sin2(70 + 36)

. '70" . (70" _ 24") [1 Sin

Sin

[ 0.883 x 0.719

1 +

,

V

0.9~ 0.866

or K" -

+

sm (36" + 24") sin (36" - 0) sin (70" _ 24°) sin (70° + 0)

1

x 0.588]' - 0.417

0.719 x 0.940

Po - ~ x 0.417 x 19 x (5.7)2 _ 128.7 kN

Therefore.

The total pressure acts inclined at 24° to the nonnal. Horizontal component,

Vertical component,

P" _ Po cos (20° + 24°) - 92.6 kN _ Po sin (20" + 24") _ 89.4 kN

p~

Calculations are shown in the tabular form. The moment are taken about toe. The clockwise moments are taken as positive. The unit weight of concrete is taken a
Forces (leN)

S.No.

..

Lever

Descriptioll

Vertical (kN)

Weight, Wt .. 1/2 )( 5 )( 0.19 )( 24 Weight, W2 .. 5x05)(24 Weight, W3 .. 112 )( 5 )( 1.71 )( 24 Weight, W" .. 3.2 )( 0.7 )( 24 Vertical componem of p .. Pay Ha-ironlal compoooni of P -PuA

11.4

053 m

60

0.84 m

50.4

102.6

1.66 m

170.3

HomonJal (kN)

Clockwise

53~

1.60 m

86.1

89.4

2.39 m

213.6

92.' 1: 317.2

175.9

1.90 m

I 92.6

Counter Clockwise

6.0

I 526.4

I 175.9 IM _ 350StN-m

Neglecting passive resistance, the factor of safety against sliding is given by Eq. 20.3 as

F~

_

~"

_ tan 24;2~6 317.2 _ 1.53 (safe)

The factor of safety against overturning is obtained from Eq. 20.4,

LMR

526.4

F" - ~M" - 175.9 - 2.99 (safe) From Sq. 20.17,

x - i~ - ~;~:;

-

1.1Om

e - b/2-x _ 1.60 - 1.10 - 0.50m From Eq. 20.18, As e < bl6, there is no tension. The pressure at the base are detcnnincd from Eq. 20.19.

540


p~

Pmin

_ ~ (1 b

~) _ 317.2 (1

+ 6

3.20

+

6 x 05) _ 192 kNl 3.2

2

m

_ ~([ _~) _ 317.2 (1 _ 6 x 05) -62kNl 2 b

6

3.20

3.2

.

m

The factor of safety against bearing capacity failure is given by Eq. 20.7 as

Fb -

p~

- :

- 3.1 (safe)

H1mtratlve Example 20.2. Check the stability of the cantilever retaining waU shown in Fig. £-20.2. TIu! 0 allowable soil pTt!Ssure is 500 kNlm2, = 34", 6 = 25 , '1 = 18 !tN1"?, i = 15°,

+'

I

f 'm

6.21m

l~

1 Fig. £..20.2.

Solution. Let us first asccctain whether Rankine's theory is applicable to the cantilever retaining wan.

From Eq. 20.9. " _ (45 0 + il2) _ CO

r.2 _sin-1 ( smcp' ~ioi )

_ (45 + 7.5) - 17 - sin-

1

(:~:) = 7.9

0

The shear ZOI!e docs oot intersect the stem. 'Therefore nie Rankine theory can be applied.

From Eq. 19.23. From Eq. 19.15,

p. -

iK.yH'

DESIGN OF RETAINING WAllS AND BULKHEADS

Pa ..

1berefore,

~

0.311

I(

541

18

I(

I(

(6.22)2 .. 108.3 kN

p .... P.. sin 15° .. 28.0kN Ph .. p .. COO 1"5° .. 104.6 kN F
S.No.

Description

1.

6.

WI - 0.41( 5.01( 24 W2 - 0.2/2 I( 5.0 I( 24 W3 _ 0.6 I( 3.50 x 24 W" _ 23x 5.0x 18 W, _ 230 x 0.62/2 I( 18 p,

7.

p,

2. 3. 4.

s.

Vt:rticaf (ON)

Moments about loe Leo,.,-

Horizon/oJ (ON)

48 12.0 50.4 207.0 12.8 28

Clockwise

1.00 0.73 1.75 235 273 3.SO

48 8.8 88.2 4865 35 98

207

104.6 I 358.2

(lcN-m)

0"" (m)

CO""", CIockwis<

216.5

I 104.6

1:764.5

I 2165

From Eq. 20.3, the factor of safct)' against sliding, .. ~Rv .. lann° I( 358.2 .. 160(--<) F' R 104.6 .:oo.Ie I1

From Eq. 20.4, the factor of safety against overturning, .IMR 764.5 F. - Hfo - 2165 - 353 (safe)

11.

i ~ . 764·;5~.i16.5

From Eq. 20.17,

i ..

From Eq. 20.18,

e .. bl2 - i .. 1.75 - 1.53 .. 0.22 m < bl6

From Eq. 20.19,

p

max

.. 1.53 m (No lension)

.. 358.2 (1 + 6 )( 0.22) .. 1419 kN/m2

3.50

350

358.2 (

Pm;n" 350

1-

6 x 0.22) 3.so..

.

63.7 kN/m

,

From Eq. 20.7, the factor of safety against bearing capacity failure,

F" ..

p~

..

1!~9

.. 3.52 (safe)

Dlustratln Example 20.3. Determine the required depth of penetration for the cantilever sheet pile shown in Fig. £-20.3. Take 'f = 16 kN/mJ. SoluUon. (Refer to Fig. 20.18 for notatioos) K. - lan' (45 - 30/2) _ 0.333, Kp .. lan 2 (45 + 3(/2) .. 3.000

Pl .. 0.333 )( 16 )( 5.0 .. 26,61cN/m2 a .. y(K

PI

p

_

26.6 K.,) .. ~- O.62m

542


PI ..

tx

26.6 )( 5 +

tx

26.60 x 0.62

" 665 + 8.2 " 74.7 kN Taking moments about 0 1 and dividing by PI'

Zl .. 66.5 )( 2'~4~7 8.2 x 0.41 .. 2.08 m l'l"y(Kp-K.J b .. 16)( 2.667 )( b .. 42.7b Pl" y{h + tf)Kp - ydK• .. 16 (5 + b + 0.62) x 3.0 - 16 (b + 0.62) )( 0.333 Pl .. 266.5 + 42.7 b 111 ..

Prom Eq. 20.26,

P2b - 2Pl P2 + PJ

42.7b 2 - 2 )( 74.7 266.5 + 85.4 b 2 P2b
From Eq. 20.27,

E ;/;;/'

~~------.

~~--t-~-1 Ag. B.20.3.

3

149.4)' " 0

74.7 (b + 2.08) _ 42.7 b + (266.5 + 85.4 b) (42.7 b' 6 6 266.5 + 85.4b 3

448.2 (b + 2.08) _ 42.7b + Solving by trwl and error,

(4;:':: ~;~:)2 . 0

b .. 4.4 m.

Alternative Method tor b

Using Eq. 20.28, where

b4 + C1b1 _ Cp2 _ C.p - C... 0 C," _ _p_'_ y (Kp - K Il)

.. 16

x 5 x 3.0 + 16 x 0.62 x 2.667 16 x 2.667

" ~6~ "6.24 BPI

C, .. y(K

p

_

8 x 74.7 K..} .. 16)( 2.667 .. 14.0

C] .. 6Ft [2"1(Kp - K .. )Zt + P.)

[y(Kp - K.ll' .. 6 x 74.7 (2 x 16 x 2.667 x 2.08 + 265.6] .. 109.07

[16 x 2.667J' and

c....

Pd6Z1P4 + 4Pd .. [y (K;, - K.Ji'

74.7 [6 x 2.08 )( 265.6 + 4 x 74.71

(16 x 2.667)'

- 148.24

Therefore. Solving by trial and error,

4

J

b + 6.24b b - 4.40 m

-

l

14.Ob

-

109.07b _ 148.24 _ 0

S43

DESIGN OF RETAINING WALLS AND BULKHEADS

d - b ... a,. 4.40 ... 0.62 -5.01m

Therefore,

D,. l.30d,. 6.s0m D1uslratlve Example 20.4. Determine the depth of penetraticn for the cantilever sheet pile wall of JIIustra/i'tl( Exomple 20.3 by the approximate method. Solution. From Eq. 20.31, taking h ::: 5 m,

(Kp - K;'Jd' - 3hK.d' - 3h'K.d - K.h' - 0 2.667d' - 5d' - 25d - 41.67 - 0 d,. 4.70m. Therefore, D - t.3d,. 6.10m Illustrative Example 20.5. Determine the depth of penetration of the cantilever sheet pile slwwn in Fig. E·20.5. The water level on both sides is the same. or Solving by trial and error,

t

Mm

r d

L.'~~:~___

---'------=-

t---- P2 - . . . . ; - - ~

---I

Fig. &20.5.

Solution.

K .. ,. 0.333 I Kp" 3.0. PI ,. 0.333 )( 16 )( 2.50 + 0.33 (9.0) )( 2.50 PI ,. 13.3 + 7.5 ,. 20.8 kN/ml

From Eq. 20.24.

"= P, =

_. _ _p,__ = 9.0 x (3.020.8- 0.333) = 0.87 y (K,. - K.. )

t

x 13.3 x

i.5 +

m.

13.3 x 2.50 + 7.5 x 2.5/2 + 20.8 x 0.8712

PI ,. 16.6 + 33.3 + 9.4 + 9.0 ,. 68.3 leN

Taking moments about 0Il

or

~~_~.~+~.=+u.~+u.~

II ,.

69.7 + 70~~ 16 ... 5.2 ,. 2.36 m

P2" y'(Kp

-

KJb ,. 9.0)( 2.667)( b,. 24.Ob

P3" [16 x 2.50 + 9.0)( 2.50] x Kp + y'd(K,. - Ka) .. (40 .. 22.5) )( 3.0 + 9.0 )( (b + 0.87) )( 2.667 P3 .. 187.5 + 24b + 21.0 .. 208.5 + 24b From Eq. 20.26,

m ..

P'P -

2 PI .. 24b

P2'tPJ .0.:

2

-

2

)C

68.3

208.5+48b


544

-

From Eq. 20.27,

Pdb + ZI) -

P2 (pz + pi! [2Ab' - 136.6]' 6 b' + -6(Pz + PJ)

24.00] (24h 1 _ 136.6)1 68.3(b + 2.36) - -6-+ 6(200.5 + 48.Ob) - 0 409.8 (b + 2.36) _ 24.Ob3 +

Solving by trial and error,

(:~-}~~}2 ..

0

b .. 5.50 m.

d - b + a .. 5.50 + 0.87 .. 6.37m D .. told -830m AltematIve method + C1 b1 -

b4

From Eq. 20.28. where

C1

wbere

..

C;,z -C,p - C" .. a

""(K:~K{I)

p" .. (16 x 2.5 + 9.0 x 2.5) Kp + 9.0 a(Kp - K.J .. (40 + 22.5)

Therefore,

CI

-

C2

-

C,-

9.0

)C

3.0 + 9

)C

0.87

)C

2.667 _ 208.4

- 8.68

8PI 8 )C 68.3 y ' {Kp _ KJ - 9.0 x 2.667 - 22.76 6PI [2y'{Kp - KJZl + P.] [y ' (K,-K.)]'

c _6

and

U:i~667

x 683 [2

)C

3

9.0 x 2.667 x 2.36 + 208.4] _ 228.81 (9.0 )C 2.667)2

PI [6 ZIP" + 4Pd and

C, -

r

[y'(K, _ K.»)'

.. 68.3 [6 x 2.36 x 208.4 + 4 )( 68.31 .. 382.21 [9.0 x 2.667)' 'Therefore, b" + 8.68bl - 22.76 b2 - 228.81 b - 382.21 .. 0 Solving by trial and error, b .. 5.50 m.

'm

1

do. b + a .. 5.50 + 0.87 .. 650m. D .. 1.3d .. 830m. Determine the depth embedment for the cantilever sheet pile shown in Fig. £-20.6. Solution. (See Fig. 20.21 for notations) In Ibis case, K" .. 0.333, Kp .. 3.0

Dlustrutive

Example

20.6.

PI .. 16.0 )( 5.0 x 0.333 .. 26.7

PI ..

~

)( 26.7 x 5 .. 66.81cN

z... 5/ 3 .. 1.67m

Fig. B-20.6.

'"

DESIGN OF RIITA1NING WALLS AND BULKHEADS

P2 - 4c -lh ., 4 )( 50 - 19 )( 5 ., 105.0 PJ - 4c + lh - 4 )( 50 + 19 )( 5 - 295

(4c -lh)d - PI (4)( 50 - 19)( 5)d- 66.8 4c 4)(50

From Eq. 20.32,.

- 0.53 d - 0.334 From Eq. 20.34,

-

PdZI +. d) - (4c - 'fh)

d' 2"

+

4

1 [(4< - yh) d c

'3 )( 16

66.8(1.67 + d) _ (200 -19 x 5)f + 11156 + 66.8d - 52Stf +

or

Simplifying and solving,

-h

Pd'

[(4 x 50 -

- 0

195~

5) d - 66.8J' _ 0

~ (105d - 66.8)2 _ 0

d-2 .60m

Alternative method

or

(4)( 50 _ 19)( 5)tf _ 2)( 66.8d -

or or

d' -

66.8(g::~: /~6: ;)66.8) _ 0

1.27d - 3.49 _ O. d _ 2.60m D _ 1.50 d _ 3.90m.

Illustrative Example 20.7. De/ermine llle deplh of embedment of the anchored sheet pile shown in Fig. £-20.7. Also determine the force in the anchor per melre of the wall. Assume free-earth support conditions. Solution. (See Fig. 20.22 for notations) K" - 0.27, Kp _ 3.69 PI - 16 )( 3.0 )( 0.27 + 9 )( 5.0 )( 0.27

or

PI - 12.96 + 12.15 - 25.11 r ~ 16 kN l rJ

• • 3':

11':9 kNlm3 1/ ~ 3,;°

i 1 _ _ _11---'-::7' T

Fig. &20.7.

..

,


P2 - 9(3.69 - 0.27)b - 30.78 b

Pz .. 30.78b x biZ .. IS.39b z PI .. 12.96 x 312 + 12.96 x 5 + 12.15 x SI2 + 25.11 x al2 in wbich the distance a is dctennincd from the equation. PI - y' (Kp - X!') a .. 0 9 x

(3~~;} ~ 0.27)

.. 0.82 m

Thus PI .. 19.44 + 64.8 + 30.38 + 25.11 x 0.82/2 .. 124.92 Taking moments about 0,

~~_~.~+~.=+3O~.~+=.=.~ or

ZI .. 132.58 ... 21\12~.;; 75.65 ... 5.63 .. 3.43 m

Taking moments of all the forces about M. using Sq. 20.38 (a), 124.92 (a + h - e - ZI) - P z ( a + h - e +

+

~)

.. 0

+

+

124.92 (0.82 8.0 - 2.5 - 3.43) - 15.39 b' (0.82 8.0 - 2.5 2b/3) - 0 361.02 - 97.26b2 - 1O.26bl .. 0 0' By trial and error, b - 1.77 m d .. b + a .. 1.77 + 0.82 .. 2.59 m Now D .. 1.3 x 2.59 - 3.40m. Illustrutlve Exumple 20.8. Determine the depth of embedment for the anchored sheet pile shown in Fig. E-20:8. Also determine the force in the QIlchor per metre run. Assume f/Xl!.d-end support conditions. Solution. (Sec Figs. 20.28 and ]1).27 for notations) KQ . . 0.27, Kp .. 3.69 PI .. 16 )( 8.0 )( 0.27 .. 34.6 leN From Eq. 20.47, From Fig. 20.26, for Therefore,

PI

*' .

34.6

a .. y (Kp _ Ko) .. 16 )( 3.42 .. 0.63 m 35°, ilh .. 0.025 i .. 0.025 )( B.O .. 0.2 m

T d

jL----~ E·20.8.

DESIGN OF RETI\JNING WALLS ANI) BULKHEADS

e; i

Po -

547

(a - I) -

~~

x 0.43 - 23.6

Talcing moments about M of all the forces acting 00 beam lB.

R, (5.5 + 0.2) -

x 34.6 x 8 x (5.33 - 2.50) + (23.6 x 0.2) x 5.60

+ (34.6 - 23.6) x 0.212 x (5.50 + 0.07) R, .. 74.4 kN 1'2 • y (K" - KJ (d - a) • 16 x 3.42 x (d - 0.65) 1'2 • 54.7 (d - 0.63)

Now

'" [Alternatively.

P2 ..

:~i

x (d - a) ..

~~4~ (d -

0.63)

or

1'2 • 54.7 (d - 0.63») Taking moments about K of all the forces acting on beam IK, R,(d _ a) .. 23.6; 0.43 (d _ a .. 0.29) _

54.7(~ -

a)2 x (d; a) _

a

74.4 (d - a) .. 5.1 (d - a .. 0.29) - 9.1 (d - ai .. 0 By trial and error,

d - a .. 2.96m or d - 2.96 + 0.63 .. 3.59m D .. 1.2 x 3.59 .., 4.31 m

From Eq. 20.50, T _ PI - R J

-

~ x

34.6 x 8 .. 23.6 x 0.2 + II x 0.1 _ 74.4

T.69.8kN PROBLEMS A. Numerical

20.1. ~ ~/:tall

stabil ity

of cantilever 100

retaining

wall sOOwo

-I ()O3~ ( 5 ) 1

fig. P-20.l.

in Fig.

~A!~:lF;:k~.~e;a::r:~~~l~;ItP~~~~i

Surcharge 50 II N/m2


548

20.2.

~~~:~;':O~~~~:~~ (~~t:CO~~ca~ ~!~~~~e~~:n:' i:i~~v~~~~pt! ~ ~~:dt~d ~~~g~~~I~ ~~ ~~~

and 4' = 30", Use approximate method. [Aos. d :: 7.35 !II) 20.3. Determine the depth of embedment Cor the cantilever pile in clay, shown in Fig. P 20.3. The water !.able is all height of 2.5 m above the dredge I~e\ on both sides. LADS. d :: 2.20 ro, D :: 3.30 mJ

J':19kNlm 3 +:30"

J'~9kNlm3 (:o;QkNlm 2 ~': 0.0

T

,.

1

,.

1 Fig. P-20.3.

Fig. P·ZO.4.

20.4. An anchored sheet-pile wall is shown in Fig. P 20.4. Find the fraction of the Ihoorelical maximum pressure IJI. the embedded length which must be mobilised for equilibrium. Usc the free· canh suppo" method. Also delennine the force in one of the anchors, assuming that they are spaced al 2.5 centres. [Aos. 0.51; 206.5 kNl 20..s. An excavation 8 m deep is to be made in c:ohesionless soil (y = 19 kN/m3, , '" 30"). The sides of the excavatitl are supported by anchored sheet piles with fixed-end support. Dctennine the minimum depth of embedment fa equilibrium. The anchors are at a depth of 1.5 m below the surface. [Ans.. 5.0 r:aJ 20.6. For the anchored sheet pile shown in Fig. P-20.6, detennine the embedment depth d and the (orce in the ancha per m run. Assume fixed-end conditions. [Ans. 5.00 , 71 kNtrai 20.7. For tbe anchored sheet pile shown in Fig. P 1JJ.7, determine the theoretical and actual depth of penetration,!hI anchor force per unit length of the wall and maximum moment. Assume frce-earth support. [Ans.. 2.40 mj 3.50 m; 39.S kN, 63.0 kN-~

B. Descriptive and Objective

'iYpe

20.8· What are differenl types of retaining walls? Discuss the methods for estimation of lateral earth pressure acti" on the walls. 20.9. Discuss the principles of the design of retaining walls. 20.10. What are different modes of failure of retaining walls? Explain with the help of sketches. 10_11. What are different types of sheet pile walls? Draw the skctche;s showing the pressure distribution. 20.12. DiSQJSS the procedure for chocking the stability of a cantilever sheet pile wall. ZO.13. How would you check the stability of an anchored sheet pile wall with free-earlh support? What is Rowe\ correction? 20.14. Describe the equivalent beam method for the analysis of an anchored sheet pile. 20.15. Discuss various methods for providing anchors for a sheet pile wall.

:RING

DESIGN OF RETAININO WALLS AND BULKHEADS

549

~~:j 35 m] sata

, m]

1 Fig P·20.6.

kN] Btion for Om]

it

"'""

'1m] , the

"-m) ~ing

we',

Fig. P-20.7.

20.16. Write whether the fol1owing statements are true or folse. (a) Gravity walls are more economj~1 lhan cantilever walls. (b) The Jatenll pressure on gravity wal1s is determined using Coulomb's theory. (e) Rankine's theory can always be used for the determination of laternl pressure on cantilever walls. (d) Counterfort retaining walls are more suitable than cantilever relaining walls for greater heights. (e,) Tension develops at the base of the retaining Willi when the resultant strikes outside the middle-third. (/) Backfill should be preferably of cohesive material. (g) In an anchored sheet pile with frce-earth support, there is no point of inflexion. (h) AI. the point of contranexure, the shenr force is zero. (I) The ultimDlC resiSlance of an anchor depends upon the passive resistance. (;) Anchors should be placed as neat the wall as possible. [An. True. (b). (
C. Muldple Choice Questions 1. The minimum allowable fBctor of safety against sliding in the case of a canlilever retaining wall is (a) 2.0 (b) 3.0 (0) 1.50 (d) 2.50 2. In the case of a counterfort retaining wall, the toe slab acts as a (a) cantilever (b) continuous slab (e) simply supported slab (d) DOlle of above. 3. In the case of fixed-earth support of a bulkhead, the stability is provided by (a) the passive resisumce of the soil (b) the force in the anchor (e) both (a) and (b) (Ii) neither (a) na (b) 4. Rowe's correction for free-enrth Support of a bulkhead depends mainly on (a) 10lial height of bulk-head (b) Young's modulus of pile material (e) moment of inertia of the pile (d) all the above. [A= 1. (0). 2. ('13. (0). 4. (d)]

21 Braced Cuts and Coffer Dams 21.1. INTRODUCUON (a) Brnced Cuts

Dccp excavations with vertical sides require lateral supports to prevent cave-in of the earth and to prolcd the adjacent areas against ground subsidence and lateral movement of the subsoil. When excavations 11ft shallow and ample space is available, the sides of the excavation can be sloped at a safe anglc to eosutt

stability. However, in deep excavations, especially in built-up areas, there may not be adequate space r(t providing safe slopeS .. Moreover, it bcoomes uneoonomical to provide safe slopes because of large quanlitio

(b) DEFORMUKN (e) PRESSURE C()M)ITION (Rotation about top)

ENVE LO PE

Fig. 21.1. Brnccd Cut.

are

of the earth involved. Excavation<; which laterally supported are called braced cuts [Fig. 21.1. (a)]. 'Ill: vertical sides of excavations are supported by a sheeting and bracing system. It consists of a relati..,qflexible sheeting placed against the excavation walls. The lateral thrust on the sheetings is resisted ~ horizontal members in compression (struts), known 115 bracing system. Bracing is provided as tbe excavati(l proceeds and the face of the sheeting becomes exposed. various types of braced sheeting. the forces actiq on them and the design of various components are discussed in this chapter. (b) Coffer Dams 1bc word 'coffer' means a casket, chest or trunk. A coffer dam is a temporary structure buill to enclm an area for excavation of foundation. Coffer dams are used when the size of excavations is very large art sheeting and bracing system becomes difficult, inconvenient or uneconomical. Coffer dams are generaI:! required for foundatiollS of structures, such as bridge piers. docks, locks, and dams, which are buill in opa water. These are also used for laying foundations on open land where there is a high ground water lab1c.A coffer dam generally consists of a reLatively impc(Viou<; wall built around the periphery of the proposa excavation to prevent the flow of water into the excavation so that the foundation may be laid in 0, condition (Fig. 21.2). Various types of coffer dams and their design aspects are considered in this mapler.

BRACED CUTS AND COFFER DAMS

~51

WATER

WATER

WORKING AREA ENCLOSED BY COFFER 0M1

.. V

COFFER

DAM

IlS PLAN Fig. 21.2. Coffer Dam.

)tcct

sure [0'

:ilies

21.2. lATERAL EARm PRESSURE ON SUEIITINGS Rankine's and Coulomb's tbeories of earth pressure cnnnot be used for the computation of lateral earth pressure on sheetings, as those theories arc applicable to rigid retaining walls rotating at base. The sheeting and bracing system is somewhat flexible and the rotation takes place al the lop of the wall. Shcclings are placed against the walls of the excavations when these are shallow. The upper strul is placed when the excavation is shallow and little lateral yield of the soil has occurred. As the excavation proceeds downwards, the lower part of the face is free to yield inward before the next strut can be placed. The inward yield of soil increases with an increase in tbe depth of excavation. Thus, the sheeting tilts about i~

top [Fig. 21.1 (b)].

The method of earth pressure calculations has been developed by Terzaghi based on observations of actual loads in struts in full - scale excavations in sand in Berlin and in sort clay in Chicago. Pressure distributions against the sheeting have been approximated on the assumption that each strut supports a sheeting area. TIle pressure distribution depends upon the speed at which excavation advances, the care '. taken in the installation of bracing, the soil type and many other factors. The effect of various factors is not yet fully understood. However, the results of field studies can be used as a basis for developing earth pressure diagram required for the design of bracing system. The pressure distribution diagram recommended for design represents an envelope which encompasses the actual pressure distribution diagrams obtained from

The ively

j by alioo Cling

dose

: and

:rally open Je.A

>OSed

, dO')'

]llili I ~·65nf(.il f1:~~ t~~~

(.a) SANO

(b) FIRM CLAY

(~':::l,) C "'"

Fig. 21.3. Apparent Pressure Diagrams.

(0;) SOFT TO '-£D!LM CLAY

(~>")


552

the field tests [Fig. 21.1 (c)]. These design pressure diagrams are also known as apparem pressure diagrams. Fig. 21.3. shows the apparent pressure diagrams suggested by Peck (1969). Fig. 21.3 (a) gives the pressure distribution for braced cuts in dry or moist sand. The pressure distribution is unifonn with a pressure (P(l) equal to 0.65 "( fI KID where K(I is Rankine's earth pressure cocITicienl, given by

K(I _ tan 2 (45

0

_

$12)

... (21.1)

Figs. 21.3 (b) and 21.3 (c) show the pressure diagrams for clay. If ("(Hlc) is less than or equal to 4, the pressure envelope shown in Fig. 21.3 (b) is used. The value of P(I varies between 0.2 y H to 0.4 y H, with an average value of 0.3 y H. If (y fllc) is greater than 4, the pressure envelope shown in Fig. 21.3 (c) is used. lbe pressure p(l is taken as (y f/ - 4c) or 0.3 y ll, whichever is greater. Non-unifonn soils 1. When the braced rut passes through layers of both sand and clay [Fig. 21.4 (a)l, an equivalent value of cohesion ce ($ = 0 concept) is detennined using the following equation suggested by Peck (1943).

rr.t

H

11

..

r~

SAND

.......

H

"2 [

l~f

CLAY

....c.qu

(a)

CLAY

'

CLAY

'2''2 CLAY LAYERS CLAY

Yn.cn (b)

fig. 21.4.

... (21.2) 2~ [y,K,H;tao$, -+ (H - H,)n'q"l where H = lotal height of the cut, "(~ = unit weight of sand, H, = height of the sand layer, K = a lateral earth pressure coefficient (- 1), $, = angle of friction of sand, q" = unconfined compressive strength of clay, n' = a Ce

-

f

coefficient of progressive failure (average value 0.75). The equivalent unit weight 'Ie of the layers is detcnnined from the following equation:

y, -

~[y,H,

+ (H - fl,)y,]

... (21.3)

where y.. = saturated unit weight of clay laycr. Once the equivalent cohesion c e and equivalenl unit weight 'Ie have been detcnnined, tile diagram in Fig. 21.3 (b) or Fig. 21.3 (c) can be used depending upon the value of ('1"lf/c,,).

2. When the braced cut passes through a numbcr of clay layers, the equivalent values of ce and Ye are detennined from tbe following equations [Fig. 21.4 (b)]. c" where

CI.

~[CIH1-+ c2H~

+ ... -+ cnHnl

... (21.4)

c2 .. c" are undrained cohesions of layer 1, 2 ... nand

HI' H2 ... H" are the thicknesses of these layers [Fig. 21.4 (b)]. Likewise.

... [215]

rG

,e 1)

,e '"d.


21.3. DIFFERENT TYI'ES OF SHEETING AND BRACING SYSTEMS The following types of shccting and bracing systems for braced ruts are commonly used. (1) Vertical nmber SheeUng. In this method, vertical timber sheeting consisting of planks about 8 (0 10 em thick are driven around the boundary of the proposed excavation 10 some depth below the base of the excavation. The soil between the shccting is then excavated. The sheeting is held in place by a system of wales and struts. The wales are horizontal beams running parallel to the excavation wall. The wales arc

(b)

(a)

Fig. 21.5. Vertical Tim1xr Shet:ling.

2)

supported by horizontal struts which extend from side to side of the excavation [Fig. 21.5 (a)J. iIowever. if the excavations arc relatively wide, it becomes economical to support the wales by inclined struts, known as rakers [Fig. 21.5 (b)J. For inclined struts to be successful, it is essential that the soil at the base of the excavation be strong enough to provide adequate reaction. Ir the soil can temporarily support itself an excavation of limited depth without all external support, the timber sheeting can be installed in Ihe open or in a partially completed excavation. Vertical timber sheetings arc cconomicIi upto a depth of 4 106m. (2) Steel Sheet ]'Iles. In this method, the steel sheet piles are driven along the sides of the proposed excavation. As the soil is excavated from the enclosure, wales and struts arc placed [Fig. 21.6(a)J. lhe wales are made of steel The struts may be of steel or wood. As the excavation progresses, another set of wales and

th

BLOCK

SHEEl

PIL E

.3

~

::D""""I~I

3)

I~I

STRUT

( b) PLAN If OET~LS OF JQINT-J

(iI)

Fig. 21.6. Stut Sheet Piles.

.4)

.5)

Slruts is inserted. The process is continued till the excavation is romplete. It is recommended that the sheet piles should be driven several metres below the bottom of excavation to prevent local heaves. If the width of a deep excavation is large, inclined bracing may be used. Fig. 21 .6 (b) shows the details of joint at J. (3) Soldier Beams. Soldier beams are H-pilcs which are driven at a spacing of 1.5 to 2.5 m around the boundary of the proposed excavation [Fig. 21.7 (a)l. As the excavation proceeds, horizontal limber planks Cllled Jaggings are placed between the soldier beams. ~n the excavation advances to a suitable depth,

,,.

SOIL MECHANICS AND FOUNDATION ENGINEERING SOLDIER

BEAMS

W4.LES

LAGGING

SOL!JIER

BEAM

(b) PlAN OF DETAILS OF JOINT.J

( 01)

Fig. 2J.7. Soldier Beams.

wales and struts arc inserted. The lagging is properly wedged between the pile flanges or behind the back !lange [Fig. 21.7 (b)]. (4) 'Ile Backs. In this method. 00 bracing in the form of struts or inclined rakers is provided. Therefore, there is no hindrance to the construction activity to be carried out inside the excavated area. The lie bade is a rod or a cable coonccted to the sheeting or lagging on one side and anchored into soil (or rock) outside the eXC!lvmion area (Fig. 21.8). Inclined holes arc drilled into the soil (or rock), and tensile SOLDIER BEAM reinforcement (tendon) is then inserted and the hole is concreted. An enlargement or a bell is LAGGING usually formed at the end of the hole. E.llCh lie back is genernlly prestressed before the depth of excavation is increased further to cope with the increased tension.

Use of Slurry Trenches An aitem..'1tive to use of sheeting and bracing system, which is being increasingly used these days, is the construction of slurry trenches nround the area to be excavated. The trench is excavated and is k.ept filled with a

~~~~~e~~sl~~rrys~~~\~~it:~~~~~t~~

BELL

fig. 21.8. Tie Il1ck.

trench. and thus the excavation can be done without sheeting and bracing. Olncrcle is tben placed through a lremie. Concrete displaces the Slurry. Reinforcement can also be placed before concreting. if required. Generally, the exterior walls of the excavation arc constructed in a slurry trench.

Z1.4. DESIGN OF VARIOUS COMPONENTS OF BRACING (a) Struts. The strut is a compression member whose load-carrying capacily depends upon slenderness ratio, fir. The effective length I of the member can be reduced by providing vertical and horizontal supportS al inlennediate points. lhe load carried by a strut can be detennined from lhe pressure envelope. The struts should have a minimum venical spacing of about 2..5 m. In the case of braced cuts in clayey soils, the depth of the first strut below the ground surface should be less than the depth of tensile aack (Z(), which is equal 10 2c/y ( see chapter 19). While calculating the load carried by various struts, it is generally assumed that the sheel piles (or soldier beams) are hinged at all the strut levels except for the top and bottom struts.

RING


'"

[Note. Some designers assume that the sheet piles (or soldier beams) are hinged al aU strut levels, exocpt (or the top J. Fig. 21.9 (0) shows a br-acing system with [our struts. Figs. 21.9 (b) and 21.9 (e), !'CSIX!Clivcly. show the pressure envelopes for sand and stiff Clays.

back

C.,

foro, ::leis

(b)

CC)

Fig. 21.9.

! the

The reactioo Rl per unit length is determined by taking moments of the forces acting 00 span a d at d. and equating them to zero. Once RI has been determined, the reaction component R2' is detenninoo from the equilibrium equatioo in the horizontal direaion for the span a d. The reaction components R z" and R]' are determined considering span df as hinged at d and

f

Thus reaction at d, Rz - Rz' + Rz" 'me reaction R4 is determined by laking momel'llS about f of the forces acting on f h. The reaction component R]" is determined from the equilibrium equation for horizontal forces acting on f II. 1be analysis is similar to that of the top strut. Thus reaction at t.

R3

..

R]' + R]"

The strut loads are then computed as under. PI .. RI )( S

LL

gh a ired.

Cru~

epili

. .. (21.6)

and P4 - R4 )( S

where s is the horizontal spacing (perpendicular to plane of paper) of struts. Proper sections for lhe struts can be chosen for the respective loads found above. (b) Wales. Wales are considered as horizontal beams pinned at the strut levels. The maximum bending moment will depend upon the span s and the loads on the struts. As the strut load<> are different at various levels, the maximum bending moments would also be different. For example.

Mrrw< ..

ness IOns

qual

P2 .. (R2' + R2") S

p] - (R3' + R]")s

R1i

8'

for the top wale

and

... [21.7(a)]

... [21.7(b)]

Once the maximum bending moments have been computed, the section modulus (S) is oomputed as

S .. Mmax a""

[dier where

O.,JI

= aUowable bending stress.

... (21.7)

5,.


(c) Sheet lines. Sheet piles act as vertical plates supported at strut levels. The maximum bending moments in various sections such as a 4 d f and f h in Figs. 21.9 (b) and 21.9 (c). are delennined. Once the maximum bending moments have been oompulcd, the section modulus of the sheet pile can be computed and the seaion chosen.

Other Criteria ror Design or Braced Cuts (I) Braced cuts in clay may become unstable due to heaving of the bottom of the excavation. Th'is is a type of bearing capacity failure and is discussed in Chapter 23. (2) Braced cuts in sand are genernUy stable against heaving failure, but these may become uJ".stable due to upward seepage of water into lhe cut if the water level ir~ide the cut is substantially }.Iwer than that outside. Piping failures are discussed later along with the design of ooffer dams.

(3) The walls of the braced cut may yield laterally and cause ground settlement in the surrounding area. This effed should be carefully assessed and suitable measure<> adopted.

COFFER DAMS 21.5. TYPES OF COFFER DAMS l11e following are different types of coffer dams common ly used in prnaice. (1) Earth Coffer dams. These are the simplest type of coffer dams weD-adapted to depths of water upto 3 m. Earth embankments are constructed around the nrea to be dcwatered. The earth coffer dams are buill of local soils, prcfernbly fine s:lnd. These usually bave a clay core or a vertically driven shcct piling in the middle. The upilrcam slope of the bank is covered with a rip rap (Fig. 21 .10). A successful coffer dam need not be completely watertigbt. For reasons of economy, it is not possible

Fig. 21.10. &tth Coffer Dam.

to make it watertight and hence some seepage of water into the excavation is usually tolerated. TIle water collected is pumped out of the excavation. The embankment should be provided with a minimum free board of 1 m to prevent overtopping by waves. Sand-bag coffer dams are used in an emergency. (2) Rockf1ll Correr dams. RocIc1iD coffer dams made of rockfill arc sometimes used 10 enclQ;C the sile 10 be dcwatered. These ~ permeable and arc usually provided with an impervious membrane of soil to reduce seepage (Fig. 21.11). The crest and the upper part of the impervious membrane arc provi:"iIXI with rip rap to provide pf'O(ection against wave aaion. Overtopping does oot cause serious damage in case of roctdiJ.l coffer dams. The slopes of a rockfill coffer dam can be made as Sleep as 1 horizontal to 1.5 vertical. (3) Single-Sheet Pile Coffer dams. Single-sheet piling Coffer dams are generally used lo enclose small foundation sites in WOller for bridges at a rcIatively...s:haliow depth. In this type of coffer dams, there is a single row of cantilever sheet piles. The piles an: someti'mes heavily braced. Joints in the sheet piles are properly sealed (Fig. 21.12). This type of coffer dams arc suitable for moderate-flow velocities of water and for depths upto 4 m. The deplh of penelration below ground surfaa: is about 0.25 h for ooarse sand and grovels, 0.50 h for fine sand and 0.85 Ii for SillS, where h is the depth of water.

55'

BRACED ClITS AND COFFER DAMS

Fig. 21.11. Rockfill Coffer Dam.

Sometimt"S, single-sheet pile coffer dams are provided with eanh fiUs on one or both sides lO increase the lateral stability.

STRUT WORKING

AREA

Fig. 21.12. Single-Sheet Pile Coffer Darn.

(4) Double-Wall Sheet Piling ColTer dams. A double-waU sheet piling coffer dam consists of two stntight, parallel vertical walls of sheet piling, lied to each other and the space between walls filled with soil. The width between the parnUcl piles is cmpirically set as (hl2 + 1.50 m), where h is height of water (Fig. 21 .13). Double-wall sheet piling coffer dams higher than 2.5 m sbould be strutted. Sometimes, an inside benn is provided to keep the phreatic line within the bcnn . The fill material should have a high coefficient of friction and unit weight so that it perfonns as a massive body lO give the colIer dam stability against sliding and ovcnuming. Suitable measures should be

1===IIr-TIE RODS

Fig. 21.13. Double-Wall Sheet Piling Colrer Dam.

5,.


..,.-WAlE adopted to reduce the uplift on the coffer dam. lbis is gCllerally done by driving the sheet ~ STEEl ROO5---,. piling on the upstream as deep as possible. lhe double·waU sheet piling coffer dam rSTRUr has the advantage of having less leakage than that in a single-wall coffer dam. 1bcse coffer darns are suitable upto a height of 10 m. WORKING (5) Braced CotTe.- dams, A braced coffer AREA RIVER 8Eo dam is formed by driving two rows of vertical "A~"V' sheeting and bracing with wales and struts. '!bese are similar to sheeting and bracing Bottom of eJl:cavat ion system discussed for braced cuts in Sed. 21.3, with one basic difference that braced cuts are '---~HEET required for excavalions in dry areas whereas PILE braced colfer dams are used to isolate a working area surrounded by water (Fig. Fig. 21.14. Braced Coercr Dam. 21.14). '!be braced coffer dams arc susceptible to' flood damage. Land Coffer dams. Braced coffer dams are sometimes used as land coffer dams to prevent ground water from entering lhe foundation pit on land and to support the soil so as to prevent cave in. Afler the pit is dewatercd, tbe structure is concreted. When concreting has been completed above the water level, the coffer dam is removed. (6) CeUula.- Coffer dams. A cellular coffer dam is constructed by driving sheet piks of special shapes to form a series of cells. 100 cells are intcrconnecte
r

559


5,"" ~rkjng(lre(l

Se~hon

A_A

Fig. 21.16. Cioc:ular Cellular ColTer Dam.

21.6. DESIGN OF CELLUlAR COFFER DAMS ON ROCKS The design of circular. cellular coffer dams am be done using one of the follOWing melhoos. (i) lVA method, (it) Cumming's method, (iii) Dnneh-Hansen method. The lVA method is more common and is discussed below. (I) Location of SutulllUon Line. For determining lhe weight of Ihc soil in the cell. it is required to locate the Salurntion line. TVA (fcnnessce Valley Authority) engineers gave the approximate location of the saturation line for diffcrent types of fill materials, as shown in Fig. 21.17. For a perfectly drnining fiU, the saturnlion line is shown in Fig. 21.17 (u). 'The lower half may be assumed saturated for analysis. Par other

(01

(bl

(cJ

Fig. 21.17.l...oc.'Ition of !i;llutfllion Line.

type of fill, the saturation line at a slope of 2 : I is assumed [Fig. 21.l7 (b)J. In ~ of a bcnn, the saturation line drops down to thc lop of the berm [Fig. 21.17 (e)J. For stability analysis, two extremc locations (marked 1 and 2) of saturation line should be Investigated in this case. (2) Average Width. 1be design of a coffer dam is made for a section 1 m long and of uniform, average width. The average width is obtained by equating the section modulus of the equivalem rectangular section to tbe actual section modulus. An approximate valuc of the average width may, however. be obtained by equating the equivalent rcaanguJar area to the actual area of the coffer dam between centre to centre. Thus Average width (b) _ areaofonemhlncell + ;~ofoneconnectingcell where 2L = distance between centre 10 centre of main cells. The lVA enginccrs gave the following relations for romputing the average width.

... (21.9)


560

and

b _ 0.785 D b _ 0.875 D

(For 90" 1"5) (For 60° ,..,)

... [2LlO(o)J ... [2LlO(b)J

where D is Ihe diameter of the main ccU.

lbc above values of the average width are assumed in the prcllminary analysis The actual width to be provided is obtained after the stability analysis. The diameter (D) of the main cells is chosen depending upon the height (II) of the coffer dam. The diameter (D) of the main cells is given by D _ l.0 II to l.UI ... [21.11(0)J

The diameter of the connocting reUs (D J ) is given by D J .0.6D

... [21.11(b)J

The cir<.:ular, cellular coffer dams are coonomical upto a total height (II) of 15

10

18 m.

(3) Sarety agairu:t sliding. The coffer dam is subjected to a horizontal sliding force due to water presSure and earth pressure. The Sliding is resisted by the frictional resistance at the base of the coffer dam. If berms are provided on the interior side, the passive earth pressure also helps in resisting the sliding (Fig. 21.18). lbus the factor of safety against sliding,

Dry

s id~

---rFig. 21.18. Sliding Analysis.

F S

-

Resisting force p( + Pp Sliding force Pd

ItW + Pp

F, - - P - -

... (21.12) ••• (21.13)

d

where It : coefficient of friction (: tan C\l). W: total weight of fill (submerged weight below saturation line). Pp :: passive resistance of the berm on dry side. P d :: driving force due to water and soil on the water side. PI = resisting force at the base. A factor of safety of at least 1.25 is generally recommended. (4) Safety against Overturning. The coffer dam should be safe agrunst failure due to overturning at toe. Neglecting the passive resistance of the berm, the factor of safety against overturning is given by F _ Resisting moment ... (21.14) ~ Overturning moment lbe factor of safely (Fi» should be greater than 2.0.

Further, as the soil canoot resist tension, the resulLant of the forces must tie within the middle third. 1be eccentricity (e) is determined after :ocating the point where the resultant strikes, as the case of retaining walls (see Chapter 20). Thus

ING

e :so b/6

(b)J be

fbe 'a)J

b)J

""

ns 3).

56'

BRACED ClITS AND COFFER DAMS

(a)J

or

b

... [21.15(a)]

-~~. :z:-V~II

...[21.15(b)J

where Z = beight of the line of action of Pd above the base. (5) Sarely lilgalnst sUpplng. As the cell tends to lip over the toe, the fill material has the tendency to run aliI (Fig. 21.19). The piles on the water side aeep upward as one unit. but the piles on the dry side sllp relative to eacb other. 1bis behaviour occurs because the frictional reslstance between the fill and the piles on the water side is smaller than the frictional resistance along the interlock. On the water side, the I force Ptt pushes tbe pile against. the fin. 1be frictional resistance between the pile and the fiU is equal to ~'P". 1be factor of safety against slipping is given by

F,p

..

Frictional rcsista~;::i70~:ward movement

... (21.16)

f) '11

g

...,,; ..-,:,-'

.... ..

Ag. 21.19. Slipping Analysis.

The value of friction ooefficienl II' is equal to tan 6, where 0 is the angle of friction between the fm and the pile. The minimum width b required can be obtained by taking moments about the toe. Thus Pd)(Z)( F,p -PdtanO)( b

b-~ tao 0

... (21.7)

The value of FIf> is usually taken ~ 1.25. If the sheet pile is embedded in the rock for a substantial depth, the effect of the active pressure and the passive pressure should also be considered when summing up the moments about the toe.

(6) Safety aX81m. vertkal shear. The cell may fall due to vertical shear deveIopcd along a plane through ilS centre line. lbe maximum shear (Vmax) is obtained by oomputing the maximum bending moment acting on the cel~ considering it as a vertical cantilever. Thus V~ • 3 M/(2b) ... (21.18) where M

= bending moment due to external forces above the base.

For stability, the shearing resistance developed must be greater than V_. The shearing resistance is equal to the sum of the shearing resistanoe of soil (S I) and the resistance due to interlock (SV obtained as follows:

1 SI _ -2

or

51 -

'{QRl

(~) 2-005 •

k

2

'{QH Ktan,

tan.

.. .[21.19(a)J .. [(21.19(b)J

where K = coefficient of earth pressure having a value greater than that for the active pressure, • = angle of internal friction of the cell fill, and lQ = aveOlge unit weight of soil. The resistance S2 is equal (0 the interlock tension T mUltiplied by the coefUcient of friction (f).

5,.

r. T

The maximum pressure is developed at a height of (3/4) HI above lhe base, where HI is the height of cell above the point of fixity. Thus

'62

SOIL MECHANICS AND FOUNDAnON eNGINEERING

T .. S2

~ 'fa

H(3/4Hf )K"

-lltJH

... (21.20)

HfKQf

The coefficient of interlock. friction f is generally taken as 0.30. 'lbe factor of safety against vertical sbear is given by SI + S2

Fv -

... (21.21)

VII,Wt

The factor of safety Fv should not be less [han 1.25. (7) StabUity ..gainst bunting. The cell should be safe against bursting. The stability against bursting depends upon the jnterlock: stresses. The interlock stresses are quite complicated and cannot be dctennincd accurately. TVA engineers gave the following expression for the ring tcnsion (Tma:<) for 9O°·Tees.

Tmu .pLsec6 Tm.u pDI2

=

where p

= horizootal

. .. (21.22) ... [21.22(a)]

pressure due to cell fill, given by

... (21.23)

P - 'YaH K" + l ... H...

= depth of soil uplO that level, H ... = water depth, S = angle which the T makes with the axis.

in which If

L = one-balf the clislance between centres of main cells. 1be maximum ring tension occurs at the lower quarter point (i.e. Hy4 above the ~ or ground surface). The computed maxintum interlock teosion should not exa:ed the allowable tension (T.../). TIle factor of safety against bursting is given by T"" F," T=

... (21.24)

The value of T dl is usually taken as 1500 kN/m . The value of F" should be at least 1.25.

21.7. DESIGN OF CE1.1.UIAR COFFER DAM ON SOIL The procedure for the design of a coffer dam embedded in deep soil is similar to that for a coffer darn resting on rock, as discu
ty

11' K. tan b

... (21.25)

where 11 = height of cell above top of the stratum, K" = coefficient of active earth pressure, 6 ::: angle of friction between fill and pile, 1 ::: unit weight of cell rut Factor of safety where

Qtdl :::

~

ultimate load capacity against bearing capacity failure (see chapter 23).

(il) If the coffer dam is embedded in clay, the ull'imate load capacity is given by Q... " (5.7c)b where c = unit cohesion. The ultimate load capacity should be greater than the fill load. 1be factor of safety is given by

... (21.26)

G

56'

8KACED CUTS AND COFFER DAMS

F _ ~_ 5.7cb

FiIlload F _ 5.7 c yH

!O)

lbH ... (21.27)

H _ 5.7c

... [21.27(0)]

yF Zl)

, A minimum factor of safety of 1.50 is recommended. (iit) If lhe coffer dam is embedded in

son to medium

clay, it should be safe against tilting c:aused by

unequal settlement. The tilting can be estimated from the compressibility cbaraaeristics of the soil. log

(iv) Cellular coffer dams on a deep sand deposit should have sufficient factor of sarety against piping

•ed

failure . Fig. 21.20 shows a coffer dam founded on deep sand bed. Water IX-rcolates under the base of the ooffer dam and rises up in front of the toe. The flow net can be drawn as shown. The flow lines are almost vertical ia front of the toe. If the seepage pressure is equal to or greater than the submerged unit weight, quick (boiling) conditions may develop. The factor of safety against boiling is given by (see chapter 10)

Z2)

:0)] 23)

I H

::e). . of

j~

24)

lam

led. ater

F · .25)

where i I

.26)

+.

Fig. 21.20. Row-Net

..

(21.28)

=hydraulic gradient at exit ( = l!. h/l), l!. h =drop between last two equipotential lines, = length of the

last flow field.

If the factor of safety is less than L50, a loaded filter is provided as shown to increase tbe downward force without increasing the seepage pressure. The factor of safety can also be increased by reducing the gradient i by driving the sheet pile deeper or by reducing the e(fed.ive head by pcnnilting some water depth on the inner side. The depth of the sheet pile below the ground surface is generally kept at least equal 10 two-thirds of the height of the coffer dam.


564

ILLUSTRATIVE EXAMPLES lIIustratlve Example 21.1. Determine Ihe loads in the three strut.! shown in Fig. E·21.1 (a). The cenlrt ttl centre spacing of the struts along the length o[ 1M CUI is 2.50 m. The soil is stiff, fissu.red clay (y = 19 2 kN1m'J, c = 40 kNlm ). Also determine the maximum bending momems in wales and sheet piles.

1' :19 kN lm) C:IoOkNlm 2

"i 0)

..!!4.

C

'0'

'"

fig- E·21.1.

Solution.

If!- . 19 :0 6.0 .. 2.85

< 4

Therefore, the pressure distribution is as given in Fig. 21.2 (b). The value of the earth pressure is given by

(a) Strut loads.

p" - 0.3yH .. 0.3 )C 19)( 6- 34.2kN/m 2 Let RIt R2 and RJ be the reactions per unit length in three struts at levels A, Band C,

respectively. Taking moments of all the forces acting above iXllni B,

R, x 2.0 .. 34.2 ; 1.50 (1.50 + 0.50) + 34.2 x 1.5 x 0.75 or

R, .. 44.9 kN

From the horizontal equilibrium of the portion AB, Rl + R2' .. 34.2 ; 1.50 + 34.2 x 1.50 .. 77.0

From symmetry, ~

R2'

..

77 - 44.9 .. 32.1 kN

R 2"

..

32.1kN and R3 .. 44.9 kN

~.~+~.~+~.~~

From Eq. 21.6, the strut loads are given as PI .. R] )( S .. 44.9 )( 2.5 .. 112.3 kN P2 .. (R2' + R2") S .. (32.1 + 32.1) )( 2.50 .. 160.5 kN p) .. 44.9 )( 2.50 .. 112.3 kN (b) Wale. 1be maximum bending moment OCOJrs in the wale at level B. From Eq. 21.7,

Mrnax ..

R~i

.. (32.1 ; 32.1) (2.5)2 .. SO kN-m

(c) Sheet pile. Fig. &21.1 (c) shows the shear force diagram for the vertical sheet pile. The shear force is zero at point D at a depth x below top point E. 'The depth x is detennined from the shear force equation.

1NO


t 'Ire

34.2 x 150 + (x - 1.5) x 34..2 - 44.9 .. O ·

Ie

or x .. 2.06 m

The maximum bending moment in the sheet pile occurs al D.

19

t

M_ ..

x 34.2 )( 1.50)( 1.e)) + (0.56

)C

34.2) )( 0.28 - 44.9 x 1.06 .. 15.0 kN-m

UlustraUve Example 21.2. [)efl!rmme the fon::es in the struts for the bracing system shown in Fig. £·21.2, Ass~ hinges at levels B, C and D. Take y = 18 IcNlmJ, c ::: 30 iN/m2 and s = 2.0 m.

1·5m

r

T

OJ

T

!

Mm

t

8m

R'

m 2..

B

" '" .,..

2·5m

R'

..1.. c

-l-

l·Sm

-.;;-

.... l8kNlrn3 C.lO~N/m2 (0)

~4J .2~

1

(d

Fig. E-21.2

e, Solution.

l.!!...18x8. 4.80>4.0 c

30.0

The prc5Sure distribution shown in Fig. 21.2 (c) is applicable in this case. Therefore.

P.. .. 1 H - 4c .. 18 x 8 - 4 )( 30 .. 24 kN/m 2 Alternatively,

p • .. 0.3 YH .. 0.3 x 18 )( 8 .. 43.2 kN/ml

Adopt larger of the two values, viz, 43.2 kN/ml The reaction in strut (1) is delennined by laking moments about B, of the forces acting above thaI level.

Rl x 2.5 ..

tx

2 x 43.2 )( 2.667 + (2 )( 43.2) x 1.0

Rl .. SO.6kN

Now PI .. 80.6 )( 2 .. 161.2 kN. From horizontal equilibrium of the portion above B,

Rl + Rl' -

i )(

2 )( 43.2 + 2 )( 43.2 - 129.6

Rl' - 129.5 - 80.6 _ 49.0 kN From horizontal equilibrium of the portioo Be.

R,l' + R)' - 43.2 )( 2.5 _ 108.0


566

Therefore,

R2

Now

P2 .. 103 x 2 .. 206 kN

..

R 2 ' + R 2"

..

49 +54 .. 103kN

Ukewise, from equilibrium of the portion CD.

R)" + Rt .. 1.5 )( 43.2 .. 64.8 R)" .. R4 .. 32.4 kN R) .. R)' + R)" .. 54.0 + 32.4 .. 86.4 kN p) .. 86.4 x 2 .. 172.SkN

Now

Dlustrative Example 21.3. Design a circular, cellular coffer dam of total height 15 m resting on ltd

(Fig. E-21.3). Tah allowable interlock tension of 1500 1eN1m, • = 300 , 6 = 25°, K = 0.60, 10 JcN/m J,

f = 0.30,1.. '

Ory side

Ag. B-21.3.

Solution. Saturation line is taken as shown in Fig. 21.17 (0), a
Preliminary section From

Eq. 21.11 (0),

From Eq. 21.10 (b),

= 18 m. = 15.75. Let us tak~

D = 1.2 H = 1.2 )( 15 b

= 0.875

x 18

b = 16 m.

Sarety against sliding Weight of fill .. 14 )( 15 )( 16 .. 3361) leN

Weight of steel shell .. 50 kN (say) Total weight W - 3360 ... 50 .. 3410kN

kK. r'lif .. ~ x kx 9.0

Active prtSSure,

PtI

Passive pressure,

Pp -

tKp,(Hi ..

Water pressure,

p ... ..

k

..

x 10

X

t 152

X

(4)2. 24kN

x 3 x 19.0 x (4)2 _ 456kN ..

From Eq. 21.13. factor or sarety against sliding,

1125 leN

!RING


567

F .. Wtan+ + PP .. 34lOtanJO° + 456 .. 211("; I Pd 1125 + 24.0 .

e,

F, ..

Neglecting Pp,

341~::~30°

.. 1.71 (safe)

Safety against overturning

roc1 Y••

Resisting moment about toe Overturning moment about toe

From Eq. 21.14,

F ..

or

F" .. 1125

"

Z ..

where

Therefore,

~!~ x+ 1~:2x

1125 )(

4/3 ..

~-r;::

.

4.82 (safe)

11i~ : ;44)( 4/3 .. 4.92m

~ .. ';6)( 1149)( 4.92 yH 14)( 15

.. 12.70m.

As b > 12.70 rn, the c:ca:ntricity is within middle third.

Safely against slipping From Eq. 21.17,

.. btan6 .. 16tan25"

F

Z

sP

4.92

.. 1.52 > 1.25 (safe) Safety agwnst vertical shear

From Eq. 21.21,

F, •

S1+ S2 --v:;-

where

SI ..

i

S2 Thking H, = 13 m, From Eq. 21.18, Therefore,

-I

'to H2 Ktan "'{d

+.. ~ )(

14 )( (15)2 )( 0.6 tan 300

..

545.6 kN

HH/Kdf

S2 -

~

)( 14 )( 15 x 13 x

Vmu ..

~

)(

¥_~

x

~!7

~

x 0.3 _ 102.4 kN

.. 530.3 kN

F" _ 545'~;;'.i02,4 .. 1.22 (slightly unsafe)

Safety against bursting Pressure at a beight of H/4 (From Eq. 21.23).

P .. 14 x (13 )( 3/4) )( 1/3 + (13 - 1) x 3/4 )( 10 .. 135.5kN/rn1 Tmu .. pL sec 9 .. p )( DI2 .. 135.5 )( 18/2 .. 1219.5 leN From Eq. 21.24,

F" ..

1~~5

.. 1.23 (slightly unsafe)


'68

PROBLEMS

~

III

T 2m

+

'"

0'

Y.18 kN/mJ

1·5m

f

,,.30 kNlml s:2m

2·5m

i-

I2l 8m

2m

",

-L

I - - 'm----i

T

Vim

1

,.,

,.,

8kNlm1

OkNlm2

,,.Z·

'm

1

0'

1.Sm 1"

I------

6m

-----<

T

Ag. P-21.2.

Fig. P-21.1.

21.2. For the braced all shown in Fig. P 21.2, draw .he eanh pressure envelope. Delennine the 10000s in SIMS (1), (2) (3) and (4)~ assuming thlll the Slruts ore hinged 01 levels 2, 3 lind 4. The spacing of the struts is 2.S m. Take 1 '" 18 kN/m and c = 30 kN/m z. [Ans. 201.8, 2575, 216.81, 81 kNJ 3 21.3. Check. the slability of the circular, cellular coffer dam SDawn in Fig. P 21.3. Toke '1 '" 18 kNhn , z 30- ,6," 25·, K ::: 0.6, f '" 0.3, TAl'" 1'500 kNIm. The diameter of the main cclls is 15 m.

+

Fig. P-ll.3.

21.4. Determine tbe factor of safety agnin.'11 piping failure of the ooffer dam founded on a deep sand bed shown In Fig- P 21.4. Take G '" 2.67 and e = 0.67.

n . Descriptive Types. 215. What is II coffer dam? Name the different types of coffer dams and discuss their relative advantages aM diSDdvimtages. 11.6. Discuss the method for the design of a circular, cellular, coffer dam on rock. 11.7. CVmpare diaphragm cellular coffer dams and ci~lar ooffer dams.

:ING

BRACED ctrrS AND COFFER DAMS

S69

Fig. P-21.4.

2UI. Sketch 11 typical section o f II brnced cut aocl show the vnriou, components. 21.9. Draw different type!> of 3ppnrent pressure diagrnm~ used in the dc~ign of braced

cut~.

What nre the factors that

affect the pressure distribution?

(2) !cy

"'J

1=

21.10. Describe the methods for the design of various ,"'Umponcnls of a braced CUI, Slating clearly the assumptions "",
C. Multiple Choice Questions. 1. For the design of braced cuts. the eanh pressurf distribution is based on ((I) Coulomb's theory (h) Rankin e's Iheory (el ilpp:lrent pressure di ugrnm (tTl none of llbove. 2. In th e design of br:u::ed cut. it is gencrnll y assumed Ihnt Ihe sheet pile is Itli hinged at all thl! strut levels. (b) Ihl!d at 1111 the ~tmt Icvels. Ie) hinged at all thc strut Icvels except the top and bottom struts. (lh none of :lOOVC. J. The diameter of the cell of a cell ular. colTer dam is usually kepi (tI) 0.5 H (h) 0.80 H Id 1.2 H (d) 2.0 H where H is the height of the coffer d:lm. 4. Single-sheet pile coffer dums ure suitublc uplo u height of fa) 5 m Ih) 10m Ie) IS III Iff) more than 15 m 5. The faclor o f safcty ag(linst piping when; 0.75 and y' 9 kN/mJ is about

=

~)l.W

(M I.OO

(e) O.KO

(d) 0.75

=

2

6. For a cellular ,""Offer dam ~~hcdded in clay. the factor of safety agninsl bearing fa ilure when c _ 60 leN/m , H

= 15mandy= 18kN/ III' ISabaut (a) 1.10

(h) 1.27

1.40

(d) O.W

(e)

(Ans. I. (el. 2. (el, 3. (e ). 4. (a), S. (a), 6. (b»)

22 Shafts, Thnnels and Underground Conduits 22.1. STRFSSFS IN SOIL IN THE VlCINqY OF A VERTICAL SIIAFr Fig. 22.1 shows the section through a vertical sbaft of radius rD' It is an axis-symmetrical case of stress distribution. The stresses at any point per, z) before the excavation of the shaft are given by (see chapter 11).

(oJ, - yZ

... [22.1(0») .. .[22.1(b»)

(0,), _ Ko y Z

and ("'), - Ko y Z ... [22.1(,») where o.p 0", au are respectively, the vertical stress., horizontal radial stress and horizontal circumferential stress (hoop stress). The suffix i indicates initial s~. 't = unit weight of soil, K" = coefficient of earth pressure at rest. 6.S.

1 Fig. 22.1. Stresses near vertical shaft.

The shear stresses 'tn are zero, as all the three stresses are principal stresses. Because the shear stresses on all cylindrical surfaces within the soil mass are zero, tbe material located within the boundaries of the. proposed shaft can be replaced by an equivalent liquid of unit weight '1, equal 10 Ko '/. without changing the state of stress in the soil. 1be horizontal pressure p due to the equivalent liquid is given by p-y,Z-KoYZ ... (22.2) As the pressure p is equal to (ar); or (oe);. there is on e£fea on the stresses in the soil in the vicinity of the shaft.

SHAFTS. TUNNELS AND UNDERGROUND CONDUITS

571

The stresses due to equivalent liquid pressure can be found using Lame's fannulae for thick. cylindcn> (sec any text on strength of materials). According to which, ... [223(a)] (oz)p - 0 (o,)p -

(a,,)p _

and

5

"). )]

)]

)] 81 Ih

p';;,.' - K, r Z (,.'.1,.') - p,}',' - -K.yZ(,.',1,.')

... [22.3(b)J ... [22.3(c)]

where suffix p indicates thai the stresses are due to pressure p of the equivalent liquid. After the shaft has been excavated, the shear stresses and the radial stresses on the interior surface are zero. The effect of excavating the shaft can be considered equivalent to that of pumping the liquid out of a cylindrical hole whose dimensions are identical with those of the shaft (BiOI, 1935). Thus the streSses at any point aftcr the excavation of the shaft can be obtained by supel1X>Sition of the initial stresses (Eq. 22.1) and those due to pressure (Eq. 22.3). The stresses due to pressure are taken as negative because the liquid is pumped out, which corresponds to a negative value of p. ... [22.4(a)J Therefore, 0: - (oz); - (oz)p -

,Z

o~ - K~ ,Z-(o,)p - K",Z(l - r;;?)

... [22.4(b)]

0, _ (0,), - (a,,)p - K. r Z (1 + ,.'.1,.') ... [22.4(c)J Fig. 22.1 also shows the variation of stresses on a horizontal plane in the soil mass. TIle vertical stress (O'~) is constant and equal to ,Z, and is independent of r. The radial stress (0,) is zero at r _ r". As r increases, 0', increases. The curve becomes asymptotic. As r tends to infmity, 0, approaches a value of K" 'Y Z. The circumferential stress (00) is equal 10 2 K", Z at r = r". As r tends to infinity, Oe also tends to a value K" 'Y Z. In olher words, at infinite radial distance, the stresses reduce to those given by Eq. 22.1 a to c. The above stresses have been derived assuming that the soil is elastic, homogeneous and isotropiC. These assumptions are seldom justified. Moreover, the circumferential stress (00) in the vicinity of the shaft is very high (= 2 K" 'Y Z) which may cause plastic flow of the soil. This may lead to a state of plastic equilibrium. Consequenlly, the actual stresses in the soil in the vicinity of the shaft may be somewhat different from those given by the elastic theory. The elastic theory is used as it is simpler than more advanced theories. 22.2. STRESSES IN SOIL AROUND TUNNELS Fig. 22.2 (a) shows a tunnel of intemal radius r" , with its centre line al a depth of Z" below the ground surface. Let a, and 00 be the radial stress and circumferential stress at any IX>int P (r. Z). Before the excavation of the tunnel, the stresses are given by Eq. 22.1 as (0,);" K",Z

and (00); - K"yZ

~'i

r

I _____ L ,-----H

es

he he

-G:j'

p(r.z)

2) of 10)

'"

Fig. 22.2. Stres>es anound a tunnel.

l"fZo

-rzo

sn


where 'i is unit weight Taking

K"

:c

of soil.

1 and

Z = ZII' (0,), ..

..,Z"

... [225(a)]

... [22.5(b)] (00)/ .. yl" As in the case of a shaft (Sed. 22.1), let us imagine tbat proposed tunnel is filled by an equivalent liquid of unit weight 'II under a pressure given by p .. 'tIZ, re the depth Z" is large compared with,,,. the liquid pressure on the tunnel m~y be taken as constant. Thus p" .. 't,Z.. .. ..,Z The stresses due to the liquid pressure on the tunnel am be dClcnnined using Lame's formulne for thick and

cylinders. (a,)". • P. (;;;,') • 1, Z. (,;;,')

... [22.6(a)]

000 (a,)".. -P.(,'.;?) • -1, Z.(;;;?) ... [22.6(b)] The stresses after the excavation of the tunnel can be obtained from the snperposition of Eqs. 22.5 and 22.6, taking the latter with negative values as it corresponds to pumping out of the liquid. Thus

0, .. 'I, Zo - 'II Z" (r;;,2) .. ,/, Z" (1 _ ;./r2)

... [22.7(a)]

aOO .[22.7(h)] Fig. 22.2(b) shows the stress distribution on a horizontal plane in the soil mass. At the interior surface, r .. TO'

Therefore, cr, .. 0 and As r lends to infinity, both cr, and Do tend to approach asymptotically a value of '11 Z" - .., Z... 'The actual stress distribution may be different, as the soil is not a purely elastic material. Further, it ha
very large compared with the radius of the tunnel (To). If the tunnel is located cinc;e to the ground surface, the result would be erroneous. The solution becomes complicated when all these aspects are considered. If the support system of the tunncl yieldS and allows an inward, or radial denection, of the walls, tbe radial stress 0, reduces but the circumferential stress CJe increases. The minimum value of 0, is attained when the circumferential stress becomes equal to the compressive strength of soil (or rock). At this stage, a statc of plastic equilibrium is established. The radial denections may further increase and 0, is also likely 10 increase:. If the maximum circumferential Slress (= 2 Ylo) is less than ttJC compressive strength of soil (or ~), theoretically no support is required for the tunnel. The deformations m~t be controlled by providing sufficient restraints to prcvent the soil from attaining a plastic equilibrium which would require Substantially greater support 1oadings.

22.3. CONSTRUCTIONS OF EARllI ruNNELS Earth tunnels are constructed by excavating from below. The procedure is similar to one used in .mining. The method of construction used would depend upon the ability of the soil to support itself temporarily during oonstrudion and also on the magnitude of the pressure ultimately developed against the support system. The fad
LINER

PLATE

Fig. 223. Constru.:lion of an eanh tunnel.

513

SHAfTS, TUNNELS AND UNDERGROUND CONDUITS

liner plme (Fig. 22.3). The liner plales are flanged steel pmles, which when boiled to similar plmes form a continuous wall. Somelimes, in place of a liner plate, l.1ggings, which arc wooden or steel hOrizontal planks braced by steel frames, are usOO. The next increment is then exc..'lVatcd and the process repeated until Ihe complcte section has been excavated. If requir~. the liner can be stiffened with curved arches. In a very soft soil, a cytiDdric..11 temporary support, known as .fMeld, is jacked through the soil. The face of the tunnel is supported by a bulkhead fitted with doors through which excavation of a limited portion of the face can be done. The pennanent lining is constructed inside the shield aftcr some excnvation has been done. The shield is then jacked ahead and the process repeated. In sand, a tunnel requires support during construction as well as pennancnt support after construction. 22.4. ARCHING IN SOlLS Arching is a phenomenon that occurs when a yielding part of a soil mass transfers pressure to adjoining soil mass which is less yielding or rigid. The action is similar to one in a structural arch which transfers the load to abulments. When a part of the soil mass yields, it has a tendency to move out of its original position. This tendency is resisted by the shearing resistance at the zone of contact between the yielding and non-yielding pmts. Coru;cquently. the pressure on the yielding part is reduced, where$ that on the non-yielding pans is increased. The soil thus arches over the yielding part and transfccs the load to the dz G non-yielding pans which act as abutmen~. Fig. 22.4 shows a long narrow section of the layer of soil having a yielding part Non yielding part Non yieldIng par t supported on a deflecting structure.

T z

1.L T

Although the actual failure surf:lccs are

Yielding

curved, for Simplicity, the analysis is

part

usually based on the assumption of vertical slip surfaces (Thrzaghi, 1943). As the structure and the soil above deflect downward, the shear resistance develops Fig. 22.4. Arching in Soils. between the soil above the yielding part and that in tbe non-yielding part. It is generally assumed that the shear stress mobilised is equal to the shear strength of the soil. An expression for the vertical stress can be determined as under. Let us consider the equilibrium of a yielding slice of thickness dz at depth z from the soil surface, The slice is of width B and unit length pecpcndicular to the plane of the paper. The slice is in equilibrium under its O'W'n weight dW, the vertical earth pressures on the upper and lower surraces, and the shear strength (s) prOOuccd by the lateral earth pressure. Thus Substituting where

dW + Boz - B(oz + do,) + 2sdz dW .. '( Bdz. and s - c + o.lan ••

o. - Koz , '(Bth. + Bo... 8(oz+ doz) + 2[c+ Ko.tan+]dz

or

'(Bth. + Boz

"

Boz !&ioz + 2cdz + 2Koz tan4ldz

Bdoz - ('IB - 2c - 2Koz lancp)dz doz _ ('I _ ~ _ 2Ko.taDCP)
...(a)

'74


At the upper surface (x = 0), the vertical stress «(J~) is equal to the surcharge q. The solution of Eq. (a) with these boundary conditions becomes. oz"

np;!I:> [1 _

e-2K(IIB)tan+ ] + qe-2K(zIB) tan

+

... (22.8)

If ... 0, Eq. (a) after integration becomes

oz"

(Y

-1f)

. .. (22.9)

z + q.

If c :::: 0 and q = 0, Eq. 22.8 becomes

At z ..

00,

Oz"

2K~~'

Oz ..

2~an cjI

[1 - e- 2K (ZIB)tan+ ]

... (22.10) ... (22.11)

Terzaghi's experimental results indicate that the value of K varies from about unity immediately above the centre of strip to a maximum of about 1.5 at a height of approximately D above the strip. It has been found by Terl.8ghi from experiments on dry sand thai the shearing resistance of sand is mobilised for a height of about 2.5 B above the yielding strip. Lei Zl be the depth of sand in which there are no shenring stresses and %2 be the depth in which shearing resistance develOps. The vertical stress 0: on the yielding strip can be obtained by making the following substitution in Eq. 22.8.

c

:=

0,

q _ lZh

and

l

-

Z:z - nE, where n

= 2.5 for sands.

Thus

. .. (22.12)

Eq. 22.12 can be used for the determination of the vertical stresses. The phenomenon of arching can be demonstrated by a simple test illustrated by Fig. 22.5. It consists of a platform fitted with a trap door AB. The lrnp door is mounted on a weighing scale (001 shown). The depth H of the dry sand layer over the trap door is very large in comparison with the width of the trap door. The pressure on the trap door and that On the adjoining platform each is equal 10 1 H when the trap 000r occupies its original position. However, when the trap door is allowed to move slightly In the downward direction, the pressure on the door dcacases considerably. as shown by the weighing scale, whereas that on the adjoining parts of the platform increases. As the prism of sarxl located above the door moves downward, the shear

~N~~': ~ ~P:;~~URE lOWERING

(b) Fig. 22.5. Dtmomtrlllion of Arching.

PRESSURE AFTER OF DOOR

'"

SHAJIl'S, TUNNELS AND UNDERGROUND CONDUITS

stresses along its lateral boundaries AD and BC resist it and, therefore, the pressure on the trap door is decreased. The pressure becomes constant after a small movement. II has been established that the ultimate pressure on the trap door is independent of the depth H of the layer of sand above it. Only a small portion of the soil prism, shown by the area ABE, contributes to the pressure on the trap door. If the sand has even a trace of cohesion, it will not drop out even jf the trap door is removed. 22.5. TYPES OF UNDERGROUND CONDUITS Underground conduits are used for sewers, drains, culverts, water mains, gas lines, electrical cables, telephone lines and many other purposes. The underground conduits should be capable of supporting aU external loads. If the conduit is under pressure, it should also be strong enough to withstand internal pressure under extreme conditions. In this text, the forces acting on the conduits due to only external loadings are considered. These forces depend upon a number of factors such as !he buried depth, rigidity of the conduit, method of installation, preparation of conduit bedding, live and dead loads acting on the conduit. For purpose of analysis, the conduits are classified into the following Iypes. (1) Ditch Conduits. (2) Positive Projecting Conduits, (3) Negative Projecting Conduits, (4) Imperfect Ditch Qmduits (5) Tunneled Conduits. The loading on different types of conduits are discussed. 22.6. nITCH CONDUITS A conduit installed in a relatively narrow ditch in

=~t~::l

a~~~:n~~i~~g~;~~~ ~t~

:n!:n type of conduits arc normally installed for sewers, drains, water mains, g'.!s mains, etc. The width Ed of the trench is generally Dot greater than 2 to 3 times the ronduit diameter B~ . The loading imposed on 10 the buried conduil can be obtained by considering the equilibrium of an elemental slice of thickness dh at a depth h below the ground surface. Let V be the vertical load on the top surface of the slice and (V + dV). on the bottoms surface. Let dW be the weight of the slice, which is equal to y Bd dh per unit length.1be horizontal pressure (ax) on the vertical sides is equal to K times the vertical pressure, where K is the coefficient of laleral pressure. Thus

j . 1

G,S. h

"i

-L dh

T

H

Fig. 22.6. Ditch Conduit

The shearing resistance (5) developed along the sides in the vertical direction is equal to Il' times the horizontal pressure, where Il' is the coefficient of Sliding friction between the backfill material and the treodl wall. Thus When the elemental slice has a tendency to move downwards, the shear resistance S acts upwards. From lhc equilibrium equation in the vertical direction, V+ dV + 2S _ V+ ylJddh dV .. yBddh - 2S or dV .. yD"dh - 2KIJ.' (VIDd ) dh . The solution of the above differential equations is

51.


l _ e-2K~' (hIBJ )

V-llf" [

-----w

2K~'

1

AI the lop of the conduit, h = II. 'Ibcrefore,

l_e-2K~' (/JIB,,)

V-llf" [

2K_'

1

.. .(22.13)

If the ditch has sloping sides. the value of Dol is taken equal to the width of the horizontal tangential pl.,nc nl the top of the conduit [Fig. 22.7{a)]. If the ditch is very wide, the wntluit is laKl in a sub-dilch 10 reduce Dol and hence to reduce the load on the cooduil [Fig. 22.7(b)].

, ----

8ockl1ll

---

,.)

'b)

Ag.22.7. The load (We) carried by the conduit depends upon the rigidity (stiffness) oC the conduit in comparisoo with that of the fill material between the sides of the conduit and the sides of the ditch. In the cnse of a vcry rigid condull, the side fills nre relatively compressible and the conduit would amy pmctically aU the load V. l_c- 2KJ,I.'(IIIB,,) From Eq. 22.13, W, - 1 2 K _' ... (22.14)

If"

If"

...(22.1S)

l_e- 2Ktl' (IIIBJ Cd 2 K _'

... (22.16)

W, - Cdl where

The coefficient Cd is known as the load coefficient for ditch conduits. The values of Cd for different values of HIBtl aDd K Il' can be obtained from Fig. 22.8. The ranges for the values of C. for a>hcsioolcss soils and cohesive soils are shown hatched. If the conduit is relatively Oel(ible and the soil is thoroughly tamped around the conduit, the stiffness of the side filb may approoch Ihot of the conduit. The load on the fiel(iblc conduit is reduced because some of lhe load is carried by the side fiUs. The lood corried by the fluible cooduit is obtained by multiplying the value given by Eq. 22.15 with the ratio (B/Bd). where B~ is the diamCler of the a:>oduit. 1berefore. ftt fJcxiblc conduits, the load is given by We " CdyBrBd ... (22.17) Eq. 22.13 has been derived a\Sumlng thai the orchlng cffect is achieved without soil cohesion. Actually, some cohesive resistance develops in the vicinity of the trench which would reduce the looding on the conduit. Therefore, Eq. 22.13 is conservnUve. Field observations indicate that the load ot the time when the fill is completed is about 80 to 90% of the finnl load. The 1000 keeps building up because of straim occurring in the soil in the zones along the vertical boundaries of the trench. This rcdua:s the shearing resistonce of the soH and causes transfer of load from the soli to the conduit. The following points are worth noting : If the trench bockfill is oompacted to blgh density, it wlU not settle relative 10 the surrounding

<,)

G

') II

SHAFfS, TUNNELS AND UNDERGROUND CONDUITS

undisturbed soi~ and the arching action would not occur. For sudl a condition, the load on the conduit is equal to the weight of prism of soil of beight H. (iI) If the surrounding undisturbed soil settles more than the fill above tbe conduit, the load on tbe conduit is computed considering it as a projecting conduit, as explained in the following section,

577

" "

"

" " , , ,

J/::

1>:""~0.1921,

-

t/:

I~

:1 1

10

_L I r-*':;~~ 1-1--

7

SO"

(2:!...)

~

22.7. POSITIVE PROJECTING CONDUITS A oonduit projecting some distance atx)Vc the natural ground surface (GS) and covered by caM fill is known as a plSitive projecting cooduit [Fig. 22.9(a)]. Common examples of positive projccting cooduits are culverts for highway, railway and airfield. In positive projective conduits, there is a differential settlement between the central zone dircctly overlying !be conduit and the side zones. Depending upon the relative settlement, the load 00 the conduit will vary, For rigid conduits, the total compression of the

RANGE OF ---..., C()-£SIONLESS

t, VI.

5

,

hr.;-

3

2

p

,

f/:

i-/J

I'

\<'; 0. 130

tt

k!<':O.l1

~~ANGE

CL AY _

FOR

I

W

1

0

Cd ----..

=~i: S~t t~~~:~1 ~c!:a~~r It::n ~h to~ Fig. 12.8. Chart for c'I. directly over the conduit. This is because greater height of fill in side woes. However, if the conduit is flexible and deflects considerably under the weight of the fill, the settlement of tbe fill in the central zone will be greater than the settlement in the side zones. The differential settlement between the two zones results in the development of shearing stresses on vertical planes passing through the sides of the oonduit. Fig. 22.9 (a) shows the direction of Shearing stresses (upward) when the central zone settles more than side zones, whereas rTOP OF EMBANKMENT4

SHEARING~t STRESSES 'I (uprNard) I 'I :

:1'I

Ii CENTRAL 4fZONE WITH ,I MORE I SETTLEMENT

:

'I

'I

t'

I

I

FlEJ(IBLE CONDUIT'

~ :

il

IYrr~~s I'

(downwacd) I LESS 'l'SETTLEMENT IN lCENTRAL ZONE RIGID CONDUIT

I

~S .

GS .

(a)

(b)

Fig. 22.9. Positive Projecting Conduit. Fig. 22.9 (b) shows the directioo. (downward) when the central zone seUles less than side zones. In the first case, the load on the oonduit is less than the weight or fill directly over the conduit, wbereas in the second case, tbe load on the conduit is more than the weight of fill. The direction of the shearing stresses can be determined from the settlement of a plane, known as a critical plane. It is a horizontal plane in the fill material at the level of tbe top of tbe conduit at the time when embankment constructioo is begun (Fig. 2-2.10). If the aiLical plane, after the coostruction, settles less than the lop of the conduil, the directioo. of shearing stresses is upward [Fig. 22.10 (a)J. However, if tbe aitical plane settles more than the top of the conduit, the direction is downward [Fig. 22.10 (b)]. A ratio, known as

5'"

SOIL MECHANICS AND FOUNDATION ENOINEERINO

(.)

Ib)

Fig. 22.10. Effect or Settlement or aitiCllI plane.

seulement ratio, is used to determine the settlement of critical plane. The settlement ratio (Tp) is given by

(6. + 6,) Tp -

(6, +

do)

... (22.18)

6...

wbere bm =: compression of the fill on the sides of the conduit in the distance pBe ~ 6, = settlement of natural ground surface adjacent to tbe conduit, 6t = settlement of conduit into its foundation. and dt = vertical deflection of the conduit, p = projection rntio. It is the ralio of the distance from the natural ground surface to the top of the conduit to its width Bt , Bt = dj.ameter or width of conduit. The setUement ratio is negative when the conduit settles more than the aitical plane. The load on the conduit is less than the weight of the overlying fill. It is similar 10 the conditions in the ditch conduit and is known as ditch condition. However, if the settlement ratio is positive, the load on the conduit is greater than the weight of the overlying fill . Il is known as projection condition. 11 is extremely difficult to determine the individual terms in the settlement ratio. However, empirically determined values given by Spangler are generally used for design purposes (Table 22.1). Table 22.1. Values fOI" St!Ulement Ratio (rp) (After Spangler) S.No. 1. 2. 3.

4.

5.

Conduit Condition Rigid conduit of foundation of rock. oc unyielding soil. Rigid conduit on foundation of ordinary soil. Rigid conduit on foundation Wt yields relative 10 adjacent ground. Fleltible conduit with poorly compacted side fills. Flexible conduit with well compacted side fills.

Settlement Ratio (rp)

+1.00 + 0.5 \0 + 0.80 0.0 to + 0.50 - 0.4 \0 0.0 - 0.2 to + 0.8 (not well~ablished)

Plane of Equal Settlement The shearing stresses acullg on the vertical plan($ do not n~rily extend 10 the top of the embankment, especially where a high fIll exists over the conduit. Differential settlements are greatest at the

'0

SIiAFfS, ruNNELS AND UNI?ERGROUNO CONDUITS

579

I C TOP OF

~MBANKMENT

I

I

-r-l(-r::~=-~b::.a~~:~,!.':EMENT

T "

Lr"<:1:,:,

tl,cRITICAL PLANE

1---

IJt

---,

I

':

I

I

I

I

I

G.S.

Fig. 22.U. Plane of equal 5elllemenl.

3)

he

Iq) surface of the conduit and progressively decrease towards the lOp of embankment. At cenain height H~ above the conduit, a borizootal plane exist at which the differential settlements arc zero. This plane is called the plane of equni seuiement (Fig. 22.11). Above the plane of equal settlement. the central zone and the side

zones settle equally. Thus 110 differential sclUcment occurs within the embankment soil above tbe plane of equal setticment. An expression for the height H,. can be obtained by equating the total vertical settlements in the central zooe and the side zooes a
Ie

;s

on ;e

+

~ 3

(!!!.)' ('!!.)+ (Ii) ('!!.) _.

(!:!... _!!"!:t: ~)l c 2KJ1'(I,f../B.-) D~

Bc

3

:t

(/J-/J,) e2K.' (II/.J _ _ 1 _,

B,

1

+!

2KI!'

2K.

B~

2

B,

B,

The upper (Plus or minus) signs apply when settlement ratio aw1y when ' " is negative.

(7,,>

B,

'pP (/JIB,)

.. (22.19)

is positive, whereas the lower signs

l)'pes of Positive Projecting Conduits The positive projecting conduits can be classified into 4 types, depending u!X>n (1) magnitudes of actual embankment height (lI) and the height of plane of equal settlement (H~) and (2) the sign of the settlement ratio (Tp), as given below. .

H < He ; H < He; H > He: H > He;

(I) Complete ditch oooduits (il) Complete projection conditions (iii) Incomplete ditch conditions (iv) Inoornplete projection conditions

Tp Tp _ Tp Tp -

negative . positive negative positive

It may be DOted that for complete ditch or complete projection oonditions, the shearing stresses extend uptO the top of the embankment; whereas (oc inoomplete ditch or iocomplcte projection oonditions, the shearing stresses extcod upto the plane of equal settlement. which is lower than the top 'o( the embankment. Marston gave the (olloY/ing expression (or tbe load on positive projeaing conduit:

We -

epyn;

... (22.20)

SOIL MECHANICS AND FQUNDXnON ENGINEERING

580

where the cocITicicnl Cp for the complete projection

Of

ditch condition is given by

e:l: 2K J,t'(HIB.) -1

Cp

... (22.21)

::t2KIJ.'

-

and ttL'll for incomplete projection or ditch condition as

Cp

_ e:!:2K\-l'(lleIBc) -1

:t:2KJ.l'

+

(!l

~) :t:KIl'(lIeIBc)

Dc - D.

e

... (22.22)

lbc positive signs apply when Tp is positive (projection conditions), whereas the negative signs apply when rp is negative (ditch conditions).

Fig. 22.12. Chart for

C,.

Values of Cp for diITcrcnt (HIB,) ratios and the products (rp x p) can be obtained from Fig. 22.12. The position of the curves is not sensitive to the variation of KIJ.' valUal expected in the field. 1be curves are fa K",' "" 0.13 and KIJ.':: 0.19.

22.8. NEGATIVE PROJEC11NG CONI)U1TS A conduit installed in a trench such that its lOp is below the original ground surface and then covered with earth embankment is known as the negative projecting conduit (Fig. 22.13). In case of negative projccting conduits, the critical plane lies below the original ground surface (G.S), and tbe setUement mlio (rII) is given by

6,-(6,,+lJ,+d() r ll

-

6"

...(22.23)

where 6, = settlement of ground surface, ad = settlement of bnckfill in the trench, b, = sculcmcnt of the conduit foundation, d t = verticnl oeOection of the conduit. TIle criticnl plnne is taken as the horizontal plane at the ground surface. Gcocrnlly. the settlcment of tbe criticnl plnne in tbc trench fill over the c<,)nduit is greater than the setUemcnt of lhc ground surface adjlJCCBt

,,"

SlIAfTS, TUNNELS AND UNDERGROUND CONDUITS

t2) .ly

Fig. 2213. Negative Projecting Conduit.

[0 the trench. As the central zone moves downward relative 10 the side zones, the shearing stresses act upward and the load on the cooduit is re
... (22.24)

c. yal

where

en

W, • is the coefficient for negative projecting conduit, given by

ell .. and

e- 2 KIl'(1II B") _ 1 -2 K~'

when H

[ 2KIl' (lf~/B.t) _ 1 +

(!!... _!l.!.) e- 2KIl' (II,/B.J

ell "

- 2KIl'

~

Dd

<

.. (22.25)

He

I \c

:~',o'31~w ",:~g E~:=-~I~P-

0 p.O , 50 ' he Or

r! 7r-ru ~

if' cd ve jo

3)

1

;;

6fSf-

'L Y'A~ /

§ I/V#,' / /

~ V,.0f7 /

flW / /

1-1- i;

0

I- I-r-§ liY;/, 1-1!//(/

i

-of/;

f/ i-+,fY /

1-1- •

/

/

/

/

p'=2·Q Te:~2

~t--,-~

f- f-I-it r- I- $ fI-

S 6 7

COFFICIENT (b)

Fig. 2214. Charts

~L

t-:lf /

/

~+4- 4-

/

/

/

/

/

/

o1/, 2 34 S 6 7 &01L , 2I 34 Cn~

I-

fl-

A/

J?i

(a)

nl

{7

1/

COFFICIENT

be

0'.' 0

1-1-- /

" f-r!~ / In 2r-:U / "J ,7'

... (22.26)

H > He

when

BJ

Cn- -

ror C..

I/, SO

,

2

3

4

5

COFFICIENT Cn- " (c)

S


582

lbc negative projection mtio p' is the ratio of the distance between the top of the conduit and the natural ground surface to the trench width. Thus the distance between the lOp of the conduit and the ground surfnce is p' Ed- h is always positive. As the settlement ratio (rft) is always negative, the product (r" x p') is a~ negative. The settlement ratio (r,,) is generally taken as -0.50 for the purpose of estimating loads on the conduit. The value of ell may be obtained for different values of r" from Fig. 22.14 for K~' - 0.13 andp' ::: 0.5, 1.0 and 2:00.

22.9. IMPERFEcr DITCH CONDUIT Imperfect ditch conduits are special type of oonduits which are oonstruded to reduce the load on I conduit under a high embankment. The construction of imperfect ditch conduits is done in two stages : C

C1ni1i
01 emban~ment

level

•. ...Q ..

.' . . ..

Top of final l'fTIbonkment

"

. ..

>

" ...·~~·G. ....·xvG_;; . ',:

4' ' . .

(0)

Fig. 22.15. Imperfect Ditch Conduit.

(I) The conduit is fm;t installed as a ~itive projecting conduit above the ground surface and it ~ covered with earthfill upto a height p' B~ where p' = 1.0 to 2.0 [Fig. 22.15 (a)J. The fln is wet compacted. (il) In the second stage, a trench is excavated directly above the conduit from the initial level ci embankment to the top of the conduit [Fig. 22.15 (b)]. The trench is backfilled with SOlllf compressible material such as hay, straw, cornstalks. The rest of the embankment is completed II usual upto the final level. Thus this consU'Uction proa:dure creates the situation like a negati\~ projection condition. As the compressible material settles, arching action deveioll'i in the overlying embankment and lilt shearing stresses act in upward direction. Thus the load on the conduit is reduced. Eqs. 22.23 to 22.26 C3I be used for the imperfect ditch conduit by substituting Be for B d • Fig. 22.14 can be used for determina· tion ofC". Imperfect ditch conduits are also known as conduits in induced trench conditwns.

22.10. TUNNELED CONDUITS Conduits passing beneath existing embankments or hilly areas arc known as tunneled conduits. Sud! conduits are constructed by tunneling or jacking methods instead of the usual open trench methods. TIt examples of the tunneled conduits are culverts or sewers installed beneath an existing highway or a railwiIj line in service (Fig. 22.16). The load on a tunneled conduit can be obtained from the following equation:

W~ - C,yn; - 2e,cB, where B, = width or diameter of the tunnel or jacked opening,

... (22.21)

'INO

T

SHAFI'S, TUNNELS AND UNDERGROUND CONDUITS c;;,TO P of >#

turnl

."

583 ~)(istinq

OX,

~ mbonkm~r'lt ( <<

,

,

lim do," '

crace also 1 the ldp'

H

l

..

~"""''''''' '' M~M'''d''"d";t

r-- BI------i Fig. 22.16. TtUlneled Conduit.

c = cohesive strength of the soil overlying conduit,

C, = coefficient for tunneled conduit. It is equal to Cm the oocfficient of ditch conduit (Eq. 22.16). The values of C, can be obtained from Fig. 22.8.

it is

we" el of

some

"'as

:ative :I the

; can

nina-

Such

TIt<

ilway

2.27)

22.11. LOADS ON CONDUITS DUE TO SURFACE LOADS Loads applied on the surface above a buried conduit are transmitted through soil to the conduit. The stresses on the conduit surface due to rona!mratoo or uniformly distnbuted loads acting on the soil surface can be detennined using the methods discussed in Cllapter 11. It is generaUy assumed that Boussinesq's soluUon is applicable. The stresses from loads applied through a highway. runway or railways depend upon the physical characteristics of the pavement (or formation below the track), the magnitude of the load and the conduit area. It is generally assumed that the loads act over some prescribed area whose size increases with depth, somewhat similar to 2 : 1 distribution. Moving loads also cause impact. Therefore, the eITect of moving loads is greater than equivalent dead loads. The effect of impact is more in the case of unpaved formation compared with that of a paved surface. Generally, an impact fador of 1.50 is taken for unpaved formation and of unity for a paved highway. Martson proposed the following equation for computing the load on the rooduit due to loads applied on the surface of the ftll

W, -

t . I~CIP

... (22.28)

where W, = average load on conduit due to wheel load (kN/m), L~ = length of conduit section, Ie = impact factor, C1 = load ooefficient, P = concentrated truck-wheel load on surface (kN). The value of the load ooefficient (e/) depends upon the length and width of the conduit section and also on the depth of soil cover over the cooduit. Its value can be determined from Boussinesq's solution. The length of rorxluit sedion (L~) is equal to the actual length of 3 precast segmental section of the conduit which is 1 m or less in length. For continuous conduits or those constructed of segmental sectioos more than 1 m in length. the effective length (L~) is the length of the conduit over which the average live load proouces the same effect as docs the actual load. An effective length of 1 m is generally taken. 22.12. CONSTRUCTION OF CONDUITS Buried ronduits are ronstructed in different shapes using different materials. The maximum load whidl a


584

conduit can lake depends upon the conduit material, shape of the conduit and the type of support. The 501fc ioc''\d which can be imposed on II rigid conduit, such as concrete pipe, is dclcnnincd from the pipe strength and the quality of foundation bedding. The strength of concrete coo.duits can be obtained from a three-edge bearing lest (Fig 22.17). As the lest simulates a very severe looding condition, which rarely develops in the actual conduits, the conduit can support even a load greater than that obtained from the test. Generally, a conduit which has a foundation shaped according to its contour can take a graltcr: lood [Fig. 22.18(b)] than thnl wbich has no bedding [Fig. 22.18(0)]. Design ch..'U1S are supplied by manufacturers which provide load faaors foc different Sh..1pcd concrete conduits based upon type of bedding, conduit

projection and settlement ratio. The factors affecting the looo-carrying capacity of flexible conduits are not well derined. Ilowcver, manufacturers of corrugated met.,1 pipes have developed some manuals which can be consulted. Such conduits fuil by excessive deflection whereas rigid conduits f:lil by the rupture of the pipe wall. 111C deflection of flexible conduits should be limited.

(a) CONDUIT WITH

NO

(b) CONDUIT

BEDDING

Load

o Fig. 22.17. Three-edgc bcaringte5t.

WITH

BEDDING Fig. 22.18.

ILLUSTRATIVE EXAMPLES Dlustmllve Example 22.1. A rigid sewer pipe with an outside diameter of 50 em is to be laid in a ditch which is 1 III wide at the top of the pipe and is to be covered with 8.0 III of clayey soil back-fill (y = 19 kN/m 3). Determine die load on the sewer. Take K Il' _ 0.12. Solution. (Sec Fig. 22.6.) In this case, 11 = 8.0 m, From Fig. 22.8, Cor I1IB/I Alternatively. Crom f-"rom Eq. 22.15,

fld

= 1.0 m, I1IBd

;;;

8.0.

= 8.0,

Eq. 22.16, Cd -

1 _ e- 2KJl'(/JIBJ ) 2KI1' - 3.56

W.. - CdyIij

_ 3.6 x 19 x (1.0)2

- 68.4kN/m lIIustmtlve Example 22.2. Determine 'he load on 0 (pcsirhY! projcclillg) square box culvert under a 16 m fill (y 19 kN/II,lj. Assume that the olltside width of tire barrel is 2.5 nI. Take projection ratio (P) OJ and the sell/ement ralio (rp) = + 0.6.

=

=

IG

:h

'9

sa,

SHAfTS, TUNNELS AND UNDERGROUND CONDUITS

Solution.

HI De - 16/2.5 - 6.4

From Fig. 22.12,

Cp

From Eq. 22.20,

W<. CpyEf,

-

and,p )( P - 0.6 )( 0.5 - + 0.3

9.0.

_ 9.0

X

19 )( (25)2

- 1068.75 kN/m. PROBLEMS A. Numerical 22.1. A rigid pipe with an outside diameter of 1 m diameter is laid in a dild:! which is 1.5 m wide at the top of the

~~1~~~ is covered with 9.0 m of sandy soil backfill (K....'.O.1S). Determine the load ~!e:rk~!~ 22.2. A reinforced concrete pipe, 2 m in diameter, is installed as nn imperfect ditch conduit, with the negative projection rutio (P') ., 1.0, The height of fill above the top of the pipe is 16 m. Determine the 1000 on the pipe. Assume K ....' = 0.13 and T" '" - 0.5. Take y = 19 kNlnl [Ans. 357.2 kNImJ

B. Descriptive and Objective 1)rpe 22.3. DiS'-"USS different types of conduits. What are the factors that affect the lood on a conduit? 22.4. What do you undcrsland by arching of soil ? Gille examples. 22.5. Explain the variation of stresses in soil in the vicinity of a venial! shnR.

22.6. Draw a sketch showing the stress distribution in soil around tunnels. U.7. 22.8. 22.9. 22.10. n.ll.

Discuss the methods of oonsuuaion of eatth tunnels. DefiDe planes of' equal settlement. What is their impooancc' ?

•

DefiDe settlement ratio. lIow is it determined ? How would you determine the load on a buried conduit due to surface loads ? Describe the methods of laying of conduits. What is the effea or foundation bedding on the load carried by the conduit? n.n. What do you understand by imperfea ditch oondition ? What is its advantage? 22.13. Write whether the following statements are true. (a) For a vertical shaft constructed in soil. the radial stress becomes infini te when the l1KI..ius tends to infinity. (b) For a tunnel in sail, as r lends to infinity, tile radial stress tends 10 approach a value of zero. (e) The StreMeS on the yielding portion of a soil mass are inaeased due to arching of soil. (d) For a ditch conduit, the load carried by a flexible ronduit is greater than that by a rigid one. (e) For a ditch conduit, the load load carried by the oonduit is reduced if the ditch is made wider. (/) The sail above the plane of equal settlement is not subjected to shear stresses. (g) lbe plane of equal seulemem cannot be above the top of embankment. (h) In imperfect ditch conduits. the excavated trench is backfilled with compact, incompressible material. (Ans. True, (f)]

C. Multiple Choice Question 1. The negative projecting conduit is installed (a) (b) (e) (d)

above the ground surracc in an embankment. in a trench below the ground surface in an embankment. in a ditch below the groW\d surface in cutting. by tunneling into an embankment.

2. in a vertical shah driven in a hornogeneoos'soil, the maximum circumferential (hoop) stress Is (a)K..a~ (b) 2K..a.. (c) o~ (d) none of above

where X. is coefficient of' earth pressure at rest and

o~

is the vertiall stress.


586

3. In a horizontal [unnel excavated in a homogeneous soil, Ihe maximum circumferential stress is (a)

2y2.,

(b)

(el 0.5 yl.,

yl.,

(il) zero

where y is Ihe unil weight of soil and 2., is the depth of lhe centre of the tunnel. 4. For ,:omplete projection condition in a posilive projccting conduit. (a) H < He nnd '1' should be positive. (b) H < Hr and rl' should be negative. (c) H > H~ and r l , should be positive. (d) H > H~ lind '1' should be negative. where H is embankment height.

H~

is height of plane of equal settlement and '1' is scl\lement ratio.

5. For a ditch comlui, of 1.0 En diameter laid in a dilCh 2 m wide III its lOp when Cd ::: 3.0, y = 18 kNlm", the load carried hy the conduit is (a) 54 kN/m (h) 216 kNfm (e) 108 kN/rn (,/)'27 kN/m 6. A ncgntivc proj!..'Cling conduit of dinmcter 1.0 m is !nid in a ditch 1.5 m wide. The load curried by the conduit when Gil = ' 5.0 nnd 20.0kNlm 2 is (al 100 kNhn (11) 150 kN/rn (c) 225 kN/m (d) 200 kN/m 7. ~,,~Si~i.~ea;,:~e~i;7) ~o;;~~~1 i:ns a diameter of 2.0 (a) 100 kN/m (c) 400 kN/m

n1.

The load carried by the condui t when

(h) 200 kN/m

(If) 300 kN/m 8. In a positive projecting conduit. the direction or the shear stres~s developed in the fill Is (a) Always upward (b) Alwnys downward (c) May be upward nr downwartl (If) Always horizontal

~_I~ ~~1~~~~~~~~~L~

INa

23 Bearing Capacity of Shallow Foundations

tho

:Iuit

23.1. INTRODUCTION A fouodalion is that part of a structure which transm its the weight of the structure to the ground. All structures constructed on land are supported on foundations. A foundation is, therefore, a connecting link: between the structure proper and the ground whim supports it. The word 'foundation' is derived from the latin word [wulaTe, meaning to sct or ground on something solid. In other words. a foundation is an artificially laid base on which a structure is set or built up. A foundation is required for distributing the loads of the superstructure 00 a large area. The foundation should be designed such thai (1) the soil below docs not fail in shear and (2) the settlement is within the safe limits. 1be pressure wbich the soil can safely wilrnrnod is known a<> the allowable bearing pressure. This

chapter gives the methods for the determination of allowable bearing pressure. e)1

Foundations may be broadly classified into two categories: (1) SbaUow foundations, (2) Decp foundations. A shallO\ll foundation transmits the loads 10 the strata al a shallow depth. A deep foundation transmits the load at ~iderable deplh below the ground surface. The distinction between a shallow foundation and a deep foundation is generally made acoording 10 Terzagbi's aiterion. According to which. a foundation is termed shaUow if it is lnid al a depth equal to or less than its width. ShaUow foundations are discussed in Chapters 23 and 24. and deep fouildatiQllS., in OUlpters 25. 26 and 27. 23.2. BASIC DEFINITIONS (1) Ultimate Bearing Capadty (q,,). The ultimate bearing capacity is the gross pressure at the base of the foundation at which the soil fails in shear. (2) Net UlUmale Bearing Capacity (q_) . It is the net increase in pressure at the base of foundation lhal causes shear failure of Ihe soil. It is equal to the gross pressure minus overburden pressure.

Thus

q~ - q. - r Df ... (23.1) where q" '" ultimate bearing capacity (gross), y unit weight of foundation soil. and Df depth of foundalioo. y Df existed even before the conslructioo of It may be noted that the overburden pressure equal foundation. (3) Net Sare Bearing Capadty (q..J. It is the net soil pressure which can be safely applied to the soil considering only shear failure. It is obtained by dividing the net ultimate bearing capacity by a suitable factor of safety. Thus

=

=

to

... (23.2)

where F = factor of safety, which is usually taken as 3.0. (4) Gross Safe Bearing Capacity (qs)' It is the maximum gross pressure which the soil can carry safely


without shear failure. It is equal (0 the oct safe bearing capacity plus the original ovaburdcn pressure. Thus q, -qlU + yD, g, •

~ + yDf

Some authors define the gross safe bearing capacity faclor of safety (I'); that is,

... (233) (q~)

as the ultimate bearing capacity divided by a

... (23.4) As the added strength due to y D, is available in full, it does not seem logical to apply a factor of safely to this term. II is, therefore, more rational 10 define the gross safe bearing ca~cily as indicated by Eq. 233. This prnctice will be followed in this text. (5) Net Safe Settlement PrtiSure (q..,). II is the oet pressure which the soil can carry without eXO::cding the allowable settlement. The maximum aUowabJc settlement generally varies between 25 mm and 40 mm for individual footings. The nct safe se!llcmcnt pressure is also known as unit soil pressure or sale bearing pressllre. (6) Net AUowable Dearing Pressure (q.) . The net allowable bearing pressure is the nct bearing pressure which can be used for the design of foundations. As the requirements for the design of foundation are that there should be no shearing failure and moreover the settlements should also be within the limits, the allowable bearing pressure is the smaller of the net safe bearing capacity (q..) and the net safe settlement pressure (q~p). Thus qnQ - q..

if qnp > qIU

... [23S(a)}

q~ • g.. if gN > g., ... [23.5(b)) The net allowable bearing pressure is also known os the allowable soil pressure or allowable bearing pressure or allowable bearing capacity.

or

2J.J. GROSS AND NEI' FOOTING I'RESSURES The gross and nel fooUng pressure at the base of a fooling can be found as follows. (1) Foundation Backfilled. Fig. 23.1 shows a footing subjected 10 a superimposed loud Q. If the weight of the footing and the soil above it is WI' the gross footing pressure is given by

q, • Q : W,

... (23.6)

G.S .

T l'f

1

FOOTING

.1 0.:

~I'~------B----~~ Fig. 21.1.

T

G

BEARING CAPACITY OF SHALLOW FOUNDATIONS

wbere A is the base of the footing.

I)


q, •

q+

589

(D'1,)A + (Dr -

~,)

1A

. .. (23.7)

where Yc '" unity weight of ooncretc, Dc = thickness of fooling. y = unit weight of soil, Q = superimposed load. The net footing pressure is equal to the gross footing pressure minus the overburden pressure. Thus q• • q, - 1 Df ... (23.8) Substituting the value of

q, from Eq. 23.7,

q• • ~ + (D'1,)A + (Df

-

~,) 1A -1 D

f

Simplifying, q" - QI A + (Yc - y)Dc If the difIereoce between the unit weight of concrete (about 24 ItN/m~ and the unit weight of soil (about 20 kN/m1 is neglected, Eq. 23.9 becomes q• • QIA ... (23.10) Thus the net footing pressure (q,,) is equal 10 the superimposed load Q divided by Ute area A. For safe design. the nct footing pressure (q,,) should be less than or equal to the net allowable bearing

pressure (q....), i.e. q,,:$ qQI A :s: q"" (2) Foundation not baddilled. 1be footing beneath the footings, the

.. (23.11) b~ment

(Fig. 23.2) are not backfilled. For such

gross pressure is given by Q

BASEMENT

1 Fig. 232.

q, - QIA + Dc 'tc + !t'Yc where If = thickness of basement floor. The net footing pressure becomes q,.-q,-lD,

... (23.12)

... (23.13)

q,. - [QIA + De le + llle] - yD/ As the thicknesses De and 1 are small in comparison with the depth D" the second and third tenns in Eq. 23.12 can be neglected. Thus q,. - QIA - yD, ... (23.14) For safe design, the net footing pressure should be equal to or less than the net allowable bearing or

pressure, i.e.,


590

q,.

'"

(QIA - yDf )

s;

qllQ

S q~

... (23.15)

... (23.16) QIA S q~ + y Df Comparing Eq. 23.11 with 23.16, it is observed that the load- carrying capacity of a foundation is considerably increased if it is not backfilled. lbe nct Cooling pressure would reduce to zero if in Eq. 23.14, QIA - yDf ... (23.17) This is the principle of compensated raft foundations in which the pressure applied is just balanced by the pressure released (see Olapter 24). The reader sbould carefully notc the diffcreoce between the footing pressure q" and the allowable bearing pressure q..... The footing pressure depends upon the superimpaiCd load Q ading on tbe footing, whereas tbe allowable bearing pressure depends upon the bearing capacity of the foundation and the allowable settlement. The allowable bearing pressure is a function of the type of soil and the footing, as discussed in later sections. Eq. 23.11 is used for the determination o( the area of footing if it is backfIlled and Eq. 23.16. if not backfilled. 23.4. RANKINE'S ANALYSIS Rankine (1885) oonsidered the plastic equilibrium of two adjaocnt soil elements, one immediately beneath the footing and the other just beyond the edge of the footing (Fig. 23.3). For the element I beneath tbe footing. the vertical stress is the major principal stress and tbe lateral stress is the minor principal stress.. However, for the clement n, the lateral stress beoomes the major stress, and the V(:rtical stress becomes the minor principal stress.

Fig. 23.3. Ranldne', Analysis.

When the footing pressure approaches the ultimate bearing capacity (q.. ). the element I attains a stale of plastic equilibrium. However, the element I can fail only when the adjacent element IT also fails. A relationship between the principal stresses aaing on the two elements can be obtained using the equations of plastic equilibrium developed in chapter 19. For the active case, 03 _ tan 2 (45 0 - .'/2) 01 For element I, Slbstituting

01 _ q.. , 2 03 - tan (45° - .'I2)q..

. ..(a)

03 - '( DI ii

1

-

~ tan'(45' - .'12)

_

'(Dr tan' (45' - .'/2)

... (b)

BEARING CAPACITY

As

or SHALLOW FOUNDATIONS

'91

oJ of element I is equaJ to 01 of element U, from Eqs. (a) and (b), lan' (45° _ "/2)

,.

q"

_

yDr 18n2 (45° _ ,'12)

q.. - y D, x tan" (450 _ .'12) 4

q.. - 1 D[fan (45 0 + .'12)

... [23.18(a)]

(~)'

... (23.18(b)]

q.. - 't D, 1 _ sin.'

Eq. 23.18 gives an approximate value of the ultimate bearing capacity q .. of the soil. As the equation does not give reliable values, it is rarely used for the determination of the ulLimate bearing capacity of the soils. It has been superseded by Tchaghi's theory and other theories which give more dependable valucs. Rankine did not consider cohesion intercept (c') of the soil. The theory gives the bearing capacity of the soil as zero if Dr • O. This is contrary to experience. These are the limitations of the theory. Eq. 23.18 is occasionally used to determine the minimum depth of foundauon (D,)min' It can be written

D . _ 2. (:,r}m," 1

(.!....=...!!!!.)' 1 + sin "

... (23.19)

where q is the intensity of loading at base.

13.5. UOGENTOGLER

AND TERZAGm'S ANALYSIS

Hogcntogler and Thrzaghi (1929) approximated the actual curved failure surfaces below the footing with II set of straight lines (Fig. 23.4) for the plastic equilibrium of a long strip footing of width B. At the lime of failure. the footing exerts a pressure q.. equal to the ultimate bearing capacity of the soiL

I--d--t

Fig. 23.4. Hogentogler and TelDghj's Analysis.

The soil in zone 1 immediately beneatb the footing is in compression'. The soil in zone I can fail only when the soil in the adjacent zone II also fails. An approximate value of the bearing capacity of the soil can be obtained by considering Ihe stresses at mid.beights of the two failure zones. The height of the failure zooe is (BI2) tan a', where a' is the angie of the failure surface, equal to (45 0 t .'(l.). The overburden pressure at the footing level is equal to ., Df . 1bis pressure is tenned as surcharge.

Zone

n. From

the equilibrium of zone II at mid-height

oJ"

YDf + (BI2) tan a' )(

t

where the second term on the right·hand side is the avtnlge vertical stress due to self weight. From the equations developed in Sect. 19.6,

... (a)

OOll MECHANICS AND FOUNDtXIlON ENGiNEERING

03 -

~

: :::: 01 -

0, - tan' ( 45' -

~~~~,

f) a, - 2c'tan ( 45' - f)

or 03_0toot2a'_2c'cota' Substituting the value of 03 from Eq. (0), Y Df +

(~

tan a' )

f - 010012 a' -

... (b)

2 c' cot a'

01- [yD,+ (yBI4)tana']lan2 a' + 2c'tana' Zone I. Now 03 of zone I is equal to 01 of zone D. Therefore,

'03

and

°1 -

_ [y D, + (y

q"

+"rt-

814) tan a'J tan2 a' +

2

c' tan a'

.•. (0)

...(Uj

tan a'

From Eq. (b). for zone I, taking 01 and OJ values from Eqs. (c) and (d), 2

(YD,+ rp-tana') tan a'+ 2c'oota' 2

- cot a'

(q" + rt-tan

0.' ) -

q"ca. a' _ yD,tan2 a' + I/-tanla' 2

2c'cota'

~cotal

+ 2c' tan a' + 2c' x

q" _ '1D,tan4a' + ('1BI4)(tanSa' - tana') + 2c'(tan3 a' + lana')

t~a'

...(23.20)

Eq. 23.20 is a general equation applicable to botb cohesive and oohesionJess soils. (0) For oohesionless soils, c' = o. Therefore, q" - '1 Df tan 4 a' + ('1814) (tanS a' - tan a') Eq. 23.21 is known as Rin~, ~ equation for bearing capacily of CQbesionless soils. (b) For purely cohesive soils, 0, c' - c.,. Thus q"-yD,+4c,,

... (23.21)

+' ..

qIW-q"-yD,-4c,,

... (23.22)

where c" is the undrained cobesioo. As the actual failure surfaces are curved and not plane as assumed in the analysis, the results obtained are approximate. However, the bearing capacity obtained is c:ooservative. 100 theory bas been superseded by

other theories as explained later. 23.6. PRANDTI1S ANALYSIS Praodtl (1921) gave a theory for the penetratioo of puocbes into metals. The theory can be used to determine the ultimate bearing capacity of soils The analysis is based 00 the assumptioo that a strip footing placed on the ground surface sinks vertically downwards into tbe soil at failure, like a punch. Fig. 23.5 shows the failure zones developed belOYf the footing. The soil in the wedge-shaped zooe I immediately under tbe footing is subjected 10 compressive stresses. As the footing sinks, zone I exer1S pressure on side zones U and Ill. The zones D are assumed to be in plastic equilibrium. The zones U push

m

zones upward. Using the theory of plasticity, PrandU developed exprtSSions fa- tbe ultimate bearing capacity for a strip footing, l&WIIing the curved part of the Slip surface of tbe shape of a logarithmic spiral For purely cobc:sive

BBARING CAPACITY OF SHAlLOW FOUNDATIONS

593

Fig. 23.5. Prandd'. Analysis.

(+

soils = 0), the spiral becomes a circular an; and Prandtl's analysis gives the following equation for the ultimate bearing capacity, ... (23.23) q" .. (It + 2) e" .. 5.14c" where e" is the undrained cohesion of soil. Eq. 23.23 indicates that tbe ultimate bearing capacity of a cohesive soil is independent of the width of the footiog (8). For cohesionless soils, Prandtl's theory shows that tbe ultimate bearing capacity increases with the width B. The theory is applicable for the footings at the surface. For the footing at a depth (D,) below the surface, an allowance can be made by increasing the bearing capacity by y Df' Hence for strip footing on cohesive

soil, q" " 5.14 c" + Y DJ .•. (23.24) PrandU's theory is valid only for footings with perfedly smooth base in contact with the soil. As the adual footings have the rough base, the theory does oot give accurate results. 23.7. TERZAGffi'S BEARING CAPACITY TIIEORY

Terzaghi (1943) gave a general theory for the bearing capacity 9f soils under a strip footing, making the following a
Derivation of Equation. As the base of the footing is rougb, the soil in the wedge ABC immediately beneath the foOling is prevented from undergoing any lateral yield (Fig. 23.6). The soil in this wedge (zone I) remains in a state of elastic equilibrium, It behaves as if it were a part of the footing itself. It is assumed that tbe angles CAB and ClM are equal to the angle of shearing resistance .' of the soiL The sloping edges AC and Be of the soil wedge CBA bear against the radial shear zones CnD and C4.F (zone III). 1be curves CD and CF are arcs of a logarithmic spiral. Two triangular zones BDE and AFG are the Rankine passive zones (zones III). An overburden pressure q - y Dr acts as a surcharge 00 the Rankine passive Wiles. The failure zones do not extend above the horizontal pLlnes pac;sing through the bao>e AD of the fOQ(ing. In other words, the shearing resistance of the soil located above the base of the fOQ(ing is neglected, and the effea of soil is taken equivalent to a surcharge of 1 D/, Because of this assumption, Terzaghi's theory is valid only for shallow foundations (Dr 5: B), in which the term 1 Dr is relativelY small.


594

Fig. 23.6. TCl'ZlIghi's Analysis.

The loading conditions are similar to that on 3 retaining wall under passive pressure case. The failure occurs when the downward pressure exerted by loads on the soil adjoirnng the inclined surfaces CD and ct of the soil wedge is equal to the upward pressure. The downward forces are due to the lood (q" )( B) and the weight of the wedge (1/4,/ If Ian $'). The upward forces are the vertical components of the resultant pao;sive pressure (Pp) and the cohesion (c') acting along the inclined surfaces. As the rcsullant passive pressure is inclined at an angle to the nannal to the surface of the wedge, it is vertical. Therefore, from the equilibrium equitation in the vert.ical direclion,

+'

i"t B'l Ian
+ q,,)( B .. '2Pp + 2c' )( L;sin
= length of the inclined surface CD [= (B{2)/ros ,'] q,,)( B .. 2Pp + Dc' tan,' -

Therefore,

~"tJiZ

tan.'

···('l

The resultant passive pressure (Pp) on the surface CB and CA constitutes the following 3 components. (1) Component (Pph which is produced by the weight of the shear zone BCDE, a (Pp)(' and (Pp)q-

q" B-2 [(Pp)T+ (Pp)(' + (Pp)q) + B c' tan q.' -

From Eq. (a),

2 (PP)T -

Substituting 2(Pp)~ +

and and

~'(JiZtao,'

.. B)(

y JiZ tanq.'

~ '(BNT

Be' tan " - B )( e' Nc 2 (Pp)., ., B)( Y D,Nq •

q" B - B)( c' Ne + B,/D,Nq + B)(

wegct

q" - c'Ne + yDfNq + 0.5 ,{BNT q" - c'Ne + q" N q + 0.5 '( BN, where. n .

~

j"

'hI' (\vrrh"mrn .......,.,,,.... _ ... n

~'{BNy ... (23.25)

.(23.25('ll

NG

,.,


Eq. 23.25 is known as Terzaghi's bearing capacity equation. The bearing capacity factors Ne,Nq and NT are lhe dimensionless numbers, depending upon the angle of shearing resistance (,') of lhe soil. These are defined by the following equations: N, -

00.,' [2<='( :;0. f) - 1]

N, - [

... [23.26(a)J

2<=' (~o f) 1

... [23.26(b)J

•

a .. e(3n/4 - .'12) tnt'

where

i (~ - 1) tan +'

~

and

he

where Kp == coefficient of passive earth pressure.

Ny ..

... [23.26(c)]

Fig. 23.7 gives the values of the bearing capacity factors. The values are also tabulated in Table 23.lo is he

These values are for general shear failure. If the ground surface on the two sides of the footing is at dificrent levels, Df is taken as the smaller of

the depths on the two sides. Eq. 23.25 gives the ultimate bearing capacity of a strip rooting. The nct ultimate bearing capacity and safe bearing capacity can be detennined as explained in Sea. 23.2. Table 23.1. Terzaghi's Bearing Capacity Fadors

:a)

5)

.)J

.', 10 15

'"25 30 35 40 4' '0

General Shear FailllFf!

Local Shur Failure

N,

N,

N,

N,

N<'

N,

Ni

5.1 1.3 9 .• 12.9 17.7 25.1 37.2 57.8 95.7 172.3 347.5

1.0

I..

0.0 05 1.2 25 5.0 9.1 19.7 42.4 100.4 2975 1153.2

5.1 •.1 8.0 9.1 11.8 14.8 19.0 25.2

1.0 1.4 1.9 2.1 3.• 5.• 8.3 12.6

0.0 0.2

2.1 4.4 1.4 12.7 225 41.4 81.3 173.3 415.1

34.'

20.5

51.2 81.3

35.1 65.6

0.' D.• 1.1 3.2 5.1 10.1 18.8 37.7 87.1

(Vesic)

1.0 1.2

I.. 2.2 3.3 5.3 .5 18.7 425 115.0 329.10

23.8. TYPES OF SHEAR FAILURE Vesic (1973) classified the bearing capacity failures into 3 categories:

(I) General Shear Failure. Fig. 23.8 (a) shows a strip footing resting on the surface of a dense sand or a stiff clay. The figure also shows the load settlement curve for the fOOling. where q is the load per unit area and s is the settlement. At a cenain load intensity equal to q... the settlement increases suddenly. A shear failure occurs in the soil al that load and the failure surfaces extend 10 the ground surfaoc. This type of failure is known as general shear failure. A heave on the sides is always observed in general shear failure. (2) Local Shear Failure. Fig. 23.8 (b) shows a strip footing resting on a medium dense sand or on a clay of medium OXlSistency. lbe figure also shows the load-settlement rurve. When the load is equal to a certain


596 00

"

50 0

II

/,

/ji

Iy

100

./

./

"c/

./"

0

V

/'

/

V

~

./

W

".~

70\

"¥

7

/

/

/

oV

/

II

/

/

!

"Of

1. 30

~

/

20!

/ o.

"o'

,"

,,"

,,"

,," An91~

of

sheoring

,,"

,,"

rt'sisl~Jr\ce

~

!

ra i

. /

60

,," ((>/)--......

Fig. 23.1. TerLaghr5 Bearing CapllCity FacIOrS.

,,"

,,"

r 50'

NS BEARING CAPACITY OF SH4LLOW FOUNDATIONS

597

~ (a) GENERAL SHEAR

FAILURE

:U

~

q

s

ulqU

~rt \ ---'..........

----- ,

':

/./~

(b) LOCAL SHEAR FAILURE

q

qu

s

(c) PUNCHING SHEAR FAILURE Fig. 23.S. Typcs of Shear Fllilure.

vulue

qo(l}l

the foundation movement is accompanied by sudden jerks. The failure surfaces gradually extend

outwards from the foundation, as shown. However, a oonsiderable movemenl of the foundation is required for tile failure surfaces to extend to the ground surface (shown dotted). The load at which this happens is equal to q... Beyond this point. an increase of load is acrompanied by a large increase in settlement. This type of failure is known as local shear failure. A heave is observed only wben there is substantial vertical seltlemenL (3) Punching Shear F8ilu~. Fig. 23.8 (c) shows a strip footing resting on a loose s.'md or a soft clay.

In this case, the failure surfaces do not extend up to the ground surface. There are jerks in foundation al II load of qo(lr The footing fails at a load of q.. at which stage the load· setUement curve becomes sleep and practically linear. This type of failure is called the punching shear failure. No heave is observed. There is only vertical movement of fooling. Vesic proposed a relationship for the mode of failure based on the relative density Dr and BO. where BO = 2B)( L/(B + L), in which B is the width of the footing and L is the length (Fig. 23.9). It is worth noting o that even for the same relative density (Dr), the mode of failure may change with a change in DIB ratio.

Dr

It is generally observe
13.9. ULTIMATE BEARING CAPACITY IN CASE OF LOCAL SIIEAR FAILURE Eq. 23.25 gives the ultimate bearing capacity of a strip footing under general shear failure. No Uteorclicai

,,.


0,

8""

3t---j---i--',II---tJ.r---j a* =!!!.L B+L

Fig. 23.9. Yule's Chart.

solution is available (or local shear failure and P9n<::hing shear failure. Terzaghi (1943) has suggested the following empirical reduction to the actual cohesion an?the angle of shearing resist.ance in case of local shear failure. Mobilised cohesion,

e'm -

~c'

Mobilised angle of shearing resistance,

... [23.27('11

,'m _tan-

t

... [23.27(bll

(2I3l3n +')

The reduced values of .' equal to ....' are used to determine bearing capacity parameters from the valllQl of we general sbear failure. lbe modified bearing capacity factors are indicated as Nt'. N q' and Ny' for loal shear failure. Fig. 23.10 gives the values of these factors for different values of .'. Thble 23.1 also gives tbe values of these factor for values of .'. It has been found from severnl mooei tests tbat the value of Nq' 81 determined above is underestimated. Sometimes, the following equation [(II' Nt,' (Vesic. 1963) is used for leal shear failure. This gives more reliable resu l[S.

N: - (e3.8 +' w.,.) tan' (45" + f)

..

(23.21)

Table 23.1 and Fig. 23.10 also give the Vesic's values of Nq'. The equation for local shear failure for strip footing can be wriuen as

q,: - ~c'N/

... (23.29)

+ 1 DIN'l ' + O.51DNT'

+m'.

where N/, Nq' and Ny' are for rOOuced values of ,', equal to Il is difficult to ascertain the limiting oonditions for which local shear failure should be assumed. the criteria given by Vesic (Fig. 23.9), the following criteria are also used.

Bcsi~

(1) For a cohcsionless soil, if ~' is greater than 36°, general shear failure is likely to occur. If ~' is lei than 29°, local shear would be more probable. For intermediate values of between 29° to 36°, the vaiUIJ of bearing capacity factors are obtained by interpolation.

+'

+'

For example, if a soil has = 35°, Eq. 23.27 (b) gives The bearing capacity factors are as under.

+m = 25°.

IG

599

BEARING CAPACrfY OF SlI~W FOUNDATIONS 00

, I

I I I I

10 a

/I

I

777

/77

,V V _______V;l:Y '

';: Dg

§

~V '0

-----

'? ,)]

..

~z;..

. .c.. ....

"'~

~/

\

'\,~

/~ /j

/

~/

/

/

17

;)]

---/

;al he

as ;al

0/

~

£

/

7

1/

!8)

f----

/

!9)

II

es,

:ss

.,,

5

,0

,

30 Aog!e

01

I rictio o.

V -----...

Fig. 23.10. Dearing Capacity faclOl'5 for Local Shear Failure.

so


600

Ne .. 57.8, Nq Ne' .. 25.2, Nq'

and

Difference (N,)d

= 32.6,

(N,),

= 28.8 and

..

..

and N, .. 42.4 12.6 and N, ' .. 10.1

41.4

(N,)d

= 323.

I\s the actual value of,' is 35" which is 6" morc than the value of " rorresponding to local shear failure

(viz. 29°). the proportiooal difference to be added to the values of Ne' Nq' and N,' is 6n times the total diITerc'pcc. Thus, the required values are Ne .. 25.2 + 617 x 32.6 .. 53.14 Nq

.. 12.6

+ 617 x 28.S .. 37.19

NT .. 10.1 + 617 x 32.3 .. 37.79

(2) If tbe failure of the specimen of the soil occurs at a relatively small strain, say less than 5%, the failure of the footing would be by general shear failure. If the slress-strain curve docs DOl show a peak and ia a rontinuously rising curve even uplO a slrain of 10 to 20%, local shear failure would occur in the footing. (3) Ir the relative density (D,) is greater than about 70%. gencml shear failure would occur. If it is less than 35%, local shear failure is more likely. (4) If the standard penetration test (SPl) value is more than 3D, the general sbear failure would ocx:ur. However, if it is less tban 5, the local shear failure is more likely. (5) If e is less tban 0.55, the general shear failure 0C0l1"S. If e is greater than 0.75, the local shear failure

occurs. 23.10. EFFECT OF WATER TABLE ON nEARING CAPACITY Eq. 23.25 for the ultimate bearing Olpneity has been developed based on the a<;Sumplion that the water table is located at a great depth. If the water table is located elose to the foundation, the bearing capacity cqu:lI.ion needs modification, as explained below. Case I Water tuble located above the base of foolIng [FIg. 23.11 (a)]

______u_YLl... __ (a)

(b)

Fig. 23.11.

The effective surcharge is reduced as the effective weight below the watet !able is equal to the submerged unit. 'Therefore,

q-D.. y+ay' where D .. = dcpthe of water table below the ground surface, a = height of water table above the base of footing. Altematively, Eq. 23.30 can be written as, substituting a - Dr D""

... (23.30)

q - y'V, + (y - y')V_ •.. (2331) Moreover, the unit weight in the third Icon of Eq. 23.25 is equal to the submerged unit weight. Thus Eq. 23.25 becomes

10

60'


q.. - c' N~ + [y' 0, ... (y - y') D ...l N q If D... - O(i.e.

(.I

-

...

0.5 'I' B N,

... (2332)

DI), .[2333(a)]

re

If a -

o(i.e. D, _

~,

D,.,), q.. - c'Hc

+.., DIN

q ...

O.5y'BN,

... [2333(b)]

Case U Water table k>cated at a depth b below brase [Flg. 1.1.11 (b)] IT the water table is located at the level of the base of footing or below it, the surcharge term is not affeded. However, the unit weight in the third term of Eq. 23.25 is modified as ne od

g.

'i -

y' +

~

(y - y')

... (23.34)

where b = depth of water table below the base, D = base width of the footing. Therefore,

q.. _ c'Ne + yDJNq + 0.58

When b - 0, i.e.

Wrr at the base,

When b - B, i.e.

wrr at depth D below the base.

['i' + ~ (y - 'i')] Ny

q.. - c:' He + yDfHq + 0.58y' Ny

er

ty

... [2335(a)]

... [23.35(b)]

q.. - c' Ne + 'I D,N., + 0.5 ByN, ... (same ~ Eq. 23.25) Heoce, when the ground water table is located at a depth b equal to or greater than B, there is no effect on the ultimate bearing capacity. General Expression Taking the submerged unit as roughly one-half of the bulk unit weight, the general equation for the ultimate bearing capacity can be written from Eqs. 23.32 and 23.35 (a) as q.. - c'N~ + yD,N'IW" + 05yB N,W, ... (2336(a)] where W'I is water table oorrection factor for tbe second term, W, - , - 0.5 aiD, < , ... [2336(b)] aod W, is water table CQCT'C(1ion factor for the third term, W, _ 0.5 + 0.5 biB < 1 •.. [23.36(c)] It may be noted that both tbe oorrections vary linearly.

13.\1. BEARING CAPACITY OF SQUARE AND CmCUiAR FOOTING Eq. 23.25 has been derived for a strip footing. The defonnations under a sttip footing are twodimensional. It is known as a plane strain case in theory of elasticity. On the other hand, the defonnations of soil under a square or a circular (ooting are three-dimensional. A rigorous analytical solutioo for a threedimensional case is extremely difficult. Based on experimental results. Terz.aghi gave the following equations for the ultimate bearing capacity for square and circular shallow footings.

0)

(a) Square Footing

q" - 1.2c'N~ + yD/N., + 0.4y BN, where B is the dimension of each side of footing.

.. (2331)

(b) Circular Footing

I) ,

q.

q" - 1.2c'N~ + yDIN., + 0.3yBN, where B is the diameter of the footing. The bearing capacity factors NnN., aDd N, are the same as that for the strip footing.

.. .(23.38)

SOIL MECHANICS AND FOUNDI\fION ENGINEERING

602

23.12. MEYERUOF'S BEARING CAPACITY TIlEORY Mcycrhof (1951) gave a general lheory of bearing capacity for a strip fOOling at any depth. Meyerhof considered the failure mechanism similar to [hat a'iSumcd by Thrnlghi, but extended the failure surfaces above foundation level. Thus the shearing strength of lhe soil above the footing base was also accounted for in the analysis. The curved rupture surfaces in the zone of radial shear were assumed to be logarithmic spirals. Meyertx>C's rupture surfaces are more general than those ~ed by Tcrzaghi.

G

c

p Terzaghi's lailure surfaces

--L

Meyerhof's surfaces

failure

Fig. 23.12. Meyerhof's Analysis.

The right-halIoe Fig. 23.12 shows the failure surfaces assumed by Meyerhof. The zone ABC is the elaslic zone but the angle which the inclined surfaces AC and BC make with horizontal was varied between .' and (45° + ,'/2). The zone BCD is the zooe of radial shear. 1be zooe BDEF is the zone of mixed shear in which shear varies between radial shear and plane shear. 1be surface BE is known as equivalent free surface. It makes an angle p with the horizontal. The resultant effect of the wedge BEF of soil is represented by the normal stress (q,,) and the shear stress ("t,,) on the surface BE. TIle angle p increases with an increase in depth Df and is equal to 9(}0 for deep foundations. The parameters j}, q" and 'to arc known as foundation depth p.1rameters. Meyeroof gave the following equation for the ultimate bearing capacity of strip footings. q" - c' N~ + qoNq + OSyBNT ... (2339) where N~. N q and NT are the general bearing capacity (actors of MeyerhoC's theory. These factors depend upon the roughness of base, depth of footing and the Shape of footing, .in addition to the angle of shearing resistance ,'. Meyerhof also gave charts for Ne>Nq and NT (Fig. 23.13) for shallow strip footings. As the equivalent free surface cannot be directly located. the normal stress q" is determined by a semi-graphical method, wbich is quite cumbersome. However, for shallow footings. q~ - y Df The ultimate bearing capacity given by Meycrhof's theory is close to the experimental values. Fa' shallow footinw>. the value Lies in-between lbe general shear value and the local shear value of 1Craghi's analysis. However, for deep footin~, Meyerhof's analysis gives values much greater than Terzaghi's analysis. The main advantage of Meyemof's tbeory is that it can also be used for deep foundations and for footings 00 slopes. . (a) Cohesionless soils For oohesionless soils (c = 0), the ultimate bearing capacity is given by a simple equation suggested by Meyerbof as

q. - 0.5 Y B N"

... (23.40)

where the parameter N yq is known as the resuitlJllt beorillg copacity factor. It depends upon the coeffICient of earth pressure (K) within the failure zone, the D';B ratio. the angle of shearing resistance " and the angle of friction b between the vertical shaft of the foundation and the sUlTOunding soil. The value of K depends upoo

603

BEARINO CAPACITY OF SHALlDW FOUNDATIONS

/

11/~ '/

///

.,

7

.• ,

/'/

~

~~

..-If

/

/'

(Aftfr

/

/'

,/ "

V

"

,,'

/ "

,,'

,

"

1'-

3~

~rho')

.,

.'" '"

Fig. 23.13. Mcycmof's Chart.

a number of factors, such as the characteristics of the soil. the stress history of the ground, the method of construction and physical characteristics of the foundation. The value of K usually lics between the active and passive earth pressure coefficients. The value of Nl'l is obtained from tbe charts for the given values of K, " and DIB ratio, assuming the base and shaft as rough. For smooth surfaces, the values of N'N are reduced to half the values. (b)~hesln

sol19 For robesive soils (+ = 0) Meyerhof's simplified equations is q.. • ' e N"l :.. y Df ... (23.41) where N"l is the bearing capacity factor depending upon the DflB ratio of the footing aod on the adhesion 00 the sides of the footing. For example, for a DIB ratio of 2.0, N~ has a maximum value of 8.30 when the adhesion is zero, and a maximum value of 8.8 when the adhesion is equal to the cohesion (e) of the soil


lbe abovc-mentioned equations are for the ultimate bearing capncily oC strip footings. The cqu.'ltions can also be used for rectangular, square and cirrular footings in conjunction with the empirical shape factors given by Mcycrhof. The values of NlfJ and N(q mny be obtained from lhe original paper. 23.13. HANSEN'S llEAIUNG CAPACITY THEORY It has been shown by Milovic (1965) that lCrzaghi's theory gives the values of the ultimate bcnring capacity higher than experimental values for cohesive soils and is unsafe. However, for cohcsionlcss soils, the theory gives conservative values of the bearing cnpncilY. For cohesive soils, Hansen (1961) gives the values of ultimate bearing capacity which arc in better agreement with experimental values. According to Hansen, the ullimate bearing capacity is given by

qlJ ..

C

N~ St de

it + q N q "., dq i., +

0.5 Y B N, ",

d,

~

... (23.42)

where N(.Nq and NT arc Ilansen's bearing Olpacity foclors (Table 23.2). and q is the effective surcharge at tbe base level ( = y' DI), and Sa Sq

and

are shape factors (lbblc 23.3).

s,

dr. d q and dy arc depth factors (fable 23.4).

in iq and i, are inclination factor (fable 23.5).

and

lbe bearing cap.'lCity factors are given by the following equations. Nfl"

um 2 (45'" +

41') (~Ian .')

.. .[23.43(a)]

... [23.43(b)]

Nt - (Nq - 1) COl " Ny" l.80(Nq -l) (nn+'

and

... [23.43(c)J

1hble 23.2. Hunsen's Dearing CapacUy Factors

.'

O·

5·

'D·

W

20·

25·

30·

35·

40·

N,

5.14

6.48

8.34

10.97

14.83

20.72

30.14

46.13

75.32

133.89

266.89

N, Ny

1.0

1.57

2.47

3.94

6.40

10.66

18.40

33.29

64.18

318.96

0.0

0.'"

0.'"

1.42

354

8.11

18.08

40.69

95.41

134.85 240.85

45·

Tobie 23.3. Shape: Factors Shope

0/ FOOlillg 1.0

1.0

"

1.0

1 + 0.2B/L

1 + 0.2 BtL

l-O.4D/L

SqUDrc Footing

1.3

1.2

0.8

Circular footing

1.3

1.2

0.6

Continuous FOOling (Slrip) RcclDngular Footing

where L = length of footing; B = width of footing.

Thble 23.4. Depth Facton 1 + 0.35 (DfllJ) 1 + 0.35 (DfIO) 1.00

where D, = depth of fooling; D [Note. Thke t1q ..

dr

for 4"

:>

= width of diameter 25'" and d'l .. 1.0 for 4" .. 0'"]

SO·

681.84

IG

605


Table 23.5. IncUnation Factors 1 - Z:BL 1 _ lotH

og

he

(;,,)'

H = horizontal component of inclined load. H:sVtan6+cBL

where

Umltation:

tan 6

where

= coefficient of friction

= length of fOOling paranel to II.

23.14. VESIC'S BEARING CAPACfIY THEORY Vesic (1973) oonfinned that the basic nature of failure surfaces in soil a<;. suggested by Terzaghi and as given in Fig. 23.6 is correct However, the anglc which the inclined surfaces AC and Be mak.e with the horizontal was found to be closer to (45 0 + 41'/2) instead of ~'. The values of bearing capacity fadors Nc,Nq and Ny for a given anglc of shearing resistance 41' change if above modification is incorporated in the analYSiS,

as under Nq

,)] »]

,)]

between footing and soil,

c = cohesion of soil between footing and soil, L

_

tan 2 (45° + q,'/2)~tan,'

... [23.44(a)]

N, - (N, - t) cot.'

... [23.44(b)]

Nl .. 2(Nq + 1) tan,'

and

... [23.44(c)] Eq. 23.44(b) was first proposed by PrandU (1921), and Eq. 23.44 (a) was given by Reissner (1924). Caquot and Kerisel (1953) and Vesic (1973) gave Eq. 23.44 (c). The values of bearing capacity factors are given in Table 23.6. The bearing capacity equation is similar in fonn to Hansen's equation, ... (23.45) qu - c' Ne Se de ic + qNq Sq d q iq +0.5 'tBNT ST d, iy where sc. Sq and ST are shape factors (fable 23.7). dc, d q and d, are depth factors, if, iq and iT are inclination factors. and As before, q is effective surcharge at the base level ( = l' Df ).

'Thble 23.6. Veslc's Bearing Capacity Factors

I

"I

N.

Nq NT

0' '.14 1.0 0.0

I

L 10'

I" I 6.49 1.57 0.45

8.35 2.47 1.22

t5' 10.98 3.94

2.65

""

14.83 6.40 5.39

25'

30'

35'

40'

4"

SO

20.72 10.66 10.88

30.14 18.40 22.40

46.12 33.30 48.03

75.31 64.20 109.41

133.88 1:>4.88 271.76

266.89 319.Q7 762.89

Table 23.7. Shape FactorS Shape of Footing

Strip

Reccangle Circle and square

10 1 + (BIL) (NqlNe) 1 + (NqINc)

"

10

'T

1 + (BIL) tan,'

10 1 - 0.4 (BIL)

1 + tan,'

0.60

The depth factors as proposed by Hansen (1970) are used for DIB :s: 1 as follows.

d. - 1 • 0.4 (DIB)

... [23.46(a)J


606

d, _ 1 + 2 tan,' (1 - sin ,')' (Dr'B)

... [23.46(b)]

d, - 1.0

... [23.46(0)] In Vesic's equation, the following inclination factors proposed by Meyerhof (1963) and Hanna and Meyerhof (1981) are generally used. i~ _ iq _ {I _ aO/900)'2 ..• [23.47(a)] and

~ _ (1 -

a"/+1'

... [23.47(b)]

where aO is the inclination of the load with vertical. 23.15. IS CODE M.O"TIJOD IS : 6403---1981 gives the equation for the net ultimate bearing capacity. which is similar to one proposed by Yesic (Eq. 23.45). q" .. - c Nt

St

dt it + q(Nq-l) Sq dq iq + O.5B yN,

51

d.y i., W'

... (23.48)

where q = effective pressure al the base. The seoond term has been changed, bccal.lSe qll" is given by (See Eq. 23.1)

q".-q.-yD,-q.-q The factor W' takes into accoWlt the effect of the water table. If the water table is at or below a depth of (Df + B), measured from the ground surface. W' '" 1.0. If the water table is lik.ely to rise to the base of the footing or above. the value of W' is taken as 0.50. If the water wble is localed at a depth D below the ground surfaoc, such that Df < D < (Df + B), the

value of W' is obtained by linear inteflXllation. A little re[lection would show that W' is the same as the factor Wy introduced in Sect. 23.10. The factor Wq is indirectly accounted for by taking q as the effective surcharge in Eq. 23.48. The bc.... ring capacity factors N~, Nq and NT are the same as those given by Vesie (rable 23.6). lbc shape factors given in Thble 233 arc used. The depth factors arc given below.

and

d, _ 1 + 0.2 (Dr'B) tan (45" + ,'/2)

... [23.49(a)]

dq -t4-1.0

... [23.49(d)]

d, _

d, -

for,'
1 + 0.1 (Dr'B)

tan (45' + ,'/2) fIX +' > 10"

... [23.49(0)]

'llie inclination factors given by Eq. 23.47 are used. Local shear failure The net ultimate bearing capacity for local shear failure is given by

q,... ..

~ c Nt' SC dcic + q (Nq '

-

l)sqd'qiq + 0.5 't B Ny' Syd,

~ ,W

.. ,(23.50)

The value of the local bearing capacity factors N/, Nq', Ny' are obtained from Table 23.6, for the angle of mobilised friction as

'.- tan-

t

(t tan+')

In case of oohesionless soils, if the relative density is greater than 70% and the void ratio is less than 0.55, the failure is considered a<; general shear failure. On the other hDnd, if the relative density is smaller than 20% and the void ratio is greater than 0.75, the failure is local shear failure (or punching failure) and Eq. 23.50 is used. For a relative density between 20% and 70% and a void ratio between 0.55 and 0.75, the bearing capacity factom are obtained by interpolation between the general shear failure and the local shear, m explained below. For a relative density between 20 to 70% (or 0.55 < e < 0.75). the value of the nct ultimate bearing


607

capacity (q.... ) can be interpolated between the general shear failure and the local shear failure conditions, depending upon the relative density. For illustration, Ict us consKler thc case when the relative density is 40% and .' is equal to 30°. The bearing capacity factors for general shear from Table 23.6 arc as under.

Nc - 30.14; Nq - IB.4O; NT - 22.40 • 21 D. The corresponding bearing capacity factln for the local shear from

Now .... _ lan- I (213 tan JO O ) Table 23.6 are

N/ - 16.01, Nq' - 7.25, NT' - 6.49. Let us find the net ultimate bearing capacity. Let us assume c' _ 0, 2 B - 2m.sq - sl - dq - ~ - 1.0,"1 - 20 kNl m and Df - 1 m, W' - 1.0. For general shear failure (Dr

:to

70%) (Eq. 23.48)

q"" - 0 + 20 X I (lB.4 - 1) + 0.5 For local shear failure (D, < 20%) ( Eq. 23.50) q,..: - 0 + 20

(7.25 - 1) + 0.5

X

X

X

20

X

2

X

22.40 • 796 kN/m 2

20 x 2 x 6.49 _ 254.B kN/m 2

For Dr = 40%, by linear interpolation,

q" .. _ 254.B +

~

(796 - 254.B) _ 471.28 kN/mz

23.16. SKEMVrON'S ANALYSIS FOR COHESIVE SOIlS Skempton (1951) showed that the bearing capacity factor Ne in Ter-.t.aghi's equation tends to increase with depth for a cohesive soil (+.. - 0, c - c..). Fig. 23.14 shows the variation of N~ with DrB ratio for Slrip and ciraJlar (or square) footings. For a Slrip footing. lhe value of Ne is equal to 5.14 for the surface footing and has a maximum value of 7.50 for Dj'B ratio Oil: 4.50. For square and circular footings, the value of N~ is equal to 6.2 for lhe surface footing. The maximum value of about 9.0 is attained for DrB ratio equal to or greater than 4.50. The UJrve for square and circular footings can also be used for rectangular footings using the following relaLion. N, (rectangle) = N, (square) [0.84 + 0.16 BIL] AltemaLivcly. the UJrvc for the strip can be used, making use of the follOWing relation.

... (23.51)

N, (rectangle) = N, (strip) [1 + 0.2 (BIL)] .. . (23.52) The following approximate relations can be used for thc determination of N~ for different DIB ratios. (a) D/O < 2.50 N, - 5.0(1 + 0.2D/O)(1 + 0.20IL)

... [23.53(a)]

Nc - 7.50(1 ... 0.2BIL)

... [23.53(b)]

(b) D/O. 2.50

Ultimate Bearing capacity For ell.. - 0,

Nq

-

1.0 and Nl - 0.0

Therefore. Eq. 23.25 gives q.. - c.. Nc .. "1 D, The net ultimate bearing capacity becomes

q_ - c.N,

... (2354)

... (2355) Eq. 23.55 is used for the detenninaLion of the net ultimate bearing capacity of footings on cohesive soils, taking N~ values given by Skempton (Fig. 23.14). It may be mentioned that Tcrzaghi's value of N~ is applicable only for shallow footings (D, < B), whcreas Skem(Xons's value can be used for all values of D,IB ratio.


608

",--,---,-----,---,-----,--,

N,

,!'-/--+----1--+--+--+--I

00

~Ratlo-Fig. 23.14. S1ccmplOn's Chart. ±

If the shear strength of the soil for a depth of 2B{3 beneath the footing docs nol. vary by more than about 50% of the average value, the value of ell in Eq. 23.55 may be taken as the average value.

23.17. IS CODE METIJOD FOR CODESIVE SOIlS 15 : 6403-1981 gives tbe following equation for the net ultimate bearing capacity of the footing immediately after construction on a cohesive soil. 1be equation is obtained by substituting N'I = l.0 and NT = 0.0 in Eq. 23.48. Thus ... (23.56) where He = 5.14 and Sco de and ic are, respectively, the shape. depth and inclination factors (see Sec. 23.15). The value of e" is obtained from unconfmed compression strength lest or it can be derived from the slatic oone lest. The stalic cone Icst gives the point resistance (qc) as explained in chapter 17. For nonnally consolidated clays, tbe point resistance qc is generally less than 2000 kN/m2 and the value of the undrained cohesion e .. varies between q/18 10 q;15. For over-consolidaled clays, the point resistance is generally 2 greater than 2000 kN/m , and the vaJue of e" varies between qJ26 and qJ22.

23.18. HEAVE OF mE BOTTOM OF mE CUT IN CLAY The lateral earth pressure acting on a braced cut h$ been discussed in Chapter 21. A braced cut may also become unstable as a result of the heaving of the bottom of the excavation (Fig. 23.15). Terzaghi (1943) has analysed the braced cut in clay and has given an expression for the factor of safety against bottom heave assuming the failure surfaces as shown in Fig. 23.16. The vertical load per unit length of the OJt at the level BD or AF aJoog the bottom of the cut is given by

no"""""""",,,,,,,,.--

Q-B11Jl-eH

... (a)

The above load acts 00 a width Bl equal to 0.7B, which can be roosidtnd as a

C()Iltinuous foundation. B is the width of !he cut.

Fig. 23.15. Heave in a Braced ClIl'

BEARING CAPACrIY Of SHALLOW FOUNDATIONS

fo°-7B G

609

8

I·

,

I

I

I

I

:(

Ct!

I

-~

I

I

H

I I

it

etl

C

I

,I

F

'D

I

0·78

-.l

Fig. 23.16.

The ultimate bearing capacity of the continuous foundation can be obtained from Eq. 23.25

~

q.. " cNf +- N'I1Df Taking Ne .. 5.7, and Nq .. 1.0,

q,," 5.7c +- yD, Therefore, I)

qfW .. 5.7c

The net ultimate load is given by The factor of safety is given by

Qnu" 5.7 cBl

Q".. 5.7cBJ F.S·-Q-BIyH_cH

ES._1-(~) H y - C/BI

... (2357)

Bjenum and Eide (1956) proposed the following equation for the factor of safety. eN,

F.S.

- 1/i

... [23.58(0»)

If there is a surcharge q, the factor of safely becomes

a

~1 •'v::;:;: :...-9

1B

sqvor~~

./

N,

'V s

1

~

~

,

tao

(Strip)

1

l! _ _ B

Fig. 23.17. Skemplon's ChIu't ror Braced Cut.

f--

,

S


610

ES. - ,

Ie;': q

... [23.58(b)J

The bearing capacity fador Nc is obtained from Skcmpton's chart (Ag. 23.11). The value of cis equal 10 the undrainoo cohesion c" which be determined as explained before. 23.19. FOUNDA110N ON lAYERED CLAY Reddy and Srinivasan (1967) derived the following equation for the bearing capacity of foundations 00 clayey soiis in two layers. q.. - CINcscde ic + q ...(2359) where CI = undrained shear strength of the uppcr·laycr. Sc = shape factor [ = 1 + (DII.) N,/NcJ de = depth factor [ = I + 0.4 (VIIl) Cor DID:s: I], ic = inclination factor. q:: surcharge (= ., Of). l11e bearing c.'lp.'lciIY factor Nr depends upon the mlio cZ/Ch where Cz is the undrained shear strength of the lower Inyc:', It also depends upon the ralio ZIB, where Z is the depth of the interface of the two layers from the bottom of foundation and B is the width of the foundation IFig. 23.18 (a)]. Fig. 23.18 (b) showS the

"r-- - - - , - - - - - =71

~I

T

t------B - - - - I

1

Nc

Up~ loy~r

~;o

Lowt'r lo~t'r

"to

4

'1

0.40

.EL0·901.01.2 _ _

'.6

1.0

c, '0)

!O)

Fig. 23.18. Layered ColK:sive DeposiL variation of Nc with C2/CI r
23.20. nEARJNG CAI'Acny FROM STANDARD I'ENETRATION TEST Method I. Ihe ultimate bC
G

BEARING CAPACITY OF SlIALLDW FOUNDATIONS

2

)J

611

q"" - ta{3N BWy +

5~100

2 + N )Df Wq ]

q".. _ 0.5N2 mVy + 0.83(100 + N 2)Df Wq where q"" = net ultimate bearing cap:lcity (kN/m1. 8 = widtb of footing. N = averngc SPT number, Df = depth of footing. If Df > 0, usc Df _ B. and WT and Wq are water table correction factor ($c(,1. 23.10). For square or circular footings, 01'

')

... (23.60)

q,.,. - io[N2 BWy + 3(100 + N2)Df Wq J q_ _ O.33N2nWy + 1.0(100 + N2 )Df Wq

or

... (23.61)

The nct allowable be.1ring capacity can be obtainccl by applying a fador of safety of 3.0. For strip footings,

'f

qns .. 0.167 N 2 BWy + 0,277 (100 +

N1 Dr Wq

kN/m 2

... [23.62(a)J

ror circular and square footings. q/U ..

0.l1N2 BlVy + 0.33(100 + N 2)Df Wq kN/m2

... [23.62(b))

13.21. ECCENTRICALLY LOADED fOUNDATIONS Foundations are sometimes sUbjeclcd to moments in addilion 10 the loads (Fig. 23.19). lbe distribution of fOOling pressure is not unifonn in this casco It is a case of bending COOIbincd with thrust, treated in the mechanics of materials. The maximum and minimum pressures are given by

q(l1a.~ and

..

qll1in '"

~

+

~-

¥

(BI2)

o/-(BI2)

=

=

where I moment of inertia ( :: Lff/12), Q lotal vertical load (gross), M = moment on the foundation, B = width of footing, L = length of fooling. Taking the eccemricity e as MIQ, Ihe above equations become

q••, and

q.o -

Ilt (I + 6c18) Ilt(1 - 6018)

... [23.63(a)J ... [23.63(b)J

The maximum pressure qllWt should be less than the safe gn:x;s bearing capacity. Mrytrbof's Method The factor of safety of eccentrically looded foundations against bearing capacity failure can be determined using the method given by Meyemof (1953), as explained below

as.,

(1) Determine the eccentricity of the load, along the widlh e'b" Mt/P

Fig. 23.19. EttentriCl'lty loaded footing.


6'2

It may by noted that the ecccrrtricity is measured from the centre of fooling. (2) Determine the effw ive width of the footing. 8' - B-2

e".

(3) Determine the effective size of the fOOling as L xB'. In the above case, the eccentricity has been assumed only along width. However. if the ea:cntricity is

also in tbe longitudinal direction along length, the eooentricity along the length is given by The effcctivc length of the footing in that case is L I _ L - 2 el and the effective size is L' x fl'. The smaller of the two dimensions B' and L' is taken as the" effective width for tbe computation of tbe ultimate bearing olpacity. (4) The ultimate bearing capacity can be obtained using Eq. 23.42 as

qll - cN.,$cd~i~ + qNqsqdqiq + O.5yB'NTs.,~i, lbc shape factors S~I Sq and s,. 3rc obtained from Table 23.7. The inclination factors arc obtained using Eq. 23.47, in which the effective width ll' and effective length L' are used. For ~pulations of depth factors, Eqs. 23.46 (a) to 23.46(c) are used, but 8' is not substituted for B. (5) The total ultimate load is computed as

Q• • q. (B' xL')

... (23.64)

(6) The factor of safety is given by

F, • Q./Q

... (23.65)

As the eccentricity causes a reduction in the load-carrying capacity of a foundation, the colum n is sometimes placed offcentre of the footing to prooucc a uniform pressure distribution (Fig. 23.20). If the rolumn is not placed oIT-<:entre. there is a possibility of tilting of the footing due to higher pressure 00 one side of the footing. This tendency of tilting can be reduced by aUO\Ving a larger faaor of safCly. If the ccccntricity is outside thc middle third, the minimum soil pressure (Eq. 23.63 b). becomes negative, indicating that [he tensile stresses develop in soil. As the soil cannot take tcnsion, there is a separation between the footing and the soil. 1be area of the footmg wbich is in tension is generally neglected. The maximum pressure in such a atse is given by

qm~' 3L(B4~2eb) where

eb

(M

--f'bl-I I

I

If r r r r r r r Fig. 23.20.

.•• (23.66[

is the eccentricity along width of footing.

23.22. SETTLEMENT OF FOUNDATION (a) Settlement under loads Foundation settlement under loads can be cl<15Sified into 3 types. (1) Immediate or elastic seillemeni (s;).. Immediate or elastic settlement takes place during (I' immediately afier the construction of the structure. It is also known as the distortion settlement is due to distortions (and not the volume change) within the foundation soil. AUoough the settlemea is not truly elastic; it is romputed using elastic theory. especially for cohesive soils. (2) Consolidation settlement (s~) .. This component of the settlement ~ due to gradual expulsioo d. water from the voids of the soil. This component is determined using Terzaghi's theory d consolidation (sec chapler 12).

as.

BEARiNG CAPACllY OF SHALLOW FOUNDATIONS

613

INO

yis

(3) Secondary Consolidation Settlement (s,..). This component of the settlement is due to secondary consolidation. This settlement occurs after oomplction of the primary consolidation. It can be determined from the roefficienl of serondary consolidation (see dlapter 12). The secondary consolidation is not significant for inorganic clays and silty soils. The total settlement (s) is given by S_S;+S~+S~

lbe

... (23.67)

(6) Settlement due to other causes In addition to settlement under loads, the settlement may also occur due to a number of other causes. (1) Underground erosion. Underground erosion may cause fonnalion of cavities in the subsoil which

when collapse: cause settlement.

Eq.

occurs.

9.

(4) Frost heave. Frost heave occurs if the structure is not founded below the depth of frost penetration. When thaw occurs, the foundation may settle. (5) Vibration and Shocks. Vibrations and shock cause large settlements, especially in loose, cohesionlcss soils. (6) Mining subsidence. Subsidence of ground may occur due to removal of minerals and other materials from mines below. (7) Land slides. If land slides occur on unstable slopes. there may be serious settlement problems. (8) Creep. The settlement may also occur due to aeep on clay Slopes. (9) Changes In die "klnity. If there are changes due to construction of a new building near the existing foundation, the settlement may occur due to ina'ease in the stresses. Suitable measures are taken to reduce the settlements due to all above causes.

'ING

1.66)

!

(2) Structural collapse of soU. Structural collapse of some soils, such as saline, non-cobcsive soils, gypsum, sillS and clays and loess, may occur due to dissolution of materials responsible for intergranular bond of grains. (3) lllermal changes. Thmpc:rature change cause shrinkage in expansive soils due to which seniemenl

IX

as it nent

"or

of

13.2J. LOADS FOR SETTLEMENT ANALYSIS Dead loads include the weight of columns, walls. footings, foundations, the overlying fill, but do not iocllKle the weight of the displaced soil. live loads depend upon the use of the structure. These loads may be taken from IS : 875. Wind loads and seismic loads should be considered wberever applicable. However, where wind (or seismic) load is less than 25% of the combined dead and live loads, it may be neglected in design, and only dead load and live loads are considered. When wind (or seismic) load is more than 25% of the combined dead and live load, the foundation is designed such that pressure due to combination of dead, live and wind (seismic) loads docs not excx:ed the allowable bearing capacity by more than 25%. For foundations resting on coarse-grained soils, the settlements should be estimQted corresponding to the full dead load, live load and wind (seismic) load. In such soils, settlements occur in a sbort time. For foundations on fine-grained SOils, the settlements are estimated corresponding to permanent loads. All dead loads and the loads due to fixed equipment are taken a'i permanent. Generally, one half of the live load is also taken $ permanent. Engineering judgment is required to ascertain the pennanent loads. 1324. IMMEDIATE SETILEMENT OF COHFSIVE SOILS The linear theory of elasticity is used to determine the elastic settlement of the footings on saturated clay. Schleidler (1926) gave the follOWing fannula for the vertical settlement under a uniformly distributed flexible ... (23.68)


614

where q = uniformly distributed load, B = characteristic length of the loaded area, Es = modulus of elasticity of the soil. fA. = Poisson's ratio ( = 0.50 for saturated clay). I = influence foctor.

The value of E. is determined from the stress-strain curve obtained from a triaxial consolidated-undrained test, with the consolidation pressure. equal to the effcctive pressure at the depth from which the sample Wal taken, as dL~ussed in chapter II, It is gencraUy taken as the initial tangent modulus or the secant modulus. For normally consolidated clays, its value varies from 250 c to 500 C, and for over-consolidated clays, from 750 c to WOO c, where c is undrained cohesion. 1:1C value of the influence factor I for a saturated day layer of scr::Ji-infinitc extent can be obtained from Tabl~ 23.8. Thble 23.8. Values of Influence Fador I. Shape

Circle

Sql.Bre Rcctagle LIB:: 1.5 LIB = 2.0 LIB = 2.0 L/B =: 5.0 LIB =: 10.0 LIB =: 100.0

I-

Flexible footillg Rigid fOOling

Cel/lre

Comer

Average

1.0 1.12

0.64 (edge) 0.56

0.85 0.95

1.36

0.68

1.20

1.06

1.53 1.78 2.10 2.52 :U8

0.77 0.89 1.05 1.26 1..69

1.31 1.52 1.83 2.2'5 2.96

1.20 l.42 1.70 2.10 3.40

0.79 0.82

Alternatively, the value of (1 - 1l1/IEr can be determined from the plate load test (Sect. 23.33).

If an incompressible 1..1yer exists at a limited depth bclow the footing, the actual settlement is less Ul3J] that given by Eq. 23.68. For such a case, Steinbrenner (1936) gave a solution. However, if the depth of the cl..'ly layer is more than 2 B. the actual settlement would not change much. If the foundation is rigid, such as a heavy beam and slab raft, the sctUement is about 0.8 times the settlement at the centre of the corresponding flexible fOWldntion. It is approximately equal to the averag~ settlem.-:nt. Table 23.8 also gives the values of / for rigid footings. Eq. 23.68 is applicable for the footing located at surface. For the footings embedded in soil, the settlement would be less than the computed values. Fox (1948) gave corrcction curves. The settlement is obtained by multiplying the computed settlements by a depth factor, which depends upon (DII~) ratio.

23.25. IMMEDIATE SE'rrLEMENT OF COIIESIONLESS SOILS As cohesionicss soils do not follow Hooke's law, immediate settlements are computed using I semi-empirical approach proposed by Schmenmann and Hartman (1978). ,. I s; = C 1 C2

(q - q)l~O t i l l

... (23.69)

where C 1 = correction factor for the depth of foundation embedment = 1 - 0.5 {ql@ - q)}. C 2 = corrcction factor for creep i!l soils [ = I + 0.2 loglo (Lime in yearslO.l)J. q -= pressure at the level of the found'l1ion, q = surCharge ( = 'Y D,), E. = modulus of elasticity, lz = stmin innuence factor. 'nlC value of the strain-influence fact.or I z varies linearly for a square or eircular foundation (Pig. 23.21~ The value of lz at depth z = 0, 0.5 Band 2IJ are respectively equal to 0.1, 0.5 and 0.0. For rcctangull

INO

IlF..ARING CAPACITY OF SHALLOW FOUNDA'nONS

615

city

ined was llus. 'rom

'rom

Ulao f the

• the ~rage

, the nt is

• B)

og a

3.69)

3.21). Igular

Fig. 23.21.

foundations, with LiD ratio equal to or greater than to.O, the values al depth Z = 0.0, B, and 48 are, respectively, 0.2, 0.5 and 0.0. For intermediate values of LIB ratio, between 1.0 and 10.0, interpolation can be made. The value of Ez can be determined from the standard penetration number .(N) using the following equations given by Schmertmann (1970). E, _ 766N (kN/ m')

... (23.70)

Alternatively, it can be estimated from the static cone 'penetration resistance (qc) as ... (23.71)

Procedure. For computation of the immediate setllement, the soil layer is divided. inlo several1ayers of thickness Az, uplO a depth z = 2B, in case of square footings and z'" 48, in ~ of rectangular footings. The immediate settlement of each layer is computed using Eq. 23.69, taking oorresponding values of EJ and J: . The required immediate settlement is equal to the sum of the settlements of all individual small layers.

23.26. CONSOUDATION SETfLEMENT IN ClAYS The consolidation settlement (s~) OCOJrs in saturated, Clayey soils when these are subjected to increased loads caused by the foundation pressure. The methods for computation of the consolidation settlement have been discussed in chapter 12. The settlement (s,) at any time t after the application of load may be estimated using Terzaghi's consolidation theory as Sf"

where U Sf

s.

s,· + U Sf

... (23.72)

= degree of consolidation, expressed as a ratio, = final consolidation settlement, = immediate settlement

The degree of consolidation (U) depends upon the time factor Tv, given by Tv - c" tiff .

. .. (23.73)


616

where c ~ = average coefficient of consolidation, t = time at which the settlement is required, d = distance of the drainage path [N&te. When considering the drainage patb, concrete of foundation is assumed as penneabJel For the final settlement, U = 1.0, and the total settlement is given by S -

Sj

+

...(23.74)

Sf

23.27. SETILEMENT OF FOUNDATION ON COllESIONLESS sons Settlements of foundations on cohesionless soils take place rather quickly after the application of the lood. The immediate settlements calculated using Schmcrtmann and Hartman method (Sect. 23.25) would also be the final sclllement in mast cases. Dccausc of the diffiOJlty of No ~'" sampling cohesionless soils, it is not __ ----~

Hj3m~~~~--~~~~~~~~~~~~~

characteristics of the insilu soils. The possible to obtain the stress-strain

sctllcmcnts arc genemlly determined

indirectly using the scmiempirical methods. (1) Stutic cone penetmtion'1 method. In Ihis method, the sand lnyer zis divided into small layers such that :!

_

-

=ta~~~~lu~ay:( lh~asco~r:r:;:~~ ~ The avemge value of the cone res~tana: E of each small layer is delennined.

n

N. 20 ~ --::::: 1_ _...'N:!o'£2"'----t_I.

is cslimatcd using the following The settlement of each smaU1957). layer I 10 equation (Dc Beer and Martens,

§

H

S -

C

00+60

log" ~o-o-

... (23.75)

where C - I.Sqc/~ in which qc = slatic cone resistaoce, 00 = mean effective overburden pressure, 6. 0 = increase in pressure al the centre of Ihe layer due to the net foundation pressure and f/ = Ihickness of layer. The lotal settlement of the entire layer is equal 10 the sum of settlements of individual layers.

~

N: l0

......-

"" . 30

___

== mt~~~§~~i~~~~~~~~§ N:t.O

N .. so

N~60

~

WIdth

a

or

lool,ng

Im)---'"

Fig. 23.22.

(2) Standurd I'enetratioll Test. Standard penetralion test can be used for the determination of the scttlement on cohesionlcss soils. IS : 8CX)9 (patt I}-1976 gives a chart for the calculation of settlement per unit pressure as a function of the width of the fOOling and the standard penetration number (Fig. 23.22). The setttemeni under any other pressure is computed assuming that the settlement is proportional to the intensity of pressure. The selllemenlS arc in metres per unit pressure in kN/m 2• If the water table is at a shaUow depth. the settlements are divided by the correction factor Wy [Eq. 23.36

(c». (3) J)lale l...ood Test The selliemcni of the footing can be estimated from the settlement of the plate in the plate load Icst (Sec. 23.33).


617

23.28. ACCURACY OF FOUNDATION SE'lTLEMENT PREDICI10N The prediction of the foundation settlements as discussed above is rnrely accurate. In [ad, it is e xtremely difficult 10 estimate the probable sctUcmcnt of a footing bccnuse of the following reasons: (1) ' 111C soil deposits are seldom isotropic and linearly e1.1s1ic. TIm deposits arc generally non-

.)

homogeneous. (2) It is not possible 10 estimate the increase in stresses caused by loads. The Boussincsq solution gives only approximate results. (3) r-or estimation of the settlement due to consolidation. it is 00( possible to locale exactly the drainage faces.

:0

(4) ror computanon of immediate settlements, it is not possible to estimate the correct value of the modulus of elasticity (5) The rigidity of the foundation is usually neglected and the pressure distribution is assumed to be

\0

be

'" be ity

36

unifonn. (6) 11 is difficult 10 obtain undisturbed samples of cohcsionlcss soils. 'The scmi..cmpirical methods do not give accurate results. (l) ScUlcments mny occur due to causes other than that due to loads. It is nol possible to estimate these settlements accurately. Despite aU the above fC.'l<;()ns. the settlements in mast cascs can be eslimatcd to an accuracy of about 25 30%. which is good enough sccing the complexity of the problem.

23.29. ALLOWABLE SE:\'fLEMENT If there is a 1.1rgc differential settlement between various parts of a structure, damage may occur due 10 additional moments developed. It is more difficult to predict differential settlement than to predict the maximum sctl1ement. TIle difIerential settlements are generally obtained indircctly from the maximum settlements. II has been found from actual observations of various existing buildings tltal the differential settlements seldom exceed 75% of the maximum settlement. In fact, in most casc.... lhc differential settlement is less than 50% of the maximum settlement. Hence, the difIerential settlement is 3ulom3lically controlled if the maximum settlement is controlled. The allowable maximum settlement depends upon the type of soil. the type of foundation and the structural framing system. The maximum setllement ranging from 20 mm to 300 mm is generally pennitted for various structures. Thcorcticnlly, no damage is done to the superstructure if the soil settles uniformly. However. settlements exceeding 150 mm may C.1USC trouble in utilities such as water pipe lines, sewers, telephone lines and also in access from streets. If sufficient precautions arc taken while designing such utilities, the maximum settlements of even uplO 2 m may be permitted in special cnscs, provided the structure setlles unifonnly. IS : 1904 (1966) permits a maximum settlement of 40 mm for isolated foondations on sand and 65 mm for those on Clay. The allowable settlement is higher for clays because progressive settlements on Clayey soils permit better strain adjustments in the structural members. ·rne maximum pennissible settlement for raft foundations on sand at\: 40 mm to 65 mm and that on clay, 65 10 '100 mm. lbe permissible settlements for rafts arc more than those for isoL.'lted foundations tx:causc the raft bridges over son patches of the soil, and the differential settlements arc reduced. If the differential settlement between two columns spaced at a distance L is 6, the angular distortion or til T is given by l' - IVL ... (23.76) Common R.C.C. buildings can sustain angu1.1r (Jistortion upto 1/150 without damage. However, to avoid the architcctural damage, it shoukl noc exceed 1/300. It corresponds to a differential settlement of 20 mm tx:twecn adjacent columns 6 m centres. According to general praaicc, the maximum differential settlement is limited to 25 mm in sandy soils and 40 mm in C1.1YCY soils. IS : 1904---1978 gives Ihe safe values of the maximum and differential settlements of different types of


618

Table 23.9. Maximum Bnd Differential Settlements (IS : 1904---1978) Plastic cloy

Sand and liard cia)'

Mar-

Diff-

Angular

Mar-

Diff-

Allgu/ar

Selliemenl

SelllemeJll

Distortion

Settlement

$ettfeme/ll

Distortion

(a) Isolated FoundaUOIIli

(i) Steel

SOmm

0.0033 L

11300

SO mm

OJXm L

1/300

SOmm

0,0015 L

1/666

75 mm

0.0015 L

1/666

75 mm

0.0033 L

1/300

IOOmm

0.0033 L

1/300

75 mm

0.002 L

1/500

100 mm

0.002 L

1/500

structure

(ii)R.C.C. slructure;s

(b) Raft Foundations (i)Stetl Structures

(ii)R.C.C. Structures

buildings. The values for isolated footings and rafts are given in Table 23.9. For further dctails, a reference may be made to the code. In actual practice, setllements of the structure at salient points such as the centre, the comer, the lightest and the heaviest column locations are determined using the methods discussed. The differential settlements are then determined from the settlements of the various points. Sometimes, the differential settlements are taken as 75% of the maximum settlement without computing the settlements otber than that at the point where the maximum settlement is expected to occur. However, for large works, thorough sub-soil invcstigations are done to determine the charaaeristics of the soil and the differential settlements are determined taking the properties of the soil into accounL

sons

23.30. ALLOWABLE SOIL PRESSURE FOR COHESIONLE'lS The allowable soil pressure (q".,) of a shallow foundation is limited either by the net safe bearing capacity (q...) or the safe settlement pressure (q" p). The design of shallow foundation on oohesionless soils is generally governed by the safe settlement pressure, as the net safe bearing capacity for a footings of usual size is quite. higb. However, in the case of narrow footings on water-logged sands, the net safe bearing capacity may be the controlling criterion for the design. Il is the normal practice for the design of footings of usual size to use empirical methods based 00 N-values for the determination of the allowable soil pressure for oohesionlcss soils. lbe plate load tests (Sect. 23.33) are also used in the case of soils having small boulders and stones which obstrua the standard penetration test. The methods using the standard penetration test are preferred to plate load tests fa homogeneous soils, as these are more economical. Footings on granular soils are generally designed using the following empirical relatinnships for the allowable soil pressure.

(1) Peck Method Terzaghi and Peck (1967) gave chans for the safe bearing pressures inducing a total seLUement of 25 mm and a differential settlement of 19 mm for different sizes of footing. Peck et 011 (1974) revised the Terzaghi and Peck. curves to take into consideration the later research, and gave the follOWing equation for the safe settlement pressure. __ .(23.7~ q"p.0.41C.. Ns where q" p = safe settlement pressure (kN/m~,

10

DEARING CAPACITY OF SHALLOW FOUNDATIONS

619

N :: average SPT number, corrected for overburden pressure and dilatancy, s = settlement (mm), C"" "" water table correction factor. For a settlement of 40 mm, q. p • 16.4 C.N Fig. 23.23 gives a chan for s = 40 mm and C", = 1.0.

(23.78)

/

600

/

~-

/ /

,.,

0

10

Its

'"'he

For a settlement of 25 mm, qll P _ to.2S C ... N

IIY ite be 00

= depth

:or he

,m

:hi

'fe

'7)

... (23.79)

of water table below the ground surface, Dr = depth of fooling, D width of footing.

(2) Teng's Equation Teng (1962) expressed the charts given by lcrzaghi and Peck (1948) in the fonn of the following formulas. Allowance was made for an increase in pressure with depth by introducing a depth factor. For a setUemeot of 25 mm,

qnP - 35.0 (N - 3)

;ts rd

(kN/m2)

The water table correction is determined using the relution C • • 0.5 + 0.5 DJ(D, + B) where D""

Uy

'" "-

Fig. 23.23.

8 + 0.3)' (--zn WT Rd

... [23.80(a)J

where qllP = safe settlement pressure (kN/m1. N = SP'T number, B = width of Cooting (m), WT = water table correction factor (Eq. 23.36 (e». Rd = depth correction factor

(

.. 1 +

B :s ~

1.20

)

The above equ..ilion can be written in general form as q.P = 1.40 (N - 3)

8 + 0.3)' . ,(----w- W, R, ,

... [23.8O(b)J

where s = tolerable settlement (mm)

(3) Meyerhof's equation Meyerhof proposed equations which are slightly different from Thng's equations. According to him, for a settlement of 25 mm,


621)

q"p - 12.2 N WTRtI for B:s; 1.2m q"p - 8.1 N

and

(

0.3)'

BB + -

... (23.81(a)J

WT R" for B ~ 1.2m

... (23.81(b)J

where all the terms are the same as in Thng's equatioo, exoept Rd. whicb is given by

R.i - 1

+ 0.33 DrB :s 1.33.

(4) Bowie's equation Bowles (19n) suggested that the Del allowable pressure given by Meyerhof's equation can be safely increased by 50%. Thus, for a settlement of 25 mm,

q. p _ 18.3N W,Rd for B • 1.2 m q" p -

and

03 12.2 N ( D +8 .

t

... [23.82(a)J

W, Rd for B :z: 1.20

.. .[23.82(b))

(5) IS , 6403-1971 equation IS: 6403-1971 gives the following equation, which is similar to Teng's equation (Eq. 23.80 (b». For a selllemeot of 40 mm, q.P - 55.4 (N - 3)

t

'1 ~

700

roo

"" "Xl

I II

I

JOO 200

(B2+D3t B w,

--- "--

... (23.83)

I

"-

~:~ ~2smm

~

100

-"0]0

N=20

N,,(\

N,,'i

F,)Uf'Idc:ltion wIdth 1m} - - . .

Fig. 23.24.

The depth factor is not considered. Fig. 23.24 gives the allowable soil pressure for a settlement of 2S mm (ferzaghi and Pede, 1967). Fig. 23.25 gives the allowable soil pressure for a settlement of 40 mm. NarTOw FootLngs on Sand The design of narrow footings on sand footings is generally governed by the net safe bearing capacity. The ultimate bearing capacity can be detennined from the theories already discussed. The value of the angle of shearing resistance (+') can be detennined from shear tests. As it is difficult to duplicate field conditions in the laboratory. 4" is usually detennined indircct.ly from standard penetration number (N). as shown in Fig. 23.7.

621

BEARING CAPACfIT OF SHALLOW FOUNDATION

23.31. ALLOWABLE SOILS PRESSURE FOR COllES1VE SOILS The allowable soil pressure for oohesive soils is generally controlled by the nct safe bearing capacity, although setllement criterion may oontrol in son clays. In the case of fouodatioas on firm to stiff clays, it is unnecessary to compute settlements, especially' when

~~t:t:~:. ~ ~:~~ ~~S:~et; ~~\se~~::tn~h: ~ IOOOr--i--"-f>-=+--

failure would generally ensure tbat the settlements '1 are within the safe limits. However, it is essential to ~ compute the consolidation settlement in all cases of : aoo heavy structures. ~

S!'ltI!'m!'flt:40rl'nI

~ N:60

~

t ~ 600f---t:::::-t':::"'i-_-!!.'-'.!'.~=::::j ~:,~ t~~ '!':r S:=~h O;:~dC': ::nm%~m~.: 1 I~___ N."____ : _

"'-

The net safe bearing capacity (qM) depends

I

computations of the ultimate bearing capacity. On .: 400f--1::--+--..:=t==~N"".3:g,0=9===1

:~~~t~dc~:e'U~~~:e ~~~~:gusi~all~la!~:l~ The undrained cohesion (COl) is determined from the vane-shear test, unconfined compression test or

~~:=li::~n:~:y t:tdc~:f~~ ~l~g ~~

'---__

N:20

200!--+--+=+===~==j N:IO

0o\----+--===1;==!==:i=,==!==i

bearing capacity theories for ell = 0 GQndilions. Width B ~ As the settlement calculation~ are involved, the . 2 2 size of the footing is first selected coll'iidering the FIg. 23.2:5. (1 tim '" 10 kN/m ) DeL safe beariog capacity aDd theo tbe settlements are checked. It may be ooted thaL the footing width has very litlIe effect on the magnitude of the settlement. The settlements depend mainly on the magnitude of the load. Consequently, if scttJements control the design, merely inq~g the size of footing may not help solve the problem. An alternative type o.f foundation such as a mat foundation, a compensating foundation or a pile foundation, would be required. 23.32. PRESUMPTIVE BEARING CAPACITY Building codes of various organisations in different countries give the allowable bearing capacity that caD be used for proportioning footings. These values, known as presumptive bearing capacity, are based on experience with other structures already buill. As the presumptive values are based only on a visual classification of the surface soils, they are not reliable. These values do noL consider important factors affecting the bearing capacity, such as the shape, width, depth of footing, location of water table. strength and compressibility of the soiL Generally, the values are conservative and can be used for preliminary design or even for the final design of small, unimportant structure. IS : 1904--1978 recommends thai safe bearing capacity should be calculated on the basis of the soil test data. However, in the absence of ~uch data, the values of the safe bearing capacity can be taken equal to the presumptive bearing capacity values given. Thble 23.10 gives the values for differeot types of soils and rocks. The table also gives the values as per New York: Building Code for comparisoo. It is further recommended that for non-cohesive soils, the values shall be reduced by 50% if the water table is above or near the base of footing. For further dctail4. a reference may be made to IS: 1904--1978. 23.33. PLATE LOAD TEST The allowable bearing prcsl)ure can be dctennined by conducting a plale load test at the site. The conduct a plate load test, a pit of the size 581' X 581" where B,. is the size of the plate, is excavated to a depth equal to the depth of foundation (DJ). The size of the plate is usually 0.3 m square. II is made of steel and is 25 mm thick. Occasionally, circular plates are also used. Sometimes, large size plates of 0.6 m square are used.


622

Tobie 23.10. Presumptive Sufe Bearing Capacity IS: 1904-1978

New York Buildillg Code

Type oj Rode/soil

2

2

kNl m2

'1m2

IeNlm

330

3240

165

1621)

90

880

4. Soft Rock II. Non-Cohes.i'l'e Soils"

45

440

77

76(J

Compact gr:lvcl. sand and gravel 2. Compact and dry coarse sand 3. Compact and dry medium sand 4. Fine sand, silt 5. l..£XX$C gravel or sand 6. l.oa$e and dy fine sand III. Cohesi'l'e SoIls I . Hard Of Sliffday, soft shalc 2. Medium day 3. Moist clay and sand day mixture 4. Soft clay 5. Very soft clay 6. Dlack cotton soil, Peat, made up soils

45 45

440 440

77-97

760-950

31- 77

300-76(J

25 15

245

25 10

245 100

36

350

19

90

45

440

49

480

25

245

19

190

15

150

10

100 50

10

95

I. Rock.. L I-lard Sound Rock 2. Laminated rock 3. Rcsiduul Deposits of Shattered and Drokcn Rocks

l.

'1m

199-591 77 112

1950-5800 76(J 1100

ISO

To be determmed after soLI m'l'CSlJgatlOM. CompadllCSS or looseness of non-cohcsi'l'e SOIls may be delennlncd by dnvmg a cone of 65 mm dIU and 60" apex angle by 11 hammer of 65 kg falling rrom 75 em. If OJrreacd number of blows (N) for 30 em pcnctr:ltion is less than 10, the soil is loose. If more than 30, it is laken WI dense. If N is between 10 and 30, il is medium dense.

A central hole of the size Bp )( Bp is excavated in the pit. 100 depth of the central hole (Dp) is oblained from the following relation :

Dp/'Bp' DIBf Dp • (Bp/'Bf ) x Df where Br is the width of the pit, and Bp is the size of plate.

... (23.84)

For conducting the plate load test, the plate is placed in the central hole and the load is applied by means o[ a hydraulic jack (Fig. 2l26). The reaaion to the jock is provided by means of a reaction beam. Sometimes, trusses are used instead of a reaction beam to take up the reaction. Alternatively, a loaded platform (kcntlc
BEARING CAPACITI OF SHALLOW FOUNDATIONS

Fig. 23.26.

P1~tc

Lond Test.

corresponding to a settlement of onc--fifth of the plate width (Bp). On the natural plot (Fig. 23.27), the ultimate lo..'Kl is obtained from thc intersection of the tangents drawn as shown. quIp)

rig. 2327.

(1) The ullimatc bearing capacity of the proposed foundalion q.. (f) can be obtained from the following relations: (a) For clayey soils, q.. (f) - q.. (P) ... (23.85)

(b) Fo, sandy soils,

q. (f) • q.(P) x

Br B,

.. (23.86)

where 0, = foundation width. (2) The plate load test can also be used to detennine the settlement for a given intensity of loading (q,,). The relations between the setllement of the plate (sp) and that of the foundation (Sf) for the same load iruensity are given below. (a) For clayey soils,

s, - sp )(

n;Br

...(23.87)

where sp is obtained from the load intensity-seulcmcnI curve for q(Y (b) For sandy soils,

Sf -

sp

0.3)]'

Br(B, • .B (Of + 0.3) [ p

...(23.88)

In above equations, Bf is the width of foundation in metres and Bp is the width of the plate also in metres. (3) For designing a shallow foundation for an allowable settlement of Sf' a trial and error procedure is adopted. First of all, a value of Of' is assumed and the value of q" is obtained as

.'"


... (23.89) 00 - QIAf where AI is the area of footing, and Q is the load. For the computed value of qo. the plate settlement (sp) is determined from the load·setUcment curve obtained from the plate load test. The value of sf is oomputed using Eq. 23.87 if tbe soil is clay and using Eq. 23.88, if sand. Thc oomputed value of is oompared with the allowable seulemcnl. The procedure is repeated till the computed value is equal to the allowable settlement. (4) The plate load lest can also be used for the determination of thc influence factor I (Sea. 23.24). From

s,

Eq.23.68.

s -

lL::J!1 ~, )( I

... (23.90)

)( qB

B

q

L-________________________

~

Fig. 23.28 shows a plot between the settlements and the lood qB. 100 slope of the line is equal [0 (I - "')IIE,.

Fig. 23.28.

Limitation of the plate load test The plate load test has the foUowing limitatiollS: (1) Size effect. The results of the plate load test reflect the strength and the settlement charncteristics of the soil within the pressure bulbs (Section 11.6). As the pressure bulb depends upon the size of the loaded area, it is mudl deeper for the actual foundation as compared to that of the plale (Fig. 23.29). The plate load test does not truly represent Ihe adual conditions if the soil is not homogeneous aoo isotropic to a large depth. (2) Scale elTed. The ultimate bearing capacity of saturated clays is independent of the size of the plate but for oohesionless soils, it increases with the size of the plate (Eq. 23.86). To reduce scale eITed, il is desirable 10 repeat the plme load test with plates of two or three different sizes and extrapolate the bearing capacity for the adual foundation and take the average of the values obtained. (3) TIme effect. A plate load test is essentially a test of short duration. For clayey soils, it docs not give the ullimate selllement. The load----&ettiement curve is not truly representative. (4) Interpretation of failure load. The failure load is not well·defincd, except in the case of a genen! shear failure. An error of personal interpretation may be involved in other types of failure. (5) Reaction load. It is not practicable to provide a reaction of more than 250 kN. Hence, the test 0118 plale of size larger than 0.6 m width is difficult. (6) Water table. The level of the water table aITeas the bearing capacily of the sandy soils. If the wilier table is above the level of the footing, it has to be lowered by pumping before placing the plate. T1r test should be performed at the water table level if it is within about 1 m below the footing.

10

625

BEARING CAPACITY OF SHALLOW FOUNDATION

A

ng ;g

FOOTING

~

>In

BULB

Ca)

>0) to

Cb)

PRESSURE BULB

Fig. 2329.

23.34. HOUSEL'S METItOn FOR DESIGN OF FOUNDATION Housel's metbod can be used for the design of a shallow foundation for a given safe settlement. 1Wa plate load tests are conducted, one with a plale of size 8 1 and the other, with a plate size 8 2- 1be load-settlement curves are obtained for both the tests. The loads corresponding to safe selLiement are obtained from the load-sclUemenl curves. Lei Ql and Q2 be the loads for the plales of size B1 and 8 2

s,

respectively. Housel expressed these loads in the following forms. and

of the

:9). ,nd

.ate :ct, !ate

~ ve

aler Ibe

..(23.91)

Q2 - A2 m + Pln

... (23.92)

where A"A2 = areas of plates of size 8 1 and 8 2 , respectively PI' P l = perimeters of plates of size Bl and 8 2 , respectively. and m, n = constants. The values of the constants m and n are determined solving equations 23.91 and 23.92. If A and P are, respectively, the area and perimeter of the given foundation, the safe load is then computed as Q - Am + Pn ., .(23.93) TIle above load Q is for a safe settlement of SI' For any other settlement 5', the safe load is given

Q' - (Qls ,) x s' ... (23.94) Eqs. 23.93 and 23.94 can also be used for the determination of the size of foundation for a given' load, as both A and P depend upon the size.

!nil

".

QI-A1m+P1n

ILLUSTRATIVE EXAMPLES Illustrative Example 23.1. Determine the ultimate bearing capacity of a strip footing, 1.20 /II wide, and having the depth o//ol/ndation 0/1,0 m. Use Terzaghi's theory and assume general shear failure. Take til' = 35", Y = 18 kN/III], and c' = 15 kN/,,? • SoluUon. From Eq. 23.25,

q" - c'

N~ + y O,Nq + 0.5 YBN,


626

For ¢I'

= 35°, Table 23.1 gives

Ne

..

57.8, N q

..

41.4

qu .. 15.0 x 57.8 + 18.0

Now

and Ny .. 42.4. )C

1.0 x 41.4 + 0.5 x 18.0 x 1.2 x 42.4

.. 2070kN/ ml lIIustraUye Example 23.2. Determine tile allowable gross load and the nel allowable load for a sqUiJTe footing of 2m side and with a depth of foundation of 1.0 m. Use Terzaghi's theory and assume local sheilr failure. Take a factor of safety of 3.0. The soil at the site has 1 = 18 kNlm3, c' :: 15 kN/m2 and if' = 25°,

Solution. From Table 23.1. for 4" = 25°' N/ .. 14.8, N q '

..

5.6

and Ny' .. 3.2

From Eq. 23.37, laking em' .. 2/3 c' .. 10 kN/m 2

qu .. 1.2 )( 1.0.0 )( 14.8 + 18 x 1.0 x 5.6 + 0.4 x 18 x 2 x 3.2 .. 325 kN/m 2

From Eq. 23.1,

q"" .. 325 - 18 x 1.0 .. 307 kN/ m2

From Eq. 23,).,

q,.. ..

Net allowable load From Eq. 23.3,

T . ~~ .

102.3 kN/rn

2

.. 102.3 x (2 x 2) .. 409.2 leN q, .. qllS + 't Df .. 102.3 + 18 )( 1.0 .. 120.3kN/ m2

Gross allowable load .. 120.3 )( (2 )( 2) .. 481.2 kN illustrative Example 23.3. A footing 2 III square is laid at a depth of 1.3 III below the ground surface. = 3D" and c' = 0. Determine the net ultimate bearing capacity using IS code method. Take y = 20 IeN/IIIJ, Solution. For = '30", Thble 23.6 gives N • .. 30.14, N q .. 18.4 and NT .. 22.4

+'

+'

From Table 23.3, From Eq. 23.49 (a),

From Eq. 23.49 (c), From Eq. 23.48,

Sc- .. 1.3, Sq .. 1.2 and Sy .. 0.80 d, _ 1 + 0.2(DI8)'.n(45' + ~'12) _ 1 + 0.2 )( (1.3/2.0) tan 60° _ 1.23

d., - d, - 1 + 0.1 (DI8) tan (45' + ~'/2) - 1.11 q"" - cN. sed. i. + q(Nq-l)sqdqiq + 0.5yBN1ST d.,~W

.. 0.0 + 1.3 )( 20 )( (18.4 - 1) )( 1.2 )( 1.11 )( 1.0 + 0.5 )( 20)( 2.0 )( 22.4 )( 0.8 )( t:Il )( 1.0

q"" _ 1000 kN/ml llIuslralive Example 23.4. Determine the net ultimate bearing capacity of the footing in Illustrative Example 23.3 if (a) the water table rises to the level of the base, (b) the water table rises to the ground surface, and (c) the water table is 1 m below the base. Solution. (a) W' = 0.50, Therefore, Eq. 23.48 gives q/IU - 1.3 )( 20.0 )( (18.4-1) )( 1.2)( 1.11 )( 1.0 + 0.5 )( 20.0)( 2.0 )( 22.4 )( 0.8 )( 1.11 )( 0.5 .. 801kN/m 1 (b) W'::: 0.50. The surcharge

q is also reduced as .the effective stress

is reduced, Thus q"" .. 1.3)( (20 - 9.81»( (18.4 - 1) )( 1.2)( 1.11 )( 1.0 + 0.5

)( 20 x 2.0 x 22.4 x 0.8 x 1.11 x 0.5 _ 586kN/m 2

/

<0


627

Eq. 23.36 (c)J.

(e) W' is obtained by linear interpolation [see

,.4

W _ 0.5 + 0.5

to

1.0 .. 0.75

q"" - 1.3 x 20.0 )( (18.4 - 1) )( 1.2 )( 1.11 )( 1.0 + 0.5 )( 20.0

Therefore,

x 2.0 x 22.4 x 0.8 x 1.11 x 0.75

_ 901 kN/m z Dlustl'lltlve Example 23.S. A square. column founda tion is to be designed for a gross allowable total load

1J:~

:;'t:; ~~es~;ea: !i ~~I:: ~~ ~s~;~l~~~:. t~ :h~;e;::~~ ~t:r;;~.e:! ;j~m5 0t:,~:(o;;;~~:

of foundation is 1.0 In. Solution. From Eq. 23.45,

2

q .... c' Nts(dtit + qNqsqdqiq + O.5'fBNTSld,~

From Thble 23.6.

Ne - 46.12, N q Sc -

1 + 33.30/46.12 - 1.72

$'1 ..

1 + Ian 35° .. 1.70, s, - 0.60

From Eq. 23.46 (a),

de - 1 + 0.4 x 1.018

From Eq. 23.46 (b),

dq _ 1 + 2 Inn 35° (1 - sin 35°)2 x 1.018 dq -I+0.255IB; d.,-1.0 i" _ iq .. (I _ aO/90)2 .. 0.694

or ceo

o.

33.30 and Ny ., 48.03

-

From Table 23.7,

From Eq. 23.47 (a),

From Eq. 23.47 (b), q.. .. 5.0

Therefore,

X

1.7

~ )C

X

46.12

a·/~)2

_ (1 )C

1.72

(1 + 0.255/8)

_ 0327

)C

(1 + 0.4/8)

X

0.694 + 05 x 19

0.694 + (19 )( 1.0)33.3

)C

)C

B x 48.03 x 0.6

X

1.0

X

0.327

.. 1022.2 + 300.7/8 + 89.5 B

From Eq. 23.1,

q"" .. q .. - 'tD, .. q .. - 19 x 1.0

3~.7

.. 1003.2 +

1.0

From Eq. 233,

qs ..

~

+ 19

.. 334.4 + Now gross load .. ql

)C

X

+ 89.58

1.0

1O~.2

+ 29.8B + 19.0

82

or 250.0 - 353.40' + 100.28 + 29.8 B' Solving by trial and error, B .. 0.7 m D1ustratlve Example 23.6. Determine the uirimtUe bearing capacity of a square footing 2 m )C 2m in a J soil with unit weight of 18 kNfm , 41' = 20°, c = 20 kNfm 2• Take the depth of [ountiLJliQn of }.50 m. Use :).5

Hansen's equation. Solution. From Eq. 23.42,

q.. .. eN" Sc d" i" + qNq Sq dq

From Table 23.2,

N e .. 14.83, Nq

From Table 23.3,

sc .. 1.2,

From Table 23.4,

Sq"

..

6.40

1.2

iq

+ 0.5 't B NT ST d, i.,

and NT .. 3.54

and s, .. 0.6

de .. 1 + 03

)C

1.512.0 .. 1.225

4,-",-1.225, d,-1.0


628

As j~ - i" ..

iy '"' 1.0, q" .. 20.0 )( 14.83 )( 1.2 )( 1.225 x l.0 ... (18 x 1.50) x 6.40 x 11 x 1.225 )( 1.0 ... 0.5 x 18 )( 2 x 3.54 )( 0.6 x 1.0 x 1.0 '"' 728.25 kN / ml

JIIustmtive Example 23.7. A strip fOOling of 2 m widtli is founded at a depth of 41/1 below the ground surface. De/ermine the ncr ultimate bearing capacity, using (a) Terzaghi's equ(lIion, (b) Skemplon's equation, alld (e) IS Code. The soil is clay (1jI = 0, C = 10 kNlm 1). The unit weight of the soj[ is 20 1rN/l1t Solution. (0) Tcr.l.'lghi's equation From Eq. 23.25, q" '"' cIIN" ... 1 DfNq + O.5yDNy Taking the values from Thblc 23.1, q" .. 10 )( 5.7 + 20 )( 4 x 1.0 + 05 x 20 x 2 )( 0.0 '"' 137

Therefore,

qlWl " q.. - yD, .. 137.0 - 20 x 4 .. S7.0kN/ m

1

(b) Skempton's equation

From Eq. 23.53 (a), for

~

.i

< 2.5 ,

N, _ 5.0 (1 + 0.2 DrB) (1 + 0.2811.) .. 5.0(1 + 0.2)( 412)(1 + 0.2)( O.O} .. 7 1

From Eq. 23.55,

qlWl .. cliNe" 10 x 7.0 .. 70.0 kN/ m

(e) IS Code

From Eq. 23.56, Taking Ne

q"" '"'

= 5.14,

clONes" ~;e

q_ .. 10

X

5.14

X

(1 + 0.2

X (

%)

tan 45°) )( 1.0

'"' 51.4 )( \.4 '"' 71.96 kN/m 2 lJIustmtive Example 23.8. A square fooling (1.5 m )( J.5 m) is located aJ a depth of 1.0 III in a clay drposit consisting of two layers. The lOp layer is 1m thick and has Cl :: 150 leN/nl and Yl :: 16 leN/mJ. The bottom layer has C2 50 kN/ml and Y 15 leN/ml. Dcrermine the net ultimate beariJ1g capacity. Solution. From Eq. 23.59, laking ie .. 1.0, q.. .. CI Ne Se de + q

=

=

From Fig. 23.18, for cl"Cj = 1/3 and Z/B ::: 1.0/ 1.5 ::: 0.67, the value of N~ is cquallo 350.

s, - I + (BIL) (N,IN,) _ 1 + 1 x 113.5 _ 1.29 de .. } + 0.4 )( 1.0/ 1.50 .. 1.27

q.. .. 150 )( 3.5 )( 1.29 )( 1.27 + 16.0 X 1.0 .. 876.1 kN/m 2

'rl'lcrcforc,

qlWl .. 876.1 - 16 .. 860.1 kN/ m 2

Illustrative Example 23.9. A square footing (1.5 m x 1.5 m) is located at a depth of 1.0 m. Tile fooling is subjected to an eccentric load of 400 kN, with an eccemriciry of 0.2 m along one of the symmetrical ares. Determine the foctor of safety against bearing failure. Use Vesic:S- cquaJion. Take y ::: 21 kNlmJ, c = 100 1eN/,,?, , '" O. Solution. Effcctive width B' .. B -2 el> .. 1.5 - 2 )( 0.2 .. l.l m From Eq. 23.45, taking Nl .. 0.0, Ne .. 5.14

and N q .. 1.0.

q.. " cNescde ie where

Sc '"

1 + (B'/L) (Nq/N e)

..

+qN~qdqiq

1 + (1.1/1.50) )( 1.0/5.14 .. 1.14

Sq " 1 + (1J' /L) tl.ln, .. 1.0 + (1.1/1.5O)13nO o

..

1.00

629

OI!ARING CAPACITY OF SHALLDW fOUNDATIONS

1.2

d~

_ I + 0.4 (DIO) .. 1 + 0.4 x LOlLS. 1.27

dq

_

1 + 2tanC/l(l - SintJI)2 (DIll]) .. 1.0

q" .. 100 x 5.14 x 1.14 x 1.27 + l.0 x (2l.0 x 1.0) x 1.0 )( 1.0

Therefore,

_ 744.2 + 21.0 _ 765.2 kN/m:

uI

n,

~ (1 +~) - 1.5~15

Feom Eq. 23.63 (a),

qmu"

0'

qmax - 320 kN/m'l

Now

(1 +6 7.5°·2) .(a)

q"" - 765.2 - 21 x 1 - 744.2. ... (b)

Therefore, f'rom Eqs. (a) and (b),

744.2/f" + 21.0 = 320

or

F .. 2.49. IIIw;trutlve Example 23.10. A square footing is required to carry a net load of 1200 leN. Determine the size of the footing if the depth of fOllndnlion ;s 2 1/1 and the tolerable settlement is 40 mm. The soil is sandy with N = 12. Take a factor of safety of 3.0. The water table is "cry deep. Use Teng ~ equarion. Solution. From Eq. 23.61. q.... _ 0.33 N1 BW, + 1.0(100 + If) DI Wq

q"" .. 0.33 (12)28 x 1.0 + 1.00 (100 + 12~ x 2 x 1.0 q".. _ 47.50 + 488.0

Ii Ii'

Total net load, Q" - (47.5 B + 488.0)/3 x 1200 • (47.58 • 488.0)13 x OJ

3

"he

Solving. by trial and error,

Fccm Eq. 23.80 (b),

1200 _ 15.8B + 162.7B B _ 2.45 m. q." 1.4O(N - 3)

q...

2

(8 ;;.3 f ,

W,Rds

1.4O(N_3)(8;;3) (1 .0.28X2)X4O

q"l''' 1.40(12 - 3) (B

;:3 f

x 40 (1 + O.4IB)

.. 126 (8 + 0.3)2 x (1 + 0.418)

·ng ·

Q• •

Now IJ()

q.,

x

Ii'

1200 • 126 (8 • 0.3)' (1 • 0.4/8) Solving. B _ 2.58 m. adopt B .. 2.60 m Illustrative Exumple 23.11. A rectangular footing (3 m x 2 m) exerts a pressure of 100 kN/m2 on a cohesive soil (E, = 5 x 104 kN/m2 and ~ = 0.50). Determine the inmlediate settlement at the cenlre, assuming (0) 'he footing is flexible,(b) the footing is rigid. Solution. From Eq.

23.~

Sj ..

q8 (

¥)

I


630

As LIB :: 3/2 '" 1.5, from Thble 23.8, I

Therefore,

= 1.36.

5," 100)( 2

(b) For rigid fooling (I

= 1.06),

(1 05') ~ 5 )( 10

)( l.36)( lcrl .. 4.08mm

s .. (1.06/1.36) )( Si

.. 1.0611.36 )( 4.08 .. 3.18 mm Illustrative Example 23.12. Fig. £-23.12 shows a square fOOling resting on a sand dePOSiL The pressure at 'he level of the foundation (q) is 200 kNlm 2, The figure also shows the variation of the elastic modulus with depth. Determine the settlement of the foundotion after 6 yeaTS of construction.

E, Fig. E-23.12.


51 ..

C 1 C 2 (q

q_ 16 x 1.5 '" 24kN/ml

-

q)

and

C, - 1 - 0.5

,.

Lo !..£.'.:l z E, q-

(~) q-q

q .. 200 _ 24 .. 176kN/m 2 _ 1 - 0.5(24/176) _ 0.932

C2 .. 1 + O.21og lO (t10.1) .. 1 + 0.210810 (6/0.1) .. 1.356

,.

Therefore,

Si ..

0.932 x 1.356 x 176

I

f if:'

dz

,. I

,.

.. 222.4

ft·

Az

The value of ~ (lzIE,) . Az is delecmined as shown in the table below. It is equal to 13.97 x 10-5,

o

BEARING CAPACIIT OF SHALLOW FOUNDATIONS

631

E, (kNlm')

0-1.00 1.0-2.0 2.0-3.0

9000 10000

3.0-4.0

11000 12300

4.0-5.0

(I,/E,)

I,

0.1 +2 0 .42 _ 0.26

8000

1.0 m

0.453 0.333 0.200 0.067

;,

3.25 x 10-5 5.03 x 10-5 3.33 x 1O-'~ 1.82 x 10- 5 0.54 x 10-5 I 13.97 x 10-5

Therefore, or

Si -

222.4 )( 13.97 x 10-5 m

31.07 mm Illustrative Example 23.13. Fig. E-23.13 shows the load- .fettlemcnt curve obtained from a plate load test conducted on a sandy soil. The size of the plate used wtu" 0.3 m x 0.3 m. Determine the size of a square column footing to carry a net load of 3000 kN with a maximwlI settlement 0/25 mm. Si _

Load

100

200

{kNlm1)--...

300

1..00

SOO

600

700

800

10

20

E ~

30

~

!

~

40

SO

Fig. E-23.l3.

BI (Be + 0.3) ]'


sf - sp [ Bp(Bf + 0.3

SI - sp

(~)' (BI ~60.3 )'

The value of Bf is fouod by trial and error, as shown in lhe table below.

BI 3.80 m 3.6 m 3.55 m

qo -

(~,

207.7 2315 238.0

Adopt a size of 3.55 m x 3.55 m.

from Fig. Ex. 21.13

5p

6mm 7rnrn

7.3mm

sf

BflBp

frrmaEq. (a)

12.67 12.00 11.83

20.62 mm 23.85 mm 24.81 mm


632

illustrative Example 23.14. Two-plate load tests at a site gave the following results. Size of plate Load Settlement 0.305 )( 0.305 In 40 kN 25 mm 0.61 x 0.61 m 40 kN 15 nun (0) Assuming Poisson's ratio as 0.1, determine the de/ormation modulus of lIIe soil. (b) If there are two columns, one of the size 2.5 m x 2.5 m, carrying a load of 2700 leN, and the other of size 3m x 3m, carrying a load of 39()() kN, determine the differential settlement. The columns are 7 In apart.

.

ql" 0.305 ~ 0.305 - 430 kN/m

Solution. (0) For the first tcst, ql

B • .. 430 x 0.305 .. 131.1 kN/m 2

~ 0.6\

lJ2 - 0.61

For tbe second (cst,

2

q'Jh -

.. 107.5 kN/m2

107.5 x 0.61 _ 65.6 kN/m 2

Fig. E-23.14 shows the plot between qB and s.

ozo .015

0.01

."",

L

/

V

SLope.~",.S2X;-o4 E,

k--

'0

/

"

""

qB (kNlm)

Fig. E-23.14.

From the plot, From Table 23.8, As Ihe plate is rigid,

Therefore,

(¥)IO

1.52 x 10-'

I .. 1.12 I .. 0.8 x 1.12 .. 0.89 2 E, .. (1 - 112) x O~96.. (1 - 03 1.52 x 10 1.52 x 10

L)( 0.896

E, .. 5364 kN/ml

(b) For the first column.

ql -

2 00 2 2.5 : 2.5 _ 4321cN/m

For the second column,

lh -

33~3

- 433 k:N/m2

633


As thc setUemcnt of the plate (0.305 m )( 0.305 m) at a load intensity of 430 kN/m~ is 25 mm, it can be

used for the dctennination of the settlement of columns.

Form Eq. 23.88,

2.5 )( (0.305 + 0.30)]' (s/h - 15 [ 0.305)( (2.5 + 0.3) - 78.42 mm 3 (0.305 + 0.30) ]'

(s;h - 2S [ 0.305 (3.0 + 03)

- 813 mm

Diffcrential seulement - 81.3 - 78.42 - 2.88 mIll mustrative ExaIllple 23.15. The results of tlW plate load tests for a settlement of 25.4 min are given. PIOte diameter Loai 0.305 m 31 \eN 0.61 m 65 \eN A square column foundo.tion is to be designed to carry a load of 800 /eN with an allowable settlement of 25.4 mm. Determine the size using Housel's melhod Solution. From Eq. 23.91 and 23.92, 31.0 - (l'tf4) x (0.305)2 )( III + l't (0.305) )( n

... (a)

65.0 - (l'tf4) x (0.61)2 )( III + l't (0.61) )( n

... (b)

Eq.(a) can be written as 62 - 2 )( 3t/4 (0.305)2 )( III + 2;c (0.305) )( n From Eqs.(b) and (e), by subtraction, 3.0 - m [;cf4 (0.372 - 0.186)] m _ 2055 From Eq. (a) 31.0 = 15 + 0.9577 n or n = 30.80 From Eq. 23.93;

Q -

n'

... (e)

x 20.55 + (30.8 x 4B)

800 _ 20.55 n' + 123.2 B

B - 3.93m say 4m x 4m. PROBLEMS A. Numericals 23.1. Determine the ultimate bearing: capacity of a square footing of size 1.2 m if the depth of foundation is 1 m. Take ~' '" 25'. y = 18 JcN/m3 and c =-. 15 IcNJm2• Use Vesic's equation. [Ans. 1050 JcN] O.Z. A circular foundation is of 2.4 m diameter. If the depth of foundation is 1m, detennine llIe net allowable load. 2 Take y = 19 kN/m3, c' = 30 leN/m , " = 15° and factor of safety as 3.0. Use Tcmghi's equation and assume [Ans. 418 kN] local shear failure. 23.3. A square footing is to be designed to carry a load of 500 kN. If the depth of foundation is 1.5 m, determine a suitable size of foundation with a factor of safety of 3Jf The water table is at foundation level. Thke $' = 25°, 3 3 [Ans. 2.10 m] Y = 16 kNJm , Ysat '" 19 kN/m . Use Terzaghi's theory. c' '" 20 IeNJm2. Assume local shear failure. 23.4. A strip footing is required to carry a net load of 1000 kN at a depth_of 1m. Thking a factor of safety of 3, 30°.'1 '" 19 JcNJm3, C:: 20 kNJm2 • Use Terzaghi's theory. determine the width of the footing. Thke Assume general shear failure .[Ans. 2.4 m] 23.5. A strip footing is 2 m wide and founded at a depth of 2m in 8 soil of unit weight 20 JcNJm 3 and a oohesion of 10 kNJm 2. Determine the increase in the bepring alpacity when is increased from zoo to 25°. Use Thmghi's equation. Assume local shear failure . [ADS. 130 kN!m1 3 Z3.6. A purely cohesive soil has a unit weight of 20 kN/m and a cohesion of 150 JcNJm2. Determine the safe bcarin$ [Ans. 448 JcNJm1 capacity for a rectangular fOOling 8m x 2m founded at a deplh of 4 m in clay. (F.S. '" 3.0) Z3.7. A strip fOOling 2 m wide is to be laid at a dCplh of 4 m in a purely cohesive soil (c '" 150 kNJm 2. y :: 19

+'"

+

634


kN/m \

Determine the uliim;l\c bcllring capacity frOlll (II) Terlaghi's thoory (b) SkcmplOn's theory. [Ans. 930 kNhn 2; 1125 kN/m 2J 23.8. E,slim:llc. [he immcdkue sen icme- llt of 11 co ncrete footing. I III X 1m size. fOllnded at a depth of J m in It soil with £ == 10" kN/m 2• 1..1 = 0.3. The foo tmg is J,ubjccted It) :1 pressure of 20() kN/rn2, Assume the footing [0 be ngid

IAns. 22.2 mmJ 23.9. A square fooli ng 2.5 m Si7.£ is founded at 11 dept h of 1.5 III in a sandy soil deposit which has the corrected N value o f 30. 1111: water table is :11 II deplh of 2 III fmm the ground surface. Find the net aHowabh! soil pressure

;f 1 ~ 3.0. (b) the permissible seUlemcnl ;s 40 mm. Use Tens's equations. rAns. 605 kN/m2, 456.5 kN/m2] 23.10. An ellcavallon 3 m wide is to he mooe to deplh of 5 m in soft clay (e u '" 15 kNJm 2, y '" 19 kN/m J ). The ground surrounding the ex.cav31ion carries a surcharge of 10 kN/nl. Determine the factor of sMety. Use Terzaghi's equations. (Ans. 1.231

(0) the deSIred (;lelor of Mlfety

8. Descriptive utld Objective Types 23,11. Define the foll owi ng l erm.~ : Net safe hearing capacity, (b) Gross safe bearing capacity, (e) Allowable ~oil prc.~~urc. 23.12. What lire the a~~umptions made in the dt:nvation of Terzaghi's bearing capacity theory? Write the equation fOf the ultimatc beari ng capacIty. 23.13. Diflerc1l1iatc hetween the ge neral shear failure and the local shc;] r failure . How the ultimate bearing capacity in local Slll!Mis dctcrmin(!(! "? 23.14. Di sc u ~~ the effect of water table on the bearing capacity of the soil. 23.15. Discuss Meyerhof's bearing C:1P[lCIt)' theory. How docs it ditrer from TeFlaghi's theory'! 23.16. Wnte shon ooles on: (a) H:msen's beanog capitClty theory, (b ) Veslc's bcanng capacity theory. 23.17. Describe Skempton's analysis for bearing ca pacity of co hesive ~uils. 23.18. How Ihe bea~ing c,lpacity of looring nn a layered cohesive mil deposit is determined '! 23.19. Discuss th e various methods uf dctemlinarion of th e 'lllowohJc soil pressure. What :u-c their limit:1rions? 23.20. What aTC different types o f settlements which ca.n occur in a foundarion '! How are these esti mated ? 23.21. Discuss the methods for estimm ing immediare settlements of foundmions on clay. .23.22. How would )'ou estimate the settlements of a found ation on cohesionlcss soils? 23.23. Describe plateiload test. Whot nrc irs lirhitation an d used '! 23.24. Write whelher the followrng ~t(lte lllents are true or (alsq. (a) A foundn.lion is cun.~ idered shallow if its deplh is less than I m. (b) Tcrzaghi 's theory is applicable when the base of rhe fooriilg is rough. (c) For footings on co hesion less soils, the bearing capacity, and not the scttlement, generally governs the design. (il) TIle plme.load tcst is more useful for cohesion1css soils than for cohesive soils. (t') The maximum v:lloe of N, is 9.0 by Skemplon's chart. (j) TIle !)TCsumprive bearing capacit)' is the same as the allowable bearing capacity. (g) '1te imm(!(!iate selliement of a rigid fnOling is nppro)( imalely equal to Ihe overage scltlement of Ihe flexible fOUling. (II) TIle allowable angular distortion in cnsc of steel structures i~ more rhan that for R,C.C. structures. (i) The differential settlement is generally three· fourths of the maximum scttlement.

«(.)

lAos. True (b), (t/). (f!), (g), (h), (I)J

C. Multiple Choice Questions 1. A shallow foundation is usuall y defined as a foundation which Ims


63'

(lI) depth Icss th:m 0.6 m

(Mdeplh Icss th.m its width (c) depth less th~ln 1.0 (d) none of above

tn,

2. The ultimate beilring .:ap".:ily of a shallow foundation on 5and is reduced to about . .. when the water table rises to the ~round surfnce. Ca) 75% (,.) 25%

(iJ) 50% (tl) none of above.

3. The :llIowable SOlI pressure for found:ulOns in cohesive soils IS gcnemlly controlled by (0) setllements (b) Ix:aring capacity (e) both (a) and (11) (tl) neither (lI) nor (b) 4. The immediale settlement or a rigid fOOling is aboul ... times the maximum settlement of an equal Oexible fOOling. (II) 0.9 (h) 0.8 (c) 0.7 (tf) 0.6 5. The bearing. c:lpncity of soil ~upporting a footing of size 3m x 3m will not be affected by the presence of water table located :Il a depth below the bn~c of footing of ((I) 1.0 (b) 1.50 m «(I) b.Om (el 3.0 m 6. A 2 m widc slnp footing rest~ nt a depth of 2 m below the ground surface where wnter table is at the ~round surface. The ultimntc IOlid which the suip can carry according to Terz.aghi's theory when Y"'I E 20 kN/m' and c = 30 kN/ml is about (h) 342 kNlm (Il) 171 kN/m (d1262 kN/m (e) 422 kN/ m 7. The pcrmis.~ible settlemcnt is the maxilllum in the casc or (CI) Isolated footing on day (h) Raft on day (el Isolate<.! fuoting on sand Ill) Ralt on sand 8, If the gross bearing C;l]lllcity of a stri~ footing 1.5 m wide I ~ated at a depth of I m in clay is 400 kN/m1 , its

~(~; ~~r~~/~'Tncity (c) 390 kN/m2

for Y:: 20

kN/~~) ~~O

kN/m2

(tl) 360 kN/m2

24 Design of Shallow Foundations 24.1. TYPES OF SUALWW FOUNDATIONS

A shallow foondation, according to TCrz.1ghi, is one whose width is greater than its depth (i.e. D/8 :s; 1). Shallow foundations are located just below the lowest pan of the walt or a column which they support. Footings are structural members, made of brick work, ma<;()nry or corx.TCtc, that arc used to lransmil the load of the wall or column such thai the lond is distributed over a large area. In fact, a footing is an enlargement of the base of the oolumn or walt it suppons. The footings arc of the following types: (1) Strip fooling. A strip footing is provided for a load·bearing wall (fig. 24.1). A strip footing is also provided for a row of columns which are SO closely sp3ccd Ihat their spread fOOings overlap or nearly touch each olher. In such a case, it is more cconomicalto provide a strip fOOling than to provide a number of spread footings in one line. A strip footing is also knOWD as continuous footing.

~ ElEVATION

",,"?~{I,,~~~ ELEVATION

on G PLAN

Fig. 24.1. Strip Footing.

PLAN

Fig. 24.2 Spread Footing.

(2) Spread or isolated footing. A sp~d (or isolated or pad) fOOling is provided to support an individual column (Fig. 24.2). A spread footing is circular, square or rectangular slab of uniform thickness. Sometimes, it is stepped or haunched to spread the load over a large area. (3) CombIned footing. A combined footing supports two C01bm1lS (Fig. 24.3). It is used when the two

DESIGN OF SHALLOW FOUNDATIONS

631

columns are so close 10 each other that their individual fOOings would overlap. A combined fOOling is also provided when the property line is so close to one column that a spre.'ld fOOling would be eccentrically loaded when kept ~ntircly within (he

~

;:':~~ i~~~e~~ :~r~:Uit~. it with that of an interior column, A combined footing may be reclanguL1r or trapezoidal in plan. ELEVATiON (4) Strap or cantilever rooLing. A slrap (or cantilever) fOOling consists of two isolated footings connected witb a structural strap or a lever, as shown in Fig. 24.4. The strap connects the two footings such that they behave as one unit. 1be strop simply acts as a connecting beam and docs not take any soil reaction. The strap L'i designed as a rigid beam . 'Illc individual footings arc so designed that Iheir combined . .. . line of action passes through the rcsult:~nt of Ihe tOial load. A FLg. 24.3. Combmcd I·oollng. simp fOOling is more economical than a combined footing when the allowable soil pressure is relatively high ,ncIlhc dis"n", betwecn the "",mns i, large. +~

0

D

~~~~ ELEVATION

@]-[QJ PLAN Fig. 24.4. Strap Footing.

(5) Milt or ran foundations. A mat or raft foundation is a large slab supponing a number of columns

and walls under the entire structure or a large part.of the strudure. A mat is required when the allowable soil pressure is low or where the columns and walls are so close that individual footings would overlap or nearty touch each other (Fig. 24.5). Mat foundations are usefu l in reducing the differential settlements on non-homogeneous soils or where 1here is a large variation in the loads on individual columns. lhc bearing capacity theories have been discussed in the preceding chapter. 'Ibe design of shallow found.1l1ons is discussed in this chapter. Ihe design is limited 10 the determination of the depth of footing, area oC foot.ing, soil pressure, shear force and bending moments. The strudural design of footings is outside the scope oC this text.

24.2, DElyrll OF FOOTINGS To perform its funclion properly. a footing must be laid at a suitable depth below the ground surface. lbe vertical distance between the ground surface and the base of fOOling is known as the depth of footing (Df ). lhe depth of footing controls the ultimate bearing capacity and the settlement, as discussed in chapter 23. While fixing the depth of COOling, the following JXlints should lIiso be considered. (1) Depth of top soil. 100 fooling should be located below the top soil consisting of organic mailers which eventually decompose. The top soil should be removed over an area slightly larger than the footing.

SOIL MECHANICS AND FOUNDATION ENGINE.ERING

638

ELEVATION

WALL

~'"

D,~~ COLUMNS

0

0

0

0

0

0

V

V- WALL

PLAN Fig. 24.5. Mat Foundation. (2) Frost Depth. The fooling should be carried below the depth of frost penetration. If the footing is located al insufficient depth, it would be subjected to the frost damage due to formation of ice lenses and consequent frost heave. Dwing summer, thawing occurs from the top downwards and the melted water is entrapped. As the soil woter freezes and melts, the footing is lifted during cold weather nnd it settles during wann weather. The shear strength of the soil is also decreased during warm weather due to an increase in water content. To prevent frost damage, the footings should be placed below the frost depth, which may be 1 m or more in cold climates. (3) Zone of Soil Volume Change. Some soils, especially clnys having high plasticity, SUdl as black cotton soil. undergo excessive volume changes. Such soils shrink upon drying aod swell upon wetting. The volume changes are generally greatest near tbe ground surface and decrease with i.ncreas.ing depth. Large volume changes beneath a footing may cause aUernate lifting and dropping. The footing should be placed below all strata that are subjected to large volume changes. (4) Adjacent footings and property lines. The Cooting should be so located that no damage is done to the existing structures. TIle adjacent structures may be damaged by construction of a new footing due 10 vibrations. undermining or lowering of the water table. The new footing may also impose additional load on the existing footings which may cause settlement. In genera~ the deeper tbe new footing and the closer to the existing structure, the greater is the potential damage to tbe existing structure. This is particularly more severe if the new fOOling is lower than the existing fOOling. As far as possible, the new footing should be placed at the same depth as the old ones, aod the si<;les of exct\vation adjacent to the existing structure shoold be suitably sUppor1ed. If the footings are plnced at different levels, the slope of line Joining the two footings should 001 be steeper than two horizootal to one vertical (Fig. 24.6), a'i per IS : 1904-1978. (5) Sloping Ground.' If a fooling is located adjacent to a sloping ground, the sloping ground surface


639

Fig. 24.6. should not encroach upon a frustum of bearing material under the footing having sides making an angle of 30° with the horizontal. Moreover, the minimum distance from the lower edge of the footing 10 the sloping ground surface should be 90 em. (Fig. 24.7). (6) Water Table. TIle footing should be placed above the ~ ground water table as far 3S possible. The presence of ground water wilhin the soil immediately around a footing in undesirable as it G. S. reduces the bearing capacity of the soil and there arc difficulties =';!=~truction. The water proofing problems also arise due to t----- O.9m-----l

is

" is

(7) Scour Depth. The footings located in streams, on water Fig. 24.7. fronts or other locations where there is a possibility of scouring, should be placed below the potential scour depth. (8) Underground Defecl
D (,)mln

_ 'l.(~)2 'I 1 + sincp'

... (24.1)

where q is the intensity of loading.

iO

;e

24.3. FOUNDATION LOADING An accurate estimation of all loads acting on the foundation should be made before it can be properly designed. A foundation may be subjeacd to one or more of the following loads. (I) Dead Lauds.. "[bc dead loads include the weight of materials pennanenUy fixed to the structure, such as beams, floolS, walls, columns, and fIXed service cquipments. The dead load can be calculated if sizes arx1 types of structural materials are known. However, there is a problem in estimating the self-weight of the structure. Hie usual procedure is to assume the self weight initially and the struaurc is designed. The weight of the structure is then found from the designed dimensions and compared with the assumed weight. 1f ncccss.1ry, the design procedure is repeated using- the revised weight. (2) Uve Loads. The live loads are the movable l
SOlL MECHANICS AND FOUNDATION ENGINEERING

It is dimwit to estimate the live loads accurately. These are specified by local building codes as uniformly distnbuted equivalent static loads. (3) Wind Loads. Wind loads act on all exposed surf".ares of structure. These loads depend upon the velocity of wind and the type of structure. Uke live loads, wind loads are also specified by building codes. (4) Snow loads. Snow loads occur due to accumulation of snow on roofs and exlerior flat surfaces in cold climates. The unit weight of snow is usually taken as 1 KN/m2. (5) Earth pressure. Earth pressures produce lateral force against the structure below the ground surface or fill surface. 1be earth pressure is delennined using the theories discussed in chapter 19. The earth pressure is nonnally treated ~ dead load. (6) Water ~~ Like earth pressure, water pressure also produces a lateral force against the structure below the water level Water pressure may also cause an upward force 00 the bottom pf the structure due 10 uplift pressure. It must be couoteracted by the dead load of the structure. (7) Earthquake loads. The force due 10 an earthquake may act vertically, laterally or torsionally on a structure in any direction. The worst oondilion should be anticipated and the relevant code consulted. The earthquake load is usually assumed as a fraction of the dead load, depending upon Ihe seismicity of the zone. Computation of Design loads. TIle dead loads and live loads on columns are usually computed by tributary area method. in which it is assumed that a oolumn carries aU the load in the floor area enclosed by lines equidistant from its adjacent columns. As live loads are temporary and transients. only a (Xlrt of it may ad for a duration thai may induce lhe settlement. especially in COhesive soils. Moreover. specified maximum live loads do not occur simultaneously at all the floors. It is usual practice to reduce !he live loads in such cases (IS : 875). It is the commoo practice to a
24.4. PRINCIPLES OF DESIGN OF FOOTINGS Before aclual design. it is essential 10 estimate the dead load, live load and other loads. The frequency and dural ion or various loads should also be available. The bending moment at the base of column (or wall) should also be ascertained if it is subjected to an eccentric load or moment. As the bearing capacity of (he soil depends upon the depth.flf footing and ilS length and width, an estimate about these dimensions is required before the actual design. For members carrying axial 10.1

DI!SIGN OF SHAlLOW FOUNDXfIONS

64J

combined with bending moments that do OOt change direction, a rectangular fOOling is more suitable than a square fooling. _. The investigation of the site should be first carried oul. 111e samplCS should be laken to determine the engineering properties of the soil. The safe bearing capacity should be calculated on the basis of soil data obtained from the tcslS using the methods discussed in chapter 23. For cohesionlcss soils, as it is difficult to obtain undisturbed samples. the bearing C3JX1city is determined from the standard peoctration test number (N) or from the plate load test The value of NIObe used is the average of the N values from the base of footings to a depth equal to width of the fOOling. The fOOling is designed using the following procedure. (1) TIlC safe bearing cap.1city is determined using the methods discussed in chapter 23. For small, unimportant structures. these values can be taken as presumptive values (fable 23.10). (2) The fOOling is proportioned making usc of the safe bearing capacity determined in Step (I). (3) The maximum settlement of the footing is determined using the methods discussed in chapler 23. An estimate of the differential sculcment between various footings is made. (4) Angular distortion is determined between various pans of the structure. (5) The maximum settlement. the differential settlement and the angular distortion oblaine
245. PROPORTIONAL FOOTINGS FOR EQUAL SETTLEMENT 'Ib reduce the differential settlement due to live load variations for fOOling; on flOe-grained soils. it is desirable [0 proportion all the footings in such a way [hat they have equal prcs.sures under the service loads. Thus all the footings would sctUe by equal amounts and the differential settlement would be considcl"::!bly reduced.1be following procedure suggcsted by Peck et nl (1974) i .. usually adopted. (I) The dead load from each column, including the weight of the footing. is determined. (2) 1he maximum live load to which each footing is SUbjected is detennincd. (3) 'Ibe ratio of the maximum live load [0 dead load for each footing is oomputcd. (4) The footing that has the lmgest live lo.1d to dead load rotio is taken as the governing looting. 1he area (Ag) of the governing footing is determined from the relation.

all~~~b:~:~~~~V~1~~~ty

'"

1C

id

Ag ... (24.2) (5) Thc service loads for all the footings are determined. (6) The design bearing capacity (qd) of aU the fooing. except the governing fooling, is determine
"ode, q, •

ley all)

on DOd

Service Lo..1<1 for governing footing

A,

(7) The area of other footings is detcnn incd as A _ Service load for tMI footing qd

... (24.3)

... (24.4)

24.6. DESIGN OF SntW FOOTINGS Plain cement concrete strip footings arc provided when the loads are light and the soil is good. If th'e


642

loads are heavy nnd the soil conditions are oot favourable, plain conaetc fooings are not eoonomical. Reinforced cemenl cooacle footings are more suitable in such cases. (a) Plain Concrete Footings. The fooling is designed so that the contact pressure on the soil docs not exceed the allowable bearing pressure. The width (B) of the fOOling is determined [rom the relation

B.

~

... (245)

where Q::: load per m run, q"o = aUowable soil pressure. If the actual width provided is different from the theorcLical width, the actual pressure is given by

qD"~

... (24.6)

lhe thickness al the edge of the footing should be at least 15 an. On cohesive soils, generally a minimum thickness of 30 em is specified in order to resist swelling pressure. The thickness of the fOOling should be adequate to minimise the development of tension on the underside of the projection acting os a cantilever.

The thickness of the projection is generally kept twice the length of the projection from the wall face. A 45 0 load distribution is also commonly used (Fig. 24.8), which gives a smalltcnsion on thc underside. For wide and thick footings, sloping of the upper surface is done to effect eronomy in the quantity oC concrete. However. the cost oC Corm work will add to the total oost. (b) Reinforced Concrete Footings. Footings carrying heavy loads on weak soils are reinforced in the transverse direction. The width of the fooling is detennined using Eq. 24.5. For computing the bending momcnt for which the footing is to be designed, the aitical section is taken as follows (IS 456 _ 1978). ~-=----

__...:!:!.DJ

(I) At the face of the monolithic wall (Fig. 24.9). Fig. 24.8. (il) Half-way between the centre line and the edge of the wall for Cootings under ma'iOnry walls.

......{41-

~I.---B ----l.1 Fig. 24.9. R.C.C. Strip FOOling.

For monolithic walls, the maximum bending moment is given by M. 'h>(B - b)'/ S where B width of footing, b

=width of wall. q(J =actual soil pressure.

... (24.~

'0

aI.

For checking the diagonal shear, the critical section is taken at a distance equal to the effective depth (d) of fOOling from the face of the wan. The diagonal shear is given by

101

.5)

643

DESIGN OF SHAu..oW FOUND,qIONS

F • qo [(B - b)12 - dJ

... (24.8)

24.7. DE'lIGN OF SPREAD FOOTINGS Spread rootin~ are u<>ed Cor distributing concentrated column loads oYcr a large area so thnl the bearing pressure is less than or equal 10 the aUowabl~ soH pressure. Fig. 24.10 shows general types of spread footings.

,6)

a

'8 (.)

(b)

(0)

Fig. 24.10.

The area of the footing is given by A • Qlq_ where Q = column load, q_ = allowable soil pressure. If the area actually provided is morc, the actual pressure is given by

qO-~

... (24.9)

... (24.10)

(a) Plain Concrete Footings. The design is similar to that of a plain concrete strip fOOling. The thickness of the footing is fixed from the consideration of preventing tension on the underside. As in the case of strip footings, the thickness is kepI equal to twice the projection or alternatively. the width of the footing is determined by the DOnnal prnctice of 45 0 distribution of loading. (b) Reinforced Concrete Footings. The area of the footing is rbtained using Eq. 24.9. The shape of the footing may be square, circular or rectangular. If the column load L~ centrally placed, the upward pressure is uniform. As the weight of the footing is directly transferred to the soil below, it does nol affect the bending moment and shear force. The critical section for bending moment is laken a<> under. (I) At the face of the column or pedestal monolithic with the footing when no metal plate is used. (if) Half-way between the face of the column or pedestal and the edge of the metal plate on which the column or pcdest1l1 rests. The maximum B.M. for the case (I) is given by (Fig. 24.11)

M. qoB(B - b)' 8

.. (24.11)

For checking the diagonal shear F, the critical section is taken al a distance equal 10 lhe effective depth Cd) of the footing from the face of the column. F. qoB[(B - b)12 - dJ ... (24.12) For punChing shear, the critical section is Iaken at a distance of d!2 from the facc of the column. GcneraUy the overall depth (do) of the footing is detennined from the punching shear considerations. do -

qo[n' - (b + d)'J 4 (b + d) O,p

••• (24.13)

where qo = actual pres<>ure, B = width of footing, b = width of square column, oq> = safe punching shear. The depth provided is checked for bending moment, shear and bond.

SOIL MECHANICS AND fOUNDA'I1ON ENGINEERING

644 Mt.t PlJNCHI NG

'f-¥

-1

Fig. 24.11. R,C.C. Spn:Ad FOOling. The m
Fb In the case moment in each contact prcs.:>ure the column. Inc fOOling

=

qo I) (LJ - /))/2

... (24.14)

of rectangular footings, the length
sqUlJfC

footing for

a circular column. the circular column is gencmlly substituted by an equivalent sqWITC column or Ihe same area. The design of the fOOling is then done using the procedure already described for a squnTC fOOling. 'JOe dcsign of the circular footing am also be done using the basic principles. I1ccausc of the curved areas, the expressions arc morc involved than those for :1 square footing. The thickness of the fOOling at the edge shall be not less th:1O 15 em for footings on soi l, nor less fh.
24.8. DESIGN OF ECCENTRICALLY LOADED SPREAD FOOTINGS A column is said to be eccentrically loaded when it is subjected 10 a load which is 001 centric or wilen it is subjected \0 a bending moment in addition to the centric lood. If the centre of gravity of the footing ti made to coincide with the centre of gravity of the centrically loodcd column, the pressure distribution ti trope-.widal, with the maximum pressure on one side and the minimum pressure on the other side. However, if the centre of gravity of the footing is placed eccentric with the centre of gravity of the column, the pressure distribution becomes unifonn (Sec Fig. 23.20). The eccentricity (e) of the footing is given by e_ MIQ when M bending moment, and Q = axial 10•.1d.

... (24.15)

DESIGN OF SIIAU ..OW FOUNDATIONS

645

If the footing is provided with the above ecccmricity. the resultant of the bearing pressure coincides with the resultant of the loads. 111e area of the fOOling is given by A •

;f

...

(24.16)

'Ine longer projcction of the footing is designed as a c
qm~ · 6(1+~)

... (24.17)

whcre e is the eccentriCity. If e > B16, the maximum pressure is given by

qm"'~

... (24.18)

The dimensions L nnd B of the footing are chosen such that the maximum pressure (qrrwr) docs not exceed thai allowable bearing pressure (qna)' The bending moment, the diagonal shear and the punching shear arc determ ined consklcring the trnpczoidal pressure distribution.

24.9. COMUINED FOOTINGS A combined foot ings supports two columns. When a foundation is built close to an existing building or the propcny line, there may not be sufficient space for equal projections on the sides of the exterior column. This results in an ccccntric loading on the fOOling. It may lead to the tilting of the foundation. To counteract this t-ilting tcndency, a combined footings is provided which joins the exterior column with an interior column. A combined footing is also required when the two individual footings overlap. In the conventional dcsign. the combined footing is assumed to be infinitely rigid and the soil pressure distribution to be planar. In the elastic line method. thc Oexibility of the footing is considered. The elastic line method is more rational but quite involvcd (see Sect. 24.1 7). 1be following discussion is limited to the conventional design. 'Ilte footing is proponioned such that the centre of gravity of the footing lies on the line of action of the resultant of the column londs. The pressure distribution thus becomes uniform. A combined fOOling is generally rcctangular in plan if sufficient space is available beyond each column. If one of the columns is near the propcny line, the rectangular footing can still be provided if the interior column is relatively hc.1vier. However. if thc interior column is lighter, a trapezoidal footing is requi red 10 keep the resultant of the column loads through the centroid of the fOOling. Thus the resultant of the soil reaction is mnde to coincide with the resultant of the column lo..1ds. ,n

!n

is is :r,

re

5)

24.10. RECfANGULAR COMIHNED FOOTINGS 111e design of a combined footing consists of selecting length and width of the fOOling such that Ihe centroid of the fOOling and the resultant of the column loads coincide, With the dimensions of the footing established, the shenr force and bending moment diagmms arc drawn. The thickness of the fOOling is selected from the bending moment Imd shenr stress considerations. The footing is designed as a continuous beam supported by two columns in Ihe longitudinal direction. The reinforcement is provided as in a continuous beam. lhe procedure consists of the following steps. (I) Determine the total column loads.

Q - QI + Q2 where QI and Q2 are the exterior and interior columns. respectively.

... (24.19)


(2) Find thc base area of the footings.

. .. (24.20)

A - QJq_ where q,... is the allowable soil pressure.

(3) Locate the line of action of the resultant of the column loads measured from onc of the column, say cxterior column (fig. 24.12).

1-----'2----1

I·

T

PLAN

1

Fig. 24.12. Rectangullir Combined FOOling.

i .. Q2)( ~Q where x2 is the distance between columns. (4) Dctcrmine thc total length of the footings. L .. 2 (i + b l l2)

... (24.21)

... (24.22)

where b J = width of the exterior column. (5) Find the width of the fooling. B .. AIL ... (24.23) (6) As thc actual Icngth and width that are provided may be slighUy more due to rounding off, the actual pressure .is given by

... (24.24) where Ao is the actual area. (7) Draw the shear force and the bending momcnt diagrams along the length of the fOOling, considering the pressure qo. For convenience, the column loads arc lakcn as oonccntrie loads acting at tbe ccntres. (8) Determine the bending moments at the face of the columns and the maximum bending moment al tile point of zero shear. (9) Find the thickness of fOOling for the maximum bending momenl. Check for diagonal shear and punching shear, as in the case of an isolated footings . Check. for bond at the point o( contraI1exure.

647


(10) Determine the longitudinal reinforcement for the maximum bending moment. For transverse reinforcement. assume a width of (b + d) to take all the bending moment in the shon diceaion, where b is the column side and d is the effective depth. 24.11. TRAI'EZOIDAL FOO"J1NG Trapezoidal combined footings are provided to avoid eccentricity of loading with respect to the ba<>e, as already mcnLioned. Trapezoidal footings are required when the space outside the exterior column is limited and the exterior column carries thc he:lVier load. The design procedure consists of the following stcp>: (1) Odermine the total column loads.

Q - Q, + Q,

... (24.25)

(2) Find the base area of the footing. ... (24.26)

A - Q/q..a (3) Locate the line of action of resultant of the column loads (Fig. 24.13).

x-

Q2~/Q

... (24.27)

INTEROR

T

1"

r

1 Fig. 24.13. Trapezoidal Combined F<:JOI.ing.

(4) Determine the distance x' of the n;<>ullant from the outer face of the exterior column. x-x+hJI2

.•. (24.28)

A trapezoidal footing is required if L/3
x

< L/3, a combined footing cannot be If x' _ LIZ, a rectangular footing is provided. However. if provided. In such a case, a stmp fOOling is suitable. (5) Detennine the widths B J and 8 2 from the following relations.


ill + 8 2

- 2 - xL .. A

!::..(8 +2B 1

and

3

2

BI + B2

...(24.29)

)_x

... (24.30)

Solving Eqs. 24.29 and 2430,

n2 _ ~ L (~L and

BI -

¥~

1)

... (24.31) ... (2432)

B2

(6) Once the dimensions BI and 8 2 have been found, the rest of the design can be done as in the case of a rcctangular combined footing. 2-1.12. STRAI' FOOTINGS 1\ strap fooling is required in the following two cases: (I) When x' < 1.13. where x' is the distance of the resultant of column loads from the exterior face of the exterior column. and I. is the length of the footing. •. (2) When the distance between the two columns is so large that a combined footing becomes excessively long and narrow. I\s mentioned before. a stmp fOOling consists of twO spread fOOlings joined by a rigid beam known a<; a .~Ir(jp. ·111e strap is nOl. subjected to any direct soil pressure from below. lIS main function is to transfer the moment from the exterior footing to the interior fOOling. ·111e following assumptions arc genernl1y made. (n) The soil pressure is uniform beneath e..1cil individual footing. (h) The stmp is perfcctly rigid. (c) The stmp is weightless. (tl) lhc interior footing is centrally loaded.

·lnc design procedure consists of the following steps. (I) I\s.<;umc a reasonable value of the eccentricity (e) between the' load Q! and the reaction R t on the exterior column (Fig. 24.14). (2) Determine the length of the footing of the exterior column. L, - 2 (e + 0.5 b,) ... (2433) where hi = width of the exterior oolumn. (3) Compute the reaction RI • by taking moments abou t the line of action of R2 •

RI _ QI x ;(2/5

... (24.34)

where.\"2 = distance between loads QI and Q2 and S = distance between reactions RI and R2 • (4) Compute areas A I and A 2·

A I'" Rllq,,,,

A 2 ... R2/qnfl The rC::lclion R2 is obviously cqunlto (Q I + Q;t} - /(1 .

and

.. .(243~ ... (24.30)

(5) Find the widths of footing;. .(24J~

ill " AIII' 1 and

Ih ...

%

(6) Dcsign the individLUIl footings
... (24J8) lh~ ca<;~

or !'.prc:uJ [(x)ting.'>.

649

DESIGN OF SHAUOW FOUNDA110NS

Fig. 24.14. Strap Fooling.

(1) Detennine the depth of the strap for diagonaJ shear and bending moment. [Note: It may be mentioned that a number of designs of the strap footing are the selected value of e.]

~ible,

depending upoo

24.13. PRINCIPLES OF DESIGN OF MAT FOUNDATIONS A mat (or raO)isfi thick reinforced oonaete slab which supports all the load-bearing walls and oolumn loads of a structure or a large portion of structure. A mat is required when the loads are heavy and the soil is very weak or highly compressible. A mat is more economical than individual footings when the total base area required for the indjviduaJ footings exceeds about one-half of the area covered by the structure. A mat is preferred to individual footings when the soil mas<> has very erratic properties and contains lenses of compressible soils. In such a case, it would be difficult to control the differential settlements if individual footings are provided. The mal spans over weak patches of the soil and thus the differential settlements are considerably reducx:d. Like all other shallow foundations, a mat must be safe against shear failure and the settlements should be within the allowable limits. As the width of a ran is very large, the pressure bulb is quite deep. Thus tbe loose soil pockelS under a raft may be more evenly distributed. 'Ibis results in a smaller differential settlement than individual footings. It is customary to assume that a differential settlement of 19 mm would occur in a raft when the maximum settlement is twice that in the individual footing. Thus a maximum settlement of 50 mm can be pennitted when the differential settlement allowed is 19 mm. As rafts are generally at some dcpth below the ground surface, a large volume of soil is excavated and, therefore, the net pressure on the soil is considerably reduced. An advant.'lge of this redUCtion in the pressure can be taken while designing a raft.

upon(~e~~h~~s ~=7: c~~~~rl~~ ::~~gw~:~'l o~: ~o~~~t:~e~nth~=~e:;~y ~~:

a and, therefore, the shear failure generally does not occur. Accordingly, the safe settlement pressure, and not tbe bearing capacity, generally governs tbe design. except for vcry loose sands (N < 5).

SOIL MECHANICS AND fOUNDA:nON ENGINEERING

6SO

Settlements depend qX'm the depth of the soil slralu;n. If a finn stratum exists at a shallow depth below the raO. the settlements orc small. However, if the sand deposit extends to a grc.'1t depth, the scUlcmcnts would be large. lhc al10wable soil pressure am be found using the following equations. The sn[c bearing capacity can be dctcnnincd as (feng, 1962),

q,.., • O.22N2 DWT + 0.67 (100 + N 2) D, Wq

... (24.39)

The safe sctlicmcnt pressure for a sctUemcnl of 2S mm is given by qnp -

17.5 (N - 3) Wy kN/m

2

... (24.40)

where B = smaller dimension of ran (III), D, = Depth of foundation, WY' Wq = waler table correction factors. 'Ihe pressures qlU and q"p arc in kN/m2, The smaller of the two values is the allowable soil pressure (q"..). As stated e.'lrlicr, Teng's equation for the safe sclliemcnl (qllp) is conservative. Using Dowie's equation [[q. 21.82 (b)]. for a settlement of 25 mm.

q"P .. 12.2 N where Rd = deptb f:Jctor = 1 In general,

(n

0 +0 .3

+ 0.33 (D/D)

q"p" 12.2N

(D +D 0.3

f fR" Wy(fs)

Rd Wy kN/m2

... (24.41)

.. .(24.42)

where s = allowable settlement (mm). As the width of a ron is very large,

03 D +0 . .. 1.0

Therefore, Taking Rd

..(24.43)

= 1.00

and

s = 50 mm, 2

. .. (24.44) q"P .. 24.4N Wy kN/m In case of rans, as the width D is very huge, and the pressure bulb is deep, the water table generally affects the safe settlement pressure. Taking Wy .. 0.5, q"P .. 12.2 N kN/m 2 ... (24.45)

Pede ct .'II (1974) express Eqs. 24.44 and 24.45) as

q"P ..

WN Wy kN/m2

...

(24.46) 2 q"P .. ION IcN/m ... (24.47) .1be above equations are ~pplic.lble for 5 s N ~ 50. If the value of (N) aftcr correctioo is less than 5, the sand is too loose for a raft foundation. 100 saod should be either rompactoo or a deep foundation, sud! as pile foundation, should be provided. For values of (N) g~\ter than 50, the above equations give unconservative results. Acoording to IS : 6403, for a settlement of 65 mm, the safe settlement pressure is given by and

Talcing W, _ 0.5,

q"P - 25.4 (N - 3) W, q"P _ 12.7 (N - 3)

... [24.48(.)J .[24.48(b)J

alpocity (q,..-) can also be determined using the equations developed in Chaptcr 23. Generally, the safe bearing c::lfXIcity is much greatcr than the safe settlement rressurc and is not of mud!. significance. The safe benring

As the raft foundations are generally used below basements, tbe foundations are not backfilled. Eq.23.16

can be used. Thus

DESIGN OF SHAllOW FOUNDATIONS

651

.(24.49) where Q = superimposed load, A = area of raft, Of = depth of foundation. (b) Rafts on Clay. The net ultimate bearing capacity is generally detennined using Skempton's equations

(Eq.23.55). q~ - 5 (1 + 0.2 DID) [1 + 0.2 (BI L)] c. where Cu = undrained cohesion. All other notations have been defined above. The safe net bearing capacity can be obtained as

. .. (24.50)

... (24.5 1)

qns - q"u lF

where F = factor of safety. Under nonna! loading conditions, the factor of safety should not be smaller than 3. IS : 6403 recommends a minimum factor of safety of 2.5. In case of rafts on clay, as the safe bearing capacity is independent of its size. it generally governs the design. But the safe settlement pressure is also important in some cases. In the case of rafts, the pressure bulb extends to a much greater depth than that for an isolated footing. The settlements of rafts on nonnaUy consolidated clays are usually very large. However, in the case of over-consolidated clays, tbe settlements are small. The setUemenlS arc calculated due to the net increaw in pressure, given by q.- QI A - YDf ... (24.52) If the soil stratum extends to a depth greater than about twice the width of the mat, the load on the mat would tend to act as a point load for the soil at large depths, and the settlements would be the same whatever be the type of foundation. In such cases, increasing the width of the mat does not help in reducing the settlements. If the settlements are large, deep foundations, such as piles or drilled caissons, would be more suitable. The factor of safety against bearing capacity failure can be written a<; F - q,.,fq".

F _ 5 (1 + O.2D/B)(1 + O.2BIL) C ... (24.53) QIA yDf The settlements of a mat foundation can be reduced by decreasing the net increase in pressure, i.e. , by U

inc,easing Df" Foe no yin~;e~~: tbe net pressure, E
Df -

fA

..

24~52 gives

I

,

~

.

"

~

o

o

(24.54)

G.5 .

A foundation that satisfies Eq. 24.54 is known as fully compensated or a floating fOlmdation. For such a foundation, tbe settlement is (theoretically) zero. The factor of safety against bc.'lring capacity failure, as per Eq. 24.53, becomes theoretically infinite. 24.14. COMMON lYPES OF I\tAT FOUNDATIONS 'Ibere are several types of mat foundations. (1) Flat Plate lYpe. In Ihis type of mat foundation, a mat of uniform thickness is provided (Fig. 24.15). This type is most suitable when the column loads are relatively light and tbe spacing of columns is relatively small and unifonn. (2) Aut Plute Thickened Under Columns. When the column loads are heavy, this type is more suitable than the flat plate type. A portion of slab under the

o

o

.-.--B---- D -

o

o

o

Plan Fig. 24.15. Flat Plate Type Mll t.

o

652

SOIL MECIIANICS AN!) FOUNDATION ENGINEERING

rl ~ ~ ~ @]

@]

@]

@]

@]

@]

@]

@]

0

0

[~J 0

@]

@]

@]

0

PLAN

r---..,

L___ j 0

0

[~J

L' __ .J,

0

0 r----;

,

,-~

0

L_j

@]

0

,

0

0

Pion

Fig. 24.16. 1'1.11 Plalc-Thicknened Under Columns

Fig. 24.17. l3cam and Slab Construction.

column is thickened to providc enough thickness for negative bending moment and diagonal shear (Fig. 24.16). Somctimes. instead of thickening thc slab, a pedcstal is provided under each ooiumn above the slab to increase the thickness. (3) Beam and Slab Construction. In this type of construction, the beams run in two perpendicular directions and ' a slab is provided betwccn the beams. l he rolumns arc located at thc intersection of beams (Pig. 24.11). This type is suitable when the bending stresses are high because of large column spacing and unequal column loods. (4) Box Structun!S. In this type of mat foundation, a box structure is provided in which the basement

.. ------.., r-----.,

i PLAN

II

r--------,

II

:

iL______ J11L_____ 1 Li _____ Ji ~

Fig. 24.19. Ma,I on Piles.

DESIGN Or: SIIAIJ.OW FOUNDATIONS

'53

walls act a... stiffeners for the mat. Ooxcs may be made of ccUular construction or rigid frames consisting of slabs and basement walls (fig. 24.18). 'This type of mat foundation can resist very high bending stresses. (5) Mats placed on Piles. "The mal is supported on piles in this type of construction (Fig. 24.19). This type of mat is used where the soil is highly compressible and the water table is high. This method of construction reduces the sett lement and also controls buoy'lOcy.

24,15. DESIGN ME'1l10DS FOR MAT FOUNDATIONS 'Ille methods of design of mnt foundations c.m be classified into 4 categories, depending upon the assumpiions made. (I) I~igid Hearn Method. In this method of design, the slab is considered to be infinitely rigid as compared with the subsoil. 'Inc nexurni deflections of the mat in this case do not influence the contact pressure distribution acting on the mal. The pressure distribution is assumed to be planar. The centroid of the soil pressure coincides with the Une of action of Ihe resultant of all the loads acting on Ihe mal. '111is method is also known as the conventional method of design. The design of combined footing discussed in Section 24. JO and 24.11 wa.<; aL'>O based on this method. (2) Simplified Elastic Mcthtlcl. '111is method is b:lscd on the assumption that the soil behaves like an infinite number of individual im!ercndcnt clastic springs. "flle springs lic solids. this method also gives approximate solutions. lbe method is complicated and rarely used in a dcsign officc. (4) Non-linear Elastic Mdhod. 111e soil is considered to be a non-linearly clastic solid. "Inc method represents Ihe behaviour of :lctual soil more closely than the clastic method and is more accurate. Numerical techniques, such as finite clement method, arc required for lhe design (Arora, 1980). The method has nol developed to a stage that this can be used in a design officc. In this text. only the first two methods are discussed.

24.16. CONVENTIONAL DESIGN OF RAFT FOUNDATIONS In the conventional method of design, Ihe raft is assumed to be infinitely rigid and the pressure distribution is taken as planar (!incnTly varying). The assumption is valid when the ran rests on a soft clay which is highly compressible, and the eccentricity of Ihe load is small. In case when the soil is stiff or when the eccentricity is large, the method does not give accurate rcsults. The elastic method, which takcs into account the stifTness of the soil and raft. is more economical and IIccurate in the I'll\er case. The Simplified el.1stic method is discussed in Sect. 24.17. According to the American Concrete Institute Committee 436 (1966), the design of mats should be daDe using the conventional method if the spacing of the columns in a strip of the raft is less than 1.75(A. metres and usi ng the simplified elaslic method when it is greater than 1.75{A, where A. is given by

A -

Blk)V. (4EJ

... (24.55)

where k = coefficient of subgrode reaction (kN/m\ 8 1 = width of Ihe strip (m). E modulus of el.1slicity of the raft material (kN/ml, I moment of inertia (m').

=

=

lhe coefficient of subgradc reaction of a soil is Ihe pressure required to proouce a unit settlement of a plate. It is given by

k _ qlz

... (24.56)


654

=

=

I

whcrc q pressure (kN/m\ z settlement (m), k ::: coefficient of subgmdc reaction (kN/m~. 'Inc cocITicient of subgrade reaction is not a constant for a given soil (ferzaghi, 1955). It depends upon a number of fadors, such as length, width, depth and shape of foundation (sec Sect. 24.20). Procedure. The procedure for tbe . conventional design consists of the foUowing stcps: (1) Dctermine the line of action of all the loads acting on the raft. The sclfweighl of the raft is not considered. as it ·is taken directly by the soil. (2) Determine. the contact pressure distribution as under. (a) If the resultant passes tbrough thc centre of the raft, the contact pressure is given by

(b) If the q -rcsuUant QIA has an eccentricity of ex and i!" in x-and y-dircctions [Fig. 24.20 (a)l.

T

1 L

n

f

~ 8,

y ... (24.57) q _ Q. (Q'eJ x. A Iyy In The maximwn contact pressure should be less than the allowable soil pressure. (3) Divide the slab into strips (bands) in x-and y-dircctions. Each strip is (a) assumed to act a'i independent beam subjected to the contact pressure and the column loads. (4) Draw the shear force and bending moment diagrams for each strip. (5) Determine the modified OOlumD loads as explained below. It is generally found that the strip (b) docs not satisfy statics, i.e. the resullant of Fig. 24.20. column loads and the resultant of contact pressure are not equal and they do not act in the same line. The rcason is that the strips do not act independently as a'iSumed and there is some shear transfer between adjoining strips. Let us consider the strip carrying column loads Qh Q2 and QJ in Fig. 24.20 (a). Let B J be the width of the strip. Let the average soil (contact) pressure on the strip be q__ Let B the length of the strip. Average load on tbe strip.,

1

Q.... -

t(

Q"" -

t

downward load

+ upward force)

(QJ + Q2 + QJ + q.... DJB)

... (24.58)

The modified average soil pressure is given by

_

qav-q""

(Qw) qlt>BJB

... (24.59)

DESIGN OF SHALLOW FOUN0XI10NS

655

The column load modification factor (F) is given by

F Q~ ... (24.60) Q\ + Q2 + Q) Allihe column loads arc multiplied by F for that strip. For tttis strip. the column loads are FQI' FQ2 and

FQ,. (6) The bending moment and shear force diagroms are drawn for the modified column loods and the modified average soil pressure W",,). (1) Design the individual strip;> for the bending moment and shear force found in step 6. The rufi is designed as an inverted floor supported at oolumns. As the analysis is IIpproximate. the actual reinforcement provided is twice the compuled valuc.

24.17. DESIGN OF COMBINED FOOTING BY ElASTIC LINE METHOD lhe design of a combined footing discussed in Sects. 24.9 to 24.1 1 is the conventional method. in which the footing is assumed 10 be rigid and the soil distribution is taken as planar. In the clastic line method. tbe stiffness of the fOOling is considered. Whcther a footing should be considered as rigid or not depends upon the soil charncteristics and the size of the footing. Hetcnyi (1946) gavc the foUowing criteria whether a beam should be considered as rigid or flcxible. As a combined footing acts like a beam on elastic founwltion. the same criteria can be used. Rigid footing AL < 0.80 Flexible fOOling ')...L > 3.0 For the value 0.8 < AL < 3.0, the footing is of intcnnediate flcxibility. The value of AL is computed from the following relation

AL _ where k E

Bid. ,)V' ( 4EJ

__ .(24.61)

=coefficicnt of subgrnde rC..1C1ion, n =width of footing, L length of footing, = modulus of footing m..1tcrial, and I = moment of inertia of fOOling.

The flexible footing is designed as infinite beam 00 elastic foundation, whereas the footings of intermediate flexibity is designed as a finite beam . According 10 the theory of beams on elastic found..11ions,

4 --

EJ yk B
of.

."__ (24.62)

= coefficient of subgrndc reaction.

BsinAx) + e-U (Coos AX + DSinAX)

_(J!"-)V' ). 4EJ

where

__ .(24.64)

s~r_._b A

___ ~, C

I

y

__ .(24.63)

I Fig. 2421.

SOIL MECHANICS AND FOUNDAllON ENGINEERING

656

Applying the boundary condition<; (Fig. 24.21), (he solution can be expressed as (Winterkorn and Fang, 1975). PkA ... [24.65 (z)] Y"kiJ

M"tr. B

... [24.65 (b)]

... [24.65 (c)] V " PC where A, Band C arc trigonometric identities, which arc quite complicated. A digital computer is required for the design of footing using above method.

24.18. FlNITE DIFFERENCE ME'mOD FOR COMBINED FOOTINGS A'S the rigorous solution for a combined footing on elastic foundation is complicated, numerical techniques arc used to obtain an approximate solution. A method proposed by Malter (1960) using finite difference is described below. TIle procedure consists of the roHowing steps (Jeng, 1962). (1) Divide the footing into 4 to 6 equal segments, each of length h. For illustration purposes, the beam in Fig. 24.22(a) is divided into 3 segments. (2) Let a, b, c and d be the deflcctions (settlements) at nodal points A, B. C and D. The soil reactions at these points are, respectively, ok,. b k#> c k, and d "" where k, - k x B. in which B is the width of the f()()(ing.

ee) S~ l

~ks

pressure

I IiI 1 et, 1

Cd) Equivalent

pressure

(e) EqlJi-..(Jtent

loads

Fig. 24.22.

1

PI

DFSIGN OF 'SHALLOW FOUNDATIONS

657

(3) The soil pressure is continuous [Fig. 24.22 (e)]. However, it can be approximated by steps [Fig. 24.22 (dJ]. For more accurate resulls, it can be represented by straight lines or parabolas. Replace the soil reactioru; by equivalent concentrated reactions R", Rs. Rc and Ro. as shown in Fig. 24.22 (e). (4) The fOOling under the applied load and the equivalent reactions should satisfy the equations of equilibrium, viz.,

IM .. O and IV ..

a

Write moment equilibrium eqU(ltions for each panel point. For example, for the point B, MB .. (a -

!:

+ e) EJ

In general,

... (24.66) ... (24.67)

Also write force equilibrium equation for the whole system. (5) Solve the simultaneous equations formed in step 4 for values of a, b, c and d. (6) Determine the actual pressure distribution from the calculated values of a, b, c and d in step 5. (7) Design the beam for the pressure distribution found in step 6.

24.19. ElASTIC PlATE METIIOD FOR MATS The design of a mat (raft) on Winkler's bed can be done using the methcKI proposed by the American Concrete Institute Committee 436 (1956). The method is based on the theory of plates 00 elastic foundations.

The step-by-step procedure is given below (Das, 1984). (1) Assume a tltickness (I) for the mat either by experience or using the conventiooal (rigid) method. (2) Determine the flexural rigidity (D) of the mat as

D.~

... (24.68)

12(1_.')

where E = modulus of footing material., jJ. = Poisson's ratio of footing material. (3) Determine the radius of effective stiIToess (L ') from the following relation.

L' • (Dlk)v,

... (24.69)

The zone of influence of any column load is of the order of 3L' to 4 L'. (4) Find the tangential aod radial momenlS at any point caused by a column load using the following equations.

[A _(I -.)A'j Q[ (I - .)A'j

Thngential moment,

M • _Q f 4

Radial moment,

M' '' -4

I

TIL'

I.I.AI+~

... (24.70)

... (24.71)

where r = radial distance from the column load, Q = column load.

as shown in Fig. 24.23. In cartesian coordinates, the above equations can be written as

AI and A2 are functions of (rIL') ,

Mx" M,sin 2 a + M,CCE.2 a M, _ M,CC62 o: + M,sin2 o: where a is the angle which the radius r makes witb x-axis. (5) Determine the shear force (\I) per unit width of the mat caused by a oolumn load as

... (24.72) ... (24.73)

6,.


v - - 4~'

... (24.74)

A]

where A] is a fundion of rl L' (Fig. 24.23).

5

,

l

1\ ~

3

Aq 2

V

VV'

'1-----1:

o

- 0'10

r---

-OJ -0-2 -D·'

O{l

I'-- ~ 0'1

0-2

0-3

0 L.

-4--A1 .A-z,A3 ---. Fig. 24.23.

(6) If the edge of the mat is located in the zone of influence of a column, dctcnninc the moment and

shear along the edge, ussuming that the mal is continuous. Moment and sbear opposite in sign to those detennined are applied at the edges to satisfy the known conditions. (7) If the zones of influence of two or more columns overlap, the method of superposition can be used to obtain the total moment and shear. 24.20. FlNITE DIFFERENCE METlIOD FOR MATS The mat is assumed 10 be a plale supponed by a bed of uniformly distributed coil springs with a spring constant k equal to the roefficicot of subgrade reaction, i.e. Winlder's bed. The differential equation for the deflection of the plate is given below. iJ" w + 2 ~ +

ax'

a;;'a;

a"w _ ~

ay'

... (24.7l)

D

where q = intensity of loading, k = coefficient of subgradc readion, w = deflection (settlement), D = flexural rigidity of plate (Eq. 24.68). Eq. 24.75 can be solved by finite difference method after dividing the mat into a square mesh (h )( Jr.). In the case of an interier point a (Fig. 24.24), the deflection wa can be expressed in the difference form as

20wa - 8(w, + w~ + w,+ wI) + 2(wt/ +W" + wb/+ woW) + (wu + whO + w/l + w") -

9!i t£ D + D

.

... (Z4.7~

where WI> w". .. . ate deflections at adjoining points. 1bc suffixes are shown in Fig. 24.24. 'The suffixes 4 ~ band r are respectively for top, left, bottom and right.

Q := concentrated load at point a. All other notations have been defined earlier. Doc difference equation can be written for each rigid point. [Note : For points ncar free edges, the difference equatioos is modified to account for boundaJy conditions.]

;)

DESIGN OF SHAlLOW FOUNDATIONS

659

It

II

t1

t

t,

l

a

,

bl

b

b,

"

bb Fig. 24.24.

By solving the simultaneous equations formed, the deflections at all points are determined. A digital computer is normally required. After the defiecl.ions have been detennined, the bending moment at any point in each direction can be found using the equations of the theory of elasticity. M, - M; + ~ My' •.. (24.71) wbere MJ(. "" total bending moment in x-direction, MJ(.' "" bending moment in x-direction without including the effed of bending moment in the transverse y-direction, My' = bending moment in y-direction without including the effect of bending moment in x-direcLion. Using the finite difference operators, the total bending moment on a strip in the I~ direction for an interior point is expressed as. M I_r

-

f [(WI -2W" + w,) + ~(w,-2w" + Wb)]

-h

... (24.78)

Thus the bending moments are computed at all points. The accuracy of the metbod depends upon the flOeness of the grid. When the squares of the grid are considerably larger than the size of columns, the results are not reliable adjacent to the columns. This diffirulty is overcome by introducing subdivisions of the grid adjacent to the columns. However, it increases the computational work considerably. 14.21. COEFFICIENT OF SUBGRADE RFACTION The coefficient of subgrnde reaction in the field can be determined by conducting a plate load test, using a square plate of the size 03 m x 03 m. A load-settlement curve is drawn, a<; explained in chapter 29 and the coefficient of subgrade reaction is detennined using Eq. 24.56. Table 24.1 gives the approximate values of the coeffident of 'ubgrade reaction (k) for different soils. As the values are for a square plate of size 03 m x 03 m, these are modified to account for the effect of size, shape and deptb as explained below. 1. meet of the size. (a) Cohesioruess soils kBxB _ «:0.3 .. 0.3

ry where B is the size of footing.

(B ~0.3f

... (24.79)


660

Thble 24.1. Coefficient of SubgOlde Reaction (k) S.No.

Value of k

COlulitiOIl of &iI

Type of Soil

MNfm

Dry or moist SOlnd

2.

8)(1031025)(10'

Mediwn

2S 10 I1D

2.S x 10 10 1.2 x loS

Do"",

120 to 350

1.2 )( loS 10-3.5 x las

a"

4

l.oo.
101015

1.0)( 105 10 1.5 )( 104

Medium

35 to 40

3.5 )( 104 to 4.0 )( 104

120 to 150

1.2)( ld' to 1.5 )( lOS

Do,,,, 3.

/cN/m]

81025

Loooe

Salurated sand

J

Stiff

101025

4 1.0 )( 10 to 2.S )( 104

Very Stiff

25 10 SO

2.S )( 104 to 5.0 )( 10'

>50

> 5 )( 104

Hard

(b) Cohesive soils kBxB - ko.3 " 0.3

where IcBx9

... (24.80)

(0.3/B)

k for size B x B metrc:s, and .to.3,,0.3 - Ie for size 0.3 m )( 0.3 m. 2. Effect 01 shape. For a rectangular foundation (L )( B), the value of k is given by -

kL

,,8 -

~kBwB (1

... (24.81)

+ BIL)

J . Effect or depth. For cohesionlcss soils, as the ronfincment pressure inaeases witb the depth. tbe modulus of elasticity increaies with depth. This causes a reduction in the settlement of the plate. Consequently, tbe value of Ie increases with depth. For cohesive soils, there is no Significant change in the value of " with depth.

[Note: Refer to chapter 29 for more details about k] ILLUSTRATIVE EXAMPLES

car;~;~ra!:: !X;;;t~/;:'~h~!t!~t;~£~~s::e~t ;;;:~:(ooting for a 1 m wide concrete Solution. From Eq. 24.5, B .. : Actual soil pressw-e

(90) ..

~

.. 4 m .. 200 kN/m 2

Maximum 8.M. per m run (Eq. 24.7), M _ q, (8 - b)' _ 200 (4 - I)' _ 225 kN-m/m

8

8

From Eq. 24.8, the diagonru shear is given by. taldllg d = 0.52 m,

F_q, [ (8; b) _d]_ 200 [(4 ; Fig. E--24.1 shows the footing . Il also shows the reinforcement.

1) _ 0.52]- 196 kN

)Wl[1


661

~ ~WAll

--'580mm

! .

20mm BARCPl &5cm lOmm BAR[.Il0cm 1520mm

fl.

4m---j Fig. E24.1.

illustrative Example 24.2. Design a square reinforced concrele fooling for the following dnta (Fig. E·24.2). Column load == 800 kN 1 Allowable soil pressure = 200kNlm Size of column =0.4mxO.4m

1. 335mm

275mm

Tl=I.=~2~",=~.II Fig. E-24.2.

Solution. From Eq. 24.9, A ..

f .: .

Size of footing.

B ..

V4 ..

Actual pressure,

qo .. 2~ 2 .. 200 kN/m 2

4 m2

2m

From Eq. 24.13, the overall depth (do) is given by

qo [8' - (b + d)'l

do· Taking

4(b + d)a,p

as 1000 kN/m2, and cover as 0.06 m, 2 d + 0.06 .. 200 (2 - (0.4 + d)21 4 (0.4 + d) x 1000 Solving. d .. 0.23 m. A~opt 0.275 m. osp

Therefore, From Eq. 24.11.

do ..

0.275 + 0.06 .. 0.335 m

qoB(B-b)' M • --8-.. 200 )( 2

~ (2 - 0.4)2 .. 118 kN-m


. .2

From Eq. 24.12, the diagonal shear is given by

F-'IoBW;b)-d] 04 _ 200 x 2 [(2 -2 . ) - 0.275]- 210kN The maximum force for bond is given by Eq. 24.14 as

Fb

..

b) .. 200 )( 2 x 2""" 1.6 B2 - boB 320kN (

llIustmtlve Example 24.3. Design a rectangular combined footing for two columns slwwn ill Fig. E·24.3 (a). Take allowable soil pressure as 100 kNl,,?

t Lt--L...JL...Lt--'.t--,-t. .Lt---11--1-t...J.t-.L-L.Jttm., L

(b) Pressure diagram

+626.2 kN

+

t!

tJBkN

~,1=:=;Clli;;:L+~:::::l2:§:.6J~mL=t:;:::;::7 _271·6kN

Ie) S.F. Ologrcm

-S61.o kN

()

~

-~kN_m

{d} 8 . M. Diogram

Fig. E·243.

Solution. Total load.

Q - 600 + 900 - 1500kN

From Eq. 24.21,

x .. Q~~ .. ~~ 5 115~

From Eq. 24.20,

A ..

From Eq. 24.22,

L .. 2

.. 3m

.. 15m2

rx + bilZ}

L - 2(3.0 + 0.3/2) - 6.3m

~

-IS1.SkN-m

kNlm


From Eq. 24.23. B

=AIL = 1516.3 = 2.38. say 2.4 m. 'lo ..

From Eq. 24.24,

~

..

6.315~2.4

.. 99.2 kN/ml

Actual pressure per metre run .. 99.2 x 2.4 .. 238.1 kN/m Fig. E-24.3 (c) shows the shear force diagram. The shear force is zero at a distance of 2.37 m from column-l. The maximum shear below column-l is 564 kN and that below oolumn-l is 626.2 kN.

Fig. B-24.3 (d) shows the 8.M. diagram. The maximum n.M. is 666 kN-m. The footing can be designed for the above values of n.M. and S.F.

OIfO!!:-::::;!:S::pI~ ;~. :;=~n a trapezoidal footing for

the

two

columns shown in Fig. £-24.4. TaU

Q .. 2000 + 1500 .. 3500 kN

Solution.

1"00

ooo kN

Q.5mXO.'m

~~1~1========~'~~~.'~m~=======:~1

kN

rl-I

~

L _ _ _ _ _ _ _ _ _ _ _I

05m".'m

I

T

T

MOm

1.6Sm

1 1~

1

~326""/m _

_

== X

1)4

(e)

Pr~~sur~

+186

(d)S.F. Diagram

-1814

V'

)42 kNlm

distribution

~

~V

-9) kN_m

-42kN_m

(~)

B. M. Diagram

Fig. E·24.4.


664

A .. 3;: .. 175m2,

L .. 6.5m

Q2~ .. 1500

From Eq. 24.27,

:r ..

From Eq. 24.28,

x' .. X + b1/2 .. 2.57 + 05/2 .. 2.82m

x 6 .. 257 m 3500 .

Q

LI2 .. 6.512 .. 3.25 m; LI3 .. 2.17 m As L/3 < x' < L/Z, a trape-L.Oidal footing is required.

From Eq. 24.31.

82"

From Eq. 24.32,

B1 ..

¥

(3{ _ 1 )

.. 2 x .!7.5 (3 )(6.;,82 _ 1) .. 1.62 m 6

¥-

.. 2 B1

Let us provide

..

82

)(6.~7.5 _ 1.62 .. 3.76 m

3.80 m and 8 2 .. 1.65 qo .. 1/2 (3.8

Pressure intensity at left C
:5~65)

)( 6.5 .. 197.6 kN/m

2

.. 197.6 )( 3.8 .. 750 kN/m .. 197.6 x 1.65 .. 326 kN/m

Pressure intensity at section X -X at a distance x from Ihe right edge, w", .. 326 + 750 6~5 326 )( x .. 326 + 65.2x

Fig.E-24 (d) shows the S.F. diagram. The maximum shear forces under the columns are - 1814 kN and 1417 leN.

B.M. atX-X .. -326t - 65.2 )( 2

~

X

~

+ 1500(x - 0.25)

.. -163J? - 10.87 x 3 + 1500x - 375 B.M. at x = 3.43 m is maximum, given by M mAA .. -1917.7 - 438.64 + 5145 - 375 .. 2413.6kN-m Fig. E-24.4. (e) shows the BM diagram . The footing can be designed for the maximum B.M. and checked for shear forces. D1ustratlve Example 24.S. Design a strap footing for the two columns shown in Fig. £·24.5 (a). The 2 allowable soil pressure is 100 kNlm • Take the eccentricity of the footing of column·1 as 1 In. Solution. From Eq. 24.33,

LI - 2(e + 0.5 hi) - 2 (1.0 + 0.5 x 0.4) _ 2.4 m

From Eq. 24.34.

RI .. 600 " 6 _ 720 kN

5

Q~ - R, - (600 + 1(00) - 7W _ 880kN

R2 - (QI + From Eq. 24.35,

AI ..

~

q_

_ 720 _ 7.2m 2

100

665


t

~_.6_00_'_N_ _ _

'm _____

-II
~r.,-:.-I--.STRAP_----I~O.4mxo.(m (a) Strop

P

(2

tooting

(b)Pressurt

o

distribution

~560 +

,20

c:::=-------

",[;:/7

440kN

(el S. F.

kN

C) td) a. M. OiogrGm

Fig. E-24.!i.

From Eq. 24.36,

A2.!!!:... OJ"

q""

880 _8.8m2 tOO

H-

3m

Intensity of pressure,

ql .. tOO)( 3 - 300kN/m

From Eq. 24.38,

82

Intemity of pressure,

tb..

..

V8.8 -

3~3

2.96 m say 3 m - 97.8kN/m

2

- 97.8 )( 3 - 293.4 kN/m Fig. E-24.5 (c) and (d) show the S.F. and B.M. diagrams. The footings can be designed. The strap should be designed for a maximum 8.M. of 480 kN-m :,nd a maximum S.F. of 120 kN.

Illustrative' Example 24.6. The plan of a mat foundation with 9 columns is shown in Fig. E-24.6 (a). Assuming tluJ( the mat is rigid, determine me soil pressure. distribution. All the columns are. of the size 0.6 In. )( 0.6 m. .

SOIL MECHANICS AND FOUNDATION ENGINEPRINO

,.)

FO,t""

'0,1'" ""

'O'j"""

I ! ! ! II! ! ! ~:~:'~:~"/m (b) Strip ABFE

+

)0'5.4kN

II

A)·/O kN

c;::::---

'561·6

(

)

+

.c::::1 42.,(/7

d

440.6kN

.r,

(t)S.F, Diagram

v~160.1~:m

6.'5

~

M

(d) a.M. Oiogrom

Fig. E-24.6.

Draw 1M S.F. and B.M. diagrams for the strip ABFE. Solution. Total loads,

Q - 400 + 500 + 300 + 1500 + 20Cl0 + 1200 + 500 + 600 + 400 ... 7400 Taking moments about lhe face AD.

i ... (400 + 1500 +

5(0) x 0.3 + (500 + ~+ 6600)6.3 + (300 + 1200 + 400) 12.3 ... 5.89m

e~ _ 1;.6 _ 5.89 ... 0.41 m (-ve) \Taking moments about the face AD.


y .. (500

667

+ 600 + 400) )( 0.3 + (1500 + ~~ 12(0) )( 8.3 + (400 + 500 + 300) )( 16.3 .. 7.98 m .,. • 1~.6 _ 7.98 • 0.32 (_"")

From Eq. 24.57,

q •

i2. _ ~ . x A

In

!?...:...2 . y

_

1JeZ.

7400 )( 0.41 )( x

7400

7400 )( 0.32 )( y1112 x 12.6 x (16.6)'

q • (12.6 x 16.6) - 1112 x (16.6)(12.6)' q-3S.4-1.lx-05y Pressure at point A. (x - - 6.3 m, y .. - 8.3 m),

q•• 35.4 - 1.1 x (-6.3) - (0.5) (-8.3). 46.4kN/m' Pressure at poiot B (x .. + 6.3 m, y .. - 83 m), q• • 35.4 - 1.1 x 6.3 - (0.5) (- 8.3) • 32.6 kN/m' Pressure at point C(x .. +6.3m, y - 8.3m), qc .. 24.4 kN/m

2

Pressure at poiDt D(x .. -6.3m.Y .. +8.3m). qD .. 38.2 kN/m 2

Fro the strip ABFE,

qw ..

46.4 ; 32.6 .. 39.S kN/m 2

From Eq. 24.5S.

Q.... ..

i

From Eq. 24.59,

-q",," qiN (Qw) q""BIB

(500 + 600 + 4(0) + 395

.. 39.5

From Eq. 24.60.

F ..

(39.5

)C

1~

)C

)C

12.6

)C

4.3 ) .. 33.6 kN/rn2

Q"", .. 500 + : QJ + Q2 + Q3

+ 400 - 1.21

Fig. E-24.6 (b) shows the modified column loads.

FQJ. 605 kN, FQ, • 726 kN, FQ, • 484 kN. Pressure intensity per metre .. 33.6 )C 4.3 .. 144.5 kN/m Figs. E-24.6 (e) and (d) show the S.F. and B.M. diagrams. S.F. at a sea.ion X-X al a dislanoe x from the edge AE Fz " 605 - 144.5x S.F. is zero at x .. 4.18 m. B.M. at sea-ion X-X

.. -1445

)C

~+

.. 1085.4 kN-m

Max. B.M. = 1085 kN-m.

Max. S.F. ;,. 561.6 kN.

4.3] .. 1820 kN

60S

)C

3.88


1000 N

,OOON

fh.,oo
~

Ie

_'

(0)

T[

ir

__

!

!

l~-:- --~m: ~j t" 'DOD"

,

.

2:~kxsl04o

(b)

f

f

~:k:l0t,b

,

C=h:~ tOt,e

f p)()()ON

2:~;()t,d

«)

Fig. E-24.7.

k, _ B )( k _ 20 )( 20 - 400 N/ cm 2, h _ 100 em. Let the settlements at points A, B, C and D be a, b, c and d respectively. Fig. &24.7 shows the seltlemeolS and the equivalent loads due to soil pressure. Eq. 24.67 can be used for writing the moment equations. Thking moments about B, Solution.

~(a

- 2 b +c) - Q,h - oh'(k,l2)

1010

""i'Ct (a -

2b + e) - 1(00 )( 100 - 20 )( lOS 0

0- 2b + e_ 0.1 - 20

.. (I)

Taking moment of forces on len of C about C,

¥Z (b -

2c + d) - 2QI h - 2 )( let )( 0 )( 200 - 4 )( Hr )( b )( 100

or b - 2 c + d _ 0.2 - 4 a - 4b Ukewise, laking moment of forces on right of C about C,

~ or

...(u)

(b - 2c + d) - Q,h - 2 x 10' d x 10'

lO' (b- 2c + d)- 10' - 2 x lO'd b - 2c +d _ 0.1 -2d

. .. (iu)

From the equilibrium of the vertical forces, 2

X

4

10

0+

4

4 )(10

b+

0+

b = e

4

4)( 10 c + 2)( 10 d_ 1000 + 1000

2b + 2c+ d- 0.1

Eqs. (l) to (iv) can be solved for a, b, e and d. In tbis case, because of symmetry, o=d and Therefore, Eq. (iv) can be written as

4

.. .(iv)

DESIGN OF SHAI.l..OW FOUNDATIONS

669

... (v)

2a + 4b - 0 .10 and Eq. (I) as 3a - b - 0.10 Solving Eq. (v), and (iv), a _ 0.0357 em Therefore, pressure at point A and D

... (v,) and b _ 0.0071 an

- 0.0357 )( 20 _ 0.714 N/em 2

and at points B, C

_ 0.0071 )( 20 _ 0.142 N/cm 2 PROBLEMS

A. Numerlcals

24.1. A raft (20 m )( 10m) exerts a gross pressure of 200 kN/m 2 at foundation level. The depth of foundation is 2.5 2 m. If the soil is clay W = 0, Cw = 80 kN/m , y = 19 kN/m~, determine the factor of safety. The raft is Cor a basement. Use Skempton's equntion. [Ans. 3.03] 24.2. Determine the depth below the ground surfDce of a roft foundmion (30 m )( 30 m) to carry a gross load of 150 2 MN. The soil is clay (cw '" 15 kN/m , y '" 19 kN/m1. Thke the factor of safety of 3. Also determine the depth for a fully compensated foundation . [Ans. 7.15 m;.8.75m] 24.3. Determine the Dllowable soil pressure for a raft (10 m )( 20 m) if the blow wunt (N) over 20 m depth is 20. Use IS code method. [AIlS. 216 kNlm1 24.4. Determine the allowable soil pressure for a raft (10 m )( 10 m) if the depth is 5 m and the undrained cohesion is 40 kN/m 2. Thke the factor of safety as 2.5. [AllS. 105.6 kNlm1 24.S. Design a rectangular combined footing to support two adjacent columns (size 40 em )( 40 cm) at a distance of 5 m and carrying loads of 3 MN nnd 4 MN. The lighter column is near the property line. The allowable soil 2 [AIlS. 6.20 )( 2.9 m] pressure is 400 kN/m . 24.6. Design a trapezoidal combined footing for twO columns (30 em )( 30 cm) carrying column lends of 1.2 MN and 0.90 MN, if the spacing between the two columns is 4 m. Take allowable soil pressure as 200 kN/m2 and the length of footing as 5 m. 24.7. Design the footing in Problem 24.6 as a strap footing if the spacing between the column is 5 m. 24.8. Design a R.C.C. footing to carry a column (50 em )( 50 em) with 2.5 MN load. The allowable soil pressure is 2 250 kN/m . 24.9. Design a R.C.C. footing for a wall 30 em wide and having a 1000 of 80 kN/m. The allowable soil pressure is 2 50 kN/m .

B. DescrlpUve and Objective 'lYpe 24.10. What are different types of shallow foundations? Explain with the help of sketches. 24.11. How would you fix the depth of foundatit)O ? Discuss Rankine's fonnula for the minimum depth. 24.12. Discuss various types of loads thai are to be considered in the design of foundations. 24.13. Describe the general procedure for the design of a shallow foundation. 24.14. Discuss the procedure for proportioning of footings for equal seUlemenl. 24.1S. Explain the procedure for the design of a (a) Strip Footing, (b) Spread FOOI:ing. 24.16. Where do you provide a combined footing ? Discuss the procedure for the design of the following types of combined footing: (a) Rectangular (b) Trapezoidal. 24.17. Describe the procedure for the design of a strap fOOling. 24.18. What are different types of raft foundmion ? Discuss the procedure for the design of a raft foundation. 24.19. Describe the procedure for the design of a combined footing by elastic line method. 24.20. Explain the method for the design of a raft on Winkler's bed. 24.21. Define the coefficient of subgrade reaction? How is it found? Discuss the various factors affecting its value. 24.2Z. Write whether the following statements are t~e or fnlse. (0) A spread footing is provided for an isolated column. (b) The critical section for bending moment in case of spread footing of a monolithic column is at the face of the column.

SOIL MECHANICS AND FOUNDATION ENGINEERING (e) A strip footing can be provided for a row of columns closely spaced. (d) A trapezoidal footing is required when the column near Ihe property line is heavier than the interior

columns. (e) The strap footings are useful when the two columns are closely spaced.

(/) The ela.,tic Jine method of II combined footing is used for the design bf a flexible footing. (8) The parameter Ai.. is dimensionless. (h) The coefficient of suh-grade reaction of a cohcsionless soil is independent or depth. lAos. True (a). (b), (e), (d), (f), (1)1

C. Multiple Choice Questions 1. Trapezoidal combined footings nre reqU1red when (a) the space outside the exterior column is limited. (b) the exterior columns is heavier. (e) both (a) and (b) (d) neither (a) nor (b)

2. For the design uf a strap fooling, the following assumption is not made (1I) The strap is perfectly rigid. (h) The suil pressure varies linearly. (e) The interior footing is centrally loaded.

(d) The strap is not suhjCf.:tcd to any direct soil pressure. 3. For conventional
(b) between 0.6 and 3.0

Ie) more than 3.0 (til None of above where AL is ch:mu;teristic parameter. 4. The coefficient of subgrude reaction depends upon (a) the size of footing (b) the shape of footing (e) the depth of rOOling

(d) allthc above. 5. According to Rankine's formula , the minimum depth of foundation when q:: 180 kNlm2,

1:: 20 kN/m 3 and 30° is ta) 0.50 m (h) 0.75 m (e) 1.0 m (e/) 2.0 m 6. When the wind load is more thlln 25% of the combined dead and live load. the safe bearing capacity is usuaUy increased hy (a) 15% (h) 20% (e) 25% (tI) 30% 1. The value of the factor Ai.. when B = 20 em, k:: 20 N/cm 3• £1 10 10 N-cm 2 and L =300 cm is (a) 1.0 (h) 2.0 (e) 3.0 (d) 4.0 [AllS, l. k), 2. (h), 3. (a), 4. (d), 5. (e), 6. (e), 1. (en

,=

=

25 Pile Foundations 25.1. INTRODUCTION

When the soil at or near the ground surface is not capable of supporting a structure, decp foundations are required to troosfer the loads to deeper stmta. Deep foundations are, therefore, used when surface soil is unsuitable for shallow foundation, and a finn stratum is so decp that it cannot be reached economically by shallow foundations. The most common types of decp foundations are piles, piers and caissons. 100 mechanism of transfer of the load to the soil is essentially the same in all 'fiX'S of deep foundations. A deep foundation is generally mudl more expensive than a shallow foundation. It should be adoplOO only when a shallow foundation is not feasible. In certain situations, a fully compensated floating raft may be more economical than a decp foundation. In some cases, the soil is improved by various methods to make it suitable for a shallow foundation. A pile is a slender strudurn) member made of steel, concrete or wood. A pile is either driven into the soil or fonned in-situ by excavating a bole and filling it with oonC"ele. A pier is a vertical ooiumn of relatively larger crms-section than a pile. A pice is installed in a dry area by excavating a cylindrical bole of large diameter to the desired depth and then backfilling it with oonaete. 1be distinction between a cast in-situ pile and a pier is rather arbitrary. A cast in-situ pile greater than 0.6 m diameter is generally tenned as a pier. A caiMon is a hollow, watertight box or chamber, which is sunk through the ground for laying found:llion under water. The caisson subsequently becomes an integral part of the foundation. A pier and a caisson differ basically only in the method of construction. Pile foundations are disrussed in this chapter. Piers and caissons are dealt with in chapter 26. Well foundations, which are special type of caissons, are discussed in chapter 27. 25.2. NECESSITY OF ]'ILE FOUNDATIONS Pile foundations arc used in the follOWing conditions: (1) When the stmta al or just below the ground surfa~ is highly compressible and very weak. to support the load transmitted by the structure. (2) When the plan of the structure is irregular relative to its outline and load distribution. It would cause non-unifonn settlement if a shallow fouodatioo b constructed. A pile foundation is required to reduce differential settlement. (3) Pile foundations are required for the transmission of structural loads through deep water to a firm stratum. (4) Pile foundations are used to resist horizontal forces in addition to support the vertical loads in earth-retaining structures and tall slruct¥res that are subjected to horizontal forces due to wind and ear1lKjuake. (5) Piles are required wbco Ihe soil conditions: are such that a wash oul, erosion or scour of soil may occur from undemealh a shallow foundatioo.

9)IL MECHANICS AND FOUNDATION ENGINEERING

(6) Piles are used for the foundatiOM of some structures, such ~ transmission towers, off-shore plateforms, which are subjected to uplift. (7) In case of expansive soils, such as black cotton soil. which swell or shrink as the water content changes. piles are used to transfer the load below the active zone, (8) Collapsible soils, such as loess, have a breakdown of structure accompanied by a sudden dcaease in void ratio when there is an increase in water cootent. Piles are U'iCd to trnnsfer the load beyond the zone of possible moisture changes in such soils.

25.3. CLASSIFICATION OF PILES Piles can be classified according to (1) the material used (2) the mode of transfer of load, (3) the method of construction, (4) the use, or (5) the displacement of soil. as described below.

(I) Classification according to materilll used There are four types of piles according to materials used.

(I) Steel Piles. Steel piles are generally either in the fonn of thick pipes or rolled steel H-sections. Pipe steel piles arc driven into the ground with their ends open or closed. Piles are provided with a driving point or shoe at the lower end. Epoxy ooatings are applied in the factOf)' during manufacture of pipes 10 reduce corrosion of the steel piles. Sometimes, concrete enc.'lSCment at sile is done as a protCClion against OOIT05ion. To take into account the corrosion, an additional thickness of the steel seaion is usually recommended. (ii) Concrete Plies. Cement concrete is used in the construction of concrete piles. Conaele piles are either precast or cast-in situ. Precast concrete piles are prepared in a factory or a casting yard. The reinforcement is provided to resist handling and driving stresses. Precast piles can also be prestressed using high strength steel pretensioned cables. A cast-in situ pile is conslruacd by making a hole in the ground and then filling it with concrete. A cast-in situ pile may be cased or uncased. A cased pile is constructed by driving a steel casing into the ground and filling it with concrete. An uncased pile is constructed by driving the casing to the desired depth and gradually withdrawing casing when fresh concrete is filled. An uncascd pile may have a pedeslal. (iii) TImber Piles. Timber piles are made from tree trunks after proper trimming. lbe timber used should be straight, sound and free from defects. Steel shoes are provided to prevent damage during driving. 10 avoid damage to the top of the pile, a metal band or a cap is provided. Splicing of timber piles is done using a pipe sleeve or metal straps and bolts. The length of the pipe sleeve should be at least five times the diameter of the pile. Timber piles below the water table have generally long life. However, above the water table, these are atlackcd by insects. '[he life of the timber piles can be increased by preservatives such as creosote oils. Timber piles should not be used in marine environment where these are attacked by various organisms. (iv) Composite piles. A composite pile is made of two materials. A composite pile may consist of the lower portion of steel and the upper portion of cast-in situ concrete. A composite pik may also have the lower ponion of timber below the permanent water table and 'the upper poction of conaete. As it is difficult to provide a proper joint bel ween two dissimilar materials, composite piles are rarely used in practice. (2) Classification Based on Mode of Transfer of Loads Based on the mooe of transfer of loads, the piles can be classified into 3 categories: (i) End-bearing plies. End-bearing piles transmit the loads through their bottom tips_ Such piles act 21 columns and transmit the load through a weak material 10 a firm stratum below. If bed red: i located within a reasonable depth, piles can be extended to the rock:. The ultimate capacity of the piIt

(1

PILE FOUNP/,1l0N

673 ,

depends upon the bearing capacity of the rock. If instead of bed rock. a fairly compact and hard stratum of soil exists at a reasonable depth, piles c.'m be extended a few metres into the hard stratum. End-bearing piles are also known as poinl- bearing piles. The ultimate load carried by the pile (Q ..) is equal to the load carried by the point or bottom end (Q,). (il) ."riction piles, Friction piles do not reach the hard stratum. These piles transfer the load through skin frictioo between the embedded surface of the pile and the surrounding soil. Friction pilcs are used when a bard stratum docs not exist at a reasonable depth. 1bc ultimate load (Q.. ) carried by the pile is equal 10 the load transferred by skin friction (Q,). [Note: The term friction pile is actually a misnomer, as in the clayey soils, the load is transferred by adhesion and not friction between the pile surface and the soil]. The friction piles are also known as floaling piles, as these do not reach me hard stratum. (iil) Combined end beuring and friction piles. 'Ibese piles transfer loads by a combination of end bearing at the bollom of the pile ~md friction along the surface of the pile shaft. The ultimate load carried by the pile is equal to the sum of the load carried by the pile point (Qp), and the load carried by the skin friction (Q,),

(3) Classification based on method of installation Based on the method of construction, Ihe piles may be classified into the following 5 categories: (I) Driven piles. These piles are driven into the soil by applying blows of a hc.wy hammer on their lOps.

(il) Driven and Cast-in-situ piles. These piles are formed by driving a casing with a closed bollom end into the soil. The casing is later fillccl with concrete. The casing mayor may not be withdrawn. (ii') Bored and Cast-in-sltu piles. These piles are fanned by excavating a bole into the ground and then

filling it with concrete. '(iv) Screw piles. 'fhese piles are screwccl into the soil. (v) Jacked piles. These piles are jackccl mto the soil by applying a downward force with the help of a

hydraulic jack. (4) Classification based on use The piles can be classified into the following 6 categories, depending upon their usc. (I) Load bearing piles. lhese piles are used to transfcr the load of the structure to a suitablc stratum

by end bearing, by (riction or by both. These are the piles mainly disrussed in this chapter. (iI) Compaction plies. These piles are dri~en into loooe granular soils to increase the relative density.

The bearing capacity of the soil is increased due to densificalion caused by vibratioos. (iii) 'Thnslon piles. These piles are in tension. These piles are used to anchor down structures SUbjected to hydrostatic uplift forces or overtuming forces. (iv) Sheet piles. Sheet piles Conn a continuous wall or bulkhead which is . used for retaining earth or water (see Chapler 20). (v) Fender piles. Fender piles are sheet piles which are used to protect water-front structures from impact of ships and vessels. (VI) Anchor piles. These piles are used 10 provide anchorage for anchored sheet piles. These piles provide resistan~ against horizootal pull for a sheet pile wall (see Olapter 20). (5) Classification based on di'iplacement or soil Based on the volume of the soil displaced during installation, Ihe piles can be classified into 2 categories:

(,) Displacement piles. All driven piles are displacement piles as the soil is diSplaced laterally when the pile is installed. The soil gets densified. 1bc installation may cause heaving of the surrounding


674

ground. Precast concrete pile and closed-end pipe piles are high displacement piles. Steel H-piles are low displacement piles. (it) Non-displacement piles. Bored piles nre non-displacement piles. As the soil is removed when the hole is bored, there is no displacement of the soil during installation. 1be installation of these piles causes very little change in the stresses in the surrounding soil

25.4. PILE DRIVING

Piles arc driven into the ground by mcam of hammers or by using a vibratory driver. Such piles are called driven piles. In some special cases, piles are installed by jetting or partial augering. The follOWing methods are commonly used. (1) Hammer Driving. Fig. 25.1 shows a pile driving rig. It consists of a hoist mechanism, a guiding frame and a hammer device. The hammers used for pile driving are of the following types: (t) Drop hammer. A drop hammer is raised by a winch and al1~wed to drop on the top of Ute pite under gravity from a certain height. During the driving operation, a cap is fixed to the top of the pile and a cushion is generally provided betwccn the pile and the cap. AnOlhcr cushion, known as hammer cushion, is placed on the pile C.1P on which the hammer causes the impact. The drop hammer is the oldest type of hammer used for pile driving. II is rarely used these days because of very slow rate of hammer blows.

HAMMER

PILE CAP PILE

Fig. 2S.l. Pile Driving Rig.

(ii) Single-acting hammer. In a single-acting hammer, the ram is raised by air (or steam) pressure to !be required beight. It is then allowed 10 drop under gravity 00 the pile cap provided with a bammcr cushion. (iiI) Double-acling hammer. In a double-acting hammer, air (or steam) pressure is used to nlisc tile hammer. When the hammer has been raised to the required height, air (or steam) pressure is applied to the other side of the piston and the hammer is pushed downward under pressure. This incrtaSCI lhe impact energy of the hammer.

PILE FOUNDATIONS

675

(il/) Diesel hammer. A diesel hammer consists of a ram and a fuel injection system. It is also provided with an anvi l block at its lower end. The ram is first raised manually and the fuel is injected near

the anvil. As soon as thc hammcr is released, it drops on the anvil and compresses the air-fuel mixture and ignition lakes place. The pressure so developed pushes the pile downward and rai ses the ram. The fuel is again injcctcd and the process is repeated. The ram lifts automatically. It has to be manually raised only 01lCC at the beginning. Diesel hammers are nor. suitable for driving piles in son soils. In such soils, the downward movement of the pile is excessive and the upward movement of the ram aner impact is small. The height achieved after the upward movement o[ the hammer may not be sufficient to ignite the air-fuel mixture. Diesel hammers are self-contained and self-activated. (2) Vibratory Pile Driver. A vibratory pile driver consists of two weights,. called exciters, which rotate in opposite directions. The horizontal components of the ccnlrifugaJ force generated by exciters cancel eadl other but the vertical romponents add. Thus a sinosoidal dynamic vertical force is applied to the pile, which forces the pile downward. 11le frequency of vibrnlion is kept equal to the natural frequency of pile-soil system for bener results. A vibrntory pile driver is userul only for sandy and gravelly soils. The speed of penetration is good. The method is used where vibrations and noise of conventional driving methods cannot be permitted. (3) Jetting Techniques, When the pile is to pcnetr:;Jte a thin hard !.1yer of sand or gravel overlying a softer soil layer, the pile can be driven through the hard layer by jetting techniques. Water under pressure is discharged al the pile bottom point by means of a pipe to wash and loosen the hard layer. (4) Partial Augering Method. Dauer piles (inclined piles) are usually advanced by partial augenng. In this method, a power auger is used to drill the hole for a part of the depth. lbe pile is then inserted in the hole and driven with hammers to the required depth. 25.5. CONSTRUCTION OF BOltED PILES (a) Drilling of holes, Bored piles arc constructed after making a hole in the ground and filling it with concrete. 'I11e following methods me used for drilling of the hole. (I) Hand auger. A hand auger Co1n be used for boring without casing in soils which are self-supporting, such as firm to stiff clays and silts and Clayey sands and gravels above the water table. 1be depth of the hole is generally limited to about 4.5 m. 'Inc diameter of the hole is usually not more than 350 mm. (2) Mechanical auger. For piles of diameter more than 350 mm or depth greater than 4.5 m, a hand auger becomes uneconomical. In such a case, a mechanical auger is used. A mechanical auger can be of rotary type or bucket type. It is power driven. The soil in this case must be self-supporting, with or without bentonite slurry. The soil should be free from tree roots, cobbles and boulders. A continuous flight auger is also used 10 drill the bore hole. (3) Boring rig, A boring rig is used \0 sink: the hole in ground whcI'C hand or mechanical augering is not possible, such as water- bearing sand or gravels, very soft clays and sills and the soils having cobbles and boulders. A specially designed boring rig, Imown as grab-type bored piling rig, is sometimes used. In this type of rig, the ca<>ing is given a continuous semi-rotary motion which causes ils sinking as the bore hole is advanced by percussion drilling. (4) Belling Bucket. Underreamcd piles are Large diameter bored piles with enlarged bases. Excavation for the undcrreamed piles is done by a special type belling bucket.

(b) Concreting

Before concrete is placed, the bored hole is bailed dry of water. Any loooe or softened soil is cleaned out and the bottom of the hole is rammed. A layer of dry concrete is placed and mmmed if the bottom of the hole is . wet. Then the concrete with a readily workable mix (7,5 to 10 em slump), not leaner than 300 kg


676

ccmcnt/m 3 of ooncrctc, is poured into a hopper placed at the mouth of the hole. If the hole cannot be bailed or pumped dry before placing the concrete, the hole is lined with a casing throughout its depth. A mass of ooncrctc is then deposited at the base of the bole by a lremie pipe. As soon as the concrete has hardened and formed a plug. the hole is pumped free of water. The casing is then gently turned aod lifted slightly to break the joint with the plug. The hole is pumped dry. The remainder of tbe coocreting is done by placing it dry upl O ulb ground surface. The casing is lhcn lined entirely (rom the bore hole. If the ground water is undcr a high pressure, there will be inllow of water between the concrete plug and the inside of the casing. The inflow should be stemmed by caulking. 1bc C$ing is cut by oxy-acetylene just above the plug. The shaH is then coocrelcd and the casing raised. The Cllt portioo of the casing around the

plug is left permanently in place. Instead of plugging the base of the pile and rona-cting, an alternative method is to concrete Ihe entire shan under waler using a tremie pipe. Coocrele should be easily workable (slump 12.5 to 17.5 cm) and the cement content should be at least 400 kg/m l . A rctDrde r, is added to the roncrcte if there is a risk of the concrete setting before the casing is lifted out. However, the quality of rona-cling done under water is not good. This method should be avoided as far as possible.

25.6. DRIVEN CAST·IN·SITU CONCRETE PILES A driven cast-in-situ roncrete pile is forme
PILE FOUNDATION

(T) A/plio piles are fonned by driving a steel tube closed with a detacoable cast iron shoe. A concrete-filled mandrel is driven inside the tube. The mandrel is gradually raised and some concrete is allowed 10 slump down in the tube. The ooncrcte is refllled in the mandrel and it is driven down as the lube is raised. Thus a pedestal is fonned. After the formaLion of the pedestal, \.he mandrel is raised and refilled with oonaete in stages. In each stage. the conaele in the pile shaft is pressed against tbe soil by the dead weight of the hammer on the mandrel 25.7. LOAD·CARRYING CAPACITY OF PILES Uke a shallow foundation, a pile foundation should be safe against shear failure and also the settlement should be within the permissible limits. The methods for cstimating the load-carrying capacity of a pile foundation can be grouped into the following 4 categories. (1) Static Methods. The static methods give the ultimate capacity of an individual pile. depending upon the characteristics of the soil. The ultimate load capacity is given by

Q. - Q, + Q, ... (25.1) where Qu ultimate failure lond, Qp point (or base or tip) resistance of the pile (Fig. 25.2), Q, = shaft resistance developed by friction (or adhesion) between the soil and the

=

=

~The methods

I

for the determination of Qp and Q, are discussed in Sects. 25.8 and ~ 25.9, respectively, for sand and Clay. The static formulas give a reasonable estimate of the pile capacity if judiciously applied. (2) Dynamic Fonnulas. The ultimate capacity of piles driven in cenain types of soils is related to the resistance against penetration developed during driving operaHon. The ultimate load capacity formulas are based on lbe principle that the resistance of a pile to further penetration by driving depends upon the energy imparted to the pile by the Fig. 25.2. hammer. It is tacitly assumed that lbe load-carrying capacity of lbe pile is equal to the dynamic resistance during driving. The dynamic formulas are not much reliable. (3) In-situ Penetration Tests. 1be pile capacity can be determined from the results of in-situ standard penetration lesl. Empirical fonnulm are used 10 determine the point resistance and the shaft resislanoe from the standard penetration number (N). Altcrnatively, the static fannulas can be used after determining the N-value, as this value is related to the angle of shearing resistance (,). Cone penetration tests are also used to estimatc the pile capacity. (4) Pile Load Tests. The most reliable melbod of estimating the pile capacity is to conduct the pile load test. The test pile is driven and loaded to failure. 1bc pile capacity is related to the ultimate load or the load at which the settlements do not exceed the permissible limits. All the above methods are discussed in detail in the following sections. 25.8 STATIC METHODS FOR DRIVEN PILES IN SAND The ultimate capacity of a single pile driven into sand is obtained using Eq. 25.1, Q.. .. Qp + Qs where

Qp .. qpAp

... (25.2)

Q, - [,A, ... (253) In above equations, qp is the ultimate bearing capacity of the soil at the pile tip and and Ap is the area of tbe pile lip; J, is.the average unit skin friction between the sand and the pile surface, and As is the effective surfaoe area of the pile in contact. with the soil. (a) Methods ror determination of Qr The ultimate bearing capacity (qp) of the soil at the pile tip can and


678

be computed from the bearing capacity equation simiL1f 10 that for a shallow foundation, as discussed in

ch..1plcr 23. For sandy soils, qp _ qNq ... O.4ylJNy where

... (25.4)

q = effective vertical pressure at the pile tip, B = pile tip width y = unit weight of Ihe soil in the zone oC the pile tip.

(or diameter).

Nq and N, = bearing capacity factors for deep foundations. In driven piles, the second Icnn of Eq. 25.4 is generally small and is, therefore, neglected. Thus qp - qNq .. ,(25.5) V~r llcClI

tq)

prt'ssvrt'

,i,M

1

Q .. ConSlont

~~~ • (b)

(a> Fig. 25.3.

of driven piles, it h..'IS been established thai the effective vertical pressure @ at the pile lip incre.'1SCS with depth only until a certain depth of pcnClr::llion. known as Ihe ailical depth (D~). I3clow the critical depth, the effective vertical pressure remains essentially constant {Fig. 25.3 (a)]. The critical depth

In

c.'lSC

00

'"

to-

'" / ~/

~ / ?'--~ ~ ----

'"

//

~, //

- --

~~ o=-

"

/ l/:/ /

~o~

Fig. 25.4.

I

>'!:-."'....

I

// /

"

/

/

o~;/

,

.5~~r---

"

1'-

01 sheoring

;1

/ / / 0)::l; ,k !5:2911 / /

3D

Angle

/

I

"',

~

0

/

10 " /

"

PILE FOUNDATION

679

upon the angle of sheming resistance (.~ of the soil and the width (or diameter) of the pile [Fig. 25.3 (b)]. lIS value can be roughly t.1ken as 10 D for loose sands and 20 D for dC .lSC sands. The bearing Co1pocity factor Nq depends upon the angie of shearing resistance (,). Vnrious investigators gave the expressions for Nq based on theoretical analysis. These values vary over a wide range because of different assumptions made in defining the shear zone near the pile lip. Fig. 25.4 shows the values of Nq given by various investigators and that given by IS : 2911. lbe values given by Ocre-,lOntzcv arc 'quite dependable,and arc generally used. In the derivation of the value of Nil' it has been assumed lhat the soil above the pile tip is similar to the soil below the pile lip. If the pile penetrates a oompact stratum only slighUy and the soil above the lip is loose, it would be more appropriate 10 use the value of Nq for a shallow foundation given in chapter 23. If the pile is of relatively large diameter, the second term in Eq. 25.4 becomes significant. The value of N, can be conservatively taken as the Nl value used for shallow foundations., given in chapter 23. Meyerhof's melhod for qP- The point bearing capacity (qp) of a pile generally increases with the depth of embedment (Db) in the bearing stratum. It reaches a maximum value at an embedment ratio of (Di/B)~,. For a homogeneous soil, Db is cqualto the actual depth D of the pile, but for a pile which has penetrated ioto a bearing stmtum for a small length. Do is less than D. Beyond the critical value of (Db/8)er' the value of qp remains COIlstant, equal to the limiting ql' The criLical ratio (Dt/B).., depends upon the soil friction angle (41) (Fig. 25.5). Once the value of (Dt/B)u has been detennined, the following procedure is used to estimate qp' (1) Determine actual (D/>/B) ratio for Ihe pile, (2) Determine Nq for (Di/o) mtio from Fig. 25.5. The value of Nq increases linearly with (Di/B) ratio and reaches a maximum value at depends

D"IB_

~(D"I8)".

(3) Detennine the point resistance Qp as Qp - Ap9 N q s Al'ql

wbere ql - 50 Nq tan •• q

... (25.6)

=vertical pressure at the pile tip (kN/m1. AI' = orca of the pile tip.

Angle 01

sh~ring

resistonce

Fig.2S.5.

;(After Meyrrhof ,1976)

680

SOIL MECIIANICS AND FOUNO,(TION ENGINEERING

If the pile initially penetrates a loose s.1nd byer and then a dense layer for a depth less than 10

n,

the

point resisulnce is given by ql(IIlDb lOB

[91(2) -

qp - q'(l )

+

... (25.7)

:S q'(2)

where 1/(1) == limiting unit point resistance of loose sand (= 50 Nql Ian +0 ql(2) = limiting unit poinl resistance of dense sand ( = 50 Nq2 1an Db == depth of penetration in dense sand. It may be mentioned that the ultimate tip resistance given by Eq. 25.2 is the gross uhimate point resistancc. The nct tip load is given by

+v

Q,(nct) - Q, - ViA,) However, in practice, the deduction of qAp is not usually made and Qp (ncl) is taken equal to Qr In case of H-pilcs and open-cnded pipe piles, the enclosed soil plug should be considered as the part of the pile for computing the area of Ihe point (Ap). (h) Methods of determination of Q•• The frictional resistance Q, is obt.'lincd from Eq. 25.3 aOcr estimming the unit skin friction (fs). lbe unit skin friction for ~ a stmight-siclcd pile depends upon thc soil pressurc acting norma] to the pilc a: surface and thc coefficicnt of friction betwccn thc soil and thc pilc material (Fig. "h & li~ ~ ~ lbc soil pressure normal to thc vcrtical pile surface is horizontal pressure (01,) and is related to the effective vcrtical soil pressure as Fig. 25.6.

j

t

0".,.

t"

Kcr"

where K = earth pressure coefficient, 0" = effective vertical pressure at that depth. Thus unit skin friction (f,) acting at any depth can be written as

J, - o"tan6

... (25.8)

[. - Ko"tan6

wherc tan 6 = coefficient of friction between sand and the pile material. Selection of suitable values of 6 and K requires good engineering judgment. Tomilson (1975) gave the values of 6 and K, as given in Tablc 25.1, based on the studies carried by Droms (1966). Tuble 25.1. Values of b anti K. Pile Material

Steet Concrete

limber

20· 0.75 oj! 0.67 oj!

K (loose salld)

K (dellsesQJuI)

0.50 1.0 15

1.0 2.0 4.0

In general, the value of 6 gcnerally varies between 0.5 4' and 0.8 iJ. In most cases, the value of K varies between 0.6 and 1.25. Meyerhof (1956) recommends that the value of K can be taken as 0.5 for loooe sand (4' = 30j and as 1.0 for dense sand (iJ = 45j. According to IS : 2911-1979, the value of () may be taken equal \0 q.. r-or driven piles in loose to medium sands, Ihe recommended value of K is between 1 and 3. Whether thc sand should be considcred as loose or dcnse depends upon not only on the initial relative density, but also on the method of installfltion. The larger the volumc of soil displacement, the higher the value of the result ing friction. For hjgh displacement driven piles, the soil is considered dellSC. For driven and cast·in place piles, the soil is oonsidered a<; medium dense if the casing is JcO in place or if the coocrcte is oompaclc.d as thc casing is wilhdmwo. The sand is considered to be loose, if the concrete is noI compacted. Tapered piles dcvelop greater unit friction lhan the stmight piles. Further, the value of K is greater if lhe pile is driven into undisturbed soil Ihan the one for installed in a predrilled hole.

PILE FOUNDATION

681

As stated earlier, tbe effcctive vertical pressure (O~) increases with depth only upto the critical depth. Below the critical depth, the value of o~ remains ronstant. The frictional resistance (a,) can be expressed as

Q, - ~ K (0,.); tan 6 (A,); j

where

".[25.9(a)]

I

~

= numbel: of layers in which

the pile is installed, (ov)j = effective nonnal stress in ith layer. 1/

(A,);

= surface area of the pile in

ith layer,

Eq. 25.9 (a) can be written ac;

Q, - I K tan () (area of Ov diagram) )( pile perimeter j

~

".[25.9(b)1

I

Eq. 25.9 (b) is useful when variation of 0 " with depth has been plotted as o.,-diagrom. The ultimate load for the pile CEq. 25.1) can be written QII - Qp+Q,

Q. -

qN""

+ I

i_I

as

K(o,,);lan6 (A,);

".(25.10)

25.9. STATIC METIIOD FOR DRIVEN I'ILES IN SATURATED ClAY Eq. 25.1 can be used for the detennination of the lo..1d-carrying capacity of driven piles in satur.ated clay. The point resistance (Qp) can be expressed as CEq. 25.2), Qp - qpAp where qp is the unit point resistance, equal to the ultimate bearing capacity (qll) of the soil For rohcsive soils ( , = 0), the ultimale bearing capacity is found from the folJowing equation, which is similar 10 that for a shallow foundation. qu - cNr + qNq As Nq

_

1.0 for , = 0, the above equation becomes

q" - cNe + q Therefore,

Qp (gross) = (cNr + q)Ap

or Qp (net) - cNrAp ... (25.11) In above equations., c is the cohesion of the clay in the zone surrounding the pile tip, and Ne is the bearing capacity factor for the deep foundation. lhe value of Ne depends upon the DIB ratio and it varies from 6 to 9.0. A value of N e =- 9.0 is generally used for the piles. In the case of short piles (DIB .s 5.0). the value of Ne is reduced to the values proposed by Skempton (see chapler 23). The skin resistance (Qs) of the pile can be expressed as (Eq. 253),

Q, - cIIAs where cII = unit adhesion (or skin friction) developed between clay and pile shaft. The unit adhesion (c..) is related \0 the unit cohesion by the relation

".(25.12)

c(J - a C ... (25.13) where a is the adhesion factor and is the average cohesion along the shaft length. 1be value of a. depends upon the consistency of the cloy. For nocmally consolidated clays, the value of a is taken as unity. According to IS : 2911-1979. the value of a. can be taken as unily for soils having soft

c

SOIL MECHANICS ANO FOUNDATION ENGINEERING

682

to very soft consistency. Fig. 25.7 shows the variation of a with the undrained cohesion c. It may be noted that for oonnally consolidated clays, with c less Ihan about 50 kN/m2, the value of a is equal to unity. As c increases, the value of a decreases. For over-consolidated sUIT to hard clays. its value is usually taken m; 0.3. For tapered piles, the value of a is generally 20% greater than that for a straight pile. For very long piles (D Ot 25 m). the above method for tttimating the skin friction is very conservative. For such soils, the unit skin friction also depends upon the elleaive overburden pressure. According 10

Vijayvcrgiya and Focht (1972). the average unit skin friction can be expressed as k \0. + 2 c)

t. •

... (25.14)

wh~ )., = friction capacity factor, Ov = mean effective vertical stress for the embedment length, c .. undrained cohesion. 0

I.OO,nIT-,---,---,----,----,

.1

J

00

0.4

/V

"

I---'

'-'

/

10

"''I----lt~_t_-_t--'L--I-_l 30

I

I I

oC

O·

Undroint!'d

Cohesion tkNlm2)---..

0 (After

Fig. 25.7.

Me Clt!'Uond ,1914)

Fig. 25.8.

The vahle of A can be obtained from Fig. 25.8 (McClelland, 1974). OnO! tbe ullit skin friction has been estimated, the shaft resistance is determined from Eq. 25.3. For cohesive soils, the ultimate load can be determined by adding the point resistance and the shaft resistance (Eq. 15.1). Thus Q.. _ cN~Ap + ncA, ... (25.15) As the clay gelS remoukled when the pile is driven, Ihis fador must be taken inlo account wben estimating the load carrying capacity. The remoulded strength is always less tilan the undisturbed strength, but because of thixotropy. the strength improves with time. 1be rote of gain of strength depends upon the consolidation characteristics of the soil and the rate of dissipation of excess pore water pressure. When using Eq. 25.15. the value of c and should be judiciously evaluated.

c

PILE FOUNDATIONS

25.10. STATIC METIIOD FOR BORED PILES Bored piles arc constructed by driUing a hole into the ground and filing it with concrete. The pile can be straight.sided for its full depth or may be constructed with a bell (or predcslal) at its basco The piles wilh a pedestal are also known a<> under·reomed piles'. The lood-carrying capacity of the bored pilcs can be determined using the procedure similar 10 lhat adopted for the drivcn piles. However. the values of the soil parameters are different,. a<> described below. (a) Bored Piles In Sand. Eq. 25.10 can be used 10 detennine the ultimate load. The equation can be written as Q• • IJiN,)A, + }; (K

a,

tan 6) (A')i

... (25.16)

i-I

"

where (j~ :: effective vertical pressure, limited to a maximum value given by the criliall depth. K = lateral earth pressure coefficient for bored foundatioo. tan 6 = coefficient of friaion between sand and concrete. 'Ibe sand in boret! piles is loosened a., a result of the boring operation, even though it may initially be in a dense or medium dense state. The value of 4' to be used to obtain Nq should be for the loose condition. An approximate value of K can be obtained from the following equation. K-I-sin4' The value of K generally varies between 0.3 and 0.75. An average value of 0.5 is usually adopted. The value of tan 6 can be taken equal to tan for bored pilcs excavated in dry soil. If a slurry has been used during excavation, the value of tan 6 should be reduced.

+

1n general, for a given initial value of 4', bored piles have a unit point resistance of

i to ~ of that of

corresponding driven piles. In driven pilcs. there is densification. cast-in-place piles with a pedcstal show about 50 10 100% greater unit point resistana:: compared wilh those without a pedestal. The impact energy of the bammer compacts tile soil during (he formation of the pedcsI.al. (b) Bored plies In Cluy. Eq 25.15 can be used to estimate the ultimate load. The equation can be written

QII -

cN~Ap + acA:

... (25.17)

where A/ = area of shaft that is effective in developing skin friction. The value of a depends upon the pile type and the method of drilling. Por straight shafts excavated dry, a is taken equal to 0.5 and that when drilled with slurry is 0.3, For belled shafts, the corresponding values are 0.3 and 0.15. For calculating the area of Sh.:1ft that is effective in developing sldn friction, the lower 1.5 m (or 2 B) of the strnight shaft and the bell section (if pr()\ljded) are neglected, because of disturbance alU$Cd. For the same reason. the top 1.5 m is also neglected. U a bored pile is installed in stiff, flSSured clay, the value of cohesion (c) should be reduced 10 75% of the value obtained from (he triaxial test. (c) Unden-eamed Piles in Clay. 1be base area of an underrearned pile is increased by underrcaming and providing a bulb [Fig. 25.9 (a)). The ultimate load is given by ... [25.17 (a)l Q.. - cN~Ab + acA: where Ab is the area of the enlarged basco The value of N r is taken as 9.0. The adhcsiQ.n factor a is taken as 0.40. When the bulb is slightly above the lip. Ab is taken equal to the area of the diameter of tile bulb and the projcacd stem below tbe bulb is ignored. 1be average value of c al the bulb ,is taken. However, if the bulb is quite high, and there is considerable difference in the value of c at the bulb level and the level of the bottom tip of the pile. the ultimate load is given by


684

BUlb

..,

(b, Fi&:.25.9.

Q... ~ (If»)( (9c) +

%(In - If»)/.

... [25.17(b))

9c' ... ac'A/

n

where = diameter of the pile shaft, DI is the diameter of the bulb. c is the unit cohesion at the tip. and c' is the unit cohesion at bulb level. the length of the shan equal to 2 n nbove Ibe bulb is usually While C1lculllling the surf,lCe area neglected. As the pile settles, there is a possibility of formation of a small gap between Ihe lop of the bulb and the overlying soil over a lenglh of 2 B, and therefoR; this length of the shaft is ncglcclcd. The little portion of the shaft projecting below the shaft is nlso neglected while computing A./.

A:.

When two or more bulbs arc provided., the Q" -

~(n2) )(

(9c) +

ultima~e

~

=

load is given by

(Of - n1 )(

(9c') + a COlA ... +

COl'

A~b

... (25.17(')J

=

where AI surface area of shaft obovc the top bulb (ignoring 2 B length). Asb surface of the cylinder circumscribing the bulbs betwccn lOp and bouom bulbs, c.. = average cohesion 00 A, and COl' = average cohesions on ASI>"

25.11. ALLOWADLE LOAD The allowable load (Qd,) is obt.:lincd from the ultimate lood (Q .. ) from the relation

Q", - Q./FS

... (25.18)

where FS is the factor of safety. FS generally vuries between 2.5 and 4.0, depending upon the uncertainties involved in the computation of the ultimate load. According to IS : 2911-1979, the minimum factor of safely on stalic Connula Sh.111 be 2.5. The fin:ll selection of the vniue of the fador oC safety should take into account the load settlement characteristics of the structure as a whole,

25.12. NEGATIVE SKIN I'lUCTION When the soil layer surrounding a portion of the pile shaft settles more than the pile, a downward drag occurs on the pile. The dr::lg is known as negalive skin friction. Negntive skin fri(.1ion develops when a soft or loose soil surrounding the pile settles after the pile has been installed. The negative skin frk"tion occurs in the soil zone which moves downward relative to the pile.

PILE .FOUNDATION

The negative friction imposes and exira downward load 00 the pile. The magnitude of the negative sk:in friction is computed using the same method as discussed in the preceding sections for the (positive) frictional resistance. However, the direction is downwards. The net ultimate lood-olnying capacity of the pile is given by the equation (Fig. 25.10).

Q.' - Q. - Q""

.. .(25.19)

""""""""""",mi '7';>7'7'7"'"'-':1 "<"<:.L..<'-U,",,-",,!

SOFT LAVER

where QIIS{ = negauve skin friction, Q,/ = net ultimate load. Where it is anticipated that negative skin friction would impose undesirable, large downward drag on a pile. it can be eliminated by providing a protective sleeve or a coating for the section which is surrounded by the settling soil. 25.13. DYNAMIC FORMULAE The load-carrying capacity of a driven pile can be Fig. 25.10. estimated from the resistance against penetration developed during driving operation. The methods give fairly good results only in tbe case of free-draining sands and hard clays in which high pore water pressures does not develop during the driving of piles. In saturated fine-grained soils, high pore water pressure develops during the driving operation and the strength of the soil is considerably changed and the methods do not give reliable results. The methods cannot be used for submerged, unifonn fine sands which may be loose enough to become quick temporarily and show a much less resistance. The dynamic formulae are based on the assumption thaI lhe kinetic energy delivered by the hammer during driving operation is equal to the work done on the pile. Thus WhTJ/,-RxS ... (25.20) where W = weight of hammer (kN). h = height of ram drop (an), ..", = efficiency of pile hammer, R = pile resistance (kN), taken equal to Q". and S = pile penetrntion per blow (em). In Eq. 25.20, no allowance has been made for tbe loss of energy. during driving operation, loss caused by elastic contraction of the pile, soil, pile cap, cushion and due to the inertia of the pile. Some energy is also lost due to generation of heat. Various formulae have been proposed, which basically differ only in the methods for accounting of the energy losses, as described below. (1) Engineering News Record Fonnula. According to Engineering News Record (ENR) formula (1888), the ultimate load is given by

Wh",

Q" -

s.c

... (25.21)

where S = penetration of pile per hammer blow. It is generally based on the average penctration obtained from the last few blows (em), C = constant (For drop hammer, C = 254 em and for steam hammer, C = 0.254 an) In Eq. 25.21, the product W x h can be replaced by tbe rated energy of hammer (E,.) in kN-an. Thus

Q. - : :~~

.. .(25.22)

The efficiency "y]" of the drop hammcr is get1erally between 0.7 and 0.9, and that for a single-ading and a double-acting hammer .is between 0.75 and 0.85. For diesel hammer. it usually lies between 0.80 and 0.90. A factor of safety of 6 is usually reoommended. However,-tbe pile load tests reveal that the actual fact

686

Modined FOrDlulli. The Engineering News Record formula has been modified recently. In the modified formula. the energy k>sscs in the hammer system and that due to impact are considered. According to this Connula.

Q _ Wh TlII .(W+e'2 p ) .. S+C W+P where P

= weight

... (25.21(a)J

of pile; e. = coefficient of restitution, and T)Jo = hammer efficiency.

The hammer efficiency (",,) depends upon vario~ factors, such as pile driving equipment, driving procedure, type of pile and the ground conditions. For drop hammers, il is usually taken between 0.75 Dnd 1.0; for single acting hammers between 0.75 and 0.85; for double-.acting or differential hammer, T)Jo = 0.85 and for diesel hammer, ll" "" 0.85 [0 1.00. The representative values of the coefficient of restitution (e) are as under.

= 0.0

Broomcd timber pile Good timber pile

= 0.25

Driving cap with timber dolly on steel pile Driving cap with plastic dollY ,on steel pile Helmet with composite plastic dolly and packing on R.C.C. pile

= 0.3 = 0.5 = 0.4

(2) IDley Fonnulu. Hiley (1925. 1930) gave a formula whkh lakes inlo account various losses.

Wh11b11h Q. - (S + e/2)

... (25.23)

=

=

where 1l1t = efficiency of hammer blow, h height of free fall of the ram or hammer (em), S final set or penetration per blow (em). C = sum of temporary elastic compression of the pile, dolly, packings and ground ( = C 1 + C 2 + C~, C 1 temporary compression of dolly and packing ( = 1.77 RIA, when the driving is without dolly, = 9.05 RIA, when the driving is with short dolly), C 2 = temporary compression of pile (= 0.657 RDIA). C 3 = temporary compression of ground ( = 3.55 RIA). D = length of the pile, A = cross-sectional area of pile. R = pile resistan<;e (Ionnes). The efficiency of hammer blow (ll,,) depends upon the weight of hammer (W). weight of pile, anvil 2nd helmet follower (P) and the coefficient of rcsistution (e). (a) Fc.r W > CPt

W ... e2 p 11" .. ~

(b) For W < eP,

W+e p (w - ep)' 11b-W-;P- W+P

... (25.24)

2

... (25.25)

The coefficient of resistutioo (e) varies [rom zero for a deteriorated condition of the head of pile to 05 for a steel ram of double-acting hammer striking on steel anvil and driving a reinforced conacte pile. For a C.1. ram of a single-acting or drop hammer striking on the head of R.C.C. pile, e = 0.4 and that striking on a well~nditioncd driving cap and helmet with hard wood on R.eC pile, e = 0.25 (IS: 2911-1979).

(3) DUDish Fonnula. According to Danish formula (1929), W x h X 1111 Q. - S + 112S. where

S. - [2'"

~hD)

r

..

. .. (25.26) (25.27)

in which So = elastic compression of pile, D = length of pile, A = cross-sectional area, E ::: modulus of elasticity of pile material The allowable load is found by taking a factor of safety of 3 to 4.

PILE FOUNDATION

687

Eq. 25.27 can also be used to determine the final set (5) per blow. Taking Q.. _ 3 Q... . .. (25.28) where Q"

=<

allowable load.

25.14. WAVE EQUATION ANALYSIS

ru the hammer strikes the top of a pile, a stress wave is transmitted through tbe length of the pile. The wave transmission theory can be used to determine the load carrying capacity of the pile and the maximum stresses that can occur within the pile during driving operation. In the wave equation analysis (Smith, 1962), the pile is represented by a series of individual spring-connectcd weights and spring damping resistance (Fig. 25.11). The weight Wl represents the weight of the ram, and W 2 represents the weight of the pile cap. Weights W3 to WlO correspond to tbe weigbts of incremental sections of the pile. The spring constant Xl represents the elasticity of the cap block; the constants K2 to Ku are for the elasticity of the pile sections. The damping springs R3 to Ru represent tbe frictional resistance of the soil surrounding the shaft; R 12 represents the soil resistancc at the pile tip.

'3 PILE

"

'" "'7" ... " ~O

!L BlOWSlcm~

~,

(.)

'"

(b)

Fig. 25.11.

The propagation of the elastic wave through the pile is analogous to that caU'ied by an impact on a long rod. A partial differential equation is written to describe the pile model shown in Fig. 25.11 (0). The equation is solved with the aid of a digital computer, and the pile capacity is determined. The pile capacity is expre.o.i>Cd as a function of penetration per blow or blows per an [Fig. 25.11 (b)J. The major drawback of the wave equation analysis for determination of the dynamiC resistance is its dependence on a computer. Moreover, tbe field, tests arc required to estimate the equivalent spring constant and soil-damping values for the pile under study. Further, tbe resuU.s obtained are valid ooly for a particular pile driven by a specified pile hammer. Despile the above shortcominw>, the wave equation analysis is a useful tool for detennining the pile capacity. The resuU.s can also be used for the selection of appropriate pile-driving equipment.


688

25.15. IN·SITU PENETRATION TFSl'S FOR .'[LE CAPACITY (Q) Standanl penetration test. TIle load-carrying capacity of a pile can be estimated from the standard penetration tcst value (N). (I) For driven piles in sand, the unit tip resistance (qp) is related 10 the uncorr"!cteti blow COUOl (N) near the pile point (Meycrhof 1976). ., - 40N (DID) < 400N ... (25.29) where qp = point resistance (kN/m\ D := length of pile, B = width (diameter) of pile. 1bc value of IIp is usually limited to 400 N. lbc average unit fridional resistance (I,) is related to the average value of the blow count

t. ..

N IcN/m 2 For low displacement piles, J.. .. 1.0 Ii kN/m2 where N i .. average of uncorrected N-values along the length of the pile. For hi8h displacement piles,

(;,) For bored piles in sand,

2.0

qp" 14 N(D"IB) kN/m 2

where Db = actual penctration into tbe granular soil. For bored piles in snnd, the unit [rictional resistance (h) is given by Is .. 0.67 N kN/m 2

(N). ... [25.30

(a)]

...{25.30 (b)] ... [25.31)]

... (25.32)

(b) Dutch cone test. Meyerhof (1965) relates the unit point resistance (qp) and the unit skin traction (h) of driven piles 10 the cone point resistance (qf)'

15

Point resistance,

qp ..

Unit skin friction

(a) f, (dense sand) .. qrl200 (b) f, (IOCfiC sand) .. qcl400 (c) f, (sill) - ./150

(D"IB)

... (25.33)

... (25.34) ... (25.35) ... (25.36)

25.16. PILE LOAD TEST 'The most rcli.'lb\e method for determining the load carrying capacity of a pile is the pile load test. The set-up generally consists of two anchor pilcs provided with an anchor girder or a reaction girder at their lop (Fig. 25.12). The test pile is instaUed between the anchor piles in the manner in which the foundation piles are to be installed. The Icst pile sbould be at least 3 B or 2.5 m clear from tbe anchor piles. ANCH(J'I GIRDER OR RE.6CTlON GlROERS

fig. 25.12. Pile Load Test.

The load is applied through a hydraulic jack resting on the reaction girder. The measurements of pile movement are taken with respect 10 a nxed reference mark. The lest is conducted after a rest period of 3 days

PILI! FOUNDATJDN

689

nner the Install:.llion in sandy soils nnd a period of one month in slils and soft clays. The load is applied in equal incremenl of about 20% of the allowable load. Settlements should be rcmroed with three dial g.1uges. Eaeh stage of the loading is maintained till the rate of movement of Ihe pil¢ LOp is not more than 0.1 mm per hour in sandy soils and 0.02 mm per hour in case of clayey soils or a maximum of two hours (IS : 2911-1979). Under each load increment. settlements are observed at 0.5. 1, 2, 4, 8, 12, 16, 20, 60 minutes. The loading should be continued upto twicc the safe load or the load lit which the total settlement reaches a specified value. 1be load is removed in. the same decrements at 1 hour interVal and the [rna) rebound is recorded 24 hours after the entire load has been removed.

Sn

-.;;;:::: . ~NET SETTLEMENT

."."'-

.

.

~ LOADlNG

Fig. 2.'i.13. Umd Settlement Curve.

Fig. 25.13 shows n typical load·selilemcnt curve (firm line) for loading as weU as unloading obtained from a pile lood test. For any given load, the nct pile sctllcmenl (SII) is given by .. .(25.37) where 5, = tOial seUIement (gross settlement), 5. = clastic settlement (rebound). Fig. 25.13 also shows the net settlement (chnin dOlted line). Fig. 25.14 shows two load·net settlement curves obtained from 8 pile load tests on two different soils. AI the ultimate land (Q...). the lo~d·net settlement curve becomes either linear as curvc (2) or there is a sharp brc.'lk as in the curve (I), ru; shown in the figure. The snfe load is usunlly tok.en os DOC, half of the uitimote ioaj. According to IS : 2911 , the sare lo.1d is taken ru; one-half of the load at which the lotal settlement is cqunl to 10 per cent of the pile di~meter (7.5 per !z cent in case of under-reamed piles) or two·lhirds of ~

~i~~:~c~~d t~~~~tr~~;t~~ :~~~~:~~:d:n~,~ ~

safe load is Inken nc; one-hIllf to two-thirds of the ~

iOOd~~iCh~~~~: n:I~~:~::~nt:~I:ri~m~ also ~ somctimes specified. Under the lond twicc the so1fe Z 100<1, the net settlement should not be more than 20 ~ mm or the gross settlement should not bc more than 2S mm.

Fig. 25.14.


690

The tcst described above is known as initial te.ft. It is carried out on a test pile to determine the ultimate load capacity and hence the safe load. lbe pile load test described in this section is a type of load-controlled test, in which the load is applied in steps. TIle test is aise> known as slow main.tained test.

25.17. OTIIER TYPES OF PILE LOAD TESTS (I) Constant rate of penetration test. In a constant-rate of penetration lest, the load 00 the pile is

continuously increased to maintain a constant rate of penetration (from 0.25 to 5 mm per minute). The force required tq achieve that rate of penetration is recorded, and a load-settlemcnt curve is drawn. The ultimate load can be determined from the curve. The lest is considerably faster than a lo.1d-controlled test. (2) Routine Load lest. This test is carried out on a working pile with a view to determine the sctUemenl corresponding to the allowable 1000. As the WOrking pile would ultimately form a pan of the foundation, the maximum load is limited to one and a half times the safe load or upto the load which gives a tOlal settlement of 12 mm. (3) Cyclic Load test. The test is carried out for scpsr'dtion of skin friction and point resistance of a pile. In the lest, an incremental load is repc.:1tcdly applied and removed. (4) Lateral Load test. The lest is conducted to determine the safe lateral load on a pile. A hydraulic jack is generally introduced between two piles to apply a lateral load. The reaction may also be suitably obtained from some other support. The test may also be Cllrricd out by applying a lateral pull by a suitable set-up. (5) 1'ull out test. The test is carried out to detcrminc the safc tC[!Sion for a pile. In the sel-up, the hydraulic jack rests against a (rame allsched to the top of the test pile such that the pile gelS pulled up. 25.18. GROUP ACTION OF PILES A pile is not used singularly beneath a column or a wall, because il is extremely difficult 10 drive the pile absolutely vertical and to place the foundation exactly over its centre line. If ccccntric loading results, the connection betwcen the pile and the column may break or the pile may fail structurally because of bending stresses. In actual practice, structural loads are supported by several piles acting as a group. For columns, a minimum of three piles in a triangular pattern are used. For walls, piles are installed in a staggered arrangement on both sides of its centre line. The loads are usually transferred to the pile group through a reinforced concrete slab, structurally tied to the pile tops such that the piles act as onc unit. TIle slab' is known as a pile cap. The load acts on the pile cap which distributes the lood to the piles (Fig. 25.15). The load carrying capacity of a pilc group is not necessarily equal to thc sum of the capacity of the individual piles. Estimation of the load-carrying capaCity of a pile group is a complicated problem. When the piles are spaced a sufficient distance apart, the group capacity may approach the sum of the individual capacities. On the other hand, if the piles . Fig. 25.15. are closely spaced, the s!ressc.s transmitted by [he piles to the soil may overlap, and this may reduce the load.carrying capacity of the piles (Fig. 25.16). For such a case, the capacity is lim ited by the group action.

t <:.~ : t ~l ;it :.:-!..:'..: t t t t tt

The efficiency (l1s) of a group of piles is defined as the ratio of the ultimate load of the group to the sum of individu.11 ultimate loads. Thus

... (25.38)

PILE FOUNDATION

691

PILE 01'

,,

(',

\

, \'!) (\'x' \\

\

\

\

\

(

\

1

/

" ---"'/

"I ' , (IY... ,) X \

(a) SINGLE PILE

---, ---,'

---'

"'",. _-------_...... (bJ PILE

"

)"

)

,/

GROUP

Fig. 25.t6.

or

11&"

Q'i"(N Q" x 100

where Q, (,, ) = ultimalc load of the group, Q"

= ultimate load of the

individual pile,

N = Number of piles in the group.

Thus the groups efficiency is equal to the ratio of the average load per pile in Ihe group al which the failure occurs 10 the ullimale load of a comparable single pile, The group efficiency depends upon the spacing of the piles. Ideally, the spacing should be such that the efficiency is 100%. Generally. the centre to centre spacing is kept between 2.5 Band 3.5 B, where B is the diameter of the pile. The methods for Ihe determination of the ultimate load of the individual piles have been discussed earlier. The methods for the estimation of Ihe ultimate load of the group are explained in the following sections. 25.19. PILE GROUPS IN SAND AND GRAVEL For piles driven in loose and medium dense cohesionlcss soils, the group efficiency is high. The soil around and between the piles is compacted due to vibration caused during the driving operation. For better results, it is essential to start driving the piles at the centre and then work. outward. The piles and the soil between them move together as a unit when subjccted to loods. The group acts as a pier foundation baving a base equal to the gross plan area contained between the piles. (0) End-bearing piles. For drivcn piles bearing on dense, compact sand with a spacing equal to or greater than 3 B, the group capacity is generally laken equal to the sum of individual capacity. Thus Q, • N Q" .. .(25.39) In this case, the load taken by the group is much grealer (TI, > 100%) than the sum of the individual capacities. and the piles fail as individual piles. For spacing less than 3 B, the group capacity is (ound for the block of piles group. (b) Friction plies. The group efficiency of friction plies in sand is obtained (rom the (ollowing expression:

~ 00 f,(P, D) 100 '1, - NQ" )( I .. NI,(PD) )( where P, = perimeter or the block. p = perimeter of the individual pile, D = length of pile,.

{, = unit frict ion resistance.

. . .(25.40)


692

If the centre-to-centre spacing is large, the group efficiency (11,) may be more than 100%. The piles will behave as individual piles, and the group capacity is obtained from Eq. 25.39. If'l1g is less than 100%. (NQ,J ... (25.41) Q,. TI, 100 lhe group efficiency can also be obtained from the Converse- Lebarre equation given below.

",

-

1

-

Cn-l)m+
mn

[

e .90

••. (25.42)

where m = number of rows of piles, n = number of pUes in a row, 8 = 180-1 (B/s). 8 ,. diameter of pile, s '" spacing of pile, centre- to-centre, '1, = group efficiency (expressed as a ratio). Bored piles. For bored piles in sand at cooventional spacing of 3 B, the group capacity is taken as 2!3 to 3/4 times the sum of individual capacities for both the cnd-bearing and the friction piles. Thus ... (25.43) Q, (u) - (2/3 to 3/4)(N Q,J In bored piles, there is limited densification of the sand surrounding the pile group. Consequently, the

efficiency is lower. 25.20. t'lLE GROUPS IN CLAY As the pile group acts as a block, its ultimate capacity is detennined by adding the b&$C resistance and the shaft resistance of the block. The capacity of the block having closely spaced piles (s :s 3 B) is often limited by the behaviour of the group acting as a block. The group capacity of the block is given by Q, (u) - qp (A,) + a c (P, D) ... (25.44) or where qp unit point resistance (Nc - 9.0).A. base area of the block, P, perimeter of tbe block, D = depth of the block, a = adhesion factor (= 1.0 for soft Clays), c = undrained cohesion. As discussed earlier. the individual pile capacity is given by Eq. 25.15. QI/ - qpAp + a C (p )( D) ••• (25.45) 'l11e group capacity core:lidering the piles as individual piles is given by Q, (u) - N Q. ...(25.46) The lower of the two values, given by Eqs. 25.44 and Eq. 25.46, is the actual capacity.

=

=

=

25.21. SETI'LEMENT OF PILE GROUPS The settlement of a pile group is due \0 elastic shortening of piles and due 10 the settlement of the soil supporting the piles. It is assumed that the pile group acts as a single large deep foundation, such as a pier

TIm 13 D

l~,·-H:

"fT"\'b.,

2y

',\

1 2

L.U.U

I

~

(a) FRICTION

PILES

(b) END BEARING PILES Fig. 25.17.

PILE FOUNDATION

693

or a mat. The tOial load is assumed to acr. at a depth equal 10 two-thirds the pile length in the case of frictional piles [Fig. 25.17 (0)1. In the case of end-bearing piles, the lotal load is assumed to act at the pile tips [Fig, 25.17 (b)J. In the case of combined action, the fricr.ional component ls assumed to act at 2/3 D and the bearing component at the tip. For determination of the settlements, the compression characteristics of the soil are required. For clayey soils, the characteristics are determined from laboratory tests 00 undisturbed samples. For oohcsionless soils, the characteristiai are obtained from empirical correlations developed from in-situ penetration tests. (a) Coheslonless soils (i) Skempton method. The settlement of the pile group is estimated from the settlement of a single pile, as determined in a pilc--lood test. 1bc settlement of the group is generally very large because the pressure bulb for the group is much deeper than that of a single pile. Skemptoll et al (1953) published curves (Fig. 25.18) relating the settlement of the pile group (S6) of a given total foundation widlh to that of a sirJgle pile (so), The curves can be used for both driven aod bored

piles. 6

2

8

'/

00

---

V --,

,

6 Width

f.----

12

olloundot!Dr\

"

\8

21

lm)~

(Aller Skempton.19S3)

Rg.25.18.

(;,) Meyerhot method. Meyerhof (1976) suggests the following empirical relation for lhe elastic settlement of a pile group in sands and gravels. "

_ 9.4

q::

I

... (25.47)

s,

where = seUlement of group (mm), q = load intensity (= Q,IA,), B, = width of the group, , = influence factor [= 1 -Dl(S B,) :t 0.5], D = length of pile, N = cocrcct.ed standard penetration number within the scat of settlement (approximately equal to B, below the tip). If statk: cone results are available, the selliement of the group can be obtained from Ihe relation,

q8,'

" - Tq;

... (25.48)

where qf = average cone penetration resistance within the seal of settlement. (6) Clayey soils The consolidation settlement of a pile group in clay can be dctennined using the procedure disrussed in chapter 12 Generally, a 2 : 1 load distribution is assumed from the level at wh;ch the load acts. Sometimes, the load is assumed to spread outwards from the edge of the block at an angle of 300 to tbe vertical. For 2 : 1 distribution, the stress inaease at the middle of each layer is calculated as (see cbaj'Jter 11),


694

q; -

Q« (8, ... Zi) (L~ ... Zi)

... (25.49)

where Zi is the distance from the level of the application of the load to the middle of clay layer i. The settlement of each layer caused by the increased stress is given by (see chapter 12).

'-'(,) - 1 ~CC~)(,) Hi

... (25.50)

where .6. e (i) = change of void ratio caused by the stress increase, e" (i) '" initial void ratio of laycr i. Hi :: thickness of laycr i. Allcmmive!y,

As(,)-

c,

(a.

IIi AOi ) -1--(-' log - -+-

+ e..

I,

... (2551)

a"

'The total consolidation settlement is equal to the sum of the settlement of all layers. 1:.6.S(I)

s, _

... (25.52)

25.22. SHARING OF LOADS IN A I)ILE GROUP All the piles in a group share equal l~ld if the load is central.

Q-

~

... (2553)

However. if the load is eccentric or if Ihe centrol Jo.1d is accompanied hy a moment, the sharing of load is computed assuming the pile cap as rigid. As the pressure distribution is planar, the pile reactions also vary linearly with lhe distance from the centroid of the cap (Fig. 25.19). 1be axial load in any pik m al a distance x from the centroid is given by

Q, (Q,' e.)x QnJ-N:I:~

... (25.54)

where ex = ccccntrkily of l~ld aboul Y.Y-(lxis, If the load is ecccntric about both thc axes.

Qg (Q.' c.)x (Q,e,)y Qm-N:I:~±IT

... (25.55)

where e y '" ccccntricity of load about X-X axis, In the above equations, the positive sign is taken for the piles on the same side as the eccentricity. If the load on any pile is negative, it indicates that the pile is in tcnsion. If the pile is not designed for tension, the lood in that pile is taken as zero, and the load between other piles is redistributed. This would cause extra compression in othcr piles. 25.23. TENSION PILES Piles supporting high structures, such as tall Chimneys., tronsmission towcrs, water towers, are required 10 resist uplifi forces due to wind. Some of the piles in these structures are required to resist tensile forces and arc known as tension piles. Resislllnce to uplift forces is provided by the friction between the pile and the surrounding soil. Thc uplift resistance of a straight-shaft pile can be computed in the same manner as the frictional resistance in rrictional piles. However. the unit skin friction (h) and adhesion (c,,) ror Ihe uplift rcsis(ancc are considerably less tban those for the compressive loads. It is

d

qJ

d

-&--f-~

d . d y (h) PLAN

crer;J:J Cd PILE LOADS Fig. 25.19.

PILE FOUNDATIONS

695

usual practice to reduce these values to one-half of the normal values if the piles arc short. For large structures, it is essential to carry out pull out tests on piles 10 determine the safe value of the unit skin friction or adhesion for uplift forces. The uplift resistance of piles can be considerably increased in the case of bored piles by under-reaming or belling oul the bollom. A bulb can also be formed in the case of driven and cast- in place piles to increase the uplift resistance. Mayerhof and Adams (1968) gave the following equations for thc pull-out resistance (P,,). (a) Shallow plies Fig. [25.20 (a)1. Pull-oul resistance,

(o}'5hollow pUe

Ip)Deep pile

Fig. 25.20.

p ..

= cohesive resislance + frictional

resistance .•. (2556)

PI' - itBtc"D + StY (iBl) LYK.. tan, + W

=

where B I = diameter of enlarged base, c" = undrained oohesion, D length of pile. ~ resistance, Sf = Shape factor (see Thble 25.3), Ku = coefficient of lateral earth pressure ( coefficient of passive earth pressurelin a cylioder of diameter BI and he;

~). 1 = bulk 1 - sm 4'

unit weight, and W

= anglc of shearing

= Kp Ian ~ ~), Kp = = weight of soil and pile

I D.

(b) Deep plies Fig. [25.20 (b)], Pull-oul resistance, P" _ rohesivc resistance + frictional resistance

P" - itBtc"H +

SfY(~Bl)

(2D-II)HK" tan

++ W

where H "" maximum height of rupture surface (see Table 25.3) (For deep piles H

:s;

.. (25.57)

D)

Thble 25.3. Values of HlDI, m and Sf

HIBI

'I

20'

25"

30"

,,"

40"

45 0

50"

2.5

3.0

4.0

5.0

7.0

9.0

11.0

0.05

0.10

0.15

0.25

0.35

0.50

0.60

t.l2

1.30

1.60

2.25

3.45

550

7.60

All other notations are the same as before.

6%

roiL MECHANICS AND FOUNDA110H ENOINEI!RINO

For purely cohesive solis, as 4' :: 0, the seoond tcnn in Eqs. 25.56 and 25.57 Is zero. for oohcsionlcss soils, as Co. = 0, the first (enn is zero. The shape factor (sf) is equal 10 1 + mD/BI for short piles, and equal \0 1 + mll/ 8\ for deep piles, where III is a cocfficJcnt depending on •• 25.24. LATERALLY LOADED PILES Piles are sometimes subjected to lateral loads due to wind pressure, water pressure, earth pressure, curthqunkcs, clc. When the horizontal component of the load is small in comparison with the vertical 1000 (say, less than 20%). it is generally assumed to be cnrricd by vertical piles and no special provision (or lateral lood is made.

If the horizontal load is large. inclined piles, known us ra/cing piles or batter piles, arc provided to take the horizontal load. These piles have a high resistance to lateral loads, as a large portion of the horizontal component of the load is carried axially by the pile. Butler piles, along with vertical piles, art provided in situntions where the horizontal loods are significant, such as wharves, jetties, bridge piers, trestles, retaining waU and tall chimneys. Daller piles arc driven at a bailer rnnging from I : 12 to 1 : 25. However, driving of batter piles is more expensive than that of vertical piles. '(be resistance to failure of vertical piles subjcded to horizontal loads is provided by the pnssivc rcsist.1oce of 0 wedge of solt in fronL of the piles. In C8S(: of baiter piles, odditlooal resistance is provided by the skin friction and the end bearing. Therefore, bauer pilcs are more effective than ven ical piles in res L~ ting horizontal lomls. Il is generally assumed that bailer piles can take the a.dolload equal to that in the ool'Tt.SpOfXling vertical pile. As the axis of thc bettcr pile is inclined, il can resist the horizontal load equal to Q cos e, where Q is the axial lCXld capncity und e is the angle which the pile makes with the horizontal When piles nre oriented in two or three direaions. Culm ann's method, as described below, is used. Steps : (1) Group the plies according to their s lopes. [In Pig. 25.21 (a), the piles are grouped in 3 directions]. (2) Draw the geomctry' ot' the pile group to some scale, and mark the directions of the inclined load Q8 and the centre line of each pile group (Ill> R2 and RJ ). (3) Dctennine the loallion of point It which is at the intersection of RI and Q" (4) Join A to the point B which is at the intersection of R2 and R). (5) Draw the force triangle [Fig. 25.21 (b)J. Select the line ab parallel to AD. From b draw a line bc pnraUeJ to Q, to some Salle. Draw a vertical at c to detennine ca which is

cqualto R I . From b draw a line parnllel 10 R3• and from a, line parallel 10 R2• to complcte the triangle abd. • (6) Detenninc forces in piles as follows. lbe magnitudes or R2 and RJ nre. respectively, given by ad and 1xI. However, R2 is compressive and RJ is tensile. The magnitude of RJ is given by co whit.'h is compressive.

(b) Fig.2S21.

PILE FOUNDATIONS

ILLUSTRATIVE EXAMPLES TIlWltrative Example 25... A concrete pile, 30 em dlamclcr, is driven into a medium dense sand (+ = 35°, 't '" 21 kNlm 2, K ::z 1.0, tan 0 • 0,70) for (J depth 018 m. Estimate the sale load, taking a fact'" 0/ safety of 2.50.

Q. -

Solulloo. From Eq. 25.10,

For • "

3~',

'iN, A,

+'~IK(O.); lon6 (A,),

D,/B _ 12.0, from Fig. 25.3. Dr'" 12)C 0.3. 3.6m

Maximum value of ov. 3.6 x 21 • 75.6 kN/m 2 The value of Nfl i.!I taken from Bcrczontzcv's

Therefore,

aJIW

(Fig. 25.4). N• • 60.

Q" ... 75.6 x 60 x Jt/4 x (0.3)2 ... K tan 6 (area ora., diagram) ... 320.5 ... 1.0 x 0.70 )( (

tx

75.6 x 3.6 + 75.6

II(

JC

4.4 ) )(

pilcs perimeter' :t

x 0.3

Q, - 320.5 + 309.2 - 629.7 kN Safe 1000, Q. _

~ _ 6~7

_ 151.9 kN

lIIuslrative Example 2S.2. Determine the safe load for the pile in IlIwtrQlive Example 25.1, i/ the table rises to 2 m below the ground surface. Take 1...... 10 kNlm J,

waler

Solution. Vertical prusure al the critical depth,

Ov ... 2 Therefore,

)C

21 + 1.6 )( (21 - to) • 59.6 kN/m 2

Q• ... 59.6 x

(j() )(

t )(

;t/4 )( (0.3)2 ... (

59.6 x 3.6 + 59.6 )(

4.4) x 0.7

)C

11.

)C

0.3

Q _ 252.6 + 243.1 - 496.3 kN Qn •

4~3

• 198.5 kN

Illustrative Example lS.3. A 30 cm diameter concret~ pile is driVttil into a homogeneous cQII.JolidoJed clay deposit (eN :;:r;. 40 kN/m2, a .. 0.7). If the em.beddt!d IDlgth is 10 m. estinuJII! the safe load (FS. = 2.5). Solution. From Eq. 25.15,

QN - cNeA, ... (leA,

Taking N, - 9.0,

QII • (40

)C

9.0) ,,/4 )( (0.3)2 + 0.7

)C

40 (,,)( 0.3) )( 10 - 289.2 kN

Q. _ ~_ ~~2 _ l1!.7kN JIIustratlye Exllmple 25.4. A square COrlCTt!t~ pile (30 em side) 10 m long is driven into coorse slJlld (1 '" 18.5 kNlmJ, N = 20). Deli!rmine the allowable load (FS. • 3.0). Solution From Eq. 25.29, In Ihlsease,

q, - 4ON(DIB) • 400N

4ON(DIB) _ 40.20 (10/0.3) _ 26666.7kN/m'

400N • 400 Adopt the lower value of 8CKX} kN/m2 From Eq. 25.30 (Q).

Therefore,

)(

J, - 2.0 •

!O • 8flXl kN/m 2

Ii _ 2.0

)( 20 • 40 kN/m 2


698

.. 8000 (0.3 )( D.3) + 40 x (4 )( 0.3 x 10) .. 1200 kN

~. I~

Q• •

_ 400kN

illustrative Example 25.5. A square concrete pile (35 an )( 35 em) is driven into Q homogenecus sand layer (+ '" 30", Y = 17 kN/mJ) for a depth of 10 In. Calculate the ultimate load Use Meyerhof's method. Take K::: 13 and 6 = 1Ir. Solution. From Fig. 25.5, (Dol D)" .. 7.0 Dc .. 7 )( 035 .. 2.45

Di/B - 10/0.35 • 28.57

Also

q .. From Fig. 25.5, From Eq. 25.6, In this case,

Nq

2.45 x 17 .. 41.7 kN/m 2

55.0

..

Qp .. ApqNq

A, ,N, -

:50

Apq/

(0.35 x 0.35) (41.7 x 55) - 280.9 kN

Apq/ .. (0.35 )( 0.35)(50 x 55 tan 30°) .. 194.5 kN Adopting the lower value, From Eq. 25.8,

Therefore,

Qp" 194.5 kN h" KO v lan6 Q, .. K tan 6 (area of Oy diagram) perimeter

.. 1.3 tan 18 0 x (41.712 x 2.45 + 41.7 )( 7.55) )( 4 x 0.35 .. 216.5 kN Q.. .. 194.5 + 216.5 .. 411 kN UlustnUve Example 25.6. A concrete pile, 4() em diameter, is driven 25 m into a soft clay (ew = 25.0 1 kN/m2, y = 19 kNlm ). Determine the allowable load using Wjayvugia and Focht method (RS. = 2.5). The Thus

water table is at the ground surface..

Solution. Taking Nc .. ' 9.0, Qp ..

From Eq. 25.14, From Fig. 25.8, for 0 = 25 m,

Therefore,

c,.NcAp

/, .. ).

(O~

..

25 )( 9 )( K/4 x (0.4}2 .. 28.3ItN

+ 2c)

). = 0.16

Is .. ,0.16 [ ~

)( 25 )( (19 - 10) + 2 )( 25] .. 26 kN/m 2

Q, .. 26 )( (l't )( 0.4) )( 2S .. 816.4 Q,. .. 28.3 + 814.4 .. 844.7 leN Q" - 844.712.5 _ 337.9 kN

Thus

llIustralive Example 25.7. A 25 m deep bored pile has a shaft of 1 m diameter and el/farged base of2.5 m diameter in the lower 1.5 m depth. The undrained cohesion of tile soil vtJries from 100 kN/m2 tJt the top 10 150 /eN/m2 aJ. the base. Determine the safe load (F.s. = 2.5). Ttike a '" 0.45. Solullon. Total depth of the shaft = 25 - 1.5 = 23.5 m Assuming no adhesion for a distance 2 B above the bell, the effective depth is 21.5 m. c,. at that depth .. 100 + (150 ;

Therefore.

100) )( 21.5 .. 143leN/m2

Q .. .. (150 )( 9) (l't/4) )( (2.5)2 + 0.45 )( (100 + 143)/2 )(

.. 10319.5 leN

Q. - lo;.~.5 _ 412.8 kN

l't )(

1 x 21.5

PILE FOUNDATIONS

lliuscratin Exumple 25.8. A precast concrete pile (35 cm x 35 cm) is driven by a single·acting steam hammer. Es(ima(e the allowable load using (a) Engineering News ,R ecord FormuUl (F.S, :: 6), (b) Hiley Forlllula (F.S. '" 4) and (c) Danish Formula (F.S. :z: 4). Use tbe following data. (I) Maximum rated energy Weight of hammer (iI) (iiI) Length of pile (ilol) Efficiency of hammer (101) Coefficient oC resistitulioE:l (loll) Weight of pile cap (viI) No. oC blows for last 25.4 mm (viii) Modulus of elasticit.y of concrete

= 3500 kN-cm :: 35 leN = 15

m

= 0.8

= 0.5 = 3 leN =6 ::: 2 x 107 kN/m2

Assume any other data, if required. Solution. (0) From Eq. 25.22,

Qu"

SE:'I1~ - 2.;;~ :~~4

.

4133'. 9kN ·:

Qu ... 4133.9/ 6 - 689 kN

Allowable load,

Weight of pile - 25 (0.35 x 0.35) x 24 .. 73.5 kN P - 735 + 3.00 .. 765 kN eP .. 0.5 )( 76.5 - 38.2 kN

W
AsW_35kN,

From Eq. 25.25,

11/1 ...

W+ e'P (W- eP)' --w-:;:-p - W + P

...

.. 35 + (0.5)2 x 76.5 _ (35 _0.5 x 76.5)' 35 + 76.5

35 + 765

'11/1 .. 0.484

J

(W1I) '11/1"'11.

From Eq. 25.23,

Q. - (5 + CI2)

Q.. ..

35~.5:/~~e;2 0.8

58 .. 254:: + el2

Assuming driving is with dolly,

.'

(9.05 + 0.657 D .. 3.55)RIA (9.05 .. 0.657 x 15 + 355) R

35 x 35 C _ 0.018 R _ 0.018Q.

0'

where Q.. is in lonnes.

e ..

e ..

For Q.. in leN,

e ..

0.0018 Q"

~3g.~Q.

Therefore, &j. (0) gives

Q. -

Solving,

Q. _ 10161eN

Allowable load.

Q• ..

0423

1~16 .. 254 kN

...(0)

SOIL MECHA.NICS AND FOUNDAnON ENGINEERING

700

(e) From Eq.

(Wh) x~.

2~.26,

Q.= 5+5,/2 c

From Eq.

_ 5" -

2~.27,

3~OO x 0.8 2800 2.54/6 + 0.5 S" ~ 0.423 + 0.5 S..

... (b)

~2~.(W")D AE

="2XO.8X3.500XI500 =1.8.5cm 3.5 x 35 x 2 x 10' x 10.... Therefore. Eq.

(b)

gi\'es

+2~~ )( 1 .~!5 = 20;.2 c S19.3 kN

QII = 0.423 Q..

Allowable load,

= 2077.2 kN

DlUitratlvt! Example 25.9. A 30 em diameter pile of length 12 m the following

~:lults Wf'tr

Wa.J

subjected

"""(kN)

Settlemem during loadinJ (em)

Settlement during unloading

'.0

(em)

Oro" senJemem (em) Rebound (em) Nd settlement (em)

a pile load test and

500

1000

1500

1000

1500

OJIS

1.05

1."

3.8

• .0

4.•

'.1

B.

'.8

•.0

Determine the dllowable load. SoIud.... Fig. &25.9 shows the load settlement curve, and the rebound calculated below. 1be figure also shows the net settlements. """(kN)

10

obtained.

I

0

ClU'VC.

!be net setdemenlS arc

1500

1000

1500

0

'00 0.85

1000

I

1.05

1.$$

3.80

•.00

I I

0

0.60

1.10

1.$0

1.80

1.00

0

0.25

0.45

1.05

1.00

4.00

0 0.0

'".0

!

J

JI

"0 Ag. &2.1.9

PlLB FOUNDATION

101

The aUowable load is determined using the following aitcria. (,) ~ of load corresponding to a gross settlement of 12 mm, Q··t)C7.50 • .500kN (;1) ~ or load com:sponding to a net settlement or 6 mm.

~

Q.(ilt)

~'Il25

_150kN

i of load roI"I'e5pondina to a gro;s settlement of 8110 ( • 3 an), Q• •

~

)C 1700 • 8.5OkN

The allowable load is the least of the Ihree values. Q. - SOD kN lIIustratlv. Example ZS.IO. A pile group C01Ul&tbtg 0/12 pile, (Fig. E·25.IO) 1& 'wbj
Ra. £·23.10

Solution. From Eq. 25.55. In this case,

Q. -

'1:-. ¥. (Q'I';i'

Ii' _ 6 • (05)' + 6 • (15)' _ 15.0

II _ 4 • The maximum load oc::cws in pile 4. Q" •

li'

(1.0)' + 4 • (1.0)' _ 8.0

+ (4.0 1~ 0.3) x 1..5 + (4.0 : 0.4) )( 1.0

_ 0.6533 MN _ 6533 kN

IIIWJtnllve Example 2$,11. A. pile group consists of 9 friction pilu of 30 em dJmw~r and 10 m length driven in clay (e,. '" 100 kNlm2, y '" 20 kNlmJ), as shown in Fig. E·25.ll. Determine the safe load for the group (FS :;:: 3, a '" 0.6). Solution. From Eq. 25.44, Q.(u). qpA. + ac(P.D).


702

Fig. E-25.lJ.

From Eq. 25.45,

- (9 )( 100)(1.8 )( 1.8) + 0.6 1100 )( (4 x 1.8 )( 10) Q, (u) • 7236 kN Q.. .. qpAp + ac(p x D)

.. (9 )( 100) )( l't/4 )( (O.3i + 0.6 )( 100 (x )( 0.3) )( 10 Q.. .. 628.8 leN From Eq. 25.46, ' Q,(u). NQ. • 9 x 628.8 • 5659.2 kN As the ultimate load Cor individual pile failure is less than the pile group load, the safe load is given by Q.. .. 56~.2 _ 1886.4 kN

Illustrative Example 25.12. A 40 em diameter pile, 11 m long, has a bell of 2 m diameter and 1 m 2 height. I{ the soil has • = 25°, ell = 20 kNlm and 'i = 19 kNlm}, estimate the allowable pull out resistance (FS = 3). Solution. From Thble 25.3, HIB, - 3.0, Therefore, H • . 3)( 2.0. 6m As D > H. the pile is deep. From Eq. 25.57, p .. - .. nB,c"H + s"t{1t/2 x B I ) (2D - H) HK.. tan $ + W where

W .. l't/4

or

W_662kN

From Table 25.3,

sf" 1.3.

X

(2.0)2 x 11 x 19 + ltI4 )( (0.4)2 )( 11 x (23 - 19)

703

PILE fOUNDATION

K .. .!....±....$..t(tan£ x,) .. I + s~n25° "I-sm, 3 l_sm25° Therefore,

p" ..

1t X

(tan~ 3

x 25

0

)

..

0.737

2.0 x 20 x 6 + 1.3 x 19 (11:12 x 2)(2 x 11 - 6) x 6 x 0.737 tan 25 0 + 662

1'" .. 3976kN p • .. 39;6 .. 1325 kN

Allowable pull,

Illustrative Example 25.13. A group of friction pilcs of 30 em diameter is subjected 10 a net load of 2000 !tN, as shown in Fig. E-25.13. Estimate the consolidation settlement.

u~ r1!t

l':t6kN/m 3

I-----

17

T

m -

L

,/ ---r---

----

,

I

10m

I Loyer

'1

----/--!::'=-=-=-11!!!.~-11 "'wt"20kN/~

"

\

~~~g.~~

/

\ \

"'so1"ZlkNf",l

, \

cc"o.zo

\

'0. 0. "

__ • __ n

_ _ _ _ _ _ _ _ _ _ _' \ ,

(Not to

'I"'"

5cot~)

Fig. E·2S.l2.

Solution. 00 at point A, middle of I layer

.. 3 x 16 + 2 )( (19 -- 10) + 8 x 10.0 .. 146 kN/m 2 00 at point B, middle of II layer .. 3 x 16 + 2 x 9.0 + 13 x 10.0 + 5 x 11 .. 251 kN/m Cross-scClionalarcaatA

.. (2.5+ 2 x 5 x

6. a ..

7.52~7.5

t) .

7.5m

.. 35.56 kN/m

2

2

2 Cross-sectional area at B .. (2.5 + ' 15 x 2 xi) .. 17.5 m do ..

17.;~7.5

1 .. 6.53kN/m

2

9'JIL MECHANICS AND FOUNDATION ENGINEERING

704

Settlement of I layer • C

~

(...lL) + 1

to

_ 0.25 x 1

Settlement of Total seuternent

n layer _

log Co + Aa CJo

+1~.7S

log 146

:~5.56

• 0.135 m

6 0.20 x 1 +1~.6S log 251 ;1 ,53 _ 0.014

• 0.135 + 0.014 • 0.149111

PROBLEMS A. Numerical 25.1.

~ ':::~IIn:~ ~~ t~ ~~9 ~~~1~ =~II)' consolidated clay depo£il

IS m thick.

2s.z. A 10 m IonS concrete pile 30 em diameter is driven into a medium dense sond (+ .. 1.0, tan 6 005). me D( IB :: 10.0. Estima$C the safe 100<1 (F.s, '" 2..5).

u.J.

JQ-,

~~: 3~; ~~i

y '" 20 kN/mJ, K:II

= lARs, 130 kN) ~p~c:a::(~)t !~c~a(=;~b~:jn: :OO~~!,7:; ~~k :~l~tn~~:pt~~::~a;~~ !~:::

dense sand (qr .. 2000 kN/m1. If the tip is embedded 1 m inlO the dense sand, cSllmalc the sate load (FS. '" 3.0) {Ans. 542 kNJ

25.4. ~~:e~;~g~ :!,::;='~r :~(:,. Lc3~~v;n. ~~~:~~I~h: !::I~~ ~b~~lrS ~n~'g~;'~rfu~

estimate the safe 1000 (F.S. '" 3). 1bkc K .. 1.0. and 6 .. 0.7S tp. [AnL 168 tNJ 2!.5. (tI) The design C8pecil)' of a pile 1& 400 kN with a factor of safely of 4. What should be the average penetration of the pile for the last few blows ? (b) HaN many blows would be required for the last one metre for the design capacity 10 be achieved ? Use collowl ng data: Energy rating .. 3500 kN-cm Ernciency of hammer • 0.80 Cro&s.seaional area 01 pile .. 100 cm'l Young's modulI» of pile material .. 2 x 10' kN/m'l Length of pile .. 10 m. Use Dani'" formul a. IAnL 9.1 mnvblow; 1101 15.6, A precast concrete pile is driven with a SO kN hammer, having 8 free (oil o( 1.0 m. U the penetration in the last blow is 0.5 em, determine the Ioad-carrying cupadlY of the pile using Engineering News Rea>td formula. (FS. '" 6.0). [Ans. 274 kN] 15.7. The pile load test on a 40 em diamelel' ~re pile in a deposit of sand indlcnles a settlement o( 4 mm under a lood of 400 kN. Estimale the seUlemcnt of. 4 x 4 pile group. The piles are driven at a spacing o( 100 em. The total 10Dd on tbe group is 6400 kN. [Ans. 2:0 mm] lS.8. A group 0( nine piles., 8 m Ions. is used as the foundJuion for a ooiumn. The piles are 30 em diameter with centre to centre spacing of 90 em The subsoil consIsts of clay with unconfined compression strength of ISO 2 kN1m . Eatimatc the sate load. (F.s. m 3.0) [Ans. 1.8 MN, individual action} 25.9. A 30 an diameter pile, 12 m Ionl. is driven lOla a sand deposit. The details of the hammer nre as under. Tbl:a1 weight of hammer .. 2:0 kN Length of stroke • 100 em Energy per blow • 2000 kN-cm Average penetration blow .. 4 mm Eaaimate ultimate rciistalKe of pllc U5ina Hiley'. formula, assuming thot driving i5 without dolly ond cushion is about 2.S an thick. [Ana. lISO kN] 25.10, A bored pile in a dayey soli failed 81 an ultimate load of 400 kN. If tho pile 15 40 em diameter and 10 m long, determine the capacity of a group of nine piles, spaoed 1 m centre 10 centre both ways. Take a ,. 0.5. {Ana. 3.6 MN, individ\l:ll aaion; 5.4 MN, group action]

PILE FOUNDATIONS

705

8. Descriptive a nd O bjective Type 25. 11. What are the condilions where. a pile. foundation 25.12. Describe various types of pile. founda110ns.

1$

more suitable than a shallow foundation '?

25. 13. Discuss different methods for the inslallnlions of piles. 25.14. How would you c.slimnle the load carrying capacity of a pile in (u) cohesionless soils, (b) cohesive soils?

25.15. What is negative skin ffl(;hOn '! What h Its effect on the pile ? 25, 16. Discuss Yariou~ dyn,nni e formulae. Whm arc Iheir limilJlions '!

25.1 7. Discuss the uses of penetration tests for the estimation of load· carrying efll);lcity of pi les. 25,18, How wou ld you estimmt! the group capacity of piles in (a) s::md (b) clay ·! 25,19. Discuss the method for the design of (a) lension piles. (b) inclined piles. 25.20. Write whether the following statements are true or fal.'iC. ((I) Pile foundations nrc more e!.:onomic:ll thnn shallow roundJtion for moderate loads. (b) TIll:: load·!.:nrrying !.:apncity of a bored pite is smaller than thm of an t:quiyalellt driven pile. (e) Negative skin friction occurs wlll.'n the surrou nding soil settles more than the pile. (tfl The most l"Clinble method for determining the load· carrying eapacHy of il pile is the load tes!. (e) TIle group <.:upJcity uf n pi le group for closely.spact.>(! piles is generally limited by the behaviour of the group as (I block. I.j) TIle group effiCiency of the piles fan be more than 100%. (g) TIle lond cmrying cupacity of tens io n piles depends upon the diameter of the bell. rA n.~. True. (I,), (el, (d). (e), (fJ. (gl]

C. Multiple Choice Questio ns 1. A pile fouodation is ust.-o when (a) the loads ure he;\Yy. (h) the soi l stratum ncar ground slIrf:lce is weak. (e) both «(I) Jnd (b) (d) neither (ll) nor (b) 2. The load.carrying cap:\city of a pile depends upon the (a) skin friction (b) point res istance (e) both (a) and (h) (ll) neither (a) nor (b). 3. The negative skill friction on a pile develops when «(I) the soil in which II is driven IS ~andy soil. (b) the soil surrounding it sc1l1es more than Ihe pile. (d the ground W,lier table riscs. Cd) the soil ncar the tip is clay. 4. The load·carrying capacity of a bored pile in SlUld is about ... times that of a driven pile. (a) 1/2 to 2/3 (b) 21.11\1 314 (C) 3/4 101.25 (til more than 1.25 S. The group cflicietll"Y uf driven piles in sand ,1\ !l !.:Iose spacing may be (a) equal to 100% (b) greater than 100% (c) well below 100% {tIl None of :looyC

6.

~c~~r ~~ls~:.~~;e~~r/~~. :~n~rie:~:~s:~:;\!r~~ ~~~~~;~u:7r~~e~~~C.~io;ls~~~~t;~s ~;~jo deposit. (a) 150.8 kN (c) 2 15.4 kN

The safc load whcn the

(h) 105.6 kN (tl) 2 11.2 kN

IA ns. I. (d. 2. (c), 3. (h), 4. (a), 5. (11), 6. (b)]

26 Drilled Piers and Caissons 26.1. INTRODUcnON (a) Drilled Pier. A drilcd pier is a large diameter concrete cylinder buill in the ground. For construction of a drilled pier, a large diameter hole is drilled in the ground and subsequenUy filled with concrete. 'The difference between a drilled pier and a bored pile is basically of the size. Generally, bored piles arc of diameter less than or equal to 0.6 m. The shafts of size larger than 0.6 m arc generally dcsignated a$ drillro piers. A driUed pier is a lype of deep foundation constructed \0 transfer heavy axi'li or lalcralloods to
DRlIl..ED PIERS AND CAISSONS

707

(b) BELLED

PIER

Fig. 26.1. Drilled Pie rs. Q~

0'

_ (qNq)Ap +

(Ka~tan6)A,

... (26.1)

a..

where qp = unit tip resistance, q = effective vertical pressure at the base, = effective vertical pressure at any dcpth, tan 6 fricllon coefficient, K = lateral carth pressure coefficient, Ap area of ba<>c, A, = surface area of shaft. As for piles, while calculating the effective vertical pressure at the base @, the limits imposed by the concept of the critical depth should be considered, if applicable. The value of K is approximately equal to the coefficient of carth pressurc at rest (K.,). Thus K-K(l .. l - sin<\l Generally, the value of K varies between 0.3 and 0.75. The value of tan 6 is takcn cqillil to tan $ whcn thc 200 excavation is donc dry. Howevcr, if sluny is used, some

=

reduction should be applied to the value of tan 6. The value of Nq for drilled pic·rs is generally lower than that for driven piles. The values given by Vesic (1963) are approximately the Jower bound, and are Nq generally recommended for drilled piers in sand (Fig. 26.2). Altematively, the values of lhe shallow foundations (Chapter 23) can be used conservatively. A factor of safety of 2.5 to 3.0 is generally applied to the ult~ate load to obtain the safe load. The allowable bearing capacity can also be obtained from N-value obtained from standard penetration test. The bearing capacity of a drilled picr is generally takcn one half of the value of an open caisson in identical conditioM, obtained from Eq. 26.5. The settlement of a pier under a given nel soil pressure is genemlly less than that of a comparable

=

v

t '~ 0 60

/

•

0

30 0 10 25

L

/

30

35

40

"'-

4S·

(Aft~r Vt-sic. ,1963)

Fig. 26.2.


708

shallow foundation becmlSoC of the confming prcs.sure of the su:rounding soil. The allowable soil pressure can be obtained from the Qlrvcs for the shallow foundations given in chapter 23, using the N-values unO)ITcctC(l for the oonfming pressure. If tbe w3Ier table is high, the water table correction is made as in shallow foundations.

As the setllcmcnt due to self weight occurs before the pier i.. completed, it is not of much practical significance. The self weight of the pier is usually subtracted from the total load when determining the aUowablc load for (he settlement. J-(owevcr, while oompuling the faaor of safety against bearing failure, the weight of the pier must be oonsidcred. lhc settlement of the pier can be computed using the procedure developed for shallow footings. Acrording to Terzaghi and Peck (1967), the settlement of a drilled pier in s.1nd al any depth is about one half the settlement of an equally loaded footing covering the same area. Eq. 23.83 can be used to determine the allowoole soil pressure. lbe unit pressure for piers on sand is generally taken twice the value for a fooling of the same size under identical oonditions, obtained from Eq. 23.83. (b) Drilled piers on cluy. 'llie analysis of a drilled pier in clay is similar to thut of bored piles in clay. The ultimate load is given by the following equmions. Qu .. qpAp + hAs

or Qu" cNtAp + neA, ... (26.2) where c = undrained cohesion, avcrdge undrained cohesion on the shaft, n adhesion factor, Nc bc.1ring capacity factor. In the case of drilled piers, the value of the Nt depends upon the DIB t ralio of the picr, whcre DI is the diameter of the bottom (see Table 26.1).

c=

=

=

Table 26.1. Vulues of Nt (After Teng, 1962) Dllll

4.0 and above 9.0

The value of a genci""dlly varies between 0.15 :md 0.50, depending upon the drilling method and the type of Ilier. An average value of 0.4 is usually takcn. If the shaft is provided with a bel~ only the straight portion is oonsidcrcd for (riction (adhesion). For bclled shaft drilled dry, the upper limit of unit adhesion is 40 kN/ml and that for the belled shuns drilled with slurry is 25 kN/m 1 . For straight shafts excavated dry. the upper limit is 100 kN/m2. lbe safe load is dclenn ined by applying a suitable [actor of safety to the ultimate load. .. .(26.3) Q, - Q,/FS Gcnerally, a factor of safety (FS) of 3 is taken. Sometimes, the safe load is obtained by applying a (actor of safety only to the tip resistance. Thus

Q, _ Q,IFS • Q,

Q,,"

(C~:SAp)

+ aeAs

... (26.4)

The settlement of a drilled pier in clay depends upon the load history of the clay. Seulemcnt analysis can be done a.<:suming the bottom of the pier as a fOOling ::Iod applying the consolidation theory (Chapter 12). Drilled piers in normally' consolidated Clays arc nOI economical. ~ the selUemenls are excessive. In actual practice, drilled piers are used only in ovcr-oonsolidated Clays. In which case, the settlements are generally

small and within the permissible limits.

26.3. CONSTRUCflON OF' DRILLED l'lERS Construction of drilled piers can be divided into 3 stages. (1) Excavation of piers. (2) Providing supports, and (3) Concreting of piers.

ORIu.rn PIERS AND

CA1~NS

709

(1) Excavation or piers Drilled picrs are gcnernlly excavated using an auger drill or some other type of driWng equipment. An auger is attached to a shaft and rotated under pressure to dig into the soil. When it is filled with soil, it is raised above the ground and emptied. For formation of a bell, the auger is replaced by an under- reaming tool. The tool usually consists of a cylinder with the cutting blades that are hinged at the top. The OJtting blades are in the folded position when the under-reamer is lowered into the hole. On reaching the bottom of the hole, the blades arc spread outward by a mechanism. As the under-reamer is rotated, a bell is formed and the loose soil falls inside the cylinder, which is raised and emptied. The process is repeated till the bell is completcly formed. TIle diameter of the bell is kept two to three times the diameter of the shaft. The angle of the bell is 30" to 45" with the vertical. The above method of drilling is oonvenient for hard clays where the hole can be left open for a few hours without a support. In cohesionlcss soils below the water table, the hole is prevented from collapsing by providing a casing or by drilling in Slurry. When rock is encountered during drilling, special machines are required. For boulders and holes socketed in hard rock, special drilliDg tools are required. When excavali01,1 is in progress, the soil is exposed at the bottom and sides. It is examined carefully 10 chec\c: that the hole is straight and has been drilled to a stmtum of adequate capacity. As the hole is of a large diameter, even a man can descend into the shaft for inspection. 2. Providing Supports (a) Chicago method. In this method, a circular hole is excavated upto the depth at which the soil will stand unsupported (about 0.5 m ror soft clays and 2 m for stiff clay). Venical boards, known as laggings, are then set in pc6ition around the C"Clvated face and are held tightly against the soil by steel ring<; [Fig. 26.3 (a)J. The shaft is tilen excavated further for 1 to 2 m and another setting of boards and rings is RINGS made. The process is repeated unlit the desired level is rc.1ched. The base of the LAGGING TELESCOPIC shaft is then belled oul. STEEL CASING (b) Gow method. In this method, excavation of the hole is done manually. Telescopic steel shells are used to support the soil [Fig. 26.3 (b)]. The telescopiC shells are extended as the hole is deepened. The shells arc removed as the concreting progresses. One section of the sheU is removed at (a) CHICAGO METHOD (b) one time. The minimum diameter of the Fig. 26.3. hole in this method is about 1.25 m.

-r

STEELtf

. , /l _______\ J

(

\

"----------.1

GOW METHOO

3. Concreting of piers. After the hole has been drilled to the required depth, the shaft is dcwatcrcd and the bottom is cleaned. The casing, if used, is removed before the concreting is done. The reinforcement cage is then inserted and concreting is done. As far as possible, concreting should be done in dry. Concreting can be done in a dry hole by gravity pouring, provided the concrete does not strike the sidcs. However. if dewatering is not possible or slurry is used to support the shaft sides, concrete is placed using a tremie. 26.4. ADVANTAGES AND DISADVANTAGES OF DRILLED PIERS Drilled piers have the following advantages and disadvantages, as compared with pile foundations; Advantages. (1) As a singlc drilled pier can takc up the load of a group of piles, it is more convenicnt . (2) Drilled picrs have higher resistance to lateral Ioods than piles. (3) Construction of drilled piers generally requires lighter equipment for drilling than thai for pile driving. There is no noise due to hammer blow in the ca<>e of drilled piers.

SOlI.. MECHANICS AND FOUNDATION ENGINEERING

710

(4) Piles driven by a hammer cause ground vibrntions and ground heaving. Such conditions do nOl exist in the case of drilled piers. (5) lhc base and the sides of the drilled pier can be inspected. This is not possible in case of driven

piles. (6) "[be base of a drilled pier enn be enlarged to provide grc;:lIcr bearing capacity and also to provide greater resistance to uplift. (7) Drilled piers can be used even when the soil contains boulders. etc. Disadvantages (1) The concreting operation requires stricter supervision. lbc quality of concn.:te obtained is generally inferior to that in prccllSt piles. (2) Deep excavation of the drilled pier, if not properly supported, can cause substantial subsidence and

damage to adjoining SlructufCS. (3) Strict supervision of drilling operation is required in drilled piers. (4) Subsurface investigntions required in the case of drilled fliers arc more than that in piles. (5) Load tests in the case of drilled piers are difficult.

26.5 DESIGN OF On:N CAISSONS Caissons are carried to a h.,rd Slr3tum, such ~ compact sand, gravel, hard clay or rock. The lood-carrying capacity can be estimated as in the ClSC of drilled piers. As a caisson also acts like a rigid mat foundation, the equ3lions for the bearing capacity given in chapter 24 for mats may also be used. The aUO'oVable soil pressure (q/W) for an open caisson in COhcsionless soils can be obtained from the following equation (ES. = 3.0). q_ _ 0.22N2 BWy + 0.67 (100 + N 2)Df Wq .. (26.5)

=

where qnQ allowable soil pressure (kN/m\ N = oorrectcd standard penetration number, B dimc,nsion of the caisson, Df = depth of foundation, measured below scour level.

= smaller

Wl and lVq are water table correction factors. According 10 IS : 3955, the allowable bearing pressure can be determined lfSing the following formula,

n

q"" .. 0.054 N2 + O. L6 000 + N2 ) Df lcN/m2 .•• [26.3(a)J If the caisson (well) rests on rock strata. the safe bearing pressure depends upon the crushing strength of rock. The crushing strength can be determined by taking the oores from the field and testing for compression. However, there may be fissures, faullS and joints in the rock which would also affect the bearing capacity and which are not detccted from the core s..,mples. Teng (1962) has suggested that the allowable bearing pressure of caissons on bed rock should not exceed that of concrete seal, which is normally taken as 3500 kN/m2 because the concretc seal is usually placed under water and the quality of concrete is poor. In case of cohesive SOils, undisturbed samples should be tested to determine the unit cohesion (c). The ultimate bearing capacity is determined as ... (266) qu - cN~ where

N~

= bearing capacity

factor.

The vertical loads acting on the c.,isson are the Jo.'lds from the superstructure Dnd the self weight. The buoyant forces should be determined for the lowest water lcvcl Dnd deducted from the downward loads. The Jatcml londs acting on the caisson are due to earth pressure, wind pressure, water pressure and earthquakes. The lateral forces may also aCI due to tractive forces from traffic, ice pressure, and currents of flow. "[be skin friction should !'Ic estimated for the most critical condition when the soil has been removed to the maximum depLh of scour. 'Ibe 10lal load is assumed to be carried by the base of the caisson if it penetralcs a relatively shallow depth of soil.

DRlLLED PIERS AND CAISSONS

7\1

G st

Ie

Iy id

Besides the above·mentioncd 10<1(15. ;1 caisson may also be subjected to large stresses during the sinking operalion. When Ule cnisson is hung up ncar the top by skin friction, Ibe lower pardon is subjected to tension. Large stresses aL'>O develop if the cais..~on is dropped suddenly during sinking or when it to; pulled to iL<; correct position from the inclined position. If the caisson is supported on one side only or on two opposite corncrs at some stage during sinking, it is subjccted to heavy stresses. 111C Clisson must be s.'lfe against nn such conditions. TIle exterior walls of the caisson llre designed to withstand the stresses due to vertlcal loads and the lateral forces. Sinking Effort. The U1issons arc designed to have sufficicnI self weight in each lift to overcome the skin friction. If the self weight is not sufficient, additional ballast is required during sinking. Occao;iOrullly, w:lIer jelling is used to reduce the friction. If it is desired to proportion a circular caisson such that no ballast is required. an expression for the unit skin friction can be obtained by equaling the weight of concrete Lo the frictional force (Fig. 26.4). lhercforc. (./4)("; - 01) Dy, _(.D.) D x f ... [26.7(a)l where Do = extemnl dinmcter of caisson. Di internal dinmcter of c..'lisson . ..,~ ;::: unit weight of concrete (; 24 kN/m 3 above water level. and 14 kN/m 3 below water level), D = deptil of penetra.tion, [= unit skin friction.

... [26.7(b) I

Therefore, te at

og

)] )f D.

tel

Fro rectangular caisson.<;,

5)

... (26.9)

where I = thickne...-.s of concrete seal. D; = internal diameter, L;.8; = internal length, width. q = unit bearing pressure at the base. a .. B/L;, 0c = allowable concrete nexural stress (} 3500 kN/m'J. If If is the depth of water above the base (Fig. 26.4). the value of q can be found from the following equation.

q - ,(.•,11 -

'(~ I

where q '" unit bearing pressure (kN/m2) • and If and t arc in metres. Taking y~ = 24 kN/m.l and ,(",:: 10 kN/m 3.

to

q=1011-241 'fhe thickness of the scal should be safe against perimetct shear,

... (26.10)


712

.

:~; .

SECTIGI

PLAN (a) CI~LAR CAISSON Fig. 26.4.

AjRy", - Ajtyc

Pit

... (26.11)

Ii!

where A j = inside area ( = n/4 for circular caissons) Pi = inside perimeter (= n: D; for circular caissons) The perimeter shear v should be less than the allowable shear. There should be overall stability of the caisson against buoyancy. For example, for circular caissons, Total upward force, F.. _ (n:/4JY")Hy,,, .. .(26.12) Total downward force, Fd - We + W, + Q.. ... (26.13) where We :: weight of empty caisson, W, ::E weight of seal, Q, '" skin friction. If Fd is greater than F... the caisson is snfe against buoyancy. However, if FJ is smaller than F .. , the thickness of the seal should be ~ to add weight. Cutting Edge. The cutting edge is provided at the base of the open caisson to facilitate penetration. It alsO protects the walls of the caisson against impact and obstacles eocounlered in its path during penetration .. The inside bevel is generally made 2 vertical 10 one horizontal. The cutting edges are usually made of anglcs and plates of structural steel or reinforced rona-ete and steel As the sharp edges are easily damaged, the blunt edges are commonly used. However, the angle of the edge should not be greater than 35"(Fig. 26.5).

DRIll..ED PIERS AND CAISSONS

" W ~

.

'. :0\" .

'

STRAP

ANCHORS

. 35·

I

(a) SHARP EDGE

713

'.,• . •..• '.: .. :

w······ .'

.#0

STRAP ....

NCHORS

.. • .- ..

<.

Cb) BLUNT EDGE

Fig.

u.s.

Th avoid tearing off the cutung edge or any lower parts of the caisson, the caisson concrete must be anchoroo or ued to the cutting edge. The lower portion of the cutting edge is provided with a 12 mm thick steel plate anchored to the concrete by means of steel straps. .

26.6. CONSTRUCTION OF OPEN CAISSONS The sinking of an open caisson is generally done by penc"trating it in the dry or from a dewatcred construction area or from an artificial island. An artificial island of sand is made for the pu~ of raising the ground surface above the water level. Thus a DREDGE WELLS dry area is obtained for sinking the caisson. The size of the sand island should be sufficient to provide working area around the caisson. (Fig; 26.6). For the construaion of a sand island, a FILL wooven wiUow mattress is first sunk to the river bottom to provide protection against scour. A timber staging is then constructed around the periphery of the intended island. Sheet piles are driven to enclose the island area. The mattress is .. ,.:' cut aloog the inside face of the shell fonnoo by

.: '.' SAND '

. r~-"---~l ,

~~t s~~

:d

th:

:~e :~~~; ~;oov~~

I

'

:

L_'-=-=--C-O:~PLETED

required level. In case it is not pa;sible to sink the caisson .'" CAISSON in dry, it is constructed in stipways or barges Fig. 26.6. and towed to its final position by floating. False bottoms are provided for this purpose. Guide piles are generally required for sinking the first few tifts of caisson. Sinking is done through open water and then penetrating it into the soiL The caisson is sunk by its own weight when the ' soil is excavated from the dredging well. As sinking progresses, additional lifts of caisson steining are installed. When a hard material is encountered, under-water blasting may be necessary, The excavauon is done by dredging with grab buckets, lbc soil near the rutting edge is removed by hand if it does not flow into the excavauon. 1be sinking operation is, of course, stopped during tbe period the concrete for the tin is cast and cured. To facilitate sinking, the exterior surface is apptied with a film of grease. Alternatively, water jets are used. When the caisson reaches the final depth, its bottom is plugged by a concrete seal. TIle concreting for the seal is done by (remie. After the concrete has m~tured, the water in the caisson is pumped out. The top of the cmcrete seal is cleaned and more COIlcrete is placed over the seal The caisson should be kept in the vertical position during the entire proctSS of sinking, However, it is extremely dilIirult to sink the caisson perfectly straight and true to its position. Corrective measure are adopted when it becomes inclined (see chapler 27).


714

26.7. PNEUl\.tATIC CAISSONS If the soil enclosed in an open caisson cannOL be excavated s..'1tisfactorily through its shaft during sinking operation, a pneumatic caisson is required. This condition occurs when the soil nows into the open caisson faster than it can be removed. A pneumatic ca isson is also used when there is a great influx of water or where difficult obstructions arc anticipated during sinking. Pneumatic caissons are suitable in soft, running soils which cannot be excavated in dry or where there is a greater danger of scour and erosion. A pneumatic caisson is a rigid. inverted box with its bottom open (Fig. AIR RELEASE--. AIR LOCK 26.7). A working chamber is provided _BLOW OUT at its bottom to keep the caisson free of COMPRESSEDPIPE water and mud by of AIR

use

compressed

...•.

';" , air. 1he design of a pneumatic caisson is similar to that of an open caisson in many respects. The ultimate load c.1rrying capacity, the design of walls. concrete seal and cutting edge are similar to tilat of open caissons. liowever, the following differences should be clearly noted. (1) Working chamber. The working chamber is made of mild steel. It is about 3 m high. It consists of a strong roof at its top. 'Ihe chamber is absolutely air tight. The air in the Fig. U.7. Pneumatic Caisson. chamber is kepi at a specified pressure to prevent entry of water and soil into it. The walls of the chamber should be thick and leak proof. 10 keep the fria.ional resistance low, tbe outside surfaces of the walls are made smooth. A cutting edge is provided 3t the bottom to facilitate the penetr'dlion of the caisson. The air pressure must be sufficient to balance the full hydrostatic pressure due to water outside. However, there is a maximum limit to air pressure. Working under a pressure of greater thun 400 kN/m2 is beyond the endurance limit of human beings. Therefore., the maxlmum depth of water through which a pneumatic caisson ~IO be sunk successfully is about 40 m. Working under a pressure greater th..1n 400 kN/m 2 may cause a special type of sickness. called caisson sickness. (2) Air Shaft. An air shaft is a vertical passage which connects the working chamber with an air lock aI the top. It provides an access to the working chamber for workmen. It is also used for the transport of the excavated materials to the ground surface. In larger caissons, two separate air Sh..1fts are provided, one for passage of workmen and one for transport of the materials. The shans arc made of steel tubes. The jOints of the tubes are provided with rubber gaskets to make them leak proof. E.'1ch shaft is provided with its own air lock at its top. As the caisson sinks, the air Sh.1ft is extended to keep the air lock always above water level. Durir-g this period, the working chamber is closed by a gate plate at the lower end of the shaft. (3) Air l...ock. An air lock is a steel chamber provided at the upper end of the air shaft above water level. The purpose of providing an air lock is to permit the workmen and materials to go in or to come out of the caisson without releasing the air pressure in the cnissoo. The steel chnmbcr of the aiilock is provided with two airtight doors, one of which opens to the shaft and Ute other opens to the outside atmosphere. When a man enters the airlock through the outside door, the pressure in the chamber is kept equal to the atmospheric pressure. The outside door is closed and the pressure in the chamber is gradually r'discd till it becomes equal to th..1t in the air shaft and the. working chamber. The door to Ihe shan is then opened and the man descends to the working chamber by a ladder provided inside the air shaft. The· procedure is reversed when a man comes out from the caisson. However, the decompression

DRIlLED PIERS AND CAISSONS

715

is done much more slowly to prevent caisson dise.1SC. A period of about 30 minutes is necessary for deromprcssion from a p~ re of 300 kN/m2 to atmospheric pressure. To prevent the air in the working chamber from beroming stale, fresh air is circulated into the worldng chamber by opening a valve in the air lock. The workers should not be kept inside the working chamber for more than two bours at a stretch. (4) Miscellaneous equipmenL MisceUaneous equipment such as motors., compressors. and pressure pumps are usually located on the shore. Pressure to the working chamber is applied through compressed air pipe. In order to cope with an emergency, al least one stand-by unit consisting of all equipment must be provided.

26.8. CONSTRUCrION OF I'NEUMAllC CAISSONS Uke open caissons., pneumatic caissons may be conslnlctcd at the site or floated and lowered from barges. Artificial sand islands may also be used. The rutting edge of the caisson is carefully positioned. Compressed air is introduced in the working chamber to expel water. Afier the working chamber has been dewatered, workmen descend through the air lock into the working chamber. 1lJe material is excavated by hand tools in dry. As the excavation progresses. the caisson gradually sinks. Concreling of the caisson is then done. The air pressure in the caisson is increased to equalise the increase in the head of water as the caisson goes down. The excavated material is. removed by buckets through the air shaft. In granular SOils, the excavated material can be removed by the blowout method through the blowo()llt pipe. When the valve in the blow pipe in opened, the granular material is blown out by high air pressure inside the working chamber. After the caisson has attained its design deplh, the working chamber is filled with ooncrcte. Precautions must be taken to ensure full contact between the ooncrele fill and the underside of the roof of the working chamber. The fresh concrete is [itSt lowered through the air shaft and a slob about 0.6 m thick is fonned on the bottom and well packed under the rutting edge. The air pressure in the chamber is kept constant till the concrete has hardened. A stiff mix of concrete is then packed into the working chamber up to the roof level. Any space left between the roof and the concrete surface is filled with cement grout. There should not be any empty space left in the chamber, as it would lead to selllement when the caisson is subjected to superimposed load. After concreling of the working chamber is completed, the shaft lUbeS are dismanlled. The shaft itself is filled up with a lean concrete. 26.9. ADVANTAGES AND DISADVANTAGES OF PNEUMATIC CAISSONS Pneumatic caissons have the following advantages and disadvantages, as compared with open caissons. Advantages (1) As there is an access to the bottom of the caisson, obstructions can be easily removed. (2) The soil can be inspected and the soil samples can be taken, if required. (3) Soil bearing capaCity can be dctennined by oonducting in-situ tests in the working chamber. (4) Excavation and pouring of concrete is done in the dry. (5) As the position of the ground water table remains unchanged, there is no flow of soil into the excavated area. (6) There is no settlement of the adjoining structures as the water table is not lowered. Dl..advantages (1) Pneumatic caissons are highly expensive. Pneumatic caissons should be used only when open caissons are not feasible. (2) The penetration deplh below water table is limited to 30 to 40 m. (3) There is a lot of inconvenience caused to the workmen while working under high pressure. The workers may develop caisson disease. (4) In pneumatic caissons, a large amount of manual work is required which inaeascs the cost. (5) Extreme care is required for the proper working of the system and to maintain the required air pressure. Any slackness may lead to an accident.


716

Z6.IO. FLOATING CAISSONS

floating caissons arc large, hollow

,

boxes with top open but bottom closed. These arc Dooted to the place where

these are to be installed. The caissons are then sunk by filling them with ballast, such ~ sand, dry concrete. gravel. Unlike the open and pnewnatic

.

CONCRETE

CAP

'

' ..

~' . . : SAND' AND .. • .:

caissons, a floating caisson does not penetrate the soil. It simply rests on a levelled surface (Fig. 26.8). The load

GR~VEl' "

.'.

carrying capacity depends solely on the base resistance, as there is no side

.

...,...

~

' .' -, .

". ~ "

~After the caissons has been sunk to i~:nal ~ition. i\ iSAcomplet~lY nl1~

:1 :: i:: I!;V~O' rece~vecres~ru~~~

..

. • . .t

. .• : .. ;,'.:' ~..: •

•

loads. To prevent soour underneath, rip rap ls placed around its base,

0 1.

.

.

Jj._..

r

'. ,'

.•. ,. .•. . ... " • .

R1PRMJ

&

Ag. 26.8. Aoaling Caissons after sinking.

floating caissons are usually oonsttuCtcd of reinforced cement concrete or steel. The plan of the caisson may be circular, square. rectangular, or elliptical. It usually contains a number of cells formed by diaphragm walls. If the caisson is to be floated in rough waters. it is designed as a ship and suitable internal strutting is provided. Z6.U. STABILITY OF FLOATING CAISSONS The caisson must be stable during flotation. According to Archimedes' principle, when a body is immersed in water, it is buoyed up by a force equal to weight of the water displaced. For equilibrium, W - U - 0 ",(26.14)

=

=

where W weight of the caisson, U buoyant fOrce. The weight W aas through the centre of gravity (0) of the body. The buoyant force U ads through. the centre of gravity of the displaced water, known as the centre 0/ buoyancy (8) (Fig. 26.9). If the caisson is lilted through a small angle e, the centre of gravity (0) remains at the same location with respect to tbe caissoo itself, but the centre of buoyancy B changes its position as the position of dbp1aced volume is changed. Some portion of the caisson wl1ich was not submerged during the verHcal position beoomes submerged in the inclined position. The point of intersection of the vertical line passing (a) VERTICAL POSITION (b) TILTED POSITION through B and the centre line of the . caisson is knO'No as metacentre (M). Fig. 26.9. The caisson would be stable if the metacen!re M is above G, i.e., the metacentric height MG is positive. The metacentric height can be detennined analyticaUy (see any text 00 Fluid Mechanics) as given belO'N. The distance BM between points Band M is given by BM _ /IV

... (26.15)

717

DRIlLED PIERS AND CAISSONS

where I ;;: second moment of area of the plan of the caisson The metacentric height is computed a<>

81.

water surface, V;;: volume of water displaced.

___ .m

..

~~

The plus sign in &j. 26.16 is used when G is below B. If the caisson is unstable, it should be either redesigned or ballast should be used to make it stable. lbe free board when 1l0o.'lting should be ancast 1 m. 16.12. ADVANTAGES AND DISADVANTAGES OF FLOAllNG CAISSONS Floating caissons have the following advantages and disadvantages as compared with open caissons.

Advantages. (1) The iffitallation of the Dooting caisson is quick and convenient. (2) As the floating cai~ are prefabricated or preciSI., the quality of mnstruCiion is good. (3) Floating caissons are less expensive than open Clissons. (4) Floating caissons can be transported by floating at a relatively low cost. Disadvantages. (1) The load<:arrying capacity of a floating caisson is much lower than that of an equivalent open caisson. (2) The foundation bed has to be levelled before installation. (3) 'The ba<>e of the floating caisson is to be protected against scour action. ILLUSTRATIVE EXAMI'LES

J

'I

lIIustrative Example 26.1. A straight-shaft drilled pier, 1.0 m in dinmeter, is constructed in a deposit consisting of loose sand overlying dense sand (Fig. £-26.1). Determine the allowable load (F.!l: := 3.). Solution. Let us take critical depth Dr _ 10 B - 10 m Q" _ (qN.,)A, + (Ka"tanb)As where q = effective vertical pressure at the base .. 8)( 17 +2)( 21 _ 178kN/m 2 From Eq. 26.1,

FromFig.26.2,N'l- 140 for, _ 40°. Therefore,

Q" .. 178 )( 140 )( ,v4 )( (1.0)2 .. 0.5 )( + 0.4 )(

1t )(

T

LOOSE SANO T.mG:O·5&

1\:0·50 ~ =30'

I

'6::111rHf",3

DENSE SAN) T.iln6=O·14 K=O·l.O

2m

,=4if 1\=21 kHfm 3

Fig. E-26.1.

(1t )(

1.0) )( 0.58

(i )(

136 )( 8.0 )

1.0 )( 0.84 [ t(l36 + 178) )( 2.0 + 178 )( 2.0]

Cloy loyer I Cu=J'S~Nlm2 0( ,,0.)0 l' .. 17 kH/mJ

Cloy loyrr \I Cu"IOO kNlm2 ,,<,,0.30 "' =18 ~Nlin3

Fig. E-26.2.


718

Q. _ 20740.00 kN Q. _ 20~40 _ 6913 kN Illustrative Example 26.2. Determine the allowable load for the drilled pier constructed in a clayey deposit, shown in Fig. £-26.2. Take. F.S. = 3.0. Solution. DIBI .. 1212 .. 6 N~

From Table 26.1,

.. 9.0.

Qu .. cN~ Ap • neA,

From Eq. 26.2,

.. 100 x 9.0 x (nI4) x (2.0)2 • (0.3 x 35) (n: x 1.0 x 8.0)

+ (0.3 x 100) (It x 1.0 x 3.0) .. 3374 kN

Q. _ 33;4 _ 1125 kN

From Eq. 26.3,

(If Eq. 26.4 is used, the allowable load = J488.4 kN) Illustrative Example 26.3. Determine tlle outside diameter of an open caisson to be sunk through 40 m of sand and water to bed rock if the allowable bearing capacity is 2000 kN/m 2• I1Ie caissol! receives a load of 50 MN from the superstruclure. The man/Ie friction is 30 kN/m 2• Test the feasibility of sinking. Also calculate the thickness of the sea/. Solution. (sec Fig. E-2.3)

" Concrete s-eol

Fig. E·26.J.

From equilibrium in the vertical direction (when imtallcd), Load from superstructure + self weight - frictional resistance -

uplift force -

base reaction

= 0.0

or 50000 + (n:/4)V;(4O) x 24 - (ruJ.,) x 40 x 30 - (;t/4)n; x 40 x 10 - (;t/4)V; x 2(0) .. 0

n; + 3.33Do -

44.23 .. 0

or D., - 5.19m

Let us adopt outside diameter as 6.0 m Feasibility of sinking From Eq. 26.7 (a), Taking

f ..

30 kN/m 2

(rt/4)

and

36 -

(D; - £if)

x 40 x 24 .. (rt Do) x 40 x

Do" 6 m, we have

df ..

30

D; _2.4m

Thickness of wall .. 6.0 ; 2.4 .. 1.8 m

f

DRILLI!D PIERS AND CAISSONS

719

11lickness oC seul

from Eq. 26.8,

1=

0.59D,..r;;;;;:

= 0.59 x 2.4 "2000/3500

= I.07m say 1.I0m Illustrative Example 26.4. Check the stability of a floating caisson 10.5 m high and having a rectangular base 20 m x 9 m. 111e weight of the caisson' is 9 MN and its centre of gravity is 4.0 m above the base. If the caisson is unstable, how WQuld you make it stahle? Take unit weight of water as 10.25 leN/mI. (b) What is the maximwlI pressure on the soil when the caisson has been fully insta/fed ? The base is at a deplll of 9 m be/ow the water level. The total .. ~ight is 50 MN, which acts' at an eccentricity of 0.1 m. Solution. (sec fig. E-26.4) 9m

I 1

IG oM

!

8

I

d

1

L - - - - . . . JA-

Immersed Volume,

Fig. £.26.4.

v .. ~ ..

878.05m l

Depth of immersion during Oo:.lting. d .. 10.25

lleight of centre of buoyancy B

~lbove

~

x 9 - 4.88m

base.

All .. 4.88/2 .. 2.44 m As the distance AG is 4 m, the distance BG is given by

Prom Eg. 26.15,

DG .. 4.0 - 2.44 .. 156m BM .. /IV

Distance AM .. AD + BM .. 2.44 + 1.38 .. 3.82 m As the metacentrc (M) is below the centre of gravity (G). the metacentric height MG is negalive and the caisson is unstable. 'Ille caisson can be made stable by filling it with ball.'1st. say sand ('r = 22 kN/m). Let the thickness of the sand L1yer be 0.5 m. The height of [he new centre of gravity (G') Ilbovc the base is given by


'''' AG' _

1lle new depth of immersion,

AB'

= height

4 + (0.5)( 9 x 20) x 22 x 0.25 .. 3.32m 9000.0.5)(9)(20)(22

00(0)(

d .. 9000 + (0.5)(9)( 20 x 22) .. 5.95m 10.25 )( 9 )( 20

of new centre of buoyancy .. 5.9512 .. 2.975 m

:0 5~9fJ

From Eq. 26.15,

B'M' .. (11;;»()(9

Metaccntric height,

M' G' .. (AB' + B'M') - AG'

.. 1.134 m

- (2.975 + 1.134) - 3.32 - 0.789 As the metacenlric height is positive (i.e. M' is above G'), the caisson is stable. (b) When the caisson has bccn fully sunk, with its base at a depth of 9 m below the water upward force is given by Ff' .. (9 x 20) )( 9 )( 10.25 .. 16605 kN

lcve~

the

As the upward force is more than the downward force of 9OClO !tN, ballast is required. Let h be the thickness of the ballast. lbercfore, (9)( 20 )( h) )( 22 + 9000 .. 16605 or h .. 1.92 m The maximum and minimum pressures when the caisson has been fully installed and subjected to design loads are given by qawc .. ;; ( 1 +

¥) . io~9

(1 + 6

)

~.d·l = 296.3 kN/m 1 .

-

... ~ (1 _ ~) _ 50000 (1 _ 6 )( 0.1 ) .. 2593 IeNI 2 qmon A b 20)(9 9 m Uplift pressure Net maximum pressure Net minimum pressure

.. 10.25 )( 9 .. 92.3 IeN/m 2 .. 296.3 - 923 _ 204 kN/m 2 .. 259.3 - 92.3 .. 167 kN/m 2

PROBLEMS A. Numericul 26.1.

~;~;I;s~a~ld::\e~: I~~ mded~hn:~e~: ;~~~e~~n~:: :Wa1=I~ (~ ~~~= =3~;' tan 6 = {AM. 16.5 MN]

26.2. A drilled pier of 1 m diameter has a tOial depth of 15 m. The diameter of the bell is 2 m and its height is 1 m. 2 J IC c.. = 80 kN/m , y = 20 kN/m and a = 0.3, detennine the allowable load (F.s. '" 3.0). {Ans. 850 WI 26.3. An open caisson, 19 m deep, has external and internal diameters of 8 m and 6 m, respeaively. If the water level

:~i~::C=I~;'I~~e~~ I~~tl~il~~~~~ :m~:I~ ::~:~e~: o7:~ ~sso~u~:~~S2~::k~y:h: ;;':i~rYj 2

and allowable perimeter shear of 650 kN/m .

26.4.

(AIls.. 1.0 m]

~~~~~; t:da~~~:~~I:~:r~~:I~!rn: J:~ ~:.d=~~ r~~1 ~u~:!g::.nd with N = [~~or4;:a~~r

8. Descriptive Types 26.5. What is the basic difJerenre between a drilled pier and a caisson ? What are the (Xmdilions io which a drilled pier is more suitable than II caisson? 26.6. Describe various methcxls for the construction of drilled piers. 26.7. How would you estimate the load carrying capacity of drilled pier in (a) sand, (b) clay?

72'

DRlLLFD PIERS AND CA ISSONS

26.8. How would you estimate the load carrymg cap:tCily of an open caisson ./ 26.9. Dmw [he ~ketch of an open caisson. How the various componcnts ,m: designed '! 26.10. Describe the various I.:ornponents o f a pneumatic caisson with the help of a sketch. 26.11. What arc the ;K.lvanlagcs and disadvantages of pneumatic caissons over open caissons? 26.12. How would you cha:k the stability of (loating caisson during Hotation ?

C. Multiple Choice Questions I. Pnt.'Umatic cnissons are used where the MJil !low into the cxeavmoo arca of an opt..'T\ caisson is (a) fastcr than II can be removed (b) slower thun it cun be removed (e) negligible (d) zero 2 . The Illuxilllum depth of fI pneul1\;llic caisson is usually limited [0 (a) 10 III (b) 20 m (c)80m (d) 40 m 3. The ad hesion factor for drilled piers on day is usually taken as ~) I n

(~QW

(c) 0.60 (d) 0.40 4. The Iloating caissons gcnernlly (a) hnlle gre;lter lnad-currying c;lpacity [han open caissons (/I) huve gre,lter dcpth below the ground surface th an open (e) are less expensivc then opcn C;lissons (d) hUlle poorer quality of construction than open caissons.

caisson~.

[Ans. I. «(I). 2. (ll", 3. (d'), 4. (dl

D. Objective Type Questions Write whether the following stalel1\~nlS are true or r;llse : (a) The d rilled piers can be of diameter less th an 0.5 m. (b) A pneummic caisson is open :u the top. te) The value of Nq for drilled piers is generall y smaller than that lor piles. (d) The angle thai the lOclincd walls of the belled pier make wilh the hori70nt
27 Well Foundations 27.1. INTRODUCTION Well foundation have their origin in India. Well foundations have been used in India for hundreds of years for providing deep foundations below the water level for monuments, bridges and aqueducts. Thj Mahal at Agra (India) has got well foundations. A well foundation is similar to an open caisson discussed in chapter 26. Well foundations can be constructed on the dry bed or after making a sand island. At locations where the depth of water is greater than 5 m to 6 m and the velocity of water is high. wells can be fabricated on the river bank and (hen floated to the final position and grounded. Great care is to be exercised while groundif'~ a well to ensure that its position is correct. Once the well bas touched the bed, sand bags are deposited around it to WEll CAP prevent scour. The well may sink into the river bed by 50 to 60 em under its own weight. Further sinking operation is similar 10 the sinking of wells on dry bed. The well is sunk into the ground to the desired level by c).cavating through the dredge holes. Fig. 27.1. shows the section of a welL A strong Culling edge is provided to facilitate sinking. The tapered portion of the well above the cut Ling edge is known as well curb. The walls of the well are known as steining. Steining is made of brick masonry, stone masonry. plain or reinforced concrete. As the steining later becomes an integral part of the structure, it should be properly designed for the imposed loods. Further, it should be heavy enough to overcome frictional resistance during sinking. After the well has been sunk to the final position, the bottom plug is fonned by concreting. The bottom plug serves as the base of the well. The well is filled with sand partly or completely. At the top of the well, a top plug is fonned by concreting. A R.c.c. well alp is provided at the top to transmit Ihe load of the superstructure (pier or abutment) 10 tl:!:e well.

TOP PLUG

CURB

Fig. 27.1. Well Found/ltion.

WEll. FOUNDATIONS

723

27.2. DIFFERENT SnAPES OF WELLS Different sbapes of wells that arc commonly used are shown in Fig. 27.2. (1) The most oommooly used shape is circular [Fig. 27.2 (a)], as it has high structural strength and is convenient in sinking. The chances of tilting are also minimum in this shape. The shape is quite suitable for piers of the single-line railway bridges and the double· lane road bridges. However, when the piers are excessively long, the circular shape becomes uneconomical. The maximum diameter of circular wells is generally limited to 9 m. (2) Double-D wells [Fig. 27.2 (b)] are generally used for the piers and abutments of (a)CIRCULAR (b) DOUBLE - 0 bridges which are too long 10 be

o CD

accommodated 00 a circular well of 9 m diameter. TIle wells of this shape can also be

sunk easily. However, considerable bending moments arc caused in the sicining due to the diffcrenre in pressure between the outside and

00 00

(c) DOUBLE OCT.AGONAL

(d) TWIN

CIRCULAR

C

the inside of the well. Further, the square comers at the partition wall offer greater resistance to sinking. (e) RECTANGULAR (I) DOUBLE RECTANGULAR (3) Double-OCUlgonal wells [Fig. 27.2 (c)] Fig. 27.2. DiCereol Shapes ol Wells. are beller than the double.D wells in many respeas. The square comers are eliminated and bending stresses are considerably reduced. However. they offer greater resistance than doublc-D wells against sinking 00 account of increased surface area. Moreover, the construction is more difficult. (4) 1Win-circular wells [Fig. 27.2 (d)] arc two independent wells placed very close to each other and having a common well cap. The wells arc sunk simultallCo~ly. These wells are suitable where the length of the pier is considerable, which cannot be accommodated on a double.D or double-octagonal well. Twin circular wells are advantageous when the depth of sinking is small and the bearing capacity of the soil is high. The disadvantage of twin circular wells is Ihat there is a possibility of the relative settlement of the two wells even if a heavy R.C.C top cap is provided unless the wells are founded on an incompressible soil. (5) Reclangular wells [Fig. 27.2 (e)] are generally used for bridge foundation having depths upto 7-8 m. For large foundations, double-rectangular wells [Fig. 27.2 (j)] are used. For piers and abutments of very large size, rectangular wells with multiple dredge holes (not shown in the figure) are used. Bending stresses in the steining are very high in rectangular wells.

E

27.3. GRIP LENGm The well should be sunk below the maximum scour level to sufficient depth such that the resistance from the sides is able to resist the lateral forces acting on the well. The depth of the bottom of the well below the maximum scour level is known as grip length. The depth of foundation should be chosen oonsidering the grip length and the bearing capacity of the soil. The maximum and minimum base pressures under worst loading conditions should be within the pennissible valucs. The depth of scour in a stream can be ascertained through aaual soundings at or near the site proposed for the bridge during or immediately after a flood. The maximum scour would be greater than the measured scour because the design discharge is greater than the flood discharge for which the soundings have been made. Moreover, there would be an increase in the velocity of water due to the obstruction of flow caused by construction of the bridge. An exira allowance should also be made in the measured scour due to the proximity of piers. In case actual soundings cannot be made, tbe normal depth of scour in alluvial soils may be calculated by Utcey's formula.

OOll MECHANICS AND FOUNDATION ENGINEERING

724

d • 0.473 ((I1f)" ... (27.1) where d :: normal scow- depth, measured below high Oood level (m), Q :: design discharge (m)/scc), f '" silt factor. lbe sill factor may be calculated from the equation

f. 1.76VJ; where dm = mean size of particle (mm) The regime width Oln be oomputed as W. 4.75 VQ

... (27.2)

.. (273)

If lhe actual water way (L) is less than the regime widlh, Ihe actual scour depth (d') is given by The maximum scour depth

(~),

from Thble 27.1.

1lle grip length for wells on railway bridges is generally taken as 50% of tbe maximum scour depth. Fa road bridges, a grip length of 30% of the maximum scour depth is generally provided. Thus the base of the well is usually taken at a depth of 2.67 d' below the high flood level. -

Table 27.1. MII)C\mum Scour Depth S. No. 1.

2 3.

.. 5. 6.

Riw:r Sectioll

Straight Reach Moderate Bend Severe Bend Right-angled Bend or lit nose of Pier Upstream Nose of Guide Banks

Severe Swirls

Maximum Scour (d..,.)

1.27 d' 1.50d' 1.75 d' 2.00 d' 2.75 tI 2.50 d'

According to IS : 3955 - 1967, the depth of foundation should 001 be less than 1.33 times the maximum scour depth. The depth of the base of the well below the maximum scour level is kept not less than 2 m for piers and abutments with arches and 1.2 m for piers and abutments supporting other Iypes of structures.

27.4. FORCES ACflNG ON THE WELL FOUNDATION The following forces should be considered in the design of a well foundation. (1) Dead loods. 1be dead loads carried by the weU include the weight of lhe superstructure aod tbe self weight. (2) Un loads. The design live loads for railway bridges fire taken according to Indian Railway Bridges Rules. For rood bridges, the live loads as specified by the Indian Road Congress Standard SpecificaLions aM Code of Practice for Road Bridges--Sect. II should be used. (3) Impact loads. Impact effect due 10 live load is considered only in the design of pier cap and the bri~gc seat on the abutment. For all other members of the well, the effed. of impact is ignored. (4) Wind loads. Wind loads on the live load,. superstructure and the pan of the substructure located above the water level are calculated according to the provisions of IS : 875, 1be wind load acts on tile exposed area in elevation and thus it acts laterally on the bridge. (5) Wilter pressure. Water pressure due to water current acts on the part of substructure which lies between the water level and the maximum scour level. On piers parallel 10 the direction of the water, the intensity of water pressure is given by

... (27.5) p • KV' where p = intensity of pressure (kN/m1, K = a constant, depending upon the shape of the well (= max. of

G

WEU- FOl!.NDAll0NS

I)

0.788 for square ended pier to a minimum of 0.237 for piers with cut and ease water), V = velocity of current (m/,.,;). It is assumed that y1 is maximum at the free surface of water and zero at the deepest scour level. The variation is assumed to be linear. The surface velocity is taken "f2 limes the average velocity. Even when the flow is pamllel to the pier, a transverse force equal to 20% of thaI acting parclUel to the pier is taken to allow for oblique flow. If the current makes an angle with the axis of the pier, the pressure along the axis of the pier and transverse to it are given by

e

PQ ::: pressure along axis ::: p cas2 (20 0 P,

= transverse pressure == p sia2 (20

0

:t:

±

9)

... (27.6)

9)

... (27.7)

(6) Longitudinal Forces. Longitudinal forces occur due to tractive and braking forces. These forces depend upOn the type of vehicles and bearings. These forces are transmitted to substruaure mainly through flXed bearings and through friction in movable bearings. According 10 IRe code, a longitudinal force: of JlW is taken on the free bearing :lOd the balance on the fixed bearing, where W is the total reaction and I-' is the coefficient of friction. (7) Centrifugal Forre. A cc~trifugal force is transmitted through bearings if the bridge is curved in plan. (8) Buoyant Forces. Buoyancy reduces the effective weight of the well. In masonry or concrete steining. 15% buoyancy is considered to acoounl for the porousness. When the well is founded 00 coarse sand or shingle, full buoyancy equal to the weight of an equivalent volume of displaced water shOuld be considered. For semi-pervioos foundatiOflS, it is suitably reduced. (9) Earth Pressure. The earth pressure is calculated acwrding to Rankine's theory or Coulomb's theory. For the stability of foundations below the soour level. the passive earth pressure of the soil i" oonsidered.

cons-:=:~~~ f~~tr':c~fftbo~ 1~:V!:. placed behind the abutment. an equivalent height of surcbarge is (10) Thmperature Stresses. Longitudinal forces are induced due 10 temperature changes. The movcmcnts due to temperature changes are partiaUy restrained in girder bridges because of friction at the moveable end. (U) Seismic (orces. For the wells constructed in the seismic zone, seismic forces should be considered. The fo~ act on all components of the structure. The force is usually specified as aW, where W is the weight of the component and a is the seismic coefficient. The value of a depends upon the seismic zone (IS : 1893). Its value is generally taken between 0.01 and 0.08. The seismic force acts through the centre of gravity of the component. Il may act in any direaion, but it is assumed to act in one direction only at a time. The seismic forces are considered separately along the axis of the pier and traosverse to it, Resultant Forces. The magnitude, direction and the point of application of all the above forces are found under the worst ~ible combinations. '£be resultant can be replaced by an 'equivalent vertical force W, and twO horizontal forces F and Q. as shown in Fig. 27.3. The horizontal force Q acting in the transverse direction gives more critical condition for the lateral stability of the welL. 17.5. TERZAGHI'S ANALYSIS Terzagbi's solution for free rigid bulkheads can be used for an

ElEVATICN

1

Fig. 27.3. Focas 0111 pier.

SlOE VIEW


approximate analysis of the well foundation. When a rigid bulkhead embedded in sand moves parallel 10 its original position, the sand on its fronl is lransfonncd into passive state whereas that on its re.1r, into active state. Assuming that both the active and passive resistances are fully mobilised, the net pressure at any depth z below the ground surface is given by p - ,z(K, - K.J ... (27.8) A free rigid bulkhead depends for its stability solely 00 the lateral resistance. Let qm;u' be the horizontal force per unit length acting on the bulkhead of total height 111 (Fig. 27.4). The pressure distribution 00 both sides of the bulkhead at the instant of failure may be represented as shown in the figure. The bulkhead rotates about the point 0 above the base B. As the soil around the well is generally Submerged, the submerged unit weight l' is used.

Fig. 27.4. Te.rl.aghi·s Analysis.

The applied force per unit length is given by

qm.".' - AreaABC - AreaFEC -

~"

d(K, - K.J -

~(2,· D)(K,

- Ko)D,

where DI is the height of point F above the bottom.

qmN<' Taking moments about the base B, q.M' fl, -

ty' k"

... (27.9)

D (Kp - K{J) (D - 2D I )

d (K, - Ko) DI3 -

k

(2,. D) (K, - K.J dj/3

where HI - H + D. Substituting the value of qrnM.' from Eq. 27.9,

k,' D(K,-Ko)(D-2D,)H, - k"(K,-K.JD'/3-k(2,· D)(K,-Ko)dj/3 (D - 2D,)H, - dl3 - 2 dj/3

.df -

3D,H I + (1.5Dll l

-

O.5d) - 0

Solving for D" we have 2 D, _ 3 HI ± V911~ - 2 D (3/i 1 - D) .. (27.10) By substituting the above value of DI in Eq. 27.9, the value of q..-' can be computed. lbe coeffICient Kp and Kd can be obtained from Rankine's theory. In the above simplified analysis, the moments due to side friction and the base reaclion are neglected. The error is on the safe side.

WEll. FOUNDATIONS

727

Heavy Wells. A beavy well embedded in cohesionless soil rotates about its base (Fig. 27.5). The force Per unit length, qmu'. can be obtained by ~k:i.ng moments about the base. qml.lJl.' HI - &"f'(Kp

or

qmn' -

~ y' (Kp -

-

IT'I~ 11

~

* .

KJd x

Ko)

(27.11)

ElTect of Sun:harge. The effect of surcharge due to tbe weight of soil above the scour line can be taken into account in the analysis. The soil below the maximum scour level is subjeded to a surcharge of beight Z of the unsroured soil (Fig. 27.6). The height Z may be taken equal to one·half the nonnal depth of scour, in case it is not possible to ascertain it by actual measurement. The pressure distribution is shown in the figure. The maximum pressure at the base is equal to "f' (Kp - KJ (D + Z). In this case, qm"",' is given by

qm.ax

Fig. Z7.5. Heavy Welt.

1 '(K K)Ii'(D+Z) -6''' p - " --H1

... (27.12)

-

,- r-----

UNseOUREO BED

" Z

J

R

, , \

w

LEVEL

'

\X~~~g~aRE \\

DISTRIBUTION

\\

\\

"

'\l __ ______

I~1D+Z)(Kp .. ~ I Fig. 27.6. Effect of

Su ~h.arge.

Allowable '1h.nsverse Load. The total resisting force for the well is equal to the product of qmrot.' and the length L of the well. The allowable load (Q,,) is equal to the resisting force divided by a suitable factor of safety. Thus

Qo _ qm;/

... (27.13)

The factor of safety (FS) should not be less than 2. The length L of the well is generally taken equal to the diameter of the well. Balwant Roo and Muthuswamy (1963) recommend that the allowable load as obtained by Eq. 27.13 should be multiplied by a suitable shape fador. The shape factor is unity in the case of rectangular wells and equal to X/4 for circular wells. However, for circular wells of the diameter larger than 4.5 m, the shape factor is taken as unity.

nIl


Maximum 1'ressure at Base. If Q is the aaual applied transverse horizontal force and Q. is the allowable equivalent resisting force. the unbalanced force (Q - Q.J aaing at a height of H above the SCOuf level would produce an overturning moment M B at the base, given by

M8 - (Q - Qo)(/J + D) The maximum and the minimum pressure at the base are W M8 qmax - A; + Z;; ~md

W

qmin -

M8

A; - Z;;

where W ::: net vertical load on the well base, aRer making allowance for buoyancy and skin friaion. Ab =
i.: [ ~Y'(Kp or

x -

K.J x2L ] - Q

[Y'(K~~!t)L

r

..

(27.14)

Taking moments about point S, Mmu - Q (H + x) - (Pressure Force) x xl3 Taking pressure force equal to Q, M~- Q(H + x) - Q(x/3)

Maus. - QH + tQx

.. (27.15)

If M is the applied moment at the scour·level, Eq. 27.15 becomes

MmilX - M +

1

Qx

... [27.15 (a)J

If the well rests squarely on rock. or an unyielding stratum, no tilt about the base or rotation about a point above the base is possible. In such a case, the moment developed upto the base is transferred to the foundation bed.

27.G. BANERJEE AND GANGOPADIIYAY'S ANALYSIS Banerjee and Gangopadhyay (1960) derived equations for the lateral load capacity of wells, making the following assumptions: • (I) The well is founded in a sandy stratum. (2) The well is acted upon by a uni-direclionai horizontal force Q in a direction across the pier. (3) The resultant unit pressure on soil at any depth is in simple proportion 10 the horizontal displacement. (4) The ratio between the contact pressure and the corresponding displacement is independent of the magnitude of pressure. (5) The coefficient of vertical subgrade reaction has the same value for every poin! of the surface acted by the contact pressure.

WELL FOUNDATIONS

The method of analysis can be divided ioto following 6 steps. (,) Expression for Horizontal Soli Reaction. When a rigid well embedded in a sandy stratum moves parallel to its original position, the sand on the front face is transfonned into passive state and that on the rear face, into active state (Fig. 27.7). The net pressure at a depths z is given by (shown by the finn line). p, • l ' (Kp - K,j ... (27.16) Let P be the load per unit area of the vertical surface and p be the corresponding displacement. If PI is the displacement required to inaease the value of the net pressure from zero to PI' we have pip .. PI Ipi

Rigid well

Substituting lJ1e value of PI from Eq. 27.16,

!!.. P

yz(Kp - K,j p,

T

P" p"(%(Kp - Kp)

p, wbcre

P .. "'f?z m .. y(Kp - K..}/pl

... (27.17)

Net pressure

... (27.18)

The constant m is known as the coefficient of horizonlal soil reaction. It is an empirical coefficient which depends upon the nature of soil, the size and shape of the loaded area.

\ __

Fossi.... pressun! /

~I

\

P,

/

AChJO pressu

\ •

!

1. _ _ _ _ _ _ _ .J~

,

\

-----.1

f--KpYO~--r KoYO ~

(u) Equations of Equilibrium. The stability of the well . can be checked assuming either no plastic flow or partial FIg. 7:7.7. plastic flow. The following analysis is based on the assumption of no plastic Dow. For the other case, a reference mny be made to the original paper. Fig. 27.8 shows a well of length L and width B. acted upon by

,, \

\

Nolt' : Rototioo Is excggt'foted

(0)

'0)

Fig. 27.8.

".


a horizontal focce P per unit length at a height H above tbe scour tine. The well is embedded to a depth D below the scour line. When the well rotates about a point 0 located al a depth Dl below the soour line, the following equations can be written from statics.

P - PI - P, - ~R •.. (27.19) PH _ M} +M2 - Ml + ~RD+ ,,(P1-Pll x 812 ... (27.20) and W - ~ (PI + Pi) + R ... (27.21) where P :::: horizontal transverse {oree, PI :::: resultant force on the front face, P2 :: resultant passive force on the rear face, "PI:::: skin friction on the front face. 1.1. P2 :::: sIcin friction on the rear face, R = resultant vertical

soil reaction at the ~ase, "R = frictional resistance of the soil at the base, MJ == momcnt at the scour line due to Ph M2 = moment at the scour line due to P2. M3 = moment due to the vertical soil reaction at its base.

Let PI :::: horizontal displacement of lbe centre line of the well at the scour level, P2 :::: horizontal displacement of the centre line of the well at ).he base level, Pl :::: downward vertical displacement of tbe well at the toe,

p'} = upward vertical displacement of the well at the beel(Pl' ... J>3}. P. = uniform vertical displacement of the well due to resultant vertical force W.

From Fig. 27.8 (a),

~

~Dl

.. D

.. ::2

... (27.22)

(UI) Evaluation oC PI and MI' Let p be the horizontal displacement at depth z below the srour Une. The given by Eq. 27.17 p P Therefore, p

pressure

~

From Fig. 27.9,

is .. mpz .. (P 1/D 1)(D1 -z) .. mz (PI/D1)(D1 - z)

... (27.23)

D,

Fora: per unit length. PI"

I"

mz (P1/D1) (D) - z) dz

PI" mplm 6

Moment,

!ttl"

- f. D'm (PI/DI) (DI or

.. (27.24)

f" DI (pdz)z z) i' dz

MI - m pdDl/12)

... (27.25) p

Fig. 2:1.9.

Pig. 2:1.10.

WEll. FOUNDAllONS

731

(;..) Evaluation of P1 and Ml From Fig. 27.10,

P p, z-DI-n; (z - D,)

P-~Pl

From Eq. 27.17,

(z - D,)

P-mz~PI

Force per unit length, P2 - fD m PI z (z - D 1) dz D,

D,

- ';;,' (2D' - 3D,D' + Moment. M2 - fD D,

~DJ z (z ,

-D 1)

Z

Dl)

... (2726)

dz

or

... (2727)

(..) Evaluation of R and M J Let

K~

be the modulus of vertical subgrade reaction.

K, - pip wbere P = vertical deflection ( = P.) of soil corresponding to the vertical reaction p. Therefore. vertical reactioo.

R _ lfll P dz _ 2 o

In. K~

.. (27.28)

p.dx

0

R _ KIIB P4 ... (27.29) or The rotation of the well is resisted by a moment M3 acting at the base due to pressure developed on account of tbe downward deflection of the toe and the upward deflection of the heel. Fig. 27.11 shows the rotation of the base, with a maximum displacement of P3 at the ends. Let P be the deflection at a distance x

-"-"""-&I~ I~ ~ f

: _x--j

I--~

"'I'

t--l

fig. 27.11.

2p,

OfP-nX From Fig. 27.28,

p _ K" P _ 2;3 K"

X

Moment M3 _ 2 { ' pxdx _ 4P3 K" 1/2 :2-dx o B Jo


... (Z1.30)

or Substituting the value of PJ from Eq. 27.22,

M) ..

ij/2) K.Ii K.Ii p, (15;- PI-6-- ~

.•. (2731)

Evaluation or m PI The maximum soil pressure at depth z below the scour line is given by ... (2732) (PJ_. y'(1(" - K.) If no plastic is allowed in the soil the horizontal soil reaction p al any depth z must not exceed (PJIIWI' 1berefore,

flo..v

...(2733) The sand sl.arts flowing as soon as the slope of the pressure parabola at scour level becomes equal to the slope of the line whose abscissa represents the value of (PJmax (see Fig. 27.8 (b»). From Eq. 27.24, From Eq. 27.23,

Therefore,

6P,

mpi ..

... (2734)

P-tnl~(DI-Z) P .. z(D I

-

(6P,) _6P

z)

D

Iif

1

(D -z)z

Dl

6P, (D, - Zz)

dp

Hence,

"iii.

AIZ" 0,

"iii •

From Eqs. 2733 and 2735,

Dr

Dl

6P,D,

tip

6P1

--nr- • Df

6P,IIif· y(Kp

-

... (27.35)

K,,)

y(K, -K,,)Iif Pi" 2 6 Substituting the above value of 6P 1/D 1 in Eq. 27.34,

... (2736)

mpi .. y(Kp - Kif}

... (2737)

Substituting the above value of m PI in Eqs. 27.25. 27.26, 27.Z7 and 27.31, M, • Y(Kp - K.)(Dl/12)

•.. (2738)

y(K -K,,).J _,--' P2" ~ (2LT - 3D1Lr + UJ)

... (27.39)

M1

..

y(K, - K , , ) . --'. 12D, (3D - 4D J LT + D 1)

M • K.y(Kp - K,,) 3 m

(-.!L) 12DI

... (27.40) .•• (27.41)

(VI) Determination or Base Pressun:. The total vertical reaction is given by (see Fig. 27.12)

R-i(p'+Ph)B

... (27.42)

WElL FOUNDATIONS

Fig. 27.12-

where PI = pressure at the toe, Plo = pressure at the heel. Moment M3 can be expressed as

1

1(2 )

B B. M3-'2(PI-Ph)x"2x'23 M) _ (PI

;2

hh) BZ

... (27.43)

1be maximum pressure should not be more tban the allowable soil pressure. TIle minimum pressure should not be negative (tensile).

27.7. SIMPLIFIED ANALYSIS FOR HEAVY WELLS The analysis given in the preceding section is considerably simplified if the well is heavy. For such wens. the reaction R at base is very high. and. therefore, the sliding at the base does not take place (i.e. P2 = 0). The well however rotates at its base (Fig. 27.13). As

pz - 0, and D J

-

D. Eqs. 27.39 and 27.40 give

Pz-O and Mz=O The equations of equilibrium (Eqs. 27.19 to 27.21) become •.. (27.44)

PH - M) - Ml + FD + and where F Sliding.

W -

~PI

~PI

+R

= frictional

SCour level

(BI2) ... (VAS) ... (27.46)

resistance at the base to prevent

Also by substituting D • D J• we have

and

P, • Y(Kp - K.)v'16

... (27.47)

M, • y(Kp - K.Jv'112

... (27.48)

M,. YK. (~ - K.) x IfD

L

The above quantities are per unit length of the well and are applicable for a rectangular well. In case of a well with a non* rectangular base, the equations, are modified as

PL - PIL - Fl PLH - MB - MIL + FID+ and

Rototion txoggerotf"d

I:-=~~-r

... (27.49)

Fig. 27.13.

~Pl

(BI2)L

... (27.50) ...(27.51) ... (27.52)

,>4


where L = maximum length of the well base, B = width of the well base, MB = total moment induced in the base due to tilting, FI '" total horizontal reaction at the base, RI ::::t total vertical reaction at the b.'lSe, WT = total vertical load of the well The pressures at the toe and heel are given by RJ

MB

RI

M8

P/'"';r;+z;

.(2753)

p,,-X;;-Z;;

and

... (2754)

where Z" is section modulus. Combining the "above Iwo equations,

2R,

... (27.55)

PI+P.-~

From Eqs. 27.41 and 27.43,

M _ 1 K, (K _ K) l

P

1/1

a

JL _ (p, - 12Ph) If 12D

The values of P, and Ph can be found from the Eqs.value 27.55 of andP,27.56. Once has been [ound, the moment MB can be found from Eq. 27.53, as

~

~

~.

(0 -'10

r

H

1

27.8. me Mh"'HOD Indian Road Congress (IRC

45-

1972) gave recommendations for estimating the resistance of the sOIl below the maximum scour level. 1be recommendations are based on the (d) PrUsur~ observed behaviour of moocls of well ~I~l~~ion foundations and research. According to the recommendations, elastic theory can be used to determine the soil pressure at the side and at the base due to design loads. However, for cstimation of the factor of s.'1.fety against shear failure, the ultimate soil resistance is computed.

1. Elastic TIleory. 100 following assumptions made in the elastic thoory : (1) The wcll behaves as a rigid

r-- -"1----' . ,. \ \

\

;"p

MB ... (P, - RJ/A")Z,, ... (2751)

I \ I

.

\1 \

p\1

e

... (27.56)

.

I

M. t--,-

(c) OeUi!'Ction Pfolill~ on 51dn

-_-_.Jj

. . . . ." \ - -___

L_-----

o

1

(c:I)Elevalion

f--.----i

It--~'I .'

bOdy. (2) The ooefficicnt oC horizontal subgrade reaction KH increases linearly with depth. (3) The unit soil reaction incrcnscs linearly with the lateral deflection.

(e)OeIlKtion at base

[! ) ! !J (fJPrtssure distribution ot be,. Ag.27.14.

WElL FOUNDATIONS (4) The well is acted upon by an external horizontal force H and a moment Mo at .the scour level Pressure Distribution on Sides. Fig. 27.14 (a) shows a rigid well with its base at a depth D below the scour level. TIle well may roLate at a point above the base, or at the base, or below the basco The base moves toward the centre of rotation if the point of rotation lies above the base and, therefore, the frictional force al the base acts in the direction of the horizontal force H. However, if the point of rotation lies below the base, the direction of tbe frictional force is in the direction opposite 10 that of II. In gcneml, the frictional force is given by ~ ~W

F _

... (27.58)

where 14 "" roefficient of friction. W III: total downward load, ~ '" a factor, which lies between -1 and + 1, depending upon the point of rotation. If tbe well rotates at JX>inl C (Fig. rT.14 (b)1 the horizontal deflection at any depth z is givenpy PII - (D -,) B ... (n.59) The horizontal soil reaction at that level is Oz - KJI(zID) {D - z)8 0 ....

where

... (27.60)

m K" (zID)(D - z) 8

m - KlflK.,

Total horizootal soil recation acting on the sides.

P·iLozdz o

-fo LmK.(vD)(D - ,)B.u _ mK;L B (0'16) or

p _ 2m:,,8 . 1"

... (27.61)

where I" _ L IY 112 '" moment of inertia about the axis passing through the c.g. of the vertical projected area. Moment of P at the base level is given by

M, -

f.o

-fo

Lo... (D - z)dz

Lmlq,ID)(D-,)'B.u

or Mp - mK"8I,, ... (27.62) Pressure Distribution at Base. The vertical deflection at a distance (x + x..) from the centre of rotation is given by [see Fig. 27.14 (c)], p - (x + Xt) 8 Therefore, vertical soil reaction 0, - 1(., (x + Xt) 8

,an Moment at the base. M8 -

M8

f-8/2

_K~e

K.J.x. + x..)8xdA

r

B

/2 -8/2

2-

dA + K"

e[812 xcxM. -812

As the reference axis is taken through the c.g. of the base, the second term is equal to zero. Therefore,

Ma - K.B1a

...(27.63)


736

where Is = moment of inertia of tbe base about an axis passing through the c.g. and perpendicular to the horizontal force H. For equilibrium, tbe sum of all the horizontal forces is zero.

H + p~W- PIlIl'P ... P P(l + P~~') _II +

Thus

p~W

P-~

... (27.64)

1 + PIlIl'

Taking moments of all the forces about the base, Mo + HD ... Ms + Mp + Il'P(aD) ...(a) where a D is equal to the distance from the axis passing through the c.g. of Ihe base to the point at whim the resultant vertical frictional force on side acts normal to the direction of the horizontal force ( = B/2 for rectangular wells aod 0.318 B for circular wells). Eq. (a) can be written as

e)

v I.l, aD x (2mK - 0 - ' Iv

+mKv9J~ +

Mo + HD ... KvB1s

Mo + liD .. Kvells + m 1~(1 + 2 Il' a)]

Kv

e ...

Kv

e-

18

Kv O ..

¥-

+mlv~ + 2~'a)

... (27.67) ... (27.67)

M-Mo+IID I .. Is + mlv{l + ZIl'a)

wb= and

~

From Eqs. 27.61 and 27.64,

P - 1 + p~~'

Using Eqs. 27.67,

P _ U;lv

where

p-

~

r ..

D '2'

-

(~ ... [27.69 (a»

I mlv

Also

If+P~W-7(I+P~~') H +

PI-' W - ~

.(27.68)

2mKv91v --D--

On simplification,

M 'J P [I-'W-~

...(27.65)

Mo + HD Is + m Iv (1 + 2 J.l'
+

~

... [27.69 (b)]

-

0/:-

(1 +

PI-' Il')

M --;:-H !;i_If

P-

or

r AI ~W-7 (~~')

As -1 <

P<

1, we have

... (27.70)

WEll.. FOUNDl\rIONS

7

(1 + I·ql') - illY < H <

7(1 -

Illl') + illY

.(27.71)

The vertical soil reaction is given by. u~ ..

lY - Il'P -

Now

W - ....' P _

Kv 0 (x + xc)

f

Kv Oxc. W Therefore,

U z ..

Oz d4 - Kv 0

f (x + x~) eM

KvOx~A

~ fl'P

' P)

w Kv 0 x + ~ (

... (27.72)

The pressures at the toe and heel are given by

and

P, • W -A IJ.' P + Kv 0 (BI2)

... [27.73(0)]

Ph - W -A IJ.' P - K v 8(DI2)

... [27.73(b)]

Substituting the value of Kv8 from Eg. 27.67,

P,-

W- .'p

A

W - p.'P Ph A

and

ME

+"2/ ME

-21

.. .(27.74) . .. (27.75)

For the soil to remain in elastic stale, the maximum soil pressure at any depth should not exceed the maxil1.lum passive pressure, i.e., O;t } Pp

mKv(~)(D AI z '" 0,

Therefore.

- z) O} y(K, - KJz

mKvO(D - z). . the term - - D - - lS a m3XllIlum. m KvO } y (K, - K,,)

m

¥- t

y (K, - Ko)

... (27.76)

Eq. 27.76 should be satisfied. Further, the maximum base pressure P, should not exceed the allowable soil pressure. The minimum pressure p" should not be negative to avoid tension. Sc.ourll'vel

1. Uilimate Solis Resistance The fridional force mobilised along the surface of rupture can be determined assuming the surface to be cylindrical (Fig. 27.15). For circular wells, the surface of rupture is assumed to be a part of sphere with its centre at the point of rotation and passing through :he periphery of the base. If W.. is the ultimate load. equal to the total vertical )oad multiplied by a suitable load factor, the load per unit width is W.,IB. It wiD also be equal 10 the upward pressure for a rectangular base. Let us consider a small arc of length Roo at an angle a from . tbe vertical axis.

Fig. 27.15. Ultimate Soil Resistance.


738

Vertical force on the element = Rda cos a (lY,.18) and normal force developed on the element

dF,. .. (lY.,.I8)Rda cos a cos a

•

2

Total normal force, F,. .. 2fo (W.l8)Rcos ada

F,. _ WiR (8 + sin 9 cos 8) wbere and

...(a)

8 .. tan- I (8/2 nO) n D is the height of point 0 above the base

"ld

Eq. (a) can be written as

R _

V(B/2)'

+ (nD)'

_!!. 2

VI

+ 4n'd

Ii'

F,.

-~VI 2 +

F

-~VI 4n'd[ -1~ ~] 2 + Iitan 2nD'" Ii + 4n2 &

,.

4n'd [ Ian -1 ~ [jl 2nD + 2Ji2

l!.Jt!Ql] ... (21.77)

The moment of resistance of the base about Ibe point of rotation is given by M. _ (F"tan~)R .. .[21.18 (a)l [Note. The right-hand side of Eq. 27.78 (0) is multiplied by a shape factor of 0.6 in the case of circular wells.] Assuming the point of rotation al a height of 0.2 D above the base. the moment of resistance of the base is given by

Mb .. eW81an'

... [21.18(b»)

where, = angle of shearing resistance, 8 = width parallel to the dircaion of forces or equal to the diameter in ciro.dar wells. C = coefficient (see Thble .2).

n

Tuble 27.2. Values of Coefficient C

DlB Recta/lgular Well Cin:ularWt'1J

Side-Resisting Moment Fig. 27.16 shows the ultimate soil pressure distribution at the front and back faces of the welL As the

Fig. 27.16.

WEll. FOUNDt\1l0NS

point of rotation is assumed at a height of 0.2 D above the base, from the similarity of triangles. it can be shown that Dl - 1/3 D. Taking moments of the forces about O.

AI: - OJ196n'[yD(Kp

AI/

=O.l l d (K,. - KJ

Taking submerged unit weigHt.,.' - O.ll'

rr (Kp -

-

K.)]

per unit length

KJ per unillcngth

M, - O.ll' d (Kp - KJ L

For a well of lcn£,th L,

.. .(27.79)

Resl"ting Moment due to rriction on front and back. races The f(idional forces on the faces ad in the vertical direction and produce resisting moment

F ..

Force/unit width,

¥

M, .

l' Jil (Kp - K,,) sin 0

10 the case of a rectangular well of length L, M,.

[!..fl'

d(Kp-XJSino]

I)(

L

AI, .. 0.183l' (Kp - KrJLB.rr sin 6 Air: 0.!8y' (Kp - K.)LBn'sm6 For circular wclls, taking Shape factor as 0.60,

... (27.80)

Air - O.l1y'(Kp - K.)Ii'n'sin6

... (27.8!)

Total Resisting Moment TIle tOl.OO resisting moment is given by M, .. Mb + M, + M, ... (27.82) IRe : 45-1972 rcoommends a reduction factor of 0.7 in tbe total resisting moment of soil. Thus M, .. 0.7 (All. + M, + M f ) ... (27.83) The applied moment AI should be less than Mr Thus

AI ~ 0.7 (AI, + AI, + Air) ... (27.84) For the computation of applied moments, the effects of moments due to tilt and shift of wells, if any, about the plane of rotation should also be considered. If Eq. 27.84 is nol s..1tisfied, increase the grip length D and revise the calculations.

Check for maximum average pressure The maximum average pressure should not exceed one half the ultimate be.1ring capaCity.

f~'¥

... (27.SS)

where q.. = ullimate bearing capacity.

27.9. DESIGN OF INDIVIDUAL COMPONENTS OF mE WELL . 1be overall dcsign of the well bas been discussed in the preceding sections. The individual romponents

are designed as explaioed below. (1) Cutting Edge. 1bc cutting edge should have a sharp angle for rutting through the soil. It should be strong enough so that it does Dot bend when penetrating through a soil containing boulders. A sharp vertical edge baving an angle of 300 with the vertical or having a slope of one horizontal to two vertical is geoerally used [Fig. 27.17 (a)]. However. if the sharp edges are likely to be damaged, a cutting edge with a Slub nose [Fig. 27.!7 (b)] is used.

SOlL MECHANICS AND FOUNDATION BNGINEERING

'40

000ERioUTER FACE

FACE

ANCHOR

ANCHOR

, (al SHARP EDGE

(b) STUB /'lOSE

Fig. 27.17. Cutting

Eage.

The cutting edge should be properly anchored to the well curb. (1) Well cum. Fig. 27.18 shows the curb of a well Curbs are generally made of reinforced concrete. During sinking operation, the curb cuts through the soil. The figure shows the forces acting on the curb when the well bas penetrated to a considerable depth below the scour level.

Fig. 27.18. Weil Curb.

Forces acting tangentially 10 the bevel surface (iMcr surface).

... (,) Q - ~p where " ooefficient of friction between soil and concrete of the curb, P = forces acting nonnal to the bevel surface.

Resolving vertically,

I4Psin8 + pease - N

or

P _

= vertical force on the curb, e :::z: Resolviog horizontally. P sin e - "p cos e

where N

. .. (b)

. N (p.sm9+ cos 8) angle which the bevel edge makes with the bc>rizontaL

_H

... (,) ...(~

where H = horizontal force on the curb per m .

Hoop tension, T - H )( t112,

H _ N (si~ 0 - ~ cos 0) (!,sme.+ cos 8) where d is the diameter.

Thus

T _ O.5N

From Eqs. (c) and (d),

(Sin.a - I'rosa) ,",smO + rose

d

... (21.86)

o

WELL FOUNDATIONS

741

Sometimes, sand-blow may cause sudden descend of the well during sinking and an inaease in the hoop tensloo. Th account for such an eventuality, the hoop tensioo is increased by 50%. Thus

T _ O.75N

(SinH - "(8) d !,sm a + rose

.. .(1:1.87)

Suitable reinforcement should be provided to resist the hoop tension T developed. When the cutting edge is not able to move downward due to reaction developed at the curb and the bottom plug, the hoop tension developed is given by

T.

(~) ~

... (1:1.88)

where q = pressure al the base _ :1 :re;~~

'--~l~~~

I-~"'"

1

Fig. 27.19.

'"

In case of granular soils, the hoop tension is relieved by active earth pressure around the curb. The net hoop tension is given by

T,.~[~-~.+p,)bl PI .)

.)

,)

S}

and

•.. (27.89)

~K.. Y' d

Pl' iK.Y' (D_b)2

in which b = height of the curb, D = depth of the curb below the scour level. At the junction of the curb and steining. a moment M .. develops due to the horizontal force H caused by bevelled action, given by M" _ H)( bl2 ... (1:1.90) Suitable reinforcement "is provided at the inner oomer to talee care of this moment and is anchored into the stcining. IRe: 21-1972 recommends a minimwn reinforcement of 72 kg/m J in a well curb. 1bc reinforcement should be properly arranged. The slope of the inner face of the curb should be such as to push forward easily. .TIle angIe with the vertical should preferably be not more than 30" in ordinary soil and 45" for sandy soils.

(3) Well Slelning. The thickness of the steining should be adequate for the stresses developed during sinking and after installation. The thickness t of the steining may be obtained from the following equation :

742


... (27.91) o O(li - h)f - 0(0 - I)Y, I Ii + P - where H = depth to which tbe well bas progressed, h = height at which it has got suspended, B = external diameter, P = weighl added for sinking. The design of steining reinforcement depends upon the skin friction and the unit weight of well. It is 3 usual practice \0 provide reinforcement of about 5 10 6 kg/m of the brick and conaete steining. About 75%

of the lolal reinforcement is in the form of vertical reinforcement and 25% in the shape of latemls or hoop rings. The vertical reinforcement is spread ncar both (the outer and inner) races. The laterals should be checked for tbe moment developed due to eocentric keotledge and half the weight of the well at an eccentricity of ooe-fourth the width of well in any direction. This COOdition is generally critical when the well ha<> sunk to about balf the designed depth. The thickness of the' siciniog is usuaUy fixed empirically. For railway bridges in India, it is generally laken as one-fourth of the outside diameter. For road bridges, it is kept as ooe.eighth of the outside diameter if it is in brick masonry and one-tenth of the outside diameter if in cement concrete. However, the thickness is increased by 12 em per 3 m of depth after the first 3 m of steining in brick masonry and 15 em per 6 m of depth after the first 6 m for cement concrete. A thumb rule commonly used is 1_ K(BI8 +1J/1OO) ... (27.92) where B = external diameter of well, H = depth below low water level, K = a constant (= l.0 for sandy soils; 1.1 for soft clay and l.25 for bard clay and boulders). (4) BoUom Plug. The bottom plug should be strong enough to traru;mit the load to the soil below. II is designed as a thick plate subjected to unit bearing pressure under the maximum vertical load, a<; already discussed in the case of open caissons in chapter 26. The bottom plug is given the Shape of an electric bulb to produce an arch action, to reduce hoop tension in the curb and to provide larg~r base area. 'The bottom plug is constructed in (1 : 2 : 4) cement concrete laid by means of a tremie or a skin box. About 10% extra cement is added because some cement is wasbed away on account of water. The water in the well must be still and at its normal level. Bouom plugging should always be done in one continuous operation. While founding the well on rock, it should be properly anchored by taking it 25 em to 30 em deep into the rock bed. Adequate dowel bars should be provided. (5) Well Cap. The bottom of the well cap is generaUy kept at the low water leveL It is designed as a slab resting on the well. The well cap may be extended as cantilevers to accommodate piers of slightly larger size tban that of the well. If the width of th..: pier is greater than the size of the dredge hole, it is assumed that the weight of a cone of concrete having an apex angle of 6()0 ~ carried by the slab and the remaining load is transmitted to tbe steining. The well cap should have a minimum reinforcement of about 80 kglm 3 . (6) Top Plug. The function of the top plug is to transmit the lood of the pier to the well steining. If a well cap is provided. there is no need of a top plug. However, it is generally provided as an extra safety precaution. Offsets are provided at the top of the steining to provide bearing to the plug. Cement concrete (1 : 2 : 4) is used for the construction of tbe top plug. (7) Sand filling. The main purpose of sand filling is to provide stability to the well by increasing its weight and to reduce the tensile stresses caused at the base by bending moment However, sand cannot be depended upon for transmitting the weight of the pier to the bottom plug and, therefore, it docs nol contribute towards the structural strength of the well. On the Indian railways, the prnctice is to do the sand filling upto the top plug. Some of the highway engineers recommend that the saod filling should be dODe upto the lowest soour level. The actual depth of the fllling should be fixed considering the requirement of the dead weight for the stability. 27.22. SINKING OF WELLS The sinking operation consists of the following steps :

WELL FOUNDIJ10NS

743

(J) Laying the Well Curb. It the river bed is dry, the cutting edge over wbich the WEll well curb is to be buill is placed at the required positioo after excavating the river bed to about 15 em. When the depth of water is uplO 5 m, a sand island is made before placing the curb. 111e size of island should be large enough to accommodate the well with adequate working space aU around (Fig. Fig. 27.'1IJ. Sand lsllllld. 27.20). 10 the case of depth of water more thaD 5 m, it is generally more economical to build the curb on dry ground at the river bank and float it to the site. It is useful to iru;crt woodeo sleepers below the cutting edge at regular intervals to distribute the load evenly over the ground. The shullering of the well curn is then erected. The outer shuttering is generally made of steel (or wood) and the inner shuttering of brick masonry. The reinforcement of the curb is then placed in proper position sum thai the vertical bars project about 2 m above the top of the curb. Concreting of the curn is done in one rontinuous operation. The curb concrete is allowed to set at least for one week before the shuttering is stripped off. The sleepers are also then removed. (2) WeD Steinlng. After sinking the well curb, the steining is raised by about 1.5 m at a time and its sinking dooe afier allowing aUesst 24 hours for setting. The steining is built using straight edges, preferably of angle iron. 'The lower portion of the straight edges is kept butted with the steining of the earlier stages. lbe steining must not be built in plumb at any intennediate stage when the well bm: tilled LO one side. Once the well hm: acquired a grip of about 6 m in ground, the steining can be raised by about 3 m at a time. The height of steining built at any stage should be such that the well does not lose stability. (3) Sinking Process. Sinking process is begun after having cast the OJrb and the first stage of steining and allowing enough time for curing. The well is sunk by excavating material from inside the curb manually KENTLEOGE or mechanically. When the depth of water Inside the well is upto 1 m, dredging can be done manually. However, beyond this depth of water, excavation is done wilh the help of 'jhams' (a type of spade). The 'jham' is tied to a rope moving over a pulley. It is pulled by the men. Every time;. a diver dives and pushes the jham into the soil and comes up. The jham ~. is then pulled out. In an improved version, the jham bas been replaced by an automatic grab operated by diesel (or steam) winches. Straight ~ chisels are used for breaking hard material so . tbal is can be taken out by grab. Undcr-cutting Fig. 27.21. Kentledge on the Well. chisels art used to loosen the material which lies under Ute steining. Explosives are used for sinking through rode. As the well sinks, the friction on the sides increases. To accelerate the proces.o; of sinking. additional looding. known as kentledgc;. is applied on the well. Kentledge is generally in the form of sand bags (Fig. 27.21) placed 00 a suitable platfonn erected on its top such Ihat it does not interfere with the excavation. Sometimes. even kenUedge is not sufficient to sink the well. In such cases, the frictional resistance developed on its outer periphery is reduced by forcing jets of water on the outer face. However, this method is effective only in the case of wells sunk in sandy strata. In some cases, pumping oul the water from inside the well is effective in well sinking. However, this method should be discouraged in early stages when the depth is shallow. It is not desirable 10 resort LO pumping out water unless the well has gone deep enough or has ~ through a clayey strata so that tbe

.\

•

744


chances of tilt and shift are reduced. Dewatering is not allowed after the well bas sunk to about 10 m. After this stage. the sinking is done by usual methods of grabbing, dlisclling, applying kcntledge or blasting. Great precaution is necessary if dewatering of the well is done when it is al a shallow depth to avoid blowing of sand from under the culling cdge. If blowing of sand occurs, it rcsulls in the loss of time and labour in removing the saDd. It also presents danger to the men working inside; as the well may get filled upto a height of a few metres if the blow is large. The well may also lill suddenly. Scrap gunny bags unci grass bundles arc placed around the periphery of the well on the outside into the funnel formed. It blocks the passage through which the blow of sand is laking place.

27.23. MEASURES FOR RECTU'ICATION OF TILTS AND SmFTS '[be well should be sunk straight and at the correct position. However, it is not an easy task to achieve this objective. Somctimes, the well tilts on one side or it shifts away from the desired position. The following precautions must be taken to avoid tillS and shifts. (1) The outer surfxe of the well curb and steining should be regular and smooth. The diameter of the curb should be kept about 4 to 8 em lager than the outer diameter of steining, and the well shoukl be symmetrically placed. (3) The cutting edge should be of uniform thickness and sharpness. (4) Dredging should be done uniformly on all sides in a circular well and in both pockelS of a twin well. TillS and shifts must be Clrefully checked and properly recorded. The correcl measurement of the tilt is an important field obscrvation required during well sinking. It is nol pnssible 10 specify Ihe permissible limits of

:::i~U:~~~S~~~~~~~~::!~:

PULLEY

PULLEY

lhat lilt should generally be limited 10 1 in 60. The Shift should be restricted to one percent of the depth sunk. In case they exceed the above limits, the following measures are taken for Ih~ir rectification: (I) Rel,,'uJuUon or Grabbing. To rectify the lilt, the higher side is grabbed more by regulating the dredging. However, Ihis method is effective in initial siages of sinking. In case of tilted wells, dredging

~C:I~~ ~~~ ~~t~:~~;!z~~n~U~e ah;~;

(a)

Fig. 27.12.

(b)

side aDd the rope of the grub is pulled through the hook [Fig. 27.22 (b)J. 1"11us dredging with hooking is done. Altcmativcly, the well may be dewalercd, if possible, and open excavation may be carried out on the higher side. R S . JOIST (2) Eccentric loading. To providc greater sinking effort on the higher side, eccentric loading is applied by adjusling the kcntlcdgc. A suitable plalfonn is ronslructed on the highcr R .5. side for this purplSc (Fig. 27.23). As JOIST the sinking progrcr.scs, heavier k.cntledge with greatcr ccccntricity would be required to rectify the till. (3) Water Jetting. If the water jets arc applicd on the outer face of thc well on the higher Side, thc side friction is reduced and the tilt is rectified.

WEU... FOUNDATIONS

(4)

Excsvatkm

'" under cultlng

-rl-~_fr ~YSLEEPER

edge. A tilted well in a hard clayey stratum docs not straighten due to unbroken hard stratum on the higher side. If dewatering of the well is possible, open excavation is done under the cutting edge. In case dewatering is not possible, divers can be sent to loosen the strata. (5) Inserting wooden Sleeper '-HOOK under the cutting edge. Sometimes sleepers arc inserted wooden (bl (al temporarily below tho rutting edge on the lower side 10 avo~d funher tilt [Pig. Fi .27.24. 27.24 (a)l. AlternatIVely, a hook is g inserted below the cutting edge on the lower side and pulled with a wire rope and kept strained [Fig. 27.24 (b)]. (6) Pulling .hl: well. If] early stages of sinking, pulling the well towards the higher side, by placing one or more steel ropes around the well with vertical sleepers packed in-between to distribute the pressure over lnrger areas of well steining, is quite eITw.ivc (Fig. 27.25). SLEEPERS

PUll STEEL RoPE UJ..c~J.L.u..JJ-.;u-j

\

,I Fig. 27.2.'5 .

(7) Strutting the well. TIle well is strutted on irs tilted side with suitable logs of wood to prevent fuMber tilt. The well steining is provided with sleepers to distribute the load from the strut. The other ends of the logs rest agaioot a finn and non-yielding base having driven piles (Fig. 27.26).

SLEEPER

Fig. 27.26.


746

Fig. 27.1:1.

(8) Pushing the wells by jocks. The till can be rectiflCd by pushing the well with a suitable arrangement through mechanical or hydraulic jacks. Fig. 27.27 shows a tilled well being pushed by a jack resting against the venically.sunk well. In aaua1 practice, a combination of several methods d~ above is gencmlly used. ILLUSTRATIVE EXAM1'LES

Dillstratl"e Example i7.1. A circular well of 6 m erlerMi dianleler and 4 m internal dimnerer is embedded to a depth of 15 m below the maxImum scour level in Q , sandy soil deposit. The well is subjected to a horizontill force of 800 kN acting at a height of 8 m above the scour level. Determine the allowable total equivalent resisting force due to earth pressure, assuming (a) the. rotation is about a point above the base, (b) the rotation is at the base. Take Y6111 ::: 20 kNlm J, 30°; factor of safety for passive resistance "" 2.0.

+: :

Use Terzaghi's analysis.

Ttr

Solution. (See Fig. E·27.1) Kp - 3.0, K(I - 0.333

P

~~==-~i: T ~~~i"'l 8m Scour

-.

(.)

II!'~I

SeD r

Tzsm

(b) Fig. E-27. 1.

(a) Rotation above the base. 1btal height, HI .. 8 + 15 .. 23 m

From Eq. 27.10,

2D, .. 3Ht ~ V9Hr - 2D(3H, - D) 2Dl .. 3 )( 23

or From Eq. 21.9,

DJ

:t

";9 )( 132 - 2)( 15,(3 x 23 - 15) .. 12.96

6.48m 1 . q~ ~ 21' D(K,. - K.J(D - 2D,) ..

747

WEll.. FOUNDATIONS

-t

x 10.0 X 15 X (3.0 - 0.333)(15 - 2 X 6.48) - 408.1 kN/m

From Eq. 27.13, the allowable transverse load.

Q,,"

q'''';sx L

(b) Rotntlon at bose. From Eqs. 27.11,

.. Therefore,

q~'

..

408.~)C

..

k

6 .. 1224.2kN

y' (Kp - KJ

jf

l.!.?t .. 652.3 kN/m

'61 )C 10 x (3.0 - 0.333) x 2J

Q" .. 652; x 6 .. 1956.9 kN

mustmtlve Example 27.2. Fig. £-27.2 shows a well. Determine the base pressure and the lateral load per unit lengtl! of tile well. The well is subjected to a net downward force of 10 MN. Assume that the horizontal deformation of the well cap at the scour level is 20 mm.

Till

H:'j:~m l~I.

I

~

1--·6~-.I ..l

Fig. E-27.2.

Use Banerjee and Gangopadhay's simplified analysis. TaJce I.l. = 0.50; y' 25000 kNlm'. Solution. Area of aoss-sectioo, A~ .. x/4 (62 _ 42) .. 15.70 m2 Moment of inertia, I .. nl64 (6' _ 44) .. 51.03 m4

Section modulus,

Z~

From &I. 27.18,

m _ l'(Kp-K.) p,

.. 51.03 .. 3.0

_ 11

X

17.0~m3

(3.0 - 0.333) _ 1466.9

20 x 10. . 3 From Eq. 27.47,

P I -

l' (Kp - K.)Ii' 6

.. 11 x (3:0 -60.333) x 8'2 .. 312.9 leN

From &I. 27.48,

M

l'(K, - K.)Ii' 1"

12

= 11 leN/m3; 4' = 30"; K~

-=


'48

.. 11.0 x (3.0 ;20.333) )( 8

- K,J m

M, _ Y'KJK

From Eq. 27.49,

11 )( From Eq. 2752,

4 ..

1251.7 kN

• IfD

25~;1.!O -

0333) )( 12 6: 8 .. 1125 kN-m

WT - flP1L +R,

10000 .. 0.5 x 312.9 x 6 + R, RI .. 9061.3kN

0'

From Eq. 27.55,

p, + PI. ..

2A~1

From Eq. 27.56,

Pt -Pit"

~ .. ~ Jr (6)

.. 2

\~1.3

.. 1154.3 IcN/m 2 .. 375kN/m1

Solving the above two equations. p, .. 764.65 IcN/m 2; PI. .. 389.65 IcN/m 2

From Eq. 27.57,

MB

From Eq. 27.50,

PL .. PIL - P,

..

(P, - R,/A b) Zb"

(764.65 - 90613) )( 17.01 .. 3189.38k:N-m ~

P )( 6.0 - 312.9 )( 6.0 - Fl

... (0)

PLH - Ms - MIL + FID + ~PI (BI2)L

From Eq. 27.51, or

P x 6.0 x 4.5 .. 3189.38 - 1251.7 x 6 + PI x 8 + 0.5 x 312.9 )( (6/2) x 6

or

27 P .. - 1504.72 + 8F,

Solving (0) and (b),

.(b)

P - 180.2 kN D1ustratlve Example 27.3. Design a well {OUIIf!tJtion for the following site conditions using IRe method. High flood level = 103.50, Bed level = 100.00, Scour level = 88.201 Base level = 78.00, External diameter = 8 In, Wuith of pier at top = 1.5 In, Total vertical load = 5000 /eN, Total horizontal load at bed level:::l: 750 leN, Moment on well at scour level 3{)()() kN-m, Horizontal WQter pressure on well = 650 liN. Talce q"", _ 600 kNlm1,. '(,.. _ 20 kNlm}; = 30°,. K = 1.25; {, = 20°,' seismic coefficient = 0.10,. Kp =

=

+

6.105, Ka _ 0.297; q.. _ 1800 kNlm

2

Solution. (See Fig. 8-273) Let us asswne that the low water level is at bed level From Eq. 27.92, t - K (8/8 + H/lOO) t _

1.25 (8/8 + 221100) _ 1.53 say 1.6 m

Internal diameter _ 8 - 2 )( 1.6 _ 4.8 m Design or Well Cap. Let us assume thai the thickness of the well cap is 75 ern. - x/4 )( (8.0)2 )( 0.75 )( 24 - 004.3 kN Weight of the well cap Assuming a buoyancy of 15%, the upward force Area of cross-section

_ 0.15 )( Jt.I4 )( (8.0)2 )( 0.75 )( 10 _ 56.5 leN _ :rtI4 )( (82 _ 4.82) _ 32.15 m2

Moment of inertia (/s)

- J't/64(8' - 4.8') -174.92m"

Section modulus

weJ.l. FOUNDATIONS

749

SeOUl"

leveL

88 .•

Bose

It'VeI

78.00

1 c:r::t=JtJtTtt 1--'.15m-..".-- l>'m-l .1 T ----z:u--t-.l-J ., T lL5.o

'252.2 kN/m 2

Fig, B-27.3.

Total vertical load, neglecting buoyancy = 5000.0 + 904.3 = 5904.3 kN The maximum and minimwn pressures are given by

itt

+

~

.. 252.2

53~53

-

~

.. 115 kN/m 2

qrmur. .. 5 3 qmin ..

Average pressure

kN/m 2

(q.) _ (252.2 ; 115.0) _ 183.6 kN/m'

Pressure at the fare of the pier [see Fig, E-27.3 (b)J.

'1J - 115.0 + (252.2 ; 115.0) x 4.75 _ 1965 kN/m'

750

SOIL MECHANICS AND FOUNDAl10N ENGiNEERING

Jotemity of pressure due to self weight, q(..

904.3 2 " 18 kN/m 2 ",/4 x (8)

In Fig. 8-27.3, let the angle AOD be 8 1 and DOC be

6:z.

cos 9 1/2 ..

~~g

or 8 1

..

158.38°

cos 8v'2 ..

~~:

or 8 2

..

143.58°

Chord

AD .. 2 x 4.0 )( sin 8 1/2 .. 7.86 m

Clard

Be ..

Area of outer segment,

2 x 2.4 )( sin

~/2

.. 4.56 m

AI_ tR2(OI-SinB1)

.. .!.2 x 402 (158.38)( ' 100 Area of inner segment,

A2 ..

~ 2

)( 242 (143.58 x

100

It

_

0.368) ..

1t

_

0.594) .. S.50ml

1916m 2

.

The distances of the centroids of these areas are obtained as under.

4 R sin) (81/2) 4)( 4 )( (0.982)3 ZI .. 3 (8 _ sin 8 ) .. 3 (2.763 _ 0.368) .. 2.1 m 1 1 7 __

~,

4 • 2.4 • (0.95)' _ 144 3 (2505 _ 0594) . m

Moment about face AD - q. [A,(l, - 0.75) - A, (z, - 0.75)J - q, [A,(Z, - 0.75)]

M - 183.6 [19.16 (2.1 - 0.75) - 5.5 (1.44 - 0.75)J - 18 • 19.16 (2,1 - 0.75) .. 3588.3 kN-m Maximum shear force

.. qQ (AI -

Au - qeAI

- 183.6 (19.16 - 5.5) - 18 • 19.16 - 2163.1 kN It can be checked that the overall thickness of 75 em of cap is sufficient from the consideration of moment and shear stresses (see any text on R.C.C. design) for M 15 concrete.

Design of Stelning Total vertical load Horizontal load

- 5000 + 9043 _ 59043 kN . _750kN

Moment at soour level - 3()(X)1cN-m Weight of steining lIplO scour level

Seismic force

00

steining

- (0/4) (8' - 4.8') • 24 • 11.05 - 8527.1 kN .. 0.1 )( 8527.1 .. 852.7 kN

Waler pressure on well .. 650 kN Total horizontal force at scour level

Q .. 750 + 852.7 + 650.0 .. ' 2252.7 leN Moment at smur level,

M .. 3()(X) + 750 x 11.8 + 852.7 )( (11.812) + 650 )( (2/3) )( 11.8

- 21994.2 kN The depth of the point of zero sbear below the soour level is given by Eq. 27.14.

WEll. FOUNDIillONS

151

Taldng F _ 2, 2 x 2 x 2252.7

x - [ 10 x (6 .105 _ 0.297) x 8.0

44 ... m

AI__ AI + 213 Qx

From Eq. 27.15 a,

.. 21994.2 + 213 x 2252.7 Let us assume a tilt of

]"

t in 60.

Ea:cntricity due to tilt at the lOp _ total

:gh1 . . ~

l(

4.4 _ 28602.1 kN-m

0.367 m

1~22 _ 0.17 m

Eca:ntricity at scour level

.. 0.367

Moment due to Ult Lei us assume a shift o[ 1%. Total shin

.. (5904.3 + 8527.1) x 0.17 .. 2453.3 kN-m

~oment due to shift Total moment

.. (5904.3 + 8527.1) x 0.22 .. 3174.9kN-m

Buoyancy

011

l(

.. 0,01 x 22 .. O.22m .. 28602.1 + 2453.3 + 3174.9 .. 34230.3 kN-rn

the well portion uplO the point of zero shear

.. (it/4) )( (8.0)2(11.80 + 4.4) x 10 x 0.15 .. 1220.8 Weight of steining upto the point of zero sbear

.. 8527.1 )(

(1 1.0:1~54.40)

.. 11922.5 kN

Thl:al vertical weight .. 5904.3 + 11922.5 - 1220.8 .. 16606 leN The maximum and minimum stresses are given by

q_ ..

;~

+

~7~~

qmill ..

~~

-

34~4~92)(

x 4 .. 1299.3 kN/m2 4 .. - 266.3 kN/m

2

Cement conacte (1 : 3 : 6) can be used for steining.

Lateral Stability of the Well 8527.1

I

Total weight of the steillillg

.. 1'l.O5 .x

Weight of water ill the well

.. (11:/4) )( (4.8)2 )( 10 )( (11.05 - 1.0) .. 1817.7 leN

21.25 .. 16398.3 kN

Weight of intermediate plug (I m) thick .. (1tI4) )( (4.8)2 x 1.0 x 24 .. 434 kN Weight of bottom plug (2.5 m thick) .. it/4 x (4.8)2 x 24 x 2.50 .. 1085.2 leN Weight of sand fill upto scour level .. :rtI4

X

(4.8)2 x 20 x 7.7 .. 2785.3 kN

Total weight at the base

.. 5904.3 + 16398.3 + 1817.7 + 434 + 2785.3 + 1085.2

Buoyancy on well

- 28424.8kN .. :rtI4 x (8.0)2 x 22 x 10 .. 11052.8 leN

752


Net downward force, W .. 28424.8 - 11052.8 .. 17372kN Weight of water and intermediate plug

.. 1817.7 + 434 _2251.7kN Horizontal seismic force .. 0.1 )( 2251.7 .. 225.2 kN Moment of seismic forces about scour level .. tSJ.n (1.75 + 10.0512) + 43.4 IC 0.50 .. 1253.2 kN Total hOrlzoolal force at scour level, II .. 2252.7 + 225.2 .. 2477.9 kN .. 21994.2 + 1253.2 _ 23247.4 kN-m Moment at scour level Moment at base level., M - 23247.4 + 2477.9 )( 10.2 + 2453.3 + 3174.9 _ 54150.2 kN-m Elastic Analysis

The design forces and moments acting al the base are as under. Vertical forces (W) = 17372 leN Horizontal force (I/) = 2477.9 kN Moment (M) = 54150.2 kN·m In this case, the grip length. D.. 10.2 m. L .. 0.9 )( 8 .. 7.2m Now

From Eq. 27.68,

Taking m ::: 1.0, where

J~

..

L/; .

7.2

xl~1O·2}l

.. 636.72m4

I .. 18 + III l~ (1 + 2 Il' a) I .. 174.92 + 1.0 x 636.72 (1 + 2 x tan 20 0

)(

a)

a .. O.3~8B .. 0.3:~.2x8 .. 0.25

Therefore,

I .. 174.92 + 636.72 (1 + 2 x tan ZOo x 0.25) .. 927.51 m4

From Eq. 27.69 (b),

D I r""2' m1v ..

From Eq. 27.71,

H > >

1~.2 ~

x 1.0

~7~.72

.. 7.43rn

(1 + f.lf.l') - f.l W

547~!~·2

(1 + tan 30" tan 20") - Ian 30" )( 17372

1/ > - 1205 As H of 2477.9 is grealer tban -1205, the above condition is stalisfted.

Also

H < <

~

(1 - !.lll') + f.l W

5471.~~.2

(1 _ tan 30" tan 20") + tan 30" )( 1737.2

2477.9 < 6760 (satisfied) From Eq. 27.76. Laldng submerged unit weight,

m¥}Y'(K,-K.)

WELL roUND(\TlONS

753

1.0 x

59~i.~;2 ~ 58.4 t

From Eq. 27.69 (a),

10 (6.105 - 0.297) 58.2 (almost satisfied)

~

p ..

54;.~~.2

.. 7288 kN

p,-~+~~

rrom Eq. 27.74,

.. (17372 - 0.364 x 7288) + 54150.2 x 8 11:/4 x (8.0)2 2 x 927.51 P, .. 526.5 kN/m 2 < 600 kN/m 2 (s.llfe)

p,,-

From Eq. 27.75.

~

A

MB -21

(17372 - 0.364 x 7288) 54150.2 x 8 .. 1(/4 x (8.0)2 - ~ PI, .. 59.5 kN/m2

Ultlmule Reslshmce

From

cq.

Tnking

q"

W

27.85,

f) .. -

'2

A

1800 kN/m 2, 17372

1800 2

,11:/4 x (8.0)2

345.8

From Eq. 27.84.

AI

t

900 (smisficd)

~ 0.7 (M.+M, +M,)

From Eq. 27.78 (b).

C =0.i;"J for DID::: 1O.m M" .. ClVlJUln$ .. 0.29 x 17372 x 8 Inn 300

From Eq. 27.79,

M, • 0.1 l' IY(K, - Ko)L

From Eq. 27.81,

- 0.1 x iO x (10.2)) x (6.105 - 0.297) x 7.2 .. 44317.3 kN-m M,_ 0.1 y' (K,. - K(J)WD 2 sin6

From Table 27.2.

= 1.23

.. 23254.9 kN-m

.. 0.1 )( 10 (6.105 _ 0.297) x 82. x (10.2)2 sin 20 0 - 13226.9

Now,

0.7 (M" + M, + M,) .. 0.7 (23254.9 + 44377.3 + 13226.9) .. 56601.4 kN-m

AI .. 54150.2 kN-m M

} 0.7 (M" + M. + M,) (satisfied)

754

SOIL MECHANICS AND FOUNDATION ENGJNE~RING

PROBLEMS A. Numencals 27.1. A well foundation has the following particulDtS. QUler diameter Inner diameter Depth below scour level

'" 5.0 m '" 3.0 m '" 12.0 m

Moment

'" 5000 kN·m

Horizontnl force acting at 8 m above the scour level Factor of safety

'" 600 kN '" 2.0

Assumi ng that the well tillS about a certain pein! above the base, compute the allowable, tOlDl equivalent resisting force due 10 earth pressure. Thke Ysflt '"' 20 kN/mJ , 30°. (Ans. 710.5 kN) 27.2. The following data refer to a well foundation: (a) Ne t downward load, including self.weight '" 12000 kN (b) Height of poim of application of horizontal force above the scour level :. 4 m (c) Depth of well below scour level '" 10 m (d) External diameter = 7.0 m (e) Internal diameter '" 5.0 m if) Vertical subgrode reaction '" 3 )( 104 kN/m J

+'"'

(g) Poisson's rmio (h) HonzonlllJ deformation of welJ cap at !he scour level (pI) (i) Allowable soi l pressure

Ihe

::::~~ ~: :~ t~~:jeB;~t=i~;SC'

D. Des
:II

0.5

• 20 mm • 600 kN/m2

determine the base pressure and lateral load per unit length of

lYpe

Discuss the situations where

0

well foundotion is more suituble than the other types of foundmiom.

What are different shapes of wells? Discuss the characteriSlics of each type. Discuss the various forces acti ng o n a well foundation. Whot do you under5U\nd by grip length ? Whot is its importance in well foundations? Describe vIlriOUS methods for the design of well foundations. What are their relative merits? What nre the various comJXlnents of II well foundation? What arc their uses ? Descri!;le the proa:dure for construction of wells. Discuss the causes and remedies for tilts and shifts. Discuss IRC method for the design of well foundntion. Explain Terz.oghi's analysis for rigid bulkhead applied to well foundation. DiSQJss Bnne~ee and Gangopadhyay',j method for the design of well foundaTions.

C. Multiple Choke Queslions 1. A well foundation is a type of (a) open caisson

(b) pneumatic caisson

(e) f100ting caisson (d) drilled pier 2. The grip length below the maximum scour level for the rnilWlly bridges is usually (a) 0.5 R (b) 0.25 R (e) R (d) 2 R where R is the max im um scour depth. 3. In some well foundation, the following is nO( provided: (a) R.C.C. well cap (b) Top plug (e) Bottom plug (e) Curb 4. The most commonly Il5Cd shope of a well foundation is (a) Double-D well (b) Circular well (e) Double octagonal well (d) Rectangular well. S. The thickness of stcining for railway bridges is usually kepc as . ..... . of outside diameter. (a) one-eight (b) one tenth (,c) one-sixth (4) one-founh. lAns. \. (a), 2. (a), 3. (b),

4. (b), S. (d)l

28 Machine Foundations 28.1. INTRODUcnON Foundations subjected to static loads have been discussed in the preceding chapters. In some cases, the foundations are subjected to dynamic loads. These loads may result from various C1USes such as vibratory motion of machines, movement of vehicles, impact of hammers, earthquakes, winds, waves, nUCic.1r blasL<;, mine explosions, and pile driving. The dynamic loads transmitted to the foundations and their effect on the SIrota below can be delennined using the principles of soil dynamics and theory of vibmtioos. 1be analysis is, however, very complex.

Machine foundations are subjected to the dynamic forces caused by the machine. lbese dynamic forces are transmitted 10 the foundation supporting the machine. Although the moving parts of the machine are genemlly balanced, there is always some unbalance in praclice which causes an eccentricity of rotating parts. This produces an oscillating force. The machine foundation must satisfy the criteria for dynamic loading, in addition to that for static loading alrc.'ldy discussed. Basically, there are three types of machines: (i) Machines which produce a periodic unbalanced force. such as reciprocating engines and compressors. The speed of such machines is genernlly less than 600 f.p.m. In these machines, the rotary molion of the crank is converted into the translatory motion. TIle unbalanced force varics sinusoidally. (i1) Machines which produce impact loads, such as forge hammers and punch presses. In these machines, the dynamic force allains a peak value in a very short time and then dies out gradually. The response is a pulsating curve. It vanishes before the next pulse. 1be speed is usually between 60 to 150 blows per minute. (iiI) High speed machines, such as turbincs and rotary compressors. The speed of such machines is very high; sometimes, it is even more than 3000 f.p,m. This chapler is devoted mainly to the design of foundation for the machines of the first type. 28.2. TYPES OF MACHINE FOUNDATIONS The following 4 types of machine foundations are commonly used. (1) Block 1)rpe. This type of machine foundation consists of a pedestal resting on a footing rfig. 28.1 (a)]. The foundation has a large mass and a smal1 natural frequency. (2) Box Type. The foundation consists of a hollow concrete block [Fig. 28.1 (b)]. The mass of lhe foundation is less than that in the block type and the natural frequency is increased. (3) Wall1)rpe. A wall type of foundation consists of a pair of walls having a top slab. The machine rests on the top slab [Fig. 28:1 (c)]. (4) Framed Type. This type of foundation consists of venical oolumns having a horizontal frame at their tops. The machine is supponcd all the frame [Fig. 28.1 (d)).

7,.


to} Block

(elWoll

tyP~

type

(b)

Bo~

type

(.:I) FromPd

type

Fig. 28.1. Type of Machille Foundation.

Suitability of various types. Machines which produce periodical and impulsive forces 3\ low speeds are gcncrnUy provided with a block type foundation. Framed type foundations are generally used for the machines working al high speeds and for those of the rotating types. Some machines which induce vcry little dynamiC forces, such a<; lathes, need not be provided with a machine foundation. Such machines may be directly bolted to the floor. 28.3. BASIC DEFINITIONS The following terms are used in the dynamic analysis of machine foundations. (1) Vibration (or oscillation). II is lhe lime-dependent, repeated motion of II"'
757

MACHINE FOUNDATIONS

(9) Resonance. When the frequency of the exciting force is equal 10 ooe of the natural frequencies of the system, the ampliludes of motion become excessively large. This condition is known as resonance. 10. Damping. The resistance to motion which develops due 10 friction and other causes is known as damping. Viscous damping is II Iype of damping in which the damping force is proportional \0 Ihe velocity. It is expressed as F - c

where c

~

... (28.2)

= damping coefficienl, and ~ = velocity

(11) Degree of }'l'eedom. The number of independent co-ordinates required to describe the motion of a system is called the degree of freedom.

(a)

(b)

fig. 28.2..

Fig. 28.2. (a) shows a system with one degree of freedom. and Fig. 28.2 (b) shows a system with two degrees of freedom. An elastic rod has an infinite degree of freedom. However, for convenience, the rod is divided inlo segmenls. The degree of freedom is made finite by coruiidering the masses of these segments. (12) Principal modes of vibrations. A system with more than one degree of freedom vibnucs in complex modes. However, if each point in the system follows a definite pattern of common natural frequency, the mode is systematic and orderly and is known as the principal mode of vibration. A system with n degrees of freedom has n principal modes and hence n natural frequencies.

·28.4. DEGREE OF FREEDOM OF A BLOCK FOUNIlATION A rigid block: foundation has 6 degree of freedom (Fig. 28.3). Any displacement can be resolved into 6 independent displacements as under. (1) Translation along X·axis, (2) Translation along Y-axis, (3) Trnru;lation along Z-axis, (4) Rotation about X-axis, (5) Rotation about Y-axis, (6) Rotation about Z-axis. Translation along Z-axes and rotation about Z-axis can occur independently of any other motion. However. translations and rotation about X-and Y-axis are coupled. as these cannot occur independent of one another. X-, y. and Z-axes are caUed, respectively. pitching, rocking and yawing axes. The discussions in Ihis elementary text are limited 10 one-degree of freedom. 29.5. GENERAL CRHERIA FOR DESiGN OF MACHINE FOUNDATIONS A good machine foundation should Satisfy the follOWing criteria. (1) Like ordinary foundations, it should be safe against shear failure caused by superimposed loads. and

758


Hg. 283. Degree of Freedom.

also the settlements should be within the safe limits. The soil pressure should normally not exceed 80% of the allowable pressure for slatic loading. (2) 'fbere should be no possibility of resonance. The natural frequency of the foundation should be either gre..1ter than or smaller than the operat ing frequency of the machine. (3) The amplitudes under service conditions should be within the permissible limits for the machine. (4) The combined centre of gravity of the machine
t

FREQUENCY

(cpm)-"

Fig. 28.4. Rchart's CharI.

MACHINE FOUNDATIONS

759

28.6. FREE VIBRATIONS

The free vibrations may be damped or undamped. (a) Undamped Vibrations. Fig. 28.5 (a) shows a rigid mass (m) resting on a spring of stiffness k. The system has olle degree of freedom. Let us assume that the system has been set in motion and it vibrates in the venical direaioll.

(a) Fig. 283.

The equation of motion can be written as under.

,p,

nld?- - kz

,p,

... (283)

nld?+kz-O

The solution of Eq. 28.3 is given by z-Acosro"t+Bsinro"t where A and B are arbitrary constants, and ro" is the natural circular frequency (radians/SeC). The solution can also be written as ... [28.4 (a)] z -A sin '(oo" t + a) where A and a are consiants. z _ A cos(ro" t - a) Alternatively, . ..[28.4 (b)] From Eq. 28.4 (a),

and

~

~

- A tu"cns(oo... t+ a) - - A 00; sin (0011 I + a)

Substituting these values in Eq. 28.3, moo~

or

00"

... (28.5)

-m A ro! sin (00" I + a) + kA sin (00" t + a) _ 0

_k

-...[k;;;;

... (28.6)

11 may be noted tbat the greater the mass m, the smaller is the frequency. If /... is the natural frequency in cycles per second,

In • ~ •

t-w;;;

If T is the time period, Eq. 28.1 gives T _ l i t. _ 2 n/-ffliii

... (28.7)

... (28.8)

Fig. 28.5 (b) shows the response OJrve of the system. As it is evident, the cycle repeats after time T.

SOIL MECHANICS AND FQUND!JION ENGINEERING

7W

(b) Damped Vibrutlon. Fig. 28.6 (a) shows a rigid mass m resting on a spring of stiffness Ie and connected to a viscous damper with a damping coefficient c. In Ibis case, there is an additional force due to damping. The equation of motion can be written as

m

d'z

dr

+

dz

C

di+

kz - 0

... (28.9)

It can be shown thai tbe general solution is of the form

Z _ C 1 eSt' + Czeszt where

... (28.10)

'I-w.(-D+~

.[28.11(a)]

'2-w.(-D-~

... [28.11(b)]

and rolf '" natural frequency, D '" damping factor ( :: c!cc-). and Cc = critical dtlmping ( :::: 2: ...rmf) If D > 1. the system is overdampcd and the motion is aperiodic, If D = I, the system is critically damped, which also gives aperiodic motion. If D < 1, the system is underdampcd and the response is periodic, as shown in Fig. z I ..·· • 0 _ 28.6 (b). Only undcrdampcd systems are of practical importance in the design of machine % ----k foundation<>. - c Eq. 28.11 can be written as (b)

rJI-f!' ~------

'I -

w.(-D + i

VI - Ii')

'2 -

w.(-D - i

~ .. 128.12(b)]

... [28.12(a)]

t

(.) Fig. 28.6.

where


z _ e-Dw.t

[c

1

d(1-d)l-J.w"t +

cze-i(l-Dllt.\!w.,]

.(28.13)

Let us make the following substitution.

w",,-w.~

... (28.14)

Therefore,

z .. e-Dw• t [(C) eiw,..,t + Cze-iw""t]

or

z .. e-Dw• t [(C) + Ci)

COS

z ._ e-Dw• t [AI cos wild'

,+ Azsin WNi']

Wild t+ i (C) - Ci) SinWNltj

... (28.15)

1be term Wild is known as damped na:u.ral frequency. Sq. 28.1 can also be written as

z .. e-Dw·t [A coo (Wild

t -

9-)

J

... (28.16)

The term e~' gives an aperiodic exponential response; whereas, the term A cos (w"a /- a) indicates a periodic sin~idal response. The net result is a periodic but gradually decreasing motion [Fig. 28.6 (b)].

For the undamped system,

$

D .. 0, z - A cos (00" t

This is the same Eq. 28.4 (b) .

-

a)

MACIIINE FOUNDA:nONS

761

28.7. FORCED VIBRATION fig. 'l1l7 shows 0 rigid mllSS with a single degree of freedom. The system is dumped and subjected Lo an exciting force F (I). llle equation of motion can be written as

m

d'z dJl

+

dz

C

di

+kz- F(t)

... (28.17)

(b)

(.) Fig. 28.7.

Lei us nssume thnt the exciting force is sinusoidul of the form F(l) - FI} sinrot where F" = magnitude of the exciting force> w = circular frequency of exciting force.

1/1';: d,

lbus

+ c

~d, + k

z - FI} sin ml

... (28.18)

lllC solution Qf Eq. 28.18 is z _ e- Dw• t [A COS (O),w 1- a)] +

F" sin(wt - 8) 2 III W~2 + c (.i

•.. (28.19)

Y(k _

The first pun of the solution is transient nnd dies Oul
responsc. Thus

z - V(k _ Substituting

(On

-

Vk7iii,

and D - 2

/11(02)2

+-

C2{J)2

&' F9 sin (wt - B)

vl1' (1- ,'2.)2 + 402,2'; where

T

= frequency rutio = w/w n

For undamped system,

=

C

=

... (28.20)

... (28.21)

•

= 0 lind D

= O. Therefore,

F"sin(OOl - ~) k(l - -,:»

... (28.22)

Whcn r 1.0,;.('., <\l 00", response is inifinilc. 'fhe condition is k.nown as As tin ideal und:tmped system is non-existent, damping always exists and the response is finite. Ilowcver, when opernting frequency (00) is close to the natural frequency (w,,), the response is vcry high. To avoid this


762

condition, the operating frequency should not be close to the natural frequeocy. For a safe design. the frequency ratio is nonnally kept outside the critical range of 0.4 to 1.50. The magnitude of the displacement is given by

1'1-

_F_._ _ - - F - . m(";-.,')

... [28.23(a)]

k(!-?)

Iz I _

In a general case,

F,Ik

... [28.23(b)J

";(1 - ?l'+ Hi'? 1be static displacement under a force F" is given by

F.

Z'f -

. .. (28.24)

k

The ratio of the magnitude of the steady-stale displacement of a forced system to the static displacement is known ~ magnification factor (M). Thus 1, 1 F,/k

0'

M -

z;; -

At_

..;(!

(F.lk) ";(1 _ ,')' + Hi'?

-y')'

(28.25)

4d,'

+

Thus z - MZ'I Fig. 28.8 shows the variation of the magnification factor M with' for different values of D. It may be noted that the magnification is high for the value of r between 04 and 15

,

t,

, , , 0

~o . o

l

~o,,~ ~~ I

0.,

,.,

10

'.0

Fr~quency

ratio

l-5

,D

,.,

4Jl

Ir}-....-

Ag.28.8.

Force Transmissibility T. A damped forced vibratory system derives its support from the foundation. The force transmitted can be expressed as FT - C ~ + Ia

Let us express

";(1_,')' where

...(28.26)

z .. (F/k)sin(wt - P) as Bsin(wt _ P), B _

+

.I),?

F,!k

";(I -

">' + 4d?

... (28.27)

7'3

MACHINE FOUNDATIONS

Eq. 28.26 can be written as Fr

.. cBwcxs(wI - j}) + k Bsin(wl -

FT - B

vI<'. +

~)

c'w' [oos(wt - ~ - y))

where 't .. 130- 1 (klcw). The magnitude of FT is given by

IFrI-B~ IFTI _

(F'/k)~ V(1 -'1' + 4li','

or IFTI- F. M~ ... (28.29) Force transmissibility (1) is defined as the rntio of the force transmitted to the applied force Thus

T ..

~

- MVl + (2D,)'

... (2830) Lik.e magnification factor (M). the trnnsmissibilily is also a function of rand D. The plot is similar in shape to that shown in Fig. 28.8. 28.8. VIBRATION ANALYSIS OF A MACHINE FOUNDATION Although a machine foundation has 6 degree of freedom, it is assumed to have a single degree of freedom for a simplified analysis. Fig. 28.9 shows a machine foundation supported on a soil mass. In this case. tbe mass of the system Jumps together the mass of the machine and the mass of foundation. lbe total acts at the centre of gravity of the system. The mass is under the supporting action of the soil. The mass elastic action can be lumped together into a single elastie spring with a stiffness k. Ukewise, aU the resistance to motion is lumped into the damping coefficient c. lous the machine foundation reduces to a single mass baving one degree of freedom, as shown in Fig. 28.7. lbe analysis of damped. forced vibration, discussed in Sect. 28.7, is, therefore, applicable to the machine foundation. Determination of Parameters. For vibration analysis of a machinC? foundation. the parameters m, c and Ie are required. lbese parameters can be determined as under. (1) Mass (m). When a machine vibrates. some portion of the supporting soil mass also vibrales. The vibrating soil mass is known as the participating mass or in-phase soil mass. 1bcrcfore, the total mass of the

m,

m,

FOUNDATION

\ \

:

I .... ______ -......... Fig. 28.9.

I

t

/~8OUNDA.RVOF VIBRATING SOIL

164


system is equal to the mass of lhe [ound:ltion block and machine (mf) and the mass (ms) of the participating soil. Thus m - /til + /tis ••• (28.31) Unfortunately, there is no rational method to determine the magnitude of m,. It is usually related to the mass of the soil in lhc pressure bulb. The value of m, generally varies between zero and m" In other words., the total mass (m) varies between m, and 2 in mast cases. (2) Spring stiffness (k). The spring stiffness depends upon the type of soil, embedment of the foundatioo block, the contact area and the contact pressure distribution. The following methods are commonly used. (0) Laboratory test. A triax.ial lest with vertical vibrations is conducted 10 dctenninc Young's modulus E. Alternatively, tho modulus of rigidity (G) is determined by conducting the tcst under torsional vibration, and E is obtained indirectly from the relation E = 2 G (1 + ~). where ~ is Poisson's ratio. The stiffness (k) is determined as

In,

k _ A£IL

=

... (28.32)

=

where A cross-sectional area of the specimen, L length of the specimen, (b) Barkan's method. The stiffness can also be obrnincd from the value of E using the foUowing relation given by Barkan. ...(28.33) where A = base area of the machine, i.e. area of contact. (e) Plate load test. A repeated plate load test is conducted and the stiffness of the soil in the test (kp) is found as the slOpe of the load-deformation curve. The spring constant k of the foundation is determined as under. (I) For cohesive soils ... (18.34) (il) For cohesionless soils

,

k _ k (B+ 0.3 ) p

... (28.35)

/J p -+ 0.3

where B is the width of foundation and Dp is the diameter of the plate. Alternatively, the spring constant can be obtained from the subgrade modulus (k.),

a
k - ks . A ... (28.36) where A = area of foundation, (d) Resonance test. The resonance frequency (f,,) is obtained using a vibrator of mass m sct up on a steel plate supported on the ground. The spring stiffness is obtained from the relation '

Ift-~ ItI· or (3)

because (I) (il)

f;..fklm

k - 41r?{"m ... (28.37) Damping constant e. Damping is duc to dissipation of vibration energy. which Q(X:urs mainly of the following reasons. Internal friction loss due to hysterisis and viscous effects. Radiational loss due to propagation of waves through soil

The damping factor D for an undcrdamped system can be determined in the laboratory. Vibration response is ploUed and the logarithmic decrement (, is found from the plot, as

MACHINE FOUNDATIONS

765

6 _ log (z,lz,)

... (2838)

where ZI and z2 are amplitudes of two consecutive cycles of an amplitude-response curve. The damping factor D and the logarithmic decrement 6 are related as

6_~ ~

... (2839) ... (28.40)

or

The damping factor D may also be obtained from the area of the hysterisis loop of the load displacement curve. as

D _

AI':

... (28.41)

where W = towl work done, 6 W .: work lost in hysterisis. The value of D for most soils genernlly v,lries between 0.01 and 0.1. 28.9. DETERMINATION Of' NATURAL FUEQUENCY The natural frequency of the foundation-soil system can be determined using the theory of vibrations discussed in Sc<.t. 28.7. The equation of motion, neglecting damping, is 111

where

III .:

~/

+ Ia: ..

F"sin{o(

, .. (28,42)

mass of machine, foundation and the participation soil, k '" equivalent spring constant of the

soiL -fbe methods for delermirwlion of k
III

h
The natuml frequency of the system is given by W il

..

.ff7m

... (28.43)

t.fk7m

... (28.44)

where wIt is in mdians per seoond.

In ..

Also where

In

is in cyclCS per second.

. .. (28.45)

Thus

where

III, = mass of machine nnd foundation.

and

III,

= mass of the participating soil mass.

Barken (1962) gave the following relation for the natural frequency.

W.# "

where

Comparing with Eq. 28.43,

... (28.46)

III

e.. '" coefficient of elastic uniform

compression, A .: oontaCl area of foundation with soil.

k .. C" )( A

... [28.46(0)]

lhe maximum amplitude is given by z",,- _ _F_·o_ _

. where F" '" exciting force .

11100;(1 -,.2)

... (28.47)

766

roiL MECHANICS AND FOUNDATION

ENGI~EERING

The coefficient of clastic unifonn compression (Cit) depends upon the type of soil. It can be obtained from the following relation. E 1 ... (28.48) C•• 1.13 (I _ 7.t:

"'>

As it is evident. the coefficient varies inversely proportional to the square root of the base area of the foundation. Thus

(A,)"

...

(C,J, (28.49) (C.),· A, Toole 28.1 gives the recommended value of CIt for A = 10 m2 for different soils (Barkao, 1962). Thble 28.1. Coefficient of Elastic UnIform Compression Coe/ficielll of elas/iC

Soil Calegory

III

uniform compression

Soil type

(C..)kNlmJ

Weak soils (clays nod silty days wi th sand in II pla<>lic Slatc; clayey sillS; soils of categories II nnd III with laminae of organic silts and of peat).

uplO 150

Soils of medium strength (clays and silty clays with days and silty dllYs with sand close 10 the plastic limit; sand) Strong soils (clays lind silty clays with SIlnd of

15~350

350-500

hard consistency; gravels and gravelly sands; I~andl~ialsoils).

IV

Rocks

>500

28.10. DESIGN CRITERIA FOR FOUNDATIONS OF RECIPROCATING MAClIINES The following design criteria for the foundation of reciprocating machines should be satisfied. (IS : 2874 Q}-I982). (1) 'The machine foundation should be isolated at all levels from the adjoining found3tions (Sec Sea. 28.13). (2) The na1ural frequency of the foundation-soi l system should be higher than the highest disturbing frequency and the frequency ratio should be less than 0.4, as far as possible. However, if it is noI possible to satisfy above criterion, the natural frequency should be lower than tbe lowest dislUrbing frequency nnd the frequency rutio should not be less than 1.50. (3) The amplitude of vibration should be within the permissible limits, as given in Fig. 28.4 or any Specified permissible limits (For example, chart given in IS : 2874 (1) - 1982). For most soils, the limiting amplitude for low speed machines is usually taken as 200 micron ( = 0.2 mm). Acrording to another criterion. the amplitude in mm is limited to 4// for frequencies less than 30 hertz and 125/t for higher frequencies, where / is frequency in hertz (cycles/sec). (4) Concrete block foundations should be used. However, when the soil is not suitable to support block foundation, cellular foundation may be used. (5) 'The size of the block: in plan should be larger than the bed plate of the machine. There should be a minimum aU-round clearance of 15 mm. The lotal width of lhe foundation measured al right angies 10 the shaft should be least equal to the distance between the centre of the shaft and the bottom of the foundation. (6) The eccentricity of the foundation system along X-X and Y-Yaxes should not exceed 5% of the length of the corresponding side of the conlact area.

MACHINE FOUNDATIONS

767

(7) 1be combined centre of gravity of machine and foundation should be as much below the top of foundation as possible. In no case, it should be above the top of foundation. (8) The depth of foundation should be sufficient to provide the required bearing capacity and to ensure stability against rotation in the vertical plane. (9) The strcsse.1O in the soil below the foundations should not exceed 80% of the allowable stresses under static loads. lbc base pressure is limited to half the nonnal allowable pressure (q"J in extreme alSCS. (10) Where it is not practicnble to design a foundation to give satisfactory dynamic response, the transmitted vibrations may be reduced by providing anti-vibration mountings either between the machine and the foundation or between the foundation and the supporting system. (11) The macpine should be anchored to the foundution block using a base plate and anchor bolts. Bolt holes should be backfiUed with concrete and the space below the plate should be filled with I : 2 cement mortar. (12) A number of similar machines can be erected on Individual pedestals on a common raft. The analysis for such machines can be made assuming thai each foundalion acts independently with an area of foundation equal to that ob1.1ined by dividing up the mft into sections corresponding to separate machines. 28.11. REINFORCEMENT AND CONSmUCnON DETAILS 3 (1) The reinforcement in the concrete block should not be less than 25 kglm • For machines requiring special design consideration of foundations, such as machine pumping explooivc 3 gases., the minimum reinforcement is 40 kglm . (2) Steel reinforcement around all pits and openings shall be at least equal to 0.5 10 0.75% of the cross-sectional area of the pit or opening. (3) The reinforcement shall run in all the three directions. lbe minimum reinforcement shall usually consist of 12 mm bars at 200 to 250 mm spacing extcnding both vertically and horizontaly near all faces of the foundation block. lbe ends of all bars should always be hooked. (4) IT the height of the foundation bkx:k exceeds one metre, sbrinkage reinforcement shaU be placed at suitable spacing in all thc three directions. (5) The cover should be a minimum of 75 mm at the bottom and 50 mm on sides and the top. (6) The concrete shall be at 1c.1st M-15 with a characteristic strength of 15 N/mm2. (7) The foundation block should be preferably cast in a single, continuous operation. In case of very thick blocks (exceeding 5 m), construction joints can be provided. 28.12. MASS OF FOUNDATION Heavy foundations eliminate excessive vibratiOll'i. Manufactures of machines sometimes recommend tbe mass of foundation required for the machincs. However, the mass recommended are generally empirical and based Largely on experience. Couzens (1938) gave the ratios of foundation mass to engine mass suitable for various types of machines (See Table 28.2). These ratios can be used for rough estimates. 28:13. VmRATION ISOLATION AND CONTROL VIbrations may cause bannful effects on the adjoining structures and machines. Besides, these vibratioos cause annoyance to the persons working in the area around the machine. However, if the frequency ratio is kept outside the aitical range of 0.4 snd 1.50, and the amplitude is within the pcnnissible limits, the harmful effects are considerably reduced, especially if the system is damped. Transmission of vibrations can be controlled and the deterimental effects considerably reduced by isolating eilhcr the source (active isolation) or by protecting the receiver (passive isolation). The following measures are generally adopted. (l) The machine toundation should be localed away from the adjoining structures. This is known as geometric isolation.


768

'flIble 28.2. Ratio or :FoundnUon

1.

2.

3. 4. 5. 6. 7.

8. 9.

mllSS

to Engine muss

Types 01 ElIgjll~

S.No. Gas Engines l---Cylindcr 2-Cylinder 4-Cylindcr 6-Cylinder 8-Cylincler Diesel Engines '2-Cylinder

R(lIio

3.0 3.0 2.75 2.25 2.0

2.75

4-Cylinder 6-Cylindcr 8-Cylincicr Rotary converter Vertical compound steam engine coupled 10 generator Vertical lriple-e"JXInsion stc;Jm cngine coupled to generator lIorizonlnl cross-compound coupled to generator liorizonlal stcllm turbine coupted to generator Vertical gas engine coupled to gellerator Vertical diesel engine coupled to generator

2.' 2.1 1.9 0.5100.75 3.8 35 3.25 3.0104.0 3.5 2.6

The amplitude of surface waves (R-waves) reduces with an increase in distance. A considerable reduction in the amplitude is achieved by locating the foundation at a great depth, as the R-waves also reduce considerably with an increase in depth. (2) Additional masses known as dampen; are attached to the foundations of high frequency machines to make it a multiple degree freedom system and to change the natuml frequency. In reciprocating machines, the vibnltions are considermly reduced by counterbalancing the exciting forces by att..-lching counterweights to the sides of the crank. (3) VibrntioM are considerably reduced by placing absort>er.>, such as rubber mountings, fellS and corks between the machine and the base. (4) If an auxiliary mass with a spring is attached to the machine foundation, the system becomes a two-degree-freedoni systcm. The ' method is especially effective when the system is in resonance. (5) If the strength of thc soil is increased by chemical or cement stabilisation, it increases the natural frequency of thc system. The mcthoo is useful for machines of low operating frequency. (6) 'Inc natund frequency of the system is modified by making struaural changes in foundation, such a') connecting the adjoining foundations, changing the base area or mass of foundation or use of attached slabs. (1) The propagation of waves can be reduced by providing sheet piles. screens or trenches. ILLUSTRATIVE EXAMPLES

DlustraU"c Example 28.1. Determine the lUllurai frequency of a machine foundation having a base area 2 m x 2 m and a mass of 15 Mg, including the mass of the mochine, Taking e" = 4 x 1(1 kN/mJ. Solution. From Eq. 28.46, Expressing

ell in

00,. -

#m

N/m J and m in kg,

MACHINE FOUNDATIONS

769

w" ..

or

V

(~ X

1

4 )( 10 X 15 x 1

2) .. 103.28 rad/sec

t-

103.28 _ 16.43 cps (hertz) or h ntustratJve Example 28.2. The natural frequency of a mochine foundtuwn is 4 hertz. Delermine its magnification at the oper(Jling frequency of 8 hertz. ToU damping factor (DJ os 0.30.

M _


1

V(1 -?)'

4Ii'.:'

+

In this case, r - 8/4 - 2. Therefore, M _

1 _ 0.31. - 2')' + 4 x (03)' x (2)' D1ustrallve Example 28.3. The exciting force of a machine is 100 kN. Determine the transmitted force if the naJural frequency of the mac/tine foundation is 3.0 Hz. Take D = 0040 and the operating frequency os

V(1

5 Hz. Solution. From Eq. 28.29.IFr l

-

F"MYI + 2Dr)2

Moo

wbere

1

4Ii'.:'

V(1 -

r')' +

V[1 -

(5/3)'f' + 4 x (0.4)' x (5/3)'

_

1

I

_ 0.45

V

Fd - 100 x 0.45 1 + (2 x 0.4 x 5/3)' _ 15 kN D1ustratlve Example 28.4. Determine the coefficient of uniform compression if a vibration tQl 011 a block 1 m X 1 m )( 1 m ga~ a resonance ~ncy of 30 Hz in the vertical directWn. ~ mass of the oscillator used wos 60 kg. Solution. Mass of foundation block. .. (l )( I x 1) x 2400 - 2400kg Total mass .. 2400 + 60 .. 2460 kg Therefore,

From Eq. 28.46,

t- -'h In this cme,

30 ..

...!... h

e.... 8.74

# Ve..

m

,-0---'----'x (1 x 1)

2460

)( 101 N/m] .. 8.74 x 10" kN/m]

D1ustrath'e Example 28.5. A 2.50 Mg vertical compressor- foundation system is operated at 40 Hz. ~ soil at the site is medium stiff clay (e.... 4 x 104 kN/mJ). Determine the natural frequency and the magnifu:alwn focwr, ossuming m, = 0.2 m" ~ base area is 2.5 m2. Take D .. O. Solution. 1bI:al mass = 2.5 + 0.2 x 25 From Eq. 28.46,

= 3.0 Mg = 3

x 1al kg

no


# . ....!.... y4

f _ ...L

m

2" M=

From Eq. 28.25.

2:t

1

X

10

X

3)(103

2.5

29.06 Hz

1

V(1

4Ii'Y'

_ ,')' +

,===:::::;::::;::=====-

V{1- (2~()5 )' r 4•O. (2~ f

1.12

+

illustrative Example 28,6. In a test block oj the size 1.5 m )( 1.0 m x 0.75 m, resonance occurs at a frequency of 20 cycles per second in the vertical vibration. Determine the coefficient of elastic uniform compression (C.J if the mass of oscillator is 70 kg and the force produ.ced by it at J 5 cycles per second is 1000 N. Also compute the maximum amplitude at 15 cycles per second Solution.

W,,'"

Mass of oscillator

:::: 70 kg

)( 1.0 x 0.75 x 24.00

= 2700 kg

'" 70 + 2700 '" 2770 kg = 1.5 x 1.0 = 1.5 m2

Total mass

Contact area From Eq. 28.46,

2x/,," 2" x 20 .. 4On;

= 1.5

Mass of block

00" ..

4Ol't ..

vc:;A7iii

ve"

x 1.5/2770

or ell .. 29.16 x lcfN/m 2 .. 29.16 )( 103 1eN/ m) From Eq. 28.47, maximum amplitude

F.

z,.,., -

moo; (1

_

Y') -

1000 2770 x (40 x)' [1 - (15120)']

.. 5.23 x 10-5 m .. 0.0523 mm mw;tmllve Examples 28.7. A foundation block of weighl 30 kN rests on a soil for which me stiffness may be assumed as 25000 kNlm. The machine is vibrated vertically by an exciting force of 3.0 sin (30 t) leN. Find the natural frequency, natural period, natural circular frequency and the amplitude of vertical displacement. l'he diJJ1Iping factor is 0.50. Solution. The exciting force F Q sin <0/ is 3.0 sin (30t). Therefore, F Q

=3.0 leN and ro =30 mdian/sccood

Now from Eq.--28.43,

roll

_ ...fk7m

(0" ..

From Eq. 28.44,

f,. ..

V

t

25000

~

1Ii

(30 x 10 /9.81)

..;vm ..

T ..

Tn ..

r ..

00/00" ..

..

90.42 rad/s

14.39 cycles/sec

0.069 second

30/90.42 .. 0.33

m

MACHINE FOUNDATIONS

AI-

From Eq. 28.25,

Static deflection, 66, Amplitude

_

!

V(I

-

V[I

- (0.33)' l' + 4 x (0.5)' x (0.33)'

(i)' +

4i:l-,' 1

.. 1.05

Folk _ 3.~OO .. O.012cm

- AI 661

-

1.05 x 0.012 .. 0.013 em

PROBLEMS

A. Numerical 28.1. [)ctermine the natural frequenc.;y of n nmchine foundation having a base area 2 m x 2 m and a mass of 10 Mg, assum ing thai the soi l mass participating in the vibration is (a) negligible (b) 20% of the mass. 1hke C u '" 104kNlm2.

[Ans. 17.43 cps, 15.92 cps]

28.2. Resonance occurs at a frequency of 20 cps in the vertical vibration of a test block (1m )( 1 m )( 1 m). Calculate the coefficient of clastic uniform compression (Cu). The mass of the oscillator is 50 kg. 4 3 [Ans.. 3.87 x 10 kN/m ] 23.3. In Problem 28.2, if the force produced by the oscillator at 10
= 5 Mg 300 r.p.m. 20 Mg

= :: = ::

25 Mg 4 60 X 10 ltN/m

Determine !he naturnl frequency and maximum amplilllde. l11ke D .. 0.1. 100 unbalanced verticil force is 12 kN. [Ails. 17.43 cps; 0.022 mm]

B. Descriptive Type 28.5. Explain the following terms: Naturnl frequency; period; resonnnce; magnification.

28.6. Discuss the use of single-degree.frccdom sySlcm in thc analysis of machine foundations. Wh!lt are its limillltions? 28.1. Describe the methods for the determination of the mass, spring constant, damping factor and the mass of participating soil. 28.8. Briefly explain the I3arknn method of machine foundation design. 28.9. What is meant by vibration isolation? How is it done ?

28.10. Discuss criteria ror the design of foundation in the follOWing cnses. (a) Free vibration without damping. (b) Free vibration with damping. (C) Focced vibration without damping. (d) Forced vibration with damping.

C. Multiple Choice Questions 1. If the circular frequency (00) is 30

11: radians per second, the time period is (a) 1 second (b) 2 soconds (c) 4 seconds (d) none of above. 2. '!be frequency of.a system increases with (a) an inaease in stiffness of system


772 (b) a decrensc in the muss of sy~tem

(c) both (a) :'lnd (h) (tf)neither(a)nOf(l,) 3. Magnilicmion factor (M) is very high if the value of frequency ratio (r) is (a) less than 0.4 (h) bctWCt!1I 0.4 and L50 (c) greater than 1.50 (d) unity. 4. For most soils. lh~ limiting ampli tude for low speed machine.~ is usually (a) 0.1 I11Ill (h) 0.2 mm (c) 0.5 mOl (J) 1.0 mm 5. For die.~eI engines. the rmio of foundmion mass 10 cngine mmt~ is (a) less than 2.0 (h) between 2.0 and 3.0 te) between 3.0 :lnU 4.0 (d) Kremer than 4.0

[Am. I.

(c).

2.

(c).

3. (b). 4. (b). 5. (b)1

Objedive type Write whethcr the following statements arc true ur fube (a) The speed o f tu rbines and rotary c(Jmpres.~ors is usually less than 600 r.p.m.

For lathes. gcner:llly block-type fOUndation is provided. te) In gencml. a rigid hlock foundat ion has 6 degrees of freedom. «(I) '111C stresses in thc soil below the machine foundation should nornlllily be less than 50% of Ihe allowable strou under static loads. (t') The minimum quantity of reinforcement in mllchine foundation is 15 kg/m). if) The minimum cover for reinforcement at the bottom of the machine foundation is 75 mm. (g) The depth of ground water table below the machine foundation should not be leu Ihan the width of foundation . (lI) The nmchine foundat ions are take n lower than the level o f the foondation of adjacent buildings. {Ans. True (e), 00. (h)1

(b)

29 Pavement Design 29.1. TYPES OF PAVEMENTS A pavement is a hard crust oonstructed over the natural soil for the purpose of providing a stable and even surface (or the vehicles. The pavement supports and distributes the wheel loads and provides an adequate wearing surface. Pavement are basically of 3 types: (1) Flexible pavements, (2) Rigid pavements and (3) Semi-flexible prIlvcments. (1) F1exlble Pavement. The flexible pavement is buill up in several layers, as shown is Fig. 29.1 (a). The natural soil beneath the pavement is known a9 subgrode. Sub-hase is built over the sub-grade and the base is constructed over the sub·base. The lop layer is known as surfacing, which is usually bitumen. The flexible pavement can resist ooly very small tensile stresses because of limited rigidity. Any derfonnation of the sub-grade results in a oom:sponding change in the surface of the pavement.

', ' -.it.. SURFACING ,,:: ~ .:.:: .::..: flOO_OOa'et°O . ' 0 · BASE • ., 0 0 0

o ~o

&9

'0.0.0:,,°

~~;;c~,.:iIt;

SUB-GRADE (a)

Fig. 29.1 (a) Flexibte PlIvemenl.

(b) (b) Rigid Pavemenl.

(2) RJgld Pavement. lbe rigid pavement is made of cement concrete. As the concrete layer is quite strong, the sub-base may not be required [Fig. 29.1 (b)). The rigid pavements have high flextural strength and can resist very high tensile stresses. The pavements are capable of bridging small depressions in the subgrade. (3) Seml-nell:lble pqnmenL A semi-flexible pavement has flexural rigidity in-between that of a rigid pavement and a flexible pavement. Such pavements are usually made of pozwlanic concrete, lean concrete, or soil-amlcot in the base cowse or sub-base. As the flexural strength of such layers is limited, the pavement can resist only moderate tensile stresses. The chapter deals with the methods of construction and the design of various types of pavements. As the space is limited, the treatment is neressarily elementary. For more details. a rererence may be made to advanced works on pavement design. 29.2. BASIC REQUIREMENT OF PAVEMENTS The basic requirement of a good pavement is to provide a stable. non-yielding surfaoe for the movement

774

SOIL MECHANICS AND FQUNO..rnON ENGINEERING

of heavy vehicles. 'lbc pavement surface should be even along the longitudinal profile to have least rolling resistance so thai fast, heavy vehicles c:m move safely and cornforwbly. Unevenness of the surface causes vertical oscillations and thus increases wear and lear and the fuel oonsurnption ~r vehiclcs. 1t ::'~.:~:) nckls \0 the discomfort and fatigue to the passengers of fast moving vehicles. 'fbe pavement carries the wheel loads and distributes it over a wide area on the subgradc. Conscqucnlly, the stresses transferred \0 the subgrndc are considerably smaller than the contact pressure. The distribution of the lond depends upon the thickness nnd the characteristics of the material used in the pavement. A pavement lnyer that distributes the wheel lood through the largest area per unit thickness of the layer is the most

efficient. The pavemcnt laycrs .should not be ovcr-stressed. Evcn for a wcll- dcsigned pavement, there would be a small, temporary deformmion when loads pass over it. These deformations must be kept within the permissible limits. If the pavement is not properly designed, rtpc.'lted applications of loa~ may cause excessive deformations, compaction and consolidation of the subgrade and even failure of the whole pnvement. . The pavement may bc constructed over an embankment or in a CUlling. Tbe pavement should be constructed above the maximum level of the ground water table to keep it dry. Moisture varimion and frost action are the main causes of deterioration of the subgrade. On the other hand, when the water content is decreased, shrinkage crackS develop, which cause differential movement in rigid p3vements and cracks in flexible pavements. The pavemenr.s should be provided wilh a suitable drainge system. In case of rigid pavemenr.s. the temperature ~md shrinkage stresses should be properly controlled. In case of flexible pavements, there should be a good bond between the individual p.1niclcs of tbe materials used and also between the surfacing and the base to check stripping or breaking up of the pavement.

29.3. FUNCTIONS OF DIFFERENT COMI)ONENTS OF A PAVEM"ENT Different components of a pavement have the foUowing characteristics and functions. (1) Subgrade. The subgrndc is a laycr of n:'ltuml soil prepared to receive the layers of the pavement. The subgrade should be strong enough to take up the stresscs imposed due to loads without shear failure Of excessive deformations. It is the general practice to compact mIeast ~op 50 em layer of the subgrade under controlled conditions of optimum water contC£lt at the maximum dry d~nsity (see chapter l4). It is essential to evaluate the strength properties of the subgradc required. As the loods are ultimatcly received by the subgrade. if it is weak. it wiU fail. The soil is gcnerally treated to inaease its strength and ID improve its properties. (2) Sub-base and Base Courses. 'Ibcsc courses provide a medium to spread the wheel loads to tbe subgrade. Tbe courses usuaUy consist of broken stones, bricks or aggregates. Boulder stones, bricks on edges and stabilised soils arc also used for sub-bases. However it is preferable to use small size graded aggregatcs bcc..'luSC large stones and bricks have a tendency to penetrate the wet soil and cause undulation and unevenness in the pavement. As the stresses in a sub-base arc much lower than those in the bast; the material used is inferior to thn! in the base. The sub-base is also known as soling. The base and sub-base in a flexible pavement improve the lood- supporting capacity of the subgrade by distributing the load over 8 large area. In a rigid pavement, the base course helps in preventing pumping OUI of the soil from below. ·lbcse also protect the subgmde against frost action. (3) Surface Course. The purpose of the surface course, also known as a wearing murse, is to give a smooth riding surface and to resist pressure exerted by wheels. 'The surface murse also provides a water·tighl barrier against the infiltration of surface water. In flexible pavements, a surface course usually consists of a bituminous surfaCing. In rigid pavements, the cement concrete may act us a base course as well as a surful-c course. . There are various types of surface tre..1tments. depending upon the availability of materials, and plants and the magnitude of the load.

PAVEMENT DESIGN

775

29.4. FACTORS AFFECTING PAVEMENT DESIGN The factors affccting the design of p..wements are discussed below. (1) Wheel Loud. The thickness of the pavement depends upon the design wheel load. The design wheel load is selected after considering the actual wheel loads of the various vehicles, contact pressure, load repetition. the dynamic effects and many other factors. The design wheel load will depend upon the multiple-wheel load mscmbly. as in the dual or multiple wheel loads. (2) Subgrade. the thickness of the pavement depends upon the propcttics of the subgrade. A thicker pavement is required over a weaker soil. The thickness also depends upon the stress-strain characteristics of the soil under static and repeated loads. As the strength and the volume change of the soil depend upon the moisture changes, the worst conditions should be considered in the design. (3) CUmate. llle climalic fadOrs. such as rainfall and temperature changes. affect the properties of the soil. The rainfall affects the moisture changes in the soil whidl affect the strcogth. 1be temperature changes affect the p:ivement. In flexible pavements, the choice of the bituminous binder depends upon the temperature. In rigid pavements. the warping stresses are caused by temperature changes. U the tcrnpernture is likely to fall to the freezing point, the possibility of the frost action is to be considered. (4) l'avement materials. The stress distribution in the pavement layers depends upon the behaviour of the materials used. 'The fatigue behaviour of the pavement materials and their durability under adverse conditions should be considered. Aggregates bear the stresses occurring in the pavement and have to resist wear due to abrasive action of traffic. '£be aggregate shouW he hard. strong and of the required size and gradation to bear the stresses. (S) Location. The design of the pavement should be done considering its location with respect 10 the ground surface. The heigh! of embankment,. the depth of cutting and the level of water table should also be considered. All these factors aITro the pcrfonnance of the pavement. (6) Miscellaneous factors. There are some special factors which affect the design of pavements. }-o.example. in case of semi-rigid pavements, the formation of shrinkage crocks. the aack pattern and the mode of their propagation should be considered in its design. Preving ring

29.5. CALIFORNIA BEARING RATIO TEST

California bearing Ratio (CBR) test is a type of test developed by the California Divisioo of Highways in 1929. The lest is used for evaluating the suitability of subgrnde and the materials used in sub-base and base courses. 'The test results have been oorrelated with the thickness of the various materials required for flexible pavements. The test may be conducted on a prepared specimen in a mould or on the soil in-situ condition. The Iaborntory CBR apparatus consists of a mould ISO mm diameter and 175 mm high. having a separate base pmte and a collar (Fig. 29.2). The load is applied by a loading frame through a plunger of 50 mm diameter. Dial gauges are used for measurement of the expansion of the specimen on soaking and for measurement of penetration. It may be noted that with the disp1a
01el

gouge ' Ptungtr

Bast plott

~150mm--..j

Fig. 29.2. C.B.R. Test AppnllL.IS.

776

SOIL MECHANICS AND

FOUNDAllO~

ENGINEERING

The test consists of causing the plunger to penetrate the specimen at the rate of 1.25 mm per minute. The loads required for a penetration of 2.5 mm aod 5.0 mm are recorded by a proving ring attached to the plunger. The load is expressed as a percentage of the standard load at tbe respective deformation level, and is known as the CBR value. Standard load values are for crushed stone and are given in Table 29.1. The CBR value is determined corresponding to both 2.5 mm and 5.0 mm penetration, and the greater value is used for the design of flexible pavement. CBR value -

St:~~r~~d

x 100

... (29.1)

Thble 29.1. Stundard Load Values S.No.

Penetration (mm)

Standard Load (kN)

1. 2. 3.

2.5 5.0 7.5

4.

10.0 125

13.44 20.16 25 .80 31.20 35.32

5.

The load-penetration rurve is drawn as silOwn in Fig. 29.3. The load corresponding to 2.5 mm and 5.0

mm penetratioo values are taken from the plOI curve (1), Hod the value of CBR detennined using Eq. 29.1.

•. of----I--+-~...,,:::< '-+--f----1 6.0+---_+~L-.-+--L--_+=,,~+------j

.~"'~ ;~+lS.------"'1_,1'.;;-----""'7.J,.'--*"."o- ---,!I1<.'-----;!ls.Q LCorretl~d

zero

Penetration

(mm) ----..

Fig. 29.3.

Generally. the CBR for 2.5 mm pene.tration is high. However, if the eBR for 5.0 mm penetration is greater than that for 2.5 mm penetration, the test is repeated. If tbe results are unchanged, the value for 5.0 mm penetration is used for defining CBR value. In some tests. there is an upward concavity of the load- penetration rurve. This may be due to the following reasons. .. (I) The plunger surface does not rome in full rontact with the top of specimen. (i!) The top layer of the soil is very soft. In the case of the initial upward concavity, the corrected zero is obtained by drawing a tangent to the curve at the point of the greatest curvature, as shown [or curve-2 in Fig. 293. The points corresponding to 2.5 mm and 5.0 mm penetration are measured from the corrected zero. To simulate worst conditions in the field, the soil specimen is kept submerged io waler for about 4 days before testing. If the test is to be conducted on an unsoaked specimen, the moulding water content should be equal to the equilibrium water content which the soil is likely to attain after tbe construction of the pavement. To simulate the effect of overlying materials, the specimen is covered with surcharge mass in the test.

PAVEMENT DESIGN

777

The minimum surcharge mass for the tests in the mould is specified

if;

45 kg. TIle annular surcharge mass

of 147 mm diameter are used for this purpose. [See O1apter 30 (Sect. 30.19) foc laboratory experiment1. 29.6. DESIGN OF FLEXIBLE PAVEMENTs A flexible pavement consists of a number of !aym in whidt the stress is transmitted by point-ta-point contact. TIle maximum intensity of stress occurs in the top layer of the pavement. 1berefore, the superior materials are used in the top layer. There is no perfect rational method for the design of flexible pavements. The design methods can be classified broodly as empirical or semi-empirical.

1. Empbicaf methods (a) Group index method. (b) California Bearing Ratio Method, (c) California R·value Methcxl, (d) McLeod Method. 2. Semi empirical methods (a) Triaxial test Method, (b) Bunnisler Method. All these methods are discussed briefly in the following sections. 29.7. GRQUl' INDEX ME'TIIOD The group index (Gl) is a parameter used in the classification of soils by AASHTO system (Chapter 5).

The group index is used in the grading of soils. The higher the value of the group index, the poorer is the subgrade. In tbe group index method (Steel, 1945), the thickness of the base and surfacing is related to the volume of the traffic. Depending up:>O the number of vehicles, the traffic volume is divided into 3 categories. Ughl volume : Less than 50 vehicles/day Medium volume : 50 to 300 yehicles/day Heavy volume : over 300 vehicles/day To determine the thickness of the pavement by this method, the group index and the t:nticipated traffic volumes are frond. The appropriate design curve (8. C or D) is used according 10 the traffic volume (Fig. 29.4) and the total thickness of the pavement (surfacing, base and sub base) is detennined. Curve A gives the thickness of the sub-base required. Curve E gives the additional base thit;irncss which may be substituted for sub-base thickness of curve A. The method is essentially empirical, wbich assumes that the soil with identical group indexes possess

EXCElLENT

(A.l_a) ~-;sfo--;tIOO;;-';'~SO"-;2~OO;"-';2S,",O--+'JO;;;O""'f..so;;-;,~oo;-;!,,"o""s;i,;o;:-o"'S~SO,...,J60'O THICI(N6S (mm) - - - . . .

Fig. 29.4.


778

icientjcal strength after rom pad ion in the field. The method assumes that the subgrade would be compacted to a density not less than 95% of the maximum Proctor dcmily. For base and sub-base materials. the density should not be less than 100% of the maximum Proctor density. 1be water table should be at least 1 m below the surface. 29.8. CRR ME11IOD In this method, the CBn values are used to determine the total thickness of the flexible pavement and tbe

thickness of various layers. Fig. 29.5 give the design curves for different wheel loads and traffic conditions. lbe design curves are based on the data CBR ,0/0 ) - " collected on a large number of pavements which

,

performed satisfactorily. The curves

give the required thickness of construction above a material of a certain CBR value. As it is evident, ibe required thickness of construction above a material decreases as the CDR value increases. The Indian Rood Congress (IRC) has

3

4

,

'567890

II III

100

Ll9h1tratfic~

131·7'5

E 200

~N

U.K. The soaked CBR value of the

~

:~~~ ~~:!~:~

~

tbe pavement

is determined using the appropriate curve. Likewise, the CBR value

1

Wheel load)

>::

:JO

40

""0 '"

-;; ~

~""i"ml'01li' T"'·81

~~~~d~a~ei:~f!~:rh~~ ~:n!:i~~ ~

:: t:~v~!:e:! ~~

10

300

kN Wh,,'lood)

,

!"o

~~Ie:ncs~~-~~ckn=~i~n'l~= ov~~ ~ 50 that material. Obviously, the thickness of +

,L-II

~avy

('5S.!"]

trattlc kN \';'heet

load)

V;

the sub-base is equal to the total thickness 600 above the subgrade minus the thickness of cooslructioo above the sub-base Li kcwise, the thickness of tbe base is determined. The CBR method is based on strength 700 parametec of the material and is, therefore. . more rational than the group index method. FIg. 29.S. The basic assumption in the method is that a layer of pavement is of superior quality than the Jayer below it. The shortcoming of the methexl is that it gives the same total thickness above a material irrespective of the quality of the overlying layers. mc : 37-1970 gives some important recommendations for the determination of CBR value and its use.

/

29.9. CALIFORNIA RESISTANCE VALDE METHOD The method uses the California Resistance value, called the R-value. The R-value is detennined by placing the specimen in tbe slabilomcter and by applying the lateral and vertical pressures as specified. The R-vaJue is given by

R _ 100 _

100 (2.5/D,)(p"/p, - 1) + 1

... (29.2)

where Pv "" vcnical pressure applied (1120 kN/m2), Pit = hori7..ontal pressure transmitted, ~ :: displaremenl of stabilomeler fluid required to increase the horizontal pressure from 35 kN/m 2 10 700 kN/ml, measured in numbec of revolutions of the calibrated pump handle. Hvccm and Carmany (1948) gave the following expression for the total thickness of the pavement

779

PAVEMENT DESIGN

1-

KT(~~-

R)

,

... (29.3)

traffic

, '6

0

thickness (em). K = numerical cornlanl ( = 0.166), T =

where I

,

CBR(·'.)~ ~

index,

R

:=

slabilometcr

resistance. value, C = cohcsiometcr value, determined from the cohesiomctcr test. The traffic index T has been empirically provi~cd to estimate the traffic volume, as

I

'0

50 . , , "

~V~ ~ ~

a

I 00 3

Iv

V

1/ 1/1/'1'<./ V !7>~ 1;: ~ V /VV ~

T _ 1.35 (EWlf" ... (29.4) .§ where EM. = equivalent wheel load. ~ 4 > 00 29.10. MCLEOD MIITlIOD McLeod gave the following formula Cor the thickness. 1- Klog,,(PIS) ... (29.5)

where I := thickness of grovel base (an), P = gross wheel load (kN), S := total subgradc support (kN), K = base course constant (varying non-linearly from 75 far bearing plate of diameter of 12.5 em to 2ID

~

"

1i. ~

The repetitive plate bearing tests conducted using plates of

Trorfic

Curn

classification

No. 01 vehicles p~r day excrrding 30kN laden weight

A

0-15

B

15_1,5

6

C

1,5_1'50

70

E

1/ V

~ [III 0"J/ ,'1 rl

SO<)

0

11/

80 a

for diameter of 125 em)

are

20

:=

ISO

+-+--

_1,50

1050 _ 1500

F

ISOO

'SOO

0

Abow

1,500

so

+-+--

(Rrcommrnd by IRC)

different sizes. The subgmde support (.5) is

Fig. 29.6.

determined from tbe unit support measured or computed [or 30 an diameter plate at 0.5 em deflection and len repetiti~. Fig. 29.7 can be used for dctennining the ratio (7) of the unit sub-grade support (SV for the Plato! 60cm

diomO!trr ........-

",m

(Ptrlm~tl'r I Arto) rotlo

Fia·29.7.

01 plolt

15cm

(em') ----..

SOIL MECHANICS AND FOUNDATION ENGINEBRINO

700

design wheel diameter to 5, that on 30 em diameter plate. The OlNes for different deflections between 0.25 em and 1.75 em are shown. Thus

S2" rS,

... (29.6)

TIle design subgrade support S is obtained by multiplying the value of S2 by the contact area of the wheel. Thus 5 .. 52 x a ... (29.7) where a = contact area of the wheel (em,.

29.U. TRIAXIAL TEST METDOD The lriaxial rompression test is conducted on a soil specimen under a lateral pressure of 140 kN/m'2 and the value of tbe mOdulus of elasticity is determined from the stress-strain curve. The thickness of the pavement in em am be dctennined using the formula given by the Kansas Highway DeptL

t

..V/(~)' _ ' 21tEtJ.

a

where P = wheel load (kN), E = modulus of elasticity (kN/cm1. oX == traffic coefficient (0.5 to 2.0), y • saturation coefficient ( 0.5 to 1.0). a area of oontac.1 (cm1, 6 = design deflection ( 0.25 em).

=

=

=

29.12. BURMISTER'S METIIOD Burmister developed a method considering the pavement as a layered system. 1bc Burmister theory is based on the following assumptions. (1) The material in each layer is homogeneous, isotropic and elastic. (2) The surface layer is infinite in the borizontal direction and finite in the vertical direction. The underlying layer is infinite in both direaions. (3) The layers are in continuous contact. (4) The top layer is free of shearing stresses and normal stresses outside the loaded area. The displacement equation given by Bunnister can be written as, assuming lA, A .. 1.5 P ~ FI for Oexible plates

tJ. _

and

=

1.18P~~2

=

... (29.9)

forrigidplates

=

lAp - 0.50,

... (29.10)

=

where p uniform pressure, a radius of plate, PI deflection factor, and E, modulus of the soiL Fig. 29.8 gives the values of the deflection factor Fl' II depends upon the ratio of the modulus of the subgrade (E,) and the modulus of pavement layer Ep- 1bc ratio (h/a) is equal to the thidalcss of the ~ layer divided by the radius of the load. 11 may be noted that for a single layer, h = 0 and E,IEp = 1.0, and hence F2 = 1.0 and the solution reduces to Boussinesq's settlement equation. The design procedure can be summarised as under. (1) Dctennine E, by conducting a plate bearing lest on a 30 em diameter plate over the subgrade.

E, _ 1.1: po

...(29.11)

(2) Calrulate FI by conducting a plate bearing test on the pavement, 6. x E, F, • 1.18 po

... (29.12)

(3) For the computed value of F2 and the given value of (hla) ratio, delennine E,IEp from Fig. 29.8.

781

PAVEMENT DESlGN 0 .S

o. 6 o.

,

."'- r---.

~ ~

E!'IEp . 1/2

---

~

1 ,~,\ ,,,,,- ~ I'-.........----- r---- ~ t-- r-~ - t--r., ~ ~ ~ ~ ~ .!LlQ.

~

1/100

11100

0.0

"'-

M l

o

10

I/SOO

i'-....

"'- "-...

I----.~ ''<00

I'-..... ~~ 20 3.0

.1J---

-- r--... F::::

--- ---'.0

-

r --

' .0

6.0

Fig. 29.8.

(4) Detennine the oontad radius of the design load, as ... (29.13) where P = design load and p = tyre pressure. (5) Compute the new value of F2 for the design deflection !J. ( = 0.5 em or 0.25 em).

F2 _ Ii)( E. ISpa (6) For the computed values of F2 and E.IEp ratio, detennine tbe

. .. (29.14)

h/a ratio from Fig. 29.8.

(7) Determine h from the hla roUo.

29.13. COEFFICIENT OF SUBGRADE REACTION For the design of rigid pavements, the coefficient of subgrade reaction is required. As discussed in chapter 24, the coefficient of subgracle readion (k) is defined as the pressure per unit deformation of tbe subgrade at a -specmcd deformation or pressure level. Thll$

k _ ~ where k = modulus of subgrode reaction (kN/an~, p

... (29.15)

= pressure (kN/cm\

and !J. = deformation (an)

[Note. In this chapter, k. is also used for k). The test sct-up for the determination of the coefficient subgrade reaction is similar to that in a plate load

test (chapter 23). It conc>ists of a loading frome which has a hydraulic jack, a proving ring and a react.ioo beam (Fig. 29.9). 1be dial gauges are used for tbe measurement of the settlement of the plate. These gauges rest on a separate datum frame. The standard size of the plate is 75 an diameter. 1be test site is levelled and the plate is properly seated 00 the prepared surface. The stiffening plates of


782

Fig. 29.9. Set-up for Detcnnination of Coefficient 0( Subgrade Reaction.

progressively deaeasing diameters are placed over the plale. The jack and proving ring assembly are litted to provide reaction. Three or four ilia] gauges are fixed on the periphery of the plate from the datum frame. A seating load of 7 kN/m2 is first applied and released after a few seconds. The load is then applied 00 the plate to cause a settlement ~f approximately 0.25 mm. When there is no perceptible increase in settlement ~ or when the rate of settlement is less than 0.025 mm per minute, the dial gauge readings are taken and the average settlement is found. The procedure is rcpcacd till the g' ~ settlement is about 0.175 an. A graph is plotted between the mean beariog pressure (P) aod the settlement (Fig. 29.10). The pressure =corresponding to a settlement of 0.125 em is read from t:. :O.12Scm the plot, and the value of k is dClcnnincd as

!1

k -

O.i25

k - P

leN/an

O~l:

l

0·'

... (29.16)

tcm)~

Fig.

kN/ml

29.10.

... [29.16(a)J

To allow for the efJcct of possible future sofiening of subgrade, the vahle of k obtained is corrected for full saturation. l\vo specimens are prepared for the consolidation lest, one without soaking and the other fully soaked. TIle consolidation tests are cooducted and the pressures required for a deformation of 0.125 em are detennined for the two specimens. The oorrected modulus of subgrade reaction is given by

k - k(.cs.) x

if;

...(29.17)

wbere p = pressure for the unsoaked specimen, and p, = pressure for the soaked specimen.

29.14. WESrERGAARD'S ANALYSIS The rigid pavements are coostructcd of cement concrete, which may or may not have reinforcement. These pavements develop slab action due to their bigh tensile strength. The stresses in the subgrade below a rigid pavement are greatly reduced due 10 the lood-spreading capacity. TIle failure of rigid pavements usually occurs by overstressing of the concrete and not by overstressing of the subgradc. TIle thickness dctcnninaLion is generally based on the calculation of stresses in the ooncrete. The stresses in the concrete slab due 10 wheel loads are dctennincd using Westergaard's theory. Westergaard oonsidered the rigid pavement as a thin el~tic plate resting on soil subgrade. The upward readion at any poiDt is assumed to be proportional to the deflection at that point. "(be slab deflection depends

PAVEMENT DESIGN

183

upon the stiffness of tbe subgrnde and !he Oexura1 strength of the slab. Thus the pressure-deformation characteristics of a rigid pavement depend upon the relative stiffness of the slab and the subgrade. Westergaard defined the radius of relative sillIness as

I

Eh.J

-

[

12(1 - ~')k

1"

... (29.18)

where I = radius of relative stiffness (an). E = modulus of elasticity of cement concrete (kN/em1. ~ = Poisson's ratio of cement ooncrete, h = slab thickness (em), k "" modulus of subgrade reaction (kN/em\

As the pavement slab bas a finite length and width, the intensity of maximum slress induced due 10 a wheel load depcD(1s upon the location of the load. The following three locations are critical. (1) Interior loading. When the load is applied in the interior of the slab at any point away from all edges. (2) Edge loading. When the load is applied on an edge of the slab at any point away from a comer. (3) Corner loading. When the centre of the load is located on the bisector of the comer angle formed by two intersecting edges of the slab, and the loaded area is at the comer touching the edges. Wcstergaard gave the following equations for the critical stresses based on the assumption thal the cement oona-ete slab is homogeneous arxl has uniform clastic properties. It is further assumed thaI the vertical subg.:adc reaction is proportional to the deflection.

O.3~~P

Interior loading

OJ"

[4Iog10(I/b) + 1.069]

... (29.19)

Edge loading

". _ o.s;;P [4 log" (lIb) + 0.359J

... (29.20)

Comer loading

", -¥> [1- (ayz]" 1

... (29.21)

where P wheel lood (kN), h ::a: slab thickness (an), I ::: radius of effective stiffness (em), a :: radius of wheel load distribution (em), b = equivalent radius of resisting action (an).

b_V1.6a2 +h 2 _O.675h for a< 1.724h

The valueofb is given by

.. .[29.22(a)J

and b_a fora
For edge loading (feller and Sutherland, 1943) ".- 0.529

~

(I +

h For corner loading (Kelley, 1939)

0.s4~)

[4 log" (lib) + log"b - 0.4048J

•

", -¥> [ (ayzr 1 -

1

... (29.23)

... (29.24)

The above equations arc presented in the fO"rm of stress charts (Figs. 29.11 and 29.12). The charts liven are applicable for a particular set of design parameters., viz; p ... 41 leN; a = 15 em; E = 3 x to) leN/em; arxl 11 .. O.IS. However, different curves are given for different value of the coefficient of subgrade reaction (k,) between 60 N/an) and 300 N/cm). The design curves are for slab thiclrness from h = 15 to 2S an.

SOIL MECHANICS AND FOUNDATION ENGINEBRING

784

~ ~°hr-+-t-t~~--~r-+-+-+-+-~ LL~~~~~~L-+"~~-L~~8 SlOb

thleknu5 (em) - - .

Slob

Fig. 29.11.

thidness

(cm)-----.

Fig. 29.12.

29.15. TEMPERATURE STRFSSES IN RIGID PAVEMENTS

A a:ment oonaelc slab expands when the temperature increases and contracts when tbe temperature decreases. IT the slab is free to change ilS length, there would be no stresses due to temperature changes. In rigid pavements, the slab is restrained against free deformation either by its own weight or by action of frictional forces, Consequently, the stresses are induced due to temperature changes. The SlreSSeS due to temperature changes can be divided into two categories.

(I) Warping stresses, (2) Frictional stresses. 1. Warping Stresses. 1bese stresses develop due to differential changes in the top and bottom surfaces of the slab. The temperature change that occurs on the top surface is not transmiued immediately throughout the entire thickness. 1be temperature difference causes different changes in the top and bottom surfaces aod a warping or bending in the slab cxx:uts during this period. Bradburry gave the following cquatioos for warping stresses.

At Interior,

a (i) .. Eat (C" + 1.1. C,)

At edges,

al(e) .. C"

At comers,

at (e) - 3 (~~ t 1.1.) ..fi7i

... (29.21)

where a coefficient of linear expansion (10 )( 10-6 per

0q and C" and C, are warping stress coefficients (Fig.

I

2

1 _

~

al

... (29.25)

1.1.2

or at(e) .. Cr~at. whicbeverishigher.

. .. (29.26)

29.13).

2. Frictional stresses. If the temperature of the concrete remains constant for a long time. tbe temperntures of the top surface and the bottom surface become equal. At that stage, there is a uniform 01 shortening of tbe slab and there is no warping. As the slab is in contact with tbe subgrade. tbe slab movements are restrained due to the friction between the bottom surface of the pavemenr: and tbe

lengthening

subgrade. The frictional stresses are given by y,Lf or - 2 x 104 where or = frictional stress (kN/an~. L coefficient of subgrade restraint (... 15).

= length of the slab (m).

... (29.28)

Yc = unit weight of concrete (kN/cm\ f =

PAVEMENT D,ESlGN

785 .2

'--l-

.0

V

0.8

1

•

:;. 0

"

II

I

.2

0 0

~

V

2

/

/

to I---L~

4

G

8

-

--------i

10

~ ( or~)_

"

4

Fig. 29,13.

To reduce temperature stresses, expansion jOints, controction joints and warping joins are provided in the pavement. 29.16. COMBINED STRESSES IN lUGID PAVEMENTS For proper design of a rigid pavement, it is necessary to consider the most critical combinations of the stresses due to loads and te!ppcrature stresses. (1) During summer, the critical combination for interior and edge regions occurs during mid-day when the slab tends to warp downward. Maximum stress at the bottom fibre:: load stress + warping stress - frictional stress . ,.(29.29) (2) However, during winter. the critical combination for the above regions occurs when the slab contracts and the slab warps downward during mid-day. Maximum stress at the bottom fibre = load stress + warping stress + frictional stres,s ... (29.30) Generally, Eq. 29.29 governs, m the differential temperatures are more in summer than in winter. (3) For comer regions. the most critical combination occurs during the midnight when the slab warps upward. Maximum stress at top fibre = load stress + warping stress .. .(2931)

ILLUSTRATIVE EXAMPLES Dlustrative Example 29.1. A sall/ple of subgrode has a group index of 8. Design the pavement for the anticipated traffic volume of over 300 commerciaJ vehicles per day. Solution. As the traffic is heavy, curve D of Fig. 29.4 is applimble. Total thickness = 500 mm From Curve A, the thickness of sub-hase = 200 mm lbcrefore, thickness of base and surracing :::: 500 - 200 = 300 mm . Illustrative Example 29.2. The CBR value of subgrade is 10%. Determine the total thickness of the pavement using IRC method. There are more than 4500 vehicles per day.


186

Solution. From Fig. 29.6, using curve G, total thickness = 360 mm

illustrative Example 29.3. Determine the stresses at interior, edge and corner regions of a rigid )C III kN/cm2, h = 20 em, J.l = 0.15, k = 4IJ

pavement using Westergaard~ method. TaJu: P = 41 leN,. E = 3 Nlcm J and a = 15 em.

SolutJon. From Eqs. 29.18, I

-

[

1_[ From Eq. 29.22 (0),

From Eq. 29.19,

From Eq. 29.20,

Eh' 12(1 _ ~')k

j"

3 x 10' x (20)'

12 (1 - 0.15 2 )

)C

0.04

j" _

84.57=

b .. ~-O.675h .. ";1.6 )( 152 + (20)2 - 0.675 ~ 20 .. 14.07 em 0, ..

O.3~~p

OJ''

~

0c"

[410g[0 (lib) + 1.069]

(20)

a.s:;p

(4

[410&10(84.57/14.07) + 1.069] .. O.13SSkN/cmz

.

10&10

(l/b) + 0.359]

.. 0.572 )C :UXJ (41oglO (84.57114.07) + 0.359] .. 0.2037 kN/cm z (20)

From Eq. 29.21,

O'-7[1-(a~r

1 0.'

_ 3 x 4l.0 [1 _ (15V2) (20)' 84.57

1_ O.1734kN/cm'

PROBLEMS A. Nlftnerical 29.1. A sample of subgrode gllve the £oIlowing results. (a) Soil portioo passing 0.074 rom sieve '" 55%. (b) Liquid Urni! '" 45 (c) Plastic Limit '" 25 Design the pavement by group index method for on Dnlicipated troffie volume of over 300 wmmercial vehicles per duy. (Aos. Total Ihickness '" 50 em, Sub-base '" 10 an) 29.2. The CBR value of subgrade is 8 percent. Determine the totol thickness of nexible pavement using design charts rerommended by IRC. There arc more than 4500 vehides per day. (Ans. 40 an)

29.3. ~t~:~~:~fS~=~~ ~~t~e;!gC Dnd comer of a'"-;":nltcrco;7~Fvement by Westergaard's cquatioos. (b) Poisson's ratio (e) Thickness of concrete pavement (d) Modulus of subgmde reaction (e) Wheel load (j) Radius of loaded area

'"' 0.15

'" 22 an

3 '" 0.03 kN/an '" 41.0 kN '" 12 an (ADS. OJ '" 0.1265, o~ '" 0.1946,

Oc =

2 0.1652 kN!an )

PAVEMENT DESIGN

781

29.4. Determine the warping stresses for a concrete pavement of thickness 20 cm with tran sverse joints :11 15 m 2 spacing and longiludimll joints at 3.5 m spacing. For concrete. f = 3 x 10~ kN/cm ,/J = 0.15. Take the modulus 3 of subgrade reaction as 30 N/cm . Assume n:mpcrnture differential for day co nditions to be l oe per cm of slab thickne ss and the cQt:fficient of lincOlr expansion of concrete as 10 x 10- 6 per °C. The radius of load area is 15cm

B. Oescl"iptivc Type and Objective Type 29.S. What arediffcrcnt typcsof puvcments ?Expluin their salient charucleristics. 29.6. What are the valious factors that anecl the de sig n of pavement ? 29.7. Describe the following methods of Ihe design of nexible pavement. (a) Group index method (b) CBR method

tel Burnl1slcr's Method .

29.S. Discus~ Westergllilrd's analysis for the rigid pavements. 29.9. Explain the lollowing terms. (a) Modulu~ of subgmde reaction. (b) Radiu s of relative sl1 ffness

29.10. Discuss the effect of tcmpemlUre stresses on the design of rigid pavements. 29.11 . Write whether the foll~wmg statements are true or false (a) Thc flexible p.wements are capahle of bridging small depressions in the subgrade.

(b) Gcnernlly. the CBR for 5.0' 111m is greater than that tor 2.5 mm. \e) The higher the value of the group index. the poorer is the subgradc. 2 (tl) 111e unitlo for the l·{lI!fficie;ll of subgnKlc rc..1ctlon arc kN/cm •

(t') For comcr

region~

\)f the rigid pavements, the entical stresses usually occur during the midnight.

[Ans. True (c), (e)l

C. Multiple Choice Questions 1. For rigid 1>3vements. generally the lollowing component is omitted (a) Surfacing

(11) Sase

(c) Slib base (tf) Subgradc 2. The minimum surcharge load for the C.S.R. \Cst is

la) 2.5 kg (e) 4.5 kg

(b) 3.5 kg (tI) 5.5 kg

3. The stand:trd size uf the pl;ltc for tlte coetTicient of ~ubgmd e reactio n test is Ill) 50 em (b) 75 em (e) 100

(tl) 125 cm

IAlis. I. (e), 2. (c). J. {b))

30 Laboratory Experiments Some of the more commonLy conducted laboratory experiments are described below.

30.1. Experiment No. 1. To detennlne the water-content or 8 soli sample by oven-drylng method. 1bcory. The water rooteol (w) of a soil sample is equal to the m~ of water divided by the mass of solids, w _ Ml - M) x 100 M)-MI where M J

= mass of empty container,

with lid,

M2 "" mass DC the container with wet soil and lid,

M3 = mass of the container with dry soil and lid. Equipment. (1) Thermostatically oontrolled oven, maintained al a temperature of llO° :!: SoC; (2) Weighing balance, with accuracy of 0.04% of the mass of the soil taken; (3) Desiccator, with any suitable desiccating agent; (4) Airtight container made of oon-cocrodible material. with lid; (5) Tongs.

Soil Specimen. The soil specimen should be representative of the soil mass. The quantity of the specimen taken would depend upon the gradation and the maximum size of particles. For more than 00% of the particles passing 425 fA IS sieve, the minimum quantity is 25 g. Procedure: (1) Clean the container, dry it and weight it with lid (M J). (2) Take the required quanlily of the wet specimen in the oontainer and close it with lid. Take the mass (Mti. (3) Place the container, with its lid removed, in the oven till its mass becomes constant (nonnally for 24 hours). (4) When the soil has dried, remove the container from the oven, using longs. Replace the lid on the container. Cool it in a desiccator.

Data Sheet for Water content by Oven-drying Method S.No.

(I) (2) (3) (4)

(5) (6) (7)

Determination No.

ObservaliotlS mid calculalions Obsen-utions Container No. Mass of empty amtainer (MI) Mass of container + soil (M2) Mass of container + dry soil (MJ) Calculations Mass of waler,M", .. M2 - MJ Mass of solids., M~ .. MJ - MI Water content, W .. (5)/(6) )( 100

ResulL 1be water content of the sample

401 20.12 g 44.32 g

41.18 g 3.14g 21.06 g 14.91

= 14.91%.

402

403

789

(5) Fmd the mass (M) of the container with lid and dry soil sample. Observations and calculations. Observations and calculations are shown in the data sheet.

30.2 Experiment No. 2. To detennlne the water content of II soil sample by pycnometer method. Theory. A pyatometer is a glass jar of about 1 litre capacity, fitted with a brass conical cap by means

of a screw-type rover. TIle cap has a small hole of about 6 mm diameter at its apex. The water cootent (w) of the sample is obtained as w -

where Ml

= mass of empty

[~Z: =Z:;

(Go 1) - 1] 100 x

pyroometer;

M2 :::: mass of PYOlometcr and wet soil; Ml = mass of pycnometer and soil, filled with water; M. = mass of PYOlometer filled with water ooly. G = specific gravity of solids. Equipment. (1) Pycnometer, (2) Weighing balance with an accuracy of 1.0 g; (3) Glass rod. Procedure. (1) 'Wash and clean the pycnometer and dry it. (2) Oetennine the mass of the pycnometer, with brass cap and washer (M .). accurate to 1 g. (3) Place about 200 to 400 g of wet soil specimen in tbe pycnometer and weigh it with its alp and

washer (MV. (4) Fill water in the pycnometer containing the wet soil specimen to about its half beight. (5) Mix the contents thorougly with a glass rod. Add more water and stir it. Fill the pycnometer with water, flush with the bole in the conical cap. (6) Dry the pycnometer from outside and take its mass (M3)' (7) Empty the pycnometer. Ceao it thoroughly. Fill it with water, flush with the hole in the oonical cap and weigh (M.).

Observations and calculations See the data sheet.

nata sheet ror water-content by pymometer method Specific gravity of solids ... 267

I.

2. 3. 4.

DetermiMtion No.

ObseNalions and calculations

S.No.

ObserwJtion.s Mass of empty pymometer (MI) Mass of pycnometer and wet soil (Mi) Mass of pycnometer soil, filled with water (M) Mass of pycnometer filled with water ooIy (M.)

580 8 844 8 1606 8 1470 g

CalculaJions S. 6. 7.

8.

M2-MI

M3-M. (G - l)/G

w-[l2

X (7)-I]XlOO

264 8 136 8 0.625 2132%

Result. Water content of the sample = 2132%.

30.3. Experiment No. J. To determine the speclflc gravity or solids by the density bottle method. Theory. The specific gravity of solid particles is the ratio of the mass density of solids to that of water. It is determined in the laboratory using the relation


790

G.. where MI

M2-MI (M, - M I ) - (M, - M.)

= mass of empty boule;

M2 = mass of the bottle and dry soil; M3 = mass of boUle, soil and water; M. = mass of bottle ftlled with wale~ only. Equipment. (1) 50 mt density bottle with stopper; (2) Oven (105 0 to 110°C); (3) Constant temperature water bath (2rC); (4) Vaalum desiccator; (5) Vacuwn pump; (6) Weighing balance, aa:uracy 0.001 g; (1) Spatula.

Procedure. (1) Wash the density bottle and dry it in an oven at 105°C to 110"C. Cool it in the desiccator. (2) Weigh the bottle, with stopper, to the nearest 0.001 g (M I ).

(3) Thke 5 to 10 g of the oven·drled soil sample and transfer it to the density bottle. Weigh the bottle with the stopper and the dry sample (M,). (4) Add de-aired distilled water to the density bottle just enough to cover the soil. Shake gently to mix soil and water. (5) Place the bottle containing the soil and water, after removing the stopper, in tbe vacuum desiccator. (6) Evacuate the desiccator gradually by operating tbe vacuum pump. Reduce the pressure to about 20 mm of mercury. Keep the bottle in the desiccator for at least I hour or until no further movement of air is notilXd. (7) Release the vacuum and remove the lid of the desiccator. Stir the soil in the boUle carefully with a spatula. Before removing the spatula from the boUle, the particles of soil adhering to it should be washed off with a few drops of air-free water. Replace the lid of the desiccator and again apply vacuum. Repeat the procedure until no more air is evolved from the specimen.

[Note. In case a varuum desiccator is not available, the entrapped air can be removed by beating the density bottle on a water bath or a sand bath.] (8) Remove the bollie from tbe desiccator. Add air-free water until the boule is full. Insert the stopper. (9) Immerse the bottie upto the neck in a constant-temperature bath for approximately I hour or until it bas attained the coostant temperature. If there is an apparent d~ in the volume of the liquid in the bottle, remove the stopper and add more water to the bottle and replace the stopper. Again place the bottle in the water bath. Allow sufficient lime to ensure that the boule and its content altain the constant temperature. (to) Take out the bottle from the water bath. Wipe it clean and dry it from outside. Fill the capillary in the stopper with drops of distilled ·water, if necessary. (11) Detennine the mass of the bottle and its contents (M~. (12) Empty the bottle and clean it thoroughly. Fill it with distilled water. Insert the stopper. (13) Immerse the bottle in the constant-temperature bath for I hour or until it has attained the oonstant temperature of the bath. If there is an apparent decrease in the volume of the liquid, remove the stopper and add more water. Again keep it in the water bath. (14) Thke out the bottle from the water bath. Wipe it dry and take the mass (M4 ).

ObservatiorL9 and calculations See the data sheet.

191

lABORATORY EXPERIMEtITS

nata Sheet for Spec1f1c Gra"ity by Density BotUe Determination No. Obserwllions and calculations

S.No.

1. 2.

3. 4.

5. 6. 7.

Observations Density Bottle No. Mass of emply density baltle (MI) Ma5s of boltle dry soil (M])

Mass of bottle, soil and water (Ml) Mass of baltic filled with water (M4) Calculations M2-MI

8.

301 41.302 g

302

30J

54.103 g 99.002 g 91.112 g 12.801 g 7.970 g 2.65

Result. Specific gravity of solids '"' 2.65. 30.4. Experiment No. 4. To detennine the specific gra"lty of solids by pycnometer method. Theory. The pymometer method can be used for the detohnination of the specific gravity of solid particles of both fine-graincd and coacse-grnined soils. The specific gravity of solids is determined using the relation G _ (M, - M,) (M, - M,) - (M, - M.) where MI = mass M2 = mass Ml "" mass M4 = mass

of empty pyrnOffieter; of pyrnometer and dry soil; of pYQ1Ollleter, soil. and water; of pycnometer filled with water only.

EqulpmenL (1) PyOlometer of about I litre capacity; (2) Weighing balance, with an acx:uracy of 1 g; (3)

Glass rod; (4) Vacuum pump. Procedure. (1) Clean and dry the pYalOmeter. lightly screw its cap. Take its mass (M I ) to the nearest

0.1 g. (2) Mark the cap and pycnometer with a vertical line parallel to the axis of the pycnometer to ensure thai the cap is screwed to the same mark each time. (3) Unscrew the cap and place about 200 g of oven-dried soil in the pYOlometer. Screw the cap. Determine the mass (Mi)' (4) Unscrew the cap and add sufficient amount of dc-aired water to the pycnometer so as to cover the soil. Screw on the cap. (5) Shake well the contents. Connect the pymometer to a vacuum pump, to remove the entrapped air, for about 20 minutes for fine- grained soils and foc about 10 minutes for marse-grained soils. (6) Disconnect the vacuum pump. Fill the pycnometer with water, about three fourths full. Reapply the vacuum for about 5 minutes, till air bubbles stop appearing on the surface of the water. (7) Fill the pymometer with water completely, upto the mark. Dry it from outside. Take its mass (Ml ). (8) Record the temperature of contenLS. (9) Empty the pymometer. Qean it and wipe it dry. (10) Fill the pycnometer with water only. Screw on the cap upto the mark. Wipe it dry. Take its mass (M4).


792

Data Sheet for Specific Gravity by Pycnometer Method Determination No.

ObseTvaJiOIl$ mui calculations

S.No.

Observaliolls I. 2. 3. 4. 5. 6.

7. 8. 9.

Pycnometer No. Room tempcnllute Mnss of empty phcnometcr (M) Mass of pycnometer and dry soil (M2) Mass or pycnometer, soil and water (M) Mass or pycnometer and water (M4) Caicu/miOlls /112 - MI

402

403

800g 1707 g 1570 g

220 g 137&

M) - M,

G -

401 U"C 580 g

(7/~\8)

2.65

ResuU. Specific gravity of solids at 26°e = 2.65.

30.5. Experiment No.5. To detennine the dry density or the soil by COI't: cutter methods. Theory. A cylindrical core cutler is a seamless steel tube. For determination of the dry density of the soil, the cutter is pressed into the soil mass so tbat it is filled with the soil. The cutter filled with the soil is lilled up. The mass of the soil in the cutter is detennincd. The dry density is obtained as

p-~-~ where M = mass of the wet soil in the cutter; V :: internal volume of the cutter; w = water content. Equipment (I) Cylindrical core cutter, 100 mm internal diameter and 130 mm long; (2) Steel rammer, mass 9 kg. overall length, with the foot and staff about 900 mm; (3) Steel dolley, 25 mm high and 100 mm internal diameter, (4) Weighing balance, accuracy I g; (5) Palette knife; (6) Straight edge, steel rule, etc. Procedure. (1) Determine the inlemul diameter and height of the core cutter to the Dearest 0.25 mm. (2) Determine the mass (M,) of the cutter to the nearest gram.

Data Sheet for Dry Density by Core Cutter Method Determination No.

ObservaJiollS olld calculatiolls

S.NO.

Observatiolls l. 2. 3. 4.

5.

Core cutter No. Intemaldiameter Internal height

Mass of empty corecuner (MI) Mass of core Cllilcr with soils (M2)

501 100 mm

129.75 mm 1130 g 3120 g

Calculatiolls

6. 7. 8. 9.

Mnssofwctsoil,M _ M2 - MI Volume of cutter, V Water content (determined as in Experiment 30-1).

"y.

DrYdensitY=~

Result. Dry density

= 1.666 glm!.

1990 g 1019.05 ml 17.75 % 1.66 gm/m1

502

503

.,.3

LABORATORY EXPERIMENTS

(3) Expose a small area of the soil mass to be tested. Level the surface, about 300 mm square in area. (4) Place the dolley over the top of the rore culter and press the rore culter into the soil mass using the rammer. Stop the process of pressing when about 1·5 mm of the doUey protrudes above the soil surfaet:. (5) Remove the soil surrounding the core cutter, and take oul the core cutter. Some soil would project from the lower end of the culter. (6) Remove the doUey. Trim the top and bottom surface of the oore cutter carefuUy using a straight edge. (7) Weigh the core cutter filled with the soil to the nearest gram (Mi). (8) Remove the core of the soil from the culler. Take a represeotalive samplc for the water content determination. Determine the water content, as described in Experiment 30.1-

30.6. Experiment No.6. To detennlne In-situ dry densily by the sand replacement method. Theory. A hole of specified dimensions is excavated in the ground. Tbe mass of the excavated soil is determined. The volume of the hole is determined by filling il with clean, uniform sand whose dry density (Ps) is determined separately by caUbration. The volume of the holc is equal to the mass of the sand filled in the hole divided by its dry density. The dry density of the excavated soil is determined as

p,.~ where M "" mass of the excavated soil; V

=volume of the hole; and w =water content.

Equipment. (1) Sand-pouring cylinder; (2) Calibrating container, 100 mm diameter and 150 mm height; (3) Soil cutting and excavating tools, such as a scraper tool, bent spoon; (4) Glass plate, 450 mm square. 9 mm thick; (5) Metal container to collect excavated soil; (6) Melaltray, 300 mm square and 40 mm deep with a hole of 100 mm in diameter aI tbe centre; (7) Weighing balance; (8) Moisture content cans; (9) Oven; (10) Desiccator. Clean, uniform sand passing I mm IS sieve aDd retained on 600 micron IS sieve in sufficient quantity. Part-I Calibration Procedure (1) Determine the internal volume of the calibrating container by filling it with water and determining the mass of water required. The mass of water in grams is approximately equal to the volume in millilitres. The volume may also be determined from the measured dimensions of the container. (2) Fill the sand-pouring cylinder with sand, within about 10 mm of its top. Determine the mass of the cylinder (M J) to the nearest gram. (3) Place the sand-pouring cylinder vertically on the calibrating conlainer. Open the shutter to allow the sand run out from the cylinder into the calibrating container till it fills the cooe oC the cylinder and the calibrating container. When there is no further movement of the sand in the cylinder, close the shutter. (4) un the pouring cylinder from the calibrating conlainer and weigh it to the nearest gram (M). (5) Again fill the pouring cylinder with sand, within 10 mm of its top. (6) Open the shutter and aUow the sand to run out of the cylinder. When the volume of the sand let out is equal to the volume of {he calibrating rontainer, close the shutter. (7) Place the cylinder over a plane surface, such as a glass plate. Open the shutter. The sand fills the oone of the cylinder. Qose the shutter when no further movement of sand takes place. (8) Remove the cylinder. CoUed the sand left on the glass plate. Determine the mass of sand (Mi) that had filled the cone by weighing lbe collected sand. (9) Determine the dry density of sand, as-sbO'Nn in the data sheet, part-I.


794

Part-II. Dry Density Procedure. (1) Expose an area of about 450 mm square on the surface of the soil mass. Trim tbe surface down to a level surface, using a scraper tool. (2) Place the metal tray on the levelled surface. (3) Excavate the soil through the central hole of the tray, using the hole in the tray as a pattern. The depth of the excavated hole should be about 150 mm. (4) Collect all the excavated soil in a metal container, and dclennine the mass of the soil (M). (5) Remove lIle metal tray from the excavated hole. (6) Fill the saoo-pouring cylinder within 10 mm of its top. Dctennine its mass (M!). (7) Place the cylinder directly over the excavated hole. Allow the sand to run out of the cylinder by opening the shutter. Close the shutter when the hole is completely filled and no further movement of sand is observed. (8) Remove the cylinder from the filled hole. Dclennine the mass of the cylinder (M4 ). (9) Take a represcnwtive sample of the excavated soil. Determine its water oontenl, as explained in the experimeot 30--1. Detennine the dry density of soil as shown in the data sheet, Part-fl. Data Sheet ror Sand Replacement Melhod Part- I. Calibration for Dry Density of Sand Determi11llfwn No.

Obsetwltions and calculations

S.Na

Obserl'Olion.s 12. 3. 4.

\-blume of calibrating cone (V<) Mass of pouring cylinder (Mt), filled with sand Mass of puring cylinder after pouring sand into the calibrating container and cone (M) Mass of sand in lhc cone (M2)

980 mI 11040 g 9120 g 450 g

Calculations

5. 6.

Mass of sand in the calibrating comBiner M, _ (2) - (3) - (4) Dry density of sand PI - M
1470 g

15 g/ml

Part-D. Dry Density of soil. Determintllion No.

Observalions and calculalions

S.No.

ObselWlions 12. 3.

4. 5.

6. 7. 8.

Mass of excavated soil (M) Mass of pouring cylinder (M.). filled with sand

Mass of pouring cyliDder after pouring inlO the hole and cone (M.) CalcuJaJlons Massofsond in the hoIeM, - MI - M. - M2 \-blume of sand in the hole V _ M,/ p, Bulk density, p _ M/V Water contenl. detennined as In experiment 30-1 (w), say,

Dry density -

rf-; - tM:Vw

ResulL Dry densIly = 1.72 g/m1.

2310 g 11040 g

8840 g 1750 g 1166.67 mI 1.98 glml 15%

1.72 gImI

LABORMORY EXPERIMENTS

30.1. Experiment No. 7. To determine the dry density of a soU sample by waterdisplacement method. Theory. A soil specimen of regular shape is coaled with paraifm wax to make it impervious to water. The total volume (V,) of the waxed specimen is found by determining the volume of water displaced by tbe specimen. The volume of the specimen (\I) Is 'given by

v_

11,- (M,-M)

p, where M, '" mass of waxed specimen; M '" mass of the specimen without wax; Pp '" density of paraffin. Dry density of specimen ..

1M::

Equipment. (I) Water-displacement apparatus; (2) Weighting balance, accuracy 1 g; (3) Paraffin wax; (4) Cutting knife; (5) Heater; (6) Oven; (7) Measuring jar; (8) Brush; (9) water amtent container. Procedure. (I) Take the soil specimen. Trim it to a regular shape. Avoid re..entrant oomers. Weigh the specimen (M). (2) Thke some paraffin wax and melt it on a heater. Apply a coat of melted paraffm wax to the specimen with a brush. When it has hardened, apply another coat. Take the mass of the waxed specimen (M,). (3) Fill the watcr-displacement apparatus with water. When the overflow occurs, cio:se the valve. (4) Place a measuring jar below the overflow tube of the apparatus. Open the valve. (5) Immerse the waxed spec'imcn slowly into the water in the apparatus. Water overflows. Conca the overflowed water in the jar.

Detennine the volume of the water collected (V,). (6) Thke out the waxed specimen from the apparatus. Dry it from outside. (7) Remove the paraffin wax by peeUng it off. (8) Cut the specimen into two pieces. Take a representative sample for the water content detennination. Determine the water content, as in Experiment 30.1.

Data sheet for Water-displacement Method Density of paraffin (Pp) :: 0.91 glm!. Determination No.

Observationr and calculations

S.No.

ObservatjOlIS 1. 2.

3.

Mass of specimen (M) Mass of waxed specimen (M,l \blume of waxed specimen by waler.
(V,)

650 g

681 g 362g

Calculatiolls 4.

Mnssofwax .. M1-M

S.

\blume of wax (Vp) .. (M, - M)/ pp

34.06 mI

6. 7.

\b[ume of specimen (\I) .. V, - Vp

327.06 mt

Wnler-O)nlenl, as in experiment 30-1, (w)

s.

Dry density (fld) ..

1M::

Result. Dry density of soil :: 1.75 g/ml.

31 g

13% 1.75 glmI

&lIL MECHANICS AND FOUNDATION ENGINEERING

796

30.8. Experiment No. 8. To detennlne lhe particle slu distribution of a soU by sieving. Theory. The soil is sieved through a set of sieves. The material retained on different sieves is determined.

lbe percentage of material retained on any sieve is given by PII _

~)(

100

where M" == mass of soil retained on sieve 'n',

aDd

M == total mass of the sample.

The cumulative percentage of the material retained.

wbere PI.h

etc.

ell-PI +P2+ .. . +p" are the percentages retained on sieve '1', '2', etc. wbich are coarser than sieve 'n', 1be

percentage finer than the sieve 'n', N. - 100 -

C.

Equipment. (1) Set of fine sieves. 2 mm, 1 mm, 600 f.l, 425 Ii> 212 Il. 150 j.l. and 75 j.l: (2) Set of roarse sieves, 100 mm. 80 mm, 40 mm, 20 mm, 10 mm and 4.75 mm; (3) Weighing balance, with accuracy of 0.1 % of the mass of sample: (4) Oven, (5) Mechanical shaker; (6) Trays; (l) Mortar, with a rubber CO'Iered pestle: (8) Brushes; (9) Rimer.

Part-I. Coarse sieve Analysis Procedure. (1) Take the required quantity of the sample, Sieve it through a 4.75 mm IS sieve. Thke the

soil Cradicn retained on 4.75 mm IS sieve for the coarse sieve analysis (part I) and that passing through the sieve for the flne sieve analysis (part II). (2) Sieve the sample through the set of coarse sieves. by hand. While sieving through ead1 sieve. the sieve should be agitated such that the sample rolls in irregular motion over the sieve. The material retained on the sieves may be rubbed with the rubber pestle in the mortar, if necessary. Care shall be taken so as not to break tbe individual particles. The quantity of the mat~l taken for sieving on each sieve shall be such that tbe maximum mass of material retained on each sieve does not exceed the specified value. (3) Detennine the mass of the material retained on eacb sieve. (4) Calculate the percentage of soil retained on eadJ sieve on the basis of the total mass of the sample, taken in step (1). (5) Detennine the percentage passing through each sieve. Part.-IL Fine sieve analysis (6) Take the portion of the soil passing 4.75 mm IS sieve. Oven dry it at 105° to 100°C. Weigh it to 0.1 % of the total mass. (7) Sieve the soil through tbe nest of flOe sieves. The sieves should be agitated so that the sample rolls in irregular motion over tbe sieves. However. no particles should be pushed through the sieve. (8) Take the material retained on various sieves in a mortar. Rub it with rubber pestle. bul do not try to break individual particles. (9) Resieve the material through the ncst of sieves. A minimum of 10 minutes of shaking is required if a med1anical shaker is used. (IO) Colled the soil fnlaion retained on each sieve in a separate coolainer. Thke the mass. (11) Detennine tbe percentage retained, cumulative percenlage retained, and the percentage finer. based on the total mass taken in step (1).

197

LABORATORY EXPERIMENTS

Data Sheet ror Sieve Analysis Total mass of dry soil = 400 g Mass of soil retained on 4.75 mm sieve:: 200 g Mass of soil passing 4.75 mm sieve = 200 g. ObservtltiOtlS

S. No.

2. 3. 4.

5. 6. 7. 8. 9. 10: 11. 12. 13. 14. 15.

ISSi~~

Calcula/ions

Masso/soil

Siuo[ Opening

retained

Coarse Fraction Par/I) lOOmm 100 mm 80mm IiOmm 40mm 40mm 20 mm 20mm

tOmm 4.75mm

tOmm 4.15 mm

Fille Fraction (Part II) 2mm 2mm lmm lmm

300" 2121A150 IA-

0.600 mm 0.425 mm 0300 mm 0.212 mm O.150mm

75"

O.D1Smm

600" 42') }l

Cumulative % retained

% {lMr

1.50

7.50 23.00 50.00

77.00 50.00

57.62 63.62

42.38 36.38

68.00 12.00 75.50 80.00 85.50 94.50 100.00

32.00

30.0 g 620 g

15.50

108.0

g

Zl.OO

30.5 24.0 11.5 16.0 14.0

g

1.62

g

6.00 4.38 4.00

g

g g

18.0 g 220 g 36.0 g 22.0 g

p~

Percentage retWned

3.50 4.50 5.50 9.00

5.50

92.50

28.00 2450 20.00 14.50

5.50

Result. Percentage finer given in the last column can be used to plot the particle s ize distribution curve with particle size as abscissa on log scale and the percentage fmer as ordinate. [Note. IT the fine fradion contains an appreciable amount of clay particles, the wet sieve analys is is requinxl. Alternatively, the following method may be used. Before conducting step (1), add the water containing sodium hexa- meta phosphate at the rate of 2 g per titre of water to the soil fraction. Stir the mix thoroughly and leave for soaking. Wash the soaked specimen on a 75.-. IS sieve until the water passing the sieve is clear. Thke the fraction retained on the s ieve and dry it in an oven. Sieve the oven dried soil through the nest of sieves as discussed in step (7). Perfonn further steps, as before. Obviously, the mass of material which would have been retained on pan is equal to the original mass of the soil before washing minus the dry mass of the soil retained on 75J! IS sieve after washing.]

30.9. Experiment No. 9. To detennlne the particle size distribution by the hydrometer method. Theor-y. Hydrometer method is used to determine the particle size distnbution of fine-grained soils passing 75J! sieve. 1be hydrometer measures the specific gravity of the soil suspension at the centre of ilS bulb. The specific gravity depends upon the mass of solids present, which, in lum, depends on the particle size. The particle size (D) is given by D - MViiJi

where M = [ g (Go~

1) p ... l ,in which, ~

YJ ,.. visoosity of water (POise), G ., specific gravity of SOlids; p ... =

density of water (gm/ml); g = 981 cm/sl:.cl, He = effective depth (em); t = time in minutes at which ot>scrvation is taken, reckoned with respect 10 the beginning of sedimentatioo. The percentage finer than tbe size D is given by

&>IL MECHANICS AND FOUNDATION ENGINEERING

798

N _

(~) G- I

x

..!i.. M.

x 100

where R = oorrected hydrometer reading; M. = mass of dry soil in 1000 ml suspension. Equipment. (1) Hydrometer; (2) Glass measuring cylinder (jar), I(XX) mli (3) Rubber bung for the cylinder (jar); (4) Mechanical stirrer; (5) Weighing balance, accuracy 0.01 g (6) Oven (7) Desiccator; (8) Evaporating dish; (9) Conical flask or beaker, 1000 mt; (10) SlOp watch; (11) Wash bottle; (12) Thermometer; (13) Glass rod; (14) Water bath; (IS) 75 ~ sieve; (16) Scale; (17) Dcflocculating agent.

Procedure. Part-I Calibration of hydrometer (1) Take about· 800 mt of water in one measuring cylinder. Place the cylinder 00 a table and observe the initial reading. (2) Immerse the hydrometer in the cylinder. Take the reading aner the immersion. (3) Determine the volume of the hydrometer (VH), which is equallO the difference between the final and initial readings. Altematively, weigh the hydrometer to the nearest 0.1 g. The volume of the hydrometer in ml is

approximately equal to its mass in grams. (4) Oetennine the area of cross·section (A) of the cylinder. It is equal to the volume indicated between any two graduations divided by the distance between them. The distance is measured with an acx:urale scale. (5) Measure the distance between the hydrometer neck and the bottom of the bulb. Record it as the height of the bulb (h). (6) Measure lhe distance (II) between the neck 10 each of the marks on the hydrometer (RII). (7) Determine the effective depth (H~). rorresponding to each of the mark (RII ). as

H·-lI+Hh-~) [Note. The fador VHIA should not be ronsidered when the hydrometer is 001 taken out when taldng readings after start of the sedimentation at 1/2, 1. 2 and 4 minutes]. (8) Draw a calibration OJrve between H~ and R~. Alternatively. prepare a table between Ho; and R~. The curve may be used for finding the effective depth H~ corresponding to reading ~.

Part-D Meniscus correction (1) Insert the hydrometer in the measuring cylinder containing about 700 mt of water. (2) Thke the readings of the hydrometer at the top and at the bottom of the menisals. (3) Determine the meniscus correction. which is equal to the difference between the two readings. (4) The meniscus correction (C",) is positive and is a constant for the hydrometer. (5) The observed hydrometer reading (R,.') is corm::ted to obtain Ihe corrected hydrometer reading (RII) as RII ~ RII ' + C", Part-m Pretreatment and Dispersion (1) Weigh accurately. to nearest 0.01 g. about 50 g air-dried soil sample passing 2 mm IS sieve. obtained by riffling from the air..
tABORA1ORY EXPERIMENTS

volume to 50 ml by boiling. Stop healing and cool the contents. (5) If the soil conlains insoluble calcium compowlds, add about 50 ml of hydrochloric acid to the cooled mixture. Stir the solution with a glass rod for a few minutes. Allow il 10 stand for one hour or so. The solution would have an acid reaction 10 litmus when the treatment is complete. (6) Filter lhe mixlure and wash it with warm water until the filtrate shows no acid reaction. (7) Transfer the damp soil on the filter and funnel to an evaporating dish, using a jet of distilled water. Use the minimum quantity of distilled water. (8) Place the evaporating dish and its contents in an oven, and dry it al 105° to llOoe. Transfer the dish to a desiccator. and allow it to cool. (9) Take the mass of the oven dried soil after pretreatment, and find tbe loss of mass due to pretreatment. (10) Add 100 ml of sodium hexa-metaphoophate solution to the oven-dried soil in the evaporating dish after pretreatment. (11) Warm the mixture gently for about 10 minutes. (12) Transfer the mixture to the cup of a mechanical mixer. Use a jet of distilled water to wash all traces of tbe soil out of the evaporating dish. Use about 150 ml of water. Stir the mixture for about 15 minutes. (13) Transfer the soil suspension to a 75 ~ IS sieve placed on a receiver (pan). Wash the soil on this sieve using a jet of distilled water. Use about 500 ml of water. (i4) Transfer the soil suspension passing 75 J1 sieve to a I(XX) ml measuring cylinder. Add more water to make the volume exactly equal to 1000 ml. (15) Collect the material retained on 75 II sieve. Dry it in an oven. Detennine its mass. If required, do the sieve analysis of Ihis fraction.

Part-IV. Sedimentation Thst (1) Place the rubber bung on the open end of the measuring cylinder containing the soil suspension. Shake it vigorously end-over-end to mix the suspension thoroughly. (2) Remove the bung aOer the shaking is complete. Place the mea
SOIL MECHANICS ANO FOUNDATION ENGINEERING

800

(13) For the detcnninalion of the compooitc correction (C). insert the hydrometer in the comparison cylinder containing 100 ml of dispersing agent solution in 1000 mt of distilled water at the same temperature. Take the reading corresponding 10 the lOp of meniscus. The negative of the reading is the composite

correction. Data Sheet for Hydrometer- Test Mass of dry soil (M,) = 50 gj Meniscus correction (C...) = +0.4 Specific gravity of solids (G) = 2.67 Calculations

Observtllions

S.

Elapsed

time(,)

No

1. 2. 3. 4.

5. 6. 7. 8. 9. 10. II. 12. 13.

1/2 minu. 1 2

Hydrometer Tempera- Composite correction keading (R.') (e) 21.80

22.0

-050

Corrt:Cted lIeig/u Reading Reading lIdcm) R = Rio" Rh' &' + C

22.2

U.8

21.3

Factor M

Particle ,iI<

D:M 1.33 x 10-

Percentage

filler

Viidi

(N)

0.065 mm

68.10

J5 30 1

h'

2 4 8

12 24

Result. Particle size distribution cUlve can be plolted using the last two columns.

30.10. Experiment No. 10. To determine the liquid limit of a soil specimen. 11100ry. The liquid limit of II soil is the water cootent at which the soil behaves prnctically like II liquid. but has n small shear strength. It flows to close the groove in just 25 blows in Casagrande's liquid limit device: As it is difficult to get exactly 25 blows in a test, 3 La 4 tests are oonducted, and the number of blows (N) required in each test is detennined. A semi·log plot is drawn between log N and the water content (w). The liquid limit is the water content corresponding to N = 25, as obtained from the plot. Equipment. (1) C4S3grnndc's liquid limit device; (2) Grooving tools of both Standard and ASTIvltype&i (3) Oven; (4) Evaporating dish or glass sheet; (5) Spatula; (6) 425 11 IS sieve; (7) Weighing balance, accuracy om g; (8) Wash boUle. Procedure. (1) Adjust the drop of the cup of the liquid limit device by relea<;ing the two screws at the top and by using the handle of the grooving tool or a gauge. The drop should be exactly 1 em at the point of contact on the base. Tighten the screw after adjustmcnt. (2) Takc about 120 g of the air-dried soil sample passing 425 "" IS sieve. (3) Mix tbe sample thoroughly with distilled water in an evaporating dish or a glass plate to fonn a unifonn paste. Mixing should be continued for about 15 to 30 minutcs, till a unifonn mix is obtained. (4) Keep the mix under humid conditions for obtaining uniform moisture distribution for sufficient period. For some fat ClaYS. this maturing lime may be uplo 24 hours. (S) Take a portion of the matured paste and remix it thoroughly. Place it in the cup of the device by a spatula and level it by a Sp.:1tula or a straight edge to have a maximum depth of the soil as 1 an at the point of the maximum thickness.

lABORATORY EXPERIMENTS

801

The excess soil, if any, should be transferred to the evaporating dish. (6) Qat a groove in the sample in the cup by using the appropriate tool. Draw the grooving tool through the paste in the cup along the symmetrical axis, along the diameter through the centre line of the cam. Hold the tool pcrpcndiClllar to the Cllp. (7) Tum the handle of the device at a rate of 2 revolutions per seoond. Count the number of blows until the two halves of the soil specimen rome in contact at lhc boltpm of the groove along a distance of 12 mm due to flow and nol by sliding. (8) Collea a representative specimen of the soil by moving spatula width-wise from one edge to the other edge of the soil cake, al righi-angles to the groove. This should include the portion of the groove in which the soil flowed 10 close the groove. Place the specimen in an aiHight container for the water oontent determination. Determine the water content. (9) Remove the remaining soil from the cup. Mix it with the soil left in Ihe evaporating dish, (10) Olange the water content of the mix in the evapornting dish. either by adding more water if the water content is to be increased, or by kneading the soil, if the water content is 10 be decreased. In no case, the dry soil should be added to reduce the waler content. (11) Repeat steps 4 to 10. and detcnnine the number of blows (N) and the water content in each case. (12) Draw the flow Cllrve between log Nand w, and determine the liquid limit corresponding to N :: 25.

Data Sheet for Liquid Umit Test S.No.

2.

3. 4.

5. 6. 7.

8.

Determination No.

Obserl'atiollS and ca/cu/atiollS Obsetvations No. of blows (N) water content can No. Mass of empcy can (MI) Mass of can + wet soil (M2) Mass of can ... dry soil (M3) Calculations Mass of waler .. M2 - 1.13 Mass of dry soil .. M3 - Ml

water content,

W ..

m)(

100

15 101

25.15 g 36.93 g 33.81 g 3.12 g 8.66 g

36%

Result. Draw a flow curve between log Nand w. Liquid limil (for N ::: 25) :: 30.11. Experiment No. 11. To detennine the plastic limit of a soli spedmen. Theory. The plastic limit of a soil is the water content of the soil below which it ceases to be plastic. 11 begins to crumble when rolled into Ihrea~ of 3 mm diameter. Equipment. (1) Porcelain evaporating dish, about 120 mm diameter or a flat glass plate, 450 mm square and 10 mm thick; (2) Ground glass plate, about 200 mm x 150 mm; (3) Metallic rod, 3 mm dia. and 100 mm long; (4) Oven; (5) Spalula or palette knife; (6) Moisture content can. Procedure. (1) Take about 30 g of air-dried soil from a thoroughly mixed sample of the soil passing 425J.l sieve. (2) Mix the soil with distilled water in an evaporating dish or on a glass plate to make it plastic enough to shape into a small ball. (3) L.eave the plastic soil mass for some time for maturing. For some fal clays, this period may be even upto 24 hours. (4) Take aboul 8 g of the plastic soil, and ~roU it with f1Oger.; on a glass plate. The rate of the rolling


8<"

should be about 80 [0 90 strokes per minute to fonn a thread of 3 mm diameter, counting one stroke when the hand moves forward and backwaid to the staring poin!. (5) If the diameter of the thread becomes less than 3 mm without cracks, it shows that the water content is more than tbe plastic limit. Knead the soil to reduce the water content, and roll it again into thread. Repeat the process of altemate rolling and kneading until the thread aumbles, and the soil can no longer

be rolled into thread. [Note. If the crumbling occurs when the thread has a diameter slightly greater than 3 mm, it may be laken as the plastic limit, provided the soil had been rolled into a thread of 3 mm diameter immediately before kneading. Do not attempt to prodlJCC failure cxaclly at 3 mm diameter.} (6) Collect the piccc::& of the crumbled soil thread in a moisture content oontaincf. (7) Repeal the procedure at least twice more with fresh samples of plastic soil each time. Data Sheet for PlasUc Umlt 'lbt Determination No.

SNo.

Obsen'aJiOIlS alia calculatiOlls ObservaliOIiS

2. 3. 4.

Moisture content container No. Mass of empcy container (MJ) Mass of container -t wet soil (M2 ) Mass of conlaincr -t dry soil (M3)

101 24.12 g 30.28 g 29.12 g

102

103

Ca/cul(lfiollS

5. 6.

Mass of waler '" M2 - 1.13 Mass of dry soil '" M3 - Mt

7.

Water Content,

W ..

~

x 100

1.16 g 5.00 g 23.2 %

Result. Plastic limit of soil '" 23.2%. 30.1:'. Experiment No. 12. To determine the shrinkage limit or a specimen or the remouldtd soiL Theory. The shrinkage limit is the water content of the soil when the water is just sufficient to fill all the pores of the soil, and the soil is just s.1turatcd. The volume of the soil does not dea'easc: when the water content is reduced below the shrinkage limit. It can be detennined from the relation W,

_

(M, - M,) - (V, - V,)P. M, )(100

where M J == inilial wet mass; VI == initial volume; M, == dry mass; V2 = volume afier drying. Equipment. (1) Shrinkage dish, having a flat bottom, 45 mm diameter and 15 mm height. (2) 1\Vo large evaporating dishes about 120 mm diameter. with a pour out and flat bottom; (3) One small mercury dish, 60 mm diameter; (4) 1\vo glass plates. one plain and one with prongs, 7S mm )( 75 mm )( 3 mm size, (5) Glass cup, 50 mm diameter and 25 mm high; (6) IS sieve 425 ~ ; (7) Oven; (8) Desiccator; (9) Weighing balance. accuracy 0.01 g; (10) Spatula; (11) Straight edge; (12) Mercury. Procedure. (I) Thke a sample of mass about 100 g from a thoroughly mixed soil passing 425 ~ sieve. (2) Thke about 30 g of the soil sample in a large evaporating dish. Mix it with distilled water to make a creamy paste which can be readily worked without entrapping the air bubbles. (3) ·fllke the shrinkage dish. CIc.1n it and detennine its mass. (4) Fill mercury in the shrinkage dish. Remove the excess mercury by pressing the plain glass plate over the lOp of the shrinkage dish. The plate should be flush with the top of the dish, and no air should be entrapped. (5) Transfer the mercury of the Shrinkage dish to a mercury weighing dish and delennine the mass of the

LADORXfORY EXPERIMENTS

803

mercury to an aocurncy of 0.1 g. The volume of the shrinlulge dish is equal to the mass of mercury in grams divided by the specific grnvity of mercury (viz. 13.6). (6) Coot the inside of the shrinkage disb with a Ihin layer of silicon grease or vaseline. Place the soil specimen in the centre of the shrinkage dish, equal 10 about one-third the volume of the Shrinkage dish. Tap the shrinkage dish on a finn, cushioned swface and aUow lhe paste to flow to the edges. (7) Add more soil paste, apprOlcimately equal to the first portion and lap the shrinkage dish as before, unlil the soil is thoroughly oompacted. Add more soil and oontinue the lapping till the shrinkage dish is completely filled, and excess soil paste projects out about its roge. Strik.e out the top surface of the ~e with a straight edge. Wipe off all soil adhering to the outside of the shrink.age dish. Determine the mass of the wet soil (M1). (8) Dry the soil in the shrinkage dish in air until the colour of the pat tums from dark to light. lben dry the pat in the oven at 105° to 110°C to constant mass. (9) Cool the dry pat in a ~iccator. Remove the dry pat from the desiccator aftcr cooling, and weigh the shrinkage dish with the dry pat-io detcrmine the dry mass of the soil (M,). (10) Place a glass cup in ~ large evaporating dish and fill it with mercury. Remove the excess mercury by pressing the gla<>s plate with prongs firmly over the top of the cup. Wipe off any mercury adhering to the outside of the cup. Remoye the glass cup full of mercury and place il in another evaporating dish, taking care not to spill any mercury from the gla<;S cup. (II) Take out the dry pat of the soil from the shrinkage dish and immerse it in the glass cup full of mercury. Thke care not to entrap air under the pat. Press the plate with prongs on the top of the cup ftrmly. (12) Colled: the meroJry displaced by the dry pat in tbe evaporating dish, and transfer it to the mercury weighing dish. Detennioe the mass of the mercury to an accuracy of 0.1 g. The volume of the dry pat (Vi) . is equal to the mass of the mercury divided by the specific gravity of mercury. (13) Repeat the test atleasl 3 times.

Data Sheet ror Shrinkage limit 'Jest Deternu'nntion No. ObserwlJions and colculaliDns

$.No.

Observotions

3.

Mass of empty mercury dish Mass of mercry dish, with mercury equal 10 volume of the shrinkage dish Mass of mercury .. (2) - (1)

4.

\t>tume of sMnlcage dish, VI_

I.

2-

~

5. 6.

Mass of empty shrinkage dish Mass of shrinkage dish + wet soil

7.

Mass of wei soil, Ml - (6) - (5) Mass of shrinkage dish + dry soil

&

9. 10.

Mass of dry soil, M, _ (8) - (5)

11.

Mass of mercury dish + mercury equal in volume of dry pal Mass of mercury displaced by dry pat. = (10) - (1)

12.

\blume of dry pal, V2 ..

m

74.2 g 361.1g 286.9 g

21.1 mI

235, 68.4g 44.9 g 57.3 g 33.8 g 304.3 g 230.1 g 16.92 mI

804


DeJerminaJionNo.

ObservoJions and caiculaJions

S.No.

CQ/culations Il.

Shrinkage limit,

WI

_

(MI-M,)-(VlM,

Vi)pw)

v7~

14.

Shrinkage r.:nio, SR ..

15.

\tllumetfic shrinkage, VS .. ( VI

20.5%

2.0

~2 Y2)

)( 100

24.70

Result. Shrinkage Limit = 20.5%.

30.13. Experiment No. 13. To determine the permeability or a soli specimen by the constant-head penneameter. Theory. "Ibe coefficient of permeability is equal to the rate of flow of water through a unit cross·sectional area under a unit hydraulic gradient. In the constanl head penncametcr, the bead causing flow through the specimen remains oonstant throughout the test. The coefficient of permeability (4-) is obtained from the relation

k-~-ffr where q ::: discharge; Q = total volwne of water, t = time period; h = head causing Dow; L = length of specimen; A '" aoss..sectiooal area. Equipment. (1) Penneameter mould, internal diameter = 100 mm, effective height = 127.3 mm, capacity::: 1000 mt; (2) Detachable collar, tOO mm tliameter, 60 mm high; (3) Dummy pJate, 108 mm dianeter, 12 mm thick; (4) Drainage base, having a porous disc; (5) Drainage cap, having a porous disc with a spring attached to the top; (6) Compaction equipment. sud:! as Proctor's mmmer or a static cxxnpactioo equipment; (7) Constant- head water-supply reservoir; (8) Vacuum pump; (9) Constant-head collecting chamber; (10) Stop watch; (11) Large funnel; (12) Thermometer; (13) Weighing balance, acruracy 0.1 g: (14) Filter papa. Procedure. (1) Remove the collar of the mould. Measure the intcmal dimensions of the mould. Weigh the mould, with dummy plate. to the nearest gram. (2) Apply a little grease on the inside to the mould. Clamp the mould between the base plate and tbe extension collar, and place the assembly on a solid base. (3) Take about 2.5 kg of the soil sample, from a thoroughly mixed wet soi~ in the mould. Compact the soil at the required dry density, using a suitable compacting device. (4) Remove the collar and base plate. Trim the excess soil level with the top ot the mould. (5) Clean the outside of the mould and the dummy plate. Find the mass of the soil in tbe mould. (6) Thke a small specimen of the soil in a container tor the water mntent determination. (7) Saturate the porous discs (stones). (8) Place a porous disc on the drainage base, and keq> a filter paper on the porous disc. (9) Remove the dummy plate, and place the mould with soil on the drainage base, after inserting a washer in between. (10) aean the edges of the mould. Apply grease in the grooves around them. (11) Place a filler paper. a porous disc and fix the drainage cap using washers. (12) Connect the water reservoir to the outlet at the base, and allow the water to Dow upwards till it has saturated the sample. Let the free water mllCCl for a depth of about 100 mm on the top of the sample. [Alternatively. the soil of low permeability can be saturated by subjecting the specimen to a gradually increasing vaaJum with bottom outlet cklsc4 so as to remove air from the voids. ~ the vacuum

"',

LABORA1ORY EXPERIMENTS

gradually to 700 mm of mercury and maintain it for 15 minutes or more, depe'uting upon the type of soil. Follow the evacuation by a process of slow saturation of the sample from th.: bottom upward under full vacuum. When the sample is saturated, close both the top and bottom oullets}. (13) Fill the empty portion of the mould with deaired water, without disturbing the soil. (14) Disconnect the reservoir from the ouUet at the bottom. (15) Connect the constant,bead reservoir to the drainage cap inlet. (16) Open the siop cock. and allow the water to flow dowoward so that all the air is removed. (17) Close the stop cock, and allow the water to flow through the soil till a steady state is attained. (18) Start the stop watch, and collect the water flowing out of the base in a measuring flask for some convenient time interval. (19) Repeat this thrice, keeping the interval the same. Cleek that the quantity of water colJected is approximately the same each time. (20) Measure the difference of bead (h) in levels between the constant bead reservoir and the outlet in the base. (21) Repeat the test for at least 2 more djfferent intcrvals. Data Sheet for Constant fiend Penneameter

Diameter

= 100 mm:

Length

= 120 mm;

\blume

= 942.48 rot. G·= 2.67: Area = 7854 mm2 , DetermillOlion No.

Observations tmd calculatiolls

S.No.

Observations

1. 2 3. 4. 5.

Mass of empty mould with base plate Mass of mould, soil and base plate Hydraulic bead (h) llrne interval (t) Quantity of flow (Q) (a) First time in period t (b) Second time in period t (e) Third time in period I

Average Q 6. 7.

Calculations Mass of soil'" (2) - (1)

BuJk density, p -

8.

Water-«>ntent,

9.

Dry density, f>d -

W,

v~ determined as in Test 30 - 1

~

10.

~dRatio,e - ~-1

11.

k

_!Jb.

5101 g 6918 g 150 mm 600 s 1210 mI 1205 m1 1215 ml 1210 mI 1210 j( 101 rom3

1817 g 1.93 gtml

14 % 1.69 glmI 058

0.205 mm/sec.

Alu

Result. Coefficient of penneability = 0.205 mm/sec. 30.14. Experiment No.. 14. To determine the penneablUty of • soli specimen by the variable-head perm~meter.

Theory. The variable-held penneameter is used to measure the penneability of relatively less pervious soils. The coefficient of penneability is given ~y

k _ 2,~:L logto(h t lh0


806

where hI = initial head; h2 = final head; t = time inlclVal; a = cross sectional area of the stand pipe, A :: cross-sectional area of the specimen, L = length of specimen. EquipmenL Alllhc equipment required for the oonst:lOt-hcad permeability lest (Experiment 30-13), and tbe following.

(1) Graduated glass stand pipe, 5 to 20 mm diameter. (2) Supporting frame for tbe stand pipe. and the clamp. Procedure. Steps 1 to 14, same as in Experiment 3Q..13. (15) Connect the sumd pipe of suitable diameter to the inlet al the lop. Fill the stand pipe with water. (16) Open the stop oock at the lop, and allow the water to flow out till all the air in the mould is rcm()\led.

(17) Oosc the stop cock, and allow the water from the stand pipe 10 flow through the soil specimen. (18) Select the heights hL and h2 measured above the centre of the outlet such that their difference is about 3(X) to 400 mm.

Mart the level corresponding to a height ~h-;. (19) Open the valve and start the SlOp watch. Record the time interval for the head to fall from hi to

Y;;:h-;

...fh;h;.,

and also from to hz. These two time intervals will be equal if the steady conditions have established. (20) Repeat the step (19) atJea<>t twice, after changing the heights hJ and h 2 • (21) Stop the flow of water. Disconnect all the P.1rts. (22) Take a small quantity of the soil specimen for the water cootent dctennination. Data Sheet ror Variable-head Permeameter Test Length of specimen

= 120 mm.

Area of specimen

= l't/4 )( (100)2 = 7853.98 mm 2.

Volume of specimen Water content Diameter of stand pipe Area of stand pipe,

Specific gravity of solids

= 1l/4 )( = 18%

Diameter", 100 mm

(100)2 )( 120 _ 942.48 )( 103 mm 3

= 10 mm a = rc/4 )( (10)2 _ 78.54 mm2

= 2.67. Detamination No.

S. No.

1. 2. 3. 4.

Obsuvations and CakulaJions Observations Mass of roould ... base plale Mass of mould ... base plate ... soil Inilial head, ht Final head, h2

5.

I-lead~

6.

TIme Inlerval hllO

V1i1Iii

V1il1i2 7.

5090 g

7120g SOOmm 200mm 316 mm

2S.

to h2

25,

hi to h2 Calc/.l/arions Mass of soil ,. (2) - (1)

50,

'"30 g

LAOORATORY IlXPERIMENTS

s.

No.

f-_ _ _---rD:.::":.::eI7liIlQliOIl No.

Obscn'OtiOIlS GIIII CQicuimiolls

8.

Bulk density, p ...

~::

2.15 glmJ

9.

Dry density. flJ ..

"I"!-;

1.83 g/mi

10.

Void mtia, e ..

11.

k ..

~

- 1

0.46

P'

2.3~: L

loglo (hllllz)

Result. Coefficicnt of permeability

0.022 mm/s

= 0.022

mm/scc.

30.15. Experiment No. 15. To detennlne the consolidation c:humcteristlcs of II soil sumple. Tbeory. Coosolidulion of a saturated soil occurs due to expulsion of water under a static, susrained lo..1d. TIlC consolidation characteristics of soils are required to predict the magnitude and the rate of settlement. The following characteristics arc obtained from the consolidation test. As per usual notations (see chapter (2), CoeffICient of comprcssiblly, Coefficient of volume change. Compression index,

Coefficient of consolidation,

c••

T. ,fl l

Equipment. (I) Coosolidometer, with a loading device; (2) Specimen ring, made of a non-corroding material; (3) Wntcr reservoir 10 salumte the sample; (4) Porous stones; (.5) Soil trimming tools, like nne wire .saw, knife, spatula, etc.; (6) Weighing balance, accuracy 0.01 g; (1) Oven; (8) Desiccator; (9) Prc:ssure pad: (10) Steel ball; (II) Dial gauge, accuracy 0.002 mOl; (12) Water content cans: (13) Large container.

Procedure. (1) Ocan and dry the metal ring. Measure its diameter and height. Take the mass of the empty ring.

(2) Press the ring into the soil snmple conlaincd in a large eoolainer at the desired density and the water contem. 1be ring is to be pressed with hands. (3) Remove the soil around the ring, The soil specimen should project about 10 mm on either side of the ring. Any voids in the specimen due to the removal of largc size particles should be filled back by pressing the soil lightly. (4) Trim the specimen flush with the top and bottom of the ring. (5) Remove any soil particles sticking to the outside of the ring. Weigh the ring with the specimen. (6) Thkc a small quantity of the soil removed during trimming for the water conteDt determination. (7) Saturate the porous stones by boiling them in distilled water for about 15 minutcs. (8) A<;.<;emble the c:onsolidomcter. Place the bottom porous stone, bottom filler paper, specimen, top filter paper and the top porous stone, one by one. (9) Position the loading block centrally on the top porous stone. Mount the mould assembly on the loading frame. Centre it such that the 1000 applied is axial. 10 the case of the levcr·looding systcm, counterbalance the system. (to) Sct the dial gauge in position. Allow sufficient margin for the swelling of the soil


808

(11) Connecl the mould assembly to the water reservoir having the water-level al about the same level as the soil specimen. Allow the water to flow into the specimen lill it is fully saturated. (12) T'dke the initial reading of the dial gauge. (13) Apply an initiaJ setting load to give a pressure of 5 kN/ml (25 kN/m2 for very son soils) to the assembly so that there is no swelling. Allow the selling load to stand lill there is no change in the dial gauge reading or for 24 hours. (14) Thke the final gauge reading under the initial setting load. (15) Apply the first load increment to apply a pressure of 10 kN/m2, and start tbe Slop walch. Record the dial gauge readings at 0, 0.25, 1.0, 2.25, 4.0, 6.25, 9.00, 12.25, 16.00, 20.25, 25.00, 36. 49, 64, 81, 100, 121 , 144, 169, 196. 225, 156, 289, 324, 361, 400, 500, 600 and 1440 minutes. (16) Increase the load 10 apply a pressure of 20 kNfm2i and repeal the step (15). Ukewise. inaease the load to apply a pressure of 40, BO, 160, 320 and 640 kN/rn or upto the desired pressure. (17) After the last load inacment had been applied and the readings taken, decrease the load 10 1/4 of the last load, and allow it to stand for 24 houl'S. Thke the dial gauge reading aner 24 houl'S. Further reduce the load to 1/4 of the previous load and repeat the above procedure. Likewise, further reduce the load to 1/4 of the previous load and repeat the procedure. FinaUy, reduce the load 10 the initial selling load, and keep it for 24 haul'S, and take the fin.'ll dial gauge reading. (18) Dismantle the assembly. Take out the ring with the specimen. Wipe out the excess surface water using a bl()(ting paper. (19) Thke the mass of the ring with the specimen. (20) Dry the specimen in the oven for 24 hours, and determine the dry mass of specimen.

Data Sheet for consolidation lest

Specific gravity of solids. G = Area of ring (A) = Mass of ring + wet soil = Mass of water = Water coolent before lest = Height of SOlids. Hs

= G ~sp",

Height of ring = Degree of saturation S =

7

Diameter of ring = Volume of ring = Mass of ring + dry soil ,. Initial height, Ho = Mass of dry soil (Ms) = Initial void ratio,

Mass of ring

eo -

!ff - 1

=

Water content after test =

(0) Coefficient of Compressibility Change in hight (.o.U)

Height H .. Ho;t:..o.H

Height ofvoith (1I~. 1I,)

Final void ratio e .. (H-H,)IH,

10 20

40 80 160 320 640

Plot a curve between

a as abscissa and final void ratio (e) as ordinate for determination of a~ and my-

Plot a graph between log 0 as abscissa and flllal void raUo as ordinate for determination of Ceo

LABORAlORY EXPERIMENTS

809

(b) Coemcient of collwlidatlon

Dial gauge readings 10

(kNlm'l

40

20

80

160

320

640

(R)

0.0 min. 0.25 1.0

1440

For each load increment, plol Vi as abscissa and the dial gauge reading (R) as ordinate. Determine the value of 19() from the plot.

Now

C, - O.84&iIt,o

30.16. Experiment No. 16. To detennine the shear parameters of a sandy soil specimen by direct shear test. Theory. Shear strength of a soil is its maximum resistance lo shearing stresses. The shear strength is expressed as s_c'+otanfjl' where c' effective cohesion; a effective stress; and fjI' effective angle of shearing resistance. The shear tests can be conducted under three different drainage conditions. The direct shear test is generally conducted on sandy soils as a consolidated-drained test. Equipment. (1) Shear box, divided inlo two halves by a horizontal plane, and fitted with locking and spacing screws; (2) Box container to hold the shear box; (3) Base plate baving cro;s grooves on its top surface; (4) Grid plates, perforated, 2 nos; (5) Porous stones, 6 mm thick, 2 nos; (6) Loading pad, (1) Loading frame; (8) Loading yoke; (9) Proving ring, capacity 2 leN: (10) Dial gauges, accuracy 0.Q1 mm, 2 nos.; (1l) Static or dynamic compaction device; (12) Spatula. Procedure. (1) Measure the internal dimensions of the shear box. Also detennine the average thickness of the grid plates. (2) Fix the upper part of the box to the lower part using the locking screws. Attach the base plate to the lower part. (3) Place the grid plate in the shear box keeping the serrations of the grid at right angles 10 Ihe direction of shear. Place a porous stone over the grid plale. (4) Weigh the shear box with base plate. grid plale and porous stone. (5) Place the soil specimen in the box. Tamp it directly in the shear box ·at the required density. When the soil in the top half of the shear box is filled upto 10 to 15 mm depth, level the soils surface. (6) Weigh the box with the soil specimen. (7) Place the box inside the box oontainer. and fix the loading pad on the box. Mount the box container on the loading frame. (8) Bring the upper half of the box in contact with the proving ring. Oleck: the contact by giving a slight movement. (9) Fill the container with water if the soil i.s to be saturatedi otherwise omit this step. (10) Mount the loading yoke on the ball placed on the loading pad. (1l) Mount one dial gauge on the loading yoke to record the vertical displacement and another dial gauge on the container to record the horizontal displacement. (12) Place the weights on the loading yoke 10 apply a normal stress of 25 kN/m 2.

=

=

=


810

Allow the sample to consolidate under the applied nonnal Stress. Note the reading of the vertical displacement dial gauge. (13) Remove locking screws. Using the spacing screws, mise the upper part slightly above the lower part such thai the gap is slightly larger than the maximum particle size. Remove the spacing screws. (14) Adju<>t aU the dial gauges to read zero. The proving ring should also read zero. (15) Apply the horizontal shear load at a constant rate of strain of 0.2 mm/minute. (16) Record readings of the proving ring. the vertical displacement dial gauge and. the horizontal displacement dial gauge at regular time intervals. Thke tbe first few readings at closer inlcr. als. (17) Continue the test till the specimen fails or till 8 strain of 20% is reached. (18) AI the end of the lest, remove the specimens from the box, and take a reprcsent.1tive sample for the water rootenl determination. (19) Repeal the lest 00 identical specimens under the normal stresses of 50, 100, ZOO, 400 kN/m 1, etc. (The range of stresses selected should correspood to the actual fiekl conditions).

Data Sheet ror Direct Shear lest Area of box = Volume of specimen = Bulk density = Dry density = . Tare mass of hanger = Total mass =

Size of box = 1bickness of specimen = Mass of soil specimen =

Water content '"

Void ratio = Mass on hanger = Nonnal stress = Mass of box + base plate + pol'OU$ stones + grid plale = Mass of box + base plate + porous stone + grid plate + soil specimen S. No.

Elapsed rim<

lIoriUHllol dial gauge

Vemcal dial gange

=

CalculaJions

Observalions

P1'rMfl8

ring

Shear Vertical ""pk=m
SMar

f-

Use separate data sheet for tests under different nonnaI stresses. Detennine the sbear stress at failure in each case. Summarise the results as follows.

Test No. 1. 2. 3. 4. 5.

N~I

51=

Shear Slress aI/ailure

Shear dlsplacmum/ aI/Dilure

Initial MXIJer

FillQilWller

25 kN/m 50 100 200 400

Plot the Coulomb envelope between the oonnal stress as abscissa and shear stress at failure as ordinate.

Result. From the plot, c' =

; .'

=

LADORAJ"QRY EXPERIMENTS

811

30.17. Experiment No. 17. To determine the unconfined compressive strength of a cohesive soli. TIleory. The unconfined compressivc strcngth (q..) is the load per unit area at which the cylindrical specimen of a Cohesive soil fails in compression. 'p

q• •

A

where P = axial load at failure; A = corrected area _ 1 A-" E' where Ao initial area of thc specimcn:

E

= axial

strain = change in length/original lenglh. The undrained shear strength (s) of the soil is equal to one half of the unoonfmed compressive strength, s - qul2. Equipment. (1) UnconflOed compression apparatus. proving ring type; (2) Proving ring, capaCity I kN, accuracy 1 N; (3) Dial gauge, mm; (4) Weighing balancc: (5) Oven; (6) Stop watch; (7) Sampling tubc; (8) Split mould, 38 mm diameter, 76 mm long; (9)Srunple extractor; (10) Knife; (11) Vernier callipers; (12) Large mould. Procedure. (I) Prepare the soil specimen at tbe desired water content and density in the large mould. (2) Push the sampling tube into the large mould, and remove the sampling tube filled with the soil For undisturbed samples, push the sampling tube into the clay sample. (3) Saturate the soil sample in the sampling tube by a suitable method. (4) Coat the split mould lightly with a thin layer of grease. Weigh the mould. (5) Extrude the &'lmple out of the sampling tube into the splil mould, using the sample extractor and the knife. (6) Trim the two ends of thc specimen in the split mould. Weigh the mould with the specimen. (1) Remove the specimen from the split mould by splitting the mould into two parts. (8) Measure the length and diameter of the specimen with a vernier callipers. (9) Place the specimen on the bollom plate of the compression machine. Adjust the upper plate to make contact with thc specimen. (10) Adjust the dial gauge and the proving-ring gauge to zero. (11) Apply the compression load 10 cause an axial slrain al the rate of 1/2 to 2% per minute. (12) Record the di
accuracy om

Duta Sheet for UnconOned Compression Test Initial length of the specimen, Lo = Initial area of the specimen, Ao = Mass of empty split mould = Mass of the specimen, M = water content, W =

Initial diameter of the specimen, Do = Initial volume of the specimen, Vo = Mass of split mould + specimen = Bulk density, P = M1Vo Dry density, Pd =

Specific gravity of SOlids, G =

Void ratio, e _ Gp... _ 1 Pd

Degree of saturation, S = weG x 100


812

Calculations

ObSD'Wllion.s

S. No.

Elapsed rim<

Proving ring

Dial gouge Reading

Deformation

(H)

Reading

Load (P)

S"''"'AL E'"

Ii)

COm!cted

Compressive

areaA.-AQI

stress

(I - t)

(0) ... PIA

Plot a curve between the compressive stress as ordinate, and axial strain. as abscissa. Results. From the plot, unoonfined compressivc strength, q.. =

Shear strength, s _

~ ...

20.18. Experiment No. 18. To determine the compaction characteristics of 8 soU specimen by Proctor's test. Theory. Compaction is the process of densification of soil by reducing air voids. The degree of compaction of a given soil is measured in terms of Its dry densily. 'The dry density is maximum at the optimum water content. A curve is drawn between the water content and the dry density to obtain the maximum dry density and the optimum water content.

Dry dcn<>ity =

1M::

where M == total mass of soil; V = volume of SOil; and w = water content. Equipment. (1) Compaction mould, capacity 10Cl0 mJ: (2) Rammer, mass 2.6 kg; (3) Detachable base pIate: (4) Collar, 60 mm high; (5) IS sieve 4.75 mm: (6) Oven; (7) Desiccator; (8) Weighing balance,

aocurncy 1 g; (9) Large mixing pan: (10) Straight edge; (11) Spatula; (12) Graduated jar; (13) Mixing tools,

spoons. trowels. etc. Procedun. (1) Thke about 20 kg of air..dried soiL Sieve it through 20 mm and 4.75 mm IS sieves. (2) Calrulate the percentage retained OIl 20 mm sieve, and 4.75 mm sieve, and the percentage passing 4.75 mm sieve. Do not use the soil retained on 20 rnm sieve. Determine the ratio of fraction retained and that passiog 4.75 mm sieve. (3) IT percentage retained OIl 4.75 rnrn sieve is greater than 2D. use the larger mould of 150 mm diameter. If it is less than 20%, the standard mould of 100 rnm diameter can be used. The following procedure is for the standard mould. (4) Mix the soil retained on 4.75 mm sieve and that passing 4.75 mrn sieve in the proportk>os determined in Step (2) to obtain about 16 to 18 kg of soil specimen. (5) Clean and dry the mould and the base plate. Grease them lighUy. (6) Weigh the mould with the base plate to the nearest 1 gram. (7) Take about 16-18 kg of soil specimen. Add water 10 it to bring tbe wattt content to about 4% if the soil is sandy and 10 about 8% if tbe soil is clayey. (8) Keep the soil in an air-tight container for about 18 to 20 hours for maturing. Mix the matured soil thoroughly. Divide the processed soil into 6 to 8 parts. (9) Attach the collar to the mould. PIaoe the mould on a solid base. (10) Thke about 2~ kg of the processed sou. and place it in the mould in 3 clqual layers.

Thk.e about one-third the quantity first, and compact it by giving 15 blows of tbe rammer. The blows

S13

should be uniformly distributed OYeC the surface of each layer. The top surface of the first layer should be saalched with a spatula before placing the scoood layer. The seoood layer should also be compacted by 25 blows of rammer. Ukewise, place the third layer and compact it. The amount of the soil used srould be just sufficient 10 fill tbe mould and leaving about 5 mm above the top of the mould to be strude off when tbe collar is removed. (11) Remove the collar, and trim off tbe excess soil projecting above the mould using a straight edge. (12) Qean the base plate and the mould from oulSide. Weigh it 10 the nearest gram. (13) Remove the soil from the mould. The soil may also be ejected out. (14) Thke the soil samples for the water content determination from the top, middle and bottom portions. Detennine the water conlent, as in Experiment 30---1. (15) Add about 3% of the water to a fresh portion of the processed SOil, aod repeat the steps 10 10 14.

Data Sheet for Compaction lest Diameter of mou.ld = 100 mm Volume of mOUld, V - x/4 )( (10.0)2 Specific gravity of solids, G = 2.67 S.

No.

Height of mould X

1. 2.

Obsuvalions Mass of empty mould. base plate Mass oC mould ... base plait ... canpaaed soil

3.

Cokulotioru Mass of compacted soil, M _ (2) - (1)

%-

4.

BulK density, P ..

5.

Water content, w (Delermined as in Sect. 30-1)

17-;

Dry density. p-' -

Voidralio,e_~_l

S.

Ory density 8t 100% saturation

yml

9%

0.80

~;;G

Degree of saturation, S ..

1605 g

1.48g1m1.

pd

(Pd) lheomax = 1

5105 g 6710 g

1.61

6.

9.

DetermiMliOll No.

ObservaliOfJS and Co/culmions

1.

= 127.3 mm

12.73 _ 1(XX) mt

:U5g1ml

7-

x 100

30%

Plot a curve between w, as abscissa, and Pd as ordinate. ResulL Max. dry density (from plot) = Optimum water cootent (from plot)

IE

30.19. Experiment No. 19. To determine the CaliIomla Bearing Ratio (CBR) of a soU specimen. Theory. The California Bearing Ratio test ~ conducted for evaluating the suitability of the subgracle and the materials used in sub-b3se and base of a flexible pavement The plunger in the CBR test penetrates the specimen in the mould at the rnte of 1.25 mm per minute. The loads required for a penetration of 2.5 mm and 5.0 mm are determined. The penetration toad is expressed as a percentage of the standard loads at the respective penetration level of 2.5 mm or 5.0 mm.

=)(


814

CBR value.

P~::~on

tOO

The CaR value is determined corresponding to both penctratioo levels. The greater of these values ~ used for the design of the pavement. Equipment. (1) CaR mould, inside diameter = 150 mm, total height = 175 mm, with detachable extensioo collar, 50 mm high, and detachable base plate, 10 mm thick.. (2) (3) (4) (5)

Spacer disc. 148 mm diameter, 47.7 mm high. Rammers, light compaction, 2.6 kg. drop 310 mmj heavy compaction, 4.89 kg, drop 450 mm. SloUed masses, annular, 2.5 kg each, 147 mm diameter, with a bole of 53 mm diameter in the centre. Cutting co~r. sleel. which can fit flush with the mould both outside and inside.

(6) Expansion measuring apparatus, cornisting of a perforated plate. 148 mm diameter, with a thread screw in tbe centre and an adjustable contact head to be screwed over the stem, and a metal tripod. (7) Penetration piston, 50 mm diameter, 100 mm long. (8) Loading deVK:c:, capacity 50 kN t equipped with a movable bead (or base) at a unifonn rate of 1.25 mm minutc. (9) 1\110 dial gauges, accuracy 0.01 mm. (to) IS sieves. 4.75 mm and 20 mm s ize. Procedure. (1) Sieve the sample through 20 mm IS sieve. Thke the material passing 20 mm IS sieve for the test. However, makc allowance for large size material by replacing plus 20 mm size material by an equal amount of material which passes 20 mm IS sieve. but is retained on 4.75 mm IS sieve. (2) Thke about 4.5 to 5.5 kg of the material, as obtained in step (1). Mix it thoroughly with the required quantity of water. If the sample is to be compacted at optimum water content aod the corresponding dry density, as found by oompactioo test (light compaction or heavy compaction), take exact quantity of water and the soil to make sure thai the water content is equal to the optimum water content. (3) Fu: the extension collar to the top of the mould. Also fix tbe base plate to the bottom. (4) Insert the spacer disc over the base, with the central bole of the disc at the lower face. Place coarse filter paper disc on the lop of the displacer disc. (5) lake the soil sample in the mould. Compact it using either the light compactioo rammer or the heavy compaction rammer, as desired. For light compact~n, the soil is to be compacted in 3 equal layers, each layer is given 56 blows by 2.6 kg rammer with drop of 310 mm. For heavy compaction, the soil is compacted in 5 equal layers, each layer is given 56 blows by 4.89 kg rammer with drop of 450 mm. (6) Remove the extensioo collar. Trim even the excess compacted soil carefully with a stnlight edge with tbe top of the mould. Any hole that may form on the surface of the compacted soil by the removal of the coarse particles should be patched with small size Plrticles and levelled. (7) Loosen the base plate. Remove the base plate and the spacer disc. (8) Weigh the mould with the oompacted soil. (9) Place a fLiter paper disc on the base plate. Invert the mould with the compacted soil Clamp the base plate. Place a perforated disc fitted with an extension stem 00 the specimen top after placing a fllter disc. (10) Place annular masses to produce a surcharge equal to the mass of the base material and wearing cost of the pavement expected. Each 2.5 kg annular mass is equivalent 10 70 mm of coostruction material. However, a minimum of two annular masses should be placed. (11) Immerse the mould assembly in a tank full of water. Allow free access of water to the lOp and bottom of tbe specimen. (12) Mount the tripod of the expansion measuring device on the edge of the mould, and take the initial reading of the dial gauge. (13) Keep the mould in the tank undisturbed for 96 hours. Take the readings of the dial gauge every 24

lABORATORY EXPERIMENTS

815

hourn, and note the time of reading. Maintain water level constant in the tank. Take the final reading at the end of 96 hours. (14) Remove the tripod. Thke out the mould from the lank. Allow the specimen to drain off for 15 minutes. Remove all the free water on the moUld, without disturbing the surface of the specimen. (15) Weigh the mould with tbe soaked specimen. (16) Place the mould containing tbe specimen, with the base plate in position but the top face exposed. 00 Lhe lower plate of the loading machine. Place the required surcharge masses on the top of !he soaked specimen. To prevent upheaval of soil into the hole of the surcharge mass, ooe 2.5 kg annular malis shall be' placed on the soil surface prior to seating the penetration plunger. After that the remaining masses are placed. (17) Seat the penetration plunger at the centre of the specimen to establish full contact between the plunger and Ihe specimen. The sec'lling lood Should be about 40 N. (18) Set the load dial gauge and the displacement dial gauge to zero. The initial load already applied to tbe plunger should be considered as zero. (19) Awly the load on the plunger. Keep the rate of penetration as 1.25 mm/minute.

Record the load corresponding to pcnetraLions of 0.0, 05, 1.0, 1.5,2.0,2.5, 3.0, 4.0, 5.0, 75, 10.0 and 12.5 mm. However, record the maximum load and the corresponding penetration jf it occurs at a penetration of less than 125 mm. (20) At the end of the test, raise the plunger, and remove the mould from the loading machine. (21) Thke about 20 to 50 g of soil specimen from the top 30 mm layer for the water content determination. If the water content of the whole specimen is required, take soil specimens from the entire depth.

Data Sheet ror California Bearing Ratio lest Optimum water cootent = Mass of mould + oompaded soil Bulk density =

=

Mass of empty mould = Mass of compacted soil = Dry density =

Soaking and Swelling Dry density before soaking = Bulk density before soaking = Bulk densily after soaking = Surcharge mass during soaking = Date and time

Dial gouge reading

Total expansion Final expansion

ratio _Final rc~!~rta;~~~~~

Penet["8tion Test Surcharge mass used = Water CQfltenl aftcr penetration test ::

reading

81.


Penl!lrDlion dial gOllge

S.NO.

Dial gauge

PeMlrOlion

reading

(mm)

1.

0.0 mm

2

O.S

3. 4.

1.0 loS 2.0

S. 6. 7. 8. 9. 10.

7S

U.

10.0

12.

12.05

[.oQd

Dial gouge Mlding

dial gauge

Load (leN)

Corfl!Cted '~d

2.S 3.0 4.0

S.O

Plot the load-penetration curve. Find the corrected loads, after zero correction, corresponding to penetrations of 2.5 mm and 5.0 mm. Result. CBR (2.5 mm) _ Corrected ;~!at 2.5 mm )( 100

CBR(S.Omrn) _

CotTeCtcd:':68t5.0mm

Ie

100

31 Introduction to Rock Mechanics 31.1. INTRODUCTION Rock mechanics is the science whiCh deals with the properties of rocks and the special methods related with the design and construction of cngin'eering works on the rock, through the rock and below the rock. The basic knowledge of rock mechanics is useful for civil engineers. Some of the oommon civil engineering applications are bridges, dams and buildings oonslructcd on lhe rock foundations, various underground installations and tunneling, deep cuts for spillways, slone quarries, etc. Rock mechanics is an inter-disciplinary subject covering various disciplines such as geology, mining, petroleum industry and civil engineering. As defined in chapter I, rock is the consolidated, coherent and relatively hard portion of the earth's crust. It is a naturaDy formed, SOlidly bonded mass of mineral matter which cannot be readily broken by hand, and which does nOI disintegrate on its first wetting and drying cycle. In general, rocks arc strong materials and can take up much more loads as compared to that by soils. Howevcr, the strength of the rocks is also limited. 1be rock may fail whcn the loads are excessively high. The aaual behaviour of a rock mass subjected to a change in stress is governed by the mechanical properties of the rock material and the geological disoontinuities such as faults, joints, fissures, etc. large bridges, high dams, tall buildings. long tunnels and deep mines induce large stress manges In the rock mass. For exact analysis and proper prediction of the rock behaviour, a complete understanding of the engineering properties and the behaviour of rock masses under different loading conditions is essentiaL This chapter gives a brief introduction to the engineering properties and behaviour of rocks. 31.2; GEOLOGICAL CLASSIFlCATION OF ROCKS According to the geological classification, the rocks can be broadly classified into 3 groups. 1. rgneous rocks 2. Sedimentary rocks 3. MetamorphiC rocks. 1. Igneous rocks. Igneous rocks are fonned by solidification of molten or liquid material called magma. lbe igneous rocks have the minerals augite, feldspar, home blende, mica, quartz, etc. Before solldificatioo. into rocks, all these mineral arc in the molten state. The igneous rocks may be aystalline or glossy (vitreous) or a combination of both. The igneous rocks may be further Slbdivided according to the grain size aDd colour.

Example of igneous rocks are grarute, basalt, diorite, etc. Z. Sedlmenlary rocks. Sedimentary rocks are formed from mineral particles wbich had been transported by water, wind, etc., or which bad been precipitated in water il:ad then solidified. Thus the sedimentary rocks are formed .by the deposition of fragments of materials like sand, clay, disintegrated rocks. etc. on the prc-existing rocks. The process of deposition OOIltinues in regular layers. In due course of time, the deposited mass becomes a sedimentary rqck. Because the sedimenlS get consolidated in horizontal or nearly horizontal layers, the sedimentary rocks show characteristic different layers distinctly. !be sedimentary rocks can be

·,,


easily split up along the bedding plnnes. Sedimentary rocks come from many sources and include distindly differeDt families. Examples of sedimcmary rocks are sand stone, shale, lime slone., etc. 3. Metamoqmlc rocks. Metamorphic rocks are fonned from igneous and sedimentary rocks by very large heat and pressure. Sometimes, metamorphic rocks are also formed from already existing metamorphic rocks. The process of change to metamorphic rock due to heat and pressure is called meuunorphism. Due to metamorphism, the original rocks change their cilarnder and the resulting ma<>s of rock change into a bard and durable foliated material. Examples of metamorphic rocks are quartzite, marble, gneiss, slate. schist, etc.

31.3. BASIC TERMINOLOGY

The following terms ' are oommonly used. l. Rock material. 1nc consolidated aggregate of mineraJ particles which Conn the solid material of the rock is called rock material. In other words, rock material is the matrix of the solid phase of the rock mass. 2. Intact rock. 'The intact rock consislS of rock material which is free from major discontinuitics. An intact rock is generally quite slrong. Moreover. it can be readily sampled and tested in the laboratory. 3. Ml\ior discontinuity. The major discontinuity is a relatively large discontinuity in the rock: which is fairly well-developed and continuous. However. the shear failure along a major discontinuity docs not involve any shearing of intact rock material. 4. Strudural discontinuity. The structural discontinuity of the rock is a fracture which makes the rock. mass a discontinuous mass and separates solid blocks of a rock: mass. The strudural discontinuities are generally is the fonn of joints, bedding planes, fissures, faulls, shear

zones, cleavage planes, and solution cavities. 111ese are planes of weakness due to which the strength or rock: mass is considerably reduced. S. Rock mass. The rock mass is defined as the aggregate of regular or irregular blocks of rock: material. These blocks are separated from one another by structural features sucb as bedding planes, joints, cavities and flssurcs. The rock: mass has anisotropy and structural discontinuity. The rock.: mass is ronccived of discrete intact blocks separated by thin joints. 111e behaviour of the rock m~ is governed by a combination of blocks and joint characteristics. 6. Block size. The block size is defined as the average diameter of a Iypical rock:. It is usually expressed as a range and a typical values. For example, the block size (1~ em), 25 cm. The ooocept of block size in rock mechanics is analogous 10 that of grain size on a microscopic scale. 7. Joint Set. A joinl set ronsists of individual joints which have similar physical and mechanical cbaractelistics and whim occur in a nearly parallel array. The joints or a rock: mass are usually subdivided into two or more such sets, which together constitute the jointing system. The characteristics of different joint sets are generally different because of diffcrenl geological origin and history. 8. Faults and shears. 1llese are joints along which there has been shearing movement in the rock. mass. 9. Rock quality designation. Rock quality designation (ROD) is frequently used to indicate the quality of rock mass. 'The rock quality designation is defined as the swn total of lengths of the cores or the Lecfgth 10 em and longer recovered from the driWng, expressed as a percentage of the total length of tbe hole drilled (Deere et ai, 1967). Thus ROD _ Total length of cores in pieces of 10 em and longer )( 100 . Length of run

... (31.1)

Eq. 31.1. is for core sticks of NX cores (57.2 mm di..1meter). The ROD of a rock mass indicates the quality of the rock:. It is a measure of the degree of fracturing and the amount of weathering in the rock mass. The rocks with ROD from 100 to 90 are classified as excellent rocks, and those with ROD from 90 to 75 as good rocks. The rocks with ROD from 75 to 50 are fait rocks, whereas those with 50 to 25 are poor rocks. The rocks with ROD less than 25 are very poor rocks.

INTRODUCTION TO ROCK MECHANICS

819

10. Joint and Bedding Description. As already mentioned, the strength of a rock mass is considerably reduced due to the presence of discontinuities. In general, the greater the number of discontinuities, the more is the reduction in strength. Moreover, the degree of such reduction depends upon the spacing, orientation and size of the discontinuities. Table 31.1. gives the general quality of rock ma<>s depending upon the spacing of discontinuities in the form of joinlS and bedding (Deere, 1968 and Singh, 1900). TABLE 31.1. Quality of rock based Average spacing of discontiltUties Greater than 3m 1 to 3 m 03 to 1 m 0.05 to 0.30 m Less than 0.05 m

00

spadng of discontinuities. &dding Description

Joinl Description

Rock q.wlity

Very wide Wide ModcrDtcly close

Thicl<

Very thick

Clooe

Thi"

Fractured rock

Very close

Yery\hin

Crushed rock

Solid rock Massive rock Blocky (seamy) rock

Medium

31.4. INDEX PROPERTIES OF ROCKS Because of a vast range in properties of rocks due to different structures, fabrics and materials, it is difficult to detennine the properties and to dcsaibe rocks quantitatively. Certain properties that are re1'ltively easy to me~ure are used to classify the rocks. These properties are known ~ the index properties of rocks. The following are the main index properties of rocks : 1. Unit weight (or mass density) 2. Porosit)' 3. Penneability 4. Point-load index 5. Slaking and durability 6. Sonic velocity All these properties are discussed in the following sections. 31.5. UNIT WEIGHT (OR MASS DENSITY) The mass density (P) of a ~ock is the mass per unit volume. It is expressed as kglm 3 • In rock mechanics. the tenn umt weigbt is commonly used. The unit weight (y) is the weight ' per unit volume. It is expressed as kN/m 1. Sometimes, the tenn density is also used for unit weight. The specific gravity of solids (G) is tbe ratio of the mass density of the rock to the mass density of water. The mass density of water is usually taken as 1000 kglm 3. The specific gravity of solids is also equal to tbe ratio of the unit weight of rock to the unit weight of water. The unit weight of water is 9.81 kN/m 3 at approximately 10 kN/m 3. Thus the rock with a specific gravity of the solids of the rock of 2.7 will have a unit weight of 27 kN/ml. 1be specific gravity of the grains of the rock: material can be determined by grinding it and using the methods discussed in chapter 3 for soil particles. Rocks show a wider range of unit weighlS ~ compared to soils. The unit weight of the rock can be detennined from tbe volume and the weigbt of a carefully dri1.led core specimen. Table 31.2. gives the average values of the dry unit weighlS of some oommon rocks. TABLE 31.2. Dry unit weights of some common \I9CU Rock

Gabbro

Diorite

Basalt

Marble

Granite

Share

Dry unit

29.4

7:1.9

27.1

27.0

26.0

22.1 to 15.7

wei~t

(kN/ml

The dry unit weight can be obtained from the wet unit weight by the relation

225

~.9

SOIL MECHANICS AND, FOUNDATION ENGINEERING

8" '( [ &.)1~bt;mt~ ~i aGm *Jo'Jl till)

~norn

!)(If .a.liJiJJnimo;)lit. to .:1(11""

r'_ Yd -

~

)\

'1'

1 +. ... w

'.t:;

t_

I"

WHl!r(!l¥.1'ts'J itf~ 4tY:lOt1it

..veigh!,rr is tqe1wet unit-.weight and w is tMl water confent... ':..' " l A k.nowledge of unit weight of rock is important for various engineering appliciltions-. Fotl'cxample; M!~hf'~t!it:f6>miKI(,) (rbm -fieavY'1eggregates would require a smaUcr volume of! concrete in a rctaining waU than that from light aggregatcs. In oil shale deposits. the unit weight indicates the:amount of the mineral commodity, as the oil yield is related directly to the unit weight. In cool deposits, the unit weight is related to the ash content.

.~iJlllnUI1(J:)o

h 10 "nt'IRq

.,0 tr.... ' -J;!

n h

V

,1.~111'"

31.6. PORQSmY "I;:" The pOt6I!!~ (n) of a ~ock: is defined -as' the ratio of the void space 10 tbe tolal volume .~~ ;~~e .rock.. It is _txprc,ssc(Las...a_pc~Dtage~ iThuL _ _ ._ . __ )"

ll'ln?

~ ,,.., ..:!;!

I

vQ . , j ; . l

n -

V )(

100

... (31.3)

where ~-fs(tW'tf"" voids (or pores) anU-V is the total volume. I" 1 l The po~'{fv depends upon a nUlJlbt r factors. such as particlc!site dlstribution.\o'r dng, gnlin ~e8'~ ] entatidn,.J:lbric, solution effects and\TlinemIogical cOmPoSitiOn. The porosity is high when ::Ill the particles arc of the same size. However i? the, F of ~guladr ~hapcfl

.9f

~ic~t~h~)~Ef~ is

INTRODUcnON ro ROCK MECHANICS

For flow of water at a temperature much different from 2Qoe or flo\\O.ofufhrids GtlbdDthall'lwa1er,llEq:u31\ 6 is modiIied as 'x ~ noonr.qrun: 10'1 .fi11'110-;:- 1)1 (
K) (

, "'1;1''''.

mm o? mm1 ,"""Oib

where K is the coefficient of absolute permenbility, with a dimension of (L J. AS alre'ady mentionea in .69fl~E ~': is i~d~~:ien\b~the ~operlics of the permeant (fluid). It is lIJirlflH~J~~~t ~I~S JJ~~¥£~io~he e arc~ - . x. em '. . :l'~it}., .:;rn· 'PiJQmlq, ~1~1p.'") 3d blu0i1a I' ~ the coeffiClc~t of VISCOSity of the pcrmcanL For water at 12lbh(r :}'f'if'l'1:kfGIW~d'S' .1rfl OJ :r"i1il.!I~ tip IS the change In pressure (Note. p = ,/11). ~ "l;l"jrni? bos )/:)rIJe _zi t;:~J ri1amlk' l,r,.,)\ The coefficicnt of permeability (k) of a rock specimen can be ~£h,epnjr~HD'b~~tflf;~cJ?)' JIl9MIDEB the vOl.ume (Q) of fluid ~<;cd t~ough ~he specimcn under a co~8MUr R~Mrn~i~~beY9"J\lC. 1Hd.t'f.f ffi' the flUId for a known penod of llme t (I.e. q :::: Qlt). 1'gn!l11;t :"1'11 ~i!m:)1:.lb OJ oot)ubfloo ~ bIuntli'. Alternatively, tbe permeability of Lhc rock can be detcrmm£fl£l¥ mCJil~M-gOi'HdR~tr)~)I\BiSJ f~~Pl1Ow

~::~:t;~~~::~:f=kt~ r;:~t~: by drilling a ~~I~f~~l hole in a cere. The coe cient of q log., (R IR ) .0 rll;;O:.I1I ... :,·... i". . . .:)"vlnJo-.> I):mitnrunu ::.dl ?i ,,\'I !)l~d''''' 2 t k -~.. rUan01l;! bl~f Jnloq ~rh II \UI')"\ .l{Ill.S) where L is the length of the specimen, IlIi is the l1iffereH'dl*n~8;5'&&$lfif!"'t~l rij18l!, )Ri .t ..l\btIINm radius of the specimen and R z is the outcr radius df(theni~)e;W ~:.ll}ff]"{ Itniq'{l :>moo 3!.Ivig \:..f( :>1(k:T In the radial permeability ICSI, a very 13~e OOI':'J\f<1'\~~!¥. 9ljrP~r~.asily generated to measure even very small permcabililics quitc accurately, Fissured rocIcs show bi~er 2Crmeability for outward flow as co~rcd to that for inward flow. It is due to thc fact that wheilt~"liB.w1S Ofwaro.tiht!to3l-e b y fOl~~ldy.,X"lopcd which opens oul the flSSurcs. On the other hand, hen the.~ is i ward, a compress e force is developed

Wbi~~I=ast~~~~:~permCability ~t Lbe~rockJs ~l£}Jri{tim in$~-O'() ~~=rf~"::~!:ic~n: ~~D=~; ~~e~c;~:;n:~~~~

water

.

;';'~~tN_~oul

tightn~~~lW~~ag~a~e:~~ .~~i~

Table 31.3. gives some typiQ.l values' oT ·th6 b:5emcia,t~ Of~ilitYOo"{J i:h~t:)~~iw81er al

20 Ge.

!Jllfl ,admun ni :.Jgn'J

)liup ~',l> *->01 :;lrU 10

'iJilidlnub

:;lrlJ 8rUI:):)110

TABLE 31.3. Typical values or the coefficient of petffi~lffil\ltYdl oon~b '(bY.:llqmoJ ,.,."."

Type

0/

roc'

8aso/1

rro' ...... f)

vlll,- 'Ii!;,,,,,,n

:>,

r!!~Il:>Jt€1bMitflil ~ I 32i~} ;:,H i:l~10~400 bI OOZ ~1.'1tJb r!("li1dm;1 .(loilllilo1;t~ ...~; n:)IM .51UJI:n ni .!mi1't~ uylmn d!9P!tv yd

J'"I SJwJe.p

Coefficient of penneability (cm/sec.)

:?'NOllol em oonirm:lJ:)b <'i el,') x,Lmi I, -

~ :L::P"1 )(

IIlUlb:)tUnih:.llliGJ:n;!/:x>}[

~')I)' .:Millo Jd~i:;lw 'no igj()T

.. \ b

.. .(31.9)

whereThP ~l' t~: ~oae~that ~~ture anedd Pfi·!i~I}h~1Vftt~fm~6Wt ,g'j(! · r~~)I?qWt£lht~©~}t:~!U)bth"i~d?;' fr,. 't~*~hha' e engtll 0 e pieces 0 n core ~ lor me es snouli:i I:)e ai' lC3st"t .. !mes c' tame er,.! \~'TJ been found that the strength depends upon the size of the specimen. 1be point load strength reduces \0 /2 to

822


the diameter of the core is increased from 10 mm to 7Omm. For comparison purposes, it is lbe standard practice to report the point load strength Coc a core of 50 1/3

a<;

mm diameter. If the diameter of the test specimen is

different from 50 mm, a suitable size..<:om:ction is applied.

For reliable resulLS. it is essential that tbe specimen should be csrefully prepared. The resuilS are highly sensitive 10 the method and style of loading. The point load strength test is quick: and simple and is fairly Fig.3!.!. reliable. It can even be conducted in the field at the site of drilling. For rocks with strong anisotropy, such as slates. shales, schists and laminated Stones, the test should be conducted to determine the strength along as well as perpendicular to the bedding planes. The point load strength is related to the unconfmed compressive strength by the relation qu -

24 /k (so)

••. (31.10)

where qu is the unconfined oompressivc strength of a cylinder of rock with a length 10 diameter ratio of 2 to 1, and 11(50) is the point load strength conected to a dia6teter of 50 mm. Eq. 31.10 docs not give oorrect results for weak rocks for whidl special correlation studies are required. Thble 31.4 gives some typical values of the point load index for different rocks.

TABLE JIA. Point Load Index Type of rod

Tertiary sand stone and clay Slone

Painl loed indeJl:

0.0:5-1 MPa

31.9. SlAKING AND DURABD...ITY To indicate the durability of a rock, some sort of index is required. However. because natural processes afIeaiog the durability of the rock are quite large in number and they are also varied, no index can completely describe the durability. Durability of the rock is essential for various engineering applications. For long life, the rock should be durable. It should not disintegrate; otherwise its life and strength would be reduced. The rock may disintegrate by various process occurring in nature, such ~ exfOliation, hydration, slaking, solulion. oxidalion, abrdsion, elc. The foUO\Ving two indexes are commOnly used 10 detennine the durability of a rock.

1. Slake durability index

2. Change in liquidity index.

1. Slake durabllUy Index. 1be slake durability index can be determined using the apparatus suggested by Franklin and Olandra (1972). It consists of a drum 140 mm in diameter and 100 mm in length. A sieve mesh fonns the cylindrical walls, with a 2 mm size openings in it. About 500 g of rock: i<; broken into 10 lumps and placed imide the drum. The drum in turned at 20 r.p.m. in a water bath for about 10 minutes. The rock retained inside the drum is then weighed. The slake durability index (/~ is determined as follows:

I d

_ Rock retained in the drum )( 100 Thtal dry weight of the rock

... (31.11)

Table 31.5 gives the classification of rocks ~ on the slake durability index. as propa:;ed by Gamble (1971).

INTRODUcnON TO ROCK MECHANICS

82J

TABLE 31.5. Gamble's Shlke Durability Classification Durability

Slake durability iltdex after aile lO-mi/lute r:ycle

Slake durability inclex after two lO-mi/lute cycles

Very high High .

Greater than 99

Medium high Medium Low Very low

95-98 85-95

Greater than 98 95-98 85-95 60-85

60-8$ Less than 60

Less than 30

98-99

30-60

2. Change in UquJdity Index. Morgenstern and Eigenbrod (1974) used a water absorption test to dctennine the amount and ratc of slaking of argillaceous materials. 'Illey came to the conclusion that non·cemented claystone or shale absorbed water faster than other materials. Moreover, all materials eventually attained a final water content equal to their liquid limits. It has been established that the material.. with high liquid limit arc more severely ..ffetied by slaking as compared to those with low liquid limit. Table 31.6 gives the amount of Slaking for different values of liquid limit. TABLE 31.6. Amount of slaklng foc- different liquid limits Liquid Limit

Greater than

Amount

Very high

140%

of slaking The rate of slaking depends upon the change in t.h c liquidity index (A It) after immersion in water for 2 hours from the relation

6h-~ wI. - wp

... (31.12)

where Aw::: Change in water content of the rock aftcr soaking for 2 hours on filter paper in a Cunnet. wI. '" water contenl at the liquid Limit, wp '" water conlent al the plastic limit. All the water contents in Eq. 31.12 must be expressed as a percentage of dry w-:::tihL TIle methods for the detennmation of dw, wL and wp are the same as those Cor SOil, discussed in elwp'..r 4. Based on the rate of slaking, the rocks are classified slow if A It. is less then 0.75. fast when dh is between 0.75 and 1.25 and very Cast when dh is greater than 1.25. 31.10. SONIC VELOCITY The sonic velocity is the vclocity with which stress waves travel through ao intact rock. The sonic velocity for an intact rock depends upon its elastic properties and density (unit weight). If there are fissures in the rock., the sonic velocity is reduced. 1berefore, the extcnt of fIssures and discontinuities in a rock mass can be assessed by comp.'lring the in·situ wave velocity with the sonic velocity of an intact rock core obtained from the same rock mass as detennincd in the laboratory. Founnainlraux (1976) gave a classification system based on longitudinal wave velocity to indicate fissuring in rock specimens. The theoretical longitudinal wave velocity (Vi) that the specimen could have if tbere were no pore or fISSUres can be estimated from the mineral composition of the rock Crom the relation 1

V, -

• C 7 vZ;

... (13.13)


824

where V,; is the longitudinal wave velocity in mineral constituent i which ha.
TABLE 31.7. Typical values of longitudinal velocity Rock

Basalt,

Gabbro

Sand stolle, quartzite

Lime slone

dolomite

Longitudinal Velocity

7000

6500 - 7000

6000 -

6500

6000

Granite 5500 -

6000

(11) (m/S) The actual longitudinal wave velocity (V,) in the rock specimen as determined in the laPoratory is less than the theoretical velocity (\Ij). The quality index (IQ) is the ratio of the actual velocity to tbe theoretical

velocity. expressed as a percentage. Thus fQ - (

f. )

x 100

... (31.14)

Founnainlraux established that IQ is affected by pores (spherical holes) in the rock., and can be expressed

as percentage where np is the porosity of the

non-fissured rock expressed as a percent. It may be noted that Eq. 31.15 docs nOI give reliable results even when there is a small fraction of flat cracks (fissures) in the rock. Based on laboratory measurements and miaoscopic observations of flSSures in the rock,

Fourmaintraux proposed a plot

between fQ and porosity n wbich can be used to determine the degree of fracturing of a rock specimen (Fig. 31.2). There are 5 zones in the plot. For example, zone I is for nOIl- fissured to sligblly flSSUred rocks;"and zone II for slightly f'issured to moderately fisSured rocks.

n (AFTER

.,.

FOURMAJNTRAUX)

Fig, 31.2

31.I'1. 'CLASSmCATION OF ROCKS FOR ENGINEERING PURPOSES Bi~niawski ('1974) gave the geomechanics classification system, which is commonly used for engineering purposes, 11 prOviCleS-' roCk~ masS' rating (RMR) values for different rocks, The RMR value depends upon the' rock quality and it may vary from 0 to 100, The RMR is derived from 5 universal parameters given below

(Gb&llnan, 1980).

INTRODUCIlON

TP ROCK

MECHANrcs

825

1. Strength of rock 2. Drill core quality 3. Joint spacing 5. Ground water conditions 4. Jqint characteristics Sometimes, orientation of joints is ,also considered as the sixth parameter, especially for specific applications in tunneling. mining and foundations. Inaements of rock mass rating corresponding to each of the above parameters are determined. 'lbcsc depend upon the characteristics of the rock as explained below. All these inaements are added up to determine the RMR value. 1. Strength or rock. The RMR value depends upon the unoonfined compressive strength of the rock. The unoonfined oompressive strength (qu) of the rock can be determined from a laboratory compression test on a prepared core of tbe rock. However for the classification of rocks, an approximate value of the compressive strength as determined from the point-load tcst on intact pieces of drill core can be used. The following equation is commonly used. . .. (31.15) wbere Is is the point load strength. Table 31.8 gives the rock mass raLing incremeDls depending upon the point load index and unconfined compressive strength of the rock. The inaements vary from 15 to O.

TABLE 31.8. Rock mass rating increments depending upon compressive strength Pain! load Index /,(MPa)

Unconfined Compressive Strength (MPa)

Rock mass raling incremenl

Greater than 8 4-8 2-4 1-2

Gretller than ZOO

15 12

100 - 200 50-100 tT 25 -50 10 - 25 3-10 Less [han 3

4, 2 1 0'

Do nOI use [his rock.

2. DriD core quality. The drill core quality is related to the rock quality designation (ROD). As already discussed, the ROD of a rock is determined from the percentage recovery of core in lengths greater than 10 em for NX cores (57.2 mm diameter). For other types of cores, the percentage recovery is detennined in lengths greater than twice the corresponding diameter. The rock mass rating increments for drill core quality depend upon the RQD value. The increments vary from 20 to 3 (see ~ble 31.9).

TABLE 31.9. Rock mass rating .increments for driD core quality ' RQD(%)

Less than 25

Rating increment

J. JOint spacing. The rock mass rating depends upon Ute spacing of joints. The rating increment should reflect the joint set which is the most critical for a particular application. If the rock mass has fewer sets of joints, the mting is increased. The rock mass mting increment depends upon the spacing of the most influential joint (fable 31.10). The increments vary from 30 to 5. TABLE 31.10. Rock mass,rating increment ror spacing or joints Joint spaci/lg(m) Ralmgillcremcllt

Less Ihan 0.005


826

4. Joint characteristics. The rock mass mUng also depends upon the characteristics of joints. The condition of joint sets most likely 10 influence a particular application should be considered foc the rating. 1be increment of rock mass rating for joint characteristics depends upon the roughness of the joint. The description of joint SUlface rougness and coaling material is weighed towards the smoothest and weakcstjoint set. Table 3 t .11 gives the rating iocrcmenlS depending upon tbe joint characteristics. 1be increments vary from 25 to O. TABLE 31.11. Rock mass radnu: Increment for joint condition

S.No.

JoinJ Description.

/lJ1ling increment

1. 2 3. 4.

Very rough surfaa:s of limited extent; hard wall rock: Slightly rough surface; aperature tess than 1 mm; hard wall rock Slightly rough surface; aperature less than 1 mm; soft wall rock Sroooth surface, or gauge filling 1-5 rom thick: or aperatUl " of 1-5 rom; joints extending more than several melers. Open joints filled with more than 5 rom of gouge, or open more than 5 mm;

25

5.

'"

12 6

joint extending more than several metres

S. Ground water condition. Ground water ronditiollS can influence the rock rna-.s behaviour to a large extent. The rock mass rating depends upon the general condition of tbe rock, which, in lum, depends upon the ground water. TIle rock: is assigned the category dry, mOist, water under moderate pressure or severe water

problems and the raling increments vary from 10 to 0 (fable 31.12). The rating increment can also be related to the inflow per 10 m length in lit/min if an exploratory adit or pilot tunnel is available at the site for measurement of waler inflow. It can also be related 10 the joint water pressure divided by the major principal stress, as given in Table 31.12. It varies from 10 to I.

TABLE 31.12. Increments of rock mass rating due to ground water condition S. No.

General

comJiti(ltl

Raling increm,mt

Joint water pre£iUTe divilkd by major

Inflow per 10 m tUllcl length (lit!min)

principal stress Completely dry

2. 3.

Moist WatcruOOerModcl1ltc

4.

Severe water problems

None 25 25 -

10 0.0-02

0.2-05

125

pressure 05

125

6. Orientation or joints. TIle orientatioo of the joints relative the the work. under consideration can have an effect on the behaviour of the rock.. Accordingly, the sum of the firsl 5 raling numbers discussed above is adjusted. If the otientation of joints is very favorable for the work under oonsideration, no points are subtracted from the sum. However, for unfavorable orientations, there are negative ratings as given in Table 31.13. For tunnels, the rating decrements are Crom 0 to 12 and Cor foundation, the rating decrements are from 10 25, depending on the orientation of joints. For assessment of influence of orientation of the joints for a particular work., the advice of nn engineering geologist is usually required.

o

TABLE 31.13. Adjuslmenl'i in rock mass rating ror joint orientation Assessment of Influence 01 ' oriOlta/ion of the work

Very favourable Favourable Fair Unfavourable Very unfavourable

Rating increments for tUIUlt:ls

Rating incrt:mt:nu for foundations

o

o

-2

-2

-s

- 7

- to

-IS

-12

-25

INTRODUcnON TO ROCK MECHANICS

827

Rock mass Rating (RMR). As already mentioned, the rock. ma..s rating is obtained by adding up the rating increments given in Tables 31.8 to 31.12 and adjusting the sum for orientation of joint as per Table 31.13. Table 31.14 gives tbe geomechanks classification of rock masses based on RMR. For example, a rock mass with the partia.ilars given below will have the RMR of 84, and the rock will be cl.a5sified as very good rock: as per Thble 31.14. 1. Point )oad index of 6 MPa = 12 (fable 31.8) 2. ROD of 80% 17 (fable 31.9) .., 25 (Thble 31.10) 3. Joint spocing 2 m 4. Very rough surface = 25 (fable 31.11) 5. Moist roodition 7 (Thble 31.12) Total = 86

= =

6. Favourable orientation for foundation RMR

= -2 (fable 31.13) _ 84

TABLE 31.14. Geomechanks classincatlon of Rock mosses

CI=

Description of Rock mass

RMR

Very good rock

81 100 61-80 41-60

Good_ Fair rock _rock Very poor rock

III IV V

21-40 0 - 20

31.12. STRENGm ClASSIFICATION OF INTACT ROCKS Intaa rocks can be classified according to the uniaxial compressive strength. Deere and Miller (1966) suggested the classifICation given in Table 31.15. The uniaxial compressive strength is determined by testing a rock. specimen with a length to diameter ratio of 2. TABLE 31.15. Strength classification of Intact rocks Uni4uial compressive strength (MPa)

CI=

Descn'ption

Greater than 224

A

Very high strength

112 - 224 56-112 28-56 Less than 28

B C D

E

High strength Medium strength Low strength Vcry low strenglh

Deere and Miller aLso classified the intact rocks according to the modulus ratio. The modulus ratio is the ratio of the modulus of elasticity to the uniaxial compressive strength. The modulus of elasticity used in the classification is the tangent modulus obtained from the stress-strain curve of the specimen in the uniaxial compression lest at a stress level equal to one half the ultimate strength of the rock. Table 31.16 gives the classification based on the modulus ratio. TABLE 31.16. Rock Classification Based on Modulus Ratio Modulus ratio

CI=

Descn'ption

Over 500 200- 500 L=1Iu>n 200

H M

High modulus ratio Medium modulus ratio Low modulus ratio


828

Combined classification based

00

strength and modulus

r'dtiO

1be classification based on compressive strength and that 011 the modulus ratio can be combined \0 indicate the quality of rock. 1\vo letters are used for classification. 'The first letter indicates the strength rating and the second leiter indicates the modulus ratio rating. For example, a high strength rock with a medium modulus ratio will be designated as BM.

Jl.IJ. LABORATORY TESTS FOR DETEID.fiNAll0N OF STRENGm OF ROCKS The following laboratory tesls are commonly used (or dctennination of the strength of rocks : 1. Unconfined compression lest

3. Spliuing tension lest 5. Ring shear lest

2. Triaxial oompression Icst

4. Beam bending test

All these tests are briefly described below (Goodman, 1980) 1. Unconfined compression lest. lbe uncon[med compression test is the most commonly used strength test on rocks. For accurate results, the test should be pcrfonncd carefully. TIle specimen should be in the Conn of a cylinder of length to width ratio varying from 2 to 2.5. The ends of the specimen should be []at. smooth and parallel. The ends should be exactly perpendicular 10 the axis of the cylinder. Cores obtained during explorations are usually trimmed for this purpose. TIle specimen is Subjected to compression between the cross-hCJd and the platen ora compression testing machine (Fig. 31.3) According to IS ': 9143-1979, the specimen should preferably have a diameter of 45 mm. In no case. the diameter shoukl be l~ than 35 mm. The load should be applied continUOUSly with a stress rate of 0.5 to 1.0 MPa ,per second. "fablc 31.17 gives some typical values of compressive strength of somc rocks. The compressive strength (q..) is determined from the relation

q.. - PIA

Fig. 313.

... (31.16) where P is the peak load aod A is the initial area of aoss-seaion of the specimen. 2. TrIaxial compression Test. The triaxial compression tcst in pririciple is similar to that used for soils (chapter 13). The cylindrical specimen is first subjeaed to the lateral pressure and then deviatoric stress is applied. As the stresses are quite large, a special type of equipment is required for conducting the test (Fig. 31.4). The usual procedure is to firsl apply the confUling pressure (P) aU around the cylinder and then to apply the deviatoric stress when the oonfining pressure is kept oonstanl. The all round pressure p increases tbe strength of the rock. However, the increase in strength is realised only when the specimen is enclosed in an impervious jacket. Normally. hydraulic oil is used as a
2P

0,

-"'itd'i

. .. (31.17) Fig. 31.4

829

INTRODUcnON 'ro ROCK MECIIANICS

where P is the '~pplied load, d is the d iameter of the cylinder aod t is the thickness of the disc (i.e. the length of the cylinder). IS 10082-1981 recommends that the diameter of the specimen should be at leasl 45 mm and the thickness of the disc shuU be approximately equal to half the diameter. lhe lood should be applied at the rate of 0.2 kN/scc. lhe load should be measured at least to an accuracy of 1%. The splitting tension test generally gives a value of tensile strength higher than iHat. l6btained~ from a direct tension test. This is probably due to the Je5S effcct of fissures in the splitting tension test as compared to that in a direct tension test. 4. Beam Bending Test. 1be beam'" berldin&.- test is al<;()

VIEW

ELEVATION

Fig. 31.5.

. .

&~~n~hc fl;:~:~'

lik'tijgg' '&,Ita,"

~~~

SIDE

flcxuOO" s(Nngth ,,,,,

~\X1ulus

p

~

\11, rupture)

~ the;'~ I@ '· '~.';)·;'/; ~i\~: :::>.:~ .~:

W.~l'.im~!1h. tCflSi1e .,§ltf'l§ j aL :ltjl \:. - l?otJ~'J s.!ll'f~ of the oore"-'S1bE "lEW " ~cs~ng !~I tI'\C~Pf:OIk load. J!/n~~ 91rfu}a!q
fJpto ~hc. ~~roe· 'fbe , flc.x%ilt~tre[lgth Jl'.t.r)'lis given by

~

:w

}fi

'fr/ 3~

..

Ffg.31.6

I"

H!1~'D T",, ~ ' 16f!i)r;;

:w~n;t.t is,Jt!9, m,~i,mum load atr!aiJl.!.fC.

ELEVATION 'P"

,~'

. ,.--:;-

.\}'

_

... (3!.lS)

4f

spWl i.&,.dislancc be~,~n ~iQns 9n ~U}.~ lqwcr sw;fficc< and d ~~IF~~.J.hC__ ~ ;.r-' 'it~JOI "" nlu·j !f~I:"':ul; :L.!.,!)' , 1 Jd i s. Ring shear test. The ring shear test is generally uscd .~~)es~ i9;.silu. rpcM. V. . givC;S tl;l~ shear strength of the rock as a function of the confining pressure. "be rock core specimen in Ihis tcst docs· nOI require perfectly square and SM'tf1~ndS.a ~}II'.r.:)1qlfllY'" ~IU 10 k ....ll l-< :1I'Ji:'l'(T ,C.If: :·t ',H lhe confining pressure is appl.iticlJby 'ttav"l00d '-psrMlel~al'Jiw.IolDelPrungcr;.;\~.~-:?,\\ffi..~~-

----

surfaces fonn ,~g) thc two planes of the lIl1pnStd( ~ (Fig. 31.7). Inc p II.OQII )-----

' shcar-strength~(,c:a\~la

whe~e

"".~,

} - 2A

OOC - 0..

1

...(31.1 9

tf

P is t~~k load and ~ is the area oG61£1SS'SCCtion of specunen. ot~J.n . 0.,. ..... As in the~ ~ of the triaxial !Ie&. there is a ;\~ntial increase bf strength ~.tfV!!§ ~9!Wing pressure. 1;'\,'~ ~-' j I' 31.14. S'J'RESSt.S'ntAIN CURVES lJ.t£ ~ m~iI~ Orelasticit,Y POisso9. [alio are !WO-im~;U;-nt - .-

-andlh'l

?cqg'Jli'cjff~Hl~horJlIrl~k~Wi! 8tHJlfc }mlpbi!I& J8e~Pib !lopo;\ iJ-hGmBepU>';!"'~ ' "'-ry~~~~~,"",,.,. of factors such a<> the size of the rock mass.. oonstituent mincC!is... tbe Fig. 31.7

degree of fracture, eiC. ..3 10 !rl!l~"Mo(i'urr&1Of ieli(Sti~i1JiebftiooirlOs!l(jr "hlat.fiblI-1 bt ! f@CIt:1 ITh'rdl!lCnn)rl&Jtl'toriYJ lhij!:~sti'am ~M obtained from a uniaxial oompressioo lest. MOSl .e1":.ttle ltodcs:fh:lIo'bfthebn~stftJim~ o».thc ~1~ ,1Qw.Jst~;l:cye~ the mtIfJulusJ of~~jcit)'; ~~rtW.ilh an .ina:easqriILsttel'uJtbedniWlb-~n1l!'tnQdutus is ,:nuih,l Is Ot.O 'to :..uw", iTll,l!Oiy't:im il.;::)lUnoodl 'Jdl 3!11"iJ1~~1 Ii 00&

rolL MECHANICS AND FOUN0.KI10N HNGlNEERING

A L,/

P

'-TANGENT MOOULUS AT P

TANGENT MODULUS

STRAIN

~

~

Fig. 31.8.

I

the Slope of the tangent drawn to the ruNe at the aigin. The secant modulus in a partirutar range is obtained by joining the stress point with tbe origin and determining the slope of that line. The tangent modulus at any stress is equal to the slope of the tangent drawn at that point. 1be initial tangent modulus, the secant modulus and the tangent mooulus are shown in Fig. 31.8. Generally. the modulus of elasticity is taken equal 10 the 50% tangent modulus, i.e. the slope of the tangent drawn at ooe-half the failure strcss. IS: 7317-1974 givcs a method for the determination of the modulus of elasticity. Fanner (1968) gave the following three types of elastic behaviour in rocks. 1. Quasi-eJastic behaviour 2. Scmi-elastic behaviour 3. NOfHlastic behaviour 1. The quasi-elastic behaviour is observed in fine-grained compact and massive rocks. The stress-strain curve in this case is almost linear right upto failure. The initial tangent modulus ranges from 6 x 104 to 11 x 1(1' MPa. 2. TIle semi-elastic behaviour is observed in coarse grained igneous rocks and also in fine-grained compacted sediments. In this case, the stress-strain curve shows a decrease in slope with increasing stress. The initial modulus ranges from 4 x 10'4 to 7 X 104 MPa. 3. 1be non-elastic behaviour is observed in rocks which are lcss cohesive and which have large pore space. The initial modulus is usually less than 5 x let Mfa. Table 31.17 gives some typical values of modulus of elasticity of some rocks.

TABLE 31.17. l)'pIcal values of the compressive strength and modulus Type of rock

or elastldty of some rocks.

Compressive strength (q..) (MPa)

Diabase QuMai1c DoIerile Granile Limestooe Shale Marble

490.0 460 - 200

330 230.0 225.0 170.0 lS3-Tl 37.0

Sand ~one

Modulus

rf elasticity (E) (Mfa)

1,00,000 78,000 84,000 56,000 Sl,OOO 68,000 48,000 - 65,000 9,700

2. Pol9son's ratio. 1be Poisson ratio (v) is tbe mtio of the lateral strain to the axial Slrain of a rock. Thus v _ ~

...

where Ed is the diametric strain (or circumferential strain) of the rock and E.. is the axial strain. The value of v geoernlly varies between 0.125 and O.34Q. for mail rocks. 10 codes subjected to uniaxial rompressioo, v remains more IX less c:oostaol with an ~ in stress, and it reacbes the tbcorttical maximum value of 0.50 at failure.

ItITRODUCTION TO ROCK MECHANICS

631

The laboratory uniaxi.1l compression test em be conducted on a rock to obtain the circumferential and axial strain. For the measurement of strains various devices, such as electrical resistance strain gauge, compressometer, optical instruments, are used. IS: 9221·1979 rcrommends that at least . two circumferential strain and Iwo moal strain measurements should be taken for each increment of lood. Moreover, the gauge length should be at least 5 times the grain size diameter.

31.15. MODFS OF FAILURE OF ROCKS The following modes of failure may OCQ.lr in rock (Goodman, 1980): 1. Aexure fa ilure 2. Shear failure 3. Direct tension failure 4. Crushing or compression failure However, in actual practice, there arc .!l large number of toad configuration and no single mode of failure predominates. The actual failure pPtlem is usually highly complex in whicb two or more modes of failure may ocx:ur simultaneously. 1. F1exure fuilure. The flexure failure OCQlrs because of the bending aaion in the rock. Dccause of bending, tensile eracks develop and propagate and cause failure. For examp~ flexure failure OCOJrs in the rock layers above the roof of a mine [fig. 31 .9 (aj l . A gap is fonned in the roof of tbe mine and a beam of . . ~ ~ ,:

" ,"

......

.,

··I~~c: .. . •

(I.)

r:= ..: ..... ',::

(b)

'"'

Ag. 31.9.

': ~:< (d)

rock: sags downward under its own weight. As the sag inacases. the tensile cracks at the bottom surface of

the beam propagate upward and the beam ultimately fail. II causes the rock to become loose and fall 2. Shear railure. This type of failure occurs when a surface of rupture is formed in the rock. because of the shear stresses bec:oming criticaL Mer !.he formation of the rupture surface, there is release of shear stresses as thc rock suffcrs a displacement along the rupture surface. This type of failure may occur in slopes cut in weak rocks sucb as weathered clay shale and
SOIL MECHAN ICS AND FOUNDATION ENqlNEERING

832

\Vhcn the rock breaks in direct tension, a surface of rupture is formed. TIle rupture surface is quite rough and free from the crushed rock particles and fragments. The surface is quite different from 111.11 Cound in shear failure. which is slick and has more powder formed Crom the crushing of rock. 4. Crushing or compression failure. Very high direct compression OOCW'S in intensely shortened rock when penetrated by a stiff punch. When the compressive stresses reach the limiting value, the crushing failure may occur,

The crushing failure mode is highly complex. It includes formation of tensile cracks and their growth and interaction through flexure. Fig. 31.9 (d) shows an example of crushing and tensile cracking which is followed by shear failure. 31.16. MOHR·COULOMB CRITERION FOR ROCKS lbe Mohr-Coulomb criterion is commonly used for detennining the Shear strength of rocks. As discussed in chapler 13, the Mohr-Coulomb criterion represents 8 linear cflvclope which louches all Mohr's circles drawn ror critical combination or princi~1 stresses (01 and OJ) at which failure occurs (fig. 3l.10). '(be criterion may be written as (Goodman. 1980).

r

"Cp _ Sj + 0 tan. . .. (31.20) where "Cp is the peak shear stress or sherif strength, Sj is the shear strength intercept, 0 is tbe normal shears on the shCc'lr plane, and q, is the angle of internal friaian. lhe Mohr-Coulomb envelope is extrapolated into the tensile stress region upto the point where 01 becomes equal to the uniaxial tensile strength (To). The minor principal Slress (01) am never be less than To. In terms of the principal stresses at peak load condition, the Mohr-Coulomb criterion can be written as

1 CJI,p _ q" + OjI3n (45" + ¥2) ... (31.21) where di:p is the major principal, stress corresponding to the peak of the stress-stroin curve and qu is the unronfincd compressivc.'strength,

,,-. The' unconfined compreSsive strength ' (q~) is given by ,,- I'

l(~,

,.,:" (~\ ~",.;J;!.t;':J" q,," 2Sj tan\1f>: + F5ctm Eqs. 31~21... t3nd 3!,22• . ~.-:. j' 'J ':1,

I.... u(!

U··f

6!; J!Jl g~ncrat.

~~; :~;,::(" (1!

,J?,:"'T.':!

or.l

"j

.-}1.

;)":1

J

OhPril.

q.

1+

,1) ,

d

01 lp I:'" ';'

'(l

... (31.22) w.

(."!.) IU"1("50I-+ q,lir q.

!r:'~ ~ ~rf' ,1',11 nI "'1.1'..

\,:;:01,

"i/~,t . ." ,I, (J;~PI:;~;'i .t~~'iA?,/qi' :I;\I, ;;1 ~:,

'

',lnt;L ~

... (31.23)

' :,!

r.

.~ J') I.;.~',

_'\1,

"1

,0,

---ci;- 10 ) .;,,. N(oJ(g..'/:. "

.. .(31.24)

833


The constants M and N in Sq. 31.24 can be detcnnined by filting a curve to lhe f~ mily of points drawn between (o)lq,,) and (o llq,, - l) obtained from the test. The maximum tension criterion must be superimpooed on the Mohr· Coulomb miction CEq. 31.21). It means that lhe failure will OCOJr because of tensile SII"CSS wherever the minor principal stress a;} becomes equal to -To> whotever may be the value of 01.

Empirical Failure ·Envelope The Mohr-Coulomb allerion is quite simplc and easy to use. But it gives 'unconscrvativc results when one of tbe principal stresses is negative (i.e. lensile) A more accurate failure ailcrion C&l be obtained by fitting an envelope Mohr circles corresponding to stresses at failure as obtained from four types of tests,

It-

TESTS(1) DIRECT TENSION (2) BRAZILIAN (]) UNCONFINED

(4)

COMPRESSION

TRIAXIAL COMPRESSON

-TO

namely, direct tension lesl, Brazilian Icst, unconfmed compression tcst and triaxial tcst. Fig. 31.11 shows the empirical failure envelope. The envelope is generally aJrved downward and the shape is in-between a straight and a parabola. It may be noted that the empirical failure envelope lies beneath the Mohr-Coulomb aitcrion with the superimposed tcnsion cutoff (shown dotted). Inside lhe hatched portion, the Mohr·Coulomb criterion with tension OJloff overestimates the strength. Therefore, it is necessary to reduce the tensile strength To and the shear slrength intercept S; when applyil1g the Mohr-Coulomb criterion to practical problems involving such conditions.

31.17. SHEAR STRENGTH OF ROCKS The shear strength of rocks is generally found from Ibe triaxial compression lest and adopting tbe Mohr-Coulomb criterion (Eq. 31.20). As already mentioned, the peak stresses are comidered for drawing the envelOpe. Hence the shear strength parameters Sj and 4t in Eq. 31.20 are for tbe peak strength. As in tbe case of dense sands discussed in chapter 13, the Mohr-Coulomb criterion can also be adopted for the residual stresses, i.e. the stresses at the ulLimate failure corx1iLioos which occur after the peak stresses. The residual shear strength is the minimum strength reached by the rocle. when subjected to defonnalions beyond the peak Stre:\S. In Ihis caSe, the subsaipt r is generally used to identify them as parameters for the residual strength. Thus Sq. 31.20 becomes 't, _ S;" + a Ian 4t, ... (31.25) Thel residual shear strength intercept (S;,,) may approach a value of zero. However, the value of the residual fridional angle 4>, usually lies Dctween zero and lhe peale. frictioo angle (+). When water is added to some rocks, the strength is considerably reduced because of the chemical deterioration of cement or clay binder of the rocks. Friable sand stone may even 10se 15% of its strength by mere saturation. A substantial redudion in shear strength occurs when pore water pressure (Pw) develops inside the pores and fissures of lhe rock. For such cases, Tcrzagbi's effective stress law (as ~ in chapter 13) can be used. Thus &j. 31.20 becomes

834


ip _ Sj + 0 tan 1$

... (31.26)

a is the effective stress ( = ° - Pw). Eq. 31.26 indicates that the IX're water pressure P... will cause the same reduction in the peak DOnna!

where

stress as that caused by a reduction of confining pressure by an equal amount of P ....

31.18. HARDNESS OF ROCKS Hardness of a material is its ability to resist scratching. The relative hardness is used as an index for the identifialtion of minerals. Mob's hardness scale is commonly used for Ihis purpose. The use of hardness for identification of rocks is however quite complex and oot well-established. In fact , rock hardness indicates the combined resistance to displacement and penetration. Il is also observed that both hardness and toughness of rock depend on the same factors.

Sometimes, the concept of hardness is applied to rocks as a synonym for streogth (e.g. soft roc:k, bard rock). Actually, hardness is a property of the rock-forming minerals and metals. Attempts have been made in the past to determine the hardness of rocks. Various tests have been suggested. Depending upon the lest, the hardness has been dcfined as scratch, indentation, abrasion, impact and rebound hardnesses. However. only the rebound tests such as Shores lest and Schmidt test have been used for the cbaracterisatioo of rocks. Deere and Miller (1966) analysed the data obtained from Shores' and Schmidt's tests. They gave correlation charts between the uniaxial oomprcssive strength and the Schmidt bardnes.... These charts can be used to predict the uniaxial compressive strength of rock with a fairly high degree of confidence limits. Abrasion hardness and impact toughness measurements are useful for the evaluation of building stones and for the suitability of the rock drilling and cutting equipment. The hardness of a rock also affects driUability. 31.19. IN·SlTU STRESSES IN ROCKS At a lX'int below the rock surface of the undisturbed rock mass, there are stresses due to weight of the overlying materials, and also due to the confinement and past stress history. These stresses are known as in-situ stresses. The in-situ stresses vary considerably from one point to the other. 1be in-situ stresses may be almost zero at some ]XIinlS, whereas at some other lX'inlS, they may be very high, even approaching the failure stress. When the in-situ stresses are almost zero, rock mass may fall from the surface and underground excavations because joints are open and weak:. On the other hand, when tile in-silu stresses are very high, any small disturbance to the stress field by tunneling or even excavation may trigger violent release of stored energy in rock and cause failure. The in-situ stresses may be approximated and the order- of their magnitude detennined by various methods, but the accuracy of estimating is always doubtful. Accurate field measurements are usually required to ~ertain tbe margin of error. Field measurements of in-situ stresses -are quite common in the mining practice. However, in the civil engineering practice. the field measuremenl are rarely done because of high cast of measurements. A basic knowledge of the in-situ stresses in quite useful in several civil engineering problems such as follows (Goodman, 1980). 1. Qrlentation ror a cavern. While selecting the orientation for 3 cavern, care is taken 10 avoid alignment along the longer dimension perpendicular to the major principal stress 0"1' 2. Selection or shape. The shape is selected to minimise the stress concentration if the initial stresses are very high. 3. Layout or complex underground works. Cracks in rocks tend to extend in the plane perp.mcUcular to the minor principal stress 0). A knowledge of the direction of the stresses is useful in selecting a layout to reduce the risk. 4. SavIng In Dning or tunnels. If the in-situ stress in the rock in which a pressure tunnel (or penstock) is constructed, is greater than the internal water pressure, lining need not be provided. 5. Large surface excavation. Substantial economy can be e ffected when making large surface excavation with pre-splitting technique by orienting the excavation perpendicular to the minor principal stress.

835

INTRODUGI10N TO ROCK MECHANIC::>

In-situ Stresses

1. Vertical StTe$S. The in-situ vertical stress at any point in the rock can generally weight of the overlying rock: per umt area. Thus

De

taken equal to tbe

Venical stress, 0" .. 1 z ... (31.27) wberez is the depth of the point and 'Y is the unit weight oflhe rock. which is usually taken as 27 kN/ml. When the ground surface is horizontal, the principal stress directions are vertical aod horizontal at points near the surface. It is generally assumed that the directions are the same even for points at large depths below the surface.. allhough the actual directions are somewhat different. However, tbis assumption is not justified in the case oC hilly terrains. Beneatb 8 valley side, there is one principal stress normal to the slope, which is equal to zero as the stresses nonnal to a free sulface cannot Fig. 31.12 exist. The other two principal stresses lie in the plane of slope (Fig. 31.12). These stresses approach zero when the rock slope is convex upward, but have high values wben the slope is ooncave upward. In tbe case of a sleep. V-shaped valleys, these stresses may even approach the failure stress or the strength of the rock. 2. Horizontal stress. The magnitude of the horizontal stress (0,,) is usually estimated from the relation

-t-

0" - K 0" ... (31.28) where K is the lateral stress coefficient. , In the case of recent sedimentation, the value of K can be taken equal to v/(1 - \f), as obtained from the theory of elasticity (chapter 11). In general, the value of K usually lies between K" and Kp , where KiJ corre
K. - ro"(45'+ +/2)_[(~)cot2(45' + +12)] x ~

... (31.29)

In the reverse faulting, the vertical stress is the minor principal stress and the failure occurs due to horizontal compression. (Ibis is similar to the passive conditions discussed in chapter 19). The value of Kp is given by

Kp _ tan2 (45 0 + +12) +

~

)(

~

.. . (31.30)

In case there is no existing fault, tbe range of possible values of K is quite wide. However, q .. can be assumed as zero near a pre-existing fault and the range of K is considerably reduced. Brown and Hoek: (1978) suggested a hyperbolic relation for the limits of K According to them,

( 0.3 +

1~)

<

"ii:

< ( 0.5_+ 15,(0)

... (3131)

where z is the depth of the point in metres and K is the ratio of the average horizontal stre5$ to the venical Sl~. The range of values of Ii: is oonsiderately less than the range of K given by Eqs. 31.29 and 31.30 even when q.. is not zero. It may be noted that in Eq. 31.31, the average horizontal stress is considered., wherea<> in Eqs. 31.29 and 31.30, Ibe ma:ltimum and minimum values of the horizontal stresses are considered.

&16

SOIL MFCHANICS AND FOUNDATION ENGINEERlNG

31.20. MEASUREMENT OF IN·srru STRESSES The following methods are commonly used for the measurement of in-situ stresses in rocks (Goodman. 1980). 1. Hydraulic fractunng method 2. flat jack method 3. Over-coring method 1. Hydraulic fracturing method. In lhis method, water pressure is used 10 create a aack in a bore hole. Water is pumped into a section of the bore hole isolated between two packers. As the water pressure is incrcascci, the initial compressive stresses existing in tbe rock at the walls of the bore hole start reducing, and at some stage become tensile. When the tensile stresses become equal to the tensile strength of the rock (-T6)' a crack is fonned in the waU. If the pumping is cootinued, the crock extends Cunher, and eventually the pressure down the hole falls to a steady value, called the shut-in pressure. Knowing the orientation oC the hydraulically induced fracture and using the equations of the theory of elasticity, the in-situ stresses can be estimated. The methcxl can be used to estimate the in-situ stresses in the rock only when the. point is at considerable uepth below the surface. 2. Flat jack method. In Ihis method. a slot is first cut in the rock. The deformation occurs in the rock: surrounding the slot. After the deformations have taken place, a fiat jnck is inserted into the slot and load is applied. 'Ibe in-situ stress is related to the pressure needed to bahmce the defonnations which occur as a resuit of slot CUlling. The first step is to marie the location of the slOl on the wall of the rock. TIlen one or more sets of the measuring points are installed such thai the Joc:ation of the slot is midway between them. The spacing d of the measuring points is kept equal to tbe gauge length of the exleOSQrneter, which is generally ,

15.24 em (6'). A deep slot is then cut perpendicular to the rock face between the mca<>uring points (Fig. 31.13). As a result of the slot cutting, the pins installed at the measuring points move and the spacing between them is reduced if the rock was under initial compression nannal to the plane of slot. The nat jack is now inserted into the slot and cemented in place. Fig. 31.13. 1be pressure in the jack is gradually increased. When the pins bave been returned to the initial spacing, the pressure in the jack is equal to the in-situ stress. The method can be used only when there is an access to 8 rock face; for example, in an underground gallery in the rock. 3. Over-coring method. In this method, a small diameter bore hole is ftrst drilled into the rock. A suitable deformation gauge is then inserted into the bore bole to measure the change in diameter. A larger diameter hole is then cored concentrically over the small diameter OEFORMATI()<.I • bore bole (Fig. 31.14). Thus a thick cylinder of rock is formed, GA~E which is detached from the rock mass and whid1 is free of stress. (f tbe rock was under initial compression, there would be enlargement of diameter, which is measured wilh the help of the deformation gauge. Analytical met~ based on the unloaded thick-walled cylinder model are used to detennine tile in-situ

stre=s. The over-coring method can be used only to measure the in-situ stresses at some distance away from the rock face. The lest is nonnally discontinued beyond about 5 m from the rock face.

Fig. 31.14.


PROBLEMS A.. Descriptive type 31.1. Define rock mechanic.~. What is its importance for n civil engillccr '! 31.2. Discuss gcologicnl clasSificntion of rocks. Give at lenst 3 e)(amplcs of each type of rock. 31.3. Explain the following term~ : (a) Rock mnteri:1I (b) Intact rock (c) Discontinuity (tl) rock mass (e)lointsct if) Rock quality designation (RQDl 31.4. What arc difFe rent index properties of rocks'! How would you determine them .! What is their importance .J 3Ui. Explain the geomct:hanics classit"iclltion of rocks. How would you determine the ruck mass raling (RMR) t,1 :1 given rock .! 31.6. Discuss the strength classification of rocks. 31.7. Explain various laboratory tcsts for the Octennination of the strength of rocks. 31.8. Draw a typical Stfess-strJIn curve of II rock. How would you detenninc the modulus of elasticity and the POI}SllI' ralio'! 31.9. What are differem modes of failure of rocks'! Give one example of each. 31.10. Discuss the Mohr-Coulomb criterion for rocks. What is an empi rical failure envelope .~ 31.11. Write shon noto;s on (a) Shear strength of rocks (b) Hardness of rock. 31 .12. Whal is the importance of in-situ stresses in rock 'J How would you measure them .,

B. Objective Type Write whether the followmg stntemcnts lire truc or falsc. (a) Slate is n sedimenwry rock. (h) Fauils are joints along which there had been shearing movemt':nlS. (c) The RQD of a rock cannot

be greater than 100%.

(c/) TIle point load strength Of a rock specimen is independent of the diameter. (e) The RMR v;lllle is reduced because of unfavourable joint oriemntion.

if) The shear stren gth of the speci men as determined by the ri ng shear test is equal to P/2A. (g) Ge nerall y, the modulu s of elasticity of the rock is take n eq ual to the 50% .~ccant modulus. on fou r different types of tests. (I) The in-silu l>tresscs in rocks arc TIOt significant for mining operations. fAns. True (b), (e), (e),

(II) The empiriclll fnillire envelope is based

lfJ. (/1)1

32 Geotechnical Earthquake Engineering 32.1. INTRODUCfIO/\ An earthquake is sudden shaking of Ihc earth surface caused by a source of disturbance inside the canh. Earthquakes occur duc to sudden mass shining in bedrock caused by forces within the canh. Shifting in bedrock occurs along faults. Such earthquakes arc called teclOnic earthquakes. Besides tectOnic earthquakes, there arc Olher types of earthquakes such as volcanic earthquakes. nuclear explosion earthquakes and mine collapse earlhquakcs. but such earthquakes aIC small. 1lle discussions herein aTC limited only to tectonic earthquakes. There is sudden release of strain energy in the ;'.One where the shifting in bedrock occurs. Shock waves arc generated duc 10 release of energy. These shock waves are propagated outward from the earthquake source. W~en these waves rench the earth surface. its shilking occOrs. The movement of the bedrock along the fault occurs at a point called the fncus (or hypocenter) (Fig. 32.1). The focus is generally 5 to 50 kill below the earth surface but may be deeper in some cases. The epicelller is the point on the earth surface vertically above the focus. 111e distance between the epicenter and the focus is C:llIed the focal depth . The distance between any po.int on the earth surface and the epicenter is known as the epicellTmldi.\"f(/l1ce. Earthquakes cause one of the most dcstrut:tive Fig. 32.1. BUSiC Tcnninology natural disasters leading to heavy losses of lile and property. Unfortunately. the earthquakes are. so far. unpredictable and unpreventable. The only course left to engineers is to design and construct the structures in such a manner that the damages caused by the earthquakes are minimum. Earthql/{/ke enRineering deals with the design and construction of structures that are earthquake resistant. TIle various structural components arc designed to withstand the earthquake forces. Geotechnical cllrlhqllllke ellgilwerillC is a young brant:h of earthquake engineering that developed in the lust two decades or so. It is concerned wi th geOiechnical aspects of earthquake engineering. such as the type and de pth of foundation soil. amplilication of earthquake's intensity by SOIl deposits. liquefaction of so ils. etc. This chapter discusses the basic concepts of the geotechnical earthquake engineering.

32.2. HISTORY OF EARTHQUAKES IN INDIA India is one or the seismic countries of the world. Some of the severe earthquakes of the world occurred in India. The earthquake of the highest magnitude, represented as M and defined later, occurred in Assam (India) on the 12th of June. 1897. The earthquake was of 11 magnitude of 8.7 on Richter's scale. 'Illat earthquake caused extensive lhlmages in Assam. . In the period from 1905 to 1950. fo~r severe earthquakes of magnitude greater than 8.0 occurred in India.

8"

GEOTECHNICAL EARTHQUAKE ENG INEERING

~

__~68~"____~71~"~~~'6~____~80~"____~'4~ " ____~8~8~__~'~i____~'~6__-, ["

~

MAP OF INDIA SHOWING SEISMIC ZONES OF INDIA

,,"

,i

"

,,"

2,

,,"

'6

0 0 0 0

ZONE

II

ZONE

ZONE

IV

.lONE

V

,,"

," ,i

"i (Redrawn after

Cuur/f'.'Y:

"

,,"

IS : 1893 .2002)

Fig. 32.2. Sci,mic wiles of Indiu The Bureau of Indian Slandurds. New Delhi-I 10002.

"'0


These earthquakes occurred in Kangra in 1905 (M = 8.6). Bihar in 1934 (M = 8.4), Andaman in 194 1 and Assam in 1950 (M = 8.6). Besides these earthq uakes: there were some less severe earthquakes. In the period from 1951 to 2000, seven moderate earthquakes of magnilUde between 6 and 7 occurred. These earthquakes occurred at Anjar in 1956 (M = 6.1), Koynn in 1967 (M = 6.5), Bihar-Nepal border in 1988 (M = 6.6), Uttarkashi in 1991 (M = 6.4), Kilian (Luur) in 1993 (M = 6.2), Jabalpur in 1997 (M = 6.0), and Chamoli in 1999 (M = 6.6). The Killari (Lal ur) earthquake was the worst, which caused about 10,000 deaths. In the beginning of this century. on 26th January, 2001, a strong eanhquake with a magnitude of 7.9 occ urred in Shuj (Gujarat), It caused heavy damages and more than 20,000 persons died and about 1.67.000 persons were injured. About 3.00.000 houses were destroyed and huge losses of property occurred. The Bhuj earthquake ha~ demonstrated the vulnerability of various Indian regions to earthquakes. It has very clearly shown that most construction in the seismic regions do nOI comply with the required safety requirements. The general public has become more concerned about earthquakes and the great damage they can cause. 11 is a major challenge to ensure that OUT constructions arc made earthquake resistanl to limit the . damages in future.

32.3. SEISMIC ZONES OF INDIA The severity of earthquakes in different parts of India is different. The seismicity at a placed is assessed by its di stance from Ihe active fault in the rock formation. It is also decided from the earthquakes already occurred in that region . In general. the seismic zoni ng of a cou ntry depends upon the seismic history of .different regions unless some sign ificam changes in tectonic features occur. As per IS : 1893 (Pan I); 2002. India has been di vided into four lones. designated as Zone II. III. IV and V (Fig. 32.2). The erstwhile zone I has now been rncrged in zone II. The zone numbers increase as the seismic severity increases. Thus zone II has the lowest seismicity and zone V has the highest seismicity. Zones IV and V arc located in the northern and north easlern regions of the country. the Kutch and Maharashtra area and Andaman . The rest of the country has only zones IT and III. Various wnes indicate the regions in which the seismicity would be approximately same . However. the severity of the earthquake in a zone may be different because of local si te conditions.

32.4. MAGNITUDE OF AN EARTHQUAKE The magnitude. (M) of an earthquake is a qUQf/tl1/ive measure of its strength in tenns of energy released at focus. It has a unique value for an earthquake. Charles Richter developed the first scale for magnitude in 1935. Magnitude is defined as the logarithm (to base 10) of the ma ximum trace amplitude (A) of the ground motion rt"corded in microns ( =IO-6 m) at a distance of 100 km from the epicenter on a Wood-Anderson Type Torsion Seismograph having damping equal to 80% of the critical damping. natural period of O.SS and magnification of 2800. Thus M = 10groA

... (32.1 )

where. M is magnitude on Richter 's scale. A is amplitude of ground motion at a distance of 100 km from the epicenter. in microns.

Since the distance of the instrument from the epicenter may nol be always 100 km. the following modified equation can be used 10 estimate the magnitude. . .. (32.2 a) M = 10g LoA - loglOAO or

M

= logro A + distance correction

... (32.2 b)

where A and Ao are usually measured in millimetres ( = IO-~ m). The quantity (-IOgIO Ao) is known as the distance correction. The distance correction can be obtained from Fig. 32.3. For a distance of 100 km . the distance correction is +3. Alternatively. the following equation may be used:

GEOTECHNICAL EARTHQUAKE ENGINEERING

841

h

.f

L--,

~

~eleSei~~

/

.!. 3

c

~-8

I ~~;t~:~e

L

~3

§ '~ 2

!

1

0

-.---

I

./

E

surface w.,es with time period 20 s

10

100 Distance

in

1000

10000

km

Fig. 32.3. Distl1llC'e Correction

... (32.3) M = 10glOA - 2.48 + 2.76 108 !J. where !J. is the epicentral distance (km) For illustration. if the amplitude of the ground motion at a distance of 100 km is 10\lm ( = 10 mm).

M", log)o 104

From Eq. 32.1,

= 4.0

From Eq. 32.2 (b), M = 10g lO 10 + 3.0

too

= 4.0

From Eq. 32.3, M = 10gLO 10 - 2.48 + 2.76 10gLo 100 = 4.04 Table 32.1 gives II typical classification of earthquakes based on their magnitudes. The table also gives annual average number of eanhquakes across the Earth in each of these groups. Table 32.1. Global occurrence of earthquakes of different magnitudes AII/II/al average mllnfMr of global occurrence

Magnitude

Group

8 and higher 7-7.9 6 - 6.9 5-5.9 4 - 4.9 3-3.9 Less than 3

Great Major Strong

IS 120

Moderate

800

Light Minor Very Minor

Abou 9000 per day ( '" 3.285)( 106 per year)

6200 (estimated) 49,000 (estimated)

Earthquakes With a magnuude of 3 or less are not felt. Earthquakes With a magnllude less than 5 nonnalty do not cause significant damage. The maximum magnitude of the earthquake occurred so far was 8.7 (Assam, 1897). The magnitude corresponding to the origina1 fonnulation proposed by Richter is also called the local magnitude (Mi). In addition to the local magnitude, three more magnitude scales have been proposed. The body wave magnitude (Mb) is based on the compressional body waves (or P-waves). The surface wave magnitude (MI) is based on the surface or Rayleigh (R) waves. The seismic moment magnitude or the wave energy magnitude (Mw) is ~ased on the seismic moment.


1142

For small earthquakes, all the four magnitude scales give almost the same value, but for large earthquakes, the values of magnitudes on different scales may be differe nt . For example, for the 1964 Alaska 6.8. M~ 8.6 and M ... 9.2. Although there are some Earthquake. the estimated magnitudes were Mh advantages in these magnitude scales. probably the Richter scale is still quite popular and is com monly used in practice.

=

=

=

Energy released in an earthquake A relmionship between the alllount of the Sltain energy released by an earthquake and its magnitude is g iven by Richter as 10glO E

= 11.4 +

1.5 M

where E is the energy released in ergs ( I crg and M is the magnitude of the earthquake.

... (32.4)

=

IO~1 1)

Earthquakes release huge amount of energy. An earthquake of magnitude 6.3 would release energy equal about 8 x 10']J.) ergs, which is equal to that of the atomic bomb dropped at Hiroshima in 1945. Fortunately, the most o f energy released by an earthquake is dissipated inside the earth into heat and fracturing of rocks. Only a small fraction of the energy goes into seismic waves that cause shaking of the ground surface. It may be noted that with an increase of one on the mtlgnilUde scale would result in an increase in amplitude of len times and an increase in energy o f 3 1.6 times. The affected area and the durmion of the earthquake increase with an increase in magnitude of the earthquake. The affected area depends upon a number of factors such as the depth of focus, type of strata, etc . The affected areas by the earthquakes of magnitudes 6. 7 and 8 respecti vely would be o f the order of 60,000, 120.000 aod 200.000 square kilometres. The corresponding dumtion of eart hquakes would be of the order of 15, 30 and 45 seconds. For example. the Koyna earthquake of 1967 (M = 6.5) affected an area of about 400 km radius. with an area of aboU! 126000 sq. km.

10

32.S. INTENSITY OF EARTHQUAKES Thc intensity of an earthquake is measured qualitatively in tcrms of damages caused by it at a particular location. In general, the larger the earthquake, the greater is the intensity. For the same earthquake, the intensi ty decreases as the epicentral distance o f the location increases. Thus the maximu'm intensity of an earthquake is ncar its epicenter. Several intensity scales have been proposed by differen t investigators. Generally, modified Mercalli Intensity (MM I) scale is used in practice. IS: 1893-2002 a lso recommends the use of Comprehensive Intensity Scale (MSK 64). Both MMI and MS J
.t.

GEOTECHNICAL EARTHQUAKE ENGIJ'lEER ING

""

Table 32.2. Modified Mcrcal1i IntensIty (MMI) Seale Imellsit.\'

Descrillli(JII

Ma.l!ldlude t"IIl Hie/I/er's set/Ie

Not felt by humans. detected by instruments only Felt only by sensitive humans; delicatcly suspended objects may swing Felt r1()liceably indoors: small vibratory motion Felt by many persons: windows and doors make cracking sounds. Felt by almost everyone; some panic: minor damages Felt by all; many frightened: some damage to ordinary structures E\'erbody run., outdoors; negligible damage to well·buih structures: considerable damge to other structures Slight damage \0 we11-built structures; serious damage \0 ordinary

1-1.9 2-2.9

Class

InSignificant Very light

IV V VI VII

Light Moderdtc Slightly strong Strong Very strong

VIII

DeSlructive

IX X

Ruinous Dis(lstrous

III

XI XII

I

Considerable damage to well-built structures Serious damage to well·buill structures: almost total destruction of non-seismic resistant structures E:uremely disastrous Almosl [Otal destruction; broad fissures in ground Catastrophic Total dmnugc: waves seen on ground; objects thrown into the air.

J

-

3.9 4.9 5.9 5.9

5-

5.9

6-

6.9

67-

6.9 7.9

7-7.9 8-8.6

observed was X. The contours of intensity IX, VITI, VII are also marked. From the MMI scale, it can be observed that the earthquake was of intensity classified as disastrous. The difference between the magnitude and lhe intensity should be carefully noted. While the magnitude of an earthquake has a u.nique val ue. the intensity decreases as the distance from the epicenter increases. In other words. for a cer1ain ear1hquake. the intensity varies with the distance. Sometimes. a sound source (or an electric bulb) analogy is used 10 demonstrate the difference between the magnitude and the intensity. While the sirength of the sound source (or electric bulb) is constant. the intensi ty of sound (or light)· decreases as the observer moves away from the soun;c. 1. Relationship between magnitude and maximum intensity The Gutenberg-Richter relation is commonly used in practice; according intensity arc approximately related as

J

3 4 5 5

M

= 1.3

10

which, the magnitude and '

+ 0.6/0

... (32 .,5)

where M is the magnitude of the eanhquake, and 10 is the maximum intensity on MMI scale (expressed in Arabic numemls). Table 32.3 gives the commonly used values of intensities for different magnitudes of earthquakes. Table 32.3_ V:llues of the maximum intensities for different magnitudes

Ma.{imulII i,l/CIU;/)' (m MMlscale

of earthquakes. XI

2. Relationship between peak ground accelerdtion and intensity The earthquake engineers require the value of the maximum ground acceleration so that they can design the structures and assess the liquefaction characteristics of soils. The exact value of the maximum seismic acceleration cannot be theoretically delennined with present stale of knowledge. The following expression is sometimes used to estimate the peak ground acceleration from the inlensity of the earthquake at any localion:


844

loglo a =

~

I - 0.5

... (32.6)

where a is the peak ground acceleration in cm/5 2, and I is I,he intensity on MMI scale (expressed in Arabic numerals).

It is the usual practil.:c to represent ground accelerations in terms of Ihe acceleration due to gravity (g). Table 32.4 gives the approximate values of the peak ground acceleration for different intensities of earthquake as given by Bolt (1993). Table 32.4. Peak ground acceleration,s for different intensities of earthquakes Intensity on MMI scalt

According to IS: 1893: 2002, the basic zone factors (z) given in the code are reasonable estimates of effective peak ground acceleration for the design of various structures. Table 32.5 gives the zone factors (or different seismic lones. The intensity as per MSK 64 broadly associated with the various zones is also given. Table 32.5. Intensity and Zone factors for different seismic Zones Seismic

Wile

fmeruity

aJ

Zmlt:

IX and· above

flU MSK 64 Jcalt:

factors

(ill terl/lS

0.36

of g)

32.6. EFFEcr OF GROUND MOTION ON STRUcrURES An earthquake produces seismic waves that cause the earth's crust to vibrate. These waves impart a momentary acceleration to the earth's crust and it starts moving in the direction in which the wave is travelling at that instant. The characteristics of vibration. such as intensity, duration, etc., at any location depend upon a number of factors, including the following: (I) the magnitude of earthquake (it) the depth of focus (iii) the epicentra[ distance (iv) the characteristics of the material through which the seismic waves ITavei.

Inertia Force When the ground motion occurs, the foundation of the structure must also move with it to avoid its rupture. When the foundation moves, the structure on it tends to stay back because of inertia. ConsequentJy. the structure is subjected to inertia forces. In the earthquake-resistant design of structures. the inertia forces due to earthquake are considered in addition to the nOTmal loads and forces. To get a clear idea about the inertia forces, let us consider thaI a person is sitting upright in a milway train not in molion. When the train suddenly starts moving. his lower body on the seal moves. with the train but his upper body tends to stay back because of inertia. An inertia force acts on his upper body in the backward direction and his head may strike the back of the seat. According to NeWlOn's second law of motion, the inertia force is equal to the product of the mass and acceleration, and it acts in a direction oppositc to (hat of the acceleration. Thus the honZOnlal inertia force is given by the expression F" Mass x acceleration

=

F/I = (

-i)

(A" g) :: WAh

where A'l is the seismic cocffidcnl in the horizontal direction, nnd W is (he seismk: weight.

... (32.7)

.•A


When the horizontal shaking of the ground occurs. the hori· ZOntal inertia forces are generated at the level of the mass of the structure. which is usually assumed to be concentrated at the floor levels (Fig. 32.S). These inertia forces arc transferred from the floor slab, through the walts (or columns), to the foundation. and finally to the strata below the foundation. In the earthquake· resistant design, it is en<;ured that each of the structural clements. including floor slabs. walls. columns and foundations, can safely transfer the inertia forces through them. Moreover. the connections between the various structural elemenls arc properly designed and constructed.

FLOOR

SLAB

INERTt,&.

FORCE

COLUMNS

Response spedrum If the building were rigid, then every point on it would move by _ _ __ the same amount as the ground, EARTHQUAKE SHAKING and consequently, the inertia force Fig. 32.S. Inenia force on a simple structure would be equal to the mass multiplied by the ground acceleration. But the bu ildings are flexible, and different parts move back·and·forth by different amounts during ground shaking. nle natural period of vibration of a structure is the time period of its undamped. free vibration. The [undamelllal lIafllral period of vibrati on is that for the first (or fundamental) mode of vibration of that structure. Each structure has a unique fundamental period of vibration at which it tends to vibrate when it is allowed to vibrate freely without any external excitation. The fundamental natural period depends upon the form and configuration of structure, the stiffness (or flexibility) of Ihe various structural members, the type and material of construction, etc. Methods for the detennination of the fundamental period of a structure are outside the scope of this lext. However. IS: 1893·2002 gives the empirical formulae for the estimation of fundamental natural period of vibrmion for huildings that can be used (see Secn.32.8). For the estimation of seismic forces in structures. the respoose spectra are commonly used in practice. The response spectrum of a structure shows the maximum response induced in the structure by the ground motion. It is usually plaited in lemlS of the maximum absolute acceleration against nalural period, although sometimes the maximum relative velocity or the maximum relative displacements arc also used. While plouing the response spectrum. it is assumed that the structure has single degree of freedom and it may have different damping. In other words, the maximum response spectrum represents the maximum acceleration of an idealised single degree freedom systems having a certain natural period of vibration and damping when it is subjected to earthquake ground motion. Fig. 32.6 shows the maximum response spectrum for a typical structura l system with a certain natural period and damping. The following points may be noted: I. The response of the system decreases a... the damping of the system increases. In buildings. usually 5% damping is a... sumed.


2. As the natural period increases. the acceleration first increases to a maximum valuc and thcn iI decreases. For the structural system shown. the greatcst acceleration occurs when natural period is ahout 0.3 s. 3. It is Ihe usual practice in the earthquakeresistant design to represent the structural response by the 0·0 '·0 2·0 J·o response factor or P~rjod (5) spectral coefficient in the normalised form Fig. 32.6. The maximum response ~pcclfum of n typical structural system as S,,/ g. 4. The response spectrum helps the earthquake engineer to predict how a panicular structure with a certain natural period will respond to the earthquake. 5. The response spectra are commonly used in Ihe estimation of seismic forces, as discussed later.

o.o~~~~

32.7. GENERAL PRINCIPLES OF EARTHQUAKE-RESISTANT DESIGN Some of the general principles of the design of earthquake-resistant design of structures arc brieny discussed below. For more details, a reference may be made 10 IS: 1893-2002. 1. Design basis earthquake. In the earthquake-resistant design, it is not the objective to make the structure absolutely eurthquake proof that will not suffer any damage during the rarest of the earthquakes. A fully earthquake-proof structure will be very huge and highly expensive. Instead. an attempt is made that the structure should be able to withstand the minor earthquakes that occur frequently in that region. Moreover, the structure should be able to resist moderate earthquakes. called design basis earthquakes (DB E), without significant structural damage. Such earthquakes could occur once during the life of the structure. Even a major earthquake. called the maximum I:onsidered earthquake (MCE). with its intensity greater than that of the design basis earthquake, would not cause complete collapse <.If a properly designed and constructed struclUre and the losses would be limited. 2. Pseudo-static analysis. Earthquakes cause dynamic loading on structures. However. for the design of earthquake-resistant structures, the dyn:unic analysi~ i~ usually not camed out. Instead, a pseudo-static analysis is used in which the earthquake forl:es arc replaced by equivalent static forces. TIle$e forces are considered in addition to the normal loads on the structure for iL.. design. II is assumed that the forces due to earthquake are not likely to occ ur simultaneously with other occasional forces such as wind loads, maximum !load forces or maximum sea waves forcc.... 3. Components of acceleration. Earthqunkes can cause acceleration in any direction. It is the usual practice to consider the components of acceleration in the vertical direction and in two perpendicular horizontal direction. Moreover, the acceleration components can be either positive or negative in these three directions. Since the three components of earthquake acceleration may not act at the same time with their maximum magnitude, the code recommends that when the maximum r~ponse from one componenl occurs. the response from lhe other two components can be 30 percent of their maximum values. All possible combinations. including plus or minus signs, should be considered in the design. Generally. the horizontal acceleration is the most predominant.

J


1W7

4. Increase in pel"missibl e s tI"CSSCS. Th~ vertical component of acceleration can incrca!'>c the normal vertical loads on the st ructure. Because of the provision of adequate factors o f safety used in the nonnal design of structures, most of the struclUres arc able to resist the additional momentary vertical loads due to earthquakes. According to the code, when em1hquake fo rces lire considered along with the normal design forces, the permissible stresses in materials in th ~ dastic method of design car. tx: increased by (llie-third. H owcv~ r, for steels having a definite yield str~ss, th~ increased stress llIay bC' limited I() the yield stress, and for steels without a definite yield point, the stress may he limited 10 110 percell! of the ultimate strength or 0.2 percent proof strain. whichever is smaller. 5. Increase in allowable bcal"ing pr-eSSUI'l'. The alJowahle hearing pressure in soils c.lO he incleased by 25 to 50%, depending on the type of :.oil and type of foundation as per details given in the code. However. for soils thai arc liable to liquefaction. :his increase is nOl applicable. Soi ls f:!lling under Indian Standard Classi fication category SP (i.e .. poorly graded sands), with standard penetration test (SPT) N-v:!lues less th:m 15 in seismic zones TIL IV and V..and less than 10 in zone II may become liquefied. Suitable

measJ.re~:;;z~:::;n~i~~ ~c~~~:~n~~~I~~i~IS f~~scc~~s~,~at~~~~~:ll~~~~tt ~~;rection

of ground motion is usually

horizontal. 111erefore. the horizontal seismic forces are most important for the earthquake-resistant design. The melllOds for the estim,ltion of horizontlll seismic forces arc discu.~sed in the next secti on. According to the code, the vertical inertia forces arc to be considered in the design unless checked and provery that they are not significant. When effects due 10 vertical earthquake loads are to be considered, the design vertical acceleration spectrum is taken ,L>; two-thirds of the design horizontal acceleration s.pectrum. Vertical inertia forces arc impon:mt for structures in which stabi lity is a criterion for desig n, such 3.<; retai ning w'llls, cant ilevered members and preslressed hori7.0nt:11 members. Rcdm:tion in grav ity forces due to vertical component of ground mot ion can be detrimental in some structures. 7. Resonance. According to the code. rcson:mce of the type as visuali zed under stelldy-state conditions will nOI occ ur because the earthquakes have irregular motion o f short dumtion in which there is not adequate time to build up the required amplitudes. However, if the structure's fundamental period is close 111 that of site, reSOnance may occur. Such conditicms have been observed for some tall buildings on deep soft soils. H. Base shear. Inert ia forces generated in the structure due to an earthquake arc assumed 10 be transferred to tnc base of the structure a... the base shear. 111e hase transfen; these forces 10 the foundation, wh ich. in tum. transfers the m to the ground. All the components of the .structure and foundations arc designed to resist these forces in 'lddit ion to the nonnal forces. The distribution of the seismic forces in different structural elements is done by the methods of structural analysis heyond the scope of this text.

32.8. DESIGN SEISMIC COEFFICIENT According 10 IS: 1893-2002. the design horizontal coeffi cient A" for a structure can be determined from

*

,'"follow;". "P::,,,~o(~ 1( 1(~ 1

... (32.8)

where. Z = Zone factor o f the seismic LUne (For seismic zones II. III, IV and V, the values of zone f!lctor are given as 0.10. 0. 16, 0.24 ~and 0.36. respectively). J = Impor1ance factor. It depends upon the functional use of the structure. The value of J is 1.5 for ~' important service community huildings. such as hospitals. schools. railway stal io ns. tire stations c inema halls , etc. ' For all other buildings, its v'llue is 1.0.

SOIL MECBANICS AND FOUNDATION ENGINEERING

1148

R = Response

reduction factor. depends on the perceived seism ic damage perfommnce of the structure. The value of R varies from I ,5 to 5.0 for different types of huildings. For ordimlry R.C. momentresisting frames , its value is 3 .0 . 111e ratio IIR should nOl be greater Ihan 1.0. S,/g = Average re.~ponse acceleration (.;oc ffi cient. II indkates the average smoothened plot of maximum a(.;celemlion. It depends upon the tyflC o f slrala and nmural period . Fig. 32.7 gives the values of (S,/,I:) for rock (or hard soil). mediulll soil and soft soil sites for 5% damping. For other valucs ()f d:unpings, the (.;odc gives the multiplying. factors. The maximum value of S,/g i!. 2.50.

z

~

J·O Typf'

8

\\

Type ',/Type

t (Rock .H.lrd Soil) II (Medium Soit) lIl{SoI!

Soil)

\ '"

.~<::.:.:.:::::.:.~:::o::..:._.:=-'"

O'O,L~_-'--_L-~_~_L-~_

0·0

0·5

t·O

1·5 2·0 2·5 PERIOD (s)

(Redrawn alter

3·0

3-5

'·0

IS : t893-2002)

NaturAl period For delemlining the value of (5./,1: ) for a FIg. 32,7. Rcsp()n.~e specll.... for rock ~.ntJ ~oi1 site~ ror S percen.' damping given structure, its natural period of vitmllion COli/u.n'; The Bureau of Indian SlomJ:ards. New Deihl. is required. The code gives Ihe following approximate fOllllulae for Ihe (';OJ1lputllti()n of the fundamental natural period of vibration (I:<). (a) Momellf-re,"hltillg frame blli/dill!:," lI';tll("" b,.ick inji/-puI/eI.\· (I) R.c. frame buildings

T, = 0 .075 "O.7~

... (32.9)

(ii) Sleel frame buildings

T., = 0.085 h01.~

... (32.10)

(b) All other Imildill!:.I". ;"c/lldi"l: lI/(mlt'lIf-resi.I·lill!: frmlle.l· with brick il/fii-panels

T<=T

... (32.11)

where.

T, = natural period of vibnllion, in seconds. II = height of building, as specified in Ihe code, in metres. d = base dimension of Ihe building .11 the plinth level along the direction of the lateral force, in melres. The code specifics that for any structure with 1'., :s; 0.1 j', the value of Air will not be taken tess than (m), whatever be the value of fiR.

32.9. DESIGN SEISMIC FORCES The design seism ic forces are computed as explained below: (a) Design lateral force

The tOlal design lateral force acting on the structure is equal 10 the product of the design horizontal coefficient (A h) and Ihe seismic weight of the structure. T~ ~eismic weight of a building is equal 10 the sum of the seismic weights of all the noors of the huilding. 'The seismic weighl of each floor is equal to its full dead load plus an appropriate fract ion of Ihe imposed IOlld, as specified in the code. While computing the

GEOTECHNICAL EARTHQUAKE ENG INEERI NG

849

seismic weight of :my floor. the dead weight of columns lind walls in any storey is equally distributed to the floors above and nclow them. The design seismic bilsc shear m.:ting on the structure is equal to the t01a1 design lilterill force and is given by lhe expression ... (32.12) where Ah = design horizontal seismic coefficient W = seismic weight of the bui ld ing VB = design hase shear The design bilse shear is di1>tributed to different floor levels, as specified in the code. The horizontal force acting 0 11 the foundation is equal to the design base shear. (b) Design vertical rorce

Since the design vertical coefficient is equal to the the design vcrtical force is given by the ~xprcssion

two~lh ird s

of the design horizontal coefficient (A h ),

F = ~ A" W " where W is the seismic weight of the building. and F" = design vertica~ force. TIle design vertil;al force is transferred from the structure to its foundations.

... (32.12 a)

(c) Design of structure

Buildings and portions thereof should be designed and constructed to resisl Ihe effect of design seismic forces. Since the seismic forces occur suddenly and without warning. it is essential to avoid construction practices that lead to brittle failure. The members should be designed to behave in a ductile manner so that complcte collapse of the structure is avoided cven during severe earthquakes. Actual design is beyond the scope of this lext. (d) Design of foundation

Since_ the eanhquake-rcsislant designs arc generally performed by pseudo-static analysis, the seismic loads on the foundation arc considered as static loads. These loads are capable of producing settlements like other loads. Thc code permi ts increase in allowable bearing pressure depending nn the soil-foundation system. The increase in allowable bearing pressure is 25 or 50 percent. In soil-foundation systems, where smal l settlements are likely to occur during earthquake, the increase in allowable bearing pressure is larger, and vice- versa. For example. for all types of foundations on rocks or hard soil. the permissible increase is 50%. For soft soils, the permissible increase for piles resting on hard rock or hard soi ls and the raft foundations is 50%, whereas that fm other types of foundations. it is 25%.

32.10. SITE-SPECIFIC RESPONSE SPECTRA The response spcctn\ of ,I structure :lre sitc~spcc ific .. as they depend upOn the type of strata at the site. Seismic waves travel diffcrently through soi ls and rocks. Therefore, the ground shaking at sites underlain by soi l is different from those underhlin by rock.

SOilsI~:~i:~~~~v~a~U~ir:~~~ea!~:~~~:n\~!e:~ ~ha;::~~~~: ~~~~t~~ ~~~St~;td~;I~~nar~ ~~t~ ~~~~~s s~rl:o: rock. The structures founded on soft soils collapsed. whereas adjacent buildings founded on stiffer soils withstood the shock without collilpsi ng. Similarly, in the 1985 Michoacan Earlhquake. Ihe damages to buildings in Mexico city, 350 km from the epicenter, were much worse than in other cities that were much closer. The reason for this difference is attributed to the difference in site conditions. The buildings in Mexico city were underlain by deep deposits of sort clay.


850

0·'

soils

0·5 0·6 0·7 Peak horizontal acceleration in bedrock (9) (Redrawn after Idriss, 1990) 0-3

O·L

Fig. 32,8. ElTed or sile conditions on the peak hori7.0nlal occcicration

Fig. 32.8 gives a simplilicd. approximate rcilitionship for different soils (Seed, et al 1916, lmd ldriss. 1990). 0.9r------~---,5:-·-,.-D-A-M-PI-NG-~

0·8 Layer)

'", 0 '6 z

Q 0·5

~

...u~ 0·'

~ 0·3

PERIOD (Seconds) (Redrawn after Seed et at, 1987) Fig. 12.':1. Ac.:det

Even in Mexico city. the spectral acceleration for different sites was dirrerenl (Eig. 32.9). The observed response at rock and hard {soils was fairly low. and the damage caused was negligible. AI the central markct site (CAO site), spectral accelerations werc amplilicd at periods of 1.3 s and within thc range of 3.5 s to 4.5 s. However. the d:l.ma!les were fairly minor. since the buildings at the CAO site did not have fundamcnlal periods within Ihese ranges and resonance did not occur. However. ncar the scr building sites. the soil deposit's fundamemal period at which amplification of spectral acceleration occurred matched with those of structures and major damages occurred due 10 resOnance.

85'

4·00

~

~

~

Soli to

m~dium ctays and sands (soil Iype 3 1

3·00

-0

zZ

Q8

~; ~~

g§ ~~

0-00 0·00

1·00 PERIOD

2·00 T

(

S~conds

3-00 1

Fig. 32.10. 1991 USC Normalised' occelcr:l.lion rcspon~e spectra

Uniform Building Code (1991) gives the normalised acceleration response spectra for different site conditions (Fig. 32.10). The spectral acceleration of a structure can be e!;timat
32.11_ HAZARDS DUE TO EARTHQUAKES lllere are many hazards due to earthquakes. The hazards rel;Jted to gtwtcchnical engineering include the following: 1. Liquefaction of soils. In satur:lled. eohcsionless soils, earthquakes may cause liquefaction (or quicksand) conditions when the shear strength is reduced (0 almost zero. Liquefaction of soils can lead to many types of failures. such as bearing capacity failures, si nking or tilting of buildings. land slides. laleml spreads. notation of underground structures. Liquefaction of soils is discussed later. 2. Ground shaking. Shaking of ground occurs during nn earthquake. The intensity. dural ion and wave form of ground shaking depend upon a number of factors such as the magnitude of earthquake. the depth of focus. ~he epicemTal distance. the energy absorhing nature of the strata. etc. The intensity of ground shaking at a given location is usually specified in terms of peak ground accelera· tion (PGA). There is an approximate re lationship between the PCA and intensity on MMl sc:lIe. as discussed earlier. To predict the response of soils ,md foundations. the characteristics of ground shaking are required. 3. Surface rupture. For earlhquakcs of small magniludes. the rupture zone occurs deep inside the earth

SOIL MECHANICS AND FOUNDATION BNCilNEERING

8S2

and il docs not extend to the ground surface. However. for the eanhquakcs of magnitudes greater than about 6.0, the ruplUre surface usually appears al the ground. Moreover. as the magnitude of the earthquake 'increases, the length of rupture surface also increases. Surface rupture creates a number of problems for buildings, bridges, railways, highways. water supply lines, etc. thai arc located directly above the fault It is the usual practice to locate the buildings and structures susceptihle to heavy damage not directly over the active faults. 4. Land slides. Eanhquakes CliO cause land slides and failure of slopes. Even when the slope fai lures do not occur. tcnsion cracks may appear in the soi l. These tension cmcks may 'lead 10 the formation of water channe ls. causing softening and saturation of soils. S. Lateral spreading. Lateral spreadi ng of soil is the massive horizontal movement of soil layers in a direction parallel to the ground slope. Lateral spreading due to earthquakes is usually observed on very gentJe slopes. If the lateral spreading occurs beneath a structure, it can tear it apart, causing heavy damages. In the design of such structures. suitable provisions
~oS~~;~It'°~slil~;lilar

h3d in nature 10 a tsunami. but il is of much smaller size and it occurs in lakes or rivers. The seiche usu3ily occurs when resonant conditions occur in the lake due to an earthquakes, i.t'.• when the nalum! frequency! of the lake and that of thc earthquake are equal. $omelimes a seiche occurs in a lake when the rupture surface is beneath the lake bed. Barly warn ing systems. should be ins1alled so Ihal suil
32.12. LIQUEFACTION PHENO'IENON Liquefaction is a state of saturllted cohesion less soil when its shear strcngth is r~uced 10 zero due to pore waler pressttrc I;aused by vibration during an earthquake. The soi l starts behaving like a liquid. To understand the liquefaction phenomenon. lei

~: :O~:i~t~ ~;0~1 :~::n~h~f ~~~::~I ~~~~~ (O.S.). Lei the dcpth of water table (w'T.) be II below t~e ground surface (Fig. 32.1 1). As discussed in Chaptcr 10. Ihc'e flec-

Fig. 32.1 1. Stresses in a soil element

...., GEOTECIINICAL EARTHQUAKE ENGINEERING

live .. tress (represcmcd as

rY

ct in this chapler) at a depth z below = yh + 1<", (z-h) - Y~. (z-h)

the G.S. is.given hy

0' = yh + y' (z-h) where y' is the submerged density (= '(,,,,-1,,,) TIle shear strength of a cohesionlcss soil is duc to internal Inclion nn9 is given by

s =

a'tnn

<\I'

If the soil deposit is subjected to ground vibrations. it lends to comp:lct and decrease in volume. However, if the drainage of porc water is prcycnled, this tendency to decrease in volume rcsu[ls in an increase in pore waler pressure. U!I UJ he the excess dynamic pore water pressure developed duc to ground vibration. The dynamic shear strength is ex pressed as SJ

= (a' - 11,/) Ian $'

For sandy soils, the angle of internal friction 41' in the dynamic conditions is almost equal to that in static conditions. The dynamic shear strength (Sd) will become zero when

cr'

= lid

~= Expressing

Il,/

I

in terms of the dynamic hydraulic head (h J ),

1~'11::

I

Thus liquefaction in a cohesionless soil will occur when Ihe dynamic pore water is equal to the effective stress. It may be noted that because of the dynamic pore water pressure, the shear strength of soil is decreased. It results in transfer of intcrgranular stress to pore water pressure. If this transfer of stress is incomplete, there is partial loss of strength and partial liquefllclion occurs. However, if the transfer of stress is complete, the shear strength becomes zero and complete liquefaction occurs. TIle sand-water mixture behaves like a viscous liquid after complete liquefaction. Large settlements occur after liquefaction and the struclUres resting on such a soil deposit start sinking. This sinking process continues till the sand remains in the liquefied stale. 11 may be summarised that for liquefaction to occur, all the following five conditions must be satisfied (Coduto, 1999). I. The soil is cohesion less. 2. The soil is loose. 3. The: soil is saturated. 4. There is shaking of ground of the required intensity and duration. 5. The undrained conditions develop in the soil due to its limited permeability. It is worth noting thai liql;lcfaction can occur in the soil dcposit at any depth whcre Ihese conditions are satisfied. Once liquefaction occurs at some depth, the flow of watcr occur.; in the upward direction, and it may cause an indirect liquefaction in the soil layers abovc. II is observed that liquefaction normally occurs in the soil cla'isificd as SP according to Indian Standard Classification when the SPT number N is less Ihan 15. However, sometimes liquefaction may also occur in the soils classified as SW, SM and ML. Fine·grained soils do not compress readily under dynamic loadings to cause a high dynamic pore water pres,sure. Moreover, they also posscss the shear strength due 10 cohesion. Therefore, liquefaction does not occur in the fine-grained soils. However, in quick clays, there is a possibility of liquefaction because their slr6cture is dcstroyed by shaking.

'54


When liquefaction occurs, sand boils (or mud SpoulS or sand foun tains) appear aI the ground surface. A large number of sand fountains were ohserved during rhe Dhuhri Earthquake in Assam ( 1934), the Bhuj Earthquake in Gujarat (2001), and nlliny other earthquakes throughout the world. Liquefaction of soil is responsible for many failures of earlh structures, slopes. foundations. II may also lead 10 land slides. laICral spreads. sinking of S\fuctures and flotation of underground structures.

32.13. FACTORS AFFECTING LIQUEFACTION TIle following factors mainly affect the liquefaction of soils in the field. 1. Soil type. As already discussed, liquefaction usually occurs in cohesion less soils, especiall y soils of type SP. On the other hand, liquefaction docs nOI occur in fine-grained. cohesive soils. However. highly sensiti ve, quick clays may liquefy. 2. Particle size and gradation. Fine. uniform sands are more prone to liquefaction than coarse, wellgraded sands. Since the penneabilily of coar~e sands is greater than that of fine sands. lhe pore water pressure is rapidly dissipated in such sands and liquefaction nOnlmlly docs not occur. Liquefaction pOiential of sands depends on percentage of fines (size < 0.075 mm) prescnt in it. For il sand of a given initial relative density, as the percentage of fine increases. the liquefaction potcntial is decreased. 3. Initial relative density. Liquefnction of sands depends to n large extent on the initial relative density (or density index). In dense sands. both porc water pressure and senlements are considcr.lbly less than those in loose sands. Hence the proneness of sand to liquefaction is reduced with an increase in relative density indicated by SPT number (or cone penetration resistance). 4. Length of drainage path. If the length of drainage path is large, a sand deposit would behave as undrained when the pore water pressure is suddenly increased due to earthquake and the liquefaction may The length of the drainage path is somet imes reduced by providing drains of highly pervious materials in sand deposits. 5. Surcharge loads. The initial e ffective stress in the sand deposit can be increased by lhe application of a surcharge load on it. With an increase in lhe eflccli\lc stress, lhe tronsfer of stress from the soi l particlcs to the pore water is delayed. Thus the sand deposit would require higher intensity vibrations for grclncr duration when the surcharge loads are applied to i1. 6. Characteristics of vibration. The m'lin characteristics of vibration are its accelcmtion. frequency. amplitude and velocity. For liquefaction of soils, the first two characteristics. namely, acceleration and frequency. are more dominant. Acceler.ltion during vibration is the most important characteristics affecting liqueraction of soils. In genera l, the greater tbe acceleration. the greater are the chances of liquefaction. Liquefaction usually occurs only aftcr a ce rtam number of vibration cycles arc re peated. Frequency of vibration is important if il is close to the natural frequency of the soil-foundation system and resonance occurs. 7. Age of soil deposit. If the soil ~eposit is very old. its proneness to liquefaction is relatively low as compared to thai of a recent soil deposit. In old deposits. somc form of cementation occurs at the contact points of sand particles and the transfer of intcrparticle stresses to pore water is delayed. 8. Trapped air.- If air is tnlpped in a sand dcposit, a part of the porc water pressure is dissipated 'due 10 its compression. Therefore, the possihilit}! of liquefaction is reduced to some extenl. 9. Miscellaneous factors. Some miscellaneous factors such as soil structure. method of ~oil deformat ions. e tc. also affect the liquefaction characteristics of soi ls but their effect is not signiticanl.

32.14. ASSESSMENT OF SUSCEPTIllILITY OF A SOIL TO LIQUEFACTION It is important to known whether the soil III the site is susceptible to liquefact ion or not, so Ihat suitable measures may be adopted, if required. Recent rcsean:h has necn directed in this dircclion. Various methods

GEOTECHNICAL EARTHQUAKE ENG1NEERlt"G

have been proposed by different investignwrs. ·nle cyclic ~are~~ npprolu:h is generull ~ u ~c d in most of the methods. The cyclic stress ratiu is derined as the mtio of Ihe cyc lic shear stress (t) to Ihe initial en·cctive slress (0"0')' llle cyclic stress ratio produced by Ihe earthquake (t/ilo')" is compared with Ihal requircd to induce liquefaction in the soil (""f/Oo'),. The faclOr of safety (F) against liquefaction is cstimated as

(32 3)

,.. "" (""f/0"0')/ (t/oo'),/

..

.1

Liqucfaction would occur if F is less Ihan unity. How!!ver, il ha.~ been observed tbat significant pore water pressures occur even at the values of F slightl y grealer than 1.0. Generally, a minimum factor of safelY of 1.25 to 1.50 is sp.::citied. The proneness of a soil deposit to liquefaction is assesscd at various depths in u soil deposit. The factors of safclY at all the points should he within the required limit. (a) Estimation of the cy~lic stress ratio produced by the CllrthqllOkc The avcmgc cyelie she:!r stress imported by the earthquake in the top 12 estimated ~lS (Seed und Idriss, 1982)

(*' ), =

065

III

of a soil deposit can be

(~'l (~ }"

..

where, (Ino.u

g

= maximum horizontal accelemtion (MHA) at the ground surfllce

= acceleration due to gravity (= 9.8 1 m/s2)

= total vertical stress UI the point of interest = effective vertical stress at the same point (""f/0"0') ,/ = cyclic stress ratio produced by the design earthquake. 0"0

00'

r,/

= stress

reduction factor.

The main problem is in the estimation of the maximum horizontal acceleration, (CXmnx). It depends upon a number o f raclors such as the magnitude of the earthquake. the rocal depth, the distance to the fault trace, the response of the sile, etc. The energy released by an earthquake :!ttcnuates as the waVeS travel away from the zone of fracture. Consequently, the maximum horizontal acceleration decreases as the epiccmral distance increases. Fig. 32. [2 shows a typical aUentualion curve for Ihe rock sites in the central U.S. regions (Nullli and ·Herrmann, 1989). The valucs of the maximum horizontal acceleration (MHA) at rock sites get further moditied due

.,

., ·5

.,

-------

"" "'\.

·3

., .\

0

~ \

10

100

EPICENTRAL DISTANCE (km) (Redrawn after Nu"Ii and Herrmann, 1984) Fig. 32.12. MHA 31tcnUlUion rct3tionslllp ror ms=7.S

(32.14)

.,.

SOIL MECHANICS AND FOUNDATION ENGINEER ING

to local soil condit ions. Fig. 32.8 (see Sec!. 32.10) shows the variation of MHA at different soil siles with respect 10 that at rock sites. both the accelerations are expressed in tenns r,f acceleration duc 10 gravity. It may be noted thUt '(he MHA aI sites on deep c lay deposits may be upt6 3 times grealer tha n that aI rock siles. The stress reduction factor ('".1) depends upon the depth of the point of interest below the ground surface. The average curve (Fig. 32.1 3) is generally use·d in practice (Seed and Idriss. 197 1). The relationship can be approximate as rJ = 1-0.008 x deplhin mclrCs ... (32.15) According 10 IS: 1893-2002. the basic zone factors (z) aTC reasonable estimate of effective peak ground acceleration in different seismic zones of India.

Sire-55

0,

... (32.16)

where

= standard

SPT Number corrected for overburden pressure (N)6l) = standard SPT Number measured in the fie ld (duly corrected for field procedures) o r' = vertical effective stress at the test section (kPa).

(NI)60

2. The particle size distribution. expressed as percentage of fines i.e., percent passing the US No. 2DO sieve ( = 0.075 Il). Alternatively, the mean gmin size (D.<,(l) can be used. Soils with less [han 5 percelll fines are 1110st susceptible to liquefaction. As the percent of fines

~:~las=~. the

liquefaction polenlial

&

"

10 4'5

Average-

:: JO 3 'Q..t.O

<

~"

Range'

"

The cyclic stress ratio (t /oo')/ required [0 induce liquefaction in a given saturated sand deposit depends upon a number of factors. In lhe simplified analysis given below. only the fo llowing three factors arc considered (Seed, e t al.. 1985). I. The corrected standard penetration [est (SPT) number indicated as (NI)6I.), obtained from the SPT number (N)oo mea... ured in the field .

~

Factor ,

10

(b) Cyclic stress ratio to induce liquefaction in a soil

(N t )60 = N w

Re-duclion

8r'-,0~"=----,,0';-'----,0:';:"=---.:,0.;:-9----:;

(Redrawn after Seed, et at, 1971)

Fig. 32. t 3. Vari:llion o r

SII'~S.~ r~duclion

roctor with dcpU!

~ 0·' ~

~l),

0"

~ ~

j

0·'

0·3

F", . .

conte!'t

.• 2

."

~~ p., American data

u

C

0.0

0

10

!

··

dat, J~~ges: d,la 20 30

~~

'iJ

... .~ ~

gg Q

0

"

(Nll60 (Redrawn after Seed el at, 1985)

"

Fig. 32.14. Cyclic strr:ss ratio to C;!USC hquefaction as:l. function of correctcd SPT (NI)fil.


3. The dUr.Ition of shaking or the magnitude of the earthquake. As the duration of the earthquake increases. the susceptibility of a soil to liquefaction also increases. In ()(her words. lhe possibility of liqucfaction is increased as the magnitude of the earthquake is increased because the duration increases with the magnitude. Fig. 32.14. gives relationship hetween the cyclic stress rJ.tio causing liquefaction (Vao'), and the SPT Number (Nt)w valucs for differem values of fines for an earthquake of magnitude 7.5. If the magnitude of the earthquake is not 7.5, the value of (t/o:o')/ obtained is to be corrected using the relation

(t-J o

M

= '"

(t-i 0

... (32.17)

J.~

where (t/oO')M is the cyctic stress ratio for the magnitude ohtained from Fig. 32.14. \jI is known as the magnitude scaling factor. Fig. 32.15 gives the value of \jI for differcnt magniludes as given by various invesl igulors. For tile earthquakes of magnitude greater than 7.5. generally Ihe value given by Ambrascys is used. as il gives the minimum value of IV in most cases. For the earthquakes of magnitudes less than 7.5. the value given by Seed and Idriss may bc used liS it gives conservative v:.lues. g It is more appropriate to use the surface wave magnitude (M.•,) for earthquakes of magnitudes greater than 6.5. Once Ihe values of (t/oa'),1 and (t/Oo'), have been obtained. the f
]

.~, I

M and (t/Oo'h.s is that for the magnitude 7.5

- -- Se~d and Idriss (198'2) -·-··Ambrast'ys (1983)

'·0 "5 '·0 0·5 0·0

5

Magnitude

32.15. PREVENTION OF LIQUEFACTION II is extremely difficull and impractical 10 Fig. 32.15. TIle vumlion of scaling f~Clof ItI with the prevent liquefaotion of soi l in the field when the magnitude or the earthquake. susceptible zonc extends to a large depth. II is generally best not to construct huildings and other structures at sites prone to liquct:'lction. Because the remedial ion or the liquefaction hazard is quite expensive, it would be cost effective on ly lit the sites where the land cost is extremely high. Remediation is sOlllelin;es necessary al sites that have neen developed and the huildings already exist and laler it has been found thllt the soi l is prone tn liquef:lclion. 111is generally h"ppens when .. dequ:lIc site investigations were not made prior to the construction of buildings. The following me
SS8


vibronotation, blasting, etc. (sec Chapter 14), The extent to which the compaclion should be done is ascertained by estimating the required SPT No. \0 prevent liq uefaction ror Ihm soil for the dc!\ign earthquake. 3. Replacing Ihe liquefiable soil. If the depth of the liquefiable soil is limited. it can be excavated and replaced with a well compacted soi l. However. if the depth o f the liqucliable s()il is brge, it OcCOIl1CS impractical to replace the soi l occausc il req ui res exte nsive dew31cring systems. This method is however sometimes adopted at thc sites of critical projects such as cnnh dams. 4. G]"<>uting Ihe soil . In this method. the soil is stabilised by inject ing chemicals or cement grOut into the soil (see Chapter 15). This method is sometimcs used at si tes where buildings already ex ist and other methods eannOl be used. 5. Ground water pumping. As already discussed. the e ffective stress at a poim increa...cs as the watrr table is lowered. By restoring to extensive ground water pum ping. the liquefaction can be prevented 10 some extent. However, this method is COSI effective on ly when Ihe water thut is pumped Cllll be used for municipal and industrial purposes. 6. Drainage of soils. 111C liqucCacli on haza rd can be reduced to some extent by providi ng coarse sand blankets and drains in the soil deposit. TIle dynamic pore water pressure is thus easily dissipaled and the cITeelive stress is increased. 7. Providing stone columns. In Ihis melhod, a number of holes are bored in the soil deposit and (iller filled wilh gravel and siones. Thus stonc columns are fanned. The stone columns have high permeability :lOd are quite eITective for rapid drainage of pore water. Thus the effective stress is increased. (TIle behavior of slOne columns is somewhat simi lar 10 thai of the sand drains used for consolidation of soils. discussed in Chapter 12) 8. Application of sUI·charge. As already discussed, when a surcharge loud is applied to II soil depusiT. the effective stress is increased. Thus the possibility of liquefaction is reduced. Monitoring of the pore water Jlrcs~ure will he required to ,lssess the magnitude of the surcharge load and its duration. ILLUSTRATIVE EXAMPLES

Illustrative Example 32. 1. The srmulartl ror.fiOIl sei.\·lI/ogml,i1 record.f {I /race amplitude (If 12.5 mm il/ £-W direcliOIl tw{/ a Imel! WII/Jlilllde of IJA 111111 ill N·S c/irl'l.: rio/l ar (/// l'piN'lIIml di.vf(llIce of /In 1011. /f rlrl! srarion correctiol/ ;s + 0./, UI;III(l/e rhe I/w{:lIil/lde of rile earthquake. Solution. From Eq. 32.2 (b). M = log,oA + distance correction The distance cOITCction for cpiccntral distance of Il v km is + 3. 1 from Fig. 32.3.

In E-W directl{ln, M = IClglO 12.5 + :U = 4.2 Applying the st:llion correction. M = 4.2 + 0. 1 = 4.3 Similarly. in N-S direction, M = lo!!:!) I 1.4 + 3. 1 + 0.1

= 4.26

Average value of M = 4.3 ; 4.26 ,; ' 4.28. s.ay 4.3

Illustrative Example 32.2. ESlimole 1i11! l!l1erg\" re/e{/.\"C(I by COli/pan' rlre n /crgy releaull

f()

w/ ' ('Orllrq/III/a'

,lrm hy 111(' Hiroshima 010111 bOil/b.

Solution. From Eq. 32.4, 10£IOE

E

=

11.4 + 15M

= 2.51

x

= 11.4

+ ' 1.5 x S.n

J O~ ,l L'l'J.!.f.

Ene rgy re leased hy the Hiroshim:1 atom hOlllh

=Kx

2lI

In l'l:t:.I".

= 23.4

of 1I/08"i1tu/e

8. 0.

85'

GEDTECHNICAL EARTHQUAKE ENGINEERING

Therefore, equivalent number of atom bombs

25 1 X I 02~

=:

~;-;oro- :=

314

Illustrative Example 32.3. E.ftimate the maxilllllll! illtellsity Oil MMI .feale lind the /II(/x/Illum acceleratioll due to an earthqlwke of magllitllde 7.0. Solution. From Eq. 32.5, the maximum intensity 10 on MMI scale is given by

KlVlllld

M:=I.3+0.6/0 7.0 1.3 + 0.6/0

=

10 = 9.5 i.e. MM[ intensity will be X. From Eq. 32.6, the maximum ground ucccleration a is given by

,i Jog lo

(J

=

t

I - 0.5 =

t

( 10.0) - 0.50 = 2.R33

= ~~.~9 If = 0.69 g

a = 681.29 cm/s2

Illustrative Example 32.4. Eflim(lle the cyclic slress r(llio pmlluce,/ by all earthquake ell a place where the maximum horiwllfal acceleraliOll is 0.15 R at (I depth of 511/ ill a s(llUly soil deposit. Assume Ille water table is at a deplh of I 1/1 be/ow lite gmwltJ miface. Take Y = Y."" 18 kN/n/

=

Solution. TOlal stress at 5

m depth,

From Eq. 32.14,

r" = ('rlao')"

=

[8

x 5 = YO kN/m1

co' = 0'0 - " = 90 - (4 x 9.8 1)

Effective stress at that depth From Eq. (32. 15),

0'0

I - 0.008 x depth

= 0.65

[* 1

=I -

(a~lV. ) ( ~ )

1',1

= 50.76 kN/m2

0.008 x 5 = 0.96

= 0.65

(7) (

5:';6) (0.96)

= 0.17

Illustrative Example 32.S. (ll) A deposit of mtllrated fille .wmd hllS 5% filles. If the corrected SPT mlntber for the .fOlld is 12. eMili/ate the cyclic .1·tn'Sl· mtio reqllired 10 ctluse liqllefactioll dlle to (Ill earthquake of mugllitllde 7.5. (b) If the magnitude of the earthquake "'ere 8.0. what wOlild have beell the required cyclic stress ratio? Assume a scalillg factor (\II) of 0.65. Solution. (a) From Fig. 32.14, the cyclic stress ratio to cause liquefaction for (N I )60 = 12. percentage of fines = 5% and for the magnitude of 7.50 is 0.12. (b) For the earthquilke of magnitude 8.0. Eq . 32.17 gives

[*1 [*) .,

= 'lI

,ls

= 0.65 (0.12) = 0.08

Illustrative Example 32.6. The maximulII horiwllraillct'elemtioll ill bed mck i.l· e.l·tinUlU(1 to be 0.20 g. What would be the maximulII IlOrit,mlfal acceleratioll al a project sill.' !ocateel ()I'er (a) a deposit of deep cohesiolliess soil. (b) a deposit of deep soft mil lleposit. Solution. (a) From Fig. 32.8, for a bed rock acccicral iQIl of 0.20 g, the maximum horizontal ground acceleration for a deep cohesion less soil deposit is 0.18 g. (b) For soft soil deposit, the maximum horizontal ground acceleration is 0.30 g. IIIust~tive Example 32.7. ((I) A jil'e.~·torey !mildiIiR. 20 III higlJ, htls .)'fee! ji'tllllel' It'il/wllt brick il/fil panels. II is corlStrllcled Oil a cleel) soil .vitt' cmui.\·tiIlR of loo.wo ..~tlfllrtlted SlIlId overlyill/: a thick .wiff clay

860


deposit. The building is to be desigllt'd 10 re.fi.w (III earthlJlwke of magllitl/de 7.5 with maximum bed mdt. horizullflll (/cct'lerl1tim: ofO.2{) R. Estill/afe rI,e I't'(/k xrolllulacceiemri(JII tit tile .file. (b) Estill/me Ihe specfml ,.e'\·IH",~·e IIf the huildillg as I'er 1991 unc Solution. (tI) From Fig. 32.8. the hcd rOl.:k horizontal acceleration of 0.20 g will be amplified to ahout

0.30 8. (b) From Eq. 32. 10, the fundnmentnl natural period for sleel framc.

1'.. := 0.085 IP·7~ = 0.085 (20)O,7~ =- 0.80 s From Fig . 32. 10, as per 1991 UBC, for a period of O.KO s and deep stiff clay, thc ratio

ClTecti:C':~~I~:~cu~:~r~~~~~cratiOn

= I.RD Therefore, the spectral acceleration = 1.80 x 0.30 g = 0.54 g Illustrative Example 32.S, Estimme 'he de.l'iRIl horiWlIllI1 seismic coeJJicient (A h ) for a buildillg from Ille fol/oll'illg dara, m per IS; 1893:2002. I. HeiR"t of buildi/lg = 30 /II 2. Base (Jimeluiol1 of th(' l)IIildil/X IIlol/X the cOII.l"ide/"et/ direclio/l of the lateral force 3. The buildillR has IIwmelll-rt'.~i.\·ting R.C/mllle with Im'ck injil fJ(l/lel.~. 4. Importal1ce focmr = 1.5 5. Re.f(WI1Se /acwr = 3.0 6. The Il/lildil1~ is ,I'illl(l1e(1 ill sei.Hl/ic ;:o/le (V of II/dia. 7. The .~lIhs".(II(J COIIS;st5 of met/illlll .mi/ (Iy('(' II). Solulion. From Eq. 32.11. the funtlamc nml nmural period (1'..) is given hy T..,

=

T

= O.O~I~

30

:=

10 III.

= 0.85 .f

Zone faclor (zl for seismic zone IV from Tahle 32.5 = r,. 24

Fmm Eq. 328.

A,

From Fig. 32 .7, for T"

1[ 1( 1

§E.

=

1.50

A/,

=

(0.;4) (hi)

g

Therefore,

=[~ ~ ~ = 0.85.1' for 111edium s(Jil.

(1.50)

= O.09g

/II of .wft to .wiff (:Iay limIer/a;" lIy 16 /II deep jille t() mlftliwl/ mlltl wit" jillf.'.r contl'"t of 15%. Tht' SPT 1"/111It'~' /II l/ijJel"lml dt'JII/u ar(' xivell ;1/ the rable. Dl'termille 'he zone of liqm1(1ctioll (Ille 10 (III etll"t"qllake with M = 7.5 callsill~ (/ maximlllll horizontal (lccelemliml 0/ O~ 18 ~ at tile xmlllll~ slIIfaCt'. • Assume 1.<", = 18 kNlllf. Tltl' wlIter t(lble 1.1' (If thl' ~ml/ll(l ~"Ulf(l(:('. Take 'Y... = 10 tNIIII" .

lIIustralive E"ample 32.9. The .nth.wi{

.ftmfll at a .~ile COIISisll' of 4

Solution. Cakulalions arc given in a tahular ftmn hclow' Rl'lIIw*

10 12 14

,

180

80

II

216

<)6

15

252

112

0.94 0.92 0.90 O.lolt)

0.25 0.24 0.24 O.2J

n.I'

0.'"

0.14 0.16 0.2J

0.58 0.67

Liquefiable Liquefiable

1.0

Juslsufe (COnl1mIL'tJ)


a,

ao'

18 2'

2H8 32'

"

360

128 144 160

Del'll!

SPT

(m)

(N)

I. 18 20 Conclusion

861

(tlao')1 Factor of safety (Eq.32.15) (Eq.32.14) (Fig. 32. 14) (Eq.32. 13)

'"

(t/OO')J

0.81

0.23 0.23 0.22

0.86

0.114

1.09

0.25 0.38 0.40

1.65

Ui2

Rnno'* Not liquefiable Not liqucfiab~ Not liquefiable

The lone between depth of 4 to 14 III is liquefiable . Below that depth, the soil is nOt lique lillble.

Illustrative Example 32.10. The c,rploraIOl)' boringl' at a project site ~'how that Ihere i.\· (l depo.\·it 5 III tllick oJ .flItllmted U/I/(I \V;th a SPT vallie of 15 fH'eriyillR a 20 III thick deposit oJ cia)' 0/1 bed mck. If the IIIILl'imlll1l /wriZ(}/IIl1lllcceieraliml ill bed rock dlle to lIlI earthquake (M = 7.5) i.~ el·timated as 0.06 g. i.\· there all)' possibility oJ liqueJaction inthe ' swut tle/1O.fi,? Assume the Witter whle i.f III 'he J:mmld mrface. perr:elllage of filles = 10% (1IIt1 Y = 19 kN/m 3. Solution. Since the sand depn.~jt overlies a thick clay depos it . the maximum horizontal acceleration will

be amplified. From Fig. 32.8, for a peak horiwllllll m;ceicration of 0.06 8 in bed rock, peak horizontal aceelerJ.tion at ground surt:ICC will be 0.14 8. Now for N 15 and Fines = 10%. the cydic stress rutio (t/Oo')/ n~quired to eause liquefaction from Fig . . 32.14 for earthquake of 1Tlllgnitude 7.5 is (1.18.

=

The tutal !>tress 00 at the depth of 5 111 is

.1

0 0 = 5 x 19 = 95kN/m! The erfective stress lit Ihal depth is

on' = 5(19-9.1:11) = 45.95kN/m 2 = 0.96 From Eq. 32 .1 4 . the cyclic stress ratio due \tI the eannquakc is given by

Struss reduction f
(* ), = x(a~",,) [~ )r" 0.65

= 0.65

x

(0.14)

(4;.~5 ) (0.96) = 0. 18

Fwm Eq. 32. J 3. the factor of s"fcty ag
F

= (t/
0.11:1

10 .

The soil is just liqueliable. However since the factor of s .. J'eIY is less than 1.25, il not safe. To prevent liquefaction. suitahle mCitSurcs should be adupted.

PROBLEMS A. NUMERICAL .. .J....

32.1.

If 1hc st(lndard torsio n scbrnogmph fCl..·urds (III ilvcmge IlOll'C llmplilude of 17 mm III all epicclllrnJ distance of

100 km, what is the pmb:lble ma[!nhude of the earthquake'?

[Ans.4.2J

{Ans. 6.31 32.2. Wh:lt .....ould he the magnilude of an earthquake to rcIC:lSC 8 x 101.\ J of energy'? 32.3. A slJl.Sloreyoo buildtng has moment-resisting R.C. rramcs wlthoui brick inlil panels and is 2S m high. What .....ould be its natural pcriurJ ()f vil'lrnlion'! (ARS. 0.8:' 5) 31.4. Estirmlle ttlt! cydic stress ratio producL': 120 kN/m2 Effective stress at that depth :=: 66 kN/ml [Ans.0.1I3] .

862


32.5. Determine the factor of safety against liquefoction at a depth of 2 m in magnilude 7.5 for the following site oonditi(lns: Maltimum horizontal occeleration = 0.18 ~ Saturated unit weight of sand = I g kN/m' Percentage of fines 5%

=

Corrected SPT number

(1,"

sand deposil due 10 an c.1rthquake of

= 20

Assume the water table is at a depth of 1.0 m below the ground surfal'C. [An.... 1.39J 32.6. Estimate the design horizontrll seismic cocfticient as per IS: 1893-2002 (or a building with the following sile co nditions: Height of building::: 28 m The building has moment-resisting R.C. frames without brick lolil panels. Importance factor 1.0 Response faclor = 3.0

=

The building is situated in seismic zone V on a subSlrala consisting of son soil (type III).

IAns. 0.102

£J

B. DESCRIPTIVE AND OBJECTIVE TYP'E 32.7. Differentiate between the magnitude and intensity of :m earthquake. How would you estimate the lUllxim um intensity if the magnilUde of the earthquake is known'! 32.8. Explain the following terms: (0) Focus (b) Epicenlre (el Focal depth (tl) EpicenlraJ distuncc 32.9. What is an isoseismal map? How ..... ould you prepare an isoseismal for a place after tnc occurrence of the earthquake. 32.10. What do you understand by Inc earthquake-resistant structure? What are the criteria for its design? 32.11. What is liquefaction of soils? What are the necessary conditions for its occurrence? 32.12. What are the factors Ihal affect liquefaction of soil? Discuss in brief. 32.13. Explain Ihe procedure for the asse.~sment of [he susceptibility of a .~alld depo~il to liquefaction. 32.14. Suggest suitable methods for prevention of liquefaction of soil~. 32.15. What are the various haznrds due to an earthqu
C. MULTIPLE CHOICE QUESTIONS 1. According 10 MSK (64) :;calc of intensity, the total numbcr 0]' intensity . ~4

(el lO

{M8 Itl) 12

c1a.~scs

(or gmdcs) is

.GEOTECHNICAL E;\RTHQUAKE ENGINEEKI NG

1163

1. For an earthquake 01 magnitude 8.0. the maximum intensity on MMI «(I) IX (II) XI

~ca1e

would be about

(c ) VIII (d) VI 3. In a sand deposit. the ellcctive str~:.~ at a point is 100 kN/m 2. The dynamic pore waler pressure nead required to cause liquefaction is about (a) 4 m (b) 8 m (c) 10 m (tI) 100 m 4. At 3 depth of 10 III below the ground surface, the stress reduction foctor is about (al 0.90 (b) 0.92 (e lO.94 (el) 0.96 5. Accnrding 10 IS : 1893-201)2. the vAlue of the horizontal seismic coefficient for any structure with T S O. I s, CAnnot be les$lh:m (a)zj2 (b)z}3 (e) z}4 (t!) z 6. According to IS : 1893-2000. the mtio (IIR) shall not be greater than (tI ) 0.50 (h) 1.0 (e) 1.50 (tl) None of above 7. TsunAmis due to enrthquakes ~lCcur in (a) Oceans (h) Rivers (e) L.'lkes «(f) Reservoirs 8. The type of foundlllion generally suitable for buildings on liquefiable soils deposit ;s (tl) Spread fOOling (M Raft foundation ((oj Well foundation [,I) Pile foundation ~_'.~~~1~~~~~&~~~'~

SELECTED REFERENCES I. Chen. W.E. (Ed.) (1995). The Cil'il EIIR;/I/!eriIlR Ha"tlbook, CRC Press, Boca Raton, New York. 2. Coduto. Donald P. (1999), Ge/}tt'Clmh~tll EIIR;lIe~riIl8, Prillci,l'e.f tIIul Prt.lclic~. PrcJltice-Uall of India, PY'I. Ltd., Ne ..... Deihl. 3. Idriss, I.M. (1990). "Response of Suft Soil Sites During E:lRhquakes'", P/Txeedblgs, H. [JolwlI Seed Memori(lf Sympo.\·;III1l. J.M . Duncan. Ed; Vol 2. IlP 27]-289, Hi Tech Vanr.:ouver, Be. 4 IS: 1893 (Part I): 2002. Cr;teritl fill' r:.:artllqllake R~s;Sltmt Desigtl "J strl/elllre, Bureau of Indian Standards, New Delhi 5. Krishna, 1., A. R. Chandr.L~ekarnn and Ch,mdra, B (1994), Elemellts oj Ea"IIqllokJ> Ellgillt'erillg, South Asia Puhllshers, New Delhi. 6. NUllli, O.W. and Hcrrmnnn, R.B. (]984), "Ground motion of Missis.~ippi Vnlley Earthquakes". J. Tech. TOJl. Civ. Ellg., ASCE 110: (54-69) 7. Richter. C.F. (l935). '"An in~trumentnl ~1rthquakc sC'Jlc", B. Scism. Soc. Am. 25 (I ): 1· 32. 8. Saran, Swami, (l99lJ) Soil D.I',ulIllie.f amI Machine Fmmtlmiolls, G:algotia Publication pvl. Ltd., New Delhi. 9. Seed, H.B .. And Idriss. I.M. (1971). 'Simplified Procedure for Evaluating Soil Liquefaction Potential'. }ol/mal oj IIII' Soil Mecltll/!;(:S tIIlll Fowl/ltt/;Ott Dh';Si(III, Vol. 107. No .. SM 9, pp 1249-1274 ASCE. 10. Seed, II . Bailon. Tokimalsu. K .. Harder. L.E, and Chung. Riley M (1985 ), "Inlluence of SIT Procedures in Soil Liqucl"action Resistance Evaluations," ASCE }ollmal oj Gt'OIt!CIUlical EII.~illt't'''iIlK, Vo l. Ill , No. 12, pp. 1425-1445. II. Seed. H.B .. Romo, M.P.. Sum. J .. Jaime. A.. and Lysnler, J. (1987) '"Relationships between so;1 conditions and earthquake gr\,'lllnd motions in Mexico city ;11 the e:1fthquake of Sept, 19, 1985." Etu'lItqtUlke Engineerillg Rt'sf!wd, Celll,-!!. Report No. UCBIEERC-87!l 5 Universily of Culifumia, Berkeley. 12. U.B.C, (1991), Uoifonn Building Code, U.S.A .

APPENDIX A GLOSSARY OF COMMON TERMS Adrorbed water. It is water bound to clay panicles because of the attraction between electrical charges existing on the clay particles lind warer molecules (dipoles). Air content. II is the mlio of !he volume of air to the volume of voids in soil. Alluvial $Oils. These are soils deposited by water. Deposits made in lakes are called lacustrine deJXlSiLS and ttme in sea (or ocean) called marine deposits. Aflowable bearing pressure. II is the net allowable bearing pressure which can be used for the design of fouOOalion. It is the smaller of the net safe bearing capacity and the nct safe seniemenl pressure. Active pressure. It is the pressure developed when the soil mass stretches due to movement of a retaining wall away from lhesoi!. Aquifer. An aquifer is a pervious SU'lllum which contains water that can be easily drained or pumped out. An aquifer is

called an unoonfined aquifer when there is an impervioos stnnum only below it and a confined aquifer when it is sandwiched between two impervious .strata. Arching. It is a pheoomeoon is which the stresses are transferred from a yielding part of a soil mDSS to an adjacent non*yielding (or tess yielding) pan of the soil mass. Al-~$t pres:w~. It is the lateral pressure in a soil mass when there is 00 movement of the mass. Atterberg Limits (Consistency limits). The liquid limit, plastic limit and shrinkage limits are known as Aueroerg's limits. The water conlent al which the soU behaviour changes from the liquid to the plnstic SUIte is called the liquid limit; from the plastic [() the semi·soIid state is the plastic limit; and from lhe semi-solid to the solid state is the shrinlcage limit. Backfill. It is the soil material which Is placed into an area that has been ex.cavated, such as againsl retaining walls and in pipe trenches. Bearing Capacity. (Ultimate bearing capacity). It is the pressure at the base of the foundation al which the soil below fails in shear. II is called the groa ultimate bearing capacity when the gross pressure is considered and the nel ultimate bearing capacity when the nct inaease in prc&Iure over the elristing overburden pressure is oonsiderC(!. The safe bearing capacity is the maximum pressure which the soil in the foundation can carry safely. lbc safe beD.ring capacity can be expressed as gross safe bearing capacity or net safe bearing capacity. Boring. It is the method of investigating subsurface conditions by drilling a hole into the eanh. Generally, soil samples are also extracted from the boring for detenninalion of the index and engineering properties. Borrow. It is soil (or rock) material obtained from another off· site source for use as fill at construction projects. Braced CUI. This is an excavation which is laterally supported. The vertical sides of excavation ore supported by sheeting and bracing system: Bullcheads. These consist of sheel.pile walls COI1SIJ'uC!OO to relain CIlnh. These are relatively flexible retaining walls c:onstrudcd for water front structures, canal loeb, coffer dams, etc. BulJcing of sand 1be phenomenon of increase in volume of sand (or 0 cohesionless soil) due 10 dampness is called the bulking of sand. The effect is predominant when the water content is between 4 and 5%. The increase in volume may be upto 20 to 25%. If the water content is increased, and the sand becomes saturated, the volume of sand mass is d = -. CaiSSotl. It is a type of foundation in which a large chamber (or box) is built above the ground level and then sunk 10 the required depth of a foundation as a single unit. The caisson may be an open caisson, a pneumatic caisson or a fleeting caisson. Open caissons are also known as well foundations.

GLDSSARY OF COMMON TERMS

'1\5

Capillarity. It is the movement of water due to surface tension and other effects but not the grovity effect. Waler moves in very smull channels beanlSe of the affinity between soil ond water. Chemical wctIfherillg. It is the process of wetlthering in which chemicol reaaions, such as Lydl'8tion, oxidation, solution, occur. When chemical weathering or chcmicol decomposition occurs. original rock mincrnls are tnlnsfonncd inlo new minerols by chemical !'t'action. aay mincrols are formed by cnemiall weathering. Cloys (clay millerals). These are very small ponicles (usunlly smaller tha n 2~) which have a crystalline stJ).laure developed as the result of the chemiall weathering of rocks. The clay pnnicles are flat or plate-like in shope. These are highly surface-aaive particles. CohesiOll. It is the atU'8ction or bonding force between the panicles of fine-grained soils UUl.: creat(1i shear strength. Compoction. It is the process of increasing the density (or unit weight) of a soil by rolling, tamping, vibrating. or OIber mechanical means. COIJSistency. The consistency ot a fine-grained soil is the physical state is which it exiscs. It is indialled by such terms as sort, firm or hard, depending upon the degree of firmness. COIuluit. It is a pipe that is usu:llly buried in a soil mass, or which passes through a soil embankment, and carries water. electriall cables, telephone albles, etc. COlISolidation. The compression of a snturnted soil under a Sleady-stnte pressure is known as oonsolidalion. It is due 10 expulsion or wafer from fhe voids. Initially. the slress imparted inlo the soil is carried by wnter. The water is gmclually forced out and the SireS> is transferred to the soil skeleton and lhe compres>ion occur:;. Critir:Oil>oid ratio. The void ratio of the soil at which 00 change in volume occurs when the soil is subjocted 10 shear strain In a drainett test Dellsity. The mass per unit volume of soil is allied the density of soil. (Sometimes, the weight per unit volume, which is the unit weight, is called density). Dewaterillg. The process of removing waler from II construction area is known as dewatering. The term dewatering is also used for lowering the wllter table to obtain a dry llre9 in the vicinity of the excavation. Deep jOlmdOliolL It is the type of found8lion which transfers the load to deep strata below the ground surface. The ccmmon types are piles, caissons, drilled piers, etc. Generally, the roundation is allied deep foundation if the depth of foundation is grenter thnn the width of footing. Dispersi1'e clays. These ate types of c1I1Ys which defiOCClJlate in still water and erode when exposed to II low~ velocity flow of water. Dispersivily is due to a high concentralion of sodium ions in a clay-pore water system. Ditch cOIlduits. These are types of conduits whicb are installed in narrow trenches (or ditches) and subsequemly backfilled with soil. Drawdown. As soon as the pumping is done from a well, the water table is lowered in its vicinity. This drop in 'Ao7lter level in the well is allied drawdown. Drilled pier. 11 is a type of deep foundation in which a large diameter hole is drilled in the ground and subsequently filled with concrete. Dynamic campactimL It is II method of compacting surface and near-surface UIOes or soil or fill by dropping a beDvy weight from a relatively great height. Multiple poundings are usually done at each location. Earth p~ssure. It is the lateral pressure exerted by a soil mass against an eanh-retaining structure (or on a fictitious vertical plane located wilhin a soil 1tlIlS». Depending upon the movement of the earth-retaining sU\lcture, the pressure ltlIly be active, passive or at-rest. When the structure moves IIWlly from the soil mass, it is aaive pressure; and when towards the soil moss, it is passive pressure. AI-rest pressure acts when there is 00 movement of structure at all. The coefficient of eanh pressure is the rolio of laleral pltSS4lre to vertical pressure existing 8t a point in the soil mass. EfJective siu. It is the size of pnrtide in a soil specimen such that 10 percent of the particles are finer than thi~ size. It is also called the cfJectil''t! diameter. Effective stress. It is the nominal stress lransmined through Ihe plrUde to particle conlllct in soil. 11 is equal to the sum of all the oormal componenlS of load divided by lhe tOlal area of cross-section. The effective stress controls the sheDr Slrenglh and compressibility of the soil. It is an abstract quantity which is obtained by subtIDCting the pore water pressure from the total stress. Electro-osmasis. It is a method of drainage or cohesive soils in which a direcl ament (IX:) is used. Pore water migrates to the calhode, which is usually a well-point. Electro-osmosis a1.so helps io inaeasing the shear strength of the cohesive soil. Expollsive clays. These are types of clays which show a large volume expansion in the presena!: of water and a large

SOIL MECHANICS AND FOUNDATION ENGINEERING volume decrease upon drying. lbe dllYs contain the mineral montmorillonite. These are highly difficult soils to work with. Fill. It is canh placed in lin excavation or otncr area to raise the eIcvlllion. It is also called eanh fill or soil fill. If fill i! required to support structural loading, it is placed in layers and compacted III a suitable water contenl to achieve I uniform 000 dense soil mass. Filles or /ine-grail/ed soils. The silt-si7.e and clay-size particles in a soil mass are called fines or fine-grained soils. Flow lIe t. A flow net consists of two sets of mutually orthogonal lines called the flow lines nnd equipotential lines. The flow lines indicate the paths of travel followed by the moving water and the equipotential lines indicate the points 01 equal poIenlial (head). A flow net is a pictOrial representation used to study the floYl of water through soil mass. FOOling. It is lin enlllrgemc'!t of the base of a column or a wall to spread the load on a large area of the stratum below. It is a type of foundation ilWalled 01 a shllllow depth. Frictioll, illJerna/. It is the friction developed in a soil due to soil-to-solid contact. It is responsible for most of the shear strength developed in a soil, espttially in a cobesionless soil. The angle of internal friction (,) is used to represent the internal friction of the soil. Frost boil. It is a phenomenon which occurs when a frozen soil mass thaws ntacl and the soil behaves lIS a liquid. Mechanical Analysis. It is the process of dctennining the sizes of various panicles in soil specimen. It is done by the sieve nnolysis for OOIlTSC particles and the sedimentation analysis for fine particles. McclulIIical weatherillg. It is the process by which physical forces, such as temperature changes and fr05t action, breakdown or reduce the rock to smaller fragments and soils. There is no chemicnl change and the properties of the soil formed are the same as those of the parent rock.

GLOSSARY OF COMMON TERMS

867

Mineral. II is a naturally formed chemical element (or compound) having a definite dH:mical composition. II usualty has a characteriSlic crySial from . Negative skin friction. II is a down drag on a pile which occurs when the soil in which lhe pile in driven senles nne than the pile. The load-carrying capacity of the. pile is reduced becaU§e of negative skin friction. Nalural frequency. A systcm under frcc conditions vibrates al a frequency calted the nalural frequency. II is the characterislic of the system. In genernl, the nalural frequency decreases as the mass increases and the spring OOIlSIanl of the system decreases. Normally consolidated soil. A soil which had not been subjected to a pressure in the past greater than the present pressure. It is also called a virgi n soil. The seulements are large in a normally consolidated soil. Net allowable bearing pressun. It is the net pressure which can be used for Ihe design of a foundation. It is !be smaller of the net safe seltlement pressure arxl the net safe bellring capacity. For oohe:sionless soils, generally the nel safe selllement pressure governs; whereas for cohesive soils, genernlly the net safe bearing capacity governs. Optimum moisture cOlllent. II is the water rontent of soil at which the rnaxil\1um dry density is achieved during compaction. Over-consofidated soil. These are Ihe soils which hod been subjeaed to a pressure in the pas! greater than the present pressure. Over-consolidated soils are also called precOIlSOfidated soils. TIle settlements are small for such soils. PQl'cmeni. It is a hard crust constructed on the subgrade (soil) for the purpose of providing a stDble Dnd even surface for the vehicles to move on. The pavement may be a flexible puvement or a rigid pavement. The rigid pavemenlS are made of cement concrete ond can take the tensile slresses. Pile. It is D relatively long, slendcr column used as D deep foundation . The pile is end-bearing (point bearing) pile whcn it obtains support from the bottom,and it is a friction pile if it develops l1!Sistance due to friction on the sides. 10 rnosl of the cases, il has resistance from bottom as well as side friction. Piping. It is a phenomenon which occurs due to erosion by sub-surfao: water moving through a soil mass. It resulls in the fonn!l1ion of continuous tunnels Of pipe-lik.e fonnations through which soil is carried by flooHing water and piping foilure may occur. Plwle slrail~ It is a state of strain in which all disptacemenl$ occur in one plane and the displacemenlS perpendicular to that plane are zero. Generally, plane strain conditions occur under a long retaining wall, strip fOOling, earth dam, etc. PltlSticiry. 11 is a propeny of fine-grained soils (panicularly clays) due to which n soil having adequote water cootenl is able 10 flow and con be remoulded without breaking apan. POissall's ratio. It is the ratio of the lateral strain to the longiludinal slmin due 10 uniaxial stress within the elastic limit. Pore pressun. II is water pre$ure developed in the voids of a soil mass. TIle shear strength of a soil is reduced due to pore pressure as the effeaive stress is decreased. Excess por..! pressun: refers to pressure greater than the normal hydr06tatic pore waler pressure. Pressure bulb. II is the zone o f the soil mass in which stresses are induced due to superimposed load. Gcnerally, it is assumed thai the pressure bulb is confined to the zone in which the stresses are more than 20% (or 10%) of the surfacelood. PressuremeJer. It is an instrument used 10 determine the in·situ strength of a soil (or rock) zooe. It is bosed on !he principle of the measurement of the pressure-related lateml expansion of a flexible cylinder inserted in a bore hole. Projtx:ling conduit. It is a type of conduit over which eonh fill or earlh embankment is placed . Quick sand condition. When the head causing upward flow in a cohcsionless soil is high, the effective stress is reduced to zero and the shear strength of the soil becomes zero. TIle condition so developed is. known as quick sand condition. The critical gradient at which a cohesiontess soil becomes quick is about unity. Rejnfo~ed earth. It is an carth mas.... strengthened by reinforcemenl. Eanh structures such as embankments, retaining walls and eanh dams constructed in I.llyers and reinforced with geofabric, metal strip! or fibres 10 inaease the strength of the soil mass are examples of reinforced earth. Retaining wall. An eanh-retaining structure constructed to resist the lateral pressure of soil is called a retaining wall. Re\>etmeli/. It is a facing buill of stone, concrete blocks, Of other durable malerial 10 protect an embankment from the wave crCfiion. It is also called rip rap. Rollers. These are types of conslruction equipment used for compaaing the soil by rolling it. The rollers are of different types. Sill/d. It is 0 Iype of coarse-grained soil whose jXlrticJe sizes range betwccn about 0.075 mm and 4.75 mm. Sand is cohesionless !lnd has high internal friction . Seepage. SeejXIge is flow through an earth mass under pressure. The term is also used 10 indicate the quomity of water


86.

flowing through a soit deposit or soil strudure or foundation. Seismic aploration. II is a method of subsurface investigation in which a shock wave is aealoo to determine tbe deptb

ofdiffcrentsuma. Secondary cOIlSo/idalioll.. This is e:onsolidatioo of a roheiive soil which occurs afier the primary consolidation is

completed. 1be SCOOfldary ronsolidalion is predominant in organic soils. SeIi/erne,lt. The downward venical movement experieoced by 0 structure Of 8 soil surface when the soil below ccmpresses. The selllemenl cx:curs mainly because of consolidation. Shear strength. II is 100 ability of a soil 10 resist sbe:lring stresses developed within a soil mass as a result of loading imposed 0010 the soil. II is the maximum resistance which develops just before the failure under shear. Sheet piling. A type of conStruction in which piles with flat aoss-section are joined 10 form 8 thin diaphragm wall or bulkhead to resist the IllIcral force of retained earth. Sieve. It i.~ a pan (ot trny) having a screen or mesh bottom. It is used 10 sepamlc panicles of a soil sample into various sizes. Silt. It is a type of (ine-grnined soil wilh the particle size smaller than 0.075 mm, but whose mineralogical oomposition remnins the same as that of the parent rock. It does not conUlin clay rninernls. Soil Stabilisation. It is method of increasing the strength of a soil and improving its properties. It Includes mixing or additives and other means of improvement sucb. as compaction and drainage. Stw/dard Penetratioll Test. (SP1) It is a type of test in which a sampler is driven into the ground by a hammer. The number of blows required for a peneltntion of 300 mm is the standard penetration number (N). The standard penctration numt>cr is correlated with shear strength. 1be lest is especially useful for cohesionlcss soils. Sump. 11 is a small pit provided to serve as a collecting basin for surface woter or near-surface under ground waler. "The water is pumped out from the sump when full. Terra.probe. It is an equipment used for COInp:!cting the soil. It consists of a vertically-vibrating tubular probe which is vibrated to the desired depth in a soil mass and then slowly withdrawn while continuing to vibrate. TIll. Soil Irnnsported by glaciers are called lilI. These usually consist of a hcterogeocous mixture of the fine-grained and coorse-grained soils. Tube well. It is II type of well in which a pipe (or tube) is driven into the soil to intercept various aquifers. The waler nows inlo the well through the strainer provided. The discharge of a lUbe well is much greater than that from an open well . Ullifonnity coefficielli. II is lhe ralio of D60 to Dlo size of the soil where D60 represents the size corresponding to 60% finer and Dlo 10 10% finer mnlerial than lhal size. For a well-graded sand, ils value is greater Ilwn 6. U"it weig"'. It is the weight per unit volume. It is expressed in kN/m1. Vibro!loatio". It is !I method of compacti ng thick deposit of sand through the use of a horizontally vibrating cylinder called a l>ibro!loat. It)id raJio. It is the ratio of the volume of voids to the volume of solids in a soil. For a fine-grained soil, the void ratio is generally greater tlwn Ihat for a coarse-grained soil. Water COllfellf. II is the ratio of the mass (or weight) of water 10 the mass (or weight) of solids in 8 soil. It is expressed as a percentage. ~Il poilli. It is the perforated end section of a well pipe installed in the ground. It permils the ground water to be drawn into the pipe for pumping.

APPENDIX B MISCELLANEOUS OBJECTIVE TYPE QUESTIONS The following arc a few Iypical questions which appeared in question papers of the G:ombined Engineering Services Examination of UPSC . J. Lists I and (I contain respectively l enns and expressions related 10 soil classification. Match the two lists and select the correci answer usi ng the codes given below the lists.

List II

List I

( I) Liquid limit - WatereQnlcn!

(A) Activity number

Plasticity index

(2)

Plasticity index Percent finerlhan 2)..1. (3) Natural water content - Plastic limit Pltl!>licilY index (4) Unconfined compressive strength of undisturbed sample

(B) Liquidity index

(C) Sensitivity index

Unconfined compressive strength of remoulded soil sample A

C

(0) (b) (c)

(,0

(C.E.S. 1993)

2. Match List I with List 11 and select the correct answer using the codes given below the lists.

List I (A) Sheep-foot roller

List II (I) Heaning of earthen dams (2) Dry sand (3) Ca~ in g of earthen dmns (4) Gravel in W.B.M. road

(8) Smooth heavy roller

(C) Pneumatic roller (D) Vibrating roller

Codes

A (0) (b) (c) (
C

o

(C.E.S. 1993)

3. Match List I (difterent types of soils) with List II (group symbols of I.S. classification) and select the correct answer using Ihe codes given below Ihe lisls.

SOIL MECHA NICS AND FOUNDATION ENGINEERING

810

List I

List II

(A) Well -graded grovel sand mixture wi th littl e or no fines (B) Poorly graded sands or gravelly sands with little or no lines (C) Inorganic silts and very fine sands or clayey silts with Jow plasticity (D) Inorganic clays of high plasticity

I . ML

2. CH 3. GW

4. SP

Codes C

A

D

(a) (b) (e)

(d)

(C.E.S . 1994)

4. Consider the la llowing statements about the properties of the fl ow nets 1. Flow linc$ are perpendicular to equ ipotential lines. 2. No two fl ow lines or equipotential lines start fronl the. same point. 3. No two flow lines cross each other

or thes~

statements :

(a) I. 2 and 3 are correct (e) I and 2 are correct

(b) 2 and 3 are correc t (d) 1 and 3 are correct

(C.E.S. 1994)

5. Matc h List I (structure) with List 11 (defonnation) and select the correct answer using the codes given below the lists. List I

List II

(A) Retaining wall

I. The wall moves abom the dredge line as a rigid structure 2. The wall moves in the form of elastic line wilh a point of I;onlrb.nexure 3. 1111: bOllom moves away from the soil 4. The lOp of the wall moves away from the soil

(B) Bridge abutment

(C) Cantilever sheet pile (0) Anchored bulkhead

Cod" A

C

D

3 3

(a)

(b) (e)

(C.E.S. 1994)

(d)

6. Malch List I with List List I (Type oj pile)

A. Friction pile B. Bauer pi le C. Tension pile D. Compaction pile

n and select the correct answer using the codes given below !.he lists List II (Situorion)

I. Stiff clay

2. Loose granular soil 3. Lateral load 4. Uplift load

MISCELLANEOUS OBJECfIVE-TYPE QUESTION

Cod", A

C

o

3

(a) (b)

(e) (d)

(C.E.S. 1994)

7. Match List I with List II and sele(;t the correct answer using the codes given below the lists List I List n (Allowable mat. ~'ettll.'lIIc!'" IS: 19(4) (Type 0/ fiJi/ndario" (llid soil fYlle) (A) 65 to 100 n11n

I. Isolated foundat ion on sand 2. Isolated foundation on clay 3. Rafts on sand 4. Rafts on clay

(B) 40 mm (C) 65 mm (D) 40 to 65 mill Codes A

C

o

I

(a)

(b)

(e) (d)

(C.E.S. 1994)

8. Consider the following statements regarding confi ned aquifer: I . The aquifer is bound at lOp and below by impervious strata. 2. 1lle pressure of water is greater than atmospheric pressure. 3. A tu be well sunk in such an aquifer starts n owing always by itself. 4. The aquifer is fully saturated

or the

above statements :

(a) I, 2 and 3 Clre correct (e) 2, 3 and 4 arc correct

(b),.,l, 2 and 4 arc correct (d) I, 3 and 4 are correct

(C.E.S. 1994)

9. Match List I with List Il and se lect the (;orrect answer using the codes given below the lisls ~I

(Type

~U

0/ suil)

(A) Lacustrine soi ls (B) Alluvial soils (C) Aeolian soi ls (D) Marine soi ls Code... A

(Mode o/transportatioll {llld depositiolt) I. Transportation by wind 2. Transportation by running water 3. Deposited at the bottom of lakes 4, Deposited in sea water

C

o

(a)

(b) (e) (d)

10. Consider the following assumptions for slope stability an;lysis: I . Friction is fully mobilised . 2. Effective stress analysis is adopted 3. Total stress analysis is used 4. Resu ltant R passes through the centre of the circle 5. Resu ltant' R is tangential to the friction circle

(C.E.S. 1995)

812


lhe assumptions necessary for fri ction circle method o f analysis ':Vou 1d include (ll) I. 3 and 4 (e) I. 3 and 5

(b) 2 and 4 Cd) 2 and 5

(C.E.S. 1995)

I J. Consider the following statements regard ing negative skin friction in piles: I. It is developed when the pile is driven through a recently deposited clay layer.

2. II is developed when the pile is driven through a layer of dense sand. 3. It is developed due 10 ,I sudden dmwdown of the water table :

or these statements

(tI) I alone is correCt (e) 2 and 3 arc correct

(b) 2 alone is correct (d) I and 3 arc correct

(C.E.S . 1995)

[2. Consider the following properties for a soi l sampler : I. Areil ratio should be low 2. CUlling edge should be thick 3. Inside c learance should be high 4. Outside c learance should be low The properties necessary for a good quality soil sampler would incl ude (a) I :rnd 4 (b) 1.2. and 3 (e) 2. 3 and 4 (d) I. 3 und 4

13. Match List

r with

(C.E.5. 1995) List II and select the correct answer using the codes given below the lisl$

List I

List

(Soil property measured) (A) (8) (C) (D)

n

(l1I·situ test) I. Cyclic pilc load test 2. Pressure metcr test 3. Plate load tesl 4. Sllindard penetration lest

Modulus of subgrade reaction Relative density and strength Skin friction lind point bearing Elastic eonSlalllS

Codes A

C

o

(a) (b) (c) (d)

(C.E.S. 1995) 14. Assertion CA) Quick sand is not u type of sand but it is condition arising in a sand mass. Reason (R) When the upward pressure becomes equal to the pressure due to submerged weight of a soil , the effe cti ve pressure becomes zero.

Codes (a) BOlh A and R arc true und R is the correct explanation of A. (b) Both A and R are true but R is not a correct explanation of A. (e) A is true but R is false (d) A is false but R is true

(C.E.S. 1996) 15. Assertion (A) The safe height (2 Zo) 10 which an unsupported vertical c ut in clay can be made is 4c/"t. Reason (R) Active earth pressure of cohesive backfi ll 'shows thai the negative pressure (tension) is developed aI depth Zo and total net pressure UplO a depth 2 Zo is zero.

Codes (a) BOlh A and R are true and R is the COrrecl explanation of A

(b) Boch A :md R arc true but R is not a correct explanation of A

MISCElLA.NEOUS OBJECTIVE-TYPE QUESTION

(c) A is true but R is fnlse (d) A is fa lse but R is true

(C.E.S. 1996)

16. Assertion (A) Negative sk in friction will act on piles in filled up soils. which shou ld be considered in design of pile foundations. Reaso n (R) 1lle lill ed up soils start consoliding and develop a drag fo rce on the pile Codes «(I) B(}(h A and R are true and R is the correct explanation of A (b) Both A and R arc true but R is not a correct explanation o f A. (c) A is tr ue but R is robe (C.E.S. 1996) (d) A is fa lse hut R is true 17 . Match LiM I with List II and select the correct answer using the codes given below the lists

List I (Flow (we)

List 11 (Flow chamcleristics) I. Seepage flow is a function of time 2. Hydraulic gradient varies wi th square of velocity 3. Flow al low velocity . .. . ifh ;P h 4. Govermng equatIon In 2D IS k$ d?- + k1· d = 0

(A) Transient n ow (8) Turbulent flow (C) Steady state flow

l

(D) Laminar !low

Cod" A

C

D

4

(a) (b) (e)

(,n

(C.E.S. 1996)

18. Consider tile fo llowmg statements In subsoil explorat ion programme, the lenn significant depth of exploration is upto 1. The width of fou ndation 2. Twice the width o f foundation 3. The depth where the additional stress intensity is less than 20% o f overburden pressure 4. The depth where the ndd itiomil stress intensity is less than 10% of the overburden pressure. 5. Hard rock level Of these stlltements : (b) 2, 3 and 5 are correci (a) I. 3 and 5 are correct (eI) 2 and 4 are COrTeel (c) I and 4 are correct (C.E.S. 1996) 19. Match List I with List II and select the correct answer using the codcs given below the lists List [ List 11 '

(Effect) (A) (8) (C) (D)

(Reasoll)

Excessive seulcmenl High expansivily Reduction in rearing capacity Acceleration o( consolidation

I. 2. 3. 4.

Rise of watcr table High compressibility MontolllOrrillon ;te Sand dmins

Codes

A

C

D

(a) (b) (e)

(d)

(C.E.S. 1997)

".


20. Match List I (property) with List II (slope of the t:urve) and select the correct answer using the codes given below the lists. Lisll

List

(A) Coefficient of compressibility (B) Compression index (C) Coeflicient of suhgrade reaction

I. 2. 3. 4.

n

Stress-deformation Stress-void ratio VOlume-pressure Log slre.ss-void ratio

Codes A

C

(a) (b) (e)

«~

(C.E.S. 1997)

21. Consider the following statements regarding undcrrcamcd piles I. TIley are used in expansive soils. 2. They are of precast reinforced concrete. 3. The ratio of bulb to shaft diameter is usually 2 to 3. 4. Minimum spacing between the piles should not be Jess than 1.5 limes lhe diameter

or these

statements :

(ll) I, 2 and 3 arc correct (c) 2. 3 and 4 arc correct

(b) [. 3 and 4 are correct (d) I, 2 and 4 are correct

(C.E.S. 1997)

22. Match List I with List II nnd select the correct answer using the codes given below the lists

List I (Fie/dlesl) (A) Plate load test (B) Standard penetration lest (C) Static Dutch cone penetration lest (D) Dynamic penetration test Cod.. A B C (a) (b) (e) (d)

List n measured)

(Param~lers

I. 2. 3. 4.

TOlal and frictional resistance Load intensity and settlement values Ncd vnlues SPT values

o

2

(C.E.s. 1997)

23. Consider the following slillcmcnls: I. Constant-head permc:II11cter is best suited for detcrminmion of coefficient of penncability of highly impenneable soils. 2. Coefficient of penneability of a soil mass decreases with increase in viscosity of the pore fluid. 3. Coefficient of pcnneability of a soil mass increases with increase in temperature of the pore fluid.

Of these statements : (lI ) I and 2 are correct (e) 2 and 3 are correct

(b) I and 3 arc correct (d) I, 2 and 3 arc correct

24. Consider the following statements : I. Relative compaction is not the ~,1111e as relative density. 2. Vibronoatalioll is 110t effective in the case of highly c()hes(ve soils. 3. 'Zero air void linc' and 100% ~
(C.E.S., 1998)

MISCELLANEOUS OBJECTIVE·TYPE QUESTION

Of UJese statements: (b) I and 3 are con-eel (d) 3 alone is com:ct. 25. Consider the following Slale!l1ents : Phrca,ic line in an earth dRm is l. elliptic in shape 2. an equipotentia l line 3. the lOp most now line with zero water pressure 4. approximately a parabola Of these statements : (b) 2, 3 and 4 arc correct (a) I, 2 and 3 are correct (e) 3 and 4 arc correct (d) I alone is con-ect (a) I and 2 arc COITeC! (e) 2 and 3 are correct

(CE.S .. 1998)

(C.E.S .. 1998)

26. Consider the following statements:

I

1

Rankine's theory and Coulomb's (heory give same values of coefficients of active and passive earth pressure when I. the retaining wall hM a venicu) hack 2. the backfill is cohesionless 3. angle of slope of backtill is equal to the angle of internal friction 4. the angle of slope of backlill is 00 5. the angle Qf wall friction 5 is 00 6. the angle of wall friction 5 is equal to 4J Of these statements : (b) I, 2, 4 and 5 are con-eel Ca) 1,2.3 and 5 are con-ect (d) I, 4 and 6 are correct (e) 2, 3 and 6 are correci (C.E.S .• 1998) 27. Consider the following statements: I. Coulomb's canh pressure theory docs not take the rOJJghncss of wall into consideration. 2. In case of non-cohes ive soils, the coefficient of active e
Of these statements : (a) 1 alone is correct (c) 2 alone is correct

....,

(b) I and 2 arc correct (d) 3 alone is correct

(C.E.S .. 1998)

28. Consider the ]allowing statements: I. The degree of saturation of a saturated soil mass subjected to pressure remains unchanged during the process of consolidation. 2. Secondary consol idation is due to the plastic defonnation of the soil when the pore fl uid is not subjected to any excess pressure 3. Primary consolidation is independent of the coeflicient of permeability of the soil but depends on the decrease in void volume due 10 air escape. Of th ese statements : (a) [ and 2 are coneet (c) 2 and 3 arc correct

(b) I and 3 arc correct (d) I, 2 and 3 arc correct

29. Consider the follow ing statements associated with local shear fai lure of soils : I. Failure is s}ldden with well-defined ultimate load 2. This fa ilure oceps in highly e()lllpressive soi ls. 3. Failure is prcc~cled by large settlerncnL".

(CES .. 1998)

SOIL MECHANICS AND FOUNDATION ENGINBERING

'" Of these statements : (a) \, 2 and 3 aTC correct (e) 2 and 3 are correct

(b) I and 2 are correct (d) I and 3 are correct

(C.E.S" 1998)

30. Consider the foll owing statements : I. Dynamic cone penetration test for site investigations is based on the principle thai elastic shock waves tra vel in different materials at different velocities. 2. Electrical resistivity method of sub-surface investigation is capable of detecti ng only the strata having

different electrical resistivity_ 3.

In~situ vane shear lest is useful fur dctcnnining the shear strength of very soft soil and sensitive clays and is unsuitable for sandy soils

or these statements

:

(a) I and 2 tire correct (c)

2 and 3 arc correct

(b) I and 3 are correct (d) 2 alone is correct

(C.E.S., 1998)

{Am. I. (d). 2. (c), J. (b), 4. (a), 5. (e), 6. (h), 7. (b), 8. (b). 9. (b), 10. (e). II . (d), 12. (a), 13. (d), 14. (a), 15. (a), 16. (a), 17. (e), 18. (,~. 19. (,~. 20. (e), 21. (h), 22. (c), 23. (e), 24. (a), 25. (e), 26. (b), 27. (d), 28. (,,), 29. (c). 30. (e)l

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62. IRC: 31, "Guidelines for the Design of Flexible Pavements", IRe, 1910. 63. IRC: 58, "Guidelines for the Design 9f Rigid Pavements for Highways", IRC, 1974. 64. IRC: 45, "Rewmmendatioll for Estimr:ltillg the Resistollce of Sofl Below the mmimum Scour Level in the Design of Well FoundoJiolls", IRC, 1910. 65. Jaeger, J.C. and N.G.W. Cook, "Fulldamentals 0{ Rock Mechanics", Olapman and Hall Ltd. and Science Paperbacks, 1969. 66. Janbu, N., "An Energy Analysis of Pile Driving with the use of DintensiofIless PartJIMters", Norwegian Geotechnical Institute, Oslo, Publicatioo No.3, 1953. 61. Jumikis, A.R., "Soil Mechanics", O. Van Nostrand Co., 1965. 68. Kasmalkat, BJ., "Foundation Engineering", hne Vidyatathi Griha Prakashan, Pune, 1983. 69. Khanna, S.K. and C.E.G. Justo, "Highway Engineering", Nem Chand & Bros.. Roorkee, 1984. 10. Lambe, TW. aod R.V. Whitman, "Soil Mechanics" , Wiley Eastern Pvt. LId., New Delhi, 1969. 11. Lambe, T.W., "Soil Testing lor Engi/leers", Wiley Easlern LId., New Delhi, 1977. 72. lambe, T.W. " Slre$ Path method". 1nl. of Soil Mech. ond Found. Engg. ASCE, \bI. 84,1958. 73. Lambe, TW., "The Structure of Compacted clay", 1nl. Soil mech. OM Found. Engg., ASCE, \bI . 84, 1958. 14. Liu, C. and lB. Evett, "SoUs olld FOUIIlMt;O/iS", Prentice Hall Inc. Englewood cliffs, New Jersey, 1981. 15. Loudon, A.G., "The Computation of Penneability for simple soil tests", Geotechllique, Vol. 3, 1952·53. 76. Malter, H., "Numerical Solutions for Beams on Elastic Foundptions", 7hms. ASCE, Vol. 125, 1960. . '7. Martson, A., "The Theory of Ex/emol Loads on closed conduits in the Light of ulest Experiments" , Bull. 96, Iowa Engg. Exp. ~alion, Amec, Iowa, 19.30. 78. McCarthy, O.F., "Esselltials of Soil mechanics alut Foundations", Reston Publishing Co. Inc., Reston, 1982. 79. McClelland, 8., "Design of Deep Penetration Pites for ocean structures", 1nL of Geotechllicol Engg. Div., ASCE, Vol. 100, No. GT. 17, 1974. SO. Means, R.E. and lV. Parcher, "Physical Properties of Soils", Prentice Hall of India Pvt. Ltd., New Delhj, 1965. 81. Meyerhof, G .G., "Bearing capacity and Settlement of Pile Foundation", 1nl. GeOl Engg. Div. ASCE, GT 3, 1976. ' 82. Meyemor, 0 .0 ., "Penetration TesIs and Bearing capacity of Cohesionless soils", 1nl Soil Meeh.. Found. Division, ASCE, SMl, 1965 . 83. Meyerhof, G.G. and Adams, J.I.,"The Ultimate Uplift capacity of Foundatioos", Conadit211 G«x, JS(4), 1968. 84. Mitchell, J.K., "Funoomelllals of Soil Behaviour", John Wiley aod Sons, New York, 1976. 85. Morgenstern, N.R. and V.E. Price, 'The Analysis of the Stability of General Slip Surfaces", Geolechnique, \bI. 15, No.1, 1965. 86. Morgenstern, N.R. and Eigenbrod, K.O. "Classifiallion of Argillaceous Soils and Rocks", J. Oear. &gg. Div, ASCE, \bl. 100, No. GT to, 1974. 87. Newmark, N.M., "Simplified Computation of Vertical Pressure in Elastic Foundation", Univ. 0/1I1inois. Engg. Exp. Sin., Circular 24, 1935. 88. Newmark, N.M., 'Influence Charts for computation of stresses in Elastic Foundations", Univ. o/Illinois. Bull. No: 338, 1942. 89. Peck, R.B., W.E. Hanson, and T.H. Thomburn, "Foundotion Enginun'ng", Wiley, New York, 1914.

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15: 1498- 1970. '"'Classijicmiml ami idemijictlliolls 0/ soiL~ for General EII),:ineering Purposes. 18:1888-1982. "Method of Load Tert 011 $Oils", IS: 1892.1979, "Code of Practice for Subsurface fI:l'esliglllions /f}r fmmdmiolls", IS: \893-\975, "Criteria for Earthquake Uesislmrl Design of S/l1ictures " IS: 1904-1986, "Desig/l omf COlIStl1iCliOIl of FoundatiollS iI' Soils, General Requiremellls". IS: 2131-1981, "Method for SI(lIId{lr(i Pel/cirrI/jOlt Test [or Soils". IS:2132· 1986, 'Code of Practice for Thill-Walled Tub e Samplillg of Soils" IS: 2720-Pan-l 1983, "Prepnratioll of DIY Somples for I'f/rious 1i'.~ IS'·. IS: 2720-Pan-2 1973, "DelerminatiO/l afWaler COll/eJrt·'. IS: 272O-Pan-3 $eel. 1-1980, "DelermiIlOlf()II of Sp«ific grQlity-Fl/le-grained .foils". IS: 272O-Pan-3 $eeI.-2-1981. " VetermIllClliOll of Specific Gravity-Nne, Medium, alld coorse-grained soils". IS: 2720-Part 4-1975, "Grain size mlo/ysis". IS: 2720-Part 5-1970, "Delermi/lalioll of Liquid alld PI(mic Lir;iu·". IS: 2720- Part 6-1972, "Determil/atioll of Sluil1kage Factors". IS: 272O-Part 7-1983, "Determil/atiol/ of Waler cOl/telll-DIY dellsity RelaliOIl usinx lighl compaclioll". IS: 272O-Part 8-1983, 'DeterminatiOIl of Wmer COillem .Dry DeIL';it)' R"fmion using Heavy Compaction. IS: 2720·P:m 9-1971, "De/ermi,rlliio;: af Dry Del'sity- MoiA/uTe n mlell/ Relatio/l by collstant weight of soil me/hod". IS: 2720 Pan 10-1973, " V etermilUltiOIl of Uncollfilled Compressir>e Strength"~. IS: 2720-P3I1 11-1971, "DctermilllllicJ/I of shear strellgth parameJers of soils from cOIJSolidated-undrained triaxial compressioll test with measuremelll of Pore-water Pressure". IS: 2720-Part 12-1981, "Determillatiol! of Shear Strength Parameters of Soils from COlISolidmed-Undrailled Triarial CompresSiOlI Test with meosureme/lt of Pore-Water Pressure". IS: 272O-Part 13-1972, "Direct Shear Test " . IS: 272O-Part 14-1983, "Determination of Density Index (Relative Density) of cohesiollless soils", IS: 272O-Part 15-1986, "DetermillatiOIl of Consolidation Properties. IS: 272O-Part 17-1977, "DetermillQlion of Lillear Shrinkage ". IS: 272O-Part 28-1974, "Determination of Dry Dellsity of Soil in-place-by the salld-replacemellt method", IS: 2720-Pnrt 29-1975, "Determinatioll of Dry Density of Soils ill place-by the core culler method". IS: 272O-Part 30-1980, "Laboratory VUlle Shear Test". IS: 2720-Part 33-1971, "DetermillatiO/1 of Ihe density ill place, by the Ring and Water Replacement Melhod", IS: 2720-Part 34-1972, "Determinatioll of the Dellsity of SoU in pIau, by the Rubber-Balloon Mehtod", IS: 272O-Part 35-1974, "Measurement of Negative Pore Water Pressure". IS: 272O-Part 36-1975, "Laboratory Determillation of Permeability of Granular Soils (Constant Head)". IS: 272O-Pan 38-1976, "Compactioll cOlI/rol Test (Hilf Method)". IS: 272O-Part 39-Scct. 1-1971, "Direcr SIUUlr Test fnr Soils eMtailling grmllC' Laboratory Test".

REFERENC~.

37. IS: 2nD-Pan 40-1977, "Deu:rmbra(iOII of F~e Swell/lldex of Soils". 38. IS: 2911-I'un l-Secl. 1·1979, "Design and COllStruction 01 Pife FOWldalions·Driven Cast in-situ cOllcrete Piles". 39. IS; 29l1-Pan I-Sect. 3·1979, --Desi8n alld cOIlS/ruction of Pile Foundation·Driven Precast Piles". 40. IS: 29U-Pan 3·1980, "Code of Practice for Design and Construction of Pile Foundation· Under-reamed Piles". 41. IS: 29l1-Pan 4-1974, "Load Test 011 Piles". 42. IS: 2950 A:I (1974), "Code oJ l'ractice lor Desigll and COIIS/ruc/ion 01 Ral/ Foundations". 43. IS: 2968·Pnn 1-1976, " Dynamic method using 50 mm Cone without Bentonite Slurry". 44. IS: 2974-Pan 1-1982, " Fountio.tiOfllor Reciprocaling Type Machint:s". 45. IS: 2974-Patt 2-1980, "Foundation lor Impact Type mochines (flarmmer Foundation )". 46. IS: 2974-Pnn 3-1975, "Foundalion lor Rotary 1}pe machines (Mt:dium and Highway F'Uiuency)". 47. IS: 2974-Pan 4-1979, "Foundations for Rotary 1ypc Machines for low Frequency". 48. IS: 2974-Part 5-1970, "FOU/lda/iOllS lor Impact Type MachillCS o/Iler tllall Harmmers (Forgin8 alld Stampi/lg Press, Pi8-breaker, Eleva/or alld f10isl Tower). 49. IS: 3764-1970, "Safely Codes for EXC3valion work". 50. IS: 3955·1967, "Indian Standard Code 01 Prac/ice for Desigll and COIlStruc/ion 01 well Foundations". 51. IS: 4434-1978, "Code of Practice for In-situ 1'{J,Ile Shear Test for soils". 52. IS: 4453-1980, "Code of Practice lor Sub-StIr/ace ExploratiOlI by Pits, Trellcli€:s, Drilts OIld Shafts." 53. IS: 496S-Pnn 2-1976, "Dynamic Method using COi le lIJId Bell/oni/e Slurry ". 54. IS: 496S-Pan 3-1976, " Static calle Pellctratioll 7es/". 55. IS: 5121-1969, "Safery code for Pi/i1l8 QJld Olher Deep Fou/uIntjons". 56. IS: 5529-Pan 1-1985, "Code of Proctice for III -situ Permeability Tests-Tests in Overburden ". 57. IS; 6403-1981, "Code of Praclice for DeterminatiOIl of Bearing Capaciry of Shallow Foundations". 58. IS: 6955-1973, "Code of Practice for Subsurface Exploration for Earth imd Rook fill Darns". 59. IS; 7317-1974, "Code of Practice l or Um·enial Jacking Test lor Deformation Modulus 01 Rock". 60. IS: 8009-P:;m 1-1976, "Shallow FoundatiO/l Subjecll:d to Symmetrical Stalic Ventea/ Loads". 61. IS: 8Q09·Parl 2-1980, "Code of Practice for calculations of settlement of FoulUIOliOlI-Deep FOUlldotiOiI subjected To Symmetrical Static Vertical Loading. 62. IS: 8763·1978, "Guide lor Ulldisturbed Samp/illg of So/Il~". 63. IS: 8764-1978, " Method lor DetermillatiOIl of PoilU-Lood Slnmgth Index oj Rocks". 64. IS: 9143· 1979, "Method for Determillatioll 01 PoilU Load Strength Index 01 Rocks". 65. IS: 9221-1980, "Method of Determination of Modulus 01 Elasticity and Poisson's ratio of Rock mmerials ill Uniaxial Compressioll. 66. IS: 9259--1979, "SpecificatiOilS for Liquid umit ApparalUS for Soils". 67. IS: 964().1981, ·'SpecificO/iOlIS for SplitSpoon Sompler". 68. IS: 10082·1981. "Method of Test for DetermilUltiOiI of Tensile Strellgth by Indirect Tests OIl Rock Specimells". 69. IS: 10379-1982, "Code of Practice for Field COlllral of Moisture alld Comptlctioo Of Soils for EmbOJlkmelll alld Sub- grade". 70. IS: 11385-1985. "Code 01 Practice l or Sub-surlace &ploratiOlI for COJUlls lIJId Crass-Drainage Works". 71. IS: 11594·1985, " SpecijicOlions for mild Steel Thill-Walled Samplill8 Tubes OJld Sampler heads".

r'

INDEX A AASHTO daS&irlCAlion, 92

Absolute specific gravity, 20 Ab60Iute permeability, 145 Abbot COOlpRctiOli 1C61, 362

Activity, 81 Adsorption, 69,114 Adsorbed waler, 116 Aeolian deposits. 6 Airconlenl,15 Air entry v~Jue, 321 A-tine, 95, 99 Air voids., 15 Alien Hazen's formula, 151 Allowable soil prcssure, 588, 618 AJluvial soil depollil. 6 Apparenl JlI"C$urec.lillgram, SSI

Anchored sheet pile. 526, 532 Angle of internal friction, 313 Angle of shearing resistance, 313 Angle of wall friction, 495 AIt:hing in soils, 573 Alomic bonds, 108 Aquiclude, 146, 402, 401 Aquifer, confined, 146, 402 Aquifer, uncon£incd, 146,402 Auger boring, 420

B B:lrbn', method, 764 Base exchange Cllpacity, 114 Base pIlrabolll, 173 Beam bending lest, 829 Bearing capacity theories, .587 Bishop's simplifi«i method. 463

Bituminous stabilisation, 381 Black cottoo soil, 7

Blanket drain, 397 Block diagram, J3 Iloiling oondilion, 201 Bonds, lOB Bored piles, 675, 685

Boring. 420 Boring log, 437 Doussinesq's solution, 221 8raced cut, 550

BulkhMds,526 Bulking of sand, 131

Bulk Density, 16

C CaiSlORS,706 California bearing IlItio, 775 Dlpillllnty-penneability test, 152 Capilllll)' potential, 12.'i Capilliit)' rise, 121 Capillary aiphoning, 131

Capillary tension, 122 Cllpillaf)' WRler, 120 CMIIgrnnde'5App;\nttus,70 CMIIgrande', phreatic line, 174 c.ap:rande', piezometer, 432 ~ral1dc:'spll\l;lidlychart,97

Cement labilisation, 377 Centriruge methcd, 127

Chcmicrd stabiliSllthn, 382 CI.,ssjfiC
CoefTkient, pcrm«lbility, 136 Coefficient, percolation, 140 Coefficient, $eCOI1tlary consolidation, 28S Coefficient, subgrade reaclion, 659, 781 CoeffICient, uniformity, 59 CoeffICient, volume change, 26.S Coffer dams, 5S6 Colloids,l08 Colluvialooils,7 Combined (OOI:ing, 645 Compaction, 357 Compm'lSion, Index, 265 Compression I1ItiOll, 279 Cone of depression, 146 Cool.' penetrometer methOO, 73 Cone penetration test, 429 Confined aqUifer, 148, 402 Consistency index, 78 Consistency limils, 69 Con~1idalion, 2.~

Contact pressure, 247

Degree of s.a\l,lr~iion , ]5 Density boule, 30 Densit)' index, 60 Denison AAmpler, 426 Depth orexpiOfation, 417 Depth or COOIillg, 637 Depth fllCior, 453 DifTusedouble layer, 114 I)ilataocy a:>rTeCtioo, 427 Dispersed struCture, 118 Dispersing IIgent, 49 J)irectshcarteSi , 314 Diteb cooduit, 575 I)raioogc, 39 1 OrawOOwn, 146, 403 Drilling methods, 420 Drilled pier, 706 Dupuii's lISSumpiioo, 146 DUl1Ibility of rock, 822 Dutch cone test, 429, 688 Dytlllmic rormul~, 68S Eanb dam, 173, 460 Earth pressure theories, 478 Eccentric footings, 611, 644 EfTective siu" 58 Effective strc.-.s, 189,313 EIT
Efficiency of pi Ie group, 690 Electrical ~nalogy, 170 EiectriC

Conbd stress, 191 Contact water, IlA Core drilling, 421 Core amel, 34 Counterfoct willi, S23 Coulomb's theory, 494 CoYalent bond, 109 Critical damping, 760 CTilical gradient, 202 Critical height, 492 Critical void IlIlio,317 Culmann', constroclion, SOl Culmann's method, 448, 696

D Damping raclof, 760 Danish fotmulll, 686 Darcy',law,135

Fadum '5 chftft, 2366 Failure. rocks, 831 Failure pI~ne, 344, 483 Failul\'.$, foulldatioos, !i96 Fail\ll1'.$, slopes, 442 Fellenius line, 457 Fenske's chan, 244 Field COIDJII.lction control, 368 Field compAction methods, 366 Field COIUOIid.,tion curve, 284 Fidd density measurement, 33, 368 FIeld identification methods, 101 Filter, 173,207 Finite difference method, 172, 6.~6 Filling methods, 2n Fixed eanb suppon, 535 Aootingcaisson, 716

'"

INDEX floccuialCd structure:, 166 Flow indell, 78 Flow net, 161 Formation of soils, 5 F()(l:Cd vibllltion 761 Foundations, 587, 636, 611, 122, 155 Free CIIntilcvcr pilt; 527 Free earth support, 532 Free vibration, 159 Friction cil'Cle method, 450, 505 Frictional Slrcs.st.~ 784 Fr06t boil, 129 Frost deplh, 638 Frost hcIIve, 128 Frost line, 128

G

"

GCOI5l3ticsttt:$$,219 General shear fllilurc, 596 Geophysical methods, 433 Gcotextilc, J8..'i GJacierdc[X)Sitoo soils, 6 Gow's method. 708 Grnded filter, 207 Grading of soils, 51 Gr.linsi7..ecurvc,57 Ground waler, 391 Gravily retaining wall, 520 Grip length, 723 Gross SlIfe bcaringcalXlcity, 587 Groe;s footing pressure, 587 Group actions, piles, 690 Growing, 384

II Ilagt:n-l'oissclliUecq .. 142 llaoo-carved sample, 426 Il~nscn's theory, 004 HIIIV~rtl como.1ction telll , 362 Heave piping, 205 Hiley's formula, 686 lIaC's method, 463 History of soU engg. II HoncY- method, 62S Hvorslcv's theory, 342 Hydraulic head, 134 Hydraulic gradient, 135 Hydrogen bond, 109 Hydrometer, 52 Hygroscopic waler, 120

1 lIIiIC,I13 Immediate settlement, 613 Indt:ll properties, 45, 819 Indian Standard classification, 98 Influence dillgram, 226 Initial C()f1solid~lion, 2."i7 In-situ 51ress, 219, 834 looic bonds, lOB Isobar diagram, 22S Isoc/tronc:!>,273 Isomorphous subslilulion, 112

Overoonsolidated soils, 261 Ovcroonsolid.1.tion IlItio, 267

Jodhpur mini COIllp!ldor, 362 Joint set, 818 Joint spacing, 825

I' . Packer test, 150 l'al1iclesize aMly~is, 45 Pal1ially saturated soils. 209, 341 Passiveslaie,418 PIIVCDJeDt design, 13 I'cnetrntion test, 427, 610 Perc:cnlllge filU!r, 55 PerCt:llIage air voids, IS Percussion drilling, 422 Permeability, 134,820 pFvalue, 125 Phreatic line, 173 Pile foundalion, 671 Pile load test, 688 Pipette method, 51

K Kaolinite, 112 KOir.Cny basic pllrabolll, 173 Kouny Carmlln eq. lSI

L umbe'sstressp.1th,339 Laminar now, 141 l...IIplllCt:·S equation, 164 LIIteral1yloa~pilC8,696

Lat.riticsoils,7 Lime stabilisation, 380 LimilS.AllCriJerg,69 Linearshrinkagc,76 Liquefaction of sa~ 343 Liquid limll, 70 Liquidity index, 78 LMd test, pl~te, 621 Load lest, pile, 690 Local shear failure, 596 Log:1rithm of time mc:thod, 278 Loudon's formllla, 152

M Machine foundalJon, 755 Mass spedficgravity, 20 MarinedeposilS,6 Mllrtson's theOf)', 583 Mat COUnd.1Iion, 649 Mccha.nicalanalysis,46 Mechanical wbilisation, 376 Mercury control ~yslcm, 320 Mcyerhof's IIl\3lysis, 602, 679 Modified failure envelope, 338 Modified Proctor test, 360 Modulus or ela.~ticity, 218 Moc.Iulus of subgrmle n:oction, 651, 181 Mohr-Coulomb criterion, 337, 832 Mobr-Coulomb theory, 312 Moistun: conlent, 15 MOnlmaillonite, 112 Multi...slagtwcll points,393

N Natural frequency, 165 Negative projecting conduit, 580 Negative skin friction, 684 Net allowable pre55ure, 587 Net footing pmosUK, !188 Ncwmllrk's c:hatt, 239 Normally consolidated soil, 2Jj7

0 OctahcdrnJ unit, 111 Open cajs.o;on~ 710 Open-end tests, 149 Open well s, 407 Optimum waler co&enl, 359 Origin of soils, 4 Origin of pllII1C5,308

•

~~::'s:~cr,

426 l'laccment watercontenl, 367 Plant ofequaJ settlement, 579 Plastic equilibrium, 338, 482 Plasticity mali. 97, 99 l'1asticity indeJ:, 78 Plastic limit, 73 Plastic models, 172 Plille 10ild 1('111, 621 Poillt-lood strength, 821 Poisson's ratio, 218 PotC~urepllNlmelc:r,333

PotC pn:s6l1fC measurement, 321 Positive proje<:ting ooodlli t, 577 I'«c:nlial function , 165 Porosity, 14, 820 Pmodtl's IIMlysis, 592 Prccompression, 371 Prcconsolidated soil, 267, 283 I'rimary oonsoJidalion, 257 Primary valence bond~, 108 Proctor lest, 358 Prodor needle, 368 Pressure bulb, 225 Pres.suremctre, 431 Presumptive bearing capacity, 621 Principal planes, 308 Pumping-in t('l;t, 148 Pumping-C>Ut 1e~1S, 146 Punching shear Cailure, 596 Py<:nome1cr, 28, 32

Q Quick. sand ooodilion, 203

R R-value,778 Rart foundatioo, 653 Rlldius ofreilltive stiffness, 7&3 Radiation method, 36 Rllnkine's 3I1:llysis, 590 Rankine'scarth pressure, 481 Recompression indu, '1h7 ReoonMis.sanoe, 417

SOIL MECHAl'lICS AND FOUNDATION ENGiNEERING Rectangular combinod footing, 645 Rc:cI3ngular plot method, 459

Recuperation test, 408 Rehbann's construction, 497 Relative density, 60 Resistivity method, 435 Resonance test, 765 Retaining walls, 517 Reynold's number, 136 Richllrt's chart, 758 Right triangle chart, 9Z Ring shear lest, 829 Rigid pavemCllts, m, 782 Rock Mc:chania., 817 Rock quality designatioo, 818 Rock mass rating, 824 Rollers, types, 366 Rolary drilling, 422 Rowe's correction,.534 Rubber balloon method, 35

S Samplers, 424, 425 Sample prt:par1Ilion, cohesionlcss soil,322 S.-l.nd bath method, 29 Sand drains, 291 Sand island, 713 Sand replacement method, 34 Saturated mass density, 16 Saturated units weight, 18 Scrap bucket sampler, 425 Secondary consolidation, 285 Second:lry valence bonds, 110 Sedimentation analysis, 47 Seepage anaIy1ois, 163 SeeP"b'C line, 173 Seepage pressu~, 197 Seepage velocity, 140 Seismic method, 433 Settlement analysis, 612 Settlement ratio, 578 Shape of partides, 59 Shear box test, 314 Shear sm:nglh, 306, &33 Shrinlulge, ]29 Shrinkage iOOcx, 76 Shrinkage limit, 74 Shrinkage ratio, 76 Shallow well, 394

Shan,57Q Sheet pile wall, 526 Shelby tube, 425 Shift, 744

Sieve analysill, 46 Significant depth, 417 Sinking or 'Wells, 742 Site inyestiglltions, 415 SUmploo-Bjenum method, 294 SltemplOll bearing capacity fllctOrs, 007 Skemptons pore prt:ssure parameters, 333, 462 Slaking, 130,822

Slip circle mdhod, 455 Slopes. stability lInalysis, 440 Slots, 396 Soil-bitumen,381 Soil-<:ement,377 Soil dassification, 89 Soil dcposits, 7 Soil formation,S Soilmodcl,171 SoIl srunpler, 424 SoillltruClure, 116 Soil suction, 125 Soil wBter, 120

Soldier beams. .553 Sonic velocity, 823 Specific gnlVity, 19 SpecifIC retention, 402 Specific surface, 107 Specific yield, 402 Splitting test, 828 Stabilisation of soils, 376 Slllbility CMrts, 453 Stability nllmbcr, 453 SUtbility or $lopes, 440 Standard pcnetrntion test, 427, 610, 688 SUtnWml Proctor ICSI, 358 SUttic oonc tCSl, 429, 616 Static formula Cor piles, 6n SlI:ining, 741 SlOkes 18w, 47 SItC.!;S--stnlin curve, 218, 829 S~p"th,339

SlI'Uctllt31 units, III Structural water, 120 Subgrade,774 Subgrndc relICtion, 659, 781 Sub...surfaceprofile, 437 Submerged density, 16 Stldd.:n drawdown, 461 Suction, 125 Suclion plale, 127 Surface tension, 120 Swedish-drcle method, 455 Swelling of !!Oils, 129

T Taylor's stability dwls, 453 Tension cmdc, 491 Tension pila. 694 Tensiometer, 127 Terminal velocity, 47 Terminology of soils, 9, 838 Terra-probe. 370 Terzaghi's analY'lis. bearing capacity,593 --cQllsoIidlllion, U,7 -rigid bulkheads, 725 Te:c:11I1l!1 clllssifiClttion, 91 Thermal slabilisation, 383 Thixotropy, 81 Three-dimensional consolid:ltion, 287 TIC backs, 554 lilt, 744

TIme factor, 273 TIme·stltJemcnt curve, 2&3 Toughness indc:c:, 79 Tonion balllnce method, 27 Trarv;Cormod 5CClion, tSO Transmissibility, 762 Trial wedge method, 503 Triaxial teSl:, 3]8, 828 True angle ol Criction, 342 True cohc.~ion, 342 Tunocls,.571 Tunocl conduits, S82

u Uitimille bellring capacity, S87 Ultimlltesoil resistance, 737 Unconfined CO!l1ptCS1 i on test, 330, 828 Unconfined aquifec, 146, 401 Underground oonduits, 575 UndistW'bcd samples, 425 Unified dRSSification, 92 Uniformity coefficient, 58 Unil weight ofsoiJ solids,S Unit weight, 17, 819

V Vacuum 'Well points, 393 VandcrWlIal"srorces,110 Vane shear test, 332, 431 Vc:sic's theory, 605 Vibration, 759 Vibration isolation, 767 Vibrofiotation, 369 Virgin curve, 267 Void ratio, 14 Volume change measureme,nl, 321 Volumetric shrinlutgll, 76 W Wales,.554 Warping stresses, 784 Wash boring, 421 Water content, 15 Waterdisplaccmentmethod,33 Wave equation, 687 Wodgefailure,447 Wedge thcory,494 Wdl roundation, m Wen hydralllics, 401 Well points, 392 Weslegaard'i soIulion, 243 Westergaard method for rigid pavements, 782 Winkler's bod, 653

y Yield ol wells, 402

Young's modulus, 218

z Zero-air voids line, 360

Roller Compacted Concrete Dams

_~ ~

v.

K. MF.HROTRA

CONTENT~

I. INTRODUCTlf):\"

II

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7.4 Or"',,):), Ilml ll!! ~onstJU~lInn 7j 1mllal ~l!n' li.kn.Hi" n_.

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77 (', , n~r~l.; !,I,.:m~ $Chrdul" 711 R~. ,·",hltll\'U<.'" anJ plal-"",en\ 7'J Wdlcr '1''P' 7 Hi Sf><'<' , ~1 jlIOC~'f11 .... I' II. t'O~.~ I Mt l(TJON t:Qlj1J'Mt:",'"

1.3 Enl.roomem Impa<.1 1..4 E.;en:~r"l'~ brmrs ~.5 OIll<:r henefi,. ',f ,,;~ ,(III~tru<;uon 2.6 Role of different orpnll.nuom

10.: IIJ

i\O:~~~Jlc I~'",~~-..;mr r1,l/l1 ~nlchl"lI JI~1 'm~mg rlanl ~"' I "U"'l'flfl:r.j: Jnd , Iu,nr'n~ ""t.hUIl:S

X ~ 'i,nfi\(.;" "k:t:u::; m...1,!,."" 11.6 SllfC..J..,.-, ~tl:f', 1I.~ V'bfmOI) j,)."t ~u[t", ~ Q V,I-r:Il,,..y .' .I!,:,

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38

SC'ep.,~col1lrol FOUnd~lion

3.9 Sl'l ll ~y.; .U O Outlel IoUlIl.:an:< 3.11 L3q;e ~:illing b;i~'ns. tna<5

9_'

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ba,·~(jlls.

fO\lndlllion of sWl;e

tanh. SI01"~~~ Inllks and Ulhtl" SIf",; wres IN\' E....,I(;ATJON AND SELECTION OF MATERIALS 41 GenC' r,1 J 2 Ce'll"'ntlliou$ mmeri~ l ~ 4.

... 3 AJ;gfej!al<:S J J Woter qU3lity

J.5

pbnl

~.II QUonl ity "I \~lu '!'lII ~ nt 9. 1'N:F.I' \N: ,H I' I" Of' SI'El:lfICATIONS 9. 1 Pmltong '1",~",~.,t,V!,~ 1M ~~ 9.2 Ekonent. vi ~1>"lIr'l~(lOlI$

Adml~lurcs

!i. MIX 1'N:OI'OK'1l0NING t'O K ROLLER COI\1I'ACTED CONCRETE 81ls,~ (or selection 01 mix proflO'"Ons

52

:;.::W::e~rec:ui~enl

SJConsislcncytllSts

proe"""'Il:Il, I('~

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1I.0A." I!'IISTN:UMENTATION II I G~r,cr.l l I I "Z T)'flc of mCa,uteme",. & ;n,trurnenls req uITed II ..' (jen~ ,.al gUI delines on the usc of instrumental Ion IIAln'lrllffiCnls ror N:CC IZ.IGeneral 12.:.1 MnlenalsllclklviOf

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S.6 Field OIdju5tment of ml~ J'lIopon ions S 7 E.t;~mple problem of ml~ design 6. I'N:OPEK'nF.S Ot' ROLtER COMPACTED CONCRETE 6.2 Workability 6.3 Oensity and yield 6.4 Ptrmc~bili ly 6.S CQmpres~ive strength 6.6 Tt nsile SlTc ngth and slrai n cnpachy 6.7 Shc.'lr strenglh .6.8 Thermal bth.lvior 6.9 Thermlll properties 6.10 Drying shrinkage 6.1 1 Autogenousvol l.Illlechanges 6.I2Crecp 6.13 Modulus or elasticily 6. 14 Poiuon's ralio 6.IS Cavitation. abl1lsion and imp;oc~

12.4 Thermallx:h.:ovlor

6. 16 Durabili~y

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INDETERMINATE STRUCTURAL ANALYSIS

lndelertX:tirww StnlCtu,aLAnalyslll

Dr. C.B. Kukreja Formerly, Head of Civil Engg .. Deon .If ond D Thapar Institute of Engg. and Technology, Patiala

First Edition

ISBN: 81-86308-59-8

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' -_ _ _01 650 diagrams and number of worked out e~amples • Fully covering the syllabus of UPSC, GATE, AMIE, and other Technical Universities.

(CONTENTS)

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Introduelion Melhod of Consistont Deformation Three Moment Equation Siope-Defloction Method Momont Distribution Method Method of Column Analogy Kini's Method Stroin Energy Methods Two Hinged and Filled Arches Influence l ines Suspension Bridges Model Anolysis Motri~ Methods Appro~imote Analysis of Framos and Trusses Secondory Stresses Appendill

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Pages: 684

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STRUCTURAL ANALYSIS

Structural Analysis

A Unified Classical and Matrix Approach A. GHALI & A.M. NEVILLE

Fourth Edition

ISBN 419-21200-0

The fourth e dition of this comprehensivo lelll book combines and develops concurrently both classical & malri~-based methods of stru~tural anolysis. The book opens wilh 0 new ~hopter on the anolysis of statically determinate stru~tures, intended to provide a beller

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Stollcolly determinote struelures Introduction to the onalysi. of statically indeterminote structures Force method of analysis Displocement method of onolysis Use of force Dnd displocement method. Siroin energyond "irtual wotk Det...."inalion of displacem ent by .... rtuol work. Furtheropplicotions of method of .... rtuol wonc Importa nt energy theorems Displaceme nt of elostic struelure5 by special methods Applications of force & d isplacement methods : column analogy & moment distributio n InflullnClllin8> for beams ond frames Inllullnce lines for grids, arches and trusses Effectli of oKiol foree on flexural stiffness Analysis of shear-wa ll structures Me'hod of finite differenclIs .. Finite-element method Furthe( devolopmtnt of finito _elemon' mothod Pla stic Clnal)'lis of continuous blloms & fromes Yield_lino & strip methods IOf .Iobs Sirudun:-' dynamics Computer onolysis of fromed strudvres Implementation of computer Clnalysis .. Nonlinoor analysis Appe ndkesA·l Index

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Indian Price Rs. 425.00

MATERIAL TESTING LABORATORY MANUAL Dr. C.B. Kukrejo & Ravi Chawla Thopor Institute of Engg. & Tech .• Patiala

( CONTENTS)

2nd Edition

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Port-l Cement Port-2 Cement Aggregates Port·3 Cement Concrote Port-4 Reinforced Concrete Port-S Bricks Port-6 Timber Part-7 Steel Port·8 Building Appendi~ Ume

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Pages: 236

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Price Is. 75.00

Dr. K.R. Arora did his B.E. (Civil Eng ineering) in 1959 in First Class from Rajasthan University Later he gOl his M. E. degree witll honours from Jodhpur University and Ph. D. Degree from 1.1. T.. DELHI. Dr. Arora worked with Irrigati on Department of Rajasthan for four years from 1959 10 1963 a Assistant Eng ineer. He was posted at Rana Pnllap Sagar OHm, Rawat Bhllla. He was associatel with design, construction and inspection of various civil engineering stmctures. such as gravity dnlll , ealtl dam, bui ldings. road~ and bridges. Dr. Arora joined M.B.M . .Eng ineeriJlg Coll ege, Faculty ofEl)gineeri ng, University of Jodhpur in 1963 as a Lecturer in eivi Engi neering. Later he was promoted as Associate Professor of Civil Engineering. He taught various Civ il Engineerin] subjects to undergraduate and postgraduate students. Dr. Arora shi ftcd to Engineering College, Kota, as Professor and Head, Department of Civil Engineering. in 1983. HI was responsib le forthecstablishmcnt orthe various laboral.oricsandotherinfraSlmcturefaci lities. He retired in 1997. Dr. Arora worked ill Iraq on a Foreign Teaching Assignment for one year 198 1-82. He a lso worked as a Professor ofC ivi Engi neering al AWl'l (Ethiopia) under a U.N.D .P. Programmc from 1997 to 1999. Dr. Arom has 36 years experience 0 teaching undergraduate and postgntduatc students. Dr. Arora has published a number of research papers in Indian and Foreign joumals and confercnces . He has written; number oflcxt books on various Civil Engineering subjects. The books are liked by students. teachers and field engineer not on ly in India bu t also in lorcigncountries. 'Soil Mechani cs and Foundation Engineering' was first publi shed in 1987 and has been revised nnd updl1lcd a number oflimcs SA LIENT FEA TURES • S I units u~ed. • Fundamentalsexplained in II simpl e, lucid language. * Subject tnalterpresented i l) a logical manner. • Objective type questions in each chaptcr. * Question from various competit ive exam inations. * Latest 13 1S codes fo llowed * Usefu l for undergraduate and poslgraduate StudcnL5 and field engineers .

IRRIGATION WATER POWER AND WATER RESOURCES ENGINEERING Or. K.R. Aror-a, Ph.D. (liTO), F.I.E.

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The book covers the syllabus of the subject usually taught al the degree level in various Indian Universities and technical institutions. The students appearing for AMIE, ICS, IES and various other competitive examinations will find the book useful. The book is written entfrely in SI units. However, useful conversion factors are given for the readers interested in M.K.S. and F.P,S. system of units.

CONTENTS

Part-I : Hydrology, Water Resources, Dam ilnd Water Power Engineering Introduction Descriptive Hydrology Stream Gauging and Hydrograph Analysis Estimation of Runoff, yield, and Flood Discharge Ground Water Hydrology River Behaviour and Training Reservoir Planning Flood Control Planning for Water Resources Development Introduction to Dam EngineerIng Gravity Dams Embankment Dams Arch Dams and Buttress Dams Spillways Dam Outtets and Sluice ways Water Power Engineering Part·1I : Diversion Headworks, Distribution works, Irrigation Practice and Irrigation Management Planning and layout of Diversion Headworks Basic Principles of Design of Hydraulic Structures Design of Diversion Headworks Water Requirements of Crops Planning and Layout of Distribution System Design of Channels Canal Regulation Works Cross·Drainage Works Canal Outlets, Escapes, Bridges and Meter Flumes lining and Maintenance of canals Well and lift Irrigation Tank, Bandhara and Inundation Irrigation System fundamental of Irrigation Practice Methods of Application of Water Water logging, Drainage and reclamation and irrigation Management Appendix A - Salient Features of Some Typical River Valley Projects Appendix B - Glossary of Common Terms Appendix C - Selected References Appendix D - Publication of Bureau of Standards.

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Price : Rs. 250.00

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Soil Mechanics & Foundation Engineering by K R Arora_6th Edition

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