published by : A. K. Jain For Standard Publishers Distributors '170S-B , Nai Sarak, Delhi-llOOO6.
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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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
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.
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.
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.
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
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,
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
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
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
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. .
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.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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" ",
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
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
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
(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.
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENINEERING
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
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,
SOIL MECHANICS AND FOUNDATION ENINEERING
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
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.
SOIL MECHANICS AND FOUNDATION ENINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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)
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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)
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
PARTICLE SIZE ANALYSIS
"
(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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
PARTICLE SIZE ANALYSIS
... (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)
,.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
PARTICLE SIZE ANALYSIS
"
(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
SOIL MECHANICS AND FOUNDATION ENGINEERING
.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
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
=
=
,.
PARTICLE SIZE ANALYSIS
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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).
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,
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
.,
PARTICLE SIZE ANALYSIS
(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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
'{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)
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
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.
.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHAN ICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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,
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
SOIL MECIIANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGiNEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
PERMEABILITY OF SOILS
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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).
PERMEABILITY OF SOILS
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERINO
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.
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
roiL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
'" 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
~
Solution. From Eq. 8.5,
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,
PERMEABILITY OF SOILS
157
Solution. From Eq. 8.6,
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. ; ;
Solution. From Eq. 8.17,
• 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)
Solution. From Eq. 8.26,
, =
~~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.
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.
PERMEABILITY OF SOILS
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._
= 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.
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?
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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). /
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,
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)
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
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
"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
;::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.
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).
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
SOIL MECHANICS AND FOUNDATION ENGINEERING
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.
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'"
SOIL MECHANICS AND FOUNDATION ENGINEERING
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)
SOIL MECHANICS AND FOUNDATION ENGINEERING
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
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)
SOIL MECHANICS AND FOUNDATION ENGINEERING
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)
SOIL MECHANICS AND FOUNDATION ENGINEERING
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 •