Boring Log Sample Computation

DVANCED FOUNDATION ENGINEERING

I. INTRODUCTION Footings are structural elements that transmit column or wall loads to the underlying soil below the structure. Footings are designed to transmit these loads to the soil without exceeding its safe bearing capacity, to prevent excessive settlement of the structure to a tolerable limit, to minimize differential settlement, and to prevent sliding and overturning. The settlement depends upon the intensity of the load, type of soil, and foundation level. Where Where poss possibi ibilit lity y of differ differen entia tiall settl settlem emen entt occu occurs rs,, the the diffe differen rentt footin footings gs should should be designed in such a way to settle independently of each other. Depending on the structure and soil encountered, various types of foundations are used. A spread footing is is simply an enlargement of a load-bearing wall or column that makes it possible to spread the load of the structure over a larger area of the soil. In soil with low load-bearing capacity, the size of the spread footings required is impracticably large. In that case, it is more economical to construct the entire structure over a concrete pad. This is called a mat foundation. Pile and drilled shaft foundations are used for heavier structures when great depth is required for supporting the load. Piles are structural members made of timber, concrete, or steel steel that that trans transmit mits s the the load load of the supers superstru tructu cture re to the lower layers layers of the the soil. soil. According According to how they transmit their load into the subsoil, piles can be divided into two cate catego gori ries es:: fric fricti tion on pile piles s and and endend-be bear arin ing g pile piles. s. In the the case case of fric fricti tion on pile piles, s, the the superstructure load is resisted by the shear stresses generated along the surface of the pile. In the end-bearing pile, the load carried by the pile is transmitted at its tip to a firm stratum.

In the the case case of drille drilled d shaft shafts, s, a shaft shaft is drill drilled ed into the subsoi subsoill and and then then is filled filled with concrete. A metal casing may be used while the shaft is being drilled. The casing may be left in place or may be withdrawn during the placing of concrete. Generally, the diameter of a drilled shaft is much larger than that that of a pile. The distinction between between piles and drilled drilled shafts becomes hazy at an approximate diameter of 1 m (3 ft), and the definitions and nomenclature nomenclature are inaccurate i naccurate.. Spread footings and mat foundations generally are referred to as shallow foundations, whereas pile and drilled-shaft foundations are classified as deep foundations. In a more general sense, shallow foundations are foundations that have a depth-of-embedment towidth ratio of approximately less than four. When the depth-of-embedment-to-width ratio of a foundation is greater than four, it may be classified as a deep foundation.

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II. PROBLEMS Note: The water table is located at depth of 2.0 m below the ground surface. Other data necessary that are not given can be assumed using the Tables and Charts.

Based on the Boring Log Data and Atterberg Limits Test Results presented, answer the following as indicated:

1. What is the ultimate and allowable bearing capacity based on soil strength up to 2B below the base using Terzhagi, Meyerhof, and Hansen equations for each given size of isolated isolated footing? Use Factor of Safety Safety = 3. a. B = 2m , L = 2m, Df =1.5m b. B = 5m , L = 5m, Df =2.0m

2. What is the allowable bearing capacity based on soil compressibility up to 2B below the base using Terzhaghi & Peck, Meyerhof, and Bowles equations for each given size of isolated footing? Use allowable total settlement = 35mm. Assume the time for creep to be 10 years. a. B = 2m , L = 2m, Df =1.5m b. B = 5m , L = 5m, Df =2.0m

3. Calculate the total settlement of each isolated footing up to 2B below the base. Use Schmertmann Schmertmann's 's method for f or immediate settlement settlement and assume all clay layers be over consolidated. Disregard the secondary consolidation. a. B = 2m , L = 2m, Df =1.5m b. B = 5m , L = 5m, Df =2.0m

4. Summarize and tabulate all your answers. answers. Compare and discuss briefly the results and make your conclusion.

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Sample Boring Log Report

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ATTERBERG LIMITS TEST RESULTS

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III. SOLUTIONS

COMPUTATION COMPUTATION OF SOIL BEARING CAPACITY CAPACITY

For the computation and determination of soil properties of the underlying foundation materials, the following equations, parameters and engineering data were used:

A. EQUATIONS EQUATIONS

A.1) for f or correcting the no. of SPT “N” values. values. Correction is carried out when water table/ ground water was encountered during drilling and potential submerged condition is likely to occur.

N’=N =15 +

for N< 15 blows (ି ଵହ )

for N> 15 blows

ଶ

Where, N’ = adjusted/ corrected No. of SPT blows. N= actual No. of SPT blows per 30 cm of penetration. A.2) for f or computation of Soil Bearing Capacity

௔=

ൣ

(−1) + 0.5

௖ ௖ +

௦ ௨ =

௖ ௖ +

௤ ௤ + 0.5

௬ ௬൧

+

௬ ௬

Where: ௔ = allowable bearing capacity in kg/m ௨ = ultimate bearing capacity in kg/m ௦ = safety factor

2

2

y= unit weight in kg/m 3 D= depth of footing in meters B= base of footing in meters

c= safety factor Page | 5


are dimensionless dimensionless

௖, ௤, ௬, ௖, ௤, ௬

parameters B. ENGINEERING ENGINEERING DATA

B.1.a) Cohesionless Soils (Sand and Sandy Silt): (Polish Code- PN-59/B-03020, 1959 Soil mechanics and Foundation Engineering by Wilun & Starzewski, v.1) Cohesion, c = 0, for sand (assumed cohesionless)

Relative Soil Condition

N

4 10 30

(Relative Density)

Approximate Angle of Internal Friction (0)

< -

< -

4 10 30 50

Very Loose Loose Medium Dense Dense

>

50

Very Dense

28 32 35

28 32 35 35 37

B.1.b) Clays (Essentials of Soil Mechanics, David F. McCarthy:


N 2 5 8 15

-

(Relative Density) 4 8 15 30

above 30

Soft Firm Stiff Very Stiff

Approximate Cohesion (kg/m^2)

1200 2400 4890 9780

Hard

-

2440 4890 9780 19300 25000

B.1.c) Mixed Soils: (Polish Code- PN-59/B-03020, 1959 Soil mechanics and Foundation Engineering by Wilun & Starzewski, v.1) 1. Slightly Clayey Sands, Sandy Silt and Silts

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N

4 8 15

< >

(Relative Density) 4 8 15 30 30

Soft Firm Stiff Very Stiff Hard

Approximate Approximate Cohesion (kg/m^2) 1200 1500 2050 3000

>

1500 2050 3000 4000 4000

Approximate Angle of Internal friction (0)

2 10 16 20

-

10 16 20 25

1. What is the ultimate and allowable bearing capacity based on soil strength up to 2B below the base using Terzhagi, Meyerhof, and Hansen equations for each given size of isolated isolated footing? Use Factor of Safety Safety = 3. a. B = 2m , L = 2m, Df =1.5m b. B = 5m , L = 5m, Df =2.0m

Depth

MC

N'

Cohesion

Gs

e

y

1.50 3.00

25.1 33.2

17 13

32 32

0 0

2.65 2.65

0.6652 0.8798

4.50

22.4

20

32

0

2.65

6.00 7.50

25 43.4

17 26

32 32

0 0

9.00

34.5

9

28

10.50

67.2

5

12.00

53

13.50 15.00

terzaghi's bearing capacity factors

Nc

Nq

Ng

19.531 18.421

44.040 44.040

28.520 28.520

26.870 26.870

0.5936

19.967

44.040

28.520

26.870

2.65 2.7

0.6625 1.1718

19.546 17.489

44 44.040 44 4 4.040

28.520 28.520

26.870 26.870

0

2.73

0.9419

18.550

31 3 1.610

17.180

13.700

28

0

2.74

1.8413

15.818

31.610

17.180

13.700

9

28

0

2.73

1. 1 .4469

16.746

31 31.610

17.180

13.700

48.4

10

28

0

2.75

1.331

17.175

31.610

17.180

13.700

33.4

10

32

0

2.75

0.9185

18.758

44.040

28.530

26.870

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meyerhof

Nc

Nq

Ng

19.531

35 35.490

23 23.180

22 22.020

0.8798 0.5936

18.421 19.967

35 35.490 35 35.490

23 23.180 23 23.180

22 22.020 22 22.020

2.65

0.6625

19.546

35.490

23.180

22.020

0

2.7

1.1718

17.489

35.490

23.180

22.020

28 28

0 0

2.73 2.74

0.9419 1.8413

18.550 15.818

25 25.800 31 31.610

14 14.720 17 17.180

11 11.190 13 13.700

9

28

0

2.73

1.4469

16.746

31.610

17.180

13.700

48.4

10

28

0

2.75

1.331

17 17.175

31 31.610

17 17.180

13 13.700

33.4

10

32

0

2.75

0.9185

18.758

35 35.490

23 23.180

22 22.020

Depth

MC

N'

Cohesion

Gs

e

unit wt.

1.50

25.1

17

32

0

2.65

0.6652

3.00 4.50

33.2 22.4

13 20

32 32

0 0

2.65 2.65

6.00

25

17

32

0

7.50

43.4

26

32

9.00 10.50

34.5 67.2

9 5

12.00

53

13.50 15.00

q = ∂Df = 1.5( 19.531) = 29.2965kN/m3 At depth of 2B of 2m x 2m footing from table 1 Ø = 31 and at depth of 2B of 5m x 5m footing from table 1Ø=0

1 .Using Terzhagi Equation ௨ =

F.S B L D Df q D/B

= = = = = = =

3 2 1.5 1.5 1.5 29.2965 0.75

௖ ௖ +

௤ ௤ + 0.5

௬ ௬

m m m m kN/m^2

Depth

q kN/m^2

FS

௨௟௧

௔௟௟.

kPa

kPa

390.3 390.374 74 657.3 657.351 51

1.50

29.296

3

3.00

55.262

3

1171. 1171.12 123 3 1972. 1972.05 053 3

4.50

89.852

3

2991. 2991.80 804 4

997.2 997.268 68

6.00

117.277

3

3764.919 3764.919

1254 1254.973 .973

The table shows the ultimate bearing capacity and allowable bearing capacity at a certain depth. Page | 8


qult =

cNcscdci c+qNqsqdqi q+0.5γBN γsγdγi γ

qall =

qu/F.S.

Kp = B/L = Nc = Nq = Nγ =

3.124035 1

2

tan (45+ φ /2) (square footing) footing)

32.8 Bearing cap. Factor 20.8 Bearing cap. Factor 18.85 18.85 Bearing cap. Factor

for Prob Problem lem 1a: sc =

1.624807 Shape Factor

1+0.2Kp B/L

dc =

1.265124 De pth Factor

1+0.2√Kp D/B

ic =

1 Incl i nati on Factor

sq =

1.312404 Shape Factor

1+0.1Kp B/L

dq =

1.132562 De pt pth Factor

1+0.1√Kp D/B

iq =

1 Incl i nati on Factor

sγ =

1.31 .312404 Sha Shape Fac Factor tor

1+0.1K .1Kp B/L

dγ =

1.13 .132562 Depth epth Fac Factor tor

1+0.1√Kp D/B

iγ =

1 Inclination Factor Factor

qult =

1679.799 kPa

qall =

559.9331 kPa

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Hansen:

qult =

cNcs cNcscd cdcic cicgc gcbc bc + qNqsqd qNqsqdqiqg qiqgqb qbq q + 0.5yB .5yB'N 'Nysy ysydyi dyiygy ygyby by

qall =

qu/F.S.

Nc = Nq = Nγ =

32.8 Be ari ng cap. Factor 20.8 Be ari ng cap. Factor 17.95 17.95 Bearing cap. Factor

B' =

B=

Eff ecti ve wi dth

L' =

L=

Eff ecti ve l ength B-2e L (assume e=0) e=0)

B'/L' =

B-2e B (assume e=0) e=0)

1

for Problem Problem 1a: k=

-1

1 tan (D/B (D/B)) for for D/B > 1 , D/B D/B for for D/B ≤ 1

0.75

sc= dc = ic = gc = bc = sq =

1.63 .634146 1.3 1 1 1 1.515038

Sha Shape Factor Depth Factor Incl i nati on Factor Ground Fa Factor Base Factor Shape Factor

dq = iq = gq = bq = sγ = dγ = iγ = gγ = bγ =

1.211973 Depth Factor 1 Incl i nati on Factor 1 Ground Fa Factor 1 Base Factor 0.6 Sha Shape Fact Factor or 1 Depth Factor 1 Inclination Factor Factor 1 Ground Ground Factor Factor 1 Base Factor

qult =

1536.772 kPa

qall =

512.2572 kPa

1.0+(Nq (Nq/Nc /Nc)*(B )*(B''/L' /L') 1.0+0.4k

1.0+(B' (B'/L')sin )sinφ 2

1+2tanφ(1-sinφ) k

1.0-0 .0-0.4 .4(B (B'/ '/LL') 1.00 for al l φ

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for Prob Problem lem 1b: F.S. = B=L = use use D = Df = q= D/B =

3 5m 2 m 2m 2

45.16 kN/m 0.4

Terzhagi:

qult =

1.3cNc+qNq+0.4γBNγ (for square square footing)

qall =

qu/F.S.

Nc = Nq = Nγ =

26 Be ari ng ca c ap. Factor 23.78 Bear earing ing cap. Factor 39 Bearing Bearing cap. cap. Factor Factor

for Proble m 1b 1b:

φ = 31

kPa

qult =

9131.924 kPa

qall =

3043.975

qult =

cNcscdci c+qNqsqdqi q+0.5γBNγsγdγi γ

qall =

qu/F.S.

Kp = B/L = Nc = Nq = Nγ =

1 1

2

32.8 Be ari ng ng cap. Factor 20.8 Be ari ng ng cap. Factor 18.85 18.85 Bearing cap. Factor

tan (45+ φ /2) (square footing) footing) φ = 31

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Hansen:

qult =

cNcs cNcsccdcic dcicgc gcbc bc + qNqsq qNqsqdq dqiqgq iqgqbq bq + 0.5y 0.5yB' B'Nys Nysydy ydyiyg iygyby yby

qall =

qu/F.S.

Nc = Nq = Nγ =

32.8 Beari ng cap. Factor 20.8 Beari ng cap. Factor 52.6 52.6 Bearing cap. Factor Factor

B' =

B=

Eff ecti ve wi dth

L' =

L=

Eff ecti ve l ength

B'/L' =

1

for Problem Problem 1b: 1b: -1

k= sc= dc = ic = gc = bc = sq =

0.4 1.63 .634146 Sha Shape Fac Factor 1.16 Depth Factor 1 Incl i nation Factor 1 Ground Factor 1 Base Factor 1.515038 Shape Factor

1 tan (D/B (D/B)) for for D/B > 1 , D/B D/B for for D/B ≤ 1

dq = iq = gq = bq = sγ = dγ = iγ = gγ = bγ =

1.113052 Depth Factor 1 Incl i nation Factor 1 Ground Factor 1 Base Factor 0.6 Shap Shape e Fac Factor tor 1 Depth Factor 1 Inclination Factor Factor 1 Ground Factor 1 Base Factor

1+2tanφ(1-sinφ) k

qult =

10212.67

qall =

3404.225

1.0+ .0+(Nq (Nq/Nc /Nc)*(B )*(B''/L') 1.0+0.4k

1.0+(B' (B'/L' /L')sin )sinφ 2

1.0-0 .0-0.4 .4(B (B''/L') /L') 1.00 f or or al l φ

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2. What is the allowable bearing capacity based on soil compressibility up to 2B below the base using Terzhaghi & Peck, Meyerhof, and Bowles equations for each given size of isolated footing? Use allowable total settlement = 35mm. Assume the time for creep to be 10 years. a. B = 2m , L = 2m, Df =1.5m b. B = 5m , L = 5m, Df =2.0m

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3. Calculate the total settlement of each isolated footing up to 2B below the base. Use Schm Schmert ertman mann' n's s metho method d for immed immedia iate te sett settle lemen mentt and and assu assume me all all clay clay layers layers be over over consolidated. Disregard the secondary consolidation. a. B = 2m , L = 2m, Df =1.5m b. B = 5m , L = 5m, Df =2.0m

Solution: A. Find total settlement settlement St = Se + Sc(p) for 2m x 2m at Depth of of 1.5m from ground

e l 0m

Ground Surf ace N=19 m=25.1%

clayey sand

Df =1.5

el -1.5m WT sil ty sand wi wi th gravel

N=13 2.0m

m=33.2%

1.5m

sil ty sand

N=24 m=22.4%

1.5m

sil ty sand

N=19 m=25.0%

1.5m

el -3.0m

el -4.5m

el -5.5m

Formula for Se & Sc Se = C1C2(q’-q) ∑(Iz/Es)∆z (q’-q) ∑(Iz/Es)∆z Sc = ∆ = ∆ eHc /(1+e)

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IMMEDIATE SETTLEMENT Step 1 : Draw the strain influence diagram to compute Iz at Figure A

0.1

l e v e l g n i t o o f w o l e b h t p e D

Influence factor (Iz) 0.2 0.3 0.4

0.5

B/2 = 1.0

B = 2.0

3.0

2B = 4.0 Figure A : Influence graph for square or circle footing (L/B = 1)

Layer boundaries are solid Layer mid-points are dashed From graph the dashed line at mid-layer at figure A we get Iz(1) = 0.400 Iz(2) = 0.292 Iz(3) = 0.083

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Step 2 : Determine the values of Elastic Modulus See Appendices for Table 5.6 for SPT Sand (Saturated) Es = 250(N + 15). N values should be estimated as N 55. Es(1) = 250(19+15)/1000 = 8.50Mpa Es(2) = 250(13+15)/1000 = 7.00Mpa Es(3) = 250(24+15)/1000 = 9.75Mpa ∆z

Layer No.

N

Es

mm

Iz middle of layer

N/mm2

(I ( Iz/Es)x ∆z

mm3/N

1

15 1500

13

7.00

0.400

85.7

2

15 1500

24

9.75

0.292

44.9

3

10 1000

19

8.50

0.083

9.8

∑(Iz/Es)x∆z =

140.4

Step 3 : Determine Embedment and Creep Factors 

Embedment Factor

C1 = 1-0.5(q/(q’-q)) Also determined the following requirement Layer No. 1 2 3

w

GS

0.332 0.2 0.3

2.70 2.70 2.70

e

0.896 0.605 0.675

ળ kN/m^2 m^2 18.6 20.2 19.8

ળ kN/m^2 18.6 20.2 19.8

Note: Gs assumed to be 2.70 Then q = ∂Df Df =1.5m, ∂w=9.81kN/m3, GS = 2.70, w = 0.251 at 1.5 below ground surface ∂ = GS∂w(1+w)/(1+e) e = wGS for S = 1 e = 0.251(2.70) = 0.6777 ∂ = 2.70(9.81 2.70(9.81)(1+0.251)/ )(1+0.251)/(1+0.6777) (1+0.6777) = 19.75kN/m3 19.75kN/m3 q = ∂Df = = 19.75(1.5) = 29.625kN/M2 Page | 16


q’ = 1500kN for 5 storey building (assumed) q’ = 1500kN /(2m x 2m) - ∂Df = = 345.375kN/M2 Getting values C1 = 1-0.5(q/(q’-q)) C1 = 1-0.5(29.625/(345.375-29.625)) = 0.9530 

Creep Factor

C2 = 1 + 0.2log(t/0.1) For end of construction C2 = 1 At end of 1 year C2 = 1 + 0.2log(1/0.1) = 1.2 se = C1C2(q’-q) ∑(Iz/Es)∆z C1C2(q’-q) ∑(Iz/Es)∆z

Immedia Immediate te sett settlement lement per elevation elevation at 1 year year (Iz/Es)x∆z C1 C2 q'-q mm3/N 0.923 1.200 0.140 191.4 0.923 1.200 0.131 191.4 0.923 1.200 0.086 191.4

Se mm 30 28 18

e l e vati on - 2.25m - 3.75m - 4.00m

PRIMARY CONSOLIDATION

e 0.880 0.594 0.663

Cc 0.1427 0.0998 0.1101

Hc

σ'o

σ'av

Sc

Se

St

mm 1500 1500 1000

kN / m2 25.32 24.45 15.6

kN/m2 220.7 220.7 220.7

mm 73 69 78

mm 30 28 18

mm 103 97 96

elevation - 2.25m - 3.75m - 4.00m

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IV.

CONCLUSION

V. REFERE REFERENCE NCE 1. 2. 3. 4.

Princi Principle pless of Foundat Foundation ion Engineer Engineering, ing, SI-7th SI-7th ed-BRAJ ed-BRAJA A M. DAS 2011 2011 Soil Soil Mechani Mechanics cs Foundat Foundation ions, s, 3rd 3rd edition edition Muni Muni Budhu Budhu Foundat Foundation ion Analys Analysis is and Design Design by Joseph Joseph Bowles Bowles 1959 Soil mechanics mechanics and Foundation Foundation Engineering Engineering by Wilun & Starzewski Starzewski

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Boring Log Sample Computation

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