(Solution Manual) Das - Principles of Geotechnical Engineering 8th SI PDF

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An Instructor’s Solutions Manual to AccompanyPRINCIPLES OF GEOTECHNICALENGINEERING, 8THEDITIONBRAJA M. DAS & KHALED SOBHANPrinted in the United States of America 1 2 3 4 5 6 7 16 15 14 13 12© 2014, 2010 Cengage LearningALL RIGHTS RESERVED. No part of this work covered by the copyright herein may...

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An Instructor’s Solutions Manual to Accompany

PRINCIPLES OF GEOTECHNICAL ENGINEERING, 8TH EDITION BRAJA M. DAS & KHALED SOBHAN

 

© 2014, 2010 Cengage Learning

ISBN-13: 978-1-133-11089-7 ISBN-10: 1-133-11089-4

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INSTRUCTOR’S SOLUTIONS MANUAL TO ACCOMPANY

PRINCIPLES OF

GEOTECHNICAL ENGINEERING Eighth Edition, SI BRAJA M. DAS KHALED SOBHAN

Contents Chapter 2 ........................................................................................................................................... 1 3 ......................................................................................................................................... 11 4 ......................................................................................................................................... 19 5 ......................................................................................................................................... 25 6 ......................................................................................................................................... 31 7 ......................................................................................................................................... 41 8 ......................................................................................................................................... 51 9 ......................................................................................................................................... 57 10 ........................................................................................................................................ 69 11 ........................................................................................................................................ 83 12 ........................................................................................................................................ 99 13 ...................................................................................................................................... 109 14 ...................................................................................................................................... 121 15 ...................................................................................................................................... 127 16 ...................................................................................................................................... 141 17 ...................................................................................................................................... 153

Page

Chapter 2 2.1

Cu =

D230 0.212 D60 0.42 = = 0.656 ≈ 0.66 = = 2.625 ≈ 2.63 ; Cc = D10 0.16 ( D60 )( D10 ) ( 0.42)( 0.16)

2.2

Cu =

D302 0.412 D60 0.81 = = 0.768 ≈ 0.77 = = 3.0 ; Cc = D10 0.27 ( D60 )( D10 ) (0.81)( 0.27)

2.3

a. Sieve no. 4 10 20 40 60 100 200 Pan

Mass of soil retained on each sieve (g) 28 42 48 128 221 86 40 24 ∑ 617 g

Percent retained on each sieve 4.54 6.81 7.78 20.75 35.82 13.94 6.48 3.89

Percent finer 95.46 88.65 80.88 60.13 24.31 10.37 3.89 0.00

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b. D10 = 0.16 mm; D30 = 0.29 mm; D60 = 0.45 mm

2.4

c. Cu =

D60 0.45 = = 2.812 ≈ 2.81 D10 0.16

d. Cc =

D302 0.29 2 = = 1.168 ≈ 1.17 ( D60 )( D10 ) (0.45)( 0.16)

a. Sieve no. 4 6 10 20 40 60 100 200 Pan

Mass of soil retained on each sieve (g) 0 30 48.7 127.3 96.8 76.6 55.2 43.4 22 ∑ 500 g

Percent retained on each sieve 0.0 6.0 9.74 25.46 19.36 15.32 11.04 8.68 4.40

Percent Finer 100.00 94.0 84.26 58.80 39.44 24.12 13.08 4.40 0.00

b. D10 = 0.13 mm; D30 = 0.3 mm; D60 = 0.9 mm 2 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.5

c. Cu =

D60 0.9 = = 6.923 ≈ 6.92 D10 0.13

d. Cc =

2 0.3 2 D30 = = 0.769 ≈ 0.77 ( D60 )( D10 ) ( 0.9)( 0.13)

a. Sieve no. 4 10 20 40 60 80 100 200 Pan

Mass of soil retained on each sieve (g) 0 40 60 89 140 122 210 56 12 ∑ 729 g

Percent retained on each sieve 0.0 5.49 8.23 12.21 19.20 16.74 28.81 7.68 1.65

Percent finer 100.00 94.51 86.28 74.07 54.87 38.13 9.33 1.65 0.00

b. D10 = 0.17 mm; D30 = 0.18 mm; D60 = 0.28 mm c. Cu =

D60 0.28 = = 1.647 ≈ 1.65 D10 0.17

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d. Cc =

2.6

D230 0.18 2 = = 0.68 (D60 )(D10 ) ( 0.28)(0.17)

a. Sieve no. 4 6 10 20 40 60 100 200 Pan

Mass of soil retained on each sieve (g) 0 0 0 9.1 249.4 179.8 22.7 15.5 23.5 ∑ 500 g

Percent retained on each sieve 0.0 0.0 0.0 1.82 49.88 35.96 4.54 3.1 4.7

Percent finer 100.00 100.00 100.00 98.18 48.3 12.34 7.8 4.7 0.00

b. D10 = 0.21 mm; D30 = 0.39 mm; D60 = 0.45 mm c. Cu =

D60 0.45 = = 2.142 ≈ 2.14 D10 0.21

d. Cc =

D302 0.39 2 = = 1.609 ≈ 1.61 ( D60 )( D10 ) ( 0.45)( 0.21) 4

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2.7

a.

b. Percent passing 2 mm = 100

GRAVEL: 100 – 100 = 0% SAND: 100 – 73 = 27% SILT: 73 – 9 = 64% CLAY: 9 – 0 = 9%

Percent passing 0.06 mm = 73 Percent passing 0.002 mm = 9

c. Percent passing 2 mm = 100 Percent passing 0.05 mm = 68 Percent passing 0.002 mm = 9

GRAVEL: 100 – 100 = 0% SAND: 100 – 68 = 32% SILT: 68 – 9 = 59% CLAY: 9 – 0 = 9%

d. Percent passing 2 mm = 100 Percent passing 0.075 mm = 80 Percent passing 0.002 mm = 9

GRAVEL: 100 – 100 = 0% SAND: 100 – 80 = 20% SILT: 80 – 9 = 71% CLAY: 9 – 0 = 9%

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2.8

a.

b. Percent passing 2 mm = 100 Percent passing 0.06 mm = 30 Percent passing 0.002 mm = 5

GRAVEL: 100 – 100 = 0% SAND: 100 – 30 = 70% SILT: 70 – 5 = 65% CLAY: 5 – 0 = 5%

c. Percent passing 2 mm = 100 Percent passing 0.05 mm = 28 Percent passing 0.002 mm = 5

GRAVEL: 100 – 100 = 0% SAND: 100 – 28 = 72% SILT: 72 – 5 = 67% CLAY: 5 – 0 = 5%

d. Percent passing 2 mm = 100 Percent passing 0.075 mm = 34 Percent passing 0.002 mm = 5

GRAVEL: 100 – 100 = 0% SAND: 100 – 34 = 66% SILT: 66 – 5 = 61% CLAY: 5 – 0 = 5%

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2.9

a.

b. Percent passing 2 mm = 100 Percent passing 0.06 mm = 84 Percent passing 0.002 mm = 28

GRAVEL: 100 – 100 = 0% SAND: 100 – 84 = 16% SILT: 84 – 28 = 56% CLAY: 28 – 0 = 28%

c. Percent passing 2 mm = 100 Percent passing 0.05 mm = 83 Percent passing 0.002 mm = 28

GRAVEL: 100 – 100 = 0% SAND: 100 – 83 = 17% SILT: 83 – 28 = 55% CLAY: 28 – 0 = 28%

d. Percent passing 2 mm = 100 Percent passing 0.075 mm = 90 Percent passing 0.002 mm = 28

GRAVEL: 100 – 100 = 0% SAND: 100 – 90 = 10% SILT: 90 – 28 = 62% CLAY: 28 – 0 = 28%

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2.10

2.11

a.

b. Percent passing 2 mm = 100 Percent passing 0.06 mm = 65 Percent passing 0.002 mm = 35

GRAVEL: 100 – 100 = 0% SAND: 100 – 65 = 35% SILT: 65 – 35 = 30% CLAY: 35 – 0 = 35%

c. Percent passing 2 mm = 100 Percent passing 0.05 mm = 62 Percent passing 0.002 mm = 35

GRAVEL: 100 – 100 = 0% SAND: 100 – 62 = 38% SILT: 62 – 35 = 27% CLAY: 35 – 0 = 35%

d. Percent passing 2 mm = 100 Percent passing 0.075 mm = 70 Percent passing 0.002 mm = 35

GRAVEL: 100 – 100 = 0% SAND: 100 – 70 = 30% SILT: 70 – 35 = 35% CLAY: 35 – 0 = 35%

Gs = 2.7; temperature = 24°; time = 60 min; L = 9.2 cm Eq. (2.5): D (mm) = K

L (cm) t (min) 8

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From Table 2.6 for Gs = 2.7 and temperature = 24°, K = 0.01282

D = 0.01282

2.12

9.2 = 0.005 mm 60

Gs = 2.75; temperature = 23°C; time = 100 min; L = 12.8 cm Eq. (2.5): D (mm) = K

L (cm) t (min)

From Table 2.6 for Gs = 2.75 and temperature = 23°, K = 0.01279

D = 0.01279

12.8 = 0.0046 mm 100

CRITICAL THINKING PROBLEM

2.C.1 a. Soil A:

Cu =

D302 52 D60 11 = = 3.78 = = 18.33 ; Cc = D10 0.6 ( D60 )( D10 ) (11)(0.6)

Soil B:

Cu =

D302 2.12 D60 7 = = 3.15 = = 35 ; Cc = D10 0.2 ( D60 )( D10 ) (7 )(0.2)

Soil C:

Cu =

D302 12 D 60 4 .5 = = 1.48 = = 30 ; Cc = D10 0.15 ( D60 )( D10 ) ( 4.5)(0.15)

b. Soil A is coarser than Soil C. A higher percentage of soil C is finer than any given size compared to Soil A. For example, about 15% is finer than 1 mm for Soil A, whereas almost 30% is finer than 1 mm in case of soil C. c. Particle segregation may take place in aggregate stockpiles such that there is a separation of coarser and finer particles. This makes representative sampling difficult. Therefore Soils A, B, and C demonstrate quite different particle size distribution.

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d. Soil A: Percent passing 4.75 mm = 29 Percent passing 0.075 mm = 1

GRAVEL: 100 – 29 = 71% SAND: 29 – 1 = 28% FINES: 1 – 0 = 1%

Soil B: Percent passing 4.75 mm = 45 Percent passing 0.075 mm = 2

GRAVEL: 100 – 45 = 55% SAND: 45 – 2 = 43% FINES: 2 – 0 = 2%

Soil C: Percent passing 4.75 mm = 53 Percent passing 0.075 mm = 3

GRAVEL: 100 – 53 = 47% SAND: 47 – 3 = 44% FINES: 3 – 0 = 3%

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Chapter 3 3.1

γ sat =

3.2

γ sat =

3.3

e γw ( G s + e) γ w G sγ w e γ w ⎛ 1 + wsat = + = + n γw = n ⎜⎜ 1+ e 1 + e 1 + e (1 + e) wsat ⎝ wsat

(Gs + e )γ w 1+ e

=

G sγ w eγ w ⎛ e ⎞ + =γd +⎜ ⎟γ w 1+ e 1 + e ⎝ 1+ e ⎠

Rearranging,

γ sat (1 + e) = γ d (1 + e) + e γ w

Therefore, e =

γ sat − γ d γ d − γ sat + γ w

⎛ 1 + w sat ⎝ 1+ e

γsat = ⎜

⎞ ⎟⎟γ w ⎠

⎞ ⎛ 1 + w sat ⎞ eγ w (1 + wsat ) nγ w ⎟G s γ w = ⎜ ⎟ = wsat ⎠ ⎝ 1 + e ⎠ wsat

Rearranging, w sat (γ sat − n γ w ) = n γ w Therefore, wsat =

3.4

a. γ =

W 12.5 = = 125 lb/ft3 V 0.1

b. γ d =

c. e =

d. n =

nγw γ sat − nγ w

γ 1+ w

G sγ w

γd

=

125 = 109.64 lb/ft3 1 + 0.14

−1 =

(2.71)(62.4) − 1 = 0.54 109.64

e 0.54 = = 0.35 1 + e 1 + 0.54

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(w )(Gs ) (0.14)( 2.71) = = 0.702 = 70.2% e 0.54

e. S =

f. Volume of water =

3.5

γw

=

(125 − 109.64)(0.1) ≈ 0.024 ft3 62.4

⎛1 + w ⎞ (1 + 0.098)( 2.69)( 9.81) a. γ = ⎜ ⎟G s γ w ; 19.2 = ; e = 0.51 1+ e ⎝ 1+ e ⎠ Gsγ w (2.69)(9.81) = = 17.48 kN/m3 1 0 . 51 1+ e +

b. γ d =

c. S =

3.6

(γ − γ d )V

a.

( w)(Gs ) (0.098)( 2.69) = = 0.517 = 51.7% e 0.51

( Gs + Se) γ w (2.69)(9.81) + (0.9)(0.51)(9.81) = 20.45 kN/m3 = 1+ e 1+ 0.51

γ =

Water to be added = 20.45 – 19.2 = 1.25 kN/m3 b. γ sat =

(Gs + e )γ w (2.69 + 0.51)(9.81) = = 20.78 kN/m3 1+ e 1 + 0.51

Water to be added = 20.78 – 19.2 = 1.58 kN/m3

3.7

π

⎛ 1 ⎞ W 9.56 (2.8) 2 (22)⎜ 3 ⎟ = 0.078ft3; γ = = = 122.56 lb/ft3 4 V 0.078 ⎝ 12 ⎠

a. V =

b. w =

W − Ws 9.56 − 8.51 = = 0.1233 = 12.33% Ws 8.51

c. γ d =

Ws 8.51 = = 109.1 lb/ft3 V 0.078

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d. e =

e. S =

3.8

a. γ =

b. e =

G sγ w

γd

(1+ 0.26)(Gs )(62.4) (1 + w) Gs γ w (1 + w )G s γ w ; Gs = 2.72 ; 108 = = (0.26)(Gs ) wGs 1 +e 1 + 1+ 0.72 S wGs (0.26)( 2.72) = = 0.98 S 0.72

a. γ d =

b. e =

3.10

(2.69)(62.4) − 1 ≈ 0.54 109.1

wGs (0.1233)( 2.69) = = 0.614 = 61.4% e 0.54

c. γ sat =

3.9

−1 =

(G s + e )γ w (2.72 + 0.98)(62.4) = = 116.6 lb/ft3 1 + 0.98 1+ e

γ 1+ w

Gs γ w

γd

=

20.6 = 17.67 kN/m3 1 + 0.166

−1 =

(2.74)(9.81) − 1 ≈ 0.52 17.67

c. n =

e 0.52 = = 0.34 1+ e 1 + 0.52

d. S =

wGs (0.166)( 2.74) = = 0.874 = 87.4% e 0.52

a. γ =

( G s + Se)γ w (2.74)(9.81) + (0.9)(0.52)(9.81) = 20.7 kN/m3 = 1+e 1 + 0.52

Water to be added = 20.7 – 20.6 = 0.1 kN/m3 b. γ sat =

(G s + e )γ w (2.74+ 0.52)(9.81) = = 21.04 kN/m3 1+ e 1 + 0.52 13

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Water to be added = 21.04 – 20.6 = 0.44 kN/m3

3.11

ρ

a. ρ d =

b.

1+ w

Gs ρ w

e=

ρd

=

1750 = 1422.76 kg/m3 1 + 0.23

−1 =

e 0.92 (2.73)(1000) = = 0.48 − 1 = 0.92 ; n = 1+ e 1 + 0.92 1422.76

wGs (0.23)( 2.73) = = 0.682 = 68.2% e 0.92

c. S =

d. ρ sat =

( Gs + e) ρ w (2.73 + 0.92)(1000) = ≈ 2967 kg/m3 1+ e 1 + 0.23

Water to be added = 2967 – 1750 = 1217 kg/m3

3.12

a. γ d =

γ 1+ w

Gs γ w

b. e =

γd

=

30.75 3 = 112 lb/ft 0.25(1 + 0.098)

−1 =

(2.66)(62.4) − 1 ≈ 0.48 112

⎛ 30.75 ⎞ − 112 ⎟(0.25)...