Principles of geotechnical engineering 8th SI (solution manual)

An Instructor’s Solutions Manual to Accompany

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

 

 


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

Page

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


Chapter 2
2.1

Cu =

0.212
D302
D60 0.42
=
= 0.656 ≈ 0.66
=
= 2.625 ≈ 2.63 ; Cc =
( D60 )( D10 ) (0.42)(0.16)

D10 0.16

2.2

Cu =

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

1
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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
0.292
=
= 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
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c. Cu =

0 .9
D60
=
= 6.923 ≈ 6.92
D10 0.13

0.32
D302
d. Cc =
=
= 0.769 ≈ 0.77
( D60 )( D10 ) (0.9)(0.13)

2.5

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

3
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D302
0.18 2
d. Cc =
=
= 0.68
( D60 )( D10 ) (0.28)(0.17)

2.6

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 =

0.39 2
D302
=
= 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%

5
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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%

6
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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%

7
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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:

D302
52
D60 11

=
= 3.78
=
= 18.33 ; Cc =
Cu =
D10 0.6
( D60 )( D10 ) (11)(0.6)
D302
2.12
D60
7
=
= 3.15
=
= 35 ; Cc =
( D60 )( D10 ) (7)(0.2)
D10 0.2

Soil B:

Cu =

Soil C:

D302
12
D60
4 .5
=
= 1.48

=
= 30 ; Cc =
Cu =
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.

9
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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%

10
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Chapter 3
3.1

γ sat =

3.2

γ sat =

3.3

⎛ 1 + wsat ⎞
(Gs + e)γ w Gsγ w e γ w
eγw

⎟⎟γ w
=
+
=
+ n γw = n⎜⎜
1+ e
1 + e 1 + e (1 + e) wsat
w
sat
⎝
⎠

(Gs + e)γ w
1+ e

=

Gs γ 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 + wsat
⎛ 1 + wsat ⎞
⎟G s γ w = ⎜
⎝ 1+ e
⎝ 1+ e ⎠

γ sat = ⎜

⎞ e γ w (1 + wsat )n γ w
=
⎟
wsat
⎠ wsat

Rearranging, wsat (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 − n γ w

γ
1+ w

Gs γ 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

11
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e. S =

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

f. Volume of water =

3.5

γw

=

(125 − 109.64)(0.1)
≈ 0.024 ft3
62.4

(1 + 0.098)( 2.69)(9.81)

⎛1+ w ⎞
a. γ = ⎜
; e = 0.51
⎟Gs γ w ; 19.2 =
1+ e
⎝ 1+ e ⎠

Gs γ w (2.69)(9.81)
=
= 17.48 kN/m3
1+ e
1 + 0.51

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

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

⎝ 12 ⎠

a. V =

b. w =

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

c. γ d =

Ws
8.51
=
= 109.1 lb/ft3
V
0.078

12
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d. e =

e. S =

3.8

a. γ =

b. e =

Gs γ w

γd

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

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 =

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

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 =

(Gs + 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 =

(Gs + 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.

e=

1+ w

Gs ρ w

ρd

=

1750
= 1422.76 kg/m3
1 + 0.23

−1 =

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

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

= 112 lb/ft3
0.25(1 + 0.098)

−1 =

(2.66)(62.4)
− 1 ≈ 0.48
112

⎞
⎛ 30.75
− 112 ⎟(0.25)
⎜
(γ − γ d )V ⎝ 0.25
⎠
c. Volume of water =
≈ 0.044 ft3
=
γw
62.4

3.13

a. e =

n
0.3
=
= 0.43
1 − n 1 − 0.3

b. ρ d =

3.14

e=

Gs γ w

γd

ρ (1 + e) 1800(1 + 0.43)
Gs ρ w
=
= 2.57
; Gs = d
1000
ρw
1+ e

−1 =

(2.69)(62.4)
− 1 ≈ 0.598
105

14
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S=

3.15

wGs (0.17)(2.69)
=
= 0.764 = 76.4%
e
0.598

a. γ =

(1 + w)Gs γ w (1 + w)Gs γ w (1 + 0.182)(2.67)(62.4)
=
=
= 122.5 lb/ft3
wGs
(0.182)(2.67)
1+ e
1+
1+
0.8
S

b. γ d =

γ
1+ w

=

122.5
= 103.6
1 + 0.182

Volume of water =

3.16

a. γ =

(γ − γ d )V

γw

=

(122.5 − 103.6)(1)
= 0.302 ft3/ft3 of soil
62.4

(G s + Se)γ w
(G + 0.55e)(62.4)
; 106 = s
1+ e
1+ e

Gs = 1.148e + 1.698

114 =

(i)

(Gs + 0.822e)(62.4)
1+ e

(ii)

From (i) and (ii): Gs = 2.73

b. Using Gs = 2.73 in Equation (i), we get e = 0.9

3.17

3.18

a.

Dr =

emax − e
0.75 − e
; 0.65 =
; e = 0.6
emax − emin
0.75 − 0.52

b.

γd =

Gs γ w (2.67)(9.81)
=
= 16.37 kN/m3
1+ e
1 + 0.6

Dr =

emax − e
0.72 − e
; 0.82 =
; e ≈ 0.51
emax − emin
0.72 − 0.46
15

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γ =

3.19

(1 + w)Gs γ w (1 + 0.11)(2.68)(9.81)
=
= 19.32 kN/m3
1+ e
1 + 0.51

γd =

γ
1+ w

=

115
= 106.48 lb/ft3
1 + 0.08

⎡ 1 ⎤ ⎡1⎤
⎡1⎤ ⎡ 1 ⎤
⎢
⎥−⎢ ⎥
⎢⎣ 92 ⎥⎦ − ⎢⎣106.48 ⎥⎦
γ
γ
d (min) ⎦
⎣ d⎦
⎣
Dr =
=
= 0.918 = 91.8%
⎡ 1 ⎤ ⎡ 1 ⎤
⎡1⎤ ⎡ 1 ⎤
⎢
⎥−⎢
⎥
⎢⎣ 92 ⎥⎦ − ⎢⎣108 ⎥⎦

γ
γ
⎣ d (min) ⎦ ⎣ d (max) ⎦

CRITICAL THINKING PROBLEMS

3.C.1 a.

e=

V
Vv
; e1 + 1 = 1
Vs
Vs

1 + 0.92 =

V1
; V1 = 1.92Vs
Vs

1 + 0.65 =

V2
; V2 = 1.65Vs
Vs

ΔV V1 − V2 1.92 − 1.65
=

=
= 0.14 = 14% (decrease)
1.92
V
V1

b. γ d (1) =

γ d ( 2) =

Gs γ w
Gγ
Gγ
= s w = s w
1 + e1 1 + 0.92 1.92
Gs γ w
1.65
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1

1

γ d ( 2) − γ d (1) 1.65 − 1.92
=
=
= 0.163 = 16.3% (increase)

1
γ d (1)

Δγ d

γ d (1)

1.92

c. S1 =

wGs wGs
wGs
=
; S2 =
0.92
e1
0.65

1
1
−
ΔS S2 − S1 0.65 0.92
=
=
= 0.415 = 41.5% (increase)
1
S1
S1
0.92

3.C.2 a. Dr =

emax − e
emax − emin

e1 = emax − Dr (emax − emin ) = 0.92 − 0.47(0.92 − 0.53) = 0.736

γd =

Gs γ w (2.65)(9.81)
=
= 14.97 kN/m3 (before compaction)
1 + e1
1 + 0.736

e2 = emax − Dr (emax − emin ) = 0.92 − 0.8(0.92 − 0.53) = 0.608

γd =

b.

(2.65)(9.81)
= 16.17 kN/m3 (after compaction)
1 + 0.608

ΔH
Δe
0.736 − 0.608
=

=
= 0.074 ; ΔH = 0.074H = (0.074)(2) = 0.148 m
H 1 + e1
1 + 0.736
Final Height = 2 – 0.148 = 1.852 m

17
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18
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Chapter 4
4.1

a. Refer to the plot
of w versus N.
LL = 29.0
b.

PI = LL – PL
= 29.0 – 13.4
= 15.6

w − PL
32 − 13.4
=
= 1.19

LL − PL
15.6

4.2

LI =

4.3

a. From the plot,
LL = 23.6.
b.

PI = LL – PL
= 23.6 – 19.1
= 4.5

19
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w − PL 21 − 19.1
=
= 0.422
LL − PL
4.5

4.4

LI =

4.5

⎛ M − M2 ⎞
⎛V −Vf
⎟⎟(100) − ⎜⎜ i
SL = ⎜⎜ 1
⎝ M2 ⎠
⎝ M2

⎞
⎟⎟( ρ w )(100)
⎠

⎛ 34 − 24 ⎞
⎛ 20.2 − 14.3 ⎞
=⎜
⎟(100) − ⎜
⎟(1)(100) = 17.08
24
⎝ 24 ⎠
⎝
⎠

SR =

4.6

M2
24

=
= 1.68
Vf ρ w (14.3)(1)

⎛ M − M2 ⎞
⎛V −Vf
⎟⎟(100) − ⎜⎜ i
SL = ⎜⎜ 1
⎝ M2 ⎠
⎝ M2

⎞
⎟⎟( ρ w )(100)
⎠

⎛ 44.6 − 32.8 ⎞
⎛ 16.2 − 10.8 ⎞
=⎜
⎟(100) − ⎜
⎟(1)(100) = 19.51
32.8 ⎠
⎝ 32.8 ⎠
⎝

SR =

32.8
M2
=
= 3.03

Vf ρw (10.8)(1)

CRITICAL THINKING PROBLEMS

4.C.1 a. From Eq. (4.26): A =

PI
(% of clay - size fraction, by weight)

The computed PI values are provided in the table on the following page.

20

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Soil
Beauharnois
Detroit I
Horten
Gosport
Mexico City
Shellhaven
St. Thuribe

% clay
(< 0.002 mm
in size)
79
36

40
55
90
41
36

A

τf-undisturbed
(kN/m2)

St

PI

τf-remolded
(kN/m2)

0.52
0.36
0.42
0.89
4.5
1.33
0.33

18
17
41
29

46
36
38

14
2.5
17
2.2
5.3
7.6
150

41.08
12.96
16.8
48.95
405
54.53
11.88

1.3
6.9
2.4
13.0
8.7
4.8
0.3

b. From Table 4.2:
Beauharnois

A = 0.52; PI = 41.08; Mineral: Illite
Detroit I
A = 0.36; PI = 12.96; Mineral: Kaolinite
Horten
A = 0.42; PI = 16.8; Mineral: Kaolinite
Gosport
A = 0.89; PI = 48.95; Mineral: Illite
Mexico City
A = 4.5; PI = 405; Mineral: Montmorillonite
Shellhaven
A = 1.33; PI = 54.53; Mineral: Illite
St. Thuribe
A = 0.33; PI = 11.88; Mineral: Kaolinite
c. Sensitivity, S t =

τ f −undisturbed
τ f −remolded

τf-remolded is calculated using the above equation and listed in the table (Part a).

21
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