Solution manual for fundamentals of geotechnical engineering 4th edition braja m. das

1
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 2
2.1 a. Sieve
No.
Mass of soil retained
on each sieve (g)
Percent retained
on each sieve
Percent
finer
4
10
20
40
60
100
200
Pan
0.0
18.5
53.2
90.5
81.8
92.2
58.5
26.5
0.0
4.4
12.6
21.5
19.4
21.9
13.9
6.3
100.0
95.6
83.0
61.5
42.1
20.2
6.3
0
Σ421.2 g
The grain-size distribution is shown.
b. From the graph, D60 = 0.4 mm; D30 = 0.22 mm; D10 = 0.12 mm
c. 3.33===
12.0
4.0
10
60
D
D
Cu
d. 1.01===
)12.0)(4.0(
)22.0(
))((
)( 2
6010
2
30
DD
D
Cc
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2
2.2 5.13===
08.0
41.0
10
60
D
D
Cu
1.48===
)41.0)(08.0(
)22.0(
))((
)( 2
6010
2
30
DD
D
Cc
2.3 7.54===
24.0
81.1
10
60
D
D
Cu
1.55===
)81.1)(24.0(
)82.0(
))((
)( 2
6010
2
30
DD
D
Cc
2.4 a. Sieve
No.
on each sieve (g)
Percent retained
on each sieve
Percent
finer
4
6
10
20
40
60
100
200
Pan
0
0
0
9.1
249.4
179.8
22.7
15.5
23.5
0
0
0
1.82
49.88
35.96
4.54
3.10
4.70
100
100
100
98.18
48.3
12.34
7.8
4.7
0
Σ500 g

3
b. From the graph, D60 = 0.48 mm; D30 = 0.33 mm; D10 = 0.23 mm.
c. 2.09==
23.0
48.0
uC
d. 0.99==
)23.0)(48.0(
)33.0( 2
cC
2.5 a. Sieve
No.
on each sieve (g)
Percent retained
on each sieve
Percent
finer
4
10
20
40
60
80
100
200
Pan
0
44
56
82
51
106
92
85
35
0
7.99
10.16
14.88
9.26
19.24
16.70
15.43
5.34
100
92.01
81.85
66.97
57.71
38.47
21.77
6.34
0
Σ 551 g
b. From the graph, D60 = 0.3 mm; D30 = 0.17 mm; D10 = 0.11 mm
c. 2.73==
11.0
3.0
uC

4
d. 0.88==
)3.0)(11.0(
)17.0( 2
cC
2.6 The grain-size distribution is shown.
From the graph, percent passing 2 mm = 100%; percent passing 0.06 mm = 58%;
percent passing 0.002 mm = 23%. See Table 2.3, so
Gravel: 0%
Sand: 100 – 58 = 42%
Silt: 58 – 23 = 35%
Clay: 23 – 0 = 23%
2.7 Refer to the graph for Problem 2.6. From the graph, percent passing 2 mm =
100%; percent passing 0.05 mm = 54%; percent passing 0.002 mm = 23%. See
Table 2.3, so
Gravel: 0%
Sand: 100 – 54 = 46%
Silt: 54 – 23 = 31%
Clay: 23 – 0 = 23%
2.8 Refer to the graph for Problem 2.6 and Table 2.3. From the graph, percent
passing 2 mm = 100%; percent passing 0.075 mm = 62%; percent passing 0.002
mm = 23%.
Gravel: 0%
Sand: 100 – 62 = 38%
Silt: 62 – 23 = 39%
Clay: 23 – 0 = 23%

5
2.9 Gs = 2.60; temperature = 24°; hydrometer reading = 43; time = 60 min.
Referring to Table 2.7, L = 9.2 cm.
Eq. (2.5): D (mm) =
(min)
cm)(
t
L
K
From Table 2.6, for Gs = 2.60 and temperature = 24°, K = 0.0132.
D = 0.0132
60
2.9
= 0.0052 mm
2.10 For Gs = 2.70 and temperature = 23°, K = 0.013 (Table 2.6), L = 12.2 (Table 2.7).
mm0.0041===
120
2.12
013.0
(min)
(cm)
(mm)
t
L
KD

6

7
Chapter 3
3.1 a. 14%=⎟
⎠
⎞
⎜
⎝
⎛ −
= )100(
9.623
9.6232.711
w
b. 3
kg/m1778===ρ
4.0
2.711
V
M
c. 3
kg/m1559.75==ρ
4.0
9.623
d
d.
e
G ws
d
+
ρ
=ρ
1
0.718=−=−
ρ
ρ
= 1
75.1559
)1000)(68.2(
1
d
wsG
e
e. 0.418=
+
=
+
=
718.01
718.0
1 e
e
n
3.2 Moisture content: 15.6%=⎟
⎠
⎞
⎜
⎝
⎛
×
×−×
== −
−−
)100(
106.153
106.153106.177
3
33
s
w
W
W
w
Moist unit weight: 3
kN/m18.99=
×
×
==γ −
−
3
3
1035.9
106.177
V
W
Dry unit weight: 3
kN/m16.43=
+
=
+
γ
=γ
100
6.15
1
99.18
1 w
d
Void ratio: 0.59=−=−
γ
γ
= 1
43.16
)81.9)(67.2(
1
d
wsG
e
Porosity: 0.37=
+
=
+
=
59.01
59.0
1 e
e
n
Degree of saturation: 70.6%=⎥⎦
⎤
⎢⎣
⎡== )100(
59.0
)67.2)(156.0(
e
wG
S s

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