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Chapter 9
9.1
Point lb/ft
σ
u
σ
′
The plot is given below.
9.2
3
1)(layer
kN/m 15.52
)81.9)(69.2(
=
+
=
+
=
γG
γ
ws
d
9.3
Point
kN/m
73
b. The water table rises to the surface. Layer 1 is saturated.
( ) (9.81)(2.69 0.7)
w s
γ G e
Decrease in
σ
′: 163.68 ‒ 134.83 = 28.85 kN/m
c. Water level rises 3 m above ground. All layers are saturated.
9.5 a.
3
(sand)
kN/m 16.2
)81.9)(66.2(
=
+
=
+
=
γG
γ
ws
d
b. Let the height of rise be h. Portions of the top sand layer will be saturated.
74
9.6
G
γ
i
s
w
=
=
=
=1
68.1
168.2
1
cr
9.7
3
sat(clay)
kN/m 19.08
)68.2)(29.0(1
)81.9)(68.2)(29.01(
1
)1( =
+
=
+
=
s
ws
wG
γGw
γ
9.8 Consider the stability of point A in terms of heaving.
E i
cr
75
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Therefore, factor of safety =
== 86.58
24.80
A
A
u
1.36
9.9 Let the maximum permissible depth of cut be H.
9.10 Let the height of water inside the cut be h (see figure)
A
86.58
u
A
9.11 a.
6.0
5.2
5.1
2
=== H
h
i
2
2
3
76
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b.
11.1
49.01
166.2
1
1
cr
=
+
=
+
=
=e
G
γ
γ
i
s
Since i < i
cr
, no boiling
c.
5.2
1.11 ;
2
cr
h
H
h
ii ===
h = 2.77 m
9.12 a.
33.0
5.4
5.1
2
=== H
h
i
9.13
(sand)
(2.69)(62.4)
s w
d
G γ
γ
= = =
+ +
77
The plot is given.
9.14
3
(sand)
kN/m 17.95
)81.9)(69.2( =
+
=
d
γ
sat(clay)
+
Depth (m)
kN/m
σ
u
σ′
9.15 From Eq. (9.22),
)(
21
HHγC
γD
FS
wo
−
=
CRITICAL THINKING PROBLEM
9.C.1 a. The flow nets for both cases are given below:
79
b. Determination of
kq
:
From Eq. (8.21):
d
f
N
N
kHq =
Case 1: N
f
= 4, N
d
= 11, H = 3.5 – 1 = 2.5 m
Case 1: Refer to the flow net and Eq. (9.24a):
643.0
)353.0)(11(
5.2
exit
=== lN
H
i
d
d. From Eq. (9.18), seepage force per unit volume is iγ
w
80
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
143.0
115.2 ==
=
H
i
3
kN/m 1.4
== )143.0)(81.9( force, Seepage iγ
w
Case 2: Refer to the flow net. At A,
136.0
135.2 ==
=
H
i