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63
Chapter 8
8.1 The flow net is shown.
k = 4
×
10
‒4
cm/sec
8.2 The flow net is shown.
/m/daym 0.518
3
=
×=
−
/m/secm 106
5
10
36
8.3 Based on the notations in Figure 8.12,
10
2
64
8.4 The flow net is shown.
8.5 Refer to the flow net given in Problem 8.4 and the figure below. The flow net has
12 potential drops. Also, H = 10 m. So the head loss for each drop = (10/12) m.
Thus,
65
© 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.
The pressure heads calculated are shown in the figure. The hydraulic uplift force
per unit length of the structure can now be calculated to be
(area of the pressure head diagram)(1)
11.67 10.84 10.84 8.754
(1.67) (1.67)
(18.32) (1.67) (1.67)
2 2 2
(9.81)(18.8 16.36 122.16 8.
w
γ
=
+ +
+
+ + +
= + + + 71 9.05)+
=1717.5 kN / m
8.6 The flow net is shown. N
f
= 3; N
d
= 5.
8.7 For this case, T
' = 8 m; S = 4 m; H = H
1
– H
2
= 6 m; B = 8 m; b = B/2 = 4 m.
4==
S; x = b – x′ = 4 – 1 = 3 m; 5.0
4
3==
b
x
66
/m/daym 2.07
≈
10
)4.0(
q
8.8
/m/daym 0.271
3
≈×=
−
/sec/mm 10139.3
36
8.9 From Problem 8.8, d = 24.2 m; H = 7 m;
α
2
= 40º
67
/m/daym 0.291
3
≈×=
×
==
−
−
/sec/mm 1037.3
)40)(sin72.2(
10
103
sin
36
2
2
4
2
2
α
kLq
8.10 For this case,
°=×==
−
57.26
2
1
tan
1
αβ
8.11 From Problem 8.10, d = 90 m, H = 25 m,
m, 3.6
25
90 ==
H
d
β
= 26.57°
© 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.
8.12 Eq. (8.34):
2
1
2
2
cotcot hH
B
H
B
h
dd
−
+−+=
αα
(a)
60
69
CRITICAL THINKING PROBLEM
8.C.1 Since ,
zx kk ≠the soil is anisotropic with respect to hydraulic conductivity.
Hence, a transformed section has to be plotted (see Section 8.5).
104
5
×
−
x
k
The following figure shows the transformed section.
70
32
1.51
1.51
22
2
2
Hdd
