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113
Chapter 12
12.1 Ko = (1 – sin 30)(2)sin30 = 0.707
1
3
12.2 Ko = 1 – sin 34 = 0.44
At z = 0 m:
ooh K
(0.44)(20) = 8.8 kN/m2; u = 0
114
12.3 a.
2
45tan2
a
K
. = 0; Ka = 1.
12.4 Eq. (12.12):
aa
a
aKcHK
K
c
HP
2
2
2
1
115
39.0
2
26
45tan
2
45tan 22
a
K
12.5
283.0
2
34
45tan2
a
K
116
kN/m 118.6
15.4407.1129.4377.832.11
)43.29)(3(
2
1
)43.1481.21)(3(
2
1
)3)(43.14()66.543.14)(2(
2
1
32.11
54321 AAAAAPa
m 1.67
6.118
)1)(15.44()1)(07.11()5.1)(29.43()67.3)(77.8()4)(32.11(
z
12.6 a. = 36; = 10; Ka = 0.2696 (Table 12.3);
aa Kz
12.7 Eq. (12.24):
m 09.2
25sin1
25sin1
18
)12)(2(
sin1
sin12
c
zc
117
12.8 a. Eq. (12.25):
)(
2
12
HKP aa
2
20
12.9 = 36;
.5.0
Table 12.7: for = 0 and = 85, Ka = 0.2718.
12.10 Eq. (12.34):
)]([
90 12 H
q
P
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118
1.37
3.3
5.2
tantan 11
2H
ba
kN/m 11.64 )]4.241.37)(3.3[(
90
25
P
12.11 Eq. (12.42):
aevae KkHP )1(
2
12
119
12.12 Eq. (12.53):
1.0098.0
)10)(115(
113
*
H
c
c
12.13 a.
2
45tan2
p
K
; = 0; Kp = 1
120
12.14 = 17.5 and = 35;
5.0
12.15 Eq. (12.72):
cos
1
2
1)(
2eppe KHP
