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Chapter 13
13.1 Refer to the diagram.
. From Table 12.3, Ka = 0.321.
Chapter 13
Chapter 13
13.2
26.0
2
45tan 1
2
a
K
H = 7.3 m
Chapter 13
(0.5)(0.3)(6.5)(c) = 22.99
m–kN 304.78
3
3.7
)25.125(
O
M
2.47 78.304
52.753
FS ng)(overturni
1.06
25.125
)30(
3
2
)4.3(15
3
2
tan)15.368(
FS(sliding)
m 0.481
15.368
78.30452.753
2
4.3
e
2
toe kN/m 200.19
4.3
)481.0)(6(
1
4.3
15.368
q
B=B – 2e = 3.4 – (2)(0.481) = 2.438 m
idqiqdqcicdcu FFNBFFNqFFNcq
22 2
1
181.1
438.2
5.1
294.01
qd
F
243.1
15tan98.10
181.11
181.1
tan
1
2
c
qd
qdcd N
F
FF
79.18
15.368
25.125
tan 1
626.0
90
79.18
12
qici FF
13.4
From Table 12.6, for
, = 0, = 71.57, Ka = 0.45, = 21.33.
Chapter 13
Refer to sections in the figure shown for Problem 13.3.
m–kN 299.13
3
8.6
)97.131(
3
H
PM vO
6.2 13.299
06.1859
FS ng)(overturni
2.35
97.131
)85.5)(40(
3
2
)66.14tan()588(
3
2
3
2
tan
FS 22
(sliding) h
P
BcV
13.5 b = 2 m; a = 1.5 m; q = 70 kN/m2
† Eq. (13.29); †† Eq. (13.30)
13.6 Eq. (13.33):
.686.0
)8)(14.0(
)2)(4.0(
4.1
14.0
4.0
4.1
H
b
M
So use M = 1.0.
Chapter 13
zz
z
b
z
ab
z
ab
2
tan
5.3
tan
2
1
tantan
2
1
2
tan
11
111
Ka = tan2(45 – 17.5) = 0.271; 1 = 17 kN/m3; q = 70 kN/m2
13.7 a.
2827.0
2
34
45tan
2
45tan 2
1
2
a
K
13.8 a. Check for overturning:
kN/m 226.16)2827.0()10)(16(
2
1
2
122
1 aa KHP
Chapter 13
13.9
= 30.
333.0
2
30
45tan2
a
K
.
)30(
3
2
tan4
tan4
F
l
13.10 Check for overturning:
kN/m 95.3)333.0()6)(9.15(
2
12
a
P
Chapter 13
)20)(tan7.3)(6)(9.15(
3
2
tan
HL
Chapter 13