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Chapter 15
15.1 Eq. (15.15):
φ
tan
′
′
c
15.2 Eq. (15.16):
15.3
γ
′
= 20.8 – 9.81 = 10.99 kN/m
3
15.4
3
sat
lb/ft 121.33
8
.
0
1
)4.62)(8.07.2(
1
)( =
+
+
=
+
+
=
e
γeG
γ
ws
144
© 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.
1.946=+=
+=
′
′
+
′
=
65
.
0
296
.
1
16tan
21tan
33.121
93.58
)16)(tan16)(cos30)(33.121(
1250
tan
tan
tancos
2
sat
2
sat
β
φ
ββ
γ
γ
Hγ
c
F
s
15.5 Eq. (15.15):
15.6 Eq. (15.15):
β
φ
ββ
tan
tan
tancos
2
′
+
′
=Hγ
c
F
s
+
=
H
413.4
25tan
23tan
)25)(tan25)(cos(
1000
)81.9)(2050(
34
35.2
2
15.7 Consider a 1-m length
of the wedge
ABC
Eq. (15.29):
15.8 Refer to the figure in the solution for Problem 15.7.
Eq. (15.29):
s
15.9 Refer to the solution for Problem 15.7.
Eq. (15.29):
146
© 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.
( ) ( )
28.38658cot32cot)22)(6)(5.0(cotcot
2
1
22
=−=−=
βθ
γHW kN
69.20432sin28.386
=
=
a
T
kN
58.32732cos28.386
=
=
a
N
kN
( )
ss
a
s
r
FF
NcCA
F
T
93.974
21tan)58.327()75(
32sin
61
tan
1
=
+
=
′
+
′
=
φ
Since
ar TT
=
,
69.204
93.974 =
s
F
Therefore,
4.76
≈
s
F
15.10 Eq. (15.42):
−−
′
−−
)1949cos(1
6.17
)cos(1
cr
φβ
γ
15.11
F
s
= 2.95;
23.14
95.2
42 ==
′
=
′
s
d
F
c
c
kN/m
2
;
°=
=
′
−
66.6
95.2
19tan
tan
1
d
φ
15.12
′
−−
′′
=)cos(1
cossin4
d
dd
γ
c
H
φβ
φβ
;
3
lb/ft 121=γ
147
15.13
ρ
= 1950 kg/m³;
3
kN/m 12.19
1000
)81.9)(1950( ==γ;
c′
= 40 kN/m²;
φ
′
= 23
°
−−
′
−−
12.19
)cos(1
d
γ
φβ
15.14 3
lb/ft 123=γ
;
c′
= 1150 lb/ft²;
φ
′
= 19
°
−−
′
−−
123
)cos(1
d
γ
φβ
15.15 Figure 15.12: For
β
= 60
°
> 53
°
,
m
≈
0.195.
)195.0)(17(
cr
mγ
15.16 Figure 15.12: For
β
= 60
°
> 53
°
,
m
≈
0.195
148
15.17 Eq. (15.50):
zacc
ozuzu
+
=
=
)0()(
15.18
°=
=
−
43.18
3
1
tan
1
β
. For
β
= 18.43
°
and
D
= 2.0,
m
≈ 0.158 (Figure 15.12).
)158.0)(122(
cr
15.20 a. 5.1
8
12 ==D;
sat
γ
= 19 kN/m
3
149
© 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.
Distance =
nH
= (1.0)(8) =
8 m
15.21 a.
°=
=
−
69.33
5.1
1
tan
1
β
15.22
°=
=
−
74.29
75.1
1
tan
1
β
;
φ
′
= 22
°
;
c′
= 25 kN/m
2
;
γ
= 19 kN/m
3
.
Refer to Table 15.3 and Figure 15.21.
d
φ
′
(deg)
d
F
φ
φ
φ
′
′
=
′
tan
tan
m
Hγmc
d
=
′
(kN/m
2
)
d
c
c
c
F′
′
=
′
150
The plot of
c
F
′
vs.
φ
′
F is
15.23
n′
= 1;
°=
=
−
45
1
1
tan
1
β
;
φ
′
= 14º;
c′
= 30 kN/m
2
;
γ
= 18 kN/m
3
;
H
= 12 m
d
φ
′
(deg)
d
F
φ
φ
φ
′
′
=
′
tan
tan
m
Hγmc
d
=
′
(kN/m
2
)
d
c
c
c
F′
′
=
′
The plot of
c
F
′
vs.
φ
′
F
is shown.
151
15.24 Refer to the figure.
80
With radius
R
= 139.47 ft, the trial surface circle has been drawn.
Now the following table can be prepared.
152
Slice Area of slices
(ft
2
)
Weight of slice
W
n
=
A
×
γ
(kip/ft)
α
n
(deg)
W
n
cos
α
n
(kip/ft)
W
n
sin
α
n
(kip/ft)
)37)(48( =
∑
′
∑
+
′
=
sin
tan)cos(
α
φ
α
θ
nn
nn
s
W
WcR
F
15.25 As in Problem 15.23, m 12
30
sin
6
sin
;sin ====
α
α
H
x
x
H
2sin 2
x
R
θ
=
Slice Area of slices
2
Weight of slice
W
n
=
A
×
γ
α
n
W
n
cos
α
n
W
n
sin
α
n
Slice
W
n
(kN/m)
α
n
(deg)
W
n
cos
α
n
(kN/m)
W
n
sin
α
n
(kN/m)
α
∆
L
(m)
Eq. (15.62):
∑
′
∑
+
′
∆
∑
=
sin
))(tancos())((
α
φ
α
α
n
s
W
WcL
F
15.27
φ
′
= 22.5º;
β
= 18.43º;
r
u
= 0.5
15.28
φ
′
= 15º;
β
= 26.56º;
r
u
= 0.5;
155
D
m′
n′
us
rnmF
′
−
′
=
15.29
β
= 30º;
φ
′
= 20º;
r
u
= 0.25;
γ
= 17 kN/m
3
;
c′
= 27 kN/m
2
;
H
= 25 m
F
s
(assumed)
c
′
d
φ
′
(deg)
s
F
φ
′
=tan
tan
d)(calculate
15.30
°=
=
−
56.26
2
1
tan
1
β
156
Figure 15.35:
F
s
(assumed)
FHγ
c
′
d
φ
′
s
F
φ
φ
′
′
=tan
tan
d)(calculate
15.31 a. Refer to Figure 15.36:
42
′
c
From Figure 15.37c and 15.38c,
157
b. In the above figure,
PO
is the radius
R
of the critical circle.
22
15.32 Michalowski’s solution (Figure 15.41)
