Chapter 13
13.113.4
φ
φ
′
′
−=
sin
))(sin1( OCRK
o
2
1
H
13.5 Given: Q = 95 kN; H = 5 m;
4.02.0
5
1≤==m
13.6 Given: H = 5 m; q = 44 kN/m.
4.02.0
5
1≤==m
203.0
n
q
124
H
z
n=
H
q203.0
22
)16.0( n
n
+
h
σ
′
(kN/m
2
)
The lateral pressure distribution with depth is shown below:
Lateral force on the wall per unit length is calculated from the area of the pressure
diagram as follows:
13.7 Given: H = 5 m;
5.1
=
′
a
m;
75.0
=
′
b
m; q = 105 kN/m
2
5
1
H
125
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°=
+
=
′
+
′
=
−−
22.24
5
75.05.1
tantan
11
2
H
ba
θ
Eq. (13.24):
[ ] [ ]
kN/m 91.52=°−°=−= )53.822.24)(5(
90
105
)(
90
12
θθ
H
q
P
13.8 Eq. (13.27):
−
′
−−+−
−= )(2
3.57)()(
12
12
2
θθ
θθ
H
HaQRH
Hz
13.913.12
(
)
245tan
2
φ
′
−=
a
K
Problem
φ
′
(deg) K
a
γHK
aHza
=
′
=)(
σ
2
2
1γHKP
aa
=
3
H
z=
()(0.282)(119)(29)
2
13.13—13.16
(
)
245tan
2
φ
′
+=
p
K
Problem
φ
′
(deg)
K
p
γHK
aHza
=
′
=)(
σ
2
2
1
γHKP
pp
=
3
H
z=
(
2.282)(99
)(
18
)
(
)(
2.282
)(
99
)(
18
)
2
13.17
.39.0
2
26
45tan
2
45tan
22
=
−=
′
−=
φ
a
K
Refer to the figure.
Area No.
Area
127
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Resultant: Taking the moment about the bottom of the wall,
ft
6.47
=
+
+
+
+
=578,15
3
17
)8.9016(
3
17
)55.3330(
2
17
)68.2890(
3
4
17)08.340(
z
13.18
333.0
2
30
45tan
2
=
−=
a
K
z = 0 m:
a
σ
′
=
q
K
a
= (16)(0.333) = 5.32 kN/m
2
; u = 0
2
Area No.
Area
Location of resultant: Taking the moment about the bottom of the wall,
m 2.49=
+
+
+
++
3
5
)
62.122
(
3
5
)
42.37
(
2
5
)
95.64
(
3
3
5
)
48.19
(
2
8
)
56.42
(
128
13.19
198.0
2
42
45tan ;259.0
2
36
45tan
2
)2(
2
)1(
=
−==
−=
aa
KK
z = 0 m:
a
σ
′
=
q
K
a(1)
= (25)(0.259) = 6.475 kN/m
2
; u = 0
2
Area No.
Area
Location of resultant: Taking the moment about the bottom of the wall,
13.20
.56.2
2
26
45tan
2
=
+=
p
K
z = 0 ft:
0
=
′
p
σ
; u = 0
2
Area No.
Area
130
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Location and Direction of Resultant: At a distance of H/3 = 7/3 = 2.33 m
above the bottom of the wall inclined at an angle α = 12°
°°
° to the horizontal.
13.22 a. Use Eq. (13.40): For α = 12°
and
°
=
′
28
φ
,
)28(cos)12(cos12cos
22
−+
13.23 a. H = 8.5 m; c
u
= 22 kN/m
2
;
γ
= 19.6 kN/m
2
;
φ
= 0
uaa
cγzK 2 ;1
45tan
2
−=
′
=
−=
σ
φ
b. Eq. (13.65):
m 2.24=== 6.19
)22)(2(2
γ
c
z
u
o
c. Eq. (13.67):
13.24 a.
aaoa
KcK 2−=
σσ
1 ;
=
+
=
ao
Kqγz
σ
132
c. Referring to the diagram in Part a,
3
13.25
.649.0 ;421.0
2
24
45tan
2
45tan
22
==
−=
′
−=
aa
KK
φ
13.26 Use Eqs. (13.77) and (13.78).
α
= 0;
θ
= 10º;
φ
′= 38º;
γ
= 21.1 kN/m
3
; H = 8 m
Part
δ
′ K
a
2
2
1γHKP
aa
=
13.27 From Eq. (13.83):
avae
KkγHP ′′
−= )1(
2
1
2
′
′
φ
δ
h
v
133
From Table 13.10, 825.0=
′
′
a
K. Therefore,
2
CRITICAL THINKING PROBLEM
13.C.1 Using Eqs. (13.42) and (13.45), the following table can be prepared:
θ = 10°
γ
(kN/m
3
)
φ
′
)( Ra
K γK
Ra )(
134
The graph of
2
5.0 HP
a
versus backfill inclination angle
α
is shown below.
The above chart shows that for any backfill inclination
α
and any wall height H,