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Chapter 14
14.1 From Figure 14.4, for
°
=
′
40
φ
and
°
=
′
20δ
, K
p
≈ 9. So,
2
p
14.2 From Table 14.1, for
°
=
′
40
φ
and
°
=
′
20δ
, K
p
= 8.30
14.3 Eq. (14.18):
RKK
δpp )0( =
′
=
14.4 Table 14.2: For
°
=
′
40
φ
and
°
=
′
20δ
, K
p
= 8.91
14.5 Table 14.3: For
°
=
′
40
φ
and
°
=
′
20δ
, K
p
= 9.68
2
p
14.6 From Table 14.1, for
°
=
′
30
φ
and
°
=
′
20δ
, K
p
= 4.68
136
© 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.
lb/ft 71,285==
2
)16)(119)(68.4(
2
1
p
P
14.7 Eq. (14.18):
RKK
δpp )0( =
′
=
14.8 Eq. (14.19):
pp
KγHP
2
1
2
1
=
14.9 Eq. (14.19):
pp
KγHP
2
1
2
1
=
)(
=
φ
δpp
137
© 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.
kN/m 692
== )58.3()077.5)(15(
2
1
2
p
P
14.10 Eq. (14.21):
pp
KγHP
2
2
1
=
2
p
14.11 Given:
°
=
′
40
φ
,
φ
′
′
δ
, 0=
′
′
cc
a
, k
v
= k
h
= 0.2
kN/m 9,823≈++=
20cos
1
)6.818,15.435,475.976,2(
Location of
pe
P
:
138
© 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.
2
( )
1 1 1
(2976.75) 3,168 kN/m
cos 2 cos 20
pγ e
γH K
δ
= =
′
From Figure (14.8), this force acts at a distance
33.2373
==H
m above the
bottom of the wall.
2. The surcharge component of
pe
P
:
[
]
kN/m 720,4)5.4435(
20
cos
1
cos
1
)(
==
′
epq
qHK
δ
From Figure (14.8), this force acts at a distance
5.3272
==H
m above the
bottom of the wall.
3. The cohesion component of
pe
P
:
14.12
4.0
7
8.2 ==
a
n
.
°
=
′
30
φ
;
°
=
′
20δ
14.13
3.0
18
4.5 ≈=
a
n
; 1.0
)18)(125(
225 ==
′
γH
c
5.0
2
γH
139
P
a
= (0.140)(0.5)(125)(18)
2
≈ 2835 lb/ft
2
2
38
2
45tan65.014.14
−=
′
−=
φ
σ
γH
a
14.15 a.
486.2
75
)12)(9.17( <==
c
γH
Use Figure (14.15c) to determine the earth pressure envelope.
b. To determine the strut loads, refer to the following diagram.
2
3
141
© 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.
Therefore, A = 150.5 kN/m
Also, sum of vertical forces,
0
=
∑
V
)3)(5.64()3)(5.64)(5.0(
1
+
=
+
BA
or
25.2905.150
1
=
+
B
Therefore,
kN/m 75.139
1
=
B
Due to symmetry,
2
139.75 kN/m
B
=
and C = 150.5 kN/m
