Civil Engineering Reference
In-Depth Information
Σ
M
O
=
0
M
M
M
M
0
+
−
−
=
Pp
1
Pp
2
Pa
1
Pa
2
(
d d
−
)(
d d
d d
2
+
+
16 5
.
)
[
]
×
0
O
3
2
2
13 5
.
d
+
40 5
.
(
d
−
d
)
+
445 5
.
(
d d
−
)
0
0
0
3
(
+
0
+
11
)
(
)
−
1 5
.
×
d
2
−
d d d
2
(
)(
2
d d
+
)
0
0
0
3
−
1 5
.
(
d
+
5 5
.
)
−
=
0
(1)
0
(
d d )
+
0
Σ
H
=
0
i e P
. .
P
P
P
0
+
−
−
=
p d
1
;
a d
2
;
a d
1
;
p d
2
;
2
2
2
2
2
2
40 5
.
d
+
4 5
.
(
d
−
d
)
−
4 5
.
(
d
+
5 5
.
)
−
[
40 5
.
(
d
−
d
)
+
445 5
.
(
d d
−
]
=
0
)
(2)
0
0
0
0
0
d
=
.
5 0
m
0
d
=
5 3
.
m
2.
Combination 2 (partial factor sets A2
+
M2
+
R1)
The calculations are the same as for Combination 1 except that this time the following partial
factors are used:
γ
G; unfav
=
1.0;
γ
φ
′
=
1 2.
.
The following expressions are then derived (K
a
=
0.409; K
p
=
2.444):
Σ
M
O
=
0
M
+
M
−
M
−
M
=
0
Pp
1
Pp
2
Pa
1
Pa
2
(
d d
−
)(
d d
d d
2
+
+
16 5
.
)
[
]
×
0
O
8 13
.
d
3
+
24 4
.
(
d
2
−
d
2
)
+
268 4
.
(
d d
−
)
0
0
0
3
(
+
0
+
11
)
(
2
2
)
−
1 36
.
×
d
−
d d d
(
)(
2
d d
+
)
0
0
0
−
1 36
.
(
d
+
5 5
.
)
3
−
=
0
(1)
0
(
d d
0
+
)
Σ
H
=
0
i e P
. .
P
P
P
0
+
−
−
=
p d
1
;
a d
2
;
a d
1
;
p d
2
;
24 4
.
d
2
+
4 09
.
(
d
2
−
d
2
)
−
4 09
.
(
d
+
5 5
.
)
2
−
[
24 4
.
(
d
2
−
d
2
)
+
268 4
.
(
d d
−
0
)]
=
0
(2)
0
0
0
0
Using
Example 8.3.xls
:
d
=
.
6 4
m
0
d
=
7 2
.
m
(b) Gross pressure method
In the gross pressure method, the net passive resistance below the point of rotation is replaced
by the horizontal force R, as shown in Fig.
8.14.
Using Rankine's theory (with
φ
′
=
30°)
K
a
=
3
; K
p
=
3.0.
Force (kN)
Lever arm (m)
Moment (kN m)
(
5
+
d
)
20
2 3
5
10
9
0
P
a
(
+
d
)
2
(
5
+
d
)
3
0
0
×
3
d
0
3
P
p
15 d
5 d
