Civil Engineering Reference
In-Depth Information
(a)
5.8
°
56.6
°
4
°
30
°
16
15
14
13
12
11
10
9
8
7
S
n
6
5
4
R
n
3
2
1
T
(
b
)
y
n
/
∆
x
M
n
L
n
P
n·t
P
n·s
P
n
R
n
S
n
S
n
/
R
n
n
y
n
Mode
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
4.0
10.0
16.0
22.0
28.0
34.0
40.0
36.0
32.0
28.0
24.0
20.0
16.0
12.0
8.0
4.0
0.4
1.0
1.6
2.2
2.8
3.4
4.0
3.6
3.2
2.8
2.4
2.0
1.6
1.2
0.8
0.4
0
0
0
0
292.5
825.7
1556.0
2826.7
3922.1
4594.8
4837.0
4637.5
3978.1
2825.6
1103.1
-1485.1
0
0
0
0
-2588.7
-3003.2
-3175.0
-3150.8
-1409.4
156.8
1300.1
2013.0
2284.1
2095.4
1413.5
472.2
0
0
0
0
292.5
825.7
1556.0
2826.7
3922.1
4594.8
4837.0
4637.5
3978.1
2825.6
1413.5
472.2
866
2165
3463
4533.4
5643.3
6787.6
7662.1
6933.8
6399.8
5872.0
5352.9
4848.1
4369.4
3707.3
2471.4
1237.1
500
1250
2000
2457.5
2966.8
3520.0
3729.3
3404.6
3327.3
3257.8
3199.5
3159.4
3152.5
2912.1
1941.3
971.8
0.577
0.577
0.577
0.542
0.526
0.519
0.487
0.491
0.520
0.555
0.598
0.652
0.722
0.7855
0.7855
0.7855
STABLE
________
17
23
29
35
36
32
28
24
20
16
12
8
4
22
28
34
35
31
27
23
19
15
11
7
3
-
T
O
P
P
L
I
N
G
________
SLIDING
(c)
Block
123456789 0 1 2 3 4 5 6
0
1
2
S
n
3
4
5
6
R
n
7
8
Figure 9.10
Limited equilibrium analysis of a toppling slope: (a) slope geometry; (b) table listing block
dimensions, calculated forces and stability mode; (c) distribution of normal
(R)
and shear
(S)
forces on base of
blocks (Goodman and Bray, 1976).
25 kN
/
m
3
. It is assumed that the slope is dry, and
that there are no external forces acting.
The stability analysis is started by examining
the toppling/sliding mode of each block, starting
at the crest. Since the friction angle on the base
of the blocks is 38.15
◦
and the dip of the base is
30
◦
, the upper blocks are stable against sliding.
Equation (9.2) is then used to assess the toppling
mode. Since cot
ψ
p
1.73, blocks 16, 15 and
14 are stable, because for each the ratio
y
n
/x
is
=
