Civil Engineering Reference
In-Depth Information
0
.01
.02
.03
.04
2.0
.05
.06
.07
.08
.09
.10
1.8
.11
.12
.13
.14
.1
.1
.1
.1
.1
.20
1.6
c
H
tan
1.4
1.2
.25
90
°
tan
FS
.30
1.0
.35
.40
.45
.50
.60
.70
.80
.90
1.0
1.5
2.0
4.0
8
0.8
Slope angle
80
°
70
0.6
°
60
°
50
°
40
°
0.4
0.2
0
0
.02
.04
.06
.08
.10
.12
.14
.16
.18
.20
.22
.24
.26
.28
.30
.32
.34
c
H
FS
Figure 8.9
Circular failure chart number 4—ground water condition 4 (Figure 8.4).
From the information given in Figure 8.14, the
value of the ratio
c/(γ H
tan
φ)
was required to check whether the cut would be
stable. The slope was in weathered and altered
material, and failure, if it occurred, would be a
circular type. Insufficient time was available for
ground water levels to be accurately established
or for shear tests to be carried out. The stability
analysis was carried out as follows:
For the condition of limiting equilibrium, FS
0.0056, and
the corresponding value of tan
φ/
FS from chart
number 2, is 0.76. Hence, the factor of safety of
the slope is 1.01. A number of trial calculations
using Janbu's method (Janbu
et al
., 1956) for the
critical slide circle shown in Figure 8.14, found
a factor of safety of 1.03.
These factors of safety indicated that the stabil-
ity of the slope was inadequate under the assumed
conditions, and steps were taken to deal with the
problem.
=
=
1 and tan
φ/
FS
tan
φ
. By reversing the pro-
cedure outlined in Figure 8.5, a range of friction
angles were used to find the values of the ratio
c/(γ H
tan
φ)
for a face angle of 42
◦
. The value of
the cohesion
c
which is mobilized at failure, for a
given friction angle, can then be calculated. This
analysis was carried out for dry slopes using chart
number 1 (line
B
, Figure 8.15), and for saturated
slopes using chart number 5 (line
A
, Figure 8.15).
=
8.5.2 Example 2—highway slope
A highway plan called for a cut at an angle of 42
◦
.
The total height of the cut would be 61 m and it
