Civil Engineering Reference
In-Depth Information
This is not the factor of safety used in slope stability, which is:
′
+
′
Shear strength
Disturbing shear
c
σ
tan
φ
F
=
i e
. .
F
=
τ
This safety factor applies equally to cohesion and to friction. F can be found by succes-
sive approximations:
′
+
′
c
F
σ
tan
φ
F
=
F
so try F
=
1.1:
tan
′
=
φ
0 364
1 1
.
=
0 33
.
=
tangent of angle of
18 3
.
°
F
.
Use this value of
φ
′
to establish a new N value from the charts:
N
=
0 019
.
⇒
c
=
0 019 18 11 3 76
.
× × =
.
kPa
m
5
3 76
′
=
⇒
F for c
(
)
=
1 33
.
.
Try F
=
1.2:
tan
′
=
φ
0 364
1 2
.
′
=
=
0 3
.
(
φ
16 9
.
°
)
F
.
5
4 75
From the charts N
=
0 024
.
⇒
F
c
=
=
1 05
.
′
.
Try F
=
1.15:
tan
φ
′
=
0 364
1 15
.
′
=
=
0 32
.
(
φ
17 6
.
°
)
F
.
5
4 36
From the charts N
=
0 022
.
⇒
F
=
=
1 15
.
(
=
F
)
c
′
φ
′
.
i.e. Factor of safety of slope
=
1.15
Example 13.7:
Taylor's charts (ii)
Slope
=
1 vertical to 4 horizontal, c
′
=
12.5 kPa, H
=
31 m,
φ
′
=
20°,
γ
=
16 kN/m
3
. Find
the F value of the slope.
Solution:
Angle of slope
=
14°, so obviously the slope is safe as it is less than the angle of shear-
ing resistance. With this case, N from the charts
=
0.
The procedure is identical with Example
13.6.
Try F
=
1.5:
tan
φ
0 364
1 5
.
′
=
=
=
0 24
.
(
φ
13 5
.
°
)
F
.
