Civil Engineering Reference
In-Depth Information
7.6.3 Example of comprehensive wedge
analysis
Table 7.5
Orientation of planes forming wedge
Plane
Dip Dip
Shear strength
The following is an example of the comprehensive
stability analysis of a wedge (Rocscience, 2001).
Consider the wedge formed by joint sets 3 and
5 in Figure 7.16. This has a friction-only factor
of safety of 1.0, so water pressures and seismic
ground motion may result in instability, depend-
ing on the cohesions on the slide surfaces. This
analysis shows the bolting force required to raise
the factor of safety to 1.5, based on the input
parameters shown in Table 7.5.
The other input parameters are as follows:
Wedge height,
H
1
direction
30
◦
,
c
3
=
3
60
360
φ
3
=
50 kPa
30
◦
,
c
5
=
5
54
118
φ
5
=
50 kPa
Slope face
76
060
Upper
surface
15
070
Tension
crack
80
060
Case (i) corresponds to the simplified analysis
using the friction-only charts for a dry, static
slope that gave a factor of safety of 1.1. With
the tension crack half-filled with water and the
seismic coefficient applied, the factor of safety
drops to 0.76 (Case (iii)). For the load condi-
tions in Case (iii), it is necessary to install a
bolting force of 27.2 MN in order to raise the
factor of safety to 1.5 (Case (iv)). In Case (iv),
the bolting force is installed in the opposite dir-
ection to the line of intersection
(α
T
=
=
28 m;
Distance of tension
crack from crest
measured along
plane 3,
L
=
9m;
26 kN
/
m
3
;
Rock density,
γ
r
=
10.0 kN
/
m
3
; and
Water density,
γ
w
=
Seismic coefficient
(horizontal),
k
H
=
0.1.
242
◦
)
and
The stability analysis of the wedge using these
parameters gave the following results:
at an angle of 8
◦
above the horizontal
(ψ
T
=
8
◦
)
—this orientation is optimal as defined by
equations (7.15) and (7.16). If the bolts are
installed at 10
◦
below the horizontal to facilit-
ate grouting
(ψ
T
−
(i) Dry, static,
c
3
=
c
5
=
0 kPa,
FS = 1.05
0,
(ii)
z
w
=
T
=
10
◦
)
and with a trend of
200
◦
, then the required bolting force to produce
a factor of safety of 1.5 increases to 38.3 MN
(Case (v)). This result illustrates the advantage
of installing anchors at, or close to, the optimal
orientation.
Case (vi) illustrates that a small amount of
cohesion can be most effective in improving
the factor of safety because the cohesion acts
over large surface areas to produce a significant
resisting force.
50%, static,
FS
=
0.93
=
c
3
=
c
5
=
0 kPa,
T
=
0
(iii)
z
w
=
50%,
k
H
=
0.1,
FS
=
0.76
c
3
=
c
5
=
0 kPa,
T
=
0
8
◦
,
(iv)
T
=
27.2 MN,
ψ
T
=−
S
=
1.5
242
◦
,
z
w
=
α
T
=
50%,
k
H
=
0.1,
c
3
=
c
5
=
0 kPa
10
◦
,
=
38.3 MN,
ψ
T
=
=
(v)
T
S
1.5
200
◦
α
T
=
(vi)
z
w
=
50%,
k
H
=
0.1,
FS
=
1.49
c
3
=
c
5
=
50 kPa,
T
=
0
