Questions and answers: underground excavation instability mechanisms 357
the direction 335/64). We are told that the strength of the fractures is
purely frictional, with 4 = 30". If we furthermore assume that there is
zero water pressure, then the stability problem reduces to that of a block
sliding on a plane, and is independent of the size and shape of the
tetrahedral wedge. The only controlling factors in this type of problem
are the friction angle and the dip angle of the fracture, and the factor of
safety is therefore given by
tan 4 tan 30 0.577
tan 64 - 2.050
- -- - 0.28.
- tan +
In the north west wall of the excavation, block 125 exhibits sliding along
the line of intersection of fracture sets 1 and 2. As with the case above,
the wedge stability problem reduces to that presented earlier for rock
slopes in which we used the concept of the wedge factor to determine
the factor of safety.
As with rock slopes, the orientations of the excavation surface and
the third plane making up the tetrahedral wedge (fracture set 4 in this
case) do not enter into the calculation, other than to ensure the kinematic
feasibility of the wedge (and this, of course, was checked in A19.5). The
geometry of the wedge and the angles required for the wedge factor are
shown in the lower hemispherical projection below. The factor of safety
is then given by
sinB tan4 sin80 tan30 0.985 0.577 - sinit tan+ sin35 tan34 0.574 0.675
Fw = k, x F - - x - =-x---x-- - - 1.47.
If the factor of safety is computed analytically, its value is found to
be 1.50. The discrepancy between this and the value given above is
due to the limited accuracy of the hemispherical projection. However,
as the error is only some 2%, we can see that hemispherical projection
techniques are a practical means of making preliminary assessments of
wedge sliding stability.