Engineering Rock Mechanics

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.