Page 235 - Rock Mechanics For Underground Mining

SUPPORT AND REINFORCEMENT OF MASSIVE ROCK

Failure domains determined in the manner discussed are not described accurately.
This can be appreciated from the observation that any rock failure causes a change
in the geometry of the elastic domain, which can cause further failure. That is, the
problem is a non-linear one, and the solution procedure suggested here examines only
the initial, linear component of the problem. For mining engineering purposes, the
suggested procedure is usually adequate.

7.6 Support and reinforcement of massive rock

Mining activity frequently takes place under conditions sufficient to induce exten-
sive failure around mine access and production openings. An understanding of the
mechanics of the techniques exploited to control the performance of fractured rock
is therefore basic to effective mining practice under these conditions. In this section,
some basic principles of rock support and reinforcement are introduced. This dis-
cussion is extended and a discussion of practical methods and procedures is given in
Chapter 11.
In evaluating the possible mechanical roles of a support system, it is instructive to
consider the effect of support on the elastic stress distribution in the rock medium.
One function of support is taken to be the application or development of a support
load at the excavation surface. It is assumed initially that this is uniformly distributed
over the boundary.
Figure 7.19a shows an elliptical opening with width/height ratio of 4, in a stress
field with vertical stress 20 MPa and horizontal stress 8 MPa. Boundary stresses at
points A and B in the sidewall and crown of the excavation may be calculated directly
from equations 7.6 and 7.7, i.e.

A = 172.0MPa, B =−8.0MPa

If a set of vertical supports is installed, sufficient to generate a vertical load of
1 MNm −2 uniformly distributed over the excavation surface, the boundary stresses
around the supported excavation can be determined from the superposition scheme
shown in Figure 7.19b. Thus

A1 = A2 + A3
8

= 1 + 19 1 − + 8
19
= 161.0MPa
B1 = B2 + B3
8 2 × 8 1

= 0 + 19 − 1 + ×
19 19 4
=−7.0MPa
From these results, it is concluded that support pressure does not modify the elastic
stress distribution around an underground opening significantly. If failure of the rock
mass is possible in the absence of support, installation of support is unlikely to modify
the stress distribution sufficiently to preclude development of failure. It is therefore
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