Engineering Rock Mechanics

280 Rock reinforcement and rock support

If we use the usual geomechanics convention of compressive stresses
and contractile strains as positive, then we can say that the relation
between and p is

where the spring stiffness, k, has units of stress. Note that the minus sign
is required to generate a compressive support pressure from an extensile
horizontal strain in the pillar.
Combining these last two equations gives
1 1
kE
--p = - [p (1 - v*) - n;v (1 + u)]

which, upon rearranging, gives

E

v(lf”):].

The case of zero horizontal strain of the pillar is represented in this

equation by an infinite spring stiffness. This condition reduces the above

equation to

Bz=p[-]

from which we see that the ratio of vertical stress to support pressure is

(1 - u)/u, as required.

@) The Hoek-Brown criterion can be written as

a1 = a3 +,/a

which, upon substitution of the numerical values given here and assum-

ing 81 = a, and a3 = p, becomes

0, = 55 = p + 4555~ + 1369.

Rearranging this to give a quadratic in p leads to

p2 - 665p + 1656 = 0

from which we find p = 2.5 MPa.

If we rearrange the equation for the pillar stiffness, we can obtain

and hence by substitution find k = 1 1.3 GPa.

To determine the area of steel bars required to provide this stiffness,

we examine the force compatibility for a unit area of pillar face. The

relation between strain and stress in the support is B = ks, and so the

force required over a unit area of face of a pillar of unit width is F = k6,

where 6 is the displacement of the support. For a bar of unit length

subjected to an axid force, the displacement is given by ti = Fb/AbEb.

Combining these equations gives