268 Rock reinforcement and rock support
terms of the angle, fl, at which the bolt is installed relative to the
plane, the angle of inclination of the plane and the Mohr-Coulomb
friction properties, c and 4, of the block/plane interface. Hence, by
differentiation, find the rockbolt angle at which the bolt tension is
minimized.
What is this angle, p, such that the
tension in the rockbolt is
minimized?A 16.2 The force, F,, acting normal to the plane is the sum of the resolved
blockweight and rockbolt tensioncomponents, F,, = W cos a!+T sin #I. The
force, F,, acting along the plane is similarly given by F, = W sin a! - T cos 6.
The limiting condition of the block/plane interface having reached
peak strength gives
Wsina!-Tcos/3=cA+(Wcosa+TsinB)tan@where c is the cohesion and A is the basal area of the block.
Rearranging gives the required answer of
T=W(sina - cosa! tan@) - cA
(cos /3 + sin B tan @)
To find the minimal rockbolt tension, we set the partial derivative
of T with respect to /3 to zero and solve the resulting equation for
B. Recognizing that the expression for T will be minimized when the
denominator is maximized, we have
a
-(cosB+sin/3tan@) =0, or -sinB+cos/3tan@=O.
aB
Rearranging gives tan @ = sin /3/ cos B = tan /3, from which /3 = @.
Thus, the rockbolt tension is minimized when the rockbolt is in-
stalled at an angle to the slope which is the same as the angle of
friction between the rock block and the slope. For a plane with a
very low angle of friction, i.e., @ M Oo, the rock block will hang from
the rockbolt, which should be installed sub-parallel to the slope. For
a very rough interlocking plane, i.e. 4 M 90°, the block is literally
bolted down to the plane, with the rockbolts being perpendicular to the
plane.
416.3 The sketch below shows a rock block resting on a slope and
subject to a stabilizing force from a rockbolt. The strength of the
interface between the block and the slope is purely frictional, with
an angle of friction 4.
Note that c is the cohesion per unit area, with units of stress.