Engineering Rock Mechanics

Question and answers: foundation and slope instability mechanisms 29 1

and use this expression to investigate the requirements for the support

force when c = 0. Rearranging gives

-- 2T (2@ - (1 + COS 2@) tan 4)

yH2 - (214 - (1 -cos2+)tan4)

and so we can plot contours of 2T/yH2 for various angles of friction and

dip of the failure plane. These are shown in the figure below.

"

0 10 20 30 40 50 60 70 80 90

dip angle of failure plane, degrees

We can see from the plot that the required bolt force reduces to zero
when the friction angle is greater than or equal to the dip angle of the
failure plane, and it approaches the weight of the block for frictionless
conditions.

Case (b) Uniform pressure

A uniform pressure is related to the horizontal force used in the above

analysis through T = p H. Making this substitution in Eq. (17.1) leads to

2c + (y H cos2 y!!f + 2p sin2 ef) tan 4

(y H - 2p) cos 1cI.f sin +f

yH (sin 2+ - (1 + cos 2+) tan4) - 4c

2 (sin2+ + (I - cos2+) tan#)

F=

Similarly, Eq. (17.2) becomes

=

Case (c) Varying pressure distribution

through T = TqH, and so we find that the factor of safety is given by

A varying ressure distribution is related to the horizontal force T

P

2c + ( y H cos2 @f + q sin2 $f) tan 4

(Y H - 9) cos l(rf sin llrf

F=