296 Foundation and slope instability mechanisms
With the limiting condition of S = CL this reduces to
N~~sin30+cLsin60+N~2sin30+cLsin60- W -pL=Oand substituting Eq. (17.7) leads to
2N12 sin 30 + 2cL sin 60 - W - pL = 0.
Substituting Eq. (17.6) and rearranging results in
10
p = -C = 5.174~.
43(17.8)For this foundation we have c = 25 kN/m3, and substituting this value
into Eq. (17.8) gives a collapse pressure of
p = 144.3 kN/m2
and with L = 6 m we obtain a collapse load of
P = 866.0 kN/m2.
Notice that the collapse pressure (and hence collapse load) is independ-
ent of the weight of the blocks. This is because the friction angle is zero,
but in general the collapse load increases as the unit weight of the rock
increases.
If we perform the above calculations for a range of values for the
friction angle and cohesion of the block boundaries, we can examine
how the collapse pressure varies with these parameters. The graph
below shows this, with the units of collapse pressure being MPa.
Notice that a logarithmic scale has been used for the curves of collapse
pressure, indicating that its magnitude increases dramatically as both
the cohesion and friction angle increase.
0 50 100 150 200 250 300 350 400 450 500
cohesion, kPa