Engineering Rock Mechanics

Question and answers: foundation and slope instability mechanisms 307

from which we obtain

(P + 2 W sin 30) = (Q -I- 2 W cos 30) tan 35.

P and Q are related through Q = P tan35 because sliding is taking
place at point X. This simplifies the above equation to

(P + 2W sin 30) = (P tan 35 + 2W cos 30) tan 35

from which we find

cos 30 tan 35 - sin 30

P=2W

1 - tan2 35

This can now be equated to Eq. (17.12) to give

(17.13)

cos 30tan 35 - sin 30 1

= - W (1.5 sin 30 - 0.375 cos 30).

2w 1-tan235 t

Rearranging this gives the value of t :

(1.5 sin 30 - 0.375 cos 30) (1 - tan2 35)

cos 30 tan 35 - sin 30

t=- 2 = 1.019 m.

Finally, in order to confirm that block A is not sliding, we must
examine the relation between the forces acting at point C. The force
equilibrium equations (parallel and normal to the plane) for block A are

FA -t P - Wsin30 = 0 and NA + Q - Wcos30 = 0

and hence

FA= Wsin30-PandNA= Wcos30-Q.

These give us a relation for the angle of friction required at A to
prevent slip:

FA Wsin30- P
tan@ = - =
NA Wc0~30-Q.
As before, because sliding is occurring at point X, this can be simplified
to

W sin30 - P

= wcos30- Ptan35'

Substituting the value of P found in Eq. (17.13) and cancelling

throughout to remove W gives the required friction angle as

C#I = tan-' (0.144) = 8.2"

As this value is much lower than the actual friction angle of 359 we

can see that sliding will not occur.

417.10 Solve Q17.5 numerically using the sector method, and com-

ment on any discrepancies between the numerical solution and the

analytical solution.