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.