306 Foundation and slope instability mechanismsabout the corner C, while block B is on the point of sliding downhill.
The shear resistance on all surfaces is purely frictional with 4 = 35O.
Given that B is twice as heavy as A, determine the thickness t of
block B. Also show that there is no tendency for block A to slip at the
corner C.A17.9 In order to solve problems of this type we start by drawing a
free body diagram, marking on all of the forces acting on the various
elements. In this case we know that, if the force due to the weight of
block A is W, then the force due to the weight of block B is 2W. Both
of these forces act through the centroid of the respective blocks. There
are also shear and normal forces acting on all of the interfaces. This
infomation leads to the diagram shown below.
As the system is in a state of limiting equilibrium with block B on
the point of sliding, we can write FB = NB tan 35. Furthermore, if block
B moves down the slope but block A does not slide, then sliding must
occur at point X and block A must rotate about point C. This information
gives us both a relation between P and Q and the position at which they
act: Q = P tan 35, acting at a distance c above the plane.
Because block A is toppling, we can write down the equation of
moment equilibrium for this block by taking moments (anticlockwise
positive) about point C:
0.75
2W sin30 x 1.5 - Wcos30 x - - Pt = 0.
Rearranging this gives
1
t
P = - W (1.5 sin30 - 0.375 cos30). (17.12)
We know that block B is sliding and hence FB = NB tan35. To de-
termine the forces in this relation, we write down the force equilibrium
equations (parallel and normal to the plane) for block B:
FB - 2Wsin30 - P = 0 and NB - WCos30 - Q = 0