Engineering Rock Mechanics

Question and answers: foundation and dope instabirity mechanisms 299

ADBC = Am - ABOC

= 4 x 17 x 17 - 4 x 17 x 17tan30

= 61.07 m2

and thus WD~C = 22.0 x 61.07 = 1343.6 kN.

The factor of safety is then

17.0

cos 30

x 10 + (1343.6 cos 30) tan 30

F= = 1.29.

1343.6 sin 30

417.6 Assume the same circumstances in Q17.5 and consider the

case when the block DBC has moved downhill a limited amount,

resulting in a considerable increase in the aperture of DB. Calculate

the factor of safety against sliding of the block DBC.

Al7.6 In the circumstance that the aperture of DB is much larger than 1
mm, far more water will enter at D than can be drained away through
BE, and as a result the pressure at B will increase towards a hydrostatic
value. If we assume that there will also be a linear pressure variation
along BC, from hydrostatic at B to zero at C, then we can compute the
new factor of safety as

ZBC * C’ + (WDBC COS 30 - U - V sin 30) tan 4’

V cos 30 + WDBC sin 30

F=

where U and V are the forces due to the water pressure along Bc and

BD, respectively. These forces are determined thus:

p~ = yw x ZBD = 9.81 x 7.185 = 70.49 kN/m2,

V = iywl&, = TPBZBD I = 253.2 kN,

U = ~~BZBC = 691.8 kN.

Hence, the factor of safety is computed as

F= = 0.44.

Clearly, increasing the aperture of DB has a significant effect on the

stability of the block, as demonstrated by performing the computation

(17/ cos 30) x 10 + (1343.6 cos 30 - 691.8 - 253.2 sin 30) tan 30

253.2 cos 30 + 1343.6 sin 30