Engineering Rock Mechanics

298 Foundation and slope instability mechanisms

We then draw up a table of heads:

Node: A BCD E

Pressurehead, plyw 0 wfyw 0 0 0

Elevation head,z 17.0 9.815 0 17.0 3.315

Total head. H 17.0 HR 0 17.0 3.315

and a table of channel conductance:

Channel: AB BC BD BE

L (m) 14.37 19.63 7.185 6.5

c (ms-’, x~O-~) 43.76 32.03 87.52 96.75

The fundamental equation for computing the head at a node is

, cij

and for node B in the network this is

CBAHA + CBCHC + CBDHD + BEH HE

CBA + @c + CBD + CBE

HB =

which is evaluated as

(43.76 x 17.0) + (32.03 x 0.0) + (87.52 x 17.0) + (96.75 x 3.315)

43.76 + 32.03 + 87.52 + 96.75

HB =

= 9.815 m.

The pressure at node B is then

p~ = yw (HB - ZB) = yw (9.815 - 9.815) = 0,

and so we can see that the drain works perfectly: the pressure is reduced

to zero at node B, and as a result the water pressure is zero everywhere

along BD and BC.

Examining the stability of block DBC, as block ABD is stable it will

play no part in the analysis. If we define a factor of safety as

resisting forces

driving forces ’

F=

then we have for block DBC

ZBC. C’ + ( WDBC COS 30) tan #’

WDBC sin 30

F=

where WDBC is the weight of block DBC. To compute this we determine

the area of DBC as follows: