Engineering Rock Mechanics

Question and answers: foundation and slope instability mechanisms 295

Block 3

fx = 0, with forces directed to the right reckoned positive,

N23 COS 30 + $3 COS 60 - S30 COS 60 - N30 COS 30 = 0. (17.3)

Limiting equilibrium tells us that, for each face of the block, S =

cL + N tan#. For this particular foundation we know that q5 = 0, and

so the limiting equilibrium condition is simply S = cL. Substituting this

into Eq. (17.3) gives us

N23 COS 30 + CL COS 60 - CL COS 60 - N30 COS 30 = 0

and hence

N30 = N23. (1 7.4)

f, = 0, with forces directed upwards reckoned positive,

N23 sin 30 + N30 sin 60 - $3 sin 60 - S30 sin 60 - W = 0.

Using the limiting condition of S = cL, this reduces to

N23 sin30 + N30sin30 - cL sin60 - cL sin60 - W = 0

and substituting Eq. (17.4) into this gives

N23 sin 30 + N23 sin 30 - cL sin 60 - cL sin 60 - W = 0

which, on rearrangement and using the identities sin30 = 1/2 and
sin60 = m, leads to

~23 = w +cL~. (17.5)

Block 2
fx = 0, with forces directed to the right being reckoned positive,
Nj2 COS 30 - $3 COS 60 - Si2 COS 60 - N23 COS 30 - S20 = 0.

With the limiting condition of S = rL, this reduces to

N12 COS 30 - CL COS 60 - CL COS 60 - N23 COS 30 - CL = 0

and upon substitution of Eq. (17.5) and rearrangement we obtain

(17.6)

Block 2
fx = 0, with forces directed to the right reckoned positive,
Nlo COS 30 + Si2 COS 60 - Slo COS 60 - N12 COS 30 = 0.

With the limiting condition of S = cL this reduces to

Nlo COS 30 + CL COS 60 - CL COS 60 - N12 COS 30 = 0

and hence

NlO = N12. (17.7)

fv = 0, with forces directed upwards reckoned positive,

Nlosin30+Slosin60+ N12sin30+S12sin60- W -pL =O.