ABSTRACT
Pp-Pa =r(H+D)+qu-yD+qu =yH+ 2qu =/5' (20.15)
(20.16)
For static equilibrium, the sum of all the horizontal forces must be equal zero, that is,
Simplifying,
Pa + 2Quh - 2QuD + rHD =0 , therefore,
2qu
Also, for equilibrium, the sum of the moments at any point should be zero. Taking moments about the base,
_ h2 (2qu - Y H)D2 Pa(y+D)+6(2qu)- 2 =0
Substituting for h in (Eq. 20.17) and simplifying,
(20.17)
(20.18)
where
Pa(6quY+Pa) C3 = (qu +r H)
The depth computed from Eq. (20.18) should be increased by 20 to 40 percent so that a factor of safety of 1.5 to 2.0 may be obtained. Alternatively the unconfined compressive strength qu may be divided by a factor of safety.
