Question

# The soil profile above the rock surface for a 25∘ infinite slope is shown in the figure, where su is the undrained shear strength and γt is total unit weight. The slip will occur at a depth of

A
8.83 m
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B
9.79 m
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C
7.83 m
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D
6.53 m
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Solution

## The correct option is A 8.83 mFOS=40+16(x)(cos2β)tanϕ16 (x) cosβsinβ but ϕ=0 ⇒FOS=40162(x)sin50∘ Even for, x = 5 FOS=1sin 50∘ i.e, F.O.S > 1 hence no slip will occur in 1st layer, For 2nd layer FOS=60[16×5+20(x−5)]cosβsinβ =60(20x−20)2sin50∘ =6010(x−1) sin50∘ FOS=6(x−1) sin50∘ for slip to occur, F.O.S = 1 ⇒6=(x−1)sin50∘ x=1+6sin 50∘=8.832m Hence slip will occur in 2nd layer of soil at a total depth of 8.832m

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