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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 m
FOS=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(x5)]cosβsinβ

=60(20x20)2sin50

=6010(x1) sin50

FOS=6(x1) sin50

for slip to occur,

F.O.S = 1

6=(x1)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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