Question

The soil profile above the rock surface for a ${25}^{\circ }$ infinite slope is shown in the figure, where $s_u$ is the undrained shear strength and ${\gamma }_t$ is total unit weight. The slip will occur at a depth of

• A. 8.83 m
• B. 9.79 m
• C. 7.83 m
• D. 6.53 m

Answer 1

The correct option is A 8.83 m

$FOS=\frac{40+16\left(x\right)\left(co{s}^2\beta \right)tan\varphi }{16\left(x\right)cos\beta sin\beta }$

$but\varphi =0$

$\Rightarrow FOS=\frac{40}{\frac{16}{2}\left(x\right)sin{50}^{\circ }}$


Even for,

x = 5

$FOS=\frac{1}{sin{50}^{\circ }}$

i.e, F.O.S > 1

hence no slip will occur in 1st layer,

For 2nd layer

$FOS=\frac{60}{[16\times 5+20(x-5\left)\right]cos\beta sin\beta }$

$=\frac{60}{\frac{(20x-20)}{2}sin{50}^{\circ }}$

$=\frac{60}{10(x-1)sin{50}^{\circ }}$

$FOS=\frac{6}{(x-1)sin{50}^{\circ }}$

for slip to occur,

F.O.S = 1

$\Rightarrow 6=(x-1)sin{50}^{\circ }$

$x=1+\frac{6}{sin{50}^{\circ }}=8.832m$

Hence slip will occur in 2nd layer of soil at a total depth of 8.832m