Question

A soil having the average properties, bulk unit weight $=19\\{kN/m}^3$, angle of internal friction =25 degree and cohesion $=15\\{kPa}$, is being formed on a rock slope at an inclination of 35 degree with the horizontal. The critical height (in m) of the soil formation up to which it would be stable without failure is (round off to one decimal place). [Assume the soil is formed parallel to the rock bedding plane and there is no ground water effect.] 

Hint:

For an infinite slope with no seepage, use $\displaystyle F=\frac{c}{\gamma z \sin\beta\cos\beta}+\frac{\tan\phi}{\tan\beta}$; set $F=1$ to get the critical thickness $z=H_c$.

Answer 1

Correct Answer: 5

For an infinite slope in a $c\mbox{-}\phi$ soil with no seepage, factor of safety is \[ F=\frac{c}{\gamma z \sin\beta\cos\beta}+\frac{\tan\phi}{\tan\beta}. \]
At critical height $z=H_c$, take $F=1$ and solve for $H_c$: \[ H_c=\frac{c}{\gamma\sin\beta\cos\beta\left(1-\dfrac{\tan\phi}{\tan\beta}\right)}. \]
Given $c=15\ \text{kPa}$, $\gamma=19\ \text{kN/m}^3$, $\phi=25^\circ$, $\beta=35^\circ$:
$\sin\beta\cos\beta=\sin35^\circ\cos35^\circ\approx0.5736\times0.8192=0.470$,
$\displaystyle 1-\frac{\tan\phi}{\tan\beta}=1-\frac{\tan25^\circ}{\tan35^\circ}\approx 1-\frac{0.4663}{0.7002}=0.334$.
Thus, \[ H_c=\frac{15}{19\times0.470\times0.334}\approx\frac{15}{2.996}\approx 5.0\ \text{m}. \] \[ \boxed{H_c \approx 5.0\ \text{m}} \]