Question

A soil having the average properties, bulk unit weight $=19\ \text{kN/m^3$, angle of internal friction $=25^\circ$ and cohesion $=15\ \text{kPa}$, is being formed on a rock slope at an inclination of $35^\circ$ with the horizontal. The critical height (in m) of the soil formation up to which it would be stable without failure is \underline{\hspace{2cm}} (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

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, \[ H_c=\frac{15}{19\times0.470\times0.334}\approx\frac{15}{2.996}\approx 5.0\ \text{m}. \] \[ \boxed{H_c \approx 5.0\ \text{m}} \]