Question

The infinite sand slope shown in the figure is on the verge of sliding failure. The ground water table coincides with ground surface. The value of effective angle of internal friction (in degrees, correct upto one decimal place) of sand is ?.
(Take ${\gamma }_w=10kN/m^3$)

• A. ${54.42}^{\circ }$
• B. ${46.12}^{\circ }$
• C. ${78.56}^{\circ }$
• D. ${89.30}^{\circ }$

Answer 1

The correct option is A ${54.42}^{\circ }$

Since the slope is on the verge of failure.

$\therefore {\gamma }_f={\gamma }_s$

${\gamma }_{sub}Hco{s}^2\beta tan\varphi ={\gamma }_{sat}Hcos\beta sin\beta$

$tan\varphi =\frac{{\gamma }_{sat}}{{\gamma }_{sub}}tan\beta$

$\varphi =ta{n}^{-1}\left\{\frac{20}{(20-10)}tan{35}^{\circ }\right\}$

$={54.42}^{\circ }$