Question

A long slope is formed in a soil with shear strength parameter: c' = 0, ${\varphi }^{\prime }={34}^{\circ }$. Firm stratum lies below the slope and it is assumed that water table may occasionally rise to the surface, with seepage taking place parallel to the slope. Use ${\gamma }_{sat}=18KN/m^3$ and ${\gamma }_w=10KN/m^3$. The maximum slope angle (in degree) to ensure the factor of safety 1.5, assuming a potential failure surface parallel to the slope, would be

• A. 45.3
• B. 44.7
• C. 12.3
• D. 11.3

Answer 1

The correct option is D 11.3

F.O.S when water surface rises to surface is given by

$F.O.S=\frac{{\gamma }_{sub}}{{\gamma }_{sat}}\cdot \frac{tan\varphi }{tan\beta }$

$\Rightarrow tan\beta =(\frac{{\gamma }_{sub}}{{\gamma }_{sat}}\cdot \frac{tan\varphi }{F.O.S})$

$\Rightarrow tan\beta =\frac{{\gamma }_{sub}}{{\gamma }_{sat}}\cdot \frac{tan\varphi }{1.5}$

$tan\beta =\frac{(18-10)}{18\times 1.5}\times tan{34}^{\circ }$

$\Rightarrow \beta ={11.30}^{\circ }$