Question

For two infinite slopes (one in dry condition and other in submerged condition) in a sand deposit having the angle of shearing resistance ${30}^{\circ }$, factor of safety was determined as 1.5 (for both slopes). The slope angles would have been.

• A. ${21.05}^{\circ }$ for dry slope and ${21.05}^{\circ }$ for submerged slope
• B. ${19.47}^{\circ }$ for dry slope and ${18.40}^{\circ }$ for submerged slope
• C. ${18.40}^{\circ }$ for dry slope and ${21.05}^{\circ }$ for submerged slope
• D. ${22.60}^{\circ }$ for dry slope and ${19.47}^{\circ }$ for submerged slope

Answer 1

The correct option is A ${21.05}^{\circ }$ for dry slope and ${21.05}^{\circ }$ for submerged slope

Angle of shearing resistance,

$\varphi ={30}^{\circ }$

FOS = 1.5

1. Dry Condition :

$FOS=\frac{ResistingForce}{DrivingForce}$

$=\frac{\gamma zco{s}^2\beta tan\varphi }{\gamma zcos\beta sin\beta }$

$=\frac{tan\varphi }{tan\beta }$

Note:

(i) There is difference between submerged condition and saturated condition. In submerged condition seepage does not take place through the soil parallel to slope, so in calculation of FOS, both in Numerator and denominator we use ${\gamma }_{sub}$.

(ii) When soil is in saturated condition, seepage takes place and water table is assumed to be parallel to the slope.

$FOS=\frac{{\gamma }_{sub}zco{s}^2\beta tan\varphi }{{\gamma }_{sat}zcos\beta sin\beta }$
(iii) FOS in submerged condition is more than that in saturated condition.