Question

The factor of safety of an infinite soil slope shown in the figure having the properties $c=0,\varphi ={35}^{\circ },{\gamma }_{dry}=16kN/m^{3}and{\gamma }_{sat}=20KN/m^3$ is approximately equal to

• A. 0.707
• B. 0.80
• C. 1.00
• D. 1.20

Answer 1

The correct option is A 0.707

Given,

C = 0

$\varphi ={35}^{\circ }$

${\gamma }_{dry}=16kN/m^3$

${\gamma }_{sat}=20kN/m^3$

${\gamma }_{sub}={\gamma }_{sat}-{\gamma }_w$

= 20 - 10

$=10kN/m^3$

$Factorofsafety=\frac{Resistingforce}{Drivingforce}$

$=\frac{C+\overline{\sigma }tan\varphi }{\tau }$

$=\frac{0+\gamma zco{s}^2\beta tan\varphi }{\gamma zcos\beta sin\beta }$

$=\frac{({\gamma }_{sub}\times 8+{\gamma }_{dry}\times 2)co{s}^2{30}^{\circ }tan{35}^{\circ }}{({\gamma }_{sat}\times 8+{\gamma }_{dry}\times 2)cos{30}^{\circ }sin{30}^{\circ }}$

$FOS=\frac{(10\times 8+16\times 2)co{s}^2{30}^{\circ }tan{35}^{\circ }}{(20\times 8+16\times 2)cos{30}^{\circ }sin{30}^{\circ }}$

= 0.707

Note : In calculation of resisting force we always take effective stress while in case of driving force or destabilizing force we always take total stress.