Question

In a drained triaxial compression test conducted on dry sand, failure occurred when the deviator stress was 218$kN/m^2$ at a confining pressure of 61 $kN/m^2$. What is the effective angle of shearing resistance and the inclination of failure plane to major principal plane?

• A. ${34}^{\circ },{62}^{\circ }$
• B. ${34}^{\circ },{28}^{\circ }$
• C. ${40}^{\circ },{25}^{\circ }$
• D. ${40}^{\circ },{65}^{\circ }$

Answer 1

The correct option is D ${40}^{\circ },{65}^{\circ }$

Given,

${\sigma }_3=61kN/m^2$

Deviator stress

${\sigma }_1-{\sigma }_3=218kN/m^2$

${\sigma }_1-61=218$

${\sigma }_1=279kN/m^2$

We know that

${\sigma }_1={\sigma }_{3}ta{n}^2(45+\frac{\varphi }{2})+2ctan(45+\frac{\varphi }{2})$

For sand, c = 0

${\sigma }_1={\sigma }_{3}ta{n}^2(45+\frac{\varphi }{2})$

$279=61ta{n}^2(45+\frac{\varphi }{2})$

$\varphi ={39.88}^{\circ }$

$\approx {40}^{\circ }$

Inclination of failure plane to major principal plane

$=45+\frac{\varphi }{2}=45+\frac{40}{2}={65}^{\circ }$