Question

In a drained triaxial compression test, a sample of sand fails at deviator stress of 150 kPa under confining pressure of 50 kPa. The angle of internal friction (in degree, round off to the nearest integer) of the sample, is .

• A. 36.87

Answer 1

The correct option is A 36.87
Major principal stress $\left({\sigma }_1\right)=$ confining stress + deviator stress

= (50 + 150)

= 200 kPa

Minor principal stress $\left({\sigma }_3\right)$ = confining stress = 50 kPa

Let, angle of internal friction is $\varphi$
$sin\varphi =\frac{{\sigma }_1-{\sigma }_3}{{\sigma }_1+{\sigma }_3}$

$=\frac{200-50}{200+50}$

$=\frac{150}{250}$

= 0.6

$\varphi =si{n}^{-1}\left(0.6\right)$

$\varphi ={36.87}^{\circ }$