Question

Following observations are made during direct shear test on sand:
Normal load at failure = 250 N
Shear load at failure = 175 N
Cross-sectional area of sample = $36c{m}^2$
The angle of internal friction (in radian) is

• A. 0.61
• B. 0.71
• C. 0.81
• D. 0.91

Answer 1

The correct option is A 0.61

Given,

$Shearstressatfailure{\tau }_f=\frac{175N}{36c{m}^2}$

$=4.86N/c{m}^2$

Normal stress on failure plane of failure,

$\sigma =\frac{250}{36}\frac{N}{c{m}^2}$

$=6.94N/c{m}^2$

Angle of internal friction
$={\tan }^{-1}\left(\frac{{\tau }_f}{\sigma }\right)$

$={\tan }^{-1}\left(\frac{4.86}{6.94}\right)={35}^{\circ }$

$=35\times \frac{\pi }{180}=0.61radians$