Question

In an unconfined compression test on stiff clay, if the failure plane made an angle of ${52}^o$ to the horizontal, what would be the angle of shearing resistance?

• A. ${16}^o$
• B. ${14}^o$
• C. ${12}^o$
• D. ${13}^o$

Answer 1

The correct option is B ${14}^o$

If the failure plane makes an angle $\alpha$ wiith the horizontal, then
$\alpha ={45}^o+\frac{\varphi }{2}$
where $\varphi$ is the angle of shearing resistance
$\therefore {54}^o={45}^o+\frac{\varphi }{2}$
$\Rightarrow \varphi ={14}^o$