Section 4: Soil and Water Conservation Engineering
Soil Mechanics
Subtopic: Shear strength
Question Statement:
A direct shear test was conducted on a dry sand sample. The normal stress applied was $150\text{ kPa}$ and the shear stress at failure was measured as $100\text{ kPa}$. The angle of internal friction $\phi$ in degrees is ________ (round off to 1 decimal place).
Answer Key:
33.7
Step-by-Step Solution & Explanation
For dry cohesionless sand ($c = 0$), the Mohr-Coulomb failure envelope is:
$\tau_f = \sigma_n \tan \phi$
Given $\sigma_n = 150\text{ kPa}$ and $\tau_f = 100\text{ kPa}$:
$\tan \phi = \frac{100}{150} = \frac{2}{3} \approx 0.6667$
$\phi = \arctan(0.6667) \approx 33.69^\circ \approx 33.7^\circ$