Question

A drained direct shear test was carried out on a sandy soil. Under a normal stress of 50 kPa, the test specimen failed at a shear stress of 35 kPa. The angle of internal friction of the sample is \underline{\hspace{3cm} degree (round off to the nearest integer).}

Hint:

For cohesionless soils, the direct shear test directly provides the angle of internal friction $\phi$ since cohesion $c = 0$. Always use $\tau = \sigma_n \tan \phi$.

Answer 1

Step 1: Recall Mohr-Coulomb failure criterion.
For sandy soil (cohesionless soil, $c=0$), the shear strength equation is: \[ \tau = \sigma_n \, \tan \phi \] where, $\tau$ = shear stress at failure, $\sigma_n$ = normal stress, $\phi$ = angle of internal friction.

Step 2: Substitute given values.
Normal stress: $\sigma_n = 50 \, \text{kPa}$
Shear stress at failure: $\tau = 35 \, \text{kPa}$
\[ \tan \phi = \frac{\tau}{\sigma_n} = \frac{35}{50} = 0.70 \]

Step 3: Find angle of internal friction.
\[ \phi = \tan^{-1}(0.70) \] Using calculator, \[ \phi = 34.99^{\circ} \approx 35^{\circ} \] \[ \boxed{\phi = 35^{\circ}} \]