Question

When a dry sand specimen tested in a triaxial test at \(50\) kPa cell pressure gave a deviation stress \(100\) kPa at failure. The angle of internal friction of sand is

• A. \(15^\circ\)
• B. \(30^\circ\)
• C. \(45^\circ\)
• D. \(56^\circ\)

Hint:

For cohesionless soil in triaxial compression, \[ \boxed{ \frac{\sigma_1}{\sigma_3} = \frac{1+\sin\phi}{1-\sin\phi} } \] Also, \[ \sigma_1=\sigma_3+\text{Deviation Stress}. \]

Answer 1

The Correct Option is B

Concept: For dry sand, \[ c=0. \] In a triaxial compression test, \[ \sigma_3=\text{Cell pressure}, \] \[ \sigma_d=\sigma_1-\sigma_3. \] The principal stress relationship is \[ \boxed{ \frac{\sigma_1}{\sigma_3} = \frac{1+\sin\phi}{1-\sin\phi} } \] where \(\phi\) is the angle of internal friction.

Step 1: Determine the principal stresses. Given, \[ \sigma_3=50\text{ kPa} \] Deviation stress, \[ \sigma_d=100\text{ kPa} \] Therefore, \[ \sigma_1 = \sigma_3+\sigma_d = 50+100 = 150\text{ kPa} \]

Step 2: Use the stress ratio equation. \[ \frac{\sigma_1}{\sigma_3} = \frac{150}{50} = 3 \] Hence, \[ 3 = \frac{1+\sin\phi}{1-\sin\phi} \] \[ 3-3\sin\phi = 1+\sin\phi \] \[ 2 = 4\sin\phi \] \[ \sin\phi=\frac12 \] \[ \phi=30^\circ \] Hence, \[ \boxed{\phi=30^\circ} \] Therefore, the correct option is \[ \boxed{(B)\;30^\circ.} \]