Question

A sandstone follows Mohr-Coulomb failure criterion. If the uniaxial compressive strength and the angle of the internal friction of the sandstone are 7 MPa and \(30^\circ\), respectively, the calculated cohesion of the rock is ................ MPa. [round off to 2 decimal places]

Hint:

For UCS problems, remember: \[ \sigma_c = \frac{2c \cos \phi}{1 - \sin \phi} \] This is the direct link between uniaxial compressive strength, cohesion, and internal friction angle.

Answer 1

Step 1: Recall Mohr-Coulomb criterion.
Shear strength of rock: \[ \tau = c + \sigma \tan \phi \] where \(c\) = cohesion, \(\phi\) = angle of internal friction.

Step 2: Relation between uniaxial compressive strength (UCS) and cohesion.
For uniaxial compression test: \[ \sigma_3 = 0, \quad \sigma_1 = \text{UCS} \] The relation is: \[ \sigma_1 = \frac{2c \cos \phi}{1 - \sin \phi} \]

Step 3: Substitute values.
Given: \(\sigma_1 = 7 \, \text{MPa}, \, \phi = 30^\circ\). \[ 7 = \frac{2c \cos 30^\circ}{1 - \sin 30^\circ} \]

Step 4: Simplify.
\[ \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2} \] \[ 7 = \frac{2c \cdot (\sqrt{3}/2)}{1 - 1/2} \] \[ 7 = \frac{c \sqrt{3}}{0.5} \] \[ 7 = 2c \sqrt{3} \]

Step 5: Solve for \(c\).
\[ c = \frac{7}{2 \sqrt{3}} = \frac{7}{3.464} \approx 2.02 \, \text{MPa} \]

Final Answer: \[ \boxed{2.02 \, \text{MPa}} \]