Question

In a uniaxial compressive strength test, a rock sample of diameter 50 mm fails at an angle of $60^\circ$ as shown. If the peak load at failure is 120 kN, the normal and shear stresses on the failure plane, in MPa, are __________________ and __________________.

• A. 15.28 and 26.46
• B. 26.46 and 15.28
• C. 57.02 and -15.28
• D. -15.28 and 15.28

Hint:

For uniaxial loading, normal stress varies as $\sigma\cos^2\theta$ and shear stress as $\sigma\sin\theta\cos\theta$ on a plane inclined at $\theta$.

Answer 1

The Correct Option is A

Step 1: Compute the axial stress.
Diameter of sample = 50 mm = 0.05 m
Radius = 0.025 m
\[ A = \pi r^2 = \pi(0.025)^2 = 1.9635 \times 10^{-3}\,\text{m}^2 \]
Peak load = 120 kN = 120,000 N
Axial compressive stress: \[ \sigma = \frac{P}{A} = \frac{120000}{1.9635 \times 10^{-3}} = 61.12\,\text{MPa} \]
Step 2: Use stress transformation for stresses on an inclined plane.
Inclination angle of failure plane = $60^\circ$
Normal stress on plane: \[ \sigma_n = \sigma \cos^2\theta \] \[ \sigma_n = 61.12 \times \cos^2(60^\circ) \] \[ \cos(60^\circ)=0.5,\quad \cos^2(60^\circ)=0.25 \] \[ \sigma_n = 61.12 \times 0.25 = 15.28\ \text{MPa} \]
Shear stress on plane: \[ \tau = \sigma \sin\theta \cos\theta \] \[ \tau = 61.12 \times (0.866)(0.5) \] \[ \tau = 61.12 \times 0.433 = 26.46\ \text{MPa} \]
Step 3: Final values.
Normal stress = 15.28 MPa
Shear stress = 26.46 MPa
Final Answer: 15.28 MPa and 26.46 MPa