Question

A shear annulus with inner and outer diameters of 240 mm and 300 mm is used to measure soil shear strength. The torque at failure is 50 N·m. Shear strength of soil in kPa is (Take $\pi = 3.14$)

• A. 14.49
• B. 18.94
• C. 21.54
• D. 28.98

Hint:

Torque–shear relation in annulus: $\tau = \dfrac{T}{A\,r_{avg}}$.

Answer 1

The Correct Option is A

Step 1: Convert to radii.
Outer radius: $R_o = 150$ mm Inner radius: $R_i = 120$ mm

Step 2: Polar moment area of annulus (shear test): \[ A = \pi(R_o^2 - R_i^2) = 3.14(150^2 - 120^2) \] \[ = 3.14(22500 - 14400) = 3.14 \times 8100 = 25434 \text{ mm}^2. \]

Step 3: Average radius: \[ r_{avg} = \frac{R_o + R_i}{2} = 135 \; \text{mm}. \]

Step 4: Shear stress from torque: \[ T = \tau \cdot A \cdot r_{avg} \] \[ 50 \times 10^3 = \tau (25434)(135) \] \[ \tau = 14.49 \text{ kPa}. \]