Question

A section of a shaft of diameter 100 mm is subjected to a moment of 4 kN-m and a torque of 3 kN-m. The ratio of maximum principal stress to minimum principal stress numerically is ............

• A. 9
• B. 2
• C. \( \frac{5}{3} \)
• D. \( \frac{4}{3} \)

Hint:

When dealing with shafts under both bending and torsion, remember that the combined effect on the principal stresses needs to be considered using both the bending stress and the torsional stress components.

Answer 1

The Correct Option is A

The maximum and minimum principal stresses can be calculated using the formulas for stress in a shaft under bending and torsion. The maximum principal stress \( \sigma_1 \) and minimum principal stress \( \sigma_2 \) are given by: \[ \sigma_1 = \frac{M}{S} + \sqrt{\left(\frac{M}{S}\right)^2 + \left(\frac{T}{J}\right)^2} \] \[ \sigma_2 = \frac{M}{S} - \sqrt{\left(\frac{M}{S}\right)^2 + \left(\frac{T}{J}\right)^2} \] Where:
- \( M \) is the moment applied (4 kN-m),
- \( T \) is the torque applied (3 kN-m),
- \( S \) is the section modulus for bending,
- \( J \) is the polar moment of inertia for the shaft's cross-section.
Given the diameter of the shaft is 100 mm, the section modulus \( S \) and the polar moment of inertia \( J \) for a circular shaft are calculated as: \[ S = \frac{\pi d^3}{32}, \quad J = \frac{\pi d^4}{32} \] Substituting the given values (d = 100 mm = 0.1 m), we calculate \( S \) and \( J \). Then we calculate the maximum and minimum principal stresses. The ratio of maximum principal stress to minimum principal stress is: \[ \frac{\sigma_1}{\sigma_2} = 9 \] Thus, the ratio is 9.