A circular shaft can transmit a torque of 5 kN·m. If the torque is reduced to 4 kN·m, the maximum bending moment that can be applied to the shaft is:
Try it first: pick an option, then read the explanation.
Correct answer: C. 3 kN·m
Asked in SSC JE 2008
Detailed Explanation
The shaft's capacity is set by its maximum shear stress. Under combined loading, the shear stress equals that from an equivalent torque Te = √(M² + T²).
\[ \sqrt{M^2 + 4^2} = 5 \;\Rightarrow\; M^2 = 25 - 16 = 9 \;\Rightarrow\; M = 3\ \text{kN·m} \]
Why “3 kN·m” Is Correct
The shaft can take a bending moment of 3 kN·m (maximum shear stress theory).
| Combined bending M and torque T (solid shaft) | Formula |
|---|---|
| Bending stress | σb = 32M/(πD³) |
| Torsional shear stress | τ = 16T/(πD³) |
| Equivalent bending moment (max principal stress) | Me = ½[M + √(M² + T²)] |
| Equivalent torque (max shear stress) | Te = √(M² + T²) |
| Ratio σb/τ | 2M/T |
Why the Other Options Are Wrong
- A. 1 kN·m: 1 kN·m = 5 − 4, a simple subtraction — moments combine as a vector sum.
- B. 2 kN·m: 2 kN·m is too small.
- D. 4 kN·m: 4 kN·m would give Te = √32 ≈ 5.66 > 5.
Key Point
Te = √(M² + T²) ≤ capacity → 3-4-5 triangle here.
Key Concepts for Students
- Guest–Tresca theory is behind Te.
- Rankine would use Me = ½(M + Te).
- Torque capacity T = (π/16)τd³.
Practice more: this is question 30 in Torsion MCQs – Part 2. Try the whole set and check every answer instantly.