A circular shaft can transmit a torque of 5 kN-m. If torque is reduced to 4 kN-m, then the maximum value of bending moment that can be applied to a shaft is
- A1 kN-m
- B2 kN-m
- C3 kN-mCorrect
- D4 kN-m
Solution
Concept
Equivalent torque under combined bending and torsion
Formula
Te = √(M² + T²)
Step-by-step solution
1. The shaft can carry a pure torque of 5 kN·m, so the equivalent torque limit is 5 kN·m. 2. With T = 4 kN·m: M² = 5² − 4² = 25 − 16 = 9. 3. M = 3 kN·m.
Explanation
Under combined bending and torsion the shaft capacity is governed by the equivalent torque √(M² + T²) (maximum shear stress theory). Reducing the torque lets the shaft carry a bending moment in a 3-4-5 relation.
Shortcut
3-4-5 triangle: T = 4 → M = 3.
Common mistake
Subtracting directly (5 − 4 = 1).
Correct answer: C) 3 kN-m