Question:medium

A circular shaft has been designed for the twisting moment of $5\text{ kN}\cdot\text{m}$. If the twisting moment is reduced to $4\text{ kN}\cdot\text{m}$, then what will be the maximum value of bending moment that can be applied for the same designed condition?

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This problem follows the standard 3-4-5 right-angle triangle relationship! Since $T_e = \sqrt{M^2 + T^2}$, if the total capacity vector length is 5 and one leg is 4, the remaining leg must be 3.
Updated On: Jul 9, 2026
  • $1\text{ kN}\cdot\text{m}$
  • $1.5\text{ kN}\cdot\text{m}$
  • $2\text{ kN}\cdot\text{m}$
  • $3\text{ kN}\cdot\text{m}$
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The Correct Option is D

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