Question

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?

• A. $1\text{ kN}\cdot\text{m}$
• B. $1.5\text{ kN}\cdot\text{m}$
• C. $2\text{ kN}\cdot\text{m}$
• D. $3\text{ kN}\cdot\text{m}$

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Find the shaft's torque capacity.
The shaft was originally sized for a pure twisting moment of 5 kN·m with no bending, so its equivalent twisting moment capacity is simply \( T_e = 5\text{ kN}\cdot\text{m} \).

Step 2: Spot the 3-4-5 triangle.
The design condition is \( T_e^2 = M^2 + T^2 \), so \( 5^2 = M^2 + T^2 \). With the new twisting moment \( T = 4\text{ kN}\cdot\text{m} \), notice that \( 3,4,5 \) is a Pythagorean triple: \( 3^2 + 4^2 = 5^2 \).

Step 3: Match the values.
Since \( T = 4 \) already matches the "4" in the triple, the bending moment must be the missing "3": \[ \boxed{M = 3\text{ kN}\cdot\text{m}} \]