Question

A solid shaft can resist a bending moment of 3 KNM and a twisting moment of 4 KNM together, then the maximum torque (in KNM) that can be applied is

• A. 7
• B. 3.5
• C. 4.5
• D. 5

Answer 1

Answer:

D. 5

For combined loading (bending and twisting); The equivalent torque is: $[T_{e}q = \sqrt{(M ^ 2 + T ^ 2)}]$ . Given: M = 3kNm and T = 4 kNm; Maximum torque applied is, $T_{e}q = \sqrt{(3 ^ 2 + 4 ^ 2)} = \sqrt{25} = 5kNm$