A shaft of diameter D is subjected to a twisting moment of T and
| A shaft of diameter D is subjected to a twisting moment of T and Bending moment of M respectively. If the maximum bending stress is equal to maximum shear stress developed. Then M is equal to
A. <span class="math-tex">\(\frac T2\)</span>
B. T
C. 2T
D. 4T
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
Maximum Bending stress due to bending moment:
\({σ _b} = \frac{{32M}}{{\pi {d^3}}}\)
Maximum Shear stress due to twisting moment/ torque:
\(τ = \frac{{16T}}{{\pi {d^3}}}\)
Calculation:
Given:
Maximum bending stress = Maximum shear stress
σb = τ
\(\frac{{32M}}{{\pi {d^3}}}\) = \( \frac{{16T}}{{\pi {d^3}}}\)
Hence, \(M = \frac T2\)