Equating maximum bending and maximum shear stresses in a circular shaft\n\nA circular shaft of diameter D is subjected simultaneously to a bending moment M and a twisting moment T. If the maximum bending stress equals the maximum torsional shear stress, what is the relation between M and T?

Introduction / Context:
Combined bending and torsion occur frequently in shafts with pulleys or gears. Comparing bending and torsional stresses helps in establishing allowable combined loading or determining an equivalent moment.

Given Data / Assumptions:

• Solid circular shaft of diameter D.
• Elastic behavior; standard bending and torsion formulas apply.
• Condition: σ_max(bending) = τ_max(torsion).

Concept / Approach:
Maximum bending normal stress is σ_max = M c / I with c = D/2 and I = π D^4 / 64. Maximum torsional shear stress is τ_max = T c / J with J = π D^4 / 32.

Step-by-Step Solution:
σ_max = M (D/2) / (π D^4 / 64) = 32 M / (π D^3).τ_max = T (D/2) / (π D^4 / 32) = 16 T / (π D^3).Set σ_max = τ_max → 32 M / (π D^3) = 16 T / (π D^3).Therefore 32 M = 16 T → M = T / 2.

Verification / Alternative check:
Dimensional and numerical consistency checks confirm the relation. If T doubles M, stresses match by the constants above.

Why Other Options Are Wrong:
Any relation other than M = T/2 violates the established proportionality between bending and torsional stress constants for a circular section.

Common Pitfalls:
Using radius instead of diameter in I and J; mixing formulas for hollow shafts; forgetting the different constants (64 vs 32) in I and J.

Final Answer:
M = T/2