Difficulty: Medium
Correct Answer: 2M / T
Explanation:
Introduction / Context:
When a circular shaft is subjected to combined loading—bending moment (M) and torque (T)—two peak stresses occur at the outer surface: maximum bending normal stress and maximum torsional shear stress. This item tests your ability to recall the fundamental stress expressions for circular shafts and to form a clean ratio independent of section dimensions.
Given Data / Assumptions:
Concept / Approach:
For a solid circular shaft of diameter d: bending stress at outer fiber is sigma_max = M*c/I and torsional shear stress at outer fiber is tau_max = T*c/J. For a circle, c = d/2, I = (pi*d^4)/64, and J = (pi*d^4)/32. Substituting and simplifying eliminates the geometric terms, leaving a simple ratio in M and T only.
Step-by-Step Solution:
1) sigma_max = (M * c) / I = (M * (d/2)) / (pi*d^4/64) = 32*M / (pi*d^3).2) tau_max = (T * c) / J = (T * (d/2)) / (pi*d^4/32) = 16*T / (pi*d^3).3) Form the ratio: (sigma_max) / (tau_max) = (32*M / (pi*d^3)) / (16*T / (pi*d^3)) = 2M / T.
Verification / Alternative check:
The pi and d^3 terms cancel, confirming that the ratio is independent of diameter for a solid circular shaft. The same ratio 2M/T also results for a thin-walled circular tube because the linear dependence on c and section properties cancels similarly when forming the ratio.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
2M / T.
Discussion & Comments