Design of solid circular shaft under torsion: A solid shaft transmits torque T. If the maximum shear stress must not exceed f_z, what diameter d is required?

Introduction / Context:
Power shafts are sized by limiting shear stress under torque. For solid circular shafts, the elastic torsion formula relates torque to maximum shear stress via the polar section modulus.

Given Data / Assumptions:

• Solid circular shaft of diameter d.
• Applied torque T (steady).
• Allowable maximum shear stress f_z (τ_allow).
• Elastic torsion; no stress concentration considered.

Concept / Approach:
For a solid circular shaft, τ_max = 16 T / (π d^3). This comes from T / J = τ / r with J = π d^4 / 32 and r = d/2, giving τ_max = 16 T / (π d^3). Solving for d yields the required diameter.

Step-by-Step Solution:
Start with τ_max = 16 T / (π d^3).Set τ_max = f_z (allowable): f_z = 16 T / (π d^3).Rearrange: d^3 = 16 T / (π f_z).Therefore, d = (16 T / (π f_z))^(1/3).

Verification / Alternative check:
Dimensional check: T has units of force*length; dividing by stress (force/area) gives length^3, consistent with d^3.

Why Other Options Are Wrong:
Square-root forms are for thin tubes or unrelated cases.Algebraic rearrangements with π misplaced do not satisfy τ_max relation.

Common Pitfalls:
Using hollow-shaft formulas or forgetting the 16/π factor for the solid section.

Final Answer:
d = (16 T / (π f_z))^(1/3).