Two shafts of the same material: Shaft A is solid with diameter D; Shaft B is hollow with outer diameter D and inner diameter D/2. Comparing torsional strength (torque capacity at the same allowable shear stress), the hollow shaft is what fraction of the solid shaft?

Introduction / Context:Torsion design often favours hollow shafts because material is more effective away from the centre. This question compares the torsional strength (torque for a given maximum shear stress) of a solid versus a hollow circular shaft with the same outer diameter and material.

Given Data / Assumptions:

• Material and allowable shear stress are the same.
• Outer diameter for both shafts is D; hollow shaft has inner diameter D/2.
• Strength comparison is based on torque at the same maximum shear stress.

Concept / Approach:For circular shafts, torsional strength T is related to polar section modulus Zp: T = τ_max * Zp. For a solid shaft, Zp_solid = (π/16) * D^3. For a hollow shaft, Zp_hollow = J/R = [(π/32) * (D^4 − d^4)] / (D/2).

Step-by-Step Solution:

Verification / Alternative check:Because most torque resistance comes from outer fibres, removing the core barely reduces strength: 15/16 ≈ 0.9375, confirming high efficiency of hollow sections.

Why Other Options Are Wrong:1/16, 1/8, 1/4 drastically underestimate strength; they ignore the outer-radius dominance in torsion.

Common Pitfalls:Comparing polar moments J (stiffness-related) instead of polar modulus Zp (strength-related); forgetting to divide J by outer radius for strength.

Final Answer:

15/16