Torsion – comparative strength of solid vs. hollow circular shafts Let a solid circular shaft of diameter D be compared with a hollow circular shaft of external diameter D and internal diameter d. What is the ratio of their torsional moments of resistance (solid : hollow)?

Introduction / Context:
Designers often select hollow shafts for weight savings while maintaining torsional capacity. Comparing the torsional “moment of resistance” (polar section modulus) clarifies why hollow shafts can be efficient.

Given Data / Assumptions:

• Solid shaft diameter = D.
• Hollow shaft external diameter = D, internal diameter = d.
• Elastic torsion (Saint-Venant) with circular sections.

Concept / Approach:
The torsional moment T that produces a given extreme shear stress tau_max is:
For solid: T = (pi / 16) * tau_max * D^3For hollow: T = (pi / 16) * tau_max * (D^4 − d^4) / DThe torsional “moment of resistance” Z_p is T / tau_max, so:
Z_p(solid) = (pi / 16) * D^3Z_p(hollow) = (pi / 16) * (D^4 − d^4) / DTherefore the ratio (solid : hollow) is:
[(pi/16) D^3] : [(pi/16) (D^4 − d^4) / D] = D^4 : (D^4 − d^4)

Step-by-Step Solution:
Write Z_p(solid) and Z_p(hollow) as above.Cancel common factor (pi/16).Multiply both sides by D to clear denominator → ratio = D^4 : (D^4 − d^4).

Verification / Alternative check:
Limiting cases: if d → 0, hollow → solid and ratio → D^4 : D^4 = 1, consistent. If d → D, hollow capacity → 0, ratio → infinite, also consistent.

Why Other Options Are Wrong:
Options with powers 1–3 confuse torsion with bending section modulus patterns; torsion depends on polar properties leading to fourth-power terms for circular shafts.

Common Pitfalls:
Using J (polar moment of inertia) directly without converting to Z_p = J / R; neglecting the outer radius R = D/2 in the hollow expression.

Final Answer:
D^4 : (D^4 − d^4)