---
title: "Torsion vs bending; hollow shaft replaced by solid"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:hollow-to-solid-shaft-replacement"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-to-solid-shaft-replacement
site: BCT Engineering
---

# Torsion vs bending; hollow shaft replaced by solid

Strength of Materials · Chapter 6: Torsion

- Asked in: 2081 Bhadra, 2+6 marks
- Video: [Comparison Between Solid and Hollow Shafts SOM Tutorial- 5](https://www.youtube.com/watch?v=3pC-U6ZTd98) — Ujjwal Kumar Sen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-to-solid-shaft-replacement
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

Clarify the basic difference between torsion and bending. A hollow shaft 20 cm internal diameter and 30 cm external diameter is to be replaced by a solid alloy shaft. If the polar modulus has the same value for both, calculate the diameter of the solid shaft and the ratio of torsional rigidities. G for steel = 2.5 times G for alloy.

## Short answer

Torsion involves twisting moment causing shear stress, zero at axis, maximum at surface; bending involves moment causing normal stress, zero at neutral axis, maximum at extreme fibre. Equal polar modulus gives solid diameter = [(30⁴-20⁴)/30]¹/³ = 27.88 cm. Torsional rigidity ratio (hollow/solid) = 2.5×J_h/J_s = 2.5×(63814/59300)= 2.69.

## Answer

#### Torsion and bending
| Point | Torsion | Bending |
|---|---|---|
| Moment | Twisting moment (torque) about the longitudinal axis | Moment acting in a plane containing the axis |
| Stress | Shear stress, zero at the axis, maximum at the surface | Normal stress, zero at the neutral axis, maximum at the extreme fibre |
| Governing equation | $\dfrac{T}{J} = \dfrac{\tau}{r} = \dfrac{G\varphi}{L}$ | $\dfrac{M}{I} = \dfrac{\sigma}{y} = \dfrac{E}{R}$ |
| Deformation | Angle of twist | Deflection and curvature |

#### Replacing the hollow shaft by a solid shaft
Equal polar modulus means $Z_p$ (hollow) $= Z_p$ (solid). Hollow shaft: $D = 30$ cm, $d = 20$ cm. Solid alloy shaft: diameter $D_s$.

$$
Z_p = \frac{\pi}{16}\,\frac{D^4-d^4}{D} = \frac{\pi}{16}\,\frac{30^4-20^4}{30} = 4254.2\ \text{cm}^3, \qquad Z_p = \frac{\pi}{16}D_s^3
$$

$$
D_s^3 = \frac{30^4-20^4}{30} = 21666.67 \Rightarrow D_s = 27.88\ \text{cm}
$$

#### Ratio of torsional rigidities
Torsional rigidity is $GJ$. Hollow (steel) $J_h = \dfrac{\pi}{32}(30^4-20^4) = 63813.6\ \text{cm}^4$; solid (alloy) $J_s = \dfrac{\pi}{32}D_s^4 = 59300.2\ \text{cm}^4$, and $G_{steel} = 2.5\,G_{alloy}$:

$$
\frac{(GJ)_{hollow}}{(GJ)_{solid}} = \frac{2.5\,G_{alloy}\,J_h}{G_{alloy}\,J_s} = 2.5\times\frac{63813.6}{59300.2} = 2.690
$$

**Answer:** solid shaft diameter $= 27.88$ cm; ratio of torsional rigidities (hollow : solid) $= 2.69$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Where is torsional shear stress zero?
   A. At the surface
   B. At the axis; maximum at the surface
   C. Everywhere
   D. Only at the neutral axis in bending
   Answer: B. Shear varies linearly with radius. (Key point: Torsion vs bending stress)

2. Which is the bending equation?
   A. T/J = tau/r
   B. M/I = tau/r
   C. M/I = sigma/y = E/R
   D. M/J = sigma/y
   Answer: C. Torsion uses T/J = tau/r = G phi/L. (Key point: Equation)

3. Which condition gives D_s?
   A. Equal areas
   B. Equal J
   C. Equal diameters
   D. Z_p hollow = Z_p solid, D_s^3 = (D^4 - d^4)/D, D_s = 27.88 cm
   Answer: D. Z_p = J/R sets equal torque capacity. (Key point: Equal polar modulus)

4. What does the ratio compare?
   A. GJ of hollow steel to GJ of solid alloy, including G = 2.5 G_alloy
   B. Only J
   C. Only G
   D. Z_p values
   Answer: A. Torsional rigidity is the product GJ; ratio = 2.5 x 63813.6/59300.2 = 2.69. (Key point: Torsional rigidity ratio)

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