---
title: "Hollow shaft stronger than solid shaft (same weight)"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:hollow-shaft-stronger-than-solid"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-shaft-stronger-than-solid
site: BCT Engineering
---

# Hollow shaft stronger than solid shaft (same weight)

Strength of Materials · Chapter 6: Torsion

- Asked in: 2075 Chaitra, 8 marks; 2074 Chaitra, 8 marks; 2072 Chaitra, 8 marks (asked 3 times)
- Video: [comparison between solid and hollow shaft/ shaft numerical](https://www.youtube.com/watch?v=vBYAOBmjm-o) — GTU MIMP
- 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-shaft-stronger-than-solid
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

Prove that the hollow shaft of same material, same weight and same length is more stronger than the solid shaft in case of torque transmission.

## Short answer

For equal weight, area equality gives D_s² = D²(1-k²). Torque ratio T_h/T_s = (1+k²)/√(1-k²) > 1, so hollow shaft transmits more torque. Material near the centre is inefficiently stressed; hollow section places it further out where stress is higher.

## Answer

**Statement:** for the same material, same length and same weight (equal cross-sectional area), a hollow shaft can transmit more torque than a solid shaft, so it is stronger in torsion.

Let the allowable shear stress be $\tau$. Let the solid shaft have diameter $D_s$ and the hollow shaft external diameter $D$ and internal diameter $d$, with $k = d/D$.

#### Equal weight, material and length means equal area
$$
\frac{\pi}{4}D_s^2 = \frac{\pi}{4}(D^2-d^2) \Rightarrow D_s^2 = D^2 - d^2 = D^2(1-k^2)
$$

#### Torque capacity from the torsion formula $T = \dfrac{\tau J}{R}$
Solid shaft:

$$
T_s = \frac{\pi}{16}\tau D_s^3
$$

Hollow shaft:

$$
T_h = \frac{\pi}{16}\tau\,\frac{D^4-d^4}{D} = \frac{\pi}{16}\tau\,\frac{(D^2-d^2)(D^2+d^2)}{D}
$$

#### Ratio
Using $D_s^2 = D^2 - d^2$:

$$
\frac{T_h}{T_s} = \frac{(D^2-d^2)(D^2+d^2)}{D\,D_s^3} = \frac{D^2+d^2}{D\,D_s} = \frac{D^2(1+k^2)}{D\cdot D\sqrt{1-k^2}} = \frac{1+k^2}{\sqrt{1-k^2}}
$$

For any $0 < k < 1$:

$$
(1+k^2)^2 - (1-k^2) = 3k^2 + k^4 > 0 \Rightarrow \frac{1+k^2}{\sqrt{1-k^2}} > 1
$$

Therefore $T_h > T_s$: the hollow shaft carries a larger torque for the same weight.

#### Numerical illustration
| $k = d/D$ | $T_h/T_s$ |
|---|---|
| 0.5 | 1.44 |
| 0.6 | 1.70 |
| 0.8 | 2.73 |

(for $k = 0.6$: $\frac{1+0.36}{\sqrt{1-0.36}} = \frac{1.36}{0.8} = 1.70$).

**Reason:** in a solid shaft the material near the axis is lightly stressed because shear stress varies linearly from zero at the centre to $\tau$ at the surface. In the hollow shaft the same material is moved to the outside, where the lever arm and the stress are greatest, so it is used more efficiently.

## Check yourself

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

1. What does equal weight, material and length imply?
   A. Equal diameters
   B. Equal cross-sectional area, D_s^2 = D^2 - d^2
   C. Equal polar moments
   D. Equal outer diameters
   Answer: B. Weight is proportional to area when length and material are the same. (Key point: Equal weight condition)

2. What is T_h/T_s?
   A. (1 - k^2)/sqrt(1 + k^2)
   B. 1 + k^2
   C. (1 + k^2)/sqrt(1 - k^2)
   D. 1/(1 - k^4)
   Answer: C. Dividing the torque expressions and using D_s^2 = D^2(1 - k^2) gives that ratio. (Key point: Ratio)

3. How is the ratio shown to exceed 1?
   A. It equals 1 for all k
   B. Because D > d
   C. Because J is larger by definition only
   D. (1 + k^2)^2 - (1 - k^2) = 3k^2 + k^4 > 0
   Answer: D. The numerator squared exceeds the denominator squared for k > 0. (Key point: Proof of greater than 1)

4. Why is the hollow shaft more efficient?
   A. Material sits at larger radius where stress and lever arm are greatest
   B. Hollow shafts have higher G
   C. It has a larger area
   D. Shear stress is uniform in a hollow shaft
   Answer: A. In a solid shaft the core is lightly stressed and carries little torque. (Key point: Physical reason)

## Related in Strength of Materials

- [Torsional rigidity; hollow shaft vs solid shaft strength](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Atorsional-rigidity-hollow-vs-solid-300rpm) — Strength of Materials, chapter 6
- [Solid shaft 300 kW; hollow shaft saving](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-percent-saving) — Strength of Materials, chapter 6
- [Torsion vs bending; hollow shaft replaced by solid](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-to-solid-shaft-replacement) — Strength of Materials, chapter 6
- [Plastic torque of solid shaft is 4/3](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplastic-torque-solid-shaft-4-3) — Strength of Materials, chapter 6
- [Hollow shaft 1670 kW at 4500 rpm: size](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-1670kw-4500rpm) — Strength of Materials, chapter 6
- [Hollow section replacing solid compression member](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-replace-solid-compression-member) — Strength of Materials, chapter 6

## Related in other subjects

- [Hollow versus solid shaft for power](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ahollow-versus-solid-shaft-power) — Strength of Materials, chapter 4
- [Hollow shaft against solid shaft](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Ahollow-versus-solid-shaft) — Machine Design I, chapter 5
- [T/F: radial more reliable than network](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-radial-more-reliable-network) — Transmission and Distribution System Design, chapter 5, asked 2×
- [T/F: more circuits rather than higher voltage](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-circuits-rather-than-voltage) — Transmission and Distribution System Design, chapter 2