---
title: "Hollow shaft against solid shaft"
subject: "Machine Design I"
chapter: 5
key: "semester_6/machine_design_i:hollow-versus-solid-shaft"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/hollow-versus-solid-shaft
site: BCT Engineering
---

# Hollow shaft against solid shaft

Machine Design I · Chapter 5: Design of shafts

- Asked in: year not given, 6 marks
- Video: [Which Shaft Wins? | Hollow vs Solid for Maximum Torque](https://www.youtube.com/watch?v=xvll7HnayN8) — Simplegyan Engineering 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/hollow-versus-solid-shaft
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-5

## Question

A solid shaft transmits 90 kW at 250 rpm, and the allowable shear stress is 40 MPa. Find its diameter. If it is replaced by a hollow shaft of the same material and the same maximum shear stress, with inside diameter equal to 0.6 times the outside diameter, find the dimensions of the hollow shaft and the percentage saving in mass.

## Short answer

Solid shaft: T=3437.7 N·m; d3=16T/(πτ) → d=75.9 mm. For hollow shaft with di/do=0.6, same torque and stress: do3 = d3/(1-0.64) → do=79.5 mm, di=47.7 mm. Mass ∝ area: save1-(do2(1-0.62)/d2) =29.8%.

## Answer

#### Torque

$$
T = \frac{60P}{2\pi n} = \frac{60 \times 90\,000}{2\pi \times 250} = 3437.7\ \text{N.m}
$$

#### Solid shaft

$$
\begin{aligned} T &= \frac{\pi}{16}\tau d^3 \\ d^3 &= \frac{16 \times 3437.7\times10^3}{\pi \times 40} = 4.377\times10^5\ \text{mm}^3 \\ d &= 75.9\ \text{mm} \end{aligned}
$$

#### Hollow shaft ($k = d_i/d_o = 0.6$)

For the same torque and stress,

$$
\frac{\pi}{16}\tau d_o^3(1 - k^4) = T
$$

$$
d_o^3 = \frac{d^3}{1 - k^4} = \frac{4.377\times10^5}{1 - 0.6^4} = \frac{4.377\times10^5}{0.8704} = 5.029\times10^5\ \text{mm}^3
$$

$$
d_o = 79.5\ \text{mm}, \qquad d_i = 0.6 \times 79.5 = 47.7\ \text{mm}
$$

#### Mass saving

Mass is proportional to cross-sectional area (same length and material):

$$
\frac{A_h}{A_s} = \frac{d_o^2(1 - k^2)}{d^2} = \frac{79.5^2 \times 0.64}{75.9^2} = 0.702
$$

Saving $= 1 - 0.702 = 0.298$, about 30 percent.

Hollow shafts save material because the inner core carries little shear stress. The outside diameter is only about 5 percent larger, but hollow shafts are costlier to make.

**Answer: Solid d = 75.9 mm; hollow $d_o$ = 79.5 mm, $d_i$ = 47.7 mm; mass saving about 29.8 percent.**

## Check yourself

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

1. The solid shaft for 90 kW at 250 rpm and 40 MPa has d =
   A. 79.5 mm
   B. 47.7 mm
   C. 75.9 mm
   D. 60 mm
   Answer: C. T = 3437.7 N.m and d^3 = 16T/(pi tau). (Key point: Solid shaft diameter)

2. For the same torque and stress, outside diameter of the hollow shaft is found from
   A. do^3 = d^3 (1 - k^4)
   B. do = d/(1 - k)
   C. do^2 = d^2/(1 - k^2)
   D. do^3 = d^3/(1 - k^4)
   Answer: D. Torque capacity is (pi/16) tau do^3 (1 - k^4). (Key point: Hollow shaft relation)

3. Mass saving is based on
   A. The ratio of cross-sectional areas, about 30 percent
   B. The ratio of diameters, about 5 percent
   C. The ratio of torques, 0 percent
   D. The ratio of lengths
   Answer: A. Ah/As = 0.702. (Key point: Mass saving)

4. Why does a hollow shaft save material?
   A. The outer layer carries little stress
   B. The inner core carries little shear stress
   C. It has higher yield strength
   D. It is easier to make
   Answer: B. Shear stress is largest at the surface. (Key point: Why hollow saves)

## Related in Machine Design I

- [Diameter of a shaft carrying a spur gear](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Agear-loaded-shaft-diameter) — Machine Design I, chapter 5
- [Shaft diameter for power and bending](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Astatic-shaft-power-bending-diameter) — Machine Design I, chapter 5
- [ASME code equation for shaft diameter](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aasme-code-shaft-diameter-derivation) — Machine Design I, chapter 5
- [Bending stress and safety factor of a spur pinion](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aagma-bending-stress-safety-factor) — Machine Design I, chapter 9

## Related in other subjects

- [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
- [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 30 kW at 240 rpm: diameters](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-30kw-240rpm) — Strength of Materials, chapter 6