---
title: "Hollow shaft 40 kW at 240 rpm: diameters"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:hollow-shaft-40kw-240rpm"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-shaft-40kw-240rpm
site: BCT Engineering
---

# Hollow shaft 40 kW at 240 rpm: diameters

Strength of Materials · Chapter 6: Torsion

- Asked in: 2069 Asar, 6 marks
- Video: [Problem on Torsion/Power Transmitted by hollow shaft/Diameter of Shaft/ Strength of Materials](https://www.youtube.com/watch?v=LT8wDQWD684) — Prof.Dr.Pravin Patil
- 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-40kw-240rpm
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

Find the external and internal diameters required for a hollow shaft which is to transmit 40 KW of power at 240 rev/minute. The shear stress is to be limited to $100\ \text{MN/m}^2$. Take external diameter to be twice the internal diameter.

## Short answer

Torque = 1.5915 kN·m. For D=2d and 100 MPa shear stress, outer diameter 44.2 mm, inner diameter 22.1 mm.

## Answer

**Data:** $P = 40$ kW, $N = 240$ rpm, $\tau_{max} = 100\ \text{MPa}$, $D = 2d$ (so $k = d/D = 0.5$ and $1-k^4 = 0.9375$).

#### Torque
$$
T = \frac{60P}{2\pi N} = \frac{60(40\times10^3)}{2\pi(240)} = 1591.5\ \text{N·m} = 1.5915\times10^{6}\ \text{N·mm}
$$

#### Diameters
$$
T = \frac{\pi}{16}\tau\,D^3(1-k^4) \Rightarrow D^3 = \frac{16T}{\pi\tau(1-k^4)} = \frac{16(1.5915\times10^{6})}{\pi(100)(0.9375)} = 8.6461\times10^{4}\ \text{mm}^3
$$

$$
D = 44.22\ \text{mm}, \qquad d = \frac{D}{2} = 22.11\ \text{mm}
$$

**Answer: external diameter $\approx 44.2$ mm, internal diameter $\approx 22.1$ mm.** (In practice adopt 45 mm and 22.5 mm.)

## Check yourself

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

1. How is the torque transmitted calculated from power P (W) and speed N (rpm)?
   A. T = 2*pi*P/(60N^2)
   B. T = 60P/(2*pi*N)
   C. T = 2*pi*N*P/60
   D. T = P/N
   Answer: B. Power = T*omega with omega = 2*pi*N/60, so T = 60P/(2*pi*N). (Key point: Torque from power)

2. Which expression gives the torque capacity of a hollow shaft of outer diameter D and k = d/D?
   A. (pi/16) tau D^3 (1 - k^2)
   B. (pi/32) tau D^4 (1 - k^4)
   C. (pi/16) tau D^3 k^4
   D. (pi/16) tau D^3 (1 - k^4)
   Answer: D. Using J = (pi/32)D^4(1-k^4) and tau = T*(D/2)/J gives T = (pi/16) tau D^3 (1-k^4). (Key point: Hollow shaft torque formula)

3. When the outer diameter is twice the inner diameter, what is the factor (1 - k^4)?
   A. 0.5
   B. 0.75
   C. 0.25
   D. 0.9375
   Answer: D. k = 0.5, so k^4 = 0.0625 and 1 - k^4 = 0.9375. (Key point: Effect of D = 2d)

4. With T about 1591.5 N·m and tau = 100 MPa, what are the diameters?
   A. D about 44.2 mm, d about 44.2 mm
   B. D about 88.4 mm, d about 44.2 mm
   C. D about 22.1 mm, d about 11.05 mm
   D. D about 44.2 mm, d about 22.1 mm
   Answer: D. D^3 = 16T/(pi*tau*0.9375) gives D about 44.2 mm and d = D/2 about 22.1 mm. (Key point: Final result)

## Related in Strength of Materials

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- [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
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- [Hollow shaft 10/7 cm: power transmitted](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-power-300rpm-13-mean) — Strength of Materials, chapter 6, asked 2×

## 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