---
title: "Hollow shaft 30 kW at 240 rpm: diameters"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:hollow-shaft-30kw-240rpm"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-shaft-30kw-240rpm
site: BCT Engineering
---

# Hollow shaft 30 kW at 240 rpm: diameters

Strength of Materials · Chapter 6: Torsion

- Asked in: 2066 Jestha (old course), 8 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-30kw-240rpm
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

Determine the internal and external diameters of a hollow shaft to transmit a power of 30 kW at 240 rpm if the shear stress must not exceed 100 MPa. The outside diameter is to be twice the inside diameter.

## Short answer

Torque T=60P/(2πN)=1.194 kN·m. Using T=(π/16)τD³(1-k⁴) with k=0.5, τ=100 MPa, gives D=40.2 mm, d=20.1 mm.

## Answer

**Data:** $P = 30$ kW, $N = 240$ rpm, $\tau_{max} = 100\ \text{MPa}$, $D = 2d$ (so $k = d/D = 0.5$ and $1-k^4 = 0.9375$).

#### Torque
$$
T = \frac{60P}{2\pi N} = \frac{60(30\times10^3)}{2\pi(240)} = 1193.7\ \text{N·m} = 1.1937\times10^{6}\ \text{N·mm}
$$

#### Diameters
$$
T = \frac{\pi}{16}\tau\,D^3(1-k^4) \Rightarrow D^3 = \frac{16T}{\pi\tau(1-k^4)} = \frac{16(1.1937\times10^{6})}{\pi(100)(0.9375)} = 6.4846\times10^{4}\ \text{mm}^3
$$

$$
D = 40.18\ \text{mm}, \qquad d = \frac{D}{2} = 20.09\ \text{mm}
$$

**Answer: external diameter $\approx 40.2$ mm, internal diameter $\approx 20.1$ mm.** (In practice adopt 41 mm and 20.5 mm.)

## Check yourself

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

1. Which formula gives torque from P in W and N in rpm?
   A. T = P N
   B. T = 60P/(2 pi N)
   C. T = 2 pi N/(60P)
   D. T = P/(2 pi N)
   Answer: B. P = T*2 pi N/60, so T = 60P/(2 pi N), about 1193.7 N·m for 30 kW at 240 rpm. (Key point: Torque from power)

2. For D = 2d, the value of 1 - k^4 is
   A. 0.5
   B. 0.75
   C. 0.875
   D. 0.9375
   Answer: D. k = 0.5, k^4 = 0.0625. (Key point: Factor 1 - k^4)

3. Which relation yields D?
   A. D^3 = 16T/(pi tau (1 - k^4))
   B. D^4 = 32T/(pi tau (1 - k^4))
   C. D^2 = 4T/(pi tau)
   D. D^3 = 16T/(pi tau)
   Answer: A. Rearranging T = (pi/16) tau D^3 (1 - k^4). (Key point: Solving for D)

4. What diameters result for 30 kW?
   A. D = 40.2 mm, d = 20.1 mm
   B. D = 20.1 mm, d = 10.05 mm
   C. D = 44.2 mm, d = 22.1 mm
   D. D = 80.4 mm, d = 40.2 mm
   Answer: A. D^3 = 6.4846e4 mm^3 gives D = 40.18 mm and d = 20.09 mm. (Key point: Result)

## Related in Strength of Materials

- [Hollow shaft 40 kW at 240 rpm: diameters](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-40kw-240rpm) — 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
- [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
- [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×
- [Solid shaft 300 kW, 120 rpm with twist limit](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-120rpm-twist-limit) — Strength of Materials, chapter 6
- [Max shear for 150 MPa stress limit; principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Adirect-stresses-100t-60c-shear-limit) — Strength of Materials, chapter 4

## 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