---
title: "Hollow shaft 1670 kW at 4500 rpm: size"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:hollow-shaft-1670kw-4500rpm"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-shaft-1670kw-4500rpm
site: BCT Engineering
---

# Hollow shaft 1670 kW at 4500 rpm: size

Strength of Materials · Chapter 6: Torsion

- Asked in: 2076 Asoj, 8 marks
- Video: [Mechanic of Materials: Lesson 24 - Power Transmission Torque Example](https://www.youtube.com/watch?v=Re_9Kxx1WEQ) — Jeff Hanson
- 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-1670kw-4500rpm
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

A 1 m long hollow cylindrical shaft is to be designed to transmit a power of 1670 KW at a rotational speed of 4500 rpm. The outer diameter is to be 1.75 times the inner diameter. The maximum shear stress of the material is to be limited to 210 MPa and the angle of twist is not to exceed 0.5 degrees. Assume maximum torque is 30% greater than the average torque. shear modulus of material is 25.5 GPa. Determine the size of the shaft.

## Short answer

Average torque 3.544 kN·m, maximum 4.607 kN·m. Strength gives outer diameter 50.0 mm, stiffness gives 124.0 mm; stiffness governs: outer diameter ≈124.0 mm, inner ≈70.8 mm.

## Answer

**Data:** $L = 1$ m, $P = 1670$ kW, $N = 4500$ rpm, $D = 1.75d$, $\tau_{max} = 210\ \text{MPa}$, $\theta \le 0.5^\circ$, $T_{max} = 1.3\,T_{mean}$, $G = 25.5\ \text{GPa}$. So $k = d/D = 1/1.75 = 0.5714$ and $1-k^4 = 0.8934$.

#### Torque
$$
T_{mean} = \frac{60P}{2\pi N} = \frac{60(1670\times10^3)}{2\pi(4500)} = 3543.9\ \text{N·m}, \qquad T_{max} = 1.3T_{mean} = 4607.0\ \text{N·m}
$$

#### Strength condition
$$
T_{max} = \frac{\pi}{16}\tau D^3(1-k^4) \Rightarrow D^3 = \frac{16(4.6070\times10^{6})}{\pi(210)(0.8934)} = 1.2506\times10^{5}\ \text{mm}^3
$$

$$
D_1 = 50.01\ \text{mm}
$$

#### Stiffness condition
$$
\theta = \frac{T_{max}L}{GJ}, \quad J = \frac{\pi}{32}D^4(1-k^4), \quad \theta = 0.5^\circ = 0.00873\ \text{rad}
$$

$$
D^4 = \frac{32\,T_{max}L}{\pi G\theta(1-k^4)} = \frac{32(4.6070\times10^{6})(1000)}{\pi(25\,500)(0.00873)(0.8934)} = 2.3605\times10^{8}\ \text{mm}^4
$$

$$
D_2 = 123.95\ \text{mm}
$$

#### Size of the shaft
The larger value governs (stiffness controls):

$$
D = 123.95\ \text{mm}, \qquad d = \frac{D}{1.75} = 70.83\ \text{mm}
$$

**Answer: external diameter $\approx 124.0$ mm, internal diameter $\approx 70.8$ mm.**

## Check yourself

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

1. What is design torque?
   A. 3543.9 N.m
   B. 1.3 x 3543.9 = 4607 N.m
   C. 1670 N.m
   D. 5000 N.m
   Answer: B. Max is 30% over average. (Key point: Torque)

2. What D does strength give?
   A. 123.95 mm
   B. 70.8 mm
   C. 50.01 mm
   D. 210 mm
   Answer: C. D^3 = 16T/(pi tau (1 - k^4)). (Key point: Strength)

3. What controls?
   A. Strength
   B. Shear modulus
   C. Neither
   D. Stiffness, D = 123.95 mm
   Answer: D. Larger value. (Key point: Stiffness)

4. What is d?
   A. D/1.75 = 70.83 mm
   B. 1.75 D
   C. D - 1.75
   D. 50 mm
   Answer: A. D = 1.75 d. (Key point: Inner)

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