---
title: "Solid shaft 300 kW; hollow shaft saving"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:solid-shaft-300kw-percent-saving"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/solid-shaft-300kw-percent-saving
site: BCT Engineering
---

# Solid shaft 300 kW; hollow shaft saving

Strength of Materials · Chapter 6: Torsion

- Asked in: 2080 Bhadra, 8 marks
- Video: [Problem on Torsion/Hollow shaft replaced by solid shaft/ Diameter of shaft/ Strength of Materials](https://www.youtube.com/watch?v=cvMGSczxEV4) — 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/solid-shaft-300kw-percent-saving
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

A solid shaft is to transmit 300 kW at 120 rpm. If the shear stress is not to exceed 100 MPa, find the diameter of the shaft. What percent saving in weight would be obtained if this shaft were replaced by a hollow one whose internal diameter equals 0.6 of the external diameter, the length, material and maximum allowable shear stress being the same?

## Short answer

Torque T = 60×300000/(2π×120) = 23.87 kNm. Solid diameter = [16T/(π×100)]¹/³ = 106.7 mm. Hollow with d_i=0.6D_o: D_o = [16T/(π×100×(1-0.6⁴))]¹/³ = 111.8 mm. Weight saving = (8947-6281)/8947 = 29.8%.

## Answer

**Data:** $P = 300$ kW, $N = 120$ rpm, $\tau_{max} = 100\ \text{MPa}$.

#### Torque
$$
T = \frac{60P}{2\pi N} = \frac{60(300\times10^3)}{2\pi(120)} = 23873.2\ \text{N·m} = 2.3873\times10^{7}\ \text{N·mm}
$$

#### Solid shaft
$$
T = \frac{\pi}{16}\tau d^3 \Rightarrow d^3 = \frac{16T}{\pi\tau} = \frac{16(2.3873\times10^{7})}{\pi(100)} = 1.2159\times10^{6}\ \text{mm}^3
$$

$$
d = 106.73\ \text{mm}
$$

#### Hollow shaft ($d_i = 0.6D_o$, same $T$ and $\tau$)
$$
T = \frac{\pi}{16}\tau D_o^3(1-0.6^4) \Rightarrow D_o^3 = \frac{16T}{\pi\tau(1-0.1296)} = 1.3969\times10^{6}\ \text{mm}^3
$$

$$
D_o = 111.79\ \text{mm}, \qquad d_i = 0.6D_o = 67.07\ \text{mm}
$$

#### Saving in weight
Weight $\propto$ area (same length and material).

$$
A_s = \frac{\pi}{4}(106.73)^2 = 8947.0\ \text{mm}^2, \qquad A_h = \frac{\pi}{4}(111.79^2 - 67.07^2) = 6281.2\ \text{mm}^2
$$

$$
\text{Saving} = \frac{A_s - A_h}{A_s}\times100 = 29.8\%
$$

**Answer:** solid shaft $d = 106.7$ mm; the hollow shaft ($D_o = 111.8$ mm, $d_i = 67.1$ mm) saves about **29.8%** of the weight.

## Check yourself

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

1. How is torque obtained from power?
   A. T = P/N
   B. T = 60P/(2 pi N) = 23873 N.m
   C. T = 2 pi N P
   D. T = P x 60
   Answer: B. P = 2 pi N T/60. (Key point: Torque)

2. What is the solid shaft diameter?
   A. 111.8 mm
   B. 67 mm
   C. 106.73 mm
   D. 53 mm
   Answer: C. d^3 = 16T/(pi tau). (Key point: Solid diameter)

3. Which factor modifies torque capacity for k = 0.6?
   A. (1 - 0.6^2)
   B. (1 - 0.6)
   C. (1 + 0.6^4)
   D. (1 - 0.6^4) = 0.8704
   Answer: D. T = (pi/16) tau D^3 (1 - k^4). (Key point: Hollow shaft)

4. How is percentage saving found?
   A. (A_s - A_h)/A_s x 100 = 29.8%
   B. (D_s - D_o)/D_s
   C. (A_h - A_s)/A_h
   D. d_i/D_o
   Answer: A. Weight is proportional to area for the same length and material. (Key point: Saving)

## Related in Strength of Materials

- [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
- [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
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- [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 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
- [Solid shaft 120 Nm: diameter from limits](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-120nm-twist-limit) — Strength of Materials, chapter 6, asked 2×

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