---
title: "Solid shaft 300 kW, 120 rpm with twist limit"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:solid-shaft-300kw-120rpm-twist-limit"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/solid-shaft-300kw-120rpm-twist-limit
site: BCT Engineering
---

# Solid shaft 300 kW, 120 rpm with twist limit

Strength of Materials · Chapter 6: Torsion

- Asked in: 2079 Bhadra, 8 marks
- Video: [Problem on Calculation of Diameter of Solid Circular Shaft - Torsion - Strength of Materials](https://www.youtube.com/watch?v=ZX7wgYDJOK8) — Ekeeda
- 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-120rpm-twist-limit
- 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. Determine the diameter if the allowable shear stress is $100\ \text{N/mm}^2$ and the allowable angle of twist is $30^\circ$ per diameter length of the shaft. Assume that the maximum torque is 1.3 times the mean torque. Take $G = 10^5\ \text{MPa}$.

## Short answer

Mean torque 23.87 kN·m, maximum 31.04 kN·m. Strength criterion gives diameter 116.5 mm, stiffness 18.2 mm; the larger governs: required diameter ≈116.5 mm.

## Answer

**Data:** $P = 300$ kW, $N = 120$ rpm, $\tau_{allow} = 100\ \text{N/mm}^2$, $G = 10^5\ \text{N/mm}^2$, allowable twist $30^\circ$ per diameter length (taken as stated, $L = d$), $T_{max} = 1.3\,T_{mean}$.

#### Torques
$$
T_{mean} = \frac{60P}{2\pi N} = \frac{60(300\times10^3)}{2\pi(120)} = 23873.2\ \text{N·m}, \qquad T_{max} = 1.3T_{mean} = 31035.2\ \text{N·m}
$$

#### Strength
$$
d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(3.1035\times10^{7})}{\pi(100)} = 1.5806\times10^{6}\ \text{mm}^3 \Rightarrow d_1 = 116.49\ \text{mm}
$$

#### Stiffness
$$
\theta = \frac{T_{max}L}{GJ} = \frac{32\,T_{max}\,d}{\pi G d^4} = \frac{32T_{max}}{\pi G d^3}, \quad \theta = 30^\circ = 0.5236\ \text{rad}
$$

$$
d^3 = \frac{32(3.1035\times10^{7})}{\pi(10^5)(0.5236)} = 6.0375\times10^{3}\ \text{mm}^3 \Rightarrow d_2 = 18.21\ \text{mm}
$$

#### Design
The larger value governs, so strength controls the design: $d = 116.49$ mm.

**Answer: diameter of the shaft $\approx 116.5\ \text{mm}$ (adopt 120 mm).**

## Check yourself

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

1. Which torque is used for design?
   A. T_mean only
   B. T_max = 1.3 T_mean = 31035 N.m
   C. T_mean/1.3
   D. 2 T_mean
   Answer: B. The peak torque governs. (Key point: Design torque)

2. What does theta become when L = d?
   A. 32T/(pi G d^4)
   B. 16T/(pi G d^3)
   C. 32T/(pi G d^3)
   D. T/(G d)
   Answer: C. d from L cancels one d from d^4. (Key point: Stiffness with L = d)

3. What d does strength give?
   A. 18.21 mm
   B. 106.7 mm
   C. 124 mm
   D. 116.49 mm
   Answer: D. d^3 = 16T_max/(pi tau). (Key point: Strength result)

4. Which governs?
   A. Strength, because it gives larger diameter
   B. Stiffness
   C. Neither
   D. Average
   Answer: A. The larger satisfies both. (Key point: Governing)

## Related in Strength of Materials

- [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×
- [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
- [Steel shaft 200 hp: torque, shear, twist](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-shaft-200hp-150rpm) — 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
- [Fixed shaft: end couples, diameter, section of no twist](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ashaft-fixed-ends-couples-no-twist) — 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

## 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
- [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
- [Stator design of 120 kW 750 rpm motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Astator-120kw-750rpm) — Electric Machine Design, chapter 4, asked 2×