---
title: "Solid shaft 120 Nm: diameter from limits"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:solid-shaft-120nm-twist-limit"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/solid-shaft-120nm-twist-limit
site: BCT Engineering
---

# Solid shaft 120 Nm: diameter from limits

Strength of Materials · Chapter 6: Torsion

- Asked in: 2073 Shrawan, 8 marks; 2072 Chaitra, 8 marks (asked 2 times)
- 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-120nm-twist-limit
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

A solid circular shaft is subjected to a torque 120 Nm. Determine the diameter if the allowable shear stress is $100\ \text{N/mm}^2$ and the allowable angle of twist is $30^\circ$ per 10 diameter length of the shaft. $G = 10^5\ \text{N/mm}^2$.

## Short answer

Strength condition: d = (16T/(πτ))¹/³ = (16×1.2×10⁵/(π×100))¹/³ = 18.28 mm. Stiffness condition: using twist 0.5236 rad over 10d gives d = (320T/(πGθ))¹/³ = (320×1.2×10⁵/(π×10⁵×0.5236))¹/³ = 6.16 mm. Use larger diameter 18.28 mm, adopt 20 mm.

## Answer

**Data:** $T = 120\ \text{N·m} = 1.2\times10^5\ \text{N·mm}$, $\tau_{allow} = 100\ \text{N/mm}^2$, $G = 10^5\ \text{N/mm}^2$, allowable twist $\theta = 30^\circ = 0.5236$ rad over a length $L = 10d$ (taken as stated in the question). Find the diameter satisfying both strength and stiffness; the larger value governs.

#### (a) Strength condition
$$
T = \frac{\pi}{16}\tau d^3 \Rightarrow d^3 = \frac{16T}{\pi\tau} = \frac{16(1.2\times10^5)}{\pi(100)} = 6111.5\ \text{mm}^3
$$

$$
d_1 = 18.28\ \text{mm}
$$

#### (b) Stiffness condition
$$
\theta = \frac{TL}{GJ} = \frac{T(10d)}{G\frac{\pi}{32}d^4} = \frac{320\,T}{\pi G d^3} \Rightarrow d^3 = \frac{320(1.2\times10^5)}{\pi(10^5)(0.5236)} = 233.44\ \text{mm}^3
$$

$$
d_2 = 6.16\ \text{mm}
$$

#### Design diameter
The larger of the two governs, so the shaft is designed for strength: $d = 18.28$ mm.

**Answer: provide a solid shaft of diameter $d = 18.28\ \text{mm}$, adopt 20 mm.**

## Check yourself

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

1. What two conditions are applied?
   A. Only strength
   B. Strength (shear stress) and stiffness (angle of twist), larger d governs
   C. Strength and bending
   D. Smaller d governs
   Answer: B. A shaft must meet both limits. (Key point: Two checks)

2. What is d from strength?
   A. d^3 = 32T/(pi tau), d = 23 mm
   B. d^3 = 16T/(pi G), d = 5 mm
   C. d^3 = 16T/(pi tau), d = 18.28 mm
   D. d = 6.16 mm
   Answer: C. T = pi tau d^3/16. (Key point: Strength)

3. How is length treated for stiffness?
   A. L = d
   B. L = 10 mm
   C. L cancels out
   D. L = 10 d, so theta = 320T/(pi G d^3)
   Answer: D. J = pi d^4/32 and L = 10 d. (Key point: Stiffness)

4. Which diameter is adopted?
   A. 18.28 mm, from strength (adopt 20 mm)
   B. 6.16 mm from stiffness
   C. 12.2 mm average
   D. 100 mm
   Answer: A. The stiffness value 6.16 mm is smaller, so strength governs. (Key point: Governing)

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