---
title: "Poisson's ratio from torsion and tension tests"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:torque-and-tension-test-poisson-ratio"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/torque-and-tension-test-poisson-ratio
site: BCT Engineering
---

# Poisson's ratio from torsion and tension tests

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2069 Chaitra, 6 marks
- Video: [Understanding Poisson's Ratio](https://www.youtube.com/watch?v=tuOlM3P7ygA) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/torque-and-tension-test-poisson-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

A steel bar of 2.5 cm diameter when subjected to a torque of 300N produces an angle of twist of 1.35 degrees in the length of 25cm. The same bar when subjected to tension elongates 0.01cm in length of 15cm under a load of 70KN. Deduce the value of poisson's ratio for the material.

## Short answer

From torsion test: G = TL/(Jθ) = 83002 N/mm², with torque 300 N·m, twist 1.35° over 250 mm. From tension test: E = PL/(Aδ) = 213904 N/mm², with load 70 kN, elongation 0.1 mm over 150 mm. Poisson's ratio ν = E/(2G) - 1 = 0.289.

## Answer

Poisson's ratio follows from $E = 2G(1+\nu)$. Find $G$ from the torsion test and $E$ from the tension test. The torque is taken as 300 N·m ($= 3\times10^5$ N·mm).

#### Modulus of rigidity (torsion test)
$$
J = \frac{\pi d^4}{32} = \frac{\pi (25)^4}{32} = 38349.5\ \text{mm}^4, \qquad \theta = 1.35^\circ = 0.02356\ \text{rad}
$$

$$
\frac{T}{J} = \frac{G\theta}{L} \Rightarrow G = \frac{TL}{J\theta} = \frac{3\times10^5 \times 250}{38349.5\times 0.02356} = 83002\ \text{N/mm}^2
$$

#### Modulus of elasticity (tension test)
$A = \dfrac{\pi}{4}(25)^2 = 490.87\ \text{mm}^2$, $\delta = 0.01\ \text{cm} = 0.1\ \text{mm}$, $L = 150$ mm.

$$
E = \frac{PL}{A\delta} = \frac{70\,000\times150}{490.87\times0.1} = 213904\ \text{N/mm}^2
$$

#### Poisson's ratio
$$
\nu = \frac{E}{2G} - 1 = \frac{213904}{2(83002)} - 1 = 0.289
$$

**Answer: $\nu \approx 0.289$.**

## Check yourself

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

1. Which relation gives Poisson's ratio from E and G?
   A. E = 2G(1 + nu)
   B. E = G(1 + 2 nu)
   C. E = 3G(1 - 2 nu)
   D. E = G/(1 + nu)
   Answer: A. nu = E/(2G) - 1. (Key point: Relation)

2. Which equation gives G?
   A. T/J = theta / (G L)
   B. T/J = G theta / L
   C. T = G J L / theta
   D. T/J = G L / theta
   Answer: B. G = T L/(J theta) from the torsion equation, about 83000 N/mm^2. (Key point: Torsion)

3. How is E found?
   A. E = P L /(A delta) = 213904 N/mm^2
   B. E = P A / L
   C. E = T L / J theta
   D. E = 2G
   Answer: A. Axial test gives E directly. (Key point: Tension)

4. What is nu?
   A. 0.5
   B. 0.35
   C. 0.289
   D. 1.58
   Answer: C. 213904/(2 x 83002) - 1 = 0.289. (Key point: Result)

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