---
title: "Pull test on bar: Poisson's ratio, moduli"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:bar-pull-test-poisson-elastic-moduli"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/bar-pull-test-poisson-elastic-moduli
site: BCT Engineering
---

# Pull test on bar: Poisson's ratio, moduli

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2076 Asoj, 6 marks
- Video: [Poisson’s ratio, Modulus of elasticity , Modulus of rigidity ( Shear Modulus), Bulk Modulus](https://www.youtube.com/watch?v=53cUUpDSr5w) — 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/bar-pull-test-poisson-elastic-moduli
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

In an experiment, a bar of 30 mm diameter is subjected to a pull of 60 kN. The measured extension on gauge length of 200 mm is 0.09 mm and the change in diameter is 0.0039 mm. Calculate the values of Poisson's ratio and three elastic moduli.

## Short answer

Compute longitudinal strain from extension and gauge length, lateral strain from diameter change. Poisson's ratio μ = 0.289. Elastic modulus E = 188.6 kN/mm², shear modulus G = 73.2 kN/mm², bulk modulus K = 148.9 kN/mm².

## Answer

#### Longitudinal quantities
$$
\begin{aligned}
A &= \frac{\pi}{4}(30)^2 = 706.86\ \text{mm}^2\\
\sigma &= \frac{P}{A} = \frac{60\times10^3}{706.86} = 84.883\ \text{N/mm}^2\\
\varepsilon_{long} &= \frac{\delta L}{L} = \frac{0.09}{200} = 0.000450\\
E &= \frac{\sigma}{\varepsilon} = \frac{84.883}{0.000450} = 188628\ \text{N/mm}^2 = 188.6\ \text{kN/mm}^2
\end{aligned}
$$

#### Poisson's ratio
$$
\varepsilon_{lat} = \frac{\delta d}{d} = \frac{0.0039}{30} = 0.000130,\qquad
\mu = \frac{\varepsilon_{lat}}{\varepsilon_{long}} = \frac{0.000130}{0.000450} = 0.2889
$$

#### Other moduli
$$
\begin{aligned}
G &= \frac{E}{2(1+\mu)} = \frac{188628}{2(1+0.2889)} = 73175\ \text{N/mm}^2\\
K &= \frac{E}{3(1-2\mu)} = \frac{188628}{3(1-2\times0.2889)} = 148917\ \text{N/mm}^2
\end{aligned}
$$

**Answer: $\mu = 0.289$; $E = 188.6$ kN/mm²; $G = 73.2$ kN/mm²; $K = 148.9$ kN/mm².**

## Check yourself

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

1. How is E found from the test?
   A. E = (P/A)/(delta L/L) = 188.6 kN/mm^2
   B. E = lateral strain / longitudinal strain
   C. E = P x delta L / A
   D. E = 2G(1 - mu)
   Answer: A. Stress 84.88 N/mm^2 divided by strain 0.00045 gives about 188.6 kN/mm^2. (Key point: Young's modulus)

2. How is Poisson's ratio obtained?
   A. Longitudinal strain / lateral strain
   B. Lateral strain / longitudinal strain = 0.289
   C. delta d / delta L
   D. E / G
   Answer: B. Lateral strain 0.00013 / longitudinal 0.00045 = 0.289. (Key point: Poisson's ratio)

3. Which formula gives G?
   A. G = E / [3(1 - 2 mu)]
   B. G = E / (1 + 2 mu)
   C. G = E / [2(1 + mu)]
   D. G = 2E(1 + mu)
   Answer: C. G = E/(2(1+mu)) gives about 73.2 kN/mm^2; the other formula is for K. (Key point: Shear modulus)

4. Which formula gives K?
   A. K = E / [2(1 + mu)]
   B. K = E / 3
   C. K = 3E(1 - 2 mu)
   D. K = E / [3(1 - 2 mu)] = 148.9 kN/mm^2
   Answer: D. K = E/(3(1-2mu)) with mu = 0.289 gives about 148.9 kN/mm^2. (Key point: Bulk modulus)

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