Stress, Strain and the Young Modulus

A spring or wire obeys Hooke's law, $F=kx$ , up to its limit of proportionality. To compare materials rather than particular objects, use stress ( $\sigma=\tfrac FA$ ) and strain ( $\varepsilon=\tfrac xL$ ). Their ratio in the linear region is the Young modulus, $E=\dfrac\sigma\varepsilon$ , a property of the material alone.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use Hooke's law and the spring constant.
Define and calculate stress and strain.
Calculate the Young modulus, including from a graph.
Describe an experiment to measure the Young modulus of a wire.

Pause and think

A thick steel cable and a thin steel wire are stretched by the same force. The thin wire stretches far more. Is the thin wire made of "stretchier" steel?

Section 2 of 14

Core idea

Start hereThick and thin wires

Hang the same load on a thick and a thin steel wire: the thin one stretches more. So stiffness depends on the sample. But divide force by area and extension by length, and every steel wire gives the same ratio: the Young modulus, a property of steel itself.

The big idea

$\sigma=\frac FA$ , $\varepsilon=\frac{\Delta L}L$ , $E=\frac\sigma\varepsilon=\frac{FL}{A\Delta L}$ in the elastic region.

1. Definitions

\sigma=\frac FA\qquad\varepsilon=\frac{\Delta L}L\qquad E=\frac{FL}{A\,\Delta L}

$\sigma$: stress (Pa).
$\varepsilon$: strain (no units).
$E$: Young modulus (Pa).

Stress–strain graph for a metal wire

Gradient of the straight part

=\dfrac{400\times10^6}{0.002}=2\times10^{11}

Pa: the Young modulus of steel.

Worked exampleSteel wire

$L=2.0$ m, diameter $0.50$ mm, load 50 N, $E=2.0\times10^{11}$ Pa.

Goal: Extension.
Strategy: Area → stress → strain → extension.

$A=\pi(0.25\times10^{-3})^2=1.96\times10^{-7}\text{ m}^2$
Radius, not diameter.
$\sigma=\frac{50}{1.96\times10^{-7}}=2.55\times10^8\text{ Pa}$
Stress.
$\varepsilon=\frac{2.55\times10^8}{2.0\times10^{11}}=1.27\times10^{-3}$
Strain.
$\Delta L=2.0\times1.27\times10^{-3}=2.5\text{ mm}$
Extension.

Answer

2.5

mm.

✓ Check

\frac{FL}{AE}=\frac{100}{39200}=2.55\times10^{-3}

m ✓.

Limit of proportionality: end of the straight part.
Elastic limit: beyond it, permanent deformation.
Ultimate tensile stress: maximum stress.

Exam tipMeasuring E Long thin wire; micrometer for diameter in several places; marker for extension; plot stress vs strain.

Common misconceptionDiameter as radius Halve the diameter before

\pi r^2

Pause and think

Double the diameter, same load and length. Extension ratio?

Summary

$\sigma=F/A$ , $\varepsilon=\Delta L/L$ .
$E$ = gradient of linear region.

Young modulus: Stress ÷ strain in the elastic region.

HistoryThomas Young, polymath. Thomas Young described the modulus that bears his name in

1807

. He also showed light is a wave (the double-slit experiment), studied how the eye focuses, and helped decipher the Rosetta Stone. He was called "the last man who knew everything".

Fun factSpider silk beats steel. Weight for weight, spider dragline silk is stronger than steel and far tougher, stretching about

30\%

before breaking. Scientists are engineering bacteria and goats to produce it for bulletproof vests and surgical threads.

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