The Young modulus: a fingerprint for a material

By the end of this lesson

By the end you can define the Young modulus, calculate it from measurements on a wire, and read it as the gradient of the straight part of a stress–strain graph.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$\sigma$	stress
$\varepsilon$	strain

From comparison to identity

Stress and strain already remove the size of the sample. Divide one by the other and you get a number that depends only on the material:

$\text{Young modulus } E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{\sigma}{\varepsilon}$

Steel comes out at about $2 \times 10^{11}\ \text{Pa}$ , copper at about $1.2 \times 10^{11}\ \text{Pa}$ , rubber at around $10^{7}\ \text{Pa}$ . Those numbers are fingerprints: they do not change when you use a longer or thicker wire.

A large $E$ means the material resists being stretched — it is stiff. Steel has a large $E$ ; rubber has a tiny one.

Definition

Young modulus

The ratio of tensile stress to tensile strain for a material, provided the limit of proportionality is not exceeded:

$E = \frac{\sigma}{\varepsilon}$

Its unit is the pascal, Pa (equivalently $\text{N m}^{-2}$ ), because strain has no unit.

Why the size cancels out

Write the definition out in terms of what you can measure:

$E = \frac{\sigma}{\varepsilon} = \frac{F/A}{x/L} = \frac{F\,L}{A\,x}$

Now look at what happens if you double the length $L$ of the wire. The extension $x$ also doubles, so $L/x$ is unchanged. Double the cross-sectional area and the extension halves, so $A\,x$ is unchanged too.

The dimensions cancel. What survives is a property of the metal, which is exactly what engineers need when they choose a material for a bridge rather than for a lab sample.

Worked example: Worked example: the full calculation

A steel wire is 2.00 m long and 0.50 mm in diameter. It extends by 2.5 mm when a load of 50 N is applied.

Calculate the Young modulus of the steel.

Have a go first?

Worked example: Worked example: using E to predict an extension

A copper wire of length 3.0 m and diameter 0.60 mm has $E = 1.2 \times 10^{11}\ \text{Pa}$ .

What extension does a 30 N load produce, assuming the wire stays within its limit of proportionality?

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

Multiple choiceQuestion 2 · Recall

Multiple choiceQuestion 3 · Apply

Answer the 3 questions above to continue.