Stress and strain: making the comparison fair

7 minAS Level6.1.5

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can define stress and strain, calculate both, and explain why a material's response is described with these ratios rather than with force and extension.

Key to the symbols

The main symbols used in this lesson:

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

The problem with force and extension

Take two steel wires and hang the same 50 N weight on each. The thick wire stretches a little; the thin wire stretches a lot. Lengthen one of them and it stretches more again.

So "it stretched 2 mm under 50 N" tells you about this wire, not about steel.

To compare materials fairly, we need to remove the effects of thickness and length. That is exactly what stress and strain do.

Stress divides the force by the cross-sectional area — removing the thickness.
Strain divides the extension by the original length — removing the length.

Definition

Tensile stress

The force per unit cross-sectional area, acting perpendicular to that area:

$\sigma = \frac{F}{A}$

where $F$ is the force in newtons and $A$ the cross-sectional area in $\text{m}^2$ . The unit is the pascal, $\text{Pa} = \text{N m}^{-2}$ .

The symbol is the Greek letter sigma, $\sigma$ ("tensile stress" or "compressive stress" depending on direction).

Definition

Tensile strain

The extension per unit original length:

$\varepsilon = \frac{x}{L}$

where $x$ is the extension and $L$ the original length, both in metres.

Strain has no unit: it is a length divided by a length. It is often written as a percentage, so a strain of $1.0 \times 10^{-3}$ is 0.10%.

Why dividing makes the comparison fair

Stress: the same force spread over twice the area is half as demanding on each square millimetre. Dividing by area turns "50 N" into "how hard is each bit of the material being pulled?", which is what the material actually feels.

Strain: stretch a 1 m wire by 1 mm and a 2 m wire of the same thickness by 1 mm, and the short wire is strained twice as much — its atoms are pulled apart proportionally more. Dividing by the original length captures that as a fraction.

Both quantities are ratios, so they describe the material at that point, not the piece of metal you happen to be holding.

Worked example: Worked example: stress in a wire

A steel wire of diameter 0.40 mm carries a tensile force of 20 N.

Calculate the tensile stress in the wire.

Have a go first?

Worked example: Worked example: strain in the same wire

The wire above has an original length of 2.50 m and stretches by 1.2 mm when the 20 N force is applied.

Calculate the tensile strain.

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

Multiple choiceQuestion 2 · Recall

Answer the 3 questions above to continue.