Physics 9702 · Paper 1

Stress & Strain

The spring constant $k$ compares whole springs. To compare materials, we first remove the effect of size and shape.
Stress $\sigma=F/A$ is force per unit cross-sectional area, in pascals (Pa).
Strain $\varepsilon=x/L$ is extension over original length: a ratio, so it has no unit.

Why we divide by area and by length

\sigma=\dfrac{F}{A} \qquad \varepsilon=\dfrac{x}{L}

Symbols

$\sigma$ = stress, force per unit cross-sectional area (Pa)
$F$ = force applied (N)
$A$ = cross-sectional area (m²)
$\varepsilon$ = strain, extension per original length (no unit)
$x$ = extension (m)
$L$ = original length (m)

A thick wire stretches less than a thin one under the same pull, so we divide the force by the area to cancel out the thickness. That gives stress $\sigma=F/A$ .
A long wire stretches more than a short one, so we divide the extension by the original length to cancel out the length. That gives strain $\varepsilon=x/L$ .
For a wire of diameter $d$ , the cross-sectional area is a circle: $A=\pi(d/2)^2$ .
Strain is a pure number, sometimes written as a percentage. Stress carries the unit, the pascal.

Stress and strain describe the material, not the particular piece. Two rods of the same metal, thick or thin, long or short, reach the same stress and strain when stretched the same way.

Top: two wires of the same length under the same load, the thinner one stretching more. Bottom: two wires of the same thickness under the same load, the longer one stretching more. — Same force on each wire, yet the thinner one (top) and the longer one (bottom) stretch more. Dividing by area (stress) and by length (strain) cancels thickness and length, so we can compare the material itself.

Worked examples: one stress, one strain

Worked example

Finding the stress

A force of 40 N pulls on a wire of cross-sectional area 2.0 × 10^-6 m².

Use $\sigma=F/A$ .
$\sigma=40/(2.0\times10^{-6})=2.0\times10^{7}\text{ Pa}$ .

Answer

\sigma=2.0\times10^{7}\text{ Pa}

Worked example

Finding the strain

A 3.0 m wire stretches by 1.5 mm.

Convert the extension: $x=1.5\text{ mm}=1.5\times10^{-3}\text{ m}$ .
$\varepsilon=x/L=1.5\times10^{-3}/3.0=5.0\times10^{-4}$ .

Answer

\varepsilon=5.0\times10^{-4}

Common mistake

Strain has no unit, so never write a pascal after it. And do not confuse

L

(original length) with

x

(extension): strain is the small extension over the full length, not the other way round.

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Stress & Strain

The spring constant $k$ compares whole springs. To compare materials, we first remove the effect of size and shape.
Stress $\sigma=F/A$ is force per unit cross-sectional area, in pascals (Pa).
Strain $\varepsilon=x/L$ is extension over original length: a ratio, so it has no unit.

Why we divide by area and by length

\sigma=\dfrac{F}{A} \qquad \varepsilon=\dfrac{x}{L}

Symbols

$\sigma$ = stress, force per unit cross-sectional area (Pa)
$F$ = force applied (N)
$A$ = cross-sectional area (m²)
$\varepsilon$ = strain, extension per original length (no unit)
$x$ = extension (m)
$L$ = original length (m)

A thick wire stretches less than a thin one under the same pull, so we divide the force by the area to cancel out the thickness. That gives stress $\sigma=F/A$ .
A long wire stretches more than a short one, so we divide the extension by the original length to cancel out the length. That gives strain $\varepsilon=x/L$ .
For a wire of diameter $d$ , the cross-sectional area is a circle: $A=\pi(d/2)^2$ .
Strain is a pure number, sometimes written as a percentage. Stress carries the unit, the pascal.

Stress and strain describe the material, not the particular piece. Two rods of the same metal, thick or thin, long or short, reach the same stress and strain when stretched the same way.

Worked examples: one stress, one strain

Worked example

Finding the stress

A force of 40 N pulls on a wire of cross-sectional area 2.0 × 10^-6 m².

Use $\sigma=F/A$ .
$\sigma=40/(2.0\times10^{-6})=2.0\times10^{7}\text{ Pa}$ .

Answer

\sigma=2.0\times10^{7}\text{ Pa}

Worked example

Finding the strain

A 3.0 m wire stretches by 1.5 mm.

Convert the extension: $x=1.5\text{ mm}=1.5\times10^{-3}\text{ m}$ .
$\varepsilon=x/L=1.5\times10^{-3}/3.0=5.0\times10^{-4}$ .

Answer

\varepsilon=5.0\times10^{-4}

Common mistake

Strain has no unit, so never write a pascal after it. And do not confuse

L

(original length) with

x

(extension): strain is the small extension over the full length, not the other way round.