Stress-strain graphs

A stress-strain graph shows how a material deforms as it is stretched: tensile stress (force per unit area) up the side, tensile strain (extension ÷ original length) along the bottom. For a metal wire it starts as a straight line, the region where the gradient is the Young modulus, then curves past the elastic limit into a plastic region, reaches the ultimate tensile stress, and ends where the wire breaks. The area under the graph is the energy stored per unit volume.

Stress and strain

Stress is in pascals (Pa, the same as N m⁻²); strain has no units because it is a length divided by a length. Plotting stress against strain, rather than force against extension, removes the size of the sample: a thick wire and a thin wire of the same metal give the same stress-strain graph. The formulas are covered in detail in the strain equation guide.

The stress-strain curve for a metal

OpenStax notes that the limits are really narrow ranges rather than sharp points, and that every material has its own curve. For a typical metal, it gives the straight section as below about 1% strain, with plastic deformation from about 1% up to about 30%: that's why the plastic region is so long on the diagram, which isn't drawn to scale. OpenStax calls the stress at the fracture point the breaking stress, or ultimate stress.

Loading past the elastic limit

If the load is removed after the elastic limit, the wire doesn't return along the curve. It unloads along a line parallel to the original straight section (the dashed line on the diagram) and is left with a permanent strain where that line meets the strain axis. OpenStax describes the material relaxing to a new shape and size.

Ductile, brittle and polymeric materials

AQA asks you to describe plastic behaviour, fracture and brittle behaviour, and to interpret simple stress-strain curves. OpenStax's table of Young moduli shows why steel's line is steeper: about 20.0 × 10¹⁰ Pa for steel, 11.0 × 10¹⁰ Pa for copper and 7.0 × 10¹⁰ Pa for aluminium. OpenStax also describes rubber-like materials, which get harder to stretch as the strain grows, so their curve bends upwards rather than flattening.

Worked examples

Example 1: the Young modulus from the graph

On the straight part of a stress-strain graph for a copper wire, the point (strain 1.2 × 10⁻³, stress 1.32 × 10⁸ Pa) lies on the line through the origin. Find the Young modulus.

Gradient = 1.32 × 10⁸ ÷ 1.2 × 10⁻³ = 1.1 × 10¹¹ Pa, matching OpenStax's value for copper. Always take the gradient from the straight section only, using a big triangle.

Example 2: energy stored, from the area

For the same wire, find the energy stored per unit volume at that point, and the total energy in a 2.0 m length of the wire whose cross-section is 2.0 × 10⁻⁷ m².

1. Area under the straight line = ½ × stress × strain = ½ × 1.32 × 10⁸ × 1.2 × 10⁻³ = 7.9 × 10⁴ J m⁻³.
2. Volume = 2.0 × 2.0 × 10⁻⁷ = 4.0 × 10⁻⁷ m³.
3. Energy = 7.92 × 10⁴ × 4.0 × 10⁻⁷ = 0.032 J.
4. Check with AQA's formula, energy = ½FΔL: F = 1.32 × 10⁸ × 2.0 × 10⁻⁷ = 26.4 N and ΔL = 1.2 × 10⁻³ × 2.0 = 2.4 × 10⁻³ m, so ½ × 26.4 × 2.4 × 10⁻³ = 0.032 J.

Example 3: breaking force from the ultimate stress

OpenStax gives the ultimate stress of aluminium as 2.2 × 10⁸ Pa. What force breaks an aluminium wire of diameter 1.0 mm?

A = π × (0.50 × 10⁻³)² = 7.85 × 10⁻⁷ m². F = stress × area = 2.2 × 10⁸ × 7.85 × 10⁻⁷ = 170 N, to 2 significant figures. Doubling the diameter would quadruple the area and the breaking force, but the breaking stress is the same: it is a property of the material.

Required practical 4: the Young modulus

AQA's required practical 4 determines the Young modulus by a simple method. The exemplar in AQA's practical handbook hangs two long steel wires from a strong support: a test wire, which is loaded, and a comparison wire beside it carrying a scale. A vernier on the test wire reads its extension against the scale on the comparison wire, which cancels out sagging of the support and thermal expansion.

1. Measure the original length L of the test wire with a metre ruler, and its diameter at several places with a micrometer screw gauge.
2. Add slotted masses one at a time and record the extension ΔL from the vernier each time.
3. Calculate stress (F ÷ A) and strain (ΔL ÷ L) and plot stress against strain.
4. The gradient of the straight section is the Young modulus.

For safety, AQA's handbook advises goggles, since the wire could break, and a sand tray beneath the masses. Use wire with no kinks, or some of the measured extension will just be kinks straightening out. The extensions are small, which is why the vernier is needed: AQA notes that 1 kg produces an extension of about 0.47 mm in a 1.5 m steel wire of diameter 0.45 mm.

Common mistakes

• Mixing up the limit of proportionality and the elastic limit. The first ends the straight line; the second ends elastic behaviour. They are close but not the same point.
• Using the diameter as the radius. A = πr², with r = d ÷ 2.
• Taking the gradient from the curved part. The Young modulus only comes from the straight section.
• Calling the plastic region elastic. Past the elastic limit, the wire is left permanently longer.
• Giving strain units. It is a ratio: no units, though it may be shown as a percentage.

Materials are in the first year of A-level Physics. Practise them on today's A-level Physics: Year 1 short timed paper, and see force against extension for springs with the Hooke's law calculator and the spring constant guide.

Read next

Physics · MechanicsStrain equation: strain = ΔL ÷ L, stress and Young modulusThe strain equation, strain = extension ÷ original length: tensile stress, the Young modulus, stress-strain graphs, strain energy and worked examples.