Stress, strain and the Young modulusEdexcel A-Level Physics: Revision notes

Tensile stress (or compressive stress) is the force applied per unit cross-sectional area:

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

Stress is measured in pascals (Pa = N m⁻²). Cross-sectional area must be in m², so convert mm² using 1 mm² = 10⁻⁶ m². For a wire, A = πd²/4 = πr².

Worked example: 80 N on a wire of area 0.50 mm² gives σ = 80 / 5.0 × 10⁻⁷ = 1.6 × 10⁸ Pa.

Strain is the extension per unit original length:

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

Strain is a ratio, so it has no unit. For compression, the strain is the decrease in length divided by the original length. A strain of 8.0 × 10⁻⁴ on a 2.50 m wire means an extension of 8.0 × 10⁻⁴ × 2.50 = 2.0 mm.

In the region where a material obeys Hooke's law, stress is proportional to strain. The Young modulus is the ratio:

$E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta x/x} = \frac{Fx}{A\,\Delta x}$

Its unit is the pascal. Unlike stiffness k, E is a property of the material, not of a particular sample, because it does not depend on the sample's length or area. For steel E ≈ 2 × 10¹¹ Pa.

Worked example: a copper wire of diameter 0.60 mm and length 1.80 m extends by 1.4 mm under 50 N. A = 2.83 × 10⁻⁷ m², stress = 1.77 × 10⁸ Pa, strain = 7.8 × 10⁻⁴, so E = 2.3 × 10¹¹ Pa.

A stress–strain graph has strain on the x-axis and stress on the y-axis. For a ductile metal under tension it is a straight line through the origin (the gradient is the Young modulus), then it curves beyond the limit of proportionality, and ends at the point of fracture.

The breaking stress is the stress at which the material breaks. The force that causes this is breaking stress × area. Under compression the graph is the same shape in the negative direction: stress and strain are negative, and the gradient of the straight section is the same for many metals.

Because stress and strain remove the effect of sample size, the stress–strain graph lets different materials be compared directly.

1. Use a long, thin wire: the extension is larger and the stress is bigger for a small load.
2. Measure the diameter with a micrometer at several places and directions, take a mean and find A.
3. Measure the original length L from the fixed point to the marker.
4. Add a small load to straighten the wire, then add masses in equal steps, recording force (mg) and extension. Unload to check that the limit of proportionality has not been passed.
5. Plot force against extension. The gradient is k = F/x, so $E = \frac{kL}{A}$

The diameter has the largest percentage uncertainty because A ∝ d², so its percentage uncertainty is doubled.