OCR OCR 2022-style practice: A nylon tie of original length 0.50 m and cross-sectional area 1.0 × 10^-5 square metres stretches by 2.0 mm under a force of 600 N. Calculate the strain, the stress and the Young's modulus of the nylon, giving units.

A Component 01 multi-step calculation. Marks for each of strain, stress and Young's modulus with units. Strain is the extension divided by the original length (it has no units): = ( L)/(L) = (0.002)/(0.50) = 0.004 (or 4.0 × 10^-3). Stress is force over area: σ = (F)/(A) = 6001.0 × 10^-5 = 6.0 × 10^7 Pa = 60 MPa. Young's modulus is stress over strain: E = (σ)/() = 6.0 × 10^70.004 = 1.5 × 10^10 Pa = 15 GPa. A common dropped mark is forgetting to convert the extension to metres (2.0 mm = 0.002 m), or mixing units; strain is dimensionless, stress is in pascals, and Young's modulus is in pascals (often gigapascals).

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How are stress, strain and Young's modulus defined and calculated, and how do they describe a material under load?

Stress, strain and Young's modulus: the definitions and formulae, their units, the stress-strain relationship and the meaning of stiffness, with worked calculations applied to product components.

A focused answer to OCR A-Level Product Design on stress, strain and Young's modulus: the definitions, formulae and units, the stress-strain relationship and the meaning of stiffness, with worked calculations of stress, strain and Young's modulus applied to product components.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Stress is the internal force per unit area in a loaded material: $\sigma = \frac{F}{A}$ , measured in pascals (Pa), where one pascal is one newton per square metre (often given in MPa). Strain is the fractional change in length: $\varepsilon = \frac{\Delta L}{L}$ , the extension divided by the original length, and it has no units. Young's modulus measures a material's stiffness, the resistance to elastic deformation: $E = \frac{\sigma}{\varepsilon}$ , also in pascals (often GPa). A high Young's modulus means a stiff material (steel) that barely stretches under load; a low one means a flexible material (rubber). To calculate, convert all lengths to metres and areas to square metres, find stress and strain, then divide stress by strain for $E$ .

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What this dot point is asking
Stress
Strain
Young's modulus and stiffness
The stress-strain relationship

What this dot point is asking

OCR wants you to define stress, strain and Young's modulus, state their formulae and units, and calculate them for a component under load. These are the quantitative measures of how a material behaves when forces act on it, and they carry calculation marks in Component 01.

Stress

Strain

The exam trap is forgetting to convert the extension and original length to the same unit (both in metres) and that strain has no units.

Young's modulus and stiffness

The stress-strain relationship

Up to a point (the limit of proportionality and the elastic limit), stress is proportional to strain, the graph is a straight line, and the material returns to shape when unloaded; Young's modulus is the gradient of that straight line. Beyond the elastic limit the material deforms permanently (plastic deformation) and eventually fails. Designers keep working stresses well within the elastic region, with a safety factor, so the product does not deform or fail in use.

Calculating stress, strain and Young's modulus

Convert to SI units and find the strain

An aluminium strut of original length $0.80$ m and cross-sectional area $5.0 \times 10^{-5}$ square metres stretches by $0.28$ mm under a tensile force of $2500$ N. First the strain: convert the extension, $\Delta L = 0.28$ mm $= 2.8 \times 10^{-4}$ m, so $\varepsilon = \frac{\Delta L}{L} = \frac{2.8 \times 10^{-4}}{0.80} = 3.5 \times 10^{-4}$ (no units).

Find the stress

Stress is force over area: $\sigma = \frac{F}{A} = \frac{2500}{5.0 \times 10^{-5}} = 5.0 \times 10^{7}$ Pa, which is $50$ MPa.

Find Young's modulus

Young's modulus is stress over strain: $E = \frac{\sigma}{\varepsilon} = \frac{5.0 \times 10^{7}}{3.5 \times 10^{-4}} = 1.43 \times 10^{11}$ Pa, about $143$ GPa.

Check the result is reasonable

State the units (Pa, dimensionless, Pa) and sense-check: aluminium's Young's modulus is about $70$ GPa, so a value of $143$ GPa here would prompt a re-check of the data, whereas a result near $70$ GPa would confirm aluminium. Always convert lengths to metres, keep strain unitless, and sense-check $E$ against known values, because a units slip is the commonest lost mark.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20204 marksA steel rod of cross-sectional area

2.0 \times 10^{-4}

square metres carries a tensile force of 8000 N. Calculate the stress, and state the unit.

Show worked answer →

A Component 01 stress calculation. Marks for the formula, the substitution and the answer with units.

Stress is force per unit area: $\sigma = \frac{F}{A} = \frac{8000}{2.0 \times 10^{-4}} = 4.0 \times 10^{7}$ pascals, which is $40$ MPa. The unit is the pascal (Pa), equal to one newton per square metre; here it is more neatly written as $40$ megapascals (MPa).

A common dropped mark is leaving the area in square millimetres or square centimetres; convert to square metres first, or give the answer in N per square mm and convert. Omitting the unit also loses a mark.

OCR 20226 marksA nylon tie of original length 0.50 m and cross-sectional area

1.0 \times 10^{-5}

square metres stretches by 2.0 mm under a force of 600 N. Calculate the strain, the stress and the Young's modulus of the nylon, giving units.

Show worked answer →

A Component 01 multi-step calculation. Marks for each of strain, stress and Young's modulus with units.

Strain is the extension divided by the original length (it has no units): $\varepsilon = \frac{\Delta L}{L} = \frac{0.002}{0.50} = 0.004$ (or $4.0 \times 10^{-3}$ ). Stress is force over area: $\sigma = \frac{F}{A} = \frac{600}{1.0 \times 10^{-5}} = 6.0 \times 10^{7}$ Pa $= 60$ MPa. Young's modulus is stress over strain: $E = \frac{\sigma}{\varepsilon} = \frac{6.0 \times 10^{7}}{0.004} = 1.5 \times 10^{10}$ Pa $= 15$ GPa.

A common dropped mark is forgetting to convert the extension to metres (2.0 mm = 0.002 m), or mixing units; strain is dimensionless, stress is in pascals, and Young's modulus is in pascals (often gigapascals).

Related dot points

Sources & how we know this

OCR A Level Design and Technology (H404-H406) specification — OCR (2017)