A wire of area 2.0 × 10^-6\ m^2 carries 40 N. Find the stress. [2 marks]

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How do we quantify how a material responds to load through stress, strain and the modulus?

Stress, strain, the Young modulus, the stress-strain graph, elastic and plastic behaviour, and factor of safety.

A CCEA A-Level Technology and Design answer on calculating stress and strain, the Young modulus, reading the stress-strain graph (elastic limit, yield, ultimate strength), elastic and plastic behaviour, and applying a factor of safety.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

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What this dot point is asking
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Examples in context
Try this

What this dot point is asking

CCEA expects you to calculate stress and strain, define and use the Young modulus, read the stress-strain graph (elastic limit, yield, ultimate strength), distinguish elastic from plastic behaviour, and apply a factor of safety. The stress/strain/modulus calculation is a regular feature.

The answer

Stress, strain and the Young modulus

The stress-strain graph

Factor of safety

Worked example: stress, strain and a safe working load

Checking a cable is safe

A cable of area $1.0 \times 10^{-4}\ \text{m}^2$ is made from a material with a yield stress of 250 MPa. It must carry 5000 N with a factor of safety of 4.

step Find the working stress

\sigma = \frac{F}{A} = \frac{5000}{1.0 \times 10^{-4}} = 5.0 \times 10^{7}\ \text{Pa} \ (50\ \text{MPa}).

step Apply the factor of safety

\text{factor of safety} = \frac{\text{yield stress}}{\text{working stress}} = \frac{250}{50} = 5.

The actual factor of safety is 5, which exceeds the required 4, so the cable is safe.

step State the meaning

A working stress of 50 MPa is well below the 250 MPa yield, so the cable stays in the elastic region and recovers after loading, with margin for overload. Computing stress and checking the factor of safety is exactly the structural calculation CCEA rewards.

Examples in context

Example 1. Lift and crane cables. Designed with a large factor of safety (often 6 or more) because failure is catastrophic, so the working stress is kept far below the cable's breaking stress.

Example 2. Choosing steel over aluminium for stiffness. Steel's high Young modulus (about 200 GPa, three times aluminium's) means it strains less under load, which is why structures needing stiffness use steel, the modulus deciding the material.

Try this

Q1. State the equation for stress. [1 mark]

Cue. $\sigma = F/A$ (force divided by cross-sectional area).

Q2. A wire of area $2.0 \times 10^{-6}\ \text{m}^2$ carries 40 N. Find the stress. [2 marks]

Cue. $\sigma = 40/(2.0\times10^{-6}) = 2.0 \times 10^{7}\ \text{Pa}$ (20 MPa).

Q3. What does the elastic limit on a stress-strain graph represent? [2 marks]

Cue. The point beyond which deformation becomes permanent (plastic); below it the material returns to its original length when unloaded.

Exam-style practice questions

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

CCEA 20206 marksA steel rod of cross-sectional area

2.0 \times 10^{-4}\ \text{m}^2

carries a tensile load of 8000 N. The rod is 2.0 m long and stretches by 0.40 mm. Calculate the stress, the strain and the Young modulus of the steel.

Show worked answer →

Stress is force per unit area:

\sigma = \frac{F}{A} = \frac{8000}{2.0 \times 10^{-4}} = 4.0 \times 10^{7}\ \text{Pa} \ (40\ \text{MPa}).

Strain is the extension divided by the original length (dimensionless):

\varepsilon = \frac{\Delta L}{L} = \frac{0.40 \times 10^{-3}}{2.0} = 2.0 \times 10^{-4}.

The Young modulus is stress over strain:

E = \frac{\sigma}{\varepsilon} = \frac{4.0 \times 10^{7}}{2.0 \times 10^{-4}} = 2.0 \times 10^{11}\ \text{Pa} \ (200\ \text{GPa}).

So the stress is 40 MPa, the strain is $2.0 \times 10^{-4}$ , and the Young modulus is about 200 GPa, the expected value for steel.

Markers reward stress = F/A, strain = extension/length (with consistent units), and E = stress/strain giving about 200 GPa.

CCEA 20214 marksUsing a stress-strain graph, explain the terms elastic limit and factor of safety.

Show worked answer →

On a stress-strain graph, the elastic limit is the point up to which the material behaves elastically - it returns to its original length when the load is removed. Below it, stress is proportional to strain (the straight, Hooke's-law region) and the deformation is recoverable; beyond the elastic limit the material deforms plastically (permanently) and does not fully return.

The factor of safety is how much stronger a component is made than it strictly needs to be for its working load:

\text{factor of safety} = \frac{\text{maximum (failure or yield) stress}}{\text{working stress}}.

A factor of safety greater than 1 (for example 2 to 5) ensures the component stays well within the elastic region in service, allowing for overloads, material flaws, fatigue and uncertainty, so it does not yield or fracture.

Markers want the elastic limit as the boundary between recoverable (elastic) and permanent (plastic) deformation, and the factor of safety as the ratio of failure/yield stress to working stress (designing with a margin).

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Sources & how we know this

CCEA GCE Technology and Design specification — CCEA (2016)