Stress, strain and Young's modulus for QCE Engineering Unit 3

Syllabus dot point

“Calculate stress, strain and Young's modulus for a material under axial load, and interpret the elastic region, proportional limit and Hooke's law from a stress-strain diagram”

QCEEngineeringUnit 3: Civil structures6 min readUpdated 2 Jun 2026

Quick answer

Stress is force per unit area, $\sigma = F/A$ , measured in pascals. Strain is the fractional change in length, $\varepsilon = \Delta L / L_0$ , and is dimensionless. Young's modulus $E = \sigma / \varepsilon$ measures stiffness: a high $E$ means the material barely stretches under load. In the linear elastic region the material obeys Hooke's law (stress proportional to strain) and returns to its original length when unloaded. The proportional limit is the stress beyond which the line stops being straight. Steel has $E \approx 200\,\text{GPa}$ , aluminium about $70\,\text{GPa}$ .

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to quantify how a material responds to an axial (pulling or pushing) load. You need to calculate stress, calculate strain, combine them into Young's modulus as a measure of stiffness, and read the elastic region and proportional limit off a stress-strain diagram. This is the bridge between the forces you found from equilibrium and whether a real material can safely carry them.

The answer

Stress

Stress is the internal force carried per unit of cross-sectional area:

\sigma = \frac{F}{A}

with $F$ in newtons and $A$ in square metres, giving stress in pascals ( $1\,\text{Pa} = 1\,\text{N/m}^2$ ). Engineering stresses are large, so megapascals ( $1\,\text{MPa} = 10^6\,\text{Pa}$ ) are usual. Tensile stress (stretching) is positive; compressive stress (squashing) is negative. A thin member under a given load carries more stress than a thick one, which is why load-bearing members are made bigger.

Strain

Strain is how much the material deforms relative to its original size:

\varepsilon = \frac{\Delta L}{L_0}

where $\Delta L$ is the change in length and $L_0$ is the original length. Strain has no units (metre divided by metre). A strain of $0.002$ means the member stretched by $0.2\%$ of its length.

Young's modulus and stiffness

Young's modulus (the modulus of elasticity) is the ratio of stress to strain in the elastic region:

E = \frac{\sigma}{\varepsilon}

Because strain is dimensionless, $E$ has the units of stress (pascals), and engineering values run to gigapascals ( $1\,\text{GPa} = 10^9\,\text{Pa}$ ). A large $E$ means a stiff material that resists deformation. Typical values: structural steel $\approx 200\,\text{GPa}$ , aluminium $\approx 70\,\text{GPa}$ , concrete $\approx 30\,\text{GPa}$ , timber $\approx 10\,\text{GPa}$ . Stiffness is not the same as strength: a material can be very stiff yet fail at a low stress, or very flexible yet strong.

The stress-strain diagram

Plotting stress (vertical) against strain (horizontal) for a tensile test produces the stress-strain curve. Key features in the early part of the curve:

Linear elastic region. A straight line through the origin. Stress is proportional to strain and the gradient equals $E$ .
Proportional limit. The point where the line stops being straight. Below it, Hooke's law holds.
Elastic limit. The greatest stress from which the material still returns fully to its original length on unloading. Beyond it, permanent (plastic) deformation begins.

Hooke's law states that within the elastic region, deformation is proportional to the applied stress:

\sigma = E\varepsilon

This is the same proportionality as a spring, where the material itself acts like a stiff spring.

Steel rod under tension

Calculate stress, strain and Young's modulus

A steel rod of original length $2.50\,\text{m}$ and cross-sectional area $4.0 \times 10^{-4}\,\text{m}^2$ carries an axial tensile force of $50\,000\,\text{N}$ . It stretches by $1.56\,\text{mm}$ . Find the stress, the strain, and Young's modulus.

Stress:

\sigma = \frac{F}{A} = \frac{50\,000}{4.0 \times 10^{-4}} = 1.25 \times 10^{8}\,\text{Pa} = 125\,\text{MPa}

Strain (convert $\Delta L$ to metres: $1.56\,\text{mm} = 1.56 \times 10^{-3}\,\text{m}$ ):

\varepsilon = \frac{\Delta L}{L_0} = \frac{1.56 \times 10^{-3}}{2.50} = 6.24 \times 10^{-4}

Young's modulus:

E = \frac{\sigma}{\varepsilon} = \frac{1.25 \times 10^{8}}{6.24 \times 10^{-4}} = 2.00 \times 10^{11}\,\text{Pa} = 200\,\text{GPa}

The result of $200\,\text{GPa}$ matches the known value for structural steel, confirming the material and the arithmetic.

Why this matters for civil structures

Engineers must keep working stresses well inside the elastic region so the structure springs back rather than deforming permanently. Young's modulus also controls deflection: a stiffer material (higher $E$ ) sags less under the same load, which matters for floors, bridges and beams where excessive deflection is a serviceability failure even before the material breaks.

Common errors

Unit slips on strain: Strain is dimensionless. If your "strain" comes out with units, you forgot to divide by the original length or mixed millimetres with metres.
Confusing stiffness with strength: Young's modulus measures stiffness (resistance to elastic deformation), not the stress at which the material breaks. A material can have a high $E$ and still fail at modest stress.
Reading the wrong gradient: Young's modulus is the gradient of the straight (elastic) part of the stress-strain curve only. Using a point in the curved plastic region gives a meaningless value.
Forgetting to convert area: A cross-section given in $\text{mm}^2$ must become $\text{m}^2$ before dividing into newtons, or the stress is wrong by a factor of $10^6$ .

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice6 marks

An aluminium tie bar of original length

1.80\ \text{m}

and cross-sectional area

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

carries an axial tensile force of

17.5\ \text{kN}

and extends by

1.80\ \text{mm}

. Determine the stress, the strain and Young's modulus for the material, and state whether the result is consistent with aluminium.

Show worked answer →

A 6 mark determine question rewards each quantity computed with correct units.

Stress: $\sigma = F/A = 17\,500 / (2.5 \times 10^{-4}) = 7.0 \times 10^{7}\ \text{Pa} = 70\ \text{MPa}$ .

Strain (convert $1.80\ \text{mm} = 1.80 \times 10^{-3}\ \text{m}$ ): $\varepsilon = \Delta L / L_0 = (1.80 \times 10^{-3})/1.80 = 1.0 \times 10^{-3}$ .

Young's modulus: $E = \sigma / \varepsilon = (7.0 \times 10^{7})/(1.0 \times 10^{-3}) = 7.0 \times 10^{10}\ \text{Pa} = 70\ \text{GPa}$ .

Markers reward unit conversion of $\Delta L$ to metres, the dimensionless strain, and recognising $70\ \text{GPa}$ as the known value for aluminium, confirming the material.

Practice4 marks

Two beams of identical dimensions carry the same load, one in steel (

E \approx 200\ \text{GPa}

) and one in timber (

E \approx 10\ \text{GPa}

). Justify which beam deflects more, referring to the stress-strain relationship.

Show worked answer →

A 4 mark justify answer needs the link between $E$ , strain and deflection.

Young's modulus is the gradient of the elastic stress-strain line, $E = \sigma/\varepsilon$ , so for a given stress a low $E$ means a large strain. The timber beam, with $E$ about twenty times smaller than steel, strains roughly twenty times more under the same stress, so it deflects far more. Deflection scales inversely with $EI$ , so the lower modulus directly produces greater sag.

Markers reward identifying timber as deflecting more and tying it to the smaller $E$ producing larger strain for the same stress.

Practise this

Sources & how we know this

Engineering General senior syllabus 2025 — QCAA