A wire of strain 0.0025 is under a stress of 5.0 × 10^8\ Pa. Find Young's modulus. [2 marks]

SQA SQA AH style-style practice: A steel rod of cross-sectional area 20\ mm^2 carries a tensile load of 4.0\ kN. The original length is 2.0\ m and it stretches by 2.0\ mm. Calculate the tensile stress, the strain and Young's modulus.

Stress is force over area; convert the area to square metres (20\ mm^2 = 20 × 10^-6\ m^2). σ = FA = 4.0 × 10^320 × 10^-6 = 2.0 × 10^8\ Pa = 200\ MPa. Strain is the extension over the original length (both in metres). = LL = 2.0 × 10^-32.0 = 1.0 × 10^-3. Young's modulus is stress over strain. E = σ = 2.0 × 10^81.0 × 10^-3 = 2.0 × 10^11\ Pa = 200\ GPa, the expected value for steel. Markers reward σ = 200\ MPa with the area converted, = 1.0 × 10^-3, and E = 2.0 × 10^11\ Pa with consistent units throughout.

ScotlandEngineering ScienceSyllabus dot point

How do we quantify how a material deforms under load, and how do we choose a safe working stress?

Stress, strain and Young's modulus, the tensile test and material properties, and the use of a factor of safety to set a safe working stress.

An SQA Advanced Higher Engineering Science answer on material behaviour, covering tensile stress and strain, Young's modulus, the stress-strain graph and material properties, and the use of a factor of safety to set a safe working stress.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-16

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

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Quick answer

Stress is the force per unit cross-sectional area, $\sigma = \dfrac{F}{A}$ , in pascals (Pa). Strain is the fractional change in length, $\varepsilon = \dfrac{\Delta L}{L}$ , a ratio with no units. While a material behaves elastically the two are proportional, and the constant of proportionality is Young's modulus $E = \dfrac{\sigma}{\varepsilon}$ , a measure of stiffness (a high $E$ , like steel, deforms little under load). The stress-strain graph reveals the elastic limit (below which the material springs back), the yield point (where it begins to deform permanently) and the ultimate tensile stress (the maximum it can bear before failing). To keep a component safely below failure, engineers set the working stress as the ultimate (or yield) stress divided by a factor of safety, allowing for material variation, overloads and deterioration.

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What this key area is asking
Stress and strain
Young's modulus
The stress-strain graph and material properties
Factor of safety and working stress
Examples in context
Try this

What this key area is asking

The SQA wants you to define and calculate tensile stress and strain, to use Young's modulus as the measure of a material's stiffness, to interpret the stress-strain graph and the material properties it reveals, and to apply a factor of safety to set a safe working stress. This is how an engineer matches a material and a section size to the forces a structure must carry.

Stress and strain

Stress measures how hard the material is being pulled per unit area, so a thin rod is more highly stressed than a thick one under the same load. Strain measures how much it stretches relative to its length. Keeping the units consistent matters: areas given in $\text{mm}^2$ must be converted to $\text{m}^2$ ( $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$ ), or the stress will be wrong by a factor of a million.

Young's modulus

Young's modulus is the gradient of the straight, elastic part of the stress-strain graph. It is a property of the material, not the specimen, so it does not depend on the rod's size. It lets you predict how much a component will deflect under load, which matters as much as strength in many designs (a floor that is strong enough but too springy is still unsatisfactory).

The stress-strain graph and material properties

A tensile test stretches a specimen and records stress against strain. The graph reveals the key properties:

Elastic region - the initial straight line; the material returns to its original length when unloaded. Its gradient is Young's modulus.
Elastic limit / yield point - beyond here the material deforms plastically (permanently) and will not fully recover.
Ultimate tensile stress - the highest stress the material reaches, its strength.
Fracture - where the specimen breaks.

The shape also shows whether a material is ductile (large plastic region, deforms before breaking, like mild steel) or brittle (little plastic deformation, snaps suddenly, like cast iron).

Factor of safety and working stress

A real component is never loaded close to its failure stress. The factor of safety divides the failure stress down to a much lower working stress, leaving a margin for things the calculation cannot fully predict: variation in material strength, unexpected or shock loads, manufacturing defects, corrosion and wear. A larger factor is used where failure would be dangerous or the loads uncertain.

Sizing a tie rod with a factor of safety

A steel tie rod must carry a working load of $24\ \text{kN}$ . The steel has an ultimate tensile stress of $480\ \text{MPa}$ , and a factor of safety of $4$ is required. Find the minimum cross-sectional area.

step 1 Find the safe working stress

$\text{working stress} = \dfrac{480}{4} = 120\ \text{MPa} = 120 \times 10^6\ \text{Pa}$ .

step 2 Rearrange the stress relationship for area

From $\sigma = \dfrac{F}{A}$ , $A = \dfrac{F}{\sigma}$ .

step 3 Substitute and evaluate

$A = \dfrac{24 \times 10^3}{120 \times 10^6} = 2.0 \times 10^{-4}\ \text{m}^2 = 200\ \text{mm}^2$ . The rod must have at least this cross-sectional area to keep the working stress within the safe limit.

Examples in context

A lift cable is sized so its working stress, after a generous factor of safety, stays well below the steel's ultimate stress, because failure would be catastrophic. A bridge member is chosen for both strength (it must not yield) and stiffness (it must not deflect too much), read from Young's modulus. A car body panel uses a ductile steel that deforms in a crash rather than shattering. A glass or ceramic part is treated cautiously because it is brittle and fails without warning. Matching the material's properties and the working stress to the duty is the engineering judgement these calculations support.

Try this

Q1. State the relationship for tensile stress. [1 mark]

Cue. $\sigma = \dfrac{F}{A}$ , force over cross-sectional area, in pascals.

Q2. A wire of strain $0.0025$ is under a stress of $5.0 \times 10^8\ \text{Pa}$ . Find Young's modulus. [2 marks]

Cue. $E = \dfrac{\sigma}{\varepsilon} = \dfrac{5.0 \times 10^8}{0.0025} = 2.0 \times 10^{11}\ \text{Pa}$ .

Q3. State one reason a factor of safety is used. [1 mark]

Cue. To allow for uncertainties such as material variation, overloads, defects or deterioration, keeping the component below its failure stress.

Exam-style practice questions

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

SQA AH style5 marksA steel rod of cross-sectional area

20\ \text{mm}^2

carries a tensile load of

4.0\ \text{kN}

. The original length is

2.0\ \text{m}

and it stretches by

2.0\ \text{mm}

. Calculate the tensile stress, the strain and Young's modulus.

Show worked answer →

Stress is force over area; convert the area to square metres ( $20\ \text{mm}^2 = 20 \times 10^{-6}\ \text{m}^2$ ).

$\sigma = \dfrac{F}{A} = \dfrac{4.0 \times 10^3}{20 \times 10^{-6}} = 2.0 \times 10^8\ \text{Pa} = 200\ \text{MPa}$ .

Strain is the extension over the original length (both in metres).

$\varepsilon = \dfrac{\Delta L}{L} = \dfrac{2.0 \times 10^{-3}}{2.0} = 1.0 \times 10^{-3}$ .

Young's modulus is stress over strain.

$E = \dfrac{\sigma}{\varepsilon} = \dfrac{2.0 \times 10^8}{1.0 \times 10^{-3}} = 2.0 \times 10^{11}\ \text{Pa} = 200\ \text{GPa}$ , the expected value for steel.

Markers reward $\sigma = 200\ \text{MPa}$ with the area converted, $\varepsilon = 1.0 \times 10^{-3}$ , and $E = 2.0 \times 10^{11}\ \text{Pa}$ with consistent units throughout.

SQA AH style4 marksA cable has an ultimate tensile stress of

400\ \text{MPa}

and must carry a working load with a factor of safety of

5

. Calculate the maximum safe working stress, and explain why a factor of safety is used.

Show worked answer →

The safe working stress is the ultimate stress divided by the factor of safety.

Relationship: $\text{working stress} = \dfrac{\text{ultimate stress}}{\text{factor of safety}}$ .

Substitution: $\text{working stress} = \dfrac{400}{5} = 80\ \text{MPa}$ .

A factor of safety is used to allow for uncertainties: variation in material strength, unexpected or shock loads, manufacturing flaws, corrosion and wear over time, so the component never approaches its failure stress in service.

Markers reward the working stress of $80\ \text{MPa}$ from dividing by the factor of safety, and a valid reason such as allowing for material variation, overloads or deterioration.

Related dot points

Sources & how we know this

Advanced Higher Engineering Science Course Specification (C823 77) — SQA (2019)
Advanced Higher Engineering Science Data Booklet — SQA (2019)