Skip to main content
Northern IrelandEngineering & ManufacturingSyllabus dot point

How do engineers test materials, and how are stress, strain and Young's modulus calculated?

Material testing and the calculations of stress, strain and Young's modulus of elasticity.

A CCEA GCSE Engineering and Manufacturing answer on testing materials and calculating stress, strain and Young's modulus, including the equations, units and a worked tensile-test calculation.

Generated by Claude Opus 4.812 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

CCEA Unit 3 expects you to know how materials are tested, and to calculate stress, strain and Young's modulus using the standard equations. These calculations let engineers compare materials and check a part will not stretch or fail under load.

The answer

Why we test materials

The tensile test is the source of the stress and strain calculations below.

Stress

Strain

Young's modulus

So a tensile test gives the load and extension, from which you find stress and strain, and their ratio is Young's modulus.

Worked example: a full tensile-test calculation

Examples in context

Example 1. A lifting cable
Engineers calculate the stress in the cable from the load and its area to check it stays below the safe limit, choosing a thicker cable (larger AA) if the stress is too high.
Example 2. Comparing two metals
A material with a higher Young's modulus stretches less under the same stress, so it is chosen where stiffness matters, such as a machine frame that must not flex.
Example 3. Quality checking a batch
A sample bar is tensile tested; if its measured stress and strain give the expected Young's modulus, the batch is accepted, linking testing to quality control.

The pattern is that the three equations always work in order: stress from force and area, strain from extension and length, then Young's modulus as their ratio.

Try this

Q1. Write the equation for tensile stress and state its unit. [2 marks]

  • Cue. σ=F/A\sigma = F/A; unit is the pascal (Pa) or N/m squared.

Q2. A bar of area 2.0×10−5 m22.0 \times 10^{-5}\ \text{m}^2 carries 3000 N3000\ \text{N}. Find the stress. [2 marks]

  • Cue. σ=3000/(2.0×10−5)=1.5×108 Pa\sigma = 3000 / (2.0 \times 10^{-5}) = 1.5 \times 10^{8}\ \text{Pa}.

Q3. Why does strain have no units? [1 mark]

  • Cue. It is a length divided by a length (extension over original length), so the units cancel.

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 style4 marksA steel rod of cross-sectional area 20 mm squared (2.0 times 10 to the power minus 5 m squared) is pulled by a force of 4000 N. Calculate the tensile stress in the rod and state its unit.
Show worked answer →

Tensile stress is force divided by cross-sectional area:

σ=FA=40002.0×10−5=2.0×108 Pa.\sigma = \dfrac{F}{A} = \dfrac{4000}{2.0 \times 10^{-5}} = 2.0 \times 10^{8}\ \text{Pa}.

So the stress is 2.0×108 Pa2.0 \times 10^{8}\ \text{Pa} (pascals), which equals 200 MPa200\ \text{MPa} or 200 N/mm2200\ \text{N/mm}^2.

Markers reward the equation σ=F/A\sigma = F/A, correct substitution with area in m2\text{m}^2, the value, and the unit (Pa or N/m squared).

CCEA style4 marksA wire 2.0 m long stretches by 1.0 mm under load. The tensile stress in it is 1.0 times 10 to the power 8 Pa. Calculate the strain and hence the Young's modulus of the material.
Show worked answer →

Strain is extension divided by original length (extension and length in the same units):

ε=ΔLL=1.0×10−32.0=5.0×10−4.\varepsilon = \dfrac{\Delta L}{L} = \dfrac{1.0 \times 10^{-3}}{2.0} = 5.0 \times 10^{-4}.

Strain has no units. Young's modulus is stress divided by strain:

E=σε=1.0×1085.0×10−4=2.0×1011 Pa.E = \dfrac{\sigma}{\varepsilon} = \dfrac{1.0 \times 10^{8}}{5.0 \times 10^{-4}} = 2.0 \times 10^{11}\ \text{Pa}.

So E=2.0×1011 PaE = 2.0 \times 10^{11}\ \text{Pa} (200 GPa), typical of steel.

Markers reward strain as a ratio with no unit, E=σ/εE = \sigma / \varepsilon, correct substitution and the value with unit Pa.

Related dot points

Sources & how we know this