Stress, strain and Young modulus

Physics· 40 min read

1. Key Definitions: Stress and Strain★★☆☆☆⏱ 10 min

When a force deforms a material, stress and strain normalise the effect of force to the size of the sample, letting us compare properties of different sized objects made of the same material.

📘 Definition

Tensile Stress

\sigma

The tensile force per unit cross-sectional area of a material, applied perpendicular to the sample face.

Example:

$σ = \frac{F}{A}$ where $F$ = tensile force, $A$ = cross-sectional area. Units: Pascals (Pa) = Nm⁻².

📘 Definition

Tensile Strain

\varepsilon

The extension of a material per unit original length. Strain is dimensionless, as it is a ratio of two lengths.

Example:

$ε = \frac{Δ L}{L _{0}}$ where $Δ L$ = extension, $L_{0}$ = original length.

📐 Worked Example

A 2.0 m long wire with diameter 0.5 mm is stretched by a 100 N force, and extends by 2.4 mm. Calculate stress and strain for the wire.

1
First calculate the cross-sectional area of the wire:
2
$A = \pi r^2 = \pi \left(\frac{0.5 \times 10^{-3}}{2}\right)^2 = 1.963 \times 10^{-7} \text{ m}^2$
3
Calculate stress using $σ = F / A$ :
4
$\sigma = \frac{100}{1.963 \times 10^{-7}} = 5.1 \times 10^8 \text{ Pa}$
5
Convert extension to metres and calculate strain:
6
$\varepsilon = \frac{\Delta L}{L_0} = \frac{2.4 \times 10^{-3}}{2.0} = 1.2 \times 10^{-3}$

2. Young Modulus: Definition and Calculation★★☆☆☆⏱ 12 min

Young modulus is an intensive material property that describes stiffness: a higher Young modulus means a stiffer material that deforms less for a given applied stress.

📘 Definition

Young Modulus

E

The ratio of tensile stress to tensile strain, valid for elastic deformation where Hooke's law is obeyed.

Example:

$E = \frac{σ}{ε} = \frac{F L _{0}}{A Δ L}$ . Units are Pa, same as stress.

📐 Worked Example

Use the values of stress and strain from the previous example to calculate the Young modulus of the wire.

1
We already have $σ = 5.1 \times 1 0^{8}$ Pa and $ε = 1.2 \times 1 0^{- 3}$ . Substitute into the formula for E:
2
$E = \frac{\sigma}{\varepsilon} = \frac{5.1 \times 10^8}{1.2 \times 10^{-3}} = 4.25 \times 10^{11} \text{ Pa} \approx 4.3 \times 10^{11} \text{ Pa}$
3
Check units: strain is dimensionless, so E has the same units as stress, which is correct.

3. Experimental Determination of Young Modulus★★★☆☆⏱ 15 min

CIE frequently asks 5-6 mark questions describing this experiment, so you need to remember the method, measurements and error reduction steps.

Clamp a long metal wire to a rigid support, with a fixed ruler alongside the wire.
Add known masses (weights) to the free end, measure extension for each weight.
Measure original length $L_{0}$ from the clamp to a marker on the wire with a metre ruler.
Measure diameter at multiple points along the wire with a micrometer, calculate average diameter.
Plot a graph of force $F$ against extension $Δ L$ , find the gradient $m = F /Δ L$ .
Calculate Young modulus with $E = \frac{m L _{0}}{A}$ , where $A = π (d /2)^{2}$ .

📐 Worked Example

A student obtains a gradient of $8.0 \times 1 0^{4}$ Nm⁻¹ from an $F$ vs $Δ L$ graph, for a 1.5 m long wire with average diameter 0.60 mm. Calculate Young modulus.

1
Calculate cross-sectional area first, converting diameter to metres:
2
$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.60 \times 10^{-3}}{2}\right)^2 = 2.827 \times 10^{-7} \text{ m}^2$
3
Substitute into the formula for E from graph gradient:
4
$E = \frac{m L_0}{A} = \frac{8.0 \times 10^4 \times 1.5}{2.827 \times 10^{-7}} = 4.2 \times 10^{11} \text{ Pa}$

4. Stress-Strain Graph Properties★★★☆☆⏱ 10 min

The gradient of a stress-strain graph in the elastic region is equal to Young modulus, since gradient = $Δ σ /Δ ε = E$ . This is a common exam question.

📐 Worked Example

The elastic region of a stress-strain curve for aluminium has a gradient of $7.0 \times 1 0^{10}$ Pa. What is Young modulus of aluminium?

1
The gradient of the stress-strain curve in the elastic region is by definition equal to Young modulus.
2
Therefore, Young modulus of aluminium is $7.0 \times 1 0^{10}$ Pa.

5. Common Pitfalls

Wrong move:

Forgetting to convert units of diameter/extension from millimetres to metres before calculation.

Why:

This leads to Young modulus values 10⁶ times too large or small, which is a common exam error.

Correct move:

Always convert all length measurements to SI units (metres) before substituting into formulas.

Wrong move:

Calculating cross-sectional area as $π d^{2}$ instead of $π (d /2)^{2}$ .

Why:

Most students measure diameter directly and forget to convert to radius for the area formula.

Correct move:

Always divide diameter by 2 to get radius before calculating area.

Wrong move:

Claiming Young modulus depends on the length or cross-sectional area of the sample.

Why:

Students confuse extension (sample-dependent) with strain, which is normalised for sample size.

Correct move:

Remember Young modulus is an intensive property of the material, not the sample.

Wrong move:

Assigning units of metres to strain.

Why:

Strain is calculated as a ratio of two lengths, so students incorrectly add units.

Correct move:

Strain is dimensionless, it has no units.

Wrong move:

Using the Young modulus formula for deformation beyond the elastic limit.

Why:

Stress is no longer proportional to strain once plastic deformation starts.

Correct move:

Only use $E = σ / ε$ for elastic deformation where Hooke's law holds.

6. Quick Reference Cheatsheet

Quantity	Symbol	Formula	Units
Tensile Stress	$σ$	$σ = F / A$	Pa (Nm⁻²)
Tensile Strain	$ε$	$ε = Δ L / L_{0}$	Dimensionless
Young Modulus	$E$	$E = (F L_{0}) / (A Δ L)$	Pa (Nm⁻²)
E from stress-strain graph	$E$	Gradient of elastic region	Pa
E from F-ΔL graph	$E$	$E = (m L_{0}) / A$ , $m = F /Δ L$	Pa

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
MCQ Young modulus calculation
2022 · 2
6 mark experiment description
2021 · 1
Stress-strain gradient question

What's Next

Mastery of stress, strain and Young modulus is the foundation for all further work on deformation of solids in CIE A-Level Physics. These concepts are used to classify the mechanical behaviour of different material types, calculate elastic potential energy in deformed solids, and answer structured questions about material properties for engineering applications. You will build on these definitions to interpret full stress-strain curves and identify key points like the elastic limit, yield point and ultimate tensile stress, which are common topics in both multiple choice and extended response questions.