Stress, Strain and Young Modulus Experiment - OCR A Level Physics
Module 3 - forces and motion

Stress, strain and Young modulus experiment

How stiff is a material? The Young modulus tells you how much a material resists stretching. This sheet covers the definitions, the wire experiment, the stress-strain graph, and how to calculate and interpret results.

Work through it in three passes

1

Learn the definitions

Distinguish stress, strain and Young modulus, and know their units.

2

Understand the experiment

See how to measure extension and why the gradient gives Young modulus.

3

Practise calculations

Use the calculator and worked examples to handle typical exam questions.

By the end you should be able to
  • Define stress, strain and Young modulus and give their units.
  • Describe an experiment to determine Young modulus for a metal wire.
  • Sketch and interpret a stress-strain graph up to breaking point.
  • Calculate Young modulus from force, extension and wire dimensions.
  • Identify the limit of proportionality, elastic limit and yield point on a graph.

Stress, strain and Young modulus

When you stretch a wire, two things control how much it extends: the force you apply and the wire's dimensions. Stress and strain normalise those effects so you can compare different materials fairly.

Stress σ

σ = F / A

Force per unit cross-sectional area. Unit: Pa or N m-2.

Strain ε

ε = ΔL / L

Fractional extension. No unit — it is a ratio of two lengths.

Young modulus E

E = σ / ε = FL / (AΔL)

Stress divided by strain in the linear region. Unit: Pa or GPa.

Why use stress and strain?

A thick wire stretches less than a thin wire under the same load. A long wire stretches more than a short wire under the same stress. Stress and strain remove those size effects, so the Young modulus depends only on the material (and its condition), not on the sample shape.

QuantitySymbolDefinitionUnit
Original lengthLLength before loadingm
ExtensionΔLIncrease in lengthm
Force / loadFTension in the wireN
Cross-sectional areaAArea of wire perpendicular to length
StressσF / APa
StrainεΔL / Lnone
Young modulusEσ / εPa or GPa

Measuring Young modulus for a metal wire

The standard school or college method uses a long wire loaded with known weights. The key is to measure very small extensions accurately.

Apparatus

Long wire (usually 2 m or more), fixed support at the top, pulley or load hanger at the bottom, micrometer or digital callipers for diameter, metre ruler for original length, travelling microscope or vernier scale for extension.

1
Measure the wire diameter. Use a micrometer at several points along the wire and take an average. Calculate A = πd²/4.
2
Measure the original length. Measure from the fixed support to the reference mark (e.g. the bottom of the wire before loading).
3
Add loads in steps. Add known weights one at a time. Record the extension after each load. Remove loads to check the wire returns to its original length.
4
Plot force against extension. In the linear region, the graph is a straight line through the origin.
5
Calculate Young modulus. Use E = FL / (AΔL), or find the gradient of a force-extension graph and use E = gradient × L / A.
Why a long wire?

Extensions are small — often less than a millimetre. A longer wire gives a larger, more measurable extension for the same strain. The control wire removes errors caused by the support sagging or the measuring scale moving.

Stress-strain graph for a metal wire

A typical stress-strain graph for a ductile metal has several distinct regions. The gradient of the initial straight part is the Young modulus.

FeatureWhat it means
Limit of proportionalityThe end of the straight-line region where σ ∝ ε and Hooke's law holds.
Elastic limitBeyond this, the material no longer returns to its original length when the force is removed.
Yield pointThe stress at which the material suddenly begins to stretch with little or no extra load.
Plastic regionThe material permanently deforms. Strain increases faster than stress.
Ultimate tensile stress (UTS)The maximum stress the material can withstand before necking begins.
Breaking stressThe stress at which the material fractures.
Young modulus from the graph

Young modulus is the gradient of the initial straight section only. Do not use a point from the plastic region, and do not use the UTS or breaking stress.

Young modulus calculator

Enter the wire dimensions and the measured extension to calculate stress, strain and Young modulus. Use SI units (metres, newtons, square metres).

Worked examples

Example 1: calculating stress and strain

A steel wire of diameter 0.40 mm and length 2.5 m is stretched by a force of 45 N. Calculate the stress and strain.

Area:

A = πd²/4 = π × (0.40 × 10-3)² / 4 = 1.26 × 10-7

Stress:

σ = F/A = 45 / 1.26 × 10-7 = 3.57 × 108 Pa

Strain:

ε = ΔL/L = extension / 2.5 (you need the extension to get a numerical answer)

Example 2: calculating Young modulus

A copper wire of length 2.00 m and diameter 0.56 mm extends by 1.2 mm when a load of 60 N is applied. Calculate Young modulus for copper.

Area:

A = π × (0.56 × 10-3)² / 4 = 2.46 × 10-7

Young modulus:

E = FL / (AΔL)
E = (60 × 2.00) / (2.46 × 10-7 × 1.2 × 10-3)
E = 4.07 × 1011 Pa ≈ 130 GPa

Example 3: from force-extension graph gradient

A force-extension graph for a metal wire has a straight-line gradient of 2.5 × 104 N m-1. The wire is 1.80 m long with cross-sectional area 3.0 × 10-7 m². Find Young modulus.

From F = (EA/L)ΔL, the gradient of a force-extension graph is EA/L.

E = gradient × L / A
E = (2.5 × 104 × 1.80) / 3.0 × 10-7
E = 1.5 × 1011 Pa = 150 GPa

Common mistakes and quick fixes

Forgetting to convert mm to m

Diameter and extension are often given in millimetres. Convert to metres before calculating area or strain.

Using diameter instead of area

Stress uses A = πd²/4, not d. A common error is to divide force by diameter.

Using the wrong part of the graph

Young modulus is the gradient of the linear region only. Do not use UTS or breaking stress.

Confusing extension with length

Strain is ΔL/L, not ΔL and not L/ΔL.

Self-check questions

1. A wire of length 3.0 m extends by 2.0 mm under load. What is the strain?

Method: Strain is the ratio of extension to original length. Convert both to metres.

ε = 2.0 × 10-3 / 3.0 = 6.7 × 10-4

2. A steel wire of diameter 0.80 mm supports a load of 120 N. Calculate the stress.

Method: First find cross-sectional area, then divide force by area.

A = π × (0.80 × 10-3)² / 4 = 5.03 × 10-7

σ = 120 / 5.03 × 10-7 = 2.39 × 108 Pa

3. A material has Young modulus 200 GPa and strain 1.5 × 10-3. What is the stress?

Method: Rearrange E = σ/ε to σ = Eε.

σ = 200 × 109 × 1.5 × 10-3 = 3.0 × 108 Pa

4. Why is a long, thin wire used in the Young modulus experiment?

Answer: A long wire gives a larger extension for the same strain, making ΔL easier to measure accurately. A thin wire gives a larger stress for the same force, so measurable extensions occur at safe loads.

Final checklist for stress and strain questions

  • Convert all lengths to metres before calculating area or strain.
  • Calculate cross-sectional area using A = πd²/4 for a circular wire.
  • Use σ = F/A and ε = ΔL/L; remember strain has no unit.
  • Calculate Young modulus from E = FL/(AΔL) or the gradient of a force-extension graph.
  • Only use the linear region of the stress-strain graph for Young modulus.
  • Label the limit of proportionality, elastic limit, yield point, UTS and breaking stress clearly on sketches.