Young Modulus from Stress-Strain Graphs Revision Notes | A Level AQA Physics
The Young modulus is the gradient of the straight-line section of a stress-strain graph. Free AQA notes with 2 worked examples and 11 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Young Modulus from Stress-Strain Graphs is part of Materials in A Level AQA Physics.
6 min
Reading time
14
Lessons in this topic
A Level AQA
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
The Young modulus is the gradient of the straight-line section of a stress-strain graph. Free AQA notes with 2 worked examples and 11 questions answered.
Explain what a stress-strain graph represents and identify its key regions
Calculate the Young modulus from the gradient of the linear region of a stress-strain graph
Describe a simple experimental method to measure the Young modulus of a wire
Analyse stress-strain graphs to compare material properties such as stiffness, strength, and ductility
Key ideas to keep in view
Use this lesson to stay inside Materials while connecting the detail back to the wider Mechanics and materials topic.
Stress: Force per unit cross-sectional area applied to a material. Measured in pascals (Pa) or N m⁻². Calculated as σ = F/A.
Strain: The fractional extension of a material under stress. A dimensionless ratio calculated as ε = ΔL/L₀, where ΔL is the extension and L₀ is the original length.
Young modulus: The ratio of stress to strain in the linear (elastic) region of a stress-strain graph. Measured in pascals (Pa). A measure of a material's stiffness.
Limit of proportionality: The point on a stress-strain graph beyond which stress is no longer proportional to strain. The linear region ends here.
Revision notes
Read the core explanations from Young Modulus from Stress-Strain Graphs before testing yourself from memory.
1. Why the Young Modulus Matters
Engineers designing a bridge, a aircraft wing, or a surgical implant all need to answer the same question: how much will this material stretch under load, and will it spring back? The Young modulus is the single number that answers that question.
What makes the Young modulus so powerful is that it is a property of the material itself, not the particular sample. A thick steel rod and a thin steel wire have the same Young modulus — because stress and strain both account for the sample's geometry. That is the insight that makes this quantity so useful in engineering and on your exam.
2. Stress, Strain, and the Graph They Produce
Before reading a stress-strain graph, you need both axes pinned down precisely. Stress ($\sigma$) is the force applied divided by the cross-sectional area: $\sigma = F/A$, measured in pascals (Pa). Strain ($\varepsilon$) is the extension divided by the original length: $\varepsilon = \Delta L / L_0$, and it is dimensionless — no units at all.
Picture the graph: stress on the vertical axis, strain on the horizontal axis. In the early part of the graph, the line is perfectly straight. This is the linear (elastic) region, and it ends at the limit of proportionality. The gradient of this straight section is the Young modulus. Beyond that point, the relationship becomes non-linear, and eventually the material yields, deforms plastically, and ultimately fractures.
The steep, straight section at the left of the graph is where the Young modulus lives. Notice how the gradient is very steep for metals — a large stress produces only a tiny strain, which is exactly what a high Young modulus tells you. The curve flattening and eventually dropping to zero represents plastic deformation and fracture.
3. Reading the Young Modulus from a Stress-Strain Graph
The key skill the specification requires is extracting the Young modulus directly from a stress-strain graph. Because $E = \sigma / \varepsilon$, and in the linear region $\sigma$ is proportional to $\varepsilon$, the Young modulus is the gradient of the straight-line portion of the graph.
To find the gradient accurately, draw the largest triangle you can along the straight section — larger triangles reduce the percentage uncertainty in your reading. Read off the change in stress ($\Delta \sigma$) and the corresponding change in strain ($\Delta \varepsilon$), then divide. Always include units: the gradient carries units of Pa (or GPa for most metals).
Worked example: Calculating Young Modulus from a Stress-Strain Graph
A stress-strain graph for a metal wire has a linear region. Two points clearly on the straight line are: Point A at strain = 0.0010, stress = 70 MPa, and Point B at strain = 0.0030, stress = 210 MPa. Calculate the Young modulus.
Point A: $\varepsilon_A = 0.0010$, $\sigma_A = 70 \times 10^6$ Pa\nPoint B: $\varepsilon_B = 0.0030$, $\sigma_B = 210 \times 10^6$ Pa — Always choose points that lie clearly on the straight line, not near the curve. Convert MPa to Pa: 1 MPa = 1 × 10⁶ Pa. [M1]
$\Delta \sigma = \sigma_B - \sigma_A = (210 - 70) \times 10^6 = 140 \times 10^6$ Pa — This is the rise of the gradient triangle. Keep values in Pa throughout. [M1]
$\Delta \varepsilon = \varepsilon_B - \varepsilon_A = 0.0030 - 0.0010 = 0.0020$ — Strain is dimensionless, so no unit conversion needed. This is the run of the gradient triangle.
$E = 7.0 \times 10^{10}$ Pa $= 70$ GPa — For metals, Young modulus values are typically in the range of tens to hundreds of GPa. 70 GPa is within the range expected for metals — a good sanity check. [A1]
Answer: $E = 7.0 \times 10^{10}$ Pa (70 GPa)
4. Interpreting Stress-Strain Graphs — Beyond the Gradient
The gradient gives you the Young modulus, but the shape of the whole graph tells a richer story about the material. Comparing two stress-strain curves on the same axes lets you analyse stiffness, strength, and ductility — all from a single diagram.
Feature of graph
What it tells you
Example material behaviour
Steep gradient in linear region
High Young modulus — stiff material
Steel has a steeper gradient than rubber
High stress at fracture
High ultimate tensile strength
High-carbon steel fractures at higher stress than copper
Large strain at fracture
Ductile material — stretches a lot before breaking
Copper shows large plastic deformation before fracture
Small strain at fracture
Brittle material — little plastic deformation
Glass fractures at low strain with almost no plastic region
Area under the linear region
Elastic strain energy stored per unit volume
Larger area = more energy stored elastically
A brittle material such as glass has a stress-strain graph that is almost entirely linear right up to the point of fracture — there is virtually no plastic region. A ductile material such as copper shows a clear yield point, a long plastic region, and fractures at a much larger strain. The Young modulus (gradient of the linear section) can be similar for two materials even if their fracture behaviours are completely different.
5. Measuring the Young Modulus — The Wire Experiment
The specification requires one simple method of measurement. The standard AQA approach uses a long metal wire under tension, measuring extension as masses are added. Using a long, thin wire is a deliberate choice — it maximises the measurable extension, reducing the percentage uncertainty in your strain calculation.
One detail that trips students up: the cross-sectional area is calculated from the diameter, not the radius. Measure the diameter with a micrometer, halve it to get the radius, then use $A = \pi r^2$. A micrometer is essential here — wire diameters are typically less than 1 mm, and a ruler cannot measure them accurately enough.
6. Worked Exam Question — Full Method
Exam questions on this topic often give you raw experimental data — a wire's dimensions and a table of load versus extension — and ask you to calculate the Young modulus. The route is always the same: convert load to stress, convert extension to strain, then find the gradient. Here is a full worked example following that route.
Worked example: Young Modulus from Experimental Data
A metal wire of original length 2.00 m and diameter 0.40 mm is loaded with a force of 60 N. The extension measured is 1.70 mm. Calculate the Young modulus of the wire.
$L_0 = 2.00$ m\n$d = 0.40$ mm $= 0.40 \times 10^{-3}$ m\n$F = 60$ N\n$\Delta L = 1.70$ mm $= 1.70 \times 10^{-3}$ m — Converting to SI base units at the start prevents unit errors later. This step alone earns a method mark in many mark schemes. [M1]
$r = d/2 = 0.40 \times 10^{-3} / 2 = 0.20 \times 10^{-3}$ m\n$A = \pi r^2 = \pi \times (0.20 \times 10^{-3})^2$\n$(0.20 \times 10^{-3})^2 = 0.20 \times 10^{-3} \times 0.20 \times 10^{-3} = 0.040 \times 10^{-6} = 4.0 \times 10^{-8}$ m$^2$\n$A = \pi \times 4.0 \times 10^{-8} = 1.257 \times 10^{-7}$ m$^2$ — Use radius, not diameter, in the area formula. Expanding the square longhand avoids the common error of squaring only the coefficient and not the power of ten. [M1]
$E = \frac{\sigma}{\varepsilon} = \frac{4.77 \times 10^{8}}{8.5 \times 10^{-4}}$\n$E = \frac{4.77}{8.5} \times 10^{8+4} = 0.5612 \times 10^{12} = 5.6 \times 10^{11}$ Pa — 4.77 ÷ 8.5 ≈ 0.561. The exponent: 10⁸ ÷ 10⁻⁴ = 10^(8−(−4)) = 10¹². So E ≈ 5.6 × 10¹¹ Pa. This value is within the range expected for metals, though it is higher than the accepted Young modulus for steel (~2 × 10¹¹ Pa) — always check your answer against known values for the specific material stated in the question. [A1]
$E \approx 5.6 \times 10^{11}$ Pa — Give the answer to 2 significant figures, matching the precision of the least precise data given (diameter was given to 2 s.f.). Always state the unit — Pa or N m⁻². [A1]
Answer: $E \approx 5.6 \times 10^{11}$ Pa
7. Exam Technique and Common Mistakes
The Young modulus calculation is reliable marks if you follow a consistent method — but there are several specific traps that appear repeatedly in AQA mark schemes. Knowing them in advance is the difference between dropping marks and collecting them.
For a 'describe' question about the experiment, the examiner wants a clear sequence: measure diameter (micrometer, multiple positions, take mean), measure original length, add known loads, record extensions, calculate stress and strain for each, plot the graph, and find the gradient of the linear region. Each of those steps is a potential mark — do not skip any.
For an 'evaluate' question about the experiment, structure your answer around sources of uncertainty and how to reduce them: the micrometer reading for diameter (random error — take multiple readings), the ruler reading for extension (parallax error — use a travelling microscope or vernier scale), and the assumption that the wire remains within its elastic limit (check by removing masses and confirming the wire returns to its original length). The Young modulus from the stress-strain graph is the specification's required method — make sure every exam answer connects back to that gradient.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
The limit of proportionality is the point on a stress-strain graph where the straight, linear region ends. Up to it, stress is proportional to strain and the gradient equals the Young modulus; beyond it, the relationship becomes non-linear, and eventually the material yields, deforms plastically and fractures.
What does the area under a stress-strain graph represent?
The area under the linear region of a stress-strain graph is the elastic strain energy stored per unit volume of the material. A larger area means more energy is stored elastically.
How do you find the Young modulus from a stress-strain graph?
The Young modulus is the gradient of the straight-line section of a stress-strain graph. The largest triangle that fits is drawn on the linear region, and the change in stress is divided by the change in strain. For points at 70 MPa (strain 0.0010) and 210 MPa (strain 0.0030), E = 140 × 10⁶ ÷ 0.0020 = 7.0 × 10¹⁰ Pa, or 70 GPa.
What does a stress-strain graph show?
A stress-strain graph shows how a material responds to load. A steep gradient in the linear region means a high Young modulus and a stiff material; a high stress at fracture means a high ultimate tensile strength; a large strain at fracture means a ductile material; and the area under the linear region is the elastic strain energy stored per unit volume.
Why is a long thin wire used to measure the Young modulus?
A long, thin wire is used because it maximises the measurable extension for a given load, which reduces the percentage uncertainty in the strain. The diameter is measured with a micrometer, because wire diameters are typically less than 1 mm and a ruler cannot measure them accurately enough.
Why draw a large gradient triangle on a stress-strain graph?
A large gradient triangle is drawn on the linear part of a stress-strain graph because larger triangles reduce the percentage uncertainty in the readings of the change in stress and change in strain. The points chosen must lie clearly on the straight line, not near the curve.
Why is a micrometer used to measure wire diameter?
A micrometer is used to measure wire diameter because wire diameters are typically less than 1 mm, and a ruler cannot measure them accurately enough. The diameter is measured at multiple positions, a mean is taken, and it is halved to give the radius for A = πr².
How do you calculate the Young modulus from experimental data?
The Young modulus from experimental data is stress divided by strain. For a 2.00 m wire of diameter 0.40 mm under 60 N with an extension of 1.70 mm, A = 1.257 × 10⁻⁷ m², stress = 4.77 × 10⁸ Pa and strain = 8.5 × 10⁻⁴, so E ≈ 5.6 × 10¹¹ Pa.
How do you convert MPa to Pa?
One MPa is 1 × 10⁶ Pa, so stresses read from a graph in MPa are converted to Pa before calculating. A stress of 210 MPa is 210 × 10⁶ Pa, and a change of 140 MPa is 140 × 10⁶ Pa.
Can two materials have the same Young modulus but break differently?
Two materials can have a similar Young modulus, the gradient of the linear section, even if their fracture behaviours are completely different. Glass is almost entirely linear up to fracture with virtually no plastic region, while copper shows a clear yield point and a long plastic region before it breaks.
What is a typical Young modulus for a metal?
Young modulus values for metals are typically in the range of tens to hundreds of GPa. A calculated value such as 70 GPa fits that range, which is a useful sanity check on an answer, and metals give a very steep stress-strain gradient.