Young Modulus from Stress-Strain Graphs Revision Notes | A Level AQA Physics

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.

1. 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]
2. $\Delta \sigma = \sigma_B - \sigma_A = (210 - 70) \times 10^6 = 140 \times 10^6$ Pa — This is the rise of the gradient triangle. Keep values in Pa throughout. [M1]
3. $\Delta \varepsilon = \varepsilon_B - \varepsilon_A = 0.0030 - 0.0010 = 0.0020$ — Strain is dimensionless, so no unit conversion needed. This is the run of the gradient triangle.
4. $E = \frac{\Delta \sigma}{\Delta \varepsilon} = \frac{140 \times 10^6}{0.0020}$\n$E = 70 \times 10^9$ Pa — Dividing: 140 × 10⁶ ÷ 0.0020 = 140 × 10⁶ × 500 = 70 000 × 10⁶ = 7.0 × 10¹⁰ Pa. [A1]
5. $E = 7.0 \times 10^{10}$ 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.

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.

1. $L_0 = 2.00$ 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]
2. $r = d/2 = 0.40 \times 10^{-3} / 2 = 0.20 \times 10^{-3}$ 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]
3. $\sigma = \frac{F}{A} = \frac{60}{1.257 \times 10^{-7}}$\n$\sigma = 4.77 \times 10^{8}$ Pa — 60 ÷ (1.257 × 10⁻⁷) = 60 × (1/1.257) × 10⁷ ≈ 47.7 × 10⁷ = 4.77 × 10⁸ Pa. [M1]
4. $\varepsilon = \frac{\Delta L}{L_0} = \frac{1.70 \times 10^{-3}}{2.00}$\n$\varepsilon = 8.5 \times 10^{-4}$ — Strain is dimensionless. 1.70 × 10⁻³ ÷ 2.00 = 0.85 × 10⁻³ = 8.5 × 10⁻⁴. [M1]
5. $E = \frac{\sigma}{\varepsilon} = \frac{4.77 \times 10^{8}}{8.5 \times 10^{-4}}$\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]
6. $E \approx 5.6 \times 10^{11}$ 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.