3.4.2.2MechanicsStretch

The Young Modulus

The spring constant tells you about one particular spring; the $Young modulus$ tells you about the $material$ itself, whatever its shape. By working with $stress$ and $strain$ instead of force and extension, we get a number that lets us compare steel with copper with bone — and read the story of how a material behaves right up to breaking.

45 min Video by Science Shorts 3.4.2 Materials

Young’s Modulus & Vernier Scales — A-Level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Define and calculate tensile stress and tensile strain
Define the Young modulus and use E = stress / strain
Interpret a stress–strain graph and its key points
Distinguish brittle, ductile and plastic behaviour
Describe how to measure the Young modulus of a wire

Tensile stress

$Tensile stress$ is the force applied per unit cross-sectional area of a sample. It captures the idea that a thin wire feels a load far more keenly than a thick one. The unit is the pascal (Pa), the same as N m⁻².

For a wire of circular cross-section, remember that the area depends on the square of the diameter, $A = π r^{2} = π (d /2)^{2}$ — so halving the diameter quarters the area and quadruples the stress.

σ = \frac{F}{A}

Tensile stress = force ÷ cross-sectional area (Pa).

Tensile strain

$Tensile strain$ is the extension expressed as a fraction of the original length. Because it is one length divided by another, strain has $no units$ — it is often quoted as a decimal or a percentage.

Strain tells you how much a material has stretched relative to its size, which is why a $1$ mm extension is dramatic for a $10$ cm sample but negligible for a $10$ m one.

ε = \frac{Δ L}{L}

Tensile strain = extension ÷ original length (no units).

The Young modulus

The $Young modulus$ is the ratio of tensile stress to tensile strain, valid while the material obeys Hooke’s law (the straight-line region). It measures $stiffness$ : a large Young modulus means a big stress is needed for a small strain. Its unit is the pascal, and for stiff materials like steel it is around $2 \times 1 0^{11}$ Pa.

Combining the definitions gives a handy working form, $E = \frac{F L}{A Δ L}$ , which is what you use when all four measured quantities are known.

E = \frac{σ}{ε} = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}

Young modulus = stress ÷ strain (Pa).

1Cross-sectional area

A = π (d /2)^{2} = π (0.5 \times 1 0^{- 3})^{2} = 7.85 \times 1 0^{- 7}

m².

2Stress

σ = F / A = 80/ (7.85 \times 1 0^{- 7}) = 1.02 \times 1 0^{8}

Pa.

3Strain

ε = Δ L / L = (1.0 \times 1 0^{- 3}) /2.0 = 5.0 \times 1 0^{- 4}

E = σ / ε = (1.02 \times 1 0^{8}) / (5.0 \times 1 0^{- 4}) = 2.0 \times 1 0^{11}

Pa.

Answer

E = 2.0 \times 1 0^{11}

Tip — Because A depends on d², an error in measuring the diameter has a doubled effect on the stress — which is why the diameter must be measured very carefully.

Stress–strain graphs

A stress–strain graph tells the whole story of a material. From the origin it rises in a straight line up to the $limit of proportionality$ , where stress and strain are proportional and the gradient is the Young modulus. Just beyond lies the $elastic limit$ — the last point from which the material returns to its original length.

A $ductile$ material such as copper then reaches a $yield point$ and stretches a lot for little extra stress (plastic behaviour) before reaching its $ultimate tensile stress$ and breaking. A $brittle$ material such as glass has almost no plastic region — it obeys Hooke’s law right up to the point where it snaps.

energy stored per unit volume = area under the stress-strain graph

In the linear region this equals ½ × stress × strain.

Tip — The area under a force–extension graph is energy (in joules); the area under a stress–strain graph is energy per unit volume (in J m⁻³). Don’t mix them up.

Measuring the Young modulus

The standard experiment uses a long, thin wire clamped at one end and loaded at the other. A long wire is chosen so the extension is large enough to measure, and a thin wire so the stress is large. Extension is read with a marker against a ruler or vernier scale; the diameter is measured with a micrometer at several points and averaged (since $A \propto d^{2}$ ).

You then increase the load in steps, record the extension each time, and plot a graph. A stress–strain graph has gradient equal to the Young modulus; equivalently, a force–extension graph has gradient $\frac{E A}{L}$ , from which $E$ is found.

Equation recap

σ = \frac{F}{A}

Tensile stress (Pa).

ε = \frac{Δ L}{L}

Tensile strain (no units).

E = \frac{σ}{ε} = \frac{F L}{A Δ L}

Young modulus (Pa).

A = π (\frac{d}{2})^{2}

Cross-sectional area of a wire from its diameter.

Common mistakes to avoid

Using the radius as the diameter (or vice versa) when finding the area.

A = π(d/2)² = πr². Halve the diameter to get the radius before squaring.

Giving the strain a unit.

Strain is a ratio of two lengths, so it has no units (it may be written as a percentage).

Calculating the Young modulus beyond the limit of proportionality.

E = stress/strain only applies in the straight-line region where Hooke’s law holds.

Reading the area under a stress–strain graph as energy in joules.

That area is energy per unit volume (J m⁻³); multiply by the volume for the total energy.

Key takeaways

Tensile stress σ = F/A (Pa); tensile strain ε = ΔL/L (no units).
Young modulus E = stress/strain = FL/(AΔL) measures a material’s stiffness.
A stress–strain graph shows the limit of proportionality, elastic limit, yield and ultimate tensile stress.
Measure E with a long thin wire, taking special care over the diameter because A ∝ d².

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