Elastic Strain Energy and Material Behaviour Revision Notes | A Level AQA Physics

1. Why Stored Energy in Materials Matters

When you stretch a spring or bend a metal rod, you are doing work on the material. That work has to go somewhere — and understanding exactly where it goes, and whether it comes back, is the heart of this topic.

This lesson connects two ideas you already know: work done equals force times displacement, and energy is always conserved. The new question is: what form does that energy take inside a deformed material, and what happens to it when the material is pushed beyond its limits?

2. Elastic Strain Energy: The Formula and the Graph

When a material obeys Hooke's law, the force-extension relationship is linear — the graph is a straight line through the origin. The work done stretching the material is not $F \times \Delta L$, because the force is not constant; it builds from zero up to the final value $F$.

The work done — and therefore the elastic strain energy stored — equals the area of the triangle under the force-extension graph. For a linear (Hookean) material, that triangle has base $\Delta L$ and height $F$, giving the result below.

The key insight is that this formula works because the graph is linear. For any material — Hookean or not — the elastic strain energy is always the area under the force-extension graph up to the point of interest. The triangular formula is a special case of that general rule.

You can also combine this with Hooke's law ($F = k\Delta L$) to write the stored energy in terms of the spring constant: substituting $F = k\Delta L$ gives $E = \frac{1}{2}k(\Delta L)^2$. This form is useful when you know $k$ but not the force directly.

Worked example: Calculating Elastic Strain Energy from a Force-Extension Graph

A metal wire is stretched by a force that increases linearly from 0 N to 24 N, producing an extension of 3.0 mm. Calculate the elastic strain energy stored in the wire. [2 marks]

1. $\Delta L = 3.0 \text{ mm} = 3.0 \times 10^{-3} \text{ m}$ — Always convert to metres before substituting. Leaving it in mm gives an answer in mJ, which is straightforward to misread. [M1] for correct unit conversion.
2. $E = \frac{1}{2} F \Delta L$ — This applies because the force-extension graph is linear (Hooke's law is obeyed), so the stored energy equals the area of the triangle under the graph.
3. $E = \frac{1}{2} \times 24 \times 3.0 \times 10^{-3}$ — F = 24 N, ΔL = 3.0 × 10⁻³ m. Both values now carry SI units.
4. $E = \frac{1}{2} \times 24 \times 3.0 \times 10^{-3} = 12 \times 3.0 \times 10^{-3} = 36 \times 10^{-3} = 0.036 \text{ J}$ — Half of 24 is 12; 12 multiplied by 3.0 × 10⁻³ gives 3.6 × 10⁻² J. [A1] for correct answer with units.
5. Area of triangle $= \frac{1}{2} \times base \times height = \frac{1}{2} \times 3.0 \times 10^{-3} \times 24 = 0.036 \text{ J}$ ✓ — Confirming via the graph-area interpretation gives the same answer, which validates the approach.

Answer: $E = 0.036 \text{ J}$ (or $36 \text{ mJ}$)

3. Energy Conservation and Elastic Strain Energy

Energy conservation applies directly here. When you do work stretching a spring or wire elastically, all of that work is stored as elastic strain energy. When you release it, that energy is returned — for example, launching a projectile or driving a mechanism.

A classic quantitative application: a catapult stores elastic strain energy in a stretched elastic band. When released, that energy converts to kinetic energy of the projectile (assuming no energy is lost to heat or sound). You can write $\frac{1}{2}F\Delta L = \frac{1}{2}mv^2$ and solve for the launch speed.

When a material is deformed plastically, the situation changes. The work done on the material is not fully recovered — some energy is permanently dissipated as heat, rearranging the internal structure of the material. For a material loaded beyond its elastic limit, the unloading curve does not return to the origin — a permanent set remains, meaning the material does not return to its original length. The area between the loading and unloading curves represents the energy dissipated during the cycle. Note that not all materials show a closed hysteresis loop: for a ductile metal deformed plastically, the unloading curve is offset from the loading curve and ends at a non-zero extension.

Worked example: Applying Energy Conservation: Elastic Strain Energy to Kinetic Energy

A spring launcher stores 0.18 J of elastic strain energy. It fires a ball of mass 0.040 kg horizontally. Assuming all stored energy converts to kinetic energy of the ball, calculate the launch speed. [3 marks]

1. $E_{\text{elastic}} = E_{\text{kinetic}}$, so $\frac{1}{2}F\Delta L = \frac{1}{2}mv^2$ — All elastic strain energy converts to kinetic energy. The assumption of no energy loss is stated in the question — always check for this. [M1]
2. $0.18 = \frac{1}{2} \times 0.040 \times v^2$ — $E_{\text{elastic}} = 0.18$ J, $m = 0.040$ kg.
3. $v^2 = \frac{2 \times 0.18}{0.040} = \frac{0.36}{0.040} = 9.0 \text{ m}^2\text{s}^{-2}$ — Multiply both sides by 2, then divide by mass. Check: 0.36 ÷ 0.040 = 9.0. [M1]
4. $v = \sqrt{9.0} = 3.0 \text{ m s}^{-1}$ — $\sqrt{9.0} = 3.0$ exactly. Always state units. [A1]

Answer: $v = 3.0 \text{ m s}^{-1}$

4. Plastic Behaviour, Fracture, and Brittle Behaviour

A force-extension graph tells a story about a material's entire life under load. Up to the elastic limit, the material returns to its original shape when the force is removed — all the work done is stored as elastic strain energy and recovered.

Beyond the elastic limit, the material enters the plastic region. The atoms within the material are permanently displaced from their equilibrium positions — the material stretches further for a given increase in force, and a permanent set remains when the load is removed. The work done in this region is not recovered; it is dissipated internally.

Eventually, the material fractures. For a ductile material such as copper or mild steel, fracture is preceded by a long plastic region — the material necks (thins locally) and then breaks. For a brittle material such as glass or cast iron, fracture occurs with little or no plastic deformation: the graph is almost entirely linear right up to the breaking point.

On the ductile graph, you can identify: the linear (Hookean) region, the elastic limit, the yield point (where plastic flow begins), a long plastic region, and finally fracture. The brittle graph shows a linear region that ends abruptly at fracture — there is no visible plastic region.

• Elastic region: force-extension is linear; material returns to original length when unloaded; elastic strain energy is fully recovered.
• Plastic region: permanent deformation occurs; energy is dissipated rather than stored elastically; the material does not return to its original length.
• Fracture: the material breaks; for ductile materials this follows significant plastic deformation; for brittle materials it follows little or none.
• Brittle fracture: sudden, with minimal energy absorbed; the area under the graph (total energy to fracture) is small compared to a ductile material.

5. Interpreting Stress-Strain Curves

A force-extension graph depends on the dimensions of the specific sample tested. Stress-strain curves remove that dependence, allowing you to compare the intrinsic mechanical properties of different materials directly.

The gradient of the linear region of a stress-strain curve is the Young modulus ($E = \sigma / \varepsilon$), a measure of the material's stiffness. A steep gradient means a stiff material; a shallow gradient means a more flexible one.

Reading a stress-strain curve requires identifying several key features. The limit of proportionality is where the graph first deviates from linearity. The elastic limit follows shortly after. The yield stress (or yield point) marks the onset of plastic flow — in mild steel, this appears as a pronounced kink. The ultimate tensile stress (UTS) is the peak of the curve, representing the maximum stress the material can withstand. Beyond the UTS, necking occurs and the engineering stress falls until fracture.

Comparing materials on the same stress-strain axes is a powerful analytical tool. A stiff material (high Young modulus) has a steep initial gradient. A strong material has a high UTS. A tough material has a large area under the entire curve — it absorbs a lot of energy before fracture. A brittle material has a small area under the curve and no plastic region.

6. Required Practical: Investigating Force-Extension for a Metal Wire

The standard AQA required practical for this topic involves loading a long metal wire and measuring its extension at each load increment. The data allows you to plot a force-extension graph, identify the elastic and plastic regions, and calculate the spring constant (stiffness) from the gradient of the linear section. To determine the Young modulus, you must also measure the wire's original length and cross-sectional area, then apply $E = \frac{F/A}{\Delta L/L_0}$ (stress divided by strain).

To extend this force-extension practical into the full AQA treatment, you also measure the original length of the wire and its diameter with a micrometer so that its cross-sectional area can be calculated. You can then convert force to stress and extension to strain, plot a stress-strain graph, and use the gradient of the initial straight-line section to determine the Young modulus.

7. Exam Technique and Common Mistakes

Exam questions on this topic test three distinct skills: correct formula use with unit handling, graph interpretation with precise language, and energy conservation reasoning. Each has its own pitfalls.

For extended-response questions asking you to compare materials using stress-strain curves, structure your answer around: gradient (stiffness/Young modulus), peak stress (strength), extent of plastic region (ductility), and total area under the curve (toughness). Covering all four properties with graph evidence earns the top mark band.

Across this lesson, every spec requirement connects back to one unifying principle: work done on a material is stored as elastic strain energy when deformation is reversible, and dissipated when it is not. The force-extension and stress-strain graphs are the visual language for reading that story — master them, and the calculations and descriptions follow naturally.

Use the next question to connect the formulae to material behaviour: the same stored elastic energy can lead to very different outcomes depending on whether the material is ductile or brittle and whether the deformation is reversible.