Elastic and Plastic Behaviour

Below its elastic limit, a material returns to its original shape when the load is removed; beyond it, it's permanently (plastically) deformed. The work done stretching a material is the area under its force–extension graph, and within the limit of proportionality the stored elastic potential energy is $E_P=\tfrac12Fx=\tfrac12kx^2$ . Ductile materials deform plastically a lot before breaking; brittle ones snap.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 30 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Distinguish elastic and plastic deformation, and define the elastic limit.
Distinguish brittle and ductile materials from their graphs.
Find work done as the area under a force–extension graph.
Use $E_P=\tfrac12Fx=\tfrac12kx^2$ .

Pause and think

Bend a paper clip slightly and it springs back. Bend it a lot and it stays bent. What changed inside the metal?

Section 2 of 14

Core idea

Start hereA catapult stores energy

Pull back a catapult and let go: the stone flies. The stretched rubber stored the work you did. That stored energy is the area under the force–extension graph.

The big idea

Elastic: reversible; plastic: permanent. Stored energy = area under $F$ – $x$ ; linear: $\frac12Fx=\frac12kx^2$ .

1. Loading and unloading

Loading and unloading past the elastic limit

On unloading, the line is parallel to the original elastic line, but it does not return to zero: the wire has been permanently stretched.

Ductile (copper): large plastic region.
Brittle (glass): little plastic deformation.
Rubber: unloading below loading; area between = energy dissipated (hysteresis).

2. Elastic potential energy

Elastic energy stored in a spring: area under F–x

extension (m) = 0.03

k=100

N/m. At 0.2 m:

E=\tfrac12(100)(0.04)=2

J. Doubling the extension stores four times the energy.

E=\frac12kx^2

$W=\text{area under the }F\text{ vs }x\text{ graph}$
Work done stretching.
$=\tfrac12Fx$
Triangle (linear region).
$=\tfrac12kx^2$
$F=kx$ .

Per unit volume:

\frac12\sigma\varepsilon

Worked exampleSpring k = 100 N m⁻¹

Energy at 0.2 m and 0.4 m.

Goal: Two energies.
Strategy: $\frac12kx^2$ .

$\tfrac12(100)(0.04)=2\text{ J}$
0.2 m.
$\tfrac12(100)(0.16)=8\text{ J}$
0.4 m.

Answer

2

8

✓ CheckDouble extension, four times energy ✓.

Worked exampleA catapult

$k=400$ N m $^{-1}$ , pulled 0.15 m, 20 g stone.

Goal: Launch speed (all energy transferred).
Strategy: Elastic PE → KE.

$\tfrac12(400)(0.15^2)=4.5\text{ J}$
Stored.
$v=\sqrt{\frac{2(4.5)}{0.020}}=21\text{ m s}^{-1}$
KE.

Answer

21

m s

^{-1}

(upper limit).

✓ CheckReal catapults lose some energy (hysteresis) ✓.

Common misconception $\frac12Fx$ beyond the limit Non-linear: estimate the area by counting squares.

Summary

Elastic vs plastic.
Energy = area; $\frac12kx^2$ when linear.

Hysteresis: Energy lost when loading and unloading curves differ.

HistoryThe Liberty ships. During the Second World War, some welded Liberty cargo ships cracked in half in cold Atlantic water. Their steel became brittle at low temperatures. The investigation founded the science of fracture mechanics, and ship steel is now tested at low temperatures.

Fun factGorilla Glass bends a little. Phone screens use chemically strengthened glass: ions in the surface are swapped for bigger ones, squeezing the surface into compression. Cracks find it harder to open, so the glass survives drops that would shatter ordinary window glass.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in