Cambridge A-Level Physics 9702 – Elastic and Plastic Behaviour
Elastic and Plastic Behaviour
Learning Objective
Understand and use the terms elastic deformation, plastic deformation and elastic limit in the context of material behaviour under load.
Key Definitions
Elastic deformation: A reversible change in shape or size of a material when a stress is applied. The material returns to its original dimensions once the stress is removed.
Plastic deformation: An irreversible change in shape or size that remains after the applied stress is removed. Occurs when the material is loaded beyond its elastic limit.
Elastic limit (or proportional limit): The maximum stress that a material can sustain while still exhibiting purely elastic behaviour. Beyond this point, permanent (plastic) deformation begins.
Stress–Strain Relationship
The behaviour of a material under tensile or compressive load is illustrated by a stress–strain curve. The curve can be divided into distinct regions:
Linear elastic region – stress and strain are proportional (Hooke’s law). $ \sigma = E \varepsilon $ where $E$ is Young’s modulus.
Elastic limit – the end of the linear region; the material still returns to its original shape if the load is removed.
Yield point – the stress at which plastic deformation begins.
Strain‑hardening region – material becomes stronger as it is deformed plastically.
Fracture point – the material finally breaks.
Suggested diagram: Typical stress–strain curve showing elastic region, elastic limit, yield point, plastic region and fracture.
Table of Typical \cdot alues
Material
Young’s Modulus $E$ (GPa)
Elastic Limit $\sigma_{e}$ (MPa)
Typical Yield Strength $\sigma_{y}$ (MPa)
Steel (mild)
200
≈250
≈350
Aluminium
70
≈70
≈95
Copper
110
≈70
≈210
Polymers (e.g., P \cdot C)
2–4
≈5
≈15
Using the Concepts in Calculations
When a material is loaded within its elastic region, the extension $\Delta L$ of a rod of original length $L$ and cross‑sectional area $A$ can be found from Hooke’s law:
$$\Delta L = \frac{F L}{A E}$$
where $F$ is the applied force. If $F$ exceeds $A \sigma_{e}$, the material will undergo plastic deformation and the above equation no longer predicts the final length.
Example Problem
Problem: A steel wire of length $2.0\ \text{m}$ and cross‑sectional area $1.0\times10^{-6}\ \text{m}^2$ is subjected to a tensile force of $300\ \text{N}$. Determine whether the deformation is elastic or plastic and calculate the extension if it is elastic. (Take $E_{\text{steel}} = 200\ \text{GPa}$ and elastic limit $\sigma_{e}=250\ \text{MPa}$.)
Compare with the elastic limit: $300\ \text{MPa} > 250\ \text{MPa}$, so the wire has exceeded its elastic limit – plastic deformation will occur.
If the force were reduced to $200\ \text{N}$ (stress $200\ \text{MPa}$), the deformation would be elastic. The extension would be:
$$\Delta L = \frac{F L}{A E} = \frac{200 \times 2.0}{1.0\times10^{-6} \times 2.0\times10^{11}} = 2.0\times10^{-3}\ \text{m} = 2.0\ \text{mm}.$$
Key Points to Remember
Elastic deformation is reversible; plastic deformation is permanent.
The elastic limit marks the boundary between the two behaviours.
Within the elastic region, stress and strain are proportional (Hooke’s law).
Beyond the elastic limit, the material yields and undergoes strain hardening before fracture.
Different materials have widely varying elastic limits and Young’s moduli, influencing their suitability for engineering applications.