Elastic and Plastic Behaviour – Cambridge IGCSE/A‑Level Physics 9702
Learning objectives
- Define load, extension and compression.
- Understand and use the terms stress, strain, limit of proportionality, elastic limit, yield point and plastic deformation.
- Apply Hooke’s law and Young’s modulus to calculate extensions in the elastic region.
- Evaluate the elastic potential energy stored in a deformed body.
- Interpret a stress‑strain graph and identify the different limits.
1. Basic quantities
- Load (F): the external force applied to a material (tension or compression).
- Extension (ΔL) / Compression (ΔL): change in length of the specimen under the load. Positive ΔL = extension, negative ΔL = compression.
- Stress (σ): load per unit cross‑sectional area.
\[
\sigma = \frac{F}{A}\qquad(\text{Pa}= \text{N m}^{-2})
\]
- Strain (ε): relative change in length.
\[
\varepsilon = \frac{\Delta L}{L}\qquad(\text{dimensionless})
\]
- Both stress and strain are independent of the size of the specimen, allowing direct comparison of different materials.
2. Hooke’s law, Young’s modulus and the limit of proportionality
- In the initial straight‑line part of a stress‑strain curve the two quantities are proportional:
\[
\sigma = E\,\varepsilon
\]
where E is Young’s modulus (Pa or GPa).
- Re‑arranging gives the familiar extension formula for a uniform rod of original length L and cross‑sectional area A:
\[
\Delta L = \frac{F L}{A E}
\]
- Limit of proportionality (sometimes called the proportional limit): the greatest stress at which the σ‑ε relationship remains strictly linear and Hooke’s law is exactly obeyed.
3. Elastic vs. plastic behaviour
- Elastic deformation: reversible change. When the load is removed the material returns to its original dimensions.
- Plastic deformation: irreversible change. A permanent strain remains after the load is removed.
4. Important limits on the stress‑strain curve
| Term |
Definition (Cambridge syllabus) |
Position on a σ‑ε graph |
| Limit of proportionality |
Maximum stress for which stress and strain are strictly proportional (Hooke’s law holds exactly). |
End of the straight‑line region. |
| Elastic limit |
Maximum stress for which deformation is completely reversible. May lie slightly beyond the limit of proportionality. |
Just before the onset of permanent (plastic) strain. |
| Yield point (yield stress) |
Stress at which permanent (plastic) deformation begins; identified by a noticeable deviation from linearity. |
Start of the curved “plastic” region. |
| Ultimate tensile strength (optional) |
Maximum stress the material can sustain before necking begins. |
Peak of the σ‑ε curve. |
5. Elastic potential energy
- When a material is deformed elastically, work done by the load is stored as elastic potential energy.
- Energy per unit volume (energy density) is the area under the linear part of the σ‑ε graph:
\[
u = \frac{1}{2}\,\sigma\,\varepsilon \qquad (\text{J m}^{-3})
\]
- Total stored energy:
\[
U = u\,V = \frac{1}{2}\,\sigma\,\varepsilon\,V
\]
where V is the volume of the specimen.
6. Typical material properties
| Material |
Young’s Modulus E (GPa) |
Elastic limit σe (MPa) |
Yield strength σy (MPa) |
| Steel (mild) |
≈ 200 |
≈ 250 |
≈ 350 |
| Aluminium |
≈ 70 |
≈ 70 |
≈ 95 |
| Copper |
≈ 110 |
≈ 70 |
≈ 210 |
| Polymers (e.g., PVC) |
2–4 |
≈ 5 |
≈ 15 |
7. Worked example (IGCSE/A‑Level style)
Problem: A steel wire 2.00 m long has a cross‑sectional area of \(1.0\times10^{-6}\,\text{m}^2\). It is subjected to a tensile load of 300 N. Determine whether the deformation is elastic or plastic and, if elastic, calculate the extension. Use \(E_{\text{steel}} = 200\;\text{GPa}\) and \(\sigma_{e}=250\;\text{MPa}\).
- Calculate the stress:
\[
\sigma = \frac{F}{A}= \frac{300}{1.0\times10^{-6}} = 3.0\times10^{8}\,\text{Pa}=300\;\text{MPa}
\]
- Compare with the elastic limit:
\[
300\;\text{MPa} \;>\; 250\;\text{MPa}
\]
The applied stress exceeds the elastic limit, so the wire will undergo plastic deformation. Hooke’s‑law formula cannot be used to find the final length.
- What if the load were 200 N?
Stress = \(\frac{200}{1.0\times10^{-6}} = 2.0\times10^{8}\,\text{Pa}=200\;\text{MPa}\) (< 250 MPa) → deformation is elastic.
Extension:
\[
\Delta L = \frac{F L}{A E}= \frac{200\times2.0}{1.0\times10^{-6}\times2.0\times10^{11}}
=2.0\times10^{-3}\,\text{m}=2.0\;\text{mm}
\]
8. Summary – key points to remember
- Load = external force; extension/compression = change in length.
- Stress = \(F/A\); Strain = \(\Delta L/L\).
- In the linear region, \(\sigma = E\varepsilon\). The limit of proportionality marks the end of exact linearity.
- The elastic limit is the greatest stress for which the material returns completely to its original shape when the load is removed.
- The yield point is where permanent (plastic) strain first appears.
- Elastic potential energy stored: \(U = \tfrac12 \sigma \varepsilon V\). It is released when the load is withdrawn.
- Materials differ markedly in Young’s modulus and elastic limits, which determines whether they are suited for springs, structural members, or flexible components.