understand and use the terms elastic deformation, plastic deformation and elastic limit

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 \(\sigma_{e}=250\;\text{MPa}\).

  1. Calculate the stress: \[ \sigma = \frac{F}{A}= \frac{300}{1.0\times10^{-6}} = 3.0\times10^{8}\,\text{Pa}=300\;\text{MPa} \]
  2. 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.
  3. 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.

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