Waves An understanding of colour from Cambridge IGCSE/O Level Physics or equivalent is assumed.

Elastic and Plastic Behaviour – Cambridge AS Physics 9702 (Topic 6.2)

1. Key Mechanical Quantities

• Load (F) – the external force applied to a specimen (N). In the syllabus the term *load* is preferred to “force” when describing the action on the material.
• Extension (ΔL) – increase in length of the specimen when a tensile load is applied (m).
• Compression – decrease in length when a compressive load is applied (m).
• Original length (L₀) – length of the specimen before any load is applied (m).

2. Limit of Proportionality vs Elastic Limit

3. Hooke’s Law

For a material that obeys Hooke’s law the restoring force is directly proportional to the extension.

F = k x

• k – spring constant (N m⁻¹) – measures the stiffness of a particular spring or rod.
• For a uniform rod of original length L₀, cross‑sectional area A and Young’s modulus E:

  k = (E A) ⁄ L₀

• Hooke’s law expressed in stress–strain form (the second required form):

  σ = E ε where σ = F⁄A and ε = ΔL⁄L₀

4. Stress and Strain

• Stress (σ) – force applied per unit area: σ = F⁄A (Pa or N m⁻²).
• Strain (ε) – relative change in length: ε = ΔL⁄L₀ (dimensionless).
• Stress has the units of pressure; strain is a pure ratio.

5. The Stress–Strain Curve

The diagram below (placeholder) summarises how a material responds to an increasing load.

6. Elastic Region

• Linear relationship: σ ∝ ε or F ∝ x.
• Hooke’s law applies: σ = E ε or F = k x.
• Young’s modulus (E) – material constant that measures stiffness (Pa). Larger E ⇒ stiffer material.
• The **limit of proportionality** marks the end of the straight‑line portion; the **elastic limit** lies at or just beyond this point, where deformation is still fully reversible.

7. Plastic Region

• Beyond the elastic limit the material deforms permanently; the curve becomes non‑linear.
• Yield point (yield stress σ_y) – stress at which noticeable plastic deformation begins.
• Ultimate tensile strength (UTS, σ_UTS) – maximum stress the material can sustain before necking.
• Fracture stress (σ_f) – stress at which the material finally breaks.

8. Key Material Properties

9. Energy Considerations

• Modulus of resilience (elastic energy)

  $$U_r=\int_{0}^{\varepsilon_y}\sigma\,d\varepsilon =\frac{1}{2}\sigma_y\varepsilon_y =\frac{\sigma_y^{2}}{2E}$$

  It is the maximum energy a material can absorb without permanent deformation.
• Modulus of toughness (total energy to fracture)

  $$U_t=\int_{0}^{\varepsilon_f}\sigma\,d\varepsilon$$

  The larger the area under the entire stress–strain curve, the tougher the material.

10. Factors Influencing Elastic and Plastic Behaviour

1. Material type – Metals usually exhibit a clear yield point; polymers often show a gradual transition.
2. Temperature – Raising temperature generally reduces E and σ_y, making materials more ductile.
3. Strain‑rate – Faster loading can increase apparent strength (strain‑rate hardening).
4. Microstructure – Grain size, dislocation density, phase composition and heat treatment all affect both elastic and plastic limits.

11. Practical Applications

• Springs – Require a high Young’s modulus and a large elastic limit; design uses F = k x with k = EA/L₀.
• Crumple zones in cars – Exploit controlled plastic deformation to absorb impact energy (high modulus of toughness).
• Structural beams – Must operate within the elastic region under service loads to avoid permanent sagging.
• Metal‑forming processes (rolling, forging, extrusion) – Use the plastic region to shape components; knowledge of yield strength and UTS is essential.

12. Summary Checklist (Exam‑style)

• Identify the limit of proportionality and the elastic limit on a stress–strain diagram.
• Write and use Hooke’s law in both forms: F = k x and σ = E ε.
• Distinguish between yield strength, ultimate tensile strength and fracture stress.
• Calculate the modulus of resilience and explain its relevance to energy‑absorbing devices.
• Explain how temperature, strain‑rate and microstructure affect elastic and plastic behaviour.

13. Worked Example – Spring Constant of a Steel Wire

Given: length L₀ = 0.50 m, cross‑sectional area A = 2.0 × 10⁻⁶ m², Young’s modulus E = 2.0 × 10¹¹ Pa.

k = (E A) ⁄ L₀ = (2.0 × 10¹¹ Pa × 2.0 × 10⁻⁶ m²) ⁄ 0.50 m = 8.0 × 10⁵ N m⁻¹

The wire obeys F = k x up to its limit of proportionality (≈ 0.2 % strain for typical steel). Beyond this point the force‑extension relationship becomes non‑linear, although the wire will still return to its original length if the elastic limit is not exceeded.