Stress, Strain and Hooke’s Law (Cambridge IGCSE / A‑Level Physics 9702 – Topic 6 Deformation of Solids)
1. Key Definitions and Units
- Load (Force, F): External force applied to a material (tension or compression). Units: newton (N).
- Original length (L₀): Length of the specimen before any load is applied. Units: metre (m).
- Extension / Compression (ΔL): Change in length caused by the load.
ΔL = L – L₀ (positive for tension, negative for compression). Units: metre (m).
- Cross‑sectional area (A): Area of the material perpendicular to the load. Units: square metre (m²).
- Stress (σ): Load per unit area.
$$\sigma = \frac{F}{A}$$
Units: pascal (Pa) = N·m⁻².
- Strain (ε): Relative change in length.
$$\varepsilon = \frac{\Delta L}{L_{0}}$$
Dimensionless (ratio of lengths).
2. Hooke’s Law and the Limit of Proportionality
Within the **elastic region** of a material, stress is directly proportional to strain:
$$\sigma = E\,\varepsilon$$
- E – Young’s modulus (elastic modulus): a measure of the material’s stiffness.
- The straight‑line portion of a stress‑strain graph ends at the limit of proportionality** (often called the **elastic limit** in the syllabus). Up to this point the material obeys Hooke’s law and will return to its original dimensions when the load is removed.
- On a stress‑strain diagram the **gradient (slope) of the initial linear region is exactly E** (units Pa). This provides a visual way to obtain Young’s modulus from experimental data.
3. Young’s Modulus (E)
Young’s modulus is the gradient of the initial straight‑line portion of the stress‑strain curve.
| Material |
Typical Young’s Modulus, E (GPa) |
| Steel | 200 |
| Aluminium | 70 |
| Copper | 110 |
| Glass | 70 |
| Rubber (soft) | 0.01 |
4. Spring Constant (k) and Its Relation to Young’s Modulus
For a uniform rod or wire that behaves like a spring:
$$k = \frac{F}{x}$$
where
x is the extension (or compression).
Using the geometry of a rod,
$$k = \frac{EA}{L_{0}}$$
- E – Young’s modulus (Pa)
- A – Cross‑sectional area (m²)
- L₀ – Original length of the rod (m)
5. Elastic and Plastic Behaviour
- Limit of proportionality / Elastic limit: Maximum stress at which the stress‑strain relationship remains linear and the material returns to its original shape when the load is removed.
- Yield point: Stress at which permanent (plastic) deformation begins; a small increase in stress produces a large increase in strain.
- Plastic region: Deformation is permanent; the material will not recover its original dimensions.
- Ultimate tensile strength (UTS) and fracture point lie beyond the plastic region (included for completeness).
6. Work Done and Elastic Potential Energy
- Work done on a material up to an extension ΔL is the area under the force‑extension graph:
$$W = \int_{0}^{\Delta L} F\,dx$$
- For a Hookean (linear) spring the graph is a right‑angled triangle, giving:
$$E_{\text{PE}} = \frac{1}{2}F\Delta L = \frac{1}{2}k(\Delta L)^{2}$$
- This energy is recoverable provided the material has not been loaded beyond the elastic limit.
7. Worked Example – Applying Hooke’s Law
- Given a steel rod:
- L₀ = 1.20 m
- A = 2.0 × 10⁻⁴ m²
- Load F = 12.0 kN
- Find stress σ, strain ε, and extension ΔL.
- Solution
- Stress:
$$\sigma = \frac{F}{A}= \frac{12.0\times10^{3}\ \text{N}}{2.0\times10^{-4}\ \text{m}^{2}} = 6.0\times10^{7}\ \text{Pa}$$
- Young’s modulus for steel: E = 200 GPa = 2.0 × 10¹¹ Pa.
- Strain (Hooke’s law):
$$\varepsilon = \frac{\sigma}{E}= \frac{6.0\times10^{7}}{2.0\times10^{11}} = 3.0\times10^{-4}$$
- Extension:
$$\Delta L = \varepsilon L_{0}= (3.0\times10^{-4})(1.20\ \text{m}) = 3.6\times10^{-4}\ \text{m}=0.36\ \text{mm}$$
8. Stress‑Strain Diagram – Key Features
When sketching or interpreting a stress‑strain graph, label the following points in order from left to right:
- Proportional limit (linear, Hooke’s law region)
- Elastic limit (same point as the proportional limit in the syllabus)
- Yield point
- Plastic region
- Ultimate tensile strength (UTS)
- Fracture point
The gradient of the initial straight line is the Young’s modulus, E.
9. Quick Revision Checklist
- Define and give units for load, stress, strain, extension, compression and cross‑sectional area.
- State Hooke’s law: σ = E ε and specify that it applies up to the **limit of proportionality (elastic limit)**.
- Remember: the gradient of the initial linear part of a stress‑strain graph equals E.
- Relate spring constant to Young’s modulus: k = EA/L₀.
- Identify the elastic limit, yield point and the different regions on a stress‑strain diagram.
- Calculate work done and elastic potential energy: Eₚₑ = ½ k x².
- Recall typical Young’s modulus values for common materials (e.g., steel ≈ 200 GPa, aluminium ≈ 70 GPa).
- Be able to rearrange σ = E ε, k = EA/L₀ and Eₚₑ = ½ k x² to solve for any unknown quantity.