Cambridge IGCSE/A‑Level Physics – Stress, Strain and Young’s Modulus (Syllabus 6.1)

Learning Objectives

• Define stress, strain and Young’s modulus and state their symbols and SI units.
• Explain the linear‑elastic region, the limit of proportionality, yield point, ultimate tensile strength and fracture point.
• Derive and use the relationships  σ = E ε,  σ = F/A,  ε = ΔL/L₀ and the elastic‑potential‑energy formulae.
• Carry out the standard practical experiment to determine Young’s modulus of a wire and evaluate uncertainties (AO3).
• Interpret a complete stress–strain diagram and solve quantitative problems.

Key Formulae

Key Definitions

Units and Dimensions

Stress–Strain Relationship (Linear‑Elastic Region)

For most engineering materials the initial portion of the stress–strain curve is linear. In this region

Since σ = F/A and ε = ΔL/L₀, Hooke’s law for a uniform rod can be written as

Elastic and Plastic Behaviour

• Elastic deformation: Reversible – the material returns to its original dimensions when the load is removed.
• Plastic deformation: Irreversible – atomic bonds are permanently rearranged.

Complete Stress–Strain Diagram (Typical for a Ductile Metal)

Elastic‑Potential‑Energy (EPE)

When a material is stretched within the elastic region, work is stored as elastic‑potential‑energy. The work done is the area under the force‑extension curve:

Because the relationship is linear (F = k x), the integral gives

For a rod, the spring constant is k = EA/L₀. Substituting k and using ε = ΔL/L₀ yields the alternative form required by the syllabus:

where V = A L₀ is the original volume of the specimen.

Worked Example 1 – Using ½ σ ε V

1. Given: steel rod, L₀ = 2.00 m, A = 5.0 × 10⁻⁴ m², tensile force F = 1.0 × 10⁴ N, E (steel) = 2.0 × 10¹¹ Pa.
2. Stress: σ = F/A = 1.0 × 10⁴ N / 5.0 × 10⁻⁴ m² = 2.0 × 10⁷ Pa.
3. Strain: ε = σ/E = 2.0 × 10⁷ / 2.0 × 10¹¹ = 1.0 × 10⁻⁴.
4. Volume: V = A L₀ = 5.0 × 10⁻⁴ × 2.00 = 1.0 × 10⁻³ m³.
5. Elastic‑potential‑energy: Uₑ = ½ σ ε V = ½ (2.0 × 10⁷)(1.0 × 10⁻⁴)(1.0 × 10⁻³) = 0.5 J.

Worked Example 2 – Using ½ k ΔL² (spring‑constant form)

A nylon thread of length L₀ = 0.50 m and cross‑sectional area A = 2.0 × 10⁻⁸ m² has Young’s modulus E = 5.0 × 10⁹ Pa. It is stretched by ΔL = 1.2 mm.

1. Spring constant: k = EA/L₀ = (5.0 × 10⁹)(2.0 × 10⁻⁸) / 0.50 = 2.0 × 10² N m⁻¹.
2. Elastic‑potential‑energy: Uₑ = ½ k ΔL² = ½ (2.0 × 10²)(1.2 × 10⁻³)² = 1.44 × 10⁻⁴ J.

Practical Determination of Young’s Modulus (AO3)

Related Topics (Cross‑References)

• Hooke’s law for springs – Topic 3 (Dynamics): F = kx.
• Vector addition of forces – needed when several loads act on the same specimen.
• Error analysis – Topic 5 (Measurements): propagation of uncertainties.
• Material properties – density and thermal expansion for comparing different solids.

Extensions for A‑Level (Optional)

• Bulk modulus (K): K = –V Δp / ΔV (resistance to uniform compression).
• Shear modulus (G): τ = G γ, where τ is shear stress and γ shear strain.
• Stress in fluids: normal stress = pressure; shear stress = η (du/dy) for Newtonian fluids.
• Non‑linear elastic materials – true stress–true strain curves for large deformations.

Common Misconceptions

• Stress ≠ pressure. Stress acts on a specific internal plane; pressure is an isotropic normal stress acting on a fluid surface.
• Strain is dimensionless; it is a ratio of lengths, not a length itself.
• Young’s modulus is a material property; it does not depend on the size or shape of the sample (provided the material is homogeneous and isotropic).
• The limit of proportionality is not always identical to the yield point – some metals exhibit a short non‑linear elastic region before yielding.

Practice Questions

1. Calculate the stress in a copper wire of diameter 2.0 mm carrying a force of 500 N.
2. A polymer has E = 1.5 × 10⁹ Pa. If a tensile load produces a stress of 3.0 × 10⁶ Pa, find the resulting strain.
3. Explain why the stress–strain curve for a brittle material (e.g., glass) differs from that of a ductile material (e.g., mild steel).
4. Using the data below, determine Young’s modulus of a nylon thread. Include a brief uncertainty analysis.

   Load (N)	Extension ΔL (mm)
   5	0.12
   10	0.25
   15	0.38

5. Derive equation (2) for elastic‑potential‑energy starting from the work‑done integral.

Summary

Stress, strain and Young’s modulus provide the quantitative framework for describing how solid materials respond to external forces. Mastery of the linear‑elastic relationship, identification of key points on a stress–strain diagram, calculation of elastic‑potential‑energy, and competence in the standard Young’s‑modulus experiment are essential for success in both IGCSE and A‑Level physics.