Stress, strain and the Young modulus

Stress and strain let us compare how different materials stretch, whatever the size of the sample. The Young modulus, stress divided by strain, is a property of the material itself: it tells engineers how stiff steel, copper or nylon is when choosing cables, bridges and wires.

Key facts

  • Tensile stress σ = F / A (Pa, or N/mm² = MPa).
  • Tensile strain ε = ΔL / L (no unit).
  • Young modulus E = stress ÷ strain = FL / (AΔL); steel ≈ 200 GPa.
  • Elastic limit: beyond it the material does not return to its original length. Ultimate tensile stress: the maximum stress before breaking.
  • Brittle materials (glass) snap with little plastic deformation; ductile materials (copper) stretch a lot plastically.
  • Elastic strain energy = ½FΔL; energy per unit volume = ½ × stress × strain.

Worked examples

Example 1

A wire of cross-sectional area 0.50 mm² carries 100 N. Find the stress.

  1. Change mm² to m². 0.50 mm² = 0.50 × 10^−6 m²
  2. Divide the force by the area of the cross-section. σ = 100 ÷ (0.50 × 10^−6) = 2.0 × 10^8 Pa = 200 MPa

Answer: 200 MPa

Example 2

A 2.0 m steel wire of area 0.40 mm² stretches 1.5 mm under 60 N. Find the Young modulus.

  1. Stress: σ = 60 ÷ (0.40 × 10^−6) = 1.5 × 10^8 Pa
  2. Strain = extension ÷ original length (exaggerated here). ε = 1.5 × 10^−3 ÷ 2.0 = 7.5 × 10^−4
  3. Young modulus = stress ÷ strain. E = 1.5 × 10^8 ÷ 7.5 × 10^−4 = 2.0 × 10^11 Pa = 200 GPa

Answer: 200 GPa

Example 3

A copper wire (E = 117 GPa) is 3.0 m long with diameter 0.80 mm. How far does it stretch under 50 N?

  1. Area of the circular cross-section (radius 0.40 mm): A = π × (0.40 × 10^−3)² = 5.03 × 10^−7 m²
  2. Rearrange E = (F/A) ÷ (ΔL/L). ΔL = 50 × 3.0 ÷ (5.03 × 10^−7 × 117 × 10^9) = 2.55 × 10^−3 m = 2.55 mm

Answer: 2.55 mm