Stress, Strain, and Young's Modulus

σ = F/A; ε = ΔL/L; E = σ/ε; stress-strain graphs; elastic and plastic deformation

# Stress, Strain, and Young's Modulus — A-Level Physics

How do materials respond to forces? Why does steel make good bridges but rubber makes good tyres? Understanding stress, strain, and Young's modulus allows us to predict how materials deform under load — essential knowledge for engineering and A-Level Physics.


1. Tensile Stress ($\sigma$)

σ=FA\boxed{\sigma = \frac{F}{A}}

  • σ\sigma = tensile stress (Pa or N/m²)
  • FF = force applied (N)
  • AA = cross-sectional area (m²)

Stress tells us the force per unit area — it's independent of the size of the sample.


2. Tensile Strain ($\varepsilon$)

ε=ΔLL\boxed{\varepsilon = \frac{\Delta L}{L}}

  • ε\varepsilon = strain (no units — it's a ratio)
  • ΔL\Delta L = extension (m)
  • LL = original length (m)

Strain tells us the fractional change in length.


3. Young's Modulus ($E$)

E=σε=F/AΔL/L=FLAΔL\boxed{E = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}}

  • EE = Young's modulus (Pa)
  • Measures the stiffness of a material
  • High EE → stiff (hard to stretch) e.g., steel (E200E \approx 200 GPa)
  • Low EE → flexible e.g., rubber (E0.01E \approx 0.01 GPa)

Typical Values

Material Young's Modulus (GPa)
Diamond 1200
Steel 200
Copper 130
Glass 70
Bone 18
Rubber 0.01–0.1

4. Stress-Strain Graphs

Key Features for a Ductile Material (e.g., copper)

  1. Linear region (O to P): Obeys Hooke's Law (σε\sigma \propto \varepsilon). Gradient = Young's modulus.
  2. Limit of proportionality (P): Beyond this, stress and strain are no longer proportional.
  3. Elastic limit (E): Beyond this, the material won't return to original length when unloaded — permanent deformation begins.
  4. Yield point (Y): Material suddenly extends significantly with little extra stress.
  5. Plastic region: Material deforms permanently.
  6. Ultimate tensile stress (UTS): Maximum stress the material can withstand.
  7. Fracture point: Material breaks.

Elastic vs Plastic Deformation

Elastic deformation: Material returns to original shape when force is removed. Energy is fully recovered. Bonds stretch but don't break.

Plastic deformation: Material is permanently deformed. Energy is NOT fully recovered (some converted to heat). Atomic planes slide past each other.


5. Hooke's Law and Springs

F=kΔLF = k\Delta L

Where kk = spring constant (N/m).

Valid up to the limit of proportionality.

Energy Stored in a Stretched Wire/Spring

E=12FΔL=12k(ΔL)2E = \frac{1}{2}F\Delta L = \frac{1}{2}k(\Delta L)^2

On a force-extension graph, energy stored = area under the curve (up to the loading line).

Energy from Stress-Strain Graph

Energy stored per unit volume: EV=12σε\frac{E}{V} = \frac{1}{2}\sigma\varepsilon

(Area under the stress-strain graph)


6. Types of Materials

Brittle Materials (e.g., glass, ceramics)

  • Break suddenly with little or no plastic deformation
  • Stress-strain graph is almost straight until fracture
  • No significant yield point

Ductile Materials (e.g., copper, mild steel)

  • Show significant plastic deformation before breaking
  • Can be drawn into wires
  • Clear yield point

Polymers (e.g., rubber, polythene)

  • Very large strains at low stress
  • Rubber: elastic but non-linear (Hooke's Law doesn't apply)
  • Loading and unloading curves differ → hysteresis (energy lost as heat)

7. Required Practical: Measuring Young's Modulus

Method

  1. Clamp a long, thin wire (e.g., copper) horizontally or vertically
  2. Measure original length LL with a tape measure
  3. Measure diameter with a micrometer at several points → calculate area A=π(d/2)2A = \pi(d/2)^2
  4. Add known masses (forces F=mgF = mg) and measure extension ΔL\Delta L with a ruler or travelling microscope
  5. Plot a graph of stress (F/AF/A) against strain (ΔL/L\Delta L/L)
  6. Gradient = Young's modulus

Improvements

  • Use a long wire (larger extension, easier to measure)
  • Use a thin wire (larger extension for same force)
  • Use a comparison wire to eliminate thermal expansion
  • Use a vernier scale or travelling microscope for accurate extension measurement
  • Take diameter measurements at multiple points and average

Worked Example: Example 1

Problem

A steel wire of length 2.5 m and diameter 0.8 mm supports a 400 N load. Find the stress, strain, and extension. (Esteel=200E_{\text{steel}} = 200 GPa)

A=π(0.4×103)2=5.03×107A = \pi(0.4 \times 10^{-3})^2 = 5.03 \times 10^{-7}

σ=F/A=400/5.03×107=7.96×108\sigma = F/A = 400/5.03 \times 10^{-7} = 7.96 \times 10^8 Pa = 796 MPa

ε=σ/E=7.96×108/200×109=3.98×103\varepsilon = \sigma/E = 7.96 \times 10^8 / 200 \times 10^9 = 3.98 \times 10^{-3}

ΔL=ε×L=3.98×103×2.5=9.95×103\Delta L = \varepsilon \times L = 3.98 \times 10^{-3} \times 2.5 = 9.95 \times 10^{-3} m ≈ 10 mm

Solution

Worked Example: Example 2

Problem

A rubber band of original length 10 cm stretches to 25 cm under a 5 N force. Cross-sectional area = 4 mm². Find Young's modulus.

σ=5/(4×106)=1.25×106\sigma = 5/(4 \times 10^{-6}) = 1.25 \times 10^6 Pa ε=0.15/0.10=1.5\varepsilon = 0.15/0.10 = 1.5 E=1.25×106/1.5=833,000E = 1.25 \times 10^6/1.5 = 833{,}000 Pa = 0.833 MPa


Solution

9. Practice Questions

    1. Define stress, strain, and Young's modulus. State units for each. (3 marks)
    1. A copper wire (E=130E = 130 GPa) of length 3 m and diameter 1.0 mm has a 200 N load applied. Calculate the extension. (4 marks)
    1. Sketch and label a stress-strain graph for a ductile material, identifying key points. (4 marks)
    1. Explain the difference between elastic and plastic deformation. (3 marks)

    Answers

    1. Stress (σ\sigma) = force/area (Pa). Strain (ε\varepsilon) = extension/original length (no units). Young's modulus (EE) = stress/strain (Pa).

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Summary

  • σ=F/A\sigma = F/A (stress), ε=ΔL/L\varepsilon = \Delta L/L (strain), E=σ/εE = \sigma/\varepsilon (Young's modulus)
  • Young's modulus = gradient of linear region of stress-strain graph
  • Elastic limit: beyond this → permanent deformation
  • Brittle: breaks with no plastic deformation; Ductile: large plastic region
  • Energy stored per unit volume = area under stress-strain graph

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