understand that deformation is caused by tensile or compressive forces (forces and deformations will be assumed to be in one dimension only)

Stress and Strain

Learning Objective

Understand that deformation of a material is caused by tensile or compressive forces. All forces and deformations are considered in one dimension only.

Key Definitions

• Tensile force: A pulling force that tries to elongate a material.
• Compressive force: A pushing force that tries to shorten a material.
• Stress (\$\sigma\$): The internal resistance of a material to an applied force, defined as force per unit area.

  \$\sigma = \frac{F}{A}\$

• Strain (\$\varepsilon\$): The relative deformation of a material, defined as the change in length divided by the original length.

  \$\varepsilon = \frac{\Delta L}{L_0}\$

• Young’s Modulus (\$E\$): The constant of proportionality between stress and strain in the linear (elastic) region.

  \$E = \frac{\sigma}{\varepsilon}\$

Units and Symbols

Behaviour of Materials Under Load

1. Elastic region: Stress and strain are proportional (Hooke’s law). The material returns to its original length when the load is removed.
2. Yield point (or proportional limit): The stress at which the linear relationship begins to break down.
3. Plastic region: Permanent deformation occurs; strain no longer returns to zero after unloading.
4. Ultimate tensile strength (UTS): Maximum stress the material can sustain before necking.
5. Fracture point: The stress at which the material breaks.

Stress–Strain Diagram (Tensile Test)

The diagram below summarises the typical relationship between stress and strain for a ductile material under tension.

Deriving Young’s Modulus from Experimental Data

When a cylindrical rod of original length \$L_0\$ and cross‑sectional area \$A\$ is subjected to a tensile force \$F\$, the extension \$\Delta L\$ is measured. Using the definitions of stress and strain:

\$\sigma = \frac{F}{A}, \qquad \varepsilon = \frac{\Delta L}{L_0}\$

In the elastic region, the slope of the straight‑line portion of the stress–strain graph gives Young’s modulus:

\$E = \frac{\sigma}{\varepsilon} = \frac{F L_0}{A \Delta L}\$

Example Calculation

Given:

• Force applied, \$F = 5.0 \,\text{kN}\$
• Original length, \$L_0 = 200 \,\text{mm}\$
• Cross‑sectional area, \$A = 10 \,\text{mm}^2\$
• Measured extension, \$\Delta L = 0.40 \,\text{mm}\$

Calculate the stress, strain, and Young’s modulus.

1. Stress:

   \$\sigma = \frac{5.0 \times 10^{3}\,\text{N}}{10 \times 10^{-6}\,\text{m}^2}=5.0 \times 10^{8}\,\text{Pa}=500\;\text{MPa}\$

2. Strain:

   \$\varepsilon = \frac{0.40 \times 10^{-3}\,\text{m}}{0.200\,\text{m}}=2.0 \times 10^{-3}\$

3. Young’s Modulus:

   \$E = \frac{5.0 \times 10^{8}\,\text{Pa}}{2.0 \times 10^{-3}}=2.5 \times 10^{11}\,\text{Pa}=250\;\text{GPa}\$

Compressive Loading

The same definitions apply for compressive forces, but the sign of \$\Delta L\$ is negative (shortening). In practice, the stress–strain curve for compression is often plotted as a mirror image of the tensile curve for the elastic region.

Key Points to Remember

• Stress is a measure of internal force per unit area; strain is a dimensionless measure of deformation.
• In the elastic region, stress and strain are linearly related by Young’s modulus.
• Beyond the proportional limit, permanent (plastic) deformation occurs.
• All calculations assume uniform cross‑section and one‑dimensional deformation.

Common Exam Questions

1. Given a material’s Young’s modulus and dimensions, calculate the extension produced by a known tensile force.
2. Interpret a stress–strain diagram to identify the elastic limit, yield point, and ultimate tensile strength.
3. Compare the stiffness of two rods of different materials and cross‑sectional areas.