define and use the terms stress, strain and the Young modulus

Stress and Strain

Learning Objective

Define stress, strain and Young’s modulus, and apply them to solve problems.

Key Definitions

• Stress (\$\sigma\$): Force applied per unit area, \$\displaystyle \sigma = \frac{F}{A}\$.
• Strain (\$\epsilon\$): Relative deformation, \$\displaystyle \epsilon = \frac{\Delta L}{L_0}\$.
• Young’s Modulus (\$E\$): Ratio of stress to strain in the linear elastic region, \$\displaystyle E = \frac{\sigma}{\epsilon}\$.

Units and Dimensions

Stress–Strain Relationship

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

\$\sigma = E \, \epsilon\$

Beyond the proportional limit the material yields and the relationship becomes non‑linear.

Example Calculation

1. Given: A steel rod of original length \$L_0 = 2.00\ \text{m}\$ and cross‑sectional area \$A = 5.0 \times 10^{-4}\ \text{m}^2\$ is subjected to a tensile force \$F = 10\,000\ \text{N}\$.
2. Calculate the stress:

   \$\sigma = \frac{F}{A} = \frac{10\,000\ \text{N}}{5.0 \times 10^{-4}\ \text{m}^2} = 2.0 \times 10^{7}\ \text{Pa}\$

3. Assuming Young’s modulus for steel \$E = 2.0 \times 10^{11}\ \text{Pa}\$, find the strain:

   \$\epsilon = \frac{\sigma}{E} = \frac{2.0 \times 10^{7}}{2.0 \times 10^{11}} = 1.0 \times 10^{-4}\$

4. Determine the extension \$\Delta L\$:

   \$\Delta L = \epsilon L_0 = (1.0 \times 10^{-4})(2.00\ \text{m}) = 2.0 \times 10^{-4}\ \text{m} = 0.20\ \text{mm}\$

Common Misconceptions

• Stress is not the same as pressure; stress acts on a specific plane within a material.
• Strain is dimensionless; it is not measured in metres.
• Young’s modulus is a material property; it does not depend on the dimensions of the sample.

Practice Questions

1. Calculate the stress in a copper wire of diameter \$2.0\ \text{mm}\$ carrying a force of \$500\ \text{N}\$.
2. A polymer has \$E = 1.5 \times 10^{9}\ \text{Pa}\$. If a tensile load produces a stress of \$3.0 \times 10^{6}\ \text{Pa}\$, what is the resulting strain?
3. Explain why the stress‑strain curve for brittle materials differs from that of ductile materials.

Summary

Stress, strain and Young’s modulus provide a quantitative framework for describing how materials respond to external forces. Mastery of these concepts enables accurate prediction of deformation and failure in engineering applications.