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

Quantity	SI Unit	Symbol
Force	newton (N)	F
Area	square metre (m²)	A
Stress	pascal (Pa) = N·m⁻²	σ
Length	metre (m)	L
Strain	dimensionless	ε
Young’s Modulus	pascal (Pa)	E

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.