How Do You Calculate Young's Modulus (Stress/Strain)?

Young's modulus is calculated by dividing tensile stress by tensile strain within the material's linear elastic region:

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

A larger Young's modulus means the material is stiffer and undergoes less elastic deformation under a given stress.

The Calculation Method

First calculate stress and strain separately.

Here, $F$ is force in newtons, $A$ is area in square metres, $\Delta L$ is extension, and $L_0$ is original length. Strain is dimensionless because it is a ratio of two lengths. Young's modulus has units of pascals, where $1\text{ Pa}=1{\text{ N m}}^{}$.

The relationship applies only before the proportional limit, where stress and strain are directly proportional and deformation is elastic. Beyond this straight-line region, dividing one stress value by its strain does not give the material's Young's modulus. Measurements must use the specimen's original dimensions.

Young's modulus is a material property rather than a property of the component's overall size or shape. A high value indicates greater resistance to stretching.

Worked Example

A steel member has an original length of $2.0\text{ m}$, cross-sectional area of $200\text{ mm}^2$, and extension of $0.60\text{ mm}$ under a tensile force of $12\text{ kN}$.

Convert to SI units:

$$A = 200 \times 10^{-6}\text{ m}^2, \quad \Delta L = 0.60 \times 10^{-3}\text{ m}, \quad F = 12\,000\text{ N}$$

Area conversion is a frequent source of error because the conversion factor must also be squared: one square millimetre equals one millionth of a square metre.

Calculate stress:

$$\sigma = \frac{12\,000}{200 \times 10^{-6}} = 60 \times 10^6\text{ Pa}$$

Calculate strain:

$$\varepsilon = \frac{0.60 \times 10^{-3}}{2.0} = 3.0 \times 10^{-4}$$

Therefore:

$$E = \frac{60 \times 10^6}{3.0 \times 10^{-4}} = 2.0 \times 10^{11}\text{ Pa} = 200\text{ GPa}$$

IB Exam Technique

For B3.2 Structural systems application and selection (HL only), show the equation, unit conversions, substitution, answer, and unit. If given a stress-strain graph, Young's modulus is the gradient of its initial straight-line region. Select two well-separated points on that linear section. A common misconception is that Young's modulus measures strength; it measures stiffness, not the stress at which failure occurs.