# Stress, strain and Young modulus

> Physics · CIE A-Level 9702
> Source: https://www.owlsprep.com/study/cie-9702-u6-stress-strain-and-young-modulus/

This module introduces core concepts for describing how materials deform under force. You will learn definitions of stress, strain and Young modulus, calculation methods, experimental technique and interpretation of results for CIE A-Level Physics.

**Prerequisites:** [Hooke's law and elastic deformation](https://www.owlsprep.com/study/cie-9702-u6-hookes-law/); SI units and uncertainty analysis

## Learning objectives

- Define tensile stress, tensile strain and Young modulus with correct units
- Calculate Young modulus from experimental force and extension data
- Describe the CIE required experiment for Young modulus measurement
- Interpret key properties of materials from stress-strain relationships

## Key Definitions: Stress and Strain

When a force deforms a material, stress and strain normalise the effect of force to the size of the sample, letting us compare properties of different sized objects made of the same material.

**Tensile Stress** — The tensile force per unit cross-sectional area of a material, applied perpendicular to the sample face.

*Notation:* \sigma

*Example:* $\sigma = \frac{F}{A}$ where $F$ = tensile force, $A$ = cross-sectional area. Units: Pascals (Pa) = Nm⁻².

**Tensile Strain** — The extension of a material per unit original length. Strain is dimensionless, as it is a ratio of two lengths.

*Notation:* \varepsilon

*Example:* $\varepsilon = \frac{\Delta L}{L_0}$ where $\Delta L$ = extension, $L_0$ = original length.

**Worked example:** A 2.0 m long wire with diameter 0.5 mm is stretched by a 100 N force, and extends by 2.4 mm. Calculate stress and strain for the wire.

1. First calculate the cross-sectional area of the wire:
2. $$A = \pi r^2 = \pi \left(\frac{0.5 \times 10^{-3}}{2}\right)^2 = 1.963 \times 10^{-7} \text{ m}^2$$
3. Calculate stress using $\sigma = F/A$:
4. $$\sigma = \frac{100}{1.963 \times 10^{-7}} = 5.1 \times 10^8 \text{ Pa}$$
5. Convert extension to metres and calculate strain:
6. $$\varepsilon = \frac{\Delta L}{L_0} = \frac{2.4 \times 10^{-3}}{2.0} = 1.2 \times 10^{-3}$$

## Young Modulus: Definition and Calculation

Young modulus is an intensive material property that describes stiffness: a higher Young modulus means a stiffer material that deforms less for a given applied stress.

**Young Modulus** — The ratio of tensile stress to tensile strain, valid for elastic deformation where Hooke's law is obeyed.

*Notation:* E

*Example:* $E = \frac{\sigma}{\varepsilon} = \frac{F L_0}{A \Delta L}$. Units are Pa, same as stress.

> **tip**
>
> Young modulus only describes elastic deformation. If the material deforms plastically (does not return to its original shape when force is removed), the relationship no longer applies.

**Worked example:** Use the values of stress and strain from the previous example to calculate the Young modulus of the wire.

1. We already have $\sigma = 5.1 \times 10^8$ Pa and $\varepsilon = 1.2 \times 10^{-3}$. Substitute into the formula for E:
2. $$E = \frac{\sigma}{\varepsilon} = \frac{5.1 \times 10^8}{1.2 \times 10^{-3}} = 4.25 \times 10^{11} \text{ Pa} \approx 4.3 \times 10^{11} \text{ Pa}$$
3. Check units: strain is dimensionless, so E has the same units as stress, which is correct.

## Experimental Determination of Young Modulus

CIE frequently asks 5-6 mark questions describing this experiment, so you need to remember the method, measurements and error reduction steps.

1. Clamp a long metal wire to a rigid support, with a fixed ruler alongside the wire.
2. Add known masses (weights) to the free end, measure extension for each weight.
3. Measure original length $L_0$ from the clamp to a marker on the wire with a metre ruler.
4. Measure diameter at multiple points along the wire with a micrometer, calculate average diameter.
5. Plot a graph of force $F$ against extension $\Delta L$, find the gradient $m = F/\Delta L$.
6. Calculate Young modulus with $E = \frac{m L_0}{A}$, where $A = \pi (d/2)^2$.

**Worked example:** A student obtains a gradient of $8.0 \times 10^4$ Nm⁻¹ from an $F$ vs $\Delta L$ graph, for a 1.5 m long wire with average diameter 0.60 mm. Calculate Young modulus.

1. Calculate cross-sectional area first, converting diameter to metres:
2. $$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.60 \times 10^{-3}}{2}\right)^2 = 2.827 \times 10^{-7} \text{ m}^2$$
3. Substitute into the formula for E from graph gradient:
4. $$E = \frac{m L_0}{A} = \frac{8.0 \times 10^4 \times 1.5}{2.827 \times 10^{-7}} = 4.2 \times 10^{11} \text{ Pa}$$

> **tip**
>
> Use a long thin wire to reduce percentage uncertainty in extension, and measure diameter multiple times to reduce random error. CIE examiners look for these points.

## Stress-Strain Graph Properties

The gradient of a stress-strain graph in the elastic region is equal to Young modulus, since gradient = $\Delta \sigma / \Delta \varepsilon = E$. This is a common exam question.

**Worked example:** The elastic region of a stress-strain curve for aluminium has a gradient of $7.0 \times 10^{10}$ Pa. What is Young modulus of aluminium?

1. The gradient of the stress-strain curve in the elastic region is by definition equal to Young modulus.
2. Therefore, Young modulus of aluminium is $7.0 \times 10^{10}$ Pa.

## Common pitfalls

- **Wrong:** Forgetting to convert units of diameter/extension from millimetres to metres before calculation.
  - Why it fails: This leads to Young modulus values 10⁶ times too large or small, which is a common exam error.
  - Correct: Always convert all length measurements to SI units (metres) before substituting into formulas.
- **Wrong:** Calculating cross-sectional area as $\pi d^2$ instead of $\pi (d/2)^2$.
  - Why it fails: Most students measure diameter directly and forget to convert to radius for the area formula.
  - Correct: Always divide diameter by 2 to get radius before calculating area.
- **Wrong:** Claiming Young modulus depends on the length or cross-sectional area of the sample.
  - Why it fails: Students confuse extension (sample-dependent) with strain, which is normalised for sample size.
  - Correct: Remember Young modulus is an intensive property of the material, not the sample.
- **Wrong:** Assigning units of metres to strain.
  - Why it fails: Strain is calculated as a ratio of two lengths, so students incorrectly add units.
  - Correct: Strain is dimensionless, it has no units.
- **Wrong:** Using the Young modulus formula for deformation beyond the elastic limit.
  - Why it fails: Stress is no longer proportional to strain once plastic deformation starts.
  - Correct: Only use $E = \sigma/\varepsilon$ for elastic deformation where Hooke's law holds.

## Cheatsheet

| Quantity | Symbol | Formula | Units |
| --- | --- | --- | --- |
| Tensile Stress | $\sigma$ | $\sigma = F/A$ | Pa (Nm⁻²) |
| Tensile Strain | $\varepsilon$ | $\varepsilon = \Delta L / L_0$ | Dimensionless |
| Young Modulus | $E$ | $E = (F L_0)/(A \Delta L)$ | Pa (Nm⁻²) |
| E from stress-strain graph | $E$ | Gradient of elastic region | Pa |
| E from F-ΔL graph | $E$ | $E = (m L_0)/A$, $m = F/\Delta L$ | Pa |

## What's next

Mastery of stress, strain and Young modulus is the foundation for all further work on deformation of solids in CIE A-Level Physics. These concepts are used to classify the mechanical behaviour of different material types, calculate elastic potential energy in deformed solids, and answer structured questions about material properties for engineering applications. You will build on these definitions to interpret full stress-strain curves and identify key points like the elastic limit, yield point and ultimate tensile stress, which are common topics in both multiple choice and extended response questions.

- [Force-extension and stress-strain graphs](https://www.owlsprep.com/study/cie-9702-u6-force-extension-and-stress-strain/)
- [Elastic and plastic deformation](https://www.owlsprep.com/study/cie-9702-u6-elastic-and-plastic-deformation/)
- [Strain energy](https://www.owlsprep.com/study/cie-9702-u6-strain-energy/)

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