The Young modulus

Welcome to the Young modulus! In the previous topic, you looked at Hooke's Law ( $\Delta L$ ), which tells us how a specific object (like a spring) behaves when you stretch it. But what if you want to compare the stiffness of copper and steel in general, regardless of their shape or thickness?

To do that, we need to strip away the dimensions of the object.

What you'll learn:

How to calculate tensile stress and tensile strain.
What the Young modulus is and how to calculate it.
How to determine the Young modulus from a stress-strain graph.
The standard practical method for measuring the Young modulus of a wire (Required Practical 4).

Moving beyond Hooke's Law

Imagine you have a thick steel cable and a thin steel guitar string. The thick cable is much harder to stretch than the thin string, so it has a much higher spring constant, $k$ . But they are made of the exact same material!

To compare the materials fairly, we need to divide the force by the cross-sectional area (to adjust for thickness), and divide the extension by the original length (to adjust for how long the wire was to begin with).

This gives us two incredibly important new physics quantities: Stress and Strain.

Tensile Stress

When you apply a pulling force to a material, you are putting it under tensile stress.

Definition

Tensile Stress

Tensile stress ( $σ\sigma$ ) is the applied force per unit cross-sectional area.

\sigma = \frac{F}{A}

Where:

$F$ is the tensile force in newtons ( $N\text{N}$ )
$A$ is the cross-sectional area in square metres ( $m2\text{m}^2$ )
$σ\sigma$ is the tensile stress in pascals ( $Pa\text{Pa}$ ) or newtons per square metre ( $m−2\text{N m}^{-2}$ )

Tip

Area of a wire

In A-Level exam questions, you are almost always dealing with round wires. Remember that the cross-sectional area is a circle. You can calculate it using the radius ( $\pi r^2$ ) or the diameter ( $\frac{\pi d^2}{4}$ ). Since micrometers measure diameter, the diameter formula is often quicker and avoids the common mistake of forgetting to halve the diameter!

Tensile Strain

When a material is put under stress, it deforms. Tensile strain is a measure of how much it has stretched compared to its original length.

Definition

Tensile Strain

Tensile strain ( $ε\varepsilon$ ) is the extension per unit original length.

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

Where:

$ΔL\Delta L$ is the extension in metres ( $m\text{m}$ )
$L$ is the original length in metres ( $m\text{m}$ )
$ε\varepsilon$ is the tensile strain (it has no units, as it is a ratio of two lengths)

Sometimes strain is expressed as a percentage. For example, if a 2 metre wire extends by 0.02 metres, the strain is $0.01$ . As a percentage, this is a $1%1\%$ strain.

If a material obeys Hooke's Law, the stress applied to it is directly proportional to the strain it experiences. The constant of proportionality is called the Young modulus.

Definition

The Young modulus

The Young modulus ( $E$ ) is the ratio of tensile stress to tensile strain in the linear region of a material's elastic deformation.

\frac{\text{tensile stress}}{\text{tensile strain}} = \frac{\sigma}{\varepsilon}

Because strain has no units, the Young modulus takes the units of stress: pascals ( $Pa\text{Pa}$ ) or $m−2\text{N m}^{-2}$ .

We can combine the equations for stress and strain to get a single, very useful equation for the Young modulus.

\begin{aligned} E &= \frac{\sigma}{\varepsilon} \\ E &= \frac{\left( \frac{F}{A} \right)}{\left( \frac{\Delta L}{L} \right)} \\ E &= \frac{F L}{A \Delta L} \end{aligned}

Key Idea

Material vs Object

This is the big takeaway: The Young modulus is a property of the material itself. All solid copper wires have the same Young modulus (roughly $1.3 \times 10^{11} \text{ Pa}$ ), regardless of whether they are long, short, thick, or thin. The spring constant ( $k$ ) is a property of the specific object.

Let's look at how this appears in an exam-style calculation.

Example

Calculating the Young modulus

A steel wire of original length 2.50 m and diameter 0.40 mm is suspended from a rigid support. A mass of 4.0 kg is hung from the free end, causing the wire to extend by 1.8 mm. Calculate the Young modulus of the steel. Give your answer to 2 significant figures. ( $\text{ m s}^{-2}$ )

First, convert all units to standard SI units. Diameter $\times 10^{-3} \text{ m}$ Extension $m\Delta L = 1.8 \times 10^{-3} \text{ m}$ Length $\text{ m}$
Calculate the cross-sectional area $A$ .

\begin{aligned} A &= \frac{\pi d^2}{4} \\ A &= \frac{\pi \times (0.40 \times 10^{-3})^2}{4} \\ A &= 1.257 \times 10^{-7} \text{ m}^2 \end{aligned}

Calculate the applied force $F$ (the weight of the mass).

\begin{aligned} F &= mg \\ F &= 4.0 \times 9.81 \\ F &= 39.24 \text{ N} \end{aligned}

Substitute all values into the Young modulus equation.

\begin{aligned} E &= \frac{F L}{A \Delta L} \\ E &= \frac{39.24 \times 2.50}{1.257 \times 10^{-7} \times 1.8 \times 10^{-3}} \\ E &= 4.33 \times 10^{11} \text{ Pa} \end{aligned}

Round to 2 significant figures (as given in the question data). The Young modulus is $4.3 \times 10^{11} \text{ Pa}$ .

Stress-Strain Graphs

Just as you plotted Force-Extension graphs for springs, you can plot Stress-Strain graphs for materials.

Because stress is proportional to strain up to the limit of proportionality, the initial part of a stress-strain graph for a metal is a straight line through the origin.

Stress-strain graph showing the linear region and gradient

Look at the equation for a straight line: $y = m x + c$ . If we plot stress ( $σ\sigma$ ) on the y-axis and strain ( $ε\varepsilon$ ) on the x-axis, the y-intercept $c$ is zero. This leaves us with $σ=mε\sigma = m \varepsilon$ .

Comparing this to our definition $\sigma / \varepsilon$ , we can clearly see that: The gradient of the linear section of a stress-strain graph is equal to the Young modulus.

Common Mistake

Watch out for axis prefixes

AQA examiners love to hide standard form prefixes on graph axes. The y-axis (stress) will almost always be in $MPa\text{MPa}$ ( $Pa10^6 \text{ Pa}$ ) or $GPa\text{GPa}$ ( $Pa10^9 \text{ Pa}$ ). The x-axis (strain) is sometimes given as $×10−3\times 10^{-3}$ or as a percentage. Always check the axis labels carefully before calculating a gradient!

Required Practical 4: Measuring the Young modulus

You need to know how to determine the Young modulus of a metal wire using a simple experimental method.

There are two common setups: a vertical setup (often using Searle's apparatus with a control wire) and a horizontal bench setup. Both rely on the same fundamental measurements.

Horizontal bench setup for determining the Young modulus of a wire

The Method

Measure the original length ( $L$ ): Clamp the wire securely. Use a metre ruler to measure the length of the wire from the clamp to the marker sticker. A longer wire reduces the percentage uncertainty in the extension measurement.
Measure the diameter ( $d$ ): Use a micrometer screw gauge to measure the diameter of the wire. Because the wire might not be perfectly cylindrical, measure the diameter at at least three different points along the wire, and at different angles, then calculate an average.
Apply a load ( $F$ ): Hang a mass hangar on the end of the wire. The tension force is $F = m g$ .
Measure the extension ( $ΔL\Delta L$ ): Record the new position of the marker against the fixed ruler (or a vernier scale/travelling microscope for higher precision). The extension is the new position minus the original position.
Repeat: Add masses one by one (e.g., in 500 g increments) and record the extension each time.

Analysing the Data

To find the Young modulus from your experimental data, you could calculate $E$ for every single mass and find an average, but AQA expects you to use a graphical method.

Example

Graphical analysis

Calculate the cross-sectional area of the wire, $\frac{\pi d^2}{4}$ .
For each mass, calculate the force applied ( $F = m g$ ).
You have two choices for the graph:
- Method A (Force-Extension): Plot $F$ on the y-axis and $ΔL\Delta L$ on the x-axis. Find the gradient of the straight line portion ( $\Delta L$ ). The Young modulus is then $\text{gradient} \times \frac{L}{A}$ .
- Method B (Stress-Strain): Calculate stress ( $F / A$ ) and strain ( $ΔL/L\Delta L/L$ ) for each point. Plot stress on the y-axis and strain on the x-axis. The gradient of the straight line is exactly equal to the Young modulus, $E$ .

Common Mistake

Confusing the gradient

If you plot Force against Extension, the gradient is not the Young modulus. The gradient is the spring constant, $k$ . You must multiply the gradient by $LA\frac{L}{A}$ to find $E$ .

Exam technique

In the exam

Hunt for prefixes: Whenever you are given a stress value, immediately look for "M" (mega, $10^6$ ) or "G" (giga, $10^9$ ). Look for "mm" in diameters and convert to metres ( $×10−3\times 10^{-3}$ ) before squaring it for area.
Micrometer justification: If asked why you use a micrometer to measure diameter instead of a ruler, state that it has a much higher resolution (typically 0.01 mm), which significantly reduces the percentage uncertainty in the measurement.
Multiple diameter readings: If asked to outline the practical method, always explicitly state that you measure the diameter at several points and average them. This is a very common marking point for Required Practical 4.
Sanity checking your answer: The Young modulus for metals is huge. Your final answer should almost always be in the region of $10^{10}$ to $Pa10^{11} \text{ Pa}$ . If you get $\text{ Pa}$ , you've likely missed a milli- or micro- conversion!

Self review

Check yourself

Can you write down the equations defining tensile stress and tensile strain from memory?
Why does a stress-strain graph allow us to compare two different materials better than a force-extension graph?
When measuring the diameter of a wire in the required practical, why should you take measurements at different points along the wire?
On a stress-strain graph, what graphical feature represents the Young modulus?

To do that, we need to strip away the dimensions of the object.

What you'll learn:

How to calculate tensile stress and tensile strain.
What the Young modulus is and how to calculate it.
How to determine the Young modulus from a stress-strain graph.
The standard practical method for measuring the Young modulus of a wire (Required Practical 4).