Young’s modulus

Young’s modulus

Narrated video. Transcript:

In the straight-line part of the stress-strain graph, the ratio of stress to strain is constant. For a wire or rod that is stretched or compressed, this ratio is called Young’s modulus, Y.

Y equals the longitudinal stress divided by the longitudinal strain: F over A, divided by ΔL over L. Rearranged, Y equals F times L, over A times ΔL.

Since strain has no unit, Young’s modulus has the same unit as stress: newtons per square metre, or pascals. Its dimensions are M L⁻¹ T⁻².

Look at a table of values. Steel is about 2 × 10¹¹ newtons per square metre. Copper is about 1.2 × 10¹¹, aluminium 7 × 10¹⁰, wood and bone around 10¹⁰. Metals have large Young’s moduli.

Notice that Y is a property of the material, not of the sample. A thick wire and a thin wire of steel need different forces for the same stretch, but both have the same Y.

Here is a puzzle: which is more elastic, steel or rubber? Most people say rubber, because it stretches easily. But in physics, steel is far more elastic.

For the same strain, steel needs a much larger stress than rubber. A larger Young’s modulus means the material resists deformation more strongly and springs back more completely.

Now a worked example. A structural steel rod of radius 10 millimetres and length 1.0 metre is pulled by a force of 100 kilonewtons. Y for steel is 2.0 × 10¹¹.

Stress is F over π r squared: 100,000 divided by 3.14 × 10⁻⁴, which is 3.18 × 10⁸ newtons per square metre.

Then ΔL equals stress times L over Y: 3.18 × 10⁸ times 1, divided by 2 × 10¹¹. That is 1.59 × 10⁻³ metres, about 1.6 millimetres. The strain is 1.59 × 10⁻³, or 0.16 percent.

Measuring Y in the laboratory

Narrated video. Transcript:

How do we measure Young’s modulus of the material of a wire? Here is the typical laboratory set-up. Two long, straight wires of the same length and the same radius hang side by side from one rigid support.

Wire A is the reference wire. It carries a millimetre main scale and a pan with a fixed weight. Wire B is the experimental wire. It carries a pan for adding weights, and a vernier scale that slides beside the main scale.

First, put a small load in both pans to keep the wires straight, and note the vernier reading. This is our starting point.

Now add slotted masses to wire B, one at a time. The wire stretches, and the vernier slides down against the main scale. Note the reading each time. The difference from the start gives the elongation, Δl.

The wire’s radius r is measured with a screw gauge, and its length L with a metre scale. The added force is M g, so Y equals M g times L, over π r squared times Δl.

Why the second wire? If the room warms up, both wires expand equally. If the support sags a little under the load, both wires drop together. The vernier only measures the stretch of B relative to A, so these errors cancel.

Let’s put in numbers. A steel wire 2 metres long and 0.5 millimetres in diameter has an area of 1.96 × 10⁻⁷ square metres. Adding 2 kilograms stretches it by about 1.0 millimetre.

Y equals 19.6 newtons times 2 metres, divided by 1.96 × 10⁻⁷ times 10⁻³. That gives 2.0 × 10¹¹ newtons per square metre, the value for steel.

For a better result, plot extension against load for several loads. The points lie on a straight line, and its slope gives Y. Keep the load small, so the wire stays well within its elastic limit.

Try it: measure Y yourself

Interactive simulation. Things to try:

• Add 0.5 kg at a time and press Record reading (or Space) after each one.
• Read the vernier: the main-scale mark just above the vernier zero, plus the vernier division that lines up × 0.1 mm.
• With four or more points, the green best-fit line gives the slope and Y.
• Tick “Warm the room”: both wires expand, but the reading does not change. That is the job of the reference wire.
• Try a 0.3 mm wire with a big load: the points turn red once the stress passes the elastic limit.

Why two wires?

In the experiment to find Young’s modulus, why is a second, reference wire hung beside the experimental wire from the same support?

1. To share the load so that the experimental wire does not break
2. To cancel changes in length caused by temperature changes and sagging of the support
3. To double the extension so it is easier to read
4. Because Young’s modulus needs two different materials to compare

Copper and steel in series

A copper wire 2.2 m long and a steel wire 1.6 m long, both 3.0 mm in diameter, are joined end to end. Under a load the total elongation is 0.70 mm. Y for steel is 2.0 × 10¹¹ N/m² and for copper 1.1 × 10¹¹ N/m². What is the load?

1. 1.8 × 10² N
2. 3.1 × 10² N
3. 6.2 × 10² N
4. 7.1 × 10² N

Summary

• Young’s modulus: Y = (F/A) / (ΔL/L) = FL / (AΔL)
• Wire experiment: Y = MgL / (πr²Δl)
• Steel: Y ≈ 2 × 10¹¹ N/m²

• Young’s modulus is longitudinal stress ÷ longitudinal strain in the elastic region. Unit N/m² (Pa).
• Y depends on the material, not on the length or thickness of the sample. Metals have large Y.
• A larger Y means a more elastic material in physics: steel is more elastic than rubber.
• In the experiment, the reference wire cancels errors from temperature changes and sagging of the support.