The Young modulus of a wire

What you are trying to do

Determine the Young modulus of the material of a wire by loading it and measuring the extension produced.

Apparatus

• A long thin wire (2 m or more) of the test material, clamped at one end
• Pulley at the bench edge, with a hanger and slotted masses
• Micrometer screw gauge, for the wire diameter
• Metre rule (or travelling microscope) and a marker or sticky-tape flag on the wire
• Second (comparison) wire alongside, if using a Searle's-type arrangement
• Safety spectacles, and a padded floor or box beneath the masses

Variables

• Independent: the load (weight) hung on the wire
• Dependent: the extension of the wire
• Control: the wire itself: same material, original length and diameter throughout, and constant temperature

Method

1. Measure the diameter of the wire with a micrometer at several places along it and in two perpendicular directions at each place, then take a mean. The wire is rarely perfectly round or uniform, and the diameter is squared in the area, so this is where most of the uncertainty lives.
2. Measure the original length from the clamp to the marker with a metre rule.
3. Add the smallest load and record the new position of the marker; the extension is the change in that reading.
4. Repeat in equal steps, recording extension against load over as wide a range as the wire will take elastically.
5. Unload in steps and check the extensions retrace the loading values. If they do, the wire stayed elastic; if the wire does not return to its original length it has passed its elastic limit, and discarding the affected points is not enough, because the wire itself is now permanently stretched. Replace the specimen, or re-clamp an undeformed section, and repeat the run within the elastic range.

Analysis

1. Plot a graph of force (y) against extension (x). Take the gradient of the straight portion through the origin.
2. Compute the cross-sectional area from the mean diameter, A = πd² / 4$A = \pi d^{2} / 4$A = πd^2 / 4.
3. The Young modulus follows from the gradient and the wire's dimensions: E = (F/A)/(ΔL/L) = gradient × L / A$E = (F/A)/(\Delta L/L) = \text{gradient} \times L / A$E = (F/A)/(ΔL/L) = gradient × L / A.
4. Plotting stress against strain instead gives the Young modulus directly as the gradient, and lets you read off the limit of proportionality.
5. What a modulus of that size supports. A gradient returning about 2 × 10¹¹ Pa, with a combined uncertainty of a few per cent, identifies the wire as steel, and that is as far as the data go. Mild steel, stainless steel and most steel alloys sit within a few per cent of each other, well inside your own uncertainty, so the result cannot name the alloy. Quote the modulus with its percentage uncertainty attached and the conclusion is defensible; quote it to four figures and it is a claim your micrometer cannot back.
6. What limits the answer, and what does not. The diameter dominates, and repeating it will not rescue you: a micrometer reading to 0.01 mm on a wire 0.40 mm across is already at its own resolution, so the improvement that helps is a finer instrument rather than more readings from the same one. Temperature is the quieter problem. Two metres of steel stretches about 24 μm for every degree it warms, and you are reading extensions to about 0.05 mm, so a lab that drifts three degrees through a long run adds a reading's worth of extension with no load applied.

A worked set of readings

Typical readings for a 2.00 m steel wire of diameter 0.40 mm:

Idealised illustrative data, chosen so the working is easy to follow. Real readings scatter about the line rather than sitting on it, and your own graph will have points either side of the best fit.

Plotting F against e gives a straight line through the origin with gradient 12566 N m⁻¹. The area is A = πd²/4 = π × (0.40 × 10⁻³)² / 4 = 1.257 × 10⁻⁷ m². Then E = gradient × L / A = 12566 × 2.000 ÷ (1.257 × 10⁻⁷) = 2.00 × 10¹¹ Pa, about 200 GPa: the right order for steel.

Evaluating the result

Idealised loading readings, with the wire allowed to slip 0.15 mm through the clamp as the 2.0 kg load went on. The third column divides extension by force, which for an elastic wire is the constant L/AE.

Idealised illustrative data, chosen so the working is easy to follow. Real readings scatter about the line rather than sitting on it, and your own graph will have points either side of the best fit.

The first three rows agree at 0.0796 mm N⁻¹, then the ratio jumps to 0.0872 and comes back down to 0.0857. That shape is the tell. A single misread extension disturbs one row, while a slip adds the same 0.15 mm to every reading after it, so the ratio steps up and then drifts back as the true extension grows around a fixed offset. No one point is the anomaly, and deleting the worst of them would leave the rest quietly wrong. Reclamp, remeasure the original length, and load again from zero.

Where the uncertainty comes from

• Wire diameter: Much the largest contributor, because the area depends on d squared, so the percentage uncertainty in the diameter is doubled in the area. Repeat measurements along the wire and average.
• Extension: Small compared with the original length, so measure it as precisely as the apparatus allows; a travelling microscope or a vernier scale beats a metre rule.
• Original length: Comparatively easy to measure well; use a long wire so the percentage uncertainty is small and the extension is big enough to read.
• Temperature and sag: Keep the temperature steady, and use a comparison wire alongside so that any sagging of the support affects both equally and cancels.

What earns the marks

Safety