Physics › Materials › Stress, strain and the Young modulus
Stress, strain and the Young modulus
The spring constant describes one object; reshape the object and k changes even though the material has not. Divide out the geometry and what remains is the Young modulus, a stiffness that belongs to the material itself.
Builds on Density and Hooke's law.
IN THIS TOPIC
- Define tensile stress and tensile strain, with their units.
- Use the Young modulus as stress over strain, and as FL/AΔL, including finding it from a graph's gradient.
- Interpret stress-strain curves, and describe one simple method for measuring the Young modulus of a wire.
WHAT YOU PROBABLY THINK
The spring constant tells you how stiff the material is.
Why k is not enough
The spring constant describes that particular object. Take the same steel and draw it into a thicker wire, and k rises; make the wire longer, and k falls. The steel has not changed at all, only its shape has, so k is hopeless as a description of the material.
To reach the material itself, divide the geometry out. Spread the force over the cross-section, and compare the extension with the length being stretched:
Stress is measured in N m−2, the pascal, the same unit as pressure and for the same reason. Strain is a length divided by a length, so it has no units at all, and small strains are often quoted as percentages.
The Young modulus
For a given material, stress and strain are proportional while the deformation stays modest, and their ratio is the Young modulus:
E is measured in pascals, and for solids the numbers are enormous: steel sits near 2 × 1011 Pa, meaning colossal stress produces only tiny strain. A large Young modulus is what “stiff material” means, independent of the shape of any sample.
Reading stress-strain curves
A full stress-strain curve tells a material's whole story, and AQA expects you to read its landmarks.
A ductile material such as copper runs straight, passes its yield point, then stretches enormously at almost constant stress, its plastic region, before fracturing. The maximum stress the specimen withstands is its breaking stress. A brittle material such as glass or cast iron has no plastic region at all: it climbs its straight line and fractures without warning, which is exactly what makes brittle failure dangerous in engineering.
A simple measurement
One simple method is required. Clamp a long, thin wire of the test material at one end of a bench, run it over a pulley at the far end, and hang masses from it. Measure the original length L from clamp to a reference marker with a metre rule, the diameter with a micrometer at several places along the wire, averaging and using A = πd2/4, and the extension ΔL by how far the marker moves against a fixed ruler as loads are added.
Plot stress against strain and take the gradient of the straight region. The wire is long and thin by design: a small area means a decent stress from bench-sized loads, and a long wire produces an extension large enough to measure with a small percentage uncertainty. Everything in this unit meets here, and so does the last one: the micrometer's resolution sets the uncertainty in d, which the power rule doubles on the way into A.
THE EXAM BIT
- Strain has no units, and saying so is often a mark in itself. If your strain comes out in metres, the lengths were not divided.
- The area comes from the diameter: A = πd2/4. Forgetting to halve the diameter, or halving it twice, are the two classic ways this line goes wrong.
- “Why long and thin?” has a precise answer: both choices enlarge the extension, and a larger ΔL carries a smaller percentage uncertainty.
- Take the Young modulus from the gradient of the straight region only; past the limit of proportionality the ratio is no longer E.
- Brittle on a graph means no plastic region: the curve fractures at, or barely beyond, the end of its straight line.
CHECK YOURSELF
A wire of length 1.8 m and diameter 0.40 mm extends by 1.8 mm under a load of 25 N. Calculate the Young modulus of the material.
Show a hint
Find the area from the diameter first, working in metres.
Show the answer
Area: A = πd2/4 = π × (0.40 × 10−3)2 / 4 = 1.26 × 10−7 m2.
Stress: F/A = 25 / 1.26 × 10−7 = 1.99 × 108 Pa. Strain: ΔL / L = 0.0018 / 1.8 = 1.0 × 10−3.
Young modulus: E = 1.99 × 108 / 1.0 × 10−3 = 2.0 × 1011 Pa, the textbook value for steel. A strain of one part in a thousand from a 25 N pull is the Young modulus of a metal doing its job.
k describes the object.
E describes the material.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.