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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.

Year 12AQA 3.4.2.2 CIE 6.1.5, 6.1.6 OCR A 3.4.2(c), 3.4.2(d)(i), 3.4.2(d)(ii), 3.4.2(e)EDEXCEL 54, 56, 57

Builds on Density and Hooke's law.

IN THIS TOPIC

Define tensile stress and tensile strain, and give the unit of each.
Use the Young modulus as stress over strain, and assemble the $FL/A\Delta L$ form yourself when you need it.
Take E from the gradient of the straight part of a stress-strain graph.
Read yield, breaking stress and brittle failure off a stress-strain curve, and describe one method of measuring E for a wire.
Tell the ductile, brittle and polymeric stress-strain shapes apart, and explain why rubber stiffens as it stretches and loses energy round its hysteresis loop.

COMMON MISCONCEPTION

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, then compare the extension against the length being stretched.

\text{tensile stress} = \frac{F}{A}

ON THE AQA DATA SHEET

\text{tensile strain} = \frac{\Delta L}{L}

ON THE AQA DATA SHEET

Stress is measured in N m⁻², the pascal, the same unit as pressure and for the same reason. Strain is a length over a length, so it has no units at all. Small strains often get quoted as percentages, which is a convenience and not a unit.

The Young modulus

For a given material, stress and strain stay proportional while the deformation is modest, and their ratio is the Young modulus.

E = \frac{\text{tensile stress}}{\text{tensile strain}}

ON THE AQA DATA SHEET

Substitute the two definitions above and the working form follows, $E = FL/(A\Delta L)$ . That combined version is not printed anywhere in the booklet, and it does not need to be. All three ingredients are printed, so you can rebuild it in the margin in about four seconds. Students who memorise it as a fourth separate formula are spending revision on something they were being given for free.

FIG. 1The four quantities the Young modulus combines: the load F, the cross-sectional area A, the original length L and the extension ΔL.

E is measured in pascals, and for solids the numbers are enormous. Steel sits near $2 \times 10^{11}$ Pa, which is to say that a colossal stress produces only a very small strain. A large Young modulus is what the phrase “stiff material” actually means, and unlike k it survives any reshaping of the sample.

FIG. 2On the linear part of a stress-strain graph, the gradient is the Young modulus.

WORKED EXAMPLE

Stress, strain, then the stretch

A brass wire (E = 1.2 × 10¹¹ Pa) of diameter 0.60 mm and length 2.5 m carries a 60 N load. Find the stress, the strain, and the extension.

Area first. A = πd²/4 = π × (6.0 × 10⁻⁴)²/4 = 2.83 × 10⁻⁷ m².

Then the stress, σ = F/A = 60/(2.83 × 10⁻⁷) = 2.1 × 10⁸ Pa.

Strain follows from ε = σ/E = 2.1 × 10⁸/(1.2 × 10¹¹) = 1.8 × 10⁻³, a bare number with no unit.

Finally ΔL = εL = 1.8 × 10⁻³ × 2.5 = 4.4 mm. Four millimetres from a two-and-a-half-metre wire under sixty newtons is exactly the scale that makes required practical 4 so fiddly.

Reading stress-strain curves

A full stress-strain curve characterises a material's tensile behaviour under the conditions of the test, and you are expected to read its landmarks.

FIG. 3A ductile material yields, stretches through a long plastic region and fractures at large strain; a brittle one climbs its straight line and snaps with little or no plastic region.

A ductile material such as copper runs straight, passes its yield point, then stretches enormously at almost constant stress through its plastic region before it fractures. The maximum stress the specimen withstands is its ultimate tensile stress, which A-level papers usually call the breaking stress; strictly, a ductile specimen necks after that maximum and finally fractures at a lower engineering stress. A brittle material such as glass or cast iron shows little or no plastic deformation. It climbs its straight line and snaps, with no warning to speak of, and that absence of warning is what makes brittle failure so feared in engineering.

A third shape completes the set. A polymeric material such as rubber or polythene is built from long molecular chains, tangled together rather than locked into a lattice, and its curve resembles neither of the other two. Stretch rubber and at first the chains merely uncoil, which takes very little stress and gives an enormous strain. Once they are pulled straight, the bonds themselves have to stretch, and the curve stiffens sharply. So a rubber graph starts shallow and steepens, the opposite way round from a metal, and it runs to strains of several hundred per cent where a steel wire has long since fractured.

FIG. 4The three characteristic shapes, each drawn on its own strain scale, because rubber outruns steel by three orders of magnitude. Brittle snaps at the end of its straight line, ductile yields and flows, and polymeric starts shallow and stiffens, its unloading curve returning below the loading one.

Two things follow from that shape. Rubber has no single Young modulus, because the gradient changes all the way along and never settles, so no one constant obeys Hooke's law over the range drawn. Yet the deformation stays elastic: let go and the sample returns to its original length. Elastic and Hooke's law are separate claims, and rubber is the material that pulls them apart.

Unloading rubber traces a lower curve than loading did, and the loop enclosed between the two is called hysteresis. Its area is work that went in and came back out as heat instead of as recovered strain energy, which is why a squash ball warms through a rally and why a tyre run hard gets hot. Polythene shows the other polymeric behaviour worth knowing: it stretches easily at low stress and stays stretched, deforming plastically where rubber recovers.

GUIDED PRACTICE

Reading E off a graph

On a stress-strain graph, a material's straight portion passes through stress 1.5 × 10⁸ Pa at strain 7.5 × 10⁻⁴. Find its Young modulus from the gradient.

Show the working

E = gradient = σ/ε = 1.5 × 10⁸/(7.5 × 10⁻⁴) = 2.0 × 10¹¹ Pa.

That is steel's value. Only the straight portion counts, because past the limit of proportionality the ratio drifts from point to point and stops being a constant of the material.

INDEPENDENT PRACTICE

Energy stored in every cubic metre

For the same material at the same point (σ = 1.5 × 10⁸ Pa, ε = 7.5 × 10⁻⁴), find the elastic energy stored per unit volume, using the area under the stress-strain line.

Show the working

The area under the straight line is a triangle, ½σε = ½ × 1.5 × 10⁸ × 7.5 × 10⁻⁴ = 5.6 × 10⁴ J m⁻³.

The units land correctly because a pascal is already a joule per cubic metre. Multiply N m⁻² by a strain that carries no unit and energy per volume is what you are left with. Fifty-odd kilojoules in every cubic metre of loaded steel is why workshops treat a tensioned cable with respect.

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 $A = \pi d^{2}/4$ , and the extension ΔL by how far the marker moves against a fixed ruler as loads are added.

FIG. 5The simple method: a long clamped wire, a marker read against a ruler, a pulley and hanging masses.

Plot stress against strain and take the gradient of the straight region. The wire is long and thin by design. A small area wrings a decent stress out of bench-sized loads, and a long wire gives an extension big enough to measure without a ruinous percentage uncertainty. The measurements unit joins in here too, because the micrometer's resolution sets the uncertainty in d, and the power rule doubles that on the way into A.

ASSESSMENT FOCUS

Strain has no units, and saying so is often a mark in its own right. A strain that comes out in metres means the lengths were never divided.
The area comes from the diameter, $A = \pi d^{2}/4$ . Forgetting to halve the diameter, or halving it twice over, are the two classic ways this line goes wrong.
“Why long and thin?” has a precise answer. Both choices enlarge ΔL, 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 not E any more.
Brittle on a graph means no plastic region. The curve fractures at, or barely beyond, the end of its straight line.
Polymeric on a graph means the gradient rises with strain, and the unloading curve sits below the loading one. Name the enclosed loop as hysteresis and the area as energy released as heat. OCR asks for all three shapes together; the other boards examine only ductile and brittle.
Rubber is elastic but not Hookean. Saying it obeys Hooke's law because it springs back is the trap in this part of the topic, and there is no single Young modulus to quote for it.
Stress, strain and E as their ratio are all printed. $FL/A\Delta L$ is not, so build it in the margin from the three that are, and do not waste a revision card on it.

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

Start with the area. $A = \pi d^{2}/4$ = π × (0.40 × 10⁻³)² / 4 = $1.26 \times 10^{−7}$ m².

Stress is $F/A$ = 25 / (1.26 × 10⁻⁷) = $1.99 \times 10^{8}$ Pa, and strain is $\Delta L / L$ = 0.0018 / 1.8 = $1.0 \times 10^{−3}$ .

Divide one by the other. $E$ = 1.99 × 10⁸ / (1.0 × 10⁻³) = $2.0 \times 10^{11}$ Pa, the textbook value for steel. One part in a thousand of stretch for a 25 N pull is what a metal's Young modulus feels like from the outside.

k describes the object.

E describes the material.

Reshape the sample and only one of them changes.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the stress, strain and the young modulus questions page.

13 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 4: the young modulus of a wire puts this topic in the lab, and the written papers ask about it.
Gradients and areas under graphs is the maths this lesson leans on, worked through from GCSE.
Uncertainty arithmetic is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Define tensile stress and tensile strain, and give the unit of each.
Use the Young modulus as stress over strain, and assemble the $FL/A\Delta L$ form yourself when you need it.
Take E from the gradient of the straight part of a stress-strain graph.
Read yield, breaking stress and brittle failure off a stress-strain curve, and describe one method of measuring E for a wire.
Tell the ductile, brittle and polymeric stress-strain shapes apart, and explain why rubber stiffens as it stretches and loses energy round its hysteresis loop.

Open the full revision checklist to track your progress across the whole unit.

Stress, strain and the Young modulus

Why k is not enough

The Young modulus

Reading stress-strain curves

A simple measurement

Question workbook

Question workbook

Answer book