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Physics for Pilot Assessments

Stretching, Squashing and Material Strength

Matter & Fluids · Lesson 6 of 6•12 min read

Every material gives a little when you load it, and the useful fact is that for a while it gives in exact proportion to the load. This lesson is about that proportionality, about where it stops, and about why a component can fail at a load it has carried safely thousands of times before. By the end you should be able to use Hooke's law in both directions, find a spring's stiffness from a single measurement, calculate a stress and a strain, and say what the elastic limit is.

Extension, and what it means

Hang a weight on a spring and it gets longer. The quantity that matters is not the new length but the change in length — the extension, meaning the amount by which the spring is now longer than its natural length, the length it has when nothing at all is pulling on it. Get this right at the outset, because items exploit it: a spring of natural length 20 centimetres that measures 28 centimetres under load has an extension of 8 centimetres, not 28. Extension is a length, so its SI unit is the metre, and every formula below wants it in metres — a measurement in centimetres must be divided by 100 first, and one in millimetres by 1000.

Hooke's law

Load a spring with 10 newtons and measure the extension; load it with 20 and the extension doubles; load it with 30 and it triples. That is the observation, and it holds for a wide range of springs and for solid bars and wires as well: within limits, the extension is proportional to the force applied. Proportional means something specific — that the ratio of the two is constant, so dividing force by extension gives the same number every time — and that is what makes a graph of force against extension a straight line passing through the origin. The constant is called the spring constant, or the stiffness, and it is the force needed per metre of extension. A stiff spring has a large one.

Hooke's law

F=kxF = kx
FF
the force stretching or compressing it, in newtons (N)
kk
the spring constant, or stiffness, in newtons per metre (N/m)
xx
the extension — the change in length, in metres (m)

The rearrangements are the usual two. Divide both sides by x and the x on the right cancels, leaving k=Fxk = \frac{F}{x}, which is how you measure a spring's stiffness from one force-and-extension pair. Divide both sides by k instead and you get x=Fkx = \frac{F}{k}, which predicts the extension a given load will produce. It works the same way in compression: squash the spring rather than stretching it and the same k relates the force to the amount by which it shortens.

forceextensionelastic limitF = kx herenot proportional here
Straight through the origin while the material behaves elastically. Past the elastic limit the same extra force buys far more extension, and the material no longer returns to its original length when the load is taken off.

The elastic limit

Keep increasing the load and at some point the straight line bends over. That point is the elastic limit, and what changes there is more fundamental than the shape of a graph. Below it the behaviour is elastic: take the load off and the material returns to exactly its original length, with nothing permanent left behind. Above it the behaviour is plastic: take the load off and some of the extension stays, so the spring is now permanently longer than it began. A spring pulled past its limit does not recover, and a paperclip bent open is plastic behaviour used deliberately. Between the elastic limit and outright fracture a material typically extends a great deal for very little extra force, which is why the graph flattens, and only then does it break.

Two words often used loosely deserve separating here. Stiffness is about how much a material deforms under a given load — a large spring constant means a small extension. Strength is about how much load it takes to break it. The two are independent: a biscuit is stiff and weak, a rubber band is floppy and surprisingly strong, and any option that offers "stronger" as the explanation for "extends less" has conflated them.

Stress and strain

Hooke's law describes a particular spring or a particular bar. It cannot tell you about a material as such, because a thick bar of a given material obviously takes more force than a thin one, and a long wire stretches further than a short one under the same pull. To talk about the material itself, the force has to be shared out over the area that carries it, and the extension has to be compared with the length that did the stretching. The first of those is stress: the force divided by the cross-sectional area it acts across, meaning the area of the slice you would see if you cut the bar straight through. It has the same unit as pressure — newtons per square metre, or pascals — for the same reason, since it too is a force spread over an area.

Stress

σ=FA\sigma = \frac{F}{A}
σ\sigma
stress, in pascals (Pa) — one pascal is one newton per square metre
FF
the force applied, in newtons (N)
AA
the cross-sectional area carrying it, in square metres (m²)

Sigma is the Greek letter s, written σ. Stress is what actually decides whether a material fails: every material has a characteristic stress it can withstand before it breaks, and that figure is a property of the material, whereas a breaking force is only a property of one particular specimen. Because real engineering stresses are large numbers of pascals, they are usually quoted in megapascals — mega- means a million, so one megapascal is 1 000 000 pascals. The companion quantity is strain: the extension divided by the original length.

Strain

ε=xL\varepsilon = \frac{x}{L}
ε\varepsilon
strain — a bare ratio, with no units at all
xx
the extension, in metres (m)
LL
the original length before loading, in metres (m)

Both quantities in that fraction are lengths, so the units cancel and strain comes out as a bare number, often quoted as a small decimal or as a percentage: a strain of 0.002 means the specimen is 0.2 per cent longer than it started. Strain is what lets you compare deformations fairly, because a 2-millimetre stretch is a great deal in a 10-centimetre specimen and almost nothing in a 40-metre cable. With stress and strain defined, Hooke's law can be restated so that it belongs to the material rather than to the specimen: while the behaviour is elastic, stress is proportional to strain. That is the same straight line as before, drawn in quantities that do not change when you swap the specimen for a thicker or longer one of the same stuff.

Why repeated loading breaks things

A component can fail at a load far below the one that would break it in a single pull, if it is loaded and unloaded enough times. The reason is that no real material is flawless: there are microscopic cracks, hard inclusions and surface scratches, and stress does not spread evenly around a flaw — it concentrates at the tip of a crack, where the local stress can be many times the average across the section. Each loading cycle drives that crack forward a minute distance, and each unloading leaves it slightly longer than it was. The crack grows, the sound cross-section that remains shrinks, and since stress is force divided by area, the stress in what is left rises with every cycle. Eventually the remaining material cannot carry one more application of the same load it has carried for years, and the part breaks. This is fatigue.

Two consequences are worth knowing. First, fatigue is progressive but its final stage is abrupt, which is what makes it dangerous — there is no gradual sagging to warn you. Second, it is why components are inspected and replaced on a schedule counted in cycles rather than simply being built strong enough: the number of load reversals matters, not only their size, and a sharp internal corner or a roughly drilled hole shortens the life of an otherwise sound part by concentrating stress at that point. Bend a paperclip back and forth and it snaps within a dozen cycles, at a force that plainly would not have broken it on the first bend.

Why it matters in aviation

An airframe is designed to deform and come back. Wings flex visibly in turbulence, the fuselage swells slightly as the cabin is pressurised, and all of that is intended elastic behaviour within the limits the structure is stressed to. The limit that governs its working life, though, is not a single large overload but repetition. Each flight applies a full cycle of loading: the cabin is pressurised in the climb and depressurised in the descent, gusts and manoeuvres load the wing up and release it again, and the landing gear absorbs one large impact at the end. Enough cycles at stresses well below the breaking stress will start a crack and then extend it, by exactly the mechanism described above.

That is why airframe and component life is tracked in flight cycles as well as in flight hours, and why an aircraft flying many short sectors accumulates fatigue faster than one of the same age flying long ones. It is also why inspections do not sample the structure evenly. Attention concentrates on the places where stress concentrates — fastener holes, skin joints, the surrounds of windows and doors, and the roots of heavily loaded members — because that is where a crack starts, long before anything would be visible elsewhere.

What the test asks

The reliable trap is extrapolation. An item gives you a force and an extension, then asks about a force ten or thirty times larger, and the arithmetically consistent answer will be sitting there among the options — 20 newtons gives 5 centimetres, so 600 newtons gives 150 centimetres. It does not, because the proportionality only holds up to the elastic limit, and the correct response is that the spring has been taken well past it. Expect a definitions item on elastic versus plastic, where the distractor simply swaps the two, and one where the extension is confused with the total length. And expect at least one item that turns on stress rather than force: two cables of the same material and different thicknesses carrying the same load are under the same force and quite different stresses, and it is the stress that decides which of them fails.

Check you understood

Three questions from this lesson — one straightforward, one that applies it, one that pushes a little further. Nothing is recorded, and you can draw a new set as often as you like.

  1. 1. Warm-up

    Strain is defined as:

  2. 2. Applying it

    Two cables of the same steel, one of cross-sectional area 1 cm² and one of 2 cm², each carry a load of 4000 N. Which is closer to failing?

  3. 3. Pushing it

    A rod of cross-sectional area 5 cm² carries a tensile load of 20 000 N. What stress is it under, in megapascals? (1 MPa = 1 000 000 Pa)

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