Stress-strain curve

Pull a steel wire slowly with an ever-increasing force and, for a long while, nothing dramatic happens — it just stretches a little more for every little bit more force, then springs straight back the moment you let go. Keep pulling, though, and something changes: past a certain load the wire starts to stretch faster than the force is rising, it stops fully springing back, and finally, with a snap, it breaks. If you had been recording the stress you applied and the strain it produced at every instant, and you plot one against the other, all of that story is compressed into a single curve — the stress-strain curve. It is one of the most information-dense graphs in all of physics: one glance tells you where a material behaves like an ideal spring, where it starts to misbehave permanently, how strong it is at its peak, and how it finally fails. This article builds directly on the stress $\sigma = F/A$ , strain $\varepsilon$ , and Hooke's law $\sigma = Y\varepsilon$ you have already met. Here those ideas get a face: the straight-line region of the curve is Hooke's law, its slope is Young's modulus, and everything that happens after the line bends is the physics of yielding, plastic flow, and fracture that no single formula from Hooke's law alone can capture. We will read the landmark points on a typical ductile-metal curve, use the area under the line to derive how much elastic energy a stretched wire stores per unit volume, work two numerical examples building on real steel data, and pin down the exact vocabulary — elastic limit versus ultimate tensile strength versus fracture point — that examiners love to test as pairs students confuse.

Key takeaways

The stress-strain curve plots stress $\sigma$ (y-axis) against strain $\varepsilon$ (x-axis) as a specimen is loaded from zero to fracture.
The initial straight-line portion is where Hooke's law holds; its slope equals Young's modulus $Y = \sigma/\varepsilon$ .
The proportional limit is where the straight line ends; the elastic limit is the stress beyond which permanent (plastic) deformation begins.
Beyond the elastic limit lies the yield region (rapid strain for little extra stress), then the ultimate tensile strength (UTS) — the maximum stress the material sustains — and finally the fracture point.
The area under the stress-strain graph up to a given strain equals the elastic potential energy stored per unit volume, $u = \tfrac12\sigma\varepsilon = \tfrac12Y\varepsilon^2$ in the linear region.
Ductile materials (steel, copper) show a long plastic region before breaking; brittle materials (glass, cast iron) fracture soon after the elastic limit; elastomers (rubber) sustain huge, non-linear, largely recoverable strains.
Key idea: the stress-strain curve is the graphical record of a material's elastic and plastic response, with its linear slope giving Young's modulus and the area beneath it giving the elastic energy stored per unit volume, up to the point where the material yields, peaks in strength, and fractures.

The core insight

A stress-strain curve tells you Young's modulus from its initial slope, the stored elastic energy per unit volume from the area beneath it, and — from where the line bends, peaks, and breaks — whether a material is brittle, ductile, or an elastomer.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 unit Mechanical Properties of Solids, right after stress, strain, and Hooke's law and just before the elastic moduli (Young's, bulk, and the qualitative idea of shear modulus) that follow it. The prescribed 2025-26 scope asks you to sketch and interpret a typical stress-strain curve, identify its landmark points, connect its slope to Young's modulus, and use the area under the linear region to get the elastic potential energy per unit volume $u = \tfrac12\sigma\varepsilon = \tfrac12Y\varepsilon^2$ . A full quantitative treatment of shear modulus, the detailed elastic-tensor description of materials, and dislocation or fatigue mathematics are explicitly beyond this board's syllabus and are left out here.

The physical picture

Imagine clamping one end of a wire and hanging steadily increasing loads on the other, measuring the extension at every step. Convert each load to stress ( $\sigma = F/A$ , force per unit original cross-sectional area) and each extension to strain ( $\varepsilon = \Delta L/L$ , extension per unit original length), and plot stress against strain. For a typical ductile metal such as annealed steel, the resulting curve rises as a straight line from the origin, gradually bends over, flattens into a nearly horizontal stretch, rises slightly to a peak, and then drops until the wire snaps.

Y = \frac{\sigma}{\varepsilon} = \tan\theta

where: $Y$ is Young's modulus (pascal, Pa), $\sigma$ is the stress on the linear part of the curve, $\varepsilon$ is the corresponding strain (dimensionless), and $\theta$ is the angle the straight portion makes with the strain axis, so $\tan\theta$ is literally the graph's slope. The physical reading is direct: a steeper initial slope means a stiffer material — more stress is needed to produce the same strain — which is exactly what a large Young's modulus means.

u = \frac{1}{2}\sigma\varepsilon = \frac{1}{2}Y\varepsilon^2

where: $u$ is the elastic potential energy stored per unit volume (joule per cubic metre, J/m³) when the material has been strained to $\varepsilon$ under stress $\sigma$ , valid while the point still lies on the straight (Hookean) part of the curve. The physical reading is that this is simply the area of the triangle under the straight-line graph up to that point — stretching a spring-like solid does work on it, and in the elastic region every joule of that work is recoverable, released back when the load is removed.

Figure

A stress on the vertical axis versus strain on the horizontal axis graph for a ductile metal wire, starting at the origin, rising as a straight line through a proportional limit point, curving through a nearby elastic limit, flattening through a yield region, rising gently to a peak marked ultimate tensile strength, then falling to a fracture point where the curve ends

Reading the curve from the origin: the segment O to A is the straight, Hookean line, where $Y = \tan\theta$ and every bit of stretch is fully recovered on unloading. Point A is the proportional limit, the highest stress at which stress and strain stay proportional. A little beyond A sits point B, the elastic limit — up to here the material still returns exactly to its original length when unloaded, but past it a small permanent (plastic) strain remains even after the load is removed. Beyond B the curve flattens into the yield region, where strain grows rapidly for very little extra stress; ductile metals show this almost-horizontal stretch clearly. The curve then climbs gently to point D, the ultimate tensile strength (UTS), the maximum stress the specimen ever bears — beyond D the cross-section starts visibly thinning (necking), so the curve can fall even though the material is still stretching, until point E, the fracture point, where it breaks. Different families of materials trace very different shapes: a brittle solid such as glass or cast iron breaks soon after its elastic limit, giving a short, almost straight curve with almost no yield region; a ductile metal like copper or mild steel shows a long plastic stretch before fracture, which is exactly why it can be drawn into wire; and an elastomer like rubber follows a strongly curved, non-Hookean path from the very start, sustaining enormous strains and yet recovering almost completely once unloaded — showing that "elastic" (returns to shape) and "obeys Hooke's law" (straight-line graph) are related but genuinely different properties.

Derivation AElastic potential energy stored per unit volume of a stretched wire

The goal is to show that when a wire is stretched within its Hookean (linear) region, the work done on it — and hence the elastic energy it stores, ready to be released the instant the load is removed — equals the area of the triangle under the stress-strain line up to that point, giving the compact result $u = \tfrac12\sigma\varepsilon = \tfrac12Y\varepsilon^2$ for the energy stored per unit volume. This uses nothing beyond the definition of work as force times displacement, applied incrementally, plus Hooke's law itself; no calculus beyond recognising a triangle's area is needed for the CBSE level, though the argument is really a small integral in disguise.

Step 1. Set up the incremental work. Consider a wire of natural length $L$ and cross-sectional area $A$ , so its volume is $V = AL$ . At some instant during stretching, the tension in the wire is $F$ and the extension so far is $x$ . Stretching it by a further tiny amount $dx$ takes a tiny amount of work $dW = F\,dx$ , exactly as for any force acting through a small displacement.

Step 2. Bring in Hooke's law for this wire. Within the linear region, $F$ is proportional to the extension already produced: $F = kx$ , where $k = \dfrac{YA}{L}$ is the wire's effective force constant (this follows directly from $Y = \dfrac{FL}{A\,\Delta L}$ rearranged for $F$ at extension $x$ ). So the tension itself grows in a straight line from zero, exactly as stress grows in a straight line with strain.

Step 3. Add up the work as the extension runs from $0$ to a final value $\Delta L$ . Because $F = kx$ rises linearly with $x$ , the total work is the area of a triangle with base $\Delta L$ and height $F_{\max} = k\,\Delta L$ , giving $W = \tfrac12 F_{\max}\,\Delta L = \tfrac12(k\,\Delta L)\,\Delta L = \tfrac12 k\,(\Delta L)^2$ . Substituting $k = YA/L$ : $W = \dfrac{1}{2}\cdot\dfrac{YA}{L}\cdot(\Delta L)^2$ .

Step 4. Convert to energy per unit volume. Divide $W$ by the wire's volume $V = AL$ : $u = \dfrac{W}{V} = \dfrac{1}{2}\cdot\dfrac{YA(\Delta L)^2}{L}\cdot\dfrac{1}{AL} = \dfrac{1}{2}Y\left(\dfrac{\Delta L}{L}\right)^2 = \dfrac{1}{2}Y\varepsilon^2$ , since strain is $\varepsilon = \Delta L/L$ . Using $\sigma = Y\varepsilon$ , this is equally $u = \tfrac12(Y\varepsilon)\varepsilon = \tfrac12\sigma\varepsilon$ — the two forms are the same result written two ways, and both are exactly the area of the triangle under the straight portion of the stress-strain graph, stress on one axis and strain on the other, scaled by volume.

Step 5. Read the result physically. This energy is genuinely elastic — stored the way a compressed or stretched ideal spring stores energy — only as long as the point $(\varepsilon,\sigma)$ lies on the straight-line part of the curve. Beyond the elastic limit, some of the work done goes into permanently rearranging the material rather than being stored recoverably, so $u=\tfrac12\sigma\varepsilon$ no longer gives the recoverable energy for points past B on the curve; it only ever describes area under a straight Hookean line.

Worked examples

Board level - Young's modulus and stored energy from a steel wire's graph

The linear portion of the stress-strain graph for a steel wire passes through the origin and the point $\varepsilon = 1.5\times10^{-4}$ , $\sigma = 3.0\times10^{7}\ \text{Pa}$ . Find Young's modulus of the steel, and the elastic energy stored per unit volume at this point.

Step 1. Read Young's modulus from the slope of the linear region. $Y = \dfrac{\sigma}{\varepsilon} = \dfrac{3.0\times10^{7}}{1.5\times10^{-4}} = 2.0\times10^{11}\ \text{Pa}$ . This matches the known order of magnitude for steel, about $2\times10^{11}\ \text{Pa}$ , confirming the graph has been read correctly.

Step 2. Use the same slope to predict stress at a nearby strain still within the linear region, say $\varepsilon = 2.5\times10^{-4}$ : $\sigma = Y\varepsilon = (2.0\times10^{11})(2.5\times10^{-4}) = 5.0\times10^{7}\ \text{Pa}$ .

Step 3. Find the elastic energy stored per unit volume at the original point, using $u = \tfrac12\sigma\varepsilon$ : $u = \tfrac12(3.0\times10^{7})(1.5\times10^{-4}) = 2.25\times10^{3}\ \text{J/m}^3$ . As a check using the other form, $u = \tfrac12Y\varepsilon^2 = \tfrac12(2.0\times10^{11})(1.5\times10^{-4})^2 = \tfrac12(2.0\times10^{11})(2.25\times10^{-8}) = 2.25\times10^{3}\ \text{J/m}^3$ — the two expressions agree exactly, as they must.

Answer: $Y = 2.0\times10^{11}\ \text{Pa}$ ; at strain $2.5\times10^{-4}$ the stress would be $5.0\times10^{7}\ \text{Pa}$ ; and the elastic energy stored per unit volume at the original point is $2.25\times10^{3}\ \text{J/m}^3$ .

Competitive level - comparing steel and an elastomer from their curve shapes and finding stored energy

A student tests two wires of equal length and area: one steel, one a rubber-like elastomer. The steel wire's curve is essentially straight up to strain $0.0015$ at stress $3.0\times10^{8}\ \text{Pa}$ (so $Y \approx 2.0\times10^{11}\ \text{Pa}$ ), then curves and finally breaks near strain $0.05$ . The elastomer's curve is visibly curved from the very start, stretches to strain $4.0$ before breaking, but returns almost fully to its original length when unloaded from strain $2.0$ . (a) Identify which wire obeys Hooke's law over most of its range and which is more ductile. (b) Estimate the elastic energy per unit volume stored in the steel wire at the end of its linear region.

Step 1. Classify the steel wire. Its stress-strain graph is a straight line up to strain $0.0015$ , so Hooke's law holds there, and it goes on to a strain of $0.05$ before fracture — a long plastic region well past its elastic limit — which is the signature of a ductile metal.

Step 2. Classify the elastomer. Because its curve is non-linear from the very first bit of loading, it does not obey Hooke's law over most of its range, even though it recovers almost completely from a huge strain of $2.0$ . This is the sharpest reminder in the whole topic: "elastic" (returns to shape) and "obeys Hooke's law" (straight-line proportionality) are related but distinct ideas — the elastomer is highly elastic yet strongly non-Hookean.

Step 3. Compute the stored energy per unit volume for the steel wire at the end of its linear region, using the values given. $u = \tfrac12\sigma\varepsilon = \tfrac12(3.0\times10^{8})(0.0015) = 2.25\times10^{5}\ \text{J/m}^3$ . Cross-check with $u=\tfrac12Y\varepsilon^2$ : $Y = \sigma/\varepsilon = (3.0\times10^{8})/(0.0015) = 2.0\times10^{11}\ \text{Pa}$ , so $u = \tfrac12(2.0\times10^{11})(0.0015)^2 = \tfrac12(2.0\times10^{11})(2.25\times10^{-6}) = 2.25\times10^{5}\ \text{J/m}^3$ , matching.

Answer: the steel wire obeys Hooke's law and is the ductile one (long plastic stretch before fracture); the elastomer is highly elastic but non-Hookean; the steel wire's stored elastic energy per unit volume at the end of its linear region is $2.25\times10^{5}\ \text{J/m}^3$ . Timing note: aim for under three minutes. The classification part (Steps 1-2) is pure reasoning from the shape of the curve, not arithmetic, so do not waste time hunting for numbers there — spend your calculation minute only on Step 3, and always sanity-check the energy figure both ways, $\tfrac12\sigma\varepsilon$ and $\tfrac12Y\varepsilon^2$ , since a units slip in either one shows up instantly as a mismatch.

Common traps

Treating "elastic" and "obeys Hooke's law" as synonyms — rubber is highly elastic (it snaps back) yet its graph is strongly curved from the start, so it disobeys Hooke's law over almost its entire range.
Confusing the elastic limit with the ultimate tensile strength — the elastic limit is the stress beyond which permanent deformation first appears; UTS is the maximum stress the material ever bears, which comes considerably later on the curve.
Assuming fracture happens exactly at the UTS point — for ductile metals, the material keeps stretching (necking) after UTS and actually breaks slightly beyond it, even as the stress reading falls.
Writing the slope as strain-over-stress instead of stress-over-strain — Young's modulus is $\sigma/\varepsilon$ , so stress must be plotted on the vertical axis and strain on the horizontal for $Y = \tan\theta$ to hold.
Forgetting that $u = \tfrac12\sigma\varepsilon$ (the triangle-area formula) is valid only in the straight-line, Hookean part of the curve — using it for a point beyond the elastic limit overstates the recoverable energy, since some work there goes into permanent rearrangement.
Mixing up units when computing $u$ — with $\sigma$ in pascal and $\varepsilon$ dimensionless, $u$ comes out directly in joule per cubic metre; no extra length or area factor is needed once you are working per unit volume.

Common trapElastic behaviour and Hooke's law are not the same thing

The single biggest misconception in this topic is assuming that if a material is elastic (it returns to its original shape), its stress-strain graph must be a straight line, or conversely that a curved graph means the material is not elastic. Both are false. A rubber band is dramatically elastic — stretch it to several times its length and it snaps back almost perfectly — yet its stress-strain curve is visibly curved from the very first bit of stretch, so it never obeys Hooke's law in any meaningful range. Meanwhile a steel wire loaded just past its elastic limit obeys Hooke's law nowhere near as well as rubber does over rubber's full range, yet fails to return to its original length at all. "Elastic" describes whether deformation is recoverable; "Hooke's law" describes whether stress and strain are proportional. Keep them as two separate questions about the same curve.

You should be able to…

Sketch a labelled stress-strain curve for a ductile metal, marking the proportional limit, elastic limit, yield region, ultimate tensile strength, and fracture point in the right order.
Read Young's modulus off the slope of the linear portion of a given stress-strain graph, and use that slope to predict stress at another strain still within the linear region.
Derive and apply $u = \tfrac12\sigma\varepsilon = \tfrac12Y\varepsilon^2$ for the elastic energy stored per unit volume in the Hookean region, and explain why it is the triangle-area under the graph.
Distinguish "elastic" (returns to original shape) from "obeys Hooke's law" (proportional stress-strain relationship), with rubber and steel as contrasting examples.
Explain, purely from curve shape, why glass is brittle, copper or steel is ductile, and rubber is an elastomer.
Explain why stress falling after the ultimate tensile strength does not mean the material is strengthening, but reflects necking measured against the original cross-section.

Ask the tutor: Want the AI tutor to sketch and compare the stress-strain curves of steel, glass, and rubber side by side and quiz you on which landmark point is which?

Why does the curve flatten out near the yield point instead of just continuing to rise?

Past the elastic limit, the material starts deforming plastically — layers of atoms are sliding past one another and rearranging permanently, not just stretching the bonds elastically the way they did in the straight-line region. That rearrangement lets the specimen extend a lot for only a small increase in applied stress, which is exactly what a nearly flat stretch on the graph means. Ductile metals like mild steel show this yield region very clearly; it is actually useful in engineering, because it gives visible warning (large deformation) before the material finally fails, rather than snapping without notice.

If the stress starts falling after the ultimate tensile strength, doesn't that mean the material is getting stronger, not weaker?

No — it is the opposite, and this is a common misreading of the graph. Stress here is being calculated using the wire's original cross-sectional area, but past the UTS the wire develops a local thin region called necking, where the true cross-section has shrunk a lot. The actual force needed to keep stretching that thinner region really is dropping, even though the true stress concentrated in the necked region is still rising right up to fracture. The falling branch after D on the curve reflects this bookkeeping choice (force divided by original area), not the material becoming easier to break.

Is the area under the whole stress-strain curve, not just the straight part, also useful?

Yes, though it is a different and more advanced quantity than what this syllabus asks you to compute. The area under just the straight (Hookean) portion, $u=\tfrac12\sigma\varepsilon$ , is the recoverable elastic energy per unit volume — release the load and you get it back. The area under the entire curve up to fracture (elastic plus plastic parts) represents the total energy absorbed per unit volume before the material breaks, called toughness, and it is why a ductile metal with a long plastic region can absorb far more energy before fracturing than a brittle material of similar strength. CBSE Class 11 only asks for the linear-region formula; toughness as a named quantity is not part of this board's prescribed scope.

Why do we plot stress and strain instead of just force and extension?

Because stress and strain strip away the specific size of the test specimen, leaving a curve that describes the material itself rather than one particular wire. A thick steel rod and a thin steel wire would give very different force-extension graphs (the thick rod needs a much bigger force for the same stretch), but once you divide force by area and extension by original length, both specimens trace the same stress-strain curve, because they are the same material. That is exactly why Young's modulus, read off this curve, is quoted as a property of steel in general, not of one particular wire.