In short

A wire of length L, area A, stretched by \Delta L with Young's modulus Y stores elastic potential energy U = \tfrac{1}{2} \cdot \sigma \cdot \varepsilon \cdot V = \tfrac{1}{2} Y \varepsilon^2 V, where V = A L is the wire's volume. The energy density (per unit volume) is u = \tfrac{1}{2}\sigma\varepsilon = \tfrac{1}{2}Y\varepsilon^2 = \tfrac{\sigma^2}{2Y} — three equivalent forms, pick the one matching the quantities you know. The stress-strain curve tells the full story of a material: its slope (in the linear region) is Y, the area under the curve is the stored energy, and the curve's shape distinguishes brittle materials (steep, short, ending in abrupt fracture) from ductile ones (long plastic region before fracture). Rubber shows elastic hysteresis: the loading and unloading curves differ, and the area between them is energy lost as heat — which is why a car tyre warms up on the highway.

Pick up a rubber band. Stretch it between your fingers. You can feel the resistance building; the further you pull, the harder the band fights back. Now let go — the band snaps back. If you release it gently, nothing much happens. But if you stretch it far and let go suddenly, the band flies across the room. The energy you put in with your fingers came back out as the band's kinetic energy.

That stored energy, waiting quietly inside a stretched elastic body, is elastic potential energy. It is the physics behind every bow and arrow that ever launched an arrow, every pole-vaulter's fibreglass pole bending and then springing back, every diving board you have ever bounced on, every spring watch that ticks through a day on a single winding. Stretch a solid, store energy. Release, recover energy. Exactly like lifting a mass against gravity, but with the "height" being strain and the "weight" being stress.

This article derives how much energy is stored in a deformed elastic body, introduces the crucial idea of energy density (so you can compare a car tyre to a bowstring on equal terms), reads every region of the stress-strain curve and what it reveals about a material, distinguishes brittle from ductile response, and explains hysteresis — the reason why rubber, unlike steel, does not return all the energy you put into it.

Deriving the stored energy

Assumptions. The wire has uniform cross-section A and original length L. It is stretched gradually (quasi-statically), so that at every instant the tension equals the applied force and kinetic energy of the wire is negligible. The wire is within its elastic limit — Hooke's law holds.

Start with a wire of the material under test. Apply a gradually increasing force F, stretching the wire from its natural length. At some point in the stretching, the wire's current extension is x, so the current strain is x/L, the stress is F/A, and — by Hooke's law — these are linked by

\frac{F}{A} = Y \cdot \frac{x}{L} \quad\Longrightarrow\quad F = \frac{YA}{L}\,x.

Why: Hooke's law \sigma = Y\varepsilon written in terms of applied force and extension. The quantity YA/L is the effective "spring constant" of the wire — call it k = YA/L. The wire behaves like a linear spring.

The work required to extend the wire by an additional tiny amount dx against this force is dW = F\,dx. The total work to stretch the wire from 0 to final extension \Delta L is

W = \int_0^{\Delta L} F\,dx = \int_0^{\Delta L} \frac{YA}{L}\,x\,dx = \frac{YA}{L} \cdot \frac{(\Delta L)^2}{2}.

Why: integrate F = (YA/L)\cdot x with respect to x, giving (YA/L) \cdot x^2/2, evaluated at the upper limit \Delta L. The quadratic behaviour is the fingerprint of a linear restoring force — the same behaviour as U = \tfrac{1}{2}kx^2 for a spring.

Because the wire's material is elastic, this work goes entirely into stored potential energy — not heat, not sound, not permanent deformation. So

U = \frac{YA(\Delta L)^2}{2L}.

This is the answer, and it can be rewritten in three equivalent forms — each useful in different problems.

Three faces of the same formula

Substitute \sigma = F/A = Y\Delta L / L and \varepsilon = \Delta L/L, and recall the wire's volume is V = AL:

Why: each form uses the Hooke's-law relation \sigma = Y\varepsilon to trade between stress and strain. Multiply/divide accordingly. All three give the same number because they are the same equation rearranged.

\boxed{\; U = \frac{1}{2}\sigma\varepsilon\,V = \frac{1}{2}Y\varepsilon^2 V = \frac{\sigma^2}{2Y}\,V \;}

Which one to use in a problem depends entirely on which quantities you have.

Energy density — energy per unit volume

Strain energy grows with the volume of the stretched material. To decouple the intensive (material-and-configuration) part from the extensive (size) part, divide both sides by V:

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

u is the elastic energy density — joules per cubic metre stored in every unit of stretched material. This is the single most useful quantity when comparing materials: how much energy can a given volume of the material hold before it breaks?

u at fracture is called the modulus of toughness — the total energy per unit volume the material absorbs before failing. A tough material has high u at fracture; a brittle material has low u at fracture even if its Y is high.

The factor of one-half — where it comes from

A common confusion: if you apply a force F and the wire stretches by \Delta L, is the work not just F \times \Delta L? Why the factor of one-half?

The answer: because the force was not constant throughout the stretching. It started at zero (unstretched wire) and grew linearly to its final value F. The average force over the stretching process is F/2, and the work done is average-force-times-displacement: W = (F/2)\Delta L = \tfrac{1}{2}F\Delta L.

Area under stress-strain curve equals energy density Linear stress-strain curve rising from origin, with the triangular area under the line shaded to represent stored elastic energy per unit volume. The final stress σ is marked on the vertical axis and the final strain ε on the horizontal axis. strain ε stress σ σ ε u = ½σε (area of triangle) slope = Y (Young's modulus)
The energy density stored in an elastically stretched solid is the area of the triangle under the stress-strain curve. Because the curve is a straight line (Hooke's law), the area is $\tfrac{1}{2}\sigma\varepsilon$. The slope of the line is $Y$.

That triangular area — half the rectangle \sigma \times \varepsilon — is the geometric origin of the factor of one-half. Every spring-like system shares this structure: U = \tfrac{1}{2}kx^2, U = \tfrac{1}{2}CV^2 for a capacitor, U = \tfrac{1}{2}LI^2 for an inductor. The "one-half-of-a-linear-thing-squared" shape is the signature of a linear restoring force.

The stress-strain curve — a material's identity card

The stress-strain curve is the single most informative plot in all of solid mechanics. One curve tells you how stiff the material is (slope), how much energy it can absorb (area under curve), whether it is brittle or ductile (shape after yield), how strong it is (maximum stress), and how far it stretches before breaking (maximum strain). Every metallurgist, civil engineer, and JEE student should be able to read one at a glance.

Four regions of the curve

Take a mild-steel sample and pull it in a tensile-testing machine. The curve rises in four distinct regions, each with a name and a physical meaning.

Four regions of the stress-strain curve for a ductile metal Stress-strain curve with four coloured regions: linear elastic region from origin to proportional limit P; elastic non-linear from P to elastic limit E; plastic region from E through yield point Y, ultimate tensile strength U, to necking region; and final fracture point F. Shaded area under elastic region represents recoverable energy; shaded area under plastic region represents dissipated plastic work. strain ε stress σ P E Y U F elastic plastic recoverable U dissipated work
A mild-steel stress-strain curve. **Elastic region** (O to E): recoverable — release the load, the sample returns to its original length. Shaded area (pink) is the stored elastic energy. **Plastic region** (E to F): non-recoverable — release the load and the sample is permanently deformed. Shaded area (grey) is the work dissipated as heat and atomic rearrangement. The key landmarks: proportional limit P, elastic limit E, yield point Y, ultimate tensile strength U, fracture F.

1. Elastic region (O → E)

Up to the elastic limit E, the sample recovers its original length when unloaded. Between the origin and the proportional limit P (usually indistinguishable from E in common metals), the curve is straight and Hooke's law applies — stress proportional to strain, slope equal to Y. A tiny non-linear segment between P and E shows elastic but non-Hookean behaviour in some materials.

Energy stored in this region is fully recoverable. The wire acts like a spring.

2. Yield region (E → Y)

Slightly past the elastic limit, the material begins to yield — it deforms permanently. For mild steel, this happens sharply: the curve drops briefly (the upper yield point spike) and then flattens at the lower yield point. This is when dislocations — line defects in the crystal lattice — start moving en masse, allowing the metal to flow.

For most non-ferrous metals (copper, aluminium), there is no sharp yield spike; the transition from elastic to plastic is gradual, and engineers define a proof stress at 0.2% plastic strain as a working substitute for "yield point".

3. Plastic region (Y → U)

Past yielding, the material continues to carry load but deforms plastically — each bit of additional stress produces far more strain than in the elastic region. Work hardening (or strain hardening) kicks in: as dislocations pile up against obstacles in the crystal, further deformation becomes harder. The curve rises gradually to the ultimate tensile strength \sigma_U, the maximum stress the material can bear.

This is the region where blacksmiths work — heating steel and hammering it at high temperatures, shaping it plastically without fracturing.

4. Necking and fracture (U → F)

At stress \sigma_U, the sample begins to narrow visibly at one location — a neck. The reduced cross-section can no longer support the full load, so the engineering stress (which uses the original area) appears to drop. The neck thins until the material tears apart at the fracture point F. For mild steel, the strain at fracture is about 0.25 — a 25% extension.

What the area under the curve means

Integrate stress with respect to strain:

\int_0^{\varepsilon} \sigma(\varepsilon')\,d\varepsilon' = \text{energy density input up to strain } \varepsilon.

For a Hookean region, this integral is \tfrac{1}{2}\sigma\varepsilon — the triangular area. Outside the Hookean region, it is just the area under whatever curve the material follows.

A spring needs high resilience (energy stored per volume, fully recoverable). A safety helmet needs high toughness (lots of energy absorbed, recoverability unimportant — the helmet is disposable).

Brittle versus ductile — the curve tells you

Not every material has all four regions. Some fracture almost as soon as they enter the plastic region. Others deform plastically for what seems like forever before failing.

Stress-strain curves for brittle and ductile materials side by side Two curves on the same axes: a brittle material curve rises steeply and then drops abruptly at a small strain, while a ductile material curve rises less steeply but continues over a much larger strain range before finally falling at fracture. strain ε stress σ brittle (e.g. glass, cast iron) fracture ductile (e.g. mild steel, copper) fracture
Brittle materials (red, like glass or cast iron) have a steep initial rise, little or no plastic region, and snap at small strain. Ductile materials (dark, like mild steel or copper) have a less steep rise and stretch through a large plastic region before failing at high strain.

Brittle materials

A brittle material has almost no plastic region. It follows Hooke's law closely, reaches a peak stress, and breaks — abruptly, with very little prior warning.

Examples: glass (window pane, bulb), cast iron (old-style water pipes, machine beds), chalk (hence the snap), concrete (under tension), ceramics (tea cups). Brittle materials often have high Young's modulus (very stiff) but low toughness — they absorb little energy before failing.

For a brittle material, the modulus of resilience essentially equals the modulus of toughness. There is no plastic reserve; once you exceed the elastic limit, you are in the danger zone.

Ductile materials

A ductile material has a large plastic region. It yields gradually, hardens as it deforms, necks down, and fails at high strain after absorbing substantial plastic work.

Examples: mild steel (structural beams, railway rails), copper (wires, coils), aluminium (foil, aircraft sheeting), lead (roofing, weights), gold (jewellery — gold is the most ductile of metals, a single gram can be drawn into a wire 3 km long). Ductile materials have high toughness.

The practical importance of ductility is warning before failure. A concrete pillar in a building might crack audibly and visibly before collapsing; a glass pane snaps without notice. Engineers prefer materials that give visual and audible hints that something is going wrong, which is why mild steel and reinforced concrete are the backbone of Indian construction. A brittle material in a load-bearing role is dangerous precisely because it fails without warning.

The brittle–ductile transition

A single material can be brittle at one temperature and ductile at another. Steel at room temperature is ductile; cool it to the temperature of liquid nitrogen and it becomes brittle. This is the reason old cast iron radiators used to crack during Delhi's colder winters — the metal's brittle-to-ductile transition temperature was not far below room temperature, and a cold night could push it over.

The same material can also appear brittle under fast loading and ductile under slow loading. A piece of clay deforms plastically if you knead it slowly, but shatters if you hit it with a hammer — the loading rate, not just the material, matters. Non-Newtonian fluids like the cornflour-paste "oobleck" exploit this dramatically.

Hysteresis — energy lost on every cycle

For a perfectly Hookean material, loading and unloading follow the same straight line on the stress-strain plot — you put in energy, you get it all back. For most real materials at moderate strains, this is approximately true.

But for rubber and many polymers, it is not. The loading curve and the unloading curve take different paths. They form a closed loop, and the area inside the loop represents energy lost as heat. This phenomenon is called elastic hysteresis.

Hysteresis loop for rubber-like materials A closed loop on the stress-strain plot for rubber: the loading curve from origin rises convex-upward to a peak stress, and the unloading curve from the peak returns to origin below the loading curve. The enclosed area is shaded to represent energy dissipated as heat. strain ε stress σ loading unloading area = energy lost as heat
Rubber's hysteresis loop. The loading curve (red, upward arrow) lies above the unloading curve (dark, downward arrow). The difference in area between the two — the shaded region inside the loop — is the energy dissipated as heat during one loading-unloading cycle. The rubber warms up.

Why rubber has hysteresis

Rubber is made of long, tangled polymer chains. When you stretch the rubber, you straighten those chains, doing work against the many weak cross-links and friction-like internal forces among them. The stretching is fast; the chains don't rearrange instantaneously. When you release the load, the chains slide back — but not along the same path they took going out. The friction-like work on the return journey is lost to heat.

Quantitatively: the area under the loading curve is the total work you did to stretch the rubber. The area under the unloading curve is the work you got back. The difference — the enclosed loop area — is energy that left the rubber as heat.

Consequences and engineering uses

Steel has almost none

High-quality spring steel has a very narrow hysteresis loop — the loading and unloading curves are nearly identical lines. This is why a well-made clock can run for a week on one winding: very little of the energy stored in the spring is wasted to hysteresis heat; it is nearly all available to turn the gears.

Worked examples

Example 1: Energy stored in a stretched steel wire

A steel wire of length L = 4.0 m and cross-sectional area A = 2.0 \times 10^{-6} m² is stretched by \Delta L = 1.0 mm by a load. Given Y_{\text{steel}} = 2.0 \times 10^{11} Pa, find the elastic potential energy stored in the wire.

Stretched steel wire with a hanging load A vertical wire of length 4 m fixed at the top and stretched by 1 mm at the bottom, with a load hanging from its free end. Energy stored is marked as U equals 0.05 joules. L = 4.0 m F A = 2×10⁻⁶ m² ΔL = 1 mm U = ½ Y ε² V ≈ 0.025 J
A 4-m steel wire stretched by a millimetre under load stores a small but non-zero elastic energy — about 25 millijoules.

Step 1. Compute the strain.

\varepsilon = \frac{\Delta L}{L} = \frac{1.0 \times 10^{-3}}{4.0} = 2.5 \times 10^{-4}.

Why: strain is the dimensionless ratio of extension to original length. Always check it is small — here 0.025%, comfortably within the elastic regime for steel.

Step 2. Compute the wire's volume.

V = A L = (2.0 \times 10^{-6}) \times 4.0 = 8.0 \times 10^{-6} \text{ m}^3.

Step 3. Apply U = \tfrac{1}{2} Y \varepsilon^2 V.

U = \frac{1}{2} \times (2.0 \times 10^{11}) \times (2.5 \times 10^{-4})^2 \times (8.0 \times 10^{-6}).

Compute in pieces:

  • (2.5 \times 10^{-4})^2 = 6.25 \times 10^{-8}.
  • \tfrac{1}{2} \times 2.0 \times 10^{11} \times 6.25 \times 10^{-8} = 6.25 \times 10^{3} J/m³ (this is the energy density u).
  • Multiply by volume: 6.25 \times 10^{3} \times 8.0 \times 10^{-6} = 5.0 \times 10^{-2} J.
U = 5.0 \times 10^{-2} \text{ J} = 50 \text{ mJ}.

Why: the energy density came out to 6.25 kJ/m³, a useful intermediate — a cubic metre of this stretched steel would hold 6.25 kJ. The wire has only 8 cm³ of material, so the total is small — 50 millijoules.

Step 4. Sanity-check with the spring-like form.

The effective spring constant is k = YA/L = (2.0 \times 10^{11})(2.0 \times 10^{-6})/4.0 = 10^5 N/m. The force at full extension is F = k\Delta L = 10^5 \times 10^{-3} = 100 N. Then

U = \tfrac{1}{2}F\Delta L = \tfrac{1}{2} \times 100 \times 10^{-3} = 5.0 \times 10^{-2} \text{ J} \checkmark.

Both methods agree.

Result: U = 50 mJ — the energy stored in the stretched wire.

What this shows: A substantial mass (100 N \approx 10 kg) hanging on a 4-m steel wire stores only about a twentieth of a joule — which is why a snapped wire is not usually dangerous on its own; it is the load falling that causes injury, not the wire's stored energy.

Example 2: Energy per unit volume — rubber band versus steel spring

A rubber band has Y_{\text{rubber}} = 5 \times 10^6 Pa, and it can be stretched to a strain of \varepsilon = 1.0 (100% extension) before failing. A steel spring has Y_{\text{steel}} = 2 \times 10^{11} Pa, and its material can strain only up to \varepsilon = 0.002 before yielding. Compare the elastic energy per unit volume each can store at maximum useful strain.

Step 1. Use u = \tfrac{1}{2}Y\varepsilon^2 for each.

Rubber: u_{\text{R}} = \tfrac{1}{2} \times 5 \times 10^6 \times (1.0)^2 = 2.5 \times 10^6 J/m³ = 2.5 MJ/m³.

Steel: u_{\text{S}} = \tfrac{1}{2} \times 2 \times 10^{11} \times (0.002)^2 = \tfrac{1}{2} \times 2 \times 10^{11} \times 4 \times 10^{-6} = 4 \times 10^{5} J/m³ = 0.4 MJ/m³.

Why: use the Y\varepsilon^2 form because both Y and \varepsilon are given. The squared strain dominates the comparison — a small-strain material, even with huge Y, may store less per unit volume than a large-strain material with small Y.

Step 2. Take the ratio.

\frac{u_{\text{R}}}{u_{\text{S}}} = \frac{2.5 \times 10^6}{4 \times 10^5} = 6.25.

Rubber stores 6.25 times more energy per unit volume than steel, at each material's maximum useful strain.

Why: even though steel's Young's modulus is 40,000 times rubber's, rubber can strain 500 times further, and strain enters as the square. The end result is that rubber wins on energy storage per kilogram — which is why archers and catapult makers have preferred elastic materials like sinew and rubber over rigid springs.

Result: Rubber's energy density at maximum strain is \approx 2.5 MJ/m³; steel's is \approx 0.4 MJ/m³. Per unit volume, rubber stores six times more elastic energy than steel.

What this shows: Young's modulus alone tells you nothing about energy storage capacity — you also need the maximum useful strain. A slingshot uses rubber because rubber can absorb a lot of strain. A railway rail uses steel because steel resists any significant strain under the wheels of a train.

Example 3: A punt of stored energy — cricket bat handle and a rebound

A cricket bat's handle contains a rubber insert of volume V = 3.0 \times 10^{-5} m³ (30 cm³). During a powerful stroke, the handle flexes under a stress of \sigma = 2.0 \times 10^6 Pa. Taking the rubber's Young's modulus as Y = 5.0 \times 10^6 Pa, find the elastic energy stored in the rubber insert at peak stress and the fraction of a typical cricket-shot kinetic energy (100 J) this represents.

Step 1. Use u = \sigma^2/(2Y).

u = \frac{(2.0 \times 10^6)^2}{2 \times 5.0 \times 10^6} = \frac{4.0 \times 10^{12}}{10^7} = 4.0 \times 10^5 \text{ J/m}^3.

Why: the stress-form is most convenient because stress is given. Using strain would require first computing \varepsilon = \sigma/Y = 0.4 — a big strain, on the edge of validity for Hooke's law in rubber.

Step 2. Multiply by volume.

U = u \cdot V = 4.0 \times 10^5 \times 3.0 \times 10^{-5} = 12 \text{ J}.

Step 3. Compare to shot KE.

\frac{U}{\text{KE}_{\text{shot}}} = \frac{12}{100} = 0.12 = 12\%.

Why: at peak stress, the rubber insert stores about a tenth of the kinetic energy of a typical powerful cricket stroke. This energy is returned (via hysteresis, most but not all of it) as the handle flexes back, contributing to the bat's springback and the ball's speed off the bat.

Result: U \approx 12 J, or about 12% of the kinetic energy of a typical cricket shot.

What this shows: Manufactured rubber inserts in cricket bat handles and tennis racket grips are not decorative — they actively store and release mechanical energy during the stroke, with hysteresis losses damping the harsh vibrations that would otherwise hurt the player's hands. The engineering trade-off: more damping (more hysteresis) means more comfort but less energy return; less damping means a livelier bat but jarring vibrations on mishits.

Common confusions

If you came here to compute stored energy, read a stress-strain curve, and understand hysteresis, you have what you need. What follows is for readers who want the energy stored in non-uniform elastic bodies, the generalisation to three-dimensional deformation, and the thermodynamic origins of hysteresis.

Energy in a non-uniform elastic body

For a wire whose stress varies with position (say, a wire hanging under its own weight, where tension is zero at the bottom and maximum at the top), you cannot use U = \tfrac{1}{2}\sigma\varepsilon V with a single value of stress. Instead, integrate the energy density over the volume:

U = \int \tfrac{1}{2}\sigma(x)\varepsilon(x)\,dV.

For a uniform wire of length L and area A hanging under its own weight (density \rho), the tension at height x above the bottom is T(x) = \rho A g x (supporting the weight below). The stress is \sigma(x) = T(x)/A = \rho g x, and the strain is \varepsilon(x) = \sigma(x)/Y = \rho g x / Y. The total energy is

U = \int_0^L \tfrac{1}{2} \cdot \rho g x \cdot \frac{\rho g x}{Y} \cdot A\,dx = \frac{A\rho^2 g^2}{2Y}\int_0^L x^2\,dx = \frac{A\rho^2 g^2 L^3}{6Y}.

Why: the stress and strain both depend linearly on height, and the energy density is quadratic in stress — so the integral is \int x^2\,dx = L^3/3. Dividing by the prefactor's denominator gives a factor of 1/6 overall.

Compare: a uniformly stressed wire of the same total extension would have U = \tfrac{1}{2}YA(\Delta L)^2/L. The self-weight case stores less energy, because a lot of the wire is under low stress (near the bottom).

The three-dimensional tensor form

For a general three-dimensional elastic deformation, the stress and strain are rank-2 tensors \sigma_{ij} and \varepsilon_{ij} (where i, j run over the three spatial directions). The elastic energy density is

u = \tfrac{1}{2}\sigma_{ij}\varepsilon_{ij},

summed over all i, j pairs. For an isotropic material (same in all directions), this simplifies to

u = \tfrac{1}{2}\lambda\,\varepsilon_{kk}^2 + \mu\, \varepsilon_{ij}\varepsilon_{ij},

where \lambda and \mu are the Lamé parameters, related to Y and \nu by \mu = G = Y/(2(1+\nu)) and \lambda = Y\nu/((1+\nu)(1-2\nu)). The first term captures volumetric-change energy; the second captures shape-change (shear) energy.

For a simple wire stretched along one direction, only one diagonal component of strain is non-zero (to leading order, before Poisson contraction), and the tensor form reduces to the scalar u = \tfrac{1}{2}\sigma\varepsilon we derived.

Why hysteresis exists — a thermodynamic view

Consider a sample of rubber at temperature T. When you stretch it, you straighten polymer chains. This decreases the configurational entropy of the chains (straight chains have fewer accessible micro-states than tangled ones). By the second law, a decrease in entropy at constant temperature must be accompanied by heat flow out of the sample — the rubber should cool as you stretch it.

This is actually measurable: rapid stretching of rubber causes a small temperature rise (the Gough-Joule effect) because the heat doesn't have time to flow out. The return stroke, when released, cools the rubber slightly. Overall, during a complete loading-unloading cycle at constant overall temperature, the net heat flow out of the sample equals the enclosed hysteresis loop area.

From a more detailed perspective, rubber's polymer chains have many low-energy configurations separated by small barriers; during deformation, some chain segments cross those barriers and relax on timescales comparable to the experiment. The loading path includes the cost of crossing barriers; the unloading path does not retrace that cost exactly — energy is lost to the thermal motion of chain segments. This is why hysteresis is larger at higher strain rates (no time to relax) and can be tuned by changing the polymer's cross-link density.

The Mullins effect and stress softening

A rubber sample that has been stretched once and then released will, upon being stretched again to the same strain, require less stress than the first time. This is the Mullins effect — a kind of "break-in" behaviour. Some weak chain connections snap during the first stretch and do not reform. The stress-strain curve for the second cycle lies below the first, until the sample is stretched beyond the previous maximum — at which point it rejoins the original curve and the next snap-points begin.

For engineering design, this means the first few loading cycles of a rubber component are non-representative. Manufacturers pre-stretch elastomeric parts ("breaking them in") before quality-testing.

The relation to simple harmonic motion

A wire stretched within its elastic limit is a linear spring with effective spring constant k = YA/L. If you hang a mass m from it and displace the mass from equilibrium by a small amount and release, it undergoes simple harmonic motion with angular frequency

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{YA}{mL}}.

This is the frequency of longitudinal oscillations of the mass on the wire — the tiny bouncing you see when you hang a weight from a wire and let go. It is also the key to Young's modulus by resonance — an alternative experimental method to Searle's, in which one excites longitudinal vibrations in a rod and measures the resonant frequency; since \omega depends on Y, the modulus can be read off. This is the method used in the resonance column experiment taught in many Indian physics labs.

Where this leads next