Elasticity: Hooke's Law, Stress and Strain

Beginner

Squeeze a sponge, stretch a rubber band or bend a plastic ruler, then let go: each one springs back.

Step 1. Elastic or not? A material is elastic if it returns to its original shape once the force is removed. By this test, steel is more elastic than rubber: it is much harder to deform, but it springs back almost perfectly. Modelling clay and dough are the opposite: they keep whatever shape you press into them.

Step 2. Hooke's rule. Hang a weight on a spring and it stretches a little. Hang a second, identical weight and it stretches by the same amount again: twice the weight, twice the stretch. Robert Hooke (1678) wrote it as $F=kx$: the pull $F$ equals a number $k$ times the stretch $x$.

Step 3. Stiffness. The number $k$ tells how stiff the spring is. A toy spiral spring is very soft (small $k$); a car's suspension spring is very stiff (large $k$), so the same load stretches it far less.

Step 4. Energy and limits. A stretched spring stores energy and gives it back when released: that is how a slingshot or a trampoline works. But pull too hard and the spring passes its elastic limit: it stays longer for good and never returns to its original length.

Check yourself: a spring stretches 2 cm under a 5 N weight. How far does it stretch under 15 N, below its elastic limit? Answer: 6 cm, three times the force, three times the stretch.

In the animation, equal weights are hung one at a time. Each one moves the red pointer down the ruler by the same step (the green marks), and the green bars grow in equal steps. When the weights are lifted off, the spring returns exactly to the dashed line where it started.

Intermediate

Load, extension and the spring constant

Hang a load on a spring and it gets longer. What matters is the change in length, the extension $x=L-L_0$, where $L_0$ is the natural length and $L$ the length under load; in the animation it is measured from the dashed line. A squeezed spring has a negative $x$.

Robert Hooke published in 1678 that, up to a limit, the extension is proportional to the force that causes it: double the load and the extension doubles.
$F=kx$

The constant $k$ is the spring constant, or stiffness, in newtons per metre (N/m): the force per metre of extension, found by dividing a load by the extension it produces:
$k=\frac{F}{x}$
A toy spiral spring has about 1 N/m, the spring in the animation 25 N/m and a car's suspension spring some 30 000 N/m.

The spring pulls back with a restoring force of the same size, directed against the extension: $F=-kx$ when the sign matters. A mass hanging at rest is in equilibrium, so $kx=mg$, which is how a spring balance weighs things. In the animation the blue $kx$ arrow and the red $F$ arrow always have the same length; while a weight is being lowered, the grey tray arrow carries the rest of its weight.

Energy and the force–extension graph

Plotted against the extension, the load gives a straight line through the origin whose slope is $k$: steep for a stiff spring, shallow for a soft one. Points stay on the line only up to the elastic limit. Below it the spring returns exactly to its natural length when unloaded; beyond it the line bends over and the spring stays longer for good, a plastic deformation.

The area under the line is the work done in stretching, stored as elastic potential energy. The force grows steadily from 0 to $kx$, so its average is half the final value and the area is a triangle:
$U=\frac{1}{2}k{x}^2=\frac{1}{2}Fx$
Twice the stretch stores four times the energy, and a bow or a slingshot gives it back when released.

Springs in series and in parallel

Hook two springs one below the other, in series, and each carries the whole load, so the extensions add: $x=F/k_1+F/k_2$. Dividing by $F$ gives
$\text{series}:\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$

Hang the load from two springs side by side, in parallel, joined by a bar that keeps their stretches equal, and they share the load, so their forces add:
$\text{parallel}:k=k_1+k_2$

Two springs like the one in the animation give 12.5 N/m in series and 50 N/m in parallel: a series pair is softer than either spring, a parallel pair stiffer. A spring is also two halves in series, so each half of the animation's spring would have 50 N/m.

Worked example: the spring in the animation

Each weight in the animation weighs 1.0 N (a mass of $1.0/9.81\approx 0.102\mathrm{kg}$) and moves the pointer 4.0 cm down the ruler. From the first row of the table,
$k=\frac{F}{x}=\frac{1.0\mathrm{N}}{0.040\mathrm{m}}=25N/m$

The other rows, from 2.0 N for 0.08 m to 4.0 N for 0.16 m, give the same ratio in the last column: that constant ratio is Hooke's law. With all four weights, $x=F/k=4.0/25=0.16\mathrm{m}$. These are the $F$ and $x$ readouts once all four weights hang, 4.0 N and 16 cm, and the green $x$ arrow spans 16 cm of the ruler.

The stored energy is then $U=\frac{1}{2}\cdot 25\cdot 0.16^2=0.32\mathrm{J}$, also $\frac{1}{2}\cdot 4.0\cdot 0.16$. After one, two and three weights it was 0.02, 0.08 and 0.18 J: each weight adds the same 4 cm, but the fourth adds 0.14 J, seven times as much as the first, because the spring already pulls with 3.0 N.

When the weights are lifted off, $x$ returns to zero and the energy comes back, since 4.0 N is below the elastic limit of 5.5 N (22 cm). A fifth weight would still be safe; a sixth would not.

Stress, strain and Young's modulus

The spring constant belongs to one object, not to its metal. A wire twice as long stretches twice as far under the same load, and a wire with twice the cross-sectional area needs twice the force for the same stretch. To describe the material, divide the force by the area and the stretch by the original length:
$\sigma =\frac{F}{A},\epsilon =\frac{\Delta L}{L_0}$

The stress $\sigma$ is in pascals ($1\mathrm{Pa}=1N/{\mathrm{m}}^2$, and $1\mathrm{MPa}=1N/m{\mathrm{m}}^2$); the strain $\epsilon$ is a pure number, often given in percent. In the linear region they are proportional, and their ratio is Young's modulus:
$E=\frac{\sigma }{\epsilon }$

It is about 200 GPa for steel, 70 GPa for aluminium and glass, 10 GPa for wood along the grain and only 0.01–0.1 GPa for rubber. Together the three relations give the stretch of any wire or rod:
$\Delta L=\frac{F{L}_0}{AE}$
A uniform wire is therefore a spring with $k=AE/L_0$: longer is softer, thicker is stiffer.

Worked example: a steel wire under load

A steel wire 2.0 m long and 1.0 mm in diameter holds a 10 kg mass. How far does it stretch? The load is the weight, $F=mg=10\cdot 9.81=98.1\mathrm{N}$, and the cross-section is a circle:
$A=\frac{\pi d^2}{4}=\frac{\pi \cdot (1.0\times 10^{-3}{)}^2}{4}=7.85\times 10^{-7}{\mathrm{m}}^2$

The stress is $\sigma =98.1/(7.85\times 10^{-7})=1.25\times 10^8\mathrm{Pa}$, that is 125 MPa. Dividing by $E=200\mathrm{GPa}$ gives the strain, $\epsilon =1.25\times 10^8/(2.0\times 10^{11})=6.25\times 10^{-4}$, about 0.06 %. The stretch is
$\Delta L=\epsilon L_0=6.25\times 10^{-4}\cdot 2.0=1.25\times 10^{-3}\mathrm{m}$
or 1.25 mm.

As a spring the wire has $k=AE/L_0\approx 7.9\times 10^{4}N/m$, some 3000 times stiffer than the animation's spring. Its stress of 125 MPa is about half the yield stress of even ordinary structural steel, roughly 250 MPa, so it springs back fully when unloaded. A wire twice as thick, with four times the area, would stretch only about 0.31 mm.

Common mistakes

• Using the length instead of the extension: a 20 cm spring stretched to 24 cm has $x=4\mathrm{cm}$, not 24 cm.
• Mixing centimetres and metres: 1.0 N over 4.0 cm gives $1.0/0.040=25N/m$, not $1.0/4.0=0.25$.
• Using the mass as the force: a 0.20 kg mass pulls with its weight, $0.20\cdot 9.81=1.96\mathrm{N}$, not with 0.20 N.
• Forgetting the one half: at 16 cm the animation's spring stores $\frac{1}{2}Fx=0.32\mathrm{J}$, not $Fx=0.64\mathrm{J}$, because the force was smaller earlier in the stretch.
• Thinking that springs in series are stiffer: two 25 N/m springs one below the other give 12.5 N/m, because each one carries the whole load.

Where you meet it

Spring balances, car suspensions, mattresses, watch hairsprings and the spring in a ballpoint pen all rely on $F=kx$. Electronic scales use strain gauges, metal foils whose resistance changes as they stretch, sensitive to strains of a few millionths. Bridges, cranes and aircraft are designed to keep every stress well inside the linear region.

A mass pulled down on a spring and released oscillates, the subject of Simple Harmonic Motion and then of Damping and Resonance. The speed of sound in a solid is set by its Young's modulus (Wave Basics), and a rail that cannot expand as it warms comes under stress (Thermal Expansion).

Advanced

Why small deformations are linear

Legend

$U$	Potential energy of the pair of atoms: the solid curve, measured from an arbitrary zero.	$\mathrm{J}$
$x$	Spacing between the two atoms.	$\mathrm{m}$
$x_0$	Equilibrium spacing, at the bottom of the well.	$\mathrm{m}$
$\frac{1}{2}k(x-x_0{)}^2$	Energy of an ideal spring with the same bottom and curvature: the dashed parabola.	$\mathrm{J}$
$k$	Stiffness of that spring, equal to the curvature $U^{\prime \prime }(x_0)$ of the well at its bottom.	$N/m$

Hooke's law says that the force needed to deform a body is proportional to the deformation. It is not a fundamental principle like the conservation of energy. It is what almost any smooth force looks like when the displacement is small, and Figure 1 shows why.

The solid curve $U$ is the potential energy of two neighbouring atoms in a solid, plotted against their spacing $x$. The atoms attract each other at long range and repel each other steeply at short range; both effects are electrical. Together they make a well whose lowest point is the equilibrium spacing $x_0$. On the left, where the atoms are pushed together, the curve climbs steeply. On the right, where they are pulled apart, it flattens out, and far enough out the bond lets go.

The dashed parabola $\frac{1}{2}k(x-x_0{)}^2$ is the energy of an ideal spring with the same bottom and the same curvature. Look first near $x_0$: there the two curves are hard to tell apart. Then look at the two dashed verticals, at equal distances either side of $x_0$. The dots on them show where the curves separate: on the squeezed side (left) the real curve lies above the parabola, on the stretched side (right) below it.

To make this precise, expand $U$ as a Taylor series about the minimum:
$U(x)=U(x_0)+U^{\prime }(x_0)(x-x_0)+\frac{1}{2}{U}^{\prime \prime }(x_0)(x-x_0{)}^2+\frac{1}{6}{U}^{\prime \prime \prime }(x_0)(x-x_0{)}^3+\cdots$

Each term carries a higher power of the displacement $x-x_0$, so for small displacements each is much smaller than the one before. The first term is a constant and exerts no force. The second is zero, because the slope of $U$ vanishes at a minimum: $U^{\prime }(x_0)=0$. Differentiating what is left gives the force $F=-dU/dx$:
$F=-k(x-x_0)-\frac{1}{2}{U}^{\prime \prime \prime }(x_0)(x-x_0{)}^2,k=U^{\prime \prime }(x_0)$

The first term is Hooke's law. For small enough displacements it beats the second, whatever the shape of the well, which is why one law covers a molecular bond, a steel cable and a bridge. The stiffness $k$ is the curvature of the well at its bottom, and $k>0$ is exactly the condition for a stable equilibrium: displaced either way, the atoms are pushed back.

The second term measures how lopsided the well is: it is the gap between the two curves of Figure 1. A real bond is harder to squeeze than to stretch, so $U^{\prime \prime \prime }(x_0)<0$, and the extra force $-\frac{1}{2}{U}^{\prime \prime \prime }(x_0)(x-x_0{)}^2$ is positive on both sides. When the bond is stretched ($x>x_0$) it works against the restoring pull, so the bond softens; when the bond is squeezed ($x<x_0$) it adds to the restoring push, so the bond stiffens. The correction is small: steel stays elastic only up to strains (fractional changes of length) of about 0.1 %, and there it is a fraction of a percent. Yet the same asymmetry is what makes solids expand when heated.

From one spring to a material

The constant $k$ belongs to one particular spring or bar. To describe a material we need a constant that does not depend on the size of the sample, and two thought experiments show what to divide by.

Put two identical blocks end to end and pull. Each block carries the same force $F$, so each stretches by the same amount, and the pair stretches twice as far as one block: this is the series rule. The extension grows in proportion to the length, so the measure of deformation that does not depend on length is the strain, the extension per unit original length, $\epsilon =\Delta L/L_0$.

Now put the two blocks side by side. To give them the same stretch takes twice the force: this is the parallel rule. The force grows in proportion to the cross-sectional area $A$, so the measure of load that does not depend on area is the stress, the force per unit area, $\sigma =F/A$.

For a uniform bar, stress is proportional to strain, and the constant of proportionality $E$, Young's modulus, belongs to the material alone. Rearranged, the same law gives the stiffness of the bar:
$\frac{F}{A}=E\frac{\Delta L}{L_0},k=\frac{AE}{L_0}$

A short, thick bar is stiff and a long, thin one is soft, yet one value of $E$ describes both if they are made of the same steel.

From atomic bonds to Young's modulus

Legend

$a$	Spacing between neighbouring atoms.	$\mathrm{m}$
$k_b$	Stiffness of one bond, drawn as a spring.	$N/m$
$L_0$	Length of the bar.	$\mathrm{m}$
$A$	Area of the end face the load acts on (the plate, seen edge-on): the cross-section of the bar.	${\mathrm{m}}^2$
$F$	Load pulling on the end face of the bar.	$\mathrm{N}$

Young's modulus can be estimated from the atoms themselves, by applying the same series and parallel counting. Figure 2 models a crystal as a cubic stack of atoms a distance $a$ apart, every bond a spring of stiffness $k_b$. The bar, of length $L_0$, is held by the wall on the left and pulled by the force $F$ on the right. The dark plate is its end face, of area $A$, seen edge-on.

First count across the bar. Each atom takes up a square of side $a$ of the end face, so $A/a^2$ chains of atoms run side by side along the bar (three in the sketch). Then count along one chain: it is a series of $L_0/a$ bonds (five in the sketch).

Now combine the two counts. In series, $L_0/a$ equal springs share the extension, so a chain is softer than one bond by that factor: its stiffness is $k_b/(L_0/a)=k_{b}a/L_0$. In parallel, the chains share the load, so their stiffnesses add:
$k=\frac{A}{a^2}\cdot \frac{k_{b}a}{L_0}=\frac{A}{L_0}\cdot \frac{k_b}{a}$

Comparing this with $k=AE/L_0$ gives $E=k_b/a$: the modulus is the stiffness of one bond divided by the spacing of the atoms.

For steel, $E=200\mathrm{GPa}$ and $a\approx 0.25\mathrm{nm}$ give $k_b=Ea\approx 50N/m$, just twice the 25 N/m of the animation's spring. A single bond is not especially stiff. The modulus is huge only because so many chains pull together: $1/a^2$, some $1.6\times 10^{19}$ of them, cross each square metre.

Stored energy as the area under the line

Back to a single spring, with $x$ now standing for its extension from the natural length. Stretching it from 0 to $x$ against its own pull $k{x}^{\prime }$ takes the work
$U={\int }_0^{x}k{x}^{\prime }d{x}^{\prime }=\frac{1}{2}k{x}^2$

On the animation's $F$–$x$ graph this is the hatched triangle under the line, with base $x$ and height $kx$. As the red dot climbs, the $U$ readout passes 0.02 J at 1 N, 0.08 J at 2 N, 0.18 J at 3 N and 0.32 J at 4 N.

Look at the increments rather than the totals: 0.02, 0.06, 0.10 and 0.14 J, in the ratio 1 : 3 : 5 : 7. Each newton of load adds 4 cm of extension, and each 4 cm strip of the triangle is a trapezium whose average height is 0.5, 1.5, 2.5 and 3.5 N. Later strips cost more because the spring is already pulling when they start. For the same reason, stretching from $x_1$ to $x_2$ costs $\frac{1}{2}k(x_2^2-x_1^2)$, not $\frac{1}{2}k(x_2-x_1{)}^2$.

For a bar the same triangle gives $U=\frac{1}{2}F\Delta L=\frac{1}{2}(\sigma A)(\epsilon L_0)$. Dividing by the volume $A{L}_0$, and using $\epsilon =\sigma /E$, gives the energy per unit volume, which is the area under the stress–strain line:
$u=\frac{1}{2}\sigma \epsilon =\frac{{\sigma }^2}{2E}$

Steel at its yield stress of about 250 MPa, the highest stress it takes elastically, holds $u=(2.5\times 10^8{)}^2/(2\cdot 2.0\times 10^{11})\approx 1.6\times 10^{5}J/{\mathrm{m}}^3$. With a density of about 7800 kg/m³, that is about 20 J per kilogram, enough to lift that kilogram by only 2 m. Springs store energy conveniently, not densely.

Sideways shrinkage, shear and coil springs

A stretched bar also gets thinner. Its width shrinks by the fraction $\nu \epsilon$, where $\nu$, Poisson's ratio, is about 0.3 for metals and close to 0.5 for rubber.

Because the laws are linear, the effects of separate loads simply add. Squeeze a cube with a pressure $p$ on all six faces: that is three compressions at once, one along each axis. Each one shortens the cube along its own axis by the strain $p/E$ and, through Poisson's ratio, lengthens it along the other two axes by $\nu p/E$. Along any one axis the net strain is therefore $-(p/E)(1-2\nu )$. The fractional change of volume is the sum of the three strains, $-3(p/E)(1-2\nu )$, and the bulk modulus, the pressure divided by the fractional loss of volume, is
$K=\frac{E}{3(1-2\nu )}$

A positive $K$ requires $\nu <1/2$. At $\nu =1/2$ a stretch leaves the volume unchanged, which is why rubber is so hard to compress.

A third modulus describes shear, a change of shape at constant volume, as when the top of a block is pushed sideways over its base. For an isotropic material, one with the same properties in every direction, it is not independent: two constants describe every small deformation. Given $E$ and $\nu$, the shear modulus is
$G=\frac{E}{2(1+\nu )}$
about 80 GPa for steel.

Shear is what matters in a coil spring. Pulling on a coil spring hardly lengthens its wire; it twists it. For wire diameter $d$, coil diameter $D$ and $n$ turns, the twisting of the wire gives
$k=\frac{G{d}^4}{8D^{3}n}$

A steel wire 0.5 mm thick, wound into 25 turns of 10 mm diameter, gives $k=8.0\times 10^{10}\cdot (5\times 10^{-4}{)}^4/(8\cdot 0.01^3\cdot 25)=25N/m$: the animation's spring. Note the strong dependence on $D$: coils 1.4 times wider would make the spring $1.4^3\approx 2.7$ times softer. So $k$ describes one spring, while $E$, $\nu$ and $G$ describe the steel it is made of.

Beyond the elastic limit

Legend

$\sigma$	Stress: the force divided by the original cross-section of the wire.	$\mathrm{Pa}$
$\epsilon$	Strain: extension per unit original length.
$E$	Young's modulus: the slope of the straight start, and of the unloading line.	$\mathrm{Pa}$
${\sigma }_y$	Yield stress, where the straight line ends and the wire starts to flow.	$\mathrm{Pa}$
${\sigma }_u$	Ultimate tensile strength: the highest stress on the curve.	$\mathrm{Pa}$
${\epsilon }_p$	Permanent strain left after unloading.
$u$	Energy absorbed per unit volume up to the break: the shaded area under the curve.	$J/{\mathrm{m}}^3$

Everything so far assumed small strains. Figure 3 shows what happens when a metal wire is pulled further: its stress $\sigma$ against its strain $\epsilon$, all the way to fracture. Here $\sigma$ is the force divided by the original cross-section of the wire.

Start at the origin. The stress rises along a straight line of slope $E$: this is the region of Hooke's law. The line ends at the dot level with ${\sigma }_y$, the yield stress (the purple dot of the animation's wire graph). Beyond it the wire flows. Dislocations, line defects in the crystal, start to glide, so planes of atoms slip past each other one row at a time instead of all at once. That is why real metals yield so early: a perfect crystal would hold out to several gigapascals, while real steels yield at 0.25–2 GPa.

The animation's $F$–$x$ graph for the spring has the same shape: its straight line ends at the elastic limit, the purple dot at 5.5 N and 22 cm, where the dashed curve bends over.

Now follow the curve past ${\sigma }_y$. It keeps rising, but more gently, because the dislocations tangle and block one another; this is called work hardening. The top of the curve is the ultimate tensile strength ${\sigma }_u$. Past it a neck forms: one short stretch of the wire thins faster than the rest. The force, and with it the plotted stress, falls, even though the true stress in the shrinking neck still rises, and the wire breaks at the cross (×).

Next, stop loading part way up the curve, at the dot on its rising part, and unload. The wire comes down along the arrow, a line parallel to the elastic one, with the same slope $E$, because only the elastic part of the strain is recovered. At zero stress it keeps the permanent strain ${\epsilon }_p$. Loaded again, it climbs the same line and rejoins the curve: it is now elastic up to a higher stress than before.

The shaded area under the whole curve, $u=\int \sigma d\epsilon$, is the energy absorbed per unit volume before fracture; in the straight part it is the triangle $\frac{1}{2}\sigma \epsilon$ met earlier. A large area means a tough material. Ductile metals absorb a great deal of energy by flowing before they break, while glass or cast iron break near the end of the straight part. The animation's spring stops at 4.0 N, 73 % of its 5.5 N limit, so on unloading it retraces the line and all 0.32 J comes back.

Worked problem: letting the load drop

Legend

$F$	Force; also the load itself, the horizontal line: the weight of the four masses hung on together.	$\mathrm{N}$
$x$	Extension of the spring from its natural length.	$\mathrm{m}$
$kx$	Pull of the spring at the extension $x$: the sloped line.	$\mathrm{N}$
$x_e$	Extension where the pull equals the load, $F/k$: the equilibrium of gentle loading.	$\mathrm{m}$
$x_L$	Extension at the elastic limit (the purple dot); beyond it the spring's line is dashed.	$\mathrm{m}$
$x_m$	Extension at the lowest point, where the weights stop: $2F/k$.	$\mathrm{m}$
$K$	Kinetic energy of the weights: the left triangle is the energy gained up to $x_e$; the right triangle, $-K$, is the same amount taken back.	$\mathrm{J}$

The tray in the animation lowers each weight gently, so the spring is in equilibrium at every instant. Suppose instead that all four weights, 4.0 N in all, were hooked on together at the natural length and let go. Figure 4 follows the fall on a graph of force against extension $x$.

Two forces act on the weights. The load $F$, their weight, is the same at every extension: the horizontal line. The pull $kx$ of the spring grows with the extension: the sloped line. The two lines cross at $x_e=F/k=4.0/25=0.16\mathrm{m}$, the 16 cm of gentle loading.

Left of $x_e$ the load is the larger force, so the weights speed up. The net force is the gap between the lines, and the work it does, the area of the left triangle, becomes their kinetic energy $K$. It is largest at $x_e$, where it equals $\frac{1}{2}F{x}_e$.

Right of $x_e$ the spring pulls harder than the load, so the weights slow down and the spring takes the energy back: the right triangle, marked $-K$. Its area equals that of the left one, because the pull climbs from $F$ to $2F$ over the same distance $x_e$ it took to climb from 0 to $F$. So the kinetic energy is zero again at $x_m=2x_e$, where the weights stop for an instant.

Energy gives the same result directly. Neglecting the spring's mass and air resistance, at the lowest point the weights are momentarily at rest, so all the energy gravity has released, $F{x}_m$, is stored in the spring:
$F{x}_m=\frac{1}{2}k{x}_m^2,x_m=\frac{2F}{k}=\frac{2\cdot 4.0}{25}=0.32\mathrm{m}$

That is twice the 16 cm of gentle loading and far past the 22 cm elastic limit $x_L$, the purple dot of Figure 4: the spring would be stretched for good. A sudden load doubles the peak deformation, so the largest load this spring survives when dropped from its natural length is half its static limit, $25\cdot 0.22/2=2.75\mathrm{N}$.

An ideal spring would oscillate between 0 and 32 cm about 16 cm. The mass is $m=4.0/9.81=0.408\mathrm{kg}$, so $\omega =\sqrt{k/m}=\sqrt{25/0.408}\approx 7.83rad/s$ and the period is $2\pi /\omega \approx 0.80\mathrm{s}$. The speed at the centre follows from energy. At 16 cm gravity has released $4.0\cdot 0.16=0.64\mathrm{J}$ and the spring holds 0.32 J, so the other 0.32 J is kinetic energy: $v=\sqrt{2\cdot 0.32/0.408}\approx 1.25m/s$, equal to $\omega \cdot 0.16\mathrm{m}$, as simple harmonic motion requires.

Gentle loading avoids all this because the tray absorbs the surplus: 0.32 J if all four weights were lowered at once. Loading in four steps, as the animation does, wastes less. In each step the tray's push starts at 1.0 N, the weight just added, and falls to zero over 4 cm, so the tray absorbs only $\frac{1}{2}\cdot 1.0\cdot 0.04=0.02\mathrm{J}$ per step. Over the four steps gravity releases 0.40 J (the weights fall 16, 12, 8 and 4 cm), the spring keeps 0.32 J and the tray takes 0.08 J. With ever smaller steps the loss vanishes.

Limits of validity

Hooke's law is a small-strain law. Mild steel stays linear only up to a strain of $250\mathrm{MPa}/200\mathrm{GPa}=1.25\times 10^{-3}$, about 0.13 %, and window glass breaks near 0.1 %.

Rubber stretches to several times its length, but past a few percent of strain its stress–strain curve is S-shaped, and its tension has a different origin. Rubber is a tangle of long chain molecules. Stretching it straightens them, and their thermal jiggling pulls them back towards tangled shapes, of which there are far more. This is an entropy effect, and it has a surprising consequence: a stretched rubber band pulls harder when heated.

Materials also depend on time. Polymers, concrete and hot metals creep: under a constant load they keep slowly deforming. Many materials also unload along a different curve from the one they loaded on. The loop between the two curves is called hysteresis, and its area is energy turned into heat in each cycle, which is why a rubber ball never bounces back to its starting height.

The spring model has limits of its own. A real spring of mass $m_s$ oscillates as if about $m_s/3$ were added to the load. Many tension springs are wound with an initial tension $F_0$: they do not stretch at all until the load exceeds $F_0$, and then follow $F=F_0+kx$. The animation's spring has no initial tension, so its line passes through the origin.

Where this leads

A mass on a spring obeys $m\ddot{x}=-kx$, the equation of Simple Harmonic Motion: the animation's 0.41 kg load would oscillate with a period of 0.80 s, and Damping and Resonance adds the internal losses. In a rod, each slice is a mass between springs, and longitudinal waves travel along it at $\sqrt{E/\rho }$, with $\rho$ the density: about 5000 m/s in steel, nearly 15 times the speed of sound in air (Wave Basics).

The lopsided bond of Figure 1 leads to Thermal Expansion. A steel rail clamped at both ends and warmed by 30 K cannot lengthen, so it comes under the stress that would squeeze it back by its free expansion: $\sigma =E\alpha \Delta T=2.0\times 10^{11}\cdot 1.2\times 10^{-5}\cdot 30\approx 7.2\times 10^7\mathrm{Pa}$, or 72 MPa, with $\alpha$ the expansion coefficient of steel. Finally, water's bulk modulus of about 2.2 GPa means that squeezing it by 1 % takes 22 MPa, some 220 atmospheres, which is why Fluid Pressure and Pascal's Principle can treat it as incompressible.