Thermal Expansion

Beginner

Step 1. Almost everything grows a little when it warms up and shrinks again when it cools, which is why concrete pavement slabs are laid with thin gaps between them. This is thermal expansion.

Step 2. Why? An object is made of tiny particles that are always jiggling. Heat it and they jiggle harder, nudging one another slightly further apart, so the whole object becomes a little longer, wider and thicker.

Step 3. How much? The growth is small but very regular: warm a rod twice as much and it grows twice as much; take a rod twice as long and it also grows twice as much. A 1 m aluminium bar heated by 200 degrees grows by less than 5 mm, yet a long steel bridge changes length by tens of centimetres between winter and summer. The pushing force is enormous, so bridges get toothed expansion joints.

Step 4. Different materials grow by different amounts. Bond a strip of brass to a strip of steel and heat them: the brass grows more, so the pair has to curl, with the brass on the outside of the bend. Such bimetallic strips switch thermostats and kettles on and off.

In the animation, flames heat a metal bar and a brass–steel strip. The bar's end moves past the dashed line (its growth is drawn 25 times bigger) while the strip curls upwards; when the flames go out, both return to their original shape.

Check yourself: a metal lid is stuck on a glass jar. Should you run hot water over the lid or over the glass? (Over the lid: metal expands more than glass, so the lid loosens.)

Water is a strange exception: between 0 °C and 4 °C it shrinks as it warms.

Intermediate

What expands, and why

Power lines sag lower on hot days, because a warm wire is slightly longer than a cold one. In a solid, atoms sit at almost fixed distances from their neighbours, held by bonds that act like stiff springs. They vibrate about those positions, and heating makes them vibrate harder. The bonds resist squeezing more than stretching, so an atom that shakes harder spends more time slightly further out, and the average spacing grows. Every length of the object grows with that spacing.

Three quantities describe the effect: the original length $L_0$, the change in length $\Delta L=L-L_0$ (negative on cooling) and the temperature change $\Delta T$. A step of 1 °C is the same size as a step of 1 K, so $\Delta T$ has the same value on both scales: warming from 20 °C to 220 °C is $\Delta T=200\mathrm{K}$.

The law of linear expansion

Measurements give a simple rule: the change in length is proportional to the original length and to the temperature change,
$\Delta L=\alpha L_0\Delta T$

Each metre of a rod grows on its own, so a rod twice as long grows twice as much. The coefficient of linear expansion $\alpha$ belongs to the material. Its unit is K⁻¹ because it is the fractional growth $\Delta L/L_0$ per kelvin: $\alpha =23\times 10^{-6}{\mathrm{K}}^{-1}$ means that each metre of aluminium gains 23 micrometres per kelvin.

Typical values, in units of $10^{-6}{\mathrm{K}}^{-1}$, are: invar (an iron–nickel alloy) 1.2, glass 9, steel and concrete about 12, copper 17, brass 19, aluminium 23. Since $\alpha \Delta T$ stays well under 1 % in everyday heating, the change is easy to overlook but hard to stop.

Areas, volumes and liquids

Heating multiplies every length of a uniform plate by $(1+\alpha \Delta T)$, so its area is multiplied by $(1+\alpha \Delta T{)}^2\approx 1+2\alpha \Delta T$; the dropped term $(\alpha \Delta T{)}^2$ is negligible. A block's volume is multiplied by the cube. To first order, area and volume therefore grow by
$\Delta A\approx 2\alpha A_0\Delta T$
$\Delta V=\beta V_0\Delta T,\beta \approx 3\alpha$

Here $\beta$ is the coefficient of volume expansion. Holes scale too: the opening of a ring grows exactly as a disc of the same material would, like a photographic enlargement.

Liquids, having no shape of their own, are described by $\beta$ alone. In units of $10^{-6}{\mathrm{K}}^{-1}$ it is about 210 for water at 20 °C, 180 for mercury and 950 for petrol, against 27 for glass. A liquid-in-glass thermometer reads this difference: the liquid outgrows its bulb and climbs the narrow tube. It is also why a tank filled to the brim with cool petrol overflows in the sun.

Worked example: the bar and the strip in the animation

The aluminium bar has $L_0=1\mathrm{m}=1000\mathrm{mm}$ and is heated from 20 °C to 220 °C, so $\Delta T=200\mathrm{K}$ and
$\Delta L=23\times 10^{-6}{\mathrm{K}}^{-1}\cdot 1000\mathrm{mm}\cdot 200\mathrm{K}=4.6\mathrm{mm}$

This is the blue $\Delta L$ readout at 220 °C: the bar is now 1.0046 m long. The growth and its arrow are drawn 25 times larger, since at true scale they would be invisible.

The strip is 10 cm of brass ($\alpha =19\times 10^{-6}{\mathrm{K}}^{-1}$) bonded to 10 cm of steel ($\alpha =12\times 10^{-6}{\mathrm{K}}^{-1}$). Free, the brass would grow by $19\times 10^{-6}\cdot 100\mathrm{mm}\cdot 200=0.38\mathrm{mm}$ and the steel by 0.24 mm, a mismatch of only 0.14 mm. Bonded, the layers cannot slide, so the strip must curve, with the longer brass on the outside of the bend. Here the brass is underneath, so the strip curls upwards.

It bends into an arc of radius $R\approx 25\mathrm{cm}$, derived at the expert level. An arc of length $L$ rises above its starting line by about $L^2/(2R)$, so the tip of the 100 mm strip lifts by $d\approx 100^2/(2\cdot 246)\approx 20\mathrm{mm}$: the green $d$ arrow and readout. A 0.14 mm mismatch has become a 20 mm movement: that is what makes the strip a good switch.

Worked example: a bridge and its expansion joint

A steel bridge deck 120 m long is fixed at one end and rests on sliding bearings at the other. Built at 15 °C, it reaches −20 °C on winter nights and +45 °C in summer sun. How much movement must the joint at the free end allow?

With $\alpha =12\times 10^{-6}{\mathrm{K}}^{-1}$ and $L_0=120000\mathrm{mm}$, each kelvin changes the length by $\alpha L_0=1.44\mathrm{mm}$. In summer, with $\Delta T=+30\mathrm{K}$, the deck grows by $1.44\cdot 30\approx 43\mathrm{mm}$; in winter, with $\Delta T=-35\mathrm{K}$, it shrinks by $1.44\cdot 35\approx 50\mathrm{mm}$. The joint must cover about 94 mm, hence its long interlocking steel teeth.

Without a joint, the abutments would have to squeeze the summer growth back out, a strain of $\alpha \Delta T=3.6\times 10^{-4}$. With Young's modulus $E=200\mathrm{GPa}$ that takes a stress $\sigma =E\alpha \Delta T=72\mathrm{MPa}$, roughly a quarter of the stress at which structural steel begins to yield, whatever the length of the deck.

Reading the animation

The dashed lines mark the cold shapes. The bar is fixed to the wall and rests on a roller, like a bridge deck on its bearings, so all its growth appears at the free end; the strip is clamped at one end only, so it is free to bend.

$\Delta L$ and $d$ are both proportional to the rise above 20 °C: halfway, at 120 °C, they read 2.30 mm and about 10 mm. Each kelvin adds 0.023 mm to the bar and about 0.1 mm to the strip's deflection. On cooling, the readouts run back through the same values to zero: thermal expansion is reversible, unless something has been stressed beyond its elastic limit.

Common mistakes

• A hole in a heated plate grows like every other length: a 20.00 mm hole in steel warmed by 100 K becomes 20.02 mm, which is how rings are shrink-fitted onto shafts.
• The atoms do not swell; their average spacing grows, by 0.46 % in the animation's bar.
• Use the change of temperature, not the temperature: 20 °C to 30 °C gives the same growth as 80 °C to 90 °C, and a change of 200 °C is 200 K, with no 273 added.
• Areas grow with $2\alpha$ and volumes with $3\alpha$: an aluminium block heated by 200 K gains 1.38 % in volume, three times the 0.46 % of each edge.
• Not everything expands when heated: water contracts as it warms from 0 °C to 4 °C, and a stretched rubber band shortens.

Where you meet it

Long structures get room to move: joints and roller bearings on bridges, gaps between concrete slabs, loops in pipelines, and power lines that sag lower on hot days. Materials that must stay joined, such as glass sealed around the wires of a lamp or a filling in a tooth, are chosen with matching coefficients, while bimetallic strips use a mismatch on purpose to switch kettles, irons and thermostats.

This lesson builds on Temperature and Heat. Water's odd behaviour returns in Phase Changes and Latent Heat: it expands by 9 % on freezing. Blocked expansion obeys the stress–strain law of Elasticity: Hooke's Law, Stress and Strain, and gases, which expand far more than solids, are explained in Kinetic Theory of Gases.

Advanced

The coefficient as a rate of change

Everything in this lesson rests on one number per material, the expansion coefficient $\alpha$. The familiar rule $\Delta L=\alpha L_0\Delta T$ is only an approximation, so it is worth seeing exactly what it approximates. The precise definition is the fractional growth in length per kelvin, taken at a single temperature:
$\alpha =\frac{1}{L}\frac{dL}{dT}$

The right-hand side is the derivative of $\ln L$ with respect to $T$. Integrating it from the starting temperature $T_0$ to $T$ therefore gives the length exactly:
$L=L_0\exp \left({\int }_{T_0}^T\alpha d{T}^{\prime }\right)$

If $\alpha$ does not depend on temperature, the integral is simply $\alpha \Delta T$, and expanding the exponential gives $L=L_0{e}^{\alpha \Delta T}=L_0(1+\alpha \Delta T+\frac{1}{2}{\alpha }^2\Delta T^2+\dots )$. The linear law keeps only the first correction, $\alpha \Delta T$, and drops the rest. For solids that costs almost nothing: in the animation $\alpha \Delta T=4.6\times 10^{-3}$, and the exponential gives 4.611 mm of growth instead of 4.600 mm.

The real weakness of the linear law is elsewhere: $\alpha$ itself changes with temperature. For real aluminium it rises by more than a tenth between 20 °C and 220 °C, so a real 1 m bar would grow by nearly 5 mm, an error far larger than the 0.011 mm lost by dropping the square term. Over wide temperature ranges, $\alpha$ must be integrated.

The animation's graph of $\Delta L$ against $T$ shows the linear law: a straight line from 0 at 20 °C to 4.6 mm at 220 °C, with slope $dL/dT=\alpha L_0=0.023mm/K$. On cooling, the dot slides back down the same line. No loop opens, because the length depends only on the present temperature, not on the history of heating and cooling. With the temperature-dependent $\alpha$ of real aluminium, the line would curve gently upwards.

The same definition applies to volume: $\beta =(1/V)dV/dT$. An isotropic solid, one that expands equally in every direction, keeps its shape, so $V\propto L^3$. Then $\ln V$ is $3\ln L$ plus a constant, and differentiating gives $\beta =3\alpha$ exactly. The familiar $\Delta V\approx 3\alpha V_0\Delta T$ is approximate only because it drops terms in $(\alpha \Delta T{)}^2$.

Why hot atoms need more room

Why should a solid expand at all? The answer lies in the shape of the bond between neighbouring atoms. Suppose the force between two neighbours were exactly $-kx$, where $x$ is the change in their separation from its resting value and $k$ is the stiffness of the bond. The potential energy would then be a parabola, every vibration would be symmetric about $x=0$, and the mean spacing would never change, however hard the atoms vibrated. A crystal with such perfectly spring-like (harmonic) bonds would not expand.

Real bonds are lopsided. Pushing two atoms together meets a steep repulsion as their electron clouds overlap, while pulling them apart meets an attraction that weakens with distance.

Legend

$x$	Change in the spacing of two neighbouring atoms, measured from the bottom of the well.	$\mathrm{m}$
$U$	Potential energy of the bond.	$\mathrm{J}$
$\frac{1}{2}k{x}^2$	Energy of an ideal spring of stiffness $k$: the symmetric parabola that the real curve follows near its bottom.	$\mathrm{J}$
$k_{B}T$	Thermal energy, Boltzmann's constant times the temperature: the height of the level at which the atoms vibrate.	$\mathrm{J}$
$\bar{x}$	Mean change of spacing: the average position of the vibrating atom, pushed to positive $x$ by the lopsided bond.	$\mathrm{m}$

Figure 1 plots the energy $U$ of one bond against the change $x$ in its spacing. Look first at the two curves. The dashed parabola $\frac{1}{2}k{x}^2$ is the ideal, symmetric spring. The solid curve is the real, lopsided bond: near the bottom of the well the two agree, but on the squeezed side (left) the real curve rises above the parabola, and on the stretched side (right) it falls below it.

Now look at the horizontal level drawn at the height $k_{B}T$, the typical thermal energy of a vibrating atom ($k_B$ is Boltzmann's constant). An atom with this energy swings back and forth between the two dots where the level meets the solid curve: these are its turning points. On the left the wall is steep, so the turning point lies close to $x=0$. On the right the wall is gentle, so the turning point lies further out. The atom therefore reaches further out than in, and its average position $\bar{x}$, the gold line, lies at positive $x$, to the right of the bottom of the well. A hotter atom rides a higher level, on a more lopsided stretch of the curve, and $\bar{x}$ moves further out. This outward drift of the average spacing, repeated in every bond of the crystal, is thermal expansion.

To make this quantitative, the simplest lopsided bond adds a cubic term to the spring, with a constant $b$ that measures the asymmetry. This is the solid curve of Figure 1, with its lopsidedness exaggerated:
$U(x)=\frac{1}{2}k{x}^2-\frac{1}{3}b{x}^3$
The minus sign puts the energy below the parabola for $x>0$ and above it for $x<0$, as in the figure. The force on the atom is $-dU/dx=-kx+b{x}^2$.

Now average over the motion. A vibrating atom stays in place on average, so the time average of the force on it must vanish: $k\bar{x}=b\overline{x^2}$, where $\overline{x^2}$ is the mean square displacement. Because $b$ is small, it is enough to take $\overline{x^2}$ from the harmonic spring. For a vibration of energy $E_{\mathrm{vib}}$, half of it is potential energy on average, so $\frac{1}{2}k\overline{x^2}=\frac{1}{2}{E}_{\mathrm{vib}}$ and $\bar{x}=b{E}_{\mathrm{vib}}/k^2$. A classical oscillator in thermal equilibrium has $E_{\mathrm{vib}}=k_{B}T$ on average, which gives
$\bar{x}=\frac{b{k}_{B}T}{k^2}$
Averaging $x$ with the Boltzmann factor $e^{-U/k_{B}T}$, to first order in $b$, gives the same result. So Figure 1, which follows one vibration at the energy $k_{B}T$, marks the same $\bar{x}$ that the thermal average gives.

The mean displacement grows in proportion to the temperature. Dividing by the resting spacing $r_0$ gives the fractional growth per kelvin, $\alpha \approx b{k}_B/(k^2{r}_0)$: constant at ordinary temperatures, and proportional to $b$. The expansion coefficient is a direct measure of how asymmetric the bonds are.

What sets the size of the coefficient

The same picture explains a striking regularity. Melting and expansion both depend on how far up the potential well the thermal energy reaches: a solid melts when its vibrations become a sizeable fraction of the atomic spacing, and it expands as the atoms ride higher up the lopsided well. A deep, stiff well therefore gives both a high melting temperature and a small $\alpha$, and for many metals their product is roughly the same, about 0.02. For aluminium, $23\times 10^{-6}\cdot 933\approx 0.021$; for tungsten, $4.5\times 10^{-6}\cdot 3695\approx 0.017$.

Invar, an iron–nickel alloy, breaks the rule: a magnetic contraction nearly cancels its normal expansion, a discovery that earned Charles-Édouard Guillaume the Nobel Prize in 1920 and makes Invar valuable in precision instruments.

At low temperatures the classical picture fails. The vibrations freeze into their quantum ground state, which hardly gains energy as the temperature rises. Expansion follows the vibrational energy, so $\alpha$ falls to zero together with the heat capacity, as the third law of thermodynamics requires. Their ratio stays nearly constant (Grüneisen's rule).

Thermal stress in a blocked rod

What happens when expansion is prevented? Take a rod held between rigid walls. It cannot change its length, so the walls must squeeze it by exactly as much as it would have expanded: they impose an elastic strain $-\alpha \Delta T$ that cancels the free expansion. By Hooke's law, stress is Young's modulus $E$ times strain, so this takes a stress of size
$\sigma =E\alpha \Delta T$
The stress is compressive on heating and tensile on cooling. Notice that the length has cancelled, because only the fractional strain matters: a 1 m rod and a 1 km rail of steel both gain 2.4 MPa of stress per kelvin.

The bending strip

A strip of two metals bonded together, such as brass on steel, is only partly blocked. Each layer tries to expand by its own amount, but the bond forces them to share one shape, and the compromise is a bend. This is how a thermostat turns a change of temperature into a movement. To find the bend, we need to know how the strain varies through the thickness of the strip.

Legend

$y$	Position across the strip, measured from the interface towards the centre of curvature (the steel side is positive).	$\mathrm{m}$
$h$	Thickness of the strip, both layers together.	$\mathrm{m}$
$\epsilon$	Strain along the strip: the fractional change of length.
${\epsilon }_0$	Strain at the interface, $y=0$: the mean of the two free strains for equal layers.
$R$	Radius of curvature of the bent strip; the blue line falls by $1/R$ per unit of $y$.	$\mathrm{m}$
${\alpha }_1\Delta T$	Strain the brass would have if it were free: its coefficient ${\alpha }_1$ times the temperature rise.
${\alpha }_2\Delta T$	Strain the steel would have if it were free: its coefficient ${\alpha }_2$ times the temperature rise.
${\sigma }_1$	Stress in the brass: $E$ times the gap between the blue line and the brass's own free strain.	$\mathrm{Pa}$
${\sigma }_2$	Stress in the steel: $E$ times the gap between the blue line and the steel's own free strain.	$\mathrm{Pa}$

Figure 2 is a cut through the thickness $h$ of the strip. The vertical axis is the position $y$ across the strip, measured from the interface towards the centre of curvature, so the steel (grey) fills $0<y<h/2$ and the brass (gold) fills $-h/2<y<0$. The horizontal axis is the strain $\epsilon$ along the strip.

Start with the two vertical lines. They are the strains each layer would have if it were free: ${\alpha }_1\Delta T$ for the brass and ${\alpha }_2\Delta T$ for the steel. Brass expands more, ${\alpha }_1>{\alpha }_2$, so its line lies further to the right.

Now the blue line. Bonded together, the layers must bend as one, to a radius $R$. In a bend, each thin slice along the strip is stretched or shortened in proportion to its distance from the centre of curvature, so the strain varies linearly across the strip:
$\epsilon (y)={\epsilon }_0-\frac{y}{R}$
Here ${\epsilon }_0$ is the strain at the interface. The line falls towards the centre of curvature, so the brass, on the outside of the bend, is stretched most.

A layer is stressed only by the part of its strain that is not free expansion, so the stress is $E(\epsilon -{\alpha }_i\Delta T)$, with ${\alpha }_1$ in the brass and ${\alpha }_2$ in the steel. In the figure this is $E$ times the horizontal gap between the blue line and the layer's own vertical line: the green wedges, ${\sigma }_1$ in the brass and ${\sigma }_2$ in the steel. Next to the interface the blue line lies left of the brass line and right of the steel line, so the brass is squeezed and the steel stretched. Near the outer faces the blue line has crossed the vertical lines, and the signs are reversed.

The blue line has two unknowns, its position ${\epsilon }_0$ and its slope $1/R$. No external load acts on the strip, so the stresses must give zero net force along it and zero net bending moment. These two conditions fix the two unknowns.

Take equal thicknesses and equal moduli. The term $-y/R$ contributes equal and opposite forces in the two halves, so the net force vanishes only if ${\epsilon }_0$ is the average of the two free strains, ${\epsilon }_0=\frac{1}{2}({\alpha }_1+{\alpha }_2)\Delta T$: the blue line crosses the interface half way between the two vertical lines. Relative to this, the brass is squeezed and the steel stretched by the same mismatch strain $\frac{1}{2}\Delta \alpha \Delta T$, where $\Delta \alpha ={\alpha }_1-{\alpha }_2$. Per unit width, these two equal and opposite forces, each $\frac{1}{4}E\Delta \alpha \Delta Th$, act $h/2$ apart and form a couple of moment $\frac{1}{8}E\Delta \alpha \Delta T{h}^2$. The bending stress $-Ey/R$ supplies the opposite moment $E{h}^3/(12R)$. Setting the two moments equal gives
$\frac{1}{R}=\frac{3\Delta \alpha \Delta T}{2h}\text{(equal thicknesses and moduli)}$
With this slope the blue line crosses each vertical line two thirds of the way from the interface to the outer face, as drawn.

For the animation, $\Delta \alpha =7\times 10^{-6}{\mathrm{K}}^{-1}$, $\Delta T=200\mathrm{K}$ and $h=0.5\mathrm{mm}$ give $1/R=4.2{\mathrm{m}}^{-1}$. The animation uses Stephen Timoshenko's general formula of 1925 with the real moduli (brass 100 GPa, steel 200 GPa) and gets $4.07{\mathrm{m}}^{-1}$, a radius of 24.6 cm.

How far does the tip move? For small angles a circle hugs a parabola, so an arc of radius $R$ falls away from its starting tangent, after a length $L$, by
$d\approx \frac{L^2}{2R}$
The 10 cm strip therefore lifts its tip by about 20.4 mm. The exact arc, which turns the tip through 23°, gives the 20.1 mm of the green readout. Combining the two formulas gives $d=3\Delta \alpha \Delta T{L}^2/(4h)$: halving the thickness doubles the movement, and doubling the length quadruples it. That is why thermostats use long, thin or coiled strips.

The water curve

Most liquids expand steadily when warmed. Water between 0 °C and about 4 °C does the opposite, and this small anomaly decides how lakes freeze.

Legend

$T$	Temperature, here in degrees Celsius.	${}^{\circ }\mathrm{C}$
$\rho$	Density of liquid water.	$kg/{\mathrm{m}}^3$
$T_m$	Temperature of the density maximum, 3.98 °C.	${}^{\circ }\mathrm{C}$
${\rho }_m$	Greatest density of liquid water, 999.97 kg/m³, reached at $T_m$.	$kg/{\mathrm{m}}^3$
$\beta$	Coefficient of volume expansion at the point of tangency: negative where the density rises with $T$, zero at the maximum, positive where it falls.	${\mathrm{K}}^{-1}$

Figure 3 shows the density $\rho$ of water from 0 to 12 °C, the right panel of the animation, with its tangent drawn at three temperatures. To read the expansion coefficient off this curve, note that for a fixed mass the density is inversely proportional to the volume, so $\beta =-(1/\rho )d\rho /dT$. The slope of the curve is therefore $-\beta \rho$, and its sign is always opposite to that of $\beta$.

Now follow the tangents from left to right. On the left the curve climbs, so the slope is positive and $\beta <0$: about $-68\times 10^{-6}{\mathrm{K}}^{-1}$ at 0 °C, which means that warming the water there makes it shrink. At the peak the tangent is level and $\beta =0$. This is the temperature $T_m=3.98$ °C of the greatest density, ${\rho }_m=999.97kg/{\mathrm{m}}^3$. Beyond it the curve falls and $\beta >0$, as for any other liquid. The vertical axis is greatly stretched: from 4 °C to 12 °C the density falls by less than 0.05 %.

This anomaly governs lakes. In autumn, cooled surface water is denser than the water below it, so it sinks, until the whole lake is at about 4 °C. Further cooling makes the surface water lighter, not heavier, so it stays on top, where it freezes, while the bottom stays near 4 °C under floating ice.

When heating makes things shrink

Water's anomaly is structural. Ice is an open, hydrogen-bonded lattice full of empty space. Melting partly collapses it, and ice-like clusters that survive in the cold liquid keep breaking up as it warms, letting the molecules pack closer. This shrinking competes with the ordinary expansion of the bonds, and it wins up to 3.98 °C.

A rubber band carrying a weight shortens when heated, for a different reason. Its tension comes mainly from long chain molecules: a straightened chain has far fewer possible arrangements than a crumpled one, so it has lower entropy, and heat drives the chains back towards shorter, disordered shapes.

Thermodynamics links this to a heat effect without any molecular model. Let $f$ be the tension and $L$ the length of the band. The free energy $F=U-TS$ then changes by $dF=-SdT+fdL$. Since $F$ is a function of $T$ and $L$, its mixed second derivatives are equal, and that gives a Maxwell relation:
${\left(\frac{\partial S}{\partial L}\right)}_T=-{\left(\frac{\partial f}{\partial T}\right)}_L$

Rubber's tension at fixed length rises with temperature, so the right-hand side is negative: stretching rubber lowers its entropy, and the band must give out heat. Stretched quickly, before that heat can escape, the band warms. A steel wire held at fixed length behaves the other way: heating it relieves its stretch, so it loses tension, by $EA\alpha$ per kelvin for a cross-section $A$. The sign reverses, and a quick stretch cools the steel, by about 0.1 K at 100 MPa.

Worked problem: the bar between two walls

Thermal stress can be enormous, but a slender bar has a way out. This problem puts numbers to both.

Legend

$L$	Length of the bar between the walls.	$\mathrm{m}$
$a$	Side of the square cross-section of the bar.	$\mathrm{m}$
$F$	Compressive force between each wall and the bar; before buckling, $F=\sigma a^2$.	$\mathrm{N}$
$\delta$	Sideways displacement of the middle of the bowed bar, exaggerated here.	$\mathrm{m}$
$L/2$	Length of the stretch between the two inflection points: half the bar.	$\mathrm{m}$

Figure 4 shows the animation's 1 m aluminium bar with the roller replaced by a second rigid wall, so that it is clamped at both ends. Its length is $L$, and its square cross-section has the side $a=10\mathrm{mm}$, so its area is $a^2=10^{-4}{\mathrm{m}}^2$; take $E=70\mathrm{GPa}$. Each wall pushes on the bar with the force $F$.

First, suppose the bar stays straight through the full 200 K. It would need the stress
$\sigma =E\alpha \Delta T=70\times 10^9\cdot 23\times 10^{-6}\cdot 200\approx 3.2\times 10^8\mathrm{Pa}$
that is 322 MPa, or $F=\sigma a^2=32\mathrm{kN}$ on each wall. Even a strong structural aluminium alloy yields at about 280 MPa, so the bar would be crushed plastically, permanently shortened.

In fact a slender bar fails much earlier, by buckling. The figure shows it after it has bowed sideways, with its middle displaced by $\delta$; the dashed outline is the straight bar, and the bowing is exaggerated. Because the clamped ends cannot turn, the bar takes the shape of one full wave, with inflection points (the dots) a quarter and three quarters of the way along.

At an inflection point the curvature of the bar passes through zero, and because the bending moment is proportional to the curvature, the moment there is zero, just as at an end that is free to turn. The stretch between the dots, of length $L/2$, therefore behaves like a bar with ends free to turn, and such a bar buckles at the force ${\pi }^{2}EI/(L/2{)}^2$. Here $I$ is the second moment of area of the section, $I=a^4/12$ for a square. So a strut with clamped ends bows sideways once the compressive force reaches $4{\pi }^{2}EI/L^2$, and dividing by the area $a^2$ gives the critical stress
${\sigma }_c=\frac{4{\pi }^{2}EI}{a^2{L}^2}=\frac{{\pi }^{2}E{a}^2}{3L^2}\approx 23\mathrm{MPa}$

The bar buckles when the thermal stress reaches this value. Setting $E\alpha \Delta T_c={\sigma }_c$, the modulus cancels:
$\Delta T_c=\frac{{\pi }^2{a}^2}{3\alpha L^2}=\frac{{\pi }^2\cdot 0.01^2}{3\cdot 23\times 10^{-6}\cdot 1^2}\approx 14\mathrm{K}$

After only 14 K, when a free bar would have grown just 0.33 mm, the clamped one bows sideways. Rails would buckle the same way on hot days if sleepers and ballast did not hold them sideways.

Limits of validity

The linear law makes two assumptions. The first, $\alpha \Delta T\ll 1$, always holds for solids: 1000 K changes steel by only about 1.2 %. The second, a constant $\alpha$, is the real limit: over wide ranges $\alpha$ must be integrated. Near absolute zero expansion dies out: copper's $\alpha$ at 50 K is only about a quarter of its room value of $17\times 10^{-6}{\mathrm{K}}^{-1}$. At a phase change the law fails abruptly: water expands by 9 % on freezing, and steel shrinks slightly as its crystal structure changes near 900 °C.

The formulas also assume a uniform temperature. Pour boiling water into a thick, cold glass, and the inner surface heats by some 50 K while the outside stays cool. The hot layer is blocked by the cold one, as in the rod between walls, and the stresses are of order $E\alpha \Delta T\approx 70\times 10^9\cdot 9\times 10^{-6}\cdot 50\approx 30\mathrm{MPa}$, comparable to the practical strength of ordinary glass. Borosilicate glass, with $\alpha \approx 3.3\times 10^{-6}{\mathrm{K}}^{-1}$, gets about a third of that, which is why it is used for ovenware and laboratory glass.

Gases follow none of this. At constant pressure an ideal gas has $\beta =1/T$, about $3.4\times 10^{-3}{\mathrm{K}}^{-1}$ at 20 °C, sixteen times the value for water.

Where this leads

Temperature and Heat supplies the kelvin; Elasticity: Hooke's Law, Stress and Strain supplies $\sigma =E\epsilon$, behind thermal stress and buckling. Kinetic Theory of Gases explains why a gas expands by $1/T$ per kelvin, and First Law of Thermodynamics and Heat Engines puts that expansion to work. The 4 °C anomaly shapes the convection of Conduction, Convection and Radiation, ice returns in Phase Changes and Latent Heat, and the entropy of a rubber band anticipates The Second Law of Thermodynamics and Entropy.

Expansion even reaches geometry. Picture a plate that is cold at its centre and hot at its rim, measured by inhabitants with rulers made of the plate's own material. Near the rim their rulers lengthen, so they count fewer ruler lengths around a circle than across it: circles around the centre measure less than $2\pi$ times their radius, and triangles have angles adding up to more than $180^{\circ }$, as on a sphere. In Gravity as Curved Spacetime, mass does to rulers and clocks what heat does here.