University Physics II · Optional extension · Institutional Extension · Temperature, Heat, and Kinetic Theory · E2.2

Thermal Expansion & Thermal Stress

Heat a solid and every length in it grows by the same small fraction — areas by twice that, volumes by three times. Block the growth and the same number comes back as stress. Water, below 4 °C, does the opposite.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Heating a solid pushes its atoms slightly further apart on average, and the effect is uniform: over a modest temperature range every length in the body grows by the same fraction, ΔL/L₀ = α ΔT. That single coefficient does all the work. Areas carry 2α and volumes 3α because they are lengths squared and cubed, holes and cavities expand with the metal around them rather than closing up, and density falls as a fixed mass spreads over more volume.

The physics gets interesting when the expansion is not allowed to happen: a bar held between rigid supports must absorb the strain α ΔT elastically instead, producing σ = −E α ΔT — a stress that does not depend on the bar's length, and the reason bridges, rails and pipework are built with somewhere to move. Two things bound the model. Water contracts as it warms from 0 °C to 3.98 °C, and α itself is a derivative that only looks constant over a window.

Simple definition: Thermal expansion is the fractional change in a body's dimensions per kelvin of temperature change. The linear coefficient α = (1/L)(dL/dT) has units K⁻¹, and under uniform heating every length grows by the same factor 1 + α ΔT.
Example: Steel has α = 12 × 10⁻⁶ K⁻¹, so a 25.0 m rail warmed by 40 K grows 12 × 10⁻⁶ × 25.0 × 40 = 12 mm. Bolt both ends to rigid supports and that 12 mm never appears; 96 MPa of compression appears instead.

Linear expansionΔL = α L₀ ΔT, so L = L₀(1 + α ΔT)

α is a fraction per kelvin, so L₀ may be any length in the body — a rod, a diameter, a gap width.

α in K⁻¹, numerically the same in °C⁻¹; steel 12 × 10⁻⁶, aluminium 23 × 10⁻⁶

α is a derivative, not a constantα(T) = (1/L)(dL/dT) → L = L₀ exp[∫α(T) dT]

Tables quote α averaged over a stated range. Use the integral when ΔT spans hundreds of kelvin.

The linear form is the first term of that exponential and needs α ΔT ≪ 1

Area and volume follow from lengthΔA = 2α A₀ ΔT · ΔV = 3α V₀ ΔT = β V₀ ΔT

First-order expansions of (1 + αΔT)² and (1 + αΔT)³. Holes and cavities scale with the same α.

Isotropic solids only; β = 3α, so steel has β = 3.6 × 10⁻⁵ K⁻¹

Density falls as the volume growsρ = ρ₀/(1 + β ΔT) ≈ ρ₀(1 − β ΔT)

At 20 °C a gas has β = 1/293 K = 3.4 × 10⁻³ K⁻¹, some ninety-five times steel's value.

The mass is fixed; for an ideal gas at constant pressure β = 1/T, not a material constant

Thermal stress in a fully constrained barσ = −E α ΔT, and F = σA = E A α ΔT in magnitude

Cancel the thermal strain α ΔT with an elastic one. Steel over 40 K: 200 GPa × 12 × 10⁻⁶ × 40 = 96 MPa.

Pa; compressive on heating, tensile on cooling. No length appears in σ.

Bimetallic strip curvature1/R = 3(α₂ − α₁) ΔT / (2t)

Brass on Invar, t = 0.50 mm, ΔT = 50 K gives R = 0.37 m, so a 30 mm strip lifts its tip 1.2 mm.

Equal thicknesses and similar moduli; t is the total thickness, tip rise ≈ L²/2R

One coefficient, quoted per kelvin

An atom in a solid sits in a potential well that is steeper on the compression side than on the stretch side. Give it more vibrational energy and its average position moves outward, so the mean atomic spacing grows with temperature. Macroscopically that appears as a fractional length change proportional to the temperature change, ΔL/L₀ = α ΔT, which defines the linear expansion coefficient α = ΔL/(L₀ ΔT). Its unit is K⁻¹, and the number is identical in °C⁻¹ because α multiplies a temperature difference rather than a temperature. Typical values, all × 10⁻⁶ K⁻¹: fused quartz 0.5, Invar 1.2, borosilicate glass 3.2, soda-lime glass 9, steel 12, copper 17, brass 19, aluminium 23, lead 29. They are small enough to ignore until either the length or the temperature range is large. A 25.0 m steel rail warmed by 40 K stretches by 12 × 10⁻⁶ × 25.0 × 40 = 1.2 × 10⁻² m, twelve millimetres. Because α is a fraction per kelvin, L₀ can be whichever length matters — a rod, a pipe's bore, the width of a gap — and one α covers them all.

Heating makes a scaled copy of the object

Uniform heating multiplies every length in an isotropic body by the same factor, 1 + α ΔT. The warm object is a scaled copy of the cold one: angles are unchanged, ratios of lengths are unchanged, and anything fixed by the positions of the material — including a hole — scales with it. If the hole troubles you, fill it with a plug cut from the same metal, heat the lot, and note that the plug still fits. The plug grew by α ΔT, so the hole grew by α ΔT. Areas and volumes then follow by squaring and cubing. A = A₀(1 + α ΔT)² = A₀(1 + 2α ΔT + α²ΔT²), and for steel over 40 K, α ΔT = 4.8 × 10⁻⁴, which makes the quadratic term only 2.4 × 10⁻⁴ of the linear one. Keep first order: ΔA = 2α A₀ ΔT. Cubing gives ΔV = 3α V₀ ΔT and defines β = 3α. Shrink fitting lives on this. A steel washer with a 20.000 mm hole heated 100 K opens to 20.000 + 12 × 10⁻⁶ × 20.000 × 100 = 20.024 mm, enough to drop over a shaft it will grip when it cools.

Block the expansion and you get stress instead

A bar free at one end expands and carries no stress. Bolt both ends to rigid supports and its length cannot change, so the thermal strain α ΔT must be cancelled by an elastic strain of −α ΔT. Hooke's law turns that into σ = −E α ΔT: compressive when heated, tensile when cooled. No length appears. Doubling the bar doubles the expansion it wants and doubles the length over which the cancelling strain is spread, and the two effects cancel, so 1 m and 100 m constrained bars load their supports to the same stress. For steel over 40 K, σ = 200 × 10⁹ × 12 × 10⁻⁶ × 40 = 9.6 × 10⁷ Pa, or 96 MPa — about 38 % of a structural steel's 250 MPa yield stress. On a 50 cm² section that is F = 9.6 × 10⁷ × 5.0 × 10⁻³ = 4.8 × 10⁵ N. At ΔT = 250 × 10⁶/(200 × 10⁹ × 12 × 10⁻⁶) = 104 K the bar yields. So structures are given room to move: expansion joints in bridge decks, roller bearings under girders, expansion loops in steam pipework.

Two materials on one strip

Bond brass (α = 19 × 10⁻⁶ K⁻¹) to Invar (1.2 × 10⁻⁶ K⁻¹) and heat the pair. The brass wants to grow about sixteen times as much, the bond forbids it, and the strip settles the argument by curving with the brass on the outside. For equal thicknesses and similar moduli, 1/R = 3(α₂ − α₁) ΔT/(2t) with t the total thickness. Take t = 0.50 mm and ΔT = 50 K: 1/R = 3 × 17.8 × 10⁻⁶ × 50/(2 × 5.0 × 10⁻⁴) = 2.7 m⁻¹, so R = 0.37 m and a 30 mm strip lifts its free tip by roughly L²/2R = 1.2 mm. Curvature is proportional to ΔT, which makes the strip a thermometer, or a thermostat once a contact is put at the tip. The same mismatch is destructive when unwanted: a glass-to-metal seal must pair materials of similar α or cooling from the sealing temperature leaves a locked-in stress. Thermal shock is that problem inside one material — boiling water against a cold outer wall. Borosilicate glass, at 3.2 × 10⁻⁶ K⁻¹, builds about a third of soda-lime's strain, which is why lab glassware is made of it.

Liquids expand more, and water reverses

A liquid has no fixed shape, so only β matters, and it runs one to two orders above a solid's: mercury 1.8 × 10⁻⁴ K⁻¹, water 2.07 × 10⁻⁴ K⁻¹ at 20 °C, petrol about 9.5 × 10⁻⁴ K⁻¹, against 3.6 × 10⁻⁵ K⁻¹ for steel. Inside a container you observe only the difference, βₐₚₚₐᵣₑₙₜ = β_liquid − β_container, so a mercury thermometer with a soda-lime bulb loses 2.7 × 10⁻⁵/1.8 × 10⁻⁴ = 15 % of the mercury's expansion to the bulb growing with it. Warm 60 L of petrol in a steel tank by 20 K: the petrol gains 60 × 9.5 × 10⁻⁴ × 20 = 1.14 L while the tank gains 0.04 L, so about 1.1 L has to go somewhere. Water is the exception that matters. Its density peaks at 3.98 °C, at 999.97 kg m⁻³, so β is negative from 0 °C up to that point — a mean near −3.3 × 10⁻⁵ K⁻¹ — and warming the water shrinks it. Hydrogen bonds hold an open, ice-like arrangement that survives melting in patches, and warming collapses those patches faster than agitation spreads the molecules apart. Ice at 0 °C, 916.7 kg m⁻³ against water's 999.84, expands 9.1 % on freezing.

Where the linear model stops

α is a derivative, α(T) = (1/L)(dL/dT), and a tabulated value is that derivative averaged over a stated range. Four things break the constant-α picture. First, α varies with temperature and falls to zero as T → 0 K, as the third law requires; over hundreds of kelvin a tabulated value can be off by tens of percent, and the honest form is L = L₀ exp[∫α(T) dT]. Second, α ΔT must stay small for ΔL = α L₀ ΔT to be the whole story — the same truncation that gave β = 3α. Third, isotropy is an assumption: calcite lengthens along its c-axis while contracting perpendicular to it, and graphite expands strongly between its layers and barely at all within them, so β = 3α fails outright for anisotropic, non-cubic crystals — cubic ones such as copper, aluminium and silicon expand isotropically and obey it exactly. Fourth, phase changes are steps, not slopes; quartz jumps in volume at its α-to-β transition near 573 °C. Invar's tiny 1.2 × 10⁻⁶ K⁻¹ is itself a near-cancellation of ordinary expansion by a magnetic contraction, and it holds only below the Curie point. A constrained slender member can also buckle sideways before σ = −E α ΔT reaches yield.

Change one variable at a time

Make the relationship visible.

Interactive model

Temperature rise ΔT40 K

Coefficient α (steel 12, aluminium 23)12 × 10⁻⁶ K⁻¹

Bar length L₀2.0 m

Drag L₀ from 0.5 m to 4.0 m: the free growth goes up eightfold while the arrows and the stress bar do not move at all, because no length appears in σ = −E α ΔT. Then set α to 12 for steel and raise ΔT until the stress bar crosses the yield mark.

GROWTH ΔL = α L₀ ΔT0.96 mm

THERMAL STRAIN α ΔT480 × 10⁻⁶

STRESS IF BLOCKED96 MPa

FORCE ON A 50 mm × 50 mm BAR240 kN

Live interpretationGROWTH ΔL = α L₀ ΔT: 0.96 mm. THERMAL STRAIN α ΔT: 480 × 10⁻⁶. STRESS IF BLOCKED: 96 MPa. FORCE ON A 50 mm × 50 mm BAR: 240 kN

Catch the common trap

Explain before calculating.

An aluminium plate (α = 23 × 10⁻⁶ K⁻¹) has a circular hole of diameter 40.00 mm at 20 °C. The whole plate is heated uniformly to 220 °C. What is the hole's diameter then?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 30.0 m aluminium bridge deck section (α = 23 × 10⁻⁶ K⁻¹) is laid at 5 °C and reaches 45 °C in summer. Find the expansion gap the designer must leave at one end, and the gap the same section would need if it were steel (α = 12 × 10⁻⁶ K⁻¹).

A Celsius interval is a kelvin interval, so ΔT = 45 − 5 = 40 K needs no conversion.
Aluminium: ΔL = α L₀ ΔT = (23 × 10⁻⁶ K⁻¹)(30.0 m)(40 K) = 2.76 × 10⁻² m = 27.6 mm.
Steel: ΔL = (12 × 10⁻⁶ K⁻¹)(30.0 m)(40 K) = 1.44 × 10⁻² m = 14.4 mm.
Same length and same ΔT, so the gaps stand in the ratio of the coefficients, 23:12 = 1.92 — the gap is set by the material, and it has to be left rather than designed away.

AnswerAluminium needs a 27.6 mm gap; the same section in steel would need 14.4 mm.

MediumAn aluminium strut (α = 23 × 10⁻⁶ K⁻¹, E = 70 GPa) of length 1.20 m and cross-section 8.0 cm² is bolted between two rigid abutments at 15 °C and heated to 65 °C. Find the compressive stress and the force on each abutment, then say what changes if the strut is 2.40 m long instead.

ΔT = 65 − 15 = 50 K. Left free, the strut would grow ΔL = α L₀ ΔT = (23 × 10⁻⁶)(1.20)(50) = 1.38 × 10⁻³ m = 1.38 mm.
Bolted at both ends it cannot, so an elastic strain must cancel the thermal one exactly: ε = −α ΔT = −(23 × 10⁻⁶)(50) = −1.15 × 10⁻³.
Hooke's law turns that into stress: σ = Eε = (70 × 10⁹ Pa)(−1.15 × 10⁻³) = −8.05 × 10⁷ Pa, i.e. 80.5 MPa in compression.
Force on each abutment: A = 8.0 cm² = 8.0 × 10⁻⁴ m², so F = σA = (8.05 × 10⁷)(8.0 × 10⁻⁴) = 6.4 × 10⁴ N = 64 kN.
At 2.40 m the free growth doubles to 2.76 mm, but σ = −E α ΔT contains no length: doubling the strut doubles the expansion it wants and doubles the length the cancelling strain is spread over. The stress stays 80.5 MPa and the force stays 64 kN.

Answerσ = 80.5 MPa compressive and F = 64 kN on each abutment; doubling the length changes neither.

HardA borosilicate flask (α = 3.2 × 10⁻⁶ K⁻¹) is filled to the brim with exactly 500.0 mL of glycerine (β = 5.1 × 10⁻⁴ K⁻¹) at 20 °C, and the pair is warmed to 80 °C. Find the volume that spills, and the fraction of the glycerine's own expansion the flask absorbs.

The cavity expands with the glass around it, so treat it as a flask-shaped lump of borosilicate: β_glass = 3α = 3(3.2 × 10⁻⁶) = 9.6 × 10⁻⁶ K⁻¹.
ΔT = 80 − 20 = 60 K. Glycerine: ΔV = β V₀ ΔT = (5.1 × 10⁻⁴)(500.0 mL)(60) = 15.3 mL.
Flask cavity: ΔV = (9.6 × 10⁻⁶)(500.0 mL)(60) = 0.288 mL.
Only the difference has nowhere to go: spill = 15.3 − 0.288 = 15.0 mL.
The flask takes up 0.288/15.3 = 1.9 % of the liquid's expansion, so a scale etched on the flask would read the apparent coefficient βₐₚₚ = β_liquid − β_glass = 5.1 × 10⁻⁴ − 0.096 × 10⁻⁴ = 5.0 × 10⁻⁴ K⁻¹, not the true 5.1 × 10⁻⁴ K⁻¹.

Answer15.0 mL spills; the flask absorbs only 1.9 % of the glycerine's expansion, so βₐₚₚₐᵣₑₙₜ = 5.0 × 10⁻⁴ K⁻¹.