Thermal expansion — linear, area, volumetric: Worked examples

This page is the workout gym for the parent topic. The parent built the three formulas from atoms up. Here we make sure that no situation can surprise you — every sign, every dimension, every degenerate case, every real-world twist gets its own fully worked example.

Before we start, one reminder of every symbol we will use, so nothing is unexplained:

The scenario matrix

Every problem this topic can throw is one (or a blend) of these cells. The examples below are labelled with the cell they hit, and together they cover the whole grid.

Cell	What makes it different	Trap it tests
C1 · Linear, heating	1-D rod, $Δ T > 0$	plain formula
C2 · Sign flip (cooling)	$Δ T < 0$ → object shrinks	remembering length drops
C3 · Area / the hole	2-D, a hole expands like metal	thinking hole shrinks
C4 · Volume + which coefficient	3-D, given $α$ not $γ$	forgetting to $\times 3$
C5 · Degenerate: $Δ T = 0$	no temperature change	answer is "nothing moves"
C6 · Differential expansion	two materials, difference of stretches	bimetallic strip / gap fit
C7 · Constrained (no room)	rod can't expand → stress instead	strain $\to$ force, links Stress and strain
C8 · Limiting / word problem	real object, engineering decision	choosing right dimension
C9 · Anomaly ( $γ < 0$ )	water 0–4 °C shrinks on heating	sign of $γ$

The worked examples

C1 — Linear expansion, straightforward heating

C2 — Sign flip: cooling makes it shrink

C3 — Area, and the famous hole

C4 — Volume, coefficient trap

C5 — Degenerate case: $Δ T = 0$

C6 — Differential expansion (bimetallic idea / fitting a gap)

Worked example Steel ring on an aluminium shaft

At $2 0^{\circ} C$ an aluminium shaft has diameter $50.00 mm$ and a steel ring has inner diameter $49.98 mm$ (too tight to slide on). Heat both equally. At what $Δ T$ do the diameters just match? $α_{Al} = 2.3 \times 1 0^{- 5}$ , $α_{steel} = 1.2 \times 1 0^{- 5} K^{- 1}$ .

Forecast: Heating both — does the gap close or open? (Which one grows faster?)

Diameters after heating: shaft $50.00 (1 + α_{Al} Δ T)$ , ring $49.98 (1 + α_{steel} Δ T)$ . Why this step? Each metal expands by its own $α$ . The ring starts smaller but expands slower — the gap actually widens the wrong way. So we must cool, or the numbers must be read for a fit-check.
Set them equal: $50.00 (1 + 2.3 \times 1 0^{- 5} Δ T) = 49.98 (1 + 1.2 \times 1 0^{- 5} Δ T)$
Expand and collect $Δ T$ : $50.00 - 49.98 = Δ T (49.98 \times 1.2 \times 1 0^{- 5} - 50.00 \times 2.3 \times 1 0^{- 5})$ $0.02 = Δ T (5.9976 \times 1 0^{- 4} - 1.15 \times 1 0^{- 3}) = Δ T (- 5.502 \times 1 0^{- 4})$ $Δ T = \frac{0.02}{- 5.502 \times 1 0 ^{- 4}} \approx - 36.4 K$

Verify: $Δ T$ is negative — you must cool by ~36 °C (down to about $- 1 6^{\circ} C$ ), because the faster-expanding aluminium shaft outruns the steel ring on heating. The sign correctly told us "heat is the wrong direction." That's the whole point of the differential cell. ✓

C7 — Constrained rod → thermal stress

Worked example Clamped steel bar

A steel bar is clamped between two rigid walls at $2 0^{\circ} C$ so it cannot lengthen. It is heated to $7 0^{\circ} C$ . Find the compressive stress. $α_{steel} = 1.2 \times 1 0^{- 5} K^{- 1}$ , Young's modulus $Y = 2.0 \times 1 0^{11} Pa$ .

Forecast: With no room to grow, where does the "wanted expansion" energy go?

If free, it would strain by $\frac{Δ L}{L _{0}} = α Δ T$ . Why this step? The walls force this strain back to zero, so the wall must impose an equal and opposite compressive strain of magnitude $α Δ T$ .
Stress $= Y \times strain = Y α Δ T$ (from Stress and strain). Why this step? Stress and strain are linked by Young's modulus $Y$ for elastic solids.
$σ = 2.0 \times 1 0^{11} \times 1.2 \times 1 0^{- 5} \times 50 = 1.2 \times 1 0^{8} Pa = 120 MPa$ .

Verify: $120 MPa$ is a real, large stress (approaching steel's yield ~250 MPa) — this is why bridges and rails need expansion joints. Notice $L_{0}$ cancels: thermal stress doesn't depend on length. ✓

C8 — Word/limiting problem: choosing the right dimension

Worked example Which coefficient does a filled tank use?

A glass tank of volume $2000 cm^{3}$ is completely full of glycerine at $1 5^{\circ} C$ . Heated to $4 5^{\circ} C$ . How much glycerine overflows? $γ_{glycerine} = 5.3 \times 1 0^{- 4}$ , $α_{glass} = 9 \times 1 0^{- 6} K^{- 1}$ .

Forecast: Does all the liquid's expansion overflow, or does the tank's own growth "make room"?

Liquid wants to grow by $Δ V_{liq} = V_{0} γ_{liq} Δ T$ .
But the container's cavity also grows, using its volume coefficient $γ_{glass} = 3 α_{glass} = 2.7 \times 1 0^{- 5} K^{- 1}$ . Why this step? The cavity is a "hole" → it expands like solid glass in 3-D. Overflow is the difference (apparent expansion).
$Δ V_{spill} = V_{0} (γ_{liq} - γ_{glass}) Δ T = 2000 (5.3 \times 1 0^{- 4} - 2.7 \times 1 0^{- 5}) (30)$ . $= 2000 \times 5.03 \times 1 0^{- 4} \times 30 = 30.18 cm^{3} .$

Verify: Positive → liquid overflows (its $γ$ far exceeds glass's). If we'd (wrongly) ignored the tank's growth we'd get $31.8 cm^{3}$ — the container correction shaves ~5 %. The difference structure mirrors C6. ✓

C9 — The anomaly: negative $γ$

Recap

Recall Which cell was which?

C1 plain linear ::: rail, $Δ L = 4.32 mm$ C2 cooling / sign flip ::: Al wire shrinks $2.875 mm$ C3 hole (area) ::: washer hole grows to $3.0204 cm^{2}$ C4 $γ$ given directly ::: mercury $Δ V = 1.8 \times 1 0^{- 3} cm^{3}$ C5 $Δ T = 0$ ::: nothing changes C6 differential ::: need to cool ~36 K for the fit C7 constrained ::: thermal stress $120 MPa$ , independent of length C8 apparent expansion ::: glycerine overflow $30.18 cm^{3}$ C9 anomaly ::: water contracts $0.272 cm^{3}$ on 0→4 °C

Connections

Thermal expansion — linear, area, volumetric — the parent that built every formula used here.
Stress and strain — the constrained-rod cell (C7) lives here.
Anomalous expansion of water — the negative- $γ$ cell (C9).
Gas laws — where much larger expansion appears and $Δ T$ must be absolute.
Pendulum and simple harmonic motion — timing errors from a lengthening rod.