A temperature change makes a free part grow or shrink by δ = α·ΔT·L, with no stress at all. Stress appears only when something stops that movement. A bar held between rigid supports develops σ = −E·α·ΔT, compressive when heated, and a partly restrained bar carries stress in proportion to the movement it is denied.
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Thermal strain
Most materials expand when heated. For moderate temperature changes the thermal strain is proportional to the change, ε_T = α·ΔT, where α is the coefficient of linear thermal expansion. Steels are around 11–13 × 10⁻⁶ per °C and aluminium alloys around 22–24 × 10⁻⁶ per °C; take the value for your actual alloy from its data sheet.
Thermal strain applies in every direction. A hole in a heated plate gets bigger, not smaller, because every dimension of the part, including the hole, scales by (1 + αΔT). That is how shrink fits work.
When thermal strain causes stress
Thermal strain by itself causes no stress. Stress comes from compatibility: if supports, bolts or a mating part limit the total deformation, the material has to carry a mechanical strain that makes up the difference. The total strain is the sum of the thermal part and the stress-related part, ε = σ/E + α·ΔT.
Fully restrained bar: total strain is zero, so σ/E = −αΔT and σ = −EαΔT. It does not depend on the bar’s length or area.
Bar with a gap: it expands freely until the gap closes, then the rest of the expansion is resisted. Only the blocked part, δ_T − gap, produces stress.
Assemblies of different materials (a steel bolt clamping an aluminium sleeve): each part wants to grow by a different amount, and equilibrium plus compatibility give the forces.
Checks that matter
A compressive thermal stress in a long, slender bar can also make it buckle, which is why rails, pipelines and long ducts have expansion joints or loops. Compare thermal stress with the yield strength, and remember that E and α both change with temperature over large ranges.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Free thermal deformation
δ_T = α·ΔT·L
α = coefficient of linear thermal expansion (1/°C), ΔT = temperature change (°C, positive when heated), L = original length (mm); δ_T in mm
Gere & Goodno, Mechanics of Materials
Total strain with temperature
ε = σ / E + α·ΔT
σ = normal stress (MPa), E = modulus of elasticity (MPa)
Beer & Johnston, Mechanics of Materials
Fully restrained bar
σ = −E·α·ΔT
negative = compression when heated (ΔT > 0)
Hibbeler, Mechanics of Materials
Partly restrained bar (gap g)
σ = −E·(α·ΔT·L − g) / L, for α·ΔT·L > g
g = initial clearance to the support (mm); no stress if the expansion is smaller than the gap
Hibbeler, Mechanics of Materials
Change in diameter (holes and shafts)
Δd = α·ΔT·d
d = original diameter (mm); applies to bores as well as solid shafts
Gere & Goodno, Mechanics of Materials
Worked examples
Example: An aluminium bar that closes a gap and then pushes on its supports
Given
An aluminium bar 1.2 m long is fixed to a rigid wall at one end. There is a 0.5 mm gap between its other end and a second rigid wall.
The temperature rises by 60 °C.
Assume α = 23 × 10⁻⁶ /°C and E = 70 GPa, typical textbook values for aluminium alloys.
Find
The free thermal expansion, whether the gap closes, and the stress in the bar.
The temperature rise at which the gap just closes.
Solution
Free expansion if nothing stopped it:δ_T = α·ΔT·L = 23 × 10⁻⁶ × 60 × 1200 mm = 1.656 mm
1.656 mm is larger than the 0.5 mm gap, so the gap closes and the walls prevent the rest:blocked deformation = 1.656 − 0.5 = 1.156 mm
The wall force must shorten the bar by that amount, so the mechanical strain is:ε = −1.156 / 1200 = −9.63 × 10⁻⁴
Stress from Hooke’s law, with E = 70 000 MPa:σ = E·ε = 70 000 × (−9.63 × 10⁻⁴) = −67.4 MPa
For comparison, with no gap the bar would be fully restrained:σ = −E·α·ΔT = −70 000 × 23 × 10⁻⁶ × 60 = −96.6 MPa
Temperature rise that just closes the gap:ΔT = g / (α·L) = 0.5 / (23 × 10⁻⁶ × 1200) = 18.1 °C
Answer: The bar would grow 1.66 mm freely. The 0.5 mm gap closes after an 18.1 °C rise, and at 60 °C the bar carries 67.4 MPa compression, against 96.6 MPa with no gap. A small clearance relieves a large share of the thermal stress.
Example: Heating a gear to shrink-fit it onto a shaft
Given
A steel gear has a 50.00 mm bore. It must fit onto a steel shaft of 50.04 mm diameter (a 0.04 mm interference).
For assembly, the heated bore must clear the shaft by at least 0.02 mm. The shop is at 20 °C.
Assume α = 12 × 10⁻⁶ /°C for both parts (typical for carbon steel).
Find
The temperature the gear must be heated to.
As an alternative, how much the shaft shrinks if it is cooled to −78 °C (dry ice) instead.
Solution
Required growth of the bore: the interference plus the assembly clearance.Δd = 0.04 + 0.02 = 0.06 mm
Rearrange Δd = α·ΔT·d for the temperature rise:ΔT = Δd / (α·d) = 0.06 / (12 × 10⁻⁶ × 50.00) = 100 °C
Resulting clearance with a room-temperature gear:50.00 − (50.04 − 0.059) = 0.019 mm
Answer: Heat the gear to about 120 °C, so its bore grows 0.06 mm. Cooling the shaft to −78 °C shrinks it about 0.059 mm, which leaves roughly the same 0.02 mm clearance. When the parts return to room temperature, the 0.04 mm interference becomes a press fit; that is a thermal-expansion problem with no thermal stress until they touch.
Common mistakes
Calculating a thermal stress in a part that is free to expand. No restraint means no thermal stress. Check the supports before writing σ = −EαΔT.
Ignoring a gap or the stiffness of the support. Subtract any clearance first. Real supports also deflect, which relieves some of the stress.
Expecting a heated hole to shrink. Holes grow with the part. A bore expands by α·ΔT·d, the same as a solid disk of that diameter.
Mixing °C and K in ΔT. A temperature difference is the same in °C and K, but not in °F. For °F, use α per °F and ΔT in °F.
Using one α and E over a very large temperature range. Both vary with temperature. For large changes, use mean values over the range from the material data.
In industry
Frequently asked questions
Why does thermal stress not depend on the length of the bar?
Both the free expansion and the deformation the supports must remove scale with L, so L cancels in the strain and the stress depends only on E, α and ΔT.
Is thermal stress tension or compression?
A restrained part that is heated goes into compression; one that is cooled goes into tension. In assemblies of different materials, one part can be in tension while the other is in compression.
What units does α use?
Per degree: 1/°C (equal to 1/K) or 1/°F. Values per °F are 5/9 of the value per °C.
Can thermal stress cause failure?
Yes. Large or repeated temperature swings can yield a restrained part, crack brittle materials through thermal shock, buckle slender members, or fatigue welds and joints over many cycles.