Thermal stress is the stress that appears when a temperature change makes a part try to expand or contract and something prevents it. A part that is free to move grows when heated and shrinks when cooled without any stress at all. When the movement is resisted, by fixings, by a neighbouring part in a different material, or by cooler material within the same part, stress results. For a bar held rigidly at both ends:
σ = EαΔT
where E is Young’s modulus, α the coefficient of thermal expansion and ΔT the temperature change. Heating a restrained bar puts it in compression; cooling it puts it in tension. The formula explains why thermal stress catches designs out: length does not appear in it. A fully restrained steel bar heated by 50 °C carries about 126 MPa whether it is 10 mm or 10 m long. That is more than a third of the yield strength of S355 steel, from temperature alone.
Thermal expansion
An unrestrained part expands by a thermal strain:
εth = αΔT, so ΔL = αLΔT
A 2 m aluminium extrusion warming by 40 °C grows by 23 × 10⁻⁶ × 2,000 mm × 40 = 1.84 mm. A steel bar of the same length grows by about 0.96 mm. Neither figure is large, but forcing either back to its original length takes a great deal of force. A temperature change of 1 °C is the same as 1 K, so α can be quoted per kelvin or per degree Celsius.
| Material | α (10⁻⁶ /K) | E (GPa) | Eα (MPa/K) |
|---|---|---|---|
| Carbon and low-alloy steel | 11 to 13 | 200 to 210 | about 2.5 |
| Austenitic stainless steel (304, 316) | 16 to 17 | 193 to 200 | about 3.2 |
| Aluminium alloys (6061, 6082) | 23 to 24 | 69 to 71 | about 1.6 |
| Copper | about 17 | 110 to 130 | about 2.0 |
| Titanium alloy Ti-6Al-4V | about 9 | about 114 | about 1.0 |
| Alumina ceramic | 7 to 8 | 300 to 380 | about 2.6 |
| ABS plastic | 70 to 100 | 2 to 2.5 | about 0.2 |
The product Eα is the stress per degree of temperature change in a fully restrained bar. The values are for temperatures near ambient; both properties change with temperature, and E in particular falls as metals get hot. Austenitic stainless steel combines high expansion with low thermal conductivity, so it tends to develop steeper gradients and higher thermal stress than carbon steel in the same duty.
The fully restrained bar
If a bar cannot change length, its mechanical strain must cancel its thermal strain:
σ / E + αΔT = 0, so σ = −EαΔT
The minus sign means a temperature rise produces compression. Two refinements cover many real cases.
- Restraint in two directions. A plate or surface layer held in its own plane in both directions carries σ = EαΔT / (1 − ν), about 43% more than the bar for a metal with Poisson’s ratio ν = 0.3.
- Partial restraint. Real supports have finite stiffness. If a bar of length L and area A is held between a rigid support and a support of stiffness k, the stress falls to σ = EαΔT / (1 + EA / (kL)). A support only as stiff as the bar itself, k = EA / L, halves the stress. That is the principle behind every expansion joint.
Worked example: aluminium on steel
An aluminium bar is fastened at both ends to a steel bar of the same section, 40 × 10 mm (A = 400 mm²), with the fixings 1 m apart. The assembly is put together at 20 °C and runs at 80 °C, so ΔT = 60 °C. Take Eal = 70 GPa and αal = 23 × 10⁻⁶ /K for the aluminium, and Est = 210 GPa and αst = 12 × 10⁻⁶ /K for the steel.
Free mismatch. Unrestrained, the aluminium would grow more than the steel by (23 − 12) × 10⁻⁶ × 60 = 660 × 10⁻⁶, or 0.66 mm over the 1 m length.
Compatibility. With rigid fixings, both bars end up the same length. The aluminium is held shorter than it wants to be and the steel is pulled longer, by an equal and opposite force F:
F (1 / EalA + 1 / EstA) = (αal − αst) ΔT
Force and stress. With E in N/mm² and A in mm², 1 / (70,000 × 400) + 1 / (210,000 × 400) = 4.76 × 10⁻⁸ per N, so F = 660 × 10⁻⁶ / 4.76 × 10⁻⁸ ≈ 13.9 kN. The stress is 13,900 N / 400 mm² ≈ 35 MPa, compressive in the aluminium and tensile in the steel.
If the steel were far stiffer than the aluminium, the aluminium would carry the full EalΔαΔT = 70,000 × 11 × 10⁻⁶ × 60 ≈ 46 MPa. Neither figure troubles a 6082-T6 bar, whose minimum 0.2% proof strength is around 250 MPa. The fixings are another matter. A shear force of 13.9 kN at each end is more than friction under a single M8 or M10 bolt can be relied on to hold, so in practice the joint slips, the bolts bear on their holes, or an adhesive bond line cracks. That is often how a mismatch problem first shows itself. If the two bars are stacked face to face rather than in line, the offset between them also bends the pair, the same effect a bimetallic strip uses on purpose.
The same mechanism works on a smaller scale in electronics. A component with an alumina ceramic body, at about 7 × 10⁻⁶ /K, soldered to an FR-4 circuit board, at about 14 to 17 × 10⁻⁶ /K in plane, loads its solder joints every time the board heats and cools.
Thermal gradients
Thermal stress does not need external restraint. If one region of a component is hotter than another, the hot region wants to expand more than the cooler material around it, and each restrains the other. When a thick part is heated quickly at its surface, the surface goes into compression and the cooler core into tension. When it is cooled quickly, as in quenching, the surface goes into tension, which is why rapid cooling can crack thick or brittle sections.
The shape of the gradient matters. A linear temperature difference through the thickness of a free plate or beam makes it bend but leaves it free of stress. If the plate is held flat, the surface stress becomes EαΔT / (2(1 − ν)), where ΔT is the difference between the two faces. Non-linear gradients, typical while a part heats up or cools down, cause stress even in a completely free part.
Brittle materials are most at risk from sudden changes. Their resistance to thermal shock rises with strength and falls with stiffness and expansion, roughly in proportion to σf(1 − ν) / (Eα), where σf is the fracture strength. Borosilicate glass, with α of about 3.3 × 10⁻⁶ /K against roughly 9 × 10⁻⁶ /K for ordinary soda-lime glass, survives temperature changes that would crack the latter.
Thermal cycling and fatigue
Every heat-up and cool-down is a stress cycle, and thermal stress is driven by imposed strain rather than imposed load. That has a useful consequence and a harmful one. If the first heat-up yields a small region, the stress there relaxes and the part often settles into elastic cycling. As a rough guide, this shakedown occurs when the elastically calculated stress range stays below about twice the yield strength. Above that, the region yields in tension and compression in every cycle, and low-cycle fatigue sets the life.
Typical examples are exhaust manifolds, turbocharger housings, brake discs, injection mould tools, heat exchanger tube sheets and power electronics, where the joints between silicon, solder, copper and ceramic layers see a cycle every time the load changes. Predicting those lives is a fatigue problem, covered in our fatigue analysis guide.
Design ways to relieve thermal stress
- Let it move. Fix the part at one point and let the others slide, using slotted holes, sliding supports or flexures. Fixing at the centre rather than at one end halves the movement each end must take.
- Add compliance. Expansion loops and bellows in pipework, flexible mounts, and thick bond lines of a low-modulus adhesive all absorb movement that would otherwise become stress.
- Match expansion. Choose materials with similar α where parts are rigidly joined, or use a low-expansion alloy where metal must be sealed to glass or ceramic.
- Reduce the gradient. Even wall thickness, better conduction paths and controlled heating and cooling rates all reduce temperature differences within a part.
- Reduce the temperature change. Cooling, insulation and heat shielding lower ΔT at source.
- Remove sharp corners. Thermal stress concentrates at the same notches as mechanical stress, so generous radii help here too.
- Mind the stress-free temperature. Parts are stress-free at the temperature at which they were assembled, bonded or cured. An adhesive that cures at 120 °C locks in stress as it cools, before the product ever runs.
What a thermal stress study needs
A thermal stress analysis has two halves: the temperature field and the structural response to it. The temperatures can come from a thermal simulation, steady-state or transient, or from measured data. The structural half then needs:
- Expansion coefficient, Young’s modulus and Poisson’s ratio for each material, ideally across the operating temperature range.
- Yield strength at temperature, since many metals lose strength as they heat.
- The stress-free reference temperature for each joint.
- How the parts are fixed and joined: bolts, slots, welds, adhesives and contacts.
- The number and range of thermal cycles, if fatigue matters.
When to bring in a specialist
Thermal stress sits between two kinds of physics, and the answer depends on getting both right together. SixGate™’s thermal stress analysis couples the temperature field to the structural model through steady operation, warm-up and repeated cycling. The result is a change you can act on, such as a slotted fixing, a different alloy pairing, a revised bond line or a better heat path, and the same team can build that change into the detailed design and its drawings, then take it into production. Where heat is coupled with flow or electromagnetics as well, see multiphysics simulation.