Thermal Stress Calculator

σ = E·α·ΔT derived rather than quoted — and the fact that it does not depend on length or on cross-section, which is why thermal stress surprises people. Plus partial restraint, where length comes back, and the bonded pair.

Last updated 24 Sep 2026

Thermal stress

Material, ΔT and restraint → stress and force

Material of the restrained memberBoth E and α come from this. The product E·α is what matters and it varies less between metals than either factor does on its own — steel is 2.34 MPa/K, aluminium 1.63, copper 1.96, titanium 0.89 — because the stiff metals tend to be the ones that expand least. Invar is the outlier at 0.23.

Temperature CHANGE ΔT (K or °C)The rise or fall from the temperature at which the member was stress-free. A rise in a restrained member gives COMPRESSION; a fall gives tension, which is usually the worse case because tension opens cracks. Note that this is a temperature DIFFERENCE, so the kelvin and the degree Celsius are the same size and no conversion is needed.

RestraintFully restrained is the worst case and the one the headline formula gives. Free is zero stress, and is what an expansion joint, a slotted hole or a sliding foot is for. Partial is everything real — and note that as soon as the restraint is finite, the member’s own length starts to matter.

Restraint stiffness (N/mm) — partial restraint onlyHow hard the surrounding structure resists the member’s expansion. Compare it with the member’s own axial stiffness EA/L, which the page computes: if the restraint is much stiffer, you are effectively fully restrained; if it is much softer, the member is effectively free. Equal stiffnesses give exactly half the full stress.

Member length L (mm)It does NOT appear in the fully restrained stress — that is the headline result of this page. It DOES appear in the free expansion, and it does appear in the partial restraint case through the member’s own stiffness EA/L. So a long member against a given restraint sees more stress than a short one, which is the honest converse of the headline.

Cross-sectional area A (mm²)Also absent from the fully restrained stress. It sets the FORCE, which is σA and which is what the surrounding structure has to react — often the number that actually matters, because it is the bolts or the brackets that fail rather than the member.

Yield strength of the member (N/mm²)For the yield comparison, which is the whole point of the 100 K example. A36 steel is 250 minimum; 304 stainless annealed 215; 6061-T6 276 typical and 240 minimum; 4140 quenched and tempered around 655. Grey cast iron has no yield point at all — use its tensile strength of about 250 and treat the answer with the brittle criteria instead.

SECOND material, for a bonded pairFor the bimetallic joint section: an aluminium housing on a steel shaft, a brazed assembly, a shrink fit that then gets hot, a bimetallic strip. What drives the stress is the DIFFERENCE in expansion coefficient, not either value.

Thickness of the first layer (mm)For the bonded pair. The two thicknesses set both the stress split and the curvature; equal thicknesses of equal modulus give the classical 1.5·Δα·ΔT/h curvature, which this page uses as a check on its own general solution.

Thickness of the second layer (mm)The other layer. Making one layer much thicker than the other reduces the curvature towards zero and pushes all the stress into the thin layer, which is why a thin coating on a thick substrate cracks or spalls rather than bending the part.

Not a circuit: this page's headline result drawn rather than asserted. Two bars of completely different length, each held between rigid walls, each taken through the same temperature change — and the fill level in the two bars is IDENTICAL, because σ = E·α·ΔT contains no length and no area. A 70 mm stub and a 250 mm bar of the same material develop exactly the same stress, and so would a 9 metre one. That is the fact that makes thermal stress surprising: it cannot be reduced by making the part shorter, and making it thicker only raises the force the walls have to react. The bar underneath is the same stress measured against the material's yield strength, with yield at the right-hand end of the scale — and for carbon steel through a hundred kelvin it arrives startlingly close to that end. Change the restraint to partial and the fill level drops while the lengths stay the same, which is the other half of the story: once the restraint is not rigid, the member's own stiffness EA/L enters, L is inside it, and length matters after all.

234.00N/mm²Example

A carbon steel bar, fully restrained, through a 100 K temperature rise

Restrained, partly restrained, and bonded

σ = E·α·ΔT · partial: σ = E·α·ΔT · k_r/(k_r + EA/L) · bonded pair: σ₁ = E₁·ΔT·A₂E₂(α₂−α₁)/(A₁E₁ + A₂E₂)

E·α: the stress per kelvin of full restraint. 2.34 N/mm² per K for carbon steel, 1.63 for aluminium, 0.23 for Invar. Note that it varies far less between metals than E or α does alone
ΔT: the temperature CHANGE from the stress-free state, so kelvin and degrees Celsius are interchangeable. A rise gives compression in a restrained member and a fall gives tension, which is usually worse
L, A: ABSENT from the fully restrained result, which is the point of this page. They reappear the instant the restraint is finite, because the member’s own stiffness is EA/L and L is in it — so a long member against a given restraint sees more stress than a short one
k_r: restraint stiffness in N/mm. Compare it with the member’s own EA/L: ten times stiffer already gives 91 per cent of the full stress, and equal gives exactly half
α₂ − α₁: for a bonded pair it is the DIFFERENCE that drives everything, not either coefficient. Identical materials give zero stress however much they expand
1/ρ: the bimetal curvature, from an exact two-layer beam solve — zero net force and zero net moment, one linear strain through both layers. At equal thickness and equal modulus it reduces to 1.5ΔαΔT/h, which this page uses as a check on itself

Worked example

A carbon steel bar, fully restrained, through a 100 K temperature rise
Derive it rather than quote it. Free to expand, the bar would lengthen by a strain of α·ΔT — for carbon steel 11.7 × 10⁻⁶ × 100 = 0.00117, which is 1,170 microstrain. That is the strain the bar WANTS
Fully restrained, it is not allowed any of it. So the restraint must impose an equal and opposite MECHANICAL strain of −0.00117 to keep the total at zero. Hooke's law turns that into a stress: σ = E × ε = 200,000 × 0.00117 = 234.0 N/mm², compressive
NOW LOOK AT WHAT IS NOT IN THAT CALCULATION. There is no length in it and no area in it. A 20 mm stub and a 9 m bar, a 5 mm² wire and a 50,000 mm² block — all of them develop 234.0 N/mm² from the same 100 K rise, if they are fully restrained. That is genuinely counter-intuitive and it is why thermal stress catches people: it cannot be reduced by making the part shorter or by making it thicker. Thicker only raises the FORCE, which is σA, and the force is usually what breaks the brackets
Is 234 N/mm² a lot? A36 structural steel's minimum yield is 250 N/mm², so this is 94 per cent of yield from a 100 kelvin rise and nothing else. No load, no pressure, no weight — just a warm day and a rigid restraint. The temperature change that would reach yield is 250/2.34 = 107 K. That is less than the difference between a cold workshop and a machine that has been running for an hour
Austenitic stainless is worse, and this is the part that surprises people twice. Its α is 17.2 against carbon steel's 11.7 and its E is about the same, so E·α is 3.32 N/mm² per K and it yields at only 65 K. ALUMINIUM, on the other hand, is BETTER: it expands twice as much as steel but is a third as stiff, so E·α is only 1.63 and a restrained aluminium member is LESS stressed than a steel one through the same temperature change. Expansion coefficient alone does not tell you who is in trouble
PARTIAL RESTRAINT, which is what everything real is. The member's own axial stiffness is EA/L = 200,000 × 500/500 = 200,000 N/mm. The free expansion, 0.585 mm, is shared between the member compressing and the restraint stretching, in inverse proportion to stiffness, so σ = E·α·ΔT · k_r/(k_r + k_m). At a restraint as stiff as the member you get exactly half the full stress; at ten times as stiff, 91 per cent; at a tenth, 9 per cent
AND HERE IS THE CONVERSE OF THE HEADLINE. The member's stiffness EA/L contains L, so as soon as the restraint is finite, LENGTH MATTERS after all — and it matters the wrong way round from intuition. A longer member is axially SOFTER, so it loses the stiffness competition against a given restraint and ends up with MORE of the full stress, not less. Doubling the length of this bar against a 1 MN/mm restraint takes the stress from 195.000 to 212.727 N/mm². Only in the fully restrained limit does length genuinely drop out
THE BONDED PAIR. An aluminium part bonded to a steel one of equal area, through the same 100 K: the two must share one strain, and the forces must balance. Solving those two conditions gives the aluminium at 61.0 N/mm² (compressive, because it wanted to grow more) and the steel at -61.0 (tensile). Note that these are much larger than either material would see alone against a rigid wall, because each is being restrained by a material of comparable stiffness AND the mismatch is the full difference in α. If the pair is free to bend instead, most of that is relieved as curvature: the exact two-layer solve gives 1/ρ = -0.0008281 /mm, a radius of -1,208 mm, and a -1.04 mm bow in a 100 mm strip. That is a bimetallic thermostat, and it is the same arithmetic

E·α by material, and the temperature change that reaches yield

The last column is the one to remember, and it is startlingly small for most engineering metals: a fully restrained carbon steel member yields at around a hundred kelvin of temperature change, and stainless steel — which expands half again as much as carbon steel — yields at under sixty. Those are not extreme temperatures; they are the difference between a workshop and a working machine. Note that E·α varies much less between metals than either factor does alone, because the stiff metals are generally the ones that expand least: steel is 2.34 N/mm² per kelvin, copper 1.96, aluminium 1.63, titanium 0.89. Aluminium expands twice as much as steel but is a third as stiff, so a restrained aluminium member is LESS stressed than a steel one through the same temperature change — which is the opposite of most people’s intuition. Invar is the real outlier at 0.23. Two published values of E are shown where the sources disagree, and grey cast iron’s 100-against-130 GPa spread is a thirty per cent spread on its thermal stress. The same designation can mean different dimensions in different standards families — ANSI against ISO, inch against metric, one national standard against another. The family used here is named beside every figure; check which one your part was made to.
Material	α (10⁻⁶/K)	E (GPa)	Second published E	E·α (N/mm² per K)	σ at ΔT = 100 K	Yield (N/mm²)	ΔT to yield if fully restrained
Carbon steel (A36, 1020)	11.7	200	215	2.340	234	250	107 K
Alloy steel (4140)	12.3	205	—	2.522	252	655	260 K
Stainless steel, austenitic (304)	17.2	193	195	3.320	332	215	65 K
Stainless steel, austenitic (316)	15.9	193	—	3.069	307	205	67 K
Grey cast iron	11.4	100	130	1.140	114	no yield point	—
Aluminium 6061-T6	23.6	69	68	1.628	163	240	147 K
Brass, cartridge (C26000)	19.9	110	102	2.189	219	100	46 K
Bronze, phosphor (C51000)	17.8	116	—	2.065	206	140	68 K
Copper (C11000)	17.0	115	117	1.955	195	70	36 K
Titanium, grade 1	8.6	103	107	0.886	89	170	192 K
Magnesium (AZ31B)	26.0	45	—	1.170	117	200	171 K
Invar (Fe-36Ni)	1.6	141	—	0.226	23	275	1,219 K

Restraint stiffness, from rigid to free, on this page’s own 500 mm × 500 mm² steel bar

The whole axis in one table, and the useful thing is how FAST it goes. A restraint ten times stiffer than the member already develops 91 per cent of the full stress, so “not quite rigid” is very nearly the same as rigid. Equal stiffnesses give exactly half. To get the stress down to a tenth you need a restraint nine times SOFTER than the member, which in practice means a designed compliance — an expansion loop, a bellows, a slotted hole, a flexure — and not merely a slightly flexible bracket. And note the sixth column: buying that compliance costs movement, and the movement is where the alignment goes. That is the real trade, and it is why the sliding foot on a long machine bed is a designed part and not an oversight. Note too that the member’s own stiffness EA/L contains L, so this entire table shifts if the member’s length changes — which is the converse of the headline result. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Restraint	k_r (N/mm)	k_r/(k_r+k_m)	Stress (N/mm²)	… of the fully restrained value	Movement (mm)	Per cent of yield
Rigid (k_r → ∞)	∞	1.0000	234.00	100.0 %	0.0000	93.6 %
100× the member	20,000,000	0.9901	231.68	99.0 %	0.0058	92.7 %
10× the member	2,000,000	0.9091	212.73	90.9 %	0.0532	85.1 %
Equal to the member	200,000	0.5000	117.00	50.0 %	0.2925	46.8 %
One tenth	20,000	0.0909	21.27	9.1 %	0.5318	8.5 %
One hundredth	2,000	0.0099	2.32	1.0 %	0.5792	0.9 %
Free (k_r = 0)	0	0.0000	0.00	0.0 %	0.5850	0.0 %

Bonded pairs through a 100 K rise, 1 mm of each layer

Two mechanisms are at work in every row and they have to be kept apart. If the pair is CONSTRAINED FROM BENDING — an aluminium housing bolted flat onto a steel base, a cladding on a thick substrate, a shrink fit — the mismatch has to be swallowed as axial stress, and the third and fourth columns are what it costs: the high-expansion material goes into compression and the low-expansion one into tension, in the ratio that balances the forces. If the pair is FREE TO BEND, most of that stress is relieved by curvature instead, and the strip bends towards the low-expansion side. The last row is the control: identical materials give zero stress and zero curvature, as they must. And the last column is the one that ends joints. The stress is DISCONTINUOUS across the interface — it has to be, because the two materials share a strain and have different moduli — and that jump is a shear the adhesive, braze or plating has to carry. It is why a plated or coated part spalls at a temperature change rather than bending, and why a large bonded assembly of dissimilar metals is a thermal problem before it is a structural one. This page sizes a part; it does not certify one. Where the answer carries a consequence — a load path, a lifting duty, a pressure boundary, a fastener holding something that can fall — confirm it against the design code that governs the application, and against the manufacturer’s own rating, before relying on it.
Pair	Δα (10⁻⁶/K)	Stress in the first (N/mm²)	… in the second	Curvature (1/mm)	Radius (mm)	Tip movement of a 100 mm strip (mm)	Interface stress jump
Aluminium on carbon steel	11.9	-61.0	61.0	0.0008281	1,208	1.035	148.5
Aluminium on Invar	22.0	-101.9	101.9	0.0015968	626	1.996	223.6
Copper on carbon steel	5.3	-38.7	38.7	0.0003898	2,565	0.487	81.9
304 stainless on carbon steel	5.5	-54.0	54.0	0.0004125	2,424	0.516	108.1
Brass on carbon steel	8.2	-58.2	58.2	0.0006012	1,663	0.751	124.2
Carbon steel on carbon steel	0.0	0.0	0.0	0.0000000	∞	0.000	0.0

Which page owns which half of the thermal problem

The split is clean and it is worth stating because the two halves feel like one problem. A free part changes size and carries no stress. A fully restrained part carries stress and does not change size. Everything real is in between, and the SAME temperature change produces some of each. The dimensional page owns the first, this page owns the second, and they use the same expansion coefficients from the same source so they cannot disagree with each other.
The question	Which page	What it gives you
My part got hot and changed SIZE. What is the new dimension, and what has happened to my fit?	Thermal effect on fit	the DIMENSIONAL half. It takes an ISO 286 fit specified at 20 °C and tells you what it becomes at temperature: the clearance, the temperature at which interference turns to clearance, the heating needed to assemble a shrink fit, and the correction for measuring parts warm
My part got hot and could NOT change size. What is the stress?	this page	the STRESS half. σ = EαΔT for full restraint, the general partial-restraint case, the force the restraint has to react, and the comparison against yield
How much interference do I need, and what pressure and torque does the fit develop?	Interference and shrink fit	the joint’s own mechanics: contact pressure, holding torque and the hoop stress in the hub, from Lamé’s equations
My restrained part yielded once and now behaves differently every cycle.	neither — and it is worth naming	thermal RATCHETING. Once a restrained member has yielded in compression on heating, it is permanently short, so on cooling it goes into TENSION. Repeat and the part walks. It needs a cyclic-plasticity treatment and is not a closed-form problem

Why length does not appear, and what happens when it comes back

σ = E·α·ΔT, and the remarkable thing is what is missing from it. Derive it in two steps. Free to expand, a member would take a thermal strain of α·ΔT. Fully restrained, it is allowed none, so the restraint imposes an equal and opposite mechanical strain, and Hooke’s law turns that into a stress E·α·ΔT. There is no length in that expression and no area. A 20 mm stub and a 9 metre bar of the same material, both fully restrained, develop exactly the same stress from the same temperature change. So does a 5 mm² wire and a 50,000 mm² block. That is genuinely counter-intuitive, it is why thermal stress surprises people, and it has a hard practical consequence: you cannot reduce a thermal stress by making the part shorter or thicker. Thicker only raises the force σA, and the force is usually what tears out the brackets.

The numbers are larger than they feel. Carbon steel’s E·α is 2.34 N/mm² per kelvin, so a 100 K rise in a fully restrained bar gives 234 N/mm² — 94 per cent of A36’s 250 N/mm² minimum yield, from a warm day and a rigid restraint with no load on the part at all. The temperature change that reaches yield is about 107 kelvin. Austenitic stainless is worse: it expands half again as much for the same modulus, so it yields at under 60 K. And aluminium is BETTER, which is the opposite of most people’s intuition — it expands twice as much as steel but is a third as stiff, so its E·α is 1.63 and a restrained aluminium member is less stressed than a steel one through the same excursion. Expansion coefficient alone does not tell you who is in trouble; the product does.

Partial restraint is the general case, and it brings length back the wrong way round. Nothing real is rigidly restrained. Put a restraint stiffness k_r against the member’s own axial stiffness k_m = EA/L and the free expansion is shared between them in inverse proportion, giving σ = E·α·ΔT·k_r/(k_r + k_m). That expression has both limits right: rigid restraint gives the full stress and zero restraint gives zero. What is worth noticing is that k_m contains L — so the moment the restraint is finite, length matters after all, and it matters in the direction intuition does not expect. A LONGER member is axially softer, loses the stiffness competition against a given restraint, and therefore develops MORE of the full stress. Only in the rigid limit does length genuinely drop out. The table on this page walks the whole axis from rigid to free, and the useful lesson is how fast it moves: a restraint only ten times stiffer than the member already produces 91 per cent of the full stress, so “fairly stiff” and “rigid” are nearly the same thing.

Two bonded materials: the difference is what matters. An aluminium housing on a steel shaft, a brazed assembly, a plated part, a bimetallic strip. If the pair cannot bend, both must take the same strain and the forces must balance; solving those two conditions gives the high-expansion material in compression and the low-expansion one in tension. If the pair CAN bend, most of that stress is relieved as curvature instead, and the assembly bows towards the low-expansion side. This page solves the exact two-layer beam — zero net force, zero net moment, one linear strain distribution through both layers — rather than quoting a published closed form, and checks it against the classical 1.5·Δα·ΔT/h for equal thicknesses of equal modulus. What both cases share is the thing that ends joints: the stress is DISCONTINUOUS across the interface, because the two materials share a strain and have different moduli, and that jump is a shear the bond has to carry. It is why a coating spalls rather than bends, and why a large bonded assembly of dissimilar metals is a thermal problem before it is a structural one.

What happens after it yields, which is where the real failures are. The linear model stops at first yield and something more interesting takes over. A restrained member heated past yield takes a permanent compressive set, so when it cools back down it is too SHORT and goes into tension. Cycle it and each pass adds a little more permanent strain: the part ratchets, and it fails after a number of cycles rather than on the first, usually in tension and usually at a weld or a fillet. That is thermal fatigue and thermal ratcheting, it is one of the commonest failure modes in anything that heats and cools in service, and it is a cyclic-plasticity problem rather than a closed-form one. The fatigue page covers the cyclic side once you have a stress range; the plastic ratcheting itself is not a closed-form calculation and this page names it rather than pretending to compute it.

This page owns the stress half only. The thermal effect on fit page owns the DIMENSIONAL half — what an ISO 286 fit specified at 20 °C becomes at operating temperature, the temperature at which interference turns to clearance, the heating needed to assemble a shrink fit, and the correction for measuring parts warm. The split is clean: a free part changes size and carries no stress, a fully restrained part carries stress and does not change size, and everything real produces some of each from the same temperature change. Both pages take their expansion coefficients from the same published source, so they cannot disagree with each other.

Frequently asked questions

Why doesn’t thermal stress depend on the length of the part?

Because both the thermal expansion and the mechanical compression that cancels it are STRAINS, and a strain is a fraction rather than a length. The thermal strain α·ΔT is the same in every millimetre of the member, so the mechanical strain needed to cancel it is the same too, and the stress that produces it is E times that. A long member wants to expand more in absolute terms, but it is also being held over a longer distance, and the two scale together exactly. The practical consequence is that you cannot design a thermal stress problem away by shortening the part.

Does making the part thicker help?

Not for the stress, and it makes the force worse. Area is absent from σ = E·α·ΔT just as length is, so a thicker fully restrained member sees exactly the same stress. What does change is the FORCE, σA, which the restraint has to react — so thickening a restrained member loads its brackets, bolts and welds harder for no gain in the member itself. If the restraint is only partial, a thicker member is axially stiffer and does win a little (it wins the stiffness competition against the restraint), but the force still rises.

Is 234 N/mm² really what a steel bar sees through 100 K?

Yes, with the properties this site uses: E = 200 GPa for A36 and α = 11.7 × 10⁻⁶/K from AmesWeb’s room-temperature column. The figure is often quoted as 250, which comes from pairing E = 207 GPa with α = 12 × 10⁻⁶/K — both defensible numbers, and the spread is about seven per cent. Either way the point is the same and it is the point worth remembering: a 100 kelvin rise takes a fully restrained mild steel member to somewhere between 93 and 100 per cent of its minimum yield strength, with no mechanical load on it at all.

Which expands more dangerously, steel or aluminium?

Steel, which surprises most people. Aluminium’s expansion coefficient is twice steel’s, so a free aluminium part moves twice as far — that is the dimensional problem and the thermal effect on fit page deals with it. But its modulus is a third of steel’s, and thermal STRESS is the product E·α: 1.63 N/mm² per kelvin for aluminium against 2.34 for steel. So a restrained aluminium member is about thirty per cent less stressed than a steel one through the same temperature change. Austenitic stainless is the worst of the common metals at 3.32, because it has steel’s modulus and half again its expansion.

Is a temperature rise or a fall worse?

A fall, usually, for three reasons. It puts a restrained member into TENSION, and tension opens and propagates cracks where compression closes them. A brittle material — cast iron, a casting, concrete, a weld with a defect — is several times weaker in tension than in compression. And a ferritic steel cooled below its transition temperature loses toughness, so a cold restrained member can fail brittly from a small flaw at a stress that would be harmless warm. It is also why cast iron machine bases crack as they cool from a hot process rather than while they heat.

What is a realistic restraint stiffness?

Compare it with the member’s own EA/L rather than trying to judge it absolutely, which this page computes for you. A short steel bracket loaded in its stiff direction has an enormous rate and is effectively rigid. A long bolt, a rubber mount, a bellows or a deliberate flexure can be genuinely soft. The useful thresholds: a restraint ten times stiffer than the member gives 91 per cent of the full stress, equal stiffness gives exactly half, and getting down to ten per cent needs a restraint nine times softer than the member. That last one is a designed compliance, not an incidentally flexible bracket.

Why does a bimetallic strip bend towards the low-expansion side?

Because the high-expansion layer wants to be longer and the bond will not let the two layers slide. The only way to accommodate a length difference between two bonded layers is to curve, with the longer layer on the outside of the curve — so the strip bows with the high-expansion material on the convex face, which means it bends TOWARDS the low-expansion side. That is a thermostat. The same mechanism warps a brazed assembly, bows a plated part and is why a weld bead pulls the plate it is laid on.

What happens if the member yields?

It stops being a linear problem and becomes a much more dangerous one. The stress stops rising at roughly yield, the excess strain becomes a permanent set, and the member is therefore permanently SHORT when it cools back down — so it goes into tension on cooling. Cycle it and each pass adds more permanent strain: the part ratchets and fails after a number of cycles, usually in tension and usually at a weld or a fillet. That is thermal ratcheting, it is one of the commonest real failure modes of restrained members, and it needs a cyclic-plasticity treatment rather than a closed form. This page flags it rather than calculating it.

Related calculators

References

AmesWeb, Thermal expansion coefficient (CTE) of metals. The source for the room-temperature expansion coefficients on this page — carbon steel 11.7, alloy steel 4140 12.3, austenitic stainless 304 17.2 and 316 15.9, grey cast iron 11.4, aluminium 6061 23.6, cartridge brass 19.9, phosphor bronze 17.8, copper 17.0, titanium grade 1 8.6, magnesium AZ31B 26.0 and Invar 1.6, all in 10⁻⁶/K. This is the SAME column Thermal effect on fit already uses in this plugin, so the two pages cannot disagree with one another.
AmesWeb, Modulus of elasticity (Young’s modulus) for metals, and The Engineering ToolBox, Metals and alloys — Young’s modulus of elasticity. Two sources for E, and they differ: grey cast iron is printed at 92 GPa by one and 130 by the other, carbon steel at 200 and at 202–215, copper at 115 and 117, brass at 110 and 102. Both readings are carried and both are shown, because E multiplies straight into the answer — a thirty per cent spread on cast iron’s modulus is a thirty per cent spread on its thermal stress.
ASTM A36/A36M, Standard Specification for Carbon Structural Steel. Cited by number, not reproduced. The two figures this page uses from it — a minimum yield of 250 MPa (36 ksi) for plate, bar and shapes under 200 mm and a tensile range of 400–550 MPa — are the specification’s headline values and are quoted here from the public summary of the standard rather than from the document.
Aluminium 6061-T6 is taken at a minimum yield of 240 MPa with typical values near 270, a minimum tensile of 290 MPa with typical values near 310, and E = 68–69 GPa. Stainless 304 and 316 in the annealed condition are taken at 215 and 205 MPa yield. Grey cast iron is carried with NO yield strength at all, deliberately: it is a brittle material with no yield point, its tensile strength is a fraction of its compressive strength, and a von Mises check on it is the wrong check. That is the whole reason the failure-criteria page asks for the material class before it asks for anything else.
S. Timoshenko, “Analysis of bi-metal thermostats”, Journal of the Optical Society of America 11(3), 1925, pp. 233–255 — the original of every bimetallic-strip curvature formula in print. The fetched summaries of it could not be trusted: one returned “1/ρ = 12ΔαΔT/h²” for the equal-thickness case, which is dimensionally impossible — a curvature is 1/length and ΔαΔT is dimensionless, so the thickness cannot appear squared. This page therefore does not quote the published closed form at all: it solves the two-layer beam from scratch (zero net force, zero net moment, one linear strain distribution through both layers), which is a 2×2 system. The result was checked three ways: against a 4,000-fibre numerical layered-beam solve, against the classical equal-thickness equal-modulus value 1.5ΔαΔT/h, and against Timoshenko’s own m–n form, with which it agrees to nine figures.
R. C. Hibbeler, Mechanics of Materials, on the superposition of axial and bending stress, on Mohr’s circle and the stress-transformation equations, and on the absolute maximum shear stress in plane stress — the point that the third principal stress is zero rather than absent, so the governing shear may be σ₁/2 and not (σ₁ − σ₂)/2.