How this instrument works
Heat a bar and, left free, it grows by ΔL = L₀·α·ΔT — ordinary thermal expansion. Clamp both ends so it cannot grow at all, and that same fractional elongation, α·ΔT, has nowhere to go except into the material as elastic strain. Hooke's law turns strain into stress by multiplying it by Young's modulus, so substituting the blocked thermal strain for ordinary mechanical strain gives σ = EαΔT directly. Notice what dropped out of the result: the original length L₀. A restrained 5 mm shim and a restrained 5 m shaft of the same steel, heated by the same 50°C, carry identical stress — length only governs how far a free part would have travelled, not how hard a trapped one pushes back.
The relation is the one-dimensional case of the Duhamel–Neumann thermoelastic law, which the mathematician Jean-Marie Duhamel worked out in 1837 by folding a temperature term into Hooke's ordinary stress–strain relation; Franz Neumann reached an equivalent result independently soon after. Process piping engineers reach for the same arithmetic when a hot line is anchored at two fixed points and an expansion loop must be sized to absorb the growth those anchors refuse to allow — skip the loop and the pipe wall itself is left carrying that growth as stress. Electronics packaging runs into a two-material version of the identical problem: a silicon die and its circuit board expand at different rates as a device warms up, and the solder joints between them absorb the mismatch, cycle after cycle, until they crack from fatigue.
Two assumptions hold the formula together. Restraint has to be essentially complete: any give in the supports lets part of the expansion happen for free, which lowers the true stress below the fully-restrained figure in rough proportion to how much motion the supports actually allowed, and a careful analysis models that with an effective restraint fraction rather than assuming zero movement. The predicted stress also has to stay under the material's yield strength for the number to describe reality, because steel does not keep climbing linearly forever as ΔT grows — past yield it deforms plastically instead, the elastic formula stops applying, and a permanent set is what remains once the part cools back down.
- Enter your material's Young's modulus in MPa — steel runs near 200,000 MPa (200 GPa), aluminium closer to 69,000 MPa.
- Enter Coefficient of thermal expansion, 1/K as a decimal — steel is about 1.2e-5, aluminium about 2.3e-5.
- Set Temperature change, °C to the difference between the final and starting temperature; use a negative number for cooling.
- Read Thermal stress in MPa, then compare it against your material's published yield strength to check the elastic assumption still holds.
- If the supports are not perfectly rigid, scale the reading down by the fraction of the free expansion they actually block.
Worked example — a steel pipe fully anchored at both ends
A run of carbon-steel process pipe sits rigidly anchored at both ends, with no expansion loop to take up any slack. Young's modulus for the steel is 200,000 MPa (200 GPa), its coefficient of thermal expansion is 1.2×10⁻⁵ per kelvin, and hot fluid raises the pipe wall by 50°C above its installed temperature. σ = EαΔT = 200,000 × 1.2×10⁻⁵ × 50 = 120 MPa: the pipe wall now carries 120 MPa of stress purely from being denied the room to grow.
That 120 MPa sits close to the yield strength of common structural and piping steel grades, which typically yield somewhere in the 235 to 355 MPa range — 120 MPa alone is not yet a failure, but it eats deeply into the margin meant for the fluid's own internal pressure, welds and any bending the run picks up elsewhere. Changing the pipe's length or diameter would not have changed that 120 MPa figure at all: length and section only matter once this stress is converted into the force or moment the anchors themselves must resist, since the stress in the restrained material depends solely on E, α and ΔT.
Questions
Is the stress from restrained heating tensile or compressive?
Compressive. A restrained bar heated above its installed temperature wants to grow, and the supports push back against that growth, squeezing the material much as a jack would. Cool a restrained bar below its installed temperature instead and the sign flips: it wants to shrink, the supports hold it at length, and the resulting stress is tensile. The formula σ = EαΔT reports magnitude; the physical sign follows from whether ΔT is a rise or a fall.
Why doesn't the length of the restrained part appear anywhere in the formula?
Because stress and displacement are different questions. A free bar's growth, ΔL = L₀αΔT, obviously scales with starting length — a longer bar moves more. But once both ends are pinned, that growth converts straight into strain, αΔT, which is already a length-independent ratio, and multiplying by E turns it into stress. A 10 mm shim and a 10 m shaft of the same restrained steel, heated by the same amount, carry identical stress; length only reenters once you ask how much force the anchors must resist.
What happens once the calculated stress goes past the material's yield strength?
The formula stops being literally true. σ = EαΔT assumes the material stays elastic, able to spring back to its original shape if the restraint were removed. Push the number past yield and the real material instead deforms plastically — it absorbs the extra thermal strain permanently rather than storing it all as recoverable stress, so the true stress plateaus near yield while the part is left slightly shorter, bowed, or otherwise changed once it cools.
Where does the σ = EαΔT relation actually come from?
It is the one-dimensional case of the Duhamel–Neumann thermoelastic law, which adds a temperature term to Hooke's ordinary stress–strain relation. Jean-Marie Duhamel worked out the extension in 1837; Franz Neumann reached an equivalent result independently soon after. Set total strain — mechanical plus thermal — to zero, since restraint forbids any net length change, and the mechanical strain that Hooke's law converts into stress must exactly cancel the thermal strain αΔT, leaving σ = EαΔT for its magnitude.
Does partial restraint just give half the stress of full restraint?
Only if the support happens to block exactly half the free expansion, which is a coincidence rather than a rule. A support of finite stiffness lets some of the bar's thermal growth actually happen, and the stress carried is EαΔT scaled by whatever fraction of that growth the support prevented — near 1 for a very stiff support, well below that for a springy one. Modelling this properly means comparing the bar's own axial stiffness to the support's stiffness, not simply halving the fully-restrained figure.
Can this formula handle a temperature gradient instead of a uniform change?
No — ΔT here is a single, uniform temperature change applied to the whole restrained member. A real gradient, hot on one face and cooler on the other, produces its own separate bending stress on top of any axial term, because each layer through the thickness wants to expand by a different amount. That case needs a thermal-gradient stress analysis, not this uniform-ΔT formula, even though both trace back to the same underlying thermoelastic idea.