Thermal Stress and Composite Bars
The one place a large stress appears with no load at all. A 50 °C rise in a restrained steel bar produces 120 MPa, and the bar's length does not enter the formula.
Skip to the animationA free body expands under heating with no stress at all; restraining that expansion produces σ = EαΔT — a stress that depends on the temperature change and the material but not on the member's length, which is why a long restrained rail is no safer than a short one.
Free expansion is free
Heating widens the mean spacing of atoms, so an unrestrained body expands by δ = αLΔT. Nothing resists the change, so no stress arises anywhere. Establishing this first matters, because the whole topic is about what happens when it is prevented.
Full restraint
- 1Imagine releasing one end and letting the bar expand freely by αLΔT.
- 2Now push it back to its original length — a compressive strain of exactly αΔT.
- 3Multiply by E: the stress is σ = EαΔT.
- 4Note what cancelled: the length L is nowhere in the answer.
For steel, Eα ≈ 2.4 MPa per °C. A 50 °C rise in a fully restrained member gives 120 MPa with nothing hung on it. A 75 °C swing reaches most of mild steel's yield stress.
Since length does not appear, a 1 m bar and a 100 m bar reach the same stress. Continuous welded rail is laid pre-tensioned for this reason, and a bridge without expansion joints would tear out its own bearings.
Partial restraint
Provide a gap g and the bar expands freely until it closes it. Only the surplus is resisted, so σ = E(αLΔT − g)/L. If the gap exceeds the free expansion, the stress is zero — the bar never touches.
Here L is back in the formula, because the gap is a fixed distance while the free expansion grows with length. That is why joint spacing is a design decision rather than a constant.
Composite bars
Bond two materials with different coefficients of expansion and heat them. Each wants a different final length and they are stuck with each other.
- 1The higher-α material is held back, ending in compression.
- 2The lower-α material is dragged out, ending in tension.
- 3Two conditions solve it: internal forces sum to zero (no external load), and both must finish the same length.
Bond them face to face instead of end to end and the mismatch resolves by bending: the higher-expansion side takes the outside of the curve. Curvature is proportional to (α₁−α₂)ΔT, so tip deflection reads temperature directly. That is a bimetallic strip, and it is how thermostats and mechanical thermometer dials work.
Where σ = EαΔT is not enough
| Situation | Why the simple formula fails |
|---|---|
| Temperature gradient through the thickness | Becomes a bending problem — this is how quenching cracks a component |
| Very large ΔT | α itself varies with temperature |
| Fire exposure | E falls steeply; structural steel keeps under half its strength at 600 °C |
| Yielding under thermal load | The stress is limited by yield, and the member deforms permanently |
Thermal stress is also self-limiting in a way mechanical stress is not: once the member yields, the strain mismatch is relieved and the stress stops growing. That is why thermal loading is treated differently from dead load in design codes.
The numbers you will be asked for
- Free expansion
δ = α · L · ΔT
- Fully restrained stress
σ = E · α · ΔT
no length term
- Partially restrained
σ = E(αLΔT − g) / L
- Composite bar
P₁ = −P₂ and δ₁ = δ₂
two equations, two unknowns
- Steel constant
Eα ≈ 2.4 MPa per °C
Watch it work
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