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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.

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A 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

  1. 1Imagine releasing one end and letting the bar expand freely by αLΔT.
  2. 2Now push it back to its original length — a compressive strain of exactly αΔT.
  3. 3Multiply by E: the stress is σ = EαΔT.
  4. 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.

  1. 1The higher-α material is held back, ending in compression.
  2. 2The lower-α material is dragged out, ending in tension.
  3. 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

SituationWhy the simple formula fails
Temperature gradient through the thicknessBecomes a bending problem — this is how quenching cracks a component
Very large ΔTα itself varies with temperature
Fire exposureE falls steeply; structural steel keeps under half its strength at 600 °C
Yielding under thermal loadThe 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

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Check yourself

question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

A steel bar is heated 40 °C with both ends free. What is the stress in it?
The same bar is now fully restrained and heated 40 °C. Does its length affect the stress?
Two bonded strips with different coefficients of expansion are heated. What happens?
When does σ = EαΔT stop being adequate?

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4 still unanswered — the dots above jump straight to them.