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Mechanical

Thermal stress in a restrained bar

Stress produced when a fully restrained bar is heated or cooled.

Ref
MEC-26
Rev
v1
Field
Mechanical
Result
σ [MPa]

Assumes full axial restraint of a uniform linearly elastic bar, uniform temperature change and free lateral contraction. Tension is positive, so heating a restrained bar gives compression. αT is shown in 10⁻⁶ K⁻¹.

A bar that cannot expand when it heats up pushes against its supports; the push is the stress given here.

Calculator

Solve for
MPa
GPa
10400
10⁻⁶ K⁻¹
130
K
−200500
  • Free thermal strain600 µε
  • Free expansion of a 1 m bar0.6 mm

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Thermal stress (MPa) current value −120 MPa
ΔTTemperature changeKey driver
−1,200480
αTExpansion coefficient
−300−10
EYoung’s modulus
−240−6
  • A steel bar heated by 50 K and fully restrained carries about 120 MPa of compression, roughly half the yield strength of mild steel.
  • Allowing expansion joints, sliding supports or gaps removes this stress.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (MPa)Local stepChange in result
ΔT Temperature change−200 to 500 K−1,2004801,680+10%+10 %
αT Expansion coefficient1 to 30 10⁻⁶ K⁻¹−300−10290+10%+10 %
E Young’s modulus10 to 400 GPa−240−6234+10%+10 %

Sweep of Temperature change

Temperature change (K)Thermal stress (MPa)
−200480
−112.5270
−2560
62.5−150
150−360
237.5−570
325−780
412.5−990
500−1,200

Sweep one variable

Thermal stress (MPa)Hover, touch, or use the arrow keys on the chart
−1000−5000500 −2000200400 Temperature change (K) now −120

Variables and how each one contributes

σ

Thermal stress

Result · MPa

What the equation gives you. Everything else feeds it.

E

Young’s modulus

GPa · typical 10 to 400

Elastic stiffness. Directly proportional.

αT

Expansion coefficient

10⁻⁶ K⁻¹ · typical 1 to 30

Linear thermal expansion coefficient (steel 12). Directly proportional.

ΔT

Temperature change

K · typical −200 to 500

Uniform change in temperature. Directly proportional; heating gives compression.

Worked example with the values above

  1. Thermal strain: αTΔT = 1.2×10−5 × 50 = 6×10−4.
  2. Elastic modulus: E = 200,000 MPa.
  3. σ = −E αT ΔT = −120 MPa (negative: compression).

Assumptions and limits

  • Full restraint of a uniform bar.
  • Linear elastic behaviour.

Common mistakes

  • Forgetting the minus sign.
  • Using the wrong α units.

Statistics and uncertainty

Stress depends on three properties each known to 5–10%, so expect about 15% uncertainty.

Where it is used

  • Rail, pipe and structure expansion design.
  • Joints between dissimilar materials.

Maths you need

  • Young’s modulus: stiffness
  • Thermal expansion: ΔL = α L ΔT

Related formulas

Practise this formula