Learn: Fully restrained thermal stress
A free bar expands when it is heated. If rigid supports prevent all axial movement, the prevented expansion becomes elastic stress instead; the restraint, not temperature alone, creates this stress.
What the formula is saying
First find the thermal strain αΔT that the bar would have developed freely. Multiplying by E gives the stress magnitude needed to suppress it. This lesson reports compression as positive: positive heating gives positive compressive stress.
Read the symbols in plain language
- E
- Young’s modulus
Elastic stiffness: the stress change needed for a unit strain in the stated material model. It is not a strength limit.
PaOne pascal is one newton per square metre. 1 MPa = 10⁶ Pa.
- α
- Thermal expansion coefficient
Thermal expansion coefficient. The coefficient converts the temperature change into a dimensionless free expansion.
1/KUse 1/K as the base unit shown here. E is in Pa, α in 1/K and ΔT in K. A temperature difference of 40 °C is also 40 K; do not add 273.15 to a difference. The result is Pa.
- ΔT
- Temperature change
Temperature change. The coefficient converts the temperature change into a dimensionless free expansion.
KUse K as the base unit shown here. E is in Pa, α in 1/K and ΔT in K. A temperature difference of 40 °C is also 40 K; do not add 273.15 to a difference. The result is Pa.
- σT
- Result to find
Fully restrained thermal stress. Elastic stiffness converts the prevented strain into compression-positive stress.
Pa
Sort out the units first
E is in Pa, α in 1/K and ΔT in K. A temperature difference of 40 °C is also 40 K; do not add 273.15 to a difference. The result is Pa.
Assumptions before calculating
Treat the material as homogeneous and linearly elastic, with small strains. Use properties for the actual temperature and loading direction; the calculation is a model of behaviour before yielding, not a failure test.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find σT and explain the result in the stated output unit.
- E · Young’s modulus
- 200000000000 Pa
- α · Thermal expansion coefficient
- 0.000012 1/K
- ΔT · Temperature change
- 40 K
Compute the prevented free strain
The coefficient converts the temperature change into a dimensionless free expansion.
(0.000012) × (40) = 0.00048Convert restrained strain to stress
Elastic stiffness converts the prevented strain into compression-positive stress.
(200000000000) × (0.00048) = 96000000 Pa
Fully restrained axial thermal stress; compression positive in this lesson.
Does this worked answer make sense?
Zero temperature change gives zero thermal stress. Cooling produces a negative answer here, meaning tension under the stated convention.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- E · Young’s modulus
- 70000000000 Pa
- α · Thermal expansion coefficient
- 0.000023 1/K
- ΔT · Temperature change
- -20 K
Compute the prevented free strain
The coefficient converts the temperature change into a dimensionless free expansion.
(0.000023) × (-20) = -0.00046Convert restrained strain to stress
Elastic stiffness converts the prevented strain into compression-positive stress.
(70000000000) × (-0.00046) = -32200000 Pa
Fully restrained axial thermal stress; compression positive in this lesson.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- E · Young’s modulus
- 200000000000 Pa
- α · Thermal expansion coefficient
- 0.000012 1/K
- ΔT · Temperature change
- 25 K
Find: Learn: Fully restrained thermal stress
A hint, not the answer
First find the thermal strain αΔT that the bar would have developed freely. Multiplying by E gives the stress magnitude needed to suppress it. This lesson reports compression as positive: positive heating gives positive compressive stress.
E is in Pa, α in 1/K and ΔT in K. A temperature difference of 40 °C is also 40 K; do not add 273.15 to a difference. The result is Pa.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Compute the prevented free strain
The coefficient converts the temperature change into a dimensionless free expansion.
(0.000012) × (25) = 0.0003Convert restrained strain to stress
Elastic stiffness converts the prevented strain into compression-positive stress.
(200000000000) × (0.0003) = 60000000 Pa
Fully restrained axial thermal stress; compression positive in this lesson.
Avoid the common trap
Do not use absolute temperature instead of temperature change. Do not apply the fully restrained formula to a freely expanding member. State the compression-positive convention when reporting a signed answer.
When this method applies — and when it does not
Assume full axial restraint, uniform temperature and constant α and E over the interval. Partial restraint, yielding, creep, support movement and thermal gradients need a different model. A tension-positive convention would reverse the reported sign.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Fully restrained thermal stress. Use the relevant elastic-response, equilibrium, constitutive-relations or shear-and-torsion module; the lesson states its particular sign convention.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.