Materials Science · Heat transport, expansion and thermal response

Thermal expansion and thermal stress

Separate free thermal strain from elastic strain, derive the stress caused by a prescribed end movement, and calculate heating and cooling cases from a stated stress-free reference.

Useful background: Energy, stability and environmental context; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Distinguish free thermal expansion from restrained thermal stress.
Explain small total axial strain as the sum of thermal and elastic strain, with tensile stress positive.
Calculate free extension, fully restrained heating or cooling stress, and stress under a prescribed end movement.
Check the reference state, units and elastic assumptions before interpreting thermal stress or a thermal cycle.

Heating changes preferred dimensions

For a small temperature interval with approximately constant linear expansion coefficient β, free extension is βLΔT. A free uniformly heated rod can expand without the axial stress predicted for full restraint. Gradients and connections change that situation.

Begin with a straight homogeneous rod that is stress-free at temperature T₀ and length L₀. Here β means the linear expansion coefficient in the rod’s axial direction, in K⁻¹. Notation varies between books: a coefficient defined for volume is a different input. All numerical cases below use a positive linear β, approximately constant over the specified interval.

Let ΔT = T − T₀. The free thermal strain is ε_th = βΔT, and the corresponding free length change is δ_free = L₀βΔT. “Free” means there is no axial restraint or applied axial force and any support friction is negligible. Uniform heating then lengthens this rod; uniform cooling shortens it. Its dimensions change without requiring elastic axial strain.

For example, L₀ = 1000 mm, β = 10⁻⁵ K⁻¹ and a rise from 20°C to 40°C give ΔT = +20 K, ε_th = +0.000200 and δ_free = +0.200 mm. The strain is a length ratio, not a displacement in millimetres. A temperature difference of 20°C has the same numerical size as 20 K; an absolute temperature of 20°C is not 20 K.

Deformation and relative strain explains the length-ratio definition. Elastic response and elastic modulus relates stress to the elastic part of strain. The distinction matters when temperature changes the length a rod would have without any axial load.

Add the strains that produce the actual length

Take tensile stress and axial extension as positive. The actual small total strain is ε_tot = (L − L₀)/L₀. In the uniaxial linear-elastic thermal model, ε_tot = ε_el + ε_th = σ/E + βΔT. E is Young’s modulus in the stated direction and temperature range; σ/E is the elastic strain relative to the thermally changed stress-free length, expressed in this small-strain approximation.

Rearranging gives σ = E(ε_tot − βΔT). First find the actual movement permitted or imposed by the surroundings, then subtract the thermal strain. Multiplying total strain by E without that subtraction would turn a free temperature change into a fictitious stress.

This is a compatibility statement: elastic and thermal contributions must add to the length change the boundary conditions permit. It is not a claim that an experiment must physically heat the rod freely and then clamp it. Imagining free expansion followed by an elastic correction is a calculation device for the same final constrained state.

Assume small strains, uniform temperature, constant cross-section and approximately constant E and β over the interval. Lateral expansion is free, so this is an axial restraint model, not a body prevented from changing size in all three directions. There is no initial stress, plastic strain or time-dependent deformation. A compressed rod is also assumed to remain straight and stable; an axial calculation alone does not check buckling.

Constraint converts mismatch into stress

In a simple fully restrained uniaxial linear-elastic rod, thermal stress magnitude is approximately EβΔT. Yield, creep, nonuniform heating or imperfect restraints can invalidate this ideal estimate. Bonded materials with different expansion responses create a related mismatch problem.

For a fully fixed case, hold the end separation at L₀ from the initial stress-free state, before changing temperature. Then ε_tot = 0, so ε_el = −βΔT and σ = −EβΔT. With β positive, heating requires a negative elastic correction and gives compression; cooling requires a positive correction and gives tension. The magnitude relation uses |ΔT| when cooling is included.

The zero total strain does not mean zero stress. At +20 K in the original 100 GPa example, +0.000200 thermal strain is cancelled by −0.000200 elastic strain. Since 100 GPa = 100000 MPa, σ = 100000 × (−0.000200) = −20.0 MPa. This is the signed version of the retained worked case.

To find the axial force carried by a uniform section, use F_ax = σA with original area A in this small-strain model. F_ax is positive in tension and negative in compression. It is not the sum of the two end forces: those forces are equal and opposite for equilibrium. Applied force and stress over an area explains why 1 MPa = 1 N/mm².

For the invented comparison below, let A = 50.0 mm² and keep the same E, β, L₀ and stress-free 20°C reference. At 40°C, the fixed rod has F_ax = −20.0 × 50.0 = −1000 N. Supports push inward on the rod. At −10°C, ΔT = −30 K instead gives σ = +30.0 MPa and F_ax = +1500 N: supports pull outward to prevent contraction. The free and fixed cases are separate experiments from the same initial state.

Original invented uniform-rod comparison: L₀ = 1000 mm, A = 50.0 mm², E = 100 GPa and linear β = 10⁻⁵ K⁻¹. Each case starts stress-free at 20°C; fixed cases are held at L₀ before their temperature change.
Final condition	Thermal strain ε_th	Elastic strain ε_el	Total movement δ	Axial stress σ
Free at 40°C	+0.000200	0	+0.200 mm	0
Fixed at 40°C	+0.000200	−0.000200	0	−20.0 MPa
Fixed at −10°C	−0.000300	+0.000300	0	+30.0 MPa

Original invented rod comparison. Every case independently starts stress-free at 20 degrees Celsius, with original length 1000 millimetres, area 50 square millimetres, modulus 100 gigapascals and positive linear expansion coefficient 10 to the minus 5 per kelvin. Fixed cases are held at the original end separation before changing temperature; the cases are not successive clamping steps. Drawn axial movements are exaggerated and the sketch is not to scale. The reference rod is blue, from x 130 to 480. A free rod heated to 40 degrees Celsius is drawn longer, ending at x 520, and its actual extension is 0.200 millimetre with zero axial stress. A fixed rod heated to 40 degrees Celsius keeps its original length; supports push inward on its ends, giving stress minus 20 megapascals and axial force minus 1000 newtons, compression. Its thermal strain plus 0.000200 is cancelled by elastic strain minus 0.000200. A fixed rod cooled to minus 10 degrees Celsius also keeps its original length; supports pull outward, giving plus 30 megapascals and plus 1500 newtons, tension. Its thermal strain minus 0.000300 is cancelled by elastic strain plus 0.000300. Arrow directions, words and signs identify the forces without relying on colour. The forces shown act on the rod and are equal and opposite at its two ends; the signed axial force is an internal tensile-or-compressive section resultant, not their vector sum.

A rod can lengthen and still be in compression

Full restraint is one boundary condition. If the end separation changes by a known δ_s relative to L₀, then ε_tot = δ_s/L₀ and σ = E(δ_s/L₀ − βΔT). Here δ_s is the change in the distance between ends, rather than a rigid translation of the whole rod.

Suppose the surroundings allow a positive length change, but less than the rod’s free thermal expansion. The elastic correction is still negative, so the rod lengthens while carrying compression. Conversely, pulling the ends farther apart than the free heated length gives tension. The sign depends on the mismatch, not just on whether the actual length increased.

A known end movement is different from an unknown compliant support. If the supports themselves deform or thermally expand, their movement must be obtained from a coupled equilibrium and compatibility model. Do not assign an arbitrary “restraint percentage” to EβΔT without showing what determines it. The prescribed-movement worked case below uses a supplied boundary displacement, not an inferred support stiffness.

Check the state and what happened before it

The reference must be the actual stress-free state used by the model. A rod allowed to expand freely from 20°C to 40°C and then secured at its new length without force is initially stress-free at that new state. It does not acquire the earlier fixed-at-20°C compression just because it is now clamped. Subsequent temperature changes must use the appropriate reference and constraints.

A rod fixed at its stress-free length, heated and returned to its initial temperature has zero stress again if the entire response was elastic and its restraints did not change. A thermal cycle alone does not prove residual stress. Plastic deformation, time-dependent strain, altered connections or an initially stressed state can change the outcome; they require additional history and material information.

If the elastic estimate exceeds a relevant stipulated yield condition at the temperature in question, it flags that the wholly elastic calculation is outside its assumed regime. It does not determine the subsequent stress or plastic strain. Plastic deformation and yield behaviour develops that distinction. Compare tensile and compressive criteria appropriately instead of assuming they are always identical.

Stress can also relax with time at a fixed length when the material develops a time-dependent strain contribution. In that case, thermal plus elastic strain alone no longer accounts for the imposed total strain. Sustained loads and creep mechanisms explains fixed-strain relaxation versus constant-load creep. This lesson supplies no relaxation time or stress-history prediction.

Bonded materials that prefer different free strains, or a body with nonuniform temperature, can develop internal stresses even when the total external force is zero. The zero-stress free-rod result required a homogeneous body at uniform temperature. Loading direction, geometry and conditions introduces why a different geometry or instability needs a different model.

Worked examples

Work through the case

An ideal restrained rod has E = 100 GPa, β = 10⁻⁵ K⁻¹ and ΔT = 20 K. Estimate stress magnitude.

Thermal strain magnitude = 2 × 10⁻⁴.
Multiply by 100 GPa.
Result = 20 MPa.

20 MPa in the stated idealisation; heating produces compression for this fully restrained rod.

Follow a different rod through cooling

A new invented rod is stress-free at 30°C, with E = 80.0 GPa, linear β = 12.0 × 10⁻⁶ K⁻¹, L₀ = 500 mm and A = 40.0 mm². It cools uniformly to −20°C. Compare free contraction with full axial restraint imposed at L₀ before cooling. A tensile yield stress of 60.0 MPa is stipulated throughout this interval solely for checking the elastic assumption.

Find the signed change: ΔT = −20 − 30 = −50.0 K. Thermal strain is ε_th = (12.0 × 10⁻⁶)(−50.0) = −0.000600.
If free, σ = 0 and ε_tot = ε_th. The length change is 500 × (−0.000600) = −0.300 mm. The rod ends at 499.700 mm without the restraint stress.
If fixed at the original end separation, ε_tot = 0 and ε_el = +0.000600. Convert E = 80000 MPa, so σ = 80000 × 0.000600 = +48.0 MPa. The force is F_ax = 48.0 × 40.0 = +1920 N, carried in tension.
The calculated tensile stress is below the stipulated 60.0 MPa yield level, consistent with that one elastic-regime check. It does not establish an assembly allowable, connection strength or safety margin for other mechanisms.

Free: contraction of 0.300 mm and zero axial stress. Fixed: zero total movement, +48.0 MPa tension and +1920 N axial force, within the problem’s stipulated yield regime.

Use the permitted movement, not a guessed restraint fraction

Use a separate copy of the new rod with E = 80.0 GPa, linear β = 12.0 × 10⁻⁶ K⁻¹ and L₀ = 500 mm. Starting stress-free, raise its temperature by 50.0 K while the apparatus increases its end separation by a prescribed 0.100 mm. Assume its response remains uniaxial and linear elastic. Is the final axial stress tensile or compressive?

Free thermal strain would be βΔT = (12.0 × 10⁻⁶)(50.0) = +0.000600, corresponding to free extension 0.300 mm.
The actual total strain is δ_s/L₀ = 0.100/500 = +0.000200. It is positive but smaller than the thermal strain.
Subtract: ε_el = ε_tot − ε_th = 0.000200 − 0.000600 = −0.000400. Then σ = 80000 × (−0.000400) = −32.0 MPa.
The final length is 500.100 mm, while its stress-free length at that temperature would be approximately 500.300 mm. Compression provides the elastic shortening relative to that preferred length. This is the stated prescribed-displacement model, not a calculation of the apparatus stiffness.

−32.0 MPa, compression. The actual rod lengthens by 0.100 mm, but it is held shorter than its free heated length.

Try the reasoning

Would the same uniform free rod have that restraint stress?
Show the answer and reasoning
No.
It can expand. For the homogeneous, uniformly heated free rod, ε_tot = βΔT, so ε_el = 0 and the axial stress is zero.
Why consider cooling too?
Show the answer and reasoning
The sign and mismatch state can change.
Cooling changes ΔT and can reverse the stress sign under unchanged restraint. Residual loads depend on the full history and response; returning to the initial temperature after a wholly elastic cycle with unchanged restraints returns this initially stress-free rod to zero stress.
The new fully fixed 80.0 GPa rod instead cools by 80.0 K from its stress-free state. Keep linear β = 12.0 × 10⁻⁶ K⁻¹ and a stipulated tensile yield stress of 60.0 MPa over this interval. Calculate the elastic estimate. May it be reported as the actual post-yield stress?
Show the answer and reasoning
The elastic extrapolation is +76.8 MPa, above 60.0 MPa. It is not an established actual post-yield stress.
ε_th = −0.000960 and ε_el = +0.000960 would give σ = 80000 × 0.000960 = 76.8 MPa. The stated yield criterion is reached within the ideal elastic model at a cooling magnitude 60.0/[80000 × (12.0 × 10⁻⁶)] = 62.5 K. That is a boundary of this stipulated model, not a universal allowable temperature drop. Beyond it, plastic constitutive and compatibility information is needed.
For the original 100 GPa rod at +20 K, first double L₀ while keeping A = 50.0 mm². Then, in a separate comparison, keep L₀ = 1000 mm and double A to 100 mm². How do free extension, fully fixed stress and fixed axial force change?
Show the answer and reasoning
Doubling length doubles free extension to 0.400 mm but leaves fixed stress −20.0 MPa and force −1000 N. Doubling area leaves free extension 0.200 mm and fixed stress −20.0 MPa, but doubles axial force to −2000 N.
δ_free = L₀βΔT contains length; σ_fixed = −EβΔT contains neither length nor area in this ideal model. F_ax = σA contains area. Changing dimensions may affect real stability or support compliance, but those effects are outside these stipulated comparisons.
A strain gauge on the original fully fixed rod indicates zero axial length change while it is heated by 20 K. A worksheet writes σ = E × 0 = 0. What contribution has it missed, and what stress follows under the stated model?
Show the answer and reasoning
It omitted thermal strain. The total strain is zero, but elastic strain is −0.000200 and stress is −20.0 MPa.
Use ε_el = ε_tot − βΔT = 0 − 0.000200 before multiplying by E. A zero actual length change establishes the compatibility condition, not zero elastic strain.
The original rod expands freely from 20°C to 40°C and is then secured without force at its new length. It cools back to 20°C with that new end separation fixed. Using the same small-strain constant E and linear β, what stress sign and approximate magnitude follow? Why is this different from the closed elastic cycle at the original fixed length?
Show the answer and reasoning
Approximately +20 MPa tension. Its new stress-free secured state is at 40°C; subsequent cooling is restrained.
Use the new stress-free reference for a −20 K change: σ ≈ −100000 × 10⁻⁵ × (−20) = +20 MPa. The supports now retain the longer separation, so the constraint history differs from holding the original L₀ throughout and returning to the initial temperature. The small difference between the two reference lengths is beyond this first-order approximation.
A rod at fixed uniform temperature and fixed end separation shows a decreasing compressive force magnitude over a long hold. Does zero total movement establish that the stress must remain −EβΔT for the entire hold?
Show the answer and reasoning
No. A changing force at unchanged area is changing stress; time-dependent strain can redistribute the strain contributions.
The thermal strain is unchanged, but an additional time-dependent contribution can allow the elastic contribution to decrease in magnitude while total strain remains fixed. The purely elastic two-part model is then incomplete. The observation alone does not determine a relaxation law, rate or remaining life.

What to carry forward

State whether expansion is free or restrained.
Use thermoelastic estimates only before their assumptions fail.
Define the stress-free reference and the linear expansion coefficient before using a temperature difference.
In this small uniaxial model, ε_tot = σ/E + βΔT; imposed movement determines the elastic correction.
Full restraint gives σ = −EβΔT, with heating compression and cooling tension for positive β.
Actual extension does not by itself determine stress sign, and an elastic closed cycle does not by itself create residual stress.

References

Teaching resources and research supporting this lesson’s explanations and examples.