Materials Science · Joining, coatings and surface engineering

Material mismatch and residual stress

Useful background: Processing routes and microstructural evolution; Composites, fibres and interfaces; Cracks, fracture and toughness. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Explain stress generated by bonded materials with different preferred strains.
Compare separate-layer free strains with the compatible strains of a bonded strip.
Explain why holding a bilayer flat and letting it bend are different mechanical boundary conditions.
Check force and moment balance in one declared elastic bilayer model, and identify evidence needed beyond its curvature.

Bonded layers cannot always expand freely

Two materials with different thermal expansion responses would change length differently if separate. Bonding forces compatible motion, creating stress and sometimes curvature. Cooling from a joining temperature can therefore leave a stressed assembly.

The useful first question is what each material would do without its neighbour. Let ε* denote this preferred, stress-free axial strain relative to a declared reference state. For a small uniform temperature change and a constant linear expansion coefficient β, ε* = βΔT. Here β means linear expansion, not volumetric expansion. The star distinguishes a preferred strain from the strain actually allowed by the assembly.

Within the one-dimensional linear elastic model used below, σ = E(ε − ε*). Tensile stress σ is positive, compressive stress negative; E is Young’s modulus and ε is actual axial strain. A layer that contracts less than it would freely is in tension even while its actual length decreases. Elastic response and elastic modulus and Thermal expansion and thermal stress develop these quantities before the two-material comparison.

Bonding constrains relative motion at the shared interface. It does not require every height in a bending strip to have the same axial strain: outer arcs can be longer than inner arcs while the interface stays joined. Equal strain throughout the thickness is an additional flat-strip assumption, not a consequence of bonding alone.

Start with two free contractions and a known bonded reference

Consider an entirely fictional pair of layers A and B. They are bonded flat and stress-free at a reference temperature T_j, with common reference length L₀ = 100 mm. Each is h = 1.00 mm thick and b = 10.0 mm wide, with E = 100 GPa = 100 000 MPa. A is the bottom layer and B the top. Their supplied constant coefficients are β_A = 20.0 × 10⁻⁶ K⁻¹ and β_B = 10.0 × 10⁻⁶ K⁻¹. Both cool uniformly by ΔT = −50.0 K. These inputs specify a teaching model, not measured alloy properties or a joining schedule.

If separate, A would have ε*_A = −0.00100 and length 99.900 mm; B would have ε*_B = −0.000500 and length 99.950 mm. The preferred-length difference is 0.050 mm. This is a mismatch between two free responses, not yet a measured opening or the extension of a bonded assembly.

Assume a long, slender, perfectly bonded elastic strip with small strains and small rotations. Plane cross-sections remain plane; temperature is uniform and there is no interlayer slip, plasticity, creep, cure shrinkage or phase change in this calculation. We model the central region away from free ends. Lateral effects and the end zones where interfacial shear transfers load are not resolved. The model predicts axial strain and stress, not an interface peel stress or a joint failure load.

Use z as height from the interface, positive into B: A occupies −h ≤ z ≤ 0 and B occupies 0 ≤ z ≤ h. Because bending changes arc length with height, write ε(z) = ε₀ + qz. The dimensionless ε₀ is strain at the interface; q is the axial-strain gradient, with units of inverse length. Positive q means the top is longer than the bottom. Its magnitude is the small-strain curvature magnitude in this strip model; the radius magnitude is 1/|q|. Defining q this way avoids leaving the bending-sign convention implicit.

Original fictional two-material dossier; all dimensions refer to the flat stress-free reference
Quantity	Bottom A	Top B
E	100 GPa	100 GPa
Layer thickness h	1.00 mm	1.00 mm
Width b	10.0 mm	10.0 mm
Linear β	20.0 × 10⁻⁶ K⁻¹	10.0 × 10⁻⁶ K⁻¹
Free strain for −50 K	−0.00100	−0.000500
Separate free length	99.900 mm	99.950 mm

Holding the bonded strip flat supplies a moment constraint

First hold the assembly flat, so q = 0, while permitting axial contraction without an external axial force. Both layers then have a common actual strain ε₀. Each layer area is bh = 10.0 mm². Axial equilibrium requires Ebh(ε₀ − ε*_A) + Ebh(ε₀ − ε*_B) = 0. Thus ε₀ is the mean of the preferred strains: −0.000750. The common length is 99.925 mm.

A has been stretched by 0.000250 relative to its shorter preferred state, giving σ_A = +25.0 MPa and axial resultant +250 N. B has been compressed by 0.000250 relative to its preferred state, giving −25.0 MPa and −250 N. The forces sum to zero; the separate stress magnitudes are not zero.

These opposite axial resultants act in different layers. In this uniform-stress flat state their lines of action are 1.00 mm apart, producing a 250 N·mm couple. Keeping the strip flat therefore requires end moments of the corresponding magnitude and sense. It is not a completely unrestrained state. Zero external axial force does not imply zero applied moment.

Stiffness and thickness enter even this flat calculation. With layer areas A_A and A_B, axial equilibrium gives ε₀ = (E_A A_A ε*_A + E_B A_B ε*_B)/(E_A A_A + E_B A_B). This is a stiffness-weighted preferred strain, not an unweighted material average. The flatness restraint must still supply whatever moment is needed. A freely bending unequal bilayer needs both extension and bending balance; this flat formula alone cannot predict its curvature.

Free bending changes the stress field without erasing it

Now remove the flatness restraint and leave both end force and end couple zero. Both total axial force N and the first moment Q of the axial stress must vanish. In this lesson Q = ∫ zσ dA, where dA is a small cross-sectional area. Q has units N·mm when z is in millimetres. Its sign is defined by this integral; requiring Q = 0 does not depend on choosing an alternative beam-moment sign convention.

For the equal-E, equal-h pair, adding the layer forces still gives ε₀ = (ε*_A + ε*_B)/2. For the first moment, integrate zE(ε₀ + qz − ε*_i) across each layer and multiply by width b. The common ε₀ terms cancel between the symmetric heights. The q term uses ∫ from −h to h of z² dz = 2h³/3, while the preferred-strain contribution is (ε*_B − ε*_A)h²/2. Consequently Q = Eb[2qh³/3 − (ε*_B − ε*_A)h²/2] = 0, giving q = 3(ε*_B − ε*_A)/(4h). The integral merely sums each small axial force times its signed distance from the interface.

The dossier gives q = +0.000375 mm⁻¹ = +0.375 m⁻¹, with radius magnitude about 2.67 m. B occupies the longer outer side and A the shorter inner side: cooling bends toward the higher-expansion A layer, because it prefers the greater contraction. On heating by the same temperature difference from the stress-free reference, all signs reverse within this elastic model.

Put q back into σ = E(ε₀ + qz − ε*). With z measured in mm, σ_A = 25.0 + 37.5z MPa and σ_B = −25.0 + 37.5z MPa. At A’s bottom outside face z = −1 mm the stress is −12.5 MPa; immediately below the interface it is +25.0 MPa. Immediately above the interface it is −25.0 MPa; at B’s outside face it is +12.5 MPa. Each layer therefore contains both tension and compression. Saying the entire A layer is tensile is now wrong even though its mean stress is positive.

The layer mean stresses are +6.25 and −6.25 MPa, giving resultants +62.5 and −62.5 N. Integrating the actual linear stresses also gives Q = 0. Using just the mean forces at the geometric layer centres would miss their internal stress gradients and incorrectly predict a remaining couple. Actual axial strain is continuous across the bonded interface, but axial stress jumps there because the preferred strains differ. This axial stress is not the traction normal to the horizontal interface; its jump is not a failure of interface force balance.

Bending has reduced the layer resultants from the held-flat case, but the maximum local stress magnitude is still 25.0 MPa at the two sides of the interface. Curvature is a deformation response, not proof that stresses have relaxed to zero. The predicted abrupt stress jump and central-region field do not resolve the free-end transfer zone.

Same fictional cooling step and materials; changing the flatness condition changes the field
Quantity	Held flat, axial force free	Force- and moment-free bending
Interface strain ε₀	−0.000750	−0.000750
Strain gradient q	0	+0.000375 mm⁻¹
A / B mean stress	+25 / −25 MPa	+6.25 / −6.25 MPa
A / B axial resultant	+250 / −250 N	+62.5 / −62.5 N
Q = ∫ zσ dA	−250 N·mm	0
Maximum \|σ\| in model	25.0 MPa	25.0 MPa

Original fictional equal-layer cooling model. Height z runs from minus one millimetre at the bottom A outside face to plus one millimetre at the top B outside face; the interface is z zero. In the upper graph, separate preferred strains are minus 1000 microstrain in A, shown short-dashed with a circle, and minus 500 microstrain in B, short-dashed with a square. The freely bent compatible actual strain is the solid green diagonal, from minus 1125 through minus 750 to minus 375 microstrain. The strain axis runs from minus 1250 to zero microstrain. In the lower graph, solid blue circles mark the free A stress from minus 12.5 MPa at its outside face to plus 25 MPa just below the interface. Long-dashed orange squares mark free B from minus 25 MPa just above the interface to plus 12.5 MPa at its outside face. Dotted violet lines are the held-flat stresses plus 25 MPa throughout A and minus 25 MPa throughout B. The stress axis runs from minus 30 to plus 30 MPa. Axial strain is continuous at the interface, but the preferred strain and axial stress jump. All graph coordinates represent the declared central-region model, not measured fields. The freely bent field has zero total axial force and stress first moment; the held-flat field requires a 250 newton millimetre end-couple magnitude.

More than one mismatch can matter

Cure shrinkage, phase changes, elastic properties and thickness ratios also influence the result. Cracking, yielding or creep can relax stresses, changing later behaviour. A complete assessment follows the process and service history.

The thermal part is referenced to the state in which the layers were bonded without stress, not automatically to room temperature or the last temperature measured. If this ideal strip returns to T_j with the bond intact and no irreversible changes, both preferred thermal strains return to zero and the elastic solution becomes straight and unstressed. If cure, plastic flow, phase transformation or damage changes the preferred dimensions or constraints, the same return temperature need not restore that original state.

An interlayer can permit relative shear deformation rather than imposing the perfect-bond kinematics used here. Its thickness, stiffness, time response and length of bonded region then affect transfer. Calling it flexible does not establish zero stress or improved durability: it can change deformation and local stress concentrations, and it introduces its own failure and environmental behaviour. The perfect-bond calculation must be replaced or tested, rather than assigned an arbitrary reduction factor.

For the equal-layer model, doubling both layer thicknesses at unchanged E, β and ΔT halves q. Yet qh stays unchanged, so stresses at corresponding fractions of the thickness remain the same. A flatter-looking thick strip is therefore not necessarily less stressed. Halving both elastic moduli instead leaves q unchanged but halves every stress. These two comparisons show why neither curvature nor expansion mismatch alone determines the local stress.

A measured curvature should be interpreted with the actual joining reference, layer geometry, local temperature history and material state. Compare it with separate-layer expansion observations and, where needed, independent strain/stress or release evidence. An equal curvature can arise with different stress magnitudes; a change after a hold could involve viscoelastic response, plasticity, bond damage or a changed thermal field. Residual stress and processing variability and Viscoelasticity, temperature and time develop those distinctions. Joint integrity needs separate interfacial and material resistance evidence.

Two bonded layers curve when their preferred strains differ. Thickness, stiffness, process history and relaxation affect actual stress and curvature. The curvature is schematic.

Worked examples

From the original bending observation to a bounded causal account

A bonded strip bends after cooling with no external force. What should be checked? Apply the declared A/B dossier and distinguish freely bending from held-flat conditions.

Compare the layers’ preferred thermal contractions.
Include their stiffnesses and thicknesses.
Assess compatible deformation and residual stress.
The independent free strains are −0.00100 and −0.000500, so A would be 0.050 mm shorter than B over the 100 mm reference length.
For the force- and moment-free equal bilayer, ε₀ = −0.000750 and q = +0.000375 mm⁻¹. B is on the longer outer arc, so the strip bends toward A.
The calculated A/B mean stresses are +6.25/−6.25 MPa, with ±62.5 N resultants and Q = 0; local interface-side magnitudes reach 25 MPa. Holding it flat would require a moment constraint and a different stress field.

Thermal mismatch can produce curvature and self-balanced stresses. In this declared cooling model the higher-expansion A layer is on the inner side. Zero net force and moment coexist with nonzero local stresses.

A contracting layer can be in tension

In the held-flat dossier, a reader calls A compressive because its actual length fell from 100.000 to 99.925 mm. Determine the stress sign from the relevant reference.

Actual strain is (99.925 − 100.000)/100.000 = −0.000750.
A’s preferred free strain is −0.00100. Its elastic strain is actual minus preferred: +0.000250.
Multiply by 100 000 MPa to obtain +25.0 MPa tension. A is longer than its own stress-free cooled length 99.900 mm.

A contracts overall but is tensile relative to its preferred cooled state. The sign of total length change alone does not identify stress.

Change one stiffness without silently removing the restraint

Keep the same geometry, coefficients and cooling step, but let E_A = 100 GPa and E_B = 50 GPa. The strip is again held flat with no external axial force. Find common strain, stresses and the required couple magnitude.

Equal areas give ε₀ = [100(−0.00100) + 50(−0.000500)]/150 = −0.00083333.
σ_A = 100 000(0.000166667) = +16.667 MPa; σ_B = 50 000(−0.000333333) = −16.667 MPa. Each area is 10 mm², so resultants are ±166.667 N.
Their geometric centre lines are 1 mm apart, giving a couple magnitude 166.667 N·mm. The flatness restraint must supply it.

Common strain ≈ −0.0008333, stresses ≈ ±16.7 MPa and couple magnitude ≈ 167 N·mm. This is a held-flat result, not the curvature of the unequal-modulus free strip.

The same curvature can conceal a different stress scale

Two otherwise identical freely bending dossier strips have E_A = E_B = 100 GPa and E_A = E_B = 50 GPa respectively. Do their curvatures distinguish their maximum axial stress magnitudes?

The equal-layer expression q = 3(ε*_B − ε*_A)/(4h) has no E because the common modulus cancels from both zero-resultant balances. Both q values are +0.000375 mm⁻¹.
Actual strain profiles are identical, but σ = E(ε − ε*) halves when both moduli halve.
Maximum magnitudes are therefore 25.0 MPa and 12.5 MPa. The corresponding layer force magnitudes are 62.5 and 31.25 N.

No. Equal curvature does not identify equal stress. One needs the material stiffness and the supported compatibility/history model as well.

Try the reasoning

Does zero applied force mean the layers are unstressed?
Show the answer and reasoning
No. Zero applied axial force constrains the sum of internal axial forces, not each local stress.
In the free dossier, +62.5 and −62.5 N sum to zero and the full stress first moment is also zero, yet interface-side axial stresses are ±25 MPa. A held-flat strip can additionally require end moments despite zero axial force.
Can a flexible interlayer change stress transfer?
Show the answer and reasoning
Yes. Interlayer deformation can change how strongly the layer motions are coupled.
The perfect-bond model then needs an interlayer transfer law and boundary conditions. Flexibility alone does not specify zero stress, the sign of every local change or a durability improvement.
Starting again from the original stress-free reference, heat the equal-layer dossier by +50 K instead of cooling by −50 K. State both free strains, q and the inner layer.
Show the answer and reasoning
ε*_A = +0.00100, ε*_B = +0.000500; q = −0.000375 mm⁻¹. B is the inner layer and A the outer.
The preferred-strain difference reverses sign. All actual elastic strains and stresses reverse within this unchanged linear model; the higher-expansion layer now needs the longer outer arc.
Double each layer thickness to 2.00 mm at unchanged width and material inputs. In the free equal-layer model, compare q, maximum |σ| and A’s resultant with the original strip.
Show the answer and reasoning
q = +0.0001875 mm⁻¹, maximum |σ| remains 25.0 MPa, and A’s resultant doubles to +125 N.
q is inversely proportional to h, while qh and the stresses at corresponding z/h remain unchanged. Each layer area doubles, so its unchanged mean stress acts over twice the area. Less curvature did not mean less local stress.
Set both β values to 10.0 × 10⁻⁶ K⁻¹, keeping the stress-free reference, −50 K step and free boundaries. What are ε₀, q and the axial stresses? What would this not prove about a real cured assembly?
Show the answer and reasoning
ε₀ = −0.000500, q = 0 and all axial stresses in this model are zero. Matching thermal coefficients does not rule out cure, transformation or other preferred-strain mismatch.
Both layers can share their same preferred contraction without elastic strain. This conclusion requires the declared absence of other strains, gradients, previous damage and restraints.
In a held-flat, axially free variant, A retains E = 100 GPa and area 10 mm², while B has E = 100 GPa and area 20 mm². The original preferred strains are unchanged. Find common strain and layer resultants; may the flat result be called fully unrestrained?
Show the answer and reasoning
ε₀ ≈ −0.0006667; σ_A ≈ +33.33 MPa and σ_B ≈ −16.67 MPa; resultants are +333.33 and −333.33 N. It is still a flatness-constrained calculation.
Weight the preferred strains by EA: [10(−0.00100) + 20(−0.000500)]/30. The areas make equal-and-opposite resultants despite unequal stresses. Without the layer positions one cannot calculate the necessary couple, and zero axial force alone does not remove it.
For the original freely bent strip, an analyst replaces each linear stress field by its mean acting at the geometric layer centre, obtains a 62.5 N·mm couple and declares the solution inconsistent. Find the error.
Show the answer and reasoning
The averaging preserved each layer force but discarded its stress-gradient moment. The actual ∫ zσ dA is zero.
For A, b∫ from −1 to 0 of z(25 + 37.5z) dz = 10(−12.5 + 12.5) = 0 N·mm; B also gives zero. A nonuniform stress resultant need not act at the geometric layer centre. Mean stress alone cannot recover a first moment.
A bonded strip has less curvature after a hot hold and cooldown. A report attributes this uniquely to a lower thermal expansion mismatch and declares the interface undamaged. Is that inference supported by curvature alone?
Show the answer and reasoning
No. Check the reference state, temperature field, layer stiffnesses and possible time-dependent or irreversible changes, plus independent interface evidence.
The same curvature can accompany different stresses, and changing plastic strain, creep, cure or bond transfer changes the model itself. Separate expansion and state measurements, controlled temperature/history observations and suitable integrity evidence are needed to discriminate causes. No life or joint-strength guarantee follows from curvature.

What to carry forward

Compare preferred strains and constraints.
Follow relaxation and damage through the history.
Stress depends on actual strain relative to the layer’s own preferred state; overall contraction can coexist with tension.
Holding a bilayer flat and leaving it force- and moment-free are different boundary conditions.
Compatible bending allows a strain gradient; zero resultant force and moment do not erase local stress.
Thickness and stiffness affect curvature and stress differently, so curvature alone is not a stress measurement.
Use the actual bonded reference and subsequent process history; a simple elastic central-region model does not resolve end transfer or qualify the joint.

References

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