Residual stress and processing variability

A cooled part can remain stressed

Uneven plastic deformation, thermal gradients or phase changes can leave incompatible preferred strains. The assembled material constrains them, producing residual stresses even without an applied external load. These stresses balance within the part and its constraints.

Residual means stress retained in the stated body and finished condition after the external processing load or transient temperature gradient has gone. Specify that boundary: a part still attached to a build plate can exchange forces with the plate, while the detached part must find a different balance. A small cut-out coupon can also relax stresses that existed in the larger component. The stress being described depends on what remains connected.

Consider a representative locally heated strip. The hot central region would expand more than its cooler surroundings, but their connection restrains it. Compressive stress can develop during heating. If the hot region deforms plastically, it acquires a permanent shortening relative to its original reference when compared later at the same temperature. On cooling, it would prefer to be shorter than the neighbouring material. The neighbours stretch it elastically, leaving central tension balanced by compression elsewhere. This is a possible sequence, not the inevitable sign pattern of every weld or printed layer.

Purely elastic, reversible thermal strain is different. For an initially stress-free homogeneous body that returns to the same uniform temperature and the same unchanged free condition, a thermal cycle alone need not leave stress. To explain a retained mismatch, identify a changed permanent strain, phase/transformation strain, different expansion response, connection or another history-dependent contribution. Thermal expansion and thermal stress develops the reversible reference case; here the new input is a retained spatial mismatch.

The strain a region would take if released from its neighbours at the comparison temperature is called its preferred or stress-free strain here. It may include plastic and transformation contributions; it is not the actual strain of the connected region. Nor does one observed final stress uniquely identify which process created that preferred strain.

Compatibility can require tension and compression together

Use an original small-strain, one-dimensional axial model of a long bonded strip. A gauge region far from the free ends has reference length L₀ = 100.00 mm, width 1.00 mm and thickness 4.00 mm. The central 2.00 mm thickness has area A_c = 2.00 mm²; each outer 1.00 mm layer has area A_o = 1.00 mm². The arrangement is symmetric about the mid-thickness plane. All zones have supplied Young’s modulus E = 200 GPa = 200000 MPa at the final common temperature.

The permanent preferred axial strain is supplied as ε*_c = −0.00200 in the centre and zero in each outer layer, all referred to L₀ at that same temperature. These are fictional model inputs, not a measured welding strain or a prediction from a thermal schedule. Assume perfect bonding, no further yielding or creep, equal axial strain in the long interior, and no external axial force or bending moment. End-transfer regions, transverse stresses and local stress concentrations are not resolved. Symmetry makes the interior bending curvature zero in this model.

Let ε be the common actual axial strain, tensile stress σ positive, and ε*_i the preferred strain in zone i. Elastic strain is ε − ε*_i, so σ_i = E_i(ε − ε*_i). Force carried by each zone is F_i = σ_i A_i. Equilibrium of the unloaded section requires the signed forces to sum to zero; it does not require each force to vanish. Thus ε = [Σ E_i A_i ε*_i]/[Σ E_i A_i], where Σ means add over the three zones. The weights are E_i A_i. With the same reference length they are proportional to the individual axial stiffnesses E_i A_i/L₀; do not give each zone an equal vote.

With the same E in all zones, ε = [2.00(−0.00200) + 1.00(0) + 1.00(0)]/4.00 = −0.00100. The centre is actually shorter than L₀ but longer than its preferred length. Its elastic strain is −0.00100 − (−0.00200) = +0.00100, giving +200 MPa tension. Each outer layer has elastic strain −0.00100 and stress −200 MPa compression. Remember that 1 MPa = 1 N/mm².

The central force is +400 N and the two outer forces are −200 N each. Their sum is zero. Their bending moments also cancel by symmetry: equal outer forces act at opposite distances from the mid-plane. The area-weighted average axial stress is zero, yet the central stress is +200 MPa. Zero net force and zero average therefore do not establish a stress-free material. The actual common length is 100.00(1 − 0.00100) = 99.900 mm; total contraction does not imply compression in every region.

Removing material changes the balance that held the shape

First remove both outer layers symmetrically from the model, introducing no new plastic or thermal strain and leaving the centre free to change length. The remaining homogeneous centre can reach its preferred strain −0.00200 with zero axial stress. Its new length is 99.800 mm, a further shortening of 0.100 mm from the original connected state. Its permanent strain did not disappear; its elastic stretching by its neighbours was released.

Removing only the top outer layer is a different problem. If the old stresses were incorrectly retained, the centre would carry +400 N and the remaining bottom layer −200 N, giving a nonzero +200 N resultant in an externally free part. The old field therefore cannot be its final equilibrium. The remaining section is asymmetric, so axial extension alone will generally be insufficient: bending can accompany redistribution.

A force-moment check makes that last point more precise without solving a beam. Take the original mid-plane as z = 0 with positive z upward. The retained centre’s centroid is at z = 0 and the bottom layer’s at −1.50 mm. The remaining section’s area centroid is −0.500 mm. Relative to that new centroid, reusing the old forces gives a signed first moment Σ zF = (0.500)(400) + (−1.00)(−200) = +400 N·mm, as well as the +200 N resultant. A free final section needs both resultants to vanish. Its actual curvature needs a bending/compatibility solution, not those old stress values pasted onto the new geometry.

Actual machining can introduce new heat, plastic deformation and surface damage as well as release existing stress. A measured shape change should therefore be compared with the removal geometry, supports, sequence and local temperature. A stress-relief treatment can also permit time-dependent strain or structural change. Neither “machined” nor “stress relieved” proves that all residual stress is zero.

They can change later performance

Residual tension can assist crack opening; compression can sometimes oppose it. Machining or stress relief may redistribute stresses and distort a part. Assess spatial distributions rather than assuming a single average describes the critical region.

For a separate elastic loading check, impose a uniform additional axial tensile stress of 80.0 MPa on the original intact strip without changing its preferred strains. Linear superposition gives central stress 200 + 80.0 = 280 MPa and outer stress −200 + 80.0 = −120 MPa. Their forces sum to +320 N, equal to the applied 80.0 MPa times 4.00 mm². The formerly unloaded self-balance has become equilibrium with an external force.

For this teaching check only, stipulate uniaxial tensile and compressive yield magnitudes of 300 MPa and no other limit. The 80.0 MPa addition remains within that assumed elastic regime. Adding 120 MPa instead would give a central elastic trial stress of 320 MPa, beyond the supplied yield condition. It is not a prediction of the actual post-yield stress: plastic redistribution changes the account. Real multiaxial yield, stress concentrations, buckling and damage require their own models.

Residual compression is not an unconditional benefit, nor is residual tension a complete failure calculation. Its direction and depth relative to a crack, the applied history, geometry and any relaxation matter. The original section model cannot supply fracture toughness, fatigue life or a safe allowable load. A useful decision follows the relevant local stress distribution in the actual finished state.

Keep the spatial map and the repeated records

Processing variability means that nominally similar runs or locations can develop different states. Local heating/cooling, constraint, starting material, dimensions and sequence can alter permanent mismatch. Separate that physical variation from measurement uncertainty and from a deliberate process change. Record the location, direction and finished condition before comparing numbers.

Use the following original six model records. All have the same three-zone geometry, E and outer preferred strains as above. Only the supplied central preferred strain varies. P and Q are invented route labels, not named heat treatments, printing settings or measured populations. The model gives central residual stress σ_c = −(E/2)ε*_c; each outer stress has the opposite sign and equal magnitude. Every record still has zero resultant force.

P has sampled central stresses 180, 200 and 220 MPa: mean 200 MPa and observed range 180–220 MPa. Q has 140, 160 and 200 MPa: mean 166.7 MPa and observed range 140–200 MPa. Q’s sampled mean is lower by 33.3 MPa, but the records overlap at 200 MPa. “Every Q part has less central tension than every P part” is contradicted by Q3 versus P1. Three chosen model records supply neither a production guarantee nor a confidence interval.

A second spatial average can hide the issue again: every specimen has zero area-average axial stress even though its central tension varies. Report the named component of stress at the critical region along with the map and repeated results. For measured data, compare the same sampled volume, direction, preparation, final constraint and measurement uncertainty; a changed surface-only measurement is not automatically a changed full-thickness field.

In an actual study of Ti-6Al-4V bridges made by laser powder-bed fusion, four scanning strategies were compared using high-energy X-ray diffraction, and two specimens were remeasured after partial release from their build plate. Higher stresses were reported near edges in the island-scanned bridges for those conditions. That result tests a particular route/material/geometry combination; it does not rank every island scan against every continuous scan. The original model records below are independent teaching data, not those experimental results.

A stress measurement needs its reference and its sampling volume

Attaching a strain gauge to the finished stressed part and setting its indication to zero does not make the part stress-free. A relief measurement instead observes strain changes when a controlled amount of constraint or material is removed, then uses a suitable elastic/geometry calibration to infer the preceding stress. The symmetric complete-release model gives a simple length change, but a real hole-drilling result is not generally one released strain multiplied by E.

Diffraction measures lattice spacings in selected directions and sampled material. Converting those to elastic strain needs an appropriate stress-free spacing; converting strain to stress needs relevant elastic and directional information. Composition, phases, texture and thermal history can change the reference or interpretation. A peak shift alone is not a complete residual-stress tensor or a whole-part map.

Specify where and at what depth the measurement averages, the stress direction/component, the current supports or build plate, section preparation, temperature, the reference/calibration and uncertainty. Check whether cutting or surface preparation has changed the state being reported. Spatial and repeat-run records are complementary: neither a smooth-looking surface, a zero net external force nor one low sampled mean establishes that the critical finished region is free of harmful stress.

Worked examples

Try the reasoning

1. Does zero external force imply zero internal stress?

   Show the answer and reasoning

   No.

   Residual stress can remain self-balanced.

2. Can residual stress interact with applied loading?

   Show the answer and reasoning

   Yes.

   The local state depends on both and any resulting plasticity.

3. In a new symmetric model keep E = 200 GPa, central preferred strain −0.00200 and outer preferred strain zero, but make all three zone areas 1.00 mm². Find the common strain and stresses. Keep the response elastic.

   Show the answer and reasoning

   Common strain is −0.0006667; central stress is +266.7 MPa and each outer stress is −133.3 MPa. The forces sum to zero.

   Weight by area because E is still common: ε = −0.00200/3. The centre’s elastic strain is +0.0013333. Each zone now has the same area, so +266.7 − 133.3 − 133.3 = 0 within rounding. Zero force does not require equal stress magnitudes when the original area proportions change.

4. Apply an additional uniform tensile stress of 120 MPa to the original intact model. Under the supplied 300 MPa uniaxial yield magnitude, may the central value from linear addition be reported as the final actual stress?

   Show the answer and reasoning

   No. The central elastic trial stress is +320 MPa, beyond the stipulated yield condition. The purely elastic prediction has exceeded its regime.

   The trial outer value is −80 MPa, but the centre would yield first in this simple axial check. Plastic redistribution and compatibility need a new constitutive calculation. The 300 MPa value is a chosen model input, not a real material allowable.

5. Return to the original 1:2:1 areas and common E, but supply central preferred strain +0.00100 with zero outer preferred strains. What happens to the signs?

   Show the answer and reasoning

   Common strain is +0.000500. The centre carries −100 MPa compression; each outer layer carries +100 MPa tension.

   The centre now prefers to be longer. Bonding shortens it relative to that preference while stretching the outer layers. Central force −200 N balances the two +100 N outer forces. A central location is not inherently tensile.

6. A separate route R has the same model and sampled central stresses 150, 200 and 250 MPa. Compare its mean and observed range with P. May the equal means justify treating the records as equivalent?

   Show the answer and reasoning

   Both sampled means are 200 MPa. R spans 150–250 MPa, a width of 100 MPa, whereas P spans 180–220 MPa, a width of 40 MPa. Equal means do not make the records equivalent.

   The spread and the largest observed local stress differ. These ranges describe the listed values only; they are not confidence bounds, tolerance limits or guarantees about unobserved parts.

7. An initially stress-free homogeneous strip returns to the same uniform temperature after a wholly elastic reversible cycle, with unchanged free connections and no phase or permanent-strain change. Must it acquire the model’s ±200 MPa residual pattern?

   Show the answer and reasoning

   No. With no retained preferred-strain mismatch, that specific pattern has no basis.

   The three-zone result required a supplied nonuniform permanent strain. A temperature history can create transient stress without necessarily leaving residual stress after the stated reversible return. Additional state or constraint changes would need their own evidence.

8. The original finished strip has zero area-average axial stress. A report therefore assigns zero to the centre before adding the 80.0 MPa load. What central value follows from the correct elastic account?

   Show the answer and reasoning

   The centre begins at +200 MPa and becomes +280 MPa under the stated elastic addition, not +80 MPa.

   Area averaging cancelled the central tension against outer compression; it did not remove either. The applied 80.0 MPa is added to the relevant local component under the unchanged linear model.

9. A strain gauge attached after processing is zeroed on the intact strip. Does its zero indication prove that the central elastic strain is zero? What would distinguish that statement from a complete symmetric-release observation?

   Show the answer and reasoning

   No. Zeroing sets an instrument reference on the already stressed body. The model instead predicts a further 0.100 mm shortening of the 100.00 mm reference gauge region when the centre is completely freed without new strain.

   The release compares two physical constraint states. Interpreting a real partial-relief measurement needs its geometry/calibration and uncertainty; neither the instrument tare nor one uncalibrated released-strain reading supplies the whole residual-stress field.

What to carry forward

• Track incompatible strains and stress redistribution.
• Measure critical regions in the finished state.
• Compatible actual strain can differ from each region’s stress-free strain, leaving balanced tension and compression.
• An unloaded section must satisfy force and moment equilibrium; zero average stress does not mean zero local stress.
• Removing material or changing constraints requires a new balance, and elastic superposition has a yield boundary.
• Keep location, repeated results, uncertainty and the final measurement reference with every process comparison.