The Gold Standard of Welding | Separating Overall Shortening from Self-Balancing Residual Stress

A ruler can establish a distance between identified points under stated conditions. That is valuable evidence about geometry. It does not directly measure the forces carried by every internal region or reveal the full history that produced the final shape.

Welding provides an important context because local heating and cooling interact with restraint and material response. TWI’s explanation of welding distortion describes why nonuniform changes and restraint matter. The actual physical process is richer than the classroom model below.

Our goal is narrower: construct two mechanically consistent hypothetical states that have the same overall length but different internal stresses. One counterexample is enough to reject a universal claim that final shortening uniquely determines residual stress.

Keep the comparison disciplined. The two hypothetical assemblies use the same member lengths, areas and elastic moduli. Their external axial resultant is zero. What differs is the supplied stress-free change assigned to each member. The example therefore isolates the information missing from the final-length observation, rather than hiding a different load or geometry inside the comparison.

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Imagine an ideal inner member and a surrounding annular member with the same original length L. Their resultant axial forces act along a common axis. Rigid end connections enforce a common end-to-end extension, and our one-dimensional model excludes bending and local connection stresses.

The coaxial arrangement matters. Opposing forces in separated off-axis bars could create a moment that needs another equilibrium account. We avoid implying that zero axial resultant alone settles every possible force and moment in an arbitrary real assembly.

Each member has a supplied positive area Aᵢ and positive elastic modulus Eᵢ. We ignore radial interaction, Poisson coupling, slip, time-dependent response, damage and changes of modulus. These exclusions make this a deliberately limited axial analogy, not a model of a specific welded section.

The common actual strain is ε = δ/L, where δ is the shared length change. Positive strain means extension and negative strain means shortening. Member i also has a supplied stress-free strain sᵢ: the length change it would prefer in the model if it were not forced to share the assembly’s length. The next step is to distinguish those two strains.

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A member’s stress-free strain is a reference change, not its actual measured strain after assembly. If the member prefers to shorten but is held longer than that preferred state, its elastic mismatch is tensile even when the whole assembly has shortened.

Use the stipulated constitutive relationship σᵢ = Eᵢ(ε − sᵢ). The difference ε − sᵢ is the elastic strain in this model. Stress follows from that difference, rather than from ε alone.

For example, if ε = −0.001 and sᵢ = −0.002, the mismatch is +0.001. The member is shorter than the original length but longer than its own stress-free length. Its calculated stress is tensile. Another member with sᵢ = 0 under the same actual strain has a compressive mismatch of −0.001.

This sign distinction is central. The words shortened and compressed are not automatically interchangeable when a material’s reference length has changed. Ask the learner to compare actual length with both the original length and the supplied stress-free length. The three-length sketch makes the stress sign understandable before any modulus is introduced.

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Compatibility says both members have the same actual axial strain because their initial lengths are equal and their end separations match. It does not say their elastic mismatches, stresses or forces are equal.

The axial force in member i is Nᵢ = Aᵢσᵢ = EᵢAᵢ(ε − sᵢ). Define wᵢ = EᵢAᵢ, a positive quantity with force units. If an external axial resultant F is applied, equilibrium requires N₁ + N₂ = F.

Combining the two statements gives (w₁+w₂)ε − w₁s₁ − w₂s₂ = F. Therefore ε = [F + w₁s₁ + w₂s₂]/(w₁+w₂). For the free axial comparison, F = 0.

The derivation uses separate jobs: compatibility connects movements, the constitutive relationship connects mismatch with stress, and equilibrium connects member forces with the external resultant. Skipping any one can produce a plausible but unsupported answer. An end displacement does not supply equilibrium by itself, and a zero external force does not impose zero force on every internal member individually.

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Supply L = 1,000 mm, E₁ = E₂ = 10,000 N/mm² and A₁ = A₂ = 100 mm². These are fictional model inputs, not measured properties of a named welding material. Each weight wᵢ is 1,000,000 N.

Set s₁ = −0.002 and s₂ = 0. With no external axial resultant, the common actual strain is the equally weighted average: ε = (−0.002 + 0)/2 = −0.001. Thus δ = −1 mm and the final assembly length is 999 mm.

The inner member’s elastic mismatch is −0.001 − (−0.002) = +0.001. Its stress is +10 N/mm² and its force is +1,000 N. The outer member’s mismatch is −0.001, giving −10 N/mm² and −1,000 N.

The forces add to zero as required. They do not individually vanish. The inner member would prefer a 998 mm stress-free length but is held at 999 mm; the outer would prefer 1,000 mm but is held at 999 mm. Their shared length reconciles two incompatible preferences by producing opposite elastic responses.

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Begin with dimensions. Modulus in N/mm² multiplied by dimensionless elastic strain gives stress in N/mm². Multiplying by area in mm² gives force in newtons. Multiplying common strain by original length gives the shared length change in millimetres.

Next check signs physically within the analogy. The inner member is one millimetre longer than its preferred length, so tension is sensible. The outer is one millimetre shorter than its preferred length, so compression is sensible. A sign error would reverse that explanation even if the force magnitudes happened to match.

Finally check equilibrium: +1,000 N and −1,000 N have zero resultant. In this coaxial axial model, their common axis avoids an unaccounted couple. Local stresses in the end connections remain excluded rather than being declared zero.

A useful independent solution starts with the three lengths, obtains the two one-millimetre elastic mismatches and divides by the original 1,000 mm length. It then recovers the same stresses and forces. Agreement between the equilibrium derivation and the length-based check is stronger than re-entering the same formula in a second spreadsheet cell.

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Keep every geometric and stiffness input unchanged. Now supply s₁ = s₂ = −0.001. Both members prefer the same 999 mm length. The equilibrium formula again gives ε = −0.001 and final length 999 mm.

This time ε − s₁ = ε − s₂ = 0. Both calculated stresses and forces are zero. The final ruler measurement is identical to the previous example, while the internal state differs.

This pair proves the limited inference result. A final shortening of one millimetre is compatible with both states under the supplied analogy. It cannot uniquely identify the stress split. The example does not claim that either history describes a particular weld. Its purpose is to expose the additional information a real stress claim would need beyond one final dimension.

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For equal weights, write the two stress-free strains as s₁ = m − d and s₂ = m + d. Their average is m. Under zero external axial force, ε = m.

The inner elastic mismatch is d and the outer mismatch is −d. If their moduli also match, their stresses are +Ed and −Ed. The average m controls the overall actual length change, while d controls the opposite mismatch in this symmetric case.

Choose m = −0.001. Taking d = 0 recovers the zero-stress case. Taking d = 0.001 recovers the first case. Other supplied differences can preserve the same final length while changing the calculated stress magnitude.

This decomposition makes the missing information explicit. The final length reveals one combination of the reference changes, not both separately. It also suggests an effective teaching question: which observation would constrain the difference? Asking for more precise measurement of the same overall length does not create a second independent relationship. Precision can narrow uncertainty in m while leaving d unresolved.

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Now keep the moduli at 10,000 N/mm² but change the areas to A₁ = 150 mm² and A₂ = 50 mm². The weights become 1,500,000 N and 500,000 N. Keep s₁ = −0.002 and s₂ = 0.

The free common strain is ε = [1,500,000(−0.002)]/2,000,000 = −0.0015. For L = 1,000 mm, the assembly shortens by 1.5 mm. The result lies closer to the inner member’s preferred change because its axial weight is larger.

The inner stress is 10,000(0.0005) = +5 N/mm², while the outer stress is 10,000(−0.0015) = −15 N/mm². Their forces are +750 N and −750 N. Equal and opposite forces do not require equal stress magnitudes when areas differ.

An unweighted average would incorrectly predict a one-millimetre shortening. A simple average is justified only by the matching weights in the earlier case. The general account uses EᵢAᵢ, not area alone or modulus alone. A useful limit check asks where ε tends as one weight becomes much larger than the other.

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Under zero external axial force, substitute the weighted-average strain into N₁ = w₁(ε − s₁). Simplifying gives N₁ = [w₁w₂/(w₁+w₂)](s₂ − s₁). The outer force is N₂ = −N₁.

The formula separates a positive stiffness-related factor from the difference in stress-free strains. If s₁ = s₂, the force is zero regardless of the common reference change. If the difference reverses sign, the member force reverses sign.

For the equal-weight case, the factor is 500,000 N. Multiplying by 0.002 gives 1,000 N. For the unequal-area case, it is 375,000 N, giving 750 N. Both agree with the earlier full calculations.

This direct form is useful after the equilibrium argument is understood. Memorising it first can conceal what F = 0 means and why a common strain still exists. Check every use against the original model: equal initial lengths, enforced common axial strain, positive linear-elastic properties and no additional coupling. A formula derived for that arrangement cannot simply be transferred to members in series, a bending plate or an arbitrary weldment.

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Add the same constant c to both supplied stress-free strains. Their weighted average increases by c, so the free actual strain also increases by c. Each elastic mismatch becomes (ε+c) − (sᵢ+c) = ε − sᵢ.

Thus the stresses and member forces stay unchanged while the overall length changes. This is the converse of the earlier counterexample: not only can the same shortening hide different stress, but different shortenings can coexist with the same stress split.

For example, begin with s₁ = −0.002 and s₂ = 0, then add c = 0.0005 to both. The new strains are −0.0015 and +0.0005. Equal weights give ε = −0.0005, a half-millimetre shortening for L = 1,000 mm. The mismatches remain +0.001 and −0.001.

These are supplied mathematical reference changes, not a proposed thermal treatment. Their value is explanatory: a visible dimensional quantity and an internal mismatch quantity need not move together. A claim that less shortening necessarily means less residual stress would fail even within this simple analogy, before the complications of a real weld are introduced.

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Suppose the final strain ε and weights are known, with F = 0. The relationship w₁s₁ + w₂s₂ = (w₁+w₂)ε constrains the two unknown stress-free strains along a line of possibilities.

For equal weights and ε = −0.001, any pair whose sum is −0.002 fits that overall measurement. The pair (−0.002,0) and the pair (−0.001,−0.001) are two examples, not an exhaustive list.

The internal force depends on s₂−s₁, which changes along that line. Without further information, the final length cannot select one member-force split. An assumed permissible range for each stress-free strain could bound the result, but those ranges would be additional inputs requiring a basis.

Do not convert this lack of identification into an unlimited real-world stress claim. Real materials have constitutive limits and physical histories that our analogy omits. The correct conclusion is that the supplied observation and model do not uniquely determine the stress. It is not that every algebraically imaginable state must be physically attainable in an actual fabricated component.

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Within the supplied model, an independently known inner stress σ₁ together with the actual strain ε would give s₁ = ε − σ₁/E₁. Equilibrium and the known areas could then determine the outer stress, and its constitutive relationship would give s₂.

In the first example, ε = −0.001 and σ₁ = +10 N/mm² imply s₁ = −0.001 − 10/10,000 = −0.002. Zero external resultant requires N₂ = −1,000 N, so σ₂ = −10 N/mm² and s₂ = −0.001 − (−10/10,000) = 0.

This algebra explains what independent information would do. It is not a recommendation to obtain a particular measurement or to treat any instrument reading as the required member stress. A real residual-stress method has a location, depth, direction, uncertainty and interpretation model.

The useful distinction is between adding a new relationship and repeating the same one. Another ruler reading of the same end separation can improve precision but does not independently reveal an internal stress. The additional observation must constrain the missing part of the state, and its meaning must match the model used to interpret it.

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Return to the first equal-weight example, but now impose ε = 0 with an external fixture in the axial model. The assembly is held at its original 1,000 mm length.

The inner member’s mismatch is 0 − (−0.002) = 0.002, giving +20 N/mm² and +2,000 N. The outer member has zero mismatch and zero force. The required external axial resultant is therefore +2,000 N.

This state is not the earlier free, self-balancing state. The external fixture supplies a nonzero resultant. Releasing that specified length constraint in the ideal elastic analogy allows the common strain to reach −0.001 and the member forces to become +1,000 N and −1,000 N.

Label the boundary condition before comparing measurements. A dimension observed while held and a dimension observed after release need not describe the same state. Nor does no visible movement under a fixture prove no stored elastic response. The mathematical comparison is useful precisely because it distinguishes imposed displacement, external reaction and internal force rather than describing all three with the single word restraint.

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For the same linear model, ε = ε_free + F/(w₁+w₂), where ε_free is the weighted stress-free average. The external load adds a common actual-strain increment to the earlier free state.

With w₁+w₂ = 2,000,000 N, an applied resultant F = 2,000 N adds strain 0.001. Starting from ε_free = −0.001, the loaded common strain becomes zero. This recovers the externally held length in the previous chapter.

The member stresses become +20 N/mm² and zero. Their forces add to the applied 2,000 N. The earlier self-balancing contribution has not been made nonexistent; the total stress is the sum of that contribution and the load-induced contribution under the stipulated linear assumptions.

This decomposition should not be extended automatically through yielding, damage, relaxation or a changed connection. Those effects can invalidate superposition. Here it teaches a limited but useful point: unloading the external resultant to zero does not necessarily unload every member to zero stress when incompatible reference changes remain. The external load history and the internal reference state must both be named.

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For a member with uniform elastic mismatch, the supplied axial model gives stored energy Uᵢ = ½LEᵢAᵢ(ε − sᵢ)². The square makes each contribution nonnegative, regardless of whether its stress is tensile or compressive.

In the first free example, each weight is 1,000,000 N, L = 1,000 mm and each mismatch magnitude is 0.001. The total is ½ × 1,000 × [1,000,000(0.001)² + 1,000,000(0.001)²] = 1,000 N mm, or one joule.

The uniform-reference-change case has the same final length but zero elastic mismatch and zero stored energy in this analogy. That provides another difference between the two states hidden by the single dimensional observation.

Do not infer a real release event or failure risk from that one-joule result. The model excludes local connection response and actual material limits. Its energy calculation is an independent consistency check: opposing member forces can coexist with positive stored elastic energy. Their zero resultant does not mean their mechanical effects or energy cancel as signed numbers.

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For the free assembly, write U(ε) = ½LΣwᵢ(ε − sᵢ)². Differentiate with respect to the common actual strain. The derivative is LΣwᵢ(ε − sᵢ), which is L times the axial resultant.

Setting that derivative to zero recovers the weighted-average strain. Because the second derivative is LΣwᵢ and every weight is positive, this stationary point is the unique energy minimum with respect to the one allowed common axial strain.

This agrees with the earlier equilibrium derivation, but it does not add omitted physical degrees of freedom. A model that excludes bending cannot use its one-variable minimum to prove that a real component will not bend or buckle. The minimum is conditional on the permitted motions and constitutive assumptions.

For learners, the dual derivation is useful. One route balances forces; the other compares elastic penalties for moving away from each member’s preferred length. Both explain why the common strain falls between the two reference strains in the free positive-weight case. Agreement helps verify the algebra while the model boundary remains visible.

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The relation between external resultant and actual strain is F = (w₁+w₂)(ε − ε_free). Its slope depends on the combined weights. Assemblies with the same weights and the same weighted reference strain share this whole ideal axial load–strain line.

Our two equal-final-length examples therefore also share the same external tangent stiffness. A small additional load test that measures only overall extension does not automatically identify the hidden residual-force difference. The internal member stresses can differ while the measured overall line remains the same.

This is stronger than saying that one isolated ruler reading is insufficient. It identifies a family of measurements that still depend on the same combined quantities. More data points help test that line and its uncertainty, but cannot necessarily recover a variable that the line does not contain independently.

The connected Wood Composite Force-Sharing Guide begins with unstressed members and develops stiffness-based sharing. The present analogy changes that reference-state assumption. Keep the two responsibilities distinct rather than using the common compatibility equation to erase their different questions.

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Use L = 500 mm, E₁ = 20,000 N/mm², A₁ = 50 mm², E₂ = 10,000 N/mm² and A₂ = 200 mm². Supply s₁ = −0.003 and s₂ = −0.0003, with zero external axial resultant.

First calculate the weights: w₁ = 1,000,000 N and w₂ = 2,000,000 N. Then ε = [1,000,000(−0.003) + 2,000,000(−0.0003)]/3,000,000 = −0.0012. The shared length change is −0.6 mm, giving final length 499.4 mm.

The inner mismatch is +0.0018, so its stress is +36 N/mm² and its force is +1,800 N. The outer mismatch is −0.0009, so its stress is −9 N/mm² and its force is −1,800 N. Force equilibrium holds despite different stress magnitudes and different moduli.

Check the preferred lengths too: 498.5 mm for the inner member and 499.85 mm for the outer. The assembly lies between them. Its position is a stiffness-weighted compromise, not the arithmetic midpoint of those lengths. The changed case tests the general derivation rather than a memorised equal-and-opposite-stress shortcut.

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Additional information can narrow the inference without identifying one exact history. In the equal-weight model, keep the measured actual strain at −0.001 and supply a hard range s₁ between −0.0025 and −0.0015. The free equilibrium relationship then requires s₂ = −0.002−s₁.

The compatible outer reference strain consequently ranges from −0.0005 to +0.0005. It is linked to s₁, not independently selectable from a separate interval. Every permitted pair still gives the same one-millimetre shortening.

The inner stress is 10,000(−0.001−s₁), so it lies between +5 and +15 N/mm². With its 100 mm² area, the inner force lies between +500 and +1,500 N. The outer force is the opposite at each matching endpoint.

Construct the endpoint pairs and check equilibrium rather than combine incompatible extremes. This is a deterministic conditional bound, not a probability statement and not an allowable stress range for a material. Its value comes from showing what the added history restriction does: a single final length still does not identify the force, but the supplied range can limit the possibilities. The credibility of an actual bound would depend on the evidence supporting that added restriction.

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The self-balancing result depends on the imposed compatibility. Imagine instead two ideal members placed in series along one axis, with no external axial resultant, no intervening axial loads and free overall ends. Their common axial force must be zero by equilibrium.

Under the same linear constitutive relationship, each member can take its own supplied stress-free strain. Its elastic mismatch is zero. The two preferred length changes add to the total assembly change without creating the opposing-force mechanism of the linked coaxial model.

For equal 1,000 mm member lengths with reference strains −0.002 and zero, the series assembly’s total length change is −2 mm. The axial model gives zero stress in both members. This is a different geometry and total original length, so it should not be compared casually with the earlier 1,000 mm assembly as if only a calculator setting had changed.

The exercise isolates the role of topology. Incompatibility creates a stress only when the connection prevents the relevant free changes from occurring independently. A real welded assembly can constrain many directions and include bending, so its compatibility cannot be inferred from the word joined alone. Draw the connections and permitted motion before choosing the equation. That step prevents a useful analogy from becoming an unsupported universal rule about every fabrication.

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Real welds experience spatially varying heating, phase changes, temperature-dependent properties, plastic deformation and complex restraint. A uniform two-member axial model cannot predict that field from a final length or from one assumed pair of reference strains.

The analogy supports a narrower claim: incompatible stress-free changes can coexist with a common actual geometry and self-balancing internal forces. It explains why visible distortion and residual stress need not be interchangeable measurements. It does not establish their numerical relationship for a real component.

Do not call sᵢ a measured plastic strain unless such a measurement and interpretation have actually been supplied. In this guide it is an imposed model input representing a preferred reference change. Its origin remains outside the calculation.

Likewise, a calculated tensile stress is not an acceptance decision. Service suitability involves material response, defects, geometry, loading, applicable requirements and competent interpretation. Return to the welding-engineering guide for that assurance chain. Preserving these limits makes the classroom result more useful: it teaches what extra evidence a claim needs instead of encouraging a reader to certify a joint from a tidy formula.

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If a learner says zero external force means zero stress, draw the two opposing member forces and add them. Equilibrium concerns their resultant. It does not require each contribution to vanish.

If they say a shortened member must be in compression, compare its actual length with its own preferred length. The inner member in the first example is shortened relative to the original reference but stretched relative to its changed stress-free state.

If they average stress-free strains without weights, ask which member has the larger EᵢAᵢ. The area-and-modulus combination determines its contribution to axial equilibrium. If they claim equal and opposite stresses with unequal areas, multiply each stress by its matching area and check the forces.

Finally, if they treat a successful model calculation as a real weld assessment, return to the missing physical inputs. Ask what represents the thermal history, material limits, spatial stress variation and measurement scope. Repairing that inference is part of the mathematics. Correct arithmetic has not completed the argument when the conclusion quietly changes from a supplied axial analogy to an actual fabricated component.

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For repair, use three drawn lengths: original, preferred and actual. Keep the modulus out initially. Ask whether the member must stretch or compress relative to its preferred state to fit the assembly.

For stabilisation, add the constitutive relationship, force equilibrium and the equal-member example. Require independent checks of dimensions, signs and resultant. Then compare the two histories with the same final length. The learner should explain which supplied quantity differs and why the ruler cannot distinguish it.

For extension, use unequal weights, add a stated external load and derive the energy minimum. Follow with the changed 500 mm case independently. A complete explanation should name both the general principle and the assumptions that prevent it from becoming a full weld simulation.

The project can be completed on paper or in a spreadsheet using fictional inputs. No heating, cutting, loading or disassembly of a real component is needed. A strong outcome is a reproducible evidence note showing what was supplied, what was calculated, which checks passed and which real-world claim still lacks support.

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Can a free assembly carry internal stress? Yes, in the supplied model its member forces can balance to zero while each member remains stressed.

Does the same final length imply the same residual stress? No. The two constructed histories have identical final length but different elastic mismatches.

Does a smaller visible shortening prove a smaller stress? No. Adding a common reference change alters overall length while leaving the mismatch stresses unchanged in this model.

Are equal and opposite forces also equal and opposite stresses? Only when the matching areas are equal. Force is stress multiplied by area.

Can a stiffness measurement recover every internal state? Not necessarily. The ideal overall load–strain line depends on combined weights and a weighted reference change, leaving some internal differences unresolved.

Is this a method for approving a weld? No. It is an explicit mechanical analogy about the limits of one dimensional observation, with real welding assessment left to the appropriate evidence and competent process.

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Check the physical context

TWI: Distortion, Types and Causes explains the relationship between nonuniform thermal changes, restraint and distortion in welding. It was checked on 23 October 2026. MIT Structural Mechanics: Effects of the Environment supplies the distinction between actual and mechanical strain and the concept of self-equilibrating stress. The coaxial-member analogy, numbers and inference checks here are original; they do not reproduce a real welding thermal cycle or an acceptance method.

Follow the distinct mechanical questions

Welding Heat Input, Travel Speed and Cooling Rates already develops energy records and thermal-history questions. Reconstructing an Ideal Fillet Throat concerns a different geometric measurement. Wood Composite Force Sharing develops compatibility for initially unstressed members.

Thermal Restraint and a Flexible Support develops an initially stress-free heated rod with an external connector. Its support reaction is a different model from the self-balancing internal force considered here.

Return to engineering evidence

Welding Engineering connects the actual joint with qualification and assurance. Return to Welding, How X Works and Learning That Transfers into Life. The final transfer task is to construct another pair of reference histories that share the same final length and explain which additional observation would distinguish their internal forces.

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