Materials Science · Processing routes and microstructural evolution

Deformation, recovery and recrystallisation

Plastic working can increase and rearrange dislocations and change grain shape and texture. Some input work becomes stored defect energy, while much is dissipated as heat. Cold working does not mean that no heating occurs.

Useful background: Processing, structure and useful properties; Energy, stability and environmental context. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Distinguish work hardening, recovery and recrystallisation.
Explain how defect storage and rearrangement connect a deformation history to later annealing.
Keep dislocation energy, hardness and a classified recrystallised area fraction distinct.
Interpret matched histories, unresolved image regions and grain growth without inferring a universal annealing schedule.

Deformation stores a changed defect state

A worked strip becomes harder to indent. After a short anneal it softens, although its visible elongated grain outlines remain. Has the strip already formed new grains, or can its internal defect structure change first? Follow the starting state, the local thermal history and two different kinds of evidence rather than using hardness as a clock.

A dislocation is a line defect whose motion permits local slip. Deformation can multiply and store these lines; their elastic fields and junctions can resist subsequent motion. This work hardening can raise resistance to continued plastic flow without adding a new element. Grains, dislocations and strengthening develops the obstacle mechanisms. Here the question is what happens when that deformed state is allowed to evolve.

Plastic working can also rotate crystal regions and produce texture: a preferred distribution of crystallographic orientations relative to the strip. Texture is different from elongated grain shape. Two grains can have similar outlines yet different orientations and ease of slip in a chosen test direction. Neither shape nor a hardness value alone supplies the orientation distribution.

The following main history concerns a representative crystalline metal whose phase and overall composition are stipulated to remain unchanged during the comparison. The new numerical records are invented teaching inputs, not alloy data or a production schedule. Recovery during the working step is assumed too limited to erase the stored state; cold working is a comparison of competing processes, not a promise of zero temperature rise.

Account for stored line energy without calling it hardness

Dislocation density ρ here means total dislocation line length divided by specimen volume. Its units are m/m³ = m⁻². It is not the number of grains per image area or a crack count. Supply an effective line energy Γ in J/m. In a deliberately simplified uniform account, total line energy is Γℓ_total, and its density is e_d = Γρ in J/m³.

For the numerical account only, stipulate Γ = χGb². G is the shear modulus in Pa, b is the Burgers-vector magnitude in m, and χ is a dimensionless effective coefficient. Pa × m² = N = J/m, as required. This captures an elastic energy scale, not an exact energy for every line: core structure, line character, elastic interactions and arrangement affect the effective coefficient. Hold Γ fixed when attributing the example’s energy change solely to a change in ρ.

Take G = 40 GPa = 4.0 × 10¹⁰ Pa, b = 0.250 nm = 2.50 × 10⁻¹⁰ m and χ = 0.50. Then Γ = 1.25 × 10⁻⁹ J/m. A reference density ρ_ref = 1.0 × 10¹³ m⁻² gives e_ref = 12,500 J/m³. A worked state at 4.0 × 10¹⁴ m⁻² gives 500,000 J/m³. The stored increase is 487,500 J/m³, or 0.4875 MJ/m³, rather than the full final line energy.

Compare that increase with an independently supplied plastic-work input of 20 MJ/m³. The accounted line-energy increase is 0.4875/20 = 0.024375, or 2.4375% of that input. Only if this example further closes the energy account to line storage and thermal energy, with other stores and kinetic changes excluded, does the remainder equal 19.5125 MJ/m³ entering the thermal account. Without heat-transfer and heat-capacity information, it is neither a temperature rise nor the heat that has already left the specimen.

At unchanged Γ, lowering ρ from 4.0 × 10¹⁴ to 1.0 × 10¹⁴ m⁻² lowers this line-energy density from 0.500 to 0.125 MJ/m³, a decrease of 75% relative to the worked state. That calculation does not imply a 75% hardness decrease or 75% recrystallisation. It contains neither an indentation law nor a boundary-growth law. Rearrangement can also lower real energy without removing the same fraction of line length, which this constant-Γ account cannot describe.

Annealing can follow different stages

Recovery rearranges or reduces parts of the defect structure; recrystallisation replaces deformed regions through growth of grains with much lower stored deformation and dislocation content. This commonly changes the grain structure without creating a different thermodynamic phase. Grain growth may continue afterward. The sequence and extent depend on alloy, deformation, temperature and duration.

During recovery, opposite dislocation segments can annihilate where their geometry and motion allow it. Other lines rearrange into lower-energy structures, including low-angle boundaries separating slightly misoriented subgrains. A subgrain is a crystal region separated from its neighbour by a relatively small orientation change. Its boundary is not automatically equivalent to the high-angle boundary between the original grains.

The outer grain outlines can therefore remain recognisably elongated while the internal state changes. Recovering dislocations may become less obstructive and local residual stresses may relax; the size of the property change depends on which features controlled that property. Recovery is not simply elastic springback after unloading, nor does it necessarily restore the original shape or every initial property.

In the conventional discontinuous recrystallisation picture used here, viable low-stored-energy regions grow into the deformed surroundings by migration of high-angle boundaries. Atoms rearrange as the boundary advances; grains are not pieces of material being translated through one another. A growing region can develop from an existing subgrain or a favourable boundary region. Calling it a new grain does not require crystallisation from a melt or a newly created chemical phase.

Recovery can proceed while recrystallised regions grow, reducing the stored-energy contrast available to drive their growth. The familiar sequence is thus a useful description, not three perfectly separate temperature boxes. During sufficiently warm deformation, recovery or recrystallisation can also occur while working continues; the present records instead concern annealing after working has stopped.

Qualitative grain and defect changes during cold work and a suitable recrystallisation anneal; not measured micrographs or a grain-size prediction.

Follow two observations through one supplied history

Use route C: a strip is reduced from 2.00 to 1.00 mm thickness, a 50% thickness reduction. This geometric measure is 100(t₀ − t_f)/t₀; it is not a universal measure of dislocation density or equivalent plastic strain. The same stock, starting grain state and chemistry are used for the comparisons below. Local specimens reach one chosen hold temperature T_a before the hold clock starts. All are then cooled to 25 °C for matched examination and hardness testing. T_a is not assigned a numerical recipe.

H denotes a conventional Vickers hardness number, reported here on one common load, dwell, surface-preparation and sampling basis. It describes indentation resistance. These HV numbers are not tensile stresses in MPa, elastic moduli or fracture toughnesses. Separate matched coupons supply the different endpoints; no hardness test is assumed to identify every structural feature.

The structural column is independently supplied: suitable orientation and boundary information, with a prior-state comparison, identifies regions that have replaced the deformed structure. X_A is the area classified as recrystallised divided by the examined area. It is an examined-section measure, not automatically a bulk volume fraction. A low internal orientation spread or an equiaxed outline by itself is insufficient: a recovered subgrain can also look relatively uniform. Keep the classification rule, resolution and unclassified area.

C0 to C5 changes H from 160 to 148, a fall of 12/160 = 7.5%. No recrystallised area is resolved in the stipulated examined fields at C5; that statement is not proof that no small region exists anywhere. The independent defect evidence supports recovery. The separate line-energy account illustrates why a large change in one defect-energy contribution need not be the same percentage change in hardness.

By C20, growing low-deformation grains occupy 60% of the classified examined area while deformed regions remain. By C60, the examined fields are fully classified as recrystallised, with a supplied mean grain size of 20 µm. At C120 they remain fully classified but the same grain-size measure is 40 µm. Continued coarsening after replacement is different from increasing the recrystallised fraction. The lower H = 94 is consistent with a changed state, but does not independently isolate a grain-size effect.

Route B begins with a smaller thickness reduction, from 2.00 to 1.50 mm, or 25%. Its post-working H is stipulated as 140. After the same T_a and 20 min hold, B20 has H = 118 and X_A = 20%. C20 has that same H but X_A = 60%. The equal hardness does not make their histories or structural states equal. Nor do the two reductions predict a universal ratio of recrystallised fractions.

Original invented, exact teaching records for one stipulated comparison basis. Time is local isothermal hold time at the same T_a after working; C and B have different prior reductions. X_A describes only the examined fields. A reported 0 means no area classified as recrystallised at the available resolution, and 100 means all examined area is classified; neither is a whole-batch certainty.
State	Hold (min)	H (common HV convention)	X_A (%)	Independent structural description
C0	0	160	0	Deformed elongated grains
C5	5	148	0	Recovered substructure; no new grains resolved
C20	20	118	60	Growing low-deformation grains and remaining deformed regions
C60	60	100	100	Replacement complete in fields; mean grain size 20 µm
C120	120	94	100	Same class fraction; mean grain size 40 µm
B20	20	118	20	Different prior deformation; incomplete replacement

Original invented route C record at one stipulated hold temperature, followed by matched examination at 25 degrees Celsius. The upper panel shows conventional Vickers hardness numbers at local hold times zero, five, twenty, sixty and one hundred twenty minutes: 160, 148, 118, 100 and 94. The lower panel shows the independently supplied percentage of examined area classified as recrystallised: zero, zero, sixty, one hundred and one hundred. Hardness uses hollow circles; classified-area percentage uses hollow squares. Panels have separate linear vertical scales and the same linear time scale. Only supplied points are drawn; no connecting curve, kinetics fit or uncertainty interval is supplied. A zero classified area does not prove absence of unresolved regions, and a fully classified field is not a whole-batch guarantee. At the early five-minute observation hardness falls before new grains are resolved; after sixty minutes the class fraction is unchanged while the supplied grain-size measure continues to change. These are teaching records, not a material treatment recipe.

A completed replacement can still evolve

Once lower-dislocation-density grains occupy the material, their boundaries still contribute energy. Reducing total boundary area can favour grain growth: some grains expand while others shrink. This can occur in an already recrystallised structure and does not require a fresh deformation step. It is distinct from growth driven mainly by the stored-energy difference between a new grain and a highly deformed neighbour.

Growth requires boundary mobility as well as a driving tendency. Solute interactions and suitable fine particles can retard boundary movement; coarse particles can instead create deformed neighbourhoods favourable for recrystallisation. Particle size, spacing, location and evolution matter. The word particle alone does not specify whether a particular population accelerates or delays the observed replacement.

A larger prior reduction can leave more stored defects and more favourable sites, but it can also change texture and heterogeneity. In a reported comparison of austenitic stainless steel, specimens rolled to different reductions and given the same subsequent thermal path had different recrystallised proportions and grain sizes. That observation demonstrates a prior-state dependence in that comparison; it does not establish a universal minimum reduction or a treatment time for another alloy.

Annealing need not restore a random orientation distribution. Some orientations may have advantages in developing viable growing regions or in boundary migration. An equiaxed final grain map can therefore retain or develop texture. Test direction matters when comparing mechanical properties; a shape-only image cannot establish isotropy.

Choose the intended final state rather than simply the longest hold. A formable intermediate state and a strong final state may require different properties. The present H and X_A records supply neither a forming limit nor a fatigue, creep or safe-service assessment. Thermal treatment and evolving microstructure distinguishes this single-phase annealing task from quenching, tempering and precipitation ageing.

Ask what observation could distinguish the alternatives

Suppose a separate segmented field contains twenty equal-area units. Six units are confidently recrystallised, twelve confidently retain deformed material, and two are unresolved. Each tile in the next accounting diagram represents 1 µm² of classified or unresolved area, not one grain or one point observation. With the assigned classes accepted, the examined-area fraction lies between 6/20 = 30% and (6 + 2)/20 = 40%. Discarding the two unresolved units would change the denominator and hide the uncertainty.

This is a deterministic classification interval under the stated labels, not a statistical confidence interval or a representative bulk fraction. Resolving one uncertain unit as recrystallised and the other as deformed yields 7/20 = 35% for this field. More fields address spatial variation; more independently treated specimens address treatment reproducibility. One cannot substitute for the other.

Different observations answer different questions. Hardness probes local indentation response. Orientation maps distinguish neighbouring lattice orientations and internal variation at their spatial and angular resolution. Diffraction profiles can respond to strain and coherent-domain size as well as instrument effects. None is an automatic total-dislocation counter. Tracking a boundary or a subgrain through a time series tests change more directly than comparing unrelated attractive images.

Individual embedded subgrains have been followed during recovery of cold-rolled aluminium by X-ray diffraction: changes in spot position and intensity were used to infer orientation and volume changes. That evidence concerns recovery within the studied specimen and method limits. In another recovered Al–Mn study, registered orientation maps and electron images correlated subgrains with smaller particles. A single contrast image would miss part of that relationship; a mapped orientation-derived dislocation measure is not every stored dislocation.

For the present hypothesis, pair a matched hardness change with evidence of internal defect rearrangement and the appearance or growth of suitable high-angle-bounded regions. Check phase and particle changes if those are possible in the actual material. Keep actual local time and temperature, prior reduction and orientation, examination locations and classification limits. A universal recrystallisation temperature, a linear hardness-to-fraction conversion or a schedule inferred from one image would exceed the evidence.

Separate original segmented-area accounting model. Twenty equal-area tiles represent one square micrometre each, for twenty square micrometres in total. These are area entries, not grains, point samples or a measured micrograph. Six tiles labelled R are confidently recrystallised, twelve labelled D confidently deformed, and two hatched tiles labelled question mark are unresolved. All twenty remain in the denominator. The smallest qualifying area is six and the largest eight square micrometres, giving an examined-area interval from thirty to forty percent. This deterministic classification bound has no statistical coverage probability and does not establish a bulk volume fraction. If one question-mark tile becomes R and the other D, seven of twenty square micrometres qualify, or thirty-five percent.

Worked examples

Work through the case

Hardness decreases before new grains are clearly observed during an anneal. Is recrystallisation the only possible explanation?

Recovery can alter the defect structure.
Check the resolution and representative sampling of images.
Combine structural and property evidence.

No. Recovery or other changes may contribute before substantial recrystallisation.

Use an energy account, then identify its missing predictions

Use the supplied G, b, χ and three dislocation densities. Find the line-energy increase from reference to worked state and its fraction of the 20 MJ/m³ plastic-work input. Then find the energy decrease to the recovered density. May these numbers be called the recrystallised fraction or heat already discharged?

Γ = 0.50 × (4.0 × 10¹⁰ Pa) × (2.50 × 10⁻¹⁰ m)² = 1.25 × 10⁻⁹ J/m. Multiplying by each density gives 0.0125, 0.500 and 0.125 MJ/m³ for reference, worked and recovered states.
Subtract the reference: Δe_d = 0.500 − 0.0125 = 0.4875 MJ/m³. Dividing by 20 gives 0.024375, or 2.4375%. Using 0.500/20 would count the reference energy as newly stored work.
From worked to recovered, the decrease is 0.500 − 0.125 = 0.375 MJ/m³, or 75% of the worked line energy. It is a change between the two model states, not an additional plastic-work input.
The assumed closed two-channel deformation account would put 19.5125 MJ/m³ into the thermal account. Actual discharge and temperature require a heat balance. Boundary-growth kinetics and indentation response are absent, so neither X_A nor H follows from the energy ratio.

The model stores an additional 0.4875 MJ/m³, or 2.4375% of the input; the later line-energy decrease is 0.375 MJ/m³. No recrystallised fraction, hardness or discharged-heat amount is determined.

Test the proposed hardness clock against a changed history

A proposed shortcut calls f_H = (H_start − H)/(H_start − H_end) the recrystallised fraction. Apply it to C20 using C0 = 160 and C60 = 100. For B20, a separate fully replaced endpoint is stipulated at H_end = 100, while H_start = 140. Compare both predictions with the independently supplied X_A values.

For C20 the numerical index is (160 − 118)/(160 − 100) = 42/60 = 0.70. Its actual stipulated examined-area fraction is 0.60, so the index exceeds it by 0.10, or 10 percentage points.
For B20 the index is (140 − 118)/(140 − 100) = 22/40 = 0.55, whereas the independent X_A is 0.20. The discrepancy is 35 percentage points. Equal current H does not even give the same normalised index when the starting states differ.
The index assumes that all softening follows a linear exchange between fixed endpoint responses. Recovery can change the remaining deformed regions; grain size, texture and other obstacles can evolve. Indentation is also a constrained local response, not an automatic linear area mixture.
The table disproves this shortcut as an exact fraction measure for these supplied records. It does not say that a material-specific calibrated hardness correlation can never be useful. Such a correlation needs independent structure evidence and a validated domain.

f_H is 70% for C20 and 55% for B20, compared with X_A = 60% and 20%. The normalised softening indices do not measure the recrystallised fractions in this record.

Resolve a threshold without deleting difficult regions

For the separate twenty-unit field, six units are recrystallised, twelve deformed and two unresolved. A declared screen requires X_A ≥ 35% for this examined field only. Can the present classification decide? What changes if one unresolved unit is then identified in each class?

Keep the full 20 µm² denominator. The smallest qualifying area is 6 µm², and the largest is 8 µm². Therefore 0.30 ≤ X_A ≤ 0.40.
The threshold 0.35 lies inside that interval. The current evidence is insufficient to declare either a pass or a failure. Reporting 6/18 = 33.3% after omitting unresolved area answers a different question.
After the stated new observations, the qualifying area is 7 µm² and the deformed area 13 µm². X_A = 7/20 = 0.35, meeting the inclusive ≥ screen exactly.
That decision applies to this stipulated field and classification. It establishes neither a whole-specimen volume fraction nor a suitable forming or service state.

Initially the 30–40% interval leaves the screen undecided. The new classification gives exactly 35%, passing only the stated examined-field screen.

Try the reasoning

Does recrystallisation necessarily change overall alloy chemistry?
Show the answer and reasoning
No.
It can change grain and defect structure at the same composition.
Why record prior deformation?
Show the answer and reasoning
It changes the starting energy and structure.
The same anneal can act differently on another state.
In a separate constant-coefficient account, G = 50 GPa, b = 0.300 nm, χ = 0.40 and ρ = 2.0 × 10¹⁴ m⁻². Find Γ, e_d and total line energy in a volume of 8.0 mm³. Does this supply a hardness?
Show the answer and reasoning
Γ = 1.80 × 10⁻⁹ J/m, e_d = 0.360 MJ/m³ and total line energy 0.00288 J. No hardness follows.
Γ = 0.40 × 5.0 × 10¹⁰ × (3.0 × 10⁻¹⁰)². Then e_d = Γρ = 360,000 J/m³. Since 8.0 mm³ = 8.0 × 10⁻⁹ m³, multiplying gives 2.88 × 10⁻³ J. An indentation model was not provided.
Keep G and b the same. State P has ρ = 2.0 × 10¹⁴ m⁻² and effective χ = 0.25; state Q has ρ = 1.0 × 10¹⁴ m⁻² and χ = 0.50. Compare their model e_d. Must their hardnesses and grain structures match?
Show the answer and reasoning
Their model line-energy densities are equal. Their hardnesses and grain structures are not determined.
Both products χρ are 5.0 × 10¹³ m⁻², so χGb²ρ is equal. Allowing χ to differ acknowledges an effective arrangement or line-character difference; equality of one scalar energy does not specify obstacle strengths, boundaries or recrystallised area.
A different twenty-unit area account has five recrystallised, twelve deformed and three unresolved units. Bound X_A. Can a requirement of at least 40% be ruled in or ruled out?
Show the answer and reasoning
25% ≤ X_A ≤ 40%. The requirement remains undecided; it passes at the upper endpoint only if all three unresolved units qualify.
The denominator stays 20. The extremes are 5/20 and 8/20. A possible upper-end pass is not demonstrated passage, while the inclusive threshold means failure is not yet demonstrated either.
C60 and C120 both have X_A = 100%, but their stated mean grain sizes are 20 and 40 µm. Has the recrystallised fraction doubled? Does the longer hold necessarily improve the intended material?
Show the answer and reasoning
No. The fraction is unchanged while the stated grain-size measure doubles. Improvement needs the actual property requirements.
Further boundary migration can coarsen a completed grain structure. Hardness, tensile ductility, creep and fracture behaviour are separate responses; the supplied numbers do not optimise them.
At one hold temperature a new prior state is 30% classified as recrystallised after 10 min. Must it be 60% after 20 min? Would a higher furnace setpoint alone establish a faster completed treatment?
Show the answer and reasoning
Neither conclusion follows.
No fraction–time law is supplied. Recovery, viable growing regions, boundary mobility, impingement and spatial heterogeneity matter. A furnace setpoint is not the specimen’s full local thermal path; compare actual states and observations.
After annealing, a map shows equiaxed grains and diffraction still identifies the same crystal phase. A report concludes that the alloy chemistry changed and every test direction now has the same strength. What is missing?
Show the answer and reasoning
Neither a chemistry change nor isotropic strength is established. Composition, orientation distribution and matched directional property evidence are missing.
Recrystallisation can replace a deformed grain structure without a phase or bulk-composition change. Equiaxed shape does not imply random crystal orientations. Texture and other state variables can preserve directional response.

What to carry forward

Name distinct structural processes.
Interpret treatment through its starting state.
Recovery can change the internal dislocation arrangement before new grains are resolved; the processes can overlap.
Stored line energy, indentation response and classified area fraction have different definitions and require different evidence.
Keep unresolved area and the prior deformation state in the inference; grain growth can continue after replacement.

References

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