Materials Science · Metals, alloys and microstructural design
Alloy composition, phases and phase diagrams
Read a bounded temperature–composition diagram, distinguish phases from chemical ingredients, and derive phase amounts by conservation. Follow a closed alloy through changed equilibrium conditions without mistaking the diagram for a cooling-rate or microstructure prediction.
Useful background: Atoms, bonding, structure and defects; Processing, structure and useful properties. Each worked case explains the quantities it uses.
Back to the chapterAfter reading, you can
- Read equilibrium phase amounts using composition balance.
- Distinguish overall composition, composition within a phase, and the amount of that phase.
- Identify single-phase and two-phase fields and read isothermal boundary intersections.
- Track changed equilibrium amounts at fixed overall composition and retain the mass, atom-count or volume basis.
- Recognise when the supplied diagram and balance do not describe the actual specimen.
An alloy can contain more than one phase
Alloying combines elements, often creating a solid solution, multiple phases or both. An equilibrium phase diagram shows which phase combinations are favoured for its composition, temperature and pressure assumptions. It does not directly show transformation speed or final grain shape.
An alloy is a material containing more than one chemical element, with metallic behaviour in the present context. The chemical ingredients are called components. A phase has a particular material state and structure; a solid solution is one phase containing more than one species in its structure. The same overall ingredients can instead divide between phases of different structures or compositions.
Keep three questions separate. What fraction of the whole alloy is ingredient B? How much B is inside phase α or phase β? What fraction of the alloy belongs to each phase? The first is overall composition, the second phase composition, and the third phase amount. The Greek labels α and β name phases, whereas A and B name the chemical components in the following invented binary system. Neither phase is assumed to be a pure component.
Grains are another distinction. Many differently oriented grains can belong to the same phase. Two-phase material need not have exactly two grains, and an overall composition does not specify particle shape. Defects, boundaries and interfaces introduces grain versus phase boundaries; Grains, dislocations and strengthening explains why those structural arrangements can affect motion.
Locate the state before drawing a tie line
The new chart is an original hypothetical, cropped solid-state binary equilibrium diagram at one fixed pressure. Its horizontal coordinate C is composition in mass% B; its vertical coordinate is absolute temperature T in kelvin. It covers only 0–80 mass% B and 500–800 K. It is not a measured alloy chart or a complete diagram up to melting. The labelled phase fields are supplied thermodynamic possibilities, rather than derived here from atomic interactions.
Within this window, stipulate straight boundary segments C_α = 10 + 0.100(T − 600) and C_β = 50 + 0.100(T − 600), where T is the numerical kelvin temperature and C is in mass% B. For example, at 700 K the endpoints are 20 and 60. These linear segments describe this local teaching diagram only; increasing solubility with this particular slope is not a law for every alloy.
Locate an alloy at its overall composition C₀ and chosen T. Left of the α-side boundary the field is single α; between the boundaries it is α + β; right of the β-side boundary it is single β. At 600 K, an alloy with C₀ = 5 lies in α and one with C₀ = 70 lies in β. A homogeneous single phase then has the overall composition of its specimen: 5 or 70 mass% B respectively. Do not replace those values with a boundary composition from another field.
For a point inside α + β, keep the temperature fixed and move horizontally to both boundaries. Those intersections give C_α and C_β, the compositions of the coexisting phases. At 600 K they are 10 and 50 mass% B. Every overall composition strictly between them shares those endpoint compositions at that temperature; its phase amounts change instead. At C₀ = 20 the α phase is not 20% B, and the β phase does not constitute 50% of the specimen just because it contains 50% B.
A point exactly on one of these simple two-phase boundaries has the limiting two-phase balance with zero amount of the other phase. For example, at 600 K and C₀ = 10, the model gives α at 10% B and zero β. No three-phase junction or additional phase is present in this cropped model.
A tie line supplies a balance
In a binary temperature–composition diagram at fixed pressure, draw a horizontal tie line through the two-phase region at the chosen temperature. Its endpoints give the equilibrium phase compositions; the overall composition lies between them. Material balance then gives the phase amounts. A mass-percent composition axis gives mass fractions, not volume fractions. An atomic-percent axis uses an atom-count basis; do not mix the bases.
Derive the amount rule from mass conservation. Let total mass be M, with phase masses M_α and M_β. Define f_α = M_α/M and f_β = M_β/M. All material belongs to one of these two phases, so f_α + f_β = 1. Fractions are dimensionless and must lie between zero and one.
The B mass is (C₀/100)M. Within the phases it is (C_α/100)M_α + (C_β/100)M_β. Equate these and divide by M/100: C₀ = f_α C_α + f_β C_β. Using f_α = 1 − f_β gives C₀ − C_α = f_β(C_β − C_α), hence f_β = (C₀ − C_α)/(C_β − C_α) and f_α = (C_β − C₀)/(C_β − C_α). The percentages can stay as 10, 20 and 50 because their common factor cancels; the answer f is still a fraction, not a percent.
This is the lever rule: a phase fraction uses the tie-line distance on the opposite side of the overall point, divided by the entire tie-line length. Near the α endpoint, only a little B-rich β is needed to raise the overall composition above C_α. At C₀ = 20 and 600 K, f_β = (20 − 10)/(50 − 10) = 10/40 = 0.25, while f_α = 0.75. Using the nearby-to-β distance 50 − 20 for f_β would reverse the answer.
For a 2.00 kg specimen at that state, M_α = 1.50 kg and M_β = 0.500 kg. The B masses are 0.10 × 1.50 = 0.150 kg and 0.50 × 0.500 = 0.250 kg. Their sum is 0.400 kg, equal to 20% of 2.00 kg. This explicit ingredient check can reveal a reversed lever arm even when both proposed phase fractions add to one.
The retained one-dimensional sketch is the equal-fraction special case at 600 K: 30 lies halfway between 10 and 50. Its horizontal axis is composition alone; it is not another temperature plot. The full chart lets us obtain its endpoint data before doing that balance.
Change the equilibrium amounts without changing the ingredients
Consider a closed 100 g alloy with C₀ = 30 mass% B. It retains 30 g of B as its temperature changes: no material enters or leaves. Its path on the chart is vertical at C₀ = 30. Suppose each listed state is allowed to reach the supplied equilibrium, with significant surface, interface, imposed-stress and other extra contributions excluded from this bulk diagram.
At 600 K, C_α = 10 and C_β = 50, so f_β = 20/40 = 0.50. At 700 K the endpoints move to 20 and 60, so f_β = (30 − 20)/40 = 0.25. At 750 K they are 25 and 65, giving f_β = 5/40 = 0.125. The amount of β decreases even though its own B concentration rises. There is no contradiction: composition within a phase and amount of that phase are different quantities.
Check the ingredient inventory at each temperature. At 600 K, B in α/β is 5 + 25 = 30 g. At 700 K it is 15 + 15 = 30 g. At 750 K it is 21.875 + 8.125 = 30 g. Redistribution changes both the phase compositions and masses while conserving total B. A phase fraction is therefore not a quantity that must remain constant merely because the specimen is closed.
The chart gives equilibrium targets under its assumptions. It gives no time needed for redistribution, and no guarantee that a rapidly heated or quenched specimen reaches them. Diffusion, nucleation, interface motion, specimen dimensions and starting state can matter. Equilibrium tendencies and transformation rates develops that separate journey; Thermal treatment and evolving microstructure follows material-specific processing sequences.
| T (K) | C_α / C_β (mass% B) | f_α / f_β | Phase masses α / β (g) | B masses in α / β (g) |
|---|---|---|---|---|
| 600 | 10 / 50 | 0.50 / 0.50 | 50 / 50 | 5 / 25 |
| 700 | 20 / 60 | 0.75 / 0.25 | 75 / 25 | 15 / 15 |
| 750 | 25 / 65 | 0.875 / 0.125 | 87.5 / 12.5 | 21.875 / 8.125 |
Keep the amount basis through the calculation
A mass-percent composition axis led to phase mass fractions because we conserved B mass. For an atomic-percent diagram, the analogous balance counts atoms: use the fraction of the total number of atoms belonging to each phase. It is not generally a mass fraction. Converting between mass and atomic composition also needs the appropriate atomic masses; the labels cannot simply be changed.
Volume is another basis. Return to the 100 g, 600 K state with 50 g in each phase. Independently supply hypothetical uniform phase densities ρ_α = 4.00 g/cm³ and ρ_β = 8.00 g/cm³ at that state. Neglect interface volume. Then V_α = M_α/ρ_α = 12.5 cm³ and V_β = M_β/ρ_β = 6.25 cm³. The β volume fraction is 6.25/(12.5 + 6.25) = 1/3, despite its mass fraction being 1/2. Equal mass occupies less volume in the denser phase.
More generally, with the same mass fractions and densities, v_β = (f_β/ρ_β)/(f_α/ρ_α + f_β/ρ_β). Use common density units and the actual state. The phase diagram did not supply those densities or prove that volume fractions add without an interface correction in every nanoscale system. The density values here are original teaching inputs, not properties of a named alloy.
Use the chart as one part of structural evidence
A diagram can specify an equilibrium phase combination and compositions without specifying grain size, particle shape, spacing or connectivity. Two specimens with the same equilibrium phase amounts can have different structural arrangements and mechanical responses. The strengthening lesson examines those arrangements rather than treating phase fraction as a complete yield-stress formula.
Check the diagram’s identity before transferring a result: components, composition basis, pressure, allowed phases and equilibrium conditions must match. An altered pressure, a third component, imposed stress or a different relevant metastable constraint can require another diagram or calculation. This cropped solid-state chart does not locate a liquidus, a eutectic or the melting point. Those cannot be read by extending its straight lines beyond the supplied window.
In an actual specimen, use suitable phase/composition observations and the processing record to assess the state reached. An average bulk composition alone cannot establish both equilibrium and the phase distribution. A quenched specimen may retain a different state; that does not change mass conservation, but it means these equilibrium endpoint values cannot be assumed without evidence. Diffusion and phase transformations develops mechanisms by which material can approach a different phase state.
Worked examples
Work through the case
An invented binary alloy contains 30 mass% B. Its two phases contain 10% and 50% B. Find their mass fractions.
- Let f be the fraction of the 50% phase.
- 30 = 50f + 10(1 − f).
- 20 = 40f, so f = 0.5.
Each phase contributes half the mass.
Use the opposite arm, then check the ingredient mass
At 600 K a 2.00 kg specimen has C₀ =20 mass% B. Read the phase compositions from the chart, find their masses and verify the total B inventory.
- A horizontal line at 600 K intersects α at 10 and β at 50 mass% B. The overall 20 lies inside that two-phase interval.
- The β fraction is(20 −10)/(50 −10)=0.25; α is 0.75. These are fractions of total mass.
- Multiply by 2.00 kg: α 1.50 kg, β 0.500 kg.
- B in α is 0.10×1.50=0.150 kg; B in β is 0.50×0.500=0.250 kg. Sum 0.400 kg agrees with 0.20×2.00. Reversing the fractions would give 0.800 kg of B, inconsistent with the specified alloy.
α at 10% B contributes 1.50 kg and β at 50% B contributes 0.500 kg. Their B inventory is 0.400 kg.
Separate changing phase amount from changing phase composition
The closed 100 g,30 mass% B alloy equilibrates first at 600 K and then at 700 K. Did B leave because β fell from 50 g to 25 g?
- At 700 K, read C_α =20 and C_β =60, rather than keeping the 600 K endpoints.
- f_β =(30 −20)/(60 −20)=0.25; α has 0.75. Phase masses are 25 and 75 g.
- B is 0.20×75=15 g in α and 0.60×25=15 g in β. The total remains 30 g.
- At 600 K, the same total was split 5 g in α and 25 g in β. Phase composition and phase amount both changed; neither alone counts total B. No time for this redistribution has been supplied.
No. β loses mass in the equilibrium balance, but the complete specimen still contains 30 g of B.
Convert an amount only after obtaining the needed density
For the 600 K equal-mass state, use 100 g total and the supplied phase densities 4.00 and 8.00 g/cm³. Is β half the material volume?
- The lever rule gives 50 g in each phase on the diagram’s mass basis.
- V_α =50/4.00=12.5 cm³; V_β =50/8.00=6.25 cm³.
- Neglecting interface volume as stipulated, total volume is 18.75 cm³ and v_β =6.25/18.75=1/3.
- The mass balance remains unchanged. Replacing f_β with this volume fraction in the mass-composition equation would mix bases.
β is half the mass and one third of the volume under these density assumptions.
Try the reasoning
Are phase compositions the same as phase fractions?
Show the answer and reasoning
No.
One describes ingredients within a phase; the other describes its amount. At 700 K and C₀ =30, β contains 60 mass% B but contributes 25% of specimen mass.
Does rapid cooling always reach the diagram’s equilibrium state?
Show the answer and reasoning
No.
Kinetics may prevent equilibration. The chart gives endpoint targets; a rate/path assessment and actual phase observations are needed for a rapidly cooled specimen.
At 650 K a 400 g alloy has C₀ =45 mass% B. Use the supplied boundary lines to find both phase masses and their B inventory.
Show the answer and reasoning
C_α =15 and C_β =55 mass% B. α contributes 100 g, β 300 g; together they contain 180 g of B.
Read the same-temperature endpoints first. f_β =(45 −15)/(55 −15)=30/40=0.75. B is 0.15×100+0.55×300=15+165=180 g, equal to 45% of 400 g. The phase’s 55% composition is not its 75% specimen mass fraction.
At 600 K, a learner inserts C₀ =5 into the 10/50 lever formula and reports f_β =−0.125. What failed, and what does this chart actually predict?
Show the answer and reasoning
The overall point is outside the two-phase field. The equilibrium model predicts single α at 5 mass% B, with no β.
The arithmetic exposes misuse of the field, not a physically negative phase amount. Locate the point before using a two-phase formula. In the single-phase field its composition equals the specimen’s overall composition; do not force it to the 10% boundary value.
A closed 200 g alloy has C₀ =30 and reaches the chart’s 750 K equilibrium. Find α/β masses and show that total B is still 60 g.
Show the answer and reasoning
α 175 g and β 25 g. B masses 43.75 and 16.25 g sum to 60 g.
Endpoints 25/65 give f_β =(30 −25)/40=0.125. Multiply by 200 g, then by the within-phase B fractions:0.25×175=43.75 and 0.65×25=16.25. A changed phase fraction does not imply loss of an ingredient.
At 700 K, C_α =20 and C_β =60. Can changing C₀ from 30 to 40 leave those endpoint compositions unchanged yet change f_β? Calculate both amounts.
Show the answer and reasoning
Yes. The endpoints stay 20/60 in this two-phase field at 700 K. f_β changes from 0.25 to 0.50.
For 30 use(30 −20)/40=0.25; for 40 use(40 −20)/40=0.50. Different overall specimens share the equilibrium phase compositions but have different amounts. This is distinct from heating one closed specimen, where C₀ is fixed.
In the equal-mass 600 K example, a learner uses β volume fraction 1/3 in C₀ =f_α 10+f_β 50 and obtains 23.33 mass% B. Why is that not a corrected alloy composition?
Show the answer and reasoning
The composition equation requires mass fractions on this axis. Volume fractions cannot replace them without the density conversion.
The true masses remain 50 g each, giving 30 mass% B. The denser β phase has smaller volume per unit mass. The 23.33 result describes an inconsistent mixture of volume weights and mass compositions, not material lost from the specimen.
Two specimens have the same phase compositions and phase mass fractions. Does this diagram establish identical particle spacing, grain size and yield stress? What extra evidence is needed?
Show the answer and reasoning
No. Examine the spatial structure, defects, processing history and relevant mechanical test state.
A phase-amount balance counts material; it does not place that material in space or determine dislocation resistance. Suitable microscopy/diffraction/composition evidence and matched property measurements address different parts of the claim. Nor can this cropped fixed-pressure chart supply an unlisted third phase or melting point.
What to carry forward
- Phase diagrams describe equilibrium under stated constraints.
- Use conservation to relate phase amounts and compositions.
- Read the field and same-temperature boundary compositions before using the two-phase balance.
- Overall composition, composition within a phase and phase amount are different quantities.
- A closed alloy can change its equilibrium phase amounts while retaining all of each ingredient.
- Mass, atom-count and volume fractions require consistent bases; volume conversion needs additional state-specific density data.
- An equilibrium phase balance does not determine rates, grain shape, particle spacing or every property.
References
Teaching resources and research supporting this lesson’s explanations and examples.
Massachusetts Institute of Technology2 sources