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Materials Science · Chapter 8 of 10 · Intermediate

Phase Diagrams

A phase diagram is a map: tell it the composition and temperature, and it tells you which phases exist, what they contain, and how much of each. It is the blueprint behind every alloy.

Readiness check

This chapter reads graphs and does ratios. Tick only what you can do closed-notes.

Read a point on a two-axis graph.
Define weight percent of a component.
Compute a ratio of two lengths.
Recall solid solutions from Chapter 4.
Recall the iron BCC-to-FCC change from Chapter 3.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview solid solutions in Chapter 4.

3 or more weak itemsRefresh weight percent and graph reading before continuing.

The core idea

At a given composition and temperature, a phase diagram names the phases present; in a two-phase region, the tie line gives their compositions and the lever rule gives their amounts.

W_L = (C_α − C₀)/(C_α − C_L)W_α = (C₀ − C_L)/(C_α − C_L)eutectoid: 0.76 wt% C, 727 °C

A phase is a region of uniform structure and composition. In a single-phase field the whole alloy is that phase. In a two-phase field, draw a horizontal tie line: its ends give the two phase compositions, and the lever rule (the opposite-arm ratio) gives the fraction of each. Cooling an alloy traces a vertical line down the diagram, and the phases that appear set the final microstructure and properties.

The skill works when: you read tie-line ends for compositions and use the opposite lever arm for each phase fraction.

The skill breaks down when: the lever arm is taken on the same side as the phase, or compositions and fractions are confused.

The concept. A binary isomorphous diagram. Above the liquidus all is liquid; below the solidus all is solid; between them the tie line's ends give the liquid and solid compositions, and the lever rule gives how much of each.

The skills, taught in order

A phase diagram is read, not memorised. Five skills cover phases, reading the map, the lever rule, eutectics, and the all-important iron-carbon system.

8.1 Phases and solubility

A phase is a physically distinct, uniform part of the system. Components dissolve in one another up to a solubility limit (the solvus line); beyond it, a second phase appears. One component can have full solubility (Cu-Ni) or limited solubility (Pb-Sn).

8.2 Reading a phase diagram

Locate the alloy by composition (x) and temperature (y). In a single-phase field, the alloy is that one phase at the overall composition. In a two-phase field, two phases coexist, and you need a tie line to resolve them.

Feature	What it tells you
Single-phase region	one phase; its composition is the overall one
Two-phase region	two phases; use a tie line and the lever rule
Liquidus	temperature above which everything is liquid
Solidus	temperature below which everything is solid

8.3 The lever rule

In a two-phase region, a horizontal tie line meets the two boundaries at the phase compositions C_L and C_α. The fraction of each phase is the length of the opposite lever arm over the total: W_L = (C_α − C₀)/(C_α − C_L). The phase you want sits at the far end from its arm, the classic point of confusion.

8.4 Eutectic systems

When solubility is limited, the liquidus dips to a eutectic point, the lowest-melting composition, where liquid transforms directly to two solids on cooling (L → α + β). Solders (Pb-Sn) exploit the low eutectic melting point. The eutectic microstructure is a fine lamellar mixture of the two solids.

8.5 The iron-carbon diagram

The Fe-Fe₃C diagram is the foundation of steel. Its eutectoid (a solid-to-two-solids reaction) sits at 0.76 wt% C and 727 °C, where austenite transforms to pearlite, alternating ferrite and cementite.

Phase or point	Detail
Ferrite (α)	BCC iron, up to 0.022 wt% C, soft and ductile
Austenite (γ)	FCC iron, up to 2.14 wt% C, stable above 727 °C
Cementite (Fe₃C)	6.70 wt% C, hard and brittle
Eutectoid	0.76 wt% C, 727 °C, gives pearlite

Engineering connection: the iron-carbon diagram and the lever rule are the starting point for all steel processing; Chapter 9 adds the time dimension to control the microstructure.

Worked example 1: the lever rule

A 35 wt% Ni copper-nickel alloy sits at 1250 °C in the liquid-plus-solid region. The tie line gives a liquid composition of 32 wt% Ni and a solid composition of 43 wt% Ni. Find the fraction of liquid and solid.

Figure 1. The tie line at 1250 °C. The overall 35% Ni lies near the liquid end, so most of the alloy is still liquid; the lever rule makes that quantitative.

ProblemFind the liquid and solid fractions for the alloy in Figure 1.
Given / findC₀ = 35, C_L = 32, C_α = 43 (all wt% Ni). Find W_L and W_α.
AssumptionsEquilibrium, tie-line compositions read from the diagram.
ModelApply the lever rule: each phase fraction is the opposite arm over the total tie-line length.
EquationsW_L = (C_α − C₀)/(C_α − C_L) W_α = (C₀ − C_L)/(C_α − C_L)
SolveW_L = (43 − 35)/(43 − 32) = 8/11 = 0.73 (73% liquid). W_α = (35 − 32)/(43 − 32) = 3/11 = 0.27 (27% solid).
CheckThe fractions sum to 1, as they must. The overall composition (35) is closer to the liquid end (32), so the larger fraction is liquid, which the numbers confirm.
ConclusionThe lever rule converts a position on a tie line into phase amounts. Note the inversion: the liquid fraction uses the arm reaching to the solid composition, the far side. This is the single most-used calculation in alloy work.

Result. 73% liquid (32% Ni) and 27% solid (43% Ni) at 1250 °C.

Worked example 2: phases in a 0.4% carbon steel

A plain-carbon steel with 0.40 wt% C is cooled to just below the eutectoid temperature (727 °C). Using ferrite at 0.022 wt% C, cementite at 6.70 wt% C, and eutectoid at 0.76 wt% C, find the fractions of total ferrite and cementite, and of pearlite and proeutectoid ferrite.

Figure 2. The eutectoid region of the iron-carbon diagram. The 0.40% C alloy (hypoeutectoid) forms proeutectoid ferrite plus pearlite just below 727 °C.

ProblemFind the phase and microconstituent fractions for the 0.40% C steel in Figure 2.
Given / findC₀ = 0.40, ferrite C_α = 0.022, cementite = 6.70, eutectoid = 0.76 (wt% C). Find total ferrite and cementite, then pearlite and proeutectoid ferrite.
AssumptionsEquilibrium cooling to just below 727 °C, lever rule on the relevant tie lines.
ModelFor total phases, apply the lever rule between ferrite and cementite. For microconstituents, apply it between ferrite and the eutectoid composition.
EquationsW_α = (6.70 − C₀)/(6.70 − 0.022) W_pearlite = (C₀ − 0.022)/(0.76 − 0.022)
SolveTotal ferrite = (6.70 − 0.40)/(6.70 − 0.022) = 6.30/6.678 = 94.3%; cementite = 5.7%. Microconstituents: pearlite = (0.40 − 0.022)/(0.76 − 0.022) = 0.378/0.738 = 51.2%; proeutectoid ferrite = 48.8%.
CheckEach pair sums to 100%. Most of the steel is soft ferrite, with a little hard cementite, consistent with a mild, formable steel. More carbon would raise the pearlite fraction and the strength.
ConclusionOne diagram and two lever-rule calculations give both the phase amounts and the microstructure. The 51% pearlite is what hardens this steel; the proeutectoid ferrite keeps it ductile, the structure-property link made concrete.

Result. 94.3% ferrite and 5.7% cementite; 51.2% pearlite and 48.8% proeutectoid ferrite.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Lever arm on the wrong side	Phase fractions swapped	"Is the arm opposite the phase I want?"	A phase's fraction uses the arm reaching to the other phase.
Composition versus fraction	Tie-line end reported as an amount	"Is this a wt% or a fraction?"	Tie-line ends give compositions; the lever rule gives fractions.
Tie line in a single-phase field	Two phases invented where one exists	"Am I in a one- or two-phase region?"	Tie lines and the lever rule apply only in two-phase fields.
Eutectic versus eutectoid	Reactions confused	"Does liquid or a solid transform?"	Eutectic: L → two solids; eutectoid: one solid → two solids.

Practice ladder

Level 1 · Direct skill

A tie line has ends at 20 and 60 wt% B, with the overall alloy at 45 wt% B. Find the fraction of the 60% phase.

Show answer

Its fraction uses the opposite arm: (45 − 20)/(60 − 20) = 25/40 = 0.625, so 62.5%. The other phase is 37.5%.

Level 2 · Mixed concept

Classify a 0.40% C steel and a 1.0% C steel as hypo- or hypereutectoid, and name the proeutectoid phase each forms.

Show answer

Below 0.76% C is hypoeutectoid (0.40% C), forming proeutectoid ferrite; above 0.76% C is hypereutectoid (1.0% C), forming proeutectoid cementite. The eutectoid composition is the dividing line.

Level 3 · Independent problem

For a 0.76% C (eutectoid) steel just below 727 °C, find the fraction of pearlite, then the total ferrite and cementite.

Show answer

At the eutectoid composition the structure is 100% pearlite. Total ferrite = (6.70 − 0.76)/(6.70 − 0.022) = 89.0%; cementite = 11.0%. All the ferrite and cementite are locked inside the pearlite lamellae.

Level 4 · Transfer to real engineering

Find a real alloy (solder, a bronze, a steel grade) and use its phase diagram to predict the phases present at room temperature and their approximate fractions.

What good work looks like

The alloy located by composition, the right region identified, tie-line compositions read, and lever-rule fractions reported with a structure-property comment.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used the opposite lever arm for each phase fraction."

"Give me five alloys; I will identify the phases present before computing fractions."

"Do the lever rule for me." Reading the tie line and inverting the arms is the skill.

"What phases are here?" Locating the region on the diagram is the point.

Portfolio task

Take one alloy through its phase diagram on cooling: identify the phases at two temperatures, compute fractions with the lever rule, and link them to expected properties.

Must include: tie-line compositions, lever-rule fractions summing to 100%, and a microstructure-property note.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What do the ends of a tie line give?

The compositions of the two coexisting phases.

2. State the lever rule for the liquid fraction.

W_L = (C_α − C₀)/(C_α − C_L), the opposite arm over the total.

3. Where is the iron-carbon eutectoid?

At 0.76 wt% C and 727 °C, where austenite becomes pearlite.

4. Name the three iron-carbon phases.

Ferrite (α, BCC), austenite (γ, FCC), and cementite (Fe₃C).

5. How do eutectic and eutectoid reactions differ?

Eutectic: liquid to two solids; eutectoid: one solid to two solids.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive both lever-rule examples from a blank page.

+3 daysOne isomorphous and one Fe-C calculation.

+7 daysAdd the time axis with transformations, Chapter 9.

+30 daysUse phase fractions to explain alloy selection, Chapter 10.

Textbook mapping

Item	Mapping
Primary source	Callister and Rethwisch, Materials Science and Engineering: An Introduction, Chapter 9 (Phase Diagrams)
Cross-reference	Askeland, Ch. 10 and 11 · Shackelford, Ch. 9
Core topics	8.1 Phases and solubility · 8.2 Reading diagrams · 8.3 Lever rule · 8.4 Eutectic systems · 8.5 Iron-carbon diagram
Engineering connection	The blueprint for every alloy and the basis of all steel processing.
Read next	Chapter 9: Phase Transformations and Heat Treatment.