Binary Phase Diagrams

Materials Science · FE Reference Handbook section

Materials Science

0 formulas

10 exam-style examples

~45 min

All Materials Science lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Binary Phase Diagrams within Materials Science. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what binary phase diagrams describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: percent versus fraction in composition and strain.

Lecture

Why this section exists. Binary Phase Diagrams is the part of Materials Science that lets you connect a steel, concrete or polymer specimen under test to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a definition, a phase-diagram read, or a one-line property calculation. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. percent versus fraction in composition and strain. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: binary phase diagrams.
Wikimedia Commons, public domain

Materials Science — Binary Phase Diagrams: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a steel, concrete or polymer specimen under test. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Allows determination of (1) what phases are present at equilibrium at any temperature and average composition,
(2) the compositions of those phases, and (3) the fractions of those phases.
Eutectic reaction (liquid → two solid phases)
Eutectoid reaction (solid → two solid phases)
Peritectic reaction (liquid + solid → solid)
Peritectoid reaction (two solid phases → solid)

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Lever rule phase fractions — Binary Phase Diagrams

An alloy of overall composition 54 wt% B lies in a two-phase field bounded by 11 wt% and 90 wt% B. What fraction of each phase is present?

Given

C₀ = 54 wt%
C_L = 11 wt%
C_S = 90 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (90 - 54) / (90 - 11) = 0.456$
Solid fraction
$W_{S} = 1 - 0.456 = 0.544$
Check
$45.6$

Answer: ≈ 45.6% liquid, 54.4% solid

Why the other options are there

54.4% liquid (arms swapped)
54% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 2

Lever rule phase fractions — Binary Phase Diagrams (2)

An alloy of overall composition 53 wt% B lies in a two-phase field bounded by 13 wt% and 74 wt% B. What fraction of each phase is present?

Given

C₀ = 53 wt%
C_L = 13 wt%
C_S = 74 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (74 - 53) / (74 - 13) = 0.344$
Solid fraction
$W_{S} = 1 - 0.344 = 0.656$
Check
$34.4$

Answer: ≈ 34.4% liquid, 65.6% solid

Why the other options are there

65.6% liquid (arms swapped)
53% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 3

Lever rule phase fractions — Binary Phase Diagrams (3)

An alloy of overall composition 59 wt% B lies in a two-phase field bounded by 20 wt% and 73 wt% B. What fraction of each phase is present?

Given

C₀ = 59 wt%
C_L = 20 wt%
C_S = 73 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (73 - 59) / (73 - 20) = 0.264$
Solid fraction
$W_{S} = 1 - 0.264 = 0.736$
Check
$26.4$

Answer: ≈ 26.4% liquid, 73.6% solid

Why the other options are there

73.6% liquid (arms swapped)
59% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 4

Lever rule phase fractions — Binary Phase Diagrams (4)

An alloy of overall composition 56 wt% B lies in a two-phase field bounded by 10 wt% and 79 wt% B. What fraction of each phase is present?

Given

C₀ = 56 wt%
C_L = 10 wt%
C_S = 79 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (79 - 56) / (79 - 10) = 0.333$
Solid fraction
$W_{S} = 1 - 0.333 = 0.667$
Check
$33.3$

Answer: ≈ 33.3% liquid, 66.7% solid

Why the other options are there

66.7% liquid (arms swapped)
56% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 5

Lever rule phase fractions — Binary Phase Diagrams (5)

An alloy of overall composition 48 wt% B lies in a two-phase field bounded by 16 wt% and 78 wt% B. What fraction of each phase is present?

Given

C₀ = 48 wt%
C_L = 16 wt%
C_S = 78 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (78 - 48) / (78 - 16) = 0.484$
Solid fraction
$W_{S} = 1 - 0.484 = 0.516$
Check
$48.4$

Answer: ≈ 48.4% liquid, 51.6% solid

Why the other options are there

51.6% liquid (arms swapped)
48% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 6

Lever rule phase fractions — Binary Phase Diagrams (6)

An alloy of overall composition 45 wt% B lies in a two-phase field bounded by 15 wt% and 82 wt% B. What fraction of each phase is present?

Given

C₀ = 45 wt%
C_L = 15 wt%
C_S = 82 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (82 - 45) / (82 - 15) = 0.552$
Solid fraction
$W_{S} = 1 - 0.552 = 0.448$
Check
$55.2$

Answer: ≈ 55.2% liquid, 44.8% solid

Why the other options are there

44.8% liquid (arms swapped)
45% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 7

Lever rule phase fractions — Binary Phase Diagrams (7)

An alloy of overall composition 39 wt% B lies in a two-phase field bounded by 11 wt% and 71 wt% B. What fraction of each phase is present?

Given

C₀ = 39 wt%
C_L = 11 wt%
C_S = 71 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (71 - 39) / (71 - 11) = 0.533$
Solid fraction
$W_{S} = 1 - 0.533 = 0.467$
Check
$53.3$

Answer: ≈ 53.3% liquid, 46.7% solid

Why the other options are there

46.7% liquid (arms swapped)
39% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 8

Lever rule phase fractions — Binary Phase Diagrams (8)

An alloy of overall composition 47 wt% B lies in a two-phase field bounded by 11 wt% and 88 wt% B. What fraction of each phase is present?

Given

C₀ = 47 wt%
C_L = 11 wt%
C_S = 88 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (88 - 47) / (88 - 11) = 0.532$
Solid fraction
$W_{S} = 1 - 0.532 = 0.468$
Check
$53.2$

Answer: ≈ 53.2% liquid, 46.8% solid

Why the other options are there

46.8% liquid (arms swapped)
47% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 9

Lever rule phase fractions — Binary Phase Diagrams (9)

An alloy of overall composition 46 wt% B lies in a two-phase field bounded by 25 wt% and 73 wt% B. What fraction of each phase is present?

Given

C₀ = 46 wt%
C_L = 25 wt%
C_S = 73 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (73 - 46) / (73 - 25) = 0.563$
Solid fraction
$W_{S} = 1 - 0.563 = 0.438$
Check
$56.3$

Answer: ≈ 56.3% liquid, 43.8% solid

Why the other options are there

43.8% liquid (arms swapped)
46% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 10

Lever rule phase fractions — Binary Phase Diagrams (10)

An alloy of overall composition 40 wt% B lies in a two-phase field bounded by 17 wt% and 86 wt% B. What fraction of each phase is present?

Given

C₀ = 40 wt%
C_L = 17 wt%
C_S = 86 wt%

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_{L} = (C_{S} - C_{0}) / (C_{S} - C_{L})$
Substituting
$W_{L} = (86 - 40) / (86 - 17) = 0.667$
Solid fraction
$W_{S} = 1 - 0.667 = 0.333$
Check
$66.7$

Answer: ≈ 66.7% liquid, 33.3% solid

Why the other options are there

33.3% liquid (arms swapped)
40% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a steel, concrete or polymer specimen under test, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Binary Phase Diagrams contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a definition, a phase-diagram read, or a one-line property calculation.
Unit rule: percent versus fraction in composition and strain.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

percent versus fraction in composition and strain
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.