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Materials

6% of exam

Phase diagrams and phase change, mechanical and thermal properties, electrical properties, and material selection.

5 concepts

A. Physical (phase diagrams) properties

Phase Diagrams and the Lever Rule

Read a binary phase diagram to find which phases exist, their compositions, and their weight fractions via the lever rule — the core materials skill on FE-Other.

A binary phase diagram is a map: feed it a composition and a temperature and it tells you three things — which phases are present at equilibrium, what each of those phases is made of, and how much of each you have by weight. That is exactly the trio the FE rewards, because a single diagram question can ask for any of the three. The Materials Science chapter of the FE Reference Handbook gives you the diagram and the lever rule but expects you to supply the reading technique, and that technique — drop a vertical at your composition, draw a horizontal tie line at your temperature, and apply the lever — is what this concept drills until it is automatic. Lose points here and you lose them on the most diagram-dense topic in the materials block.

What the regions and lines mean

On a temperature-versus-composition plot, single-phase regions (liquid

L

, or a solid solution like

\alpha

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or

\beta

) are open fields, while two-phase regions (

L+\alpha

,

\alpha+\beta

) sit between them. The boundary above which everything is liquid is the liquidus; the boundary below which everything is solid is the solidus; the line bounding the limit of solid solubility is the solvus. Inside any two-phase region the system is genuinely a mixture — two distinct phases coexist, each with its own fixed composition set by the phase boundaries, not by the overall alloy composition.

The tie line locates phase compositions

To find what each phase is made of, draw a horizontal line — the tie line (isotherm) — across the two-phase region at your temperature. Where it strikes the left boundary gives the composition

x_\alpha

of the

\alpha

phase; where it strikes the right boundary gives

x_\beta

. The overall alloy composition

x

lies somewhere on that same line between the two ends. Crucially, the phase compositions are read at the ends of the tie line and are not equal to

x

— that is the most common conceptual error on these problems.

x_\alpha \;\le\; x \;\le\; x_\beta \quad\text{(read }x_\alpha,\,x_\beta\text{ at the tie-line ends)}

The lever rule gives phase fractions

Once the tie line is drawn, the weight fraction of each phase comes from balancing it like a seesaw pivoted at the overall composition

x

. The fraction of a phase is proportional to the length of the tie-line segment on the opposite side of

x

— the far arm, not the near arm. So the

\beta

fraction uses the

\alpha

-side segment and vice versa, and the two fractions always sum to one.

W_\alpha = \dfrac{x_\beta - x}{x_\beta - x_\alpha}, \qquad W_\beta = \dfrac{x - x_\alpha}{x_\beta - x_\alpha}

Eutectic systems and invariant reactions

Many binary alloys form a eutectic: a single point — the lowest melting composition — where liquid transforms directly into two solid phases on cooling. The eutectic reaction

L \rightarrow \alpha + \beta

happens at a fixed temperature (the eutectic isotherm) and a fixed composition, producing the characteristic fine two-phase microstructure. The handbook lists the four invariant reactions you should recognize on sight: eutectic (

L \rightarrow \alpha+\beta

), eutectoid (one solid

\rightarrow

two solids), peritectic (

L+\alpha \rightarrow \beta

), and peritectoid (two solids

\rightarrow

one solid). They are distinguished by what is on each side of the arrow, not by the temperature.

L \xrightarrow{\;T_E\;} \alpha + \beta \quad\text{(eutectic)}

Cooling curves

A cooling curve plots temperature against time as a melt solidifies. A pure metal shows a flat thermal arrest at its single melting point — heat of fusion leaves at constant temperature. An off-eutectic alloy instead shows two slope changes: the curve bends at the liquidus (solidification begins) and bends again at the solidus (solidification ends), with a sloped two-phase region between. An alloy of exactly eutectic composition behaves like a pure metal — one flat arrest at the eutectic temperature. Reading the number and shape of these breaks tells you whether a composition is pure, eutectic, or in between. (Note: these solidification cooling curves are added context to read the diagram; the only 'cooling curves' the handbook itself prints are Jominy quench cooling-RATE curves for hardenability — a different topic, not solidification thermal arrests.)

The iron-carbon diagram

The iron-iron-carbide diagram is the most important phase diagram in engineering and a frequent FE target. Its key invariant point is the eutectoid at about

0.76\text{–}0.77\%

carbon (this concept uses

0.76\%

throughout) and

727^\circ\text{C}

, where austenite (FCC

\gamma

iron) transforms on cooling into pearlite — alternating lamellae of ferrite (BCC

\alpha

iron, very low carbon) and cementite (

\text{Fe}_3\text{C}

,

6.67\%

C). A hypoeutectoid steel (below

0.76\%

C) forms proeutectoid ferrite first, then the remaining austenite becomes pearlite at the eutectoid line; a hypereutectoid steel forms proeutectoid cementite first. Rapid quenching of austenite bypasses these equilibrium products and traps the carbon as hard, brittle martensite — a non-equilibrium phase the diagram does not show.

\gamma\;(0.76\%\,\text{C}) \xrightarrow{\;727^\circ\text{C}\;} \alpha + \text{Fe}_3\text{C}\ \text{(pearlite)}

Exam strategy

Build a fixed reflex for every diagram problem: (1) mark the overall composition

x

on the horizontal axis, (2) draw the horizontal tie line at the stated temperature, (3) read

x_\alpha

and

x_\beta

where it hits the boundaries, (4) apply the lever rule with the OPPOSITE arm in each numerator. If the point lands in a single-phase field, that phase is

100\%

and there is no lever to compute. Keep all compositions in the same units (wt% B) and remember the two fractions must add to one — if they do not, you swapped an arm. For iron-carbon, memorize the eutectoid coordinates

0.76\%

C and

727^\circ\text{C}

; many questions hinge on them alone.

Key equations

Lever rule — fraction of α

W_\alpha = \dfrac{x_\beta - x}{x_\beta - x_\alpha}

Weight fraction of the α phase. $x$ = overall composition, $x_\alpha,\,x_\beta$ = phase compositions at the tie-line ends (all in same units, e.g. wt% B). Uses the FAR (β-side) arm.

Lever rule — fraction of β

W_\beta = \dfrac{x - x_\alpha}{x_\beta - x_\alpha}

Phase-fraction closure

W_\alpha + W_\beta = 1

Mass-balance check for any two-phase tie line; if your two fractions do not sum to 1 you swapped an arm or misread a boundary.

Liquid/solid fractions

W_L = \dfrac{x_s - x}{x_s - x_L}, \quad W_S = \dfrac{x - x_L}{x_s - x_L}

Eutectic reaction

L \rightarrow \alpha + \beta

Liquid transforms to two solid phases at the fixed eutectic temperature and composition.

Eutectoid reaction

\gamma \rightarrow \alpha + \text{Fe}_3\text{C}

One solid phase transforms to two solid phases; in steel: austenite → pearlite (ferrite + cementite) at 0.76% C, 727°C.

Peritectic reaction

L + \alpha \rightarrow \beta

Liquid plus one solid transform to a second solid on cooling.

Pearlite fraction in hypoeutectoid steel

W_{pearlite} = \dfrac{C_0 - 0.022}{0.76 - 0.022}

Gibbs phase rule (condensed)

F = C - P + 1

Degrees of freedom $F$ for a binary ( $C=2$ ) system at fixed pressure; $P$

True/engineering composition basis

x = \dfrac{\text{mass of B}}{\text{total mass}}\times 100\%

Weight-percent basis used throughout; do not mix wt% with at% on the same tie line.

Worked examples

Lever rule in a two-solid region

Problem. A lead-tin alloy of overall composition $40\%$ Sn is held at a temperature where the $\alpha$ phase boundary is at $18\%$ Sn and the $\beta$ phase boundary is at $97\%$ Sn. Find the weight fraction of each phase.

Solution. Draw the tie line:

x_\alpha = 18\%

,

x_\beta = 97\%

, overall

x = 40\%

W_\alpha = \dfrac{97-40}{97-18} = 0.722, \qquad W_\beta = \dfrac{40-18}{97-18} = 0.278

Liquid and solid fractions during freezing

Problem. A binary alloy of overall composition $40\%$ B is cooled into its $L+\text{solid}$ region. At the current temperature the solidus reads $20\%$ B and the liquidus reads $60\%$ B. What fraction of the alloy has solidified?

Solution. Tie-line ends:

x_s = 20\%

Pearlite and proeutectoid ferrite in a 0.40% C steel

Problem. A plain-carbon steel contains $0.40\%$ carbon. Just below the eutectoid temperature ( $727^\circ\text{C}$ ), find the weight fraction of pearlite and of proeutectoid ferrite. Take the ferrite solubility limit as $0.022\%$ C and the eutectoid composition as $0.76\%$ C.

Solution. Pearlite forms from the austenite that was still present at the eutectoid line, so its lever arm runs from the proeutectoid ferrite composition to the eutectoid:

Common pitfalls

•Reading phase compositions as the overall composition $x$ . The phase compositions are $x_\alpha$ and $x_\beta$ at the tie-line ENDS; $x$ only sets where the lever pivots.
•Using the near arm instead of the far arm. The fraction of a phase is proportional to the tie-line length on the OPPOSITE side of $x$ — $W_\alpha$ uses the β-side segment.
•Confusing eutectic with eutectoid. Eutectic starts from LIQUID ( $L\to\alpha+\beta$ ); eutectoid starts from a SOLID ( $\gamma\to\alpha+\text{Fe}_3\text{C}$ ). Wrong reaction = wrong tie line.
•Forgetting that a eutectic-composition alloy cools like a pure metal — one flat arrest, not two slope breaks — when reading cooling curves.
•Mixing wt% and at% on one diagram, or mixing the steel carbon fractions; keep every composition on the same basis.
•Treating martensite as if it appears on the equilibrium iron-carbon diagram. Martensite is a non-equilibrium quench product and is not on the diagram.
•Computing a lever in a single-phase field. If the point lands in $\alpha$ , $L$ , or $\beta$ alone, that phase is 100% and there is no second phase to balance.

References

NCEES FE Reference Handbook — Materials Science/Structure of Matter (Binary Phase Diagrams, Lever Rule, Iron-Iron Carbide)
Callister & Rethwisch, Materials Science and Engineering: An Introduction — phase diagrams and microstructure development chapters
Van Vlack, Elements of Materials Science and Engineering — classic source for the iron-iron-carbide phase diagram

B. Mechanical properties

Mechanical Properties of Materials

Extract modulus, yield, ultimate strength, ductility, resilience and toughness from the engineering stress-strain curve, and connect them to fatigue, creep and hardness.

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C. Chemical properties

Chemical Properties and Material Selection

Corrosion and oxidation mechanisms, the galvanic series, chemical compatibility, and the strength-weight-cost trade-offs that drive engineering material selection.

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D. Thermal properties

Thermal Properties of Materials

Coefficient of thermal expansion, conductivity, specific heat and diffusivity, plus the thermal stress that constrained expansion drives — and melting and glass-transition temperatures.

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E. Electrical properties

Electrical Properties of Materials

Resistivity and conductivity, the temperature coefficient of resistance, semiconductors and doping, band gaps, and dielectric permittivity — how materials carry, block, and store charge.

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Weight fraction of the β phase, using the α-side arm. Always $W_\alpha + W_\beta = 1$ .

Lever rule in the $L+\text{solid}$ region; $x_L$ = liquidus composition, $x_s$ = solidus composition at temperature $T$ .

At just below 727°C, fraction of pearlite for a steel of carbon content $C_0$ (wt%); ferrite solubility limit ≈ 0.022% C, eutectoid = 0.76% C.

= number of phases.

F=0

at an invariant (e.g. eutectic) point.

. Fraction of

\beta

(uses the

\alpha

-side arm):

W_\beta = \frac{x - x_\alpha}{x_\beta - x_\alpha} = \frac{40 - 18}{97 - 18} = \frac{22}{79} = 0.278.

Fraction of

\alpha

(uses the

\beta

-side arm):

W_\alpha = \frac{x_\beta - x}{x_\beta - x_\alpha} = \frac{97 - 40}{79} = \frac{57}{79} = 0.722.

So

W_\alpha = 0.722

and

W_\beta = 0.278

. Sanity check:

0.722 + 0.278 = 1.000

, and the alloy is closer to the

\alpha

end, so

\alpha

rightly dominates.

(solid),

x_L = 60\%

(liquid), overall

x = 40\%

. Solid fraction uses the liquid-side arm:

W_S = \frac{x_L - x}{x_L - x_s} = \frac{60 - 40}{60 - 20} = \frac{20}{40} = 0.500.

Liquid fraction:

W_L = \frac{x - x_s}{x_L - x_s} = \frac{40 - 20}{40} = 0.500.

Half solid, half liquid:

W_S = 0.500

. Sanity check: the overall composition sits exactly midway between the two boundaries, so a 50/50 split is expected.

W_S = \dfrac{60-40}{60-20} = 0.500

W_{pearlite} = \frac{C_0 - 0.022}{0.76 - 0.022} = \frac{0.40 - 0.022}{0.738} = \frac{0.378}{0.738} = 0.512.

Proeutectoid ferrite is the remainder:

W_{\alpha,pro} = \frac{0.76 - 0.40}{0.738} = \frac{0.36}{0.738} = 0.488.

So

W_{pearlite} = 0.512

and

W_{\alpha} = 0.488

. Sanity check: they sum to

1.000

, and a

0.40\%

C steel sits roughly midway to the eutectoid, giving the near-even split observed.

W_{pearlite} = \dfrac{0.40 - 0.022}{0.76 - 0.022} = 0.512