1. Chapter 9: Phase Diagrams
ISSUES TO ADDRESS...
• When we combine two elements...
what is the resulting equilibrium state?
• In particular, if we specify...
-- the composition (e.g., wt% Cu - wt% Ni), and
-- the temperature (T )
then...
How many phases form?
What is the composition of each phase?
What is the amount of each phase?
What is the structure of each phase
Phase Behavior
Laughlin p. 67
Phase A
Phase B
Nickel atom
Copper atom

2. Components and Phases • Components: • Phases: Optical Mettalography α
The elements or compounds which are present in the alloy
(e.g., Al and Cu)
• Phases:
The physically and chemically distinct material regions
that form (e.g., α and β).
Optical Mettalography α
(lighter
phase) β
(darker
1µm
phase)
Aluminum-Copper Alloy

3. Solubility Limit
• Solution – solid, liquid, or gas solutions, single phase
• Mixture – more than one phase
• Solubility Limit:
Maximum concentration for
which only a single phase solution exists.
Sugar/Water Phase Diagram
Temperature (ºC) 20 40 60 80
100
Composition (wt% sugar) L
(syrup)
Solubility
Limit
(liquid) + S
(solid
sugar) 4 6 8 10
Question: What is the
solubility limit for sugar in
water at 20ºC?
Answer: 65 wt% sugar.
At 20ºC, if C < 65 wt% sugar: syrup
At 20ºC, if C > 65 wt% sugar:
syrup + sugar 65
Problem 9.2: At 170°C, what is the maximum solubility (a) of Pb in Sn? (b) of Sn in Pb?
The lead-tin phase diagram is shown in the Animated Figure 9.8.

4. Isomorphous Binary Phase Diagram
Cu-Ni phase diagram
wt% Ni 20 40 60 80
100
1000
1100
1200
1300
1400
1500
1600
T(ºC)
L (liquid) α
solid solution L +
liquidus
solidus
• System is:
-- binary
2 components: Cu and Ni.
-- isomorphous
i.e., complete solubility of
one component in another

5. Criteria for Solid Solubility: Hume–Rothery rules
Same crystal structure
Similar electronegativities
Similar atomic radii
Crystal
Structure
electroneg
r (nm) Ni
FCC
1.9
0.1246 Cu
1.8
0.1278
Ni and Cu are totally soluble in one another for all proportions.

6. Structure
BCC
FCC

7. Periodic Table

8. Determination of phases present
• If we know T and Co, then we know which phases are present.
Cu-Ni phase diagram
wt% Ni 20 40 60 80
100
1000
1100
1200
1300
1400
1500
1600
T(ºC) L α +
• Examples: B
(1250ºC,35)
A(1100ºC, 60 wt% Ni):
1 phase: α B
(1250ºC, 35 wt% Ni):
2 phases: L + α
A(1100ºC,60)

9. Determination of phase compositions: Lever Rule
Tie line –– also sometimes called an isotherm
wt% Ni 20
1200
1300
T(ºC) L α 3 4 5 + B T
tie line C0 CL Cα S R
What fraction of each phase? Think of the tie line as a lever
(teeter-totter) ML Mα R S

10. Slow Cooling of a Cu-Ni Alloy
C0 = 35 wt% Ni alloy
T(ºC) L
L: 35wt%Ni
α: 46 wt% Ni
L: 35 wt% Ni
130 46 35 43 32
a: 43 wt% Ni
L: 32 wt% Ni 24 36
L: 24 wt% Ni
a: 36 wt% Ni
120 α
110 20 3 35 4 5 C0
wt% Ni

11. Cored vs Equilibrium Structures
Uniform Ca:
35 wt% Ni
T(ºC) L
130 46 35 43 32 24 36
120
First α to solidify:
46 wt% Ni α
Last a to solidify:
< 35 wt% Ni
110 20 3 35 4 5 C0
wt% Ni

12. Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Tensile strength (TS)
-- Ductility (%EL)
Tensile Strength (MPa)
Composition, wt% Ni Cu Ni 20 40 60 80
100
200
300
400
TS for
pure Ni
TS for pure Cu
Elongation (%EL)
Composition, wt% Ni Cu Ni 20 40 60 80
100 30 50
%EL
for
pure Ni
for pure Cu

13. Binary-Eutectic Systems
=low melting
2 components
Cu-Ag system
Ex.: Cu-Ag system
1200
• 3 single phase regions
(L, α, β)
L (liquid)
1000 α L + α L + β
800
779ºC β TE
8.0
71.9
91.2
600 α + β
400
200 20 40 60 CE 80
100
wt% Ag
Eutectic Phase Reaction:
cooling
heating

14. Microstructural Development in Eutectic Systems IL + α
200
T(ºC)
wt% Sn 10 2 20 C0
300
100 30 β
400
(room T solubility limit) TE
Pb-Sn
• For alloys for which
C0 < 2 wt% Sn
• Result: at room temperature
-- polycrystalline with grains of
a phase having
composition C0
L: C0 wt% Sn α L
α: C0 wt% Sn

15. Microstructural Development in Eutectic Systems II
L: C0 wt% Sn
• For alloys for which
2 wt% Sn < C0 < 18.3 wt% Sn
• Result:
at temperatures in α + β range
-- polycrystalline with a grains
and small β-phase particles L + α
200
T(ºC) C ,
wt% Sn 10
18.3 20 C0
300
100 30 β
400
(sol. limit at TE) TE 2
(sol. limit at T
room ) L α
a: C0 wt% Sn α β

16. Microstructural Development in Eutectic Systems III
• For alloy of composition C0 = CE
• Result: Eutectic microstructure (lamellar structure)
-- alternating layers (lamellae) of α and β phases.
200
T(ºC)
wt% Sn 20 60 80
100
300 L α β +
183ºC 40 TE
160 μm
Micrograph of Pb-Sn eutectic microstructure
L: C0 wt% Sn CE
61.9
18.3
α: 18.3 wt%Sn
97.8
β: 97.8 wt% Sn

17. Lamellar Eutectic Structure

18. Microstructural Development in Eutectic Systems IV
• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn
• Result: α phase particles and a eutectic microconstituent WL
= (1- W α )
= 0.50 Cα
= 18.3 wt% Sn CL
= 61.9 wt% Sn S R + Wα =
• Just above TE :
200
T(ºC)
wt% Sn 20 60 80
100
300 L α + 40 TE
L: C0 L α
18.3
61.9 S R
97.8 S R
primary α
eutectic β
• Just below TE : C α
= 18.3 wt% Sn β
= 97.8 wt% Sn S R + W =
= 0.73
= 0.27

19. Hypoeutectic & Hypereutectic
300 L
T(ºC) L + α α β
200 L + β TE
System) α + β
100
175 mm α
hypoeutectic: C0 = 50 wt% Sn
hypereutectic: (illustration only) β
C, wt% Sn 20 40 60 80
100
eutectic
61.9
eutectic: C0 = 61.9 wt% Sn
160 mm
eutectic micro-constituent

20. Intermetallic Compounds
Mg2Pb
Note: intermetallic compound exists as a line - not an area – because stoichiometry (i.e. composition of a compound) is fixed.

21. Eutectic, Eutectoid, & Peritectic
Eutectic - liquid transforms to two solid phases
L S1+S3 (For Pb-Sn, 183ºC, 61.9 wt% Sn)
cool
heat
Eutectoid –all solid phases
S S1+S3
ϒ α+ Fe3C (For Fe-C, 727ºC, 0.76 wt% C)
intermetallic compound - cementite
cool
heat
Peritectic - liquid and one solid phase transform to a second solid phase
S1 + L S2
δ + L ϒ (For Fe-C, 1493ºC, 0.16 wt% C)
cool
heat
Peritectoid – all solid phases
S1 + S S3

22. Eutectoid & Peritectic
Cu-Zn Phase diagram
Peritectic transformation γ + L δ
Eutectoid transformation δ γ + ε

23. Iron-Carbon (Fe-C) Phase Diagram
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L γ
(austenite) +L
+Fe3C α + δ
(Fe)
wt% C
1148ºC
T(ºC)
727ºC = T
eutectoid
- Eutectic (A): L → γ +
Fe3C A
L+Fe3C
- Eutectoid (B): γ → α +
Fe3C γ
Result: Pearlite =
alternating layers of α
and Fe3C phases
120 mm
0.76 B
Fe3C (cementite-hard) α
(ferrite-soft)
4.30

24. Hypereutectoid Steel a T(ºC) δ L +L γ γ (austenite) γ γ α
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L γ
(austenite) +L
+Fe3C α
L+Fe3C δ
(Fe)
C, wt%C
1148ºC
T(ºC) a
proeutectoid Fe3C γ γ
Fe3C γ
pearlite
60 μm
pearlite C0
0.76

25. Hypoeutectoid Steel T(ºC) δ L +L γ (austenite) pearlite
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L γ
(austenite) +L
+ Fe3C α
L+Fe3C δ
(Fe)
C, wt% C
1148ºC
T(ºC)
727ºC C0
0.76 γ
100 μm
pearlite γ α
proeutectoid ferrite α
pearlite
For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following:
The compositions of Fe3C and ferrite (α).
The amount of cementite (in grams) that forms in 100 g of steel.
The amounts of pearlite and proeutectoid ferrite (α) in the 100 g.

26. Solution to Example Problem
The compositions of Fe3C and ferrite (α).
RS tie line just below the eutectoid
Cα = wt% C CFe3C = 6.70 wt% C
pearlite
Wight Fraction of cementite
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L γ
(austenite) +L
+ Fe3C α
L+Fe3C δ
C , wt% C
1148ºC
T(ºC)
727ºC C0 R S
CFe C 3 Cα
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g

27. The amounts of pearlite in the 100 g.
c) Using the VX tie line just above the eutectoid and realizing that
C0 = 0.40 wt% C Cα = wt% C Cpearlite = Cγ = 0.76 wt% C α
pearlite
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L γ
(austenite) +L
+ Fe3C α
L+Fe3C δ
C, wt% C
1148ºC
T(ºC)
727ºC C0 V X Cγ Cα
Amount of pearlite in 100 g
= (100 g)Wpearlite
= (100 g)(0.512) = 51.2 g

28. Alloying with Other Elements
• Teutectoid changes: T
Eutectoid
(ºC)
wt. % of alloying elements Ti Ni Mo Si W Cr Mn
• Ceutectoid changes:
wt. % of alloying elements C
eutectoid
(wt% C) Ni Ti Cr Si Mn W Mo

29. VMSE: Interactive Phase Diagrams
Microstructure, phase compositions, and phase fractions respond interactively
Change alloy composition

30. Summary • Phase diagrams are useful tools to determine:
-- the number and types of phases present,
-- the composition of each phase,
-- and the weight fraction of each phase
given the temperature and composition of the system.
• The microstructure of an alloy depends on
-- its composition, and
-- whether or not cooling rate allows for maintenance of equilibrium.
• Important phase diagram phase transformations include eutectic, eutectoid, and peritectic.

32. Effect of Temperature & Composition
• Altering T can change # of phases:
path A to B.
Altering C can change # of phases:
path B to D.
B (100ºC,C = 70)
1 phase
D (100ºC,C = 90)
2 phases 70 80
100 60 40 20
Temperature (ºC) C
= Composition (wt% sugar) L (
liquid solution
i.e., syrup)
(liquid) + S
(solid
sugar)
A (20ºC,C = 70)
2 phases

33. EX 1: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC
Cα = 11 wt% Sn
Pb-Snsystem L + α β
200
T(ºC)
18.3
wt% Sn 20 60 80
100
300
L (liquid)
183ºC
61.9
97.8
Cβ = 99 wt% Sn
-- the relative amount of each phase
150 R S 11 Cα 40 C0 99 Cβ W α =
Cβ - C0
Cβ - Cα 59 88
= 0.67 S
R+S W β =
C0 - Cα
Cβ - Cα R
R+S 29 88
= 0.33

34. EX 2: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC L + β a b
200
T(ºC)
wt% Sn 20 60 80
100
300 α
183ºC
Ca = 17 wt% Sn
CL = 46 wt% Sn
220 17 Cα R 40 C0 S 46 CL
the relative amount of each phase W a =
CL - C0
CL - Cα 6 29
= 0.21 WL =
C0 - Cα
CL - Cα 23 29
= 0.79