1. Introduction to Materials Science and Engineering
Chapter 9: Phase Diagrams
Textbook Chapter 11: Phase Diagrams

2. Content Introduction Solubility Limit Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy

3. Introduction ISSUES TO ADDRESS... • When we combine two elements...
what equilibrium state do we get?
• In particular, if we specify...
--a composition (e.g., wt%Cu – wt%Ni), and
--a temperature (T)
then...
How many phases do we get?
What is the composition of each phase?
How much of each phase do we get?

4. Introduction ISSUES TO ADDRESS...
Phase diagram is a graphical illustration of a thermodynamic equilibrium state of a material system which is defined by the composition, temperature, and pressure, namely, Gibbs free energy of a system, and illustrates,
The number of phases at equilibrium
The composition of each phases
The amount of each phases
at equilibrium.

5. Content Introduction Solubility Limit Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy

6. THE SOLUBILITY LIMIT • Solubility Limit: + L S • Ex: Phase Diagram:
Max. concentration for which only a solution occurs.
Pure
Sugar
Temperature (ºC) 20 4 6 8 10 C o
= Composition (wt% sugar) L
(liquid solution
i.e., syrup)
Solubility
Limit
(liquid) + S
(solid
sugar) 65
Water
• Ex: Phase Diagram:
Water-Sugar System
Question: What is the
solubility limit at 20C?
Answer: 65wt% sugar.
If Co < 65wt% sugar: sugar  syrup
If Co > 65wt% sugar: syrup + sugar.
Adapted from Fig. 9.1,
Callister 6e.
• Solubility limit increases with T:
e.g., if T = 100C, solubility limit = 80wt% sugar.

7. COMPONENTS AND PHASES • Components: • Phases:
The elements or compounds which are mixed initially
(e.g., Al and Cu)
• Phases:
The physically and chemically identical material regions
that result (e.g., a and b).
Aluminum-
Copper
Alloy
Adapted from Fig. 9.0,
Callister 3e.

8. EFFECT OF T & COMPOSITION (Co)
• Changing T can change # of phases: path A to B.
• Changing Co can change # of phases: path B to D. 70 8 10 6 4 2
Temperature (ºC) C o
=Composition (wt% sugar) L (
liquid solution
i.e., syrup) A
(20,70 )
2 phases B
(100,70)
1 phase 20 D
(100,90)
(liquid) + S
(solid
sugar)
• water-
sugar
system
Adapted from Fig. 9.1,
Callister 6e.

9. Content Introduction Solubility Limit Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy

10. PHASE DIAGRAMS • Tell us about phases as function of T, Co, P.
Gibbs Phase Rule:
• Tell us about phases as function of T, Co, P.
• For this course:
--binary systems: just 2 components.
--independent variables: T and Co (P = 1 atm is always used).
P+F=C+2
2 phases: L
(liquid) a
(FCC solid solution)
3 phase fields:
L +
wt% Ni 20 4 6 8 10
100
110
120
130
140
150
160
T(C)
L (liquid)
(FCC solid
solution) +
liquidus
solidus
• Phase
Diagram
for Cu-Ni
system
Adapted from Fig. 9.2(a), Callister 6e.
(Fig. 9.2(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

11. PHASE DIAGRAMS: # and types of phases
• Rule 1: If we know T and Co, then we know:
--the # and types of phases present.
wt% Ni 20 4 6 8 10
100
110
120
130
140
150
160
T(ºC)
L (liquid) a
(FCC solid
solution) L +
liquidus
solidus
A(1100,60) B
(1250,35)
• Examples:
Cu-Ni
phase
diagram
Adapted from Fig. 9.2(a), Callister 6e.
(Fig. 9.2(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991).

12. PHASE DIAGRAMS: composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
Cu-Ni system
• Examples:
wt% Ni 20
120
130
T(ºC)
L (liquid) a
(solid) L +
liquidus
solidus 3 4 5 T A D B
tie line 35 32 C o C o
= 35wt%Ni
At T A
: ?
Only Liquid (L) C L
= C o
( = 35wt% Ni)
At T D
: ?
Only Solid ( a ) C a
= C o
( = 35wt% Ni )
At T B
: ?
Both a
and L
Adapted from Fig. 9.2(b), Callister 6e.
(Fig. 9.2(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.) C L
= C
liquidus
( = 32wt% Ni here) C a
= C
solidus
( = 43wt% Ni here)

13. PHASE DIAGRAMS: weight fractions of phases
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
Cu-Ni
system
• Examples:
T(ºC) C o
= 38wt%Ni A T A
At T
tie line A
: Only Liquid (L)
L (liquid)
130 W L
= 100wt%, W a
= 0 a B + L
At T a D :
Only Solid ( ) T B R S W L
= 0, W a
= 100wt% a + a L
At T :
Both a B
and L
120 D T
(solid) D
What would be WL and W? 20 3 32 38 4 4 4 5 W L = S R + C L C C o a
wt% Ni W a = R + S

14. PHASE DIAGRAMS: weight fractions of phases
Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%). •
Cu-Ni
system
• Examples:
T(ºC) A T A
tie line
L (liquid)
130 a B + L T B R S a + a L
120 D T
(solid) D
What would be WL and W? 20 3 32 35 4 4 4 5 W L = S R + C L C C o a
wt% Ni W a = R + S

15. PHASE DIAGRAMS: weight fractions of phases
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
Cu-Ni system
wt% Ni 20
120
130
T(ºC)
L (liquid) a
(solid) L +
liquidus
solidus 3 4 5 T A D B
tie line 35 32 C o R S
What would be WL and W? W L = S R +
= 27wt% W a = R + S
Adapted from Fig. 9.2(b), Callister 6e.
(Fig. 9.2(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)

16. THE LEVER RULE: A PROOF • Sum of weight fractions:
• Conservation of mass (Ni):
• Combine above equations:
• A geometric interpretation:

17. Content Introduction Solubility Limit Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy

18. EX: COOLING IN A Cu-Ni BINARY
• Phase diagram:
Cu-Ni system.
T(ºC)
L (liquid)
L: 35wt%Ni
Cu-Ni
system
• System is:
--binary
i.e., 2 components:
Cu and Ni.
--isomorphous
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100wt% Ni.
130 a A +
L: 35wt%Ni L a :
46wt%Ni B 35 46 C 32 43 D 24
L: 32wt%Ni 36 a a + :
43wt%Ni
120 L E
L: 24wt%Ni a a :
36wt%Ni
(solid)
• Consider
Co = 35wt%Ni.
What would be the microstructure?
110 20 3 35 4 5 C
wt% Ni o

19. EX: COOLING IN A Cu-Ni BINARY
wt% Ni 20
120
130 3 4 5
110
L (liquid) a
(solid) L +
T(ºC) A D B 35 C o
L: 35wt%Ni :
46wt%Ni E 46 43 32 24 36
43wt%Ni
L: 32wt%Ni
L: 24wt%Ni
36wt%Ni
Cu-Ni
system
• Consider
Co = 35wt%Ni.
Adapted from Fig. 9.3, Callister 6e.

20. If the alloy (65%Cu-35%Ni) is cooled from A to D at 10oC/sec,
what would be the details inside the solid?
wt% Ni 20
120
130
T(ºC)
L (liquid) a
(solid) L +
liquidus
solidus 3 4 5 T A D B
tie line 35 32 C o

21. CORED VS EQUILIBRIUM PHASES
• Ca changes as we solidify.
• Cu-Ni case:
First a to solidify has Ca = 46wt%Ni.
Last a to solidify has Ca = 35wt%Ni.
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure

23. MECHANICAL PROPERTIES: Cu-Ni System
• Effect of solid solution strengthening on:
--Tensile strength (TS)
--Ductility (%EL,%AR)
Elongation (%EL)
Composition, wt%Ni Cu Ni 20 4 6 8 10 3 5
%EL
for
pure Ni
for pure Cu
Tensile Strength (MPa)
200 00
TS for
TS for pure Cu ?

24. MECHANICAL PROPERTIES: Cu-Ni System
• Effect of solid solution strengthening on:
--Tensile strength (TS)
--Ductility (%EL,%AR)
--Peak as a function of Co
--Min. as a function of Co

25. Content Introduction Solubility Limit Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy

26. BINARY-EUTECTIC SYSTEMS
Ex.: Cu-Ag system
Gibbs Phase Rule:
P+F=C+2
3 single phase regions (L, a, b)
Limited solubility:
L (liquid) a L + b C o ,
wt% Ag 20 4 6 8 10 2
120
T(ºC) 00 E T
8.0 71 .9
91.2
779ºC a
: mostly Cu b
: mostly Ag
TE: no liquid below TE
CE: minimum melting Temp.
composition

27. EX: Pb-Sn EUTECTIC SYSTEM (1)
• For a 40wt%Sn-60wt%Pb alloy at 150C, find...
--the phases present:
a + b
--the compositions of
the phases:
T(ºC) 3 00
L (liquid) L + a a L + b b 2
183ºC
18.3
61.9
97.8
150 a + b 1 00 20 4 6 8 10 C o C ,
wt% Sn o

28. EX: Pb-Sn EUTECTIC SYSTEM (2)
• For a 40wt%Sn-60wt%Pb alloy at 150C, find...
--the phases present: a + b
--the compositions of
the phases:
Ca = ?
Cb = ?
--the relative amounts
of each phase:
T(ºC) 3 00
L (liquid) L + a a L + b b 2
183ºC
18.3
61.9
97.8
150 R S 1 00
Wa = ? wt%
Wb = ? wt% a + b 11 20 4 6 8 99 10 C o C ,
wt% Sn o

29. EX: Pb-Sn EUTECTIC SYSTEM (2)
• For a 40wt%Sn-60wt%Pb alloy at 150C, find...
--the phases present: a + b
--the compositions of
the phases:
Ca = 11wt%Sn
Cb = 99wt%Sn
--the relative amounts
of each phase:
T(ºC) 3 00
L (liquid) L + a a L + b b 2
183ºC
18.3
61.9
97.8
150 R S 1 00 a + b 11 20 4 6 8 99 10 C o C ,
wt% Sn o

30. MICROSTRUCTURES IN EUTECTIC SYSTEMS-I
• Co < 2wt%Sn
• Result: ?
T(ºC)
L: C o
wt%Sn 4 00 L 3 00 L + a a 2
(Pb-Sn T E
System) 1 00 a + b 1 2 3 C o C o ,
wt% Sn 2
(room T solubility limit)

31. MICROSTRUCTURES IN EUTECTIC SYSTEMS-I
• Co < 2wt%Sn
• Result:
--polycrystal of a grains. L + a 2
T(ºC) C o ,
wt% Sn 1 3 00
L: C
wt%Sn
: C b 4
(room T solubility limit) T E

32. MICROSTRUCTURES IN EUTECTIC SYSTEMS-II
• 2wt%Sn < Co < 18.3wt%Sn
• Result:?
T(ºC)
L: C o
wt%Sn 4 00 L L 3 00 a L + a a : C o
wt%Sn a 2 T E a b 1 00 a + b
Pb-Sn
system 1 2 3 C C o ,
wt% Sn 2 o
(sol. limit at T
room )
18.3
(sol. limit at T E )

33. MICROSTRUCTURES IN EUTECTIC SYSTEMS-II
• 2wt%Sn < Co < 18.3wt%Sn
• Result:
--a polycrystal with fine
b crystals. a : C o
wt%Sn L + 2
T(ºC) ,
wt% Sn 1
18.3 3 00
L: C b 4 T E )
Pb-Sn
system
(sol. limit at T
room
(sol. limit at T E )

34. MICROSTRUCTURES IN EUTECTIC SYSTEMS-III
• Co = CE
• Result: ? L + a 2
T(ºC) C o ,
wt% Sn 4 3 00 1 6
L: C
wt%Sn b T E
: 18.3wt%Sn 8
100
18.3
97.8
61.9
183ºC
: 97.8wt%Sn
Pb-Sn
system

35. MICROSTRUCTURES IN EUTECTIC SYSTEMS-III
• Co = CE
• Result: Eutectic microstructure
--alternating layers of a and b crystals.
T(ºC)
L: C o
wt%Sn 3 00 L
Pb-Sn
system L + a L + b 2 a b
183ºC T E 1 00 a + b b
: 97.8wt%Sn a
: 18.3wt%Sn 2 4 6 8
100
18.3 C
97.8 E C o ,
wt% Sn
61.9

36. Formation of Eutectic Lamellar Structure

37. MICROSTRUCTURES IN EUTECTIC SYSTEMS-IV
• 18.3wt%Sn < Co < 61.9wt%Sn
• Result: a crystals and an eutectic microstructure
Just above TE: L + a 2
T(ºC) C o ,
wt% Sn 4 3 00 1 6
L: C
wt%Sn b T E 8
100
18.3
61.9
97.8 S R C a
= 18.3wt%Sn C L
= 61.9wt%Sn S W a =
=50wt% R + S W L
= (1- W a )
=50wt%
Just below TE: C a
= 18.3wt%Sn C b
= 97.8wt%Sn S W a =
=73wt% R + S W b
= 27wt%

38. MICROSTRUCTURES IN EUTECTIC SYSTEMS-IV
• 18.3wt%Sn < Co < 61.9wt%Sn
• Result: a crystals and an eutectic microstructure
Just above TE : W L
= (1- a )
=50wt% C
= 18.3wt%Sn
= 61.9wt%Sn S R + =
L: C
wt%Sn L a o
T(ºC) L 3 00 L a L + a b 2 R S L + b a T E R S
Just below TE : C a
= 18.3wt%Sn b
= 97.8wt%Sn S R + W =
=73wt%
= 27wt% 1 00 a + b
primary a
eutectic a
eutectic b 2 4 6 8
100
18.3 C o
61.9
97.8 C o ,
wt% Sn

39. HYPOEUTECTIC & HYPEREUTECTIC
T(ºC) 3 00 L L + a 2 a L + b
(Pb-Sn T E b
System) a + b 1 00 C o C o
hypoeutectic
hypereutectic C 2 4 6 8
100 o ,
wt% Sn
eutectic
97.8
18.3
175 m a
hypoeutectic: C o
=50wt%Sn
61.9
hypereutectic: (illustration only)
160 m o
=61.9wt%Sn
eutectic: C b b b b b b
eutectic micro-constituent

40. OTHER EXAMPLES

41. OTHER EXAMPLES

42. Content Introduction Solubility Limit Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy

43. IRON-CARBON (Fe-C) PHASE DIAGRAM3
C (cementite)
1600
1400
1200
1000
800 6 00 4 1 2 5
6.7 L g (
austenite) +L
+Fe C a +
L+Fe d
(Fe) o ,
wt% C
0.77
4.30
727ºC = T
eutectoid
1148ºC
T(ºC) A B S R
Two important points
-Eutectic (A):
-Eutectoid (B):
L + g + Fe 3 C
g + a

44. IRON-CARBON (Fe-C) PHASE DIAGRAM3
C (cementite)
1600
1400
1200
1000
800 6 00 4 1 2 5
6.7 L g (
austenite) +L
+Fe C a +
L+Fe d
(Fe) o ,
wt% C
0.77
4.30
727ºC = T
eutectoid
1148ºC
T(ºC) A B S R
C (cementite-hard)
ferrite-soft)
Two important points
-Eutectic (A):
-Eutectoid (B):
L + g + Fe 3 C
g + a
Result: Pearlite =
alternating layers of a
and Fe3C
phases.
120 m

45. Formation of Pearlite Lamellar Structure

46. HYPOEUTECTOID STEEL Fe 3 C (cementite) L g ( austenite) +L +Fe a L+Fe
1600
1400
1200
1000
800 6 00 4 1 2 5
6.7 L g (
austenite) +L
+Fe a
L+Fe d ,
wt% C
0.77
727ºC
1148ºC
T(ºC) R S r s

47. HYPOEUTECTOID STEEL Fe 3 C (cementite) L g ( austenite) +L +Fe a L+Fe
1600
1400
1200
1000
800 6 00 4 1 2 5
6.7 L g (
austenite) +L
+Fe a
L+Fe d ,
wt% C
0.77
727ºC
1148ºC
T(ºC) R S r s w = /( + )
(1-
Fe3C
pearlite
100 m

48. HYPEREUTECTOID STEEL a T(ºC) d L (Fe-C g g +L System) L+Fe C ( 3
1600 d L
1400
(Fe-C g g +L
System)
1200
L+Fe C (
1148ºC 3
austenite)
1000 g
+Fe C
Adapted from Figs and 9.29,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) 3
C (cementite)
800 r s a R S 6 00 3 a
+Fe 3 C Fe 4 00 1
0.77 C 2 3 4 5 6
6.7 o C o ,
wt% C

49. HYPEREUTECTOID STEEL (Fe-C System) Fe 3 C (cementite) L g ( austenite)
1600
1400
1200
1000
800 6 00 4 1 2 5
6.7 L g (
austenite) +L
+Fe a
L+Fe d ,
wt% C
0.77
1148ºC
T(ºC) R S s w
Fe3C = r /( + )
=(1-
(1-
pearlite m
Hypereutectoid
steel

50. SUMMARY • Phase diagrams are useful tools to determine:
--the number and types of phases,
--the wt% of each phase,
--and the composition of each phase
for a given T and composition of the system.
• Alloying to produce a solid solution usually
--increases the tensile strength (TS)
--decreases the ductility.
• Binary eutectics and binary eutectoids allow for
a range of microstructures.