Introduction to Materials Science and Engineering
Chapter 9: Phase Diagrams
Textbook Chapter 11: Phase Diagrams
Content
1.
2.
3.
4.
5.
6.
Introduction
Solubility Limit
Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy
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?
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,
(a) The number of phases at equilibrium
(b) The composition of each phases
(c) The amount of each phases
at equilibrium.
Content
1.
2.
3.
4.
5.
6.
Introduction
Solubility Limit
Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy
THE SOLUBILITY LIMIT
Water-Sugar System
Question: What is the
solubility limit at 20C?
Answer: 65wt% sugar.
Solubility
Limit
80
60
L
40
(liquid solution
i.e., syrup)
20
0
Pure
Water
• Ex: Phase Diagram:
Temperature (ºC)
Max. concentration for which only
a solution occurs.
100
20
40
L
(liquid)
+
S
(solid
sugar)
6065 80
C o = Composition (wt% sugar)
If Co < 65wt% sugar: sugar  syrup
If Co > 65wt% sugar: syrup + sugar.
• Solubility limit increases with T:
e.g., if T = 100C, solubility limit = 80wt% sugar.
100
Pure
Sugar
• Solubility Limit:
Adapted from Fig. 9.1,
Callister 6e.
COMPONENTS AND PHASES
• Components:
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).
AluminumCopper
Alloy
Adapted from
Fig. 9.0,
Callister 3e.
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.
B(100,70) D(100,90)
1 phase
• watersugar
system
Adapted from
Fig. 9.1,
Callister 6e.
Temperature (ºC)
100
L
80
60
40
20
0
0
2 phases
(liquid)
L
(liquid solution
i.e., syrup)
+
S
(solid
sugar)
A(20,700)
2 phases
20
40
60 70 80
100
Co=Composition (wt% sugar)
Content
1.
2.
3.
4.
5.
6.
Introduction
Solubility Limit
Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy
PHASE DIAGRAMS
• Tell us about phases as function of T, Co, P.
• For this course:
Gibbs Phase Rule:
P+F=C+2
--binary systems: just 2 components.
--independent variables: T and Co (P = 1 atm is always used).
T(C)
• Phase
Diagram
for Cu-Ni
system
1600
1500
2 phases:
L (liquid)
a (FCC solid solution)
L (liquid)
1400
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
80
3 phase fields:
L
L+a
a
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).
100
wt% Ni
PHASE DIAGRAMS:
# and types of phases
• Rule 1: If we know T and Co, then we know:
--the # and types of phases present.
1600
1500
L (liquid)
B(1250,35)
• Examples:
T(ºC)
1400
1300
1200
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).
1100
1000
0
Cu-Ni
phase
diagram
a
(FCC solid
solution)
A(1100,60)
20
40
60
80
100
wt% Ni
PHASE DIAGRAMS:
composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
Co = 35wt%Ni
At TA: ?
Only Liquid (L)
CL = Co ( = 35wt% Ni)
At TD: ?
Only Solid (a)
Ca = Co ( = 35wt% Ni)
At TB: ?
Both a and L
CL = Cliquidus ( = 32wt% Ni here)
Ca = Csolidus ( = 43wt% Ni here)
Cu-Ni
system
T(ºC)
TA
1300
TB
1200
TD
20
A
L (liquid)
tie line
B
D
a
(solid)
3032 35 4043
CLCo
50
Ca wt% Ni
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.)
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:
Co = 38wt%Ni
At TA: Only Liquid (L)
WL = 100wt%, Wa = 0
At TD: Only Solid (a)
WL = 0, Wa = 100wt%
At TB: Both a and L
What would be WL and Wa?
WL 
S
44  38

 50wt %
+
R S 44  32
Wa 
38  32
R

 50wt %
+
R S 44  32
Cu-Ni
system
T(ºC)
TA
1300
TB
A
L (liquid)
R
B
S
D
1200
TD
20
tie line
3032
a
(solid)
50
38 4044
CL Co Ca
wt% Ni
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
T(ºC)
TA
1300
TB
A
L (liquid)
WL 
S
44  35

 75wt %
+
R S 44  32
Wa 
35  32
R

 25wt %
+
R S 44  32
S
D
TD
20
B
R
1200
What would be WL and Wa?
tie line
303235
CLCo
a
(solid)
4044
50
Ca
wt% Ni
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:
T(ºC)
A
TA
B
R S
TB
WL 
S
43  35

 73wt %
+
R S 43  32
Wa 
R
R+S
= 27wt%
tie line
L (liquid)
1300
What would be WL and Wa?
Cu-Ni system
1200
D
TD
20
a
(solid)
3032 35 4043
CLCo
50
Ca
wt% Ni
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.)
THE LEVER RULE: A PROOF
• Sum of weight fractions: WL + Wa  1
• Conservation of mass (Ni): Co  WL CL + WaCa
• Combine above equations:
• A geometric interpretation:
moment equilibrium:
WLR  WaS
1 Wa
solving gives Lever Rule
Content
1.
2.
3.
4.
5.
6.
Introduction
Solubility Limit
Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy
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.
1300
L: 35wt%Ni
a: 46wt%Ni
A
32
35
B
C
46
43
D
24
L: 32wt%Ni
36
1200
a: 43wt%Ni
E
L: 24wt%Ni
a: 36wt%Ni
a
(solid)
• Consider
Co = 35wt%Ni.
1100
20
What would be the microstructure?
30
35
Co
40
50
wt% Ni
EX: COOLING IN A Cu-Ni BINARY
T(ºC) L (liquid)
1300
L: 35wt%Ni
a: 46wt%Ni
L: 35wt%Ni
Cu-Ni
system
A
32
35
B
C
46
43
D
24
L: 32wt%Ni
36
1200
a: 43wt%Ni
E
L: 24wt%Ni
a: 36wt%Ni
a
(solid)
• Consider
Co = 35wt%Ni.
1100
20
30
Adapted from Fig. 9.3,
Callister 6e.
35
Co
40
50
wt% Ni
If the alloy (65%Cu-35%Ni) is cooled from A to D at 10oC/sec,
what would be the details inside the solid?
T(ºC)
TA
1300
TB
1200
TD
20
A
L (liquid)
tie line
B
D
a
(solid)
3032 35 4043
CLCo
50
Ca wt% Ni
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
MECHANICAL PROPERTIES:
Cu-Ni System
• Effect of solid solution strengthening on:
--Ductility (%EL,%AR)
60
400
300
?
TS for
pure Ni
TS for pure Cu
200
0 20 40 60 80 100
Cu
Ni
Composition, wt%Ni
Elongation (%EL)
Tensile Strength (MPa)
--Tensile strength (TS)
%EL for pure Cu
50
40
30
?
%EL for
pure Ni
20
0 20 40 60 80 100
Cu
Ni
Composition, wt%Ni
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
Content
1.
2.
3.
4.
5.
6.
Introduction
Solubility Limit
Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy
BINARY-EUTECTIC SYSTEMS
Ex.: Cu-Ag system
Gibbs Phase Rule:
P+F=C+2
3 single phase regions (L, a, b)
Limited solubility:
a: mostly Cu
b: mostly Ag
T(ºC)
120 0
L (liquid)
1000
TE: no liquid below TE
T E 800
CE: minimum melting Temp.
composition
600
a
L+a
8.0
71.9 91.2
a+b
400
200
0
L+ b b
779ºC
20
40
60 C E 80
100
C o, wt% Ag
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)
300
L (liquid)
a
200
L+a
18.3
150
183ºC
61.9
97.8
a+b
100
0
L+b b
20
40
Co
60
80
100
Co, wt% Sn
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:
Wa = ? wt%
Wb = ? wt%
T(ºC)
300
L (liquid)
a
200
L+a
18.3
150
100
183ºC
61.9
R
0 11 20
L+b b
97.8
S
a+b
40
Co
60
80
99100
Co, wt% Sn
EX: Pb-Sn EUTECTIC SYSTEM (2)
• For a 40wt%Sn-60wt%Pb alloy at 150C, find...
T(ºC)
--the phases present: a + b
--the compositions of
300
the phases:
L (liquid)
Ca = 11wt%Sn
L+a
a
Cb = 99wt%Sn
L+b b
200
183ºC
18.3
--the relative amounts
61.9
97.8
150
of each phase:
R
S
100
0 11 20
a+b
40
Co
60
80
99100
Co, wt% Sn
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-I
• Co < 2wt%Sn
• Result: ?
T(ºC)
400
L: Cowt%Sn
L
300
200
TE
100
a
L+a
(Pb-Sn
System)
a+b
0
10
20
30
Co
Co, wt%
2
(room T solubility limit)
Sn
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-I
• Co < 2wt%Sn
• Result:
--polycrystal of a grains.
T(ºC)
400
L: Cowt%Sn
L
a
L
300
200
TE
100
a
L+a
a: Cowt%Sn
a+b
0
10
20
30
Co
Co, wt%
2
(room T solubility limit)
Sn
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-II
• 2wt%Sn < Co < 18.3wt%Sn
• Result:?
L: Cowt%Sn
T(ºC)
400
L
L
a
300
L+a
a
200
TE
a
b
100
0
a: C owt%Sn
a+b
10
20
Pb-Sn
system
30
Co Co, wt%
2
(sol. limit at Troom)
18.3
(sol. limit at TE)
Sn
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-II
• 2wt%Sn < Co < 18.3wt%Sn
• Result:
--a polycrystal with fine
b crystals.
L: Cowt%Sn
T(ºC)
400
L
L
a
300
L+a
a
200
TE
a
b
100
0
a: C owt%Sn
a+b
10
20
Pb-Sn
system
30
Co Co, wt%
2
(sol. limit at Troom)
18.3
(sol. limit at TE)
Sn
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-III
• Co = CE
• Result: ?
T(ºC)
300
L
Pb-Sn
system
200
TE
a
L+a
20
18.3
L+b b
183ºC
a+b
100
0
0
L: Cowt%Sn
40
b: 97.8wt%Sn
a: 18.3wt%Sn
60
CE
61.9
80
100
97.8
Co, wt% Sn
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-III
• Co = C E
• Result: Eutectic microstructure
--alternating layers of a and b crystals.
T(ºC)
300
L
Pb-Sn
system
200
TE
a
L+a
20
18.3
L+b b
183ºC
a+b
100
0
0
L: Cowt%Sn
40
b: 97.8wt%Sn
a: 18.3wt%Sn
60
CE
61.9
80
100
97.8
Co, wt% Sn
Formation of Eutectic Lamellar Structure
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-IV
• 18.3wt%Sn < Co < 61.9wt%Sn
• Result: a crystals and an eutectic microstructure
Just above TE:
L: Cowt%Sn
T(ºC)
L
a
L
300
L+a
200
TE
R
R
a
100
0
0
L+b
S
b
S
Just below TE:
a+b
20
18.3
40
Co
60
61.9
Ca = 18.3wt%Sn
CL = 61.9wt%Sn
Wa = S =50wt%
R+S
WL = (1-Wa) =50wt%
80
100
97.8
Co, wt% Sn
Ca = 18.3wt%Sn
Cb = 97.8wt%Sn
Wa = S =73wt%
R+S
Wb = 27wt%
MICROSTRUCTURES
IN EUTECTIC SYSTEMS-IV
• 18.3wt%Sn < Co < 61.9wt%Sn
• Result: a crystals and an eutectic microstructure
a L
L: Cowt%Sn
T(ºC)
L
a
L
300
L+a
200
TE
R
R
a
100
0
0
L+b
S
S
40
Co
60
61.9
Ca = 18.3wt%Sn
CL = 61.9wt%Sn
Wa = S =50wt%
R+S
WL = (1-Wa) =50wt%
Just below TE :
a+b
20
18.3
b
Just above TE :
80
primary a
eutectic a
eutectic b
100
97.8
Co, wt% Sn
Ca = 18.3wt%Sn
Cb = 97.8wt%Sn
Wa = S =73wt%
R+S
Wb = 27wt%
HYPOEUTECTIC & HYPEREUTECTIC
T(ºC)
L
300
200
TE
a
L+a
L+b
a+b
100
Co
Co
0
0
hypoeutectic hypereutectic
20
40
o =50wt%Sn
60
61.9
eutectic: C
o =61.9wt%Sn
a
a
a
a
100
97.8
Co, wt% Sn
hypereutectic: (illustration only)
b
b
a
a
80
eutectic
18.3
hypoeutectic: C
(Pb-Sn
System)
b
b
b b
160 m m
175 m m
eutectic micro-constituent
b
OTHER EXAMPLES
OTHER EXAMPLES
Content
1.
2.
3.
4.
5.
6.
Introduction
Solubility Limit
Phase Diagrams
Microstructure Evolution during Cooling
Eutectic Systems
Fe-C Alloy
IRON-CARBON (Fe-C) PHASE DIAGRAM
T(ºC)
-Eutectic (A):
L + g + Fe3C
1600
d
1200
-Eutectoid (B):
g + a + Fe 3C
L
1400
g +L
g
A
1148ºC
( austenite)
L+Fe 3 C
S
R
1000
Fe 3 C (cementite)
Two important points
g +Fe 3 C
800
B
a
727ºC = T eutectoid
R
S
a +Fe 3 C
600
400
0
(Fe)
1
0.77
2
3
4
4.30
5
6
6.7
C o , wt% C
IRON-CARBON (Fe-C) PHASE DIAGRAM
T(ºC)
-Eutectic (A):
L + g + Fe3C
-Eutectoid (B):
1600
d
L
1400
1200
g + a + Fe 3C
g +L
g
1148ºC
( austenite)
800
B
a
g +Fe 3 C
727ºC = T eutectoid
R
S
a +Fe 3 C
600
120 m m
Result: Pearlite =
alternating layers of
a and Fe3C phases.
400
0
(Fe)
L+Fe 3 C
S
R
g g
g g
1000
A
Fe 3 C (cementite)
Two important points
1
0.77
2
3
4
4.30
5
6
6.7
C o , wt% C
Fe 3 C (cementite-hard)
a (ferrite-soft)
Formation of Pearlite Lamellar Structure
HYPOEUTECTOID STEEL
T(ºC)
1600
d
L
1400
( austenite)
1000
800
g +Fe 3 C
r s
727ºC
aR
S
6 00
a +Fe 3 C
Co
0.77
4 00
0
L+Fe 3 C
1148ºC
Fe 3 C (cementite)
1200
g +L
g
1
2
3
4
5
6
6.7
C o , wt% C
HYPOEUTECTOID STEEL
T(ºC)
1600
d
L
g g
g g
g g
g g
1200
g +L
g
( austenite)
1000
a
g
g
a
g ag
800
g +Fe 3 C
r s
727ºC
aRS
a
a
a
pearlite
Co
0.77
w a = s/(r+ s) 600
w g = (1- w a )
400
0
w pearlite = w g
w a = S/( R+ S)
w Fe3C = (1- w a )
L+Fe 3 C
1148ºC
Fe 3 C (cementite)
1400
a +Fe 3 C
1
2
3
4
5
6
6.7
C o , wt% C
100 m m
HYPEREUTECTOID STEEL
T(ºC)
1600
d
L
g+L
g
1200
(austenite)
L+Fe3C
1148ºC
1000
g+Fe3C
600
400
0
R
0.77
a
s
r
800
S
1
Co
a+Fe3C
2
3
4
5
(Fe-C
System)
Fe3C (cementite)
1400
6
6.7
Co, wt% C
Adapted from Figs.
9.21 and 9.29,Callister
6e. (Fig. 9.21 adapted
from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B.
Massalski (Ed.-inChief), ASM
International, Materials
Park, OH, 1990.)
HYPEREUTECTOID STEEL
T(ºC)
1600
d
L
g+L
g
1200
(austenite)
L+Fe3C
1148ºC
1000
g+Fe3C
a
wFe3C =r/(r+s)600
wg =(1-w Fe3C)
400
0
pearlite
R
wpearlite = wg
wa =S/(R+S)
wFe3C =(1-w a )
s
r
800
0.77
g g
g g
g g
g g
Fe 3C
g g
g g
S
1
Co
a+Fe3C
2
3
4
5
(Fe-C
System)
Fe3C (cementite)
1400
6
6.7
Co, wt% C
60mm Hypereutectoid
steel
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.