Structure
Properties
Process
1
CHAPTER 9: PHASE DIAGRAMS
ISSUES TO ADDRESS...
• When we combine two elements in an alloy...
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 phases are we going to have?
What is the composition of each phase?
How much of each phase do we get?
Phase B
Phase A
Nickel atom
Copper atom
2
Importance of Phase Diagrams
• The phase diagram is an important tool for Materials Scientists
and Engineers
• A phase diagram can be used to predict the equilibrium
compositions and proportion of the phases in a material for a
given chemical composition at a particular temperature
• A material’s microstructure can be predicted using information
from the phase diagram and its processing route
• A strong correlation exists between microstructure and
mechanical properties of materials; thus the mechanical
behaviour of materials can be predicted from phase diagrams
3
Definitions and Basic Concepts
• COMPONENTS: The pure metals and/or compounds of which
an alloy is composed (or elements which are mixed to form an
alloy). For example: in brass; the components are copper and
zinc)
• SYSTEM: System has two meanings; it refers to either the
particular body of material being considered, or the possible
alloys of a combination of same alloying components (E.g: the
iron-carbon system)
• PHASES: A phase is a homogeneous part of the system with
uniform physical and chemical characteristics
•
SOLUTE and SOLVENT; SUBSTITUIONAL and INTERSTITIALS
positions
4
Solid solutions
Substitutional
Solid Solution
Interstitial
• Solid solution of B in A (i.e., random dist. of point
defects)
OR
Substitutional alloy
(e.g., Cu in Ni)
Interstitial alloy
(e.g., Steel: C in Fe)
5
THE SOLUBILITY LIMIT
• Solubility Limit of a component in a phase is the maximum
amount of the component that can be dissolved in it.
- e.g. alcohol has unlimited solubility in water, sugar has a limited
solubility, oil is insoluble.
- The same concepts apply to solid phases: Cu and Ni are soluble in
any amount (unlimited solid solubility), while C has a ……….
solubility in Fe.
Question: What is the
solubility limit at 20C?
Answer: 65wt% sugar.
Solubility
Limit
80
60
40
20
0
Pure
Water
If Co < 65wt% sugar: syrup
If Co > 65wt% sugar: syrup + sugar.
L
(liquid solution
i.e., syrup)
L
(liquid)
+
S
(solid
sugar)
20
40
6065 80
100
Co=Composition (wt% sugar)
Pure
Sugar
• Ex: Phase Diagram:
Water-Sugar System
Temperature (°C)
100
6
EFFECT OF Temp. & COMPOSITION (Co)
• Solubility limit increases with T:
e.g., if T = 100 º C, solubility limit = 80wt% sugar.
• 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
Temperature (°C)
100
L
80
60
40
20
0
0
2 phases
(liquid)
L
(liquid solution
i.e., syrup)
+
S
(solid
sugar)
A(70,20)
2 phases
20
40
60 70 80
100
Co=Composition (wt% sugar)
7
Phases
• Phase must have the following characteristics:
– Homogeneous in crystal structure or atomic arrangement, have same
chemical and physical properties throughout, have a definite interface and
able to be mechanically separated from its surroundings.
– Do not confuse phase with grain. A single phase material may contain
many grains, however, a single grain consists of only one phase. A
phase may contain one or ……….. components.
•
A pure material can be considered to be a phase: H2O has three different
phases possible: Ice, Water and Steam
•
A single phases system are called HOMOGENEOUS and multi-phase
system are HETEROGENEOUS systems
8
Microstructure
• Microstructure is observed by OPTICAL and ELECTRON
microscopy techniques using appropriate polishing and etching
• Microstructure of an alloy is characterized by (1) number of phases
present, (2) their proportions and (3) manner in which they are
distributed or arranged
• Microstructure of an alloy depend on the elements present, their
concentration, and the heat treatment (I.e. heated to what
temperature, time kept at that temperature and the rate of cooling)
AluminumCopper
Alloy
 (lighter
phase)
(darker
phase)
9
1.
2.
3.
4.
5.
6.
7.
8.
9.
needle like
Islands (continues or separated)
Spherical
Fine dispersion
Lamellar
Banded
Coarse and blocky
Globular
Networking
10
Equilibrium Phase Diagrams
• Graphical representation of the number and quantities of phases
that can exist in an alloy at any given composition, temperature
and Pressure
• The presence of the different phases under the above conditions is
dictated by PHASE EQUILIBRIA, that is, the system minimizes
its energy to reach an equilibrium state (that is the reason phase
diagrams are termed as Equilibrium Diagrams)
• Stable (equilibrium) configuration of a system, when a sufficient
amount of time has elapsed that ..transformation ... occur.
• In the present discussion we will focus on Binary Phase Diagram
(for an alloy with two components) with temperature and
compositions as variables (Pressure is kept constant at 1atm)
• We will focus on two simple types of phase diagram Binary
Isomorphous and Binary Eutectic
11
12
Binary Isomorphous System
• Simplest phase diagram to understand and interpret,
characterized by Copper Nickel System
•
Important parts of the phase
diagram are:
– Temperature Axis
– Composition Axis
– Liquidus Line
– Solidus Line
Liquidus and Solidus Line
Constitute Phase Boundaries Lines
13
Cu-Ni Phase Diagram
The phase diagram is characterized
by:
• Cu and Ni are completely soluble
in both liquid and solid solutions
(Isomorphous behavior)
•The complete solubility is explained
by the fact that both have same
crystal structure (FCC) and satisfy
other requirements
•The solid solution is called  and is
substitutional solid solution with
FCC crystal structure
•For any composition other than pure
components the melting takes place
over a range of temperatures
T(°C)
• 2 phases:
1600
1500
L (liquid)

L (liquid)
1400
• 3 phase
fields:
L
L+

1300

(FCC solid
solution)
1200
1100
1000
0
20
40
60
80
100
wt% Ni
14
What are the # and types of phases present?
• Rule 1: If we know T and Co, then we know:
--the # and types of phases present.
1600
1500
L (liquid)
B(1250,35)
Method: Locate the
temperature-composition point on
the diagram and note the phase(s)
with which the corresponding
phase field is labeled
T(°C)
1400
1300
• Examples:
 
+
L
1200
A (1100°C, 60 wt% Ni-40wt% Cu):
1100
1 phase: 
B (1250°C, 35 wt% Ni-65wt% Cu):1000
0
2 phases: L + 
s
u
d
i
u
s
liq
u
lid
o
s
(FCC solid
solution)
Cu-Ni
phase
diagram
A(1100,60)
20
40
60
80
100
Adapted from Phase Diagrams of Binary
Nickel Alloys, P. Nash (Ed.), ASM
International, Materials Park, OH, 1991).
wt% Ni
15
Determination of Phase Compositions
Steps to compute the composition includes:
For Single Phase Region:
– The composition of the phase is the same as overall composition of alloy
For Two Phase Region:
– A TIE LINE or isotherm is constructed across the two phase region at
the temperature of the alloy
(A tie line at a given temperature extend across the two-phase region and
terminate (end) at the phase boundary lines on either side)
– The intersections of the tie line and the phase boundaries on either side
are noted
– Perpendiculars are dropped from these intersections to the horizontal
composition axis and the composition of each respective phase is noted
16
What are the composition of phases present?
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
T(°C)
TA
Cu-Ni
system
A
Co = 35wt%Ni-65wt%Cu
1300 L (liquid)
At TA:
Only Liquid (L)
B
CL = Co ( = 35wt% Ni-65wt%Cu) TB

At TD:
+
L
1200
D
Only Solid ()
TD
C = Co ( = 35wt% Ni-65wt%Cu)
20
3032 35
CLCo
At TB:
Both  and L
CL = Cliquidus ( = 31.5 wt% Ni-68.5wt% Cu here)
C = Csolidus ( = 42.5 wt% Ni-57.5wt% Cu here)
tie line dus
i
liqu

+
L
s
so l
idu

(solid)
4043
50
C wt% Ni
17
Determination of the Phase Amounts
The relative amount as fractions or as percentage can be calculated
by following steps:
For Single Phase Region:
– The alloys is composed entirely of one phase that is the phase fraction is
1 and percentage 100%
For Two Phase Region Tie Line is constructed and Lever rule used:
– A tie line is constructed across two-phase region at the given temperature
– The overall composition is located on the tie line
– The fraction of one phase is computed by taking the length of tie line
from the overall composition to the phase boundary for the OTHER
phase, and dividing by the total tie line length (This is called Lever Rule
or Inverse Lever Rule)
– The fraction of other phase is determined in the same manner
– Tie Line segments length can be determined by direct measurement (by a
scale) or by subtracting compositions as taken from the composition axis
18
What is the weight fractions of each phase present?
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
Co = 35wt%Ni-65wt%Cu
At TA: Only Liquid (L)
WL = 100wt%, W = 0
At TD:
At TB:
Only Solid ()
WL = 0, W = 100wt%
Both  and L
S
43  35


 73wt %
WL
R  S 43  32
R
W  
R S
= 27wt%
Cu-Ni
system
T(°C)
A
TA
1300
L (liquid)
TB
1200
TD
20

+
L
tie line dus
i
liqu

L + dus
B
R S
i
sol

(solid)
D
3032 35 4043
Please note:
CLCo
50
C
wt% Ni
C  C0  S
C0  C L  R
C  C L  R  S
19
THE LEVER RULE: A PROOF
• Sum of weight fractions: WL  W  1
• Conservation of mass (Ni): Co  WL CL  W C
• Combine above equations:
C  Co  S
Co  CL  R

WL
W


C  CL R  S
C  CL R  S
• A geometric interpretation:
Co
CL
C
R
S
WL
W
moment equilibrium:
WLR  WS
1 W
solving gives Lever Rule
20
Summary of Reading a Phase
Diagram
Given: Composition and Temperature
•
What Phases are Present?
– Read off phases from the field
at a given composition and
temperature
•
What Are Their Compositions?
– In a two-phase region, use a
horizontal tie-line
•
How Much of Each Phase is
Present (or percent fraction of
each phase)?
– In a two-phase region, use the
Lever Rule and the tie-line
C  C0  S
C0  C L  R
21
Development of Microstructure in
Isomorphous Alloys
EQUILIBRIUM COOLING:
• Cooling occurs very slowly
• Phase equilibrium is continuously maintained
• Consider binary Cu-Ni system with isomorphous behavior, i.e.,
complete solubility of one component in another;  phase field
extends from 0 to 100 wt% Ni
• Consider cooling the alloy of composition 35 wt% Ni-65 wt%
Cu as it is cooled from 1300°C, that is, moving down a vertical
line at the above composition
• See Figure on next page
22
• Point A (1300C): alloy
completely liquid of composition
35 wt% Ni
• Point B (~1260C): The first
solid begins to form with the
composition dictated by tie line
Cu-Ni
system
T(°C) L (liquid)
1300
• The fraction of liquid and 
phase will change as we go from B
to D (fraction of  will increase)
• Point D (~1220C): Last
remaining liquid solidify.
• Final Product: Polycrystalline
solid solution with uniform
composition of 35 wt% Ni
32
35
B
C
46
43
D
24
1200

+
L
A
• With cooling the composition of L: 35wt%Ni
liquid and  phase will follow the : 46wt%Ni
liquidus and solidus lines,
respectively
• The overall composition of
alloy remains the same as it is
cooled
L: 35wt%Ni
L: 32wt%Ni
36

+
L
: 43wt%Ni
E
L: 24wt%Ni
: 36wt%Ni

(solid)
1100
20
30
35
Co
40
50
wt% Ni
23
MECHANICAL PROPERTIES: Cu-Ni System
• Solid Solution Strengthening: An increase in the strength
and hardness of an alloy by addition of other component
400
TS for
pure Ni
300
TS for pure Cu
200
0 20 40 60 80 100
Cu
Ni
Composition, wt%Ni
--Passes through maximum
as a function of Co
--Ductility (%EL,%AR)
Elongation (%EL)
Tensile Strength (MPa)
--Tensile strength (TS)
60
%EL for pure Cu
%EL for
pure Ni
50
40
30
20
0 20 40 60 80 100
Cu
Ni
Composition, wt%Ni
--Show a minimum as a
function of Co
24
BINARY-EUTECTIC SYSTEMS (1)
2 components
Example: Cu-Ag system
• 3 single phase regions
(L, ,)
• Three two phase regions
Means easily melted, has a special
Composition with a minimum melting
Temperature
T(°C)
1200
L (liquid)
1000
 L +  779°C
TE 800
• Limited solubility:
8.0
: mostly Cu
600
: mostly Ag
• TE: No liquid below TE
400
• CE: Min. melting T
200
0
20
composition
L+
71.9 91.2

40
60 CE 80
100
Co, wt% Ag
25
BINARY-EUTECTIC SYSTEMS (2)
•
Important Feature: Limited
Solubility
•
-Phase:
– A solid solution rich in Cu
– Silver is the solute component
– Only a limited concentration of Ag will
dissolve in Cu
•
Liquidus
Eutectic Point
Solidus
-Phase:
– A solid solution rich in Ag
– Copper is the solute component
•
Both  and  have FCC Crystal Structure
•
Solubility limit of -phase correspond
to the boundary line CBA (max.
solubility is at 779°C at point B)
•
Limited solubility is possibly due to
different crystal structure of pure
elements, atom size, valency,
electronegativity…... )
Eutectic Isotherm
Solvus
26
Eutectic Point/Reaction
• As Silver (Ag) is added to Copper (Cu) the liquidus line temperature
decreases (Line AE), that is, the melting point of Cu is lowered by Ag
addition. The same is true for Ag (Line FE)
• The two liquidus lines meet at point E; Point E is called the Eutectic
Point or Invariant Point (Eutectic Temperature TE=779°C and
Eutectic Composition CE=71.9 wt% Ag)
• Eutectic means easy melting
• Eutectic Reaction takes place at point E: L(CE) =  (C E) +  (CE)
• Upon cooling, liquid phase of concentration CE transformed in two
solid phases  and  at the temperature TE
• Thus for Cu-Ag system Eutectic reaction is:
L (71.9% Ag)=  (8% Ag) +  (91.2% Ag) at 779°C
• Eutectic reaction is similar to solidification and melting of pure
components
27
EX 9.2: Pb-Sn EUTECTIC SYSTEM (1)
T(°C)
• -phase: A solid solution of
Tin in Lead
Pb: Solvent Sn: Solute
• -phase: A solid solution of
Lead in Tin
Sn: Solvent Pb: Solute
• Eutectic Invariant Point:
61.9 wt% Sn and 183°C
• The solder materials commonly
used in electronics is 60 wt% Sn
+ 40 wt% Pb , the composition is
chosen because it melts at very
low temperature 185°C (very
close to the eutectic point)
300
L (liquid)
 L + 
200
18.3
150
183°C
61.9
97.8

100
0
L+ 
20
40
Co
60
80
100
Co, wt% Sn
For a 40wt%Sn-60wt%Pb alloy at 150C, find...
--the phases present
--the compositions of the phases
--the relative amount of each phase interms of
mass fraction
28
EX 9.2: Pb-Sn EUTECTIC SYSTEM (2)
• For a 40wt%Sn-60wt%Pb alloy at 150C, find...
--the phases present:  + 
Pb-Sn
T(°C)
--the compositions of
system
the phases:
300
C = 10wt%Sn
L (liquid)
C = 98wt%Sn
--the relative amounts 200  L +  183°C
L+ 
of each phase:
18.3
61.9
97.8
W
W
 58  66wt%
88
 30  34wt%
88
C   C0  S  98  40  58
150
100
R
0 10 20
S

40
Co
60
80
98100
Co, wt% Sn
C0  C  R  40  10  30
C   C  R  S  98  10  88
29
Pb-Sn EUTECTIC SYSTEM
30
Microstructures in Eutectic Systems-I
• Different microstructures are
possible depending on
composition even for slow rate
of cooling
T(°C)
400
L: Cowt%Sn
L

L
300
• Consider Pb-Sn Phase Diagram
• for Composition:
0 < Co < 2wt%Sn
(pure to maximum solid solubility
at room temperature)
• Microstructure development
similar to the binary
isomorphous system
• Result:
200
TE
100

L+
: Cowt%Sn
(Pb-Sn
System)
+
0
10
20
30
Co
Co, wt%
2
(room T solubility limit)
Sn
--polycrystal of  grains
31
Microstructures in Eutectic Systems-II
Pb-Sn
system
• for Composition:
2wt%Sn < Co < 18.3wt%Sn
(solubility limit at RT to maximum solid
solubility at the eutectic temperature)
400
L
• Alpha phase solidifies first and then
precipitates beta phase; as the alpha
solid solubility is exceeded
300
• On further cooling the beta particles
grow in size because fraction of beta
phase increases slightly
200
TE
• Result:
L: Cowt%Sn
T(°C)
L+

L

: C owt%Sn


100
+
-- polycrystal with fine
 crystals
0
10
20
30
Co Co, wt%
2
(sol. limit at Troom)
18.3
(sol. limit at TE)
Sn
32
Microstructures in Eutectic Systems-III
• for Composition Co = CE (The Eutectic
Composition)
• Eutectic Reaction:
L (61.9% Sn)=  (18.3% Sn) +  (97.8% Sn) at 183°C
all composition are different so atomic diffusion should
take place for redistribution of Pb and Sn
• Result: Eutectic microstructure
Alternating layer or Lamellar
--alternating layers of  and  crystals
T(°C)
Pb-Sn
system
configuration
300
200
TE
L

L+
20
18.3
L+ 
183°C
+
100
0
0
L: Cowt%Sn
40
: 97.8wt%Sn
: 18.3wt%Sn
60
CE
61.9
80
100
97.8
Co, wt% Sn
160mm
Micrograph of Pb-Sn
eutectic microstructure
33
Microstructures in Eutectic Systems-IV
• for Composition: 18.3wt%Sn < Co < 97.8wt%Sn
(along the eutectic isotherm except eutectic point)
• The primary alpha phase solidifies first and the last liquid solidifies as
the eutectic structure (alpha in eutectic structure is called eutectic
alpha)
• Result:  crystals and a eutectic microstructure
• Just above TE :
L
L: Cowt%Sn

T(°C)
C = 18.3wt%Sn
L
300
L

CL = 61.9wt%Sn
S
W
=
=50wt%
L+
R+S

R
S
200 
L+
WL = (1-W) =50wt%
TE
R
• Just below TE :
100
0
0
S
+
20
18.3
40
Co
60
61.9
80
primary 
eutectic 
eutectic 
100
97.8
Co, wt% Sn
C = 18.3wt%Sn
C = 97.8wt%Sn
W = S =73wt%
R+S
34
W = 27wt%
HYPOEUTECTIC & HYPEREUTECTIC
T(°C)
L
300
200
TE

L+
L+
 + 
100
Co
Co
0
0
hypoeutectic hypereutectic
20
40
hypoeutectic: Co=50wt%Sn
Metallography and
Microstructures,
American Society
for Metals,
Materials Park,
OH, 1985.)



60
80
eutectic
18.3
(Figs. 9.12 and
9.15 from Metals
Handbook, 9th ed.,
Vol. 9,
61.9
100
Co, wt% Sn
97.8
hypereutectic: (illustration only)
eutectic: Co=61.9wt%Sn

 


175mm
(Pb-Sn
System)


 

160mm
eutectic micro-constituent
35
IRON-CARBON (Fe-C) PHASE DIAGRAM
Four one phase
regions:
, ,  and L
A compound Fe3C
Two phases
regions?
36
Important Concepts to Understand Fe-C Phase
Diagram (also called Fe-Fe3C Phase Diagram)
• Compounds:
– Compounds have distinct chemical formula (fixed composition) as
opposed to a range of compositions observed in single phases
– Compounds appear as straight line on phase diagrams e.g: Fe3C
(Cementite or Iron Carbide)
• Fe-C phase diagram show three distinct invariant reactions:
– Eutectic reaction: LiquidTwo Solid Phases
• L =  + Fe3C at 4.3 wt% Carbon at 1147°C
– Eutectoid reaction: One solid phase Two other solid phases
•  =  + Fe3C at 0.76 wt% Carbon at 727°C
– Peritectic reaction: One solid phase One Liquid +another solid phase
•  =  + L at 1493 °C
• Fe-C phase diagram only extends from 0 to 6.7 wt %Carbon
because all steels and cast irons have carbon content below it
37
Important Features of Fe-C Phase Diagram
•
Fe C phase diagram very important since Steels and Cast Iron, the most
important structural materials, are both alloys of Iron and Carbon
•
Pure Iron goes through the following transformations as it is heated (see the
left vertical axis on phase diagram):
– At room temperature it is -iron or Ferrite (a BCC crystal structure)
– At 912°C -iron  -iron or Austenite (a FCC crystal structure)
a Polymorphic Transformation
– At 1394°C -iron -iron (a BCC crystal structure)
– At 1538 °C melting of Iron takes place -iron Liquid
•
Carbon is interstitial impurity in iron and forms a solid solution with ,  and
-iron
•
Solubility of C is very high in -iron (2.14%) as compared to  (0.022%) and
-iron due to the different crystal structures (BCC)
•
Above the solubility limit of C in  and  iron, Cementite or Fe3C is produced
•
Cementite is very hard and brittle so increases the strength of steels
38
Types of Steels and Cast Iron
• Classification of Ferrous Alloys Based on Carbon Content:
• Pure Iron: Containing less than 0.008 wt% C. Not used
commonly because it is soft
• Steels: Containing 0.008 to 2.14 wt% C. Composition rarely
goes above 1%. Types are:
– Low Carbon Steels: <0.25 wt%C, used in car body panels, beams,
sheet, tubes
– Medium Carbon Steels: 0.25-0.6 wt%C, used in shafts, gears, bolts
– High Carbon Steels: 0.6-1.4wt%C, used in wire, springs, knives, chisels
• Cast Iron: Containing 2.14 to 6.7 wt% C. Composition
normally 4.5 wt% C. Used in making machine body, cylinder
blocks
39
Development of Microstructure in Fe-C alloys
• Consider slow cooling, i.e., equilibrium is continuously maintained
• Consider the eutectoid reaction at eutectoid composition of 0.76 wt% C
Result: Pearlite
alternating layers of  and Fe3C phases
40
Carbon Steels
•
The important phases in carbon steels
Pearlite
Ferrite
Austenite
(BCC)
(FCC)
(Ferrite+Carbide Eutectoid)
Pearlie has mechanical properties
between soft ductile ferrite and
hard brittle cementite
(Grains are also called colonies)
41
Carbon Steels
•
Carbon controls the amount of
pearlite
42
HYPOEUTECTOID STEEL
Consider a composition Co to the left of eutectoid point, i.e., hypoeutectoid (or
less than eutectoid) alloy between 0.022 - 0.76 wt %;
T(°C)
1600

L
 
 
 
 




 
1200
+L

austenite)
1000
800
+Fe3C
r s
727°C
RS
400
0 Co
pearlite
0.77
w =s/(r+s) 600
w =(1-w )



L+Fe3C
1148°C
wpearlite = w
w =S/(R+S)
wFe3C =(1-w )
+Fe3C
1
2
3
4
5
(Fe-C
System)
Fe3C (cementite)
1400
6
6.7
Co, wt% C
100mm Hypoeutectoid
steel
Ferrite present in pearliete is called
eutectoid ferrtie and the one formed
above 727C is called proeutectoid 43
ferrite
HYPEREUTECTOID STEEL
Composition between 0.76-2.14 wt %; Cementite produced above 727C
is called proeutectoid cementite
T(°C)
1600

L
 
 
 
 
Fe3C
 
 
1200
+L

austenite)
L+Fe3C
1148°C
1000
+Fe3C

wFe3C =r/(r+s)600
s
r
800
R
S
+Fe3C
400
0
pearlite
0.77
w =(1-w Fe3C)
wpearlite = w
w =S/(R+S)
wFe3C =(1-w )
1
Co
2
3
4
5
(Fe-C
System)
Fe3C (cementite)
1400
6
6.7
Co, wt% C
60mm Hypereutectoid
steel
44
SUMMARY
• Phase diagrams are useful tools to determine
for a given T and composition of the system:
-- the number and types of phases,
-- the composition of each phase,
-- and the wt% or fraction of each phase present
• 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.
45