ME 3101 Material Science Fall 2016

1. 1
Chapter 9 - 1
Ch 9: Phase Diagrams
• Read: Ch 9 – general reading sections 1-6, careful study
sections 7-12, and 18-20. Plus all example problems. .
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Chapter 9 -
Ch9:Do we know these terms?
– Gibb’s Phase Rule
– γ-Austenite,
– Cementite
– Pearlite
– Isomorphous
– Lever Rule
– α-Ferrite
– Eutectic
– Eutectoid
– Liquidus, Solidus,
Solvus lines etc.
2
– Terminal solid
Solution
– Tie line
– Martensite
– Hypoeutectic
– Hypereutectic
– Proeutectoid
– Hypoeutectoid
– hypereutectoid,
etc.
Chapter 9 - 3
Phase Equilibria: Solubility Limit
Introduction
– Solutions – solid solutions (alloys), single phase
– Mixtures – more than one phase
• Solubility Limit:
Max concentration for
which only a single phase
solution occurs.
Question: What is the
solubility limit at 20°C?
Answer: 65 wt% sugar.
If Co < 65 wt% sugar: syrup
If Co > 65 wt% sugar: syrup + sugar.
65
Sucrose/Water Phase Diagram
Pure
Sugar
Temperature(°C)
0 20 40 60 80 100
Co =Composition (wt% sugar)
L
(liquid solution
i.e., syrup)
Solubility
Limit L
(liquid)
+
S
(solid
sugar)20
40
60
80
100
Pure
Water

2. 2
Chapter 9 - 4
• Components:
The elements or compounds which are present in the mixture
(e.g., Al and Cu) there are 2.
• Phases:
The physically and chemically distinct material regions
that result (e.g., α and β). So, there are how many phases?
Aluminum-Copper Alloy
( a type of Bronze)
Components and Phases
α (darker
phase)
β (lighter
phase)
Adapted from
chapter-opening
photograph, Chapter
9, Callister 3e.
Chapter 9 - 5
Phase Equilibria
Crystal
Structure
electroneg r (nm)
Ni FCC 1.9 0.1246
Cu FCC 1.8 0.1278
• Both have the same crystal structure (FCC) and have
similar electronegativities and atomic radii (W. Hume –
Rothery rules) suggesting high mutual solubility.
Simple solution system (e.g., Ni-Cu solution)
• Result Ni and Cu are totally miscible in ALL proportions.
= ISOMORPHOUS ALLOY
Chapter 9 - 6
Binary Phase Diagrams: e.g. Cu-Ni.
• Indicate phases as function of T, Co, and P.
• For this course:
-binary systems: just 2 components.
-independent variables: T and Co (P = 1 atm is almost always used).
• Phase
Diagram
for Cu-Ni
system
Adapted from Fig. 9.3(a), Callister 7e.
(Fig. 9.3(a) is adapted from Phase
Diagrams of Binary Nickel Alloys, P. Nash
(Ed.), ASM International, Materials Park,
OH (1991).
• 2 phases:
L (liquid)
α (FCC solid solution)
• 3 phase fields:
L
L + α
α
wt% Ni20 40 60 80 1000
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
α
(FCC solid
solution)

3. 3
Chapter 9 - 7
wt% Ni20 40 60 80 1000
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
α
(FCC solid
solution)
Cu-Ni
phase
diagram
Phase Diagrams:EXPL
# and types of phases
• Rule 1: If we know T and Co, then we know:
--the # and types of phases present.
• Examples:
A(1100°C, 60):
1 phase: α
B(1250°C, 35):
2 phases: L + α
Adapted from Fig. 9.3(a), Callister 7e.
(Fig. 9.3(a) is adapted from Phase
Diagrams of Binary Nickel Alloys, P. Nash
(Ed.), ASM International, Materials Park,
OH, 1991).
B(1250°C,35)
A(1100°C,60)
Chapter 9 - 8
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)
30 40 50
Cu-Ni
system
Phase Diagrams:EXPL
composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
TA
A
35
Co
32
CL
At TA = 1320°C:
Only Liquid (L)
CL = Co ( = 35 wt% Ni)
At TB = 1250°C:
Both α and L
CL = Cliquidus ( = 32 wt% Ni here)
Cα = Csolidus ( = 43 wt% Ni here)
At TD = 1190°C:
Only Solid ( α)
Cα = Co ( = 35 wt% Ni)
Co = 35 wt% Ni
Adapted from Fig. 9.3(b), Callister 7e.
(Fig. 9.3(b) is adapted from Phase Diagrams
of Binary Nickel Alloys, P. Nash (Ed.), ASM
International, Materials Park, OH, 1991.)
B
TB
D
TD
tie line
4
Cα
3
Chapter 9 - 9
• Tie line – connects the phases in equilibrium with
each other - essentially an isothermal line
The Lever Rule
How much of each phase?
Think of it as a lever (teeter-totter)
ML Mα
R S
RMSM L ⋅=⋅α
L
L
LL
L
L
CC
CC
SR
R
W
CC
CC
SR
S
MM
M
W
−
−
=
+
=
−
−
=
+
=
+
=
α
α
α
α
α
00
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)
30 40 50
B
TB
tie line
CoCL Cα
SR
Adapted from Fig. 9.3(b),
Callister 7e.

4. 4
Chapter 9 - 10
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
At TA: Only Liquid (L)
WL = 100 wt%, Wα = 0
At TD: Only Solid ( α)
WL = 0, Wα = 100 wt%
Co = 35 wt% Ni
Adapted from Fig. 9.3(b), Callister 7e.
(Fig. 9.3(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
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)
30 40 50
Cu-Ni
system
TA
A
35
Co
32
CL
B
TB
D
TD
tie line
4
Cα
3
R S
At TB: Both α and L
%73
3243
3543
wt=
−
−
=
= 27 wt%
WL
= S
R +S
Wα
= R
R +S
Chapter 9 - 11
wt% Ni
20
1200
1300
30 40 50
110 0
L (liquid)
α
(solid)
T(°C)
A
35
Co
L: 35wt%Ni
Cu-Ni
system
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
--isomorphous
i.e., complete
solubility of one
component in
another; α phase
field extends from
0 to 100 wt% Ni.
Adapted from Fig. 9.4,
Callister 7e.
• Consider
Co = 35 wt%Ni.
Development of microstructure upon cooling
4635
43
32
α: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
α: 36 wt% Ni
Bα: 46 wt% Ni
L: 35 wt% Ni
C
D
E
24 36
Chapter 9 - 12
• Cα changes as we solidify.
• Cu-Ni case:
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure
First α to solidify has Cα = 46 wt% Ni.
Last α to solidify has Cα = 35 wt% Ni.
Cored vs Equilibrium Phases
First α to solidify:
46 wt% Ni
Uniform Cα:
35 wt% Ni
Last α to solidify:
< 35 wt% Ni
Provided the 35% soluton
is mantained..
How’s it possible
that average is
same as the
outer comp. Of
the cored grain?

5. 5
Chapter 9 - 13
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
Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.
TensileStrength(MPa)
Composition, wt% Ni
Cu Ni
0 20 40 60 80 100
200
300
400
TS for
pure Ni
TS for pure Cu
Elongation(%EL)
Composition, wt% Ni
Cu Ni
0 20 40 60 80 100
20
30
40
50
60
%EL for
pure Ni
%EL for pure Cu
Chapter 9 - 14
Review
Isomorphous Alloy
Lever Rule
Phase Analysis 3 main items
phases present
?
?
Solidus, Liquidus, Vapodus
Tie Line
Microstructure development
Cored structure ?
Chapter 9 - 15
: Min. melting TE
2 components
has a special composition
with a min. melting T.
Adapted from Fig. 9.7,
Callister 7e.
Binary-Eutectic Systems
• Eutectic transition
L(CE) α(CαE) + β(CβE)
• 3 single phase regions
(L, α, β ) total regions=?
• Limited solubility:
α: mostly Cu
β: mostly Ag
• TE : No liquid below TE
• @CE
composition
Ex.: Cu-Ag system
Cu-Ag
system
L (liquid)
α L + α
L+ββ
α + β
Co , wt% Ag
20 40 60 80 1000
200
1200
T(°C)
400
600
800
1000
CE
TE 8.0 71.9 91.2
779°C

6. 6
Chapter 9 - 16
L+α
L+β
α + β
200
T(°C)
18.3
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
α 183°C
61.9 97.8
β
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present: Pb-Sn
system
EX: Pb-Sn Eutectic System (1)
α + β
--compositions of phases:
CO = 40 wt% Sn
--the relative amount
of each phase:
150
40
Co
11
Cα
99
Cβ
SR
Cα = 11 wt% Sn
Cβ = 99 wt% Sn
Wα=
Cβ - CO
Cβ - Cα
=
99 - 40
99 - 11
=
59
88
= 67 wt%
S
R+S
=
Wβ =
CO - Cα
Cβ - Cα
=
R
R+S
=
29
88
= 33 wt%=
40 - 11
99 - 11
Adapted from Fig. 9.8,
Callister 7e.
Chapter 9 - 17
L+β
α + β
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
α β
L+α
183°C
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find...
--the phases present: Pb-Sn
system
Adapted from Fig. 9.8,
Callister 7e.
EX: Pb-Sn Eutectic System (2)
α + L
--compositions of phases:
CO = 40 wt% Sn
--the relative amount
of each phase:
Wα =
CL - CO
CL - Cα
=
46 - 40
46 - 17
=
6
29
= 21 wt%
WL =
CO - Cα
CL - Cα
=
23
29
= 79 wt%
40
Co
46
CL
17
Cα
220
SR
Cα = 17 wt% Sn
CL = 46 wt% Sn
Chapter 9 - 18
• Co < 2 wt% Sn
• Result:
--at extreme ends
--polycrystal of α grains
i.e., only one solid phase.
Adapted from Fig. 9.11,
Callister 7e.
Microstructures
in Eutectic Systems: I
0
L+ α
200
T(°C)
Co, wt% Sn
10
2
20
Co
300
100
L
α
30
α+β
400
(room T solubility limit)
TE
(Pb-Sn
System)
α
L
L: Co wt% Sn
α: Co wt% Sn

7. 7
Chapter 9 - 19
• 2 wt% Sn < Co < 18.3 wt% Sn
• Result:
Initially liquid + α
then α alone
finally two phases
α polycrystal
fine β-phase inclusions
Adapted from Fig. 9.12,
Callister 7e.
Microstructures
in Eutectic Systems: II
Pb-Sn
system
L + α
200
T(°C)
Co , wt% Sn
10
18.3
200
Co
300
100
L
α
30
α+ β
400
(sol. limit at TE)
TE
2
(sol. limit at Troom)
L
α
L: Co wt% Sn
α
β
α: Co wt% Sn
Chapter 9 - 20
• Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of α and β crystals.
Adapted from Fig. 9.13,
Callister 7e.
Microstructures
in Eutectic Systems: III
Adapted from Fig. 9.14, Callister 7e.
160µm
Micrograph of Pb-Sn
eutectic
microstructure
Pb-Sn
system
L+β
α + β
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L
α β
L+α
183°C
40
TE
18.3
α: 18.3 wt%Sn
97.8
β: 97.8 wt% Sn
CE
61.9
L: Co wt% Sn
Chapter 9 - 21
• 18.3 wt% Sn < Co < 61.9 wt% Sn e.g. 40% Sn
• Result: α crystals and a eutectic microstructure
Microstructures
in Eutectic Systems: IV
18.3 61.9
SR
97.8
SR
primary α
eutectic α
eutectic β
WL = (1-Wα) = 50 wt%
Cα = 18.3 wt% Sn
CL = 61.9 wt% Sn
S
R + S
Wα= = 50 wt%
• Just above TE :
• Just below TE :
Cα = 18.3 wt% Sn
Cβ = 97.8 wt% Sn
S
R + S
Wα= = 73 wt%
Wβ = 27 wt%
Adapted from Fig. 9.16,
Callister 7e.
Pb-Sn
system
L+β200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
α β
L+α
40
α+β
TE
L: Co wt% Sn Lα
L
α

8. 8
Chapter 9 - 22
L+α
L+β
α + β
200
Co, wt% Sn20 60 80 1000
300
100
L
α β
TE
40
(Pb-Sn
System)
Hypoeutectic & Hypereutectic
Adapted from Fig. 9.8,
Callister 7e. (Fig. 9.8 adapted from
Binary Phase Diagrams, 2nd ed.,
Vol. 3, T.B. Massalski (Editor-in-
Chief), ASM International, Materials
Park, OH, 1990.)
160 µm
eutectic micro-constituent
Adapted from Fig. 9.14,
Callister 7e.
hypereutectic: (illustration only)
β
β
β
β
β
β
Adapted from Fig. 9.17, Callister 7e.
(Illustration only)
(Figs. 9.14 and 9.17 from
Metals Handbook, 9th ed.,
Vol. 9, Metallography and
Microstructures, American
Society for Metals,
Materials Park, OH, 1985.)
175 µm
α
α
α
α
α
α
hypoeutectic: Co = 50 wt% Sn
Adapted from
Fig. 9.17, Callister 7e.
T(°C)
61.9
eutectic
eutectic: Co =61.9wt% Sn
Chapter 9 - 23
Review
Binary Eutectic System
Hypo and hyper eutectic alloys
Eutectic grain structure
Eutectic system phase analyses
Eutectic system microstructure formation
Cored structure of grains
Phase Analysis
phases present
?
?
Eutectoid (not Eutectic system)
Composition of phases
Wt% of phases
Chapter 9 - 24
Other Types:Eutectoid & Peritectic
Cu-Zn Phase diagram
Adapted from
Fig. 9.21, Callister 7e.
Eutectoid transition δ γ + ε
Peritectic transition γ + L δ
Identify
Eutectoid
Pt.
Now identify
a Peritectic
Pt.
Another ?
P

9. 9
Chapter 9 - 25
Source: www.substec.com
Chapter 9 - 26
Iron-Carbon (Fe-C) Phase Diagram
• 2 important
points
-Eutectoid (B):
γ ⇒ α +Fe3C
-Eutectic (A):
L ⇒ γ +Fe3C
Adapted from Fig. 9.24,Callister 7e.
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ+Fe3C
α+Fe3C
L+Fe3C
δ
(Fe) Co, wt% C
1148°C
T(°C)
α 727°C = Teutectoid
A
SR
4.30
Result: Pearlite =
alternating layers of
α and Fe3C phases
120 µm
(Adapted from Fig. 9.27, Callister 7e.)
γ γ
γγ
R S
0.76
Ceutectoid
B
Fe3C (cementite-hard)
α (ferrite-soft)
Chapter 9 - 27
Hypoeutectoid Steel
Adapted from Figs. 9.24
and 9.29,Callister 7e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ + Fe3C
α+ Fe3C
L+Fe3C
δ
(Fe) Co, wt% C
1148°C
T(°C)
α
727°C
(Fe-C
System)
C0
0.76
Adapted from Fig. 9.30,Callister 7e.
proeutectoid ferritepearlite
100 µm
Hypoeutectoid
steel
R S
α
wα =S/(R+S)
wFe3
C
=(1-wα)
wpearlite = wγ
pearlite
r s
wα =s/(r+s)
wγ =(1- wα)
γ
γ γ
γα
α
α
γγ
γ γ
γ γ
γγ

10. 10
Chapter 9 - 28
Hypereutectoid Steel
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ +Fe3C
α +Fe3C
L+Fe3C
δ
(Fe) Co, wt%C
1148°C
T(°C)
α
Adapted from Figs. 9.24
and 9.32,Callister 7e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
(Fe-C
System)
0.76
Co
Adapted from Fig. 9.33,Callister 7e.
proeutectoid Fe3C
60 µmHypereutectoid
steel
pearlite
R S
wα =S/(R+S)
wFe3C =(1-wα)
wpearlite = wγ
pearlite
sr
wFe3C =r/(r+s)
wγ =(1-w Fe3C )
Fe3C
γγ
γ γ
γγ
γ γ
γγ
γ γ
Chapter 9 - 29
Example: Phase Equilibria
For a AISI1040 ( how much wt% Fe) steel at a temperature just
below the eutectoid temperature(727°C), determine the
following:
a) What are the phases?
b) composition of phases Fe3C and ferrite (α)
c) the amount of carbide (cementite) in grams that forms per 100
g of steel
d) the amount of pearlite and proeutectoid ferrite (α)
Solutions next page
Use a ruler on the handout diagram and
show all lines and %C.
Chapter 9 - 30
Chapter 9 – Phase Equilibria
Solution:
g3.94
g5.7CFe
g7.5100
022.07.6
022.04.0
100x
CFe
CFe
3
CFe3
3
3
=α
=
=
−
−
=
−
−
=
α+ α
α
x
CC
CCo
b) the amount of carbide
(cementite) in grams that
forms per 100 g of steel
a) composition of Fe3C and ferrite (α)
CO = 0.40 wt% C
Cα = 0.022 wt% C
CFe C = 6.70 wt% C
3
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ + Fe3C
α+ Fe3C
L+Fe3C
δ
Co, wt% C
1148°C
T(°C)
727°C
CO
R S
CFe C3Cα

11. 11
Chapter 9 - 31
Chapter 9 – Phase Equilibria
c. the amount of pearlite and proeutectoid ferrite (α)
note: amount of pearlite = amount of γ just above TE
Co = 0.40 wt% C
Cα = 0.022 wt% C
Cpearlite = Cγ = 0.76 wt% C
γ
γ + α
=
Co −Cα
Cγ −Cα
x 100 = 51.2 g
pearlite = 51.2 g
proeutectoid α = 48.8 g
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ + Fe3C
α+ Fe3C
L+Fe3C
δ
Co, wt% C
1148°C
T(°C)
727°C
CO
R S
CγCα
Chapter 9 - 32
Alloying Steel with More Elements
• Teutectoid changes: • Ceutectoid changes:
Adapted from Fig. 9.34,Callister 7e. (Fig. 9.34
from Edgar C. Bain, Functions of the Alloying
Elements in Steel, American Society for Metals,
1939, p. 127.)
Adapted from Fig. 9.35,Callister 7e. (Fig. 9.35
from Edgar C. Bain, Functions of the Alloying
Elements in Steel, American Society for Metals,
1939, p. 127.)
TEutectoid(°C)
wt. % of alloying elements
Ti
Ni
Mo
Si
W
Cr
Mn
wt. % of alloying elements
Ceutectoid(wt%C)
Ni
Ti
Cr
Si
Mn
W
Mo
Chapter 9 - 33
Practice Problem 9.66
• Teutectoid changes:
TEutectoid(°C)
wt. % of alloying elements
Ti
Ni
Mo
Si
W
Cr
Mn
6% Mn
So:
TE=700F
Contd.
Alloy Steel has 6% Mn,
and .25% C
TE=?,
%Eutectoid C=?
ProEutectoid Phase = ?

12. 12
Chapter 9 - 34
Eutectoid %C
• Ceutectoid changes:
wt. % of alloying elements
Ceutectoid(wt%C)
Ni
Ti
Cr
Si
Mn
W
Mo
From figure Eutectoid
Mixture happens at 0.44% C
If mixture has .25% C –
Then it is left of eutectoid
point, i.e. proeutectoid
phase is Ferrite
If %C in mixture is >
Eutectoid Mixture then
proeutectoid phase would
have been Cementite
Chapter 9 - 35
• Phase diagrams are useful tools to determine:
--the number and types of phases in an alloy,
--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.
Summary