Phase Diagrams- Introduction
ES-312
DRAFT
1
Phase Diagrams- Objectives
• Identify the phases in a phase diagram
• In binary phase diagrams, given composition and temperature,
identify the phases present, composition of phases, weight
fraction of phases
• Read an Iron-carbon phase diagram and identify the phases and
their respective quantities present
• Development of microstructures in alloy systems
• Prediction of properties of the final microstructure
• Last but not least, design and process alloys
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2
What is a Solution ?
DRAFT
3
Imperfections in Solids
Conditions for substitutional solid solution: Hume – Rothery
rules
–1.
–2.
–3.
–4.
r (atomic radius) < 15%
Proximity in periodic table (i.e., similar electronegativities)
Same crystal structure for pure metals
Valency equality
All else being equal, a metal will have a greater tendency to dissolve a
metal of higher valency than one of lower valency
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4
Alloys/Compounds/Intermediates
Intermetallics
Ni, Al  FCC
Ni-Al Alloy  NiAl (Nickel
Aluminide Intermetallic)  Disordered at high
temperature – Individual lattice is BCC - Room
temperature shows Simple Cubic Crystal Lattice
[Light, 5 X stronger than SS, Refractory: Has best of both
metals & ceramics – but workability is less due to high
brittleness]
AKA Electron Compounds/Hume Rothery Compounds:
SC
[ORDERED ALLOYS ?????]
DRAFT
Motif: 1Ni + 1Al
5
Definition f Phase/Component/State
Phases of Materials
What is the difference between a phase and state ?
A phase of matter is uniform (homogeneous) with respect to its
physical and chemical properties –
“Primary” phases of matter is also known as States of Matter – Solid,
Liquid, Gas, Plasma & BEC (or degenerate matter)
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6
Components/States/Phases
DRAFT
7
Difference between Purely Physical Terms:
Components/States/Phases
Phase of matter: A region of space in a material where intensive properties are uniform
[Density/Composition/Temperature/Pressure] -
Salt + Water = Salt solution (2C/1P)
Salt + Water + Ice = Salt Solution+ Ice (2C/2P)
Salt + Water + Olive Oil = Salt Solution+ Olive Oil (3C/2P)
Salt + Water + Olive Oil+ Ice = Salt Solution+ Olive Oil + Ice (3C/3P)
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8
Phase Diagram of Pure Substances
The lines themselves represent the combinations of T and
P at which two states are in equilibrium.
Invariant Triple Point
DRAFT
9
Iron
Same Line (Assume Transposed)
DRAFT
10
Components and Phases
• Components:
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., a and b).
b (lighter
phase)
Al-Cu Alloy
a (darker
phase)
Adapted from chapteropening photograph,
Chapter 9, Callister,
Materials Science &
Engineering: An
Introduction, 3e.
DRAFT
11
Titanium
Room Temperature
1 atm
DRAFT
12
Gibbs Phase Rule
[DOF = (# of Components) – (# of Phases) + Environmental Variables]
F=C-P+E
For most processes, E= Temp, Pressure etc.
For equilibrium diagrams, pressure is assumed constant at 1atm
Gibbs’ Phase Rule:
F=C-P+2
F=C-P+1
[Callister, Sec 9.17]
For one component with p and T and variables
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13
Phase rule applied to a Pure Substance
F=C-P+2
Degree of Freedom => Thermodynamic
Variables that can independently be changed
without changing phases present
If any change is made in one of the variables,
another one must change itself
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14
H2O
Triple Point of H2O
Where liquid water, water vapor and ice coexist, there is a
'triple point' where both the boiling point of water and melting
point of ice are equal.
A 'critical point' phase boundary (liquid/vapor) vanishes
L/V phases become indistinguishable.
Below critical temperature (Tc ) a vapor can be liquefied by
increasing pressure only.
liq. H2O is hot enough and gaseous H2O is under sufficient
pressure that their densities are identical (0.322 g cm-3 )
Triple Point: 273.16K/4.58 mm Hg (0.00603659 atm) [−22 °C/2070 atm ]
Critical Point: 373.99 oC/217.75 atm [liq H2O hot enough & H2O Vapor is under sufficient pressure that their densities are identical ]
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15
What is the Difference ???
DRAFT
16
Stillinger-Weber Potential
Liquified noble gases (Ar, Kr, Xe) typify such systems, for which the famous
Lennard-Jones (LJ) interaction is well defined/discussed
No usual pair potential describes general structures involving tetrahedral
coordination/diamond structure (Si) as UL-J stabilizes the close-packed
crystals characteristic of the noble gases.
Molecular Dynamics Simulation- using Stillinger-Weber Potential
Quadruple Point
The crystalline phases are BCC, β-tin, DC and SC
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18
Phase Equilibria:
Solubility
Limit
Sugar/Water
Phase Diagram
Adapted from Fig. 9.1,
Callister & Rethwisch 8e.
• Solubility Limit:
Answer: Approx 65 wt% sugar.
At 20ºC, if C < 65 wt% sugar: syrup
At 20ºC, if C > 65 wt% sugar:
L2
(liquid)
60
40
20
0
L1
(liquid solution
i.e., syrup)
+
s
(solid
sugar)
20
40
6065 80
100
C = Composition (wt% sugar)
Sugar
for sugar in water at 20ºC?
Solubility
Limit
80
Water
Question: What is the solubility limit
Temperature (ºC)
Maximum concentration for
which only a single phase
solution exists.
100
syrup + sugar
DRAFT
19
Moving
the Phase Equilibria
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) D (100ºC,C = 90)
100
Temperature (ºC)
Water-Sugar System
Adapted from Fig. 9.1,
Callister & Rethwisch 8e.
1 phase
2 phases
L
80
(liquid)
60
L
(liquid solution
40
i.e., syrup)
+
S
(solid
sugar)
A (20ºC,C = 70)
20
2 phases
0
0
20
40
60 70 80
100
C = Composition (wt% sugar)
DRAFT
20
Criteria for Solid Solubility
Simple system (e.g., Ni-Cu solution)
Crystal Structure
electroneg
r (nm)
Ni
FCC
1.91
0.1246
Cu
FCC
1.9
0.1278
• Both have the same crystal structure (FCC) and have similar electronegativity values and
atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility.
• Ni and Cu are totally soluble in one another in all proportions.
DRAFT
21
Cooling curves of Alloys
Cu-Ni at different compositions
time
Flat region indicates
latent heat of fusion
(transformation of
phase)
22
Binary Phase Diagrams
• Indicate phases as a function of T, C, and P.
• For this course:
- only binary (2 components) systems:
- independent variables: T and C (P = constant = 1 atm).
T(ºC)
• 2 phases:
1600
Phase Diagram
Cu-Ni system
1500
L (liquid)
1400
1300
a
(FCC solid
solution)
1200
1100
1000
L (liquid)
a (FCC solid solution)
0
20
40
60
DRAFT 80
• 3 different phase fields:
L
L+a
a
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys,
P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
100
wt% Ni
23
Binary Isomorphous System
• Phase diagram:
Cu-Ni system.
• System is:
T(ºC)
1600
1500
L (liquid)
-- 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 100 wt% Ni.
Cu-Ni
phase
diagram
1400
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
80
100
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys,
P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
DRAFT
wt% Ni
24
Phase Determination
• Rule 1: If we know T and Co, then we know:
-- which phase(s) is (are) present [Coordinates in phase diagram].
Point A(1100ºC, 60 wt% Ni):
1 phase: a
B(~ 1250ºC, 35 wt% Ni):
2 phases: L + a
1600
L (liquid)
B (1250ºC,35)
• Examples:
T(ºC)
1500
1400
1300
a
(FCC solid
solution)
1200
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys,
P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
Cu-Ni phase diagram
A(1100ºC,60)
1100
1000
0
20
DRAFT
40
60
80
100
wt% Ni
25
Composition Determination: Tie Rule
• Rule 2: We know T and C0, we can determine: Composition of each phase.
T(ºC)
• Examples:
= 1320ºC:
B
TB
a
= 1190ºC:
Only Solid (a) present
Ca = C0
1200
TD
( = 35 wt% Ni)
= 1250ºC:
Both
(solid)
D
20
At TB
tie line
L (liquid)
1300
Only Liquid (L) present
( = 35 wt% Ni)
CL = C0
At TD
A
TA
Consider C0 = 35 wt% Ni
At TA
Cu-Ni
system
30
40
32 35
a and L present
CL C0
50
wt% Ni
43
Ca
CL = C liquidus
( = 32 wt% Ni)
Ca = C solidus
( = 43 wt% Ni)  Comp. at “tie-line hitting solidus”
26
Between 33% to 42% (wt%) Ni and a fixed temp of 1250oC - the CL & Ca are fixed
The Inverse-Lever Rule
T(ºC)
What fraction of each phase?
Think of the tie line as a lever (teeter-totter)
tie line
1300
L (liquid)
TB
a
(solid)
1200
R
20
Ma
ML
B
30CL
S
R
C0 40 Ca
wt% Ni
Ca  C0
ML
S
WL 


ML  Ma R  S Ca  CL
50
Adapted from Fig. 9.3(b),
Callister & Rethwisch 8e.
C0  CL
R
Wa 

R  S Ca  CL
DRAFT
S
We can find both the weight
fraction and the composition
of SS formed at a given
temperature for any alloy
during solidification
27
Determination of weight fractions
• Rule 3: Knowing T and C0, we can determine the weight fraction
• Examples:
Consider C0 = 35 wt% Ni
T(ºC)
S
R +S
Wa 
R
= 0.27
R +S

WL=Red/Red+Blue
WS=Blue/Red+Blue
tie line
B
R S
TB
1200
TD
D
20
30
35
32 C0
CL
a
(solid)
40 43
50
Ca wt% Ni
Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig.
9.3(a) is adapted from Phase Diagrams of Binary Nickel
Alloys, P. Nash (Ed.), ASM International, Materials Park, OH
(1991).
DRAFT
S
L (liquid)
1300
43  35
 0.73
43  32
WL 
A
TA
At TA : Only Liquid (L) present
WL = 1.00, Wa = 0
At TD : Only Solid ( a) present
WL = 0, Wa = 1.00
At TB : Both a and L present
L
Cu-Ni
system
28
Equilibrium Cooling
• Phase diagram:
Cu-Ni system.
• Consider
microstuctural
changes that
accompany the
cooling of a
C0 = 35 wt% Ni alloy
T(ºC) L (liquid)
130 0
L: 35 wt% Ni
a: 46 wt% Ni
L: 35wt%Ni
Compositional readjustments are
accomplished by diffusional
processes
A
35
32
B
C
46
43
D
24
L: 32 wt% Ni
36
120 0
a: 43 wt% Ni
E
L: 24 wt% Ni
a: 36 wt% Ni
a
(solid)
110 0
20
30
Adapted from Fig. 9.4,
Callister & Rethwisch 8e.
Cu-Ni
system
Diffusion is slower with reducing
temp
35
C0
DRAFT
40
50
wt% Ni
29
Non-equilibrium cooling
~1240 deg C – diffusion is slow and
composition doesn’t reach 40% Nickel but
rather is ~ 42%
CL=29%Ni-71%Cu (Cα should be 40%)
α-phase formed at b’ should lose Ni with
lowering of temp but slow diffusion inhibits the
process and hence has a higher value than
predicted from tie-line
As if the solidus line on the phase diagram has been
shifted to higher Ni contents due to fast cooling
30
Cored vs Equilibrium Structures
• Ca changes as we solidify.
• Cu-Ni case: First a to solidify has Ca = 46 wt% Ni.
• Slow rate of cooling:
• Fast rate of cooling:
Equilibrium structure
Cored structure
Uniform Ca:
35 wt% Ni
First a to solidify:
46 wt% Ni
DRAFT
31
Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Ductility (%EL)
400
TS for
pure Ni
300
TS for pure Cu
200
0 20 40
Cu
60 80 100
Ni
Composition, wt% Ni
Adapted from Fig. 9.6(a),
Callister & Rethwisch 8e.
Elongation (%EL)
Tensile Strength (MPa)
-- Tensile strength (TS)
60
%EL for pure Cu
%EL for
pure Ni
50
40
30
20
0 20
Cu
40
60
80 100
Ni
Composition, wt% Ni
Adapted from Fig. 9.6(b),
Callister & Rethwisch 8e.
DRAFT
32
Binary-Eutectic Systems
Cu-Ag & Pb-Sn systems
DRAFT
33
Violation of Hume-Rothery Rules
Crystal Structure
electroneg
r (nm)
Ag
FCC
1.93
0.172
Cu
FCC
1.9
0.127
DRAFT
34
Copper-Silver: Binary-Eutectic Systems
has a special composition
with a min. melting T.
Cu-Ag
system
2 components
Ex: Cu-Ag system
T(ºC)
1200
• 3 single phase regions
(L, a, b)
L (liquid)
1000
a : mostly Cu or Cu-rich phase
b : mostly Ag or Ag-rich phase
TE 800
• TE : No liquid below TE
• CE : Composition at temp TE
• Eutectic reaction
L+a
cooling
heating
779ºC
8.0
71.9
L+b b
91.2
600
ab
400
200
0
20
a(CaE) + b(CbE)
L(CE)
L(71.9 wt% Ag)
a
40
60 CE 80
100
wt% Ag
Adapted from Fig. 9.7,
Callister & Rethwisch 8e.
a (8.0 wt% Ag)  b (91.2 wt% Ag)
35
Solder
Solder Alloys
Solder alloys can be elemental metals such as tin
and indium;
binary alloys such as Sn–Ag or Sn-Pb;
complex ternary and quaternary compositions
Of interest are the solidus temperature, at which
melting begins, and the liquidus temperature, at
and above which the alloy is fully liquid.
The temperature range between the solidus and
liquidus temperatures is identified as the pasty
range or the mushy zone, in which the solder is a
mixture of liquid and solid particles.
P.T. Vianco, Y Feng, Electronic Packaging: Solder Mounting Technologies,
Reference Module in Materials Science and Materials Engineering, Elsevier, 2016
DRAFT
36
Lead-Tin (Pb-Sn) System (solder)
• For a 40% Sn (60% Pb) alloy at 150ºC, determine:
Pb-Sn
system
-- the phases present and weight fractions
T(ºC)
Answer: a + b
-- the phase compositions
Answer: Ca = 11 wt% Sn
Cb = 99 wt% Sn
-- the relative amount
of each phase
Answer:
S
=
W =
R
+S
a
99 - 40
=
99 - 11
Wb = R =
R+S
40 - 11
=
99 - 11
300
200
Cb - C0
Cb - Ca
59
= 0.67
88
C0 - Ca
Cb - Ca
=
=
29
= 0.33
88
L (liquid)
a
L+ a
100
L+b b
183ºC
18.3
150
Tie-line
Isotherm @150 degC
61.9
R
97.8
S
a+b
0 11 20
Ca
40
C0
60
80
99100
Cb
wt% Sn
Adapted from Fig. 9.8,
Callister & Rethwisch 8e.
DRAFT
37
Example 2: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine:
-- the phases and compositions:
T(ºC)
Answer: a + L
-- the phase compositions
Answer: Ca = 17 wt% Sn
CL = 46 wt% Sn
-- the relative amount
of each phase
Pb-Sn
system
300
a
220
200
L (liquid)
L+a
R
L+b
S
b
183ºC
Answer:
CL - C0
46 - 40
=
Wa =
CL - Ca
46 - 17
6
=
= 0.21
29
C0 - C a
23
=
= 0.79
WL =
CL - C a
29
100
a+b
0
17 20
Ca
40 46 60
C0 CL
80
100
C, wt% Sn
Adapted from Fig. 9.8,
Callister & Rethwisch 8e.
DRAFT
38
Microstructure Development Binary-Eutectic Systems
DRAFT
39
Microstructural Developments Pb-Sn System
Limited solubility – completely soluble upto 2% Sn –
T(ºC)
L: C0 wt% Sn
400
• For alloys where
C0 < 2 wt% Sn
L
a
L
300
200
TE
• Result: at room temperature
-- polycrystalline with grains of
a-phase having composition C0
100
Adapted from Fig. 9.11,
Callister & Rethwisch 8e.
DRAFT
a
L+ a
a: C0 wt% Sn
(Pb-Sn
System)
a+ b
0
10
20
30
C0
C , wt% Sn
2
(room T solubility limit)
40
Microstructure Developments:
Intermediate Composition
L: C0 wt% Sn
T(ºC)
400
• For alloys for which
2 wt% Sn < C0 < 18.3 wt% Sn
L is low-Sn (High-Pb)
L
300
• Result:
at temperatures in a + b range
-- polycrystalline with a grains
and small b phase particles
L +a
a
200
TE
a: C0 wt% Sn
a
b
100
Adapted from Fig. 9.12, Callister & Rethwisch 8e.
L
a
a+ b
0
10
20
Pb-Sn
system
30
C0
C, wt%
2
(sol. limit at T room )
18.3
(sol. limit at TE)
DRAFT
Sn
41
Microstructural Developments:
Eutectic Composition
Eutectic Mixture
• For alloy of composition C0 = CE
• Result: Eutectic microstructure (lamellar structure)
-- alternating layers (lamellae) of a and b phases.
T(ºC)
L : C0 wt% Sn
300
Pb-Sn
system
a
200
L+ a
L
Lb b
183ºC
TE
α - Pb-rich
100
ab
β - Sn-rich
Tie-line gives
composition
0
20
18.3
Adapted from Fig. 9.13,
Callister & Rethwisch 8e.
Micrograph of Pb-Sn
eutectic microstructure
40
b: 97.8 wt% Sn
a: 18.3 wt%Sn
60
CE
61.9
80
Adapted from Fig. 9.14,
Callister & Rethwisch 8e.
100
97.8
C, wt% Sn
DRAFT
160 m
42
Lamellar Eutectic Structure
Alternate layered system with two alloy types - α and β
α and β phases have different compositions
neither of which is the same as that of the liquid
Redistribution is accomplished by atomic diffusion
Interdiffusion of Pb & Sn occurs in the liquid just
ahead of the “eutectic solid–liquid” interface.
Lamellar configuration preferred as atomic
diffusion of Pb and Sn need only occur over
relatively short distances.
Adapted from Figs. 9.14 & 9.15, Callister & Rethwisch 8e.
DRAFT
43
Liquidus: Line separating pure Liquid
(L) phase field from a mixture of
{Liquid (L) + solid(α)} phase field
Solidus: ?
Solvus line: a line on a phase diagram
that separates a solid phase (α) from a
mixed solid (β) + solid(α) phase
Two phase fields have different
microstructures.
DRAFT
44
Hypoeutectic & Hypereutectic
300
T(ºC)
Adapted from Fig. 9.8,
Callister & Rethwisch 8e.
200
(Fig. 10.8 adapted from
TE
Binary Phase Diagrams,
2nd ed., Vol. 3, T.B.
Massalski (Editor-in-Chief),
100
ASM International,
Materials Park, OH, 1990.)
a
L+ a
20
a
L+b b
40
hypoeutectic: C0 = 50 wt% Sn
a
Eutectic: Lamellar μ-structure
Hypo:
α-rich + eutectic
Hyper: β-rich +eutectic
a+b
0
(Figs. 9.14 and 9.17
from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
(Pb-Sn
System)
L
a
60
80
eutectic
61.9
C, wt% Sn
hypereutectic: (illustration only)
eutectic: C0 = 61.9 wt% Sn
b
a a
b
a
175 m
100
b
b b
b
160 m
eutectic micro-constituent
45
Intermetallic Compounds
Intermetallic compounds are defined as solid phases involving two
or more metallic or semimetallic elements with an ordered
structure
Usually has well-defined and fixed stoichiometry
Non-stoichiometric Intermetallics: Ordered but have a variable
composition
DRAFT
46
Intermetallic Compounds in Phase Diagrams
Adapted from
Fig. 9.20, Callister &
Rethwisch 8e.
Mg2Pb
Note: intermetallics usually exists as a line on the diagram
(i.e. composition of a compound is a fixed value).
Non-stoichiometric Intermetallics
usually have a variable composition
Invariant Reactions
Syntectic:
L1+L2α
Catatectic:
β  L+α
DRAFT
48
Tin 63% / Lead 37% solder which melts
and freezes at 183 °C. This melting
point is much lower than the melting
points of either pure metal which are
232 °C (Sn) and 327 °C (Pb)
DRAFT
49
Eutectoid & Peritectic
Peritectic transformation  + L

Cu-Zn Phase diagram (Brass)
Eutectoid transformation 
+
DRAFT
Adapted from Fig. 9.21,
Callister & Rethwisch 8e.
50
Allotropes of Iron and various phases
DRAFT
51
Fe-C Phase Diagram: Invariant reactions
Eutectic - liquid transforms to two solid phases
L cool a + b
(For Pb-Sn, 183ºC, 61.9 wt% Sn)
heat
• Eutectoid – one solid phase transforms to two
other solid phases
compound - cementite
S2
S1+S3
 cool a + Fe3C (For Fe-C, 727ºC, 0.76 wt% C)
heat
• Peritectic - liquid and one solid phase transform
to a second solid phase
S1 + S2
 +L
cool
heat

(For Fe-C, 1493ºC, 0.16 wt% C)
DRAFT
52
Iron-Iron carbide (Fe-Fe3C) Phase Diagram
L   + Fe3C
T(ºC)
1600

1200
- Eutectoid (B):
  a + Fe3C
L
1400
 +L

(austenite)
 
 
1000
a
800
A
1148ºC
 +Fe3C
727ºC = T eutectoid
B
a+Fe3C
600
120 m
Result: Pearlite =
alternating layers of
a and Fe3C phases
(Adapted from Fig. 9.27,
Callister & Rethwisch 8e.)
400
0
(Fe)
L+Fe3C
1
0.76
2
3
4
4.30
5
6
Fe3C (cementite)
• 2 important
points
- Eutectic (A):
6.7
C, wt% C
Fe3C (cementite-hard)
a (ferrite-soft)
Adapted from Fig. 9.24,
DRAFT
Callister & Rethwisch 8e.
53
Voids in Octahedron/Tetrahedron
FCC: 8 OV and 4 TV per unit cell
BCC: 12 OV and 6TV
Each of 8 OVs in FCC >> Each of 12 OVs in BCC
DRAFT
54
Hypoeutectoid Steel
T(ºC)
1600

L
 
 
 
 
a


a
 a
 +L

1200
(austenite)
1000
 + Fe3C
800
727ºC
a
a + Fe3C
600
pearlite
1
0.76
a
400
0
(Fe)C0
L+Fe3C
1148ºC
2
3
4
5
6
6.7
Adapted from Figs. 9.24
and 9.29,Callister &
Rethwisch 8e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
C, wt% C
100 m
pearlite
(Fe-C
System)
Fe3C (cementite)
1400
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 9.30,
Callister & Rethwisch 8e.
DRAFT
55
Hypoeutectoid Steel
T(ºC)
1600

L
a
 +L

1200
(austenite)
a

 a
1000
 + Fe3C
Wa = s/(r + s)
800 r s
W =(1 - Wa)
727ºC
aRS
a + Fe3C
600
pearlite
Wpearlite = W
400
0
(Fe)C0
1
0.76
a
L+Fe3C
1148ºC
Wa’ = S/(R + S)
WFe3C =(1 – Wa’)
pearlite
2
3
4
5
6
(Fe-C
System)
Fe3C (cementite)
1400
6.7
Adapted from Figs. 9.24
and 9.29,Callister &
Rethwisch 8e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
C, wt% C
100 m
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 9.30,
Callister & Rethwisch 8e.
DRAFT
56
Hypereutectoid Steel
T(ºC)
1600

L
Fe3C


 +L

1200
(austenite)


1000
 
 
L+Fe3C
1148ºC
 +Fe3C
800
a
a +Fe3C
600
400
0
(Fe)
pearlite
0.76




(Fe-C
System)
1 C0
2
3
4
5
6
Fe3C (cementite)
1400
Adapted from Figs. 9.24
and 9.32,Callister &
Rethwisch 8e. (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.)
6.7
C, wt%C
60 mHypereutectoid
steel
pearlite
proeutectoid Fe3C
Adapted from Fig. 9.33,
Callister & Rethwisch 8e.
DRAFT
57
Hypereutectoid Steel
T(ºC)
1600

L
 +L

1200
(austenite)
 
 
W =x/(v + x)
WFe3C =(1-W)
 +Fe3C
v x
800
a V
600
pearlite
400
0
(Fe)
Wpearlite = W
X
1 C0
Wa = X/(V + X)
WFe
3C’
L+Fe3C
1148ºC
1000
0.76
Fe3C
(Fe-C
System)
a +Fe3C
2
3
4
5
6
Fe3C (cementite)
1400
Adapted from Figs. 9.24
and 9.32,Callister &
Rethwisch 8e. (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.)
6.7
C, wt%C
60 mHypereutectoid
steel
=(1 - Wa)
pearlite
proeutectoid Fe3C
Adapted from Fig. 9.33,
Callister & Rethwisch 8e.
DRAFT
58
Alloying with Other Elements
Ti
Mo
Si
W
Cr
Mn
Ni
• Ceutectoid changes:
Ceutectoid (wt% C)
TEutectoid (ºC)
• Teutectoid changes:
Ni
Cr
Si
Ti Mo
W
Mn
wt. % of alloying elements
wt. % of alloying elements
Adapted from Fig. 9.34,Callister & Rethwisch 8e.
(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 & Rethwisch 8e.
(Fig. 9.35 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
DRAFT
59
Example Problem
For a 99.6 wt% Fe-0.40 wt% C steel at a temperature
just below the eutectoid, determine the following:
a) The compositions of Fe3C and ferrite (a).
b) The amount of cementite (in grams) that forms in
100 g of steel.
c) The amounts of pearlite and proeutectoid ferrite (a)
in the 100 g.
DRAFT
60
Solution to Example Problem
a) Using the RS tie line just below the eutectoid
Ca = 0.022 wt% C
CFe3C = 6.70 wt% C
WFe 3C =
=
R
C - Ca
= 0
R + S CFe 3C - Ca
0.40 - 0.022
= 0.057
6.70 - 0.022
1600

L
1400
T(ºC)
1200

 +L
1000
 + Fe3C
800
727ºC
R
Amount of Fe3C in 100 g
S
a + Fe3C
600
= (100 g)WFe3C
400
0
= (100 g)(0.057) = 5.7 g
Ca C0
DRAFT
L+Fe3C
1148ºC
(austenite)
Fe3C (cementite)
b) Using the lever rule with
the tie line shown
1
2
3
4
C, wt% C
5
6
6.7
CFe
3C
61
Solution to Example Problem (cont.)
c) Using the VX tie line just above the eutectoid and realizing that
C0 = 0.40 wt% C
Ca = 0.022 wt% C
Cpearlite = C = 0.76 wt% C
1600

Wpearlite =
=
V
C - Ca
= 0
V + X Cg - Ca
0.40 - 0.022
= 0.512
0.76 - 0.022
Amount of pearlite in 100 g
= (100 g)Wpearlite
L
1400
T(ºC)
1200

 +L
L+Fe3C
1148ºC
(austenite)
1000
 + Fe3C
800
727ºC
VX
a + Fe3C
600
400
0
1
Ca C0 C
= (100 g)(0.512) = 51.2 g
2
3
4
5
6
Fe3C (cementite)
•
6.7
C, wt% C
DRAFT
62
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.
DRAFT
63