Microstructure of a lead-tin alloy
1
Example: Building Structures
Burj Dubai under construction
2007; completed 2010
Example: Aerospace Industry
F/A22 Raptor
Single-Crystal Titanium panels
Example: Aerospace Industry
Pratt and Whitney F119 Engine
Why study Phase Diagrams?
• Provide valuable information about:
– Melting
– Casting
– Microstructure
– Crystallization
– Temperature
– Pressure
– Composition
– Phase transformations
Phase Diagrams
• Component: a chemical constituent of an
alloy, which may be used to specify its
composition
– Example: In a copper-zinc brass, the
components are Cu and Zn
• Solid solution (Chapter 4):
– Consists of atoms of at least two different types
– Solute atoms occupy either substitutional or
interstitial positions
– Crystal structure of solvent is maintained
Solubility Limit
• Maximum concentration of solute atoms
that may dissolve in the solvent, at a specific
temperature, to form a solid solution
• Adding solute past the solubility limit yields
another solid solution (different crystal
structure) or compound that has a different
composition
Phases
• Phase: a homogeneous portion of a system that
has uniform physical and chemical characteristics
• The following are considered phases:
– Solid solution
– Liquid solution
– Gaseous solution
• If more than one phase is present in a system, each
will have its own distinct properties
• Phases exist over a range of compositions,
temperatures and pressures
• Example: Water and Ice  two phases (physically
dissimilar, but chemically the same)
8
Stability of Phase
• Phase stability depends on temperature, pressure and
composition,
9
Phases
For each of the following situations how many
phases are present?
1. Ice cubes floating in a glass of water.
2. Water at room temperature containing 20
wt% sugar.
3. A bottle of 7-Up.
10
Components vs. Phases
• Components:
– The elements (or compounds)
that are mixed to make the
alloy
• Al and Cu
• MgO and Al2O3
• styrene and butadiene
– The composition is the
concentration of each
component present in an alloy
• Phase:
Al-Cu alloy
Phase α
– Physically and/or chemically
distinct regions in the
material
Phase β
Aluminum atom
Copper atom
11
Sugar- Water Demo
• Initially, sugar is added to water
and there is a sugar- water
solution or syrup
• Solution becomes more
concentrated with sugar until
the solubility limit is reached
(saturated)
• System cannot dissolve any
more sugar (at the specified
temperature) and further
additions settle to the bottom of
the container
• System now contains two
phases; syrup and solid
(undissolved) sugar crystals
Solid Solubility
• We define solubility
as the maximum
equilibrium
concentration of a
solute in a host
material.
ALS:
What is the solubility of sugar in water at 20oC?
Answer: 65 wt% sugar
Solid Solubility
• We define solubility as
the maximum
equilibrium
concentration of a
solute in a host
material
Previous Answer:
65 wt% sugar
ALS:
If you have 1 L (1000 g) of water, how much sugar can you dissolve in it?
a) 650 g of sugar
b) 2000 g of sugar
c) 1350 g of sugar
Solid Solubility
• We define solubility as
the maximum
equilibrium
concentration of a
solute in a host
material
Previous Answer:
65 wt% sugar
ALS:
If you have 1 L (1000 g) of water, how much sugar can you dissolve in it?
Answer: The solution is 65% (about 2/3) sugar and 1/3 water. The weight of the
sugar is twice that of the water, so 2000 g is the answer.
Solubility Limits
If you leave a saturated water/sugar
solution open to the air and the water
starts to evaporate, what will happen?
Answer:
At some point there won’t be enough water to
hold all of the sugar. The solubility for sugar will
be exceeded, and some of it will come out of the
solution in crystal form (i.e. precipitate) to
compensate.
Phase Diagrams
17
•
•
•
•
One-Component (Or Unary) Phase
Diagrams
3 externally controllable parameters that affect phase structure:
– Temperature
– Pressure (usually scaled logarithmically)
– Composition
Unary phase diagrams only deal with pressure-temperature graphs 
composition remains constant
Along the phase boundaries, the phases on either side are in equilibrium
Triple point: all three
states are in equilibrium
(at 273.16 K at a
pressure of 6.04x10-3
atm in the example)
(H20) Each of the phases
will exist in equilibrium
conditions over the
corresponding temperature
18
Binary Phase
Diagrams
19
Binary Phase Diagrams
• Temperature and composition are variable
parameters
• Binary alloys – contain two components
• Pressure held constant  normally ≈ 1 atm
• Maps that represent the relationship
between temperatures, compositions and
quantities of phases at equilibrium
• Accurately predict phase transformations
and resulting microstructures
Binary Isomorphous Systems
• Easiest binary phase diagram to understand
– Example: copper-nickel system
• Temperature is plotted along the vertical axis
• Composition of the alloy is plotted along the
horizontal axis
– Bottom- weight %
– Top – atom %
– Composition ranges from 0 wt% Ni (100 wt% Cu)
on the left of the horizontal axis to 100 wt% Ni (0
wt% Cu) on the right of the horizontal axis
Binary Isomorphous Systems
• Three phase regions or fields:
– Alpha(α)
• Substitutional solid solution
consisting of Cu and Ni and
both have an FCC crystal
structure
– Liquid (L)
• Homogeneous liquid solution
composed of both copper and
nickel
– Two phase  α + L
• The copper-nickel system is
termed isomorphous due to the
complete liquid and solid
solubility of the two
components
Phase Boundary
Interpretation of Phase Diagrams
• For a binary system of known composition and
temperature that is at equilibrium, you can
identify:
1. Phases present
2. Compositions of phases
3. Percentages or fractions of phases
•
Phases Present:
– Locate the temperature-composition point on the
diagram and one can identify the phase(s) within
the phase field that are/ is present
Interpretation of Phase Diagrams
• Phase compositions:
– One phase present:
• composition of phase is the same as the overall
composition of the alloy
– Two phases present:
• A Tie line is constructed across two-phase region at
a specified temperature of the alloy
• Intersections of tie lines and phase boundaries on
both sides are noted
• The compositions corresponding to these points are
read from the x-axis.
Interpretation of Phase Diagrams
• Determining Phase Amounts:
– One phase present:
• Alloy is composed entirely of that phase; phase fraction is 1.0 or
percentage is 100%
– Two phases present:
• Use the lever rule; tie line is constructed across the two-phase
region at a specified temperature of the alloy
• Overall alloy composition is located on the tie line
• Fraction of one phase is determined by taking the length of the tie
line from the overall alloy composition to the phase boundary for
the other phase, and dividing by the total lie line length
• The fraction of the other phase is determined in the same manner
• Segment lengths can be determined either by direct measurement
from the phase diagram using ruler or by subtracting compositions
from the composition axis
Alternate Approach - Phase
diagrams
• Liquid mass
fraction:
C  C0
S
WL 

R  S C  CL
• α-Phase mass
fraction:
C0  CL
R
W 

R  S C  CL
Developing Microstructures –
Isomorphous Alloys
• Equilibrium cooling:
– Cooling occurs very
slowly
Mechanical Properties of
Isomorphous Alloys
• Solute Strengthening
TS for pure Ni
%EL for pure Ni
%EL for pure Cu
TS for pure Cu
Binary Eutectic Systems
One common type of alloy system is called the eutectic system.
Binary eutectic systems occur when 2 elements completely dissolve into
each other in the liquid state, but have only limited solubility in the solid
state.
More than one solid phase exists in
these systems.
There are multiphase regions
between single-phase regions.
29
Which phases are present
depends, as before, on the
temperature and the composition
of the material.
Binary Eutectic Systems
• Consider Cu-Ag alloys:
The copper-silver alloy system is binary
eutectic.
− How many phases are there
in the system?
3 (L, , b)
− What are the two solid
phases?
: mostly Cu
b: mostly Ag
− Where is the eutectic point?
eutectic temperature: TE
eutectic composition: CE
30
ALS: Pb-Sn Eutectic System
The lead-tin system is also binary eutectic
• Consider an alloy which contains 40 wt%
Sn at 150oC.
• What phases are present? What are their
compositions?
•
Answer:
There are 2 phases:
a and b.
Ca = 11 wt% Sn
Cb = 99 wt% Sn
31
ALS: Pb-Sn Eutectic System
• How much of each phase is present?
•
Answer:
From the lever rule:
59
W   67%
88
Wb  1 W  33%
32
Microstructures in Eutectic
Systems
• The result is...
– … a single phase
polycrystal with 
grains.
Adapted from Fig. 9.9,
33
Callister 6e.
L
Temperature oC
• Consider a eutectic
system (e.g. Pb-Sn)
with an initial
concentration (Co)
less than 6 wt% Sn.
400
300

L 
200
 b
100
0%
Co
10%
%Sn
20%
30%
Microstructures in Eutectic
Systems
• Consider a eutectic
system with an initial
concentration (Co)
between 2 and 18.3 wt%
Sn.
• The result is...
– … a 2-phase
polycrystal with...
• …  and β grains.
• … a fine grained
structure.
Adapted from Fig. 9.10,
34
Callister 6e.
L: Cowt%Sn
T(°C)
400
L
300
200
TE
100
L+

L

: Cowt%Sn

b
+b
Pb-Sn
system
0
10
20
30
Co Co, wt%
2
(sol. limit at Troom)
18.3
(sol. limit at TE)
Sn
Microstructures in Eutectic Systems
• When the initial concentration (Co) equals the eutectic
composition (CE) the result is...
– … alternating layers of  and β crystals
Pb-Sn
system
Adapted from Fig. 9.12, Callister 6e. (Fig.
9.12 from Metals Handbook, Vol. 9, 9th ed.,
Metallography and Microstructures,
American Society for Metals, Materials
Park, OH, 1985.)
Adapted from Fig. 9.11,
35
Callister 6e.
Microstructures in Eutectic
Systems
• Co between 18.3 and 61.9 wt% Sn results in...
– … a combination of  crystals and eutectic regions
L

L
300
Pb-Sn
system
L+ 
2 00
TE
R
R

100
0
0
b
S
40
Co
Callister 6e.
C L = 61.9 wt% Sn
W =
+ b
20
18.3
Just above TE:
C  = 18.3 wt% Sn
L+ b
S
60
61.9
Adapted from Fig. 9.14,
36
 L
L: Cowt%Sn
T(°C)
80
primary 
eutectic 
eutectic b
100
97.8
wt% Sn
S
= 50 wt%
R+S
WL = (1-W) = 50 wt%
Microstructures in Eutectic
Systems
• Co between 18.3 and 61.9 wt% Sn results in...
– … a combination of  crystals and eutectic regions
Just below TE:
Pb-Sn
system
C  = 18.3 wt%Sn
C b = 97.8 wt%Sn
W =
S = 73 wt%
R+S
Wb = 27 wt%
Adapted from Fig. 9.14,
37
Callister 6e.
Eutectoid and Peritectic Reactions
• Eutectoid reaction: a reaction wherein upon cooling, one
solid phase transforms isothermally and reversibly into
two new solid phases that are intimately mixed

cooling





heating
 
• Peritectic reaction: a
reaction wherein upon
cooling, a solid and a
liquid phases
transform
isothermally and
reversibly to a solid
phase having a
different composition
 L
Copper-zinc phase diagram
cooling





heating

38
The Iron-Carbon
System
39
The Iron-Iron Carbide (Fe-Fe3C)
Phase Diagram
• When heated, pure iron experiences two
changes in its crystal structure before it
reaches its melting temperature:
– At room temperature, exists as ferrite (α iron) and
has a BCC structure
– Ferrite experiences a polymorphic transformation
to FCC austenite (γ iron) at 912°C
– Experiences a second transformation to BCC phase
known as δ ferrite at 1394°C, which finally melts at
1538°C
The Iron-Iron Carbide (Fe-Fe3C)
Phase Diagram
• Cementite,
(Iron Carbide
- Fe3C). It is
25at% C or
6.7 wt% C.
The Iron-Iron Carbide (Fe-Fe3C)
Phase
Diagram
Eutectic reaction for
the iron-iron carbide
system:
L
cooling





heating
Euctectic Point
  Fe3C
Eutectoid Point
Eutectoid reaction
for the iron-iron
carbide system:
 (0.76wt %C)
cooling





heating
 (0.022wt %C)  Fe3C(6.7wt %C)
Development of Microstructure in
Iron-Carbon Alloys
•
The microstructure depends on:
– Concentration of carbon
– Heat treatment
•
Pearlite: a two-phase microstucture
results from the transformation of
austenite and consists of alternating
layers ( or lamellae) of -ferrite and
cementite
Pearlite
43
Hypoeutectoid Alloys
• Hypoeutectoid alloy : (less
than eutectoid) between
0.022 and 0.76wt% C
• Proeutectoid ferrite: that
formed above the eutectoid
temperature
44
Hypereutectoid Alloys
• Hypereutectoid alloys: 
containing between 0.76 and
2.14 wt% C
• Proeutectoid cementite: that
which forms before the
eutectoid reaction  cementite
composition remains constant
as temperature changes
45
Non-Equilibrium Cooling
• In most situations, the cooling rates are impractically slow and
unnecessary
• Nonequilibrium effects of practical importance:
– Phase changes or transformations at temperatures other than those
predicted in phase diagrams
– The existence of non-equilibrium phases that do not appear on the phase
diagram
46
Chapter 9 Review
What did you learn?
•
•
•
•
•
•
47
Chapter 9
Practice Problems
Practice Problems
1. Salt is often spread on roads during the winter in order
to depress the freezing point of water. A binary phase
diagram for water and salt (NaCl) is shown below.
Liquid (Brine)
Field “Z”
Practice Problems
a) What are the two phases that exist in equilibrium in
the field “Z” region.
b) At -10°C determine the overall composition for which
equal amounts of these phases exist in equilibrium.
c) What are the compositions of each phase under these
conditions?
d) What is the lowest temperature at which adding salt
can prevent water from freezing completely?
Practice Problems
Answer:
a)Liquid (Brine) + Ice
b)8 wt% NaCl
c)Ice – 0.5 wt% NaCl
Salt + Brine – 16 wt% NaCl
d) -21°C
Practice Problems
2. Using the Fe-Fe3C phase diagram, compute
the weight fractions of proeutectoid ferrite
and pearlite, respectively, that form in an
iron-carbon alloy containing 0.25 wt% C if
the alloy is slowly cooled to a temperature
just below the eutectoid temperature.
Practice Problems
a) Wα = 0.31,
b) Wα = 0.69,
c) Wα = 0.56,
d) Wα = 0.44,
Wp = 0.69
Wp = 0.31
Wp = 0.44
Wp = 0.56
Practice Problems
Answer:
0.76  0.25
W 
 0.69
0.76  0.022
0.25  0.022
Wp 
 0.31
0.76  0.022
Therefore, the answer is b)
54
Practice Problems
c)
Before eutectic reaction :
61.9  30
W 
 0.7317
61.9  18.3
After eutectic reaction :
97.8  30
W 
 0.8528
97.8  18.3
 formed during eutectic reaction :
W  0.8528 0.7317  0.1211or 12.11%
55