Introduction to Materials Science and Engineering
Eun Soo Park
Office: 33-313
Telephone: 880-7221 Email: espark@snu.ac.kr
Office hours: by appointment
2020 Fall
10. 27. 2020
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• Sharp corners produce large stress concentrations and premature failure.
SUMMARY
• Engineering materials not as strong as predicted by theory
• Flaws act as stress concentrators that cause failure at stresses lower than theoretical values.
• Failure type depends on T and σ :
- For simple fracture (noncyclic σ and T < 0.4T m ), failure stress decreases with:
- increased maximum flaw size, - decreased T,
- increased rate of loading.
- For fatigue (cyclic σ ):
- cycles to fail decreases as Δσ increases.
- For creep (T > 0.4T m ):
- time to rupture decreases as σ or T increases.
Contents for previous class
Chapter 8: Failure
3
ISSUES TO ADDRESS...
• When we combine two elements...
what is the resulting equilibrium state?
• In particular, if we specify...
-- the composition (e.g., wt% Cu - wt% Ni), and -- the temperature (T )
then...
How many phases form?
What is the composition of each phase?
What is the amount of each phase?
Chapter 9: Phase Diagrams
Phase B Phase A
Nickel atom
Copper atom
I. Component, Phase, Equilibrium
→ Phase diagram (Gibb’s phase rule)
II. one component phase diagram two component phase diagram
: solubility limit (Hume-Rothery Rule)
III. Isomorphous Binary Phase Diagram
: tie line, lever rule
IV. Binary-Eutectic Systems V. Binary invariant reaction
: Eutectic, Eutectoid, & Peritectic - Ternary, Quarternary phase diagram - Phase transformation
What we will learn about
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5
• 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., α and β).
Aluminum- Copper Alloy
I-a. Components and Phases
α (darker phase)
β (lighter phase)
Adapted from chapter- opening photograph, Chapter 9, Callister, Materials Science &
Engineering: An
Introduction, 3e.
b. Equilibrium
Mechanical equilibrium
: total potential energy of the system is a minimum.
Thermal equilibrium
: absence of temperature gradients in the system
Chemical equilibrium
: no further reaction occurs between the reacting substances i.e. the forward and reverse rates of reaction are equal.
Thermodynamic equilibrium
: the system is under mechanical, thermal and chemical equilibrium
The properties of the system-P, T, V, concentrations-do not change with time.
matchbox
6
= 0 dG
2 1 0
G G G
∆ = − <
Equilibrium Lowest possible value of Gibb’s Free Energy No desire to change ad infinitum
Phase Transformation
metastable
unstable
Activation E
stable
7
phase diagram: graphical representation of the phases present and the ranges in composition, temperature,
and pressure over which the phases are stable
Gibbs phase rule: F=C-P+2
C: # components, P: # phases in equilibrium
F: degree of freedom (temp., press., composition)
ex) H 2 O, C=1, F=C-P+2=3-P 1 phase F=2 2 phase F=1
3 phase F=0 (invariant)
* pressure or temperature constant F=C-P+1
c. Phase diagram & Gibbs phase rule
8
II-a. Single component system
One element (Al, Fe) One type of molecule (H 2 O)
Fe C
Gibbs Free Energy as a Function of Temp.
TS H
G = −
- Allotropic forms?
- How is phase stability measured?
9
One Component Phase Diagram
Pressure-temperature diagram for H 2 O. Notice the solid-liquid line
sloping to the left. At normal pressure (1 atm or 760 torr), the melting temperature is 273 K. A possible scheme for freeze drying is shown as starting with point S and following the dashed line to the left.
Equilibrium and non-equilibrium states by heating
10
b: What are binary systems?
“Mixture vs. Solution vs. Compound”
Multi-component system:
11
* Binary system (two components) A, B
- Mixture ; A – A, B – B ; the physical combination of two or
more substances on which the identities and boundaries are retained.
A B
* Single component system One element (Al, Fe), One type of molecule (H
2O)
: Equilibrium depends on pressure and temperature.
: Equilibrium depends on not only pressure and temperature but also composition.
Alluvial mining
Winnowing
wash rice Select egg
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- Solution ; A – A – A ; atomic scale mixture/ Random distribution A – B – A Solid solution : substitutional or interstitial
- Compound ; A – B – A – B ; fixed A, B positions/ Ordered state
B – A – B – A
MgCNi3
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14
Mg 2 Pb
Note: intermetallic compound exists as a line on the diagram - not an area - because of stoichiometry (i.e. composition of a compound
is a fixed value).
Fig. 11.19, Callister &
Rethwisch 9e.
[Adapted from Phase Diagrams of Binary Magnesium Alloys, A. A.
Nayeb-Hashemi and J. B.
Clark (Editors), 1988.
Reprinted by permission of ASM International, Materials Park, OH.]
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c. Solubility
• Unlimited Solubility
– Hume Rothery’ Conditions
• Similar Size
• Same Crystal Structure
• Same Valance
• Similar Electronegativity
– Implies single phase
• Limited Solubility
– Implies multiple phases
• No Solubility
- oil and water region
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Solubility Limit
Question: What is the
solubility limit for sugar in water at 20°C?
Answer: 65 wt% sugar.
At 20°C, if C < 65 wt% sugar: syrup
At 20°C, if C > 65 wt% sugar:
syrup + sugar
65
• Solubility Limit:
Maximum concentration for which only a single phase solution exists.
Sugar/Water Phase Diagram
S uga r
T em p er at u re ( ° C)
0 20 40 60 80 100
C = Composition (wt% sugar)
L
(liquid solution i.e., syrup)
Solubility
Limit L
(liquid)
+ S
(solid sugar) 20
40 60 80 100
W at er
Adapted from Fig. 11.1,
Callister & Rethwisch 9e.
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70 80 100
60 40
20 0
Tem per atur e ( ° C)
C = Composition (wt% sugar)
L
( liquid solution i.e., syrup)
20 100
40 60 80
0
L
(liquid)
+ S
(solid sugar)
Effect of Temperature & Composition
• Altering T can change # of phases: path A to B.
• Altering C can change # of phases: path B to D.
water- sugar system
Fig. 11.1, Callister &
Rethwisch 9e.
D (100°C,C = 90)
2 phases
B (100°C,C = 70)
1 phase
A (20°C,C = 70)
2 phases
Solubility Limit
(c) In copper-zinc alloys containing more than 30%
Zn, a second phase forms because of the limited solubility of zinc in copper.
(a) Liquid copper and liquid nickel are
completely soluble in each other.
(b) Solid copper-nickel alloys display complete solid
solubility, with copper and nickel atoms occupying random lattice sites.
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Empirical rules for substitutional solid-solution formation were identified from experiment that are not exact, but give an expectation of formation.
Briefly,
1) Atomic Size Factor The 15% Rule
If "size difference" of elements are greater than ±15%, the lattice distortions (i.e. local lattice strain) are too big and solid-solution will not be favored.
DR%= < ±15% will not disallow formation.
2) Crystal Structure Like elemental crystal structures are better
For appreciable solubility, the crystal structure for metals must be the same.
3) Electronegativity DE ~ 0 favors solid-solution.
The more electropositive one element and the more electronegative the other, then "intermetallic compounds" (order alloys) are more likely.
4) Valences Higher in lower alright. Lower in higher, it’s a fight.
A metal will dissolve another metal of higher valency more than one of lower valency.
r solute − r solvent
r solvent x100%
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d. Criteria for Solid Solubility: Hume-Rothery Rule
Phase Equilibria
Crystal Structure
Electro-
negativity 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 solid solution system (e.g., Ni-Cu solution)
• Ni and Cu are totally miscible in all proportions.
Complete mixture
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Cu-Ag Alloys Cu-Ni Alloys
complete solid solution limited solid solution
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e. Complete immiscibility of two metals does not exist.
: The solubility of one metal in another may be so low (e.g. Cu in Ge <10
-7at%.) that it is difficult to detect experimentally, but there will always be a measure of solubility.
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• Indicate phases as a function of T, C, and P.
• For this course:
- binary systems: just 2 components ( Cu and Ni).
- independent variables: T and C (P = 1 atm is almost always used).
Phase Diagram for Cu-Ni system
Fig. 11.3(a), Callister & Rethwisch 9e.
(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)
• 2 phases:
L (liquid)
α (FCC solid solution)
• 3 different phase fields:
L
L + α α
wt% Ni
20 40 60 80 100 1000 0
1100 1200 1300 1400 1500 1600
T(°C)
L (liquid)
α
(FCC solid solution)
III. Isomorphous Binary Phase Diagram
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wt% Ni
20 40 60 80 100 1000 0
1100 1200 1300 1400 1500 1600
T(°C)
L (liquid)
α
(FCC solid solution)
Cu-Ni phase diagram
a. Determination of phase(s) present
• Rule 1: If we know T and C o , then we know:
-- which phase(s) is (are) present.
• Examples:
A(1100°C, 60 wt% Ni):
1 phase: α
B(1250°C, 35 wt% Ni):
2 phases: L + α
B (1250 ºC ,35)
A(1100ºC,60)
Fig. 11.3(a), Callister & Rethwisch 9e.
(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)
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wt% Ni
20 1200 1300
T(°C)
L (liquid)
α (solid)
30 40 50
Cu-Ni system
b. Determination of phase compositions
• Rule 2: If we know T and C 0 , then we can determine:
-- the composition of each phase.
• Examples:
T A A
35 C 0 32 C L At T A = 1320°C:
Only Liquid (L) present C L = C 0 ( = 35 wt% Ni)
At T B = 1250°C:
Both α and L present
C L = Cliquidus ( = 32 wt% Ni) C α = Csolidus ( = 43 wt% Ni) At T D = 1190°C:
Only Solid ( α) present C α = C 0 ( = 35 wt% Ni) Consider C 0 = 35 wt% Ni
T D D
tie line
4 C 3 α
Fig. 11.3(b), Callister & Rethwisch 9e.
(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)
T B B
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• Rule 3: If we know T and C 0 , then can determine:
-- the weight fraction of each phase.
• Examples:
At T A : Only Liquid (L) present W L = 1.00, W α = 0 At T D : Only Solid ( α) present
W L = 0, W α = 1.00
c. Determination of phase weight fractions
wt% Ni
20 1200 1300
T(°C)
L (liquid)
α (solid)
30 40 50
Cu-Ni system
T A A
35 C 0 32 C L T B B
T D D
tie line
4 C 3 α
R S
At T B : Both α and L present
= 0.27
W L = S R + S W α = R
R + S
Consider C 0 = 35 wt% Ni
Fig. 11.3(b), Callister & Rethwisch 9e.
(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)
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• Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm
The Lever Rule
M L M α
R S
wt% Ni
20 1200 1300
T(°C)
L (liquid)
α (solid)
30 40 50
TB B
tie line
C 0
C L
C α S
R
Adapted from Fig. 11.3(b), Callister & Rethwisch 9e.
What fraction of each phase?
Think of the tie line as a lever
(teeter-totter)
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wt% Ni
20 120 0 130 0
30 40 50
110 0
L (liquid)
α (solid) T( ° C)
A
35 C 0
L: 35 wt%Ni
Cu-Ni system
• Phase diagram:
Cu-Ni system.
Adapted from Fig. 11.4, Callister & Rethwisch 9e.
• Consider
microstuctural changes that accompany the cooling of a
C 0 = 35 wt% Ni alloy
d. Cooling of a Cu-Ni Alloy
35 46 32 43
α: 43 wt% Ni L: 32 wt% Ni α: 46 wt% Ni L: 35 wt% Ni B
C
E L: 24 wt% Ni
α: 36 wt% Ni
24 D 36
α: 35 wt% Ni
• C α changes as we solidify.
• Cu-Ni case:
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure First α to solidify has C α = 46 wt% Ni.
Last α to solidify has C α = 35 wt% Ni.
e. Cored vs Equilibrium Phases
First α to solidify:
46 wt% Ni
Uniform C α:
35 wt% Ni Last α to solidify:
< 35 wt% Ni
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f. Relationship Between Properties and the Phase Diagram
• Solid-solution strengthening - Increasing the
strength of a metallic material via the formation of a solid solution.
• Dispersion strengthening - Strengthening, typically used in metallic materials, by the
formation of ultra-fine dispersions of a second phase.
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Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Tensile strength (TS) -- Ductility (%EL)
Adapted from Fig. 11.5(a), Callister & Rethwisch 9e.
T ens ile S tr engt h (M P a)
Composition, wt% Ni
Cu Ni
0 20 40 60 80 100 200
300 400
TS for pure Ni
TS for pure Cu
E longat ion (% EL )
Composition, wt% Ni
Cu 0 20 40 60 80 100 Ni 20
30 40 50 60
%EL for pure Ni
%EL for pure Cu
Adapted from Fig. 11.5(b),
Callister & Rethwisch 9e.
Copper is strengthened by up to 60% Ni and
nickel is strengthened by up to 40% Cu.
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™is a trademark used herein under license.
Mechanical Properties: Cu-Ni System
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• 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.
Summary
Introduction to Materials Science and Engineering
Eun Soo Park
Office: 33-313
Telephone: 880-7221 Email: espark@snu.ac.kr
Office hours: by appointment
2020 Fall
10. 27. 2020
1
Contents for previous class
- Binary System mixture/ solution / compound
Chapter 9
= 0 dG
Phase Diagrams - Equilibrium
2 1 0
G G G
∆ = − <
- Phase Transformation
Lowest possible value of G
No desire to change ad infinitum
3
- Solubility
• Unlimited Solubility
– Hume Rothery’ Conditions
• Similar Size
• Same Crystal Structure
• Same Valance
• Similar Electronegativity
– Implies single phase
• Limited Solubility
– Implies multiple phases
• No Solubility
- oil and water region
3
Cu-Ag Alloys Cu-Ni Alloys
complete solid solution limited solid solution
4
5
wt% Ni
20 120 0 130 0
30 40 50
110 0
L (liquid)
α (solid) T( ° C)
A
35 C 0
L: 35 wt%Ni
Cu-Ni system
• Phase diagram:
Cu-Ni system.
Adapted from Fig. 11.4, Callister & Rethwisch 9e.
• Consider
microstuctural changes that accompany the cooling of a
C 0 = 35 wt% Ni alloy
- Cooling of a Cu-Ni Alloy
35 46 32 43
α: 43 wt% Ni L: 32 wt% Ni α: 46 wt% Ni L: 35 wt% Ni B
C
E L: 24 wt% Ni
α: 36 wt% Ni
24 D 36
α: 35 wt% Ni
I. Component, Phase, Equilibrium
→ Phase diagram (Gibb’s phase rule)
II. one component phase diagram two component phase diagram
: solubility limit (Hume-Rothery Rule)
III. Isomorphous Binary Phase Diagram
: tie line, lever rule
IV. Binary-Eutectic Systems V. Binary invariant reaction
: Eutectic, Eutectoid, & Peritectic - Ternary, Quarternary phase diagram - Phase transformation
What we will learn about
6
Chapter 9: Phase Diagrams
Eutectic reaction L α + β
2 components has a special composition with a min. melting T.
IV. Binary-Eutectic Systems
7
2
2 3 2 1
1
1
1 single phase F = C - P + 1
= 2 - 1 + 1
= 2
can vary T and composition independently
2 two phase
F = C - P + 1
= 2 - 2 + 1
= 1
can vary T or composition
3 eutectic point F = C - P + 1
= 2 - 3 + 1
= 0
can’t vary T or composition For Constant Pressure,
P + F = C + 1
The Gibbs Phase Rule
8
9 Fig. 11.6, Callister & Rethwisch 9e
[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.].
• 3 single phase regions (L, α, β)
• Limited solubility:
α: mostly Cu β: mostly Ag
• T E : No liquid below T E : Composition at
temperature T E
• C E
Cu-Ag system
L (liquid)
α L + α L+ ββ
α + β
C, wt% Ag
20 40 60 80 100
200 0 1200
T(°C)
400 600 800 1000
C E
T E 8.0 779° C 71.9 91.2
cooling heating
• Eutectic reaction
L(C E ) α(C αE ) + β(C βE )
EX 1: Cu-Ag Eutectic System
10
L+ α
L+ β
α + β
200
T(°C)
18.3
C, wt% Sn
20 60 80 100
0 300
100
L (liquid)
α 183°C
61.9 97.8 β
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, determine:
-- the phases present Pb-Sn
system
EX 2: Pb-Sn Eutectic System
Answer: α + β
-- the phase compositions
-- the relative amount of each phase
150
40 C 0 11
C α
99 C β
R S
Answer: C α = 11 wt% Sn C β = 99 wt% Sn
W α = C β - C 0 C β - C α
= 99 - 40
99 - 11 = 59
88 = 0.67 S
R+S =
Wβ = C 0 - C α C β - C α R =
R+S
= 29
88 = 0.33
= 40 - 11 99 - 11 Answer:
Fig. 11.7, Callister & Rethwisch 9e.
[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
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Answer: C α = 17 wt% Sn -- the phase compositions
L+ β
α + β
200
T(°C)
C, wt% Sn
20 60 80 100
0 300
100
L (liquid)
α β
L+ α
183°C
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:
-- the phases present: Pb-Sn
system
EX 2: Pb-Sn Eutectic System
-- the relative amount of each phase
W α = C L - C 0
C L - C α = 46 - 40 46 - 17
= 6
29 = 0.21 W L = C 0 - C α
C L - C α = 23
29 = 0.79
40 C 0
46 C L 17
C α
220 R S
Answer: α + L
C L = 46 wt% Sn
Answer:
Fig. 11.7, Callister & Rethwisch 9e.
[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
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• For alloys for which C 0 < 2 wt% Sn
• Result: at room temperature
-- polycrystalline with grains of α phase having
composition C 0
Microstructural Developments in Eutectic Systems I
0
L + α
200
T( ° C)
C, wt% Sn
10 2
20 C
0300
100
L α
30
α + β
400
(room T solubility limit)
T
E(Pb-Sn
System)
α L L: C
0wt% Sn
α : C
0wt% Sn
Fig. 11.10, Callister &
Rethwisch 9e.
13
• For alloys for which
2 wt% Sn < C 0 < 18.3 wt% Sn
• Result:
at temperatures in α + β range -- polycrystalline with α grains
and small β-phase particles
Fig. 11.11, Callister &
Rethwisch 9e.
Microstructural Developments in Eutectic Systems II
Pb-Sn system L + α
200
T(°C)
C, wt% Sn
10
18.3 20 0
C
0300
100
L
α
30
α + β
400
(sol. limit at T
E) T
E2
(sol. limit at Troom)
L α L: C
0wt% Sn
α β
α : C
0wt% Sn
14
• For alloy of composition C 0 = C E
• Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of α and β phases.
Fig. 11.12, Callister &
Rethwisch 9e.
Microstructural Developments in Eutectic Systems III
Fig. 11.13, Callister & Rethwisch 9e.
(From Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, 1985.
Reproduced by permission of ASM International, Materials Park, OH.)
160 μm
Micrograph of Pb-Sn eutectic
microstructure
Pb-Sn system
L + β
α + β
200
T(°C)
C, wt% Sn
20 60 80 100
0
300
100
L
α β
L+ α
183°C
40 T
E18.3
α : 18.3 wt%Sn
97.8 β: 97.8 wt% Sn
C
E61.9
L: C
0wt% Sn
15
Lamellar Eutectic Structure
Figs. 11.13 & 11.14, Callister & Rethwisch 9e.
(Fig. 11.13 from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.)
16
• For alloys for which 18.3 wt% Sn < C 0 < 61.9 wt% Sn
• Result: α phase particles and a eutectic microconstituent
Microstructural Developments in Eutectic Systems IV
18.3 61.9
S R
97.8 R S
primary α eutectic α
eutectic β
W L = (1- W α) = 0.50 C α = 18.3 wt% Sn C L = 61.9 wt% Sn
S R + S
W α = = 0.50
• Just above T E :
• Just below T E : C α = 18.3 wt% Sn C β = 97.8 wt% Sn
S R + S
W α = = 0.73 W β = 0.27
Fig. 11.15, Callister &
Rethwisch 9e.
Pb-Sn system
L+ β
200
T(°C)
C, wt% Sn
20 60 80 100
0
300
100
L
α β
L+ α
40
α + β
T
EL: C 0 wt% Sn α L
L α
17
L+ α
L+ β α + β
200
C, wt% Sn
20 60 80 100
0 300
100
L
α β
T
E40
(Pb-Sn System)
Hypoeutectic & Hypereutectic
Fig. 11.7, Callister & Rethwisch 9e.
[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B.Massalski (Editor-in-Chief), 1990.
Reprinted by permission of ASM International, Materials Park, OH.]
160 μm eutectic micro-constituent
Fig. 11.13, Callister &
Rethwisch 9e.
hypereutectic: (illustration only)
β β β
β β
β
Adapted from Fig. 11.16, Callister & Rethwisch 9e.
(Illustration only)
(Figs. 11.13 and 11.16 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, 1985.
Reproduced by permission of ASM International, Materials Park, OH.)
175 μm
α α
α α α
α
hypoeutectic: C
0= 50 wt% Sn
Fig. 11.16, Callister &
Rethwisch 9e.
T(°C)
61.9 eutectic
eutectic: C
0= 61.9 wt% Sn
Metatectic reaction: β ↔ L + α Ex. Co-Os, Co-Re, Co-Ru 18
19
Syntectic reaction Liquid 1 +Liquid 2 ↔ α
L 1 +L 2 α
K-Zn, Na-Zn,
K-Pb, Pb-U, Ca-Cd
19
20
21
• Eutectoid – one solid phase transforms to two other solid phases
S 2 S 1 +S 3
γ α + Fe 3 C (For Fe-C, 727°C, 0.76 wt% C)
intermetallic compound - cementite
cool heat
V. Eutectic, Eutectoid, & Peritectic
• Eutectic - liquid transforms to two solid phases
L cool α + β (For Pb-Sn, 183°C, 61.9 wt% Sn)
heat
cool heat
• Peritectic - liquid and one solid phase transform to a second solid phase
S 1 + L S 2
δ + L γ (For Fe-C, 1493°C, 0.16 wt% C)
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a. Eutectoid & Peritectic
Cu-Zn Phase diagram
Fig. 11.20, Callister & Rethwisch 9e.
[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 2, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
Eutectoid transformation δ γ + e
Peritectic transformation γ + L δ
b. Congruent vs Incongruent
Congruent phase transformations: no compositional change associated with transformation
Incongruent phase transformation:
at least one phase will experience change in composition
Examples:
• Allotropic phase transformations
• Melting points of pure metals
• Congruent Melting Point
Examples:
• Melting in isomorphous alloys
• Eutectic reactions
• Pertectic Reactions
• Eutectoid reactions
Ni Ti
23
α iron Ferrite
(BCC)
soft & ductile γ iron
austenite (FCC)
Cementite (Fe 3 C) hard & brittle
A; eutectic B; eutectoid
B
A
c. Fe-C phase diagram
C concentration 0.008w% 2.14w% 6.7w%
iron steel cast iron
24
25
Iron-Carbon (Fe-C) Phase Diagram
• 2 important points
- Eutectoid (B):
γ ⇒ α +Fe 3 C
- Eutectic (A):
L ⇒ γ +Fe 3 C
Fig. 11.23, Callister
& Rethwisch 9e.
[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
Fe 3 C ( c em ent it e)
1600 1400 1200
1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite)
γ +L
γ +Fe 3 C
α+Fe 3 C δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C = T eutectoid
4.30
Result: Pearlite = alternating layers of α and Fe
3C phases
120 μm
Fig. 11.26, Callister & Rethwisch 9e.
(From Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, 1985.
Reproduced by permission of ASM International, Materials Park, OH.)
0.76
B
γ γ γ γ
A L+Fe
3 C
Fe 3 C (cementite-hard)
α (ferrite-soft)
Pearlite microstructure
26
27
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite) γ+L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
0 .7 6
Hypoeutectoid Steel
Adapted from Figs. 11.23 and 11.28, Callister &
Rethwisch 9e.
[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.
Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
Adapted from Fig. 11.29, Callister & Rethwisch 9e.
(Photomicrograph courtesy of Republic Steel Corporation.)
proeutectoid ferrite pearlite
100 μm Hypoeutectoid steel α
pearlite
γ γ γ α γ
α α
γ γ γ γ
γ γ γ γ
28
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite) γ+L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
0 .7 6
Hypoeutectoid Steel
γ γ γ α γ
α α
s r
W α = s/(r + s) W γ =(1 - W α )
R S
α
pearlite
W pearlite = W γ W α ’ = S/(R + S) W Fe =(1 – W α’ )
3 C
Adapted from Fig. 11.29, Callister & Rethwisch 9e.
(Photomicrograph courtesy of Republic Steel Corporation.)
proeutectoid ferrite pearlite
100 μm Hypoeutectoid steel
Adapted from Figs. 11.23 and 11.28, Callister &
Rethwisch 9e.
[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.
Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
Hypoeutectoid Steel
Proeutect ferrite oid
Pearlite
dark – ferrite light - cementite
29
30
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite) γ+L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
Hypereutectoid Steel
0. 76
C
0Fe
3C
γ γ γ γ γ γ γ γ γ γ γ γ
Adapted from Fig. 11.32, Callister & Rethwisch 9e.
(Copyright 1971 by United States Steel Corporation.)
proeutectoid Fe 3 C
60 μm Hypereutectoid steel
pearlite
pearlite
Adapted from Figs. 11.23 and 11.31, Callister &
Rethwisch 9e.
[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.
Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
31
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite) γ+L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
Adapted from Fig. 11.32, Callister & Rethwisch 9e.
(Copyright 1971 by United States Steel Corporation.)
proeutectoid Fe 3 C
60 μm Hypereutectoid steel
pearlite
Adapted from Figs. 11.23 and 11.31, Callister &
Rethwisch 9e.
[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.
Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
Hypereutectoid Steel
0. 76 C
0pearlite Fe
3C
γ γ γ γ
v x
V X
W pearlite = W γ W α = X/(V + X)
W Fe =(1 - W α )
3
C’
W =(1-W γ ) W γ =x/(v + x)
Fe
3C
Hypereutectoid Steel
Proeutectoid cementite Pearlite
32
Evolution of microstructure of hypoeutectoid and hypereutectoid steels during cooling
33
34
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 Fe 3 C and ferrite ( α).
b) The amount of cementite (in grams) that forms in 100 g of steel.
c) The amounts of pearlite and proeutectoid
ferrite ( α) in the 100 g.
35
Solution to Example Problem
b) Using the lever rule with the tie line shown
a) Using the RS tie line just below the eutectoid
C α = 0.022 wt% C C Fe
3 C = 6.70 wt% C
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800 600
400 0 1 2 3 4 5 6 6.7
L
γ
(austenite) γ+L
γ + Fe
3C α + Fe
3C
L+Fe
3C
δ
C, wt% C 1148°C
T(°C)
727°C
C 0
R S
C Fe C C α 3
Amount of Fe 3 C in 100 g
= (100 g)W Fe
3 C
= (100 g)(0.057) = 5.7 g
Fig. 11.23, Callister & Rethwisch 9e.
[From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T.
B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
36
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800 600
400 0 1 2 3 4 5 6 6.7
L
γ
(austenite) γ+L
γ + Fe
3C α + Fe
3C
L+Fe
3C
δ
C, wt% C 1148°C
T(°C)
727°°C
Solution to Example Problem (cont.)
c) Using the VX tie line just above the eutectoid and realizing that
C 0 = 0.40 wt% C C α = 0.022 wt% C
C pearlite = C γ = 0.76 wt% C
C 0 V X
C γ C α
Amount of pearlite in 100 g
= (100 g)W pearlite
= (100 g)(0.512) = 51.2 g
Fig. 11.23, Callister & Rethwisch 9e.
[From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T.
B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]
37
Alloying with Other Elements
• T eutectoid changes:
Fig. 11.33, Callister & Rethwisch 9e
.(From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.)
T Eu te c to id (º C)
wt. % of alloying elements
Ti
Ni
Mo Si
W
Cr Mn
• C eutectoid changes:
Fig. 11.34,Callister & Rethwisch 9e.
(From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.)