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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

(2)

2

• 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)

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

(4)

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

4

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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.

(6)

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

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= 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

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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

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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

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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

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b: What are binary systems?

“Mixture vs. Solution vs. Compound”

Multi-component system:

11

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* 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

2

O)

: Equilibrium depends on pressure and temperature.

: Equilibrium depends on not only pressure and temperature but also composition.

Alluvial mining

Winnowing

wash rice Select egg

12

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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

13

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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.]

(15)

15

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

15

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16

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.

(17)

17

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

(18)

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.

18

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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%

19

d. Criteria for Solid Solubility: Hume-Rothery Rule

(20)

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

20

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Cu-Ag Alloys Cu-Ni Alloys

complete solid solution limited solid solution

21

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22

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

-7

at%.) that it is difficult to detect experimentally, but there will always be a measure of solubility.

22

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23

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24

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25

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26

• 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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27

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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28

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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29

• 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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30

• 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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31

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

(32)

• 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

32

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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.

33

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34

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.

(35)

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 Learningis a trademark used herein under license.

Mechanical Properties: Cu-Ni System

35

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36

• 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

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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

(38)

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

(39)

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

(40)

Cu-Ag Alloys Cu-Ni Alloys

complete solid solution limited solid solution

4

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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

(42)

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

(43)

Eutectic reaction L  α + β

2 components has a special composition with a min. melting T.

IV. Binary-Eutectic Systems

7

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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

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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

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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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11

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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12

• 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

0

300

100

L α

30

α + β

400

(room T solubility limit)

T

E

(Pb-Sn

System)

α L L: C

0

wt% Sn

α : C

0

wt% Sn

Fig. 11.10, Callister &

Rethwisch 9e.

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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

0

300

100

L

α

30

α + β

400

(sol. limit at T

E

) T

E

2

(sol. limit at Troom)

L α L: C

0

wt% Sn

α β

α : C

0

wt% Sn

(50)

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

E

18.3

α : 18.3 wt%Sn

97.8 β: 97.8 wt% Sn

C

E

61.9

L: C

0

wt% Sn

(51)

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.)

(52)

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

E

L: C 0 wt% Sn α L

L α

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17

L+ α

L+ β α + β

200

C, wt% Sn

20 60 80 100

0 300

100

L

α β

T

E

40

(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

(54)

Metatectic reaction: β ↔ L + α Ex. Co-Os, Co-Re, Co-Ru 18

(55)

19

Syntectic reaction Liquid 1 +Liquid 2 ↔ α

L 1 +L 2 α

K-Zn, Na-Zn,

K-Pb, Pb-U, Ca-Cd

19

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20

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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)

(58)

22

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

(59)

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

(60)

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

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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

3

C 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)

(62)

Pearlite microstructure

26

(63)

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

(64)

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.]

(65)

Hypoeutectoid Steel

Proeutect ferrite oid

Pearlite

dark – ferrite light - cementite

29

(66)

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

0

Fe

3

C

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.]

(67)

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

0

pearlite Fe

3

C

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

3

C

(68)

Hypereutectoid Steel

Proeutectoid cementite Pearlite

32

(69)

Evolution of microstructure of hypoeutectoid and hypereutectoid steels during cooling

33

(70)

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.

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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

3

C α + Fe

3

C

L+Fe

3

C

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.]

(72)

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

3

C α + Fe

3

C

L+Fe

3

C

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.]

(73)

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.)

wt. % of alloying elements C eut ec toi d (w t% C )

Ni

Ti

Cr Si

W Mn

Mo

(74)

Ternary Eutectic System

(with Solid Solubility)

38

(75)

Ternary Eutectic System

(with Solid Solubility)

39

(76)

40

Ternary Eutectic System

(with Solid Solubility)

40

(77)

Ternary Eutectic System

(with Solid Solubility)

41

(78)

Ternary Eutectic System

(with Solid Solubility)

42

(79)

43

Ternary Eutectic System

(with Solid Solubility)

43

(80)

44

Ternary Eutectic System

(with Solid Solubility)

44

(81)

Ternary Eutectic System

(with Solid Solubility)

45

(82)

Ternary Eutectic System

(with Solid Solubility)

T= ternary eutectic temp.

L+ β+γ

L+ α+γ L+ α+β

46

(83)

Ternary Eutectic System

(with Solid Solubility)

http://www.youtube.com/watch?v=yzhVomAdetM 47

(84)

Isothermal section (T A > T > T B )

THE EUTECTIC EQUILIBRIUM (l = α + β + γ)

(a) T A > T >T B (b) T = e 1 (c) e 1 > T > e 3 (d) T = e 3

(g) T A = E

(e) T = e 2 (f) e 2 > T > E (h) E = T 48

(85)

Vertical section Location of vertical section

THE EUTECTIC EQUILIBRIUM (l = α + β + γ)

49

(86)

Vertical section Location of vertical section

THE EUTECTIC EQUILIBRIUM (l = α + β + γ)

50

(87)

Vertical section Location of vertical section

THE EUTECTIC EQUILIBRIUM (l = α + β + γ)

(88)

52

Vertical section Location of vertical section

THE EUTECTIC EQUILIBRIUM (l = α + β + γ)

(89)

< Quaternary phase Diagrams >

Four components: A, B, C, D

A difficulty of four-dimensional geometry

→ further restriction on the system Most common figure:

“ equilateral tetrahedron “ Assuming isobaric conditions, Four variables: X A , X B , X C and T

4 pure components 6 binary systems 4 ternary systems

A quarternary system

(90)

Microstructure-Properties Relationships

Microstructure Properties

Alloy design &

Processing Performance

“ Phase Transformation ”

“Tailor-made Materials Design”

down to atomic scale

54

(91)

(Ch1) Thermodynamics and Phase Diagrams (Ch2) Diffusion: Kinetics

(Ch3) Crystal Interface and Microstructure

(Ch4) Solidification: Liquid → Solid

(Ch5) Diffusional Transformations in Solid: Solid → Solid (Ch6) Diffusionless Transformations: Solid → Solid

Background to understand

phase

transformation

Representative Phase

transformation

Contents _Phase transformation course

55

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56

• 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

참조

관련 문서

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