lecture 4

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Phase Diagrams

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2

Phase Equilibria: Solubility Limit

Question: What is thesolubility limit for sugar inwater at 20ºC?

Answer: 65 wt% sugar.At 20ºC, if C < 65 wt% sugar: syrupAt 20ºC, if C > 65 wt% sugar:

syrup + sugar

65

• Solubility Limit:Maximum concentration forwhich only a single phasesolution exists.

Sugar/Water Phase Diagram

Suga

r

Tem

pera

ture

(ºC

)

0 20 40 60 80 100C = Composition (wt% sugar)

L(liquid solution

i.e., syrup)

SolubilityLimit L

(liquid)+S

(solidsugar)20

40

60

80

100

Wat

er

Adapted from Fig. 9.1,Callister & Rethwisch 8e.

• Solution – solid, liquid, or gas solutions, single phase• Mixture – more than one phase

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3

• Components:The elements or compounds which are present in the alloy

(e.g., Al and Cu)

• Phases:The physically and chemically distinct material regionsthat form (e.g., α and β).

Aluminum-CopperAlloy

Components and Phases

α (darkerphase)

β (lighterphase)

Adapted from chapter-opening photograph,Chapter 9, Callister,Materials Science &Engineering: AnIntroduction, 3e.

4

70 80 1006040200

Tem

pera

ture

(ºC

)

C = Composition (wt% sugar)

L(liquid solution

i.e., syrup)

20

100

40

60

80

0

L(liquid)

+S

(solidsugar)

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-sugarsystem


D (100ºC,C = 90)2 phases

B (100ºC,C = 70)1 phase

A (20ºC,C = 70)2 phases

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Cu-Ni phase diagram

Isomorphous Binary Phase Diagram

• Phase diagram:Cu-Ni system.

• System is:

Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adaptedfrom Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASMInternational, Materials Park, OH (1991).

-- binaryi.e., 2 components:Cu and Ni.

-- isomorphousi.e., completesolubility of onecomponent inanother; α phasefield extends from0 to 100 wt% Ni.

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(ºC)

L (liquid)

α(FCC solidsolution)

6

Phase Diagrams• Indicate phases as a function of T, C, and P.• For this course:

- binary systems: just 2 components.- independent variables: T and C (P = 1 atm is almost always used).

PhaseDiagramfor Cu-Nisystem

Adapted from Fig. 9.3(a), Callister &Rethwisch 8e. (Fig. 9.3(a) is adapted fromPhase Diagrams of Binary Nickel Alloys,P. Nash (Ed.), ASM International,Materials Park, OH (1991).

• 2 phases:L (liquid)α (FCC solid solution)

• 3 different phase fields:LL + αα

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(ºC)

L (liquid)


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wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(ºC)

L (liquid)


Cu-Niphase

diagram

7

Phase Diagrams:Determination of phase(s) present

• Rule 1: If we know T and Co, then we know:-- which phase(s) is (are) present.

• Examples:A(1100ºC, 60 wt% Ni):

1 phase: α

B(1250ºC, 35 wt% Ni):2 phases: L + α

B(1

250º

C,3

5)A(1100ºC,60)

Adapted from Fig. 9.3(a), Callister &Rethwisch 8e. (Fig. 9.3(a) is adapted fromPhase Diagrams of Binary Nickel Alloys,P. Nash (Ed.), ASM International,Materials Park, OH (1991).

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wt% Ni20

1200

1300

T(ºC)

L (liquid)

α(solid)

30 40 50

Cu-Nisystem

Phase Diagrams:Determination of phase compositions

• Rule 2: If we know T and C0, then we can determine:-- the composition of each phase.

• Examples:

TAA

35C0

32CL

At TA = 1320ºC:Only Liquid (L) presentCL = C0 ( = 35 wt% Ni)

At TB = 1250ºC:Both and L presentCL = C liquidus ( = 32 wt% Ni)C = Csolidus ( = 43 wt% Ni)

At TD = 1190ºC:Only Solid (α) presentC = C0 ( = 35 wt% Ni)

Consider C0 = 35 wt% Ni

DTD

tie line

4C3

Adapted from Fig. 9.3(a), Callister & Rethwisch 8e.(Fig. 9.3(a) is adapted from Phase Diagrams ofBinary Nickel Alloys, P. Nash (Ed.), ASMInternational, Materials Park, OH (1991).

BTB

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• Rule 3: If we know T and C0, then can determine:-- the weight fraction of each phase.

• Examples:

At TA : Only Liquid (L) presentWL = 1.00, Wα = 0

At TD : Only Solid () presentWL = 0, Wα = 1.00

Phase Diagrams:Determination of phase weight fractions

wt% Ni20

1200

1300

T(ºC)

L (liquid)

α(solid)

30 40 50

Cu-Nisystem

TAA

35C0

32CL

BTB

DTD

tie line

4Cα3

R SAt TB : Both and L present

73.032433543

= 0.27

WL = SR +S

Wα = RR +S

Consider C0 = 35 wt% Ni

Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) isadapted from Phase Diagrams of Binary Nickel Alloys, P. Nash(Ed.), ASM International, Materials Park, OH (1991).

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Tie line – connects the phases in equilibrium with each other –also sometimes called an isotherm

The Lever Rule

What fraction of each phase?Think of the tie line as a lever(teeter-totter)

ML M

R S

M x S ML x R

L

L

LL

LL CC

CCSR

RWCCCC

SRS

MMMW

00

wt% Ni20

1200

1300

T(ºC)

L (liquid)

α(solid)

30 40 50

BTB

tie line

C0CL C

SR

Adapted from Fig. 9.3(b),Callister & Rethwisch 8e.

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11wt% Ni

20

1200

1300

30 40 50110 0

L (liquid)

α(solid)

T(ºC)

A

35C0

L: 35wt%NiCu-Ni

system

• Phase diagram:Cu-Ni system.


• Considermicrostucturalchanges thataccompany thecooling of aC0 = 35 wt% Ni alloy

Ex: Cooling of a Cu-Ni Alloy

46354332

α: 43 wt% NiL: 32 wt% Ni

Bα: 46 wt% NiL: 35 wt% Ni

C

EL: 24 wt% Ni

α: 36 wt% Ni

24 36D

• Slow rate of cooling:Equilibrium structure

• Fast rate of cooling:Cored structure

First α to solidify:46 wt% NiLast α to solidify:< 35 wt% Ni

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• Cα changes as we solidify.• Cu-Ni case: First α to solidify has Cα = 46 wt% Ni.

Last α to solidify has Cα = 35 wt% Ni.

Cored vs Equilibrium Structures

Uniform Cα:35 wt% Ni

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2 componentshas a special compositionwith a min. melting T.


Binary-Eutectic Systems

• 3 single phase regions(L, α, β)

• Limited solubility:α: mostly Cuβ: mostly Ag

• TE : No liquid below TE

: Composition attemperature TE

• CE

Ex.: Cu-Ag systemCu-Agsystem

L (liquid)

α L + α L+β β

α + β

C, wt% Ag20 40 60 80 1000

200

1200T(ºC)

400

600

800

1000

CE

TE 8.0 71.9 91.2779ºC

Ag) wt%1.29( Ag) wt%.08( Ag) wt%9.71( Lcooling

heating

• Eutectic reactionL(CE) (CE) + (CE)

14

L+αL+β

α + β

200

T(ºC)

18.3

C, wt% Sn20 60 80 1000

300

100

L (liquid)

α 183ºC61.9 97.8

β

• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, determine:-- the phases present

Pb-Snsystem

EX 1: Pb-Sn Eutectic System

Answer: α + β-- the phase compositions

-- the relative amountof each phase

150

40C0

11C

99C

SR

Answer: Cα = 11 wt% SnCβ = 99 wt% Sn

W

=C - C0C - C

= 99 - 4099 - 11 = 59

88 = 0.67

SR+S =

W =C0 - CC - C

=RR+S

= 2988

= 0.33= 40 - 1199 - 11

Answer:


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Answer: Cα = 17 wt% Sn-- the phase compositions

L+β

α + β

200

T(ºC)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

α βL+α

183ºC

• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine:-- the phases present: Pb-Sn

system

EX 2: Pb-Sn Eutectic System

-- the relative amountof each phase

Wα =CL - C0

CL - C=

46 - 4046 - 17

=6

29 = 0.21

WL =C0 - CCL - C

=2329 = 0.79

40C0

46CL

17C

220SR

Answer: α + L

CL = 46 wt% Sn

Answer:


16

• For alloys for whichC0 < 2 wt% Sn

• Result: at room temperature-- polycrystalline with grains of

α phase havingcomposition C0

Microstructural Developmentsin Eutectic Systems I

0

L+ α200

T(ºC)

C , wt% Sn10

2

20C0

300

100

L

α

30

α+β

400

(room T solubility limit)

TE(Pb-SnSystem)

αL

L: C0 wt% Sn

α: C0 wt% Sn


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• For alloys for which2 wt% Sn < C0 < 18.3 wt% Sn

• Result:at temperatures in α + β range-- polycrystalline with α grains

and small β-phase particles


Microstructural Developmentsin Eutectic Systems II

Pb-Snsystem

L + α

200

T(ºC)

C, wt% Sn10

18.3

200C0

300

100

L

α

30

α+ β

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

Lα

L: C0 wt% Sn

αβ

α: C0 wt% Sn

18

• For alloy of composition C0 = CE• Result: Eutectic microstructure (lamellar structure)

-- alternating layers (lamellae) of α and β phases.

Adapted fromFig. 9.13,Callister &Rethwisch 8e.

Microstructural Developmentsin Eutectic Systems III


160m

Micrograph of Pb-Sneutecticmicrostructure

Pb-Snsystem

L

α

200

T(ºC)

C, wt% Sn20 60 80 1000

300

100

L

α βL+α

183ºC

40

TE

18.3

: 18.3 wt%Sn

97.8

: 97.8 wt% Sn

CE61.9

L: C0 wt% Sn

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Lamellar Eutectic Structure

Adapted from Figs. 9.14 & 9.15, Callister & Rethwisch 8e.

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• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn• Result: α phase particles and a eutectic microconstituent

Microstructural Developmentsin Eutectic Systems IV

18.3 61.9

SR

97.8

SR

primary αeutectic α

eutectic β

WL = (1- Wα) = 0.50

Cα = 18.3 wt% Sn

CL = 61.9 wt% SnS

R + SWα = = 0.50

• Just above TE :

• Just below TE :C = 18.3 wt% Sn

C = 97.8 wt% SnS

R + SW= = 0.73

W = 0.27Adapted from Fig. 9.16,Callister & Rethwisch 8e.

Pb-Snsystem

L+200

T(ºC)

C, wt% Sn

20 60 80 1000

300

100

L

α βL+

40

+

TE

L: C0 wt% Sn LL

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L+αL+β

α + β

200

C, wt% Sn20 60 80 1000

300

100

L

α βTE

40

(Pb-SnSystem)

Hypoeutectic & Hypereutectic

Adapted from Fig. 9.8,Callister & Rethwisch 8e.(Fig. 10.8 adapted fromBinary Phase Diagrams,2nd ed., Vol. 3, T.B.Massalski (Editor-in-Chief),ASM International,Materials Park, OH, 1990.)

160 µmeutectic micro-constituent


hypereutectic: (illustration only)

β

βββ

β

β

Adapted from Fig. 9.17,Callister & Rethwisch 8e.(Illustration only)

(Figs. 9.14 and 9.17from MetalsHandbook, 9th ed.,Vol. 9,Metallography andMicrostructures,American Society forMetals, MaterialsPark, OH, 1985.)

175 µm

α

α

α

αα

α

hypoeutectic: C0 = 50 wt% Sn

Adapted fromFig. 9.17, Callister &Rethwisch 8e.

T(ºC)

61.9eutectic

eutectic: C0 = 61.9wt% Sn

22

• Eutectoid – one solid phase transforms to two other solidphasesS2 S1+S3

+ Fe3C (For Fe-C, 727ºC, 0.76 wt% C)

intermetallic compound- cementite

coolheat

Eutectic, Eutectoid, & Peritectic

• Eutectic - liquid transforms to two solid phasesL + (For Pb-Sn, 183ºC, 61.9 wt% Sn)cool

heat

coolheat

• Peritectic - liquid and one solid phase transform to asecond solid phaseS1 + L S2

+ L (For Fe-C, 1493ºC, 0.16 wt% C)

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Eutectoid & Peritectic

Cu-Zn Phase diagram

Adapted from Fig. 9.21,Callister & Rethwisch 8e.Eutectoid transformation +

Peritectic transformation + L

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Iron-Carbon (Fe-C) Phase Diagram• 2 important

points

- Eutectoid (B):γ α +Fe3C

- Eutectic (A):L γ +Fe3C


Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ(austenite)

γ+L

γ+Fe3C

α+Fe3C

δ

(Fe) C, wt% C

1148ºC

T(ºC)

α 727ºC = Teutectoid

4.30Result: Pearlite =alternating layers ofα and Fe3C phases

120 µm

(Adapted from Fig. 9.27,Callister & Rethwisch 8e.)

0.76

Bγ γ

γγ

A L+Fe3C

Fe3C (cementite-hard)α (ferrite-soft)

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25

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ(austenite)

γ+L

γ + Fe3C

α+ Fe3C

L+Fe3C

δ

(Fe) C, wt% C

1148ºC

T(ºC)

α727ºC

(Fe-CSystem)

C0

0.76

Hypoeutectoid Steel

Adapted from Figs. 9.24and 9.29,Callister &Rethwisch 8e.(Fig. 9.24 adapted fromBinary Alloy PhaseDiagrams, 2nd ed., Vol.1, T.B. Massalski (Ed.-in-Chief), ASM International,Materials Park, OH,1990.)

Adapted from Fig. 9.30, Callister & Rethwisch 8e.proeutectoid ferritepearlite

100 µm Hypoeutectoidsteel

α

pearlite

γγ γ

γααα

γγγ γ

γ γγγ

26

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ(austenite)

γ+L

γ + Fe3C

α+ Fe3C

L+Fe3C

δ

(Fe) C, wt% C

1148ºC

T(ºC)

α727ºC

(Fe-CSystem)

C0

0.76

Hypoeutectoid Steel

γγ γ

γααα

srWα = s/(r + s)Wγ =(1 - Wα)

R Sα

pearlite

Wpearlite = Wγ

Wα’ = S/(R + S)W =(1 – Wα’)Fe3C

Adapted from Figs.9.24 and9.29,Callister &Rethwisch 8e.(Fig. 9.24 adaptedfrom Binary AlloyPhase Diagrams, 2nded., Vol. 1, T.B.Massalski (Ed.-in-Chief), ASMInternational,Materials Park, OH,1990.)

Adapted from Fig. 9.30, Callister & Rethwisch 8e.

proeutectoid ferritepearlite

100 µm Hypoeutectoidsteel

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27

Hypereutectoid Steel

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ(austenite)

γ+L

γ +Fe3C

α +Fe3C

L+Fe3C

δ

(Fe) C, wt%C

1148ºC

T(ºC)

α

Adapted from Figs.9.24 and 9.32,Callister& Rethwisch 8e. (Fig.9.24 adapted fromBinary Alloy PhaseDiagrams, 2nd ed., Vol.1, T.B. Massalski (Ed.-in-Chief), ASMInternational, MaterialsPark, OH, 1990.)

(Fe-CSystem)

0.76 C0

Fe3C

γγγ γ

γγγ γ

γγγ γ


proeutectoid Fe3C

60 µmHypereutectoidsteel

pearlite

pearlite

28

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ(austenite)

γ+L

γ +Fe3C

α +Fe3C

L+Fe3C

δ

(Fe) C, wt%C

1148ºC

T(ºC)

α

Hypereutectoid Steel

(Fe-CSystem)

0.76 C0

pearlite

Fe3C

γγγ γ

xv

V X

Wpearlite = Wγ

Wα = X/(V +X)

W =(1 - Wα)Fe3C’

W =(1-Wγ)Wγ =x/(v + x)

Fe3C


proeutectoid Fe3C

60 µmHypereutectoidsteel

pearlite

Adapted from Figs.9.24 and9.32,Callister &Rethwisch 8e. (Fig.9.24 adapted fromBinary Alloy PhaseDiagrams, 2nd ed.,Vol. 1, T.B.Massalski (Ed.-in-Chief), ASMInternational,Materials Park, OH,1990.)

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Alloying with Other Elements

• Teutectoid changes:

Adapted from Fig. 9.34,Callister & Rethwisch 8e.(Fig. 9.34 from Edgar C. Bain, Functions of theAlloying Elements in Steel, American Society forMetals, 1939, p. 127.)

T Eut

ecto

id(º

C)

wt. % of alloying elements

Ti

Ni

Mo SiW

Cr

Mn

• Ceutectoid changes:

Adapted from Fig. 9.35,Callister & Rethwisch 8e.(Fig. 9.35 from Edgar C. Bain, Functions of theAlloying Elements in Steel, American Society forMetals, 1939, p. 127.)

wt. % of alloying elements

Ceu

tect

oid

(wt%

C)

Ni

Ti

Cr

SiMnWMo

30

• 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 phasegiven 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 includeeutectic, eutectoid, and peritectic.

Summary

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Phase Transformations

32

Phase TransformationsNucleation

nuclei (seeds) act as templates on which crystals grow for nucleusto form rate of addition of atoms to nucleus must be faster than rateof loss once nucleated, growth proceeds until equilibrium isattained

Driving force to nucleate increases as we increase T– supercooling (eutectic, eutectoid)– superheating (peritectic)

Small supercooling slow nucleation rate - few nuclei - large crystals

Large supercooling rapid nucleation rate - many nuclei - small crystals

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Solidification: Nucleation Types

• Homogeneous nucleationnuclei form in the bulk of liquid metal requiresconsiderable supercooling (typically 80-300ºC)

• Heterogeneous nucleation– much easier since stable “nucleating surface” is

already present — e.g., mold wall, impurities inliquid phase

– only very slight supercooling (0.1-10ºC)

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r* = critical nucleus: for r < r* nuclei shrink; for r >r* nuclei grow (to reduce energy)Adapted from Fig.10.2(b), Callister & Rethwisch 8e.

Homogeneous Nucleation & Energy Effects

∆GT = Total Free Energy= ∆GS + ∆GV

Surface Free Energy - destabilizesthe nuclei (it takes energy to makean interface)

24 rGS

γ = surface tension

Volume (Bulk) Free Energy –stabilizes the nuclei (releases energy)

GrGV3

34

volumeunitenergyfreevolume

G

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Solidification

THTr

f

m

2*

Note: Hf and are weakly dependent on T

r* decreases as T increases

For typical T r* ~ 10 nm

Hf = latent heat of solidificationTm = melting temperatureγ = surface free energy

∆T = Tm - T = supercooling

r* = critical radius

36

Transformations & Undercooling

• For transf. to occur, must cool tobelow 727ºC(i.e., must “undercool”)

• Eutectoid transf. (Fe-Fe3C system):γ α + Fe3C

0.76 wt% C0.022 wt% C

6.7 wt% C

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ(austenite)

γ+L

γ +Fe3C

α +Fe3C

L+Fe3C

δ

(Fe) C, wt%C

1148ºC

T(ºC)

αferrite

727ºC

Eutectoid:Equil. Cooling: Ttransf. = 727ºC

∆TUndercooling by Ttransf. < 727C

0.76

0.02

2

Adapted from Fig.9.24,Callister &Rethwisch 8e. (Fig. 9.24adapted from BinaryAlloy Phase Diagrams,2nd ed., Vol. 1, T.B.Massalski (Ed.-in-Chief),ASM International,Materials Park, OH,1990.)

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• Eutectoid composition, C0 = 0.76 wt% C• Begin at T > 727ºC• Rapidly cool to 625ºC• Hold T (625ºC) constant (isothermal treatment)

Adapted from Fig.10.14,Callister & Rethwisch8e. (Fig. 10.14 adaptedfrom H. Boyer (Ed.) Atlas ofIsothermal Transformationand Cooling TransformationDiagrams, AmericanSociety for Metals, 1997, p.28.)

Austenite-to-Pearlite Isothermal Transformation

400

500

600

700

Austenite (stable) TE (727ºC)Austenite(unstable)

Pearlite

T(ºC)

1 10 102 103 104 105

time (s)

γ γ

γγ γ

γ

38

10 103 105

time (s)10-1

400

600

800

T(ºC

)

Austenite (stable)

200

P

B

TEA

A

Bainite: Another Fe-Fe3C Transformation Product• Bainite:

-- elongated Fe3C particles inα-ferrite matrix

-- diffusion controlled• Isothermal Transf. Diagram,

C0 = 0.76 wt% C

Adapted from Fig. 10.18, Callister &Rethwisch 8e.

Adapted from Fig. 10.17, Callister &Rethwisch 8e. (Fig. 10.17 from MetalsHandbook, 8th ed., Vol. 8, Metallography,Structures, and Phase Diagrams, AmericanSociety for Metals, Materials Park, OH,1973.)

Fe3C(cementite)

5 µm

a(ferrite)

100% bainite

100% pearlite

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• Martensite:-- γ(FCC) to Martensite (BCT)

Adapted from Fig. 10.21, Callister &Rethwisch 8e. (Fig. 10.21 courtesyUnited States Steel Corporation.)


Martensite:A Nonequilibrium Transformation Product

Martensite needlesAustenite

60m

40

(FCC) (BCC) + Fe3C

Martensite Formationslow cooling

temperingquench

M (BCT)

Martensite (M) – single phase– has body centered tetragonal (BCT)

crystal structure

Diffusionless transformationBCT if C0 > 0.15 wt% CBCT few slip planes hard, brittle

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Tempered Martensite

• tempered martensite less brittle than martensite• tempering reduces internal stresses caused by quenching

Adapted from Fig.10.33, Callister &Rethwisch 8e. (Fig.10.33 copyright byUnited States SteelCorporation, 1971.)

• tempering decreases TS, YS but increases %RA• tempering produces extremely small Fe3C particles surrounded by α.

Adapted fromFig. 10.34,Callister &Rethwisch 8e.(Fig. 10.34adapted fromFig. furnishedcourtesy ofRepublic SteelCorporation.)

9µm

YS(MPa)TS(MPa)

800

1000

1200

1400

1600

1800

30405060

200 400 600Tempering T (ºC)

%RA

TS

YS

%RA

Heat treat martensite to form tempered martensite

42

Phase Transformations of AlloysEffect of adding other elementsChange transition temp.

Cr, Ni, Mo, Si, Mn

retard + Fe3Creaction (and formation of

pearlite, bainite)


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43


Continuous CoolingTransformation Diagrams

Conversion of isothermaltransformation diagram tocontinuous coolingtransformation diagram

Cooling curve

44

Summary of Possible Transformations


Austenite (γ)

Pearlite(α + Fe3C layers + aproeutectoid phase)

slowcool

Bainite(α + elong. Fe3C particles)

moderatecool

Martensite(BCT phasediffusionless

transformation)

rapidquench

TemperedMartensite(α + very fine

Fe3C particles)

reheat

Stre

ngth

Duc

tility

MartensiteT Martensite

bainitefine pearlite

coarse pearlitespheroidite

General Trends

lecture 4

Documents

altering c

c liquidus

composition wt

ll wt

cu phases

callister rethwisch

single phase mixture

different phase fields