Ch 10 Lecture

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Chapter 10 - 1

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 10: Phase Diagrams

Phase BPhase A

Nickel atom

Copper atom

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Chapter 10 - 2

Phase Equilibria: Solubility Limit

Question: What is the

solubility limit for sugar inwater at 20C?

Answer: 65 wt% sugar.At 20C, if C< 65 wt% sugar:syrup

At 20C,if C> 65 wt% sugar:syrup + sugar

65

Solubility Limit:Maximum concentration forwhich only a single phase

solution exists.

Sugar/Water Phase Diagram

S

ugar

Temperature(C)

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

L (liquid solution

i.e., syrup)

SolubilityLimit L

(liquid)

+S

(solidsugar)20

406080

100

W

ater

Adapted from Fig. 10.1,

Callister & Rethwisch 4e.

Solutionsolid, liquid, or gas solutions, single phase Mixturemore than one phase

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Chapter 10 - 3

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., aand b).

Aluminum-

Copper

Alloy

Components and Phases

a (darkerphase)

b(lighter

phase)

Adapted from chapter-

opening photograph,

Chapter 9, Callister,

Materials Science &

Engineering: An

Introduction, 3e.

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Chapter 10 - 4

70 80 1006040200

Temperature(C

)

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 Tcan change # of phases: pathAto B.

Altering Ccan change # of phases: path Bto D.

water-

sugar

system



D(100C,C= 90)2 phases

B(100C,C= 70)1 phase

A(20C,C= 70)2 phases

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Chapter 10 - 6

Phase Diagrams

Indicate phases as a function of T, C, and P.

For this course:- binary systems: just 2 components.

- independent variables: Tand C (P= 1 atm is almost always used).

PhaseDiagram

for Cu-Ni

system

Adapted from Fig. 10.3(a), Callister &

Rethwisch 4e.(Fig. 10.3(a) is adapted

from Phase Diagrams of Binary Nickel

Alloys, P. Nash (Ed.), ASM International,

Materials Park, OH (1991).

2 phases:L(liquid)a(FCC solid solution)

3 different phase fields:LL+

a

a

wt% Ni20 40 60 80 10001000110012001300140015001600T(C)

L(liquid)

a(FCC solidsolution)

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Chapter 10 - 7

Cu-Ni

phase

diagram

Isomorphous Binary Phase Diagram

Phase diagram:

Cu-Ni system.

System is:


Rethwisch 4e.(Fig. 10.3(a) is adapted

from Phase Diagrams of Binary Nickel



-- binaryi.e., 2 components:

Cu and Ni.

-- isomorphousi.e., complete

solubility of one

component in

another; aphase

field extends from

0 to 100 wt% Ni.wt% Ni20 40 60 80 10001000

110012001300140015001600T(C)

L(liquid)

a(FCC solidsolution)

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Chapter 10 -

wt% Ni20 40 60 80 1000100011001200

1300140015001600T(C)

L(liquid)

a(FCC solid

solution)

Cu-Ni

phase

diagram

8

Phase Diagrams:Determination of phase(s) present

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

Examples:

A(1100C, 60 wt% Ni):

1 phase: a

B(1250C, 35 wt% Ni):2 phases: L+ a

B(12

50

C,35)

A (1100C,60)Adapted from Fig. 10.3(a), Callister &Rethwisch 4e.(Fig. 10.3(a) is adaptedfrom Phase Diagrams of Binary Nickel



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Chapter 10 - 9

wt% Ni201200

1300

T(C)L(liquid)

a(solid)

30 40 50

Cu-Ni

system

Phase Diagrams:Determination of phase compositions

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

Examples:TA A

35C032CL

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

At TB = 1250C:Both aand L presentCL = Cliquidus ( = 32 wt% Ni)Ca = Csolidus ( = 43 wt% Ni)

At TD = 1190C:Only Solid (a) presentCa= C0 ( = 35 wt% Ni)

Consider C0= 35 wt% Ni

DTD

tie line

4Ca3Adapted from Fig. 10.3(a), Callister &

Rethwisch 4e.(Fig. 10.3(a) is adapted from

Phase Diagrams of Binary Nickel Alloys, P.

Nash (Ed.), ASM International, Materials

Park, OH (1991).

BTB

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Chapter 10 - 10

Rule 3: If we know Tand C0, then can determine:-- the weight fraction of each phase. Examples:

At TA : Only Liquid (L) present

WL = 1.00, Wa= 0

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

Phase Diagrams:Determination of phase weight fractions

wt% Ni201200

1300

T(C)L(liquid)

a(solid)

30 40 50

Cu-Ni

system

TAA

35C0

32CL

BTB

DTD

tie line

4Ca3

R SAt TB : Both a and L present

73.03243

3543=

=

= 0.27

WL= SR+SWa= RR+S

Consider C0= 35 wt% Ni


Rethwisch 4e.(Fig. 10.3(a) is adapted from

Phase Diagrams of Binary Nickel Alloys, P.

Nash (Ed.), ASM International, Materials

Park, OH (1991).

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Chapter 10 - 11

Tie lineconnects the phases in equilibrium with

each otheralso sometimes called an isotherm

The Lever Rule

What fraction of each phase?

Think of the tie line as a lever

(teeter-totter)

ML Ma

R S

M x S=MLx R

L

L

LL

L

LCC

CC

SR

RW

CC

CC

SR

S

MM

MW

=

+

=

=

+

=

+

=

a

a

a

a

a

00

wt% Ni20

1200

1300T(C)

L(liquid)

a(solid)

30 40 50

BTB

tie line

C0CL CaSR

Adapted from Fig. 10.3(b),


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Chapter 10 -

Slow rate of cooling:Equilibrium structure

Fast rate of cooling:Cored structure

First ato solidify:46 wt% Ni

Last ato solidify:

< 35 wt% Ni

13

Cachanges as we solidify.

Cu-Ni case: First ato solidify has Ca= 46 wt% Ni.Last ato solidify has Ca= 35 wt% Ni.

Cored vs Equilibrium Structures

Uniform Ca:35 wt% Ni

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Chapter 10 - 14

Mechanical Properties:Cu-Ni System

Effect of solid solution strengthening on:

-- Tensile strength (TS) -- Ductility (%EL)

Adapted from Fig. 10.6(a),


TensileStrength

(MPa)

Composition, wt% NiCu Ni

0 20 40 60 80 100200

300

400

TSforpure Ni

TSfor pure Cu

Elongation

(%EL)

Composition, wt% NiCu Ni

0 20 40 60 80 10020

30

40

50

60

%ELfor

pure Ni

%ELfor pure Cu

Adapted from Fig. 10.6(b),


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Chapter 10 - 15

2 components

has a special composition

with a min. melting T.



Binary-Eutectic Systems

3 single phase regions

(L, a, b)

Limited solubility:a: mostly Cu

b: mostly Ag

TE : No liquid below TE

: Composition at

temperature TE

CE

Ex.: Cu-Ag system

Cu-Ag

system

L(liquid)

a L+ a L+bba + b

C, wt% Ag20 40 60 80 1000

200

1200T(C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779C

Ag)wt%1.29(Ag)wt%.08(Ag)wt%9.71( b+L

cooling

heating

Eutectic reaction

L(CE) a(CaE) + b(CbE)

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Chapter 10 - 16

L+a L+b

a+b

200

T(C)

18.3

C, wt% Sn

20 60 80 1000

300

100

L(liquid)

a 183C61.9 97.8

b

For a 40 wt% Sn-60 wt% Pb alloy at 150C, determine:

-- the phases present Pb-Snsystem

EX 1: Pb-Sn Eutectic System

Answer:a+ b-- the phase compositions

-- the relative amountof each phase

150

40C0

11Ca

99Cb

SR

Answer:Ca= 11 wt% SnCb= 99 wt% Sn

Wa= Cb- C0Cb- Ca=

99 - 4099 - 11

=5988

= 0.67

SR+S

=

Wb=C0 - CaCb - Ca

=R

R+S

=29

88= 0.33=

40 - 11

99 - 11

Answer:



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Chapter 10 - 17

Answer:Ca= 17 wt% Sn

-- the phase compositions

L+b

a+b

200

T(C)

C, wt% Sn

20 60 80 1000

300

100

L(liquid)

a bL+a

183C

For a 40 wt% Sn-60 wt% Pb alloy at 220C, determine:

-- the phases present: Pb-Snsystem

EX 2: Pb-Sn Eutectic System

-- the relative amountof each phase

Wa= CL- C0CL- Ca

=46 - 40

46 - 17

=6

29= 0.21

WL=C0- CaCL- Ca

=23

29= 0.79

40C0

46CL

17Ca

220 SR

Answer:a+ L

CL= 46 wt% Sn

Answer:



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Chapter 10 - 18

For alloys for which

C0< 2 wt% Sn

Result: at room temperature-- polycrystalline with grains of

aphase havingcomposition C0

Microstructural Developments

in Eutectic Systems I

0

L+a200

T(C)

C , wt% Sn10

220

C0

300

100

L

a

30

a+b

400

(room Tsolubility limit)

TE(Pb-SnSystem)

aLL: C0wt% Sn

a: C0wt% Sn



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Chapter 10 - 19

For alloys for which

2 wt% Sn < C0< 18.3 wt% Sn

Result:

at temperatures in a+ brange

-- polycrystalline with agrains

and small b-phaseparticles




in Eutectic Systems II

Pb-Snsystem

L+ a200

T(C)

C, wt% Sn10

18.3

200C0

300

100

L

a

30

a+b

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

L

a

L: C0wt% Sn

ab

a: C0wt% Sn

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Chapter 10 - 20

For alloy of composition C0= C

E

Result: Eutectic microstructure (lamellar structure)-- alternating layers (lamellae) of aand bphases.

Adapted from Fig. 10.13,Callister & Rethwisch 4e.


in Eutectic Systems III



160m

Micrograph of Pb-Sneutectic

microstructure

Pb-Sn

system

L+b

a + b

200

T(C)

C, wt% Sn

20 60 80 1000

300

100

L

a bL+a

183C

40

TE

18.3

a: 18.3 wt%Sn

97.8

b: 97.8 wt% Sn

CE

61.9

L: C0wt% Sn

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Chapter 10 - 21

Lamellar Eutectic Structure

Adapted from Figs. 10.14 & 10.15,


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Chapter 10 - 22

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


in Eutectic Systems IV

18.3 61.9

SR

97.8

SR

primary aeutectic a

eutectic b

WL= (1-Wa)= 0.50

Ca = 18.3 wt% Sn

CL = 61.9 wt% SnS

R+ SWa= = 0.50

Just above TE :

Just below TE:

Ca = 18.3 wt% SnCb = 97.8 wt% Sn

S

R+ SWa= = 0.73

Wb= 0.27Adapted from Fig. 10.16,


Pb-Sn

systemL+

b200

T(C)

C, wt% Sn

20 60 80 1000

300

100

L

a bL+a

40

a+bTE

L: C0wt% Sn LaL

a

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Chapter 10 - 23

L+aL+b

a+b200

C, wt% Sn20 60 80 1000

300

100

L

a bTE

40

(Pb-SnSystem)

Hypoeutectic & Hypereutectic



(Fig. 10.8 adapted from

Binary Phase Diagrams,

2nd ed., Vol. 3, T.B.

Massalski (Editor-in-Chief),

ASM International,

Materials Park, OH, 1990.)

160 m

eutectic micro-constituentAdapted from Fig. 10.14,


hypereutectic: (illustration only)

bbb

bbb



(Illustration only)

(Figs. 10.14 and

10.17 from Metals

Handbook, 9th ed.,

Vol. 9,

Metallography and

Microstructures,

American Society for

Metals, Materials

Park, OH, 1985.)

175 m

aaa

aaa

hypoeutectic: C0= 50 wt% Sn

Adapted from

Fig. 10.17, Callister &

Rethwisch 4e.

T(C)

61.9

eutectic

eutectic: C0=61.9wt% Sn

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Chapter 10 - 24

Intermetallic Compounds

Mg2Pb

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).

Adapted from

Fig. 10.20, Callister &

Rethwisch 4e.

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Chapter 10 - 25

Eutectoidone solid phase transforms to two other

solid phases

S2 S1+S3

a+ Fe3C (For Fe-C, 727C, 0.76 wt% C)

intermetallic compound

- cementite

cool

heat

Eutectic, Eutectoid, & Peritectic

Eutectic- liquid transforms to two solid phases

L a+ b (For Pb-Sn, 183C, 61.9 wt% Sn)coolheat

cool

heat

Peritectic- liquid and one solid phase transform to a

second solid phaseS1 + L S2

+ L (For Fe-C, 1493C, 0.16 wt% C)

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Chapter 10 - 26

Eutectoid & Peritectic

Cu-Zn Phase diagram



Eutectoid transformation +

Peritectic transformation + L

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Chapter 10 - 27

Ceramic Phase Diagrams

MgO-Al2O3 diagram:

Adapted from Fig.

10.24, Callister &

Rethwisch 4e.

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Chapter 10 - 28

Iron-Carbon (Fe-C) Phase Diagram

2 important

points

- Eutectoid (B):

a+Fe3C

- Eutectic (A):L +Fe3C

Adapted from Fig. 10.28,Callister & Rethwisch 4e.

Fe3C(cementite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a+Fe3C

(Fe) C, wt% C

1148C

T(C)

a 727C = Teutectoid

4.30

Result: Pearlite =alternating layers ofaand Fe3C phases

120 m

(Adapted from Fig. 10.31,Callister & Rethwisch 4e.)

0.76

B

AL+Fe3C

Fe3C (cementite-hard)

a(ferrite-soft)

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Chapter 10 - 30

Fe3C(cem

entite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a+Fe3C

L+Fe3C

(Fe) C, wt% C

1148C

T(C)

a727C

(Fe-C

System)

C0

0.76

Hypoeutectoid Steel

a

aa

srWa= s/(r+s)W=(1 - Wa)

RSa

pearlite

Wpearlite=W

Wa= S/(R+S)

W =(1Wa)Fe3C

Adapted from Figs. 10.28

and 10.33,Callister &Rethwisch 4e.

(Fig. 10.28 adapted from

Binary Alloy Phase

Diagrams, 2nd ed., Vol.

1, T.B. Massalski (Ed.-in-

Chief), ASM International,

Materials Park, OH,

1990.)

Adapted from Fig. 10.34, Callister & Rethwisch 4e.

proeutectoid ferritepearlite

100 mHypoeutectoid

steel

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Chapter 10 - 31

Hypereutectoid Steel

Fe3C(cem

entite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a+Fe3C

L+Fe3C

(Fe) C, wt%C

1148C

T(C)

a


and 10.36,Callister &Rethwisch 4e. (Fig.

10.28 adapted from

Binary Alloy Phase




Materials Park, OH,

1990.)

(Fe-C

System)

0.7

6 C0

Fe3C


proeutectoid Fe3C

60 mHypereutectoidsteel

pearlite

pearlite

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Chapter 10 - 32

Fe3C(cem

entite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a+Fe3C

L+Fe3C

(Fe) C, wt%C

1148C

T(C)

a

Hypereutectoid Steel

(Fe-C

System)

0.7

6 C0

pearlite

Fe3C

xv

V X

Wpearlite=W

Wa=X/(V+X)

W =(1 - Wa)Fe3C

W =(1-W)

W =x/(v+x)

Fe3C


proeutectoid Fe3C

60 mHypereutectoidsteel

pearlite


and 10.36,Callister &Rethwisch 4e. (Fig.

10.28 adapted from

Binary Alloy Phase




Materials Park, OH,

1990.)

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Chapter 10 - 33

Example Problem

For a 99.6 wt% Fe-0.40 wt% C steel at atemperature just below the eutectoid,

determine the following:

a) The compositions of Fe3C and ferrite (a).

b) The amount of cementite (in grams) thatforms in 100 g of steel.

c) The amounts of pearlite and proeutectoid

ferrite (a) in the 100 g.

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Chapter 10 - 34

Solution to Example Problem

WFe3C=

R

R+S=

C0 CCFe3C

C

= 0.40 0.022

6.70 0.022 = 0.057

b) Using the lever rule with

the tie line shown

a) Using the RStie line just below the eutectoid

Ca= 0.022 wt% CCFe3C= 6.70 wt% C

Fe3C(cementite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a+Fe3C

L+Fe3C

C, wt% C

1148C

T(C)

727

C

C0

R S

CFe C3C

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) =5.7 g

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Chapter 10 - 35

Solution to Example Problem (cont.)

c) Using the VXtie line just above the eutectoid and

realizing thatC0= 0.40 wt% C

Ca= 0.022 wt% C

Cpearlite= C= 0.76 wt% C

Fe3C(cementite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a+Fe3C

L+Fe3C

C, wt% C

1148C

T(C)

727

C

C0

VX

CC

Wpearlite =V

V+X=C0 CC C

= 0.40 0.022

0.76 0.022 = 0.512

Amount of pearlite in 100 g

= (100 g)Wpearlite

= (100 g)(0.512) =51.2 g

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Chapter 10 -

VMSE: Interactive Phase Diagrams

36Change alloy composition

Microstructure, phase compositions, and phase fractions respond interactively

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Chapter 10 - 37

Alloying with Other Elements

Teutectoidchanges:

Adapted from Fig. 10.38,Callister & Rethwisch 4e.(Fig. 10.38 from Edgar C. Bain, Functions of the

Alloying Elements in Steel, American Society for

Metals, 1939, p. 127.)

TEutectoid(C)

wt. % of alloying elements

Ti

Ni

MoSi

W

Cr

Mn

Ceutectoidchanges:

Adapted from Fig. 10.39,Callister & Rethwisch 4e.(Fig. 10.39 from Edgar C. Bain, Functions of the

Alloying Elements in Steel, American Society for

Metals, 1939, p. 127.)

wt. % of alloying elements

Ceutectoid(wt%C

)

Ni

Ti

Cr

Si MnW

Mo

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Chapter 10 - 38

Phase diagramsare useful tools to determine:-- the number and types of phases present,

-- the compositionof 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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Ch t 10 39

Core Problems:

Self-help Problems:

ANNOUNCEMENTS

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