Ch09 phase diagm-sent-fall2016

1

Chapter 9 - 1

Ch 9: Phase Diagrams

• Read: Ch 9 – general reading sections 1-6, careful study sections 7-12, and 18-20. Plus all example problems. .

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Chapter 9 -

Ch9:Do we know these terms?

– Gibb’s Phase Rule

– γ-Austenite,

– Cementite

– Pearlite

– Isomorphous

– Lever Rule

– α-Ferrite

– Eutectic

– Eutectoid

– Liquidus, Solidus, Solvus lines etc.

2

– Terminal solid Solution

– Tie line

– Martensite

– Hypoeutectic

– Hypereutectic

– Proeutectoid

– Hypoeutectoid

– hypereutectoid, etc.

Chapter 9 - 3

Phase Equilibria: Solubility LimitIntroduction

– Solutions – solid solutions (alloys), single phase

– Mixtures – more than one phase

• Solubility Limit:Max concentration for

which only a single phase

solution occurs.

Question: What is the

solubility limit at 20°C?

Answer: 65 wt% sugar.If Co < 65 wt% sugar: syrup

If Co > 65 wt% sugar: syrup + sugar.65

Sucrose/Water Phase Diagram

Pu

re

Su

ga

r

Te

mp

era

ture

(°C

)

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

L

(liquid solution i.e., syrup)

Solubility

Limit L

(liquid)

+ S

(solid

sugar)20

40

60

80

100

Pu

re

Wa

ter

2

Chapter 9 - 4

• Components:

The elements or compounds which are present in the mixture

(e.g., Al and Cu) there are 2.

• Phases:

The physically and chemically distinct material regions

that result (e.g., α and β). So, there are how many phases?

Aluminum-Copper Alloy ( a type of Bronze)

Components and Phases

α (darker

phase)

β (lighter

phase)

Adapted from

chapter-opening

photograph, Chapter

9, Callister 3e.

Chapter 9 - 5

Phase Equilibria

CrystalStructure

electroneg 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 solution system (e.g., Ni-Cu solution)

• Result �Ni and Cu are totally miscible in ALL proportions.

= ISOMORPHOUS ALLOY

Chapter 9 - 6

Binary Phase Diagrams: e.g. Cu-Ni.

• Indicate phases as function of T, Co, and P.

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

-independent variables: T and Co (P = 1 atm is almost always used).

• Phase

Diagram

for Cu-Ni

system

Adapted from Fig. 9.3(a), Callister 7e.

(Fig. 9.3(a) is adapted from Phase

Diagrams of Binary Nickel Alloys, P. Nash

(Ed.), ASM International, Materials Park,

OH (1991).

• 2 phases:

L (liquid)

α (FCC solid solution)

• 3 phase fields:

L

L + α

α

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

α

(FCC solid solution)

3

Chapter 9 - 7

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

α

(FCC solid solution)

Cu-Ni

phase

diagram

Phase Diagrams:EXPL # and types of phases

• Rule 1: If we know T and Co, then we know:--the # and types of phases present.

• Examples:

A(1100°C, 60): 1 phase: α

B(1250°C, 35): 2 phases: L + α

Adapted from Fig. 9.3(a), Callister 7e.

(Fig. 9.3(a) is adapted from Phase

Diagrams of Binary Nickel Alloys, P. Nash

(Ed.), ASM International, Materials Park,

OH, 1991).

B(1

25

0°C

,35

)

A(1100°C,60)

Chapter 9 - 8

wt% Ni

20

1200

1300

T(°C)

L (liquid)

α

(solid)

30 40 50

Cu-Ni

system

Phase Diagrams:EXPLcomposition of phases

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

• Examples:TA

A

35Co

32CL

At TA = 1320°C:

Only Liquid (L) CL = Co ( = 35 wt% Ni)

At TB = 1250°C:

Both α and L

CL = C liquidus ( = 32 wt% Ni here)

Cα = Csolidus ( = 43 wt% Ni here)

At TD = 1190°C:

Only Solid ( α)

Cα = Co ( = 35 wt% Ni)

Co = 35 wt% Ni

Adapted from Fig. 9.3(b), Callister 7e.

(Fig. 9.3(b) is adapted from Phase Diagrams

of Binary Nickel Alloys, P. Nash (Ed.), ASM

International, Materials Park, OH, 1991.)

BTB

DTD

tie line

4Cα3

Chapter 9 - 9

• Tie line – connects the phases in equilibrium with

each other - essentially an isothermal line

The Lever Rule

How much of each phase?

Think of it as a lever (teeter-totter)

MLMα

R S

RMSM L ⋅=⋅α

L

L

LL

LL

CC

CC

SR

RW

CC

CC

SR

S

MM

MW

−

−=

+=

−

−=

+=

+=

α

α

α

α

α

00

wt% Ni

20

1200

1300

T(°C)

L (liquid)

α

(solid)

30 40 50

BTB

tie line

Co

CL Cα

SR

Adapted from Fig. 9.3(b),

Callister 7e.

4

Chapter 9 - 10

• Rule 3: If we know T and Co, then we know:--the amount of each phase (given in wt%).

• Examples:

At TA: Only Liquid (L)

WL = 100 wt%, Wα = 0

At TD: Only Solid ( α)

WL = 0, Wα = 100 wt%

Co = 35 wt% Ni

Adapted from Fig. 9.3(b), Callister 7e.

(Fig. 9.3(b) is adapted from Phase Diagrams of

Binary Nickel Alloys, P. Nash (Ed.), ASM

International, Materials Park, OH, 1991.)

Phase Diagrams:

weight fractions of phases

wt% Ni

20

1200

1300

T(°C)

L (liquid)

α

(solid)

30 40 50

Cu-Ni

system

TAA

35Co

32CL

BTB

DTD

tie line

4Cα3

R S

At TB: Both α and L

% 733243

3543wt=

−

−=

= 27 wt%

WL= S

R +S

Wα= R

R +S

Chapter 9 - 11

wt% Ni20

1200

1300

30 40 50110 0

L (liquid)

α

(solid)

T(°C)

A

35Co

L: 35wt%Ni

Cu-Ni

system

• Phase diagram:Cu-Ni system.

• System is:--binary

i.e., 2 components:

Cu and Ni.

--isomorphousi.e., complete

solubility of one

component in

another; α phase

field extends from

0 to 100 wt% Ni.

Adapted from Fig. 9.4,

Callister 7e.

• ConsiderCo = 35 wt%Ni.

Development of microstructure upon cooling

4635

4332

α: 43 wt% Ni

L: 32 wt% Ni

L: 24 wt% Ni

α: 36 wt% Ni

Bα: 46 wt% NiL: 35 wt% Ni

C

D

E

24 36

Chapter 9 - 12

• Cα changes as we solidify.

• Cu-Ni case:

• Fast rate of cooling:Cored structure

• Slow rate of cooling:Equilibrium structure

First α to solidify has Cα = 46 wt% Ni.

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

Cored vs Equilibrium Phases

First α to solidify:

46 wt% Ni

Uniform Cα:

35 wt% Ni

Last α to solidify:

< 35 wt% Ni

Provided the 35% soluton

is mantained..

How’s it possible

that average is

same as the

outer comp. Of

the cored grain?

5

Chapter 9 - 13

Mechanical Properties: Cu-Ni System

• Effect of solid solution strengthening on:

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

--Peak as a function of Co --Min. as a function of Co

Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.

Tensile S

trength

(M

Pa)

Composition, wt% NiCu Ni0 20 40 60 80 100

200

300

400

TS for

pure Ni

TS for pure CuE

longation (

%E

L)

Composition, wt% NiCu Ni0 20 40 60 80 100

20

30

40

50

60

%EL for pure Ni

%EL for pure Cu

Chapter 9 - 14

Review

Isomorphous Alloy

Lever Rule

Phase Analysis �3 main items

phases present

?

?

Solidus, Liquidus, Vapodus

Tie Line

Microstructure development

Cored structure ?

Chapter 9 - 15

: Min. melting TE

2 componentshas a special composition

with a min. melting T.


Callister 7e.

Binary-Eutectic Systems

• Eutectic transition

L(CE) α(CαE) + β(CβE)

• 3 single phase regions

(L, α, β ) total regions=?

• Limited solubility:

α: mostly Cu

β: mostly Ag

• TE : No liquid below TE

• @CE

composition

Ex.: Cu-Ag system

Cu-Ag

system

L (liquid)

α L + αL+ββ

α + β

Co , wt% Ag20 40 60 80 1000

200

1200T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779°C

6

Chapter 9 - 16

L+αL+β

α + β

200

T(°C)

18.3

C, wt% Sn

20 60 80 1000

300

100

L (liquid)

α183°C

61.9 97.8

β

• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...--the phases present: Pb-Sn

system

EX: Pb-Sn Eutectic System (1)

α + β--compositions of phases:

CO = 40 wt% Sn

--the relative amount

of each phase:150

40Co

11Cα

99Cβ

SR

Cα = 11 wt% Sn

Cβ = 99 wt% Sn

Wα=Cβ - CO

Cβ - Cα

=99 - 4099 - 11

=5988

= 67 wt%

SR+S

=

Wβ =CO - Cα

Cβ - Cα

=R

R+S

=29

88= 33 wt%=

40 - 11

99 - 11


Callister 7e.

Chapter 9 - 17

L+β

α + β

200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

L (liquid)

α β

L+α

183°C

• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find...--the phases present: Pb-Sn

system


Callister 7e.

EX: Pb-Sn Eutectic System (2)

α + L--compositions of phases:

CO = 40 wt% Sn

--the relative amount

of each phase:

Wα =CL - CO

CL - Cα

=46 - 40

46 - 17

=6

29= 21 wt%

WL =CO - Cα

CL - Cα

=23

29= 79 wt%

40Co

46CL

17Cα

220SR

Cα = 17 wt% Sn

CL = 46 wt% Sn

Chapter 9 - 18

• Co < 2 wt% Sn

• Result:

--at extreme ends

--polycrystal of α grains

i.e., only one solid phase.


Callister 7e.

Microstructures

in Eutectic Systems: I

0

L+ α

200

T(°C)

Co, wt% Sn10

2

20Co

300

100

L

α

30

α+ β

400

(room T solubility limit)

TE

(Pb-SnSystem)

αL

L: Co wt% Sn

α: Co wt% Sn

7

Chapter 9 - 19

• 2 wt% Sn < Co < 18.3 wt% Sn

• Result:� Initially liquid + α

� then α alone

� finally two phases

� α polycrystal� fine β-phase inclusions


Callister 7e.

Microstructures

in Eutectic Systems: II

Pb-Sn

system

L + α

200

T(°C)

Co , wt% Sn10

18.3

200Co

300

100

L

α

30

α+ β

400

(sol. limit at TE)

TE

2(sol. limit at T room)

L

α

L: Co wt% Sn

αβ

α: Co wt% Sn

Chapter 9 - 20

• Co = CE

• Result: Eutectic microstructure (lamellar structure)

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


Callister 7e.

Microstructures

in Eutectic Systems: III

Adapted from Fig. 9.14, Callister 7e.

160µm

Micrograph of Pb-Sn

eutectic microstructure

Pb-Sn

system

L + β

α + β

200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

L

α β

L+α

183°C

40

TE

18.3

α: 18.3 wt%Sn

97.8

β: 97.8 wt% Sn

CE61.9

L: Co wt% Sn

Chapter 9 - 21

• 18.3 wt% Sn < Co < 61.9 wt% Sn e.g. 40% Sn

• Result: α crystals and a eutectic microstructure

Microstructures

in Eutectic Systems: IV

18.3 61.9

SR

97.8

SR

primary α

eutectic αeutectic β

WL = (1-Wα) = 50 wt%

Cα = 18.3 wt% Sn

CL = 61.9 wt% SnS

R + SWα= = 50 wt%

• Just above TE :

• Just below TE :

Cα = 18.3 wt% Sn

Cβ = 97.8 wt% SnS

R + SWα= = 73 wt%

Wβ = 27 wt%Adapted from Fig. 9.16,

Callister 7e.

Pb-Sn

systemL+β200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

α β

L+α

40

α+β

TE

L: Co wt% Sn Lα

Lα

8

Chapter 9 - 22

L+αL+β

α + β

200

Co, wt% Sn20 60 80 1000

300

100

L

α βTE

40

(Pb-Sn System)

Hypoeutectic & Hypereutectic


Callister 7e. (Fig. 9.8 adapted from

Binary Phase Diagrams, 2nd ed.,

Vol. 3, T.B. Massalski (Editor-in-

Chief), ASM International, Materials

Park, OH, 1990.)

160 µm

eutectic micro-constituentAdapted from Fig. 9.14,

Callister 7e.

hypereutectic: (illustration only)

β

ββ

ββ

β

Adapted from Fig. 9.17, Callister 7e.

(Illustration only)

(Figs. 9.14 and 9.17 from

Metals Handbook, 9th ed.,

Vol. 9, Metallography and

Microstructures, American

Society for Metals,

Materials Park, OH, 1985.)

175 µm

α

α

α

αα

α

hypoeutectic: Co = 50 wt% Sn

Adapted from

Fig. 9.17, Callister 7e.

T(°C)

61.9

eutectic

eutectic: Co =61.9wt% Sn

Chapter 9 - 23

Review

Binary Eutectic System

Hypo and hyper eutectic alloys

Eutectic grain structure

Eutectic system phase analyses

Eutectic system microstructure formation

Cored structure of grains

Phase Analysis �

phases present

?

?

Eutectoid (not Eutectic system)

Composition of phases

Wt% of phases

Chapter 9 - 24

Other Types:Eutectoid & Peritectic

Cu-Zn Phase diagram

Adapted from

Fig. 9.21, Callister 7e.

Eutectoid transition δ γ + ε

Peritectic transition γ + L δ

Identify

Eutectoid

Pt.

Now identify

a Peritectic

Pt.

Another ?

P

9

Chapter 9 - 25

Source: www.substec.com

Chapter 9 - 26

Iron-Carbon (Fe-C) Phase Diagram

• 2 important points

-Eutectoid (B):γ ⇒ α +Fe3C

-Eutectic (A):L ⇒ γ +Fe3C

Adapted from Fig. 9.24,Callister 7e.

Fe

3C

(cem

entite

)

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) Co, wt% C

1148°C

T(°C)

α727°C = Teutectoid

A

SR

4.30

Result: Pearlite = alternating layers of α and Fe3C phases

120 µm

(Adapted from Fig. 9.27, Callister 7e.)

γ γ

γγ

R S

0.76

Ce

ute

cto

id

B

Fe3C (cementite-hard)

α (ferrite-soft)

Chapter 9 - 27

Hypoeutectoid Steel

Adapted from Figs. 9.24

and 9.29,Callister 7e.

(Fig. 9.24 adapted from

Binary Alloy Phase

Diagrams, 2nd ed., Vol.

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

Chief), ASM International,

Materials Park, OH,

1990.)

Fe

3C

(cem

entite

)

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) Co , wt% C

1148°C

T(°C)

α727°C

(Fe-C

System)

C0

0.7

6


proeutectoid ferritepearlite

100 µmHypoeutectoid

steel

R S

α

wα =S/(R+S)

wFe3

C

=(1-wα)

wpearlite = wγpearlite

r s

wα =s/(r+s)

wγ =(1- wα)

γ

γ γ

γα

αα

γγγ γ

γ γ

γγ

10

Chapter 9 - 28

Hypereutectoid Steel

Fe

3C

(cem

entite

)

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) Co , wt%C

1148°C

T(°C)

α

Adapted from Figs. 9.24

and 9.32,Callister 7e.

(Fig. 9.24 adapted from

Binary Alloy Phase

Diagrams, 2nd ed., Vol.

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

Chief), ASM International,

Materials Park, OH,

1990.)

(Fe-C

System)

0.7

6 Co


proeutectoid Fe3C

60 µmHypereutectoid steel

pearlite

R S

wα =S/(R+S)

wFe3C =(1-w α)

wpearlite = wγpearlite

sr

wFe3C=r /(r+s)

wγ =(1-w Fe3C)

Fe3C

γγ

γ γ

γγγ γ

γγγ γ

Chapter 9 - 29

Example: Phase Equilibria

For a AISI1040 ( how much wt% Fe) steel at a temperature just below the eutectoid temperature(727°C), determine the following:

a) What are the phases?

b) composition of phases Fe3C and ferrite (α)

c) the amount of carbide (cementite) in grams that forms per 100 g of steel

d) the amount of pearlite and proeutectoid ferrite (α)

Solutions � next page

Use a ruler on the handout diagram and

show all lines and %C.

Chapter 9 - 30

Chapter 9 – Phase Equilibria

Solution:

g 3.94

g 5.7 CFe

g7.5100 022.07.6

022.04.0

100xCFe

CFe

3

CFe3

3

3

=α

=

=−

−=

−

−=

α+ α

α

x

CC

CCo

b) the amount of carbide

(cementite) in grams that

forms per 100 g of steel

a) composition of Fe3C and ferrite (α)

CO = 0.40 wt% C

Cα = 0.022 wt% C

CFe C = 6.70 wt% C3

Fe

3C

(cem

entite

)

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

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CFe C3Cα

11

Chapter 9 - 31

Chapter 9 – Phase Equilibriac. the amount of pearlite and proeutectoid ferrite (α)

note: amount of pearlite = amount of γ just above TE

Co = 0.40 wt% C

Cα = 0.022 wt% C

Cpearlite = Cγ = 0.76 wt% C

γ

γ + α=

Co −Cα

Cγ −Cα

x 100 = 51.2 g

pearlite = 51.2 g

proeutectoid α = 48.8 g

Fe

3C

(cem

entite

)

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

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CγCα

Chapter 9 - 32

Alloying Steel with More Elements

• Teutectoid changes: • Ceutectoid changes:

Adapted from Fig. 9.34,Callister 7e. (Fig. 9.34

from Edgar C. Bain, Functions of the Alloying

Elements in Steel, American Society for Metals,

1939, p. 127.)

Adapted from Fig. 9.35,Callister 7e. (Fig. 9.35

from Edgar C. Bain, Functions of the Alloying

Elements in Steel, American Society for Metals,

1939, p. 127.)

TE

ute

cto

id(°

C)

wt. % of alloying elements

Ti

Ni

MoSi

W

Cr

Mn


Ceute

cto

id(w

t%C

)

Ni

Ti

Cr

SiMn

WMo

Chapter 9 - 33

Practice Problem 9.66

• Teutectoid changes:

TE

ute

cto

id(°

C)


Ti

Ni

MoSi

W

Cr

Mn

6% Mn

So:

TE=700F

Contd.

Alloy Steel has 6% Mn,

and .25% C

TE=?,

%Eutectoid C=?

ProEutectoid Phase = ?

12

Chapter 9 - 34

Eutectoid %C

• Ceutectoid changes:


Ceute

cto

id(w

t%C

)

Ni

Ti

Cr

Si

MnW

Mo

From figure Eutectoid

Mixture happens at 0.44% C

If mixture has .25% C –

Then it is left of eutectoid

point, i.e. proeutectoid

phase is Ferrite

If %C in mixture is >

Eutectoid Mixture then

proeutectoid phase would

have been Cementite

Chapter 9 - 35

• Phase diagrams are useful tools to determine:

--the number and types of phases in an alloy,

--the wt% of each phase,

--and the composition of each phase

for a given T and composition of the system.

• Alloying to produce a solid solution usually

--increases the tensile strength (TS)

--decreases the ductility.

• Binary eutectics and binary eutectoids allow for

a range of microstructures.

Summary

Ch09 phase diagm-sent-fall2016

Engineering

Transcript of Ch09 phase diagm-sent-fall2016