Introduction to Engineering Materials ENGR2000 Chapter 9 ...engineering.armstrong.edu/cameron/ENGR2000_phasediagrams1.pdf · •Tie lines and lever rule still apply. ... calculate

Introduction to Engineering Materials

ENGR2000

Chapter 9: Phase Diagrams IChapter 9: Phase Diagrams I

Dr. Coates

Definitions and Basic Concepts

• A component is a chemically distinct and

essentially indivisible substance.

– Fe, Si, C are single component systems

– H2O is a single component system

– window glass (SiO2, Na2O & CaO) is a three-– window glass (SiO2, Na2O & CaO) is a three-

component system

9.3 Phases

• A phase is a homogeneous portion of a system

that has uniform physical and/or chemical

characteristics.

– Liquid– Liquid

– Solid

– Gaseous

– Crystal structure

– Pure metal

Phase and Component

• Pure FCC Al is a

– single component, single phase system

• A mixture of pure ice and pure water is

– a single component, two phase system– a single component, two phase system

• A mixture of BCC Fe and FCC Fe is a

– single component, two phase system.

• A solid solution of Cu and Ni is a

– two component, single phase system

• A solid solution of NiO and MgO is a

– two component, single phase system

9.2 Solubility Limit

• Maximum concentration of solute atoms that may

dissolve in the solvent to form a solid solution

– At a specific temperature– At a specific temperature

– For a specific alloy system

• How would temp. affect solubility limit?

• Why?

– Atoms in solvent are able to vibrate with greater

amplitude and frequency, allowing more spaces per unit

time for solute atoms to occupy

Sugar-water solution + solid sugar at 20 C

in equilibrium

9.4 Microstructure

(Review 4.10 polishing and etching)

• Microstructure

– subject to direct observation using optical or electron

microscopes

• Metal alloys characterized by

– # of phases present– # of phases present

– Proportions of phases

– Distribution or arrangement of phases

• Microstructure depends on

– Alloying elements

– Concentrations

– Heat treatment

9.5 Phase Equilibrium

• Free energy = function (internal energy,

randomness or disorder of the atoms)

• A system is at equilibrium if its free energy is at a

minimum under some specified combination ofminimum under some specified combination of

temperature, pressure and composition.

• Phase equilibrium is reflected by a constancy with

time in the phase characteristics of a system.

(applies where more than one phase exists in the

system)

Sugar-water solution + solid sugar at 20 C

in equilibrium

If suddenly

increase temp

from 200 C to

800 C, what

happens?

9.5 Phase Equilibria

• Free energy considerations provide information

about the equilibrium characteristics of a

particular system– No indication of time periods needed for attainment of new– No indication of time periods needed for attainment of new

equilibrium state

• Often equilibrium never achieved because rate of

approach to equilibrium is extremely slow-System

state is said to be in non-equilibrium or metastable

9.8 Binary Isomorphous Systems

• Binary

– 2 components

• Isomorphous• Isomorphous

– Complete liquid and solid solubility

Phase diagrams are maps that present the relationship between

temperature and composition and the quantities of phases at

equilibrium

•presented here are for a constant pressure of 1 atm

Substitutional solid solutions

• Cu & Ni solution - Isomorphous

– Atomic radii for Cu & Ni are 0.128 nm & 0.125 nm.

– Both have FCC crystal structure

– EN values are 1.9 & 1.8– EN values are 1.9 & 1.8

– Valences are +1 & +2.

• The liquid L is a homogenous liquid solution

composed of both copper and nickel

• The α phase is a substitutional solid solution

consisting of both Cu and Ni atoms, having an

FCC crsytal structure.FCC crsytal structure.

• At temp. below 10800 C, Cu and Ni are mutually

soluble in each other in the solid state for all

compositions. Why? (See section 4.3)

pure

com

ponen

tC

u

pure

com

ponen

tN

i

Melting

Temperature of

pure Cu

9.8 Interpretation of Phase Diagrams

Phases Present

35 wt % Ni – 65 wt % Cu alloy

at 1250 C


at 1100 C

at 1250 C

Determination of Phase Compositions


at 1100 Cat 1100 C

NiwtC .%60=α

The composition at (11000 C)

is given by

SINGLE PHASE!

• Within a two phase system, how is the

composition of each phase calculated?

• Ex. Suppose we wanted to know what % nickel

was present in the α portion within the α +L phase

NiwtC .%5.42=α

NiwtC

NiwtC

L .%5.31

.%5.42

=

=α

Note-these are the

equilibrium concentrations

of the two phases

Determination of Phase Amounts

%100=αW

What (mass fraction)

percentage of the α phase

is present

1. in the α region

2. in the L region

3. in the α+L region?

%0=αW


at 1100 C

3. in the α+L region?

Determination of Phase Amounts

- Lever rule


at 1250 C

.%5.31

.%5.42

=

=

NiwtC

NiwtCα

1 :Note

68.0

32.0

.%35

.%5.31

0

0

0

=+

=−

−=

=−

−=

=

=

L

L

L

L

L

L

WW

CC

CCW

CC

CCW

NiwtC

NiwtC

α

α

α

α

α

Reading Assignments

• Section 9.8

– Tie line

– Lever rule

– Example 9.1– Example 9.1

– Volume fractions & mass fractions

9.9 Development of Microstructure in

Isomorphous Alloys

• Equilibrium cooling

– Cooling occurs very slowly

– Phase equilibrium is continuously maintained


Isomorphous Alloys

• Non-equilibrium cooling

– Rapid cooling rates (in most practical situations)

– Sufficient time must be allowed at each temperature for

compositional readjustment, otherwisecompositional readjustment, otherwise

– Equilibrium is not always maintained

Because diffusion in

the solid α phase is

relatively slow!

At a give temp in the two phase

region, will more or less liquid be

present for nonequilibrium cooling vs

equilibrium cooling?

How does Ni content compare along

solidus line for nonequilibrium vs

equilibrium cooling?

How much solidus deviation would

you expect with increasing cooling

rate?

• Distribution of grains will be non-uniform

(segregation)

• Center of each grain is rich in high melting

element/concentration, the concentration of low

melting element increases from this region

towards grain boundary-coring

• CORING will affect temperature response/sudden• CORING will affect temperature response/sudden

loss of mechanical integrity as structure is

reheated

9.10 Mechanical Properties of

Isomorphous Alloys

• Each component experiences solid-solution strengthening

– At temp below lowest-melting component

•Three phases, α, β, liquid

•α -Solid solution rich in Cu, silver as solute

FCC crystal structure

•β-Solid solution, FCC, Cu is solute

9.11 Binary Eutectic Systems

•L-liquid

•Three two phase regions

•α+β•α+L

•α+β•Tie lines and lever rule still apply

9.11 Binary Eutectic Systems

Melting

temperature

Pure copper

Max

solubility

Ag in

α phase

Max

solubility

Cu in

β phase

Invariant point

• Which line represents lowest temperature at which

a liquid phase may exist for any Cu-Ag alloy at

equilibrium?

• What happens to temp at which alloys become

totally liquid (melting temp)as Ag is added to Cu?

Vice versa?

BEG

AE EF

• What is special about point E?Eutectic Reaction-upon

cooling, liquid is

transformed to two solid αand β phases

Eutectic (easily melted) point

( ) ( ) ( )

re temperatueutectic -

ncompositio eutectic -

:n compositio ofalloy an for reaction Eutectic

E

E

EEE

E

T

C

CCCL

C

βα βα +⇔

re temperatueutectic - ET

L(71.9 wt% Ag)

α(8.0wt% Ag)+β(91.2 wt% Ag)

60-40 Solder

(60 wt. % Sn – 40 wt. % Pb)

• Alloy is

completely

molten at

~185 C~185 C

• Easily melted

• Low-temperature

solder

Example 9.2

• For a 40 wt % Sn – 60 wt % Pb alloy at 150 C

– What phases are present?

– What are the compositions of the phases?

SnwtC %11=α

βα

:nscompositio Phase

&

:Bpoint at Phases

SnwtC %98=β

α

Example 9.3

• For a 40 wt % Sn – 60 wt % Pb alloy at 1500 C,

calculate the relative amount of each phase present

in terms of

– mass fraction and– mass fraction and

– volume fraction.

At 1500 C take the densities of Pb and Sn to be

11.23g/cm3 and 7.24 g/cm3 respectively.

.%40

%98

%11

1

−

=

=

=

β

α

CC

SnwtC

SnwtC

SnwtC

:fractions) (mass

amounts Phase

:nscompositio Phase

33.0

67.0

1

1

=−

−=

=−

−=

αβ

αβ

αβ

β

α

CC

CCW

CC

CCW

( )

( )

3

3

3

/59.108911

100100

/24.7

/23.11

cmgCC

cmgSn

cmgPb

===

=

=

ρ

ρ

ρ

α

:phase each of Density

:densities Given

( )

( )( )

( )

( )

( )( )

( )

3/29.7

23.11

2

24.7

98

100100

/59.10

23.11

89

24.7

11

cmg

Pb

C

Sn

C

cmg

Pb

C

Sn

C

PbSn

PbSn

=

+

=

+

=

=

+

=

+

=

ρρ

ρ

ρρ

ρ

βββ

ααα

58.0=

+

=βα

α

α

α

ρρ

ρ

WW

W

V

:fractions) (volume amounts Phase

43.0=

+

=

+

β

β

α

α

β

β

β

βα

ρρ

ρ

ρρ

WW

W

V


Eutectic Alloys

• Eutectic alloys

– Eutectic composition

• Hypoeutectic alloys

– Alloy composition < eutectic composition

• Hypereutectic alloys

– Eutectic composition < alloy composition

Several different types of microstructures are possible

for the slow cooling of eutectic alloys!

C1 ≤≤

component of solubility solidmax

component pure

SnwtCSnwt %2%0 1 ≤≤

:system Pb-Sn theFor

C) 20(~ etemperatur roomat

20

C

component of solubilitymax

re temperaturoomat


2 ≤≤

SnwtCSnwt %3.18%2

:system Pb-Sn For the

re temperatueutecticat


2 ≤≤

( ) ( ) ( )SnwtSnwtSnwtL

SnwtC

C

.%8.97.%3.18.%9.61

:reaction Eutectic

%9.61


ncompositio eutectic

3

3

βα +⇔

=

=

Eutectic structure

• Microstructure of solid that results from the

eutectic reaction

• Alternating layers (lamellae) of α and β phases

that form simultaneouslythat form simultaneously

SnwtCSnwt

C

%9.61%3.18


isotherm. eutectic thecross

cooled, when that,eutectic the

other than nscompositio all

4

4

≤≤

∈

1

3.189.61

9.61

3.189.61

3.18

'4

'4

'

=+

−

−=

+=

−

−=

+==

PeutecticL

eutecticLe

WW

:Check

C

QP

QW

:primary the of fraction) (mass amount Phase

C

QP

PWW

:ituentmicroconst eutectic the of fraction) (mass amount Phase

α

α

α

The eutectic microconstituent always

forms from the liquid having the

eutectic composition

We, Wα’, are mass fractions of eutectic

microconstituent and primary α

1

3.188.97

3.18

3.188.97

8.97

'4

'4

=+

−

−=

++=

−

−=

++

+=

totaltotal

total

total

WW :Note

C

RPQ

PW

C

RQP

RQW

: and total the of fraction) (mass amounts Phase

βα

β

α

βα

Intermediate Phases or Compounds

• Terminal Solid Solution

– The two solid phases exist over composition ranges

near the concentration extremes

• Intermediate Solid Solution• Intermediate Solid Solution

– Intermediate phases found at other than the two

composition extremes

• Intermetallic Compound

– Discrete intermediate compounds exist (rather than

solid solutions) in metal-metal solutions

9.12 Equilibrium Diagrams having

Intermediate Phases or Compounds

Intermediate solid solutions

Terminal solid solution

(commercial brass)(commercial brass)

Intermetallic Compounds

Adapted from

Fig. 9.20, Callister &

Rethwisch 8e.

Mg-Mg2Pb

eutectic system

Mg2Pb-Pb

eutectic system

51

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).

• Mg2Pb (magnesium plumbide) exists by itself

only at the precise composition (81 wt% Pb)

• What is the melting temperature of Mg2Pb?

• How does the solubility of magnesium in lead

compare to the solubility of lead in magnesium?

• Note that the diagram could be separated into two

eutectic diagrams at Mg2Pb line

• Eutectoid – one solid phase transforms to two other solid

phases

S2 S1+S3

intermetallic compound -cementite

Eutectic, Eutectoid, & Peritectic

• Eutectic - liquid transforms to two solid phases

L α + β (For Pb-Sn, 183ºC, 61.9 wt% Sn)cool

heat

53

S2 S1+S3

γ α + Fe3C (For Fe-C, 727ºC, 0.76 wt% C)cool

heat

cool

heat

• Peritectic - liquid and one solid phase transform to a

second solid phase

S1 + L S2

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

Eutectoid & Peritectic

Cu-Zn Phase diagram

Peritectic transformation γ + L δ

54

Adapted from Fig. 9.21,

Callister & Rethwisch 8e.

Eutectoid transformation δ γ + ε

A eutectoid reaction is found in the iron-carbon system, very important in heat treating of

steels!

9.15 Congruent Phase Transformations

• Transformations with no change in composition• Melting of pure materials

• Allotropic transformations

• Incongruent• Incongruent

– At least one of the phases will experience a change in

composition

• Eutectic, eutectoid

• Melting of an alloy that belongs to an isomorphous system

Congruent melting for γ solid solution of 44.9 wt.% Ti

Where is peritectic

reaction?

Congruent/incongruent?

Portion of Ni-Ti Phase Diagram

Iron-Carbon (Fe-C) Phase Diagram

• 2 important points

- Eutectoid (B):

γ ⇒ α +Fe3C

- Eutectic (A):L⇒ γ + Fe3C

C (

cem

enti

te)

1600

1400

1200

1000

L

γ

(austenite)

γ+L

γ+Fe3C

δ

1148ºC

T(ºC)

γ γγγ

AL+Fe3C

57Adapted from Fig. 9.24,

Callister & Rethwisch 8e.

Fe 3

C (

cem

enti

te)

800

600

4000 1 2 3 4 5 6 6.7

γ+Fe3C

α+Fe3C

(Fe) C, wt% C

α 727ºC = T eutectoid

4.30

Result: Pearlite =

alternating layers of α and Fe3C phases

120 µm

(Adapted from Fig. 9.27,

Callister & Rethwisch 8e.)

0.76

B

γ γγγ

Fe3C (cementite-hard)

α (ferrite-soft)

C (

cem

enti

te)

1600

1400

1200

1000

800

L

γ

(austenite)

γ+L

γ + Fe3C

L+Fe3C

δ

1148ºC

T(ºC)

(Fe-C

System)

Hypoeutectoid Steel

Adapted from Figs. 9.24

and 9.29,Callister &

Rethwisch 8e. γ γα

γγγ γ

γ γγγ

58

Fe 3

C (

cem

enti

te)

800

600

4000 1 2 3 4 5 6 6.7

α+ Fe3C

(Fe) C, wt% C

α727ºC

C0

0.7

6

Rethwisch 8e.

(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.)

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

proeutectoid ferritepearlite

100 µmHypoeutectoid

steel

α

pearlite

γ

γ γ

γα

αα

C (

cem

enti

te)

1600

1400

1200

1000

800

L

γ

(austenite)

γ+L

γ + Fe3C

L+Fe3C

δ

1148ºC

T(ºC)

(Fe-C

System)

Hypoeutectoid Steel

γ

γ γ

γα

αα

srWα’ = s/(r + s)



Rethwisch 8e.

The eutectoid microconstituent

pearlite will form from the solid γhaving the eutectic composition

59

Fe 3

C (

cem

enti

te)

800

600

4000 1 2 3 4 5 6 6.7

α+ Fe3C

(Fe) C, wt% C

α727ºC

C0

0.7

6

srWγ =(1 – Wα’)R T

α

pearlite

Wpearlite = Wγ

Wαtotal = T/(R +T)

W =(1 – Wαtotal)Fe3C

Rethwisch 8e.

(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.)


proeutectoid ferritepearlite

100 µmHypoeutectoid

steel

Hypereutectoid Steel

C (

cem

enti

te)

1600

1400

1200

1000

800

L

γ

(austenite)

γ+L

γ +Fe3C

L+Fe3C

δ

1148ºC

T(ºC)



Rethwisch 8e. (Fig. 9.24

(Fe-C

System)

Fe3Cγ γ

γγγ γ

γγγ γ

60

Fe 3

C (

cem

enti

te)

800

600

4000 1 2 3 4 5 6 6.7

α +Fe3C

(Fe) C, wt%C

α


adapted from Binary Alloy

Phase Diagrams, 2nd ed.,

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

in-Chief), ASM

International, Materials

Park, OH, 1990.)

0.7

6 C0

Fe3C

γγ

γ γ


proeutectoid Fe3C

60 µm Hypereutectoid

steel

pearlite

pearlite

C (

cem

enti

te)

1600

1400

1200

1000

800

L

γ

(austenite)

γ+L

γ +Fe3C

L+Fe3C

δ

1148ºC

T(ºC)Hypereutectoid Steel

(Fe-C

System)Fe3C

γγ

γ γ

xvWγ =x/(v + x)




61

Fe 3

C (

cem

enti

te)

800

600

4000 1 2 3 4 5 6 6.7

α +Fe3C

(Fe) C, wt%C

α

0.7

6 C0

pearlite

xv

V X

Wpearlite = Wγ

Wαtotal = X/(V +X)

W =(1 - Wαtotal)Fe3C’

W =(1-Wγ)Fe3C


proeutectoid Fe3C

60 µm Hypereutectoid

steel

pearlite


adapted from Binary Alloy

Phase Diagrams, 2nd ed.,

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

in-Chief), ASM

International, Materials

Park, OH, 1990.)

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 Fe3C and ferrite (α).

62

3

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.

Solution to Example Problem

W =R

=C0 −Cα

b) Using the lever rule with

the tie line shown

a) Using the RS tie line just below the eutectoid

Cα = 0.022 wt% C

CFe3C = 6.70 wt% C

1600

1400 L

γ

δ

T(ºC)

63

WFe3C =R

R + S=

C0 −Cα

CFe3C −Cα

= 0.40 − 0.022

6.70 − 0.022 = 0.057

Fe 3

C (

cem

enti

te)1200

1000

800

600

4000 1 2 3 4 5 6 6.7

γ (austenite)

γ+L

γ + Fe3C

α+ Fe3C

L+Fe3C

C, wt% C

1148ºC

T(ºC)

727ºC

C0

R S

CFe C3Cαααα

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 g

Solution to Example Problem (cont.)

c) Using the VX tie line just above the eutectoid and realizing

thatC0 = 0.40 wt% C

Cα = 0.022 wt% C

Cpearlite = Cγ = 0.76 wt% C1600

1400 L

γ

δ

T(ºC)=

V=

C −Cα

64

Fe 3

C (

cem

enti

te)1200

1000

800

600

4000 1 2 3 4 5 6 6.7

γ (austenite)

γ+L

γ + Fe3C

α+ Fe3C

L+Fe3C

C, wt% C

1148ºC

T(ºC)

727ºC

C0

V X

CγγγγCαααα

Wpearlite =V

V + X=

C0 −Cα

Cγ −Cα

= 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

Alloying with Other Elements

• Teutectoid changes:

Eu

tect

oid

(ºC

) TiMo

SiW

Cr

• Ceutectoid changes:

eute

cto

id(w

t% C

)

Ni

Cr

SiMn

65

Adapted from Fig. 9.34,Callister & Rethwisch 8e.

(Fig. 9.34 from Edgar C. Bain, Functions of the

Alloying Elements in Steel, American Society for

Metals, 1939, p. 127.)

TE

ute

cto

id

wt. % of alloying elements

Ni

Mn

Adapted from Fig. 9.35,Callister & Rethwisch 8e. (Fig.

9.35 from Edgar C. Bain, Functions of the Alloying

Elements in Steel, American Society for Metals, 1939,

p. 127.)

wt. % of alloying elements

Ceu

tect

oid

Ti

MnW

Mo

• 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.

Summary

66

• 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.

9.11, 9.13, 9.17, 9.20, 9.32

Core Problems:

ANNOUNCEMENTS

Reading:

9.1 thru 9.20, will not be tested on 9.16, 9.17

67

Self-help Problems:

Documents

Introduction to Engineering Materials ENGR2000 Chapter 9 ...engineering.armstrong.edu/cameron/ENGR2000_phasediagrams1.pdf · •Tie lines and lever rule still apply. ... calculate