Chapter 10: Phase Diagrams and Microstructures

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Chapter 10: Phase Diagrams and Microstructures. Temperature vs. composition behavior of the various alloy constitutes the phases. Allows design and control of heat treatment by controlling equilibrium and non-equilibrium structure.

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Chapter 10: Phase Diagrams and Microstructures

• Temperature vs. composition behavior of the various
• alloy constitutes the phases.
• Allows design and control of heat treatment by controlling
• equilibrium and non-equilibrium structure.
• Structure(phases and microstructure) often controls properties
• of materials, and, therefore, depends on thermal history.
• Objectives:
• Read and evaluate phases present at T.
• Determine composition of phases and phase fraction.
• Know difference between types of reactions, e.g., eutectic, eutectoid, proeutectoid.
• Make a schematic diagram of microstructure.

Phase B

Phase A

Nickel atom

Copper atom

• When we combine two elements, what equilibrium state do we get?

• In particular, if we specify...

--a composition (e.g., wt%A - wt%B), and

--a temperature (T)

then...

How many phases do we get?

What is the composition of each phase?

How much of each phase do we get?

Ordered

Phase

Solid-solution

Phase

Example Phase (T-c) Diagram : Cu-Zn

• Solution – solid, liquid, or gas solutions, single phase

• Mixture – more than one phase

liquid

bcc-ss

fcc-ss

2-phase

hcp-ss

hcp

bcc

CsCl

Sugar/Water Phase Diagram

10

0

Solubility

L

Limit

8

0

(liquid)

6

0

+

L

Temperature (°C)

S

(liquid solution

4

0

i.e., syrup)

(solid

20

sugar)

0

20

40

60

80

100

C

= Composition (wt% sugar)

Sugar

Water

Solubility Limit in Phase Diagram

• Solubility Limit:

Max concentration for which only a solution occurs.

• Ex: Water-Sugar

Q: What is solubility limit at 20oC?

If Co < 65wt% sugar:syrup

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

Callister & Rethwisch 3e.

• Solubility limit increases with T:

e.g., if T = 100C, solubility limit = 80wt% sugar.

Components and Phases

• Components: elements or compounds that are mixed initially (e.g., Al and Cu)

• Phases: physically and chemically distinct material regions that result (e.g., a and b).

Aluminum-Copper Alloy

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

B(100°C,C = 70)

1 phase

D(100°C,C = 90)

2 phases

100

L

80

(liquid)

+

60

L

S

Temperature (°C)

(

liquid solution

(solid

40

i.e., syrup)

sugar)

A(20°C,C = 70)

2 phases

20

0

0

20

40

60

70

80

100

C

= Composition (wt% sugar)

Temperature (T) and Composition (C) Effects

• Changing T can change # of phases: path A to B.

• Changing Co can change # of phases: path B to D.

water- sugar

system

Callister & Rethwisch 3e.

Criteria for Solid Solubility: recall Hume-Rothery rules

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

• Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility.
• Ni and Cu are totally soluble in one another for all proportions.

T(°C)

• 2 phases:

1600

(liquid)

1500

L (liquid)

(FCC solid solution)

• 3 phase fields:

1400

L

a

liquidus

+

1300

L +

a

L

solidus

a

a

1200

(FCC solid

1100

solution)

1000

wt% Ni

0

20

40

60

80

100

Cu-NiPhase Diagram: T vs. c (wt% or at%)

• Tell us about phases as function of T, Co, P.

• For this course:

--binary systems: just 2 components.

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

-- isomorphous

i.e., complete solubility of one component in another; a phase field extends from 0 to 100 wt% Ni.

Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

T(°C)

1600

1500

L (liquid)

1 phase: a

B

(1250°C, 35 wt% Ni):

(1250°C,35)

1400

liquidus

2 phases: L + a

a

+

1300

solidus

L

B

a

(FCC solidsolution)

1200

A(1100°C,60)

1100

1000

wt% Ni

0

20

40

60

80

100

Phase Diagrams: # and types of phases

• Rule 1: If we know T and Co, then we know:

--the # and types of phases present.

• Examples:

Cu-Ni

phase

diagram

A(1100°C, 60 wt% Ni):

Cu-Ni system

T(°C)

A

TA

tie line

liquidus

L (liquid)

At TA

= 1320°C:

1300

a

+

L

Only Liquid (L) present

B

TB

solidus

CL = C0

( = 35 wt% Ni)

a

a

At TD

= 1190°C:

+

L

(solid)

1200

D

Only Solid (a) present

TD

C = C0

( = 35 wt% Ni)

32

35

4

3

20

30

40

50

At TB

= 1250°C:

CL

C0

C

wt% Ni

Both

and L present

CL

= C

( = 32 wt% Ni)

liquidus

C

= C

( = 43 wt% Ni)

solidus

Phase Diagrams: composition of phases

• Rule 2: If we know T and Co, then we know:

--the composition of each phase.

• Examples:

Consider C0 = 35 wt% Ni

Cu-Ni system

T(°C)

A

TA

liquidus

tie line

L (liquid)

At TA

: Only Liquid (L) present

1300

a

+

L

B

WL= 1.00, Wa = 0

TB

solidus

At TD

:

Only Solid (

) present

S

R

a

a

+

WL

= 0, W

= 1.00

L

a

(solid)

1200

D

TD

At TB

:

Both

and L present

32

35

4

3

20

3

0

4

0

5

0

S

=

WL

CL

C0

Ca

wt% Ni

R

+

S

R

=

= 0.27

Wa

R

+

S

Phase Diagrams: wt. fraction of phases

• Rule 3: If we know T and C0, then can determine:

-- the weight fraction of each phase.

• Examples:

Consider C0 = 35 wt% Ni

W = wt. fraction of phase out of whole.

T(°C)

tie line

liquidus

L (liquid)

1300

a

+

L

B

solidus

T

B

a

a

+

L

(solid)

1200

S

R

20

3

0

4

0

5

0

C0

CL

C

wt% Ni

The Lever Rule

• Sum of weight fractions:

• Conservation of mass (Ni):

• Combine above equations:

T(°C)

tie line

liquidus

L (liquid)

1300

a

+

M

ML

L

B

solidus

T

B

a

a

+

L

(solid)

1200

R

S

S

R

20

3

0

4

0

5

0

C0

CL

C

wt% Ni

The Lever Rule: an interpretation

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

What fraction of each phase?

Think of tie line as a lever

T(°C)

L: 35wt%Ni

L (liquid)

a

130

0

A

+

L

L: 35 wt% Ni

B

a: 46 wt% Ni

35

46

C

32

43

D

L: 32 wt% Ni

24

36

a

a: 43 wt% Ni

+

120

0

E

L

L: 24 wt% Ni

a: 36 wt% Ni

a

(solid)

110

0

35

20

3

0

4

0

5

0

wt% Ni

C0

Phase and Microstructure (equilibrium)

Example: Cooling in Cu-Ni Binary

• Consider microstuctural changes that accompany the cooling of a

C0 = 35 wt% Ni alloy

• From liquid, solid phase nucleates.
• From solid, other phases can nucleate.
• Like ice, many grains of solid form.
• wt% of SOLUTE given by line dropped from boundaries
• Fraction of PHASES present given by the “lever rule”.
• Microstructure different depending on cool slowly or quench.

Phases and Microstructure (non-equilibrium)

• Upon cooling quickly (i.e. nonequilibrium), microstructure has range of composition depending on when it was formed.
• Inside nucleus of solid phase higher composition (in Cu-Ni case) due to its creation at higher T.
• Outside part of growing solid phase nucleus has lower composition due to its forming at lower T.

Core-like development

Cored versus Equilibrium Phases

• Ca changes as we solidify.

• Cu-Ni case:

First a to solidify has Ca = 46 wt% Ni.

Last a to solidify has Ca = 35 wt% Ni.

• Fast rate of cooling:

Cored structure

• Slow rate of cooling:

Equilibrium structure

60

%EL

for pure Cu

400

%EL

for

50

pure Ni

TS for

Elongation (%EL)

40

pure Ni

Tensile Strength (MPa)

300

30

TS for pure Cu

200

20

0

20

40

60

80

100

0

20

40

60

80

100

Cu

Ni

Cu

Ni

Composition, wt% Ni

Composition, wt% Ni

Recall: Mechanical Properties of Cu-Ni

• Effect of solid solution strengthening on:

--Ductility (%EL,%AR)

--Tensile strength (TS)

-Peak as a function of Co

-Minimum as a function of Co

Cu-Ag

system

T(°C)

1200

L (liquid)

1000

a

L

+

a

b

L

+

779°C

b

800

TE

8.0

91.2

71.9

600

a

+

b

400

200

80

100

0

20

40

60

CE

C

,

wt% Ag

cooling

heating

Binary-Eutectic Systems

has a special composition

with a minimum melting T.

• Ex: Cu-Ag
• 3 single-phase regions
• (L, , )
• Limited solubility
• : mostly Cu
• : mostly Ag
• TE: no liquid below TE.
• cE: composition for min. melting T (Eutectic).

Eutectic: direct from liquid to 2-phase solid upon cooling: L   + 

T(°C)

300

L (liquid)

a

L

+

a

b

b

L

+

200

183°C

18.3

61.9

97.8

C- C0

150

S

R

S

=

W

=

R+S

C- C

100

a

+

b

99 - 40

59

=

=

= 0.67

99 - 11

88

100

0

11

20

60

80

99

40

C0- C

R

W

C

C

C0

=

=

C, wt% Sn

C - C

R+S

40 - 11

29

=

= 0.33

=

99 - 11

88

Solder for electronics

Example 1: Lead-Tin (Pb-Sn) Eutectic Diagram

• For a 40wt%Sn-60wt%Pb alloy at 150oC, determine...

--phases present: a + b

--compositions of phases:

Ca = 11 wt% Sn

Cb = 99 wt% Sn

-- relative amount of each phase:

Use the “Lever Rule”

Adapted from Fig. 10.8, Callister & Rethwisch 3e.

T(°C)

300

L (liquid)

a

L

+

220

a

b

b

R

L

+

S

200

CL - C0

46 - 40

183°C

=

W

=

= 0.21

a

CL - C

46 - 17

100

a

+

b

100

17

46

0

20

40

60

80

C

CL

C0

C, wt% Sn

C0 - C

23

=

WL

=

= 0.79

CL - C

29

Example 2: Pb-Sn Eutectic System

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

-- phases present:

a + L

-- phase compositions

Ca = 17 wt% Sn

CL = 46 wt% Sn

-- relative amt of phases

T(°C)

L: C0wt% Sn

400

L

a

L

300

L

a

+

a

200

(Pb-Sn

a: C0wt% Sn

TE

System)

100

b

+

a

0

10

20

30

C

,

wt% Sn

C0

2

(room T solubility limit)

Microstructure for Eutectic Diagrams: Pb-Sn

• For alloys with C0 < 2 wt% Sn

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

a phase having C0

• Note: liquid-to-solid creates grain boundaries.

L: C0 wt% Sn

T(°C)

400

L

L

300

a

L

+

a

a: C0wt% Sn

a

200

TE

a

b

100

b

+

a

Pb-Sn

system

0

10

20

30

C

,

wt% Sn

C0

2

(sol. limit at T

)

18.3

room

(sol. limit at TE)

Microstructure for Eutectic Diagrams: Pb-Sn

• For alloys with

2 wt% Sn < C0< 18.3 wt% Sn

• Result:

at T’s in a + b range

-- polycrystalline with agrains

and small b-phase particles

• Note: Solubility depends on T.

Micrograph of Pb-Sn

T(°C)

eutectic

L: C0 wt% Sn

microstructure

300

L

Pb-Sn

system

a

L

+

a

b

L

200

183°C

TE

100

160m

a

: 97.8 wt% Sn

Adapted from Fig. 10.14, Callister & Rethwisch 3e.

: 18.3 wt%Sn

0

20

40

60

80

100

97.8

18.3

CE

C, wt% Sn

61.9

Microstructure at Eutectic

• For alloy of composition C0 = CE

• Result: Eutectic microstructure (lamellar structure)

-- alternating layers (lamellae) of a and b phases.

Light: Sn-rich

Dark: Pb-rich

Adapted from Fig. 10.13, Callister & Rethwisch 3e.

Lamellar Eutectic Structure

Diffusion local

Adapted from Figs. 10.14 & 10.15, Callister & Rethwisch 3e.

• Just above TE :

L

T(°C)

L: C0

wt% Sn

Ca

= 18.3 wt% Sn

L



CL

= 61.9 wt% Sn

300

L

Pb-Sn

system

S

Wa

L

+

= 0.50

=

R

+

S

a

b

WL

= (1-

W

)

= 0.50

L

+

a

R

S

200

TE

S

R

• Just below TE :

C

= 18.3 wt% Sn

100

+

a

primary

C

= 97.8 wt% Sn

a

eutectic

S

b

eutectic

W

= 0.73

=

R

+

S

0

20

40

60

80

100

W

= 0.27

18.3

61.9

97.8

Adapted from Fig. 10.16, Callister & Rethwisch 3e.

C, wt% Sn

Microstructure “below” Eutectic (hypoeutectic)

• For alloys for 18.3wt%Sn < Co < 61.9wt%Sn

• Result: a crystals and a eutectic microstructure

• For 50 wt% Pb alloy:
• Lamellar eutectic structure
• of Sn-rich  phase (light).

L

L + 

 + 

* Why is Liquid-phase ~62.9wt%Sn and -phase ~16.3wt%Sn at 180 C?

* For fraction of total  phase (both eutectic and primary), use the Lever Rule.

Hypoeutectic & Hypereutectic

Callister & Rethwisch 3e.

(Figs. 10.14 and 10.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)

Mg2Pb

Intermetallic Compounds

Fig. 10.20, Callister & Rethwisch 3e.

Intermetallic compounds exists as a line on the phase diagram - not an area - because of stoichiometry (i.e. c of a compound is a fixed value).

cool

cool

cool

heat

heat

heat

• Eutectoid – one solid phase transforms to two other solid phases
• S2S1+S3
•  + Fe3C (For Fe-C, 727C, 0.76 wt% C)

intermetallic compound - cementite

• Peritectic - liquid and one solid phase transform to a second solid phase
• S1 + LS2
•  + L (For Fe-C, 1493°C, 0.16 wt% C)

Eutectic, Eutectoid, & Peritectic

• Eutectic - liquid transforms to two solid phases
• L +  (For Pb-Sn, 183C, 61.9 wt% Sn)

Peritectic transformation  + L

Eutectoid transformation  + 

Eutectoid, & Peritectic

Cu-Zn Phase diagram

Adapted from Fig. 10.21, Callister & Rethwisch 3e.

Ceramic Phase Diagrams

MgO-Al2O3 diagram:

Adapted from Fig. 10.24, Callister & Rethwisch 3e.

Iron-Carbon (Fe-C) Phase Diagram

T(°C)

1600

Adapted from Fig. 10.28,Callister & Rethwisch 3e.

d

L

1400

g

+L

A

- Eutectic (A):

g

L+Fe3C

1200

1148°C

(austenite)

Þ

g

+

L

Fe3C

A

- Eutectoid (B):

1000

g

+Fe3C

Fe3C (cementite)

a

g

Þ

a

+

Fe3C

g

g

+

800

g

a

g

g

B

727°C = T

eutectoid

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

C, wt% C

(Fe)

Result: Pearlite is

4.30

0.76

alternating layers of and Fe3C phases

120 mm

a

Fe3C (cementite-hard)

a

(ferrite-soft)

• 2 points

T(°C)

1600

d

L

1400

(Fe-C

g

+L

g

g

g

System)

1200

L+Fe3C

1148°C

(austenite)

g

g

g

1000

g

g

+Fe3C

g

g

Fe3C (cementite)

a

g

g

800

727°C

a

a

a

g

g

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

a

C, wt% C

(Fe)

C0

0.76

pearlite

Hypoeutectoid

100 mm

steel

pearlite

proeutectoid ferrite

Adapted from Fig. 10.34, Callister & Rethwisch 3e.

Hypoeutectoid Steel

Adapted from Figs. 10.28 and 10.33

T(°C)

1600

d

L

1400

(Fe-C

g

+L

g

System)

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Wa= s/(r+s)

Wa’= S/(R+S)

Fe3C (cementite)

a

r

s

g

g

800

Wg=(1 - Wa)

W=(1 – Wa’)

727°C

a

a

a

g

g

R

S

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

a

C, wt% C

(Fe)

C0

0.76

Wpearlite=Wg

pearlite

Hypoeutectoid

100 mm

steel

Fe3C

pearlite

proeutectoid ferrite

Adapted from Fig. 10.34, Callister & Rethwisch 3e.

Hypoeutectoid Steel

T(°C)

1600

d

L

1400

g

+L

g

g

g

1200

L+Fe3C

1148°C

g

g

(austenite)

g

1000

g

g

+Fe3C

g

g

Fe3C (cementite)

Fe3C

800

g

g

a

g

g

600

a

+Fe3C

400

C0

0

1

2

3

4

5

6

6.7

C, wt%C

0.76

(Fe)

pearlite

Hypereutectoid

60mm

steel

pearlite

proeutectoid Fe3C

Adapted from Fig. 10.37, Callister & Rethwisch 3e.

Hypereutectoid Steel

T(°C)

1600

d

L

1400

g

+L

g

Fe3C

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Wg=x/(v + x)

Fe3C (cementite)

x

v

800

g

W=(1-Wg)

g

Fe3C

a

g

g

V

X

600

a

+Fe3C

400

C0

0

1

2

3

4

5

6

6.7

C, wt%C

0.76

(Fe)

pearlite

Wpearlite=Wg

Wa= X/(V+X)

Hypereutectoid

60mm

W=(1 - Wa)

steel

Fe3C’

pearlite

proeutectoid Fe3C

Adapted from Fig. 10.37, Callister & Rethwisch 3e.

Hypereutectoid Steel

Example Problem Steel

• For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following:
• The compositions of Fe3C and ferrite ().
• The amount of cementite (in grams) that forms in 100 g of steel.
• The amounts of pearlite and proeutectoid ferrite () in the 100 g.

1600

d

L

1400

T(°C)

g

+L

g

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Fe3C (cementite)

800

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 g

727°C

R

S

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

C0

C

C, wt% C

CFe C

3

Solution to Problem

a) Use RS tie line just below Eutectoid

Ca = 0.022 wt% CCFe3C = 6.70 wt% C

• Use lever rule with the tie line shown

1600

d

L

1400

T(°C)

g

+L

g

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Fe3C (cementite)

Amount of pearlite in 100 g

= (100 g)Wpearlite

= (100 g)(0.512) = 51.2 g

800

727°C

V

X

600

a

+Fe3C

400

C0

0

1

2

3

4

5

6

6.7

C

C

C, wt% C

Solution to Problem

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

C0 = 0.40 wt% CCa = 0.022 wt% CCpearlite = C = 0.76 wt% C

Alloying Steel With More Elements

• Teutectoid changes:

• Ceutectoid changes:

(wt% C)

Ti

Si

Mo

(°C)

Ni

W

Cr

Cr

eutectoid

Eutectoid

Si

Mn

W

Mn

Ti

Mo

C

T

Ni

wt. % of alloying elements

wt. % of alloying elements

Adapted from Fig. 10.38,Callister & Rethwisch 3e. (Fig. 10.38 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

Adapted from Fig. 10.39,Callister & Rethwisch 3e. (Fig. 10.39 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

Summary

Phase Diagrams and Microstructures

• Phase (T vs c) diagrams are useful to determine:

- the number and types of phases,

- 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.

- Alloy composition and annealing temperature and time

determine possible microstructure(s). Slow vs. Fast.

- In 2-phase regions, use “lever rule” to get wt% of phases.

- Varying microstructure affects mechanical properties.