ISSUES TO ADDRESS...

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?

What is the structure of each phase

Chapter 9: Phase Diagrams

Phase B Phase A

Nickel atom Copper atom

Phase Behavior Laughlin p. 67

• 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., α and β).

Aluminum-Copper Alloy

Components and Phases

β (darker phase)

α (lighter phase)

1µm

Optical Mettalography

Solubility Limit

Question: What is the

solubility limit for sugar in water at 20ºC?

65

• Solubility Limit:

Maximum concentration for which only a single phase solution exists.

Sugar/Water Phase Diagram

Temperature (ºC)

0 20 40 60 80 100

Composition (wt% sugar)

L

(syrup)

Solubility

Limit L

(liquid)

+ S

(solid sugar) 20

40 60 80 100

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

• Mixture – more than one phase

Problem 9.2: At 170°C, what is the maximum solubility (a) of Pb in Sn? (b) of Sn in Pb?The lead-tin phase diagram is shown in the Animated Figure 9.8.

Isomorphous Binary Phase Diagram

• System is:

-- binary

2 components: Cu and Ni.

-- isomorphous

i.e., complete solubility of one component in another

Cu-Ni phase diagram

wt% Ni

20 40 60 80 100 10000

1100 1200 1300 1400 1500 1600

T(ºC)

L (liquid)

α solid solution

L + ^α

liquidus solidus

Criteria for Solid Solubility: Hume–Rothery rules

Crystal Structure

electroneg r (nm)

Ni FCC 1.9 0.1246

Cu FCC 1.8 0.1278

• Same crystal structure

• Similar electronegativities

• Similar atomic radii

• Ni and Cu are totally soluble in one another for all proportions.

Structure

BCC

FCC

Periodic Table

Cu-Ni phase diagram

wt% Ni

20 40 60 80 100 10000

1100 1200 1300 1400 1500 1600

T(ºC)

L

α L + ^α

Determination of phases present

• If we know T and C_o, then we know which phases are present.

• Examples:

A(1100ºC, 60 wt% Ni):

1 phase: α

B(1250ºC, 35 wt% Ni):

2 phases: L + α ^{B (}

1250ºC,35)

A(1100ºC,60)

Determination of phase compositions: Lever Rule

What fraction of each phase?

Think of the tie line as a lever (teeter-totter)

M_L M_α

R S



M_ x S  M_L x R

L L L

L L L

C C

C C

S R WR C

C

C C

S R

S M

M WM

























0

0

wt% Ni

20 1200 1300

T(ºC) L

α

30 40 50

L + α TB B

tie line

C₀

C_L C_α

S R

Tie line –– also sometimes called an isotherm

wt% Ni

20 1200 1300

30 40 50

1100

L

α T(ºC)

35 C₀

L: 35wt%Ni

C₀ = 35 wt% Ni alloy

Slow Cooling of a Cu-Ni Alloy

35 46 32 43

: 43 wt% Ni L: 32 wt% Ni α: 46 wt% NiL: 35 wt% Ni

L: 24 wt% Ni

: 36 wt% Ni

24 36

Last  to solidify:

< 35 wt% Ni

Cored vs Equilibrium Structures

Uniform C_: 35 wt% Ni

First α to solidify:

46 wt% Ni

wt% Ni

20 1200 1300

30 40 50

1100

L

α T(ºC)

35 C₀

35 46 32 43

24 36

Mechanical Properties: Cu-Ni System

• Effect of solid solution strengthening on:

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

Tensile Strength (MPa)

Composition, wt% Ni

Cu0 20 40 60 80 100Ni 200

300 400

TS for pure Ni TS for pure Cu

Elongation (%EL)

Composition, wt% Ni

Cu0 20 40 60 80 100Ni 20

30 40 50 60

%EL for pure Ni

%EL for pure Cu

2 components =low melting

Binary-Eutectic Systems

• 3 single phase regions (L, α, β)

Ex.: Cu-Ag system

L (liquid)

α L + α _L+

β β

α + β

wt% Ag

20 40 60 80 100

2000 1200

400 600 800 1000

C_E

T_{E 8.0} ^779ºC _{71.9 91.2}

Ag) wt%

1.2 9 ( Ag)

wt%

.0 8 ( Ag)

wt%

9 . 71

(   

L ^cooling

heating

Eutectic Phase Reaction:

Cu-Ag system

• For alloys for which C₀ < 2 wt% Sn

• Result: at room temperature

-- polycrystalline with grains of a phase having

composition C₀

Microstructural Development in Eutectic Systems I

0

L+ α

200

T(ºC)

wt% Sn

10 2

20 C₀

300

100

L α

30

α+β

400

(room T solubility limit)

T_E

Pb-Sn

αL L: C₀ wt% Sn

α: C₀ wt% Sn

• For alloys for which

2 wt% Sn < C₀ < 18.3 wt% Sn

• Result:

at temperatures in α + β range -- polycrystalline with a grains and small β-phase particles

Microstructural Development in Eutectic Systems II

L + α

200

T(ºC)

C, wt% Sn

10

18.3 20

0 C₀

300

100

L

α

30

α+ β

400

(sol. limit at T_E) T_E

(sol. limit at Troom)2

L α L: C₀ wt% Sn

αβ

a: C₀ wt% Sn

• For alloy of composition C₀ = C_E

• Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of α and β phases.

Microstructural Development in Eutectic Systems III

160μm

Micrograph of Pb-Sn eutectic microstructure

200

T(ºC)

wt% Sn

20 60 80 100

0

300

100

L

α β

L+ α

183ºC

40 T_E

18.3

α: 18.3 wt%Sn

97.8

β: 97.8 wt% Sn

C_E 61.9

L: C₀ wt% Sn

Lamellar Eutectic Structure

• For alloys for which 18.3 wt% Sn < C₀ < 61.9 wt% Sn

• Result: α phase particles and a eutectic microconstituent

Microstructural Development in Eutectic Systems IV

18.3 61.9

S R

97.8 R S

primary α eutectic α

eutectic β

W_L = (1- Wα) = 0.50 C_α = 18.3 wt% Sn C_L = 61.9 wt% Sn

S R+ S

W_α = = 0.50

• Just above T_E :

• Just below T_E :

Cα = 18.3 wt% Sn Cβ = 97.8 wt% Sn

S R+ S

Wα = = 0.73 Wβ = 0.27

200

T(ºC)

wt% Sn

20 60 80 100

0

300

100

L α

L+

α

40 T_E

L: C₀ wt% Sn L

α

L+α

L+β α +β

200

C, wt% Sn

20 60 80 100

0 300

100

L

α β

T_E

40

System)

Hypoeutectic & Hypereutectic

160 mm eutectic micro-constituent

hypereutectic: (illustration only)

β β β

β β

β

175 mm

α α

α α α

α

hypoeutectic: C₀ = 50 wt% Sn

T(ºC)

61.9 eutectic

eutectic: C₀=61.9wt% Sn

Intermetallic Compounds

Mg

Pb

Note: intermetallic compound exists as a line - not an area – because stoichiometry (i.e. composition of a compound) is fixed.

• Eutectoid –all solid phases S₂ S₁+S₃

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

intermetallic compound - cementite

cool heat

Eutectic, Eutectoid, & Peritectic

• Eutectic - liquid transforms to two solid phases

L S_heat^cool ₁+S₃ (For Pb-Sn, 183ºC, 61.9 wt% Sn)

cool heat

• Peritectic - liquid and one solid phase transform to a second solid phase

S₁ + L S₂

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

• Peritectoid – all solid phases S₁ + S₂ S₃

Eutectoid & Peritectic

Eutectoid transformation δ γ + ε

Peritectic transformation γ + L δ

Cu-Zn Phase diagram

Iron-Carbon (Fe-C) Phase Diagram

- Eutectoid (B):

γ → α + Fe³C

- Eutectic (A):

L → γ + Fe³C

Fe 3C (cementite)

1600 1400 1200

1000 800

600

4000 1 2 3 4 5 6 6.7

L γ

(austenite)

γ+L

γ^+Fe3C

α+Fe₃C α+γ

δ

(Fe) wt% C

1148ºC

T(ºC)

α 727ºC = Teutectoid

4.30

Result: Pearlite = alternating layers of

α and Fe₃C phases

120 mm 0.76

B

γ γγ γ

A _L+Fe

3C

Fe₃C (cementite-hard) α (ferrite-soft)

Hypereutectoid Steel

1600 1400 1200 1000 800

600

4000 1 2 3 4 5 6 6.7

L γ

(austenite)

γ+L

γ ^+Fe₃^C

α +Fe₃C

L+Fe₃C

δ

(Fe) C, wt%C

1148ºC

T(ºC)

a

0.76 ^C0

Fe₃C

γ γ γ γγ γγ

proeutectoid Fe₃C

60 μm

pearlite

pearlite

1600 1400 1200 1000 800

600

4000 1 2 3 4 5 6 6.7

L γ

(austenite)

γ+L

γ ^+Fe₃^C

α ^+Fe₃^C

L+Fe₃C

δ

(Fe) C, wt% C

1148ºC

T(ºC)

α

727ºC

C₀

0.76

Hypoeutectoid Steel

proeutectoid ferrite pearlite

100 μm

γ γγ γ

α

pearlite

γ γ γ

α γ

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).

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_Fe

3C  R

R  S  C₀ C_ C_Fe

3C C_  0.40 0.022

6.70 0.022  0.057

b) Wight Fraction of cementite

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

RS tie line just below the eutectoid

C_α = 0.022 wt% C C_Fe

3C = 6.70 wt% C

Fe 3C (cementite)

1600 1400 1200 1000 800 600

4000 1 2 3 4 5 6 6.7

L

γ

(austenite) γ+L

γ +Fe₃C

α ^+Fe₃^C

L+Fe₃C

δ

C, wt% C

1148ºC

T(ºC)

727ºC

C₀

R S

C_{Fe C} C_α 3

Amount of Fe₃C in 100 g

= (100 g)W_Fe3C

= (100 g)(0.057) = 5.7 g

pearlite

The amounts of pearlite in the 100 g.

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

C₀ = 0.40 wt% C C_α = 0.022 wt% C

C_pearlite = C_γ = 0.76 wt% C

Fe 3C (cementite)

1600 1400 1200 1000 800 600

4000 1 2 3 4 5 6 6.7

L (austenite)γ

γ+L

γ +Fe₃C

α^+Fe3C

L+Fe₃C

δ

C, wt% C

1148ºC

T(ºC)

727ºC

C₀ V X

C_γ C_α



W_pearlite  V

V  X  C₀ C_ C_ C_  0.40 0.022

0.76 0.022  0.512

Amount of pearlite in 100 g

= (100 g)W_pearlite

= (100 g)(0.512) = 51.2 g

α

pearlite

Alloying with Other Elements

• Teutectoid changes:

T Eutectoid (ºC)

wt. % of alloying elements

Ti

Ni

Mo Si

W

Cr Mn

• Ceutectoid changes:

wt. % of alloying elements C eutectoid (wt% C)

Ni

Ti

Cr

Si Mn

Mo W

VMSE: Interactive Phase Diagrams

Change alloy composition

Microstructure, phase compositions, and phase fractions respond interactively

• 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.

• 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

70 80 100 60

40 20

0

Temperature (ºC)

C = Composition (wt% sugar)

L

(liquid solution i.e., syrup)

20 100

40 60 80

0

L

(liquid)

+ S

(solid sugar)

Effect of Temperature & Composition

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

• Altering C can change # of phases: path B to D.

D (100ºC,C = 90)

2 phases

B (100ºC,C = 70)

1 phase

A (20ºC,C = 70)

2 phases

L+α

L+β

α + β

200

T(ºC)

18.3

wt% Sn

20 60 80 100

0 300

100

L (liquid)

α _183ºC

61.9 97.8 β

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

Pb-Snsystem

EX 1: Pb-Sn Eutectic System

-- the relative amount of each phase

150

40 C₀ 11

C_α

99 C_β

R S

C_α = 11 wt% Sn C_β = 99 wt% Sn

Wα = C_β - C₀ C_β - C_α

= 99 - 40

99 - 11 = 59

88 = 0.67 S

R+S =

Wβ = C₀ - C_α C_β - C_α R =

R+S

= 29

88 = 0.33

= 40 - 11 99 - 11

C_a = 17 wt% Sn

L+β

a + b

200

T(ºC)

wt% Sn

20 60 80 100

0 300

100

L

α β

L+ α

183ºC

• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC

EX 2: Pb-Sn Eutectic System

the relative amount of each phase

Wa = C_L - C₀

C_L - C_α = 46 - 40 46 - 17

= 6

29 = 0.21 W_L = C₀ - C_α

C_L - C_α = 23

29 = 0.79

40 C₀

46 C_L 17

C_α

220 R S

C_L = 46 wt% Sn