MET assignment 2

Moh-993

MET180-Chapter11.pdf

Home >Physics homework help >Mechanics homework help >MET assignment 2

MET 180 Structure and Properties of Materials

Week 15

CHAPTER 11:

Phase Diagrams

Content

1. Components and Phases

2. Phase Diagrams

3. Eutectic, Eutectoid and Peritectic Systems

4. Iron - Carbon (Fe-C) System

5. Summary

ISSUES TO ADDRESS...

• When we combine two elements...

what is the resulting equilibrium state?

• In particular, 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 11: Phase Diagrams

Phase B

Phase A

Nickel atom Copper atom

Phase Equilibria: Solubility Limit

Question: What is the solubility limit for sugar in

water at 20°C?

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

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

• Solubility Limit: Maximum concentration for

which only a single phase

solution exists.

Sugar/Water Phase Diagram

S u

g a

T e

m p

e ra

tu re

(° C

) 0 20 40 60 80 100

C = Composition (wt% sugar)

L (liquid solution

i.e., syrup)

Solubility Limit L

(liquid)

+ S

(solid sugar)20

100

W a

te r

Fig. 11.1,

Callister & Rethwisch 9e.

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

• Mixture – more than one phase

• Components: The elements or compounds which are present in the alloy

(e.g., Copper-zinc brass: the components are Cu and Zn)

• Phases: The physically and chemically distinct material regions

that form (e.g., α and β).

Aluminum-

Copper

Alloy

Components and Phases

α (darker phase)

β (lighter phase)

70 80 1006040200

T e

m p

e ra

tu re

( °

C )

C = Composition (wt% sugar)

L (liquid solution

i.e., syrup)

100

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.

water-

sugar

system

Fig. 11.1, Callister &

Rethwisch 9e.

D (100°C,C = 90) 2 phases

B (100°C,C = 70) 1 phase

A (20°C,C = 70) 2 phases

Criteria for Solid Solubility

Crystal Structure

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

electro-negativities and atomic radii suggesting high mutual

solubility.

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

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

• A phase is homogenous portion of a material system that has uniform physical and chemical characteristics.

• Every pure material is considered to be a phase; so also is every solid, liquid and gaseous solution.

Phase Diagrams

• System is an alloy consisting of the same components (e.g., the iron–carbon system)

• The understanding of phase diagrams for alloy systems is extremely important because there is a strong correlation between microstructure and mechanical properties.

• The development of microstructure of an alloy is related to the characteristics of its phase diagram.

• In addition, phase diagrams provide valuable information about melting, casting, and crystallization.

Phase Diagrams

• The physical and mechanical properties of a material depend on the microstructure.

• In metal alloys, microstructure is characterized by the number of phases present, their proportions, and the manner in which they are distributed.

Phase Diagrams

• The microstructure of an alloy depends on alloying elements present, their concentrations, and the heat treatment of the alloy

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

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

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

Phase

Diagram

for Cu-Ni

system

Fig. 11.3(a), Callister & Rethwisch 9e.

• 2 phases:

L (liquid) α (FCC solid solution)

• 3 different phase fields:

L + α

wt% Ni20 40 60 80 1000 1000

1100

1200

1300

1400

1500

1600 T(°C)

L (liquid)

(FCC solid solution)

Cu-Ni

phase

diagram

Isomorphous Binary Phase Diagram

• Phase diagram: Cu-Ni system.

• System is:

Fig. 11.3(a), Callister & Rethwisch 9e

-- binary i.e., 2 components:

Cu and Ni.

-- isomorphous i.e., complete solubility of one

component in another;

α phase field extends from

0 to 100 wt% Ni. wt% Ni20 40 60 80 1000 1000

1100

1200

1300

1400

1500

1600 T(°C)

L (liquid)

(FCC solid solution)

wt% Ni20 40 60 80 1000 1000

1100

1200

1300

1400

1500

1600 T(°C)

L (liquid)

α (FCC solid

solution)

Cu-Ni

phase

diagram

Phase Diagrams: Determination of phase(s) present

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

• Examples:

A(1100°C, 60 wt% Ni): 1 phase: α

B(1250°C, 35 wt% Ni):

2 phases: L + α

B (1 2 5 0 ºC ,3 5 )

A(1100ºC,60)

Fig. 11.3(a), Callister & Rethwisch 9e

wt% Ni

1200

1300

T(°C)

L (liquid)

(solid)

30 40 50

Cu-Ni

system

Phase Diagrams: Determination of phase compositions

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

• Examples:

TA A

35 C0

32 CL

At TA = 1320°C:

Only Liquid (L) present CL = C0 ( = 35 wt% Ni)

At TB = 1250°C:

Both α and L present

CL = Cliquidus ( = 32 wt% Ni)

Cα = Csolidus ( = 43 wt% Ni)

At TD = 1190°C:

Only Solid (α) present

Cα = C0 ( = 35 wt% Ni)

Consider C0 = 35 wt% Ni

D TD

tie line

4 Cα 3

Fig. 11.3(b), Callister & Rethwisch 9e.

B TB

• Rule 3: If we know T and 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 (α ) present

WL = 0, Wα = 1.00

Phase Diagrams: Determination of phase weight fractions

wt% Ni

1200

1300

T(°C)

L (liquid)

(solid)

30 40 50

Cu-Ni

system

TA A

35 C0

32 CL

B TB

D TD

tie line

4 Cα 3

R S

At TB : Both α and L present

73.0 3243

3543 =

- =

= 0.27

WL = S

R +S

Wα = R

R +S

Consider C0 = 35 wt% Ni

Fig. 11.3(b), Callister & Rethwisch 9e.

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

The Lever Rule

What fraction of each phase?

Think of the tie line as a lever

(teeter-totter)

ML Mα

R S

wt% Ni

1200

1300

T(°C)

L (liquid)

(solid)

30 40 50

B TB

tie line

C0 CL Cα

• Phase diagram: Cu-Ni system.

Fig. 11.4, Callister &

Rethwisch 9e.

• Consider

microstuctural

changes that

accompany the

cooling of a

C0 = 35 wt% Ni alloy

Example: Cooling of a Cu-Ni Alloy

wt% Ni 20

1200

1300

30 40 50 110 0

L (liquid)

(solid)

T(°C)

35 C0

L: 35 wt%Ni

Cu-Ni

system

4635

43 32

α: 43 wt% Ni

L: 32 wt% Ni

Bα: 46 wt% Ni L: 35 wt% Ni

E L: 24 wt% Ni

α: 36 wt% Ni

24 36 D

α: 35 wt% Ni

Mechanical Properties: Cu-Ni System

• Effect of solid solution strengthening on:

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

Fig. 11.5(a), Callister & Rethwisch 9e.

T e

n s ile

S tr

e n

g th

( M

P a

)

Composition, wt% Ni Cu Ni 0 20 40 60 80 100

200

300

400

TS for pure Ni

TS for pure Cu

E lo

n g

a ti o

n (

% E

L )

Composition, wt% Ni Cu Ni 0 20 40 60 80 100

%EL for pure Ni

%EL for pure Cu

Fig. 11.5(b), Callister & Rethwisch 9e.

Eutectic Systems

Cu-Ag

system

L (liquid)

α L + α L+ββ

α + β

C, wt% Ag 20 40 60 80 1000

200

1200 T(°C)

400

600

800

1000

TE 8.0 71.9 91.2 779°C

• Eutectic reaction

L α + β cooling

heating

• Eutectic means “easily melted”.

• Upon cooling, a liquid phase is transformed into the two solid phases at the temperature TE; the opposite reaction occurs upon heating.

2 components has a special composition

with a min. melting T.

Fig. 11.6, Callister & Rethwisch 9e

Binary - Eutectic Systems

• 3 single phase regions

(L, α, β)

• Limited solubility:

α: mostly Cu

β: mostly Ag

• TE : No liquid below TE

: Composition at

temperature TE

• CE

Ex.: Cu-Ag system

Cu-Ag

system

L (liquid)

α L + α L+ββ

α + β

C, wt% Ag 20 40 60 80 1000

200

1200 T(°C)

400

600

800

1000

TE 8.0 71.9 91.2 779°C

cooling

heating

• Eutectic reaction

L(CE) α(CαE) + β(CβE)

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

Ex. 1: Pb-Sn Eutectic System

Answer: α + β

-- the phase compositions

-- the relative amount of each phase L+α

L+β

α + β

200

T(°C)

18.3

C, wt% Sn

20 60 80 1000

300

100

L (liquid)

α 183°C

61.9 97.8 β

Pb-Sn

system

150

40 C0

11 Cα

99 Cβ

Answer: Cα = 11 wt% Sn Cβ = 99 wt% Sn

Wα = Cβ - C0 Cβ - Cα

= 99 - 40

99 - 11 =

88 = 0.67

S R+S

Wβ = C0 - Cα Cβ - Cα

= R

R+S

= 29

88 = 0.33=

40 - 11

99 - 11

Answer:

Fig. 11.7, Callister & Rethwisch 9e.

Answer: Cα = 17 wt% Sn

-- the phase compositions

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

Ex. 2: Pb-Sn Eutectic System

-- the relative amount

of each phase

Wα = CL - C0

CL - Cα =

46 - 40

46 - 17

= 6

29 = 0.21

WL = C0 - Cα

CL - Cα =

29 = 0.79

L+β

α + β

200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

L (liquid)

α β

L+ α

183°C

Pb-Sn system

40 C0

46 CL

17 Cα

220 SR

Answer: α + L

CL = 46 wt% Sn

Answer:

Fig. 11.7, Callister & Rethwisch 9e.

• For alloys for which

C0 < 2 wt% Sn

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

α phase having

composition C0

Microstructural Developments

in Eutectic Systems I

L+ α 200

T(°C)

C, wt% Sn 10

20 C0

300

100

α +β

400

(room T solubility limit)

TE (Pb-Sn System)

α L

L: C0 wt% Sn

α: C0 wt% Sn

Fig. 11.10, Callister &

Rethwisch 9e.

• For alloys for which

2 wt% Sn < C0 < 18.3 wt% Sn

• Result:

at temperatures in α + β range

-- polycrystalline with α grains

and small β-phase particles

Fig. 11.11, Callister &

Rethwisch 9e.

Microstructural Developments

in Eutectic Systems II

Pb-Sn system

L + α

200

T(°C)

C, wt% Sn 10

18.3

200 C0

300

100

α + β

400

(sol. limit at TE)

2 (sol. limit at Troom)

L: C0 wt% Sn

α β

α: C0 wt% Sn

• For alloy of composition C0 = CE • Result: Eutectic microstructure (lamellar structure)

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

Fig. 11.12, Callister &

Rethwisch 9e.

Microstructural Developments

in Eutectic Systems III

Fig. 11.13, Callister & Rethwisch 9e.

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

α β

L+ α

183°C

18.3

α: 18.3 wt%Sn

97.8

β: 97.8 wt% Sn

CE 61.9

L: C0 wt% Sn

Lamellar Eutectic Structure

Figs. 11.13 & 11.14,

Callister & Rethwisch 9e. a eutectic micro-constituent

Photomicrograph showing the microstructure of a lead–tin alloy of eutectic composition.

This microstructure consists of alternating layers of a lead- rich α-phase solid solution, and a tin-rich β-phase solid solution.

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

• Result: α phase particles and a eutectic micro-constituent

Microstructural Developments

in Eutectic Systems IV

WL = (1-Wα) = 0.50

Cα = 18.3 wt% Sn

CL = 61.9 wt% Sn S

R + S Wα = = 0.50

• Just above TE :

• Just below TE :

Cα = 18.3 wt% Sn

Cβ = 97.8 wt% Sn S

R + S Wα = = 0.73

Wβ = 0.2718.3 61.9

97.8

primary α

eutectic α

eutectic β

Fig. 11.15

Pb-Sn

system

L+β200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

α β

L+α

α+ β

L: C0 wt% Sn Lα L

L+α L+β

α + β

200

C, wt% Sn20 60 80 1000

300

100

α β TE

(Pb-Sn System)

Hypoeutectic & Hypereutectic

Fig. 11.7, Callister & Rethwisch

160 μm

eutectic micro-constituent Fig. 11.13

hypereutectic: (illustration only)

β β

Fig. 11.16 175 μm

α α

hypoeutectic: C0 = 50 wt% Sn

Fig. 11.16

T(°C)

61.9

eutectic

eutectic: C0 =61.9wt% Sn

Intermetallic Compounds

• For some metal–metal systems, discrete compounds rather than solid solutions may be found on the phase diagram, and these compounds have distinct chemical formulas.

• For the magnesium–lead phase system, the compound Mg2Pb has a composition of 81 wt% Pb.

• It is represented as a vertical line on the diagram, rather than an area.

Intermetallic Compounds

Intermetallic compound exists as a line on the diagram - not an area –

because composition of a compound is a fixed value).

Mg2Pb Fig. 11.19, Callister

& Rethwisch 9e.

The magnesium–lead phase diagram

• Eutectoid – one solid phase transforms to two other

solid phases

S2 S1+S3

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

intermetallic compound - cementite

cool

heat

Eutectic, Eutectoid, & Peritectic • Eutectic - liquid transforms to two solid phases

L α + β (For Pb-Sn, 183°C, 61.9 wt% Sn) 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, 1493°C, 0.16 wt% C)

Eutectoid & Peritectic

Cu-Zn Phase diagram

Fig. 11.20, Callister & Rethwisch 9e.

Eutectoid transformation δ γ + e

Peritectic transformation γ + L δ

The Iron – Carbon System

• Of all binary alloy systems, the one that is possibly the most important is that for iron and carbon.

• Both steels and cast irons are essentially iron– carbon alloys.

• Steels are primary structural materials in every technologically advanced culture.

The Iron – Carbon System

The Iron – Carbon System • Pure iron, upon heating, experiences two changes in

crystal structure before it melts.

• At room temperature the stable form, called α-ferrite, has a BCC crystal structure.

• α-ferrite transforms to FCC γ-austenite, at 912°C.

• This austenite persists to 1394°C, at which temperature the FCC austenite reverts back to a BCC phase known as δ-ferrite, which finally melts at 1538°C.

The Iron – Carbon System

• Cementite (Fe3C): iron carbide

• At 6.70 wt% C; the intermediate compound iron carbide, or cementite (Fe3C), is formed.

• Cementite (Fe3C) forms when the solubility limit of carbon in α-ferrite or γ-austenite is exceeded.

• Mechanically, cementite is very hard and brittle; the strength of some steels is greatly enhanced by its presence.

The Iron – Carbon System

• Eutectic point exists for the iron–iron carbide system, at 4.30 wt% C and 1147°C . This eutectic reaction:

• Eutectoid point exists at a composition of 0.76 wt%C and a temperature of 727°C. This eutectoid reaction:

The Iron – Carbon System

• Pure iron contains less than 0.008 wt% C.

• Iron–carbon alloys that contain between 0.008 and 2.14 wt% C are classified as steels.

• In most steels the microstructure consists of both α and Fe3C phases.

• Cast irons are classified as ferrous alloys that contain between 2.14 and 6.70 wt% C.

Iron-Carbon (Fe-C) Phase Diagram • 2 important

points

- Eutectoid (B): γ  α +Fe3C

- Eutectic (A): L  γ +Fe3C

Fig. 11.23

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

(austenite)

γ+L

γ+Fe3C

α+Fe3C

(Fe) C, wt% C

1148°C

T(°C)

α 727°C = Teutectoid

4.30 Result: Pearlite = alternating layers of α and Fe3C phases

120 μm

Fig. 11.26

0.76

γ γ

γγ

A L+Fe3C

Fe3C (cementite-hard)

α (ferrite-soft)

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

(austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

(Fe) C, wt% C

1148°C

T(°C)

a 727°C

(Fe-C System)

0 .7

Hypoeutectoid Steel

Figs. 11.23 and 11.28

Fig. 11.29

proeutectoid ferritepearlite

100 μm Hypoeutectoid steel

pearlite

γ γ γ

γ α

α α

γγ γ γ

γ γ

γγ

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

(austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

(Fe) C, wt% C

1148°C

T(°C)

α 727°C

(Fe-C System)

0 .7

Hypoeutectoid Steel

γ γ

γ α

α α

sr Wα = s/(r + s)

Wγ =(1 - Wα) R S

pearlite

Wpearlite = Wγ

Wα’= S/(R + S)

W =(1 – Wα’)Fe3C

Fig. 11.29

proeutectoid ferritepearlite

100 μm Hypoeutectoidsteel

Figs. 11.23 and 11.28

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

(austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

(Fe) C, wt% C

1148°C

T(°C)

α 727°C

(Fe-C System)

Hypereutectoid Steel

0 .7

6 C0

Fe3C

γγ

γ γ

γγ γ γ

Fig. 11.32

proeutectoid Fe3C

60 μm Hypereutectoid steel

pearlite

Figs. 11.23 and 11.31

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

(austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

(Fe) C, wt% C

1148°C

T(°C)

α 727°C

(Fe-C System

Fig. 11.32

proeutectoid Fe3C

60 μm Hypereutectoid steel

pearlite

Figs. 11.23 and 11.31

0 .7

6 C0

pearlite

Fe3C

γγ

γ γ

V X

Wpearlite = Wγ

Wα = X/(V + X)

W =(1 - Wα)Fe3C’

W =(1-Wγ)

Wγ =x/(v + x)

Fe3C

Hypereutectoid Steel

Example 11.4 on page 372

For a 99.65 wt% Fe–0.35 wt% C alloy at a

temperature just below the eutectoid,

determine the following:

a)The fractions of total ferrite and cementite

phases?

b) The fractions of the proeutectoid ferrite and

pearlite?

c)The fraction of eutectoid ferrite?

Example 11.4 on page 372

Figure 11.30

The fraction of pearlite, Wp, may be determined according to

The fraction of proeutectoid α, Wpro_α , is computed as follows

Example 11.4 on page 372 (a) The fractions of total ferrite and cementite phases? Application of the lever rule expressions employing a tie line that extends all the way across the α and Fe3C phase field. Thus, C0 is 0.35 wt% C:

(b) The fractions of proeutectoid ferrite and pearlite? Using the lever rule and a tie line that extends only to the eutectoid composition:

Example 11.4 on page 372 (c) The fraction of eutectoid ferrite?

All ferrite is either as proeutectoid or eutectoid (in the pearlite). Therefore, the sum of these two ferrite fractions will equal the fraction of total ferrite, where Wαe denotes the fraction of the total alloy that is eutectoid ferrite:

Example Problem 1

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

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

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

C, wt% C

1148°C

T(°C)

727°C

R S

CFe C3Cα

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 g Fig. 11.23, Callister & Rethwisch 9e.

99.6 wt% Fe - 0.40 wt% C steel

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

C, wt% C

1148°C

T(°C)

727°°C

Solution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and

realizing that

C0 = 0.40 wt% C

Cα = 0.022 wt% C

Cpearlite = Cγ = 0.76 wt% C

V X

Amount of pearlite in 100 g

= (100 g)Wpearlite

= (100 g)(0.512) = 51.2 g Fig. 11.23, Callister & Rethwisch 9e.

The amounts of pearlite and proeutectoid ferrite (α) in the 100 g?

F e

3 C

( c e

m e

n ti te

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

C, wt% C

1148°C

T(°C)

727°°C

Solution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and

realizing that

C0 = 0.40 wt% C

Cα = 0.022 wt% C

Cpearlite = Cγ = 0.76 wt% C

V X

Amount of proeutectoid ferrite in 100 g

= (100 g)Wpro_α

= (100 g)(0.488) = 48.8 g Fig. 11.23, Callister & Rethwisch 9e.

The amounts of pearlite and proeutectoid ferrite (α) in the 100 g?

𝑊𝑝𝑟𝑜_𝛼 = 𝑋

𝑉 + 𝑋 =

0.76 − 0.4

0.76 − 0.022

𝑊𝑝𝑟𝑜_𝛼 = 0.488

• 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