1

Chapter 9: PHASE DIAGRAM

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?

Phase B Phase A

Nickel atom Copper atom

2

IRON-CARBON (Fe-C) PHASE DIAGRAM

(EXAMPLE 1) • 2 important points

- Eutectoid (B):

g a + Fe3C

- Eutectic (A):

L g + Fe3C

Adapted from Fig. 9.24,

Callister & Rethwisch 8e.

Fe

3C

(cem

entite

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

L

g

(austenite)

g +L

g +Fe3C

a +Fe3C

d

(Fe) C, wt% C

1148ºC

T(ºC)

a 727ºC = T eutectoid

4.30

Result: Pearlite = alternating layers of a and Fe3C phases

120 mm

(Adapted from Fig. 9.27,

Callister & Rethwisch 8e.)

0.76

B

g g

g g

A L+Fe3C

Fe3C (cementite-hard)

a (ferrite-soft)

3

EXAMPLE 1

• An alloy of eutectoid composition (0.76 wt% C) as it is

cooled down from a temperature within the g-phase

region (e.g., at 800 ºC).

• Initially the alloy is composed entirely of the austenitic

phase having a composition of 0.76 wt% C

• As the alloy is cooled, no changes will occur until the

eutectoid temperature (727 ºC).

• Upon crossing this temperature to point B, the austenite

transforms according to:

Eutectoid (B):

g (0.76 wt% C) a (0.022 wt% C) + Fe3C (6.7 wt% C)

4

EXAMPLE 1 (cont.)

• The microstructure for this eutectoid steel is slowly

cooled through the eutectoid temperature consists of

alternating layers or lamellae of the two phases (a and Fe3C) that form simultaneously during the

transformation.

• Point B is called pearlite.

• Mechanically, pearlite has properties intermediate

between the soft, ductile ferrite and the hard, brittle

cementite.

5

EXAMPLE 1 (cont.)

• The alternating a and Fe3C layers in pearlite form as

such for the same reason that the eutectic structure

forms because the composition of austenite (0.76 %wt

C) is different from either of ferrite (0.022 wt% C) and

cementite (6.70 wt% C), and the phase transformation

requires that there be a redistribution of the carbon by

diffusion.

• Subsequent cooling of the pearlite from point B will

produce relatively insignificant microstructural changes.

6

Fe

3C

(ce

me

ntite

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

L

g

(austenite)

g +L

g + Fe3C

a + Fe3C

L+Fe3C

d

(Fe) C, wt% C

1148ºC

T(ºC)

a 727ºC

(Fe-C

System)

C0

0.7

6

Hypoeutectoid Steel (EXAMPLE 2)

Adapted from Figs. 9.24

and 9.29,Callister &

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 ferrite pearlite

100 mm Hypoeutectoid

steel

a

pearlite

g

g g

g a

a a

g g g g

g g

g g

7

EXAMPLE 2 (cont.)

• Within the a + g region, most of the a particles will form

along the original g grain boundaries.

• The particles will grow larger just above the eutectoid line. As the temperature is lowered below Te, all the g

phase will transform to pearlite according to:

• There will be virtually no change in the a phase that existed just above the Te.

• This a that is formed above Te is called proeutectoid

(pro=pre=before eutectoid) ferrite.

g a + Fe3C

8

EXAMPLE 2 (cont.)

• The ferrite that is present in the pearlite is called

eutectoid ferrite.

• As a result, two microconstituents are present in the last micrograph (the one below Te): proeutectoid ferrite

and pearlite

9

Fe

3C

(ce

me

ntite

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

L

g

(austenite)

g +L

g + Fe3C

a + Fe3C

L+Fe3C

d

(Fe) C, wt% C

1148ºC

T(ºC)

a 727ºC

(Fe-C

System)

C0

0.7

6

EXAMPLE 2

g

g g

g a

a a

s r Wa = s/(r + s)

Wg =(1 - Wa) R S

a

pearlite

Wpearlite = Wg

Wa’ = S/(R + S)

W =(1 – Wa’) Fe3C

Adapted from Figs. 9.24

and 9.29,Callister &

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 ferrite pearlite

100 mm Hypoeutectoid

steel

10

HYPEREUTECTOID STEEL (EXAMPLE 3)

Fe

3C

(ce

me

ntite

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

L

g

(austenite)

g +L

g +Fe3C

a +Fe3C

L+Fe3C

d

(Fe) C, wt%C

1148ºC

T(ºC)

a

Adapted from Figs. 9.24

and 9.32,Callister &

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.)

(Fe-C

System)

0.7

6

C0

Fe3C

g g

g g

g g g g

g g g g

Adapted from Fig. 9.33, Callister & Rethwisch 8e.

proeutectoid Fe3C

60 mm Hypereutectoid steel

pearlite

pearlite

11

Fe

3C

(ce

me

ntite

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

L

g

(austenite)

g +L

g +Fe3C

a +Fe3C

L+Fe3C

d

(Fe) C, wt%C

1148ºC

T(ºC)

a

EXAMPLE 3 (cont.)

(Fe-C

System)

0.7

6

C0

pearlite

Fe3C

g g

g g

x v

V X

Wpearlite = Wg

Wa = X/(V + X)

W =(1 - Wa) Fe3C’

W =(1-Wg)

Wg =x/(v + x)

Fe3C

Adapted from Fig. 9.33, Callister & Rethwisch 8e.

proeutectoid Fe3C

60 mm Hypereutectoid steel

pearlite

Adapted from Figs. 9.24

and 9.32,Callister &

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.)

12

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 (a).

b) The amount of cementite (in grams) that forms in 100 g

of steel.

c) The amounts of pearlite and proeutectoid ferrite (a) in

the 100 g

13

SOLUTION TO PROBLEM

WFe3C R

R + S

C0 Ca

CFe3C Ca

0.40 0.022

6.70 0.022 0.057

b) Using the lever rule with

the tie line shown

a) Using the RS tie line just below the eutectoid

Ca = 0.022 wt% C

CFe3C = 6.70 wt% C

Fe

3C

(ce

me

ntite

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

L

g (austenite)

g +L

g + Fe3C

a + Fe3C

L+Fe3C

d

C , wt% C

1148ºC

T(ºC)

727ºC

C0

R S

CFe C 3

Ca

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 g

14

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

C0 = 0.40 wt% C

Ca = 0.022 wt% C

Cpearlite = Cg = 0.76 wt% C

Fe

3C

(ce

me

ntite

)

1600

1400

1200

1000

800

600

400 0 1 2 3 4 5 6 6.7

L

g (austenite)

g +L

g + Fe3C

a + Fe3C

L+Fe3C

d

C, wt% C

1148ºC

T(ºC)

727ºC

C0

V X

Cg Ca

Wpearlite V

V + X

C0 Ca

Cg Ca

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

SOLUTION TO PROBLEM (CONT.)

15

ALLOYING WITH OTHER ELEMENTS

• Teutectoid changes:

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.)

T E

ute

cto

id (ºC

)

wt. % of alloying elements

Ti

Ni

Mo Si

W

Cr

Mn

• Ceutectoid changes:

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

C e

ute

cto

id (w

t% C

)

Ni

Ti

Cr

Si Mn

W Mo