Iron Carbide (Fe-Fe C) Phase Diagram, Lecture notes of Engineering

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The

Iron

Iron

Carbide

(Fe

Fe

3

C)

Phase

Diagram

Peritectic L

• δ = γ

at

T=

o

C

and

0.18wt%C

Eutectic

L

= γ +

Fe

3

C

at

T=

o

C

and

4.3wt%C

Eutectoid

γ = α +

Fe

3

C

at

T=

o

C

and

0.77wt%C

Phases

Present

L

Reactions

δ

ferrite

delta

Bcc

structure

Paramagnetic

γ

austenite

Fcc structure Non

magnetic

ductile

α

ferrite

Bcc

structure

Ferromagnetic

Fairly

ductile

Fe

3

C

cementite

Orthorhombic

Hard,

brittle

Max.

solubility

of

C

in

ferrite=0.022%

in

austenite=2.11%

Comments

on

Fe–Fe

3

C

system

C

is

an

interstitial

impurity

in

Fe.

It

forms

a

solid

solution

with

α, γ,

δ

phases

of

iron

Maximum

solubility

in

BCC

α

ferrite is

wt%

at

C.

BCC:

relatively

small

interstitial

positions

Maximum

solubility

in

FCC

austenite

is

wt%

at

C

FCC

has

larger

interstitial

positions

Mechanical

properties:

Cementite

(Fe

3

C

is

hard

and

brittle:

strengthens

steels).

Mechanical

properties

also

depend

on

microstructure:

how

ferrite

and

cementite are

mixed.

Magnetic

properties:

α

ferrite

is

magnetic

below

C,

austenite

is

non

magnetic

Classification. Three

types

of

ferrous

alloys:

Iron:

wt

C in

α

ferrite

at

room

T

Steels:

wt

C

(usually

wt

α

ferrite

Fe

3

C

at

room

T

Cast

iron:

wt

(usually

wt

Eutectoid

steel

Alloy

of

eutectoid

composition

wt

C)

is

cooled

slowly:

forms

pearlite,

layered

structure

of

two

phases:

α

ferrite

and

cementite

(Fe

3

C)

Microstructure

of

eutectoid

steel

Mechanically,

pearlite has

properties

intermediate

to

soft,

ductile

ferrite

and

hard,

brittle

cementite.

Hypoeutectoid alloys contain

proeutectoid

ferrite

(formed

above

the

eutectoid

temperature)

plus

the

eutectoid

perlite

that

contain

eutectoid

ferrite

and

cementite.

Hypereutectoid

steel

How

to

calculate

the

relative

amounts

of

proeutectoid phase

α

or

Fe

3

C)

and

pearlite?

Use

the

lever

rule

and

a

tie

line

that

extends

from

the

eutectoid

composition

wt%

C)

to

α

wt%

C)

for

hypoeutectoid

alloys and

to

Fe

3

C

wt%

C)

for

hypereutectoid

alloys.

Example:

hypereutectoid

alloy,

composition

C

1

(

)

(

)

(

)

(

)

(

)

(

)

Cementite

id

Proeutecto

of

Fraction

Pearlite

of

Fraction

1

3

1

C

X

V

V

W

C

X

V

X

W

C

Fe

P

The

problem

is

to

solve

for

compositions

at

the

phase

boundaries

for

both

α

and

β

phases

(i.e.,

C

and

C

We

may

set

up

two

independent

lever

rule

expressions,

one

for

each

composition,

in

terms

of

C

and

C

as

follows:

α

β

β

α

β

o

β

α

C

C

C

=

C

C

C C =. = W

− −

− −

60

57

0

1

1

α

β β

α

β

o

β

α

C

C C

=

C

C

C C =. = W

− −

− −

30

14

0

2

2

In

these

expressions,

compositions

are

given

in

weight

percent

A

Solving

for

C

α

and

C

β

from

these

equations,

yield

C

α

(or

wt%

A

wt%

B)

C

β

(or

wt%

A

wt%

B)

Heat

Treatment

of

Steels

On

slowly

cooling

the

steels,

the

properties

of

the

steel

are

dependent

mainly

on

the

percentage

carbon.

Different

percentage

carbon

implies

different

percentage

of

microconstituents and

phases

•pearlite and

ferrite

pro

eutectoid

for

the

hypo

eutectoid

steels •pearlite and

cementite pro

eutectoid

for

the

hyper

eutectoid

steels.

The

temperature

is

high

enough

and

the

time

at

high

temperature

is

long

enough,

for

the

atoms

to

diffuse

and

attain

equilibrium

conditions