fe-c diagram

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The Iron–Carbon Phase Diagram

Iron–Carbon Phase Diagram

• In their simplest form, steels are alloys of Iron (Fe) and Carbon (C). The Fe-C phase diagram is a fairly complex one, but we will only consider the steel and cast iron part of the diagram, up to 6.67% Carbon.

Fe – C Equilibrium Diagram

Phases in Fe–C Phase Diagram• α-ferrite - solid solution of C in BCC Fe

• Stable form of iron at room temperature.• The maximum solubility of C is 0.025 % at eutectoid temperature• Transforms to FCC γ-austenite at 910°C

• γ-austenite - solid solution of C in FCC Fe• The maximum solubility of C is 2.14 % at eutectic temperature• Transforms to BCC δ-ferrite at 1395 °C• Is not stable below the eutectoid temperature(727 ° C) unless cooled rapidly

• δ-ferrite solid solution of C in BCC Fe• The same structure as α-ferrite• Stable only at high Temp., above 1394 °C• Melts at 1538 °C

Phases in Fe–C Phase Diagram

• Fe3C (iron carbide or cementite)

It is an intermetallic compound between iron and carbon. It contains 6.67 % carbon and is hard and brittle.This intermetallic compound is a metastablephase and it remains as a compound indefinitely at room temperature.

A few comments on Fe–C system

• Carbon occupies interstitial positions in Fe. It forms a solid solution with α, γ, δ phases of iron

• Maximum solubility in BCC α-ferrite is limited (max. 0.025 % at 727 °C) - BCC has relatively small interstitial positions

• Maximum solubility in FCC austenite is 2.14 % at 1147 °C - FCC has larger interstitial positions

A few comments on Fe–C system

• Mechanical properties– Cementite is very hard and brittle - can strengthen

steels.– Mechanical properties depend on the

microstructure, that is, how ferrite and cementite are distributed.

• Magnetic properties: α -ferrite is magnetic below 768 °C, austenite is non-magnetic

Classification

• Three types of ferrous alloy• Iron: less than 0.008% C in α−ferrite at room temp.• Steels: 0.008 - 2.14% C (usually < 1%) α-ferrite + Fe3C at room temp.

• Cast iron: 2.14 - 6.7% (usually < 4.5%)

Eutectic and eutectoid reactions in Fe–C

Eutectic: 4.30 wt% C, 1147 °C L (4.30% C) ↔ γ (2.14% C) + Fe3C

Eutectoid: 0.76 wt%C, 727 °Cγ(0.76% C) ↔ α (0.022% C) + Fe3C

• Eutectic and eutectoid reactions are very important in heat treatment of steels

Development of Microstructure in Iron - Carbon alloys

• Microstructure depends on composition (carbon content) and heat treatment. In the discussion below we consider slow cooling in which equilibrium is maintained.

12

Iron-Carbon (Fe-C) Phase Diagram

• 2 points

- Eutectoid (B):

g Þ a +Fe3C

- Eutectic (A):

L Þ g + Fe3C

Result: Pearlite is alternating layers of and Fe3C phases a

120 mm

A

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g+Fe3C

a+Fe3C

a+g

d

(Fe) C, wt% C

1148°C

T(°C)

a 727°C = Teutectoid

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

4.30

g ggg

AL+Fe3C

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

0.76

B

Microstructure of eutectoid steel

Microstructure of eutectoid steel

• When alloy of eutectoid composition (0.76 wt % C) is cooled slowly it forms perlite, a lamellar or layered structure of two phases: α-ferrite and cementite (Fe3C).

• The layers of alternating phases in pearlite are formed for the same reason as layered structure of eutectic structures: redistribution C atoms between ferrite (0.022 wt%) and cementite (6.7 wt%) by atomic diffusion.

• Mechanically, pearlite has properties intermediate to soft, ductile ferrite and hard, brittle cementite.

Microstructure of eutectoid steelIn the micrograph, the dark areas areFe3C layers, the light phase is α-ferrite

Microstructure of hypoeutectoid steel

Compositions to the left of eutectoid (0.022 - 0.76 wt % C) is hypoeutectoid (less than eutectoid -Greek) alloys.

γ → α + γ → α + Fe3C

17

Hypoeutectoid Steel

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

proeutectoid ferritepearlite

100 mm Hypoeutectoidsteel

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g + Fe3C

a + Fe3C

L+Fe3C

d

(Fe) C, wt% C

1148°C

T(°C)

a727°C

(Fe-C System)

C0

0.76

a

pearlite

gg g

ga

aa

ggg g

g ggg

Adapted from Figs. 10.28 and 10.33

18

Hypoeutectoid Steel

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g + Fe3C

a + Fe3C

L+Fe3C

d

(Fe) C, wt% C

1148°C

T(°C)

a727°C

(Fe-C System)

C0

0.76

gg g

ga

aa

srWa = s/(r + s)

Wg =(1 - Wa)R S

a

pearlite

Wpearlite = Wg

Wa’ = S/(R + S)

W =(1 – Wa’)Fe3C

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

proeutectoid ferritepearlite

100 mm Hypoeutectoidsteel

Microstructure of hypoeutectoid steelHypoeutectoid alloys contain proeutectoid ferrite (formedabove the eutectoid temperature) plus the eutectoid perlitethat contain eutectoid ferrite and cementite.

Microstructure of hypereutectoid steel

Compositions to the right of eutectoid (0.76 - 2.14 wt % C) is hypereutectoid (more than eutectoid -Greek) alloys.

γ → γ + Fe3C → α + Fe3C

Microstructure of hypereutectoid steel

22

Hypereutectoid Steel

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g +Fe3C

a +Fe3C

L+Fe3C

d

(Fe) C, wt%C

1148°C

T(°C)

a

0.7

6 C0

Fe3C

ggg g

ggg g

ggg g

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

proeutectoid Fe3C

60 mmHypereutectoid steel

pearlite

pearlite

23

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g +Fe3C

a +Fe3C

L+Fe3C

d

(Fe) C, wt%C

1148°C

T(°C)

a

0.7

6 C0

pearlite

Fe3C

ggg g

xv

V X

Wpearlite = Wg

Wa = X/(V + X)

W =(1 - Wa)Fe3C’

W =(1-Wg)

Wg =x/(v + x)

Fe3C

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

proeutectoid Fe3C

60 mmHypereutectoid steel

pearlite

Hypereutectoid Steel

Relative amounts of proeutectoidphase (α or Fe3C) and pearlite?

Application of the lever rule with tie line that extends from the eutectoid composition (0.76 wt% C) to α – (α + Fe3C) boundary (0.022 wt% C) for hypoeutectoid alloys and to (α + Fe3C) – Fe3C boundary (6.7 wt% C) for hypereutectoid alloys.

Fraction of α phase is determined by application of the lever rule across the entire (α + Fe3C) phase field.

Example for hypereutectoid alloy with composition C1

Fraction of pearlite: WP = X / (V+X) = (6.7 – C1) / (6.7 – 0.76)Fraction of proeutectoid cementite: WFe3C = V / (V+X) = (C1 – 0.76) / (6.7 – 0.76)

26

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 ().

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

c) The amounts of pearlite and proeutectoid ferrite () in the 100 g.

Example Problem Steel

27

Solution to Problem

WFe3C R

R S

C0 C

CFe3C C

0.40 0.0226.70 0.022

0.057

b) Use lever rule with the tie line shown

a) Use RS tie line just below Eutectoid

Ca = 0.022 wt% C

CFe3C = 6.70 wt% C

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 g

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g + Fe3C

a + Fe3C

L+Fe3C

d

C , wt% C

1148°C

T(°C)

727°C

C0

R S

CFe C3C

28

Solution to Problem

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

C0 = 0.40 wt% C

Ca = 0.022 wt% C

Cpearlite = C = 0.76 wt% C

Wpearlite V

V X

C0 C

C C

0.40 0.0220.76 0.022

0.512

Amount of pearlite in 100 g = (100 g)Wpearlite = (100 g)(0.512) = 51.2 g

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g + Fe3C

a + Fe3C

L+Fe3C

d

C, wt% C

1148°C

T(°C)

727°C

C0

V X

CC

29

• 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.

SummaryPhase Diagrams and Microstructures

30

Alloying Steel With More Elements

• Teutectoid changes:

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.)

TE

ute

cto

id (

°C)

wt. % of alloying elements

Ti

Ni

Mo SiW

Cr

Mn

• Ceutectoid changes:

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.)

wt. % of alloying elements

Ce

ute

cto

id (

wt%

C)

Ni

Ti

Cr

SiMn

WMo

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