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.
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%) αferrite + 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 °C
γ(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.
Iron-Carbon (Fe-C) Phase Diagram
T(°C)
1600
- Eutectic (A):
L
Adapted from Fig. 10.28,
Callister & Rethwisch 3e.
L
1400
+ Fe3C
- Eutectoid (B):
+ Fe3C
1200
+L
AA
(austenite)
1000
800
+Fe3C
B
727°C = T eutectoid
600
120 m
Result: Pearlite is
alternating layers of
400
0
(Fe)
L+Fe3C
1148°C
Fe3C (cementite)
• 2 points
+Fe3C
1
0.76
and Fe3C phases
2
3
4
4.30
5
6
6.7
C, wt% C
Fe3C (cementite-hard)
(ferrite-soft)
12
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 steel
In the micrograph, the dark areas are
Fe3C layers, the light phase is αferrite
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
Hypoeutectoid Steel
T(°C)
1600
Adapted from Figs. 10.28 and 10.33
L
1400
1000
+ Fe3C
800
727°C
600
pearlite
+ Fe3C
1
0.76
400
0
(Fe)C0
L+Fe3C
1148°C
(austenite)
2
3
4
5
6
6.7
C, wt% C
100 m
pearlite
(Fe-C
System)
Fe3C (cementite)
1200
+L
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 10.34, Callister & Rethwisch 3e.
17
Hypoeutectoid Steel
T(°C)
1600
L
1400
(austenite)
1000
800
600
pearlite
Wpearlite = W
+ Fe3C
r s
727°C
RS
400
0
(Fe)C0
+ Fe3C
1
0.76
W = s/(r + s)
W =(1 - W )
L+Fe3C
1148°C
W ’ = S/(R + S)
WFe3C
=(1 – W ’)
pearlite
2
3
4
5
6
(Fe-C
System)
Fe3C (cementite)
1200
+L
6.7
C, wt% C
100 m
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 10.34, Callister & Rethwisch 3e.
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Microstructure of hypoeutectoid steel
Hypoeutectoid alloys contain proeutectoid ferrite (formed
above the eutectoid temperature) plus the eutectoid perlite
that contain eutectoid ferrite and cementite.
T(°C)
Hypereutectoid Steel
1600
L
1400
1200
+L
1000
+Fe3C
W =x/(v + x)
v x
800
600
pearlite
400
0
(Fe)
Wpearlite = W
V
X
+Fe3C
0.76
WFe3C =(1-W )
1 C0
W = X/(V + X)
WFe
3C’
L+Fe3C
1148°C
(austenite)
2
3
4
5
6
Fe3C (cementite)
Fe3C
6.7
C, wt%C
60 mHypereutectoid
steel
=(1 - W )
pearlite
proeutectoid Fe3C
Adapted from Fig. 10.37, Callister & Rethwisch 3e.
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Relative amounts of proeutectoid
phase (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)
Example Problem Steel
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.
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Solution to Problem
a) Use RS tie line just below
Eutectoid
b)
Use lever rule with
the tie line shown
WFe 3C
R
R S
1600
T(°C)
1200
C0 C
CFe 3C C
0.40 0.022
6.70 0.022
L
1400
+L
1000
+ Fe3C
800
727°C
R
0.057
S
+ Fe3C
600
400
0
Amount of Fe3C in 100 g
L+Fe3C
1148°C
(austenite)
Fe3C (cementite)
C = 0.022 wt% C
CFe3C = 6.70 wt% C
C C0
1
2
3
4
C, wt% C
5
6
6.7
CFe
3C
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g
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Solution to Problem
c) Using the VX tie line just above the eutectoid
and realizing that
C0 = 0.40 wt% C
C = 0.022 wt% C
Cpearlite = C = 0.76 wt% C
V X
T(°C)
C0 C
C C
0.40 0.022
0.76 0.022
L
1400
1200
+L
L+Fe3C
1148°C
(austenite)
1000
+ Fe3C
0.512
800
727°C
VX
Amount of pearlite in 100 g
= (100 g)Wpearlite
= (100 g)(0.512) = 51.2 g
600
400
0
+ Fe3C
1
C C0 C
2
3
4
5
6
Fe C (cementite)
Wpearlite
V
1600
6.7
C, wt% C
28
Summary
Phase Diagrams and Microstructures
• 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.
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Alloying Steel With More Elements
Ti
Mo
• Ceutectoid changes:
Si
W
Cr
Mn
Ni
wt. % of alloying elements
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.)
Ceutectoid (wt% C)
T Eutectoid (°C)
• Teutectoid changes:
Ni
Cr
Si
Ti Mo
W
Mn
wt. % of alloying elements
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.)
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