1. Phase Diagram Fe3C

2. 100X

4. 500X

5. Ferrite and Austenite
Ferrite
Austenite

18. Hypoeutectoid Steel T(°C) d L +L g (austenite) Fe3C (cementite) a
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
(Fe)
Co , wt% C
1148°C
T(°C)
727°C
(Fe-C
System) C0
0.76 g g r s w a = /( + ) g
(1- R S a w = S /( R + )
Fe3C
(1-
pearlite g
proeutectoid ferrite
pearlite
100 mm
Hypoeutectoid
steel

20. Hypereutectoid Steel Fe3C (cementite) L g (austenite) +L a d T(°C) g g
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+Fe3C a
L+Fe3C d
(Fe)
Co , wt%C
1148°C
T(°C)
(Fe-C
System) g g s r w
Fe3C = /( + ) g
=(1-
proeutectoid Fe3C
60 mm
Hypereutectoid
steel
pearlite R S w a = /( + )
Fe3C
(1- g Co
0.76

22. Iron-Carbon (Fe-C) Phase Diagram
• 2 important
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+Fe3C a +
L+Fe3C d
(Fe)
Co, wt% C
1148°C
T(°C)
727°C = T
eutectoid
points
-Eutectic (A): L Þ g +
Fe3C A S R
4.30
-Eutectoid (B): g Þ a +
Fe3C g
120 mm
0.76 C
eutectoid B R S
Fe3C (cementite-hard) a
(ferrite-soft)
Result: Pearlite =
alternating layers of a
and Fe3C phases

23. Example: Phase Equilibria
For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following
composition of Fe3C and ferrite ()
the amount of carbide (cementite) in grams that forms per 100 g of steel
the amount of pearlite and proeutectoid ferrite ()

24. Solution: a) composition of Fe3C and ferrite () 3
the amount of carbide (cementite) in grams that forms per 100 g of steel
CO = 0.40 wt% C Ca = wt% C CFe C = 6.70 wt% C 3
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
Co , wt% C
1148°C
T(°C)
727°C CO R S
CFe C 3 C

25. Co = 0.40 wt% C Ca = 0.022 wt% C Cpearlite = C = 0.76 wt% C
the amount of pearlite and proeutectoid ferrite ()
note: amount of pearlite = amount of g just above TE
Co = 0.40 wt% C Ca = wt% C Cpearlite = C = 0.76 wt% C
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
Co , wt% C
1148°C
T(°C)
727°C CO R S C C
pearlite = 51.2 g proeutectoid  = 48.8 g

26. Alloying Steel with More Elements
• Teutectoid changes:
• Ceutectoid changes: Ti
(wt%C) Si
(°C) Mo W Ni Cr Cr
Eutectoid
eutectoid Si Mn Mn W Ti Mo T C Ni
wt. % of alloying elements
wt. % of alloying elements

32. Summary • Phase diagrams are useful tools 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.

34. End