Phase Diagram
MSE 201 Lab III
Overview
 Examine microstructures of the selected
carbon steels
 Correlate chemical composition and
temperature with microstructure of the
steel samples using equilibrium Fe-C
phase diagram
 Hardness testing of the steels of
equilibrium and non-equilibrium cooling
Samples Conditions
 AISI-SAE 1018 – 0.18 % C
 AISI-SAE 1045 – 0.45 % C
 AISI-SAE 1095 – 0.95 % C
 Austenitized at 870°C for 2 hours followed
by
 Slow cooling (furnace cool)
 Fast cooling (water quench)
Iron-Carbon Phase Diagram:
It’s All Greek To Me
 Alpha
 “Ferrite”, BCC Iron
 Room Temperature
 Gamma
 “Austenite”, FCC Iron
 Elevated Temperatures
 These are PHASES of iron. Adding carbon
changes the phase transformation temperature.
Where Does the Carbon Go?
Interstitial Sites:
FCC (Austenite)
BCC (Ferrite)
BCT (Martensite)
Solubility Limits
BCC ( or
Ferrite)
Iron can’t
hold much
Carbon, it
has a low
solubility
limit
(0.022%)
But, FCC ( or Austenite)
Iron can hold up to 2.14%
Carbon!
P 2.14
E
4.30
L + Fe3C
F
G
x
M
O
N
H
0.76
0.022
Cementite Fe3C
C
x’
6.70
Eutectoid Reaction (Pearlite Formation)
 Austenite precipitates
Fe3C at Eutectoid
Transformation
Temperature (727°C).
Cooling

 + Fe3C
Heating
 When cooled slowly,
forms Pearlite, which is a
microcontituent made of
ferrite () and Cementite
(Fe3C), looks like Mother
of Pearl.
Microstructure of Pearlite
Photomicrographs of (a) coarse pearlite and (b) fine pearlite. 3000X
Hypo-Eutectoid
“Proeutectoid” means it
formed ABOVE or BEFORE
the Eutectoid Temperature!
Microstructure of Hypo-Eutectoid
Hyper-Eutectoid
“Proeutectoid” means it
formed ABOVE or BEFORE
the Eutectoid Temperature!
Microstructure of Hyper-Eutectoid
Exercise with Lever Rule
• Figure out what type of reaction of the transformation and the
phases/microconstituents involved
• Construct a tie line at the temperature of alloy
• Project the intersections to determine % concentration
• C0 = C  f + (1- f ) C Liq
Exercise with Lever Rule
– Hypo-eutectoid
C0 = 0.1% C

1. Phase(s) present: , Fe3C
2. % C:  - 0.022%, Fe3C – 6.7%
3. Amount of each phase:
f - fraction of 
Then C0 = C  f + (1- f ) C Fe3C
f = (C0 - C Fe3C)/(C  - C Fe3C)
f Fe3C = 1- f
4. Microconstituents: Primary Ferrite,
Pearlite
5. % C in each microconstituent:
Ferrite - 0.022%
Pearlite – 0.76%
6. Amount of each microconstituent:
f - fraction of 
Then C0 = C  f + (1- f ) C Pearlite
f = (C0 - C Pearlite)/(C  - C Pearlite)
f Pearlite = 1- f
You will figure out how to
work it out for hypereutectoid compositions …
What if cooled really fast?
 Faster cooling gives “non-equilibrium
microconstituents” … Martensite, and more!
 Let’s measure the hardness of slow and fast
cooled samples and compare…
NEXT LAB : 11/03/06
NO GROUP REPORTS FOR THIS LAB!!!!