Title: Phase Diagram
1Phase Diagram
2Overview
- 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
3Samples Conditions
- AISI-SAE 1018 0.18 C
- AISI-SAE 1045 0.45 C
- AISI-SAE 1095 0.95 C
- Austenitized at 870C for 2 hours followed by
- Slow cooling (furnace cool)
- Fast cooling (water quench)
4Iron-Carbon Phase Diagram
5Its 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.
6Where Does the Carbon Go?
FCC (Austenite)
BCC (Ferrite) BCT (Martensite)
7Solubility Limits
BCC (? or Ferrite) Iron cant hold much Carbon,
it has a low solubility limit (0.022)
But, FCC (? or Austenite) Iron can hold up to
2.14 Carbon!
8Eutectoid Reaction (Pearlite Formation)
- Austenite precipitates Fe3C at Eutectoid
Transformation Temperature (727C). -
- When cooled slowly, forms Pearlite, which is a
microcontituent made of ferrite (?) and Cementite
(Fe3C), looks like Mother of Pearl.
9Microstructure of Pearlite
Photomicrographs of (a) coarse pearlite and (b)
fine pearlite. 3000X
10Hypo-Eutectoid
Proeutectoid means it formed ABOVE or BEFORE
the Eutectoid Temperature!
11Microstructure of Hypo-Eutectoid
12Hyper-Eutectoid
Proeutectoid means it formed ABOVE or BEFORE
the Eutectoid Temperature!
13Microstructure of Hyper-Eutectoid
14Exercise 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
15Exercise with Lever Rule Hypo-eutectoid
- 4. Microconstituents Primary Ferrite, Pearlite
- 5. C in each microconstituent
- Ferrite - 0.022
- Pearlite 0.76
- 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?
-
- Phase(s) present ?, Fe3C
- C ? - 0.022, Fe3C 6.7
- 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?
You will figure out how to work it out for
hyper-eutectoid compositions
16What if cooled really fast?
- Faster cooling gives non-equilibrium
microconstituents Martensite, and more! - Lets measure the hardness of slow and fast
cooled samples and compare
NEXT LAB 11/03/06 NO GROUP REPORTS FOR THIS
LAB!!!!