Chapter 9: Phase Diagrams - me.emu.edu.trme.emu.edu.tr/behzad/meng286/Fe-C diagram MENG286.pdf ·...
Transcript of Chapter 9: Phase Diagrams - me.emu.edu.trme.emu.edu.tr/behzad/meng286/Fe-C diagram MENG286.pdf ·...
1
Chapter 9: PHASE DIAGRAM
ISSUES TO ADDRESS... • When we combine two elements...
what is the resulting equilibrium state?
• In particular, if we specify...
-- the composition (e.g., wt% Cu - wt% Ni), and
-- the temperature (T )
then...
How many phases form?
What is the composition of each phase?
What is the amount of each phase?
Phase B Phase A
Nickel atom Copper atom
2
IRON-CARBON (Fe-C) PHASE DIAGRAM
(EXAMPLE 1) • 2 important points
- Eutectoid (B):
g a + Fe3C
- Eutectic (A):
L g + Fe3C
Adapted from Fig. 9.24,
Callister & Rethwisch 8e.
Fe
3C
(cem
entite
)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
g
(austenite)
g +L
g +Fe3C
a +Fe3C
d
(Fe) C, wt% C
1148ºC
T(ºC)
a 727ºC = T eutectoid
4.30
Result: Pearlite = alternating layers of a and Fe3C phases
120 mm
(Adapted from Fig. 9.27,
Callister & Rethwisch 8e.)
0.76
B
g g
g g
A L+Fe3C
Fe3C (cementite-hard)
a (ferrite-soft)
3
EXAMPLE 1
• An alloy of eutectoid composition (0.76 wt% C) as it is
cooled down from a temperature within the g-phase
region (e.g., at 800 ºC).
• Initially the alloy is composed entirely of the austenitic
phase having a composition of 0.76 wt% C
• As the alloy is cooled, no changes will occur until the
eutectoid temperature (727 ºC).
• Upon crossing this temperature to point B, the austenite
transforms according to:
Eutectoid (B):
g (0.76 wt% C) a (0.022 wt% C) + Fe3C (6.7 wt% C)
4
EXAMPLE 1 (cont.)
• The microstructure for this eutectoid steel is slowly
cooled through the eutectoid temperature consists of
alternating layers or lamellae of the two phases (a and Fe3C) that form simultaneously during the
transformation.
• Point B is called pearlite.
• Mechanically, pearlite has properties intermediate
between the soft, ductile ferrite and the hard, brittle
cementite.
5
EXAMPLE 1 (cont.)
• The alternating a and Fe3C layers in pearlite form as
such for the same reason that the eutectic structure
forms because the composition of austenite (0.76 %wt
C) is different from either of ferrite (0.022 wt% C) and
cementite (6.70 wt% C), and the phase transformation
requires that there be a redistribution of the carbon by
diffusion.
• Subsequent cooling of the pearlite from point B will
produce relatively insignificant microstructural changes.
6
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
g
(austenite)
g +L
g + Fe3C
a + Fe3C
L+Fe3C
d
(Fe) C, wt% C
1148ºC
T(ºC)
a 727ºC
(Fe-C
System)
C0
0.7
6
Hypoeutectoid Steel (EXAMPLE 2)
Adapted from Figs. 9.24
and 9.29,Callister &
Rethwisch 8e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
Adapted from Fig. 9.30, Callister & Rethwisch 8e.
proeutectoid ferrite pearlite
100 mm Hypoeutectoid
steel
a
pearlite
g
g g
g a
a a
g g g g
g g
g g
7
EXAMPLE 2 (cont.)
• Within the a + g region, most of the a particles will form
along the original g grain boundaries.
• The particles will grow larger just above the eutectoid line. As the temperature is lowered below Te, all the g
phase will transform to pearlite according to:
• There will be virtually no change in the a phase that existed just above the Te.
• This a that is formed above Te is called proeutectoid
(pro=pre=before eutectoid) ferrite.
g a + Fe3C
8
EXAMPLE 2 (cont.)
• The ferrite that is present in the pearlite is called
eutectoid ferrite.
• As a result, two microconstituents are present in the last micrograph (the one below Te): proeutectoid ferrite
and pearlite
9
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
g
(austenite)
g +L
g + Fe3C
a + Fe3C
L+Fe3C
d
(Fe) C, wt% C
1148ºC
T(ºC)
a 727ºC
(Fe-C
System)
C0
0.7
6
EXAMPLE 2
g
g g
g a
a a
s r Wa = s/(r + s)
Wg =(1 - Wa) R S
a
pearlite
Wpearlite = Wg
Wa’ = S/(R + S)
W =(1 – Wa’) Fe3C
Adapted from Figs. 9.24
and 9.29,Callister &
Rethwisch 8e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
Adapted from Fig. 9.30, Callister & Rethwisch 8e.
proeutectoid ferrite pearlite
100 mm Hypoeutectoid
steel
10
HYPEREUTECTOID STEEL (EXAMPLE 3)
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
g
(austenite)
g +L
g +Fe3C
a +Fe3C
L+Fe3C
d
(Fe) C, wt%C
1148ºC
T(ºC)
a
Adapted from Figs. 9.24
and 9.32,Callister &
Rethwisch 8e. (Fig. 9.24
adapted from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B. Massalski
(Ed.-in-Chief), ASM
International, Materials
Park, OH, 1990.)
(Fe-C
System)
0.7
6
C0
Fe3C
g g
g g
g g g g
g g g g
Adapted from Fig. 9.33, Callister & Rethwisch 8e.
proeutectoid Fe3C
60 mm Hypereutectoid steel
pearlite
pearlite
11
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
g
(austenite)
g +L
g +Fe3C
a +Fe3C
L+Fe3C
d
(Fe) C, wt%C
1148ºC
T(ºC)
a
EXAMPLE 3 (cont.)
(Fe-C
System)
0.7
6
C0
pearlite
Fe3C
g g
g g
x v
V X
Wpearlite = Wg
Wa = X/(V + X)
W =(1 - Wa) Fe3C’
W =(1-Wg)
Wg =x/(v + x)
Fe3C
Adapted from Fig. 9.33, Callister & Rethwisch 8e.
proeutectoid Fe3C
60 mm Hypereutectoid steel
pearlite
Adapted from Figs. 9.24
and 9.32,Callister &
Rethwisch 8e. (Fig. 9.24
adapted from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B. Massalski
(Ed.-in-Chief), ASM
International, Materials
Park, OH, 1990.)
12
PROBLEM
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 (a).
b) The amount of cementite (in grams) that forms in 100 g
of steel.
c) The amounts of pearlite and proeutectoid ferrite (a) in
the 100 g
13
SOLUTION TO PROBLEM
WFe3C R
R + S
C0 Ca
CFe3C Ca
0.40 0.022
6.70 0.022 0.057
b) Using the lever rule with
the tie line shown
a) Using the RS tie line just below the eutectoid
Ca = 0.022 wt% C
CFe3C = 6.70 wt% C
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
g (austenite)
g +L
g + Fe3C
a + Fe3C
L+Fe3C
d
C , wt% C
1148ºC
T(ºC)
727ºC
C0
R S
CFe C 3
Ca
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g
14
c) Using the VX tie line just above the eutectoid and realizing that
C0 = 0.40 wt% C
Ca = 0.022 wt% C
Cpearlite = Cg = 0.76 wt% C
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
g (austenite)
g +L
g + Fe3C
a + Fe3C
L+Fe3C
d
C, wt% C
1148ºC
T(ºC)
727ºC
C0
V X
Cg Ca
Wpearlite V
V + X
C0 Ca
Cg Ca
0.40 0.022
0.76 0.022 0.512
Amount of pearlite in 100 g
= (100 g)Wpearlite
= (100 g)(0.512) = 51.2 g
SOLUTION TO PROBLEM (CONT.)
15
ALLOYING WITH OTHER ELEMENTS
• Teutectoid changes:
Adapted from Fig. 9.34,Callister & Rethwisch 8e.
(Fig. 9.34 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
T E
ute
cto
id (ºC
)
wt. % of alloying elements
Ti
Ni
Mo Si
W
Cr
Mn
• Ceutectoid changes:
Adapted from Fig. 9.35,Callister & Rethwisch 8e.
(Fig. 9.35 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
wt. % of alloying elements
C e
ute
cto
id (w
t% C
)
Ni
Ti
Cr
Si Mn
W Mo