Introduction to Engineering Materials ENGR2000 Chapter 9: Phase Diagrams II Dr. Coates
Introduction to Engineering Materials
ENGR2000
Chapter 9: Phase Diagrams II
Dr. Coates
The Iron-Carbon System
• Ferrous alloys (iron + C & other elements)
– Iron: less than 0.008 wt. % C
– Steel: between 0.008 and 2.14 wt. % C
– Cast iron: between 2.14 and 6.70 wt. % C
Terminology for microstructures in
Fe-C alloys
9.l7 The Iron-Iron Carbide Phase Diagram
Eutectoid
reaction
Peritectic
reaction Eutectic
reaction
Properties of the different phases
• α-ferrite is relatively soft
• δ-ferrite is very similar but is stable over a higher
range of temperatures.
• γ-austenite is non-magnetic
• Cementite is hard & brittle
Eutectoid
reaction
( ) ( ) ( )CwtCFeCwtCwt .%7.6.%022.0.%76.0
:reaction Eutectoid
3+⇔ αγ
CwtC %76.0
:ncompositio Eutectoid
0 =
pearlite
Eutectoid alloys
CwtC %76.0
:ncompositio eutectoid-Hypo
0 <
Hypo-eutectoid alloys
Eutectoid ferrite and Pro-eutectoid ferrite
• α-ferrite is present in the pearlite (eutectoid
ferrite) and
• Is formed above the eutectoid temperature as well
(pro-eutectoid ferrite)
022.076.0
76.0
: idproeutecto of fraction) (massamount Phase
022.076.0
022.0
:pearlite of fraction) (massamount Phase
0
0
−
−=
−
−=
CW
CW
idproeutecto
pearlite
α
α
CwtCCwt %14.2%76.0
:ncompositio eutectoid-Hyper
1 <<
Hyper-eutectoid alloys 10-18-10
76.070.6
76.0
:C idproeutecto of fraction) (massamount Phase
76.070.6
70.6
:pearlite of fraction) (massamount Phase
1
3
1
3 −
−=
−
−=
CW
Fe
CW
CFeidproeutecto
pearlite
Example 9.4
• For a 99.65 wt. % Fe-0.35 wt. % C alloy at a
temperature just below the eutectoid, determine
the following:
(a) Fractions of total ferrite and cementite phases
(b) Fractions of the proeutectoid ferrite and pearlite
(c) Fraction of eutectoid ferrite
Fractions of total ferrite and cementite phases
Cwt.%35.0C
:ncompositioalloy Given
0 =
05.0022.070.6
022.035.0
:cementite totalofFraction
95.0022.070.6
35.070.6
:ferrite totalofFraction
3=
−
−=
=−
−=
CFeW
Wα
Fractions of the proeutectoid ferrite & pearlite
44.0022.076.0
022.035.0
:pearlite ofFraction
56.0022.076.0
35.076.0
:ferrite idproeutecto ofFraction
=−
−=
=−
−=
pearlite
idproeutecto
W
W α Remember the γ becomes
eutectoid pearlite
Fraction of eutectoid ferrite
39.056.095.0
:ferrite eutectoid ofFraction
=−=
=+
α
ααα
eutectoid
totaleutectoididproeutecto
W
WWW
The eutectoid α will be 89% of the pearlite that is present at ANY
composition
89.0022.070.6
76.070.6=
−
−=αeutectoidW
:reaction eutectoid the from ferrite eutectoid of Fraction
39.044.089.089.0 =×=×=
=
pearliteeutectoid WW
ferrite eutectoid of fraction The
α
At a composition of 0.35%, the amount of pearlite present is 44%,
therefore:
Fraction of eutectoid ferrite
(another method)
Problems
9.12 A 50 wt% Pb-50 wt% Mg alloy is slowly cooled from 700°C (1290°F) to 400°C (750°F).
(a) At what temperature does the first solid phase form?
(b) What is the composition of this solid phase?
(c) At what temperature does the liquid solidify?
(d) What is the composition of this last remaining liquid phase?
Solution in
class!
9.33 The microstructure of a lead-tin alloy at 180°C (355°F) consists of primary β
and eutectic structures. If the mass fractions of these two microconstituents are
0.57 and 0.43, respectively, determine the composition of the alloy.
Solution
in class!
9.51 Consider 2.5 kg of austenite containing 0.65 wt% C, cooled to below 727°C (1341°F).
(a) What is the proeutectoid phase?
(b) How many kilograms each of total ferrite and cementite form?
(c) How many kilograms each of pearlite and the proeutectoid phase form?
(d) Schematically sketch and label the resulting microstructure.
The influence of other alloying elements