Student Name ………………………… Student Number …………………
DEPARTMENT OF MECHANICAL ENGINEERING
MECH 473 Ferrous and Non-Ferrous Materials
1 hour and 15 minutes Mid-term Test No.2 Thursday July 15, 2005
1. Pearlite (total – 10 marks)
Describe in metallurgical terms the nucleation, growth and morphology of pearlite in
hypereutectoid and hypoeutectoid steels. To do so, answer the following.
a) Why type of nucleation is involved in the formation of pearlite? (1) The nucleation of
pearlite is heterogeneous.
b) If the austenite is homogeneous, where will the nucleation occur? (1) If the austenite is
homogeneous, the pearlite will nucleate at grain boundaries.
c) If the austenite is not homogeneous where else can the pearlite nucleate? (1)
If the austenite contains concentrations gradients or residual carbide particles because of a
lack of time to achieve equilibrium in the –phase region, then pearlite can nucleate at these
particles and defect sites within the austenite grains.
d) How will the pearlite grow into the austenite at the grain boundary? (1) A nucleated
pearlite colony will normally grow into only one of the two austenite grains at the boundary.
It has an orientation relationship with one grain, 1, and grows into the other grain, 2.
e) What is the orientation relationship of the planes and directions of the ferrite, , that is
nucleating the pearlite with respect to the austenite, , if the grain boundaries are “clean”, i.e.,
contain no proeutectoid cementite. (1) If the grain boundaries are “clean”, then the ferrite in
the pearlite will have an orientation related to the austenite grain, 1 of,
111  110a
110

111 a
f) If the austenite grain boundaries contain proeutectoid cementite as found in hypereutectoid
steels, what will be the orientation relationship of the cementite and ferrite in the pearlite
with respect to the proeutectoid cementite? (1) The cementite in the pearlite will have the
same orientation as the Fe3C in the boundary, i.e., the proeutectoid cementite, which has a
complicated relationship to the 1 grain but the orientation of the ferrite is unrelated to that of
the proeutectoid cementite or the 1 grain. In both of these cases, the orientation of the
cementite and the ferrite in the pearlite is unrelated to the orientation of the other austenite
grain, 2, into which the pearlite grows.
g) Why does pearlite not grow into one austenite grain? (1) Pearlite does not grow into the
austenite grain, 1, with which it has an orientation relationship because it involves a lowenergy, low-mobility interface.
Pearlite does grow into the other austenite grain, 2, because the ends of the cementite and
ferrite lamellae, which penetrate this grain, have high energy, high mobility interfaces.
h) How will the growth of a ferrite plate affect the carbon concentration in its vicinity and
what will it stimulate to nucleate at the ferrite-austenite interface? (1) The growth of a ferrite
plate will increase the carbon concentration in its vicinity and thus favour the nucleation of
an adjacent cementite plate at the ferrite-austenite interface.
i) The growth of pearlite nodules usually have what kind of shape and why? (1) The pearlite
nodules are usually spherical in shape because the rate of growth is approximately equal in
all directions
j) Before impinging on other colonies what is the growth rate of pearlite? (1) Before
impingement the rate of pearlite growth is linear.
2. Martensite (total – 10 marks)
a) What are three unique characteristics of the martensite transformation process from
austenite? (1.5)
The unique characteristics of the martensite transformation are:
1) it is an athermal or diffusionless transformation process, (1/2)
2) it is a displacive or shear-like transformation and (1/2)
3) it involves a change in crystal structure from fcc to bct. (1/2)
b) Explain the significance of each characteristic. (1.5)
The diffusionless characteristic signifies two things: 1) the amount, and 2) the composition of
the old or parent phase is the same as the martensite. (1/2)
The displacive or shear-like transformation signifies that a large volume of the old phase may
transform at one time. (1/2)
The change in crystal structure, i.e., bcc to bct, signifies that the characteristics of the
martensite are different from the old phase, which makes martensite to be the primary source
of hardening in steels. (1/2)
c) How does the hardenability depend on carbon content? (1)
The hardenability or the ability to form martensite increases with increased carbon content.
d) How does the hardness depend on carbon content? (1)
The hardness or strength of the martensite solely depends on the carbon content.
e) How does the hardenability depend on most of the other alloying additions containing <
5% total alloy content? (1)
The hardenability increases with increased alloy content for most of the alloy additions.
f) How does the hardness depend on the other alloying additions? (1)
The hardness or strength of the martensite does not depend on the other alloys.
g) What type of martensite forms at high carbon concentrations, above what carbon content
does it form, and how brittle is this martensite (does it have to be tempered in order to be
used?)? (1.5)
Lenticular martensite occurs in high carbon steels with > 0.4% C. They are brittle and have
to be tempered in order to be used.
h) How do the alloying elements other than C affect the softening of the martenistic steel
during tempering? (1)
The alloying elements will slow down the softening of the martenistic steel during tempering.
i) In a low carbon, low alloy steel solution treated above the eutectoid temperature, how
much of the austenite will transform to martensite when quenched? (1/2)
100% of the austenite in a low carbon, low alloy steel will transform to martensite when
quenched.
3. Use of Phase Diagram – (total – 10 marks)
a) Consider a steel with 0.2 wt%C given the phase diagram below. a) Describe all the
changes in its structure as it is heated from room temperature to 850 oC. Start with its
room temperature structure. (4.5)
At room temperature, the 1020 steel has a ferrite plus pearlite structure (1/2). At 727 oC,
the austenite phase, , forms (1/2) as the pearlite dissolves. (1/2) The iron carbide dissolves
completely (1/2) and there is retained ferrite. (1/2) In the range of 727 oC to 830 oC, the ferrite and the –austenite phases co-exist. (1/2) As the temperature increases, the amount
of austenite increases at the expense of the ferrite. (1/2) At 830 oC, all the ferrite
transforms to austenite (1/2) and the structure at 850 oC is 100% austenite. (1/2)
b) at 750 oC, write the equation to determine the equilibrium compositions and relative
amounts of ferrite and austenite. Calculations of the percentages are not required. (1)
At 750 oC,  = 0.62%C;  = 0.02%C
At 750 oC and 0.2 wt%C
0.62 - 0.2
% 
 100  70%
0.62 - 0.02
0.2 - 0.02
% 
 100  30%
0.62 - 0.02
c) at 727 oC, write the equation to determine the total amount of ferrite and cementite
present (1). d) write the equation to determine the amount of ferrite in the pearlite. (1) e)
write the equation to determine the amount of ferrite in the specimen that is not in the pearlite
(1).
At the eutectic temperatu re of 727 o C
 has 0.02% C; Fe 3 C has 6.67% C
For an 1020 steel of 0.2 wt%C
6.67 - 0.2
 100  97%
6.67 - 0.02
30 - 19
(% Fe 3 C 
 100  3%)
6.67 - 0.02
d) The eutectoid compositio n, ie., pearlite is 0.77% C.
% 
The fraction of  (and Fe 3 C) in the pearlite is :
6.67 - 0.77
 100  88.7%
6.67 - 0.02
0.77 - 0.02
(% Fe 3 C 
 100  11.3%)
6.67 - 0.02
e) The fraction of ferrite of the specimen and not in the pearlite is.
% 
0.77 - 0.2
 100  76%
0.77 - 0.02
0.2 - 0.02
(% pearlite 
 100  24%)
0.77 - 0.02
% 
f) What is the relationship between the yield strength of pearlite and its interlamellar
spacing? (1/2)
The yield strength is an inverse linear function of the interlamellar spacing.
g) How does the pearlite interlamellar spacing vary with transformation temperature from
austenite. (1/2)
The lower the transformation temperature from austenite, the smaller the pearlite spacing.
h) Briefly describe the method used in practice to determine the pearlite interlamellar
spacing. (1/2)
In practice the interlamellar spacing is determined by measuring the distance between
adjacent lamellae along a number of random lines marked on a polished surface. The
true interlamellar spacing is taken as the mean value measured divided by 2.
4. Low Alloy Steels and TTT Diagrams – (total – 10 marks)
a) What is the “ideal quench diameter” (DI)of a steel specimen? (1)
The “ideal quench diameter” is the diameter of a sample that would have 50/50 P/M at its
center (1/2) when cooled by an ideal quench (1/2).
b) Write the equation to calculate DI for a steel consisting of 0.4%C, 0.8%Mn, 0.55%Ni,
0.1%Si, 0.55%Cr having a grain size of ASTM 7 and using the Table provided. (1)
D1 = 0.213 x 3.667 x 1.201 x 1.07 x 2.188 = 2.196 inches
c) If a steel with a particular quench diameter is required, rather than order one custom
made, what is better to do instead? (1)
If a steel is required having a particular quench diameter, it is better to examine the
output from various mills and order the one, which manufactures to the required tolerance
with the specification. (Consider other answers that seem reasonable although this one
was given in class).
d) What is microstructure produced by the cooling rates indicated on the TTT diagram given
below? Indicate the any difference in morphology between c) and d) (2)?
a)
b)
c)
d)
a) martensite, b) martensite and pearlite, c) fine pearlite d) course pearlite
e) What does the dashed line in the TTT figure given below indicate and what microstructure
is produced? (1)
The dashed line refers to the critical cooling rate resulting in 100% martensite.
f) What effects on the TTT curves do the alloy additions of 1% Cr and 3.5% Ni give a 0.4%
C steel? (1)
1% Cr changes the TTT curves into two C-curves where the high temperature curve
refers to pearlite and the low curve refers to bainite. (1/2)
3.5% Ni retards the start of the pearlite start curve. A fast water quench will give
martensite, while a slow oil quench will give fine pearlite. (1/2)
g) What does all this mean? Briefly describe how the tensile strength and % elongation
(ductility) vary as a function of transformation temperature for the three metastable
phases existing in a eutectoid composition steel. What is required of the martensite? Use
a graph if it helps. (2)
The graph below shows the relationship between the tensile strength and % elongation
(ductility) of the steel as a function of transformation temperature.
Martensite has high tensile strength but no ductility (1/2). It needs to be tempered as it is
rarely used “as quenched” (1/2).
Pearlite can be made to have reasonably high tensile strength (600 oC) with reasonably good
ductility. (1/2)
Bainite at 400 oC is in between. (1/2)
h) Why is Mn an indispensible addition to hot rolled steels? (1)
Mn is an indispensable addition because it reacts with the sulfur remaining in the steel to
form MnS, which remains solid and deforms with the steel during hot rolling, whereas
FeS is a liquid and will induce the steel to split during hot rolling.
5. Stainless Steels (1 mark each – total 10 marks)
a) What is the definition of a stainless steel that differentiates it from all the other types of
steel? A stainless steel is defined as a steel having 12%Cr or more as an alloying
addition.
b) What is the principle reason why one would want to use a stainless steel? For its
corrosion resistance.
c) What form of chromium carbide is normally found in stainless steels? (FeCr)4C
d) Concerning the Fe-Cr phase diagram, how does Cr affect the ferritic phases (1/2) and
austenite phase (1/2)?
Cr is a strong ferritic phase stabilizer of the bcc structure. The delta ferrite and alpha
ferrite become indistinguishable. The austenite phase is condensed into a small -loop
over the temperature range of 900 – 1400 oC.
e) Concerning the Fe-Ni phase diagram, how does Ni affect the austenite phase (1/2) and
ferritic phases (1/2) down to room temperature?
Ni – Ni is a strong gamma phase stabilizer of the FCC structure down to room
temperature in Ni-Cr stainless steels such as Fe-8%Ni-18%Cr. The ferritic phases are
very restricted.
f) How does the gamma phase affect the brittleness of the steel at low temperatures (1/2)
and why (1/2)? The gamma phase maintains the ductility of the stainless steel at low
temperatures because it has the FCC structure.
g) If a FCC stainless steel were to corrode, say in a marine environment, what would be
the morphology of the corrosion (1/2) and why (1/2)? Steels having the FCC structure
tend to form pits if and when they corrode. Dislocations form grain boundaries as a
network and act as pipes for diffusion, which is required for corrosion. Dislocations in
FCC materials react with each other to form extended networks that can extend deep into
the material from the surface, which enables the pit morphology to be created.
h) Stainless steels can suffer from the precipitation of chromium-iron carbides at grain
boundaries. What do they suffer (1/2) and what are two methods that can be used to
eliminate it (1/2)? Precipitation of (CrFe)4C at grain boundaries can reduce the Cr
content to below 12%, which degrades the corrosion resistant property of the steel. It can
be eliminated by lowering the carbon content to 0.03% or alloy the steel with Ti or Nb,
which remove the carbon at TiC or NbC, without lowering the Cr content of the
austenite.
i) What is the predominant microstructure of the 400 series of heat treatable stainless
steels (1/2)? What are the high-carbon grades used for containing 0.6-1.2%C and 16-18
% Cr, which have much harder martensite upon quenching (1/2)? The 400 series stainless
steels have a martensitic structure. The high carbon grades are used for surgical
instruments.
j) In the thermomechanical treatment of steels, what is ausforming (1/2)? What
transformation property of the steel, which is provided by its composition, is necessary
for ausforming (1/2). Ausforming is the deformation of the austenite prior to the
formation of martensite. It can only be used if the composition of the steel results in a
bay between the pearlite and bainite reactions so that pearlite and bainite will not form
during the deformation process. Afterwards the steel is quenched to room temperature to
form martensite.
6. Cast Irons (total – 10 marks)
What are the three constituents of cast iron? Iron, carbon and silicon. (1)
What is its eutectic temperature, eutectic reaction and compositions? (1.5)
eutectic
1154 oC L4.26%C 2.08%C + Cgraphite
What is its eutectoid temperature, eutectoid reaction and compositions? (1.5)
eutectoid
738 oC 0.68%C  0.02%C + Cgraphite
Why is silicon necessary to produce cast iron? (2)
Silicon is used to stabilize the formation of graphite by increasing the undercooling of the
eutectoid reaction (1), which encourages the formation of ferrite and graphite phases from
austenite (1). Alternatively - Si retards the formation of cementite, Fe3C, which allows more
time to form graphite (1).
Write the equation to convert silicon into an equivalent carbon based on the eutectic
composition. (1)
%Si
Carbon Equivalent  %C 
 4.3%
3
How is the addition of phosphorus treated in the carbon equivalent equation? (1)
%Si  %P
Carbon Equivalent  %C 
 4.3%
3
How does white cast iron obtain its name (1/2) and why is it used (1/2)?
When white cast iron fractures its surface appears white. It is very hard and resistant to wear.
How is the strength and ductility of gray cast iron and what is it most commonly used for?
(1) Gray cast iron has low strength and low ductility and it mostly used for engine blocks.
Bonus 1 mark
Calculate the percentages of phases present using the Lever Rule in question 3 and the Ideal
Quench Diameter in question 4.