MECH 473 Ferrous and Non-Ferrous Materials
Student Name ………………………… Student Number …………………
Assignment #1
Marks shown in brackets ( ) at end of question. Total marks are ~89.
Due date to be determined in class.
1. Give three good reasons why we need to know the properties of materials? (3)
a) Cost - Cost of materials can significantly affect the final product cost
with 50% being the rule. For example, automobile materials typically
are 70% of the manufacturing costs. For electronic devices, such as
computers, materials can be 75% of the manufacturing costs.
b) Competitiveness - The manufacturer with the best materials and
processes has the best product and an advantage over the competitors.
To remain competitive, companies hire engineers for the required
research and development (R&D).
c) Defense - Throughout the history of mankind, the strongest countries
produced the best materials and tended to dominate the world, e.g., the
UK thumped a lot of competitors using steel ships and ended up
controlling 1/5 or its land mass. Thus, materials are of national
importance to a country
2. What are the three pillars of the materials selection process that affect a product’s
cost and performance and briefly describe them? (3)
a) Composition – composition determines the material’s cost and the
materials properties such as strength, magnetic, electrical and optical
output, etc
b) synthesis and processing - determines the material’s cost of putting the
material into a form useful for its application
c) microstructure – determines many of the properties of the material
such as its strength, corrosion resistance, etc., necessary for the product
to be competitive with other similar products.
3. List 10 functional classifications of materials, give a brief description and an
example. (5)
1) Biomedical materials – materials such as cobalt, titanium,
stainless steel, teflon, calcium-based ceramics, etc that replace
bones, organs, teeth, etc.
2) Electronic materials – semiconductors such as Si, Ge, GaAs, etc.,
used in computers, ceramics used as sensors, metals used as
conductors, superconductors for powerful magnets.
3) Energy and Environmental materials – The nuclear industry use
uranium for fuel, zirconium to hold the uranium and high-strength,
low-corrosion steel in the nuclear reactors. The fuel cell industry
uses many types of materials such as zeolites, alumina, etc as
catalysts. Solar panels use materials such as amorphous silicon.
4) Magnetic materials – computer hard disks and audio and video
cassettes use many types of ceramics, metals, and polymers.
5) “Smart” materials – can sense and respond to an external
stimulus such as temperature, stress, humidity or chemical
environment consisting of a sensors and actuators and read change
and initiate an action such as Lead Zirconium Niobate (PZT).
6) Aerospace Materials – such as aluminum, carbon composites,
titanium, gallium arsenide and silicon, which are strong, light
weight, resistance to radiation damage
7) Automotive Materials such as carbon composites, aluminum, and
steels that are light-weight, high-strength materials, high-strength,
high-temperature materials
8) Structural materials such as steels, aluminum, concrete and
composites are designed for carrying some kind of stress such as in
buildings and bridges with combinations of strength, stiffnes and
toughness are needed under different conditions of temperature and
loading. These are the most common use of materials.
9) Optoelectronic materials such as silicon, GaAs, organics, etc for
solar panels having multiple layers of material to collect a large
range of wavelength from the sun’s light. Optoelectronic
materials such as GaAs, Indium Phosphide, GaSb, CdTe, silicon,
germanium, etc used for the communication industry for detectors
and lasers. Silica is used for fiber optics and sensors. Polymers are
used to make Liquid Crystal Displays, being used in the projector
for this lecture, etc.
10) Defense materials such as uranium and tungsten for projectiles,
steels for shields, tanks, ships, etc.
4. What is an amorphous metal? Who discovered it? Why are they special and what
will they enable us to do in the future? (2)
Amorphous Metals such as Zr41.2Be22.5Ti13.8Cu12.5Ni10.0 were discovered by
a Canadian named Moe Rainville. These materials have a super-high elastic
constant making them extremely strong. They will be used to make buildings and
structures in cities that reach the clouds. They will be used to make things that we
can only imagine and sometime see in science fiction movies.
5. What is a metamaterial? Give an example of the materials used and describe
how they work. (2)
Metamaterials have a negative refractive index, which will make them extremely
useful as perfect lenses for optical communication devices and for microscopes,
as well as, for cloaking devices. For long wavelengths such as microwaves,
metamaterial consists of fiberglass panels having copper loops and wires on its
surface. The electromagnetic waves resonate with the material’s structure to
obtain a negative refractive index.
6. Even though iron was the first “really hard” material to be used by man, why did
it take the longest to be produced into useful tools? (1)
It took a very long time to learn how to make “clean” iron, low in carbon and
impurities, which required high temperature furnaces. Early iron was brittle as a
result of being too high in carbon and impurities.
7. Who were the first people to learn how to make and use iron and where did they
live? (1)
The Hittites in Ur (Middle East) were the first to make and use iron before 1,200
BC.
8. What type of metallurgical processing in the manufacture of steel swords did the
Romans learn from the Greeks that enabled them to defeat the Gauls? This metal
processing is still in use today in the making of steel. It is being used to help
protect our peace-keeper in Iraq. How did this processing affect the swords of the
Romans so that they became superior to those of the Gauls? (1)
The Romans quenched and tempered their swords, which made them harder,
stronger and less brittle.
9. What significant design contribution did the Arabs bring to iron and steel making
between 700 – 1200 AD? What did this design enable? What type of modern
furnace is now based on their contribution? What two types of iron did it enable
to be produced? (2)
The Arabs developed furnaces with very high chimneys (Philosophers Egg) (½)
that produced higher temperatures and thus purer iron (½). These furnaces
evolved into the blast furnace of today (½) and enabled wrought iron and cast
iron to be produced (½).
10. In the beginning what did metallurgists use for their source of carbon and what
was it replaced by? (1)
In the beginning, iron makers used wood for their source of carbon (charcoal)
made preferably from hard wood, which was eventually replaced by coal (coke)
as the availability of hard wood diminished.
11. Slowly the Iron age turned into the Steel Age as impurities were removed from
the iron and iron alloys began being produced, however, where and when was
steel first produced, called Wootz steel, and why was it not continued? (1)
Wootz steel was produced in clay pots (crucible) in India around 500 BC but it
wasn’t continued because it wasn’t reproducible.
12. The UK, USA and most of the developed world produced steel using a crucible
process that dominated steel making right up until the 1980s. What is this method
and why did it work so well? (1)
It’s the Open Hearth Method, which was able to be well controlled to make
high-quality steel.
13. Hot-shortness brittled the steel so it could not be forged when hot. How was the
hot shortness problem solved (2 changes were necessary)? (1)
Hot shortness was solved by first reducing the Carbon content, which purified the
steel, and then by adding Mn (spiegelsen, a C-Mn alloy) to re-introduce the C,
which strengthened the steels. Mn adds ductility to carbon-based steels and is
used extensively today.
14. What steel making method first conceived in Austria and developed in Canada
has made Canada one of the leaders, if not the leader, in producing high-quality,
low cost steel used for buildings and automobiles? Briefly describe this method.
(1)
The BOF method was brought to Canada by the Austrians in the 1950s and
developed by Dofasco. It became the leading method during the 1990s. It uses
liquid oxygen blown into the pig iron for it to reach high temperatures in a short
time to clean and produce high-quality steel.
15. When austenitization of the steel is complete and it cools down below the
eutectoid temperature where will the pearlite nucleate? (1)
The pearlite will nucleate at the austenite grain boundaries.
16. In general, describe in terms of the 1 grain and the 2 grain how pearlite grows
from austenite and state what phase will form first in hypereutectoid and
hypoeutectoid steels. (4)
(any one of the following is worth ½ mark)
The pearlite grows into only one of the two austenite grains at the boundary.
It nucleates on one grain that it can establish a crystallographic relationship (1 grain),
and it then grows into the other grain (2) that it has no crystallographic relationship.
Pearlite does not grow into the one the austenite grain, 1, with which it has an
orientation relationship because it involves a low-energy, low-mobility interface.
Pearlite grows 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.
In a hypereutectoid steel the nucleation of pearlite will first produce cementite followed
by ferrite in the formation of pearlite.
In a hypoeutectoid steel, ferrite will first nucleate followed by cementite in the formation
of pearlite.
The austenite grain that is used as a substrate by the ferrite has the following orientation
relationship with 1:
111 110a

110

111 a
17. Draw the Kurjumov-Sachs crystallographic relationship indicating the
planes and directions for ferrite grain growth into an austenite grain. (4)
18. Many materials produce eutectic and eutectoid microstructures. For pearlite how
does the local carbon concentration and its diffusion affect the nucleation and
growth of the ferrite and cementite platelets? (1)
The growth of a ferrite plate will increase the carbon concentration in its vicinity
and thus stimulate the nucleation of a cementite plate at the ferrite-austenite
interface.
The growth of a cementite plate will decrease the carbon concentration in its
vicinity and thus stimulate the nucleation of a ferrite plate at the cementiteaustenite interface.
19. When we look at pearlite under a microscope, why is the spacing not
representative of the pearlite’s true spacing? Briefly describe 3 methods to
measure the true interlamellar spacing of pearlite. (2)
The interlamellar spacing seen by the microscope is not representative of the true
interlamellar spacing because the ferrite and pearlite platelets can approach the
surface at an angle which makes them appear wider than they actually are. (1/2)
1) Given a large enough representative cross-section, the “minimum” observed
spacing can be taken as the true spacing, lo, defined as the spacing between
lamellae normal to the surface.
2) In practise,  is obtained by measuring the distance, “l” between adjacent
lamellae along a number of random lines marked on the polished surface. If  is
the mean of a number of such measurements, then the true value of the lamellar
spacing is given by:
m
 
2
3) The Electron Microscopy (EM) method involved protecting ½ of the surface of
a steel specimen that contained pearlite using a rubber compound, then chemically
etching the unprotected surface to created a height difference between the two
surfaces, then removing the rubber compound and evaporating a thin layer of
Aluminum (Al) and Carbon (C) on the two surfaces. The Al-C thin film
replicates the surface and is removed and examined in the EM to measure the
required parameters of the true interlamellar spacing, ttrue.
20. What is the crystal structure of austenite, ferrite and martensite? (1 ½)
FCC, BCC and BCT
21. What is the molecular composition of cementite and Hagg carbide? (1)
Fe3C and Fe2..2C
22. Where does carbon reside in the crystal structure of austenite, ferrite and
martensite? (1 ½)
In austenite, the carbon resides in octahedral sites at the center of each edge of
the unit cell (½, 0, 0) and in the cube center (½, ½, ½).
In ferrite, the carbon is randomly located in tetrahedral sites ((1/4, ½, 0).
In martensite, the carbon is restricted to octahedral sites (center of crystal faces)
and unit cells edges that lie parallel to the c-axis
23. What are the differences between lath martensite and lenticular martensite in
terms of their carbon concentration and required tempering? (1)
Lath martensite is formed at carbon concentrations < 0.4 C.
Lenticular martensite is formed at carbon concentration > 0.4 C.
Lath martensite does not necessarily have to be tempered for its use.
Lenticular martensite is too brittle and must be tempered for its use.
24. Define hardenability of a steel. (1)
Hardenability is the ability of a steel to form martensite.
25. In general, how does the hardenability of steel depend on most of the alloy
additions? (1)
The hardenability increases with increasing alloy content for most of the alloy
additions.
26. How does the hardness of the martensite depend on carbon and the other alloying
additions? (1)
The hardness of martensite depends solely on the carbon content and does not
depend on the other alloys.
27. 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 martensite during
tempering.
28. From your phase diagram lecture, briefly explain how an equilibrium phase
diagram is created. At the eutectic composition and temperature, how many
degrees of freedom are there? (3)
The phase diagram is created by measuring the temperature of solidification of a
metallic liquid of many alloys of different compositions while noting that the rate
of change of the temperature changes at the liquidus and solidus points for each
alloy. The liquidus lines and solidus lines are drawn by connecting all of the
liquidus and solidus points of the alloys. At the eutectic there is zero degrees of
freedom.
29. What can happen to the liquidus and solidus lines if the alloy is quenched rapidly,
i.e., non-equilibrium, and what can be the consequences in terms of the phase
structures that can be produced? Give an example. (2)
If the alloy is quenched rapidly, the solidus and liquidus lines can shift to the right
or left resulting in non-equilibrium phases being formed, for example, such as the
eutectic structure being formed last in the solidification of the Pb-15%Sn alloy.
30. Consider a 1020 steel consisting of ferrite and pearlite given the phase diagram
below. a) Describe the changes in its constituents as it is heated from room
temperature to 850 oC. (2) b) at 750 oC, determine the relative amounts of ferrite
and austenite. (1) c) at 727 oC, determine the total amount of ferrite and
cementite present (1), d) determine the amount of ferrite in the pearlite. (1). From
the TTT diagram provided, draw the expected microstructures of the 1020 steel
clearly indicating the phases present, given their heat treatments provided by 1, 2,
3, and 4 (4). (9)
a) At room temperature, the 1020 steel has a ferrite plus pearlite structure. At 727 oC, the
austenite phase, , forms as the pearlite dissolves (1/4). As the temperature increases in
the range of 727 oC to 830 oC, the pearlite dissolves completely and there is retained
ferrite co-existing with –austenite (1/4). As the temperature increases, the amount of
austenite increases at the expense of the ferrite. At 830 oC, all the ferrite transforms to
austenite (1/4) and the structure at 850 oC is 100%. (2)
b) Relative amounts of austenite and ferrite 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) The total amount of ferrite is:
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
0.2 - 0.02
(% 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
% 
Time in seconds
31. What are the four assumptions used to derive the Johnson-Mehl equation? (2)
1) The nuclei are randomly distributed throughout the austenite
2) The average rate of nucleation, N, is constant
3) The nodules remain spherical in shape, and
4) The growth rate, G, is constant
32. Determine the constants c and n for the sigmoidal equation used to establish a
time-temperature-transformation plot, given the plot provided. (6) b) Determine
the activation energy for crystallization. (2) Hint: the rate of transformation is the
reciprocal of the time required for half of the transformation to occur. Use the
Arrhenius equation. c) For an isothermally transformed eutectoid steel having
yield strength of ~500 MPa, determine the approximate interlamellar spacing and
transformation temperature of the pearlite. (2)
From your notes:
For an isothermally transformed eutectoid steel having yield strength of ~500 MPa,
determine the approximate interlamellar spacing and transformation temperature of the
pearlite
1/ = ~75,000;  = ~1.3 x 10-5 cm
The transformation temperature is ~575 oC
33. What is the “ideal quench diameter” of a steel specimen? Calculate it for steel
consisting of 0.6%C, 1%Mn, 0.55%Ni, 0.65%Cr having a grain size of ASTM 6
using the Table provided. (1)
The “ideal quench diameter” is the diameter of a sample, which would have 50/50
P/M at its center when cooled by an ideal quench. (1/2)
D1 = 0.28 x 4.3 x 1.2 x 2.4 = 3.5 inches (1/2)
(see chart below)
34. What is the effect of austenite grain size on hardenability? Give a brief
explanation. (1)
Because pearlite nucleates heterogeneously at austenite grain boundaries, the
number of pearlite nuclei that form per unit time, ie., nucleation rate, is directly
proportional to the grain boundary surface area, which is inversely related to the
austenite grain size. The rate of nucleation of pearlite will be much more rapid in
fine grained steel, which will consequently have a lower hardenability.
35. Why are low amounts of Cr important for a HSLA steel’s microstructure? What
physical properties does Cr enhance? (1)
(Any one of these)
Low amounts of Cr, such as in the HSLA steels enables martensite and bainite to
form upon quenching (1/4), which can be later tempered to improve strength
(1/4).
Upon cooling from austenite, Cr dissolves in both the ferrite and in the cementite
(1/4) where it substitutes for Fe atoms in the ’s bcc and Fe3C crystal structures
(1/4).
This redistribution of Cr among the ferrite + cementite phases slows down the
eutectoid reaction (1/4) so that the TTT curves are displaced towards higher times
to the right (1/4), thereby increasing hardenability (1/4).
36. Why is low amounts of Ni important for a HSLA steel’s microstructure? What are
the changes, or not, to the physical properties of HSLA steels by the addition of
Ni? (1)
(Any one of these)
The presence of 3.5 %Ni does not improve the strength (1/4) above that of a hot
worked plain carbon steel because Ni does not significantly improve the
hardenability (1/4)
but after tempering, the ductile-brittle transition temperature of this steel is
lowered (1/4) to below -20 oC.
The increase of Ni content to 9% gives higher ductility (1/4) so that its ductile to
brittle transition temperature is lowered to below –200 oC.
Solid solution strengthening (1/4) is achieved by the addition of Ni, which
partitions to the ferrite phase rather than the carbide (1/4).
The redistribution of Ni among the ferrite phase slows down the eutectoid
reaction (1/4) so that its TTT curves are displaced towards higher times to the
right (1/4), thereby slightly increasing hardenability.
37. Why is Mn such an important alloying addition for the low carbon steels and
HSLA steels (1/2)? Why is Mn an indispensable alloying addition the formability
properties of low carbon steels and HSLA steels (1/2)? (1)
Mn was the first alloying addition used to significantly enhance steel’s ductility
(1/4). The redistribution of Mn among the ferrite + cementite phases is the most
effective alloying addition to slow down the eutectoid reaction (1/4) so that the
TTT curves are displaced towards higher times to the right (1/4), thereby
increasing hardenability (1/4).
Solid solution strengthening is achieved by the Mn that partitions to the ferrite
phase (1/4).
Manganese is an indispensable addition in steels because it reacts with the sulfur
remaining in the steel to form MnS. Without Mn, the sulfur reacts with FeS,
which is a liquid at the normal hot-rolling temperature, which will induce the steel
to split during hot-rolling. MnS remains solid and deforms with the steel during
hot rolling.
38. What is interphase precipitation and why is it important for HSLA steels? (1)
HSLA steels can be strengthened by “interphase precipitation” in which other
carbides such as VC, WC, etc form in preference to Fe3C. The other carbides
are less brittle than Fe3C so the steel is tougher.
39. Normally we use alloying additions and temperature processing to produce
martensitic steels from austenite, however, there is another means by which
engineers have used eutectoid steels to produce hard wearing surfaces for steel
rails. Describe this process (1/2) and what alloying addition and how much is
being used to make it happen (1/2)? (1)
The means is stress induced transformation of  to martensite. 14 %Mn is
added to rails intended to be used for switching, which creates pockets of retained
austenite. Under the heavy load of a train, the retained austenite at the surface
transforms to martensite to give an extremely hard wearing surface layer, which is
progressively replenished by successive transits.
40. What is killed steel and why are steels killed? (1)
A steel, which has been deoxidized using Al, which enables control of the
microstructure and grain size. It gives lower strength but better ductility.
41. What is the definition of a stainless steel that differentiates it from all the other
types of steel? (1)
A stainless steel is defined as a steel having 12%Cr or more as an alloying
addition.
42. What is the principle reason why one would want to use a stainless steel? (1)
For its corrosion resistance.
43. How does Cr affect the “delta” ferrite phase (1/2) in the Fe-Cr phase diagram and
why do we want to avoid the sigma () phase? (1)
Cr is a strong ferritic phase stabilizer of the bcc structure so the delta ferrite
phase and alpha ferrite phase become indistinguishable. The sigma phase is
brittle and should be avoided.
44. In a Fe-Cr-Ni stainless steel, how does Ni affect the austenite phase (1/2) and
ferrite phase (1/2)? (1)
Ni is a strong gamma (austenite) 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.
45. How is the gamma phase affected by low temperatures (1/2) and why (1/2)? (1)
The gamma phase maintains the ductility of the stainless steel at low
temperatures because it has the FCC structure.
46. If a delta ferrite stainless steel were to corrode, say in a marine environment, what
would be the morphology of the corrosion product (1/2) and why (1/2)? (1)
Stainless steels having the delta ferrite structure tend to form scales (continuous
thin layer) when they corrode. Dislocations enhance diffusion, which is required
for corrosion. Dislocations in BCC materials do not interact with each other so
tend to localize the corrosion at the surface of the stee, which enhances the scale
morphology to be created instead of a pit morphology.
47. How can precipitation of chromium-iron carbides occurring at grain boundaries
create a problem for a 12%Cr stainless steel (1/2) and what are two methods that
can be used to eliminate this problem (1/2)? (1)
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 1) lowering the carbon content to 0.03% or by 2) alloying the
stainless steel with Ti or Nb, which removes the carbon by precipitation of TiC or
NbC, without lowering the Cr content.
48. What are the three constituents and their relative concentrations that make cast
iron? (1)
Carbon (2-4%) and silicon (0.5 – 3%) balance iron.
49. Give at least two reasons why silicon is necessary to produce cast iron? (1)
Silicon is used to 1) stabilize the formation of graphite by increasing the
undercooling of the eutectoid reaction (1/2), which encourages the formation of
ferrite and graphite phases from austenite (1/2). Alternatively – 2) Si retards the
formation of cementite (1/2), Fe3C, which allows more time to form graphite
(1/2).
50. In the TTT diagram of the graphitic cast iron shown below, for each numbered
curve, state the type of heat treatment used and state its resulting microstructure.
(2)
2
4
1
1)
2)
3)
4)
3
Quenching (Martensite + graphite) (1/2)
Annealing (Ferrite + graphite) (1/2)
Normalizing (Pearlite + graphite) (1/2)
Austempering (Bainite + graphite) (1/2)