MECH 473 Ferrous and Non-Ferrous Materials
Student Name ………………………… Student Number …………………
Assignment #2
Marks shown in brackets ( ) at the end of each question.
Due Friday March 30, 2012
1. What are the principle means of strengthening Copper and its alloys and
give an example of a material using these means? (5)
1) Solid solution strengthening, eg. Commercially pure Cu, Cu-Ni alloys have complete
solid solubility, Cu-35%Zn, Cu-10%Zn
2) Grain size strengthening – pure Cu
3) Strain hardening – pure Cu, Cu-35%Zn
4) Age hardening – Cu-2%Be
5) Phase change hardening – martensite (Cu-Al) and eutectoid reaction (Cu-Mn)
2. When copper and its alloys strengthen by cold working how are the crystal
and its dislocation process(es) involved in the strengthening? (2)
(Describe the dislocation mechanism involved in cold working Cu and its alloys)
Copper and its alloys have the FCC structure. It work hardens by the dislocations
having B = ao/3<110>{111} being able to interact with each other to form a dense,
tangled network where their strain fields resist their movement thereby providing
strengthening.
3. What copper alloy is used for thermocouples and why? (1)
Cu 45%Ni, constantan is used for precision standard resistors and rheostats to
measure temperature as thermocouples because it has the highest electrical resistivity
combined with a very low temperature dependence of resistivity.
4. What Copper alloy is able to make stiff wires and springs? What are the
current and a modern application of it at a very small scale? (1)
An alloy containing 1.8 % Be- 0.25% Co has high hardness and excellent stiffness
– so it is used for miniature springs and for current and future MEMS
applications.
5. What are two Cu-based shaped memory alloys and how do they work? (1)
The most effective and widely used Cu alloys include CuAlNi and CuZnAl.
The shape change involves a solid state phase change involving a molecular
rearrangement between Martensite and Austenite upon a temperature change of only
about 10 oC.
6. What is coring and why does it occur in Cu-Ni alloys? (1)
Coring is segregation of the Ni concentration from the center to edge of the grains
produced during solidification due to the low diffusivity of Cu and Ni.
7. What is the most widely used cast Cu alloy and why? (1)
The three-fives bronze – 5%Sn 5%Zn and 5%Pb – is the most widely used casting
alloy – it is a combined brass and Sn bronze – with good mechanical casting and
corrosion properties and excellent machinability (due to Pb).
8. Describe the three steps that would be required to age harden the Cu-2%Be
alloys. (3)
1. Solution Treatment – the alloy is heated above the solvus temperature (above 576
C and usually up to 800 C) into a single phase region of the phase diagram to
dissolve any secondary phases such as precipitates. The material is held at this
temperature until a homogeneous solid solution is produced.
2. Quench – the alloy is rapidly cooled so the atoms do not have enough time to
diffuse to potential nucleation sites. The alloy remains as a single phase material
that is supersaturated with Be. If the material is work hardened, the increase in
dislocations density can be used as nucleation sites during aging.
3. Aged – The alloy is heated to a temperature below the solvus (below 576 C and
usually at 400 C) so the atoms can diffuse to numerous nucleation sites to produce
precipitates. Ideally, uniform highly dispersed, ultrafine  precipitates form to
give the best effect in age hardening or precipitate strengthening.
9. What four conditions must be meet to age harden the Cu-2%Be alloys? (2)
The four conditions that must be satisfied for an alloy to have an age-hardened
response during heat treatment.
1) The Cu-2%Be alloy system must display a decreasing solid solubility with
decreasing temperature. In other words, the alloy must form a single phase on
heating above the solvus line, then enter a two-phase region on cooling.
2) The matrix should be relatively soft and ductile, and the  precipitate must be
hard and brittle.
3) The alloy must be quenchable.
4) The  precipitate must have coherence with the Cu matrix.
10. Other than Cu alloys, what other alloys are age hardenable? (1)
Many important alloys are age-hardenable including stainless steels and alloys based
on aluminum, magnesium, titanium, nickel, chromium, and iron.
11. What alloying addition to Al has very low solubility and how is it used by the
electronics industry. (1)
Si has very low solubility in Al. There is no Si-Al phase at 100%Si on the phase
diagram. This makes it good for an interconnect of components in devises for the
electronics industry.
12. Why must Mg be added to Al-Si alloys to make them age hardenable? What
can be added to Al-Mg alloys to make them age hardenable? (1)
Mg needs to be added to Al-Si alloys to make them age hardenable because an Al-Si
precipitate is not formed whereas a Mg-Si precipitate will form enabling aging. Zn
can be added to Al-Mg alloys to make them age hardenable.
13. What is added to Al-Si alloys and how are they heat treated to enhance their
strength and toughness. (1)
The addition of Na just before casting allows a hypereutectic alloy to become
supercooled, which shifts the eutectic from 11 to 13% Si and causes the Si to form as
very fine particles giving high strength and ductility. Rapid cooling gives smaller,
more rounded precipitates with greater ductility.
14. What is the strengthening phase in Al-Cu alloys? Why is it so effective in
strengthening these alloys? On what close-packed atomic planes does this
phase block the motion of dislocations? Why do the strength of these alloys
fall off if they are aged too long? (2)
The  phase is the strengthening phase of the precipitates. They are effective because
they are coherent with the matrix. These precipitates block the motion of dislocations
on the close-packed {111} planes. Their strength falls off if they are aged too long
because of ripening (precipitates get round and large).
15. Why do Al alloys initially soften during aging and why does prior stretching
of an Al alloy enhance its aging response? (1)
Al alloys initial soften because the dislocations are annealed out at the aging
temperature. Prior stretching enhances the aging response because the dislocations are
nucleation sites for precipitate formation.
16. What Al alloy is known for superplasticity and why? What can be added to
enhance its precipitation hardening (1)
Al-Zn alloys are good for superplasicity because Al can dissolve a high percentage of
Zn. Mg, Cu and Cr can be added for precipitate strengthening in Al-Zn alloys.
17. Why is pure Al used to clad Al alloys? Why is it used to clad steel cables used
by the power industry? (1)
Pure Al helps prevent corrosion of Al alloys. It’s used to clad steel cables used for
the transmission of electricity.
18. If an Al alloy corrodes (other than forming its protective oxide), what is/are
the mechanism(s) by which it corrodes? (1)
(Describe the dislocation mechanism involved in corroding Al and its alloys.)
Thermal fluctuations or mechanical deformation cause precipitates existing on the
surface of an alloy to crack the protective oxide enabling corrosion products to
penetrate and corrode the Al alloy.
19. What is the structure of the corrosion product on the surface of the Al
material and why. (1)
The corrosion product is usually a white pit on the surface caused by the
dislocations of FCC Al to be able to interact (join each other), which enables the
corrosion product to be able to penetrate deep within the material rather than
forming a scale like rust similar to a plain carbon steel.
20. Explain why aluminum alloys containing more than about 17% Mg are not
used. Assume that the  phase is an intermetallic compound and a eutectic
structure is produced. (1)
When more than 17% Mg is added to Al, a eutectic microconstituent consisting of the
 phase and a brittle  phase is produced during solidification as seen in its phase
diagram. This eutectic contains
% 
35.0  17.4
 97.2%
35.5  17.4
Most of the eutectic is thus the brittle  intermetallic compound and it will likely
embrittle the eutectic. The brittle eutectic, which is the continuous microconstituent,
will make the entire alloy brittle.
21. What is a common problem of cast aluminum alloys occurring during
solidification and what are two methods by which it can be reduced. (1)
Shrinkage occurs between the dendrite that create pores, which reduces Al’s density
and act as defects for failure. They can be reduced by controlling the solidification
process to produce smaller dendrites and by compressing the Al part that will squish
the pores to reduce their size and increase the density.
22. The company that you are working for welds aluminum parts together using
the GMAW (MIG) process. Currently, they are using CO2 gas to protect the
aluminum during the welding process. However, the CO2 gas is too cold and
leaves a rough finish on the welds. What gas would you choose to improve
the finish on the welds? Why would you choose this particular gas? (1)
Answer: Choose argon gas because it does not cool the weld as quickly. This allows
the weld puddle to smooth out more before freezing.
23. Stainless steels can suffer from “weld decay”. What is it? (1)
Weld decay occurs in the stainless steels and is the precipitation of (CrFe)4C, which
contains 70 % Cr, at grain boundaries causing the concentration of Cr in the adjacent
grains to fall below 12%, which degrades the corrosion resistance properties of the
stainless steels. The optimum temperature for precipitation of (CrFe)4C is around
650 oC, which is attained in the heat affected zone adjacent to a fusion weld.
24. What are three methods that can be used to eliminate weld decay and why?
(1)
The thee methods to eliminate weld decay are to use a low carbon steel, use a higher
Cr steel or add W and Nb as additives. The first two methods reduce the C in the steel
to help prevent the formation of (CrFe)4C. The second method has sufficient Cr to
protect the steel even if (CrFe)4C forms.
25. Submerged-arc welding is used for the high-output welding of large steel
panels (such as ship decks) in the flat position. Describe the Submerged-Arc
Welding process. (1)
Submerged-arc welding is both fast and automatic. Heat is supplied by an arc
between a bare metal electrode and the work piece while the shielding is provided by
a blanket of granular flux. The electrode and the flux are both fed continuously as the
arc is traversed along a pre-machined vee interface. The flux melts in the vicinity of
the arc and then solidifies to form a protective cover over the cooling weld. The flux
is an insulator in the solid state but become highly conductive on melting and thus
maintains the submerged arc. High quality welds can be produced at high rates of
weld metal deposition. The high deposition rates are achieved with narrow electrodes
passing high currents, which results in a relatively coarse columnar structure and an
enlarged heat affected zone causing poorer low temperature ductility.
26. How should the welding rod match the base steel metal that is heat treatable
and non-heat treatable in terms of Ni, Mn, Cr and Mo? (1)
In non-heat treatable steels the filler usually has slightly more Mn and Ni.
In heat treatable steels, the Ni, Cr, and Mo content of the filler is usually increased to
counteract evaporation losses when passing through the arc.
27. Draw the microstructure of steel in the fusion zone and heat-affected zone in
a weld showing in a) the initial structure at the maximum temperature, b) the
structure after cooling of a low-carbon steel and c) the structure after cooling
of a high hardenability steel. (3)
28. By what means does the FIB (Focus Ion Beam) weld materials at the
nanoscale? (1)
The metal-organic gas that is introduced into the vacuum chamber is broken down by
the focused ion beam, which deposits the metal atoms onto the surfaces being welded.
29. What is the crystal structure of Mg? What are its close-packed planes and
directions? (not required for the assignment but may be required for the 2nd
mid-term)
Mg is HCP up to its melting point. Its close packed planes are the (0001) and its
close-packed directions are the <1120>.
30. Mg being a highly reactive metal, how is it protected for open and closed
crucibles during melting? (not required for the assignment but may be
required for the 2nd mid-term)
It must be covered with a flux during melting.
For covered crucibles the flux is 20% KCl, 50% Mg2Cl, and 15% CaF2.
For open pots the flux is 55% KCl, 34% Mg2Cl, 9% BaCl2 and 2% CaF2.
31. At what temperatures are Mg alloys solution treated and what happens if the
temperature is too high? (not required for the assignment but may be
required for the 2nd mid-term)
To dissolve magnesium alloy precipitates, it is solution treated at 390 – 410 C.
If the solution temperature is too high, it will “burn” where low melting grain
boundary phases are exuded at the surface.
A grey-black powder appears on the surface
Internal voids form due to evolution of gaseous phases
32. On a stiffness basis, how does Mg match up with Al and steel? On a mass
basis, how does it match? (not required for the assignment but may be
required for the 2nd mid-term)
Mg has a modulus of elasticity of 45 GPa – compared to 71 GPa for Al and 200 GPa
for steel.
Mg density is 1.8 g/cm – compared to 2.8 g/cm for Al and 7.9 g/cm for steel – on a
mass basis, Mg has the greatest stiffness – and steel the least.
33. By what means can Mg alloys be welded and what’s possible for temporary
repairs in the field? (not required for the assignment but may be required for
the 2nd mid-term)
Mg must be protected from the atmosphere by an inert gas using a tungsten arc or
consumable Mg
It can be welded using a gas torch with suitable flux for temporary repairs in the field.
34. In Mg-Zn alloys show how much alpha phase exists in the eutectic assuming
the maximum solubility of Al in the delta phase is 73%. Is the eutectic alloy
ductile? (not required for the assignment but may be required for the 2nd
mid-term)
73  54
 100  29%
73  8.4
The Mg-Zn alloy is not ductile but brittle because of the delta phase is brittle.
%a 
35. What are the two means to strengthen Mg-Al alloys and at which
compositions do they apply? (not required for the assignment but may be
required for the 2nd mid-term)
Alloys containing up to 3 wt% Al are solution strengthened
Alloys with 6-9 wt% Al can be precipitation hardened
36. At what temperatures can wrought Mg alloys be forged into shapes? By what
means and temperature are they strengthened and why? (not required for
the assignment but may be required for the 2nd mid-term)
All solid solution Mg alloys can be hot forged at 300 – 400 C. Precipitation hardening
during subsequent aging at room temperature is required as they are not heat
treatable.
37. In sand cast Mg alloys, what problems can arise due to the reactive nature of
Mg and how is it mitigated in a similar way as in Al alloys? (not required for
the assignment but may be required for the 2nd mid-term)
The reactive nature of Mg also means that sand cast alloys are subject to
microporosity caused by evolution of hydrogen with a consequent deterioration of its
mechanical properties. Insoluble gases such as He and Cl are bubbled through the
melt before casting to remove reactive gases such as H.
38. Why are die cast Mg alloys significantly stronger than sand cast alloys and
give three uses for them? (not required for the assignment but may be
required for the 2nd mid-term)
Die cast alloys are significantly stronger than sand cast alloys as they are not
susceptible to microporosity.
They are used for auto wheels, for crankcases for air cooled engines like VWs and for
dash boards in GM trucks.
39. What are the low-temperature and high-temperature phases of Titanium and
how can the high temperature phase be partially and fully stabilized to low
temperatures? (2)
The low-temperature phase is hcp. The high temperature phase is bcc, which can be
stabilized by alloying with alloying additions of Mn Fe Cr Cu Ni H for partially
stabilized and Mo V Ta Nb for fully stabilized bcc phase.
40. What are the two mechanisms of deformation in Ti and what are the slip
plane and directions active in the hcp material? (2)
Twinning and dislocations are the deformation mechanisms. The slip plane and
_
direction in the hcp material are the (0001) and 11 2 0 .
41. What impurities easily dissolve in Ti and how are they expressed? (2)
O N and C easily dissolve in both hcp and bcc Ti as interstitials and are conveniently
expressed in terms of its yield strength at 0.2% offset and in terms of the percent
oxygen equivalent, i.e., % O equivalent = (%O) + 2(%N) + 0.67(%C).
42. What alloying addition to the + Ti alloys is added for high temperature
creep resistance and why is it effective? (2)
Small amounts of Si are added to the + alloys to increase high temperature creep
resistance due to insoluble silicide phases.
43. What similar alloys to the Ti alloys are used in the nuclear industry and
why? (2)
Zr and its alloys, i.e., Zircaloys (Zircaloy-1. Zircaloy-2 and Zircaloy-4) are similar to
Ti and are used as cladding of fuel rods in nuclear reactors because they have very
low absorption cross-section of thermal neutrons.
44. Why is Ta a special material and what new method of processing enables it to
be used commercial purposes. Give an example of an application. (2)
Pure Ta is one of the most corrosion resistant materials. It is now being deposited as
Ta gas onto material surfaces that require extra-strong protection to form a thin,
protective film of Ta.
Its applications include the production of H, biomass to ethanol, oil & gas, etc that
involve a highly corrosive environment.
45. What is the structure of Nb over its entire temperature range and give two
methods of how it is strengthened for high-temperatures applications? (2)
Nb is bcc. Nb is solid solution strengthened using the additions of Zr, Mo W or Ti
and dispersioned strengthened using ZrO2 ThO2 ZrC and HfC, which strengthen by
preventing recrystallization and grain boundary sliding.
46. What methods (there are 2) and alloying additions are used to enhance the
high-temperature strength of Mo and its alloys and to what temperature are
they effective? (2)
Strain hardening is used in unalloyed Mo and Ti and Zr are added to form a fine
dispersion of carbides in order to restrict the recrystallization of Mo so that strain
hardening can be retained to high temperatures effective to 1300 C.
47. What is the basis for Tungsten (W) being used for filaments for incandescent
light bulbs, as cathodes in electronic and X-ray tubes and as electrodes for
welding and other arc sources? What is unique about W that makes it
different than the other refractory metals? (2)
W has the highest melting temperature (3695 C). It doesn’t absorb or react with
hydrogen so it can be treated or worked in a hydrogen or reducing atmosphere
required for welding whereas the other refractory metals readily absorb H so can’t be
used in a reducing environment.
48. What crystal orientation of W penetrated a steel target the best and why? (2)
The [100] orientation because all 4 slip systems are active in the deformation process
whereas the [111] has just three orientations, which resulted in mushrooming of the
head, and the [110] had just two, which resulted in the penetrator corkscrewing
through the steel as in penetrated.
49. Would you expect W to work harden during its extrusion from the tip of the
penetrator to its sides? Justify your answer by showing the dislocation
mechanism(s) responsible for deformation. (2)
W does not work harden because it has the bcc structure, which does not allow the
dislocations to interact to form complex, tangled, interconnected dislocations like fcc
metals. This is why refractory metals only have medium strength.
For bcc metals, if two dislocations try to interact their resultant dislocation will
generate a larger Burgers vector. Since the energy of a dislocation is proportional to
the square of the Burgers vector, these dislocations tend not to form, i.e.,
ao
1110 1 1  ao 1 11 0 1 1  ao 022 0 1 1
2
2
2
Or,
ao
1110 1 1  ao 1 1 1 011  ao 2021 01
2
2
2

 

50. How many cracks would it take to make a single crystallite in the W
penetrator? What would be the crystal planes operative to make the
crystallites? (2)
It would take 6 cracks on the {100} planes to make the crystallite. These planes
include 1001 000100 1 000100 1  .
51. Why are Zirconium Alloys used in nuclear reactors? What are the three
Zirconium alloys being used today for the CANDU nuclear reactor? (2)
They are used because of their low neutron capture cross-section. The three alloys
being used are:
1) Zircaloy-2 consisting of 98.25 wt % Zr, 1.20-1.70% Sn, 0.07-0.20% Fe, 0.050.15% Cr, and 0.03-0.08% Ni.
2) Zircaloy-4 consisting of 98.23 wt % Zr, 1.45% Sn, 0.21% Fe, 0.1% Cr, and
0.01% Hf.
3) Zr-2.5 % Nb
52. When a thermal neutron collides with a Zr atom what two types of energy
dissipation mechanisms occur? What two types of point defects are created?
(2)
Both a displacement spike and thermal spike are created, which dissipate the
energy of the thermal neutron. Both self-interstitial atoms (SIAs) and vacancies
are created in the process.
53. What types of dislocations do the point defects create? How do these defects
go on to created an extended dislocation network? (2)
The point defects create interstitial loops and vacancy loops. The loops coalesce
to form an extended dislocation network.
54. What type of material deformation limiting the life-time of the CANDU
reactor is created by the annihilation of point defects at the extended
dislocation networks? When single crystals of Zr are tested for their
deformation, what is the direction dependency of the deformation? (2)
Irradiation growth is created by the annihilation of point defects at the extended
defects. The material expands in the <a> axis, shrinks in the <c> axis and remains
unchanged in the <a+c> axis.
55. What is the term used for the high-rate of material deformation at the high
neutron doses and what two mechanisms are responsible for it? (2)
Breakaway growth occurs at high neutron doses, which is caused by the formation
of high-energy dislocations having B = [0001](0001) and by the glide of network
dislocations preferentially interacting with interstitial loops.
56. In general, what happens to Fe, Ni and Cr precipitates and grain boundaries
during neutron irradiation of Zr alloys? (2)
The precipitates dissolve. Their elements (Ni, Fe, Cr) segregate to the grain
boundaries where they re-precipitate with a slightly different composition. The
grain boundaries move due to their absorption of point defects.