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:

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 cementite-austenite 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:

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

c) The total amount of ferrite is:

31. What are the four assumptions used to derive the Johnson-Mehl equation? (2)

1) The nuclei are randomly distributed throughout the austenite2) The average rate of nucleation, N, is constant3) The nodules remain spherical in shape, and4) The growth rate, G, is constant

Time in seconds

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 cmThe 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)

1) Quenching (Martensite + graphite) (1/2)2) Annealing (Ferrite + graphite) (1/2)3) Normalizing (Pearlite + graphite) (1/2)4) Austempering (Bainite + graphite) (1/2)

1

2

4

3