INDUSTRIAL MATERIALS

1. INDUSTRIAL MATERIALS Instructed by: Dr. SajidZaidi PhD in Advanced Mechanics, UTC, France MS in Advanced Mechanics, UTC, France B.Sc. in Mechanical Engineering, UET, Lahore B.TECH Mechanical Technology IQRA COLLEGE OF TECHNOLOGY (ICT) INTERNATIONAL ISLAMIC UNIVERSITY, ISLAMABAD

2. Formation of Alloys • For most manufacturing applications, metals are not used in their pure form. • Engineering materials tend to be ALLOYS, materials composed of two or more different elements, and they tend to exhibit their own characteristic properties. • An ‘‘alloy’’ can also be defined as a material that exhibits properties of a metallic material and is made from multiple elements. • There are different ways in which a metal might respond to the addition of another element. • The most popular way is when the two elements exhibit some degree of solubility in the solid state. The two materials form a solid solution, where the alloy element dissolves in the base metal. INDUSTRIAL MATERIALS

3. Formation of Alloys • The solid solution can be • Substitutional or Interstitial • In the substitutional solutions, some atoms of the alloy elements occupy lattice sites normally filled by atoms of the base metal. The replacement is totally random in nature, with the alloy atoms being distributed throughout the base lattice. • In the interstitial solutions, the alloy element atoms squeeze into the empty spaces between the atoms of the base metal lattice. INDUSTRIAL MATERIALS Interstitial Substitutional

4. Phases • Alloys can be classified as: • Single – Phase Alloys • Multiple – Phase Alloys • A phase can be defined as a homogeneous portion of a system that has uniform physical and chemical characteristics. • A phase has the following characteristics: • same structure or atomic arrangement throughout • roughly the same composition and properties throughout • definite interface between the phase and any surrounding or adjoining phases. INDUSTRIAL MATERIALS Formation of Alloys

5. Phases • Water has three phases: liquid water, solid ice, and steam. • Phases do not always have to be solid, liquid, and gaseous forms of a material. • An element, such as iron (Fe), can exist in FCC and BCC crystal structures. These two solid forms of iron are two different phases of iron that will be stable at different temperatures and pressure conditions. • Similarly, ice, itself, can exist in several crystal structures. • Carbon can exist in many forms (e.g., graphite or diamond). These are only two of the many possible phases of carbon. INDUSTRIAL MATERIALS Formation of Alloys

6. Phase Rule • Josiah Willard Gibbs developed the phase rule in 1875–1876. It describes the relationship between the number of components and the number of phases for a given system and the conditions that may be allowed to change (e.g., temperature, pressure, etc.) • The general form of Phase Rule is: 2 + C = F + P C = number of chemically independent components F = number of degrees of freedom P = number of phases present 2 = both temperature and pressure are allowed to change INDUSTRIAL MATERIALS Formation of Alloys

7. Phase Rule • Unary phase diagram • Melting point • Boiling point • Phase Rule • Point A • Point B • Point X • Triple Point (point X) • Condition of sublime • P-T diagram INDUSTRIAL MATERIALS Formation of Alloys Schematic unary phase diagram for magnesium

8. Solubility • How much of each material or component we can combine without producing an additional phase i.e., the solubility of one material into another (e.g., sugar in water, copper in nickel, phosphorus in silicon, etc.). INDUSTRIAL MATERIALS Formation of Alloys No Solubility Unlimited Solubility Limited Solubility

9. Unlimited Solubility • Water and Alcohol have unlimited solubility and when combined only one phase is produced. • Similarly, if we were to mix any amounts of liquid copper and liquid nickel, only one liquid phase would be produced. This liquid alloy has the same composition and properties everywhere, because nickel and copper have unlimited liquid solubility. INDUSTRIAL MATERIALS Formation of Alloys

10. Unlimited Solubility • If the liquid copper-nickel alloy solidifies and cools to room temperature while maintaining thermal equilibrium, only one solid phase is produced. After solidification, the copper and nickel atoms do not separate but, instead, are randomly located within the FCC crystal structure. Within the solid phase, the structure, properties, and composition are uniform and no interface exists between the copper and nickel atoms. Therefore, copper and nickel also have unlimited solid solubility. The solid phase is a solid solution of copper and nickel. INDUSTRIAL MATERIALS Formation of Alloys

11. Unlimited Solubility • A solid solution is not a mixture. A mixture contains more than one type of phase whose characteristics are retained when the mixture is formed. In contrast to this, the components of a solid solution completely dissolve in one another and do not retain their individual characteristics. Conditions of Unlimited Solubility In order for an alloy system to have unlimited solid solubility, certain conditions must be satisfied. These conditions, the Hume-Rothery rules, are as follows: • Size factor: The atoms or ions must be of similar size, with no more than a 15% difference in atomic radius, in order to minimize the lattice strain INDUSTRIAL MATERIALS Formation of Alloys

12. Unlimited Solubility • Crystal structure: The materials must have the same crystal structure; otherwise, there is some point at which a transition occurs from one phase to a second phase with a different structure. • Valence: The ions must have the same valence; otherwise, the valence electron difference encourages the formation of compounds rather than solutions. • Electronegativity: The atoms must have approximately the same electronegativity. Electronegativity is the affinity for electrons. If the electronegativities differ significantly, compounds form as when sodium and chlorine atoms combine to form sodium chloride. INDUSTRIAL MATERIALS Formation of Alloys

13. Limited Solubility • Salt or sugar have a limited solubility in water. • If we add a small amount of liquid zinc to liquid copper, a single liquid solution is produced. When that copper-zinc solution cools and solidifies, a single solid solution having an FCC structure results, with copper and zinc atoms randomly located at the normal lattice points. • However, during the solidification of the liquid solution containing more than about 30% Zn, some of the excess zinc atoms combine with some of the copper atoms to form a CuZn compound. • Two solid phases now coexist: a solid solution of • copper saturated with about 30% Zn plus a CuZn compound. • The solubility of zinc in copper is limited. INDUSTRIAL MATERIALS Formation of Alloys

14. Limited Solubility INDUSTRIAL MATERIALS Formation of Alloys

15. Solid – Solution Strengthening • In metallic materials, one of the important effects of solid-solution formation is the resultant solid-solution strengthening. • This strengthening is caused by increased resistance to dislocation motion. • This is one of the important reasons why brass (Cu-Zn alloy) is stronger than pure copper. • Similarly small levels of carbon strengthen iron. • Jewelry could be made out of pure gold or silver. However, pure gold and pure silver are extremely soft and malleable and the jewelry pieces made will not retain their shape. That’s why jewelers add copper to gold or silver. INDUSTRIAL MATERIALS Formation of Alloys

16. Solid – Solution Strengthening The effects of several alloying elements on the yield strength of copper. Nickel and zinc atoms are about the same size as copper atoms, but beryllium and tin atoms are much different from copper atoms. Increasing both atomic size difference and amount of alloying element increases solid-solution Strengthening. INDUSTRIAL MATERIALS Formation of Alloys

17. Solid – Solution Strengthening Degree of Solid-Solution Strengthening • The degree of solid-solution strengthening depends on two factors. • First, a large difference in atomic size between the original (host or solvent) atom and the added (guest or solute) atom increases the strengthening effect. • A larger size difference produces a greater disruption of the initial crystal structure, making slip more difficult. • Second, the greater the amount of alloying element added, the greater the strengthening effect. A Cu-20% Ni alloy is stronger than a Cu-10% Ni alloy. INDUSTRIAL MATERIALS Formation of Alloys

18. Solid – Solution Strengthening Effect of Solid-Solution Strengthening on Properties • The yield strength, tensile strength, and hardness of alloys are greater than those of the pure metals. For example, small concentrations of Mg are added to aluminum to provide higher strength to the aluminum alloys used in making aluminum beverage cans. • Almost always, the ductility of the alloy is less than that of the pure metal. Only rarely, as in copper-zinc alloys, does solid-solution strengthening increase both strength and ductility. INDUSTRIAL MATERIALS Formation of Alloys

19. Solid – Solution Strengthening Effect of Solid-Solution Strengthening on Properties • Electrical conductivity of the alloy is much lower than that of the pure metal. This is because electrons get more scattered off the atoms of the alloying elements. Solid-solution strengthening of copper or aluminum wires used for transmission of electrical power is not recommended because of this pronounced effect. Electrical conductivity of many alloys, although lower than that of pure metals, is often more stable as a function of temperature. • The resistance to creep, or loss of strength at elevated temperatures, is improved by solid-solution strengthening. High temperatures do not cause a catastrophic change in the properties of solid-solution-strengthened alloys. Many high-temperature alloys, such as those used for jet engines, rely partly on extensive solid-solution strengthening. INDUSTRIAL MATERIALS Formation of Alloys

20. Solid – Solution Strengthening INDUSTRIAL MATERIALS Formation of Alloys The effect of additions of zinc to copper on the properties of the solid-solution strengthened Alloy.

21. Phase Diagram • A phase diagram shows the phases and their compositions at any combination of temperature and alloy composition. • When only two elements or two compounds are present in a material, a binary phase diagram can be constructed. • Isomorphous binary phase diagrams are found in a number of metallic and ceramic systems. In the isomorphous systems, which include the copper-nickel systems, only one solid phase forms; the two components in the system display complete solid solubility. • In phase diagram, x-axis represents mole%, weight%, atomic% or mole fraction of one of the components. The y-axis represents the temperature. • A ternary phase diagram is a phase diagram for systems consisting of three components. INDUSTRIAL MATERIALS Formation of Alloys

22. Phase Diagram • Transition in structure appear as characteristic points in the temperature – time plot of the cooling history, known as cooling curve. It is obtained when a fixed composition material is heated and subsequently cooled at a uniformly slow rate INDUSTRIAL MATERIALS Formation of Alloys Liquidus

23. Phase Diagram Cooling curves for Cu-Ni alloy form different Ni composition INDUSTRIAL MATERIALS Formation of Alloys

24. Phase Diagram • Liquidus temperature is the temperature above which a material is completely liquid. The upper curve represents the liquidus temperatures for copper-nickel alloys for different compositions. • We must heat an alloy above the liquidus temperature to produce a completely liquid alloy that can then be cast into a useful shape. The liquid alloy begins to solidify when the temperature decreases to the liquidus temperature. For the Cu-40%Ni alloy in figure the liquidus temperature is 1280ºC. INDUSTRIAL MATERIALS Formation of Alloys

25. Phase Diagram • Solidus temperature is the temperature below which the alloy is 100% solid. The lower curve in figure represents the solidus temperatures for Cu-Ni alloys for different compositions. • An alloy is not completely solid until the material cools below the solidus temperature. • If we use a copper-nickel alloy at high temperatures, we must be sure that the service temperature is below the solidus so that no melting occurs. For the Cu-40% Ni alloy in figure, the solidus temperature is 1240ºC. INDUSTRIAL MATERIALS Formation of Alloys

26. Phase Diagram • Alloys melt and freeze over a range of temperatures between the liquidus and the solidus. The temperature difference between the liquidus and the solidus is the freezing range of the alloy. • Within the freezing range, two phases coexist: a liquid and a solid. For Cu-Ni, the solid is a solution of copper and nickel atoms and is designated as the αphase. • For the Cu-40% Ni alloy in figure, the freezing range is 1280 – 1240 = 40ºC. • Note that pure metals solidify at a fixed temperature i.e., the freezing range is zero degrees. INDUSTRIAL MATERIALS Formation of Alloys

27. Phase Diagram • Often we are interested to know the phases present in an alloy at a particular temperature. • If we plan to make a casting, we must be sure that the metal is initially all liquid. • If we plan to heat treat an alloy component, we must be sure that no liquid forms during the process. • Different solid phases have different properties. For example, BCC Fe (indicated as α phase on the iron carbon phase diagram) is magnetic. However, FCC iron (indicated as γ phase on the Fe-C diagram) is not. • The phase diagram can be treated as a road map; if we know the temperature and alloy composition we can determine the phases present. INDUSTRIAL MATERIALS Formation of Alloys

28. Phase Diagram • From the phase diagram for the NiO-MgO binary system [figure below], describe a composition that can melt at 2600ºC but will not melt when placed into service at 2300ºC. Often we are interested to know the phases present in an alloy at a particular temperature. INDUSTRIAL MATERIALS Formation of Alloys

29. Phase Diagram • One method to improve the fracture toughness of a ceramic material is to reinforce the ceramic matrix with ceramic fibers. A materials designer has suggested that Al2O3 could be reinforced with 25% Cr2O3 fibers, which would interfere with the propagation of any cracks in the alumina. The resulting composite is expected to operate under load at 2000ºC for several months. Criticize the appropriateness of this design. INDUSTRIAL MATERIALS Formation of Alloys

30. Phase Diagram Gibbs Rule for Isomorphous Phase Diagram • We keep the pressure fixed (e.g., one atmosphere), which is normal for binary phase diagrams. • The phase rule can be rewritten as: 1 + C = F + P (for constant pressure) • Determine the degrees of freedom in a Cu-40% Ni alloy at (a) 1300ºC, (b) 1250ºC, and (c) 1200ºC. • At 1300ºC, P = 1, C = 2, 1 + C = F + P → F = 2 • At 1250ºC, P = 2, C = 2, 1 + C = F + P → F = 1 • At 1200ºC, P = 1, C = 2, 1 + C = F + P → F = 2 INDUSTRIAL MATERIALS Formation of Alloys

31. Phase Diagram Gibbs Rule for Isomorphous Phase Diagram • As there is only one degree of freedom in a two-phase region of a binary phase diagram, the compositions of the two phases are always fixed when we specify the temperature. Therefore, we can use a tie line to determine the composition of the two phases. • A tie line is a horizontal line within a two-phase region drawn at the temperature of interest (see figure). In an isomorphous system, the tie line connects the liquidus and solidus points at the specified temperature. The ends of the tie line represent the compositions of the two phases in equilibrium. • Tie lines are not used in single-phase regions because we do not have two phases to ‘‘tie’’ in. INDUSTRIAL MATERIALS Formation of Alloys

32. Phase Diagram Use of Tie Line • Determine the composition of each phase in a Cu-40% Ni alloy at 1300ºC, 1270ºC, 1250ºC, and 1200ºC (see fig.). INDUSTRIAL MATERIALS Formation of Alloys

33. Phase Diagram Lever Rule (amount of each phase) • To calculate the amounts of liquid and solid, we construct a lever on our tie line, with the fulcrum of our lever being the original composition of the alloy. The leg of the lever opposite to the composition of the phase, whose amount we are calculating, is divided by the total length of the lever to give the amount of that phase. • The lever rule in general can be written as: Phase percent = opposite arm of lever x 100 total length of tie line INDUSTRIAL MATERIALS Formation of Alloys

34. Phase Diagram Lever Rule (amount of each phase) • Fraction of the solid phase = (Wo – Wl)/(Ws-Wl) INDUSTRIAL MATERIALS Formation of Alloys

35. Phase Diagram Lever Rule (amount of each phase) • Determine the amount of each phase in the Cu-40% Ni alloy at 1300ºC, 1270ºC, 1250ºC, and 1200ºC. at 1300ºC: There is only one phase, so 100% L. at 1270ºC: %L = 50 – 40 x 100 = 77% 50 – 37 %α = 40 – 37 x 100 = 23% 50 – 37 INDUSTRIAL MATERIALS Formation of Alloys

36. Eutectic • A eutectic or eutectic mixture is a mixture of two or more phases at a composition that has the lowest melting point. • It is where the phases simultaneously crystallize from molten solution. • The proper ratios of phases to obtain a eutectic is identified by the eutectic point on a binary phase diagram. • The eutectic point is the point where the liquid phase borders directly on the solid (α + β) phase. INDUSTRIAL MATERIALS Formation of Alloys Eutectic point

37. Eutectic INDUSTRIAL MATERIALS Formation of Alloys • The eutectic point represents the minimum melting temperature of any possible A-B alloy. • The temperature that corresponds to this point is known as the eutectic temperature. • Not all binary system alloys have a eutectic point: those that form a solid solution at all concentrations, such as the Cu-Ni system, have no eutectic. • An alloy system that has a eutectic is often referred to as a eutectic system, or eutectic alloy.

38. Binary – Eutectic Phase Diagram Cu – Ag Ectectic Phase Diagram INDUSTRIAL MATERIALS Formation of Alloys

39. Binary – Eutectic Phase Diagram INDUSTRIAL MATERIALS Formation of Alloys Cooling curve of a eutectic alloy Cu –Ag Eutectic Phase Diagram • Three single-phase regions can be found on the diagram: α, β, and liquid. • The α phase is a solid solution rich in copper; it has silver as the solute component and an FCC crystal structure. • The β phase solid solution also has an FCC structure, but copper is the solute. • No liquid below TE • CE is the composition at TE • The intersection of CE and TE is called the eutectic point. If the alloy has the eutectic composition, it will simultaneously crystallize from molten solution i.e., without having any freezing range.

40. Binary – Eutectic Phase Diagram INDUSTRIAL MATERIALS Formation of Alloys Cu –Ag Eutectic Phase Diagram • The solubility in each of the solid phases is limited. • The solubility limit for the α phase corresponds to the boundary line, CBA, between the α/(α+β) and α/(α+L) phase regions. • It increases with temperature to a maximum [8.0 wt% Ag at 779ºC] at point B, and decreases back to zero at the melting temperature of pure copper, point A [1085ºC]. • At temperatures below 779ºC, the solid solubility limit line separating the α and α+β phase regions is termed as solvus line (CB). • The boundary AB between the α and α+L regions is the solidus line.

41. Binary – Eutectic Phase Diagram INDUSTRIAL MATERIALS Formation of Alloys Cu –Ag Eutectic Phase Diagram • For the β phase, both solvus and solidus lines also exist, HG and GF, respectively. • There are also three two-phase regions found for the copper–silver system: α+L, β+L, and α+β. • The α and β phase solid solutions coexist for all compositions and temperatures within the α+β phase field. • The α+liquid and β+liquid phases also coexist in their respective phase regions. • An important reaction occurs for an alloy of composition CE as it changes temperature in passing through TE; this reaction may be written as follows:

42. Binary – Eutectic Phase Diagram INDUSTRIAL MATERIALS Formation of Alloys Cu –Ag Eutectic Phase Diagram • Upon cooling, a liquid phase is transformed into the two solid α and β phases at the temperature TE; the opposite reaction occurs upon heating. This is called a eutectic reaction; CαE and CβE are the respective compositions of the α and β phases at TE. Thus, for the copper–silver system, the eutectic reaction may be written as follows: • The eutectic reaction, upon cooling, is similar to solidification for pure components, however, the solid product of eutectic solidification is always two solid phases, whereas for a pure component only a single phase forms.

43. Binary – Eutectic Phase Diagram INDUSTRIAL MATERIALS Formation of Alloys Pb –Sn Eutectic Phase Diagram

44. (Ex) Pb-Sn Eutectic System T(°C) 300 L (liquid) a L + a b b L + 200 183°C 18.3 61.9 97.8 C- C0 150 W  = C- C 100 a + b 99 - 40 59 = = = 0.67 99 - 11 88 100 0 11 20 60 80 99 40 C0- C W C C  C0 = C, wt% Sn C - C 40 - 11 29 = = 0.33 = 99 - 11 88 • For a 40 wt% Sn-60 wt% Pb alloy at 150°C, determine: -- the phases present Pb-Sn system Answer: a + b -- the phase compositions INDUSTRIAL MATERIALS Formation of Alloys Answer: Ca = 11 wt% Sn Cb = 99 wt% Sn -- the relative amount of each phase Answer:

45. (Ex contd. ) T(°C) CL - C0 46 - 40 = W = a CL - C 46 - 17 300 L (liquid) 6 a L + = = 0.21 29 220 a b L + 200 b 183°C 100 a + b 100 17 46 0 20 40 60 80 C CL C0 C, wt% Sn C0 - C 23 = WL = = 0.79 CL - C 29 • For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine: -- the phases present: Answer: a + L -- the phase compositions INDUSTRIAL MATERIALS Formation of Alloys Answer: Ca = 17 wt% Sn CL = 46 wt% Sn -- the relative amount of each phase Answer:

46. (Ex contd.) INDUSTRIAL MATERIALS Formation of Alloys • Determine (a) the solubility of tin in solid lead at 100ºC, (b) the maximum solubility of lead in solid tin, (c) the amount of β that forms if a Pb-10% Sn alloy is cooled to 0ºC. • (a) Determine the amount and composition of each phase in a lead-tin alloy of eutectic composition. (b) Calculate the mass of phases present. (c) Calculate the amount of lead and tin in each phase, assuming you have 200 g of the alloy.

47. Hypoeutectic and Hypereutectic Alloys INDUSTRIAL MATERIALS Formation of Alloys • A hypoeutectic alloy is an alloy whose composition will be between that of the left-hand-side end of the tie line defining the eutectic reaction and the eutectic composition. As a hypoeutectic alloy cools, the liquid begins to solidify at the liquidus temperature, producing solid α. However, solidification is completed only after going through the eutectic reaction. • An alloy composition between that of the right-hand-side end of the tie line defining the eutectic reaction and the eutectic composition is known as a hypereutectic alloy.

48. Hypoeutectic and Hypereutectic Alloys INDUSTRIAL MATERIALS Formation of Alloys The cooling curve for a hypoeutectic Pb-30% Sn alloy