REsolução de exercicios, Exercícios de Física

Baixe REsolução de exercicios e outras Exercícios em PDF para Física, somente na Docsity! Instructors' Solution Manual THE SCIENCE AND ENGINEERING Or MATERIALS Fourth Edition 2 The Science and Engineering of Materials Instructor's Solution Manual Tails, with the polymer holding the broken pieces of glass together until the canopy can be replaced. Another approach might be to use a transparent, “glassy” polymer material such as polycarbonate. Some polymers have reasonably good impact properties and may resist failure. The polymers can also be toughened to resist impact by introducing tiny globules of a rubber, or elastomer, into the polymer; these globules improve the energy- absorbing ability of the composite polymer, while being too small to interfere with the optical properties of the material. 1-7 Coiled springs ought to be very strong and stiff. Si,N, is a strong, stiff material. Would you select this material for a spring? Explain. Solution: Springs are intended to resist high elastic forces, where only the atomic bonds are stretched when the force is applied. The silicon nitride would satisfy this requirement. However, we would like to also have good resistance to impact and at least some ductility (in case the spring is overloaded) to assure that the spring will not fail catastrophically. We also would like to be sure that all springs will perform satisfactorily. Ceramic materials such as silicon nitride have virtually no ductility, poor impact properties, and often are difficult to manufacture without introducing at least some small flaws that cause to fail even for relatively low forces. The silicon nitride is NOT recommended. 1-8 Temperature indicators are sometimes produced from a coiled metal strip that uncoils a specific amount when the temperature increases. How does this work; from what kind of material would the indicator be made; and what are the important properties that the material in the indicator must possess? 1-9 Solution: Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change. In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough modulus of elasticity so that no permanent deformation of the material occurs. You would like to design an aircraft that can be flown by human power nonstop for a distance of 30 km. What types of material properties would you recommend”? What materials might be appropriate? Solution: Such an aircraft must possess enough strength and stifíness to resist its own weight, the weight of the human “power source”, and any aerodynamic forces imposed on it. On the other hand, it must be as light as possible to assure that the human can generate enough work to operate the aircraft. Composite materials, particularly those based on a polymer matrix, might comprise the bulk of the aircraft. The polymers have a light weight (with densities of less than half that of aluminum) and can be strengthened by introducing strong, stiff fibers made of glass, carbon, or other polymers. Composites having the strength and stifíness CHAPTER 1 Introduction to Materials Science and Engineering 3 of steel, but with only a fraction of the weight, can be produced in this manner. 1-10 You would like to place a three-foot diameter microsatellite into orbit. The satellite will contain delicate electronic equipment that will send and receive radio signals from earth. Design the outer shell within which the electronic equipment is contained. What properties will be required and what kind of materials might be considered? Solution: The shell of the microsatellite must satisfy several criteria. The material should have a low density, minimizing the satellite weight so that it can be lifted economically into its orbit, the material must be strong, hard, and impact resistant in order to assure that any “space dust” that might strike the satellite does not penetrate and damage the electronic equipment; the material must be transparent to the radio signals that provide communication between the satellite and earth; and the material must provide some thermal insulation to assure that solar heating does not damage the electronics. One approach might be to use a composite shell of several materials. The outside surface might be a very thin reflective metal coating that would help reflect solar heat. The main body of the shell might be a light weight fiber-reinforced composite that would provide impact resistance (preventing penetration by dust particles) but would be transparent to radio signals. 1-11 What properties should the head of a carpenter"s hammer possess? How would you manufacture a hammer head? 1-12 Solution: The head for a carpenter's hammer is produced by forging, a metal- working process; a simple steel shape is heated and formed in several steps while hot into the required shape. The head is then heat treated to produce the required mechanical and physical properties. The striking face and claws of the hammer should be hard —the metal should not dent or deform when driving or removing nails. Yet these portions must also possess some impact resistance, particularly so that chips do not flake off the striking face and cause injuries. The hull of the space shuttle consists of ceramic tiles bonded to an aluminum skin. Discuss the design requirements of the shuttle hull that led to the use of this combi- nation of materials. What problems in producing the hull might the designers and manufacturers have faced? Solution: The space shuttle experiences extreme temperatures during re-entry into earth's atmosphere; consequentIy a thermal protection system must be used to prevent damage to the structure of the shuttle (not to mention its contents!). The skin must therefore be composed of a material that has an exceptionally low thermal conductivity. The material must be capable of being firmly attached to the skin of the shuttle and to be easily repaired when damage occurs. The tiles used on the space shuttle are composed of silica fibers bonded together to produce a very low density ceramic. The thermal conductivity is so low that a person can hold on to one side of the tile while the opposite surface is red hot. The tiles are attached to the shuttle 4 The Science and Engineering of Materials Instructor's Solution Manual skin using a rubbery polymer that helps assure that the forces do not break the tile loose, which would then expose the underlying skin to high temperatures. 1-13 You would like to select a material for the electrical contacts in an electrical switch- ing device which opens and closes frequently and forcefully. What properties should the contact material possess? What type of material might you recommend? Would ALO, be a good choice? Explain. Solution: The material must have a high electrical conductivity to assure that no electrical heating or arcing occurs when the switch is closed. High purity (and therefore very soft) metals such as copper, aluminum, silver or gold provide the high conductivity. However the device must also have good wear resistance, requiring that the material be hard. Most hard, wear resistant materials have poor electrical conductivity. One solution to this problem is to produce a particulate composite material composed of hard ceramic particles embedded in a continuous matrix of the electrical conductor. For example, silicon carbide particles could be introduced into pure aluminum; the silicon carbide particles provide wear resistance while aluminum provides conductivity. Other examples of these materials are described in Chapter 16. ALO, by itself would not be a good choice-—alumina is a ceramic material and is an electrical insulator. However alumina particles dispersed into a copper matrix might provide wear resistance to the composite. 1-14 Aluminum has a density of 2.7 g/cm?. Suppose you would like to produce a com- 1-15 posite material based on aluminum having a density of 1.5 g/cm?. Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.95 g/cm?, into the aluminum be a likely possibility? Explain. Solution: In order to produce an aluminum-matrix composite material with a density of 1.5 g/em?, we would need to select a material having a density considerably less than 1.5 g/cm?. While polyethylene”s density would make it a possibility, the polyethylene has a very low melting point compared to aluminum; this would make it very difficult to introduce the polyethylene into a solid aluminum matrix —processes such as casting or powder metallurgy would destroy the polyethylene. Therefore polyethylene would NOT be a likely possibility. One approach, however, might be to introduce hollow glass beads. Although ceramic glasses have densities comparable to that of aluminum, a hollow bead will have a very low density. The glass also has a high melting temperature and could be introduced into liquid aluminum for processing as a casting. You would like to be able to identify different materials without resorting to chemical analysis or lengthy testing procedures. Describe some possible testing and sorting techniques you might be able to use based on the physical properties of materials. Solution: Some typical methods might include: measuring the density of the material (may help in separating metal groups such as aluminum, copper, steel, magnesium, etc.), determining the electrical conductivity Atomic Structure 2-6(a) 2-6(b) 2a) 2-7(b) Aluminum foil used for storing food weighs about 0.3 g per square inch. How many atoms of aluminum are contained in this sample of foil? Solution: In a one square inch sample: (0.3 86.02 x 1023 atoms/mol) number = =6.69 x 10?! atoms 26.981 g/mol Using the densities and atomic weights given in Appendix A, calculate and com- pare the number of atoms per cubic centimeter in (a) lead and (b) lithium. Solution: (a) In lead: (11.36 g/em*)(1 cm?)(6.02 x 1023 atoms/mol) 207.19 g/mol =3.3x 1022 atoms/cm? (b) In lithium: (0.534 g/em?)(1 cm?)(6.02 x 10? atoms/mol) 6.94 g/mol = 4.63 x 102? atoms/cm? Using data in Appendix A, calculate the number of iron atoms in one ton (2000 pounds). Solution: (2000 Ib)(454 g/1b)(6.02 x 1023 atoms/mol) 55.847 g/mol =9.79 x 107 atoms/ton Using data in Appendix A, calculate the volume in cubic centimeters occupied by one mole of boron. Solution: (1 mol(10.81 g/mol) — 3 23 gem *7em 8 The Science and Engineering of Materials Instructor's Solution Manual 2-8 In order to plate a steel part having a surface area of 200 in.? with a 0.002 in. thick layer of nickel, (a) how many atoms of nickel are required and (b) how many moles of nickel are required? Solution: Volume = (200 in.2)(0.002 in.)(2.54 cm/in.)? = 6.555 cm? 6.555 cm3)(8.902 g/em?)(6.02 x 10? atoms/mol (a) (6.555 cm')(8.902 g/em” (6.02 x 10 atomsímol) (5. nos oms 58.71 g/mol b) (6.555 cm?)(8.902 gem? (b) (6.555 cm(8.002 gfem) 6 994 mo Ni required 58.71 g/mol + , 0.002 in T CO 2-9 Suppose an element has a valence of 2 and an atomic number of 27. Based only on the quantum numbers, how many electrons must be present in the 3d energy level? Solution: We can let x be the number of electrons in the 3d energy level. Then: 152 2522pº3523pº3d"4s? (must be 2 electrons in 4s for valence = 2) Since 27-(24246+246+2) = 7 =x there must be 7 electrons in the 3d level. 2-10 Indium, which has an atomic number of 49, contains no electrons in its 4f energy level. Based only on this information, what must be the valence of indium? Solution: We can let x be the number of electrons in the outer sp energy level. Then: 1522522p3523p'3d 04524 po4d AF OS(sp)" 49-(24246+2+6+10+2+6+10+0) = 3 Therefore the outer 5sp level must be: 5s25p! or valence = 3 2-1 Without consulting Appendix C, describe the quantum numbers for each of the 18 electrons in the M shell of copper, using a format similar to that in Figure 2-9. Solution: For the M shell: n=3;(=0,1,2:m,=2/+1 2-12 214 215 CHAPTER2 Atomic Structure 9 3d!O Electrical charge is transferred in metals by movement of valence electrons. How many potential charge carriers are there in an aluminum wire 1 mm in diameter and 100 m in length? Solution: Aluminum has 3 valence electrons per atom; the volume of the wire is: Volume = (1/4)d?! = (7/4)0.1 cm)?(10,000 cm) = 78.54 cm? (78.54 em?)(2.699 g/cm?)(6.02 x 102? atoms/mol)(3 electrons/atom) n= 26.981 g/mol n=1.42x 10% carriers Bonding in the intermetallic compound Ni;Al is predominantly metallic. Explain why there will be little, if any, ionic bonding component. The electronegativity of nickel is about 1.8. Solution: The electronegativity of Alis 1.5, while that of Ni is 1.8. These values are relatively close, so we wouldn't expect much ionic bonding. Also, both are metals and prefer to give up their electrons rather than share or donate them. Plot the melting temperatures of elements in the 4A to 8-10 columns of the periodic table versus atomic number (i.e., plot melting temperatures of Ti through Ni, Zr through Pd, and Hf through Pb. Discuss these relationships, based on atomic bonding and binding energy, (a) as the atomic number increases in each row of the periodic table and (b) as the atomic number increases in each column of the periodic table. Solution: Ti -1668 Zr 1852 Hf-2227 V —1900 Nb-2468 Ta-2996 Cr -1875 Mo-2610 W-3410 Mn-1244 Te -2200 Re-3180 Fe -1538 Ru -2310 0s-2700 Co -1495 Rh-1963 Tr -2447 Ni -1453 Pd -1552 Pt -1769 12 | The Science and Engineering of Materials Instructor's Solution Manual 2-W Calculate the fraction of bonding of MgoO that is ionic. Solution: Eme= 1.2 Eç=3.5 Loovaten = EXPI(-0.25)(3.5 — 1.292] = exp(-1.3225) = 0.266 Sonic = 1 — 0.266 = 0.734 .:. bonding is mostly ionic 2-29 Beryllium and magnesium, both in the 2A column of the periodic table, are light- weight metals. Which would you expect to have the higher modulus of elasticity? Explain, considering binding energy and atom radii and using appropriate sketches of force versus interatomic spacing. Solution: 4Be 152252 E=42x10psi rg =LI43 À 12Mg 15252p632 E=6xX106 psi ryç= 604 À Force The smaller Be electrons are held closer to the core .:. held more tightly, giving a higher binding energy. 2-30 Boron has a much lower coefficient of thermal expansion than aluminum, even though both are in the 3B column of the periodic table. Explain, based on binding energy, atomic size, and the energy well, why this difference is expected. Solution: 5B 1522522p! rg =0.46 À 13 Al 252p33p! ry=1432Á B AI 3 distance “a” 2 di =m-sa a Aa a Electrons in Al are not as tightly bonded as those in B due to the smaller size of the boron atom and the lower binding energy associated with its size. 2-3 232 2-33 2-34 2-35 CHAPTER 2 Atomic Structure 13 Would you expect MgO or magnesium to have the higher modulus of elasticity? Explain. Solution: Mgo has ionic bonds, which are strong compared to the metallic bonds in Mg. A higher force will be required to cause the same separation between the ions in MgO compared to the atoms in Mg. Therefore, MgO should have the higher modulus of elasticity. In Mg, E = 6x 10º psi; in Mgo, E = 30 x 10º psi. Would you expect AL,O, or aluminum to have the higher coefficient of thermal expansion? Explain. Solution: ALO, has stronger bonds than Al; therefore, ALO, should have a lower thermal expansion coefficient than Al. In Al, a = 25 x 10% cm/emºC; in ALO,, a =6.7x 10% em/emºc. Aluminum and silicon are side by side in the periodic table. Which would you expect to have the higher modulus of elasticity (E)? Explain. Solution: Silicon has covalent bonds; aluminum has metallic bonds. Therefore, Si should have a higher modulus of elasticity. Explain why the modulus of elasticity of simple thermoplastic polymers, such as polyethylene and polystyrene, is expected to be very low compared with that of metals and ceramics. Solution: The chains in polymers are held to other chains by Van der Waals bonds, which are much weaker than metallic, ionic, and covalent bonds. For this reason, much less force is required to shear these weak bonds and to unkink and straighten the chains. Steel is coated with a thin layer of ceramic to help protect against corrosion. What do you expect to happen to the coating when the temperature of the steel is increased significantly? Explain. Solution: Ceramics are expected to have a low coefficient of thermal expansion due to strong ionic/covalent bonds; steel has a high thermal expansion coefficient. When the structure heats, steel expands more than the coat- ing, which may crack and expose the underlying steel to corrosion. CHAPTER 3 Atomic and lonic Arrangements 17 3-32 Bismuth has a hexagonal structure, with a, = 0.4546 nm and c, = 1.186 nm. The density is 9.808 g/cm” and the atomic weight is 208.98 g/mol. Determine (a) the volume of the unit cell and (b) how many atoms are in each unit cell. Solution: (a) The volume of the unit cellis V= a Ze c0s30. V = (0.4546 nm)*(1.186 nm)(cos30) = 0.21226 nm? =2.1226x 102 cm? (b) If “x” is the number of atoms per unit cell, then: (x atoms/cell)(208.98 g/mol) 9.808 gem'= — >> E 8 (2.1226 x 102 em?)(6.02 x 103 atoms/mol) x=6 atoms/cell 3-33 Gallium has an orthorhombic structure, with a, = 0.45258 nm, b, = 0.45186 nm, and c, = 0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 gem? and the atomic weight is 69.72 g/mol. Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell. Solution: The volume of the unit cellis V= a b,c, or V= (0.45258 nm)(0.45 186 nm)(0.76570 nm) = 0.1566 nm? = 1.566x 1022? cm? (a) From the density equation: (x atoms/cell)(69.72 g/mol) 5.904 glem' = ————>>—> > E 8 (1.566 x 1022 cm?)(6.02 x 1023 atoms/mol) x= 8 atoms/cell (b) From the packing factor (PF) equation: (8 atoms/cell)(47/3)(0.1218 nm)? PF 0.1566 nm? = 0.387 3-34 Beryllium has a hexagonal crystal structure, with a, = 0.22858 nm and c, =0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 gem, and the atomic weight is 9.01 g/mol. Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell. Solution: V= (0.22858 nm)*(0.35842 nmjcos 30 = 0.01622 nm? = 16.22 x 104 cm” (a) From the density equation: (x atoms/cell)(9.01 g/mol) 1.848gkm'= 1 E 8 (16.22 x 102º cm?)(6.02 x 1023 atoms/mol) x =2 atoms/cell (b) The packing factor (PF) is: pr= (2 atoms/cell)(47/3)(0.1143 nm)? — 0.77 0.01622 nm? 18 | The Science and Engineering of Materials Instructor's Solution Manual 3-39 Above 882ºC, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion? Solution: We can find the volume of each unit cell. Two atoms are present in both BCC and HCP titanium unit cells, so the volumes of the unit cells can be directly compared. Vacc = (0.332 nm)? = 0.03659 nm? Vuc = (0.2978 nm)*(0.4735 nm)cos30 = 0.03637 nm? Vncr — Vece — 0.03637 nm? — 0.03659 nm x 100 x 100=-0.6% Voce 0.03659 nm? AV= Therefore titanium contracts 0.6% during cooling. 3-40 Mn has a cubic structure with a, = 0.8931 nm and a density of 7.47 glem?. B-Mn has a different cubic structure, with a, = 0.6326 nm and a density of 7.26 gem. The atomic weight of manganese is 54.938 g/mol and the atomic radius is 0.112 nm. Determine the percent volume change that would occur if o-Mn transforms to B-Mn. Solution: First we need to find the number of atoms in each unit cell so we can determine the volume change based on equal numbers of atoms. From the density equation, we find for the o-Mn: 747 ejom? - (x atoms/cell)(54.938 g/mol) (8.931 x 108 cm)3(6.02 x 1023 atoms/mol) x=58 atoms/cell V. = (8.931 x 108 cm) = 7.12 x 102 cm? o For B-Mn: 7.26 g/emê (x atoms/cell)(54.938 g/mol) .26 gem?=——— E (6.326 x 108 cm)3(6.02 x 1023 atoms/mol) x=20 atoms/cell V. mn = (6.326 x 108 cm)? = 2.53 x 1022 cm? The volume of the B-Mn can be adjusted by a factor of 58/20, to account for the different number of atoms per cell. The volume change is then: (58/20)V sam — Va S80)(2.53) = 7.12 av CEE era “oca (o 68202.53) — 7.12 x 100=+3.05% Von 712 The manganese expands by 3.05% during the transformation. 3-35 Atypical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the num- ber of unit cells and (b) the number of iron atoms in the paper clip. (See Appendix A for required data) Solution: The lattice parameter for BCC iron is 2.866 x 108 cm. Therefore V, = (2.866 x 108 em) = 2.354 x 1023 cm? unit cell (a) The density is 7.87 glem?. The number of unit cells is: 059 g > >> E — = 3.185 x 10?! cells (7.87 g/em?)(2.354 x 10? cm?fcell) number = CHAPTER 3 Atomic and lonic Arrangements 19 (b) There are 2 atoms/cell in BCC iron. The number of atoms is: number = (3.185 x 10?! cells)(2 atoms/cell) = 6.37 x 102! atoms 3-36 Aluminum foil used to package food is approximately 0.001 inch thick. Assume that 3-51 3-52 3-53 all of the unit cells of the aluminum are arranged so that a, is perpendicular to the foil surface. For a 4 in. x 4 in. square of the foil, determine (a) the total number of unit cells in the foil and (b) the thickness of the foil in number of unit cells. (See Appendix A) Solution: The lattice parameter for aluminum is 4.04958 x 1078 cm. Therefore: V, unit cell = (4.04958 x 108)? = 6.6409 x 10? cm? The volume of the foil is: Vioy = (4in (4 in.)(0.001 in.) = 0.016 in3 = 0.262 em? (a) The number of unit cells in the foil is: 0.262 emê number="——— OM 3945x 10% cells 6.6409 x 102 cm?Ycell (b) The thickness of the foil, in number of unit cells, is: — (0.001 in.)(2.54 cmi.) number = =6.27x 10º cells 4.04958 x 10-8 cm Determine the Miller indices for the directions in the cubic unit cell shown in Figure 3-48. Solution: A:0,1,0-0,1,1=0,0-1 = [007] B:4,00-0,10=/4-1,0 =[120] C:0,1,1-1,0,0=-1,1,1 =[TI] D: 1,0,4- 0,6,1=1,-4—% = [277] Determine the indices for the directions in the cubic unit cell shown in Figure 3-49. Solution: A:0,0,1-1,00=-1,0,1 =[T01] B:10,1-410=h-11 =[122] C:100-041=1,24-1 = [437] D:0,1,/4-0,00=0,1,4 =[021] Determine the indices for the planes in the cubic unit cell shown in Figure 3-50. Solution: A x=1 lx = 1 y 1y = AT) z=1 lz =1 B: x lx =0 y 1y =3 (030) z 1 =0 22 The Science and Engineering of Materials Instructor's Solution Manual PN 213 2 h Ee ' : 113) 4 1 3 i | k ' | z 2 3-60 Sketch the following planes and directions within a cubic unit cell. Solution: (a) [TO] (b) [221] (e) [410] (a) [072] (e) [321] (DO INTI] QD MOD (030) () 2 (13) (041) 7 12 a INS c d y X a f h 9 112 114 Atomic and lonic Arrangements 23 CHAPTER 3 3-61 Sketch the following planes and directions within a hexagonal unit cell. (d) (0003) (e) (1010) (f) (O1TI) Solution: (a) [OITO] (b) [1120] (e) [To1] c c (001) c (io) c (om) Dl DD íC IL IDO O g- a RN Í TT Ê HR / TS % . & Cf & en & A oria a Sl a a a 3-62 Sketch the following planes and directions within a hexagonal unit cell. Solution: (a) [2110] (b) [1121] (e) [1OTO] (d) (1210) (e) (TT22) (1) (1230) nm a E c (ão) c (iz c (tão) ; ão j ão - % A o, a 8 a a é (2110) “foto 3-63 What are the indices of the six directions of the form <110> that lie in the (11T) [011] plane of a cubic cell? [oTT] Solution: [Tio] [101] [TO] [Tot] z 3-64 What are the indices of the four directions of the form <111> that lie in the (TO1) plane of a cubic cell? [1] TT) Solution: TI) (mt) z 24 - The Science and Engineering of Materials Instructor's Solution Manual 3-65 3-66 3-67 Determine the number of directions of the form <110> in a tetragonal unit cell and compare to the number of directions of the form <110> in an orthorhombic unit cell. Solution: Tetragonal: [110], [TTO], [T10), [1TO] = 4 Orthorhombic: [110], [TIO] =2 Note that in cubic systems, there are 12 directions of the form <110>. Determine the angle between the [110] direction and the (110) plane in a tetragonal unit cell; then determine the angle between the [011] direction and the (011) plane in a tetragonal cell. The lattice parameters are a, = 4 À and c=s À. What is responsible for the difference? Solution: [110] + (110) 25 N tan(9/2)=2.5/2=1.25 92=51.34º 0= 102.68º The lattice parameters in the x and y directions are the same; this allows the angle between [110] and (110) to be 90º. But the lattice parameters in the y and z direc- tions are different! Determine the Miller indices of the plane that passes through three points having the following coordinates. Solution: (a) 0,0,1; 1,0,0; and !4,/5,0 (b) 14,0,1: 44,0,0: and 0,1,0 (c) 1,0,0; 0,1,/4; and 1,44 (d) 1,0,0; 0,0,/4; and 14,1,0 (a) (111) (b) Q10) (c) (0T2) (d) (218) CHAPTER 3 Atomic and lonic Arrangements 27 For (10): 2 points (3.5167 x 10-8 em) (2) (3.5167 x 108 em) =0.1144 x 10719 points/cm? planar density = 2 packing fraction = —2m 0.555 (ar DX COS — Va <a> For (11): From the sketch, we can determine that the area of the (111) plane is (N2a9/2)(N3a,/n/2) = 0.86602. There are (3)(15) + (3)(4%6) = 2 atoms in this area. 2 points 0.866(3.5167 x 10-8 cm)? = 0.1867 x 10!º points/em? 27( NBast4)? packing fraction= —————+— = 0.907 086642 The (111) is close packed. planar density = 3-72 Determine the planar density and packing fraction for BCC lithium in the (100), (110), and (111) planes. Which, if any, of these planes is close-packed? Solution: a, =3.5089 À For (100): 1 planar density = -——>>—" 3 = 0.0812 x 1016 points/cm? (3.5089 x 108 cm)? 2 . co MNBaoid] packing fraction = += 0.589 a? o O OO + p—— 28 The Science and Engineering of Materials Instructor's Solution Manual 3-73 3-74 For (JO): itv= 2 - 16 noi 2 planar density= — 2 =0.1149x 10!º points/cm 2 «/2(3.5089 x 108 em) 27[N3açi4 packing fraction = 2m[N3aota) 0.833 N2a2 (5 O ! = v2a, > For (11): There are only (3)(4%) = !4 points in the plane, which has an area of 0.866a7. hj planar density = — = 0.0469 x 10!º points/cm? 0.866(3.5089 x 108 cm 2 4 m)N3a/4] vo mN3ata] as 0.86642 There is no close-packed plane in BCC structures. packing fraction = A=0.866 a8 Suppose that FCC rhodium is produced as a 1 mm thick sheet, with the (111) plane parallel to the surface of the sheet. How many (111) interplanar spacings d,,, thick is the sheet? See Appendix A for necessary data. a, IT Á VP+P+ro 8 (Ii mm/10 mm/em) thickness =———— = 4.563 x 10º d, 2.1916x 108 cm Solution: dn = =2I9IGà 111 Spacings In a FCC unit cell, how many d,,, are present between the 0,0,0 point and the 1,1,1 point? Solution: The distance between the 0,0,0 and 1,1,1 points is 3a. The interplanar spacing is dn =aNP+P+O =aN3 Therefore the number of interplanar spacings is number of d,,, spacings = Naa JA) =3 CHAPTER 3 Atomic and lonic Arrangements 29 Point 11,1 Point 0,0, 3-79 Determine the minimum radius of an atom that will just fit into (a) the tetrahedral interstitial site in FCC nickel and (b) the octahedral interstitial site in BCC lithium. Solution: (a) For the tetrahedral site in FCC nickel (a, = 3.5167 Ã): 2(3.5167 À) n=——— = 1,243 À N 4 rlrw; = 0.225 for a tetrahedral site. Therefore: r= (1.243 ÀY(0.225) = 0.2797 À (b) For the octahedral site in BCC lithium (a, = 3.5089 Â): /3(3.5089) n=— =1519Á “ 4 rir, ; = 0.414 for an octrahedral site. Therefore: r= (1.519 ÀYX0.414) = 0.629 À 3-86 What is the radius of an atom that will just fit into the octahedral site in FCC copper without disturbing the lattice? Solution: re,=1278à rlrey = 0.414 for an octahedral site. Therefore: r= (1.278 ÀY(0.414) = 0.529 À 3-87 Using the ionic radii given in Appendix B, determine the coordination number expected for the following compounds. Solution: (a) Y,0, (b) UO, (c) Bao (d) Si N, (Geo, (DMO (MS (WKBr 0.89 0.53 On Uot= 1597067 CND (0) raNro2= 55040 0.97 O nftio?= 159703 CN=6 (OD rutlio 132 O Ara 547099 ENS8 (O) ni (O nt 0097036 CENS4 (na! = 56 = 068 CN=6 32 The Science and Engineering of Materials Instructor's Solution Manual 3-93 E = (RM | -— v2a, —— MgoO, which has the sodium chloride structure, has a lattice parameter of 0.396 nm. Determine the planar density and the planar packing fraction for the (111) and (222) planes of MgO. What ions are present on each plane? Solution: As described in the answer to Problem 3-71, the area of the (111) plane is 0.866 2. a,= 214524219? = 2(0.66 + 1.32) = 3.96 À 2Mg (0.866)3.96 x 108 cm)? — mO6 OB6OG9P (111): PD.= = 0.1473 x 101º points/cm? (222): PD. = 0.1473 x 10!$ points/cm? MAP “OB6OG9P (222) (11) IN, 3-100 Polypropylene forms an orthorhombic unit cell with Iattice parameters of a, = 3-101 1.450 nm, b, = 0.569 nm, and c, = 0.740 nm. The chemical formula for the propy- lene molecule, from which the polymer is produced, is C,H,. The density of the polymer is about 0.90 g/cm. Determine the number of propylene molecules, the number of carbon atoms, and the number of hydrogen atoms in each unit cell. Solution: MWpp=3 C+6H=3(12) + 6=42 g/mol 090 glemê = CHA? glmol) . (14.5 cm)(5.69 em)(7.40 em)(10-24)(6.02 x 1023 molecules/mol) x=8C,H, molecules or 24 C atoms and 48 H atoms The density of cristobalite is about 1.538 g/cm?, and it has a lattice parameter of 0.8037 nm. Calculate the number of SiO, ions, the number of silicon ions, and the number of oxygen ions in each unit cell. (x SiO,)[28.08 + 2(16) g/mol] Solution: 1.538gkm'= ED — E —— 8.037 x 108 cm)3(6.02 x 1023 ions/mol) x=8 SiO, or 8 Si'* ions and 16 02 ions 3-105 3-106 3-107 CHAPTER 3 Atomic and lonic Arrangements 33 A diffracted x-ray beam is observed from the (220) planes of iron at a 20 angle of 99.1º when x-rays of 0.15418 nm wavelength are used. Calculate the lattice parame- ter of the iron. Solution: sin 0=/2d, 2 2 2 sin(99.1/2) = 0-15418V2º + 2º +07 2a, 01541848 à => —— = 0.2865 nm 2sin(49.55) A diffracted x-ray beam is observed from the (311) planes of aluminum at a 20 angle of 78.3º when x-rays of 0.15418 nm wavelength are used. Calculate the lattice parameter of the aluminum. Solution: sin 0=A/d,, — OIS4ISVP AL +I? = = 0.40497 nm 2sin(78.32) lo Figure 3-56 shows the results of an x-ray diffraction experiment in the form of the intensity of the diffracted peak versus the 20 diffraction angle. If x-rays with a wavelength of 0.15418 nm are used, determine (a) the crystal structure of the metal, (b) the indices of the planes that produce each of the peaks, and (c) the lattice parameter of the metal. Solution: The 20 values can be estimated from Figure 3-56: Planar 20 sin?) sin?0/00077 indices d=/2sin0 a=dÃR+RA+P 1 17.5 0.023 3 (1) 0.5068 0.8778 2 205 0.032 4 (200) 0.4332 0.8664 3 285 0.061 8 (220) 0.3132 0.8859 4 335 0.083 q (311) 0.2675 0.8872 5 355 0.093 12 (222) 0.2529 0.8761 6 4 0.123 16 (400) 0.2201 0.8804 7 4 0.146 19 (331) 0.2014 0.8779 8 465 0.156 20 (420) 0.1953 0.8734 The sin?9 values must be divided by 0.077 (one third the first sin29 value) in order to produce a possible sequence of numbers) (a) The 3,4,8,11, .. . sequence means that the material is FCC (c) The average a, = 0.8781 nm 34 The Science and Engineering of Materials Instructor's Solution Manual 3-108 Figure 3-57 shows the results of an x-ray diffraction experiment in the form of the intensity of the diffracted peak versus the 20 diffraction angle. If x-rays with a wavelength of 0.0717 nm are used, determine (a) the crystal structure of the metal, (b) the indices of the planes that produce each of the peaks, and (c) the lattice parameter of the metal. Solution: The 20 values can be estimated from the figure: Planar 20 sin?) sin20/0047 | indices d=1/2sin6 a, = daN rRAP 1255 0047 1 (III) 01610 0.2277 2 36 0095 2 (200) 0.1150 0.2300 3 445 0143 3 Qi) 00938 0.2299 4515 0189 4 (220) 00818 0.2313 5 58 0235 5 (10) 00733 0.2318 6 645 0285 6 (22) 00666 0.2307 7/7 0329 7 (321) 006195 0.2318 8 755 0375 8 (400) 0.0580 0.2322 (a) The sequence 1,2,3,4,5,6,7,8 (which includes the “7”) means that the material is BCC. (c) The average a, = 0.2307 nm CHAPTER 4 Imperfections in the Atomic and lonic Arrangements 37 There are 2 atoms per cell in BCC metals. Thus: S= 0.69/2 = 0.345 4-8 Tin atoms are introduced into a FCC copper crystal, producing an alloy with a lat- tice parameter of 3.7589 x 10-8 cm and a density of 8.772 g/cm?. Calculate the atomic percentage of tin present in the alloy. Solution: 877 glem? = (isa (118.69 g/mol) + (4 — xs (63.54 g/mol) — (3.7589x 108 cm)*(6.02 x 1023 atoms/mol) 280.5 =55.15xç, + 254.16 or xs = 0.478 Sn atoms/cell There are 4 atoms per cell in FCC metals; therefore the at% Sn is: (0.478/4) = 11.95% 4-9 We replace 7.5 atomic percent of the chromium atoms in its BCC crystal with tanta- lum. X-ray diffraction shows that the lattice parameter is 0.29158 nm. Calculate the density of the alloy. Solution: — (2)0.925)(51.996 g/mol) + 2(0.075)(180.95 g/mol) =8.265 glem? (2.9158 x 10-8 cm)3(6.02 x 1023 atoms/mol) a 4-10 Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find (a) the density and (b) the packing factor. Solution: There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 € per unit cell: 6) 5 — ISSA gfmol) + (15X plo) — = =7.89 glem? (2.867 x 108 cm)?(6.02 x 1023 atoms/mol) 8 3 3 (b) Packing Factor = A47/3)(1.241)? + (1/50)(470/3 0.77) =0.681 (2.867) 4-11 The density of BCC iron is 7.882 g/em? and the lattice parameter is 0.2866 nm when hydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms and (b) the number of unit cells required on aver- age to contain one hydrogen atom. Solution: 2(55.847 1) + x(1.00797 L (a) 7.882 glemê = — 2605-847 e/mol) + x(1.00797 emo (2.866 x 108 cm)3(6.02 x 1023 atoms/mol) x=0.0081 H atoms/cell The total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus: 0.0081 hs = 0.004 2.0081 (b) Since there is 0.0081 H/cell, then the number of cells containing H atoms is: cells = 1/0.0081 = 123.5 or 1 Hin 123.5 cells 38 The Science and Engineering of Materials Instructor's Solution Manual 412 413 414 422 Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate (a) the number of anion vacancies per em? and (b) the density of the ceramic. Solution: In 10 unit cells, we expect 40 Mg + 40 O ions, but due to the defect: 40Mg-1=39 400 —-1=39 (a) 1 vacancy/(10 cells)(3.96 x 108 cm)? = 1.61 x 10?! vacancies/em? (b) — (39/40)(4)(24.312 g/mol) + (39/404)(16 g/mol) = 4.205 gem”? (3.96 x 10-8 cm)?(6.02 x 1023 atoms/mol) x ZnS has the zinc blende structure. If the density is 3.02 g/cm? and the lattice param- eter is 0.59583 nm, determine the number of Schottky defects (a) per unit cell and (b) per cubic centimeter. Solution: Let x be the number of each type of ion in the unit cell. There normally are 4 of each type. x(65.38 g/mol) + x(32.064 g/mol) (5.9583 x 108 em)*(6.02 x 10? ions/mol) 4 3.9465 = 0.0535 defects/u.c. (a) 3.02 gem? = x=3.9465 (b) * of unit cells/em? = 1/(5.9683 x 108 cm)? = 4.704 x 10?! Schottky defects per cm? = (4.704 x 102!)(0.0535) = 2.517 x 1020 Suppose we introduce the following point defects. What other changes in each structure might be necessary to maintain a charge balance? Explain. (a) Mg? ions substitute for yttrium atoms in Y,0, (b) Fe3+ ions substitute for magnesium ions in MgO (e) Li!* ions substitute for magnesium ions in MgO (d) Fe?+ ions replace sodium ions in NaCl Solution: (a) Remove 2 Y** and add 3 Mg?* — create cation interstitial. (b) Remove 3 Mg* and add 2 Fe? — create cation vacancy. (c) Remove 1 Mg?* and add 2 Li* — create cation interstitial. (d) Remove 2 Na* and add 1 Fe?+ — create cation vacancy. What are the Miller indices of the slip directions (a) on the (111) plane in an FCC unit cell (b) on the (011) plane in a BCC unit cell? Solution: [0T1), [01T] nto, TT) [Tio], [ITO] (TI), DT] [Toi], [107] z z CHAPTER 4 Imperfections in the Atomic and lonic Arrangements 39 4-23 What are the Miller indices of the slip planes in FCC unit cells that include the [101] slip direction? Solution: (UT, TI) (TD, (TT) z 4-24 What are the Miller indices of the (110) slip planes in BCC unit cells that include the [111] slip direction? Solution: (ITO), (TIO) (OTI), (01T) (107), (TO1) 4-25 Calculate the length of the Burgers vector in the following materials: (a) BCC niobium (b) FCC silver (c) diamond cubic silicon Solution: (a) The repeat distance, or Burgers vector, is half the body diagonal, or: b= repeat distance = (14) (3) (3.294 À) = 2.853 À (b) The repeat distance, or Burgers vector, is half of the face diagonal, or: b=(14) (2a) = (6) (N2) (4.0862 À) = 2.889 À (c) The slip direction is [110], where the repeat distance is half of the face diagonal: b=(4) (N2) (5.4307 À) = 3.840 À 4-26 Determine the interplanar spacing and the length of the Burgers vector for slip on the expected slip systems in FCC aluminum. Repeat, assuming that the slip system is a (110) plane and a [1T1] direction. What is the ratio between the shear stresses required for slip for the two systems? Assume that k = 2 in Equation 4-2. Solution: (a) For (111)/[110], b=(1) (2) (4.04958 À) = 2.863 À dy = AOWSSA o 338 VI+1+1 (b) If (11O[111], then: 4.04958à b= 3 (404958 À)=7014Á dm= =2.863Á VP+2+0? 42 The Science ai nd Engineering of Materials Instructor's Solution Manual Stress 4-45 Our discussion of Schmid's law dealt with single crystals of a metal. Discuss slip 4-49 4-50 and Schmid's law in a polycrystalline material. What might happen as the grain size gets smaller and smaller? Solution: With smaller grains, the movement of the dislocations is impeded by frequent intersections with the grain boundaries. The strength of metals is not nearly as low as might be predicted from the critical resolved shear stress as a consequence of these interactions. The strength of titanium is found to be 65,000 psi when the grain size is 17 x 109 m and 82,000 psi when the grain size is 0.8 x 104m. Determine (a) the con- stants in the Hall-Petch equation and (b) the strength of the titanium when the grain size is reduced to 0.2 x 10º m. Solution: 65,000 = o, + K =6,+2425K 1 N17x 10% 1 0.8 x 104 82,000 = 0, +K = 0,+11180K (a) By solving the two simultaneous equations: K=19.4psi/Ym o, = 60,290 psi (b) 6 = 60,290 + 19.4 / 40.2 x 10º = 103,670 psi A copper-zinc alloy has the following properties: grain diameter (mm) strength (MPa) dr 0.015 170 MPa 8.165 0.025 158 MPa 6.325 0.035 151 MPa 5.345 0.050 145 MPa 4472 Determine (a) the constants in the Hall-Petch equation and (b) the grain size required to obtain a strength of 200 MPa. 451 4.52 4.53 CHAPTER 4 Imperfections in the Atomic and lonic Arrangements 43 Solution: The values of d-“ are included in the table; the graph shows the relation- ship. We can determine K and 6, either from the graph or by using two of the data points. (a) 170= 6, + K(8.165) 145=0, + K(4.472) 25=3.693K K=677MPa/Ymm o, = 114.7 MPa (b) To obtain a strength of 200 MPa: 200 = 114.7+ 6.77 1d 85.3 = 6771 d = 0.0063 mm 180 160 Strenght (MPa) 140 d112 For an ASTM grain size number of 8, calculate the number of grains per square inch (a) ata magnification of 100 and (b) with no magnification. Solution: ()N=2"! N=28!=27= 128 grains/in.? (b) No magnification means that the magnification is “1”: (27)(100/12 = 1.28 x 10º grains/in.? Determine the ASTM grain size number if 20 grains/square inch are observed at a magnification of 400. Solution: (20)(400/100)2 = 2"! 10g(320) = (n-1)log(2) 2.505 = (n-1)(0.301) or n = 9.3 Determine the ASTM grain size number if 25 grains/square inch are observed at a magnification of 50. Solution: 25(50/100)? = 2"! log(6.25) = (n-1)log(2) 0.796 = (n-1)(0.301) or n = 3.6 44 The Science and Engineering of Materials Instructor's Solution Manual 4-54 Determine the ASTM grain size number for the materials in (a) Figure 4-18 (b) Figure 4-23 Solution: (a) There are about 26 grains in the photomicrograph, which has the dimensions 2.375 in. x 2 in. The magnification is 100, thus: —26 gm -0738= (nr - Tosa *? log(5.47) = 0.738 = (n-)log(2) n=3.5 (b) There are about 59 grains in the photomicrograph, which has the dimensions 2.25 in. x 2in. The magnification is 500, thus: 59(500/100) mol l0g(328) = 2.516= (n-1log(2) n=94 (2.252) There are about 28 grains in the photomicrograph, which has the dimensions 2 in. x 2.25 in. The magnification is 200, thus: 28(200/1002 21 1og(24.889) = 1.396 = (n-1)log(2) n=5.6 220) og( ) (n-Dlog(2) n 4-58 The angle 0 of a tilt boundary is given by sin(9/2) = b/2D (see Figure 4-19). Verify the correctness of this equation. Solution: From the figure, we note that the grains are offset one Burgers vector, b, only for two spacings D. Then it is apparent that sin(9/2) must be b divided by two D. b Pos lh po o 4-59 Calculate the angle O of a small angle grain boundary in FCC aluminum when the dislocations are 5000 À apart. (See Figure 4-19 and equation in Problem 4-58.) Solution: b=(15) (42) (4.04958) = 2.8635 À and D= 5000 À 28635 (25000) 92 =0.0164 sin(0/2) = 000286 0=0.0328º 4-60 For BCC iron, calculate the average distance between dislocations in a small angle grain boundary tilted 0.50º. (See Figure 4-19.) y Solution: sin(0.50)=- + (N32.866) 2D 0.004364 = 1.241/D D=284Á CHAPTER 5 Atom and lon Movements in Materials 47 0.0167 x 102º — 0.0209 x 107 AclAx => >>> = 1.68 Zn atoms/em?-cm 0.0025 cm 5-44 A 0.001 in. BCC iron foil is used to separate a high hydrogen gas from a low hydro- gen gas at 650ºC. 5 x 108 H atoms/cm? are in equilibrium with the hot side of the foil, while 2 x 10? H atoms/cm? are in equilibrium with the cold side Determine (a) the concentration gradient of hydrogen and (b) the flux of hydrogen through the foil. 2x 10º—5x 108 —> >>> —— = 1969 x 108 H atoms/cm?-cm (0.001 in.)(2.54 em/in.) (b) J=D(AciAx) = —0.0012 exp[-3600/(1.987)(923)-1969 x 108) J=0.33 x 108 H atoms/cm?-s Solution: (a) AclAx = 5-45 A I-mm sheet of FCC iron is used to contain nitrogen in a heat exchanger at 1200ºC. The concentration of N at one surface is 0.04 atomic percent and the con- centration at the second surface is 0.005 atomic percent. Determine the flux of nitro- gen through the foil in atoms/cm?-s. Solution: (a) Acihr= (0.00005 — 0.0004)(4 atoms per cell)/(3.589 x 10 cm)? (1 mm)(0.1 cm/mm) = 3.03 x 1020 N atoms/cm?-cm (b) J=—D(AciAx) = —0.0034 exp[-34,600/(1.987)(1473)](-3.03 x 102º) =7.57x 102 N atoms/em?—s 5-46 A4 cm-diameter, 0.5 mm-thick spherical container made of BCC iron holds nitro- gen at 700ºC. The concentration at the inner surface is 0.05 atomic percent and at the outer surface is 0.002 atomic percent. Calculate the number of grams of nitrogen that are lost from the container per hour. Solution: AclAx = [0.00002 — 0.0005](2 atoms/cell)/(2.866 x 105 cm)? (0.5 mm)(0.1 em/mm) =-8.16x 102º N/cm-cm J = —0.0047 exp[-18,300/(1.987)(973)][-8.16 x 102º] = 2.97 x 10!4 N/cm?-s A =4m2=47(2 cm? = 50.27 cm? t= 3600 s/h sphere N atoms/h = (2.97 x 1014)(50.27)(3600) = 5.37 x 10!º N atoms/h 2 (537x 10!º atoms)(14.007 g/mol) - 3 Nloss (6.02 x 10? atoms/mo 1285 x 10 gh 5-47 ABCC iron structure is to be manufactured that will allow no more than 50 g of hydrogen to be lost per year through each square centimeter of the iron at 400º. If the concentration of hydrogen at one surface is 0.05 H atom per unit cell and is 0.001 H atom per unit cell at the second surface, determine the minimum thickness of the iron. Solution: c, = 0.05 H/(2.866 x 108 em)? = 212.4 x 10!º H atoms/cm? 0.001 H/(2.866 x 108 cm)? = 4.25 x 10!º H atoms/cm? 4.25 x 1019 — 212.4 x 1019] —2.08 x 102! Ax A AclAx= 48 | The Science and Engineering of Materials Instructor's Solution Manual (50 g/em? y)(6.02 x 102? atoms/mol) — (1.00797 g/mol)31.536 x 109 s/y) J=9.47x 10" H atoms/cm?-s =(-2.08 x 102!/Ax)(0.0012)exp[-3600/((1.987)(673))] Ax= 0.179 em =947x 1017 H atoms/cm?-s 5-48 Determine the maximum allowable temperature that will produce a flux of less than 2000 H atoms/cm?-s through a BCC iron foil when the concentration gradient is —5 x 1016 atoms/cm3-cm. (Note the negative sign for the flux.) Solution: 2000 H atoms/cm?-s = —0.0012 exp[-3600/1.987T][-5 x 1016 atoms/em3-cm] In(3.33 x 1071!) = -3600/1.987T T=3600/(-24.12)(1.987)) = 75 K = —1989€ 5-53 Explain why a rubber balloon filled with helium gas deflates over time. Solution: Helium atoms diffuse through the chains of the polymer material due to the small size of the helium atoms and the ease at which they dif- fuse between the loosely-packed chains. 5-59 The electrical conductivity of Mn,O, is 8 x 10-18 ohm-!-cm”! at 140ºC and is 1 x 107 ohm”!-em”! at 400ºC. Determine the activation energy that controls the tem- perature dependence of conductivity. Explain the process by which the temperature controls conductivity. 8x 10-18 = Cexpl-O/(1.987X413)] 1x 107 Cexpl-O/(1.987)(673)] 8x 101! =exp(-0.0004710) or —23.25 =-0.0004710 OQ = 49,360 cal/mol Solution: Electrical charge is carried by the diffusion of the atoms; as the temperature increases, more rapid diffusion occurs and consequently the electrical conductivity is higher. 5-60 Compare the rate at which oxygen ions diffuse in ALO, with the rate at which alu- minum ions diffuse in AL,O, at 1500ºC. Explain the difference. Solution: Do? = 1900 exp[-152,000/(1.987)(1773)] = 3.47 x 10-16 cm?/s D = 28 exp[-114,000/(1.987)(1773)] = 2.48 x 10-13 em?/s The ionic radius of the oxygen ion is 1.32 Á, compared with the aluminum ionic radius of 0.51 A; consequently it is much easier for the smaller aluminum ion to diffuse in the ceramic. 5-61. Compare the diffusion coefficients of carbon in BCC and FCC iron at the allotropic transformation temperature of 912ºC and explain the difference. Solution: Dycc = 0.011 exp[-20,900/(1.987)(1185)] = 1.51 x 10% cm?/s Dpcc= 0.23 exp[-32,900/(1.987)(1185)] = 1.92 x 107 cm?/s CHAPTER 5 Atom and lon Movements in Materials 49 Packing factor of the BCC lattice (0.68) is less than that of the FCC lattice; consequently atoms are expected to be able to diffuse more rapidly in the BCC iron. 5-62 Compare the diffusion coefficients for hydrogen and nitrogen in FCC iron at 1000ºC and explain the difference in their values. Solution: D, = 0.0063 exp[-10,300/(1.987)(1273)] = 1.074 x 104 em?/s D, = 0.0034 exp[-34,600/(1.987)(1273)] = 3.898 x 10º cm?/s Hin BCC Nin FCC Nitrogen atoms have a larger atoms radius (0.71 Â) compared with that of hydrogen atoms (0.46 À); the smaller hydrogen ions are expected to diffuse more rapidly. 5-66 A carburizing process is carried out on a 0.10% C steel by introducing 1.0% C at the surface at 980ºC, where the iron is FCC. Calculate the carbon content at 0.01 cm, 0.05 cm, and 0.10 em beneath the surface after 1 h. Solution: D= 0.23 exp[-32,900/(1.987)(1253)] = 42 x 108 cm?/s TE = ertlx / (Q(42 x 103)(3600)] = exflx / 0.0778] 1 1-c x= 0.01: erf70.01/0.0778] = erf(0.1285) = ET =0.144 c,=0.87%C 1-c x= 0.05: erfT0.05/0.0778] = erf(0.643) = Co 0.636 c,=0.43%C 1- x= 0.10: erf70.10/0.0778] = erf(1.285) = rey 0914 c=0.18%C Surface 0.05 x 010 0.15 5-67 Iron containing 0.05% C is heated to 912ºC in an atmosphere that produces 1.20% Cat the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if (a) the iron is BCC and (b) the iron is FCC. Explain the dif- ference. Solution: t= (24 hX(3600 s/h) = 86,400 s Dpcc = 0.011 exp[-20,900/(1.987)(1185)] = 1.54 x 10 em?/s Dycc= 0:23 exp[-32,900/(1.987)(1185)] = 1.97 x 107 em?/s 12-c BCC: TE = erf0.05/ (2/(1.54 x 10-4)(86, 400) )] = erf[0.0685] = 0.077 c,=1.1%C 52 | The Science and Engineering of Materials Instructor's Solution Manual 5-86 5-87 5-88 t 1h 1200 6 Dot D.= (3.019 x 10-(1) 1200 11200950 = Tg =9.95h t9s0 = During freezing of a Cu-Zn alloy, we find that the composition is nonuniform. By heating the alloy to 600ºC for 3 hours, diffusion of zinc helps to make the composi- tion more uniform. What temperature would be required if we wished to perform this homogenization treatment in 30 minutes? Solution: Dego= 0.78 exp[-43,900/(1.987/(873)] = 7.9636 x 102 1oo=3h 1,=05h D,= Doo teod't, = (79636 x 102)3)/0.5 D,=4.778x 10"! = 0.78 exp[-43,900/1.9877] In (6.1258 x 107!) = 23.516 =—43,900/1.987 T T=940K = 667 A ceramic part made of MgoO is sintered successfully at 1700ºC in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500ºC. Which will limit the rate at which sintering can be done: diffusion of mag- nesium ions or diffusion of oxygen ions? What time will be required at the lower temperature? Solution: Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen. D,100 = 0.000043 expf-82,100/(1.987)(1973)] = 3.455 x 10714 cm?/s D,500 = 0.000043 exp[-82,100/(1.987)(1773)] = 3.255 x 10-15 cm?/s (3.455 x 10-1490) . fiso= Dino !rmodDisoo =" a ass Tqis 955 min = 15.9h A Cu-Zn alloy has an initial grain diameter of 0.01 mm. The alloy is then heated to various temperatures, permitting grain growth to occur. The times required for the grains to grow to a diameter of 0.30 mm are Solution: Temperature (ºC) Time (min) 500 80,000 600 3,000 700 120 800 10 850 3 Determine the activation energy for grain growth. Does this correlate with the diffu- sion of zinc in copper? (Hint: Note that rate is the reciprocal of time.) Solution: Temperature WT Time Rate (C) (K) (Ko!) (min) (min-!) 500 773 0.00129 80,000 1.25x 105 600 873 0.00115 3,000 3.33x 104 700 973 0.001028 120 8.33x 102 800 1073 0.000932 10 0.100 850 1123 0.000890 3 0.333 CHAPTER 5 Atom and lon Movements in Materials 53 From the graph, we find that Q = 51,286 cal/mol, which does correlate with the acti- vation energy for diffusion of zinc in copper. 101 Q/R=25,810 Q=51,286 102 o Ê 103 104 0.00123 - 0.009 0.0010 0.0012 vT 5-91 A sheet of gold is diffusion-bonded to a sheet of silver in 1 h at 700ºC. At 500ºC, 440 h are required to obtain the same degree of bonding, and at 300ºC, bonding requires 1530 years. What is the activation energy for the diffusion bonding process? Does it appear that diffusion of gold or diffusion of silver controls the bonding rate? (Hint - note that rate is the reciprocal of time.) Solution: Temperature YT Time Rate CO K) (K!) (s) (sec-!) 700 973 0.001007 3600 0.278 x 103 500 773 0.001294 1.584x 10º 0.631 x 106 300 573 0.001745 4.825x10!º 0.207 x 10-10 0.278 x 10? expl-Q/(1.987/973)] exp[-0.00051720] 0.207 x 10-19. expl-0/(1.987)/(573)] “ exp[-0.00087830] In(1.343 x 107) = 16.413 = 0.0003611 O Q = 45,450 cal/mol. The activation energy for the diffusion of gold in silver is 45,500 cal/mole; thus the diffusion of gold appears to control the bonding rate. 54 - The Science and Engineering of Materials Instructor's Solution Manual 102 108 10-10 0.0010 0.0012 0.0014 0.0016 0.0018 vT CHAPTER 6 Mechanical Properties and Behavior 57 Load GageLengh Siess Stnin | (b) (in.) (psi) (in./in.) 0 2.00000 0 0.0 3,000 2.00167 15,000 0.000835 6,000 2.00333 30,000 0.001665 7,500 2.00417 37,500 0.002085 9,000 2.0090 45,000 0.0045 10,500 2.040 52,500 0.02 12,000 2.26 60,000 0.13 12,400 2.50 (max load) — 62,000 0.25 11,400 3.02 (fracture) 57,000 0.51 a 8 0.2% ofíset Stress (ksi) o a 8 8 9 [= 5 0.001 0.02 0.01 Strain (in.4n.) After fracture, the gage length is 3.014 in. and the diameter is 0.374 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (a) 0.2% offset yield strength = 45,000 psi (b) tensile strength = 62,000 psi (c) E = (30,000 — 0) / (0.001665 — 0) = 18 x 10º psi (d) %Elongation = cmo x 100 = 50.7% 2. 2 (e) %Reduction in area = (74X 0.505)" — (m/4)(0.374)! x 100 = 45.2% (m/4)(0.505)? (f) engineering stress at fracture = 57,000 psi (g) true stress at fracture = 11,400 lb / (1/4)(0.3742 = 103,770 psi (h) From the graph, yielding begins at about 37,500 psi. Thus: (yield strength)(strain at yield) = 14(37,500)(0.002085) = 39.1 psi 58 The Science and Engineering of Materials Instructor's Solution Manual 6-34 The following data were collected from a 0.4-in. diameter test specimen of polyvinyl chloride (/, = 2.0 in.): Solution: o =FIm/4)(0.4) = F/0.1257 U-2)12 Load GageLengh — Suess Stan - (lb) (in) (psi) (in.fino) 0 2.00000 o 00 300 2.00746 2387 000373 600 2.01496 4773 000748 900 202374 7,160 001187 1200 2.032 9,547 0016 1500 2.046 11,933 0.023 1660 2.070 (max load) 13206 0.035 1600 2.094 12.729 0047 1420 2.12 (fracture) 11297 006 0.2% ofíset Yielding o o 3 Stress (ksi) Pos 0.002 0.01 0.02 0.03 Strain (in./in.) After fracture, the gage length is 2.09 in. and the diameter is 0.393 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (a) 0.2% offset yield strength = 11,600 psi (b) tensile strength = 12,729 psi (c) E= (7160 — 0) / (0.01187 — 0) = 603,000 psi . (2.09 — 2) (d) %Elongation = 2. x 100 = 4.5% (e) %Reduction in area = (MAN OA? (m/40.393)7 x 100 = 3.5% (m/4X04? (1) engineering stress at fracture = 11,297 psi (g) true stress at fracture = 1420 Ib / (7/4)(0.393)2 = 11,706 psi (h) From the figure, yielding begins near 9550 psi. Thus: (yield strength)(strain at yield) = 4(9550)(0.016) = 76.4 psi CHAPTER 6 Mechanical Properties and Behavior 59 6-35 The following data were collected from a 12-mm-diameter test specimen of magnesium (£, = 30.00 mm): Solution: o=FI(u4(2mm)?=F/1131 (4 — 30)/30 Load GageLengh — Suess Stan - (N) (mm) (MPa) (mm/mm) 0 30.0000 0 0.0 5,000 30.0296 44.2 0.000987 10,000 30.0592 88.4 0.001973 15,000 30.0888 132.6 0.00296 20,000 30.15 176.8 0.005 25,000 30.51 221.0 0.017 26,500 30.90 234.3 0.030 27,000 31.50 (max load) 238.7 0.050 26,500 32.10 234.3 0.070 25,000 32.79 (fracture) 221.0 0.093 250 8 8 s 0.2% offset a 8 Stress (Mpa) 8 0.001 0.01 0.02 0.03 Strain (mmimm) After fracture, the gage length is 32.61 mm and the diameter is 11.74 mm. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (a) 0.2% offset yield strength = 186 MPa (b) tensile strength = 238.7 MPa (c) E = (132.6 — 0) / (0.00296 — 0) = 44,800 MPa = 44.8 GPa . (32.61 — 30) (d) %Elongation = 30 x 100 = 8.7% 2. 2 (e) %Reduction in area = MAYID — (411 7A] x 100 = 4.3% (m/4)(12)? (f) engineering stress at fracture = 221 MPa (g) true stress at fracture = 25,000 N / (m/4)(1 1.742 = 231 MPa 62 The Science and Engineering of Materials Instructor's Solution Manual The applied stress is much less than the yield strength; therefore Hooke's law can be used. The strain is e = o/E = 3,979 psi / (16 x 109 psi) = 0.00024868 in./in. LL 4, 12in. «e = CTN = 000024868 inn. in. o 4,= 12.00298 in. From Poisson's ratio, yu =— =03 Ei! Etong ja =— (0.3)(0.00024868) = — 0.0000746 in./in. d-d, d-04in. ão - 40ÊM 00000746in./in. d, 04 d,= 0.39997 in. 6-40(b) When a tensile load is applied to a 1.5-cm diameter copper bar, the diameter is reduced to 1.498-cm diameter. Determine the applied load, using the data in Table 6-3. Solution: From Table 6-3, u =! Elong = 0.36 1498- 1.5 Eu 157 = 001333 Eong = — Ea / 12 =— (-0.001333) / 0.36 = 0.0037 in.fin. o = Ee = (124.8 GPa)(1000 MPa/GPa)(0.0037 in./in.) = 462 MPa F=0A = (462 MPa)(x/4)(15 mm) = 81,640 N 6-41 A three-point bend test is performed on a block of ZrO, that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 1b is applied, the specimen deflects 0.037 in. and breaks. Calculate (a) the flex- ural strength and (b) the flexural modulus, assuming that no plastic deformation oceurs. (3)(400 Ib)(4 in.) Solution: (a) flexural strength = 3FL/2wh? = — >" (20.5 in.)(0.25 in.)? = 76,800 psi (b) flexural modulus = FI?/4wh?ô o (400 Ib)(4 in.)? “(AXO in.)(0.25 in.)*(0.037 in.) =22.14x 10º psi 6-42 A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 em thick and is resting on two supports 7.5 cm apart. The sam- ple breaks when a deflection of 0.09 mm is recorded. Calculate (a) the force that caused the fracture and (b) the flexural strength. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs. Solution: (a) The force F required to produce a deflection of 0.09 mm is F = (flexural modulus)(4wl36/L? F=(480,000 MPa)(4)(15 mm)(6 mm)(0.09 mm) / (75 mm)? F=1327N CHAPTER 6 Mechanical Properties and Behavior o3 (b) flexural strength = 3FL/2wh? = (3)(1327 N)(75 mmy/(2X15 mm)(6 mm? = 276 MPa 6-43(a) | A thermosetting polymer containing glass beads is required to deflect 0.5 mm when a force of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs. Solution: The minimum distance L between the supports can be calculated from the flexural modulus. IL? = 4wh? S(flexural modulus)/F IL? = (4)(20 mmJ(5 mm)*(0.5 mm)(6.9 GPA)(1000 MPa/GPa) / 500 N 1 =69,000mm? o L=4mm The stress acting on the bar when a deflection of 0.5 mm is obtained is o =3FL/2wh? = (3)(500 NJ(41 mm) / (2/20 mm)(5 mm)? = 61.5 MPa The applied stress is less than the flexural strength of 85 MPa; the poly- mer is not expected to fracture. 6-43(b) The flexural modulus of alumina is 45 x 109 psi and its flexural strength is 6-52 6-53 6-55 46,000 psi. A bar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs. Solution: The force required to break the bar is F = 2wh(flexural strength)/3L F= (2X in.(0.3 in.)(46,000 psi / (37 in.) = 394 Ib The deflection just prior to fracture is 8 = FI3/4wh?(flexural modulus) 3 = (394 1b)(7 in MAYA in.)(0.3 in.)(45 x 10º psi) = 0.0278 in. A Brinell hardness measurement, using a 10-mm-diameter indenter and a 500 kg load, produces an indentation of 4.5 mm on an aluminum plate. Determine the Brinell hardness number HB of the metal. 500 kg =D DD > > — É > = 298 (x 1 2)10 mm)[LO — 10? — 4.52] Solution: HB When a 3000 kg load is applied to a 10-mm-diameter ball in a Brinell test of a steel, an indentation of 3.1 mm is produced. Estimate the tensile strength of the steel. 3000 kg B="———— E = 388 (x 1 2)(10 mm)[LO — 410? — 3.12] Tensile strength = 500 HB = (500)(388) = 194,000 psi Solution: The data below were obtained from a series of Charpy impact tests performed on four steels, each having a different manganese content. Plot the data and determine (a) the transition temperature (defined by the mean of the absorbed energies in the 64 | The Science and Engineering of Materials Instructor's Solution Manual ductile and brittle regions) and (b) the transition temperature (defined as the temper- ature that provides 50 J absorbed energy). Plot the transition temperature versus manganese content and discuss the effect of manganese on the toughness of steel. What would be the minimum manganese allowed in the steel if a part is to be used at 09C? Solution: Test temperature Impact energy (J) SC 0.30% Mn 0.39% Mn 1.01% Mn 1.55% Mn —100 2 5 5 15 —75 2 5 7 25 —50 2 12 20 45 -25 10 25 40 70 0 30 55 75 Ho 25 60 100 no 135 50 105 125 130 140 75 130 135 135 140 100 130 135 135 140 7 G 20 1204 c Average 5 $ r 5 80d a 5% 5 g J r g E -20 564 E 40d õ s d F -s0 —HO o 100 Temperature (ºC) 0.3 0.6 0.9 1.2 15 % Mn (a) Transition temperatures defined by the mean of the absorbed energies are: 0.30% Mn: mean energy = 2 + (130 +2)/2 = 68]; T=27C 0.39% Mn: mean energy = 5 + (135 +5)/2 = 75 J; T=10º€ 1.01% Mn: mean energy = 5 + (135 +5)/2=75]; T= 0€ 1.55% Mn: mean energy = 15 + (140 + 15)/2 = 92.5 J, T=—12ºC (b) Transition temperatures defined by 50 J are: 0.30% Mn: T=15ºC 0.39% Mn: T=-—50€ 1.01% Mn: T=—15ºC 1.55% Mn: T=—45ºC 668 669 6-70 [rá CHAPTER 6 Mechanical Properties and Behavior 87 interface in an effort to continue propagating. In addition, the fibers may begin to pull out of the matrix, particularly if bonding is poor; the fiber pull-out requires energy, thus improving toughness. Finally, the fibers may bridge across the crack, helping to hold the material together and requiring more energy to propagate the crack. A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the composite is 45 MPam and the tensile strength is 550 MPa. Will the flaw cause the composite to fail before the tensile strength is reached? Assume that f= 1. Solution: Since the crack is internal, 2a = 0.001 em = 0.00001 m. Therefore a =0.000005 m Kw = folma or o=kKl za o = (45 MPaym) / (1),/7(0.000005 m) = 11,354 MPa The applied stress required for the crack to cause failure is much larger than the tensile strength of 550 MPa. Any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws. An aluminum alloy that has a plane strain fracture toughness of 25,000 psi fails when a stress of 42,000 psi is applied. Observation of the fracture surface indi- cates that fracture began at the surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f= 1.1. Solution: Kw=folta or a=(UMDKcl foR a = (11 7)[25,000 psivin. / (1.142, 000 psi)? = 0.093in. A polymer that contains internal flaws 1 mm in length fails at a stress of 25 MPa. Determine the plane strain fracture toughness of the polymer. Assume that f= 1. Solution: Since the flaws are internal, 2a = 1 mm = 0.001 m; thus a = 0.0005 m Ku = foma = (1)(25 MPa)/7(0.0005 m) = 0.99 MPavm A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain fracture toughness of 5,000 psivin. To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05 in. long. Assuming that f= 1.4, does our nondestructive test have the required sensitivity? Explain. Solution: The applied stress is o = (/4)(75,000 psi) = 25,000 psi a=(UMDIK, fo]? = (1/m)[5,000 psivin. 1 (1.4)(25,000 psi)? a=0.0065 in. The length of internal flaws is 2a = 0.013 in. Our nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size required for failure. Thus our NDT test is not satisfactory. 68 The Science and Engineering of Materials Instructor's Solution Manual 6-86 6-87 6-88 6-89 690 A cylindrical tool steel specimen that is 6 in. long and 0.25 in. in diameter rotates as a cantilever beam and is to be designed so that failure never occurs. Assuming that the maximum tensile and compressive stresses are equal, determine the maximum load that can be applied to the end of the beam. (See Figure 6-50.) Solution: The stress must be less than the endurance limit, 60,000 psi. o = 10.18LFId? or F= (endurance limit)d2/10.18L F = (60,000 psi(0.25 in.)2 / (10.186 in.) = 15.35 Ib A 2 cm-diameter, 20-cm-long bar of an acetal polymer (Figure 6-61) is loaded on one end and is expected to survive one million cycles of loading, with equal maxi- mum tensile and compressive stresses, during its lifetime. What is the maximum permissible load that can be applied? Solution: From the figure, we find that the fatigue strength must be 22 MPa in order for the polymer to survive one million cycles. Thus, the maximum load is F = (fatigue strength)d?/10.18L F= (22 MPa)(20 mm)? / (10.18)(200 mm) = 86.4 N Acyclical load of 1500 lb is to be exerted at the end of a 10-in. long aluminum beam (Figure 6-50). The bar must survive for at least 109 cycles. What is the mini- mum diameter of the bar? Solution: From the figure, we find that the fatigue strength must be 35,000 psi in order for the aluminum to survive 10º cycles. Thus, the minimum diam- eter of the bar is d= 4iO.18LF / fatigue strength d = 3f10.18)(10 in.)(1500 1b) / 35,000 psi = 1.634 in. A cylindrical acetal polymer bar 20 cm long and 1.5 cm in diameter is subjected to a vibrational load at a frequency of 500 vibrations per minute with a load of 50 N. How many hours will the part survive before breaking? (See Figure 6-61) Solution: The stress acting on the polymer is o = 10.18LF/d? = (10.18)(200 mm)(50 N) / (15 mm)? = 30.16 MPa From the figure, the fatigue life at 30.16 MPa is about 2 x 105 cycles. Based on 500 cycles per minute, the life of the part is life = 2 x 10º cycles / (500 cycles/min)(60 min/h) = 6.7 h Suppose that we would like a part produced from the acetal polymer shown in Figure 6-61 to survive for one million cycles under conditions that provide for equal compressive and tensile stresses. What is the fatigue strength, or maximum stress amplitude, required? What are the maximum stress, the minimum stress, and the mean stress on the part during its use? What effect would the frequency of the stress application have on your answers? Explain. Solution: From the figure, the fatigue strength at one million cycles is 22 MPa. The maximum stress is +22 MPa, the minimum stress is —-22 MPa, and the mean stress is O MPa. CHAPTER 6 Mechanical Properties and Behavior 69 A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time. 6-91 The high-strength steel in Figure 6-52 is subjected to a stress alternating at 200 revolutions per minute between 600 MPa and 200 MPa (both tension). Calculate the growth rate of a surface crack when it reaches a length of 0.2 mm in both m/cycle and m/s. Assume that f= 1.0. Solution: For the steel, C= 1.62 x 10-22 and n = 3.2. The change in the stress intensity factor AK is AK — fãoN'ma = (1.2)(600 MPa — 200 MPa),/7(0.0002 m = 12.03 MPavm. The crack growth rate is daldN = 1.62 x 10-!(AK)2 daldN = 1.62 x 10-!(12.03)*2 = 4.638 x 10º m/cycle datdt = (4.638 x 10º m/cycle)(200 cycles/min)/ 60 s/min daldt = 1.55 x 108 m/s 6-92 The high-strength steel in Figure 6-52, which has a critical fracture toughness of 80 MPavm, is subjected to an alternating stress varying from -900 MPa (compres- sion) to +900 MPa (tension). It is to survive for 10 cycles before failure occurs. Calculate (a) the size of a surface crack required for failure to occur and (b) the largest initial surface crack size that will permit this to happen. Assume that f= 1. Solution: (a) Only the tensile portion of the applied stress is considered in Ac. Based on the applied stress of 900 MPa and the fracture toughness of 80 MPavm, the size of a surface crack required for failure to occur is K= fora. ox a=(07MIKIfof a, = (1/ 7)[80 MPavm / (1)(900 MPa)P = 0.0025m = 2.5mm (b) The largest initial surface crack tolerable to prevent failure within 10º cycles is 21(0.0025 m) 23282 — q 3212] 2(36.41 — (a)-069] (-1.2)1.62 x 10P)(12.84 x 10º)(6.244) (a 98 = 1760 a,=3.9 x 10% m = 0.0039 mm N= I& cycles = 10 = 6-93 The acrylic polymer from which Figure 6-62 was obtained has a critical fracture toughness of 2 MpPavm. Itis subjected to a stress alternating between —10 and +10 MPa. Calculate the growth rate of a surface crack when it reaches a length of 5 x 10%miff=1.0. Solution: Ao = 10 MPa — 0 = 10 MPa, since the crack doesn't propagate for compressive loads. 72 | The Science and Engineering of Materials Instructor's Solution Manual Determine (a) the load applied to the specimen during the test, (b) the approximate length of time during which linear creep occurs, (c) the creep rate in in./in.h and in %/h, and (d) the true stress acting on the specimen at the time of rupture. (a) The loadis F = o A = (10,000 psi)(7/4)(0.6 in.)2 = 2827 Ib (b) The plot of strain versus time is linear between approximately 500 and 6000 hours, or a total of 5500 hours. (c) From the graph, the strain rate is the slope of the linear portion of the curve. 0.095 — 0.03 a Ag/At= "= 1.44 x 10% in./inch = 1.44 x 10? %h 6000 — 1500 (d) At the time of rupture, the force is still 2827 1b, but the diameter is reduced to 0.52 in. The true stress is therefore 0,= FIA = 2827 1 / (W/4)0.52 in.) = 13,312 psi 6-105 A stainless steel is held at 705ºC under different loads. The following data are obtained: Solution: Applied Stress (MPa) Rupture Time (A) Creep Rate (%/h) 106.9 1200 0.022 128.2 710 0.068 147.5 300 0.201 160.0 no 0.332 Determine the exponents “n” and “m” in Equations 6-40 and 6-41 that describe the dependence of creep rate and rupture time on applied stress. Plots describing the effect of applied stress on creep rate and on rupture time are shown below. In the first plot, the creep rate is given by As/At= Co" and the graph is a log-log plot. In the second plot, rupture time is given by t,=A o”, another log-log plot. The exponents “n” and “m” are the slopes of the two graphs. In this case, n=6.86 m=-—6.9 2000: o10 1000) ê E 7 0.06 É eo s e $ 004 É «0 º 008 = 00217 Noope -6.86 200 Me Coste dE Cof 001 100 100 200 300 100 200 300 Stress (MPa) Stress (MPa) CHAPTER 6 Mechanical Properties and Behavior 73 6-106 Using the data in Figure 6-59(a) for an iron-chromium-nickel alloy, determine the activation energy Q, and the constant “m” for rupture in the temperature range 980 to 1090ºC. Solution: The appropriate equation is 1,= Ko”exp(O /RT). From Figure 6-59(a), we can determine the rupture time versus tempera ture for a fixed stress, say o = 1000 psi: t,=º 2400h at 1090ºC = 1363K t,= 14,000h at 1040ºC =1313K t,= 100,000h at 980ºC =1253K From this data, the equation becomes 1, = Kexp(Q, |/RT) and we can find O, by simultaneous equations or graphically. O, = 117,000 cal/mol We can also determine the rupture time versus applied stress for a con- stant temperature, say 1090ºC: 1,=106h — foro= 450 psi 1,=10th — foro = 800 psi 1,=10h — foro = 1200 psi t 102h — foro =2100 psi With this approach, the equation becomes 1, = K“o”, where “ni” is obtained graphically or by simultaneous equations: m=3.9 105 10º Q/R = 59,000 Q=117,000 calímol Rupture time (h) Rupture time (h) 10º 0.00074 0.00076 0.00078 0.00080 vT (K?) Stress (psi) 6-107 A 1-in. diameter bar of an iron-chromium-nickel alloy is subjected to a load of 2500 Ib. How many days will the bar survive without rupturing at 980ºC? [See Figure 6-59(a).] Solution: The stress is o = F/A = 2500 Ib/ (m/4)(1 in.) = 3183 psi From the graph, the rupture time is 700 h/ 24 h/day = 29 days 74 | The Science and Engineering of Materials Instructor's Solution Manual 6-108 6-109 6-110 6111 6-112 A 5 mm x 20 mm bar of an iron-chromium-nickel alloy is to operate at 1040 for 10 years without rupturing. What is the maximum load that can be applied? [See Figure 6-59(a).] Solution: The operating time is (10 years)(365 days/year)(24 h/day) = 87,600 h From the graph, the stress must be less than 500 psi. The load is then F= A =(500 psi)(S mm/25.4 mm/in.)(20 mm/25.4 mmyin.) = 77.5 lb An iron-chromium-nickel alloy is to withstand a load of 1500 Ib at 760ºC for 6 years. Calculate the minimum diameter of the bar. [See Figure 6-59(a).] Solution: The operating time is (6 years)(365 days/year)(24 h/day) = 52,560 h From the graph, the stress must be less than 7000 psi. The minimum diameter of the bar is then d=(4/ nXF Io = J(41 7)(1500 lb / 7000 psi) = 0.52 in. A 1.2-in.-diameter bar of an iron-chromium-nickel alloy is to operate for 5 years under a load of 4000 Ib. What is the maximum operating temperature”? [See Figure 6-59(a).] Solution: The operating time is (5 years)(365 days/year)(24 h/day) = 43,800 h The stress is o = F/A = 4000 1b/ (m/4)(1.2 in.)2 = 3537 psi From the figure, the temperature must be below 850ºC in order for the bar to survive five years at 3537 psi. Alin.x2in. ductile cast iron bar must operate for 9 years at 650ºC. What is the maximum load that can be applied? [See Figure 6-59(b).] Solution: The operating time is (9 year)(365 days/year)(24 h/day) = 78,840 h. The temperature is 650 + 273 = 923 K LM = (923/1000)[36 + 0.78 In(78,840)] = 41.35 From the graph, the stress must be no more than about 1000 psi. The load is then F=0A = (1000 psi)(2 in.?) = 2000 Ib A ductile cast iron bar is to operate at a stress of 6000 psi for 1 year. What is the maximum allowable temperature? [See Figure 6-59(b).] Solution: The operating time is (1 year)(365 days/year)(24 h/day) = 8760 h From the graph, the Larson-Miller parameter must be 34.4 at a stress of 6000 psi. Thus 34.4=(T/ 1000)[36 + 0.78 In(8760)] = 0.0437 T=800K =527ºC CHAPTER 7 Strain Hardening and Annealing 77 7-9 A true stress-true strain curve is shown in Figure 7-22. Determine the strain harden- ing exponent for the metal. Solution: o,=Ke; e o, 0.05 in./in. 60,000 psi 0.10in./in. 66,000 psi 0.20 in./in. 74,000 psi 0.30 in./in. 76,000 psi 0.40 in./in. 81,000 psi From graph: K = 92,000 psi n=0.15 100 += 92,000 for e,=1 True stress (ksi) 0.05 0.10 0.20 0.40 True strain (in./in.) 7-10 ACu-30% Zn alloy bar has a strain hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm and an initial gage length of 3 cm, fails at an engi- neering stress of 120 MPa. After fracture, the gage length is 3.5 cm and the diameter is 0.926 cm. No necking occurred. Calculate the true stress when the true strain is 0.05 em/cm. Solution: e,= In(G,/t,) = In(3.5/3.0) = 0.154 o p= 120 MPa = RR (m4X10mm) F=9425N 9425 N = 139.95 MPa TE ODE (m/4)(9.26 mm); o,= K(0.154)º5 = 139.95 MPa or K=3566 The true stress at &, = 0.05 cm/cm is: 0,= 356.6 (00525 or o,=79.7 MPa The Frank-Read source shown in Figure 7-5(e) has created four dislocation loops from the original dislocation line. Estimate the total dislocation line present in the photograph and determine the percent increase in the length of dislocations pro- duced by the deformation. Solution: If the length of the original dislocation line is 1 mm on the photograph, then we can estimate the circumference of the dislocation loops. The loops are not perfect circles, so we might measure the smallest and largest diameters, then use the average: 78 | The Science and Engineering of Materials Instructor's Solution Manual first loop: D, = 10 mm, D, mau = arge circumference = 12.07 = 14 mm; DagT 12 mm second loop: Dn= 18 mm; Doasge circumference = 19.07 =20 mm; Dag= 19 mm third loop: D, =28 mm; D mau = large circumference = 29.07 =30 mm; Das =29 mm fourth loop: Dna =42 mm; Dasge circumference = 43.57 =45 mm; Dae =43.5 mm Therefore in the photograph itself: total length = 1 + (12.0 + 19.0 + 29.0 + 43.5)x = 326 mm The magnification in the photograph is 30,000. Therefore: total length = 326 / 30,000 = 0.0109 mm The original dislocation line is 1 mm / 30,000 = 3.33 x 105 mm Yo increase = (0.0109 —0.0000333) / 3.33 x 105 mm) x 100 = 32,630% 7-19 A 0.25-in.-thick copper plate is to be cold worked 63%. Find the final thickness. 0.25 -1,. Solution: (See Figure 7-7.) 63 “o x 100% or 1,=0.0925 in. 7-2 AQ.25-in.-diameter copper bar is to be cold worked 63%. Find the final diameter. (0.252 — d? Solution: 63= (0.257 — df x 100% or d2=0.023 or d=0.152in. (0.25% f 7 7-21 A2-in.-diameter copper rod is reduced to 1.5 in. diameter, then reduced again to a final diameter of 1 in. In a second case, the 2-in.-diameter rod is reduced in one step from 2 in. to a 1 in. diameter. Calculate the %CW for both cases. . »2-a Solution: acw= Sn Dr x 100 = 75% in both cases y 7-22 A3105 aluminum plate is reduced from 1.75 in. to 1.15 in. Determine the final properties of the plate. (See Figure 7-23.) Solution: JCW = ETEEO x 100% = 34.3% TS = 26 ksi YS = 22 ksi elongation = 5% 7-23 A Cu-30% Zn brass bar is reduced from 1-in. diameter to a 0.45-in. diameter. Determine the final properties of the bar. (See Figure 7-24.) 2 2 “ (0.45) x 100 = 79.75% TS=105ks YS= 68ksi elongation = 1% Solution: %CW = 7-24 7-25 7-26 CHAPTER 7 Strain Hardening and Annealing 79 A 3105 aluminum bar is reduced from a 1-in. diameter, to a 0.8-in. diameter, to a 0.6-in. diameter, to a final 0.4-in. diameter. Determine the %CW and the properties after each step of the process. Calculate the total percent cold work. (See Figure 7-23.) Solution: If we calculated the percent deformation in each step separately, we would find that 36% deformation is required to go from 1 in. to 0.8 in. The deformation from 0.8 in. to 0.6 in. (using 0.8 in. as the initial diame- ter) is 43.75%, and the deformation from 0.6 in. to 0.4 in. (using 0.6 in. as the initial diameter) is 55.6%. If we added these three deformations, the total would be 135.35%. This would not be correct. Instead, we must always use the original 1 in. diameter as our starting point. The following table summarizes the actual deformation and properties after each step. TS Ys % ksi ksi elongation 12- (0.82 Dir 0.8 x 100 = 36% 26 23 6 az 12 — (0.62 Mim 0.6! x 100 = 64% 30 27 3 «2 12- (0.42 E x 100 = 84% 32 29 2 The total percent cold work is actually 84%, not the 135.35%. We want a copper bar to have a tensile strength of at least 70,000 psi and a final diameter of 0.375 in. What is the minimum diameter of the original bar? (See Figure 7-7.) Solution: CW > 50% to achieve the minimum tensile strength 2. 2 50 = Ay (0.375 Y (0.375) x 100 o 0.5d2=0.140625 or d,=0.53in. We want a Cu-30% Zn brass plate originally 1.2-in. thick to have a yield strength greater than 50,000 psi and a %Elongation of at least 10%. What range of final thicknesses must be obtained? (See Figure 7-24.) Solution: YS > 50,000 psi requires CW > 20% E > 10% requires CW < 35% 12-t,=0.20 12-1,.=0.35 12 12 1- 0.96 in. 1- 0.78 in. 4,=0.78 to 0.96 in. 82 The Science and Engineering of Materials Instructor's Solution Manual (c) Hot working temperature = 900ºC (d) 04 Tp = 7509C = 1023 K Tap E 1023/04 = 2558 K = 2285ºC Properties 400 7-56 The following data were obtained when a cold worked metal was annealed. (a) Estimate the recovery, recrystallization, and grain growth temperatures. (b) Recommend a suitable temperature for obtaining a high strength, high electrical conductivity wire. (c) Recommend a suitable temperature for a hot-working process. Electrical 'conductivity 600 800 Temperature (ºC) 1200 (d) Estimate the melting temperature of the alloy. Annealing Residual Temperature Stresses (e) (psi) 250 21,000 275 21,000 300 5,000 325 0 350 0 375 0 400 0 425 o Solution: (a) recovery temperature = 280ºC Tensile Strength (psi) 52,000 52,000 52,000 52,000 34,000 30,000 27,000 25,000 recrystallization temperature = 330ºC grain growth temperature = 380ºC (b) For a high strength, high conductivity wire, we want to heat into the Grain Size (in) 0.0030 0.0030 0.0030 0.0030 0.0010 0.0010 0.0035 0.0072 recovery range. A suitable temperature might be 320º. (c) Hot working temperature = 375ºC CHAPTER 7 Strain Hardening and Annealing 83 (d) 0.4 Tp = 3309C = 603 K Top E 603/04 = 1508 K = 1235ºC Tensile strength Properties Residual Grain size stress 250 300 350 400 450 Temperature (ºC) 7-58 Determine the ASTM grain size number for each of the micrographs in Figure 7-16 and plot the grain size number versus the annealing temperature. Solution: The approximate number of grains per square inch in each photomicro- graph at 75x is: = log(14.6) = 2.683 = (n-1)0.301) n=49 650ºC: N = (3 grains/in?)(75/1007 “7 grains/in? = 2-1 log(1.7) = 0.23 = (n-1)(0.301) n=1.8 800ºC: N = (0.7 grains/in.?)(75/100)2 =04 grainsin.? = 271 log(0.4) = —0.40 = (n—1)(0.301) n=-0.3 84 | The Science and Engineering of Materials Instructor's Solution Manual » mn ASTM Grain Size Number (n) o 400 600 800 Temperature (ºC) 7-66 Using the data in Table 7-4, plot the recrystallization temperature versus the melting temperature of each metal, using absolute temperatures (Kelvin). Measure the slope and compare with the expected relationship between these two temperatures. Is our approximation a good one? Solution: Converting the recrystallization and melting temperatures to Kelvin, we can obtain the graph shown. The relationship of T, = 0.47, (K) is very closely followed. Tm z, Al 933K 423K Mg 923K - 43K Ag 1235K — 4BK Cu 1358K — 43BK Fe I811K | 723K Ni 1726K 8BK Mo 2883K 1173K wW 3683K 1473K Recrystalization temperature (K) 1000 2000 3000 4000 Melting temperature (K) Principles of Solidification 8-10 Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required, and (b) the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm. Solution: From Table 8-1, AT, ,, = 480ºC é QU25S x 107 Memê)(1453 + 273) pts DEDA O SMA IIS = 6.65 x 108 cm (2756 em?)(480) a,=3.56À V=45.118x 102% cm? Vaucleus = (4M3)(6.65 x 108 em)? = 1232 x 104 cm? nucleus number of unit cells = 1232/45.118 = 27.3 atoms per nucleus = (4 atoms/cell)(27.3 cells) = 109 atoms 8-11 . Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required, and (b) the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is 2.92 À. Solution: «(04 x 107 Jlem?)(1538 + 273) rf=T[[[["]DD DDD (1737 Jlem?)(420) V= (47/3)(10.128)? = 4352 À? = 4352 x 10-24 cm? Vos (2.92 À = 24.897 À? = 24.897 x 10% cm? number of unit cells = 4352/24.897 = 175 = 10.128 x 108 cm atoms per nucleus = (175 cells)(2 atoms/cell) = 350 atoms 8-12 Suppose that solid nickel was able to nucleate homogeneously with an undercooling of only 22ºC. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid FCC nickel is 0.356 nm. 87 88 The Science and Engineering of Materials Instructor's Solution Manual 813 s-14 8-28 8-29 = (BSS x 107 Hemêy(14s3 +273) . (2756 Jlem?)(22) Ve= 45.118 x 102 cm? (see Problem 8-10) Vac = (4U3)(145.18 x 108 cm) = 1.282 x 107 cm? number of unit cells = 1.282 x 10-17 / 45.118 x 1024 = 2.84 x 109 Solution: = 145.18 x 10% cm atoms per nucleus = (4 atoms/cells)(2.84 x 109 cell) = 1.136 x 109 Suppose that solid iron was able to nucleate homogeneously with an undercooling of only 15ºC. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid BCC iron is 2.92 À. Solution: «(204 x 107 Jlem?)(1538 + 273) f=-T]]]|[D[DDwD>———— (1737 Jlem?)(15) Vic = 24.897 x 1024 cm? (see Problem 8-10) Vuc = (47/3)(283.6 x 108 cm)? = 95,544,850 x 10-24 cm? number of unit cells = 95,544,850/24.897 = 3.838 x 109 atoms per nucleus = (2 atoms/cells)(3.838 x 10º cell) = 7.676 x 109 =283.6x 108 cm Calculate the fraction of solidification that occurs dendritically when iron nucleates (a) at 10ºC undercooling, (b) at 100ºC undercooling, and (c) homogeneously. The specific heat of iron is 5.78 J/em?.ºC. ion: a 30 Solution: CAT (5.38 Vem PO(OC) 0.0333 AH, 1737 Jem? ? 30 CAT - (5.78 J/em” ºC)(100ºC) = 0333 AH, 1737 Jem? cAT (5.78 Jlem3.ºC)(420ºC) nn — = 2, therefore, all dendritically AH, 1737 Jem? Calculate the fraction of solidification that occurs dendritically when silver nucle- ates (a) at 10ºC undercooling, (b) at 100ºC undercooling, and (c) homogeneously. The specific heat of silver is 3.25 J/cm?ºC. lution: "AT 3.25 Jlem?*C)(10ºC Solution CAT 62 Vem COMO 37 AH, 965 J/cmê ) 3. CAT - (3.25 J/em*-“C)(100ºC) = 0337 AH, 965 J/em? ) 3. CAT - (3.25 J/em*-“C)(250ºC) = 0842 AH, 965 J/em? Analysis of a nickel casting suggests that 28% of the solidification process occurred in a dendritic manner. Calculate the temperature at which nucleation occurred. The specific heat of nickel is 4.1 J/cm?.ºC. Solution: T (4 Jlem?.ºC)(AT) 56 em? =0.28 A AT=188C or T=1453- 188=1265ºC sa 832 833 CHAPTER 8 Principles of Solidification 89 A 2-in. cube solidifies in 4.6 min. Calculate (a) the mold constant in Chvorinov's rule and (b) the solidification time for a 0.5 in. x 0.5 in. x 6 in. bar cast under the same conditions. Assume that n = 2. Solution: (a) We can find the volume and surface area of the cube: V=0P=8in? A=602=24in? 1=46=B(8242 B=4.6/(0.333)2 = 41.48 minin.? (b) For the bar, assuming that B = 41.48 min/in.2: V=(0.5/0.5/6) = 1.5 in.2 A=20.5)0.5) + 4(0.5)(6) = 12.5 in.? t= (41.48)(1.5/12.5)2 = 0.60 min A 5-cm diameter sphere solidifies in 1050 s. Calculate the solidification time for a 0.3 em x 10 em x 20 cm plate cast under the same conditions. Assume that n = 2. Solution: (47 132.5) t=1050s=B 4m(2.5)? 2 | =B[2.53P or B=15129cm? (1512)(0.3 x 10 x 20)? 1= > STO = 5I2[60MI8P=31.155 [2(0.3)(10) + 2(0.3)(20) + 2(10)(20)? Find the constants B and n in Chvorinov's rule by plotting the following data on a log-log plot: “Casting | Solidification dimensions time (in.) (min) 0.5x8x12 3.48 2x3x 10 15.78 2.5 cube 10.17 Ix4x9 8.13 Solution: V(in?) Ain?) VIA (in.) 48 212 0.226 s0 12 0.536 15.6 37.5 0416 36 98 0.367 From the graph, we find that B=48 minin? and n = 1.72 92 The Science and Engineering of Materials Instructor's Solution Manual (b) 1.5 = 0.070/7 — 0.3 fuer = (1.8/0.077 =661 8 center (c) The mold gets hot during the solidification process, and consequentIy heat is extracted from the casting more slowly. This in turn changes the constants in the equation and increases the time required for complete solidification. 15 center E 2 g10 Due to heating 5 of mold 3 E e 205 E E B a o 10 20 30 vt (vs) 8-36 Figure 8-9(b) shows a photograph of an aluminum alloy. Estimate (a) the secondary dendrite arm spacing and (b) the local solidification time for that area of the casting. Solution: (a) The distance between adjacent dendrite arms can be measured. Although most people doing these measurements will arrive at slightly different numbers, the author”s calculations obtained from four different primary arms are: 16 mm/ 6 arms = 2.67 mm 9 mm/5 arms = 1.80 mm 13 mm/7 arms = 1.85 mm 18 mm/9 rms = 2.00 mm average = 2.08 mm = 0.208 cm Dividing by the magnification of x50: SDAS = 0.208 cm / 50 = 4.16 x 10% cm (b) From Figure 8-10, we find that local solidification time (LST) = 90 s 8-37 Figure 8-31 shows a photograph of FeO dendrites that have precipitated from a glass (an undercooled liquid). Estimate the secondary dendrite arm spacing. Solution: We can find 13 SDAS along a 3.5 cm distance on the photomicro- graph. The magnification of the photomicrograph is x450, while we want the actual length (at magnification x 1). Thus: SDAS = (13 SDAS/3.5 cm)(1/450) = 8.25 x 10? cm 8-38 Find the constants c and m relating the secondary dendrite arm spacing to the local solidification time by plotting the following data on a log-log plot: CHAPTER 8 Principles of Solidification 93 Solidification Time SDAS. (8) (em) 156 0.0176 282 0.0216 606 0.0282 1356 0.0374 Solution: From the slope of the graph: m= 34/100 = 0.34 We can then pick a point off the graph (say SDAS = 0.0225 cm when LST =300s) and calculate “c”: 0.0225 = c(300)9:4 = 6.954c 0032 SDAS (cm) 100 300 500 1000 3000 Time (s) 8-39 Figure 8-32 shows dendrites in a titanium powder particle that has been rapidly solidified. Assuming that the size of the titanium dendrites is related to solidification time by the same relationship as in aluminum, estimate the solidification time of the powder particle. Solution: The secondary dendrite arm spacing can be estimated from the pho- tomicrograph at several locations. The author”s calculations, derived from measurements at three locations, are 1 mm/8 arms = 1.375 mm 13 mm/ 8 arms = 1.625 mm 13 mm/ 8 arms = 1.625 mm average = 1.540 mm Dividing by the magnification of 2200: SDAS = (1.540 mm(0.1 em/mm) / 2200 = 7 x 105 em The relationship between SDAS and solidification time for aluminum is: SDAS = 8x 1041042=7x 105 t= (0.0875)"0-42 = 0.003 s 94 | The Science and Engineering of Materials Instructor's Solution Manual 8-40 The secondary dendrite arm spacing in an electron beam weld of copper is 8-45 8-46 8-47 9.5 x 104 cm. Estimate the solidification time of the weld. Solution: From Figure 8-10, we can determine the equation relating SDAS and solidification time for copper: n=19/50=0.38 c=4x 10% cm Then for the copper weld: 9.5 x 104=4 x 104LST)8 (Note: LST is local solidification time) 0.2375 = (LST)º38 or -—1438=0.38 InLST In LST=-3.783 or LST=0.023s A cooling curve is shown in Figure 8-33. Determine (a) the pouring temperature, (b) the solidification temperature, (c) the superheat, (d) the cooling rate just before solidification begins, (e) the total solidification time, (f) the local solidification time, and (g) the probable identity of the metal. (h) If the cooling curve was obtained at the center of the casting sketched in the figure, determine the mold constant, assum- ing thatn=2. Solution: (a) Toon = 475€ (e)1,=470s (b) Toy = 3200€ (D LST = 470 — 130 = 340 s (c) AT, = 475 — 320 = 155€ (g) Cadmium (Cd) 475 = 320 () 1,=470=B[38.4/121.6? (d) ATIA =" = 1.2.0C)s 5 130-0 B=4713s/em? A cooling curve is shown in Figure 8-34. Determine (a) the pouring temperature, (b) the solidification temperature, (c) the superheat, (d) the cooling rate just before solidification begins, (e) the total solidification time, (f) the local solidification time, (g) the undercooling, and (h) the probable identity of the metal. (i) If the cooling curve was obtained at the center of the casting sketched in the figure, determine the mold constant, assuming n = 2. Solution: (a) Tour = 900ºC (e) 1,=9.7 min (b) Too = 420ºC (DLST=9.7- 1.6=8.I min (c) AT, = 900 — 420 = 480ºC (g) 420 — 360 = 60ºC 900 — 400 . (h)Zn (d) ATIA = 600” 312 ºC/min (D1,=9.7=BI8/24P or B=87.5 minin? Figure 8-35 shows the cooling curves obtained from several locations within a cylindrical aluminum casting. Determine the local solidification times and the SDAS at each location, then plot the tensile strength versus distance from the cast- ing surface. Would you recommend that the casting be designed so that a large or small amount of material must be machined from the surface during finishing? Explain. Solution: The local solidification times can be found from the cooling curves and can be used to find the expected SDAS values from Figure 8-10. The SDAS values can then be used to find the tensile strength, using Figure 8-11.