Fundamentals of Materials Science and Engineering Solutions Ch 17-20

CH A PTE R 17 THERMA L PROPERTIES 17.4 (a) For aluminum, Cv at 50 K may be approximated by Equation (17.2), since this temperature is significantly below the Debye temperature. The value of Cv at 30 K is given, and thus, we may compute the constant A as A= 0.81 J/mol-K Cv = 3 × 10−5 J/mol-K 4 = (30 K) 3 T3 Therefore, at 50 K C v = AT 3 = (3 × 10−5 J/mol-K 4 )(50 K) 3 = 3.75 J/mol-K and cv = (3.75 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 139 J/kg-K (b) Since 425 K is above the D ebye temperature, a good approximation for Cv is C v = 3R = (3)(8.31 J/mol-K) = 24.9 J/mol-K A nd, converting this to specific heat cv = (24.9 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 925 J/kg-K 17.14 This problem asks for us to determine the temperature to which a cylindrical rod of tungsten 10.000 mm in diameter must be heated in order for it to just fit into a 9.988 mm diameter circular hole in a plate of 316 stainless steel, assuming that the initial temperature is 25◦ C. This requires the use of E quation (17.3a), which is applied to the diameters of the rod and hole. That is df − do = ␣l (T f − T o ) do Solving this expression for df yields d f = d o [1 + ␣l (T f − T o )] Now all we need do is to establish expressions for df (316 stainless) and df (W), set them equal to one another, and solve for Tf . A ccording to Table 17.1, ␣l (316 stainless) = 16.0 × 10−6 ( ◦ C) −1 and ␣l (W) = 4.5 × 10−6 ( ◦ C) −1 . Thus d f (316 stainless) = d f (W) (9.988 mm)[1 + {16.0 × 10−6 ( ◦ C) −1 }(T f − 25◦ C)] = (10.000 mm)[1 + {4.5 × 10−6 ( ◦ C) −1 }(T f − 25◦ C)] Now solving for Tf gives Tf = 129.5◦ C. 62 17.24 This problem asks that we treat a porous material as a composite wherein one of the phases is a pore phase, and that we estimate upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a 0.30 volume fraction of pores. The upper limit of k (kupper) may be determined using E quation (15.1) with thermal conductivity substituted for the elastic modulus, E. From Table 17.1, the value of k for MgO is 37.7 W/m-K, while for still air in the pore phase, k = 0.02 W/m-K. Thus k upper = V p k air + V MgO k MgO = (0.30)(0.02 W/m-K) + (0.70)(37.7 W/m-K) = 26.4 W/m-K For the lower limit we employ a modification of E quation (15.2) as k lower = = k air k MgO V p k MgO + V MgO k air (0.02 W/m-K)(37.7 W/m-K) (0.30)(37.7 W/m-K) + (0.70)(0.02 W/m-K) = 0.067 W/m-K 17.29 We want to heat the copper wire in order to reduce the stress level from 70 MPa to 35 MPa; in doing so, we reduce the stress in the wire by 70 MPa − 35 MPa = 35 MPa, which will be a compressive stress (i.e., ␴ = −35 MPa). Taking a value for E of 110 G Pa (Table 7.1) and solving for Tf from E quation (17.8) Tf = To − ␴ E ␣1 = 20◦ C − −35 MPa (110 × 103 MPa)[17 × 10−6 ( ◦ C) −1 ] = 20◦ C + 19◦ C = 39◦ C (101◦ F) 17.D 1 This problem stipulates that 1025 steel railroad tracks are laid at a temperature of 10°C. We are asked to determine the hottest possible temperature that can be tolerated without the introduction of thermal stresses if the rails are 11.9 m lon g, and for a joint space of 4.6 mm. For these railroad tracks, each end is allowed to expand one-half of the joint space distance, or the track may expand a total of this distance (4.6 mm). E quation (17.3a) is used to solve for Tf , where ␣l for the 1025 steel is found in Table 17.1. Thus, Tf = = l + To ␣l lo 4.6 × 10−3 m + 10◦ C [12.0 × 10−6 ( ◦ C) −1 ] (11.9 m) = 32.2◦ C + 10◦ C = 42.2◦ C (108◦ F) 63 CH A PTE R 18 MAGNETIC PROPERTIES 18.1 This problem concerns a coil of wire 0.20 m long that has 200 turns and carries 10 A . (a) We may calculate the magnetic field strength generated by this coil using E quation (18.1) as NI l (200 turns)(10 A ) = 10,000 A -turns/m = 0.2 m H = (b) In a vacuum, the flux density is determined from E quation (18.3). Thus, B o = ␮o H = (1.257 × 10−6 H /m)(10,000 A -turns/m) = 1.257 × 10−2 tesla (c) When a bar of titanium is positioned within the coil, we must use an expression that is a combination of Equations (18.5) and (18.6) in order to compute the flux density given the magnetic susceptibility. Inasmuch as ␹ m = 1.81 × 10−4 (Table 18.2), then B = ␮o H + ␮o M = ␮o H + ␮o ␹ m H = ␮o H (1 + ␹ m ) = (1.257 × 10−6 H /m)(10,000 A -turns/m)(1 + 1.81 × 10−4 ) ∼ = 1.257 × 10−2 tesla which is essentially the same result as part (b). This is to say that the influence of the titanium bar within the coil makes an imperceptible difference in the magnitude of the B field. (d) The magnetization is computed from E quation (18.6): M = ␹ m H = (1.81 × 10−4 )(10,000 A -turns/m) = 1.81 A /m 18.4 For this problem, we want to convert the volume susceptibility of silver (i.e., 2.38 × 10−5 ) into other systems of units. For the mass susceptibility ␹ m (kg) = = ␹m ␳ (kg/m 3 ) −2.38 × 10−5 = −2.27 × 10−9 10.49 × 103 kg/m 3 For the atomic susceptibility ␹ m (a) = ␹ m (kg) × [atomic weight (in kg)] = (−2.27 × 10−9 )(0.10787 kg/mol) = −2.45 × 10−10 18.6 This problem stipulates that the magnetic flux density within a bar of some material is 0.435 tesla at an H field of 3.44 × 105 A /m. 64 (a) We are first of all asked to compute the magnetic permeability of this material. This is possible using E quation (18.2) as ␮= B 0.435 tesla = 1.2645 × 10−6 H /m = 3.44 × 105 A /m H (b) The magnetic susceptibility is calculated as ␹m = 1.2645 × 10−6 H /m ␮ −1= −1 1.257 × 10−6 H /m ␮o = 6 × 10−3 18.27 (a) The B-H data provided in the problem are plotted below. 1.6 B (tesla) 1.2 0.8 0.4 0.0 0 200 400 600 800 1000 1200 H (A/m) (b) This portion of the problem asks for us to determine values of the initial permeability and initial relative permeability. The first four data points are plotted below. 0.3 B (tesla) 0.2 0.1 0.0 0 i 10 20 30 H (A/m) 65 40 50 60 The slope of the initial portion of the curve is ␮i (as shown), is ␮i = (0.15 − 0) tesla B = 3.0 × 10−3 H /m = (50 − 0) A /m H A lso, the initial relative permeability [E quation (18.4)] is just ␮ri = 3.0 × 10−3 H /m ␮i = = 2400 ␮o 1.257 × 10−6 H /m (c) The maximum permeability is the tangent to the B-H curve having the greatest slope; it is drawn on the plot below, and designated as ␮(max) . 1.6 (max) B (tesla) 1.2 0.8 0.4 0.0 0 200 400 600 800 1000 1200 H (A/m) The value of ␮(max) is ␮(max) = B (1.3 − 0.3) tesla = 8.70 × 10−3 H /m = (160 − 45) A -m H 18.32 (a) G iven E quation (18.12) and the data in Table 18.7, we are asked to calculate the critical magnetic fields for tin at 1.5 and 2.5 K. From the table, for Sn, TC = 3.72 K and B C (0) = 0.0305 tesla. Thus, from E quation (18.3) H C (0) = = B C (0) ␮o 0.0305 tesla = 2.43 × 104 A /m 1.257 × 10−6 H /m Now, solving for H C (2.5) using E quation (18.12) yields H C (T) = H C (0) 1 − 2 TT2  2 C  H C (2.5) = (2.43 × 104 A /m) 1 − 66 (2.5 K) (3.72 K) 2  = 1.33 × 104 A /m (b) Now we are to determine the temperature to which lead must be cooled in a magnetic field of 20,000 A /m in order for it to be superconductive. The value of H C (0) must first be determined using B C (0) given in the table (i.e., 0.0803 tesla); thus H C (0) = B C (0) 0.0803 tesla = 6.39 × 104 A /m = ␮o 1.257 × 10−6 H /m Since TC = 7.19 K we may solve for T using E quation (18.12) as  T = TC 1 − C (0) HHC (T)  = (7.19 K) 1 − 63900AA 20000 /m/m 67 = 5.96 K CH A PTE R 19 OPTICAL PROPERTIES 19.9 We want to compute the velocity of light in calcium fluoride given that r = 2.056 and ␹ m = −1.43 × 10−5 . The velocity is determined using E quation (19.8); but first, we must calculate the values of and ␮ for calcium fluoride. A ccording to E quation (12.30) = r o = (2.056)(8.85 × 10−12 F/m) = 1.82 × 10−11 F/m Now, utilizing E quations (18.4) and (18.7) ␮ = ␮o (␹ m + 1) = (1.257 × 10−6 H /m)(1 − 1.43 × 10−5 ) = 1.257 × 10−6 H /m A nd, finally 1 v= √ ␮ 1 = (1.82 × 10−11 F/m)(1.257 × 10−6 H /m) = 2.09 × 108 m/s 19.11 This problem asks for us, using data in Table 19.1, to estimate the dielectric constants for silica glass, soda-lime glass, PTFE , polyethylene, and polystyrene, and then to compare these values with those cited in Table 12.4 and briefly explain any discrepancies. From E quation (19.10) r = n2 Thus, for fused silica, since n = 1.458 r = (1.458) 2 = 2.13 When we compare this value with that given in Table 12.4 at a frequency of 1 MH z (i.e., r = 3.8) there is a significant discrepancy. The reason for this is that, for this material, an ionic component to the dielectric constant is present at 1 MH z, which is absent at frequencies within the visible electromagnetic spectrum, frequencies which are on the order 109 MH z. This effect may be noted in Figure 12.32. 19.19 In this problem we are asked to calculate the fraction of nonreflected light transmitted through a 20 mm thickness of transparent material, given that the fraction transmitted through a 10 mm width is 0.90. From E quation (19.18), the fraction of nonreflected light transmitted is just I′T /I′ o . U sing this expression we must first determine the value of ␤ as ␤=− =− Now, solving for IT′ I′o 1 x   ln T II′′ o 1 10 mm when x = 20 mm   ln(0.90) = 1.05 × 10−2 mm −1 I ′T = exp( −␤x) I ′o exp[−(1.05 × 10−2 mm −1 )(20 mm)] = 0.81 68 19.30 This problem asks for the difference in energy between metastable and ground electron states for a ruby laser. The wavelength of the radiation emitted by an electron transition from the metastable to ground state is cited as 0.6943 ␮m. The difference in energy between these states, E, may be determined from E quation (19.6), as E = h␯ = = hc ␭ (4.13 × 10−15 eV-s)(3 × 108 m/s) 6.943 × 10−7 m = 1.78 eV 69 CH A PTE R 20 MATERIA LS SELECTION A ND D ESIGN CONSID ERATIONS 20.D 3 (a) This portion of the problem asks that we derive a performance index expression for strength analogous to E quation (20.9) for a cylindrical cantilever beam that is stressed in the manner shown in the accompanying figure. The stress on the unfixed end, , for an imposed force, F, is given by the expression [E quation (20.24) in the textbook] = FLr I (20.D 1) where L and r are the rod length and radius, respectively, and I is the moment of inertia; for a cylinder the expression for I is provided in Figure 7.18: I= r 4 4 (20.D 2) Substitution for I into E quation (20.D 1) leads to = 4FL r 3 (20.D 3) Now, the mass m of some given quantity of material is the product of its density ( ) and volume. Inasmuch as the volume of a cylinder is just r2 L, then m = r 2 L (20.D 4) From this expression, the radius is just r= (20.D 5) L m Inclusion of E quation (20.D 5) into E quation (20.D 3) yields = 4F 1/2 L 5/2  3/2 m 3/2 (20.D 6) A nd solving for the mass gives m = (16F 2 L 5 ) 1/3  2/3 (20.D 7) To ensure that the beam will not fail, we replace stress in E quation (20.D 7) with the yield strength ( y ) divided by a factor of safety (N) as m = (16F 2 L 5 N 2 ) 1/3  y 70 2 /3 (20.D 8) Thus, the best materials to be used for this cylindrical cantilever beam when strength is a consideration are those having low ␳2/3 ratios. Furthermore, the strength performance index, P, is just the y reciprocal of this ratio, or 2/3 y P= ␳ (20.D 9) The second portion of the problem asks for an expression for the stiffness performance index. Let us begin by consideration of E quation (20.25) which relates ␦, the elastic deflection at the unfixed end, to the force (F), beam length (L), the modulus of elasticity (E), and moment of inertia (I) as FL 3 3E I ␦= (20.25) A gain, E quation (20.D 2) gives an expression for I for a cylinder, which when substituted into E quation (20.25) yields ␦= 4FL 3 3 E r 4 (20.D 10) A nd, substitution of the expression for r [E quation (20.D 5)] into E quation (20.D 10), leads to 4FL 3 = 4  m 3E 5 2L 4FL  = 3E m 2 (20.D 11) Now solving this expression for the mass m yields 1/2√ (20.D 12) 4FL 5  E 3 O r, for this cantilever situation, the mass of material experiencing a given deflection produced by a  specific force is proportional to the √ ratio for that material. A nd, finally, the stiffness performance E index, P, is just the reciprocal of this ratio, or  m= √ P= E  (20.D 13) (b) H ere we are asked to select those metal alloys in the database that have stiffness performance indices greater than 3.0 (in SI units). (Note: for this performance index of 3.0, density has been taken in terms of g/cm 3 rather than in the SI units of kg/m 3 .) Seventeen metal alloys satisfy this √ criterion; they and their E/ values are listed below, and ranked from highest to lowest value. 71 √ A lloy E ␳ Condition A Z 31B Mg A Z 31B Mg A Z 91D Mg 356.0 A l 356.0 A l 356.0 A l 6061 A l 6061 A l 6061 A l 2024 A l 2024 A l 2024 A l 1100 A l 1100 A l 7075 A l 7075 A l 7075 A l Rolled Extruded A s cast A s cast, high production A s cast, custom T6 O T6 T651 O T3 T351 O H 14 O T6 T651 3.790 3.790 3.706 3.163 3.163 3.163 3.077 3.077 3.077 3.072 3.072 3.072 3.065 3.065 3.009 3.009 3.009 √ (c) We are now asked to do a cost analysis on the above alloys. Below are tabulated the ␳ / E ratio, the relative material cost ( c̄), and the product of these two parameters; also those alloys for which cost data are provided are ranked, from least to most expensive. A lloy A Z 91D Mg 6061 A l 356.0 A l 6061 A l A Z 31B Mg 1100 A l A Z 31B Mg 7075 A l 2024 A l 356.0 A l 356.0 A l 2024 A l 1100 A l 2024 A l 6061 A l 7075 A l 7075 A l ␳ Condition A s cast T6 A s cast, high production T651 E xtruded O R olled T6 T3 A s cast, custom T6 T351 H 14 O O O T651 √ E 0.2640 0.3250 0.3162 0.3250 0.2640 0.3263 0.2640 0.3323 0.3255 0.3162 0.3162 0.3255 0.3263 0.3255 0.3250 0.3323 0.3323 c̄ 5.4 7.6 7.9 8.7 12.6 12.3 15.7 13.4 14.1 15.7 16.6 16.2 – – – – –   ␳ c̄ √ E 1.43 2.47 2.50 2.83 3.33 4.01 4.14 4.45 4.59 4.96 5.25 5.27 – – – – – It is up to the student to select the best metal alloy to be used for this cantilever beam on a stiffness-per-mass basis, including the element of cost, and other relevant considerations. 72 20.D 6 (a) This portion of the problem asks that we compute the maximum tensile load that may be applied to a spring constructed of a 14 hard 304 stainless steel such that the total deflection is less than 5 mm; there are 10 coils in the spring, whereas, its center-to-center diameter is 15 mm, and the wire diameter is 2.0 mm. The total spring deflection ␦s may be determined by combining E quations (20.14) and (20.15); solving for the load F from the combined equation leads to F= ␦sd4G 8N c D 3 H owever, it becomes necessary to determine the value of the shear modulus G. This is possible using E quation (7.9) and values of the modulus of elasticity (193 G Pa) and Poisson’s ratio (0.30) as taken from Tables B.2 and B.3 in A ppendix B. Thus G = E 2(1 + ␯ ) 193 G Pa = = 74.2 G Pa 2(1 + 0.30) Substitution of this value and values of the other parameters into the above equation for F leads to (5 × 10−3 m)(2 × 10−3 m) 4 (74.2 × 109 N/m 2 ) (8)(10 coils)(15 × 10−3 m) 3 = 22.0 N (5.1 lb f ) F= (b) We are now asked to compute the maximum tensile load that may be applied without any permanent deformation of the spring wire. This requires that we combine E quations (20.12) and (20.13), and then solve for F. H owever, it is first necessary to calculate the shear yield strength and substitute it for ␶ in E quation (20.12). The problem statement stipulates that ␶y = 0.6 y . From Table B.4 in A ppendix B, we note that the tensile yield strength for this alloy in the 1/4 hardened state is 515 MPa; thus y = 309 MPa. Thus, solving for F as outlined above F= = y d 3  −0.140 D (1.6)(8)(D ) d (309 × 106 N/m 2 )(2 × 10−3 m) 3  (1.6)(8)(15 × 10−3 m) = 53.6 N (12.5 lb f ) 73 15 × 10−3 m 2 × 10−3 m −0.140