Fundamentals of Materials Science and Engineering Solutions Ch 7-8

CH A PTE R 7 MECHA NICAL PROPERTIES 7.4 We are asked to compute the maximum length of a cylindrical titanium alloy specimen that is deformed elastically in tension. For a cylindrical specimen 2 do 2 where do is the original diameter. Combining E quations (7.1), (7.2), and (7.5) and solving for lo leads to Ao = ␲ lo = = E ␲d 2o l 4F (107 × 109 N/m 2 )( ␲)(3.8 × 10−3 m) 2 (0.42 × 10−3 m) (4)(2000 N) = 0.25 m = 250 mm (10 in .) 7.9 This problem asks that we calculate the elongation l of a specimen of steel the stress-strain behavior of which is shown in Figure 7.33. First it becomes necessary to compute the stress when a load of 23,500 N is applied as ␴= F = Ao F ␲ 23500 N 2 = 2d o 2 = 300 MPa (44,400 psi) 10 × 10−3 m ␲ 2 R eferring to Figure 7.33, at this stress level we are in the elastic region on the stress-strain curve, which corresponds to a strain of 0.0013. Now, utilization of E quation (7.2) yields l = lo = (0.0013)(75 mm) = 0.10 mm (0.004 in.) 7.14 (a) We are asked, in this portion of the problem, to determine the elongation of a cylindrical specimen of aluminum. U sing E quations (7.1), (7.2), and (7.5) F  =E o 2 ␲ l lo d4 Or l= = 4Flo ␲do2E (4)(48,800 N)(200 × 10−3 m) = 0.50 mm (0.02 in .) (␲)(19 × 10−3 m) 2 (69 × 109 N/m 2 ) (b) We are now called upon to determine the change in diameter, ␯=− x z =− 22 d/d o l/lo d. U sing E quation (7.8) From Table 7.1, for A l, ␯ = 0.33. Now, solving for d yields ␯ ld o (0.33)(0.50 mm)(19 mm) =− d=− lo 200 mm = −1.6 × 10−2 mm (−6.2 × 10−4 in .) The diameter will decrease. 7.16 This problem asks that we compute Poisson’s ratio for the metal alloy. From E quations (7.5) and (7.1) Since the transverse strain ␴ F/A o = = E E = z x F = 2E 4F ␲do2E do ␲ 2 is just x = d do and Poisson’s ratio is defined by E quation (7.8) then ␯=− x z d /d o =−  =− d o d␲E 4F 4F2 ␲d oE (8 × 10−3 m)( −5 × 10−6 m)( ␲)(140 × 109 N/m 2 ) =− (4)(15,700 N) = 0.280 7.21 (a) This portion of the problem asks that we compute the elongation of the brass specimen. The first calculation necessary is that of the applied stress using E quation (7.1), as ␴= F = Ao F ␲ 5000 N 2 = d2o 2 = 177 MPa (25,000 psi) 6 × 10−3 m ␲ 2 From the stress-strain plot in Figure 7.12, this stress corresponds to a strain of about 2.0 × 10−3 . From the definition of strain, E quation (7.2), l = lo = (2.0 × 10−3 )(50 mm) = 0.10 mm (4 × 10−3 in .) (b) In order to determine the reduction in diameter d, it is necessary to use E quation (7.8) and the definition of lateral strain (i.e., x = d/do ) as follows: d = do x = −d o ␯ z = −(6 mm)(0.30)(2.0 × 10−3 ) = −3.6 × 10−3 mm (−1.4 × 10−4 in .) 7.27 This problem asks us to determine the deformation characteristics of a steel specimen, the stressstrain behavior of which is shown in Figure 7.33. (a) In order to ascertain whether the deformation is elastic or plastic, we must first compute the stress, then locate it on the stress-strain curve, and, finally, note whether this point is on the elastic 23 or plastic region. Thus, ␴= F = Ao 44500 N 2 = 565 MPa (80,000 psi) 10 × 10−3 m 2 ␲ The 565 MPa point is past the linear portion of the curve, and, therefore, the deformation will be both elastic and plastic. (b) This portion of the problem asks us to compute the increase in specimen length. From the stress-strain curve, the strain at 565 MPa is approximately 0.008. Thus, from E quation (7.2) l = lo = (0.008)(500 mm) = 4 mm (0.16 in.) 7.29 This problem calls for us to make a stress-strain plot for aluminum, given its tensile load-length data, and then to determine some of its mechanical characteristics. (a) The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for the second, the curve extends just beyond the elastic region of deformation. 300 Stress (MPa) 200 100 0 0.000 0.002 0.004 0.006 Strain 24 0.008 0.010 0.012 (b) The elastic modulus is the slope in the linear elastic region as E= ␴ = 200 MPa − 0 MPa = 62.5 × 103 MPa = 62.5 G Pa (9.1 × 106 psi) 0.0032 − 0 (c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at approximately 285 MPa (41,000 psi). (d) The tensile strength is approximately 370 MPa (54,000 psi), corresponding to the maximum stress on the complete stress-strain plot. (e) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by onehundred. The total fracture strain at fracture is 0.165; subtracting out the elastic strain (which is about 0.005) leaves a plastic strain of 0.160. Thus, the ductility is about 16% EL. (f) From E quation (7.14), the modulus of resilience is just Ur = ␴y2 2E which, using data computed in the problem yields a value of Ur = (285 MPa) 2 = 6.5 × 105 J/m 3 (93.8 in.-lb f /in .3 ) (2)(62.5 × 103 MPa) 7.32 This problem asks us to calculate the moduli of resilience for the materials having the stress-strain behaviors shown in Figures 7.12 and 7.33. A ccording to E quation (7.14), the modulus of resilience U r is a function of the yield strength and the modulus of elasticity as Ur = ␴y2 2E The values for ␴y and E for the brass in Figure 7.12 are 250 MPa (36,000 psi) and 93.9 G Pa (13.6 × 106 psi), respectively. Thus Ur = (250 MPa) 2 = 3.32 × 105 J/m 3 (47.6 in.-lb f /in .3 ) (2)(93.9 × 103 MPa) 7.41 For this problem, we are given two values of T and ␴T , from which we are asked to calculate the true stress which produces a true plastic strain of 0.25. E mploying E quation (7.19), we may set up two simultaneous equations with two unknowns (the unknowns being K and n), as log(50,000 psi) = log K + n log(0.10) log(60,000 psi) = log K + n log(0.20) From these two expressions, n= log(50,000) − log(60,000) = 0.263 log(0.1) − log(0.2) log K = 4.96 or K = 91,623 psi Thus, for T = 0.25 ␴T = K( T) 2 = (91,623 psi)(0.25) 0.263 = 63,700 psi (440 MPa) 25 7.45 This problem calls for us to utilize the appropriate data from Problem 7.29 in order to determine the values of n and K for this material. From E quation (7.32) the slope and intercept of a log ␴T versus log T plot will yield n and log K, respectively. However, E quation (7.19) is only valid in the region of plastic deformation to the point of necking; thus, only the 7th, 8th, 9th, and 10th data points may be utilized. The log-log plot with these data points is given below. 2.60 2.58 log true stress (MPa) 2.56 2.54 2.52 2.50 2.48 2.46 -2.2 -2.0 -1.8 -1.6 -1.4 -1.2 log true strain The slope yields a value of 0.136 for n, whereas the intercept gives a value of 2.7497 for log K, and thus K = 562 MPa. 7.50 For this problem, the load is given at which a circular specimen of aluminum oxide fractures when subjected to a three-point bending test; we are then are asked to determine the load at which a specimen of the same material having a square cross-section fractures. It is first necessary to compute the flexural strength of the alumina using E quation (7.20b), and then, using this value, we may calculate the value of Ff in E quation (7.20a). From E quation (7.20b) ␴fs = = FfL ␲R 3 (950 N)(50 × 10−3 m) = 352 × 106 N/m 2 = 352 MPa (50,000 psi) (␲)(3.5 × 10−3 m) 3 Now, solving for Ff from E quation (7.20a), realizing that b = d = 12 mm, yields Ff = = 2␴fs d 3 3L (2)(352 × 106 N/m 2 )(12 × 10−3 m) 3 = 10,100 N (2165 lb f ) (3)(40 × 10−3 m) 7.54* (a) This part of the problem asks us to determine the flexural strength of nonporous MgO assuming that the value of n in E quation (7.22) is 3.75. Taking natural logarithms of both sides of 26 E quation (7.22) yields ln ␴fs = ln ␴o − nP In Table 7.2 it is noted that for P = 0.05, ␴fs = 105 MPa. For the nonporous material P = 0 and, ln ␴o = ln ␴fs . Solving for ln ␴o from the above equation gives and using these data gives ln ␴o = ln ␴fs + nP = ln(105 MPa) + (3.75)(0.05) = 4.841 or ␴o = e 4.841 = 127 MPa (18,100 psi) (b) Now we are asked to compute the volume percent porosity to yield a ␴fs of 62 MPa (9000 psi). Taking the natural logarithm of E quation (7.22) and solving for P leads to ln ␴o − ln ␴fs n ln(127 MPa) − ln(62 MPa) = 3.75 = 0.19 or 19 vol% P= 7.65 This problem calls for estimations of Brinell and R ockwell hardnesses. (a) For the brass specimen, the stress-strain behavior for which is shown in Figure 7.12, the tensile strength is 450 MPa (65,000 psi). From Figure 7.31, the hardness for brass corresponding to this tensile strength is about 125 H B or 70 H R B. 7.70 The working stresses for the two alloys, the stress-strain behaviors of which are shown in Figures 7.12 and 7.33, are calculated by dividing the yield strength by a factor of safety, which we will take to be 2. For the brass alloy (Figure 7.12), since ␴y = 250 MPa (36,000 psi), the working stress is 125 MPa (18,000 psi), whereas for the steel alloy (Figure 7.33), ␴y = 570 MPa (82,000 psi), and, therefore, ␴w = 285 MPa (41,000 psi). 27 CH A PTE R 8 D EFORMATION A ND STRENGTHENING MECHA NISMS 8.7 In the manner of Figure 8.6b, we are to sketch the atomic packing for a BCC {110} type plane, and with arrows indicate two different 111 type directions. Such is shown below. 8.10* We are asked to compute the Schmid factor for an FCC crystal oriented with its [100] direction parallel to the loading axis. With this scheme, slip may occur on the (111) plane and in the [110] direction as noted in the figure below. z [111] y x _ [110] [100] The angle between the [100] and [110] directions, ␭, is 45◦ . For the (111) plane, the angle √ between its normal (which is the [111] direction) and the [100] direction, ␾, is tan−1 ( a 2 ) = 54.74◦ ; a therefore cos ␭ cos ␾ = cos(45◦ ) cos(54.74◦ ) = 0.408 8.20 We are asked to determine the grain diameter for an iron which will give a yield strength of 205 MPa (30,000 psi). The best way to solve this problem is to first establish two simultaneous expressions of E quation (8.5), solve for ␴o and ky , and finally determine the value of d when ␴y = 205 MPa. The data pertaining to this problem may be tabulated as follows: 28 ␴y 135 MPa 260 MPa d (mm) d −1/2 (mm) −1/2 5 × 10−2 8 × 10−3 4.47 11.18 The two equations thus become 135 MPa = ␴o + (4.47)k y 260 MPa = ␴o + (11.18)k y which yield the values, ␴o = 51.7 MPa and ky = 18.63 MPa(mm) 1/2 . A t a yield strength of 205 MPa 205 MPa = 51.7 MPa + [18.63 MPa(mm) 1/2 ]d −1/2 or d−1/2 = 8.23 (mm) −1/2 , which gives d = 1.48 × 10−2 mm. 8.25 This problem stipulates that two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas. For one specimen, the initial and deformed radii are 16 mm and 11 mm, respectively. The second specimen with an initial radius of 12 mm is to have the same deformed hardness as the first specimen. We are asked to compute the radius of the second specimen after deformation. In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to the same percent cold work. For the first specimen % CW = 2 2 A o − A d × 100 = ␲r o − ␲r d × 100 Ao ␲r 2 o = ␲(16 mm) 2 − ␲(11 mm) 2 ␲(16 mm) 2 × 100 = 52.7% CW For the second specimen, the deformed radius is computed using the above equation and solving for rd as rd = ro 1 − 100 % CW = 8.25 mm 52.7% CW 100 = (12 mm) 1 − 8.27 This problem calls for us to calculate the precold-worked radius of a cylindrical specimen of copper that has a cold-worked ductility of 25% E L. From Figure 8.19(c), copper that has a ductility of 25% E L will have experienced a deformation of about 11% CW. For a cylindrical specimen, E quation (8.6) becomes  % CW = 2 o − ␲r d  2 ␲r 2␲r Since rd = 10 mm (0.40 in.), solving for ro yields rd ro = 1− % 100 CW = × 100 o 10 mm = 10.6 mm (0.424 in .) 1− 100 11.0 29 8.35 In this problem, we are asked for the length of time required for the average grain size of a brass material to increase a specified amount using Figure 8.25. (a) A t 500◦ C, the time necessary for the average grain diameter to increase from 0.01 to 0.1 mm is approximately 3500 min. (b) A t 600◦ C the time required for this same grain size increase is approximately 150 min. 8.45* This problem gives us the tensile strengths and associated number-average molecular weights for two polymethyl methacrylate materials and then asks that we estimate the tensile strength for Mn = 30,000 g/mol. E quation (8.9) provides the dependence of the tensile strength on Mn . Thus, using the data provided in the problem, we may set up two simultaneous equations from which it is possible to solve for the two constants TS∞ and A . These equations are as follows: 107 MPa = TS∞ − A 40000 g/mol 170 MPa = TS∞ − A 60000 g/mol Thus, the values of the two constants are TS∞ = 296 MPa and A = 7.56 × 106 MPa-g/mol. Substituting these values into an equation for which Mn = 30,000 g/mol leads to A 30000 g/mol TS = TS∞ − = 296 MPa − 7.56 × 106 MPa-g/mol 30000 g/mol = 44 MPa 8.54 This problem asks that we compute the fraction of possible crosslink sites in 10 kg of polybutadiene when 4.8 kg of S is added, assuming that, on the average, 4.5 sulfur atoms participate in each crosslink bond. G iven the butadiene mer unit in Table 4.5, we may calculate its molecular weight as follows: A (butadiene) = 4(A C ) + 6(A H ) = (4)(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol which means that in 10 kg of butadiene there are 10000 g = 184.9 mol. 54.09 g/mol For the vulcanization of polybutadiene, there are two possible crosslink sites per mer—one for each of the two carbon atoms that are doubly bonded. Furthermore, each of these crosslinks forms a bridge between two mers. Therefore, we can say that there is the equivalent of one crosslink per mer. Therefore, let us now calculate the number of moles of sulfur (nsulfur) that react with the butadiene, by taking the mole ratio of sulfur to butadiene, and then dividing this ratio by 4.5 atoms per crosslink; this yields the fraction of possible sites that are crosslinked. Thus n sulfur = 4800 g 32.06 g/mol = 149.7 mol A nd fraction sites crosslinked = 30 149.7 mol 184.9 mol 4.5 = 0.180 8.D 1 This problem calls for us to determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and a ductility of at least 12% E L. A ccording to Figure 7.31, a Brinell hardness of 225 corresponds to a tensile strength of 800 MPa (116,000 psi). Furthermore, from Figure 8.19(b), in order to achieve a tensile strength of 800 MPa, deformation of at least 13% CW is necessary. Finally, if we cold work the steel to 13% CW, then the ductility is reduced to only 14% E L from Figure 8.19(c). Therefore, it is possible to meet both of these criteria by plastically deforming the steel. 8.D 6 This problem stipulates that a cylindrical rod of copper originally 16.0 mm in diameter is to be cold worked by drawing; a cold-worked yield strength in excess of 250 MPa and a ductility of at least 12% E L are required, whereas the final diameter must be 11.3 mm. We are to explain how this is to be accomplished. Let us first calculate the percent cold work and attendant yield strength and ductility if the drawing is carried out without interruption. From E quation (8.6)  2  2 do dd −␲ % CW = ␲ 2 × 100 2 2  do ␲ 2 2 2  16 mm 11.3 mm − ␲ 2 × 100 = 50% CW 2 2  16 mm 2 ␲ A t 50% CW, the copper will have a yield strength on the order of 330 MPa (48,000 psi), Figure 8.19(a), which is adequate; however, the ductility will be about 4% E L, Figure 8.19(c), which is insufficient. Instead of performing the drawing in a single operation, let us initially draw some fraction of the total deformation, then anneal to recrystallize, and, finally, cold work the material a second time in order to achieve the final diameter, yield strength, and ductility. R eference to Figure 8.19(a) indicates that 21% CW is necessary to give a yield strength of 250 MPa. Similarly, a maximum of 23% CW is possible for 12% E L [Figure 8.19(c)]. The average of these two values is 22% CW, which we will use in the calculations. If the final diameter after the first drawing is d′o , then  = ␲  o d′ 22% CW = ␲ 2 2 2  11.3 −␲ × 100 o 2 2  ′ d ␲ 2 A nd, solving for d′o yields d′o = 12.8 mm (0.50 in.). 31