Chap 8,9 solutions.pdf

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Materials Science & Engineering
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MSE101H1
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Scott Ramsay

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CHAPTER 8 DEFORMATION AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS Slip Systems 8.3 (a) Compare planar densities (Section 3.15 and Probl em W3.46 [which appears on the book’s Web site]) for the (100), (110), and (111) planes for FCC. (b) Compare planar densities (Problem 3.44) for the (100), (110), and (111) planes for BCC. Solution (a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.12 as PD (FCC) = 1 = 0.177 110 4R 2 2 R2 Furthermore, the planar densities of the (1nd (111) planes are calculated in Homework Problem W3.46, which are as follows: 1 0.25 PD100(FCC) = 2 = 2 4R R 1 0.29 PD111(FCC) = 2 = 2 2R 3 R (b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in Homework Problem 3.44, which are as follows: PD (BCC) = 3 = 0.19 100 16R 2 R 2 3 0.27 PD110(BCC) = 2 = 2 8R 2 R Excerpts from this work may be reproduced bdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textboAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Below is a BCC unit cell, withinwhich is shown a (111) plane. (a) The centers of the three corner atoms, denoted by A, B, and C lie on this plan e. Furthermore, the (111) plane does not pass through the center of atom D, which is located at th e unit cell center. The atomic packing of this plane is presented in the following figure; the corresponding atom positions from the Figure (a) are also noted. (b) Inasmuch as this plane does not pass through the center of atom D, it is not included in the atom count. One sixth of each of the three atoms labeled A, B, a nd C is associated with this plane, which gives an equivalence of one-half atom. In Figure (b) the triangle with A, B, and C at its corners is an equilateral triangle. And, from Figure ( b), xy the area of this triangle is . The triangle edge length, x, is equal to the length of a face diagonal, as indicated in 2 Figure (a). And its length is related to the unit cell edge length,a, as Excerpts from this work may be reproduced by instructorsdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has been aAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2 2 2 2 x = a + a = 2a or x = a 2 4R For BCC, a = (Equation 3.3), and, therefore, 3 4R 2 x = 3 Also, from Figure (b), with respect to the length y we may write 2 y2 + ⎜x ⎟ = x 2 ⎝2 ⎠ x 3 which leads toy = . And, substitution for the above expression for x yields 2 x 3 ⎛4R 2 ⎞⎛ 3⎞ 4R 2 y = = ⎜ ⎟⎜ ⎟ = 2 ⎝ 3 ⎠⎝ 2 ⎠ 2 Thus, the area of this triangle is equal to ⎛ ⎞⎛ ⎞ 2 1 ⎛1⎞ 4R 2 4R 2 8R AREA = 2 xy = ⎝2⎠ ⎜ 3 ⎟⎜ 2 ⎟ = 3 ⎝ ⎠⎝ ⎠ And, finally, the planar density for this (111) plane is 3 PD (BCC) = 0.5 atom = = 0.11 111 8R 2 16R 2 R2 3 Excerpts from this work may be reproduced by indistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.4 One slip system for the BCC crystal structure is { 110}111 . In a manner similar to Figure 8.6b, sketch a {110}-type plane for the BCC structure, represen ting atom positions with circles. Now, using arrows, indicate two different 111 slip directions within this plane. Solution Below is shown the atomic packing for a BCC {110} -type plane. The arrows indicate two different 111 type directions. Excerpts from this work may be reproduced by instrudistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has beAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Slip in Single Crystals 8.6 Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60° and 35°, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa (900 psi), will an applied stress of 12 MPa (1750 psi) cause the single crystal to yield? If not, what stress will be necessary? Solution This problem calls for us to determine whether or not a metal single crystal having a specific orientation and of given critical resolved shear stress will yield. We are given that φ = 60°, λ = 35°, and that the values of the critical resolved shear stress and applied tensile stress ar e 6.2 MPa (900 psi) and 12 MPa (1750 psi), respectively. From Equation 8.2 τR = σ cos φ cos λ = (12 MPa)(cos 60°)(cos 35°) = 4.91 MPa (717 psi) Since the resolved shear stress (4.91 MPa) is less that the critical resolved shear stress (6.2 MPa), the single crystal will not yield. However, from Equation 8.4, the stress at which yielding occurs is τ σ = crss = 6.2 MPa = 15.1 MPa (2200 psi) y cos φ cos λ (cos 60° )(cos 35°) Excerpts from this work may be reproduced by instrudistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has beAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.8 (a) A single crystal of a metal that has the BCC crys tal structure is oriented such that a tensile stress is applied in the [100] direction. If the magnitude of this stress is 4.0 MPa, compute the resolved shear stress in the [11 1] direction on each of the (110), (011), and(101 )planes. (b) On the basis of these resolved shear stress va lues, which slip system(s) is (are) most favorably oriented? Solution (a) This part of the problem asks, for a BCC metal, that we compute the resolved shear stress in the1 1] direction on each of the (110), (011), and (101 ) planes. In order to solve this problem it is necessary to employ Equation 8.2, which means that we first need to solve for the for angles λ and φ for the three slip systems. For each of these thr ee slip systems, theλ will be the same—i.e., the angle between the direction of the applied stress, [100] and the slip directi[11 1]. This angle λ may be determined using Equation 8.6 ⎡ ⎤ −1⎢ u1 2+ v v1 2 + w1 2 ⎥ λ= cos ⎢ 2 2 2 2 2 2 ⎥ ⎢ ()1 + v1 + w1 ()2 + v 2 + w 2 ⎥ ⎣ ⎦ where (for [100]) u = 1 ,v = 0, w = 0, and (for [1 1 1) u = 1, v = –1, w = 1. Therefore, λ is determined as 1 1 1 2 2 2 ⎡ ⎤ −1⎢ (1)(1)+ (0)(−1) + (0)(1) ⎥ λ= cos ⎢ 2 2 2 2 2 2 ⎥ ⎢ []) + (0) + (0) [ ]) + (−1) + (1) ⎥ ⎣ ⎦ −1 ⎛ 1 ⎞ = cos ⎜ ⎟= 54.7° ⎝ 3 ⎠ Let us now determine φ for the angle between the direction of the a pplied tensile stress—i.e., the [100] direction— and the normal to the (110) slip plane—i.e., the [110] direction. Again, using Equation 8.6 where u = ,v = 0, w 1 1 1 = 0 (for [100]), and u = 1, v = 1, w = 0 (for [110]), φ is equal to 2 2 2 ⎡ ⎤ −1⎢ (1)(1)+ (0)(1) + (0)(0) ⎥ φ[100]−[110] = cos ⎢ ⎥ ⎢ []1) + (0) + (0) 2 []) + (1) + (0) 2 ⎥ ⎣ ⎦ ⎛ 1 ⎞ = cos−1 ⎜ ⎟= 45° ⎝ 2 ⎠ Excerpts from this work may be reproduced by instrdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Now, using Equation 8.2 τR = σ cosφ cosλ we solve for the resolved shear stress for this slip system as τ = (4.0 MPa) cos(45°)cos(54.7°) = (4.0 MPa)(0.707)(0.578) = 1.63 MPa R(110)−[11 1] [ ] Now, we must determine the value of φ for the (011)– [11 1]slip system—that is, the angle between the direction of the applied stress, [100], and the normal to the (011) plane—i.e., the [011] direction. Again using Equation 8.6 ⎡ ⎤ −1⎢ (1)(0)+ (0)(1) + (0)(1) ⎥ λ [100]−[011] = cos ⎢ 2 2 2 2 2 2 ⎥ ⎣ []1) + (0) + (0) []0) + (1) + (1) ⎦ = cos−1 (0) = 90° Thus, the resolved shear stress for this (011[11 1] slip system is τR (011)−[11]= (4.0 MPa) c[s(90°)cos(54.7°) = ]4.0 MPa)(0)(0.578) = 0 MPa And, finally, it is necessary to determine the value of φ for the (101 )–[1 1 1]slip system —that is, the angle between the direction of the app lied stress, [100], and the normal to the (101 )plane—i.e., the [101 ] direction. Again using Equation 8.6 ⎡ ⎤ −1 ⎢ (1)(1)+ (0)(0) + (0)(−1) ⎥ λ[100]−[10 1 ] cos ⎢ ⎥ ⎢ []1) + (0) + (0) 2 [ ]) + (0) + (−1) 2 ⎥ ⎣ ⎦ ⎛ 1 ⎞ = cos −1⎜ ⎟ = 45° ⎝ 2⎠ Here, as with the (110)–[11 1] slip system above, the value of φ is 45°, which again leads to Excerpts from this work may be reproduced by instrdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has bAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. τR = (4.0 MPa) c[ ]45°)cos(54.7°) = (4.0 MPa)(0.707)(0.578) = 1.63 MPa (10 )−[11 1] (b) The most favored slip system(s) is (a re) the one(s) that has (have) the largest τ value. Both (110)– R [1 1 1]and (101 )−[11 1] slip systems are most favored since they have the same τ (R.63 MPa), which is greater than the τRvalue for (011)−[11 1] (viz., 0 MPa). Excerpts from this work may be reproduced by instrudistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has bAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.9 The critical resolved shear stress for copper is 0. 48 MPa (70 psi). Determine the maximum possible yield strength for a single crystal of Cu pulled in tension. Solution In order to determine the maximum possible yield strength for a single crystal of Cu pulled in tension, we simply employ Equation 8.5 as σ =2 τ = (2)(0.48 MPa) = 0.96 MPa (140 psi) y crss Excerpts from this work may be reproduced by instructodistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has been Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Strengthening by Grain Size Reduction 8.10 Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries. Answer Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographicmisalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction. Excerpts from this work may be reproduced by instructordistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has been aAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Strain Hardening 8.14 Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross ons). For one specimen, the initial and deformed radii are 15 mm and 12 mm, respectively. The second specimen, with an initial radius of 11 mm, must have the same deformed hardness as the first specimen; compute the second specimen’s radius after deformation. Solution 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 the percent cold workis computed using Equation 8.8 as A − A π r2 −π r 2 %CW = 0 d × 100 = 0 d × 100 A0 πr0 2 2 π (15 mm) −π (12 mm) = π (15 mm) 2 × 100 = 36%CW For the second specimen, the deformed radius is computed using the aboveequation and solving for r d %CW rd = r0 1 − 100 36%CW = (11 mm) 1 − = 8.80 mm 100 Excerpts from this work may be reproduced by indistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Factors That Influence the Mechanical Properties of Semicrystalline Polymers Deformation of Elastomers 8.22 Briefly explain how each of the following influences the tensile or yield strength of a semicrystalline polymer and why: (a) Molecular weight (b) Degree of crystallinity (c) Deformation by drawing (d) Annealing of an undeformed material Solution (a) The tensile strength of a semicrystalline polymer increases with increasing molecular weight. This effect is explained by increased chain entanglements at higher molecular weights. (b) Increasing the degree of crys tallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Again, this is due to enhanced inte rchain bonding and forces; in response to applied stresses, interchain motions are thus inhibited. (c) Deformation by drawing increases the tensile stre ngth of a semicrystalline polymer. This effect is due to the highly oriented chain structure th at is produced by drawing, which give s rise to higher interchain secondary bonding forces. (d) Annealing an undeformed semicrystalline polymer produces an increase in its tensile strength. Excerpts from this work may be reproduced by instructdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has beenAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.23 The tensile strength and number-average molecular weight for two poly(methyl methacrylate) materials are as follows: Tensile Strength Number-Average Molecular (MPa) Weight (g/mol) 50 30,000 150 50,000 Estimate the tensile strength at a number-average molecular weight of 40,000 g/mol. Solution This problem gives us the tensile strengths aassociated number-average molecular weights for two poly(methyl methacrylate) materials and then as ks that we estimate the tensile strMnn= 40,000 g/mol. Equation 8.11 cites the dependen ce of the tensile strenMtn. Thus, using the data provided in the problem statement, we may set up two simultaneous equations fro m which it is possible to solve for the two constants TS ∞ and A. These equations are as follows: A 50 MPa = TS −∞ 30,000 g/mol 150 MPa = TS − A ∞ 50,000 g/mol 6 Thus, the values of the two constants aTS ∞ 300 MPa and A = 7.50 × 10 MPa-g/mol. Substituting these values into Equation 8.11 fon = 40,000 g/mol leads to A TS = TS ∞ 40,000 g/mol 7.50 × 10 MPa - g/mol = 300 MPa − 40,000 g/mol = 112.5 MPa Excerpts from this work may be reproduced bydistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbooAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.24 For each of the following pairs of polymers, do the fo llowing: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for y our choice; and (3) if it is not possible to decide, then state why. (a) Branched and atactic poly(vinyl chloride) with a we ight-average molecular weight of 100,000 g/mol; linear and isotactic poly(vinyl chloride) having aweight-average molecular weight of 75,000 g/mol (b) Random styrene-butadiene copolymer with 5% of po ssible sites crosslinked; block styrene-butadiene copolymer with 10% of possible sites crosslinked (c) Branched polyethylene with a number-average mo lecular weight of 100,000 g/mol; atactic polypropylene with a number-average molecular weight of 150,000 g/mol Solution (a) Yes, it is possible. The linear and isotactic pol y(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, te nsile modulus is independent of molecular weight--the atactic/branched materialhas the higher molecular weight. (b) Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material. A higher degree of crystallinity favor s larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amountof crosslinking also enhances the tensile modulus. (c) No, it is not possible. Branched polyethylen e will tend to have a low de gree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene proba bly also has a relatively low degree of crystallinity; atactic structures al so don't tend to crystallize, and polypropy lene has a more complex repeat unit structure than does polyethylene. Tens ile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight. Excerpts from this work may be reproduced by instructordistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has been Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.25 For each of the following pairs of polymers, plot and label schematic stress–strain curves on the same graph (i.e., make separate plots for parts a, b, and c). (a) Polyisoprene having a number-average molecular weight of 100,000 g/mol and 10% of available sites crosslinked; polyisoprene having a number-average molecular weight of 100,000 g/mol and 20% of available sites crosslinked (b) Syndiotactic polypropylene having a weight-average molecular weight of 100,000 g/mol; atactic polypropylene having a weight-average molecular weight of 75,000 g/mol (c) Branched polyethylene having a num ber-average molecular weight of 90,000 g/mol; heavily crosslinked polyethylene having a number-average molecular weight of 90,000 g/mol Solution (a) Shown below are the stress-strain curves for th e two polyisoprene materials, both of which have a molecular weight of 100,000 g/mol. These two materials are elastomers and will have curves similar to curve C in Figure 7.22. However, the curve for the material having the greater number of crossli nks (20%) will have a higher elastic modulus at all strains. (b) Shown below are the stress-strain curves for th e two polypropylene materials. These materials will most probably display the stress-strain behavior of a normal plastic, curve B in Figure 7.22. However, the syndiotactic polypropylene has a hi gher molecular weight and will also undoubtedly have a higher degree of crystallinity; therefore, it will have a higher strength. Excerpts from this work may be reproduced by instructodistribution on a not-for-profit ba sis for testing or instructiona l purposes only to students enrolled in courses for which the textbook has beenAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (c) Shown below are the stress-strain curves for the two polyethylene mate rials. The branched polyethylene will display the behavior of a normal
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