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7-6 7.6 (a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.11 as PD110(FCC) =14R22=0.177R2 Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.53, which are as follows: PD100(FCC) = 14R2=0.25R2 PD111(FCC) =12R23=0.29R2 (b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in Homework Problem 3.54, which are as follows: PD100(BCC) =316R2=0.19R2 PD110(BCC) =38R22=0.27R2 Below is a BCC unit cell, within which is shown a (111) plane. (a) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 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.7-7 The centers of the three corner atoms, denoted by A, B, and C lie on this plane. Furthermore, the (111) plane does not pass through the center of atom D, which is located at the 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, and 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), the area of this triangle is xy2. The triangle edge length, x, is equal to the length of a face diagonal, as indicated in Figure (a). And its length is related to the unit cell edge length, a, as x2= a2+ a2= 2a2 or x=a 2 For BCC, a =4R3 (Equation 3.3), and, therefore, x =4R 23 Also, from Figure (b), with respect to the length y we may write Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 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.7-29 7.23 (a) Perhaps the easiest way to solve for σ0 and ky in Equation 7.7 is to pick two values each of σy and d-1/2 from Figure 7.15, and then solve two simultaneous equations, which may be set up. For example d-1/2 (mm) -1/2 σy (MPa) 4 75 12 175 The two equations are thus 75 =σ0+4ky 175 =σ0+12ky Solution of these equations yield the values of ky= 12.5 MPa (mm)1/21810 psi(mm)1/2[] σ0 = 25 MPa (3630 psi) (b) When d = 2.0 x 10-3 mm, d-1/2 = 22.4 mm-1/2, and, using Equation 7.7, σy= σ0+ kyd-1/2 = (25 MPa) + 12.5 MPa (mm)1/2⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (22.4 mm-1/2) = 305 MPa (44,200 psi) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 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.7-16 7.13 We are asked to compute the critical resolved shear stress for Zn. As stipulated in the problem, φ = 65°, while possible values for λ are 30°, 48°, and 78°. (a) Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the largest cos λ. Cosines for the possible λ values are given below. cos(30°) = 0.87 cos(48°) = 0.67 cos(78°) = 0.21 Thus, the slip direction is at an angle of 30° with the tensile axis. (b) From Equation 7.4, the critical resolved shear stress is just τcrss=σy(cos φ cos λ)max = (2.5 MPa) cos(65°) cos(30°)[]= 0.90 MPa (130 psi) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 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.7-9 7.7 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 instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 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


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