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# HW8 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 08 DISLOCATIONS AND STRENGTHENING MECHANISMS 7.6 (a) Compare planar densities (Section 3.11 and Problem 3.54) for the (100), (110), and (111) planes for FCC. (b) Compare planar densities (Problem 3.55) 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 3.11 as PD 110 (FCC) 1 4 R 2 2 0.177 R 2 Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.54, which are as follows: PD 100 (FCC) = 1 4 R 2 0.25 R 2 PD 111 (FCC) 1 2 R 2 3 0.29 R 2 (b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in Homework Problem 3.55, which are as follows: PD 100 (BCC) = 3 16 R 2 0.19 R 2 PD 110 (BCC) 3 8 R 2 2 0.27 R 2 Below is a BCC unit cell, within which is shown a (111) plane.

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( a ) 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 xy 2 . 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