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
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x 2 a 2 a 2 2 a 2 or x a 2 For BCC, a 4 R 3 (Equation 3.3), and, therefore, x 4 R 2 3 Also, from Figure ( b ), with respect to the length y we may write y 2 x 2 2 x 2 which leads to y x 3 2 . And, substitution for the above expression for x yields y x 3 2 4 R 2 3 3 2 4 R 2 2 Thus, the area of this triangle is equal to AREA 1 2 xy 1 2 4 R 2 3 4 R 2 2 8 R 2 3 And, finally, the planar density for this (111) plane is PD 111 (BCC) 0.5 atom 8 R 2 3 3 16 R 2 0.11 R 2
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7.7 One slip system for the BCC crystal structure is 110 111 . In a manner similar to Figure 7.6b, sketch a 110  -type plane for the BCC structure, representing 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.
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7.9 Equations 7.1a and 7.1b, expressions for Burgers vectors for FCC and BCC crystal structures, are of the form b a 2 uvw where a is the unit cell edge length. Also, since the magnitudes of these Burgers vectors may be determined from the following equation: b a 2 u 2 v 2 w 2
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This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.

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

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