Chap 07 Solns-6E

# Chap 07 Solns-6E - CHAPTER 7 DISLOCATIONS AND STRENGTHENING...

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CHAPTER 7 DISLOCATIONS AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS 7.1 The dislocation density is just the total dislocation length per unit volume of material (in this case per cubic millimeters). Thus, the total length in 1000 mm 3 of material having a density of 10 5 mm -2 is just 10 5 mm -2 ( 29 1000 μμ 3 ( 29 = 10 8 μμ = 10 5 μ = 62 μι Similarly, for a dislocation density of 10 9 mm -2 , the total length is 10 9 mm -2 ( 29 1000 μμ 3 ( 29 = 10 12 μμ = 10 9 μ = 6.2 ξ 10 5 μι 7.2 When the two edge dislocations become aligned, a planar region of vacancies will exist between the dislocations as: 7.3 It is possible for two screw dislocations of opposite sign to annihilate one another if their dislocation lines are parallel. This is demonstrated in the figure below. 172

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7.4 For the various dislocation types, the relationships between the direction of the applied shear stress and the direction of dislocation line motion are as follows: edge dislocation--parallel screw dislocation--perpendicular mixed dislocation--neither parallel nor perpendicular 7.5 (a) A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs. (b) All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure. 7.6 (a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation (3.12) as PD 110 (FCC) = 1 4 Ρ 2 2 = 0.177 Ρ 2 Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.47, which are as follows: PD 100 (FCC) = 1 4R 2 = 0.25 Ρ 2 PD 111 (FCC) = 1 2 3 = 0.29 Ρ 2 (b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in Homework Problem 3.48, which are as follows: PD 100 (BCC) = 3 16 R 2 = 0.19 Ρ 2 PD 110 (BCC) = 3 2 2 = 0.27 Ρ 2 173
Below is a BCC unit cell, within which is shown a (111) plane. (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.

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Chap 07 Solns-6E - CHAPTER 7 DISLOCATIONS AND STRENGTHENING...

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