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Resolução Callister Capítulo 3 Engenharia dos Materiais

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And, solving for v from Equation 3.6c leads to 
 
 v = − (u + t) 
 
which, when substituted into the above expression for u’ yields 
 
 u' = v + 2u = − u − t + 2u = u − t 
 
 In solving for an expression for v’, we begin with the one of the above expressions for this parameter—i.e., 
 
 v' = 2u' − 3u 
 
Now, substitution of the above expression for u’ into this equation leads to 
 
 vÕ= 2uÕ− 3u = (2)(u − t) − 3u = − u −2t 
 
And solving for u from Equation 3.6c gives 
 
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 3-43 
 u = − v − t 
 
which, when substituted in the previous equation results in the following expression for v’ 
 
 vÕ= − u −2t = − (− v − t) − 2t = v − t 
 
And, of course from Equation 3.6d 
 
w’ = w 
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 3-44 
 Crystallographic Planes 
 
 3.37 (a) We are asked to draw a (021 ) plane within an orthorhombic unit cell. First remove the three 
indices from the parentheses, and take their reciprocals--i.e., ∞, 1/2, and -1. This means that the plane parallels the 
x-axis, intersects the y-axis at b/2, and intersects the z-axis at -c. The plane that satisfies these requirements has 
been drawn within the orthorhombic unit cell below. (For orthorhombic, a ≠ b ≠ c, and α = β = γ = 90°.) 
 
 
 
 (b) A (200) plane is drawn within the monoclinic cell shown below. We first remove the parentheses and 
take the reciprocals of the indices; this gives 1/2, ∞, and ∞,. Thus, the (200) plane parallels both y- and z-axes, and 
intercepts the x-axis at a/2, as indicated in the drawing. (For monoclinic, a ≠ b ≠ c, and α = γ = 90° ≠ β.) 
 
 
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 3-45 
 3.38 This problem calls for specification of the indices for the two planes that are drawn in the sketch. 
 Plane 1 is a (211) plane. The determination of its indices is summarized below. 
 
 x y z
 Intercepts a/2 b c 
 Intercepts in terms of a, b, and c 1/2 1 1 
 Reciprocals of intercepts 2 1 1 
 Enclosure (211) 
 
 Plane 2 is a (02 0) plane, as summarized below. 
 
 x y z
 
 Intercepts ∞a -b/2 ∞c 
 Intercepts in terms of a, b, and c ∞ -1/2 ∞ 
 Reciprocals of intercepts 0 -2 0 
 Enclosure (02 0) 
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 3-46 
 3.39 The planes called for are plotted in the cubic unit cells shown below. 
 
 
 
 
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 3-47 
 3.40 For plane A we will leave the origin at the unit cell as shown. If we extend this plane back into the 
plane of the page, then it is a (111 ) plane, as summarized below. 
 
 x y z
 Intercepts a b – c 
 Intercepts in terms of a, b, and c 1 1 – 1 
 Reciprocals of intercepts 1 1 – 1 
 Reduction not necessary 
 Enclosure (111 ) 
 
[Note: If we move the origin one unit cell distance parallel to the x axis and then one unit cell distance parallel to 
the y axis, the direction becomes (1 1 1) ]. 
 
 For plane B we will leave the origin of the unit cell as shown; this is a (230) plane, as summarized below. 
 
 x y z
 Intercepts 
 
a
2
 
 
b
3
 ∞c 
 Intercepts in terms of a, b, and c 
 
1
2
 
 
1
3
 ∞ 
 Reciprocals of intercepts 2 3 0 
 Enclosure (230) 
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 3-48 
 3.41 For plane A we will move the origin of the coordinate system one unit cell distance to the right along 
the y axis; thus, this is a (11 0) plane, as summarized below. 
 
 x y z
 Intercepts 
 
a
2
 – 
 
b
2
 ∞ c 
 Intercepts in terms of a, b, and c 
 
1
2
 – 
 
1
2
 ∞ 
 Reciprocals of intercepts 2 – 2 0 
 Reduction 1 – 1 0 
 Enclosure (11 0) 
 
 For plane B we will leave the origin of the unit cell as shown; thus, this is a (122) plane, as summarized 
below. 
 
 x y z
 Intercepts a 
 
b
2
 
 
c
2
 
 Intercepts in terms of a, b, and c 1 
 
1
2
 
 
1
2
 
 Reciprocals of intercepts 1 2 2 
 Reduction not necessary 
 Enclosure (122) 
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 3-41 
 
t = − (u+ v) = − 4
3
−
2
3
⎛ 
⎝ 
⎜ 
⎞ 
⎠ 
⎟ = −
2
3
 
 
 w = w' = 1 
 
Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result 
that the direction D is a [42 2 3] direction. 
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted 
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 3-49 
 3.42 For plane A since the plane passes through the origin of the coordinate system as shown, we will 
move the origin of the coordinate system one unit cell distance vertically along the z axis; thus, this is a (211 ) 
plane, as summarized below. 
 
 x y z
 Intercepts 
 
a
2
 b – c 
 Intercepts in terms of a, b, and c 
 
1
2
 1 – 1 
 Reciprocals of intercepts 2 1 – 1 
 Reduction not necessary 
 Enclosure (211 ) 
 
 For plane B, since the plane passes through the origin of the coordinate system as shown, we will move the 
origin one unit cell distance vertically along the z axis; this is a (021 ) plane, as summarized below. 
 
 x y z
 Intercepts ∞ a