Solutions Manual for Physical Metallurgy: Principles and Design, 1st Edition - Gregory Haidemenopoulos | ISBN 9781498778558

SOLUTIONS

, Chapter 2 q




Problem 2.1 In FCC the relation between the lattice parameter and the atomic radius is
q q q q q q q q q q q q q q




4R
 , then α=4.95 Angstroms. On the cube phase (100) correspond 2 atoms (4x1/4+1). Then
q

q q q q q q q q q q q q q q q



2
the density of the (100) plane is
q q q q q q




2
(100)   8.2x1012 atoms/mm2
4.95x107
q q




In the (111) plane there are 3/6+3/2=2 atoms. The base of the triangle is 4R and the height 2 3R
q q q q q q q q q q q q q q q q q q q q




After some math we get ρ(111)=9.5x1012 atoms/mm2. We see that the (111) plane has higher density t
q q q q q q q q q q q q q q q q



han the (100) plane, it is a close-packed plane.
q q q q q q q q




Problem 2.2 The (100)-type plane closer to the origin is the (002) plane which cuts the z axis at
q q q q q q q q q q q q q q q q q q



½. This has
q q




a 2R a
  d(002) 
q

q q


002 2
q
2 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
q q q q q q




In the same way
q q q



a a 4R
d(111)   
q

q q
q

1 1 1 3 6
and d(111)=2.85 Angstroms. We see that the close-packed planes have a larger interplanar spacing.
q q q q q q q q q q q q q




Problem 2.3. The structure of vanadium is BCC. In this structure, the close-packed direction is
q q q q q q q q q q q q q q



[111] , which corresponds to the diagonal of the cubic unit cell where there is a consecutive contact of s
q q q q q q q q q q q q q q q q q q q




pheres (in the model of hard spheres). Furthermore, the number of atoms per unit cell for the BCC stru
q q q q q q q q q q q q q q q q q q




cture is 2. The first step is to find the lattice parameter α. The density is
q q q q q q q q q q q q q q q





2 q



 q



3 q




Where  is the Avogadro’s number. Therefore the lattice parameter is
q q q q q q q q q q




250.94
3   a  3.08108 cm  3.081010 m
q q
q
q q q q q q q q
23
5.8 6.02310 q




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, The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2
q q q q q q q q q q q q q q q q q




atoms. Hence the atomic density of the close-packed direction of vanadium (V) is
q q q q q q q q q q q q




2 2
[111]   10
 3.75109 atoms / m q q q q


 3
q

3.0810 3
The aforementioned atomic density result translates to 3750 atoms/μm or 3.75 atoms/nm.
q q q q q q q q q q q




4R

q


Problem 2.4. The lattice parameter for the FCC structure is
q q
. The (100) plane is the q q q q q q q
q q q q q q q



2
face of the unit cell. The face comprises ¼ of atoms at each corner plus 1 atom at the center of
q q q q q q q q q q q q q q q q q q q q



the face. Hence the face consists of 4(1/ 4) 1 2 atoms. The atomic density of the (100)
q q q q q q q q q q q q q q q q q q q q q q




plane is q




2 2 1
(100)   2 
a  4R 
q
2
4R2
q

q


 
q




 2 
The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this triangle i
q q q q q q q q q q q q q q q q q q




s 4R . Using the Pythagorean Theorem, we can calculate the height of the triangle which
q q q q q q q q q q q q q q q




is 2 3R . Thus the area of the triangle is (base height / 2)  4 3R2 . The equilateral triangle
q q q q q q q q q q q q q q q q q q q q q q




comprises 6 of the atoms at each corner and ½ of the atoms at the middle of each side. Thus the
q q q q q q q q q q q q q q q q q q q q



equilateral triangle consists of 3(1/ 6)  3(1/ 2)  2 atoms. The atomic density of the (111)
q q q q q q q q q q q q q q q q q q q q q




plane is q




2 1
(111)  
4 3R2 2 3R2
q




The ratio of the atomic densities is
q q q q q q




(111) 2
 1.154 1 q q




(100)
q q q




Therefore (111)  (100) and specifically the (111) plane has 15% higher atomic density than the
q
q
q
q
q q q q q q q q q q q




(100) plane. This is important since the plastic deformation of metals (Al, Cu, Ni, γ-
q q q q q q q q q q q q q q




Fe, etc.) is accomplished with dislocation glide on the close-packed planes.
q q q q q q q q q q




Problem 2.5. The ideal c/a ratio in HCP structure results when the atoms of this structure have an arra
q q q q q q q q q q q q q q q q q q




ngement as dense as the atoms of the FCC structure. The distance between the (0001) bases of the H
q q q q q q q q q q q q q q q q q q




CP structure is c. Using the fact that the (0001) planes of HCP structure
q q q q q q q q q q q q q




correspond to the (111) planes of the FCC structure, we get
q q q q q q q q q q




c  2d(111) FCC
q q q
q




Where d(111) is the distance between the (111)close-packed planes. We find that
q
q
q q q q q q q q q q




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, a
d  a a 
 
q




 3 
q q

(111)
h2  k 2  l 2 12  12  12  4R
 (111) FCC 
  d
q



q

q
q



q



 4R 6
a
q




q q


2

Thus,

c  c 8R 4
   1.63
q
q q
q q q



6 
a 6
HCP: a  2 R
 q q q




Therefore, the ideal ratio c/a for close packing in HCP structure is equal to 1.63. The c/a ratio for zinc (
q q q q q q q q q q q q q q q q q q q q



Zn) is 1.86 while for titanium (Ti) is 1.59 (see Table 7.1, Book). This means that the distance between
q q q q q q q q q q q q q q q q q q



qthe (0001) planes is longer in Zn than in Ti. This fact affects the plastic
q q q q q q q q q q q q q q




deformation in these metals, since the slip on (0001) planes is easier in Zn than in Ti. Indeed the critic
q q q q q q q q q q q q q q q q q q q




al shear stress of Zn is only 0.18 MPa, while of Ti is 110 MPa. Due to this, the plastic deformation in Ti
q q q q q q q q q q q q q q q q q q q q q q q




is performed on (1010) plane, where the critical shear stress is approximately
q q q q q q q q q q q q q




49 MPa. Thus in Ti the slip is not performed on the close-
q q q q q q q q q q q q



packed planes of the crystal structure. For more details look at the 7.3 paragraph of the book (plastic de
q q q q q q q q q q q q q q q q q q



formation of single crystals with slip). q q q q q




Problem 2.6. The cell volume of HCP structure is the product of the base area (hexagon)
q q q q q q q q q q q q q q q


8R
multiplied by the height c. The base of hexagon is A  6 R2 3 and the height is c 
q q q q q q q q q q q q
q q q q q q q
. As a
q q


6
result, the cell volume is
q q q q




V  24 q q R3

The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing factor is
q q q q q q q q q q q q q q q q q q




4
6  R3
q q



 q q



APF  3   0.74 q
q q q
q

HCP 3
24 R

Regarding the BCC structure, the number of atoms per unit cell is 2 and the cell volume is
q q q q q q q q q q q q q q q q q a3 , q


4R
where a 
q q

. Therefore the atomic packing factor of BCC structure is
q q q q q q q q q q q q



3
4
2  R3
q q

q q




APFBCC  3  3  0.68
3 
q q q
q
q q


 4R  8 q q q


 
 



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