Numerical and Analytical Methods with MATLAB (1st Ed) Solutions Manual | William Bober | Complete Step-by-Step Solutions

Chapters 2 - 10 Covered
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SOLUTIONS

, Chapter 2 w




Problem 2.1 In FCC the relation between the lattice parameter and the atomic radius is
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4R
 , then α=4.95 Angstroms. On the cube phase (100) correspond 2 atoms (4x1/4+1). Then
w

w w w w w w w w w w w w w w w



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




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




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




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



y than the (100) plane, it is a close-packed plane.
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Problem 2.2 The (100)-type plane closer to the origin is the (002) plane which cuts the z axis at
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½. This has
w w




a a 2R
d(002)   
w

w w


002 2
w

w w w w 2 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
w w w w w w




In the same way
w w w



a a 4R
d(111)   
w

w w
w

1 1 1 3 6 w w




and d(111)=2.85 Angstroms. We see that the close-packed planes have a larger interplanar spacing.
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Problem 2.3. The structure of vanadium is BCC. In this structure, the close-packed direction is
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[111] , which corresponds to the diagonal of the cubic unit cell where there is a consecutive contact
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of spheres (in the model of hard spheres). Furthermore, the number of atoms per unit cell for the BC
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C structure is 2. The first step is to find the lattice parameter α. The density is
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
2 w



 w



3 w




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




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




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, The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2
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atoms. Hence the atomic density of the close-packed direction of vanadium (V) is
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2 2
[111]   10
 3.75109 atoms / m w w w w


 3 3.0810
w

w w 3 w w




The aforementioned atomic density result translates to 3750 atoms/μm or 3.75 atoms/nm.
w w w w w w w w w w w




4R

w


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



2
face of the unit cell. The face comprises ¼ of atoms at each corner plus 1 atom at the center of
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the face. Hence the face consists of 4  () 1  2 atoms. The atomic density of the (100)
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plane is w




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

w


 
w




 2 
The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this triangl
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e is 4R . Using the Pythagorean Theorem, we can calculate the height of the triangle which
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is 2 3R . Thus the area of the triangle is (base  height / 2)  4 3R2 . The equilateral triangle
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comprises 6 of the atoms at each corner and ½ of the atoms at the middle of each side. Thus the
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equilateral triangle consists of 3 ()  3 ()  2 atoms. The atomic density of the (111)
w w w w w w w w w w w w w w w w w w w w w




plane is w




2 1
(111)  
4 3R2 2 3R2
w


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




(111) 2
  1.154  1 w w




(100)
w w w




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




(100) plane. This is important since the plastic deformation of metals (Al, Cu, Ni, γ-
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Fe, etc.) is accomplished with dislocation glide on the close-packed planes.
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Problem 2.5. The ideal c/a ratio in HCP structure results when the atoms of this structure have an a
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rrangement as dense as the atoms of the FCC structure. The distance between the (0001) bases of t
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he HCP structure is c. Using the fact that the (0001) planes of HCP structure
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correspond to the (111) planes of the FCC structure, we get
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c  2d(111) FCC
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w




Where d(111) is the distance between the (111) close-packed planes. We find that
w
w
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, a
d  a a 
 
w




 3 
w w

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



w w w w w w w w w w w w
w w
w



w



 4R 6
a
w




w w


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



6 
a 6
HCP: a  2 R
 w w w




Therefore, the ideal ratio c/a for close packing in HCP structure is equal to 1.63. The c/a ratio for zin
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c (Zn) is 1.86 while for titanium (Ti) is 1.59 (see Table 7.1, Book). This means that the distance bet
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ween the (0001) planes is longer in Zn than in Ti. This fact affects the plastic
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deformation in these metals, since the slip on (0001) planes is easier in Zn than in Ti. Indeed the crit
w w w w w w w w w w w w w w w w w w w




ical shear stress of Zn is only 0.18 MPa, while of Ti is 110 MPa. Due to this, the plastic deformation
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in Ti is performed on (10 1 0) plane, where the critical shear stress is approximately
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49 MPa. Thus in Ti the slip is not performed on the close-
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packed planes of the crystal structure. For more details look at the 7.3 paragraph of the book (plastic
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deformation of single crystals with slip). w w w w w




Problem 2.6. The cell volume of HCP structure is the product of the base area (hexagon)
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8R
multiplied by the height c. The base of hexagon is A  6 R2 3 and the height is c 
w w w w w w w w w w w w
w w w w w w w
. As a
w w


6
result, the cell volume is
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V  24 w w R3

The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing factor is
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4
6  R3
w w



 w w



APF  3   0.74 w
w w w
w

HCP 3
24 R

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


4R
where a 
w w

. Therefore the atomic packing factor of BCC structure is
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3
4
2  R3
w w

w w




APFBCC  3  3  0.68
3 
w w w
w
w w


 4R  8 w w w


 
 



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