Solutions Manual for Physical Metallurgy Principles and Design 1st Edition by Haidemenopoulos

Chapters 2 - 10 Covered




SOLUTIONS

, Chapter 2 b




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




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


b b b b b b b b b b b b b b b




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




2
(100) = = 8.2x1012 atoms/mm2
4.95x10−7
b b




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




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




than the (100) plane, it is a close-packed plane.
b b b b b b b b b




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




½. This has b b




a a 2R
d(002) = = =
b


b b




0 + 0 + 22 2 2
b




Setting R=1.749 Angstromswe get d(002)=2.745 Angstroms.
b b b b b b




In the same way
b b b




a = 4R
d(111) = =
b


b
b
a 6
1+ 1+ 1 3 b




and d(111)=2.85 Angstroms. We see that the close-packed planes have a larger interplanar spacing.
b b b b b b b b b b b b b




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




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




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




structure is 2. The first step is to find the lattice parameter α. The density is
b b b b b b b b b b b b b b b b





2 b




= b



3 b




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




250.94
3 =  a = 3.0810−8cm = 3.0810−10m
b b
b

b b b b b b b b

23
5.8 6.02310 b




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




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




2 2
[111] = = = 3.75109 atoms / m
 3 3.0810−10 3
b b b b
b




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




4R
Problem 2.4. The lattice parameter for the FCC structure is  =
b




b
. The (100) plane is the b b b b b b b b
b b b b b b b




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




the face. Hence the face consists of 4(1/ 4)+1= 2 atoms. The atomic density of the (100)
b b b b b b b b b b b b b b b b b b b b b b




plane is b




2 2 1
(100) = = 2 =
a  4R 
b
2
4R2
b




 
b b




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




is4R . Using the Pythagorean Theorem, we can calculate the height of the triangle which
b b b b b b b b b b b b b b b b




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




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




equilateral triangle consists of 3(1/ 6)+3(1/ 2) = 2 atoms. The atomic density of the (111)
b b b b b b b b b b b b b b b b b b b b b




plane is b




2 1
(111) = =
4 3R2 2 3R2
b




The ratio of the atomic densities is
b b b b b b




(111) 2
= =1.154 1
b
b




(100)
b b b




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




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




accomplished with dislocation glide on the close-packed planes.
b b b b b b b b




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




arrangement as dense as the atoms of the FCC structure. The distance between the (0001) bases of
b b b b b b b b b b b b b b b b b




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




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




c = 2d(111) FCC
b b b
b




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




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, a
d = a a 
= =
b




3
b b


(111)
h2 + k 2 + l 2 12 + 12 + 12 4R
  d =
b




 (111)
b


b




FCC
b



4R 6
a=
b
b





b b




2

Thus,

c=  c 8R 4
 = = 1.63
b

b b
b b b




6 
a 6
HCP: a = 2R b b b




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




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




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




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




critical shear stress of Zn is only 0.18 MPa, while of Ti is 110 MPa. Due to this, the plastic
b b b b b b b b b b b b b b b b b b b b




deformation in Ti is performed on (1010) plane, where the critical shear stress is approximately
b b b b b b b b b b b b b b b b b




49 MPa. Thus in Ti the slip is not performed on the close-packed planes of the crystal structure. For
b b b b b b b b b b b b b b b b b b




more details look at the 7.3 paragraph of the book (plastic deformation of single crystals with slip).
b b b b b b b b b b b b b b b b b




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




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




6
result, the cell volume is
b b b b




V = 24 b b R3

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




4
6  R3
b b




3  b b




APF = = = 0.74 b
b b b

b



HCP 3
24 R

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




4R
where a =
b b


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




3
4
2  R3
b b


b b




APFBCC = 3  3 = 0.68
3 = b
b b b
b
b




 4R  8
 
b b b




 




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