Solutions Manual for Physical Metallurgy: Principles and Design (1st Edition, 2018) by Haidemenopoulos – Covers Chapters (2 - 10)

Chapters 2 - 10 Covered




SOLUTIONS

, Chapter 2


Problem 2.1 In FCC the relation between the lattice parameter and the atomic radius is
4R
 , then α=4.95 Angstroms. On the cube phase (100) correspond 2 atoms (4x1/4+1). Then
2
the density of the (100) plane is
2
r(100) = = 8.2x1012 atoms/mm 2
4.95x10 -7
In the (111) plane there are 3/6+3/2=2 atoms. The base of the triangle is 4R and the height 2 3R
After some math we get ρ(111)=9.5x1012 atoms/mm2. We see that the (111) plane has higher
density than the (100) plane, it is a close-packed plane.


Problem 2.2 The (100)-type plane closer to the origin is the (002) plane which cuts the z axis at
½. This has
a a 2R
d(002) = = =
0 + 0 + 22 2 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.

In the same way

a a 4R
d(111) = = =
1+1+1 3 6
and d(111)=2.85 Angstroms. We see that the close-packed planes have a larger interplanar spacing.



Problem 2.3. The structure of vanadium is BCC. In this structure, the close-packed direction is
[111] , which corresponds to the diagonal of the cubic unit cell where there is a consecutive
contact of spheres (in the model of hard spheres). Furthermore, the number of atoms per unit cell
for the BCC structure is 2. The first step is to find the lattice parameter α. The density is

2


3
Where   is the Avogadro’s number. Therefore the lattice parameter is

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




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,The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2
atoms. Hence the atomic density of the close-packed direction of vanadium (V) is
2 2
[111]    3.75 109 atoms / m
 3 3.08 10 10
3
The aforementioned atomic density result translates to 3750 atoms/μm or 3.75 atoms/nm.
4R
Problem 2.4. The lattice parameter for the FCC structure is   . The (10 0) plane is the
2
face of the unit cell. The face comprises ¼ of atoms at each corner plus 1 atom at the center of
the face. Hence the face consists of 4  (1/ 4)  1  2 atoms. The atomic density of the (10 0)
plane is
2 2 1
 (100)  2
 2

a  4R  4R2
 
 2
The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this
triangle is 4R . Using the Pythagorean Theorem, we can calculate the height of the triangle which
is 2 3R . Thus the area of the triangle is (base  height / 2)  4 3R 2 . The equilateral triangle
comprises 6 of the atoms at each corner and ½ of the atoms at the middle of each side. Thus the
equilateral triangle consists of 3  (1/ 6)  3  (1/ 2)  2 atoms. The atomic density of the (111)
plane is
2 1
(111)  2

4 3R 2 3R 2
The ratio of the atomic densities is
(111) 2
  1.154  1
(100) 3

Therefore  (111)   (100) and specifically the (111) plane has 15% higher atomic density than the
(10 0) plane. This is important since the plastic deformation of metals (Al, Cu, Ni, γ-Fe, etc.) is
accomplished with dislocation glide on the close-packed planes.
Problem 2.5. The ideal c/a ratio in HCP structure results when the atoms of this structure have
an arrangement as dense as the atoms of the FCC structure. The distance between the (0 0 01)
bases of the HCP structure is c. Using the fact that the (0 0 01) planes of HCP structure
correspond to the (111) planes of the FCC structure, we get

c  2 d(111) 
FCC


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



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, a a a 
d (111)   
h k l
2 2 2
1 1 1
2 2 2
3
 4R
   d (111)  FCC 
a
4R  6
2 

Thus,
8R 
c  c 4
6    1.63
 a 6
HCP : a  2 R 

Therefore, the ideal ratio c/a for close packing in HCP structure is equal to 1.63. The c/a ratio for
zinc (Zn) is 1.86 while for titanium (Ti) is 1.59 (see Table 7.1, Book). This means that the
distance between the (0 0 01) planes is longer in Zn than in Ti. This fact affects the plastic
deformation in these metals, since the slip on (0 0 01) planes is easier in Zn than in Ti. Indeed the
critical shear stress of Zn is only 0.18 MPa, while of Ti is 110 MPa. Due to this, the plastic
deformation in Ti is performed on (10 1 0) plane, where the critical shear stress is approximately
49 MPa. Thus in Ti the slip is not performed on the close-packed planes of the crystal structure.
For more details look at the 7.3 paragraph of the book (plastic deformation of single crystals with
slip).


Problem 2.6. The cell volume of HCP structure is the product of the base area (hexagon)
8R
multiplied by the height c. The base of hexagon is A  6 R 2 3 and the height is c  . As a
6
result, the cell volume is

V  24 2 R 3

The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing factor is
4
6  R3 
APFHCP  3 3
  0.74
24 2 R 3 2

Regarding the BCC structure, the number of atoms per unit cell is 2 and the cell volume is a 3 ,
4R
where a  . Therefore the atomic packing factor of BCC structure is
3
4
2  R3  3
APFBCC  3   0.68
3
 4R  8
 
 3



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