****iNSTANT DOWNLOAD****PDF***Solutions Manual for Physical Metallurgy: Principles and Design (1st Edition) by Gregory N. Haidemenopoulos

Chapters 2 - 10 Covered
v v v v




SOLUTIONS

, Chapter 2 v




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




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


v v v v v v v v v v v v v v v




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




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




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




After some math we get ρ(111)=9.5x1012 atoms/mm2. We see that the (111) plane has higher density
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than the (100) plane, it is a close-packed plane.
v v v v v v v v v




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




½. This has v v




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


v v




0+0+ 22
v


v v v v
2 2
Setting R=1.749 Angstromswe get d(002)=2.745 Angstroms.
v v v v v v




In the same way
v v v




a = 4R
d(111) = =
v




a
v



6
v




1+1+1 3 v v
v




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




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




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




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




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





2 v




= v



3 v




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




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



23
v v v v v v v v



5.8 6.02310 v




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




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




2 2
[111] = = = 3.75109 atoms / m
 3
v v v v




3.0810−10 3
v


v v v v




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




4R
Problem 2.4. The lattice parameter for the FCC structure is  = . The (100) plane is the
v


v v v v v v v
v v v v v v v v v




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




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




plane is v




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




 
v v




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




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




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




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




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




plane is v




2 1
(111) = =
4 3R2 2 3R2
v




The ratio of the atomic densities is
v v v v v v




(111) 2
= =1.154 1
v
v




(100)
v v v




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




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




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




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




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




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




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




c = 2d(111) FCC
v v v
v




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




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, a
d = a a 
= =
v




3
v v


(111)
h2 + k2 + l2 12 + 12 + 12 4R
  d =
v
v





v v v v v v v v v v v

v
(111) v




FCC 6
v



4R
a=
v
v





v v




2

Thus,

 c 8R 4
c=  =
v v
v


v v = 1.63 v




6 
a 6
HCP: a = 2R v v v




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




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




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




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




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




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




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




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




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




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




6
result, the cell volume is
v v v v




V = 24 v v R3

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




4
6  R3
v v




3  v v




APF = = = 0.74 v
v v v

v



HCP 3
24 R

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




4R
where a = . Therefore the atomic packing factor of BCC structure is
v v


v v v v v v v v v v v v




3
4
2  R3
v v


v v




APFBCC = 3  3 = 0.68
3 = v v v
v



8
v v




 4R 
 
v v v




 




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