Physical Metallurgy: Principles and Design (1st Edition) by Gregory N. Haidemenopoulos — A comprehensive exploration of the processing‑structure‑properties triangle for metals and alloys, covering crystallography, phase transformations, plastic deformatio

Chapters 2 - 10 Covered
c c c c




SOLUTIONS

, Chapter 2 cb




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




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


c b c b cb cb cb cb cb cb cb cb cb cb cb cb cb




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




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




In the (111) plane there are 3/6+3/2=2 atoms. The base of the triangle is 4R and the height 2
cb cb cb cb cb cb cb cb cb cb cb cb cb cb cb cb cb c b c b c b 3R
After some math we get ρ(111)=9.5x1012 atoms/mm2. We see that the (111) plane has higher density
cb cb cb cb cb cb cb cb cb cb cb cb cb cb cb




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




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




½. This has cb cb




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


cb c b




0+0+22
c b


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




In the same way
cb cb cb




a = 4R
d(111) = =
cb




a
cb



6
c b




1+1+1 3 c c
c b




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




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




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




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




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





2 cb




= cb



3 cb




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




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



23
cb c b c b cb cb c b cb cb



5.8 6.02310 cb




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@SSeeisismmicicisisoolalatitoionn

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




2
c b




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




2 2
[111] = = = 3.75109 atoms / m
 3
cb cb cb cb
c b


c c 3.0810 − 10
3 c c




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




4R
 = . The (100) plane is
c b




Problem 2.4. The lattice parameter for the FCC c b c b c b c b c b c b c b
c b c b c b c b c b c b




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




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




c (100)
b




plane is cb




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




 
cb cb




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




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




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




triangle
c b




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




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




plane is cb




2 1
(111) = =
4 3R2 2 3R2
c b




The ratio of the atomic densities is
cb cb cb cb cb cb




(111) 2
= =1.154 1
c b
c b




(100)
cb cb cb




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




the
cb




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




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




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




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




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




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




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, c = 2d(111) FCC
cb cb cb
cb




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




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@SSeeisismmicicisisoolalatitoionn