14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 08
DISLOCATIONS AND STRENGTHENING MECHANISMS
7.6
(a)
Compare planar densities (Section 3.11 and Problem 3.54) for the (100), (110), and (111) planes for FCC.
(b) Compare planar densities (Problem 3.55) for the (100), (110), and (111) planes for BCC.
Solution
(a)
For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.11 as
PD
110
(FCC)
1
4
R
2
2
0.177
R
2
Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.54,
which are as follows:
PD
100
(FCC) =
1
4
R
2
0.25
R
2
PD
111
(FCC)
1
2
R
2
3
0.29
R
2
(b)
For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in
Homework Problem 3.55, which are as follows:
PD
100
(BCC) =
3
16
R
2
0.19
R
2
PD
110
(BCC)
3
8
R
2
2
0.27
R
2
Below is a BCC unit cell, within which is shown a (111) plane.

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
(
a
)
The centers of the three corner atoms, denoted by A, B, and C lie on this plane.
Furthermore, the (111) plane does
not pass through the center of atom D, which is located at the unit cell center.
The atomic packing of this plane is
presented in the following figure;
the corresponding atom positions from the Figure (
a
) are also noted.
(
b
)
Inasmuch as this plane does not pass through the center of atom D, it is not included in the atom count.
One sixth of
each of the three atoms labeled A, B, and C is associated with this plane, which gives an equivalence of one-half
atom.
In Figure (
b
) the triangle with A, B, and C at its corners is an equilateral triangle.
And, from Figure (
b
), the
area of this triangle is
xy
2
.
The triangle edge length,
x
, is equal to the length of a face diagonal, as indicated in
Figure (
a
).
And its length is related to the unit cell edge length,
a
, as