CHAPTER 4 – INTRODUCTION TO DISLOCATIONS
4.1
A single crystal of copper yields under a shear stress of about 0.62 MPa. The shear modulus of
. With this data, compute an approximate value for the ratio of
copper is approximately
the theoretical to the experimental shear stresses in copper.
Solution:
Assuming that the theoretical yield shear stress equals one half the shear modulus, i.e.
and
4.4
slip systems are there in the fcc lattice?
(a) How many equivalent {111}
(b) Indentify each system by writing out its slip plane and slip direction indices.
Solution:
(a) The various slip systems are listed below.
(111)[
]
(
)[110]
(
)[110]
(
)[
(111)[
]
(
)[
]
(
)[011]
(
)[011]
(111)[
]
(
)[101]
(
)[
(
)[101]
]
]
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27
4.5
Assume that the triangle in the drawing below lies on the (111) plane of a face‐centered cubic
crystal, and that its edges are equal in magnitude to the Burgers vectors of the three total
dislocations that can glide in this plane. Then, if
lies at the centroid of this triangle, lines
accordingly correspond to the three possible partial dislocations of this plane.
(a) Indentify each line
with its proper Burgers vector expressed in the vector
notation.
(b) Demonstrate by vector addition that:
z
B
C
y
A
x
Solution:
(a) The direction vectors of the lines in question are as follows:
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28
involves the vector
(b) Note that the equation
negative of
4.10
whose direction indices are the
Thus,
(a) Write a simple computer program that gives the shear stress of a screw dislocation as a
function of the perpendicular distance from the dislocation (see Eq. 4.8). Assuming that the
shear modulus is 86 GPa, and the Burgers vector is 0.248 nm, use the program to obtain the
shear stress at the following values of r: 50, 100, 150, and 200 nm, respectively. Plot the
resulting versus r data, and with the aid of this curve, determine the distance from the
dislocation where is 27.6 MPa, the shear stress at which an iron crystal will begin to undergo
slip.
(b) To how many Burgers vectors does this distance correspond?
Solution:
(a) Equation 4.8 is:
This becomes, on substituting the given values of and as well as the value of ,
where r is given in nm. This equation is plotted in the figure below. Now as may be seen in the
plot, the critical resolved shear stress of iron, 27.6 MPa, corresponds to a distance of 122.9 nm.
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29
80
70
60
t, MPa
50
40
27.6 MPa
30
20
122.9 nm
10
0
0
50
100
150
200
250
r, nm
(b) The distance r, at which the stress of the dislocation becomes equal to the critical resolved shear
stress is 122.9 nm. The length of the iron Burgers vector is 0.248 nm. Thus, the number of
Burgers vectors to the point where the shear stress of the dislocation become equal to the
critical resolved shear stress is 122.9/0.248 = 496 Burgers vectors.
4.14
The strain energy of a dislocation normally varies as the square of its Burgers vector. One may
see this by examining Eqs. 4.19 and 4.20. This relationship between the dislocation strain energy
and the Burgers vector is known as Frank’s rule. Thus, if
where a is a numerical
factor, then:
Show that in an fcc crystal the dissociation of a total dislocation into its two partials is
energetically feasible. See Eq. 4.4.
Solution:
Equations 4.4 is:
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30
The corresponding energies of the three dislocations are for the total dislocation,
for both partials,
, and
. Thus,
The dissociation is energetically feasible.
4.15
The
ratio of the hcp zinc crystal is 1.866. Determine the ratio of the strain energy of a
dislocation with a
Burgers vector, to that of a basal slip dislocation in a zinc crystal.
Solution:
The Burgers vector of a basal slip dislocation is equal to a, and to that of the pyramidal slip
dislocation is proportional to the vector sum of an a dislocation and a c dislocation. The ratio of
the energies of these two Burger vectors is accordingly:
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31