CHAPTER 4 – INTRODUCTION TO DISLOCATIONS
4.1
A single crystal of copper of copper yields under a shear stress of about of about 0.62 MPa. The shear modulus of copper is approximately
. With this data, compute an approximate value for the ratio of
the theoretical to the experimental shear stresses in copper. Solution:
Assuming that the theoretical yield shear stress equals one half the half the shear modulus, i.e.
and
4.4
(a) How many equivalent {111}
slip systems are there in the fcc lattice?
(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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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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(b) Note that the equation negative of
4.10
involves the vector
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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80 70 60 50 a P
M 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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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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