280
26 Eigen Stresses and Dislocations
−1, along a closed path in the crystal is equal to the total Burgers vector b of the
dislocation enclosed by this path. This is the result of Kröner’s analysis.
For the illustration of these facts, we turn to Figs. 7.7 and 7.8, in Chap. 7. We
consider the case that a straight dislocation is created by the removal of the half of a
lattice plane. In the ideal experiment, one can depict this as follows. The crystal is
cut open along the line of dislocation that is to be created. One removes one half of
the lattice plane from this cross section and puts the crystal back together so that the
cross section can no longer be seen, see Fig. 26.1 and compare with Kröner [47, 48].
The experimental proof of dislocations is possible in many ways. Today, dislocations can be made visible using the electron microscope, see Kittel [46], Chap. 19
foll.
The rest is just mathematics. We do not just approximate the crystal using a
continuum, but also the large number of contain dislocations using a continuous
dislocation density α. The second-order tensor of dislocation density α is defined
according to
A i α ik = b k
that is
A · α = b
(369)
due to the sum b k of the Burgers vectors of all lines of dislocation that perpendicularly penetrate the area A i .
Equations (367) and (369) together make up the equation of Kröner and Ney that
describes the complicated relationship between dislocations and elastic distorsions
of the lattice in a most simplistic manner, Kröner [47] 1958, in English Kröner [48]
1980,
irs β sk , r = α ik
that is
curl β = α .
(370)
Figures 26.1 and 26.2: Dislocation and plastic deformation. Only the position of
the dislocation line pointing perpendicularily out of the drawing plane marked with
the sign ⊥ for an edge dislocation, can be experimentally verified. The dislocation
in Fig. 26.2 is created by removing the shaded half lattice plane of Fig. 26.1 from
the crystal and then putting the two halves back together again. The dotted lines in
Fig. 26.2 show which lattice atoms received new next neighbours after this procedure.
Here, the atoms were plastically displaced by δs
p
o . The final position of the atoms
is achieved by an additional elastic displacement. The partially filled atom has been
labelled to allow better orientation in the lattice. On the thick lined orbit, only this
atom has experienced a plastic displacement and thus alone adds δs
p
o to the sum
δs
p . All other atoms in this orbit were only displaced elastically. This results in
b = −
δs
p
= −δs
p
o for the Burgers vector b of the dislocation enclosed in
the orbit.
From (370), we come to an important conclusion. With the help of (358), one can
easily calculate that always
∂
∂x i
irs β sk , s = irs β sk , si = 0 .
Hence, for an arbitrary distribution of dislocations, the following is always valid,
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