DISLOCATION MOVEMENT
• Because a dislocation represents the boundary b/w the
slipped an unslipped region of a crystal, therefore it
must:
– be a closed loop, or
– end at the free surface of a crystal or
– end at a grain boundary.
• As shown in Fig. 12-1, dislocations will ordinarily take
the form of curves or loops, which in three dimensions
form an interlocking dislocation network.
• In considering a dislocation line, it can be:
– edge,
– screw or
– mixed type, which can be resolved into edge and screw
components.
• For example, in Figure 12-1, the dislocation loop is:
– pure screw at point A
– pure edge at point B,
– along most of its length it has mixed edge and screw
components.
Figure 12-1. Dislocation loop lying in a slip plane
• The process of cross slip illustrated in Fig 12-2, will serve as
an example of dislocation loops.
• In Fig. 12-2(a), a small loop of dislocation line with
b  a / 2[ 1 01] is moving on a (111) plane in an fcc crystal.
o
• The dislocation loop is pure positive edge at “w” and pure
negative edge at “y”.
• At “x” the dislocation is a right-handed screw while at “z”
the dislocation loop is a pure left-handed screw dislocation.
• At some stage (Fig. 12-2b), the shear causing expansion of
the loop tends to move the dislocation on the intersecting
(1 1 1) plane.
• Since the dislocation is pure screw at “z”, it is free to move
on this second (1 1 1) plane.
• In Fig. 11-2(d), double cross slip has taken place as the loop
glides back onto the original (111) plane.
Cross-Slip Mechanism
Figure 12-2. Cross slip in a face-centered cubic crystal.
Figure 12-3. Movement (glide) of an edge dislocation.
Figure 12-4. (a) Macroscopic deformation of a cube produced by glide
of an edge dislocation. (b) Macroscopic deformation of a cube by glide
of a screw dislocation. Note that the end result is identical for both
situations.
• The macroscopic slip produced by the motion of an edge
dislocation is shown in Fig. 12-4a and by a screw
dislocation in Fig.12-4b.
• Both the shear stress and final deformation are identical for
both situations, but for an edge dislocation line moves
parallel to the slip direction while the screw dislocation
moves at right angles to it.
• Recall the geometric properties of dislocations. This is
summarized in Table 10-1.
Table 10-1 Geometric properties of dislocations
Dislocation property
Relationship b/w dislocation
line l and b
Type of dislocation
Edge
Screw
perpendicular
parallel
Slip direction
parallel to b
parallel to b
Direction of dislocation line l
movement relative to b
(slip direction)
parallel
perpendicular
Process by which dislocation
may leave slip plane
climb
cross-slip
• Aside from the shear stress associated with the glide of
edge, screw, and mixed dislocations, it is possible for edge
dislocations to produce tensile strains by moving in a
direction perpendicular to the Burgers vector [Fig. 125(a)].
• This type of motion is called dislocation climb and
involves the transport of atoms (or vacancies) away from
or to the dislocation (edge of the extra half plane) by
diffusion.
• Because dislocation climb is diffusion-controlled, it is
important only at relatively high temperatures where the
diffusion rates are fairly rapid.
• Motion of an edge dislocation showing both glide and
climb motion is illustrated in Fig. 12-5(b).
Figure 12-5(a) . Illustration of climb of edge dislocation
Figure 12-5(b) . Illustration of glide and climb of
edge dislocation.
Relation Between Dislocation Movement and
Plastic Flow
• To demonstrate how we can relate the motion of
individual dislocations to macroscopic plastic flow, let
us consider the strain produced by the motion of a
single dislocation.
• Suppose we have a crystal in the shape of a cube ,
where the edge length is L.
• Under the application of an appropriate shear stress, a
dislocation is somehow produced on the left-hand side
of the crystal and is allowed to move all the way
through the crystal as shown in Fig 12-6.
• The plastic shear strain associated with the passage of
this dislocation had not passed all the way through the
crystal but had gone only a distance x , where x<L.
• Then the overall shear strain would somewhere between 0
and b/L, the exact value being proportional to the fraction of
the slip plane area that the dislocation has traversed. In
moving a distance x, the dislocation has swept put x/L of its
slip plane and the resulting shear strain is given by
xb
p 
LL
(12.1)
• If instead of one dislocation we had considered N
dislocation, all moving an average distance x [ Fig 12-7],
the resulting strain would be
Nx b
p  2
L
(12.2)
Figure 12-6. Relation between shear strain and dislocation
movement in glide.
Figure 12-7 . Relation between shear strain and dislocation
movement in glide.
• Which can be rewritten
NLx b
 b  3  bx
L
(12.3)
• where NL is the total length of the dislocation line and  is
the dislocation density, NL/L-3. In units of length per unit
volume. The shear strain is then the product of the
dislocation density (the number of dislocations that have
moved), the Burgers vector, and the average distance each
dislocation has moved. If the shear strain b occurs over a
time t, then we can write
b
t
or

bx
t
 p  bv
(12.4)
(12.5)
• Where p  is the shear strain rate and v is the average
dislocation velocity. Equation (12.5) is a type of transport
equation that often occurs in physics.
• It simply states that the strain rate is equal to the density of
defects producing strain (dislocations) times the strength of
each defect (the amount of strain associated with each
defect), times the average velocity of the defects.
• Later we shall show that an identical equation exists for
describing electrical conduction. For electrical conduction
we write
J  nqv
(12.6)
Stress Required for Slip
• The flow stresses of crystals are highly anistropic.
• For instance, the yield stress of zinc under uniaxial tension
varies by a factor of at least 6.
• Consequently, it is very important to specify the orientation
of the load.
• In shear or torsion tests, the shear plane and directions are
precisely known.
• Because dislocations can glide only under the effect of shear
stresses, these shear stresses have to be determined.
• In uniaxial tensile and compressive test ( the most common
tests), one has to determine mathematically the shear
component of the applied stress acting on the plane in which
slip is taking place.
• Figure 12-8 shows a crystal with a normal cross-sectional
area A upon which a tensile load acts, generating a uniaxial
stress P/A.
Figure 12-8. Relationship between stress axis and slip
plane and direction.
• The slip plane and direction are indicated, respectively,
by the angles  and  that they make with the tensile
axis.
• We take the normal n of the plane that makes an angle
 with the loading direction .
• The areas A1 and A are related by the angle . Area A is
the projection of A1 onto the horizontal plane; thus, we
can write
A  A1 cos 
(12.7)
• The shear stress  acting on the slip plane and along the
slip direction S is obtained by dividing the resolved load
along the slip direction ( P cos  ) by A1:
P cos  P

 cos  cos 
A1
A
(12.8)
• But P / A   is the normal stress applied to the
specimen. Hence,
   cos  cos 
• Note that cos   sin 
(12.9)
• This equation shows that  will be zero when either  or 
is equal to 0o.
• When both  and  are equal to 45o, the shear component
is maximum and we have, in this case,
 max   cos 45 cos 45 
o
o

2
(12.10)
• The angle between any two directions a and b can be
obtained from the scalar product of these vectors:
a  b  a b cos
(12.11)
a b
cos  
ab
(12.12)
or
• For cubic crystals, planes and directions with the same
indices are perpendicular, and the angle is determined
from the coefficients, h, k, and l. For two vectors
and
a  h1i  k1 j  l1k
(12.13a)
b  h2i  k2 j  l2 k
(12.13b)
the angle  is given by
cos 
h1h2  k1k2  l1l2
(h12  k12  l12 )1/ 2 (h22  k22  l22 )1/ 2
(12.14)
• If the two directions are perpendicular, their dot product is
zero; and the same is true for a direction that is contained in
a plane.
• From Eq. 12.11, it is possible to obtain the cos  and cos 
terms for all desired crystallographic directions of a crystal.
• For instance, if the loading directions is [123] for an FCC
crystal, then the Schmid factors of the various slip systems
are found by obtaining the angles of [123] with <111>
(perpendicular to slip planes) and [123] with (110) (in slip
directions).
• Note that each slip plane contains three slip directions
and that 12 values (4 x 3) have to be obtained.
• Schmid and coworkers used the variation in the
resolved shear stress to explain the great differences in
the yield stresses of monocrystals of certain metals.
• They proposed the following rationalization, known as
the Schmid law: Metals flows plastically when the
resolved shear stress acting in the plane and along the
direction of slip reaches the critical value
 c   o sin  cos   M o
(11.15)
M  sin  cos   cos  cos 
(11.16)
• where the factor M is usually known as the Schmid
factor
• A single crystal subjected to a shear stress can deform
extensively with slip on only a single slip system.
However, this is not the case for polycrystals.
• Since each grain in a polycrystal has a different
crystallographic orientation, each will respond differently
(slip in a different direction) when subjected to a shear
stress.
• Thus, unless the region around grain boundaries can be
undergo arbitrary shape changes, voids at grain
boundaries will be opened up and the material will
fracture at low strains.
• Example: An NaCl single crystal is loaded in uniaxial
compression with the [001] direction parallel to the
compression axis. Dislocation motion is first observed
on the (101)[10 1 ] slip system when the applied tensile
stress is 500 psi. Calculate the inherent lattice resistance
to dislocation motion.
• Solution: The resolved shear stress on the (101)[10 1 ]
slip system is given by Eq.(6.7), where
   cos cos 
• or    / 2 for (101)[10 1 ] slip since     45o. The
shear stress necessary to overcome the inherent lattice
resistance to dislocation motion is therefore 250 psi.