M sc sem iv u i

Dr. WS Barde, Shri Shivaji Science College, Amravati 1
4PHY-3(ii) : CONDENSED MATTER PHYSICS-II
Unit-I
Imperfections in Crystal: Mechanisms of plastic deformation in solid, Dislocations, stress &
strain field of screw dislocation, elastic energy of dislocations, Slip, Cross slip, climb,
Dislocation Multiplications, stress needed to operate Frank Read Source.
Book Referred :
Dislosations in Crystals
By W. T. Read
&
Introdustion to Dislosations
Fourth Edition
by D. Hull and D. J. Bacon

1.1 Dislocations in Crystals
Taylor, Orowan, and others introduced dislocations into physics in the 1930's; they were
interested in understanding what the atoms do when a crystal deforms. The understanding of
crystals in terms of atoms has two parts: (1) the theory of perfect crystals, (2) the theory of
imperfections.
Dislocations come under the theory of imperfections. By an imperfection we mean a small
region(at least of few atomic diameter) where the regular pattern breaks down and some atoms
are not properly surrounded by neighbors. A lattice vacancy is a simple example of an
imperfection.
1.2 Line defects or Dislocations:
Line imperfections (one-dimensional defects) are also called Dislocations. They are abrupt
changes in the regular ordering of atoms along a line (dislocation line) in the solid. The abrupt
changes in ordering are occurred because of the relative slip along the slip plane between two
regions in the crystal. The line separating slipped and unslipped regions on a slip plane is called
dislocation line. Dislocations occur in high densities and strongly influence the mechanical
properties of material. They are characterized by the Burgers vector (b), whose direction and
magnitude can be determined by constructing a loop around the disrupted region and noticing
the extra inter-atomic spacing needed to close the loop. The Burgers vector in metals points in a
close packed lattice direction.
Dislocations occur when an extra incomplete plane is inserted. The dislocation line is at the end
of the plane. Dislocations can be best understood by referring to two limiting cases - Edge
dislocation and Screw dislocation.
1.3 Edge dislocation or Taylor-Orowan dislocation
The geometry of an edge dislocation is relatively easy to visualize when the crystal has a simple
cubic crystal structure. An edge dislocation in a simple cubic structure is drawn in Figure 1,
which shows both a two-dimensional view and a three-dimensional section along the
dislocation line. The dislocation can be created by making a cut in the crystal on the dashed
plane that terminates at
the dislocation line,
displacing the material
above the cut plane to
the left of the
dislocation by one
lattice spacing, and
allowing the atoms to
re-bond across the slip
plane. This slip
recreates the simple
cubic unit cell
everywhere except on
the dislocation line
itself. Hence the Burgers vector, b, of the dislocation that is drawn in the figure is b = a[100],
where a is a vector along the edge of the cubic unit cell.

In edge dislocation a Burger’s vector is perpendicular to the dislocation line. The dislocation line
is an edge of an extra plane of atoms within a crystal structure. Thus regions of compression and
tension are associated with an edge dislocation. Because of extra incomplete plane of atoms, the
atoms above the dislocation line are compressed together whereas atoms below are pulled
apart and experience tensile stresses. Edge dislocation is considered positive when compressive
stresses present above the dislocation line, and is represented by ┴. If compressive stresses
exist below the dislocation line, it is considered as negative edge dislocation, and represented
by ┴.
A pure edge dislocation can glide or slip in a direction perpendicular to its length i.e. along its
Burger’s vector in the slip plane. The dislocation ( extra plane of atoms) may move vertically
by a process known as climb, if diffusion of atoms or vacancies can take place at
appropriate rate. When atoms are added to the extra plane, it may move downward by a
process called negative climb.
1.4 Screw dislocation or Burger’s dislocation
Screw dislocation or Burgers dislocation has its dislocation line parallel to the Burger’s vector. A
screw dislocation is like a spiral ramp with a dislocation line down its axis. Screw dislocations
result when displacing planes relative to each other through shear. Screw dislocation is
considered positive if Burger’s vector and t-vector are parallel, and vice versa. A positive screw
dislocation is represented by a dot surrounded by circular direction in clock-wise
direction”, whereas the negative screw dislocation is represented by a dot surrounded by a
circular direction in anti-clock-wise direction”. A schematic view of a negative screw dislocation
is shown in figure.
Like edge dislocation, a screw
dislocation does not have a
preferred slip plane. Therefore the
motion of a screw dislocation is
less restricted than the motion of
an edge dislocation. As there is no
preferred slip plane, screw
dislocation can cross-slip on to
another plane, and can continue
its glide under favorable stress
conditions. However, screw
dislocation cannot move by climb
process, whereas edge dislocations
cannot cross-slip.
A screw dislocation allows easy crystal growth because additional atoms can be added to the
‘step’ of the screw.
1.5 Mixed dislocation
A dislocation that contains partly edge and partly screws dislocation is called as mixed
dislocation. In mixed dislocation boundary separating slipped and unslipped regions is curved
in shape as shown in figure 1 or it is a closed loop inside the crystal as shown in figure 2.

A closed dislocation loop is shown in figure. The left and right parts of the dislocation loop
which are perpendicular to the Burger vector ‘b’ show positive and negative edge dislocations,
while front and back parts of the loop which are parallel
to Burger vector show screw anti-clockwise and
clockwise dislocations. Rest of the curved part ( for
example XY) of the loop shows mixed dislocation.
Burger vector b for mixed dislocation XY is resolved into
corresponding edge and screw components, which is
shown in figure. A pure edge vector b1 of length bsinθ
at right angle to XY and a pure screw vector b2 of length
bcosθ parallel to XY, such that; b=b1+b2.
Dislocations can be removed by heating to high temperatures where they cancel each other or
move out through the crystal to its surface. The density of dislocations in a crystal is measures
by counting the number of points at which they intersect a random cross-section of the crystal.
These points, called etch-pits, can be seen under microscope. In an annealed crystal, the
dislocation density is the range of 10
8
-10
10
m
-2
.
1.6 The Burger’s Circuit and Burger's Vector:
To describe the size and the direction of the main lattice distortion caused by a dislocation we
should introduce so called Burger’s vector b. To find the Burger’s vector, we should make a
circuit , called Burger’s circuit.
The procedure to find the Burger’s vectors of any dislocation is as follow.
• The circuit is traversed in the same manner as a rotating right-hand screw advancing in
the positive direction of the dislocation.
• The circuit must close in a perfect crystal and must go completely round the dislocation
in the imperfect (real) crystal.
• The vector that closes the circuit in the imperfect crystal (by connecting the end point to
the starting point) is the Burger’s vector.
Characteristics of Burger’s vector
• The Burger’s vector b represents the magnitude and direction of the lattice distortion
associated with a dislocation.

• Even though a dislocation changes direction and nature within a crystal (e.g. from edge
to mixed to screw), the burger’s vector will be the same at all points along its line.
• The vector sum of Burger’s vectors of dislocations meeting at a point, called node and
must be zero.
• The t-vectors of all the dislocations meeting at a node must either point towards it or
away from it.
• A dislocation line cannot end abruptly within the crystal. It can close on itself as a loop,
or ends either at a node or at the surface.
• As dislocations in the real crystal can be either full dislocations or partial dislocations.
- Partial dislocation’s Burger’s vector will be a fraction of a lattice translation and
- Full dislocation’s Burger’s vector is an integral multiple of a lattice translation
• Elastic energy associated with a dislocation is proportional to square of its Burger’s
vector.
The Burger’s circuit for Edge and Screw dislocations are shown in figures.
Burger's vector b denotes actually the dislocation-displacement vector.
• The Burger’s vector of a screw dislocation is parallel to the dislocation line.
• The Burger’s vector of an edge dislocation is perpendicular to the dislocation line.
• The Burger’s vector lies in the slip plane.
• In general cases, the Burger's vector may have other directions with respect to the
dislocation and for these cases the dislocation is a mixture of both edge and screw types.
Thus the mixed dislocation is defined in terms of the direction of the Burger's vector.
• The point where several dislocations meet is
called a node.
• The sum of the Burgers vectors of the
dislocations meeting at a node must vanish if
each dislocation is considered to go into the
node.
• If dislocation is going into a node and
branching into the other dislocations then the
vector sum of the Burgers vectors of the
branches is equal to the Burgers vector of the
original dislocation. Dislocations forming a
node is shown in figure.
• The Burger vector is described by the indices
of crystallographic direction along which
Burger vector lies. For example a Burger vector defined as b =
1
2
[111] in BCC indicates
displacement of a/2 in X-direction, a/2 in Y-direction and a/2 in Z-direction.

1.7 Motion of dislocation
There are two kinds of motion of a dislocation, Glide and Climb.
Glide :
It is relatively easy to move
dislocations to produce shear
deformation in the crystal. It is
very difficult to shear a crystal by
forcing the rigid planes to glide
over one another. But the slip is
more easily achieved by moving
dislocations stepwise through the
crystal. Stepwise dislocation
motion requires a much smaller
force because in each step reconfiguration of the bonds take place. The stepwise motion of an
edge dislocation in a simple cubic crystal is illustrated in Fig.
In order for the dislocation to move one lattice spacing to the right it is only necessary to break
the bond indicated by the long dash in fig. a and establish the bond indicated by the short dash.
The new configuration is shown in Fig. b. If the dislocation moves through the crystal in a
sequence of individual steps like that shown in Fig. a and b , it causes a net slip of the material
above its plane of motion by the Burgers vector, b, and hence causes a rigid displacement of the
whole upper part of the crystal. The type of motion that is illustrated above is called dislocation
glide, and is relatively easy to accomplish. However, an edge dislocation cannot glide in an
arbitrary direction. It can only glide in a particular plane, the slip plane or glide plane, which
contains both the Burgers vector and the dislocation line.
Climb:
When an edge dislocation moves out of its glide plane its motion is called climb. The climb of a
dislocation is difficult at ordinary temperatures since it requires that atoms be absorbed on or
liberated from the extra half-plane of atoms. The climb of an edge dislocation is illustrated in
following Figures.
The mechanism is slightly different depending on whether the dislocation moves up, which
contracts the extra half-
plane, or moves down,
which extends it. If the
dislocation climbs up atoms
must be liberated from the
edge of the extra half-
plane. Since the number of
atoms is conserved, this
requires the absorption of
vacancies from the lattice. If the dislocation climbs down it must add atoms to the extra half-
plane, and can only do this by liberating one vacancy per plane into the matrix.
That is Climb involves addition or subtraction of a row of atoms below the half plane .
► +ve climb = climb up → removal of a plane of atoms
► −ve climb = climb down → addition of a plane of atoms.
Both processes are possible at high temperature.

1.8 Plastic Behaviour and Concept of Slip
Plastic deformation takes place in a crystal due to the
sliding of one part of a crystal with respect to the
other. This results in slight increase in the length of
the crystal ABCD under the effect of a tension FF
applied to it as shown in fig. The 'process of sliding is
called slip. The direction and place in which the sliding
takes place are called respectively the slip direction as shown by the arrow P and slip plane. The
outer surface of the single crystal is deformed and a slip band is formed, as is seen in the figure,
which may be several thousand Angstroms wide. This can be observed by means of an optical
microscope, but when observed by an electron microscope a slip band is found to consist of
several slip planes. The examination of slips by an electron microscope reveals that these extend
over several tens of lattice constants. The slip lines do not run throughout the crystal but end
inside it, showing that slips do not take place simultaneously over the whole Slip planes but
occur only locally.
Slip Planes and Slip Systems
The slip planes are normally the planes with highest atomic density and hence are widely
spaced.
The slip directions in the slip plane are the directions on the slip plane along which atoms are
closely spaced and hence lattice translation vectors along the directions are shortest.
• In case of FCC crystal slip occurs on the planes {111} in the 〈110〉 directions. The shortest
lattice vectors in slip directions are
1
2
〈110〉.
• While in case of BCC crystals slip occurs along the directions 〈111〉 having shortest lattice
vectors
1
2
〈111〉 in the slip directions. But slip planes are not well defined, observations
suggest that in BCC crystals slip planes are {321} , {211} and {110} and therefore slip lines
are wavy. At low temperature the slip along {110} planes are preferred.
• In case of HCP (hexagonal closed packed) crystal slip occurs on basal planes {0001} in the
〈112̅0〉 directions. The shortest lattice vectors in slip directions are
1
2
〈112̅0〉.
A slip plane and a slip direction in the plane constitute a slip system.
• The FCC crystals have four {111} planes with three 〈110〉 directions in each plane and
therefore have 12 {111} 〈110〉 slip systems. (example; Cu, Al, Ni, Au, Ag …)
• The BCC crystals have six {110} planes with two 〈1̅11〉 directions, twelve {211} planes with
one 〈1̅11〉 direction and 24 {321} planes with one 〈1̅11〉 direction constituting slip systems.
Thus in all there are 48 slip systems. (examples; Fe, Mo, K, Na….)

• The HCP ( hexagonal closed packed) crystals have one {0001} plane with three 〈112̅0〉
directions, three {101̅0} planes with one 〈112̅0〉 direction and six {101̅1} planes with one
〈112̅0〉 direction. Thus in all there are 12 slip systems. (examples; Cd, Zn, Mg, Be, Ti….)
1.9 Cross Slip
In general screw dislocation tends to move in certain crystallographic planes. For example in
case of FCC metals a screw
dislocation moves in {111}
planes but can switch from one
{111} type plane to another if it
contain direction of vector b.
This process is called as cross
slip.
The process of cross slip is
illustrated in following figure.
In this figure the dislocation
line and Burger vector
1
2
[1̅01]
are gliding towards left in (111)
plane under the external shear
stress. The other plane containing this slip vector is (11̅1) plane. Under the shear stress the slip
may preferred to switch from in (111) plane to (11̅1) plane. So cross slip occurs at S. In this case
only pure screw dislocation gets cross slipped while edge dislocation continues to slip over
same plane. The glide of dislocation remain continue over on (11̅1) plane and then double cross
slip occurs on the plane parallel to (111) plane as shown in figure.
1.10 Shear Strength of Crystals: Theoretical Critical Stress required for slip (plastic
deformation) Or Peierls- Nabarro Stress
J. Frenkel in calculating the theoretical shear strength of a perfect crystal. The model proposed
by him is given in figure, showing a cross-section through two adjacent atomic planes separated
by a distance d. The full line circles indicate the equilibrium positions of the atoms without any
external force.
Let us now apply a shear stress in the direction shown in figure. All the atoms in the upper
plane are thus displaced by an amount x from the original positions as shown by the dotted
circles. In fig. the shear stress  has been plotted as a function of the relative displacement of the
planes from their
equilibrium positions
and this gives the
periodic behavior of  as
supposed by Frenkel.
'  is found to become
zero for x = 0, a/2, a etc.,
where a is the distance

between the atoms in the direction of the shear. Frenkel assumed that this periodic function is
given by
a
x
c


2
sin= (1)
where the amplitude c denotes the critical shear stress which we have to calculate For x << a,
we have as usual,
a
x
c


2
= (2)
In order to calculate the force required to shear the two planes of atoms, we from the definition
of shear modulus
a
x
dxyStrain
Stress
G c


 2
/
===
where G is the shear module and
Gd
x
y

== is the elastic strain






=
d
x
G (3)
Comparing it with equation (2), we have












=





b
a
x
G
or
d
x
G
a
x
cc .
2
2



or daif
G
x
G
c  ,
62
 (4)
This gives the maximum critical stress above which the crystal becomes unstable. It is about one
sixth of the shear modulus. In a cubic crystal, G = 1011 dynes per cm. for a shear in the <100>
direction. Hence the theoretical value of the critical shear stress on Frenkel's model is c = 1010
dynes per sq. cm. which is much larger than the observed values for pure crystals.
However, the experimental values for the maximum resolved shear stress required to start the
plastic flow in metals were of the order of 10-3 to 10-4 G at that time and it was not a agreement
with the theoretical results .
The disagreement in values is due to the supposition of perfect lattices. We have , therefore, to
consider the presence of imperfections which act as sources of mechanical weakness in actual
crystals.
1.11 Stress Fields around Screw Dislocations : Screw
Dislocation Energy
We know that the core of dislocations is a region within a few
lattice constants of the centre of dislocation and that it is a
"bad" region where the atomic arrangement of the crystal is
severely changed from the regular state.
Let us have a cylindrical shell of a material surrounding an
axial screw dislocation. Let the radius of the shell be r and the
thickness dr. The circumference of the shell is r2 and let it
be sheared by an amount b, so that the shear strain
𝛾 =
b
2πr
(1)
and the corresponding shear stress in the good region is,
𝜏 = Gγ = G
b
2πr
(2)

where G is the shear modulus or modulus of rigidity of the material. A distribution of forces is
exerted over the surface of the cut for producing a displacement b and the work done by the
forces to do it gives the energy dEs per unit volume of the screw dislocation for spherical shell is
𝑑𝐸
𝑑𝑉
=
1
2
𝑆𝑡𝑟𝑒𝑠𝑠 𝑋 𝑠𝑡𝑟𝑎𝑖𝑛 =
1
2 r
bG
2
. b
2πr
[3]
The volume of the annular region is 𝑑𝑉 = 2πrldr, this gives us,
𝑑𝐸 =
1
2 r
bG
2
. b
2πr
2πrldr =
lGb2
4π
dr
r
Therefore screw dislocation energy for the strained region is given by,
𝐸𝑠 = ∫ 𝑑𝐸 = ∫
lGb2
4π
dr
r
R
r0
𝐸𝑠 =
lGb2
4π
ln (
R
r0
) ……………….[4]
where R and r0 the proper upper and lower limits of r. The energy depends upon the values
taken for R and r0 is suitable when it is equal to about the Burger's vector b or equal to one or
two lattice constants and the value of R is not more than the size of the crystal.
Stress field of an edge dislocation: Edge Dislocation energy
Let us consider a cylindrical material whose axis is along the z-axis and a cut has been made
along a slip plane. Let R be the radius of cross-section , G be the shear modulus and ν be
Poisson's ratio of the cylindrical material. The portion on one side of the cut is now slipped
relative to the portion on other by an amount b, the Burger's vector as shown in fig. Thus, a
positive edge dislocation has been produced along the z-axis.
Let σrr be the radial tensile stress and let σθθ be the circumferential tensile stress. Let τrθ denote
the shear stress acting in a radial direction.
Without giving the details of calculations here, the stress field of the edge dislocation in terms of
r and  are given by the following : .
rv
Gb
rr


 
sin
.
)1(4 −
== ……(1)
and
rv
Gb
r


 
cos
.
)1(2 −
= …..(2)
It may be noted that for r = 0, the stresses become infinite and so a
small cylindrical region of radius ro around the dislocation must be
excluded. This is necessary because in this region the Hooks law of
elasticity does not hold. To know the value of ro, let us put ro = b,
the magnitude of the strain there is then of the order of
1/2π(1-v). We shall now calculate the energy of formation of an
edge dislocation of unit length. The final shear stress in the plane y
= 0 is given by (2) by putting θ = 0 will be,
rv
Gb
r
1
.
)1(2 −
=

 
Also corresponding strain is, 𝛾 = 𝑏/2𝜋𝑟
Thus the strain energy for edge dislocation will be given by
𝐸𝑒 = ∫ strainxstress
2
1𝑅
𝑟0
=
1
2
∫
rv
Gb 1
.
)1(2 −
𝑅
𝑟0
𝑏
2𝜋𝑟
𝑑𝑟 =
0
2
log
)1(4 r
R
v
Gb
−
…[3]

1.13 Critical Resolved Shear Stress in slip direction
Consider the crystal as shown in figure which is being
deformed by applied force F along the axis of the cylindrical
crystal. If A be the cross sectional area then tensile stress
parallel to the force is 𝜎 = 𝐹/𝐴.
Let λ be the angle between force F and slip direction. Then
the component of the force is 𝐹𝑐𝑜𝑠𝜆
This force 𝐹𝑐𝑜𝑠𝜆 acts on slip surface which has an area
𝐴/𝑐𝑜𝑠𝜙, where 𝜙 is the angle between force F and the
normal to the slip plane.
Thus the shear stress resolved on slip plane and in slip
direction is given by,
𝝉 𝑹 =
𝑭𝒄𝒐𝒔𝝀
𝑨/𝒄𝒐𝒔𝝓
=
𝑭
𝑨
𝒄𝒐𝒔𝝀 𝒄𝒐𝒔𝝓
Or,
𝝉 𝑹 = 𝝈 𝒄𝒐𝒔𝝀 𝒄𝒐𝒔𝝓 , where the quantity 𝒄𝒐𝒔𝝀 𝒄𝒐𝒔𝝓
is called Schimid factor.
1.14 Forces on Dislocations : stress needed to operate Frank Read Source
The forces acting on dislocation may be different types for example; glide forces, climb forces
and image forces. The illustration for glide force is given below.
Consider a dislocation moving in a slip plane under the action of resolved shear stress 𝝉. If dl
be the small element of the dislocation line of Burger
vector b moves forward through a displacement of
ds. The average shear displacement of the crystal
surface produced by glide of an element dl is
𝒅𝒔 𝒅𝒍
𝑨
𝒃.
Where A is the area of the slip plane. The external
force due to 𝝉 acting on this area is Aτ . Thus the
work done during the slip is given by,
dW = 𝝉𝑨 (
𝒅𝒔 𝒅𝒍
𝑨
𝒃) …………….[1]
The glide force on unit area can be written as,
F =
dW
𝒅𝒔 𝒅𝒍
= 𝝉𝒃 ……………………….[2]
In addition to this force a dislocation line has a line tension T. The line tension is defined as the
increase in energy per unit increase in the length of dislocation line.
That is , T = αGb2
………….[3]
Consider a curved dislocation as shown in following figure.

The line tension T tends to make the element dl straight. The resultant tension directs along AO
towards the center of curvature O. Thus to keep the element curved, a resultant shear stress
should be acting on dl opposite to the resultant tension. If the shear stress 𝝉 is required to keep
the radius of curvature R as it is, then,
the corresponding force along OA on an element dl = 𝝉𝒃𝒅𝒍 ………………..[4]
From figure we have, angle subtended by the element at center of curvature is 𝒅𝜽 = 𝒅𝒍/𝑹…..[5]
Thus resultant tension towards AO from two ends of dl is given by, 𝟐𝑻𝒔𝒊𝒏 (
𝒅𝜽
𝟐
).
Now, if 𝒅𝜽 is small then
Tension towards AO = 𝟐𝑻 (
𝒅𝜽
𝟐
) = 𝑻𝒅𝜽 ……………………..[6]
For equilibrium of an element dl we have,
𝑻𝒅𝜽 = 𝝉𝒃𝒅𝒍……………………..[7], using equations [3] and [5] we get,
αGb2
𝒅𝒍
𝑹
= 𝝉𝒃𝒅𝒍
or, 𝝉 =
𝛂𝐆𝐛
𝐑
………..[7]
This gives the stress required to bend the dislocation.
Bending of dislocation, in this way, results in Frank-Read type multiplication of dislocation.
Thus the equation[7] is the stress needed to operate the Frank-Read type multiplication.
1.15 Forces between dislocations
Consider two parallel edge dislocations lying in the same plane. They can either both positive or
of opposite sign as shown in figure.
When dislocations are separated by large
distance, then total dislocation energy per unit
length is given by,
𝐺𝑏2
+ 𝐺𝑏2
………1
If dislocations are very close (fig a) , then they
can be considered as a approximately single
dislocation of Burger vector of magnitude 2b.
Then total dislocation energy per unit length is
given by,
𝐺(2𝑏)2
………2, which is twice the dislocation,
than that for the situation when they are
separated by large distance. Thus the
dislocations repel each other to reduce the
dislocation energy.

When dislocations are of opposite sign close to each other (fig.b), then resultant Burger vector is
zero. Thus dislocations of opposite sign attract each other to reduce their dislocation energy.
Similar effect occurs when dislocations are not in same plane, but in such cases conditions for
repulsion and attraction are more complicated.
1.15 Multiplication of Dislocations
There are two main mechanisms of multiplication of dislocations.
1. Multiplication by Frank-Read Source
2. Multiplication by Multiple cross glides
1. Multiplication by Frank-Read Source
The first model for dislocation multiplication was proposed by Frank and Read. This can be
illustrated with the help of following
figure. In this figure the dislocation is
lying in two different planes. The
length BC of the dislocation ABC is in
the slip plane while length AB is not in
the slip plane. So AB is non-moveable.
Thus under the stresses dislocation
BC will anchored and revolves about B
as shown in second figure. The
revolution about B causes the upward
displacement of the slip plane. If displacement reaches to the surface then a large slip step is
produced on the surface.
The Frank-Read source is illustrated as follow-
1. The segment AB is the dislocation with
Burger vector lies in the slip plane as
shown in figure (a).
2. Both ends of AB are held fixed by some
unspecified barriers ( dislocation
intersection, jogs, precipitate etc).
3. An applied stress exert the force of
magnitude 𝝉𝒃 on the line AB and tends to
make the dislocation bow as shown in figure b.
4. As 𝝉 increases radius of curvature R decreases.
The line bows continuously till the radius of
curvature become minimum( R equals to half of
AB) as shown in figure c.
5. If the stress is sufficient, then the line continues
to expand and here R increases ( figures d and
e) which makes the dislocation unstable.
6. The dislocation forms the kidney shaped loop,
as shown in figure (f) With regenerated
dislocation line AB which repeat the process.

2. Multiplication by Multiple cross glides
This type of multiplication is
associated with initiation and
broadening of a slip band. Such
dislocations are observed by etch
pit technique.
Initially the slip band started with
single loop of dislocation. The slip
band can be widen by multiple
cross glide on different slip planes
as shown in figure. Screw
dislocation lying along AB can
cross glide to new position CD on
another parallel glide plane. If stress is grater on primary plane then long jogs AC and BD
are relatively immobile.

M sc sem iv u i

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