Lectures 07-12, 2023.pdf

Friday, February 24,
2023 (IUB)
1
Lectures 07-12
DISLOCATIONS & PLASTIC DEFORMATION
Materials Simulation Group
(MSG)
By
Dr. Altaf Hussain
(Professor)
Institute of Physics (IoP),
The Islamia University of Bahawalpur

2023 (IUB)
2
DISLOCATIONS & PLASTIC DEFORMATION
(MSG)
▪ Early materials studies led to the computation of the theoretical
strengths of perfect crystals, which were many times greater than
those actually measured.
▪ During the 1930s it was theorized that this discrepancy in
mechanical strengths could be explained by a type of linear
crystalline defect that has since come to be known as a
dislocation.
▪ Not until the 1950s, however, was the existence of such
dislocation defects established by direct observation with the
electron microscope.
▪ Since then, a theory of dislocations has evolved that explains
many of the physical and mechanical phenomena in metals [as
well as crystalline ceramics].

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3
(MSG)
Section-1
Basic Concepts

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4
BASIC CONCEPTS
(MSG)
▪ Edge and screw are the two fundamental dislocation
types.
▪ In an edge
dislocation,
localized lattice
distortion exists
along the end of an
extra half-plane of
atoms, which also
defines the
dislocation line.

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5
Cont.
(MSG)
▪ A screw
dislocation may
be thought of as
resulting from
shear distortion.
▪ In this case, the
dislocation line
passes through
the center of a
spiral, atomic
plane ramp.

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6
Cont.
(MSG)
▪ Many
dislocations in
crystalline
materials have
both edge and
screw
components.
▪ These are known
as mixed
dislocations.

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7
Mechanism of Plastic Deformation
(MSG)
▪ Plastic deformation corresponds to the motion of large numbers of
dislocations.
▪ An edge dislocation moves in response to a shear stress applied in a
direction perpendicular to its line.
▪ The mechanics of dislocation motion are represented in Figure
below.

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8
Cont.
(MSG)
▪ Let the initial extra half-plane of atoms be plane A.
▪ When the shear stress is applied as indicated (Figure a), plane A is forced to the
right; this in turn pushes the top halves of planes B, C, D, and so on, in the same
direction.
▪ If the applied shear stress is of sufficient magnitude, the interatomic bonds of plane B
are severed along the shear plane, and the upper half of plane B becomes the extra
half-plane as plane A links up with the bottom half of plane B (Figure b).

2023 (IUB)
9
Cont.
(MSG)
▪ This process is subsequently repeated for the other planes, such that the extra
half-plane, by discrete steps, moves from left to right by successive and repeated
breaking of bonds and shifting by interatomic distances of upper half-planes.
▪ Before and after the movement of a dislocation through some particular region
of the crystal, the atomic arrangement is ordered and perfect; it is only during
the passage of the extra half-plane that the lattice structure is disrupted.
▪ Ultimately this extra half-plane may emerge from the right surface of the crystal,
forming an edge that is one atomic distance wide; this is shown in Figure c.

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10
(MSG)
Section-2
Slip & Slip Systems

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11
Slip
(MSG)
▪ The process by which plastic deformation is produced by dislocation
motion is termed as slip.
▪ The crystallographic plane along which the dislocation line traverses
is the slip plane, as indicated in Figure (on previous slide).
▪ Macroscopic plastic deformation simply corresponds to permanent
deformation that results from the movement of dislocations, or slip,
in response to an applied shear stress, as represented in Figure
below.

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12
Cont.
(MSG)
▪ The motion of a screw dislocation in response to the applied shear stress
is shown in Figure below.
▪ The direction of movement is perpendicular to the stress direction.
▪ For an edge dislocation, motion is parallel to the shear stress.
▪ However, the net plastic deformation for the motion of both dislocation
types is the same.
▪ The direction of motion of the mixed dislocation line is neither
perpendicular nor parallel to the applied stress, but lies somewhere in
between.

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13
Dislocation Density
(MSG)
▪ All metals and alloys contain some dislocations that
were introduced during solidification, during plastic
deformation, and as a consequence of thermal stresses
that result from rapid cooling.
▪ The number of dislocations, or dislocation density in a
material, is expressed as the total dislocation length per
unit volume or, equivalently, the number of dislocations
that intersect a unit area of a random section.
▪ The units of dislocation density are millimeters of
dislocation per cubic millimeter or just per square
millimeter.

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14
Cont.
(MSG)
▪ Dislocation densities as low as 103 mm-2 are typically
found in carefully solidified metal crystals.
▪ For heavily deformed metals, the density may run as
high as 109 to 1010 mm-2.
▪ Heat-treating a deformed metal specimen can diminish
the density to on the order of 105 to 106 mm-2.
▪ By way of contrast, a typical dislocation density for
ceramic materials is between 102 and 104 mm-2.
▪ For silicon single crystals used in integrated circuits the
value normally lies between 0.1 and 1 mm-2.

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15
Characteristics of Dislocations
(MSG)
▪ Several characteristics of dislocations are important with
regard to the mechanical properties of metals.
▪ These include strain fields that exist around dislocations,
which are influential in determining the mobility of the
dislocations, as well as their ability to multiply.
▪ When metals are plastically deformed, some fraction of
the deformation energy (approximately 5%) is retained
internally; the remainder is dissipated as heat.
▪ The major portion of this stored energy is as strain
energy associated with dislocations.

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16
Cont.
(MSG)
▪ Consider the edge dislocation
represented in Figure.
▪ As already mentioned, some atomic
lattice distortion exists around the
dislocation line because of the presence
of the extra half-plane of atoms.
▪ As a consequence, there are regions in
which compressive, tensile, and shear
lattice strains are imposed on the
neighboring atoms.
▪ For example, atoms immediately above and adjacent to the dislocation line
are squeezed together.
▪ As a result, these atoms may be thought of as experiencing a compressive
strain relative to atoms positioned in the perfect crystal and far removed
from the dislocation; this is illustrated in Figure.
Regions of compression (green) and tension
(yellow) located around an edge dislocation.

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17
Cont.
(MSG)
▪ Directly below the half-plane, the effect
is just the opposite; lattice atoms sustain
an imposed tensile strain, which is as
shown.
▪ Shear strains also exist in the vicinity of
the edge dislocation.
▪ For a screw dislocation, lattice strains
are pure shear only.
▪ These lattice distortions may be
considered to be strain fields that
radiate from the dislocation line.
▪ The strains extend into the surrounding
atoms, and their magnitude decreases
with radial distance from the
dislocation.
Regions of compression (green) and tension
(yellow) located around an edge dislocation.

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18
Repulsive Interaction
(MSG)
▪ The strain fields surrounding dislocations in close proximity to one
another may interact such that forces are imposed on each dislocation
by the combined interactions of all its neighboring dislocations.
▪ For example, consider two edge dislocations that have the same sign
and the identical slip plane, as represented in Figure a.
▪ The compressive and tensile strain fields for both lie on the same side of
the slip plane; the strain field interaction is such that there exists
between these two isolated dislocations a mutual repulsive force that
tends to move them apart.
(a) Two edge dislocations of the
same sign and lying on the same
slip plane exert a repulsive force
on each other; C and T denote
compression and tensile regions,
respectively.

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Attractive Interaction
(MSG)
▪ On the other hand, two dislocations of opposite sign and having the
same slip plane will be attracted to one another, as indicated in Figure b.
▪ Dislocation annihilation will occur when they meet.
▪ That is, the two extra half-planes of atoms will align and become a
complete plane.
▪ Dislocation interactions are possible between edge, screw, and/or
mixed dislocations, and for a variety of orientations.
(b) Edge dislocations
of opposite sign and
lying on the same slip
plane exert an
attractive force on
each other. Upon
meeting, they
annihilate each other
and leave a region of
perfect crystal.

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20
Cont.
(MSG)
▪ These strain fields and associated forces are important in
the strengthening mechanisms for metals.
▪ During plastic deformation, the number of dislocations
increases dramatically.
▪ We know that the dislocation density in a metal that has
been highly deformed may be as high as 1010 mm-2.
▪ One important source of these new dislocations is existing
dislocations, which multiply; furthermore, grain boundaries,
as well as internal defects and surface irregularities such as
scratches and nicks, which act as stress concentrations, may
serve as dislocation formation sites during deformation.

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SLIP SYSTEMS
(MSG)
▪ Dislocations do not move with the same degree of ease on all
crystallographic planes of atoms and in all crystallographic directions.
▪ Ordinarily there is a preferred plane, and in that plane there are
specific directions along which dislocation motion occurs.
▪ This plane is called the slip plane; it follows that the direction of
movement is called the slip direction.
▪ This combination of the slip plane and the slip direction is termed the
slip system.
▪ The slip system depends on the crystal structure of the metal and is
such that the atomic distortion that accompanies the motion of a
dislocation is a minimum.
▪ For a particular crystal structure, the slip plane is the plane that has the
most dense atomic packing—that is, has the greatest planar density.
▪ The slip direction corresponds to the direction, in this plane, that is
most closely packed with atoms—that is, has the highest linear density.

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22
Cont.
(MSG)
▪ Consider, for example, the FCC crystal structure, a unit cell of which is
shown in Figure a below.
▪ There is a set of planes, the {111} family, all of which are closely
packed.
▪ A (111)-type plane is indicated in the unit cell; in Figure b below, this
plane is positioned within the plane of the page, in which atoms are
now represented as touching nearest neighbors.
(a) A {111} slip system shown
within an FCC unit cell. (b) The
(111) plane from (a) and three slip
directions (as indicated by arrows)
within that plane constitute
possible slip systems.

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23
Cont.
(MSG)
▪ Slip occurs along <110>-type directions within the {111} planes, as
indicated by arrows in Figure shown before.
▪ Hence, {111}<110> represents the slip plane and direction
combination, or the slip system for FCC.
▪ Figure b shown on previous slide demonstrates that a given slip plane
may contain more than a single slip direction.
▪ Thus, several slip systems may exist for a particular crystal structure;
the number of independent slip systems represents the different
possible combinations of slip planes and directions.
▪ For example, for face-centered cubic, there are 12 slip systems: four
unique {111} planes (i.e. (111), (ഥ
𝟏𝟏𝟏), (ഥ
𝟏ഥ
𝟏𝟏) and (ഥ
𝟏ഥ
𝟏ഥ
𝟏) planes) and,
within each plane, three independent directions.
▪ The possible slip systems for BCC and HCP crystal structures are listed
in Table on next slide.

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24
Cont.
(MSG)

Body Centered Cubic (𝛂 − 𝐅𝐞, 𝐖, 𝐌𝐨 )
2023 (IUB)
Materials Simulation Group (MSG) 25
▪ 𝟏𝟏𝟎 𝐒𝐥𝐢𝐩 𝐩𝐥𝐚𝐧𝐞 𝐟𝐚𝐦𝐢𝐥𝐲 = 12 following planes
➢ 𝟏𝟏𝟎 , 𝟏𝟎𝟏 , 𝟎𝟏𝟏 ,
➢ ഥ
𝟏𝟏𝟎 , ഥ
𝟏𝟎𝟏 , 𝟎ഥ
𝟏𝟏 ,
➢ 𝟏ഥ
𝟏𝟎 , 𝟏𝟎ഥ
𝟏 , 𝟎𝟏ഥ
𝟏 ,
➢ ഥ
𝟏ഥ
𝟏𝟎 , ഥ
𝟏𝟎ഥ
𝟏 , 𝟎ഥ
𝟏ഥ
𝟏
▪ Out of these 12 planes
▪ 𝟏𝟏𝟎 and ഥ
𝟏ഥ
𝟏𝟎 are similar and will be counted as 1 plane
▪ 𝟏𝟎𝟏 and ഥ
𝟏𝟎ഥ
𝟏 are similar and will be counted as 1 plane
▪ 𝟎𝟏𝟏 and 𝟎ഥ
𝟏ഥ
𝟏 are similar and will be counted as 1 plane
▪ And rest of 6 planes ഥ
𝟏𝟏𝟎 , ഥ
𝟏𝟎𝟏 , 𝟎ഥ
𝟏𝟏 , 𝟏ഥ
𝟏𝟎 , 𝟏𝟎ഥ
𝟏 , 𝟎𝟏ഥ
𝟏 are similar and
will be treated as single
❖ So, at the end we will be having four (04) planes.
▪ ഥ
𝟏𝟏𝟏 𝐒𝐥𝐢𝐩 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐟𝐚𝐦𝐢𝐥𝐲 = 03 following directions
➢ ഥ
𝟏𝟏𝟏 , 𝟏ഥ
𝟏𝟏 , 𝟏𝟏ഥ
𝟏
❑ So the number of slip systems are 𝟎𝟒 × 𝟎𝟑 = 𝟏𝟐

2023 (IUB)

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28
Cont.
(MSG)
𝟏𝟏ഥ
𝟏
ഥ
𝟏𝟏𝟏
𝟏ഥ
𝟏𝟏
y-axis
z-axis
x-axis

Body Centered Cubic (𝛂 − 𝐅𝐞, 𝐖)
2023 (IUB)
▪ 𝟐𝟏𝟏 𝐒𝐥𝐢𝐩 𝐩𝐥𝐚𝐧𝐞 𝐟𝐚𝐦𝐢𝐥𝐲 = 24 following planes
➢ 𝟐𝟏𝟏 , 𝟏𝟐𝟏 , 𝟏𝟏𝟐 ,
➢ ഥ
𝟐𝟏𝟏 , ഥ
𝟏𝟐𝟏 , ഥ
𝟏𝟏𝟐 ,
➢ 𝟐ഥ
𝟏𝟏 , 𝟏ഥ
𝟐𝟏 , 𝟏ഥ
𝟏𝟐 ,
➢ 𝟐𝟏ഥ
𝟏 , 𝟏𝟐ഥ
𝟏 , 𝟏𝟏ഥ
𝟐 ,
➢ 𝟐𝟏𝟏 , 𝟏𝟐𝟏 , 𝟏𝟏𝟐 ,
➢ ഥ
𝟐𝟏ഥ
𝟏 , ഥ
𝟏𝟐ഥ
𝟏 , ഥ
𝟏𝟏ഥ
𝟐 ,
➢ 𝟐𝟏𝟏 , 𝟏𝟐𝟏 , 𝟏𝟏𝟐 ,
➢ ഥ
𝟐ഥ
𝟏ഥ
𝟏 , ഥ
𝟏ഥ
𝟐ഥ
𝟏 , ഥ
𝟏ഥ
𝟏ഥ
𝟐
❖ So, at the end we will be having four (04) planes.
03 Unique Planes (03 colors)
01 Unique Plane
▪ ഥ
𝟏𝟏𝟏 𝐟𝐚𝐦𝐢𝐥𝐲 = 03 following unique directions: ഥ
𝟏𝟏𝟏 , 𝟏ഥ
𝟏𝟏 , 𝟏𝟏ഥ
𝟏
❑ So the number of slip systems are 𝟎𝟒 × 𝟎𝟑 = 𝟏𝟐

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32
Cont.
(MSG)
▪ For each of these structures, slip is possible on more than one family of
planes (e.g., {110}, {211}, and {321} for BCC). For metals having these
two crystal structures, some slip systems are often operable only at
elevated temperatures.
▪ Metals with FCC or BCC crystal structures have a relatively large
number of slip systems (at least 12).
▪ These metals are quite ductile because extensive plastic deformation is
normally possible along the various systems.
▪ Conversely, HCP metals, having few active slip systems, are normally
quite brittle.
▪ With regard to the process of slip, a Burgers vector’s direction
corresponds to a dislocation’s slip direction, whereas its magnitude is
equal to the unit slip distance (or interatomic separation in this
direction).

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33
Cont.
(MSG)
▪ Of course, both the direction and the magnitude of 𝒃 will
depend on crystal structure, and it is convenient to specify a
Burgers vector in terms of unit cell edge length (a) and
crystallographic direction indices.
▪ Burgers vectors for face-centered cubic, body-centered cubic,
and hexagonal close-packed crystal structures are given as
follows:
𝐛 FCC =
𝑎
2
< 110 >
𝐛 BCC =
𝑎
2
< 111 >
𝐛 HCP =
𝑎
3
< 11ത
20 >

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34
(MSG)
Section-3
Slip in Single Crystals

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35
SLIP IN SINGLE CRYSTALS
(MSG)
▪ Though an applied stress
may be pure tensile (or
compressive), shear
components exist at all but
parallel or perpendicular
alignments to the stress
direction.
▪ These are termed resolved
shear stresses, and their
magnitudes depend not
only on the applied stress,
but also on the orientation
of both the slip plane and
direction within that plane.

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36
Cont.
(MSG)
▪ Let 𝝓 represents the angle between the
normal to the slip plane and the applied
stress direction, and 𝝀 be the angle between
the slip and stress directions.
Slip direction
F
Component of force
along slip direction
F𝐜𝐨𝐬 𝝀
𝝀
▪ The resolved shear stress 𝝉𝑹
𝝉𝑹 =
𝐑𝐞𝐬𝐨𝐥𝐯𝐞𝐝 𝐟𝐨𝐫𝐜𝐞 𝐚𝐜𝐭𝐢𝐧𝐠 𝐚𝐥𝐨𝐧𝐠 𝐬𝐥𝐢𝐩 𝐩𝐥𝐚𝐧𝐞
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐬𝐥𝐢𝐩 𝐩𝐥𝐚𝐧𝐞
Area of slip plane
A
𝜙 Base
Hyp
⊥ 𝐚𝐫
𝐜𝐨𝐬 𝝓 =
𝐀
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐬𝐥𝐢𝐩 𝐩𝐥𝐚𝐧𝐞 =
𝐀
cos 𝝓

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37
Cont.
(MSG)
▪ 𝝉𝑹 =
𝐑𝐞𝐬𝐨𝐥𝐯𝐞𝐝 𝐟𝐨𝐫𝐜𝐞 𝐚𝐜𝐭𝐢𝐧𝐠 𝐚𝐥𝐨𝐧𝐠 𝐬𝐥𝐢𝐩 𝐩𝐥𝐚𝐧𝐞
▪ 𝝉𝑹 =
𝐅 cos λ
𝐀
cos ϕ
=
𝐅
𝐀
cos λ cos ϕ
▪ 𝝉𝑹 = 𝝈 cos ϕ cos λ
▪ 𝝉𝑹 = 𝝈 cos ϕ cos λ
▪ where 𝝈 is the applied stress.
▪ Schmid’s Law: The value of 𝝉𝑹 at which slip occurs in a given material
with specified dislocation density and purity is a constant, known as
the critical resolved shear stress 𝝉𝒄𝒓𝒔𝒔.
▪ 𝝉𝑹 𝒎𝒂𝒙 = 𝝈 cos ϕ cos λ 𝒎𝒂𝒙
▪ 𝝉𝒄𝒓𝒔𝒔 is the minimum shear stress required to initiate the slip and is the
property of material.
Schmid factor

2023 (IUB)
38
Cont.
(MSG)
▪ In general, 𝝓 + 𝝀 ≠ 𝟗𝟎°, as the tensile axis, the slip plane normal, and
the slip direction all need not lie in the same plane.
▪ A metal single crystal has a number of different slip systems that are
capable of operating.
▪ The resolved shear stress normally differs for each one because the
orientation of each relative to the stress axis (𝝓 and 𝝀 angles) also
differs.
▪ However, one slip system is generally oriented most favorably—that is,
has the largest resolved shear stress, 𝝉𝑹(𝐦𝐚𝐱):
▪ 𝝉𝑹 𝒎𝒂𝒙 = 𝝈 (cos 𝝓 cos 𝝀)𝐦𝐚𝐱
▪ In response to an applied tensile or compressive stress, slip in a single
crystal commences (starts or initiates) on the most favorably oriented
slip system when the resolved shear stress reaches some critical value,
termed the critical resolved shear stress 𝝉𝒄𝒓𝒔𝒔.

2023 (IUB)
39
Cont.
(MSG)
▪ 𝝉𝒄𝒓𝒔𝒔 represents the minimum shear stress required to initiate slip and
is a property of the material that determines when yielding occurs.
▪ The single crystal plastically deforms or yields when
▪ 𝝉𝑹 𝒎𝒂𝒙 = 𝝉𝐜𝐫𝐬𝐬
▪ 𝝈𝒚 (cos 𝝓 cos 𝝀)𝐦𝐚𝐱 = 𝝉𝐜𝐫𝐬𝐬
▪ The magnitude of the applied stress required to initiate yielding (i.e.,
the yield strength 𝝈𝒚) is
▪ 𝝈𝒚 =
𝝉𝐜𝐫𝐬𝐬
(cos 𝝓 cos 𝝀)𝐦𝐚𝐱
(i)
▪ The minimum stress necessary to introduce yielding occurs when a
single crystal is oriented such that 𝝓 = 𝝀 = 𝟒𝟓° ; under these
conditions
▪ 𝝈𝒚 = 𝟐𝝉𝐜𝐫𝐬𝐬 using cos 45°
=
2
2
in equation (i)

2023 (IUB)
40
Cont.
(MSG)
▪ For a single-crystal specimen that is stressed in
tension, deformation will be as in Figure ------->,
where slip occurs along a number of equivalent
and most favorably oriented planes and
directions at various positions along the
specimen length.
▪ This slip deformation forms as small steps on the
surface of the single crystal that are parallel to
one another and loop around the circumference
of the specimen as indicated in Figure ------->.
▪ Each step results from the movement of a large
number of dislocations along the same slip plane.
▪ On the surface of a polished single-crystal
specimen, these steps appear as lines, which are
called slip lines.

2023 (IUB)
41
Zinc Single Crystal and Slip
(MSG)
▪ A zinc single crystal that has been plastically deformed to
the degree that these slip markings are discernible
(visible or apparent) is shown in Figure on right.
▪ With continued extension of a single crystal, both the
number of slip lines and the slip step width will increase.
▪ For FCC and BCC metals, slip may eventually begin along
a second slip system, the system that is next most
favorably oriented with the tensile axis.
▪ Furthermore, for HCP crystals having few slip systems, if
the stress axis for the most favorable slip system is either
perpendicular to the slip direction (𝝀 = 𝟗𝟎°) or parallel
to the slip plane (𝝓 = 𝟗𝟎°
), the critical resolved shear
stress will be zero.
▪ For these extreme orientations the crystal ordinarily
fractures rather than deforming plastically.

2023 (IUB)
42
Cont.
(MSG)
Figure: Schematic representation
showing normal ( 𝝈′ ) and shear ( 𝝉′ )
stresses that act on a plane oriented at
an angle 𝜃 relative to the plane taken
perpendicular to the direction along
which a pure tensile stress (𝜎) is applied.
𝜏′
= 𝜎 sin θ cos θ = 𝜎
sin 2θ
2
Equation 6.4b

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43
Problems for Students
(MSG)
Problem Book Page
No.
Example Problem 3.8 Materials Sci & Eng by WD Callister (2010)
8th Edition
61
8th Edition
63
8th Edition
68
8th Edition
93
8th Edition
113
8th Edition
207

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44
Materials Simulation Group (MSG)

Lectures 07-12, 2023.pdf

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