Defects in Crystals

1. DEFECTS IN CRYSTALS

2. DEFECTS IN CRYSTALS
• All materials however pure will have defects
especially vacancies.
• There are three main categories of defects in
crystals defined by size and dimensions
• a) Point Defects
– Can be a vacancy site with one atom missing or
– an interstitial atom not occupying its normal position
or
•

3. POINT DEFECTS IN CRYSTALS
• Substitutional impurities.
–These are impurities that substitute
themselves for one of the host atoms
in the lattice or
• Interstitial impurities.
–These are impurities that are
squeezed at points other than normal
lattice points.

4. DEFECTS IN CRYSTALS
b) Line defects
• These are one dimensional or lines of
imperfections in a material. They are known as
dislocations
c) Sheet/ surface or plane defects
• Examples of these are stacking faults, grain
boundaries, interphase boundaries
d) Volume Defects
these could be cavities in a material (voids or
bubbles)
• or even Precipitates

5. EM 310 – Material Science II
Crystallographic Structure – Volume/point Defects

7. DISLOCATIONS
• Dislocation geometry - A dislocation is a line of
defects that separates a region of a crystal that
has slipped from a region that has not
• Between the slipped and unslipped regions, the
structure will be dislocated; this boundary is
referred to as a dislocation line, or dislocation
• Dislocations affects electrical, chemical and in
particular mechanical properties
• Slip is a process by which materials undergo
deformation

9. DISLOCATION
• It is a line of discontinuity,
• it forms a closed loop in the interior of
the crystal or emerges at the surface
and
• the difference in the amount of slip
across the dislocation line is constant.

10. BURGER’S VECTOR
• a dislocation is characterized by the magnitude and
direction of the slip movement associated with it.
• This is called the Burgers vector, b, which for any
given dislocation line is the same all along its
length.
• In a crystal, the burger’s vector b is usually the
shortest lattice translation vector.
• Perfect dislocations are translation vectors.
• It is possible that b may not be a lattice translation
vector in other words may not join two lattice sites.
What results is an imperfect dislocation or a partial
dislocation.
• Typically the magnitude of a burgers vector is of the
order of a few Å (0.2-0.4 nm).

11. DISLOCATIONS
• There are two basic types of dislocations.
These are the edge and the screw dislocations
• Edge dislocations
– The burgers vector is perpendicular to the
dislocation line in an edge dislocation.

12. • an example of a positive edge dislocation which is
designated by symbol .
• The perpendicular line represents the extra half
plane, while the horizontal line represents the
direction of the slip plane.
• A negative edge dislocation has the opposite sign
and the extra half plane is below the slip plane.
• Positive and negative edge dislocations are physically
opposite dislocations and combining them results in
a perfect crystal.

13. SCREW DISLOCATION
• In the screw dislocation the Burgers vector is
parallel to the dislocation line.
• After slip has taken place, the planes which
were parallel before the dislocation form a
helicoidal surface i.e a screw thread with a
pitch of the magnitude of a burgers vector.
• A screw dislocation can be left handed or right
handed.
• Left and right handed screws are equal and
opposite.

14. SCREW DISLOCATION

15. MIXED DISLOCATIONS
In the figure above, the dislocation loop is pure screw at point A and pure edge at
point B, while along most of its length it has mixed edge and screw components.
A dislocation line is rarely pure edge or pure screw
Any dislocation can be resolved into edge and screw components.

16. BURGER’S CIRCUIT
• start at a lattice point and imagine a clockwise
path traced from atom to atom an equal
distance in each direction
• At the finish of the path the circuit does not
close.
• The closure failure from finish to start is the
Burgers vector b of the dislocation

17. BURGER’S CIRCUIT

18. DEFORMATION BY SLIP
• The usual method of plastic deformation in
metals is by the sliding of blocks of crystals over
one another along definite crystallographic
planes called slip planes.
• Slip is the process by which metals deform
plastically
• During slip the atoms move an integral number of
atomic distances along the slip plane.
• the crystal structure is perfectly restored after slip
has taken place provided that the deformation
was uniform.
• Each atom in the slipped part moves forward the
same integral number of lattice spacings.

19. DEFORMATION BY SLIP
• Slip occurs more readily in specific directions on
certain crystallographic planes.
• Generally the slip plane is the plane of the
greatest atomic density and
• the slip direction is the closest-packed direction
within the slip plane.
• Since the planes of greatest atomic density are
also the most widely spaced planes in the crystal
structure, the resistance to slip is generally less
for these planes than for any other set of
planes.

20. SLIP SYSTEMS
• The slip plane together with the slip direction
establishes the slip system.
The limited number of slip systems in HCP is the reason for
the extreme orientation dependence and low ductility of HCP
crystals.
BCC materials have no definite single slip plane. It has three
planes and one closest packed direction
Since planes are not as close packed as FCC, higher shearing
stresses are required to cause slip in BCC

21. SLIP SYSTEMS
• Example: Determine the slip systems for slip on the (111) in the
FCC crystal structure and sketch the result.
• The slip directions in the FCC are the <110> type directions. Slip
directions are most easily established from a sketch of the
(111) plane.
• To prove that these slip directions lie in the plane, a dot
product between the plane and the direction should result in a
value of zero.

22. SLIP IN A PERFECT LATTICE
• If slip is assumed to occur by the translation of one
plane of atoms over another, it is possible to make a
reasonable estimate of the shear stress required for
such a movement in a perfect lattice.
• Consider two planes of atoms subjected to a
homogeneous shear stress.
• The shear stress is assumed to act in the slip plane
along the slip direction.
• The distance between atoms in the slip direction is b
• the spacing between adjacent planes is a.
• The shear stress causes a displacement x in the slip
direction between the pair of adjacent lattice planes.

24. SLIP IN A PERFECT LATTICE
• The shearing stress is initially zero when the two
planes are coincident,
• and it is also zero when the atoms of the top plane
is over point 2 on the bottom plane.
• The shearing stress is also zero when the atoms of
the top plane are midway between those of the
bottom plane, since this is a symmetry position.
• Between these positions each atom is attracted to
wards the nearest atom of the other row,
• so that the shearing stress is a periodic function of
the displacement.

25. As a first approximation, the relationship between shear stress and
displacement can be expressed as a sine function
𝜏 = 𝜏𝑚𝑠𝑖𝑛
2𝜋𝑥
𝑏
(1)
Where 𝜏𝑚 is the amplitude of the sine wave and 𝑏 is the period.
At small values of displacements, Hook’s law should apply.
𝜏 = 𝐺𝛾 =
𝐺𝑥
𝑎
(2)
For small values of 𝑥
𝑏 the sine function can be written as
𝜏 ≈ 𝜏𝑚
2𝜋𝑥
𝑏
(3)
Combining equations (2) and (3) provides an expression for the
maximum stress at which slip should occur
𝜏𝑚 =
𝐺
2𝜋
𝑏
𝑎
(4)
As a rough approximation, 𝑏 can be taken as equal to 𝑎, with the
result that the theoretical shear strength of a perfect crystal is
approximately equal to the shear modulus divided by 2𝜋.
𝜏𝑚 =
𝐺
2𝜋

26. SLIP IN A PERFECT LATTICE
• The shear modulus for metals is in the range 20 to
150 GPa.
• therefore equations (3-5) predicts that the
theoretical shear stress will be in the range (3 to 30)
GPa)
• actual values of shear stress required to produce
plastic deformation in metal single crystals are in the
range 0.5 MPa to 10 MPa.
• Since the shear strength of metal crystals is
approximately 1000 times greater than the observed
shear strength, it must be concluded that a
mechanism other than bodily shearing of planes of
atoms is responsible for slip.

27. SLIP BY DISLOCATION MOVEMENT
• The concept of the dislocation was first
introduced to explain the discrepancy between
the observed and the theoretical shear strength
of metals.
• For the dislocation concept to be valid, it is
necessary to show the following
– that the motion of a dislocation through a crystal
lattice requires a stress far smaller than the
theoretical shear stress
– that the movement of the dislocations produces a
step, or a slip band at the free surface.

28. SLIP BY DISLOCATION MOVEMENT
• In a perfect lattice atoms above and below the
slip plane are in minimum energy positions
• When a shear stress is applied to a crystal, the
same force opposing the movement acts on all
the atoms.
• When there is a dislocation in the crystal the
atoms well away from the dislocation are still in
the minimum energy positions but at a
dislocation only a small movement of atoms is
required.

29. SLIP BY DISLOCATION MOTION

30. • In figure (a) the extra plane of atoms at the edge dislocation
initially is at 4.
• Under the action of the shear stress, a very small movement of
atoms to the right will allow this half plane to line up with the
half plane 5’, at the same time cutting the half plane 5 from its
neighbours below the slip plane.
• the edge dislocation line has moved from its initial position
between planes 4’ and 5’ to a new position between planes 5’
and 6’
• Since the atoms around the dislocations are symmetrically
placed on opposite sides of the extra half plane, equal and
opposite forces oppose and assist the motion.
• To a first approximation there is no net force on the dislocation
and the stress required to move the dislocation is zero
. The continuation of this process under the stresses moves the
dislocation to the right.
When the extra half plane of atoms reaches a free surface figure
(b) it results in a slip step of one Burger’s vector, or one atomic
distance for the simple cubic lattice.

31. SLIP BY DISLOCATION
MOVEMENT

32. SLIP BY DISLOCATION MOVEMENT -
COTTRELL
• Consider that plastic deformation is the
transition from an unslipped state to a slipped
• the process is opposed by an energy barrier ∆E

33. SLIP BY DISLOCATION MOVEMENT -
COTTRELL
• Consider that plastic deformation is the
transition from an unslipped state to a slipped
• The process is opposed by an energy barrier ∆E

34. SLIP BY DISLOCATION MOVEMENT -COTTRELL
• The slipped material will grow at the expense
of the unslipped region by an advance of an
interfacial region
• the interfacial region is a dislocation.
• To minimise the energy for the transition, we
expect the interface thickness w to be
narrow

35. SLIP BY DISLOCATION MOVEMENT -
COTTRELL
• The distance w is the width of the dislocation
• The smaller the width of the dislocation, the
lower is the interfacial energy,
• But the wider the dislocation, the lower is the
elastic energy of the crystal because then the
atomic spacing in the slip direction is closer to
its equilibrium spacing.
• The equilibrium width of the dislocation is
determined by a balance between these
opposing energy changes.

36. SLIP BY DISLOCATION MOVEMENT -
COTTRELL
• The dislocation width is important because
it determines the force required to move a
dislocation through the crystal lattice
• This force is called the Peierls-Nabarro
force.
• The Peierls stress is the shear stress
required to move a dislocation through a
crystal lattice in a particular direction

37. SLIP BY DISLOCATION MOVEMENT -
COTTRELL
• Where a is the distance between slip planes and b is the
distance between atoms in a slip direction.
• Note that the dislocation width appears in the exponential
term so that the Peielrs stress will be sensitive to the
atomic position at the core of the dislocation.
• The equation is accurate enough to show that the stress
required to move a dislocation in a metal is quite low.

38. SLIP BY DISLOCATION MOVEMENT -
COTTRELL
• materials with wide dislocations will require a low stress to move the
dislocation.
• Physically this is so because when the dislocation is wide, the highly
distorted region at the core of the dislocation is not localized on any
particular atom in the crystal lattice.
• In ductile materials, the dislocation width is of the order of 10 atomic
spacings .
• However, in ceramic materials with directional covalent bonds, the
interfacial energy is high and the dislocations are narrow and
relatively immobile.
• This fact combined with the restriction on the slip systems imposed by
the requirements of electrostatic forces result in the low degree of
plasticity of ceramic materials.
• Ceramics become more ductile at high temperatures because thermal
activation helps the dislocations overcome the energy barrier.

39. SLIP BY DISLOCATION MOVEMENT -
COTTRELL
• The fact that slip occurs in close packed directions (b is
minimized),the Peierls stress will be lower.
• If a<b, as would occur for closely spaced but loosely
packed planes, the Peierls stress would be high.
• Thus the equation provides a basis for the observation
that slip occurs most readily on close packed planes in
the close packed directions.
• When the crystal structure is complex without highly
closely packed planes and directions, the dislocations
tend to be immobile.
• This causes the brittleness and high hardness of
intermetallic compounds.

40. SHEAR STRAIN

41. SHEAR STRAIN

42. • consider the case where a dislocation has moved
part way through the crystal along the slip plane.
Since b is very small compare to L or h,

47. CRITICAL RESOLVED SHEAR STRESS
• The extent of slip in a single crystal depends on
• the magnitude of the shearing stress produced by
the eternal loads
• the geometry of the crystal structure, and
• the orientation of the active slip plane with
respect to the shearing stresses.

48. CRITICAL RESOLVED SHEAR STRESS
• Slip begins when the shearing stress on the slip
plane in the slip direction reaches a threshold
value called the critical resolved shear stress.
• This value is really the single crystal equivalent of
the yield stress of an ordinary stress strain curve.
• The value of the critical resolved shear stress
depends chiefly on the composition and the
temperature.

49. CRITICAL RESOLVED SHEAR STRESS
• The fact that different tensile loads are required to
produce slip in a single crystal of different orientation
can be rationalized by a critical resolved shear stress;
• this was first realized by Schmid.
• To calculate the critical resolved shear stress from a
single crystal tested in tension, it is necessary to
know, from x ray diffraction, the orientation, with
respect to the tensile axis, of the plane on which slip
first appears and the slip direction.

50. CRITICAL RESOLVED SHEAR STRESS
• Consider a cylindrical single crystal with cross sectional area A.
• The angle between the normal to the slip plane and the tensile axis
is 𝜙
• and the angle which the slip direction makes with the tensile axis
is 𝜆
• The area of the slip plane inclined at the angle 𝜑 will be
𝐴
𝑐𝑜𝑠𝜙
• and the component of the axial load acting in the slip plane in the
slip direction is 𝑃𝑐𝑜𝑠𝜆
• Therefore, the critical resolved shear stress is given by
• 𝜏𝑅 =
𝑃𝑐𝑜𝑠𝜆
𝐴
𝑐𝑜𝑠𝜙
=
𝑃
𝐴
𝑐𝑜𝑠𝜙𝑐𝑜𝑠𝜆.

51. CRITICAL RESOLVED SHEAR STRESS

52. CRITICAL RESOLVED SHEAR STRESS
• This equation gives the shear stress resolved on the slip plane in
the slip direction.
• This shear stress is a maximum when 𝜙 = 𝜆 = 45𝑜
• so that 𝜏𝑅 =
1
2
𝑃
𝐴
• If the tension axis is normal to the slip plane (𝜆 = 90𝑜)
• or if it is parallel to the slip plane (𝜙 = 90𝑜), the resolved shear
stress is zero.
• Slip will not occur for these extreme orientations since there is
no shear stress on the slip plane.
• Crystals close to these orientations tend to fracture rather than
slip.

53. EXAMPLE
• Determine the tensile stress that is applied along the 11
¯
0 axis of a silver
crystal to cause slip on the 111
¯
01
¯
1 system. The critical resolved shear
stress is 6 Mpa.
• The angle between the tensile axis 11
¯
0 and the normal to (111
¯
) is
• 𝑐𝑜𝑠𝜙 =
1 1 + −1 −1 + 0 (−1)
(1)2+(−1)2+(0)2 (1)2+(−1)2+(−1)2
=
2
2 3
=
2
6
• The angle between the tensile axis 11
¯
0 and the slip direction 011
¯
is
• 𝑐𝑜𝑠𝜆 =
1 0 + −1 −1 + 0 (−1)
2 (0)2+(−1)2+(−1)2
=
1
2 2
=
1
2
• 𝜎 =
𝑃
𝐴
=
𝜏𝑅
𝑐𝑜𝑠𝜙𝑐𝑜𝑠𝜆
=
6
2
6
1
2
= 6 6 = 14.7𝑀𝑃𝑎

54. CRITICAL RESOLVED SHEAR STRESS
• The magnitude of the critical resolved shear stress of a
crystal is determined by the interaction of its
population of dislocations with each other and with
defects such as vacancies, interstitials, and impurity
atoms.
• This stress is greater than the stress required to move a
single dislocation, but it is appreciably lower than the
stress required to produce slip in a perfect lattice.

55. CRITICAL RESOLVED SHEAR STRESS
• On the basis of this reasoning, the critical
resolved shear stress should decrease as the
density of defects decreases, provided that the
total number of imperfections is not zero.
• When the last dislocation is eliminated, the
critical resolved shear stress should rise abruptly
to the high value predicted for the shear strength
of a perfect crystal.

56. GENERATION OF DISLOCATIONS
• Before plastic deformation can happen, there must be
some dislocations present.
• If there were no sources generating dislocations, cold
work should decrease rather than increase the density
of dislocations in a single crystal.
• The fact that cold worked metals have high dislocation
densities indicates that there must be a mechanisms
that generate dislocations in the first place!

57. GENERATION OF DISLOCATIONS
• The most important mechanism dislocation
generation is the Frank-Read mechanism.
• Consider a segment of dislocation firmly anchored at
two points (red circles).
• The force F = b · τ res is shown by a sequence of
arrows

58. DISLOCATION GENERATION
• The dislocation segment responds to the force by
bowing out.
• If the force is large enough, the critical configuration
of a semicircle may be reached.
• This requires a maximum shear stress of τmax = Gb/R

59. DISLOCATION GENERATION
• If the shear stress is higher than Gb/R, the radius of
curvature is too small to stop further bowing out.
• The dislocation is unstable and the following process
now proceeds automatically and quickly.

60. DISLOCATION GENERATION
• Since the two line vectors at the point of
contact have opposite signs (or, if you only look
at the two parts almost touching: the Burgers
vectors have different signs for the same line
vectors), the segments in contact will annihilate
each other.
• The configuration shown is what you have
immediately after contact; it is totally unstable
(think of the rubber band model!).

61. DISLOCATION GENERATION
• It will immediately form a straight segment and a
"nice" dislocation loop which will expand under the
influence of the resolved shear stress.

62. DISLOCATION GENERATION
• The regained old segment will immediately start
to go through the whole process again, and
again, and again, as long as the force exists. A
whole sequence of nested dislocation loops will
be produced.

63. • The loop is free to move, i.e. grow much larger under the
applied stress.
• It will encounter other dislocations, form knots and become
part of a network.
• The next loop will follow and so on - as long as there is enough
shear stress.
• The process can be repeated over and over again at a single
source each time producing a dislocation loop which produces
slip of one burgers vector along the slip plane.
• The source does not however, continue indefinitely.
• The back stresses produced by the dislocation pile up along the
slip plane opposes the applied stresses and when this is equal
to the critical stress Tau is equal to GB/l the source ceases to
operate.

65. MECHANISMS OF SLIP AND CLIMB
• If the resolved shear stress on the slip plane is τ
and the Burgers vector of the dislocation b, the
force on the dislocation, i.e. force per unit length
of dislocation, is F = τb.

66. DISLOCATION GLIDE
• A dislocation is able to glide in that slip plane which contains
both the line of the dislocation and its Burgers vector.
• The edge dislocation is confined to glide in one plane only.
• An important difference between the motion of a screw
dislocation and that of an edge dislocation arises from the fact
that the screw dislocation is cylindrically symmetrical about its
axis with its b parallel to this axis.
• To a screw dislocation all crystal planes passing through the
axis look the same and therefore the motion of the screw
dislocation is not restricted to a single slip plane, as is the case
for a gliding edge dislocation.

67. CROSS SLIP
• The process whereby a screw dislocation
glides into another slip plane having a slip
direction in common with the original slip
plane is called cross-slip.
• This is illustrated in the figure below. Usually,
the cross-slip plane is also a close-packed
plane, e.g. {111} in fcc crystals.

68. • The mechanism of slip illustrated above
shows that the slip or glide motion of an
edge dislocation is restricted, since it can
only glide in that slip plane which
contains both the dislocation line and its
Burgers vector

69. DISLOCATION CLIMB
• However, movement of the dislocation line in a direction normal to
the slip plane can occur under certain circumstances; this is called
dislocation climb.
• To move the extra half-plane either up or down, as is required for
climb, requires mass transport by diffusion and is a non-
conservative motion.
• For example, if vacancies diffuse to the dislocation line it climbs up
and the extra half-plane will shorten.
• However, since the vacancies will not necessarily arrive at the
dislocation at the same instant, or uniformly, the dislocation climbs
one atom at a time and some sections will lie in one plane and other
sections in parallel neighboring planes.

70. DISLOCATION CLIMB
• Where the dislocation deviates from one plane to
another it is known as a jog, and from the
diagrams above it is evident that a jog in a
dislocation may be regarded as a short length of
dislocation not lying in the same slip plane as the
main dislocation but having the same Burgers
vector.
• Jogs may also form when a moving dislocation
cuts through intersecting dislocations, i.e. forest
dislocations, during its glide motion.

71. DISLOCATION JOGS AND CLIMB
• Where the dislocation deviates from one plane to
another it is known as a jog, and from the diagrams
above it is evident that a jog in a dislocation may be
regarded as a short length of dislocation not lying in
the same slip plane as the main dislocation but
having the same Burgers vector.
• Jogs may also form when a moving dislocation cuts
through intersecting dislocations, i.e. forest
dislocations, during its glide motion.