What you have to learn from this chapter

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Chapter 4. Introduction to Dislocations
What you have to learn from this chapter? ?
Definition of Dislocations ? Crystallographic of
Dislocations ? Dislocation Motions ?
Dislocation Interactions ? Stress Field of
Dislocations ? Energy of Dislocations ?
Interaction of Dislocations with Point Defects
2
? Discovery of dislocation discrepancy between
the theoretical and observed yield stresses of
crystal e.g. a Mg single crystal strain in
tension gt easily stretches out to a narrow
ribbon much longer than the original crystal!
gt examine the surface of the deformed crystal,
a series of fine steps have been formed
(on basal plane) gt the phenomena is
called slip, visible marking on the
surface is called slip lines, the
crystallographic plane on which the shear has
occurred is called slip plane. See.
Fig.4.1, 4.2, 4.3.
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The shear stress required to produce plastic
flow ltlt cohesive force between atoms! Q how do
atoms move to produce steps?
Force
elastic deformation
0.7MPa
Mg
Stress
plastic deformation
Force
Strain
t
t
a
Assume the radius of the atom is a!
2a
Shear strain ?? a/2a 0.5
Shear stress (?) shear modulus (?) ? Shear
strain (?) For Mg ? 17.2 GPa gt ? ? 8-9 ?103
MPa (8-9 ?103 MPa) / (0.7 MPa) ? 104. gt
discrepancy gt dislocations
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(observable with TEM, dislocation lines on the
slip plane), see. Fig. 4.7-4.10. gt evidence
etch pits and slip plane
See Fig. 4.11 for a 3D view of a edge dislocation
Large distortion only occurs around core region,
the shear force only have to move atoms in the
core region to have the extra plane moving gt to
the edge of crystal gt slip step
a
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See Fig. 4.13-4.14
Two basic orientations a dislocation may take
edge and screw dislocation
Edge dislocation positive and
negative Screw dislocation left-hand and
right hand See Fig. 4.15. (Is the definition
absolute?)
Screw dislocation represents the
boundary between slipped and unslipped area
Dislocations could be a combination of pure edge
and screw dislocations! The combinations include
different directions and See, Fig. 4.16-18.
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screw
screw
edge
below the slip plane
above the slip plane
edge
Sharp transition from a pure edge to a pure
screw! Fig. 4.19 is a dislocation with part of
edge component and part of screw component. gt
dislocations could be curved.
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Fig. 4.20 dislocation loop (not necessary any
shape) under stress gt open or close. ?
Dislocations is a boundary between a slipped area
and unslipped area gt can not end in a crystal,
but could end at the surface!
? The Burgers Vector
local Burgers vector
Extra plane
Starting point
End point
Edge dislocation
Screw dislocation
Convention right hand circuit finish point to
start point (RHFS)
true Burgers vector
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? RH or LH (left hand) no universal convention ?
If the Burgers vector and the orientation of the
dislocation are known gt Dislocation is
completely described!
? Summarize (a) For a given dislocation, there
is only one Burgers vector. (b) A
dislocation must end on itself (forming a loop)
or on other dislocations, (forming a
network), or on surface such as an
external surface or a grain boundary.
(c) The sum of the Burgers vectors at the node or
point of junction of the dislocation is
zero. (d) The slip (or glide) plane of a
dislocation contains the dislocation line
and its Burgers vector.
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(e) Edge dislocation the line of an
edge dislocation ? Burgers vector an
edge dislocation moves in the direction of
Burgers vector (slip direction) under a shear
stress (f) screw dislocation the line
of a screw dislocation ?? Burgers vector
a screw dislocation moves in the direction
perpendicular to the Burgers vector under a
shear stress (g) When dislocation
slips, the atoms move in the direction
of the Burgers vector, but line moves
normal to itself on the slip plane.
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? Vector Notation for Dislocation ? In
real crystal, the arrangement of atoms is much
more complex than previously mentioned
cases gt difficult to visualize the
geometrical appearance of a complicated
dislocation gt dislocation is defined by
its Burgers vector and dislocation line. ?
In most crystal, the slip plane for a
dislocation typically happens to be the
close-packed plane! ?smallest shear
distance involved during a slip
movement gt dislocation with Burgers vector
smallest shear vector, energetically
favored ? Burgers vector notation include
direction and length
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z
z
z
y
y
y
x
x
x
Closed-packed plane in simple cubic is 100 gt
the smallest shear vector 100
Closed-packed plane in BCC is 110 gt the
smallest shear vector 1/2111
Closed-packed plane in FCC is 111 gt The
smallest shear vector 1/2110
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? Dislocations in the FCC Lattice Slip
plane 111
b Burgers vector 1/2lt110gt
b
b
b
The kinetic path for this movement
is Necessarily energetically favorable!
Some other paths may involve even lower energy!
b
c
d
b
b
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b 1/2-110
b cd gt
z
b total dislocation c, d partial dislocation
y
x
c 1/6-12-1
d 1/6-21-1
The strain energy of the system is lowered when a
total dislocation breaks down into two partial
dislocations. (otherwise no reason to do that!) ?
Dislocation energy is proportional to the square
of Burgers vector ? The break down is favored
when b2 gt c2 d2. (kinetics seems to be related
to the least action rules!)
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A
An extended dislocation!
B
C
C
B
C
C
B
C
C B A C B
C B A C B
A C A C B
Stacking faults
The stacking fault occurs on the slip plane and
is bounded at its ends by the partial
dislocations (Shockley). gt If a stacking fault
terminates inside a crystal, its boundaries will
form a partial dislocation.
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101
121
Open circles atoms in the (10-1) plane Closed
circles atoms Immediately above or below (10-1)
plane
111
1-21
C
B
M
A
C
B
Slip vector
A
M location of Shockley partial dislocation
? Intrinsic and Extrinsic Stacking Faults in FCC
metals
Intrinsic SF condensation of vacancies Extrinsic
SF condensation of interstitials gt See Fig.
4.29 AB gt Frank partial dislocation 1/3lt111gt
(edge) loop
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B
B
A
A
111
C
C
B
A
A
C
C
B
B
Frank partial dislocation the Burgers vector is
not contained in the close-packed plane gt it
cannot glide
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101
121
111
1-21
C
B
A
C
Burgers Vector
B
A
Removal (or addition) of a close pack layer of
atoms gt formation of a 1/3lt111gt Frank partial
dislocation
Shockley partial Frank partial gt Perfect
dislocation
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B
A
A
C
C
B
A
A
C
C
B
B
Extrinsic Frank needs two Shockley partials to
eliminate the faulted area!
19
? Extended Dislocations in Hexagonal Metals
Close packed plane basal plane a total
dislocation gt dissociated partials
FCC equivalent
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? Climb of Edge Dislocations
Slip plane the plane contains both the
dislocation line and its Burgers vector. Screw gt
any plane ? dislocation line Burgers
vector Edge gt only one possible plane
?dislocation line ? Burgers vector Climb is
another way for edge dislocation to (in the
direction ? slip plane)
Positive Climb ?
Negative Climb ?
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Compressive stress gt promote positive
climb Tensile stress gt promote negative
climb Shear stress gt slip normal stress gt
climb Any processes that produce lots of
vacancies gt promote positive climb e.g.
high temperature . Any processes that
produce lots of interstitials gt promote
negative climb e.g. oxidations . There
are other examples that show the interaction
of defects with vacancies and interstitials!
22
? Dislocation Intersections
? In real crystal gt 3D network of linear faults
gt any given plane a certain number of
dislocations on it as well as many other
dislocations that intersect it at various
angles gt require works for a dislocation to
cut through other dislocations ? Two basic
types (1) kinks steps that lie in the slip
plane of a dislocation kink in a screw
dislocation is edge type kink in a edge
dislocation is screw type
edge
screw
p
p
o
screw
o
edge
n
n
See Fig. 4.34
b
b
m
m
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b
b
Kinks could be easily removed
(2) jog steps normal to the slip plane of a
dislocation
Edge
screw
b
b
Can not glide along the vertical plane can
only move by climb
Edge
edge
mobile
immobile
b
b
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? The Stress Field of a Screw Dislocation
b
Dislocation line
Rotate 2pr gt displacement b
y
er
2pr
q
gt shear strain
r
eq
z
x
b
For isotropic materials, G(?)E/2(1u) E
Youngs modulus u Poissons ratio
y
?, ? strain ? normal stress, ? shear stress
(x, y)
?
x
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displacement in z direction from 0
for ? 0 to b for ? 2?
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Similarly,
Not a correct solution for a finite rod!
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gt elastic strain around a dislocation 1/r (r gt
b) near the core, the strain are no longer
elastic. The elastic distortion contains no
tensile or compressive components only shear
component.
? The Stress Field of an Edge Dislocation
The displacement in x-direction is b in
z-direction is 0. Plane strain displacement
ux(x,y) uy(x,y) uz0 ? /?z0 Airy stress
function (see J.P. Hirth and J. Lothe, Theory of
Dislocations, Krieger Publishing Company, 1992.)
y
r
?
x
b
z
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in Cartesian coordinate
in polar coordinate
? Force on a dislocation
Consider a single edge dislocation The work done
by shear stress t which causes the crystal to
slip by a distance b is Wt Fb t lD
b
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Assume dislocation has a force Fd (force per
unit length) gt when dislocation move, the
work done by the dislocation is Fd lD gt
dislocation move across the crystal gt crystal
upper part shift b relative to lower part gt Wt
t lD b Fd lD gt Fd t b similar
argument could be applied to screw
dislocations as well gt Fd t b. Force acts
normal to the dislocations lines for both
types of dislocations.
F
H
l
D
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z
The climb forces on an edge dislocations
similar argument gt Fc -? b ? normal
stress compressive gt climb force points
toward the z
direction tensile gt climb force points
toward the -z direction!
F
?
?
y
b
x
In the general case, dislocations are curved
gt general treatment, consider the dislocation
line vector (? ?x, ?y, ?z), Burgers vector
(b bx, by, bz), and a general stress field
(? ?x, ?y, ?z), where ? b?G (G stress
tensor, typical symbol is ?)
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Only stress components that are parallel to
Burgers vector moves dislocations and the
force on the dislocation is ???. (? unit
length) with direction normal to dislocation
line gt F(b?G )??
Difficult to understand?! Look at a simple case!
z
y
x
Lets look at the effect of ?x on a ?y. Three
term in ?x (1)stress ?xx, Burgers vector
bx (2) stress ?xy, Burgers vector by (3)
stress ?xz, Burgers vector bz.
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z
y
z
?xx
?xx
?xy
?xz
?xy
?xz
(1)
(2)
(2)
x
x
x
bx
by
bz
Glide force of a screw component
Glide force of a edge component
Climb force of a dislocation
gt In general, the force per unit length is
F(b?G )??
with a glide component
and a climb component
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? Energy of a screw dislocations Summation
of elastic energy of the stress field.
Total strain energy S (1/2) ? stress ?
strain. Differential strain energy of a
screw dislocation at a distance r from the
dislocation core is
dV2pr dr ? unit length
G energy per unit length UC core energy
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UC could be estimated from heat of
fusion. Typical range
Assume r0b and
Typically, r0b/? ? 2-4
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? Energy of an edge dislocations
For an edge dislocation, the energy is modified
by Poissons ratio u. GS for screw dislocation
Ge for edge dislocation
In general,G aGb2 0.5 lt a lt 1
For Si, G 7.5 x 1010 N/m2, b (a/2)1 1 0,
and a 0.543 nm gt dislocation energy 10.4 eV
per 0.1 nm. Very large energy! Difficult to
produce by random thermal fluctuations! Could be
produced by stresses during thermal
and mechanical processing including growth!
? 4 gt Textbook, Eq. 4.20
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? Interaction of Dislocations with Point Defects
RS atomic size of solute atom Rh atomic size of
host atom
Compressive good for smaller atom
Solute atom with size gt host atom
vacancy
Tensile good for larger atom
RS gt Rh
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Example Dislocations have only one Burgers
vector for the whole line! The
loop will expand or contract in response
to a shear stress.
?
Dislocation loop
b
a positive edge dislocation b left screw
dislocation c negative edge dislocation d
right edge dislocation
?
a
c
?
b
Expand
d
Axiom reversing the sense of the dislocation
line by reversing the direction ? causes b to
reverse its direction
gt In a loop, the positive and negative
dislocations have the same Burgers vector