Forces Acting on Dislocations
¾ Peach - Koehler equation
¾ External force - constant Peach - Koehler force
¾ Forces acting between dislocations
¾ Interactions between dislocations
¾ Energy of dislocation configurations
¾ Climb and chemical forces
¾ Image force
¾ Self-force (or line tension)
References:
Hull and Bacon, Ch. 4.5-4.8
Kelly and Knowles, Ch. 8
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Force acting on dislocations: Peach - Koehler equation
Application of external stresses and stresses generated by other crystal defects may cause
movement of a dislocation on its glide plane. As a result, work may be done by the applied
stresses. Let’s consider dislocation motion in a sample with dimensions Lx×Ly×Lz due to the
shear stress τ :
τ
y
τ
b
τ
b
Lx
τ
x
τ
τ
b
The work done by the shear stress τ in changing the system from the initial to the final state is
equal to τSb = τLxLzb. If we consider the process as movement of a dislocation under the action
of force F acting on a unit length of the dislocation, the same work can also be written as
( force FLz ) × ( distance Lx) = FLzLx
Thus F = τb
The force acting on a dislocation line is not a physical force (like mechanical force of a spring or
electrostatic force acting on a charged particle) but a way to describe the tendency of dislocation
to move through the crystal when external or internal stresses are present.
This work done at the slip plane is dissipated into heat (similar to work done by friction forcers)
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Force acting on dislocations
r
F
F = τb
F is always perpendicular to the dislocation
line even though τ is constant on the glide
plane
r
F
τ
r
b
r
F
r
F
to produce the same deformation, the
same τ generates force on a screw
dislocation that is perpendicular to the
force on an edge dislocation
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Force acting on dislocations: Peach - Koehler equation
F = τb
τ is the shear stress in the glide plane resolved in the direction of b and F acts
normal to the dislocation line
Since the dislocation moves on its glide plane, we only need to consider the shear stress on this
plane. Stress components normal to the glide plane do not contribute to the dislocation movement
Moreover, only the shear stress components in the direction of b (called the resolved shear
stress τ) are contribution to the movement of the dislocation.
F is perpendicular to the dislocation line
τ is constant on the glide plane
σ yy
y
r
F
In general:
r
b
( )
r r
r
F = σ ⋅b × l
Peach-Koehler equation
r
F - force per unit length at an arbitrary point P
along the dislocation line
τ
σ yy
x
σ - local stress field
r
l - local line tangent direction at point P
r
σ ⋅ b - local force per unit length acting on a plane
(of area b) normal to the Burgers vector
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Force acting on dislocations: Peach - Koehler equation
Example: edge dislocation runs along y-axis and Burgers vector is in the negative direction of x-axis
glide plane is (001)
z
( )
r r
r
F = σ ⋅b × l
r
g = (gx, g y , gz )
y
σ xx
r
b
σ yx
σ zx
r
b = (b x , b y , b z ) = ( − b ,0,0 )
r
l = (l x , l y , l z ) = ( 010 )
x
g x = σ xx b x + σ xy b y + σ xz b z = − σ xx b
g y = σ yx b x + σ yy b y + σ yz b z = − σ yx b
g z = σ zx b x + σ zy b y + σ zz b z = − σ zx b
traction on the surface normal to b is defined by stress components σxx, σyx, σzx
g x = − σ xx b is ⊥ to l ⇒ σxx produces |F| = σxxb downwards, perpendicular to the glide plane
r
F
if σxx < 0 (compression) ⇒ F is directed upwards ⇒ climb force
r
r
g
l
g y = − σ yx b is || to l ⇒ σyx makes no contribution to F
g z = − σ zx b is ⊥ to l ⇒ σzx produces |F| = σzxb along the x-axis ⇒ glide force
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Force acting on dislocations: Peach - Koehler equation
Example: screw dislocation runs along x-axis and Burgers vector is in the negative direction of x-axis
glide plane is (001)
r
b
( )
r
l
x
r
b = (b x , b y , b z ) = ( − b ,0,0 )
r
l = (l x , l y , l z ) = (100 )
r r
r
F = σ ⋅b × l
r
g = (gx, g y , gz )
g x = σ xx b x + σ xy b y + σ xz b z = − σ xx b
g y = σ yx b x + σ yy b y + σ yz b z = − σ yx b
g z = σ zx b x + σ zy b y + σ zz b z = − σ zx b
traction on the surface normal to b is defined by stress components σxx, σyx, σzx
g x = − σ xx b is || to l ⇒ σxx makes no contribution to F
g y = − σ yx b is ⊥ to l ⇒ σyx produces |F| = σyxb up- or down-wards (depending on the sign
of σyx), perpendicular to the xy slip plane - can induce cross-slip
g z = − σ zx b is ⊥ to l ⇒ σzx produces |F| = σzxb along the y-axis ⇒ glide force along xy slip
plane
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Force acting on dislocations: Peach - Koehler equation
Example: screw dislocation runs along y-axis and Burgers vector is in the positive direction of y-axis
z
glide plane is (001)
y
r
l
r
b
( )
σ xy
σ zy
σ yy
x
r
b = (b x , b y , b z ) = ( 0, b ,0 )
r
l = (l x , l y , l z ) = ( 010 )
r r
r
F = σ ⋅b × l
r
g = (gx, g y , gz )
g x = σ xx b x + σ xy b y + σ xz b z = σ xy b
g y = σ yx b x + σ yy b y + σ yz b z = σ yy b
g z = σ zx b x + σ zy b y + σ zz b z = σ zy b
traction on the surface normal to b is defined by stress components σxy, σyy, σzy
g x = σ xy b
is ⊥ to l ⇒ σxy produces |F| = σxyb up- or down-wards (depending on the sign
r
of σxy), perpendicular to the xy slip plane - can induce cross-slip
r
l
F
g y = σ yy b
is || to l ⇒ σyy makes no contribution to F
g z = σ zy b
is ⊥ to l ⇒ σzy produces |F| = σzyb along the x-axis ⇒ glide force along xy slip
plane
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
r
g
Forces between dislocations
Force due to the interaction with other dislocations = sum of all Peach–Koehler forces between
the segments of all other dislocations in the system
Example: interaction between two
parallel straight edge dislocations
(2)
y
(1)
x
z
(1) (2)
( )
r r
r
F = σ ⋅b × l
r
g = (gx, g y , gz )
r
b = (b x , b y , b z )
r
l = (l x , l y , l z )
disl. (2)
g x = σ xx b x + σ xy b y + σ xz b z
g y = σ yx b x + σ yy b y + σ yz b z
g z = σ zx b x + σ zy b y + σ zz b z
Edge dislocation (1) produces a stress field that dislocation (2) responds to
Peach–Koehler gives force acting on dislocation (2) due to the presence of dislocation (1)
Must use consistent convention to describe dislocations (direction of the Burgers circuit,
line direction into page, start-finish)
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Forces between dislocations
Example: interaction between two parallel straight edge dislocations
(2)
y
(1)
r
(1) b = (b1 ,0 ,0 )
r
l1 = ( 0,0,1)
(1) (2)
r
(2) b = (b2 ,0,0 )
r
l 2 = ( 0,0,1)
(
x
Step 1:
Step 3:
g x = σ xx b x + σ xy b y + σ xz b z = σ xx b2
use expressions for stresses generated
by dislocation (1):
Gb1
y (3 x 2 + y 2 )
σ xx = −
2 π (1 − ν ) (x 2 + y 2 )2
g y = σ yx b x + σ yy b y + σ yz b z = σ yx b2
g z = σ zx b x + σ zy b y + σ zz b z = 0
r
b = (b2 ,0 ,0 ) for disl. (2)
σ from disl. (1)
r
g = ( σ xx b2 , σ yx b2 ,0 )
Step 2:
r r r
F = g × l 2 = ( σ yx b2 , − σ xx b2 ,0 )
)
r
r
r
F = σ ⋅ b2 × l 2
r
g
σ xy
r
F =
(
(
)
Gb1
x x2 − y2
=
2 π (1 − ν ) x 2 + y 2 2
(
)
)
(
(
)
Gb 1b2 x x 2 − y 2
Gb1b2 y 3 x 2 + y 2
xˆ +
yˆ
2
2
2
2
2
2
2 π (1 − ν ) x + y
2 π (1 − ν ) x + y
(
)
a × b = c = c1i + c2j + c3k
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
)
Forces between dislocations
Example: interaction between two parallel straight edge dislocations
(2)
r
F =
y
(1)
y
)
(
(
)
(
)
)
x
glide plane
(2)
attraction
r
Fx = 0
r
Fy =
x= y
45 °
(1)
Gb1b2 1
2 π (1 − ν ) y
x=0
r
Fy =
90 °
- climb force (need diffusion)
attraction
r
Fx = 0
repulsion
- unstable equilibrium (Fx < 0 when x < y and Fx > 0 when x > y )
x
y (2)
(1)
(
Gb 1b2 x x 2 − y 2
Gb1b2 y 3 x 2 + y 2
xˆ +
yˆ
2 π (1 − ν ) x 2 + y 2 2
2 π (1 − ν ) x 2 + y 2 2
attraction
- stable equilibrium (Fx > 0 when x < 0 and Fx < 0 when x > 0 )
Gb1b2 1
2 π (1 − ν ) y
- climb force (need diffusion)
x
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Forces between dislocations
Example: interaction between two parallel straight edge dislocations
(2)
y
(1)
x
r
(1) b = (b1 ,0 ,0 )
r
l1 = ( 0,0,1)
(1) (2)
r
(2) b = ( − b2 ,0 ,0 )
r
l 2 = ( 0,0,1)
(
same as before, but opposite sign of (2)
Step 1:
g x = σ xx b x + σ xy b y + σ xz b z = − σ xx b2
g y = σ yx b x + σ yy b y + σ yz b z = − σ yx b2
g z = σ zx b x + σ zy b y + σ zz b z = 0
r
b = ( − b2 ,0,0 ) for disl. (2)
σ from disl. (1)
r
g = ( − σ xx b2 , − σ yx b2 ,0 )
Step 2:
r r r
F = g × l 2 = ( − σ yx b2 , σ xx b2 ,0 )
)
r
r
r
F = σ ⋅ b2 × l 2
r
g
Step 3:
use expressions for stresses generated
by dislocation (1):
Gb 1
y (3 x 2 + y 2 )
σ xx = −
2 π (1 − ν ) (x 2 + y 2 )2
σ xy
(
(
)
Gb1
x x2 − y2
=
2 π (1 − ν ) x 2 + y 2 2
(
)
)
(
(
)
r
Gb 1b2 x x 2 − y 2
Gb 1b2 y 3 x 2 + y 2
F =−
xˆ −
yˆ
2
2
2
2
2
2
2 π (1 − ν ) x + y
2 π (1 − ν ) x + y
(
)
the sign is reversed when the
dislocations are of opposite sign
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
)
Forces between dislocations
Example: interaction between two parallel straight edge dislocations
r
F =±
(
)
(
(
)
Gb 1b2 x x 2 − y 2
Gb 1b2 y 3 x 2 + y 2
xˆ ±
yˆ
2 π (1 − ν ) x 2 + y 2 2
2 π (1 − ν ) x 2 + y 2 2
(
)
)
y (2`)
(2)
(2*) (2#)
(2º)
45 °
x
(1)
(2)
force in units of
Gb1b2
2π(1 − ν ) y
y (2`) (2*) (2#)
from Hull and Bacon
45 °
(1)
x
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
(2º)
Forces between dislocations
Example: interaction between two parallel straight screw dislocations
(2)
(1)
y
r
(1) b = ( 0 ,0 , b1 )
r
l1 = ( 0,0,1)
(1) (2)
r
(2) b = ( 0 ,0 , b2 )
r
l 2 = ( 0,0,1)
(
x
Step 1:
)
r
r
r
F = σ ⋅ b2 × l 2
r
g
Step 3:
g x = σ xx b x + σ xy b y + σ xz b z = σ xz b2
use expressions for stresses generated
by dislocation (1):
g y = σ yx b x + σ yy b y + σ yz b z = σ yz b2
σ xz = σ zx = −
g z = σ zx b x + σ zy b y + σ zz b z = 0
r
b = ( 0 ,0 , b2 ) for disl. (2)
σ from disl. (1)
r
g = ( σ xz b2 , σ yz b2 ,0 )
Step 2:
r r r
F = g × l 2 = ( σ yz b2 , − σ xz b2 ,0 )
σ yz = σ zy =
Gb
y
Gb sin θ
=
−
2π x 2 + y 2
2π r
Gb
x
Gb cos θ
=
2π x 2 + y 2
2π r
r Gb1b2
x
Gb1b2
y
ˆ
F =
x
yˆ
+
2
2
2
2
2π x + y
2π x + y
r Gb1b2
(cos θ xˆ + sin θ yˆ )
F =
2 πr
repulsive for screws of the same sign and
attractive for screws of opposite sign
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Summary on interactions between dislocations
General basic rules:
The superposition of the stress fields of two dislocations as they move towards each other can
result in
(1) larger combined stress field as compared to a single dislocation (e.g., overlap of the
regions of compressive or tensile stresses from the two dislocations) ⇒ increase in the
energy of the configuration ⇒ repulsion between dislocations.
(2) lower combined stress field as compared to a single dislocation (e.g., overlap of regions of
compressive stress from one dislocation with regions of tensile stress from the other
dislocation) ⇒ attraction between dislocations.
Arbitrarily (curved) dislocations on the same glide plane:
Dislocations with opposite b will attract each other and annihilate
Dislocations with the same b will always repel each other
Edge dislocations with identical or opposite Burgers vector b on neighboring glide planes
may attract or repulse each other, depending on the precise geometry.
The force between screw dislocations is repulsive for dislocations of the same sign and attractive
for dislocations of opposite sign.
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Energy of dislocation configurations
stable configuration for parallel straight edge dislocations
y
dislocation wall
σ xx
superposition of compressive and tensile stresses of
dislocations in the wall ⇒ screening radius R ~ h
Wdisl = Wel + Wcore
h
for h = 75b,
x
this configuration has strong
long-range stress field
ln
Rmax
≈ 16
r0
⎞
Gb 2 L ⎛ R
⎜⎜ ln + Z ⎟⎟
=
4πK ⎝ r0
⎠
h
h
≈ 50, and ln ≈ 4
r0
r0
energy of dislocation in the wall
is up to 4 times lower than energy
of an individual dislocation
dislocation walls form during recovery, when stored internal
energy accumulated during plastic deformation decreases as the
dislocations form low-energy (stable) configurations
a wall of edge dislocations corresponds to a low-angle tilt grain
boundary ⇒ the rearrangement of dislocations into low-angle
grain boundaries can lead to the formation of cellular subgrain structure
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Energy of dislocation configurations
stable configuration for parallel straight edge dislocations
pile-ups of dislocations of opposite sign
chessboard structure (Taylor lattice)
dislocation dipoles
τ
1
2
3
4
L
these configurations have weak long-range stress field
.... n
dislocation pile ups can be generated at the initial
stage of plastic deformation (unstable configuration)
at large distances from pile-up (r >> L), the stress field created by the pile up is analogous to a
super-dislocation with B = nb
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Low-energy of dislocation configurations
2D computer simulation of a low-energy dislocation configuration
(“Taylor lattice”): (a) relaxed quadrupole configuration and (b)
configuration under critical stress for disintegration into dipole walls
P. Neumann, Mater. Sci. Eng. 81, 465, 1986
Doris Kuhlmann-Wilsdorf, in
Dislocations in Solids, Vol. 11, Ch. 59
Model for formation of a Taylor lattice: (a) activation of a
dislocation source in a volume element, (b) coordinated
shape change of the volume elements as a similar sequence
of dislocations arrives from a neighboring element, (c)
formation
dipoles
and (d) formation
of a Taylor
lattice.
Formation of in
Taylor
lattice in Leonid
α-brass Zhigilei
University
of Virginia,
MSE 6020:
Defects
and Microstructure
Materials,
1μm
Climb force
glide plane
y
climb force
(2)
r
F = ( σ yx b2 , − σ xx b2 ,0 )
climb force originates from normal stress, e.g., σxx in the figure to the
left, acting to squeeze the extra half plane of dislocation from the crystal
(1)
x
normal stress can also be created by external forces, line tension, etc.
edge or mixed dislocation can move away from its slip plane only with the help of point defects
(vacancies or interstitials) - such motion is called non-conservative motion or climb (in contrast to
the conservative motion within the slip plane)
motion of a screw dislocation is always conservative (never involves point defects)
prismatic loop of partial dislocation
“conservative climb” is also possible for small prismatic
loops of edge dislocations. The loop can move in its
plane without shrinking or expanding at low T, when
bulk diffusion of point defects is negligible.
the motion is due to the pipe diffusion of vacancies
produced at one side of the loop and moving along the
dislocation core to another side.
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Chemical force
y
x
r
Fclimb
σ xx
vacancies
σ xx
r
Fchem
r
Fclimb = − σ xx b
r
Fclimb
σ xx
σ xx
let’s consider a volume of material where dislocation is the only sink and source of vacancies
without external stress, the equilibrium concentration of vacancies is maintained by absorption
or birth of vacancies on the dislocation
when σxx is applied, |Fclimb| = σxxb acts downwards and dislocation moves by emitting vacancies
vacancy concentration increases above the equilibrium concentration c0 and it is increasingly
difficult to create new vacancies
eventually, at vacancy concentration c, the movement stops; we can consider a chemical force
r
acting against the climb force and opposite in direction, i.e., r
Fchem = − Fclimb at c > c 0
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Chemical force
the work done when a segment l climbs distance s in response to Fclimb is Wclimb = Fclimbls and
the number of vacancies emitted is nclimb = bls/Ωa, where Ωa is volume per atom
the effective change in the vacancy formation energy is Wclimb/nclimb = FclimbΩa/b
⎛ Δs
c = exp ⎜⎜ v
⎝ kB
⎛ Δh
⎞
⎟⎟ exp ⎜⎜ − f
⎝ k BT
⎠
r
r
since Fchem = − Fclimb
⎛ F Ω /b⎞
⎞
⎛ F Ω /b⎞
⎟⎟ exp ⎜⎜ − climb a ⎟⎟ = c0 exp ⎜⎜ − climb a ⎟⎟
k BT
k BT
⎠
⎠
⎝
⎠
⎝
Fchem =
bk BT ⎛ c ⎞
ln ⎜⎜ ⎟⎟
Ωa
⎝ c0 ⎠
r
Fclimb
r
Fclimb
negative climb (Fclimb < 0) ⇒ vacancy emission ⇒ c > c0
positive climb (Fclimb > 0) ⇒ vacancy absorption ⇒ c < c0
supersaturation of vacancies ⇒ chemical force ⇒
r
Fchem
dislocation climb until Fchem is not compensated by
Fclimb due to external/internal stresses or line tension
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
r
Fchem
Image forces
The stress field generated by a dislocation is modified near a free surface, leading to extra
forces acting on the dislocation (dislocation-surface interaction).
The normal and shear stress at a free surface are zero (there is “nothing” on one side of the
boundary to provide reaction forces ⇒ there must be no normal or shear stress on the inside)
image
dislocation
let’s consider a straight screw dislocation parallel to z axes located at
a distance d from a free surface at x = 0
y
d
d
x
to satisfy the condition of zero traction on plane x = 0,
i.e., σxx = σyx = σzx = 0, we can add stress field of an
imaginary screw dislocation of opposite sign at x = -d
the stress field inside the body (x > 0) is then
force acting on the screw dislocation
from the surface = force acting from the
image dislocation:
Gb 2
Fx = σ zy ( x = d , y = 0 ) b = −
4 πd
σ zx =
⎤
Gb ⎡
y
y
−
⎢
⎥
2 π ⎣ ( x + d )2 + y 2 ( x − d )2 + y 2 ⎦
σ zy = −
⎤
Gb ⎡
x+d
x−d
−
⎢
⎥
2 π ⎣ ( x + d )2 + y 2 ( x − d )2 + y 2 ⎦
only stress due to the image dislocation is accounted for
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
image
dislocation
Image forces
y
d
For edge dislocation, adding an image dislocation at x = -d
cancels σxx and x = 0, but not σyx. Thus, an extra term
should be added to match the boundary condition.
x
d
The shear stress is then:
σ yx
Gb
=−
2 π (1 − ν )
⎡
(x + d )2 − y 2
(x − d )2 − y 2
2 d ( x − d )( x + d ) 3 − 6 x ( x + d ) y 2 + y 4 ⎤
− (x − d )
+
⎢( x + d )
⎥
3
2
2 2
2
2 2
(x + d ) + y
(x − d ) + y
(x + d )2 + y 2
⎢⎣
⎥⎦
(
)
(
)
(
)
extra term to ensure σyx = 0 at x = 0
force acting on the edge dislocation from the surface = force due to the 1st and 3rd terms
Gb 2
Fx = σ yx ( x = d , y = 0 ) b = −
4 π (1 − υ ) d
(contribution from the 3rd term is zero)
dislocations are attracted to the surface ⇒ image forces can remove dislocations from the surface
regions given that slip planes are oriented at large angles to the surface
Interactions of curved dislocations, dislocation loops and dipoles, etc. can result in very complex
stress fields that are often difficult to evaluate analytically
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Line tension
Recall the result of our analysis of the dislocation energy: Wdisl = Wel + Wcore
or, for energy per unit length,
Wdisl
Gb 2
=
L 4πK
⎞
⎛ R
⎜⎜ ln + Z ⎟⎟ ≈ αGb 2
⎠
⎝ r0
α ≈ 0 .5 − 1 .5
The line energy (energy per length) has the
same dimension as a force and corresponds to
line tension, i.e., a force in the direction of the
line vector which tries to shorten the dislocation
r
FT
The resulting force FT acting on an element dl
of a dislocation line is related to the line tension
T at the ends of the element and is perpendicular
to the dislocation line r
FT = 2T sin( dθ / 2) ≈ Tdθ
T=
Wdisl
L
≈ αGb 2
dθ ≈ dl
R
for small dθ
force Fτ acting on the same element dl due to the external shear stress is
balance of forces to maintain
radius R of the curved dislocation:
⎞
Gb 2 L ⎛ R
⎜⎜ ln + Z ⎟⎟
=
4πK ⎝ r0
⎠
Td θ = τbdl
τ=
r
Fτ = τbdl
α Gb
Td θ
T
=
=
bdl
bR
R
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei
Line tension
What happens when τ > αGb/Rmin ?
τ
2 R min R
min
dislocation segment pinned at its ends
How the dependence of the energy of dislocation on its type
(edge vs. screw) affects the shape of the dislocation loop?
τ
b
University of Virginia, MSE 6020: Defects and Microstructure in Materials, Leonid Zhigilei