Forces on Dislocations
Plastic Deformation in Crystalline Materials
Kamyar Davoudi
Lecture 9
Lemma: Energy of Inclusions
•  Consider a finite body D, containing inclusions Ω.
Material is either isotropic or anisotropic. Assume that
eigenstrain ε*ij (inelastic strains such as plastic or thermal
strains) is prescribed in Ω.
Ω
D
εij*
Lemma: Energy of Inclusions
•  First let D be free of any external forces and surface constraints. The
elastic strain energy is:
Eel =
1
σijε(e)
dV
∫
ij
2 D
(e)
*
•  Remember that εij = εij − εij
and
εij =
1
ui, j + u j,i
2
(
)
•  Integration by parts reveal that:
∫σu
ij i, j
D
dV =∫ σij ui n j dS − ∫ σij, j ui dV = 0
$
!"#
S
D
σij n j =Ti =0
=0
•  Therefore:
Eel =
1
1
(e)
*
σ
ε
dV
=
−
σ
ε
dV
∫
∫
ij ij
ij ij
2 D
2 Ω
Exercise
•  Now assume that D is subjected to surface traction. If
ui0 , σij0 : elastic fields if traction act alone in the absence of
eigenstrain
ui , σij : elastic fields due to the eigenstrain prescribed in Ω,
then prove that:
Eel =
Hint: Note that
Eel =
1
1
0 0
*
σ
u
dV
−
σ
ε
dV
∫
∫
ij i, j
ij ij
2 D
2 Ω
1
0
0
*
(σ
+
σ
)(u
+
u
−
ε
)dV
∫
ij
ij
i, j
i, j
ij
2 D
Example: Calculation of Elastic Strain Energy of
an Edge Dislocation
1
1
(e)
Eel = ∫ σijεij dV = − ∫ σijε(ijp) dV
2V
2V
Elastic strain energy per unit length
⎞⎟
1
µb
x2 − y 2 ⎛⎜ b
( p)
⎜⎜ δ( y)H(−x)⎟⎟ dx dy
Eel = − ∫∫ 2σxyεxy dx dy = −∫∫
x
4
⎟⎠
⎜⎝ 2
2 sphere
r
sphere 2π 1−ν
(
)
⎛ R ⎞⎟
−r0
1
µb 2
µb 2
⎜
= −∫
dx = −
ln x =
ln ⎜⎜ ⎟⎟⎟
−R
⎜⎜⎝ r ⎟⎠
x
4π 1−ν
4π 1−ν
−R 4π 1−ν
0
−r0
µb 2
(
)
(
)
(
)
Variational Approach
•  Remember that* !
! !
βij( r ) = −b j niδ(S − r )
•  It can also be written
as
( p)
β ji = −bi n jδ (ζ )H (η0 − η )
•  The ζ axis is along the normal vector n
•  the η-axis is along ξ×n.
•  η is taken from the dislocation line, where the
position is η0=η0 (l) .
Variational Approach
•  Now consider:
Δx: displacement of the dislocation.
dβ ij( p) = −bj niδ (ζ )dH (η0 − η )
•  Because
then
dH = (dH
/ dη0 )dη0 = δ (η0 − η )dη0
! !
!
Δx × ξ
Δ
η
=
Δ
x
sin φ
n̂ = !
0
| Δx | sin φ
Δβ ij( p) = −bjεilmΔxlξ mδ (ζ )δ (η0 − η )
The variation in the elastic energy is:
ΔEel = − ∫ σ ij Δβ ij( p)dV = σ ij bjεilmξ mΔxl Δl
Force per unit length is: F = − ∂(E / Δl) = −σ b ε ξ = ε σ b ξ
l
ij j ilm m
lim ij j m
∂xl
Force on Dislocations
•  The Peach-Koehler force on a dislocation with line direction ξ
that is under the external stress σ including the stress due to
other dislocations reads as
F = ( · b) ⇥ ⇠.
•  Note that Peach-Koehler forces are not really Newtonian
forces. In fact they are thermodynamic driving forces
•  It has glide component Fg and climb component Fc
Fg = F · (⇠ ⇥ m) ,
Fc = F · m
where m = b ⇥ ⇠/|b ⇥ ⇠| is the unit vector perpendicular to
the glide plane of the dislocation
Glide and Climb Components of P-K Force for
an Edge Dislocation
If b=[b 0 0] and ξ=[0 0 1] as shown
above. Then and
2
b xy
F~ = 4 b xx 5 .
0
[Bulatov&Cai,ComputerSimula3onsofDisloca3ons]
The glide component
3
F x = Fg = b
xy
urges the dislocation to glide in the x-direction and the climb
component
F = F = b
y
c
xx
urges the dislocation to climb downward. In this case m = -j .
Exercise
•  Find the Peach-Koehler force on a screw
dislocation