An interesting and often not appreciated result is that since the strain energy of a body is independent of the position of the dislocation, the presence of the dislocation does not alter the elastic constants of the material. causes an elastic displacement and the strain energy in the body is just the work done by the force, ( ) 1 1 1 2 3 , , u u x x x = 1 T Consider an unstrained body with no dislocation present. Apply and external force per unit area over the surface of the body. 1 T 1 T S ds
Transcript
An interesting and often not appreciated result is that since the strain energy of a body is independent of the position of the dislocation, the presence of the dislocation does not alter the elastic constants of the material.
causes an elastic displacement and the strain energyin the body is just the work done by the force,
( )1 1 1 2 3, ,u u x x x=1T
Consider an unstrained body with no dislocation present. Apply and external forceper unit area over the surface of the body.
1T
1TS ds
1TS ds
( )1 1 11/ 2S
U T u dS= •ò
2TNow form a dislocation by cutting and applying to the cut faces a force causing 2.u
( )2 2 21/ 2S
U T u dA= •ò
Also, there is work done by as is being applied. The total strain energy is 1T 2u
( ) ( )1 2 1 2 1 2S A
U U U T u dS T u dA= + + • + •ò ò
2TA
Now we can do exactly the same thing in reverse. First form the dislocationby applying resulting in and then apply 2T 2u 1.T
Alternatively we could remove and then in the manner of a thermodynamic cycle.
1T 2T
Since does not act on ( )2 2 1, 0 ; 0S
S T on S T u dS= • =ò2T
Also, is single valued whereas is equal and opposite on the two sides of thecut surface; the work gained on one face of the cut is exactly balanced by the work lost on the other face,
1u 2T
( )2 1 0A
T u dA• =ò
( ) ( )1 2 2 1 2 1S A
U U U T u dS T u dA= + + • + •ò ò ( ) ( )1 2 1 2 1 2S A
U U U T u dS T u dA= + + • + •ò ò
So, therefore is independent of the position of the dislocation.1 2U U U= +
This means that the elastic constants are insensitive to the presence orlocation of stationary dislocations. *This is for an “infinite” solid as thepresence or effect of surfaces is not taken into account.
The “force” on the dislocation will be due entirely to the external work that isdone when it moves,
( )1 2A
W T u dA= - •ò
( ) ( )
( ) ( )
1 2 1 2
1 2 1 2
0S A
A S
T u dS T u dA
T u dA T u dS
• + • =
- • = •
ò ò
ò ò
Also, this means that the work done by over S and A as the dislocation is formed must sum to zero.
1T
and 1 21W W
T uA l x
¶ ¶= - • º
¶ ¶
This may be considered as the force per unit length acting on a dislocation line
1 21 W
F T u bl x
s¶= - = • =
¶
where is the stress in the slip plane in the slip direction.s
Edge dislocation S Screw dislocation
glide
S
only one glide plane(mixed dislocations also)
any crystallographically allowable glide plane
climb
S
edge & mixed dislocations
screw dislocations don’t climb (unless they have a small edge component)
. and both containing plane aon motion :motion) ive(conservat Glide
b!!
x
. direction in motion :motion) veconservati-(non Climb
b!!
´±x
coren dislocatio fromaway or toed transportbemust matter :motion veconservati-non
Dislocation Climb
Figure showing edge dislocation climb
Jogged edge dislocation indicating annihilation and removal of atomic planes.
Climb: important in high temperature creep (helps in dislocation annihilation and circumventing obstructions)
dislocation emitting or absorbing vacancies?
dislocation emitting or absorbing vacancies?
dislocations are important sinks and sources for vacancies
adsorbingvacancies
emittingvacancies
Mixed & edge dislocations have only one glide plane and can’t cross slip
S
another toplane glide one from change :slip-cross
Cross slip is important as a means of dislocations 1) rearranging themselves, 2) getting around obstacles and 2) annihilating (dynamic recovery)
+S
-S
Screw dislocations can cross-slip
b!
r!dx
Consider a mixed dislocation of line direction x and Burgers vector b. A stress s acts on the crystal.
Question: What is the force (per unit length) acting on the dislocation?
We know that, in principle, if the dislocation moves by a displacement dr, then the work done on the crystal will be dW.
rFW !! dd ×=
Force on a Dislocation: Peach Koehler Equation
glide component
climb component
Force on a Dislocation: Peach Koehler Equation
Let us calculate dW. To do so we need to move the dislocation by an amount dr. This is the way we move it: We first make a cut of length L(the length of the dislocation) and width dr. The area of this cut we denote |dA|, where the vector dA points perpendicular to the plane of the cut:
( )rLA !!!dxd ´=
b!
r!dx
LA!
d
Peach Koehler Equation (continued)Note: While making the cut we don’t want the faces of the cut to move in the presence of the applied stress. In order to prevent that from happening, we have to apply tractions to the surfaces of the cut that are equal and opposite to the applied stress.
Next, while the tractions are applied, we shift the two faces with respect to each other by an amount b. In order to do this it might be necessary to take material away from the cut and place it on the surface of the crystal (or vice versa) so that the dislocation can climb. (This will be necessary if dr lies outside the glide plane).
While we are moving the two surfaces, two things are happening:1) The external stress is doing work on the crystal,2) The dislocation is effectively moving by an amount dr to the new location.
b!
r!dx
LA!
d
Peach Koehler Equation (continued)The amount of work done is given by:
! "#$%&
forcentdisplaceme
AbW dsd ××=
This can be rearranged to give:
( )rLA !!!dxd ´=
( )[ ]xsdd ˆ´××= brLW !!
The force per unit length acting on the dislocation is therefore:
( ) xs ˆ´×= bf!! Peach-Koehler
Equation
with
The component of force per unit length acting in the direction of climb is:
The glide force (the most important component low temps) is:
where trss is the resolved shear stress, that is, the component of stress acting on the glide plane and in the direction of the Burgers vector. The glide force acts perpendicular to the line of the dislocation in such a way as to expand the slipped area.
( )[ ] ( )x
xxsˆ
ˆˆ
´
´×´×=
bbbfcl !
!!!
bfff rssclgl t=-=!!
Climb and glide forces
Examples
x
y
z
Mixed dislocation with a tangent vector in the negativez-direction and a Burgers vector
Dislocation with a tangent vector in the +x direction and a Burgers vector
x
Examples
!b = by j
Apply stressesAll other stresses are zero.
σ xy and σ yzb
0 σ xy 0
σ xy 0 σ yz
0 σ yz 0
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
0by0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟=
σ xyby0
σ yzby
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟=
Gx
0Gz
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
i j kGx 0 Gy
1 0 0
=σ yzby j
Examples
y
z
x b
x
0 σ xy 0
σ xy 0 σ yz
0 σ yz 0
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
0by0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟=
σ xyby0
σ yzby
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟=
Gx
0Gz
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
i j kGx 0 Gy
1 0 0
=σ yzby j
Note that if we had chosen a coordinate system like thisthen
x
x b
z
y
x
y
z
Examples
x
y
z
rA B
We want to calculate the force (per unit length) that dislocation A exerts on dislocation B.
z
Since the stress field assumes that the dislocation is at the origin of the coordinate system,we place A there and note that the tangent vector is parallel to the z-direction. Then the PKEquation takes the form ( ) ˆ
A on B AB B A Bf f b s x= = ´! "i
ˆ
ˆˆ ˆA B x
A B
b b b i
kx x
= =
= =
! !
( )0
0 0 00 0 0 0
ˆ ˆ00 0 1
Bxx xy xx x x
Bx xy yy xy x y
zz
x y y x
b Gb b G
i j kG G G i G j
s s ss s s
s
æ öæ ö æ öç ÷ç ÷ ç ÷= =ç ÷ç ÷ ç ÷
ç ÷ ç ÷ç ÷è ø è øè ø
= -
( )( )( )
( )( )( ) ( )
2 2
22 2
2 2
22 2
0
12 1
32 1
2 1
xx
xxy
A
y x yb
x y
x x ybry
b
x
µsp n
µs µn np p
=+
=- +
-=
--=
+
Since
( ) ( )21 1ˆ ˆ
2 1 2 1
A Bx x x
AB yb b b
f G i ir r
µ µp n p n
= = =- -
0; 0xx yGs = =
Examples
x
y
zr
A
B
We want to calculate the force (per unit length) that dislocation A exerts on dislocation B.
