Lauren Ayers 2271 Outline Partial Dislocations Why Partials Stacking Faults LomerCottrel Lock Force on a Dislocation Line Tension Model Dislocation Density Partial Dislocations b1 Single unit dislocation can break down into two Shockley partials ID: 273054
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Slide1
Partial Dislocations
Lauren Ayers
22.71 Slide2
Outline
Partial Dislocations
Why Partials?
Stacking Faults
Lomer-Cottrel
Lock
Force on a Dislocation
Line Tension Model
Dislocation DensitySlide3
Partial Dislocations
b1
Single unit dislocation can break down into two Shockley partialsSlide4
Partial Dislocations
b2
Single unit dislocation can break down into two Shockley partials
b3Slide5
Why Partials?
Frank’s Rule:
|b
1
|
2
>|b2|2
+|b3|2Energy of a dislocation
is proportional to |b
|
2
Partial dislocations decrease strain energy of the latticeSlide6
Stacking Faults
Movement of partial dislocations generate discontinuity in stacking planes, ex: ABCAXCABC
Two separated partials have smaller energy than a full dislocation
Reduction in elastic energy proportional to:
Equilibrium splitting distanceSlide7
What does this mean?
Wide vs. narrow ribbon affects cross slip
Cu, s = 2nm: high constriction energy barrier
Al, s= 4 A: cross slip occurs more easilySlide8
Lomer-Cottrel Lock (LC)
2 Dislocations on primary slip planes combine
Formed by:
Slip by
b
LC
creates a high energy stacking faultNo {111} plane which the LC can move as an edge dislocation
“Lock”: Once the state is formed, hard to leaveActs as a barrier against other dislocationsSlide9
Force on a Dislocation
Climb: “Non-conservative”
Glide: “Conservative”Slide10
Line Tension Model
Assume:
Line tension
η
(total elastic energy per length) independent of line direction
ξ
Dislocations do not interact with each other elastically
Local modelFrom dF, derive critical external stress:Slide11
Dislocation Density
Total length of all dislocations in a unit volume of material
1/m
2
or m/m
3Slide12
Thanks!