85. 10 Forces on dislocate i on ◆|.| ntr i on 1. Definition Fis a fictitious force which is supposed to be responsible for the motion of dislocation Criterion The work done on crystal by stress must be equal to that done on dislocation by force F 2. Principle of derivation of F Macroscopic work=microscopic work (or deformation work)
§5.10 Forces on dislocation ◆ Ⅰ.Introduction 1. Definition: is a fictitious force which is supposed to be responsible for the motion of dislocation. ∥ Criterion: The work done on crystal by stress must be equal to that done on dislocation by force F. 2. Principle of derivation of : Macroscopic work＝microscopic work (or deformation work) F F F
I Der ivat ion 1. Force on screw dislocation F B Criterion The work done on crystal by stress must be equal to that done on dislocation by force F
◆ Ⅱ.Derivation: 1. Force on screw dislocation Criterion: The work done on crystal by stress must be equal to that done on dislocation by force F. A L B F b
B 小< for screw dislocation F·dS=(A)·()·(LdS) A microscopic work=-macroscopic work F=cbL
A A B B F dS L b for screw dislocation d ( ) ( ) (LdS) A b F S = A microscopic work＝macroscopic work F =bL
2. Force on edge dislocation Give: stress field [ o l caused by applies force or other f x dislocation microscopic work=macroscopic wok determine. the slip forcef s, climb force fc, and total force Fon edge dislocation
2. Force on edge dislocation x y s f c f ⊥ F Give: stress field caused by applies force or other dislocation [ ] ij determine: the slip force f s , climb force fc , and total force F on edge dislocation. microscopic work＝macroscopic wok
f: let dislocation slip a distance dS L FOS ds ∫ SP fs dS=txy A). ().I-ds fs make the dislocation move along the contrary direction
x y s f c f ⊥ F b S.P. 1= L ( )s F f dS f b S A b f S A xy xy = = s s d ( ) ( ) 1 d if f s make the dislocation move along the contrary direction f b s xy = − ① f s : let dislocation slip a distance dS
2 f: let dislocation climb a distance dS b·1 f·dS"=-(x·A Jf b b tot
② fc : let dislocation climb a distance dS' S A b f S x A = − d 1 d ( ) c c c f b c = − x ③ f tot 2 c 2 tot s f = f + f b dS ⊥
3. Force on mixed dislocation B C X given: D a mixed dislocation AB of unit length with b//OX (AB,b=I-a determine: force on AB for slip and climb
3. Force on mixed dislocation given: D a mixed dislocation of unit length with AB b // OX ( , = −) AB b determine: force on for slip and climb AB x y z B A s f b v
Solution ①f f s=ob L x b directs to that part of crystal which contains extra-half plane I x directs to that part of crystal which moves along b b
Solution： ① f s : f s =b directs to that part of crystal which contains extra-half plane. L b directs to that part of crystal which moves along . l v f s = xz b b
y A to determine fc, AB is B resolved into Ac and CB X For: CB, F=o bcos. For: AC F=-obsin a AB→AB AC→A fe=F+F2=b(tx cos a-ox sin a)
For : , ; For : , ② fc : x z y o b C A B C A B to determine fc , is resolved into and AB → AB AB AC CB CB F1 = xybcos AC F2 = − x bsin AC → AC ( cos sin ) f c = F1 + F2 = b xy − x
4. General case Peach-Kochler formula gIven dF dA ① an mixed dislocation ds with arbitrary shape determine: force on d/ segment
4. General case － Peach-Kochler formula. l d s d dA F d given: ① an mixed dislocation with arbitrary shape ② [ ] determine: force on segment l d
