Chap. 7. Intersection of Dislocations

Transcription

1 Chap. 7. Intersection of Dislocations Plastic Deformation of Crystal - Slip starts at a slip system having the largest Schmid factor. - A dislocation moving in slip plane will intersect other dislocations crossing the slip plane. "Forest Dislocation" - Slip proceeds at other slip planes having lower Schmid factors Intersection of Dislocations Intersection of Disl.1 (b 1) with Disl. (b) - Each disl. gives displacement of its Burgers vector to other dislocation Disl.1 : Displacement b Disl. : Displacement b1 1. Intersection of Two Edge Disl. with Perpendicular Burgers Vectors

2 Edge Disl. XY : No change Edge Disl. AB : Formation of jog (PP') - length & small edge disl. Burges vector b ) ( direction b 1. Intersection of Two Edge Disl. with Parallel Burgers Vector Edge Disl. XY : Formation of kink (QQ') - length b Edge Disl. AB : Formation of kink (PP') - length b 1 small screw disl.( Burgers vector b )

3 3. Intersection of Edge Disl. with Perpendicular Screw Disl. Edge Disl. AB : Formation of jog(pp') Screw Disl. XY: Formation of kink(qq') or jog(qq') If QQ' have same slip plane with screw, QQ' is kink. If QQ' have different slip plane with screw, QQ' is jog. 4. Intersection of Two Perpendicular Screw Disl. Screw Disl. AB : Formation of jog or kink Screw Disl. XY : Formation of jog or kink

4 Elementary Jog (or Unit Jog): Length of jog is equal to one Burgers vector. Superjog : Length of jog is multiples of Burgers vector. 7- Movement of Dislocation containing Jogs and Kinks 1. Edge Disl. with Jog Jog PP' : small edge disl. Jog has same slip plane with edge disl. Jog in edge disl. does not affect the glide of edge disl.. Edge Disl. with Kink

5 Kink PP' : small screw disl. Kink in edge disl. does not resist the glide of edge disl., in fact, the kink assists the glide of edge disl. 3. Screw Disl. with Kink (Fig. 7.5) PP' : small edge disl. Slip plane of kink PP'(P'PRR') is same with slip plane. Screw Disl.(P'PRR') Kink does not affect the glide of screw disl. 4. Screw Disl. with Jog (Fig. 7.5) Slip plane of jog PP'(P'PRR') is different from slip plane of screw disl.(q'p'r'b'). As screw disl. glide(ap A'Q, P'R' Q'B'), jog can move only by climb(pp' QQ'). Jog resists the movement of screw disl. and the climb process requires thermal activation thus the movement of the screw disl. will be temperature dependent.

6 7-3. Movement of Screw Disl. with Jogs Screw Disl. with Array of Jogs Screw disl. glide Screw disl. bows out between the jogs to a radius of curvaturer = αgb/ τ Induce glide force on the jogs due to the difference in line tension Two relatively close-spaced jogs glide together and resulting in annihilation or formation of a superjog The remaining jogs will be of approximately uniform spacing, x. Critical Stress to Move a Screw Disl. with Uniform Jogs Let, τ x : : applied shear stress uniform spacing between jogs Forward force on jog due to applied stress = τbx

7 As jog moves forward b, one point defect is formed. When jog moves forward b, Work done by the applied stress = ( τ bx)b = τb Formation energy of a point defect = E f x Jog willmove whenτb x E f Critical stress for jog movement (or dragging), τ o If jog moves to a direction to produce vacancies If jog moves to a direction to produce interstitials vacancy jog interstitial jog E f, interstitial > E f, vacancy Plastic Deformation Increase in disl. density & vacancy It is more likely that vacancies are formed by jog dragging. concentration Motion of Screw Disl. with Superjogs Movement depends on the magnitude of superjog. 1. Small Jog (height = nb, n = small) As jog climbs b, n point defects are formed. Critical stress for movement of superjog If τ > τo, screw disl.glidesby dragging jog.(fig.7. 8a)

8 . Longer Jog (height = nb, n = large) Critical Stress As n increases, τo increases. Bending Stress of Screw Disl. τ = αgb R If τmax < τ < τo, screw disl. glides without formation of disl. dipole. dragging jog. NP, MO - edge disl. of opposite sign two dislocations are fixed due to the interaction stress between them

9 Interaction stress between dislocation dipole NP, MO 1) For Medium Jog If τ <τ o ne b x f ( τ0 = ) αgb 0.5G & <τ< x πn(1- ν) formation of dislocation dipole since the edge disl. can not overcome the interaction force.(fig. 7.8b) ) For Long Jog If τ max 0.5G <τ & <τ< n(1-) π ν τ o αgb ( τmax = ) x two edge disl. can move independent of each other. (Fig. 7.8c) ex)if τ = 10-3 when y 60b. G, the transition between 1) and ) is found

10 7-5. Jogs and Dislocation Loops Formation of Superjog in Crystal 1) Double Cross Slip ) Formation of Dipole due to Disl. Interaction TT' on Disl. NM' 3) Repeated Disl. Intersection

11 Formation of Dislocation Loop in Crystal 1) Diffusion and Coalescence of Vacancies ) Cross Slip at Dislocation Dipole 3) Movement of Dislocation past Obstacles Oxide Dispersion Strengthened(ODS) Materials (a) : Orowan Mechanism - without cross slip (b),(c),(d) : Hirsch Mechanism - with cross slip

12 7-6 Intersection of Extended Dislocations The extended dislocation can only cut through each other by constricting to form perfect unextended disl. in the region close to the intersection. The jog also can dissociate to have stacking fault.

13 Intersection of Partial Disl. Disl. BA on plane ABD Bα and αd on plane BCD by a Jog - line direction BA b = BD slip plane ABD Dissociation of Jog BD Bγ + γd onplaneabd bi at New Junction? B α - Bγ - γα = Bα + γb + αγ = 0 partialdisl.: γd, Bγ stair - rod partial disl. : γα, αγ - opposite sign

14 7-7 Attractive and Repulsive Junctions Intersection of Disl. A and Disl. B - Attractive, repulsive or neutral interactions depending on the stress fields of disl. A and disl. B Interaction between Two Parallel Screw Disl. Let, Spacing = d Interaction Force on Disl. A - glide force on z plane F = σ = σ zx zy b b Gb = π x x + σ zy x + y b At Position (x = d, y) Gb = π d d F + y y + σ zz b Interaction Force on Disl. B - glide force on y plane F = σyxbx + σyyby + σyzbz (bx= by= 0, bz= b) = σyzb Gb x = π x + z z (bx= bz= 0, by= b)

15 At Position (x = d, z) Gb = π d d + z F Disl. A, Disl. B - opposite sign Disl. A, Disl. B - same sign repulsive "repulsive junction" attractive "attractive junction" 7-8. Formation of New Disl. Segment by Combination of Disl. A and Disl. B on Intersection Line Thomson's Notation DC on Plane BCD + CB on Plane ABC (Forest Disl.) New Disl. : line direction CB b = DC + CB = DB

16 7-9 Stress to Move a Disl. through Forest Disl. of Spacing x FormationE.of a jog= Ej = coreenergy length = 0.Gb b Work done by external load when a disl. of length x moves Minimum stress to move a dislocation through forest dislocations 0.Gb 3 Gb τ 5x τb 1 x x Strengthening of materials due to forest dislocations.

17 7-10. Extended Stacking Fault Nodes Dissociation of a New Disl. Segment formed by Disl. Intersection of Disl. DC and Disl. CB (Fig ) Disl. DC Disl. CB Dα + αc on Cδ + δb on New Disl. at Junction Disl.DB plane BCD plane ABC Dα + αb on planebcd αb,dα, αc : 3 Shockley partialson planebcd Equil. shape of 3 partial disl. due to the surface tension of stacking fault "Extended Stacking Fault Nodes" At position x on partial disl. αc, Stacking Fault Energy, γ equil. Line Tension, T bending of disl. line with radius of curvature, R T Straigteni ng Force / Unit Length due to Line Tension = R Bending Force/Unit Length due to Stacking Fault E. = γ Equilibrium T αgb γ = = R R Measurement of R calculation of γ

18 Equil. Spacing of Shockley Partial Disl., d Gb d = 4πγ Gb γ = 4πd Measurement of d Calculation of γ Equil. Shape of Extended Stacking Fault Nodes γ αgb = Gb R 4πd R 4παd 6d ( α 0.5) Measurement of R Calculation of γ Measurement of R is more accurate to calculate γ, because R is about 6 times larger than d.