Stress Fields Around Dislocations

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1 Stress Fields Around Dislocations The crystal lattice in the vicinity of a dislocation is distorted (or strained). The stresses that accompanied the strains can e calculated y elasticity theory eginning from a radial distance aout 5, or ~ 15 Å from the axis of the dislocation. The dislocation core is universally ignored in calculating the consequences of the stresses around dislocations. The stress field around a dislocation is responsile for several important interactions with the environment. These include: 1. An applied shear stress on the slip plane exerts a force on the dislocation line, which responds y moving or changing shape.. Interaction of the stress fields of dislocations in close proximity to one another results in forces on oth which are either repulsive or attractive. 3. Edge dislocations attract and collect interstitial impurity atoms dispersed in the lattice. This phenomenon is especially important for caron in iron alloys.

2 Screw Dislocation Assume that the material is an elastic continuous and a perfect crystal of cylindrical shape of length L and radius r. Now, introduce a screw dislocation along AB. The Burger s vector is parallel to the dislocation line ζ. Now let us, unwrap the surface of the cylinder into the plane of the paper A πr B L γ tanθ πr G τ Gγ πr

3 Then, the strain energy per unit volume is: τ γ Strain ene rgy We have identified the strain at any point with cylindrical coordinates (r,θ,z) A z r θ Slip plane B τ Zθ z r θ τ θz Slip plane B G 8π r A G τ θ Z Gγ The elastic energy associated with an element is its πr energy per unit volume times its volume. The volume of a pipe is πrδr 1 Energy per unit Length G r1 r0 πr δr 1 G ( ) πr 4π r0 r ln To otain the total energy locked in the crystal due to the screw dislocation we need to integrate the aove equation for all values of r with a r o (minimum) of 5. 1

4 Shear plane Energy per unit length of screw dislocation (integrating from r 0 to r): Elasticity theory reaks down for r 0 ~5 so core energy is ignored here. Roughly, due to r dependence E screw G 4π ln r G r 0 1 4π ln r 1 r 0 The total strain energy of a dislocation is the sum of the elastic strain energy plus the energy of the core of the dislocation (aout 1/15 th of the total energy quantum mechanical calculations). We have shown the distortion of a cylindrical element y a screw dislocation and the equivalent to a simple shear type of distortion. When translated to a coordinate system, the only shear possile are those with a z component.

5 The strains given in cartesian and cylindrical coordinates are: All the other strains should e zero in an isotropic material. The γ γ 13 3 γ γ 31 3 π π G y ( x + y ) x π sinθ r cosθ ( x + y ) π r ( ) associated stresses are given y: π x + y The strain field surrounding the core of a screw dislocation can e represented as: ε 0 0 ε 31 ε ε ε σ σ σ 11 3 σ σ σ 3 σ 33 G π σ 1 x σ y 1 0 G π G cosθ ( x + y ) π r Where each value ε ij depends on the x y position for dislocation lying along the z axis (or 3 axis). sin θ r

6 Energy of Edge Dislocations For idealized edge component, one entire plane has een pushed into the other planes aove the glide plane ut not elow (tensile + compressive stresses). Hence, there is Poisson Effect along length of line, which yields a (1 v) in denominator for strain. compression Idealized Edge E edge G 4π(1 ν) ln r r 0 G (1 ν) For many metals, ν~1/3, so E edge 3 E screw Elasticity theory reaks down for r 0 ~5 so core energy is ignored here. Slides presentations taken from

7 Dislocation Stress Fields: Edges Edges with z axis line direction est descrie in x y plane in cartesian coordinates. Use: See ook, Hirth and Loath. +Edge at center with u(001) and (100) interacting with +edge somewhere neary x x x 1 3 x r cosθ y r sinθ z yx Other stresses are zero! General trend: Aove the edge (x0, y>0), pure compression. Below the edge (x0, y<0), pure tension. Along the slip plane (y0), pure shear. All other locations, compression + tension + shear. y x

8

9 compression σ σ σ τ xx yy zz xy Gy 3x + y π (1 ν) ( x + y ) + Gy x π (1 ν) ( x ν( σ τ yx xx + σ yy y + y ) Gyν ) π(1 ν) ( x + Gx x π (1 ν) ( x y + y 1 G y p ( σ xx + σ yy + σ zz) (1 + ν) 3 3 π (1 ν) ( x + y The elastic displacements around edge dislocations in isotropic materials include all three normal strains ε xx, ε yy and ε zz, and the shear strains in the x y plane γ xy. For an edge dislocation with a core along the z axis and the Burger s vector in the positive x direction ) ε 1 + y ε ε ) ε ε 1 0 ε )

10 Energy and Forces etween Edge dislocations Idealized E edge G (1 ν) Roughly, your expectation should e (as found from intuition): Energy efore: G Energy after: G tot 0 0 Should attract Energy efore: G Energy after: G tot G() 4G Should repel

11

12 Mixed Dislocations Mixed dislocations are dislocation segments wherein the angle etween the Burgers vector and the line direction is neither 90 o (edge) or 0 o (screw). Each mixed dislocation can e resolved into edge and screw components. Energy has component from oth types: E mixed G( ) 4π(1 ν) ln r + G( ) ln r r 0 4π r 0 Edge Screw

13 Comining (Screw, Mixed, Edge): E mixed G 4π(1 ν) ln r (1 ν cos θ) r 0 Where θ is the angle etween the Burger s vector and the line direction. E total E core + E mixed (θ) Screw mixed u sinθ θ π θ cosθ 1 4π ln r ~ 1 r 0 (core energy < 10% of E ) The elastic energy of a dislocation can e generalized as: Edge E dislocation αg Where α is a dimensionless factor ( ) and l is the dislocation length. It can e noted that smaller values of lead to smallest energy. l

14 (111) fcc plane FCC Partial Dislocations and Stacking Faults Partial Dislocations 1 + a 1 01 [ ] a 6 11 [ ]+ a [ ] 1 1y y 1x x 1y and y are attractive screw segments 1x and x are repulsive edge segments

15 If energy is favorale, G > G 1 + G, then partial dislocation form. (We need to show: Ga / > Ga /3) Energies of Full and Partials are E 1 G E 1+ G 1 + G Ga Ga E G G ( ) 4 > E 1+ Ga 36 ( ) Ga 3 Favorale for partials to form, i.e. dislocation disassociate. Dislocations may e sessile if not on the correct slip plane. due to ABC stacking Here partials form, edge repulsion wins out, which creates stacking faulted region in etween.

16 Separation of partials Motion of partials FCC partial Partial Dislocations 1 + a [ 101 ] a [ 6 11 ]+ a [ ] In FCC, due to ABC stacking, if partials form, edge repulsion wins out, which creates stacking faulted region in etween. Green Partials Separate. Stacking Faults are defects that cost energy Energy alance etween separating partials to lower elastic energy and creation of more SF. A B C A B HCP C Partial dislocations move apart. As they move apart leave hcp SF rion. ABC 3 layers AB layers ABCABC converts to ABABAB

17 Non conservative Motion for Edge Dislocation: Vacancy assisted Clim Vacancy: Missing atom Edge clims up Swapped with atom at ottom of edge Clim is non conservative in work. Only a part of the dislocation line clims up, hence it will generate jogs. Edge can clim down too!

18 Vacancy assisted Clim creates jogs! n is the slip plane normal. is the Burger s vector of edge dislocation. τ is shear stress (could e applied or from other dislocation lines). Jogs will create sessile edge dislocation segments Moves Glissile Does not move Sessile With jogs, an edge dislocation will have sections that are sessile! Why? Because segments are not laying in possile slip plane. n xn n, τ Slip plane Not a slip plane xu, τ Slip plane τ xn xn

19 Conservative Motion for Screws: Cross Slip For a FCC n 1 ( 111) n ( 111) n 1 ( 111) u The Burger s vector is : This is a Screw dislocation It moves in the direction of n n d ( 111) ( 111) [ 101] 1 n ( 111) ( 101) [ 11] 1 Cross Slip of Screw Component n 1 ( 111) n ( 111) n 1 ( 111) u

20 Separation of Partials: Stacking Fault Here partials are favorale, G > G 1 + G, since Ga / > Ga /3. In FCC crystals, the magnitude of the Burgers vector is: a u for edge Partial Dislocations 1 + a 1 01 [ ] a 6 11 [ ]+ a [ ] 1 screw 1x edge 1 y a [ ] y x edge a [ ] As the edge components have the same direction, the 1x and x components of the partials will repel. As the screw components of the partial dislocations 1y and y are in opposite direction these will attract. 1y y a 1 a 1 [ 1 1] [ 1 1]

21 As the partials separate, the energy increase y d*sfe, where SFE is the stacking fault energy and d is the fault separation distance. Thus, the chemical force resisting separation is SFE (dimensions of joules per square meter or force per unit length). The attractive force etween the parallel screw dislocations separated y a distance d is: Gs FScrew ± τ ± Gγs ± πd The repulsive force etween the edge components of the dislocations is: + G x x y F edge π (1 ν) ( x + y ) Since the dislocations are in the same plane and separated y a distance d, then y0 and xd F G edge π (1 ν ) d Since the units of SFE are in force per unit length, and so are those of the forces aove, we can just equate them directly.

22 SFE + F Screw F Edge SFE+ G π 1 d G π (1 ν) 1 d G 1 d π ( SFE) (1 ν) 1 a For a perfect dislocation 110 For a partial dislocation a 11 6 a 6 d Ga 1 1π ( SFE) (1 ν) 1 High SFE low separation, low SFE large separation in TEM.

23 Stacking Faults and Energy Partial dislocation repel and leave stacking faults Stacking Faults are defects that cost energy Energy alance etween separating partials to lower elastic energy and creation of more SF. Partial Dislocations 1 + For ideal case: d sf fcc G1 π ( SFE) hcp fcc 1 d SF TEM image In TEM, you see contrast etween faulted and unfaulted regions, hence, you can measure d SF and get SFE.

24 A Full (Mixed) Dislocation Recomination: Lomer Cottrell Locks β γ C D δ α B Consider two slip planes in FCC crystals, δ and γ, namely (111) and (11 1) planes. The three perfect dislocations in the (111) plane are a [ 110] a [ 101] a [ 011] 1 3 The three perfect dislocations in the (11 1)plane are a [ 110] a a 4 5 [ 101] 6 [ 011] The direction 1 and 4 are in opposite direction and they will cancel each other. The comination and 5 will result in a higher energy (not possile). The sole comination that result in a energy reduction is the 3 and 5. a [ ] a [ ] a [ ] 3 a + 5 This dislocation is not moile in either (111) or (11 1) planes. a + a >

25 Full (Mixed) Dislocation Recomination: Cottrell Locks (MIXED) full dislocation reaction: 1 + Check slip planes? 1 a [ 101 ] n 1 ( 11 1 ) 1 n 1 0 a [ 01 1 ] n ( 111 ) n 0 n(001) a [ 110 ] motion u [ 1 1 0] [ ] 1 a 101 ( 11 1 ) u 1 [ 1 1 0] [ ] u motion [ ] a 01 1 ( 111) Mixed dislocations: u s are all parallel to intersection, and s are not to u s.

26 Burger s vector,? (See figure aove in cue) Favorale to recomine? Yes, G 1 + G > G 1 + a [ 101 ]+ a [ 01 1 ] a [ 110 ] 1 + a > a Slip Plane? does not lie in either of the two slip planes, ut does lie in n x u (001). a [ 110 ] Glissile or Sessile? Sessile, not {111} fcc slip plane It impedes slip and therefore is called a lock. Unless lock (sessile dislocation) is removed, dislocation on same plane cannot move past. Going ack > 1 + would allow other dislocation to glide again.

27 Schockley Partial Dislocations Recomination: Locks Partial dislocations reaction: p1 + p 1 a [ 101 ] p p a [ 6 11 ]+ a n 1 0 n 1 ( 111 ) n 0 n ( 111) [ ] a [ 01 1 p1 ] + p a [ ]+ a 6 11 [ ] n motion 1 p p 1 u [ 1 1 0] 1 p 1 u motion ( 11 1 ) u 1 [ 1 1 0] p u ( 111) Lormer Cottrell lock. But if full s comine, it is Cottrell lock.

28 Burger s vector? Leading partials comine 1 p + p1 a [ ]+ a [ ] a [ ] Favorale to recomine? Check G 1 + G > G 1 + a 3 > a 18 Line Direction? u n 1 n i j k i j + k0 [1 1 0] Slip Plane? n u [00 1 ] Glissile or Sessile? Sessile, not {111} fcc slip plane Unless lock (sessile dislocation) is removed, dislocation on same plane cannot move past. Other possile cominations give: a [ ]

29 Edge Edge Interactions: creates edge jogs **Dislocations each acquire a jog equal to the component of the other dislocation s Burger s vector that is normal to its own slip plane. Energy cost of jog: G / (Energy/length) x (length of jog) G 3 / This dislocation got a jog in direction of 1e. efore after 1e e e 1e Dislocation 1 got a jog in direction of e of the other dislocation; thus, it got longer. Extra atoms in half plane increases length. What happens when dislocations are extended, i.e. composed of two trailing partials?

30 Screw Edge Dislocation Interactions: creates edge jogs Direction of screw dislocation motion Time snap shots s line edge (later) s edge e e B of screw e This end of edge goes underneath creates jog. Edge jog is in direction of s. Jogs slow motion of dislocation. Energy cost of jog is G 3 / Why does screw also have jog?

31 Screw Edge Dislocation Interactions: creates edge jogs Direction of screw dislocation motion s line Edge jog is in direction of s. Jogs slow motion of dislocation. B of screw e edge (early) s edge (later) e Screw jog is in direction of e. Why is there a jog in screw?

32 Screw Screw Dislocation Interactions: creates edge jogs s line screw (later) s screw Time snap shots s e s Energy cost of jog G 3 / Jogs slow motion of dislocation. In screw screw case, jog has to move via CLIMB, or generate a row VACANCIES or INTERSTITIALS. Clim is non conservative, and point defects costs more energy.

33 Multiplication of Dislocations To account for large plastic strain that can e produced in crystals, it is necessary to have regenerative multiplication of dislocations. Of course, there are many variants that lead to many effects. Two important mechanisms for this are: Frank Read sources and multiple cross glide Marked pts could e from cross slip Fig Cross slip of single crystal of Fe 3.5%Si From Hall and Bacon 4th Ed

34 Dislocations Generation: single ended Frank Read source Single ended Frank Read source leads to regenerative multiplication. This mechanisms can e attained from a superjog, where an extended line is out of the slip plane and thus sessile. Sessile segment Segment BC is edge anchored at one end. (a) Moves y rotating. () Each revolution around B displaces the crystal aove slip plane y, so n revolutions gives n slip. Spiraling around B increases line length. CEF For super jogs, see ook and Gilman and Johnston, Solid State Physics 13, 147 (196).

35 Small jog Dislocation Generation: Frank Read Source After effects of dislocation dislocation interaction τ applied shear stress Unstale position: loop expands Shear owing of line τ rθ L Tsinθ/ r θ/ θ/ T Tsinθ/ Line Tension (E/L) T sinθ/ ~ Tθ θg / What type of dislocations? What can happen? Generated a dislocation in place of old one, which is now a loop. (Shaded area has 1 unit of slip.) Larger density. Back stresses hinder motions. Screws annihilate Shape due to Si directional onding opposes owing via shear τ: F/L * owing arc τ rθ So, τ rθ θg / τ G/r Radius of curvature r smallest for semicircular arc of r L/. Larger L easier to deform. τ max G/L

36 A more likely mechanism for dislocation loop formation is the Frank Read Source dislocation pinned at oth ends. Radius of curvature depends on applied shear stress. Critical ow out for R L/ (L AB) Further steps are the formation of a kidney shaped loop and the annihilation of dislocation segments with the same vector ut opposite line sense.

37 Dislocation Generation: Frank Read Source via Cross Slip What type of dislocation is in (a)? τ applied shear stress can e parallel and perpendicular to. Looks like two pints on (111) plane, as in Si case jog τ applied shear stress Bowing of cross slipped dislocation line is similar to jogged dislocations. L

38 Summary Dislocations (line defects) give rise to complicated interactions in a crystal. Dislocation multiplication is responsile for the very large increases in YS. Dislocation dislocation interactions, or dislocations interacting with other defects, lead to higher stresses required to move the dislocations further (work hardening). For example, dislocation pile up, jogs, trasnfer across grain oundaries, etc., all contriute to YS increases. Dislocations interacting with anything lead to other defects (point, planar, volumetric). Consequences are found in the allowed slip and strengthening of materials. Be familiar/conversant with how dislocations interact and the consequences. Are these consequences ale to e mathematically descried?

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