Stress Fields Around Dislocations The crystal lattice in the vicinity of a dislocation is distorted (or strained). The stresses that accompanied the strains can be calculated by elasticity theory beginning from a radial distance about 5b, 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 responsible 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 by moving or changing shape. 2. Interaction of the stress fields of dislocations in close proximity to one another results in forces on both 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 carbon in iron alloys. 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 b A 2πr GL b γ = = tanθ 2πr G bG τ = Gγ = B 2πr 2 Then, the strain energy per unit volume is: τ× γ b G Strain energy = = 2π 82r 2 We have identified the strain at any point with cylindrical coordinates (r,θ,z) τ τZθ θZ B r B θ r θ Slip plane z Slip plane A z G A b G τ=G γ = The elastic energy associated with an element is its θZ 2πr energy per unit volume times its volume. The volume of a pipe is 2πrδ r 1 r1 2πr 1 ⎛ r ⎞ Energy unit per Leng= th2 Gbδr = Gb2 ln⎜ 1 ⎟ ∫r0 2 ⎜ ⎟ 2 ()2πr 4π ⎝ r0 ⎠ To obtain the total energy locked in the crystal due to the screw dislocation we need to integrate the above equation for all values of r with a ro (minimum) of 5b. Energy per unit length of screw dislocation Roughly, due to r dependence (integrating from r0 to r): S 1 ⎛ r ⎞ 1 hea ln⎜ ⎟ ≈ Elasticity theory breaks down for r0~5b so 4π ⎝ r0 ⎠ 2 r p la core energy is ignored here. ne Gb2 ⎛ r ⎞ Gb2 Escrew = ln⎜ ⎟ ≈ 4π ⎝ r0 ⎠ 2 The total strain energy of a dislocation is the sum of the elastic strain energy plus the energy of the core of the dislocation (about 1/15th of the total energy – quantum mechanical calculations). We have shown the distortion of a cylindrical element by a screw dislocation and the equivalent to a simple shear type of distortion. When translated to a coordinate system, the only shear possible are those with a z‐component. − b y − b sinθ γ13= γ 31 = = The strains given in cartesian and 2π x()2+ y 2 2π r cylindrical coordinates are: b x b cosθ γ23= γ 32 = = 2π x()2+ y 2 2π r σ = σ = σ = σ = σ = 0 All the other strains should be zero11 22 33 1221 in an isotropic material. The − Gb y − Gb sinθ σ13= σ 31 = 2 2 = associated stresses are given by: 2π x()+ y 2π r Gb x Gb cosθ σ23= σ 32 = = The strain field surrounding the 2π x()2+ y 2 2π r core of a screw dislocation can be represented as: Where each value εij depends on the x‐y ⎡0 0 ε13 ⎤ position for dislocation lying along the z‐ ε = ⎢0 0 ε ⎥ axis (or 3‐axis). ⎢ 23 ⎥ ε⎣⎢31 ε 32 0 ⎦⎥ Energy of Edge Dislocations For idealized edge component, one entire plane has been pushed into the other planes above the glide plane but not below (tensile + compressive stresses). Hence, there is Poisson Effect along length of line, which yields a (1‐v) in denominator for strain. c om Idealized Edge pre s sion Gb2 ⎛ r ⎞ Gb2 Eedge = ln⎜ ⎟ ≈ 4π(1−ν) ⎝ r0 ⎠ 2(1−ν) For many metals, ν~1/3, so 3 Eedge = Escrew 2 Elasticity theory breaks down for r0~5b so core energy is ignored here. Slides presentations taken from http://web.mse.uiuc.edu/courses/mse406/Handouts/index.html Dislocation Stress Fields: Edges Edges with z‐axis line direction See book, Hirth and Loath. best describe in x‐y plane in +Edge at center with u=(001) and b=(100) cartesian coordinates. Use: interacting with +edge somewhere nearby y=x x1 = x= cos r θ x2= y = sin r θ x3 = z Other stresses are zero! y=–x General trend: • Above the edge (x=0, y>0), pure compression. • Below the edge (x=0, y<0), pure tension. • Along the slip plane (y=0), pure shear. • All other locations, compression + tension + shear. Gby– 3 2+ x 2 y σ = xx2π ( 1− ν()x2 ) + y2 2 c 2 2 om Gby+ – x y pre s σ yy = 2 2 2 sion 2π (− 1 ν()x ) + y – Gbyν 1 σ= ν() σ + σ = zz xx yy π( 1− ν()x )2 + y 2 Gbx+ 2– x 2 y τ= τ = xy2 yx π (− 1 ν()x2 ) + y2 2 1 2 Gb y = −p σ( + σ + σ) ( =1ν ) + 3 xx yy zz 3 2π ( 1− νx )2 ( + y 2 ) The elastic displacements around edge dislocations in isotropic materials include all three normal strains εxx, εyy and εzz, and the shear strains in the x‐y plane γxy. ε⎡ ε 0 ⎤ For an edge dislocation with a core along the z‐ 11 12 ε = ε⎢ ε 0 ⎥ axis and the Burger’s vector in the positive x‐ ⎢21 22 ⎥ direction ⎣⎢0 0 ε33 ⎦⎥ Energy and Forces between Gb2 Edge dislocations Idealized Eedge ≈ 2(1−ν) Roughly, your expectation should be (as found from intuition): Energy before: 2Gb2 b Should attract 2 –b Energy after: Gbtot = 0 b=0 2 b Energy before: 2Gb b Should repel 2 2 2 b=2 Energy after: Gbtot = G(2b) = 4Gb Mixed Dislocations Mixed dislocations are dislocation segments wherein the angle between the Burgers vector and the line direction is neither 90o (edge) or 0o (screw). Each mixed dislocation can be resolved into edge and screw components. Energy has component from both types: 2 ⎛ ⎞ 2 ⎛ ⎞ G(b⊥) r G(b||) r Emixed = ln⎜ ⎟ + ln⎜ ⎟ 4π(1−ν) ⎝ r0 ⎠ 4π ⎝ r0 ⎠ Edge Screw Combining (Screw, Mixed, Edge): Where θ is the angle between 2 ⎛ ⎞ Gb r 2 the Burger’s vector and the line Emixed = ln⎜ ⎟(1 −ν cos θ) 4π(1−ν) ⎝ r0 ⎠ direction. E total = Ecore + Emixed (θ) b Screw 1 ⎛ r ⎞ 1 mixed ln⎜ ⎟ ~ (core energy < 10% of E ) u 4 π ⎝ r0 ⎠ 2 θ b π −θ b bsin 2 || = θ The elastic energy of a dislocation b ⊥ = bcosθ can be generalized as: b 2 Edge Edislocation = αGb l Where α is a dimensionless factor (0.5‐1.0) and l is the dislocation length. It can be noted that smaller values of b lead to smallest energy. FCC Partial Dislocations and Stacking Faults (111) fcc plane Partial Dislocations b = b1 + b2 b a a a []1 01 = []2 11 + [1 1 2] 2 6 6 b2 b1 b1y b2y b1x b2x b1y and b2y are attractive screw segments b and b are repulsive edge segments b 1x 2x 2 2 2 If energy is favorable, Gb > Gb1 + Gb2 , then partial dislocation form. (We need to show: Ga2/2 > Ga2/3) 2 2 2 E1 = Gb E1+2 = Gb1 + Gb2 Energies of Full and Partials are 2 2 2 2 2 Ga Ga Ga Ga E= Gb = • Gb =( 1 b 0 + 1 + ) => E1+2 = 2 (4 + 1+ 1) = 4 2 36 3 Favorable for partials to form, i.e. dislocation disassociate. Dislocations may be sessile if not on the correct slip plane. due to …ABC… tackings Here partials form, edge repulsion wins out, which creates stacking faulted region in between. Partial Dislocations b = b1 + b2 a a a Motion of []1 01 = []2 11 + [1 1 2] partials 2 6 6 In FCC, due to …ABC… stacking, if partials b form, edge repulsion wins out, which creates stacking faulted region in between. Green Partials Separate. Separation of Stacking Faults are defects that cost energy partials Energy balance between separating partials to lower elastic energy and creation of more SF. FCC partial A A B B Partial dislocations move C C apart. As they move apart leave hcp SF ribbon. ABC = 3 layers AB = 2 layers HCP ABCABC converts to ABABAB Non‐conservative Motion for Edge Dislocation: Vacancy‐assisted Climb Edge climbs up ⊥ Swapped with atom at bottom of edge Vacancy: Missing atom Climb is non‐conservative in work. Only a part of the dislocation line climbs up, hence it will generate jogs. Edge can climb down too! n Vacancy‐assisted Climb creates jogs! • n is the slip plane normal. b, τ • b is the Burger’s vector of edge dislocation. • τ is shear stress (could be applied or from other dislocation lines). Slip plane Jogs will create sessile edge dislocation segments Moves = Glissile Does not move = Sessile bxn With jogs, an edge dislocation will have sections that are sessile! Why? Because segments are not laying in possible slip plane. n b, τ τ Not a slip plane Slip plane bxu2 b bxn bxn Conservative Motion for Screws: Cross‐Slip For a FCC n1 = (1 1 1) n2 = (1 1 1) n1 = ()1 1 1 b u The Burger’s vector is : This is a Screw dislocation b = n1 × n2 = ()1 1 1 × (1 1 1)= [1 0 1] It moves in the direction of d = n1 × b = ()1 1 1 × (1 0 1)= [1 2 1] Cross‐Slip of Screw Component n1 = (1 1 1) n2 = (1 1 1) n1 = ()1 1 1 b u Separation of Partials: Stacking Fault Partial Dislocations b = b1 + b2 Here partials are favorable, 2 2 2 a a a Gb > Gb1 + Gb2 , []1 01 = []2 11 + []1 1 2 since Ga2/2 > Ga2/3.