Non-Equilibrium Continuum Physics TA session #10 TA: Yohai Bar Sinai Dislocations

Transcription

1 Non-Equilibrium Continuum Physics TA session #0 TA: Yohai Bar Sinai Dislocations References There are countless books about islocations. The ones that I recommen are Theory of islocations, Hirth & Lothe. Crystals, efects an micro-structures, Rob Philips (Chap. 8). Elementary islocation theory, Weertman & Weertman. 2 Continuum theory of islocations (a) Schematic of a islocation (From Rob Philips book) (b) Yiel stress as a function of islocation ensity (Kocks & Mecking, 200). Figure Dislocations are line efects in crystals. Uner normal conitions, they are the main carriers of plastic eformation, an therefore are crucial in its escription. The efining property of a islocation is that preforming a line integral over the isplacement fiel aroun the islocation line results in a non-zero value: u i = j u i x j b i 0 () This value, b, is calle the Burger s vector of the islocation.

2 2. Volterra moel, elastic fiels The Volterra moel for a islocation consists of cutting a bulk along a half plane, an then shifting the relative parts (Fig a). The bounary of the half plane is calle the islocation line. If the isplacement is parallel to the islocation line, the islocation is calle screw islocation an is similar to multi-stories parking lots. If the eformation is perpenicular to the islocation line, it is calle an ege islocation. When both components are present the islocation is calle a mixe islocation. The iscrete analogue of the continuum contour integral () can be thought of as walking along the crystalline irections in what is suppose to be a close circle (Fig a, 0 atoms ownwars, 0 to the right, 0 upwars an 0 left ). If the loop oes not close, you have encircle a islocation. Note that very close to the islocation core it is har to tell what are the crystalline axes, but far away there s no problem. You also see that the irection an magnitue of the Burgers vector is relate to the crystal structure. The elastic fiels of such a straight islocation can be calculate. In fact, we ve alreay preforme this calculation in class for screw islocations (starting aroun Eq (5.45) in the lecture notes). If the islocation line is efine as the ẑ irection, then the fiels are u z = b z θ 2π, (2) ɛ θz = r θu z = b z 2πr, (3) σ θz = µɛ θz µb Z 2πr, (4) an all other components vanish. b z is the z component of the burgers vector. A similar but a tiny bit less elegant result can be obtaine for the case of an ege islocation. We ll quote only the stresses: σ rr = σ θθ = µb sin θ, 2π r (5) σ zz = 2ν µb sin θ, ν 2π r (6) σ rθ = µb cos θ ν 2π r (7) where θ is measure from the extra plane of the islocation. Since the theory is linear, for mixe islocations we can simply a their screw an ege components. 2.2 Dislocation energy Note that the fiels iverge at r 0. This is clearly unphysical, an where the stress is too high linear elasticity breaks own an something else happens. Also, as always we have a short-length cutoff at length comparable to the lattice spacing. The region where the elastic solution is no longer vali is calle the islocation core, an is of the orer of the lattice spacing. Note that since the total energy goes like log(r max /r min ) we nee both upper an lower cutoffs - the system size an the islocation core size. The islocation energy iverges with the system size! 2

3 In the core, aitional energy is store which is not escribe by linear elasticity. As a crue estimate, we can say that the stress there is fixe at its value on the core raius r c, where r c is the core s raius (something like elastic-perfect-plastic behavior). The energy (per unit length) of the core is thus E core = µ ɛ rθ σ rθ (πrc). 2 If we estimate r c b, the total energy per unit length of the islocation reas, roughly, ( E tot /L µb 2 log r ) max + (8) b For metals, typically µ 50GPa, b Å. Even for very small systems, say r max 0nm, an log( ), the energy is µb ev per nm. This is much larger than thermal energies at R.T. (k B T /40eV), so this raises serious questions about the nature of islocations. They can not be create by thermal fluctuations - they are purely out-ofequilibrium creatures. 2.3 Deformation Plastic eformation occurs when islocations travel to the bounary of the material. In fact, the islocation line can be thought of the bounary between the region along the slip plane that has slippe an the region that in t slip yet (again, Fig. a). When the islocation reaches the bounary, an atomic step is create. Each islocation carries with it a eformation charge equal to its Burgers vector. This is, in a way, a topological charge, that is conserve. Dislocation lines can split, coalesce, form junctions etc. but the Burgers vector is always conserve. A way to see this is that the eformation fiels are continuous an therefore when eforming the integration contour () the result cannot change. 2.4 Forces, configurational forces an interactions In orer to investigate the interaction between islocations an other stuff (external forces, other islocations, free bounaries, point efects...) we nee to look at the energy. For example, let s consier two parallel ege islocations with two parallel Burger vectors b, b 2 at istance apart. The total energy of the system is ( ) ( ) E tot = Ecore + Ecore 2 + ɛ ij + ɛ 2 ij σ ij + σij 2 3 x (9) Ω ( ) = Eself + Eself 2 + ɛ ij σij 2 + ɛ 2 ijσij 3 x (0) Ω }{{} The self energies are inepenent of. However, the interaction term oes epen on, an its erivative with respect to is the force that the islocation exert on one another. These kin of forces that is, erivatives of the energy with respect to a translationalsymmetry-breaking parameter are calle configurational forces. The interaction energy (per unit length square) is evaluate to be E int E int = µb b 2 () 2π( ν) x x 3

4 The interaction is repulsive if the b an b 2 have the same sign, an vice versa. This is a simple case where the islocations are parallel an so is their Burgers vector - in the general case the interaction is highly anisotropic. We see that islocations are very strongly interacting (interaction energy ecays as r ). This is usually thought to inuce strain harening, etc (Fig. b). Following the same proceure, we can calculate the interaction energy of a islocation an an external stress fiel σ. Differentiating with respect to the position of the islocation, we get the Peach-Koheler force: F = E ijk σ il b l ξ j = (σ b) ˆξ (2) where F is the force per unit length an ˆξ is the irection of the islocation line. 3 Discrete effects 3. A continuous islocation Figure 2: Setup for the calculation of the Peierls-Nabarro islocation. From Hirth & Lothe. Note that here we assume that the magnitue of the Burgers vector is one lattice spacing. We ve seen that there are configurational forces acting on islocations (an other efects), but in orer to get a notion about the plastic flow, we still nee to say something about the ynamics of islocation movement. When two islocations interact an the energy is not at its minimum (i.e., there are forces), will the islocation always move? An what about external forces? For this, we nee some moel of the islocation core. Probably the simplest moel that offers goo quantitative preictions is the Peierls-Nabarro moel. The setup is sketche in Fig. 2: Assume we have two half-infinite lattices with the bottom lattice having an extra plane of atoms. Now glue the half lattices together. What will be the isplacement fiel that minimize the energy? To each point along the slip plane we can assign a local 4

5 misregistry value (x) which measures to what extent the crystals above an below the plane are misaligne. We know that ( ) = () = b/2, an knowing the profile of (x) is the goal. The energy of the islocation is compose of the elastic energy of the bulk, erive earlier, an an aitional term, E mis which is calle the misfit energy an which stems from the inter-planar potential φ( ). The potential has the same perioicity as the lattice. The misfit energy might be thought of as the core energy mentione before. We write E mis = φ( (x))x (3) We now remin ourselves of Clapeyron s theorem, which says that in static conitions the elastic energy store in the bulk of a general soli is E bulk = σ ij u j n i, (4) 2 In our case, we ll integrate on the two half-crystals above an below the slip plane. The normal is therefore ŷ, an we have two surfaces as y = ±η with η 0. The energy is E bulk = 2 Ω σ xy (u x (η) u x ( η))x = 2 For the Volterra islocation, we use the stresses from (7): σ xy (x)x (5) σ xy (y = 0) = µb 2π( ν) x (6) an (x) is simply so the energy is (x) = E bulk = 2 0 { b x < 0 0 x > 0 µb 2 2π( ν) x an it is infinite. The misfit energy, however, is zero. Thus, we expect that the lattice will fin a way to reuce the energy by smoothing out the islocation core, whose size is zero in the Volterra moel, thus increasing the misfit energy but reucing the bulk energy. The key point in our moeling of the core is consiering the slip plane as compose of a continuous istribution of infinitesimal parallel ege islocations, with their magnitue given by b(x) = ( x )x, or ensity ρ = x. We ll see that for a islocation with core ensity ρ(x) the stress fiel will not iverge: it is the convolution of Eq. (6) with ρ, i.e., σ xy (x) = µ 2π( ν) (7) (8) ρ(x ) x x x. (9) In orer to fin the minimzer of the energy we can write the total energy an use the Euler-Lagrange Equations. This is a somewhat technical calculation which we will not 5

6 follow here. However, it s final result is very intuitive: it simply states that the stresses inuce by the islocations shoul exactly cancel the effective stresses inuce by the misfit energy, i.e. φ µ x (x ) ( ) = x (20) 4π( ν) x x This is an integro-ifferential equation an to solve it, we nee an explicit form for φ. Note that φ must have the same symmetry of the lattice, an in particular in must be a perioic function with perio. The simplest function that oes that is a cosine, φ( ) = φ 0 2 ( cos 2π ), (2) where is the inter-planar spacing. This form was also use in the erivation of the ieal shear strength (Eq. (.5) in Eran s notes). The constant φ 0 can be obtaine by emaning compliance with Hooke s law: for small isplacements the restoring stress must be σ xy = 2µε xy = 2µ + O( 2 ) (22) φ = πφ ( ) 0 2π sin = 2π2 φ 0 + O( 2 ) (23) 2 an equating the two gives φ 0 = µb2. With this form for the potential we arrive at the π 2 famous Peierls-Nabarro integro-ifferential equation µ 4π( ν) (x ) x x x = φ 0π sin 2π (24) The solution of integro-ifferential equation is a whole story that we re not getting into. The solution to this particular equation was given by Peierls to be (x) = b π tan ( x ζ ) b2, ζ = µb 2 4π 2 φ 0 ( ν) (25) Note that the core half-with ζ represents the competition between the elastic stiffness µ, which tens to sprea the islocations out, an the non-linear misfit potential φ 0, which tens to localize the islocation core. Plugging in our value for φ 0 yiels ζ = this confirms our estimation that the core with scales with the lattice spacing. an 2( ν) 6