Non-Equilibrium Continuum Physics TA session #5 TA: Yohai Bar Sinai 13.04.2016 Linear elasticity I This TA session is the first of three (at least, maybe more) in which we'll dive deep deep into linear elasticity theory. Linear elasticity by itself can be the topic of a year-long course, and in these lectures we'll try to convey a significant fraction of the richness of the theory. While Eran will continue to explore more complex constitutive behaviors (non-linear elasticity, visco-elasticity, and so on), in the next few TA sessions we'll stay in the linear realm and try to acquire the knowledge we believe is fundamental for a well-educated physicist. 1 Hooke's law, stiffness, and compliance Hooke's law is 2 1 σ = λε δ + 2µε = λ + µ tr(")δ + 2µ " − tr(")δ (1) ij kk ij ij 3 ij ij 3 ij Let's write it explicitly, to get a better feel of what's going on σxx = 2µεxx + λ ("xx + "yy + "zz) = (2µ + λ)"xx + λ ("yy + "zz) σ = 2µε + λ (" + " + " ) = (2µ + λ)" + λ (" + " ) yy yy xx yy zz yy xx zz (2) σzz = 2µεzz + λ ("xx + "yy + "zz) = (2µ + λ)"zz + λ ("xx + "yy) σij = 2µεij; i 6= j The off-diagonal terms (shear terms) have a simple dependence, while the diagonal ones are a bit more complicated. Let's try to invert these relations to find "(σ) - that is, let's find the compliance matrix for an isotropic linear elastic material. The same considera- tions that we used to derive Eq. (1) (i.e. that " and σ are related by a 4-rank isotropic tensor) allow us to write "ij = a σkkδij + b σij (3) So finding the compliance reduces to finding a; b. If tr " = 0, then Eq.(1) and (3) reduce to, respectively, σij = 2µ "ij (4) "ij = b σij (5) So we immediately find b = (2µ)−1. Taking the trace of Eq. (1)and (3) gives, respectively tr(σ) = (3λ + 2µ) tr(") (6) tr(") = (3a + b) tr(σ) (7) This tells us that 1 1 −λ 3a + b = (3λ + 2µ)−1 ) a = − b = (8) 3 3λ + 2µ 2µ(3λ + 2µ) λ 1 " = − tr(σ)δ + σ (9) ij 2µ(3λ + 2µ) ij 2µ ij 1 Writing explicitly, we have λ 1 λ " = − + σ − (σ + σ ) (10) xx 2µ(3λ + 2µ) 2µ xx 2µ(3λ + 2µ) yy zz λ 1 λ " = − + σ − (σ + σ ) (11) yy 2µ(3λ + 2µ) 2µ yy 2µ(3λ + 2µ) xx zz λ 1 λ " = − + σ − (σ + σ ) (12) zz 2µ(3λ + 2µ) 2µ zz 2µ(3λ + 2µ) xx yy 1 " = σ i 6= j (13) ij 2µ ij As discussed in class, the term in parentheses in Eq.(10) is the inverse of the Young's modulus, and for uniaxial stress it reads λ 1 −1 3λ + 2µ E = σ =" = − + = µ : (14) xx xx 2µ(3λ + 2µ) 2µ λ + µ It is the microscopic analogue of the spring constant. If the uniaxial stress is in the x direction, then we have λ λ λ " = − σ = − E" = − " ; (15) yy 2µ(3λ + 2µ) xx 2µ(3λ + 2µ) xx 2(λ + µ) xx λ and as discussed in class, −"yy="xx is known as the Poisson ratio ν = 2(λ+µ) . Rewriting Eqs. (10) with these quantities yields a much nicer expression: 1 " = [σ − ν(σ + σ )] ; xx E xx yy zz 1 "yy = [σyy − ν(σxx + σzz)] ; E (16) 1 " = [σ − ν(σ + σ )] ; zz E zz xx yy 1 + ν " = σ for i 6= j : ij E ij 2 Green's function for an infinite medium It seems that the time is ripe to fully and completely solve a problem, with all the 2π's and everything, without resorting to hand waving and scaling arguments. While the emphasis will still be on the structure of the problem, we think it will be instructive, at least once, to write down a problem and solve it exactly. A nice problem to consider is the response of an infinite linear isotropic homogeneous ~ elastic medium to a localized force f = Fiδ(~r), i.e. finding the Green's function of an infinite medium. We define the Green function (matrix) Gij(~r1; ~r2) as the displacement in the i direction at the point ~r1 as a response to a localized force in the j direction applied at ~r2. For homogeneous materials we know that Gij(~r1; ~r2) = Gij(~r1 − ~r2). We therefore denote 2 ~r = ~r1 − ~r2. You all know well that, within the linear elastic theory, this will allow us to solve the problem of an arbitrary force distribution f(~r) by convolving f(~r) with the Green function. Conceptually, the structure is the following. We want to find a displacement field ui(~r), from which we calculate ui ) "ij ) σij ) @jσij = δ(~r) Of course, we will want to do the whole thing backwards. I stress (no pun intended1) that we already know how to express div(σ) in terms of ui - this is what we called the Navier-Lam´eequation. But for completeness, let's do it again: σij = λεkk δij + 2µ "ij = λ∂kukδij + µ (@iuj + @jui) @jσij = @j (λ∂kukδij + µ (@iuj + @jui)) = (λ + µ)@i@juj + µ∂j@jui ; which is nothing but the u-dependent term of the Navier-Lam´eequation. Since the equation is linear, it seems right to solve the problem by Fourier transform. We use the conventions Z 3 i~q·~r ui(~q) = d ~xe ui(~r) ; (17) 1 Z u (~r) = d3~qe−i~q·~ru (~q) : (18) i (2π)3 i The equation we want to transform is (λ + µ)@j@iui + µ∂j@jui = −Fiδ(~x) ; (19) which readily gives − (λ + µ)qjqi uj − µ qjqj ui = −Fi : (20) This is a matrix equation: [(λ + µ)qjqi + µ qkqkδij] uj = Fi : (21) Or even more explicitly: 0 2 2 1 0 1 0 1 (λ + µ) q1 + µj~qj (λ + µ) q1q2 (λ + µ) q1q3 u1 F1 2 2 @ (λ + µ) q1q2 (λ + µ) q2 + µj~qj (λ + µ) q2q3 A @u2A = @F2A (22) 2 2 (λ + µ) q1q3 (λ + µ) q2q3 (λ + µ) q3 + µj~qj u3 F3 This matrix can be inverted by using any of your favorite methods, giving 1 δij 1 qiqj ui = − 2 Fj : (23) µ qkqk 2(1 − ν) (qkqk) λ Where we used ν = 2(λ+µ) . In other words, we have found the Fourier representation of the Green function: 1 δij 1 qiqj Gij(~q) = − 2 : (24) µ qkqk 2(1 − ν) (qkqk) 1Just kidding, of course it's intended. 3 We now need to preform the inverse Fourier transform. We'll begin with the first term, and do it in spherical coordinates with the qz direction parallel to ~r: 1 Z e−i~q·~x 1 Z e−iqr cos θ 1 Z d3~q = q2 sin θ dθ dq dφ = e−iqr cos θ sin θ dθ dq (2π)3 q2 (2π)3 q2 (2π)2 1 Z eiqr − e−iqr 2 Z 1 sin(qr) 1 = 2 dq = 2 dq = (2π) iqr (2π) 0 qr 4πr If this result surprises you, maybe you should remind yourself of the first linear field theory that you met in your life - Poisson's equation for a point charge r2φ = δ(~r). I'll let you complete the analogy by yourselves. For the second term, we use a dirty trick: −1 qiqj 1 −1 @ 1 i @ −1 @ 1 F 2 = − F qi = F (qkqk) 2 @qj qkqk 2 @xi @qj qkqk i @ −1 1 1 @ xj = −ixjF = (25) 2 @xi qkqk 2 @xi 4πr 1 δ x x = ij − i j 2 4πr 4πr3 Plugging in (24) we get 1 h x x i G (~r) = (3 − 4ν)δ + i j : (26) ij 16(1 − ν)πµr ij r2 A more elegant way to go, is to write Gij as gradients of r (I mean jrj, not ~r): 1 1 1 rrr G (~r) = @ @ r δ − @ @ r = I r2r − (27) ij 8πµr k k ij 2(1 − ν) i j 8πµr 2(1 − ν) 2.1 Notes about the solution 1. The displacements go as 1=r, which means that the strain/stress go as 1=r2. There- fore the elastic energy density, which goes like "2, goes like 1=r4 and its integral diverges. This is much like the case of electrostatics, where the total energy of the electrical field of a point charge diverges. 2. The scaling u ∼ 1=r could also have been obtained from simple dimensional anal- ysis. It is quite common that dimensional considerations in elasticity take the \dimension" from the shear modulus µ, and then there's an unknown (and usually uninteresting) function of ν. 3. You might have noticed that Gij is a symmetric matrix. This might look at first glance as a trivial property that stems from the translational symmetry or rotational symmetry (=isotropy), but this is not the case. This symmetry property does not stem from any simple argument (that I can think of). Instead, this symmetry is a special case of a more general property that is called reciprocity. For a general 4 Figure 1: Top: Displacement field lines in the x − y plane for point F~ = F x^ in the horizontal direction. From left to right, with ν = 0, 0.33, 0.5.. Bottom: deformation of a regular mesh under this motion. Note that crossing of two lines of the same color is physically forbidden (why?). linear elastic solid, and by general I mean that Cijkl(r) can have any symmetry and can even depend on space, the static Green function satisfies 0 0 Gij(r; r ) = Gji(r ; r) (28) The proof of this general property is given in Sec.4 of this file.