Lecture 11 Notes: Elasticity

18.354J Nonlinear Dynamics II: Continuum Systems 11 Lecture 11 Elasticity What shape does a piece of paper take when we push it in at the ends? To answer this question let’s acquaint ourselves with another continuum approximation, used to describe the deformation of elastic solids (we might actually have studied this before our work on fluids, as it is conceptually simpler). We first need to find a way to describe stress and strain within the solid, and then determine the relation between the two. Then we can derive the equations of elasticity and apply them to the buckling of a thin plate. 11.1 Strain If a solid is deformed, then points within the solid will move. We fix our attention on a single point, whose coordinates are (x1 , x2 , x3 ), and the close neighborhood of this point. We suppose that in the strained state the Cartesian coordinates of the same point have become (x01 , x02 , x03 ). The displacement of this point due to the deformation is denoted by u = u(x1 , x2 , x3 ), where ui = x0i − xi . (259) The vector u is called the displacement vector. When a body is deformed the distance between its points change. Let’s consider two points very close together. If the vector joining them before is dxi , the vector joining them in 0 the i . This distance between the points was originally dl = p deformed body is dxi = dxi + dup 0 2 02 2 02 2 dx1 + dx2 + dx3 and is now dl = dx02 1 + dx2 + dx3 . Using the summation convention, which tells us to sum over repeated indices (i.e., ai bi = a1 b1 + a2 b2 + a3 b3 ), and substituting in that dui = (∂ui /∂xk )dxk , we get dl02 = dl2 + 2 ∂ui ∂ui ∂ui dxk dxl . dxi dxk + ∂xk ∂xk ∂xl (260) We shall neglect the last of these terms, as we consider the ui to be small, so that dl02 = dl2 + 2eij dxi dxj (261) where eij = 1 2  ∂uj ∂ui + ∂xi ∂xj  (262) are the components of the strain tensor e. This is called linear elasticity (even though it is not really linear). It is often very useful to separate pure shear from pure compression effects, which can be achieved by rewriting12     δij δij δij δij eij = eij − ell + ell = eij − ∇·u + ∇ · u. (263) 3 3 3 3 The first part in parentheses has a vanishing trace and therefore represents pure shear. 11.2 Stress tensor When a body is deformed, the arrangement of molecules within is changed, and forces arise that want to restore the body to its equilibrium configuration. These are called internal stresses, represented by a stress tensor σ = (σij ), and when there is no deformation the stress is zero σik = 0. (264) The three components of a force on a volume element V can be obtained from stresses by transforming a surface integral into a volume integral Z Z Z σik dAk = (∂k σik )dV = fi dV (265) ∂V V V Hence, the vector fi must be given by the divergence of the stress tensor σik . fi = ∂σik . ∂xk (266) We recognize that σik dAk is the force per unit area in the i-direction on the surface element with outward normal dA. One thing we know about the stress tensor is that it is symmetric 12 In d dimensions one would simply replace δij /3 by δij /d everywhere. (σij = σji ). If, for example, the body is in a gravitational field then the internal stresses must everywhere balance gravity, in which case the equilibrium equations are ∂σik + ρgi = 0. ∂xk (267) Additional external forces applied to the surface of the body will enters as boundary conditions that complement the equilibrium conditions (267). For instance, if there is an external force per unit area, fˆ, acting over the surface, then we require σik nk = fˆi , (268) where n is the outward unit normal on the surface. 11.3 Hooke’s law In general, we would like to use Eqs. (267) to predict the deformation of a solid body under a given force distribution. That is, we have to express the stress tensor σij in terms of the displacement field u. The main body of the mathematical theory of elasticity rests on the assumption of a linear homogeneous relation between the elements of the stress tensor and the strain tensor. This is just the continuum version of Hooke’s Law. To simplify matters, let’s focus on materials that are isotropic (i.e., the elastic properties are independent of direction). In this case σij = λδij (e11 + e22 + e33 ) + 2µeij = λδij Tre + 2µeij (269) where δij is the Kronecker delta and λ and µ are positive elastic constants of the material, called Lame coefficients. The corresponding (free) energy density E of the body associated with deformation, obtained from the relation σij = ∂E , ∂eij (270) is therefore 1 (271) E = λe2ii + µe2ij . 2 As stated above, the sum eii = Tre is related to the change in volume associated with a deformation. If this is zero, only the shape of the body is altered, corresponding to pure shear. Recalling our above decomposition 1 1 eij = (eij − δij ell ) + δij ell . (272) 3 3 we can obtain a general expression for the energy density of a deformed isotropic body, by replacing (271) with 1 1 E = Ke2ll + µ(eik − δik ell )2 (273) 2 3 where K and µ positive constants, respectively called the modulus of compression and the modulus of rigidity. In 3D, K is related to the Lame coefficients by13 2 K =λ+ µ 3 13 In 2D, this relation becomes K = λ + µ. (274) 11.4 A simple problem Consider the simple case of a beam. Let the beam be along the z-axis, and let us pull it a both ends to stretch it. The force per unit area p is uniform over each end. The resulting deformation is uniform throughout the body and, hence, so is the stress tensor. It therefore follows that all components σik are zero except for σzz , and from the forcing condition at the end we have that σzz = p. From the general expression that relates the components of the stress and strain tensors, we see that all components eik with i 6= k are zero. The equilibrium equations are therefore   1 1 1 p (275) exx = eyy = − − 3 2µ 3K and ezz  1 = 3 1 1 + µ 3K  p. (276) The component ezz gives the lengthening of the rod, and the coefficient of p is the coefficient of extension. Its reciprocal is Young’s modulus Y = 9Kµ . 3K + µ (277) The components exx and eyy give the relative compression of the rod in the transverse direction. The ratio of the transverse compression to the longitudinal extension is called Poisson’s ratio, ν: exx = −νezz , (278) where 1 ν= 2  3K − 2µ 3K + µ  . (279) Since K and µ are always positive, Poisson’s ratio can vary between -1 and 21 . Note that a negative value corresponds to pulling on the beam and it getting thicker! Now we see why we use Y and ν; they are easier to measure. Inverting these formulae, we get µ= Y , 2(1 + ν) K= Y . 3(1 − 2ν) (280) The free energy then becomes Y E= 2(1 + ν)  e2ik +  ν 2 e . 1 − 2ν ll The stress tensor is given in terms of the strain tensor by   Y ν σik = eik + ell δik . 1+ν 1 − 2ν (281) (282) Conversely eik = 1 [(1 + ν)σik − νσll δik ]. Y (283) 11.5 Bending of a thin beam Now we are in a position to try and calculate the shape of a bent beam. Our analysis requires that the thickness be much smaller than the lateral dimension. The deformations must also be small, such that the displacements are small compared with the thickness. Although the general equilibrium equations are greatly simplified when considering thin plates, it is more convenient not to derive our result from these. Rather, we shall use our knowledge of variational calculus to calculate afresh the energy of a bent plate, and set about varying that energy. When a plate is bent, it is stretched at some points and compressed at others: on the convex side there is evidently an extension and on the concave side there is compression. Somewhere in the middle there is a neutral surface, on which there is no extension or compression. The neutral surface lies midway through the plate. We take a coordinate system with the origin on the neutral surface and the z-axis normal to the surface. The xy-plane is that of the undeformed surface. The displacement of the neutral surface is given by uz = w(x, y). For further calculations we note that since the plate is thin, comparatively small forces on the surface are needed to bend it. These forces are always considerably less than the internal stresses caused in the deformed beam by the extension and compression of its parts. Thus we have on both surfaces of the plate σxz = σyz = σzz = 0. (284) Since the plate is small, these quantities must be small within the plate if they are zero on the surface. We therefore conclude that σxz = σyz = σzz are small everywhere, and equate them to zero. From our general formulae relating stress and strain, we have σzx = σzz = Y ezx , 1+ν σzy = Y ezy , 1+ν (285) Y [(1 − ν)ezz + ν(exx + eyy )]. (1 + ν)(1 − 2ν) (286) Substituting in our expression for the the strain tensor and equating to zero, we get ∂uy ∂uz =− , ∂z ∂y ∂ux ∂uz =− , ∂x ∂z ezz = − (287) ν(exx + eyy ) . (1 − ν) (288) In the first two of these equations uz can be replaced by w(x, y). Thus, integrating the above relations gives ∂w ∂w ux = −z , uy = −z , (289) ∂x ∂y where the constants of integration were chosen so as to make ux = uy = 0 for z = 0. Knowing ux and uy we can now determine all the components of the strain tensor: exx = −z ∂2w , ∂x2 eyy = −z ∂2w ∂y 2 exy = −z ∂2w , ∂x∂y (290) exz = eyz = 0, ezz zν = 1−ν  ∂2w ∂2w + ∂x2 ∂y 2  . We now calculate the free energy of the plate, using our general formula, " #  2 2  2 2 ∂ w ∂2w 1 ∂ w ∂2w ∂2w 2 Y E=z + + − . ∂y 2 ∂x∂y ∂x2 ∂y 2 1 + ν 2(1 − ν) ∂x2 (291) (292) Integrating from − h2 to h2 , where h is the thickness of the plate, then integrating again over an area element gives the free energy per unit area " #) 2 2 Z Z ( 2 ∂ w ∂2w ∂2w ∂2w ∂2w Y h3 + + 2(1 − ν) − dxdy. EA = 24(1 − ν 2 ) ∂x2 ∂y 2 ∂x∂y ∂x2 ∂y 2 (293) where the element of area can be written dxdy since the deformation is small. We now derive the equilibrium equation for a plate from the condition that it’s free energy is a minimum. To simplify things, let’s just ignore any y-dependence and consider a 2D problem. Using the calculus of variations we have that the energy of the distorted beam is δE Y h3 d4 w . (294) = δw 12(1 − ν 2 ) dx4 This expression must equal the force f (x) applied to deform the plate: Y h3 d4 w = f (x). 12(1 − ν 2 ) dx4 (295) The simplest boundary conditions are if the edges are clamped, in which case w = 0, dw =0 dx (296) at the edges. The first of these expresses the fact that the edge of the plate undergoes no deformation, and the second that it remains horizontal. For more details, see chapter 2 in Theory of Elasticity, Landau & Lifschitz.