Feynman Physics Lectures V2 Ch39 1963-04-01 Elastic Materials

Feynman Physics Lectures V2 Ch39 1963-04-01 Elastic Materials

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Unformatted text preview: 39 Elastic Materials 39—1 The tensor of strain In the last chapter we talked about the distortions of particular elastic objects. In this chapter we want to look at what can happen in general inside an elastic material. We would like to be able to describe the conditions of stress and strain inside some big glob of Jello which is twisted and squashed in some complicated way. To do this, we need to be able to describe the local strain at every pomt in an elastic body; we can do it by giving a set of six numbers—which are the components of a symmetric tensor—for each point. Earlier, we spoke of the stress tensor (Chapter 31); now we need the tensor of strain. Imagine that we start with the material initially unstrained and watch the motion of a small speck of “dirt” embedded in the material when the strain is applied. A speck that was at the point P located at r = (x, y, 2) moves to a new position P’ at r’ = (x’, y?. 2’) as shown in Fig. 39‘1. We will call u the vector displacements from P to I Then u = r’ — r. (39.1) The displacement u depends, of course, on which point P we start with, so u is a vector function of r—or, if you prefer, of (x, y, 2). Let’s look first at a simple situation in which the strain IS constant over the material—so we have what is called a homogeneous strain. Suppose, for instance, that we have a block of material and we stretch it uniformly. We just change its dimensions uniformly in one direction—say, in the x-direction, as shown in Fig. 39—2. The motion ux of a speck at x is proportional to x. In fact, ux _ Al 7 T We will write u, this way: ux = eux. BEFORE 39—1 The tensor of strain 39—2 The tensor of elasticity 39-3 The motions in an elastic body 39—4 Nonelastic behavior 39—5 Calculating the elastic constants Reference: C. Kittel, Introduction to Solid State Physics, John Wiley and Sons, Inc., New York, 2nd ed., 1956. BEFORE Fig. 39—1 . A speck of the material at the point P in an unstrained block Fig. 39—2. A homogeneous stretch-type strain. moves to P' where the block is strained. 39—1 The proportionality constant em is, of course, the same thing as Al/l. (You will see shortly why we use a double subscript.) If the strain is not uniform, the relation between um and x will vary from place to place in the material. For the general situation, we define the cm by a kind of local Al/l, namely by em = aux/8x. (39.2) This number—which is now a function of x, y, and z—describes the amount of stretching in the x-direction throughout the hunk of jello. There may, of course, also be stretching in the y- and z—directions. We describe them by the numbers 6u 6L1 6”” = 5711’ e“ = 622' (39.3) We need to be able to describe also the shear-type strains. Suppose we imagine a little cube marked out in the initially undisturbed JCllO. When the jello is pushed out of shape, this cube may get changed into a parallelogram, as sketched in Fig. 39—3.* In this kind of a strain, the x-motion of each particle is proportional to its y-coordinate, um = y. (39.4) qu> And there is also a y-motion proportional to x, x. (39.5) NIQ: My: So we can describe such a shear-type strain by writing uz = exyy, uy = eyxx with 6‘“, "—' 6111: NW: Now you might think that when the strains are not homogeneous we could describe the generalized shear strains by defining the quantities e“, and egg, by Fig. 39—3. A homogeneous shear strain. 81y — a e“ I ——- (39.6) But there is one difliculty. Suppose that the displacements um and L4,, were given by 0 0 ut=§y, lly:_‘i‘ * We choose for the moment to split the total shear angle 6 into two equal parts and make the strain symmetric With respect to x and y. 39—2 BEFORE L Fig. 39—4. A homogeneous rotation—there is no strain. They are like Eqs. (39.4) and (39.5) except that the sign of uy is reversed. With these displacements a little cube in the jello simply gets shifted by the angle 0/2, as shown in Fig. 39—4. There is no strain at alljust a rotation in space. There is no distortion of the material; the relative positions of all the atoms are not changed at all. We must somehow make our definitions so that pure rotations are not included in our definitions of a shear strain. The key point is that if auy/ax and aux/6y are equal and opposite, there is no strain; so we can fix things up by defining A f ”‘7' em, = em = flaw/ax + aux/6y). For a pure rotation they are both zero, but for a pure shear we get that e“, is equal to em, as we would like. In the most general distortion—which may include stretching or compression as well as shear—we define the state of strain by giving the nine numbers aux err = (9X ’ em, = (:1: (39.7) 93y = flatly/8x + aux/0y), These are the terms of a tensor of strain. Because it is a symmetric tensor—our definitions make em: = eyx. always—there are really only six different numbers. You remember (see Chapter 31) that the general characteristic of a tensor is that the terms transform like the products of the components of two vectors. (If A and B are vectors, C” = A,B, is a tensor.) Each term of e“ is a product (or the sum of such products) of the components of the vector u = (um, uy, 14,), and of the operator V = (El/ax, 6/6y, 6/62), which we know transforms like a vector. Let’s let x1, x2, and x3 stand for x, y, and z and ul, u2, and 113 stand for uz, uy, and uz; then we can write the general term ei, of the strain tensor as e” = gait/ax. + aut/axJ), (393) where i andj can be 1, 2, or 3. When we have a homogeneous strain—which may include both stretching and shear—all of the 9w are constants, and we can write ux = emx + ezyy + e122. (39.9) (We choose our origin of x, y, 2 at the point where u is zero.) In this case, the strain tensor e” gives the relationship between two vectors: the coordinate vector r = (x, y, z) and the displacement vector u = (uz, uy, uz). 39—3 When the strains are not homogeneous, any piece of the jello may also get somewhat twisted——there will be a local rotation. If the distortions are all small, we would have Au. = Z (em — w,,)Ax,, (39.10) J where a)” is an antisymmetric tensor, 0),] = %(6u,/6x, — 6u1/6xj), (39.11) which describes the rotation. We will, however, not worry any more about rota- tions, but only about the strains described by the symmetric tensor e”. 39-2 The tensor of elasticity Now that we have described the strains, we want to relate them to the internal forces—the stresses in the material. For each small piece of the material, we assume Hooke’s law holds and write that the stresses are proportional to the strains. In Chapter 31 we defined the stress tensor S” as the ith component of the force across a umt area perpendicular to the j—axis. Hooke’s law says that each component of S” is linearly related to each of the components of strain. Since S and e each have nine components, there are 9 X 9 = 81 possible coefficients which describe the elastic properties of the material. They are constants if the material itself is homogeneous. We wr1te these coefficients as C2,“ and define them by the equation S1] = Z Cuklekla (39.12) k,l where i, j, k, I all take on the values 1, 2, or 3. Since the coefficients CU“ relate one tensor to another, they also form a tensor—a tensor of the fourth rank. We can call it the tensor of elasticity. Suppose that all the C’s are known and that you put a complicated force on an object of some peculiar shape. There will be all kinds of distortion, and the thing will settle down with some twisted shape. What are the displacements? You can see that it is a complicated problem. If you knew the strains, you could find the stresses from Eq. (39.12)—or vice versa. But the stresses and strains you end up with at any point depend on what happens in all the rest of the material. The easiest way to get at the problem is by thinking of the energy. When there is a force F proportional to a displacement x, say F = kx, the work requ1red for any displacement x is kx2/2. In a similar way, the work w that goes into each unit volume of a distorted mater1al turns out to be w = % Z Cultleuekl- (39.13) zjkl The total work W done in distorting the body is the integral of w over its volume: W = / a Z kalewekl dVol. (39.14) 11101 This is then the potential energy stored in the internal stresses of the material. Now when a body is in equilibrium, this internal energy must be at a minimum. So the problem of finding the strains in a body can be solved by finding the set of displacements u throughout the body which will make W a minimum. In Chapter 19 we gave some of the general ideas of the calculus of variations that are used in tackling minimization problems l1ke this. We cannot go into the problem in any more detail here. What we are mainly interested in now is what we can say about the general properties of the tensor of elasticity. First, it is clear that there are not really 81 different terms in Cm. Since both S” and e” are symmetric tensors, each with only six dlfferent terms, there can be at most 36 different terms in Cu“. There are, however, usually many fewer than this. 39—4 1 Let’s look at the special case of a cubic crystal. In it, the energy density w starts out like this: _ 1 2 W '— §{CIII$eZI + CZnyeIZefiby + Crxxzexxexz + szyxexzexy + Cxxyyegjxeyy . . . etc . . . + nywefm +...etc...etc...}, (39.15) with 81 terms in all! Now a cubic crystal has certain symmetries. In particular, if the crystal is rotated 90°, it has the same physical properties. It has the same stiffness for stretching in the y-direction as for stretching in the x-direction. There- fore, if we change our definition of the coordinate directions x and y in Eq. (39.15), the energy wouldn’t change. It must be that for a cubic crystal Cum = nyyy = szzz- (39.16) Next we can show that the terms like Cmy must be zero. A cubic crystal has the property that it is symmetric under a reflection about any plane perpendicular to one of the axes. If we replace y by — y, nothing is different. But changing y to — y changes em, to — erg—a displacement which was toward + y is now toward — y. If the energy is not to change, Cmy must go into — Cami, when we make a reflec- tion. But a reflected crystal is the same as before, so Cum, must be the same as — Cxxxy- This can happen only if both are zero. You say, “But the same argument will make Cm”, = 0!” No, because there are four y’s. The sign changes once for each y, and four minuses make a plus. If there are two or four y’s, the term does not have to be zero. It is zero only when there is one, or three. So, for a cubic crystal, any nonzero term of C will have only an even number of identical subscripts. (The arguments we have made for y ob- viously hold also for x and z.) We might then have terms like Cm”, Cam, CHM, and so on. We have already shown, however, that if we change all x’s to y’s and vice versa (or all 2’s and x’s, and so on) we must get—for a cubic crystal—the same number. This means that there are only three diflerent nonzero possibilities: szxz (= nyyy = C2222), Cur/y (2 Cut/m Cuzz, etc-L (39.17) Czyzy (z Cmcyx szxz: etc-)- H For a cubic crystal, then, the energy density will look like this: W : %{Czrxx(egx + 912,1, + 83:) +2Cmyy(euem, + ewe“ + ezzem) (39.18) +4Cryxy(ea20y + 932 + 931)}. For an isotropic—that is, noncrystalline—material, the symmetry is still higher. The C’s must be the same for any choice of the coordinate system. Then it turns out that there is another relation among the C’s, namely, that Cream: = Cxxyy + Czyxy- (39.19) We can see that this is so by the following general argument. The stress tensor S”- has to be related to e” in a way that doesn’t depend at all on the coordinate directions—it must be related only by scalar quantities. “That’s easy,” you say. “The only way to obtain 5,, from 6,, is by multiplication by a scalar constant. It’s just Hooke’s law. It must be that S“ = (const)e,,.” But that’s not quite right; there could also be the unzt tensor 5,] multiplied by some scalar, linearly related to e”. The only invariant you can make that 13 linear in the e’s IS Zen (It transforms like x2 + y2 + Z2, which is a scalar.) So the most general form for the equation relating S” to eu—for isotropic materials—is 5,, = 2M6” + x (Z em.) 5,]. (39.20) [c (The first constant is usually written as two times ,u; then the coefficient M is equal 39—5 Fig. 39—5. A small volume element V bounded by the surface A. to the shear modulus we defined in the last chapter.) The constants u and )\ are called the Lamé elastic constants. Comparing Eq. (39.20) with Eq. (39.12), you see that szyy = h, )7 Cum, 2,u, ”.5 (39.21) Cxxxx : 2M + )\. So we have proved that Eq. (39.19) is indeed true. You also see that the elastic properties of an isotropic material are completely given by two constants, as we said in the last chapter. The C’s can be put in terms of any two of the elastic constants we have used earlier—for instance, in terms of Young’s modulus Y and Poisson’s ratio a. We Will leave it for you to show that Y a CxLL£—]+O’(1+l_20>a Y a Cm... — 1 + a (I _ 20> , (39.22) Y CW = (175' 39—3 The motions in an elastic body We have pointed out that for an elastic body in equilibrium the internal stresses adjuSt themselves to make the energy a minimum. Now we take a look at What happens when the internal forces are not in equilibrium. Let’s say we have a small piece of the material inside some surface A. See Fig. 39—5. If the piece is in equilibrium, the total force F acting on it must be zero. We can think of this force as being made up of two parts. There could be one part due to “external” forces like graVIty, which act from a distance on the matter in the piece to produce a force per unit volume fem. The total external force Fe,“ is the integral of fm over the volume of the piece: FCXI: = [1%.th (39.23) In equilibrium, this force would be balanced by the total force Fm from the neigh- boring material which acts across the surface A. When the piece is not in equili- brium—if it is moving—the sum of the internal and external forces is equal to the mass times the acceleration. We would have Foxt ’1‘ Fint : f P” dV: (3924) where p is the density of the material, and r IS its acceleration. We can now com- bine Eqs. (39.23) and (39.24), writing F1111; = / (_fext + pr) (IV (3925) We will simplify our writing by defining f: _fext. + Pk (39.26) Then Eq. (39.25) is written Fm. = fde. (39.27) What we have called Fm, is related to the stresses in the material. The stress tensor 5., was defined (Chapter 31) so that the x-component of the force dF across a surface element da, whose unit normal is n, is given by dF. = (sun. + szynz, + sznz)da. (39.28) 39—6 The x-component of Fm on our little piece is then the integral of sz over the surface. Substituting this into the x-component of Eq. (39.27), we get / ($1:an + Sari/n11 + 5.22712) dd 2 / f1; dV. (39.29) A - ‘1) We have a surface integral related to a volume integral—and that reminds us of something we learned in electricity. Note that if you ignore the first subscript x on each of the S’s in the left—hand side of Eq. (39.29), it looks Just like the integral of a quantity “S” - n—that is, the normal component of a vector—over the surface. It would be the flux of “S” out of the volume. And this could be written, using Gauss law, as the volume integral of the divergence of “S”. It is, in fact, true whether the x-subscript is there or not—it is just a mathematical theorem you get by integrating by parts. In other words, we can change Eq. (39.29) into ' as” 63,, as... _ / /.<ax + 6y + az>dV 0)"de (39.30) Now we can leave off the volume integrals and write the differential equation for the general component of f as f. = Z 65% (39.31) This tells us how the force per unit volume is related to the stress tensor S”. The theory of the motions inside a solid works this way. If we start out know— ing the initial displacements—given by, say, u—we can work out the strains 6”. From the strains we can get the stresses from Eq. (39.12). From the stresses we can get the force densityfin Eq. (39.31). Knowingf, we can get, from Eq. (39.26), the acceleration r of the material, which tells us how the displacements Will be changing. Putting everything together, we get the horrible equation of motion for an elastic solid. We Will just write down the results that come out for an isotropic material. If you use (39.20) for S”, and write the e” as fine/ax] + BuJ/ax“ you end up with the vector equation f = (x + 9) WV ' u) + W214. (39.32) You can, in fact, see that the equation relating f and u must have this form. The force must depend on the second derivatives of the displacements u What second derivatives of u are there that are vectors? One is V(V - u); that’s a true vector. The only other one is V211. So the most general form is I f: aV(V-u)+bV2u, which is just (39.32) with a different definition of the constants. You may be wondering why we don’t have a third term usmg V X V X u, which is also a vector. But remember that V X V X u is the same thing as V2u — V(V ‘ u), so it is a linear combination of the two terms we have. Adding it would add nothing new. We have proved once more that isotropic material has only two elastic constants. For the equation of motion of the material, we can set (39.32) equal to p 6211/ 6t2—neglecting for now any body forces like graVitye—and get 2 p g}; = (>. + ,u) v(v ~ u) + ”V2... (39.33) It looks something like the wave equation we had in electromagnetism, except that there is an additional complicating term. For materials whose elastic proper- ties are everywhere the same we can see what the general solutions look like in the following way. You will remember that any vector field can be written as the sum of two vectors: one whose divergence is zero, and the other whose curl is zero. In 39—7 POLAROIDS UNDER STRESS Fig. 39—6. Measuring internal stresses with polarized light. Fig. 39—7. A stressed plastic model as seen between crossed polaroids. [From F. W. Sears, Optics, Addison- Wesley Publishing Co., Reading, Moss., 1949.] other words, we can put u = u] + “2, (39.34) where V 'u1 = 0, V X 1‘2 = 0. (39.35) Substituting ul + 112 for u in (39.33), we get P 62/6t2lul + “2] = 0‘ + I!) V(V ‘ "2) + MV2(1‘1 + u2). (3936) We can eliminate u] by taking the divergence of this equation, P 32/612(V ' "2) = 0‘ + M)V2(V “142) + MV ' V2112. Since the operators (V2) and (V-) can be interchanged, we can factor out the di- vergence to get v - {p 62u2/6t2 — (A + 2M)V2u2} = 0. (39.37) Since V X u; is zero by definition, the curl of the bracket {} is also zero; so the bracket itself IS identically zero, and p62u2/612 = (x + 2M)V2u2. 3 (39.38) This is the vector wave equation for waves which move at the speed C2 = \WOT—F 2p.)/p. Since the curl of u2 lS zero, there is no shearing associated with this wave; this wave 18 just the compressional—sound-type———wave we discussed in the last chapter, and the velocity is just what we found for Clong. In a similar way—by taking the curl of Eq. (39.36)—we can show that ul satisfies the equation p62u1/612 = szul. (39.39) This is again a vector wave equation for waves With the speed C2 = u/p. Since V - ul is zero, ul produces no changes in density; the vector ul corresponds to the transverse, or shear-type, wave we saw in the last chapter, and C 2 = Cm”. If we wished to know the static stresses in an isotropic material, we could, in princrple, find them by solvmg Eq. (39.32) with fequal to zero—0r equal to the static body forces from gravity such as pg—under certain conditions which are related to the forces acting on the surfaces of our large block of material. This is somewhat more dilficult to do than the corresponding problems in electromagne- tism. It is more difficult, first, because the equations are a little more difficult to handle, and second, because the shape of the elastic bodies we are likely to be interested in are usually much more complicated. In electromagnetism, we are often interested in solving Maxwell’s equations around relatively simple ...
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