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 $\FLPF$ 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 $\FLPf_{\text{ext}}$. The total external force $\FLPF_{\text{ext}}$ is the integral of $\FLPf_{\text{ext}}$ over the volume of the piece: \begin{equation} \label{Eq:II:39:23} \FLPF_{\text{ext}}=\int\FLPf_{\text{ext}}\,dV. \end{equation}  In equilibrium, this force would be balanced by the total force $\FLPF_{\text{int}}$ from the neighboring material which acts across the surface $A$. When the piece is not in equilibrium—if it is moving—the sum of the internal and external forces is equal to the mass times the acceleration. We would have \begin{equation} \label{Eq:II:39:24} \FLPF_{\text{ext}}+\FLPF_{\text{int}}= \int\rho\ddot{\FLPr}\,dV, \end{equation}  where $\rho$ is the density of the material, and $\ddot{\FLPr}$ is its acceleration. We can now combine Eqs. (39.23) and (39.24), writing \begin{equation} \label{Eq:II:39:25} \FLPF_{\text{int}}=\int_v(-\FLPf_{\text{ext}}+\rho\ddot{\FLPr})\,dV. \end{equation}  We will simplify our writing by defining \begin{equation} \label{Eq:II:39:26} \FLPf=-\FLPf_{\text{ext}}+\rho\ddot{\FLPr}. \end{equation}  Then Eq. (39.25) is written \begin{equation} \label{Eq:II:39:27} \FLPF_{\text{int}}=\int_v\FLPf\,dV. \end{equation}  What we have called $\FLPF_{\text{int}}$ is related to the stresses in the material. The stress tensor $S_{ij}$ was defined (Chapter 31) so that the $x$-component of the force $dF$ across a surface element $da$, whose unit normal is $\FLPn$, is given by \begin{equation} \label{Eq:II:39:28} dF_x=(S_{xx}n_x+S_{xy}n_y+S_{xz}n_z)\,da. \end{equation}  The $x$-component of $\FLPF_{\text{int}}$ on our little piece is then the integral of $dF_x$ over the surface. Substituting this into the $x$-component of Eq. (39.27), we get \begin{equation} \label{Eq:II:39:29} \int_A(S_{xx}n_x+S_{xy}n_y+S_{xz}n_z)\,da=\int_vf_x\,dV. \end{equation}  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 $\unicode{x201C}\FLPS\,\unicode{x201D}\cdot\FLPn$—that is, the normal component of a vector—over the surface. It would be the flux of $\unicode{x201C}\FLPS\,\unicode{x201D}$ out of the volume. And this could be written, using Gauss law, as the volume integral of the divergence of $\unicode{x201C}\FLPS\,\unicode{x201D}$. 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 \begin{equation} \label{Eq:II:39:30} \int_v\biggl( \ddp{S_{xx}}{x}+\ddp{S_{xy}}{y}+\ddp{S_{xz}}{z} \biggr)dV=\int_vf_x\,dV. \end{equation}  Now we can leave off the volume integrals and write the differential equation for the general component of $\FLPf$ as \begin{equation} \label{Eq:II:39:31} f_i=\sum_j\ddp{S_{ij}}{x_j}. \end{equation}  This tells us how the force per unit volume is related to the stress tensor $S_{ij}$. The theory of the motions inside a solid works this way. If we start out knowing the initial displacements—given by, say, $\FLPu$—we can work out the strains $e_{ij}$. From the strains we can get the stresses from Eq. (39.12). From the stresses we can get the force density $\FLPf$ in Eq. (39.31). Knowing $\FLPf$, we can get, from Eq. (39.26), the acceleration $\ddot{\FLPr}$ 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 $S_{ij}$, and write the $e_{ij}$ as $\tfrac{1}{2}(\ddpl{u_i}{x_j}+\ddpl{u_j}{x_i})$, you end up with the vector equation \begin{equation} \label{Eq:II:39:32} \FLPf=(\lambda+\mu)\,\FLPgrad{(\FLPdiv{\FLPu})}+\mu\,\nabla^2\FLPu. \end{equation}  You can, in fact, see that the equation relating $\FLPf$ and $\FLPu$ must have this form. The force must depend on the second derivatives of the displacements $\FLPu$. What second derivatives of $\FLPu$ are there that are vectors? One is $\FLPgrad{(\FLPdiv{\FLPu})}$; that’s a true vector. The only other one is $\nabla^2\FLPu$. So the most general form is \begin{equation*} \FLPf=a\,\FLPgrad{(\FLPdiv{\FLPu})}+b\,\nabla^2\FLPu, \end{equation*}  which is just (39.32) with a different definition of the constants. You may be wondering why we don’t have a third term using $\FLPcurl{\FLPcurl{\FLPu}}$, which is also a vector. But remember that $\FLPcurl{\FLPcurl{\FLPu}}$ is the same thing as $\FLPgrad{(\FLPdiv{\FLPu})}-\nabla^2\FLPu$, 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 $\rho\,\partial^2\FLPu/\partial t^2$—neglecting for now any body forces like gravity—and get \begin{equation} \label{Eq:II:39:33} \rho\,\frac{\partial^2\FLPu}{\partial t^2}= (\lambda+\mu)\,\FLPgrad{(\FLPdiv{\FLPu})}+\mu\,\nabla^2\FLPu. \end{equation}  It looks something like the wave equation we had in electromagnetism, except that there is an additional complicating term. For materials whose elastic properties 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 other words, we can put \begin{equation} \label{Eq:II:39:34} \FLPu=\FLPu_1+\FLPu_2, \end{equation}  where \begin{equation} \label{Eq:II:39:35} \FLPdiv{\FLPu_1}=0,\quad \FLPcurl{\FLPu_2}=\FLPzero. \end{equation}  Substituting $\FLPu_1+\FLPu_2$ for $\FLPu$ in (39.33), we get  \begin{equation} \label{Eq:II:39:36} \rho\,\partial^2/\partial t^2[\FLPu_1+\FLPu_2]= (\lambda+\mu)\,\FLPgrad{(\FLPdiv{\FLPu_2})}+ \mu\,\nabla^2(\FLPu_1+\FLPu_2). \end{equation} 
 \begin{align} \rho\,\partial^2/&\partial t^2[\FLPu_1+\FLPu_2]=\notag\\[1ex] \label{Eq:II:39:36} &(\lambda+\mu)\,\FLPgrad{(\FLPdiv{\FLPu_2})}+ \mu\,\nabla^2(\FLPu_1+\FLPu_2). \end{align}    We can eliminate $\FLPu_1$ by taking the divergence of this equation,  \begin{equation*} \rho\,\partial^2/\partial t^2(\FLPdiv{\FLPu_2})= (\lambda+\mu)\,\nabla^2(\FLPdiv{\FLPu_2})+ \mu\,\FLPdiv{\nabla^2(\FLPu_2)}. \end{equation*} 
  \begin{align*} \rho\,\partial^2/&\partial t^2(\FLPdiv{\FLPu_2})=\notag\\[1ex] &(\lambda+\mu)\,\nabla^2(\FLPdiv{\FLPu_2})+ \mu\,\FLPdiv{\nabla^2(\FLPu_2)}. \end{align*}   Since the operators ($\nabla^2$) and ($\FLPdiv{}$) can be interchanged, we can factor out the divergence to get \begin{equation} \label{Eq:II:39:37} \FLPdiv{\{\rho\,\partial^2\FLPu_2/\partial t^2- (\lambda+2\mu)\,\nabla^2\FLPu_2\}}=0. \end{equation}  Since $\FLPcurl{\FLPu_2}$ is zero by definition, the curl of the bracket $\{\}$ is also zero; so the bracket itself is identically zero, and \begin{equation} \label{Eq:II:39:38} \rho\,\partial^2\FLPu_2/\partial t^2= (\lambda+2\mu)\,\nabla^2\FLPu_2. \end{equation}  This is the vector wave equation for waves which move at the speed $C_2=\sqrt{(\lambda+2\mu)/\rho}$. Since the curl of $\FLPu_2$ is zero, there is no shearing associated with this wave; this wave is just the compressional—sound-type—wave we discussed in the last chapter, and the velocity is just what we found for $C_{\text{long}}$. In a similar way—by taking the curl of Eq. (39.36)—we can show that $\FLPu_1$ satisfies the equation \begin{equation} \label{Eq:II:39:39} \rho\,\partial^2\FLPu_1/\partial t^2=\mu\,\nabla^2\FLPu_1. \end{equation}  This is again a vector wave equation for waves with the speed $C_1=\sqrt{\mu/\rho}$. Since $\FLPdiv{\FLPu_1}$ is zero, $\FLPu_1$ produces no changes in density; the vector $\FLPu_1$ corresponds to the transverse, or shear-type, wave we saw in the last chapter, and $C_1=C_{\text{shear}}$. If we wished to know the static stresses in an isotropic material, we could, in principle, find them by solving Eq. (39.32) with $\FLPf$ equal to zero—or equal to the static body forces from gravity such as $\rho\FLPg$—under certain conditions which are related to the forces acting on the surfaces of our large block of material. This is somewhat more difficult to do than the corresponding problems in electromagnetism. 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 geometric shapes such as cylinders, spheres, and so on, since these are convenient shapes for electrical devices. In elasticity, the objects we would like to analyze may have quite complicated shapes—like a crane hook, or an automobile crankshaft, or the rotor of a gas turbine. Such problems can sometimes be worked out approximately by numerical methods, using the minimum energy principle we mentioned earlier. Another way is to use a model of the object and measure the internal strains experimentally, using polarized light.  It works this way: When a transparent isotropic material—for example, a clear plastic like lucite—is put under stress, it becomes birefringent. If you put polarized light through it, the plane of polarization will be rotated by an amount related to the stress: by measuring the rotation, you can measure the stress. Figure 39–6 shows how such a setup might look. Figure 39–7 is a photograph of a photoelastic model of a complicated shape under stress. 