The tensor of stress
The symmetric tensors we have described so far arose as coefficients in relating one vector to another. We would like to look now at a tensor which has a different physical significance—the tensor of stress. Suppose we have a solid object with various forces on it. We say that there are various “stresses” inside, by which we mean that there are internal forces between neighboring parts of the material. We have talked a little about such stresses in a two-dimensional case when we considered the surface tension in a stretched diaphragm in Section 12–3. We will now see that the internal forces in the material of a three-dimensional body can be described in terms of a tensor. Consider a body of some elastic material—say a block of jello. If we make a cut through the block, the material on each side of the cut will, in general, get displaced by the internal forces. Before the cut was made, there must have been forces between the two parts of the block that kept the material in place; we can define the stresses in terms of these forces. Suppose we look at an imaginary plane perpendicular to the $x$-axis—like the plane $\sigma$ in Fig. 31–5—and ask about the force across a small area $\Delta y\,\Delta z$ in this plane. The material on the left of the area exerts the force $\Delta\FLPF_1$ on the material to the right, as shown in part (b) of the figure. There is, of course, the opposite reaction force $-\Delta\FLPF_1$ exerted on the material to the left of the surface. If the area is small enough, we expect that $\Delta\FLPF_1$ is proportional to the area $\Delta y\,\Delta z$. You are already familiar with one kind of stress—the pressure in a static liquid. There the force is equal to the pressure times the area and is at right angles to the surface element. For solids—also for viscous liquids in motion—the force need not be normal to the surface; there are shear forces in addition to pressures (positive or negative). (By a “shear” force we mean the tangential components of the force across a surface.) All three components of the force must be taken into account. Notice also that if we make our cut on a plane with some other orientation, the forces will be different. A complete description of the internal stress requires a tensor. We define the stress tensor in the following way: First, we imagine a cut perpendicular to the $x$-axis and resolve the force $\Delta\FLPF_1$ across the cut into its components $\Delta F_{x1}$, $\Delta F_{y1}$, $\Delta F_{z1}$, as in Fig. 31–6. The ratio of these forces to the area $\Delta y\,\Delta z$, we call $S_{xx}$, $S_{yx}$, and $S_{zx}$. For example, \begin{equation*} S_{yx}=\frac{\Delta F_{y1}}{\Delta y\,\Delta z}. \end{equation*}  The first index $y$ refers to the direction force component; the second index $x$ is normal to the area. If you wish, you can write the area $\Delta y\,\Delta z$ as $\Delta a_x$, meaning an element of area perpendicular to $x$. Then \begin{equation*} S_{yx}=\frac{\Delta F_{y1}}{\Delta a_x}. \end{equation*}  Next, we think of an imaginary cut perpendicular to the $y$-axis. Across a small area $\Delta x\,\Delta z$ there will be a force $\Delta\FLPF_2$. Again we resolve this force into three components, as shown in Fig. 31–7, and define the three components of the stress, $S_{xy}$, $S_{yy}$, $S_{zy}$, as the force per unit area in the three directions. Finally, we make an imaginary cut perpendicular to $z$ and define the three components $S_{xz}$, $S_{yz}$, and $S_{zz}$. So we have the nine numbers \begin{equation} \label{Eq:II:31:23} S_{ij}=\begin{bmatrix} S_{xx} & S_{xy} & S_{xz}\\ S_{yx} & S_{yy} & S_{yz}\\ S_{zx} & S_{zy} & S_{zz}\\ \end{bmatrix}. \end{equation}  We want to show now that these nine numbers are sufficient to describe completely the internal state of stress, and that $S_{ij}$ is indeed a tensor. Suppose we want to know the force across a surface oriented at some arbitrary angle. Can we find it from $S_{ij}$? Yes, in the following way: We imagine a little solid figure which has one face $N$ in the new surface, and the other faces parallel to the coordinate axes. If the face $N$ happened to be parallel to the $z$-axis, we would have the triangular piece shown in Fig. 31–8. (This is a somewhat special case, but will illustrate well enough the general method.) Now the stress forces on the little solid triangle in Fig. 31–8 are in equilibrium (at least in the limit of infinitesimal dimensions), so the total force on it must be zero. We know the forces on the faces parallel to the coordinate axes directly from $S_{ij}$. Their vector sum must equal the force on the face $N$, so we can express this force in terms of $S_{ij}$. Our assumption that the surface forces on the small triangular volume are in equilibrium neglects any other body forces that might be present, such as gravity or pseudo forces if our coordinate system is not an inertial frame. Notice, however, that such body forces will be proportional to the volume of the little triangle and, therefore, to $\Delta x\,\Delta y\,\Delta z$, whereas all the surface forces are proportional to the areas such as $\Delta x\,\Delta y$, $\Delta y\,\Delta z$, etc. So if we take the scale of the little wedge small enough, the body forces can always be neglected in comparison with the surface forces. Let’s now add up the forces on the little wedge. We take first the $x$-component, which is the sum of five parts—one from each face. However, if $\Delta z$ is small enough, the forces on the triangular faces (perpendicular to the $z$-axis) will be equal and opposite, so we can forget them. The $x$-component of the force on the bottom rectangle is \begin{equation*} \Delta F_{x2}=S_{xy}\,\Delta x\,\Delta z. \end{equation*}  The $x$-component of the force on the vertical rectangle is \begin{equation*} \Delta F_{x1}=S_{xx}\,\Delta y\,\Delta z. \end{equation*}  These two must be equal to the $x$-component of the force outward across the face $N$. Let’s call $\FLPn$ the unit vector normal to the face $N$, and the force on it $\FLPF_n$; then we have \begin{equation*} \Delta F_{xn}=S_{xx}\,\Delta y\,\Delta z+S_{xy}\,\Delta x\,\Delta z. \end{equation*}  The $x$-component $S_{xn}$ of the stress across this plane is equal to $\Delta F_{xn}$ divided by the area, which is $\Delta z\sqrt{\Delta x^2+\Delta y^2}$, or \begin{equation*} S_{xn}=S_{xx}\,\frac{\Delta y}{\sqrt{\Delta x^2+\Delta y^2}}+ S_{xy}\,\frac{\Delta x}{\sqrt{\Delta x^2+\Delta y^2}}. \end{equation*}  Now $\Delta x/\sqrt{\Delta x^2+\Delta y^2}$ is the cosine of the angle $\theta$ between $\FLPn$ and the $y$-axis, as shown in Fig. 31–8, so it can also be written as $n_y$, the $y$-component of $\FLPn$. Similarly, $\Delta y/\sqrt{\Delta x^2+\Delta y^2}$ is $\sin\theta=n_x$. We can write \begin{equation*} S_{xn}=S_{xx}n_x+S_{xy}n_y. \end{equation*}  If we now generalize to an arbitrary surface element, we would get that \begin{equation*} S_{xn}=S_{xx}n_x+S_{xy}n_y+S_{xz}n_z \end{equation*}  or, in general, \begin{equation} \label{Eq:II:31:24} S_{in}=\sum_jS_{ij}n_j. \end{equation}  We can find the force across any surface element in terms of the $S_{ij}$, so it does describe completely the state of internal stress of the material. Equation (31.24) says that the tensor $S_{ij}$ relates the stress $\FLPS_n$ to the unit vector $\FLPn$, just as $\alpha_{ij}$ relates $\FLPP$ to $\FLPE$. Since $\FLPn$ and $\FLPS_n$ are vectors, the components of $S_{ij}$ must transform as a tensor with changes in coordinate axes. So $S_{ij}$ is indeed a tensor. We can also show that $S_{ij}$ is a symmetric tensor by looking at the forces on a little cube of material. Suppose we take a little cube, oriented with its faces parallel to our coordinate axes, and look at it in cross section, as shown in Fig. 31–9. If we let the edge of the cube be one unit, the $x$- and $y$-components of the forces on the faces normal to the $x$- and $y$-axes might be as shown in the figure. If the cube is small, the stresses do not change appreciably from one side of the cube to the opposite side, so the force components are equal and opposite as shown. Now there must be no torque on the cube, or it would start spinning. The total torque about the center is $(S_{yx}-S_{xy})$ (times the unit edge of the cube), and since the total is zero, $S_{yx}$ is equal to $S_{xy}$, and the stress tensor is symmetric. Since $S_{ij}$ is a symmetric tensor, it can be described by an ellipsoid which will have three principal axes. For surfaces normal to these axes, the stresses are particularly simple—they correspond to pushes or pulls perpendicular to the surfaces. There are no shear forces along these faces. For any stress, we can always choose our axes so that the shear components are zero. If the ellipsoid is a sphere, there are only normal forces in any direction. This corresponds to a hydrostatic pressure (positive or negative). So for a hydrostatic pressure, the tensor is diagonal and all three components are equal; they are, in fact, just equal to the pressure $p$. We can write \begin{equation} \label{Eq:II:31:25} S_{ij}=p\delta_{ij}. \end{equation}  The stress tensor—and also its ellipsoid—will, in general, vary from point to point in a block of material; to describe the whole block we need to give the value of each component of $S_{ij}$ as a function of position. So the stress tensor is a field. We have had scalar fields, like the temperature $T(x,y,z)$, which give one number for each point in space, and vector fields like $\FLPE(x,y,z)$, which give three numbers for each point. Now we have a tensor field which gives nine numbers for each point in space—or really six for the symmetric tensor $S_{ij}$. A complete description of the internal forces in an arbitrarily distorted solid requires six functions of $x$, $y$, and $z$.