Uniform strains
As our first example let’s find out what happens to a rectangular block under uniform hydrostatic pressure. Let’s put a block under water in a pressure tank. Then there will be a force acting inward on every face of the block proportional to the area (see Fig. 38–2). Since the hydrostatic pressure is uniform, the stress (force per unit area) on each face of the block is the same. We will work out first the change in the length. The change in length of the block can be thought of as the sum of changes in length that would occur in the three independent problems which are sketched in Fig. 38–3. Problem 1. If we push on the ends of the block with a pressure $p$, the compressional strain is $p/Y$, and it is negative, \begin{equation*} \frac{\Delta l_1}{l}=-\frac{p}{Y}. \end{equation*}  Problem 2. If we push on the two sides of the block with pressure $p$, the compressional strain is again $p/Y$, but now we want the lengthwise strain. We can get that from the sideways strain multiplied by $-\sigma$. The sideways strain is \begin{equation*} \frac{\Delta w}{w}=-\frac{p}{Y}; \end{equation*}  so \begin{equation*} \frac{\Delta l_2}{l}=+\sigma\,\frac{p}{Y}. \end{equation*}  Problem 3. If we push on the top of the block, the compressional strain is once more $p/Y$, and the corresponding strain in the sideways direction is again $-\sigma p/Y$. We get \begin{equation*} \frac{\Delta l_3}{l}=+\sigma\,\frac{p}{Y}. \end{equation*}  Combining the results of the three problems—that is, taking $\Delta l=\Delta l_1+\Delta l_2+\Delta l_3$—we get \begin{equation} \label{Eq:II:38:6} \frac{\Delta l}{l}=-\frac{p}{Y}\,(1-2\sigma). \end{equation}  The problem is, of course, symmetrical in all three directions; it follows that \begin{equation} \label{Eq:II:38:7} \frac{\Delta w}{w}=\frac{\Delta h}{h}=-\frac{p}{Y}\,(1-2\sigma). \end{equation}  The change in the volume under hydrostatic pressure is also of some interest. Since $V=lwh$, we can write, for small displacements, \begin{equation*} \frac{\Delta V}{V}=\frac{\Delta l}{l}+\frac{\Delta w}{w}+ \frac{\Delta h}{h}. \end{equation*}  Using (38.6) and (38.7), we have \begin{equation} \label{Eq:II:38:8} \frac{\Delta V}{V}=-3\,\frac{p}{Y}\,(1-2\sigma). \end{equation}  People like to call $\Delta V/V$ the volume strain and write \begin{equation*} p=-K\,\frac{\Delta V}{V}. \end{equation*}  The volume stress $p$ is proportional to the volume strain—Hooke’s law once more. The coefficient $K$ is called the bulk modulus; it is related to the other constants by \begin{equation} \label{Eq:II:38:9} K=\frac{Y}{3(1-2\sigma)}. \end{equation}  Since $K$ is of some practical interest, many handbooks give $Y$ and $K$ instead of $Y$ and $\sigma$. If you want $\sigma$ you can always get it from Eq. (38.9). We can also see from Eq. (38.9) that Poisson’s ratio, $\sigma$, must be less than one-half. If it were not, the bulk modulus $K$ would be negative, and the material would expand under increasing pressure. That would allow us to get mechanical energy out of any old block—it would mean that the block was in unstable equilibrium. If it started to expand it would continue by itself with a release of energy. Now we want to consider what happens when you put a “shear” strain on something. By shear strain we mean the kind of distortion shown in Fig. 38–4. As a preliminary to this, let us look at the strains in a cube of material subjected to the forces shown in Fig. 38–5. Again we can break it up into two problems: the vertical pushes, and the horizontal pulls. Calling $A$ the area of the cube face, we have for the change in horizontal length \begin{equation} \label{Eq:II:38:10} \frac{\Delta l}{l}=\frac{1}{Y}\,\frac{F}{A}+ \sigma\,\frac{1}{Y}\,\frac{F}{A}= \frac{1+\sigma}{Y}\,\frac{F}{A}. \end{equation}  The change in the vertical height is just the negative of this. Now suppose we have the same cube and subject it to the shearing forces shown in Fig. 38–6(a). Note that all the forces have to be equal if there are to be no net torques and the cube is to be in equilibrium. (Similar forces must also exist in Fig. 38–4, since the block is in equilibrium. They are provided through the “glue” that holds the block to the table.) The cube is then said to be in a state of pure shear. But note that if we cut the cube by a plane at $45^\circ$—say along the diagonal $A$ in the figure—the total force acting across the plane is normal to the plane and is equal to $\sqrt{2}G$. The area over which this force acts is $\sqrt{2}A$; therefore, the tensile stress normal to this plane is simply $G/A$. Similarly, if we examine a plane at an angle of $45^\circ$ the other way—the diagonal $B$ in the figure—we see that there is a compressional stress normal to this plane of $-G/A$. From this, we see that the stress in a “pure shear” is equivalent to a combination of tension and compression stresses of equal strength and at right angles to each other, and at $45^\circ$ to the original faces of the cube. The internal stresses and strains are the same as we would find in the larger block of material with the forces shown in Fig. 38–6(b). But this is the problem we have already solved. The change in length of the diagonal is given by Eq. (38.10), \begin{equation} \label{Eq:II:38:11} \frac{\Delta D}{D}=\frac{1+\sigma}{Y}\,\frac{G}{A}. \end{equation}  (One diagonal is shortened; the other is elongated.) It is often convenient to express a shear strain in terms of the angle by which the cube is twisted—the angle $\theta$ in Fig. 38–7. From the geometry of the figure you can see that the horizontal shift $\delta$ of the top edge is equal to $\sqrt{2}\,\Delta D$. So \begin{equation} \label{Eq:II:38:12} \theta=\frac{\delta}{l}=\frac{\sqrt{2}\,\Delta D}{l}=2\,\frac{\Delta D}{D}. \end{equation}  The shear stress $g$ is defined as the tangential force on one face divided by the area, $g=G/A$. Using Eq. (38.11) in (38.12), we get \begin{equation*} \theta=2\,\frac{1+\sigma}{Y}\,g. \end{equation*}  Or, writing this in the form “stress${}={}$constant times strain,” \begin{equation} \label{Eq:II:38:13} g=\mu\theta. \end{equation}  The proportionality coefficient $\mu$ is called the shear modulus (or, sometimes, the coefficient of rigidity). It is given in terms of $Y$ and $\sigma$ by \begin{equation} \label{Eq:II:38:14} \mu=\frac{Y}{2(1+\sigma)}. \end{equation}  Incidentally, the shear modulus must be positive—otherwise you could get work out of a self-shearing block. From Eq. (38.14), $\sigma$ must be greater than $-1$. We know, then, that $\sigma$ must be between $-1$ and $+\tfrac{1}{2}$; in practice, however, it is always greater than zero. As a last example of the type of situation where the stresses are uniform through the material, let’s consider the problem of a block which is stretched, while it is at the same time constrained so that no lateral contraction can take place. (Technically, it’s a little easier to compress it while keeping the sides from bulging out—but it’s the same problem.) What happens? Well, there must be sideways forces which keep it from changing its thickness—forces we don’t know off-hand but will have to calculate. It’s the same kind of problem we have already done, only with a little different algebra. We imagine forces on all three sides, as shown in Fig. 38–8; we calculate the changes in dimensions, and we choose the transverse forces to make the width and height remain constant. Following the usual arguments, we get for the three strains:  \begin{align} \label{Eq:II:38:15} \frac{\Delta l_x}{l_x}&=\frac{1}{Y}\,\frac{F_x}{A_x}- \frac{\sigma}{Y}\,\frac{F_y}{A_y}- \frac{\sigma}{Y}\,\frac{F_z}{A_z}= \frac{1}{Y}\biggl[\frac{F_x}{A_x}-\sigma\biggl( \frac{F_y}{A_y}+\frac{F_z}{A_z} \biggr)\biggr],\\[3pt] \label{Eq:II:38:16} \frac{\Delta l_y}{l_y}&=\frac{1}{Y} \biggl[\frac{F_y}{A_y}-\sigma\biggl( \frac{F_x}{A_x}+\frac{F_z}{A_z} \biggr)\biggr],\\[5pt] \label{Eq:II:38:17} \frac{\Delta l_z}{l_z}&=\frac{1}{Y} \biggl[\frac{F_z}{A_z}-\sigma\biggl( \frac{F_x}{A_x}+\frac{F_y}{A_y} \biggr)\biggr]. \end{align} 
 \begin{align} \frac{\Delta l_x}{l_x}&=\frac{1}{Y}\,\frac{F_x}{A_x}- \frac{\sigma}{Y}\,\frac{F_y}{A_y}- \frac{\sigma}{Y}\,\frac{F_z}{A_z}\notag\\[.75ex] \label{Eq:II:38:15} % ebook insert: \label{Eq:II:0:0} &=\frac{1}{Y}\biggl[\frac{F_x}{A_x}-\sigma\biggl( \frac{F_y}{A_y}+\frac{F_z}{A_z} \biggr)\biggr],\\[1ex] \label{Eq:II:38:16} \frac{\Delta l_y}{l_y}&=\frac{1}{Y} \biggl[\frac{F_y}{A_y}-\sigma\biggl( \frac{F_x}{A_x}+\frac{F_z}{A_z} \biggr)\biggr],\\[1ex] \label{Eq:II:38:17} \frac{\Delta l_z}{l_z}&=\frac{1}{Y} \biggl[\frac{F_z}{A_z}-\sigma\biggl( \frac{F_x}{A_x}+\frac{F_y}{A_y} \biggr)\biggr]. \end{align} 
 Now since $\Delta l_y$ and $\Delta l_z$ are supposed to be zero, Eqs. (38.16) and (38.17) give two equations relating $F_y$ and $F_z$ to $F_x$. Solving them together, we get that \begin{equation} \label{Eq:II:38:18} \frac{F_y}{A_y}=\frac{F_z}{A_z}=\frac{\sigma}{1-\sigma}\,\frac{F_x}{A_x}. \end{equation}  Substituting in (38.15), we have  \begin{equation} \label{Eq:II:38:19} \frac{\Delta l_x}{l_x}=\frac{1}{Y}\biggl( 1-\frac{2\sigma^2}{1-\sigma} \biggr)\frac{F_x}{A_x}=\frac{1}{Y}\biggl( \frac{1-\sigma-2\sigma^2}{1-\sigma} \biggr)\frac{F_x}{A_x}. \end{equation} 
 \begin{align} \frac{\Delta l_x}{l_x}&=\frac{1}{Y}\biggl( 1-\frac{2\sigma^2}{1-\sigma} \biggr)\frac{F_x}{A_x}\notag\\[1ex] \label{Eq:II:38:19} &=\frac{1}{Y}\biggl( \frac{1-\sigma-2\sigma^2}{1-\sigma} \biggr)\frac{F_x}{A_x}. \end{align}  Often, you will see this turned around, and with the quadratic in $\sigma$ factored out, it is then written \begin{equation} \label{Eq:II:38:20} \frac{F}{A}=\frac{1-\sigma}{(1+\sigma)(1-2\sigma)} \,Y\,\frac{\Delta l}{l}. \end{equation}  When we constrain the sides, Young’s modulus gets multiplied by a complicated function of $\sigma$. As you can most easily see from Eq. (38.19), the factor in front of $Y$ is always greater than $1$. It is harder to stretch the block when the sides are held—which also means that a block is stronger when the sides are held than when they are not.