The bent beam
We want now to look at another practical matter—the bending of a rod or a beam. What are the forces when we bend a bar of some arbitrary cross section? We will work it out thinking of a bar with a circular cross section, but our answer will be good for any shape. To save time, however, we will cut some corners, so our theory we will work out is only approximate. Our results will be correct only when the radius of the bend is much larger than the thickness of the beam. Suppose you grab the two ends of a straight bar and bend it into some curve like the one shown in Fig. 38–11. What goes on inside the bar? Well, if it is curved, that means that the material on the inside of the curve is compressed and the material on the outside is stretched. There is some surface which goes along more or less parallel to the axis of the bar that is neither stretched nor compressed. This is called the neutral surface. You would expect this surface to be near the “middle” of the cross section. It can be shown (but we won’t do it here) that, for small bending of simple beams, the neutral surface goes through the “center of gravity” of the cross section. This is true only for “pure” bending—if you are not stretching or compressing the beam at the same time. For pure bending, then, a thin transverse slice of the bar is distorted as shown in Fig. 38–12(a). The material below the neutral surface has a compressional strain which is proportional to the distance from the neutral surface; and the material above is stretched, also in proportion to its distance from the neutral surface. So the longitudinal stretch $\Delta l$ is proportional to the height $y$. The constant of proportionality is just $l$ over the radius of curvature of the bar—see Fig. 38–12: \begin{equation*} \frac{\Delta l}{l}=\frac{y}{R}. \end{equation*}  So the force per unit area—the stress—in a small strip at $y$ is also proportional to the distance from the neutral surface \begin{equation} \label{Eq:II:38:34} \frac{\Delta F}{\Delta A}=Y\,\frac{y}{R}. \end{equation}  Now let’s look at the forces that would produce such a strain. The forces acting on the little segment drawn in Fig. 38–12 are shown in the figure. If we think of any transverse cut, the forces acting across it are one way above the neutral surface and the other way below. They come in pairs to make a “bending moment” $\bendingMom$—by which we mean the torque about the neutral line. We can compute the total moment by integrating the force times the distance from the neutral surface for one of the faces of the segment of Fig. 38–12: \begin{equation} \label{Eq:II:38:35} \bendingMom=\underset{\substack{\text{cross}\\\text{sect}}}{\int} y\,dF. \end{equation}  From Eq. (38.34), $dF=Yy/R\,dA$, so \begin{equation*} \bendingMom=\frac{Y}{R}\int y^2\,dA. \end{equation*}  The integral of $y^2\,dA$ is what we can call the “moment of inertia” of the geometric cross section about a horizontal axis through its “center of mass”;1 we will call it $I$: \begin{align} \label{Eq:II:38:36} \bendingMom&=\frac{YI}{R}\\[2ex] \label{Eq:II:38:37} I&=\int y^2\,dA. \end{align}  Equation (38.36), then, gives us the relation between the bending moment $\bendingMom$ and the curvature $1/R$ of the beam. The “stiffness” of the beam is proportional to $Y$ and to the moment of inertia $I$. In other words, if you want the stiffest possible beam with a given amount of, say, aluminum, you want to put as much of it as possible as far as you can from the neutral surface, to make a large moment of inertia. You can’t carry this to an extreme, however, because then the thing will not curve as we have supposed—it will buckle or twist and become weaker again. But now you see why structural beams are made in the form of an I or an H—as shown in Fig. 38–13. As an example of the use of our beam equation (38.36), let’s work out the deflection of a cantilevered beam with a concentrated force $W$ acting at the free end, as sketched in Fig. 38–14. (By “cantilevered” we simply mean that the beam is supported in such a way that both the position and the slope are fixed at one end—it is stuck into a cement wall.) What is the shape of the beam?  Let’s call the deflection at the distance $x$ from the fixed end $z$; we want to know $z(x)$. We’ll work it out only for small deflections. We will also assume that the beam is long in comparison with its cross section. Now, as you know from your mathematics courses, the curvature $1/R$ of any curve $z(x)$ is given by \begin{equation} \label{Eq:II:38:38} \frac{1}{R}=\frac{d^2z/dx^2}{[1+(dz/dx)^2]^{3/2}}. \end{equation}  Since we are interested only in small slopes—this is usually the case in engineering structures—we neglect $(dz/dx)^2$ in comparison with $1$, and take \begin{equation} \label{Eq:II:38:39} \frac{1}{R}=\frac{d^2z}{dx^2}. \end{equation}  We also need to know the bending moment $\bendingMom$. It is a function of $x$ because it is equal to the torque about the neutral axis of any cross section. Let’s neglect the weight of the beam and take only the downward force $W$ at the end of the beam. (You can put in the beam weight yourself if you want.) Then the bending moment at $x$ is \begin{equation*} \bendingMom(x)=W(L-x), \end{equation*}  because that is the torque about the point at $x$, exerted by the weight $W$—the torque which the beam must support at $x$. We get \begin{equation} W(L-x)=\frac{YI}{R}=YI\,\frac{d^2z}{dx^2}\notag \end{equation}  or \begin{equation} \label{Eq:II:38:40} \frac{d^2z}{dx^2}=\frac{W}{YI}\,(L-x). \end{equation}  This one we can integrate without any tricks; we get \begin{equation} \label{Eq:II:38:41} z=\frac{W}{YI}\biggl( \frac{Lx^2}{2}-\frac{x^3}{6} \biggr), \end{equation}  using our assumptions that $z(0)=0$ and that $dz/dx$ is also zero at $x=0$. That is the shape of the beam. The displacement of the end is \begin{equation} \label{Eq:II:38:42} z(L)=\frac{W}{YI}\,\frac{L^3}{3}; \end{equation}  the displacement of the end of a beam increases as the cube of the length. In deriving our approximate beam theory, we have assumed that the cross section of the beam did not change when the beam was bent. When the thickness of the beam is small compared to the radius of curvature, the cross section changes very little and our result is O.K. In general, however, this effect cannot be neglected, as you can easily demonstrate for yourselves by bending a soft-rubber eraser in your fingers. If the cross section was originally rectangular, you will find that when it is bent it bulges at the bottom (see Fig. 38–15). This happens because when we compress the bottom, the material expands sideways—as described by Poisson’s ratio. Rubber is easy to bend or stretch, but it is somewhat like a liquid in that it’s hard to change the volume—as shows up nicely when you bend the eraser. For an incompressible material, Poisson’s ratio would be exactly $1/2$—for rubber it is nearly that.