Bending vibration of a slender beam can be understood via the simplest approximation, Euler-Bernoulli beam theory. Suppose the beam is made of material with Young’s modulus $E$ and density $\rho$, and that it has a uniform cross-section with area $A$. We are interested in free vibration with small transverse displacement $w(x,t)$. Consider the forces acting on a small element of the beam, as sketched in Fig. 1.
If you imagine a cut across the cross-section of the beam, there will be an equal and opposite pair for forces acting on the two faces of the cut, and also an equal and opposite pair of moments. The force is called the shear force, $S(x,t)$, and the moment is called the bending moment, $M(x,t)$. This force and moment act on both ends of the small element, at positions $x$ and $x + \delta x$. Within this simplest theoretical framework, the rotational inertia of the element is ignored so that the element is approximately in overall moment balance. This requires
$$S(x,t) \delta x \approx M(x+ \delta x,t) – M(x,t). \tag{1}$$
In the limit as $\delta x \rightarrow 0$ this implies
$$S \approx \dfrac{\partial M}{\partial x} . \tag{2}$$
Now apply Newton’s law to the transverse motion of the element:
$$\rho A ~\delta x \dfrac{\partial^2 w}{\partial t^2} \approx S(x,t) – S(x+\delta x,t) \tag{3}$$
so that as $\delta x \rightarrow 0$,
$$\rho A \dfrac{\partial^2 w}{\partial t^2} = -\dfrac{\partial S}{\partial x}. \tag{4}$$
For linearised bending theory, the bending moment $M$ is proportional to the curvature of the beam:
$$M(x,t) = EI\dfrac{\partial^2 w}{\partial x^2} . \tag{5}$$
The constant of proportionality is called the bending rigidity, and is the product of the Young’s modulus $E$ with a quantity $I$ which is called the second moment of area of the cross-section. For a rectangular section, appropriate for our xylophone bar, there is a simple formula in terms of the width $b$ and the thickness $h$:
$$I=\dfrac{1}{12} bh^3. \tag{6}$$
Now combining eqs. (2), (4) and (5) we find the equation of motion for free vibration of the beam:
$$m \dfrac{\partial^2 w}{\partial t^2}+EI \dfrac{\partial^4 w}{\partial x^4}=0 . \tag{7}$$
Because this is a fourth-order differential equation in $x$, we will need two boundary conditions at each end of the beam. We are interested in the case with free ends, so we obtain boundary conditions from the fact that both the bending moment and the shear force must be zero: this requires
$$\dfrac{\partial^2 w}{\partial x^2}=0 \mathrm{~~and~~} \dfrac{\partial^3 w}{\partial x^3}=0~~\mathrm{(free~boundary).} \tag{8}$$
Now we are ready to find the modes of a free-free beam, the case we are interested in with both ends free. As usual, we start by looking for solutions of the form
$$w(x,t) = u(x) e^{i \omega t} \tag{9}$$
so that eq. (7) requires
$$EI\dfrac{d^4 u}{d x^4}= \rho A \omega^2 u. \tag{10}$$
The general solution to this equation can be written
$$u=C_1 e^{ikx} + C_2 e^{-ikx} + C_3 e^{kx} + C_4 e^{-kx} \tag{11}$$
where $C_1 – C_4$ are arbitrary constants and the wavenumber $k$ satisfies
$$k^4 = \dfrac{\rho A}{EI}\omega^2 . \tag{12}$$
The terms in eq. (11) can be described physically. The terms $e^{i \omega t} e^{\pm i k x}$ describe sinusoidal waves which travel along the beam, to the left or to the right, at a speed $\omega/k$. These are directly similar to the waves found on the stretched string in section 2.1.1. The terms $e^{i \omega t} e^{\pm k x}$ describe something new. These are disturbances on the beam which do not travel, but simply decay exponentially in one direction or the other along the beam. These are known as evanescent waves or near fields.
Equally well, instead of (11) we could write the general solution in the equivalent form
$$u=K_1 \cos kx + K_2 \sin kx + K_3 \cosh kx + K_4 \sinh kx \tag{13}$$
with a different set of constants $K_1 – K_4$. This form turns out to be more convenient for calculating the mode shapes. Applying the two boundary conditions (8) at $x=0$ leads immediately to $K_1=K_3$ and $K_2=K_4$. Enforcing the same two conditions at $x=L$ then requires
$$K_1 (-\cos kL +\cosh kL) +K_2 (-\sin kL +\sinh kL)=0 \tag{14}$$
and
$$K_1 (\sin kL +\sinh kL) + K_2(-\cos kL +\cosh kL)=0. \tag{15}$$
These two simultaneous equations for $K_1$ and $K_2$ both have zero on the right-hand side, so they can only have non-zero solutions if the determinant of the $2 \times 2$ matrix of coefficients is zero. This requires
$$(\cosh kL -\cos kL)^2 = (\sinh kL + \sin kL)(\sinh kL – \sin kL) \tag{16}$$
which simplifies after a little algebra to the requirement
$$\cos kL \cosh kL =1. \tag{17}$$
The roots of this equation give the only possible values of $k$, which then lead to corresponding values of $\omega$ via eq. (12). These are the natural frequencies. Having found the value of $k$ for a given mode, the mode shape can be found by substituting back in the equations for the four constants $K_1 – K_4$ from the earlier calculation.
We cannot solve eq. (17) exactly to obtain the natural frequencies. However, we can easily obtain an approximate answer using a graphical approach. Rearrange eq. (17) in the form
$$\cos kL =1/\cosh kL. \tag{18}$$
We can easily sketch the left- and right-hand side functions; intersections of the two plots will give the roots. Figure 2 shows the result. As the value of $kL$ increases, $\cosh kL$ increases rapidly so the intersections get closer and closer to the zeros of $\cos kL$. This gives an approximate expression for $n$th intersection:
$$kL \approx (n +1/2) \pi \tag{19}$$
and the corresponding natural frequency is
$$\omega_n \approx \left[ \dfrac{(n+1/2)\pi)}{L}\right]^2 \sqrt{\dfrac{EI}{\rho A}} = \left[ \dfrac{(n+1/2)\pi)}{L}\right]^2 \sqrt{\dfrac{Eh^2}{12 \rho}} \tag{20}$$
for $n=1,2,3…$, where the final expression is for the rectangular section.
Looking at Fig. 2, you may notice that the red and blue curves meet in a tangential manner at $kL=0$. Is there an additional mode associated with this condition? We know from equation (12) that it would correspond to a natural frequency of zero, which would mean a “rigid-body mode”: a possible motion of the beam involving no bending deformation. The easiest way to spot rigid-body modes like this is via physical intuition rather than mathematics.
As soon as you ask yourself whether there are ways that a free-free beam can move without bending deformation, the answer is obvious. Any unconstrained body can move in six different ways: there are three degrees of freedom of lateral motion in the three directions $x,y,z$, and another three degrees of freedom to rotate about each of the coordinate axes. In the case of our beam we have formulated the problem in terms of motion confined to a plane, in which each point on the beam is only allowed to move perpendicular to the beam’s axis. Two of the six degrees of freedom comply with those conditions: the beam could fall, or it could rotate about an axis perpendicular to the upper sketch of Fig. 1.
We conclude that our beam should have two modes at zero frequency. The simplest way to verify this mathematically is to substitute the functions $w=\mathrm{constant}$ and $w=x$ into the governing equation (7) and the boundary conditions (8): both are trivially satisfied.
Finally, there is another type of bending beam that is sometimes used for musical purposes. You can make a kind of musical note by holding a ruler firmly against the edge of a table and twanging the overhanging end. The same principle is used in the kalimba or “thumb piano”, in which a number of cantilever beams of different lengths are tuned to give different notes of a scale: you can play tunes by plucking them with your thumbs.
A cantilever beam like these examples will satisfy the same governing equation (7), and at the free end it will still satisfy the boundary conditions (8). But at the other end, the beam is clamped: it has no displacement and no rotation. So at $x=0$ we must now satisfy
$$w=0 \mathrm{~~~and~~~} \dfrac{\partial w}{\partial x}=0 . \tag{21}$$
The steps of the calculation described above can be followed with the new boundary condition, and after a little algebra we find that in place of eq. (17) the solutions must now satisfy
$$\cos kL \cosh kL =-1. \tag{22}$$
We can visualise the solutions by a graphical construction like Fig. 2, but incorporating the new minus sign: the result is plotted in Fig. 3. This time there is no intersection of the curves at $kL=0$, and indeed it is obvious that a cantilever beam has no rigid body degrees of freedom. Most of the intersections are rather similar to those in Fig. 2, close to $kL=3 \pi/2, 5 \pi/2, 7 \pi/2…$ But the lowest mode now has a position which is not well approximated by the simple formula. Numerical solution reveals that it occurs at $kL \approx 1.875$.
