Fundamentals

Identical natural frequencies of beams with different boundary conditions

One somewhat surprising result in the vibration of beams is that the natural frequencies of a free-free beam are identical to those of a clamped-clamped beam. Our intuition may tell us that the fixed-fixed beam is stiffer and should have higher natural frequencies but while the fixed beam is indeed statically stiffer, the natural frequencies are identical. As we show below, fixed and free boundary conditions result in the equivalent characteristic equations to find the natural frequency because they differ simply by two differentiations. Similarly, pinned and sliding boundary conditions also have the same equivalence. As a result many pairs of boundary conditions can result in identical sets of natural frequencies but with clearly different mode shapes.

Fixed-fixed and free-free beams

We begin with a very brief refresher of Euler-Bernoulli beam theory as described in one of my favorite texts \emph{Mechanical Vibration} by S.S. Rao (link to book).  The governing equation for a vibrating beam is

    \begin{equation*} EI \frac{\partial^4 w}{\partial x^4}+\rho A \frac{\partial^2 w}{\partial t^2} = f(x,t) \end{equation*}

For free vibration,f(x,t)=0 and separation of variables is used to obtain two equations

    \begin{equation*} \frac{d^4 W}{d x^4}-\beta^4 W=0 \end{equation*}

and

    \begin{equation*} \frac{d^2 T}{d t^2}+\omega^2 T=0 \end{equation*}

where

    \begin{equation*} \beta^4=\omega^2 \frac{\rho A}{EI} \end{equation*}

The mode shape of the beam is

    \begin{equation*} W(x)=C_1 \cos\beta x + C_2 \sin \beta x + C_3 \cosh \beta x +C_4 \sinh \beta x \end{equation*}

The boundary conditions of the beam provide four equations which are used to solve for the C_i. Fixed boundary conditions result in w=0 and \partial w/\partial x = 0. Free boundary conditions result in zero moment and shear force at the ends \partial^2 w/\partial x^2 = 0 and \partial^3 w/\partial x^3 = 0. Notice here that the free boundary conditions are two additional derivatives of the fixed boundary conditions. Pinned boundary conditions have zero displacement and zero moment at the ends w=0 and \partial^2 w/\partial x^2 = 0. Sliding restrained boundary conditions have zero slope and zero shear force at the ends \partial w/\partial x = 0, \partial^3 w/\partial x^3 = 0. Again these two boundary conditions are related by two differentiations.

For the fixed-fixed beam, substitution of the boundary conditions results in a four equations

    \begin{equation*} \begin{bmatrix} 1 & 0 & 1 & 0 \\ \cos \beta L & \sin \beta L & \cosh \beta L & \sinh \beta L \\ 0 & 1 & 0 & 1 \\ -\sin \beta L & \cos \beta L & \sinh \beta L & \cosh \beta L \\ \end{bmatrix} \begin{pmatrix} C_1 \\ C_2 \\ C_3 \\ C_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{equation*}

Similarly for the free-free beam

    \begin{equation*} \begin{bmatrix} -1 & 0 & 1 & 0 \\ -\cos \beta L & -\sin \beta L & \cosh \beta L & \sinh \beta L \\ 0 & -1 & 0 & 1 \\ \sin \beta L & -\cos \beta L & \sinh \beta L & \cosh \beta L \\ \end{bmatrix} \begin{pmatrix} C_1 \\ C_2 \\ C_3 \\ C_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{equation*}

Comparing the two equations, the only differences are that the first two columns of the free-free case are the negative of the first two columns of the fixed-fixed case. To solve for the non-trivial solution, one must set the determinant of the matrix equal to zero to obtain the characteristic equation. One property of the determinant is that \det(A)=\det(A^T). When we compare the transpose of the matrices we see that the first two rows of the free-free case and the fixed-fixed case differ only by a negative sign and can be made identical. Therefore, the characteristic equation will have the same solution for \beta in each case and the natural frequencies are identical. When one goes back to solve for the mode shapes, it is quite clear that the negative sign results in different shapes.

The result of the characteristic equation is

    \begin{equation*} cos \beta_n L \cosh \beta_n L=1 \end{equation*}

The mode shape for the fixed-fixed beam is

    \begin{equation*} W_n(x)=C_n \left( \sin \beta x - \sinh \beta x + \alpha_n \left( \cos\beta x - \cosh \beta x \right)\right) \end{equation*}

where

    \begin{equation*} \alpha_n=\frac{\sin \beta_n L - \sinh \beta_n L}{\cosh \beta_n L - \cos \beta_n L} \end{equation*}

The mode shape for the free-free beam is

    \begin{equation*} W_n(x)=C_n \left( \sin \beta x + \sinh \beta x + \alpha_n \left( \cos\beta x + \cosh \beta x \right)\right) \end{equation*}

where \alpha_n is the same as for the fixed-fixed case given above.

Below we plot the first mode shape as well as its derivatives for the fixed-fixed and free-free boundary conditions. The maximum displacement has been normalized to 1 and EI and \rho A and L have also been set to 1 without loss of generality.

Comparison of first mode shape for fixed-fixed and free-free beams

Another useful insight is to compare the strain and kinetic energy densities. Because the natural frequencies are the same, we know that the ratio of potential energy and strain energy are the same, but the energy density shows how the beams are different.

Strain (potential) and kinetic energy densities for fixed-fixed and free-free beams

A similar procedure can also be followed to compare pinned and sliding boundary conditions. Furthermore, one also finds that this holds true for natural frequencies of plates.