Eigenfunctions for a Uniform Free Beam

Introduction

We show here how to find the eigenfunction for a beam with free edge conditions.

Equations

We can find a the eigenfunction which satisfy

[math]\displaystyle{ \partial_x^4 w_n = \lambda_n^4 w_n \,\,\, -L \leq x \leq L }[/math]

plus the edge conditions of zero bending moment and shear stress

[math]\displaystyle{ \begin{matrix} \partial_x^3 w_n= 0 \;\;\;\; \mbox{ at } z = 0 \;\;\; x = \pm L, \end{matrix} }[/math] [math]\displaystyle{ \begin{matrix} \partial_x^2 w_n = 0 \;\;\; \mbox{ at } z = 0 \;\;\; x = \pm L. \end{matrix} }[/math]

Solution

General solution of the differential equation for [math]\displaystyle{ \lambda \neq 0 }[/math] is

[math]\displaystyle{ w_n(x) = C_1 \sin(\lambda_n x) + C_2 \cos(\lambda_n x) + C_3 \sinh(\lambda_n x) + C_4 \cosh(\lambda_n x)\, }[/math]

Due to symmetry of the problem, dry natural vibrations of a free beam can be split into two different sets of modes, symmetric (even) modes and skew-symmetric (odd) modes.

Modes for [math]\displaystyle{ \lambda = 0 }[/math]

There are two modes for [math]\displaystyle{ \lambda = 0 }[/math] which are the two rigid body motions they are given by

[math]\displaystyle{ w_0 = \frac{1}{\sqrt{2L}} }[/math]

and

[math]\displaystyle{ w_1 = \sqrt{\frac{3}{2L^3}} x }[/math]

Symmetric modes

[math]\displaystyle{ C_1 = C_3 = 0 \Rightarrow w_n(x) = C_2 \cos(\lambda_n x) + C_4 \cosh(\lambda_n x) }[/math]

By imposing boundary conditions at [math]\displaystyle{ x = L }[/math] :

[math]\displaystyle{ \begin{bmatrix} - \cos(\lambda_n L)&\cosh(\lambda_n L)\\ \sin(\lambda_n L)&\sinh(\lambda_n L)\\ \end{bmatrix} \begin{bmatrix} C_2\\ C_4\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} }[/math]

For a nontrivial solution one gets:

[math]\displaystyle{ \tan(\lambda_n L)+\tanh(\lambda_n L)=0\, }[/math]

The first three roots are :

[math]\displaystyle{ \lambda_0 L = 0, \lambda_2 L = 2.365, \lambda_4 L = 5.497\, }[/math]

Symmetric natural modes can be written in normalized form as :

[math]\displaystyle{ w_{2n}(x) = \frac{1}{\sqrt{2L}}\left( \frac{\cos(\lambda_n x)}{\cos(\lambda_n L)}+\frac{\cosh(\lambda_n x)}{\cosh(\lambda_n L)} \right ) \,\,\,n\geq 1 }[/math]

where the The symmetric modes have been normalised so that their inner products equal the Kronecker delta.

Anti-symmetric modes

[math]\displaystyle{ C_2 = C_4 = 0 \Rightarrow w_n(x) = C_1 \sin(\lambda_n x) + C_3 \sinh(\lambda_n x) }[/math]

By imposing boundary conditions at [math]\displaystyle{ x = l }[/math] :

[math]\displaystyle{ \begin{bmatrix} - \sin(\lambda_n L)&\sinh(\lambda_n L)\\ -\cos(\lambda_n L)&\cosh(\lambda_n L)\\ \end{bmatrix} \begin{bmatrix} C_1\\ C_3\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} }[/math]

For a nontrivial solution one gets:

[math]\displaystyle{ -\tan(\lambda_n L)+\tanh(\lambda_n L)=0\, }[/math]

The first three roots are :

[math]\displaystyle{ \lambda_1 L = 0, \lambda_3 L = 3.925, \lambda_5 L = 7.068\, }[/math]

Anti-symmetric natural modes can be written in normalized form as :

[math]\displaystyle{ w_{2n+1}(x) = \frac{1}{\sqrt{2L}}\left( \frac{\sin(\lambda_n x)}{\sin(\lambda_n L)}+\frac{\sinh(\lambda_n x)}{\sinh(\lambda_n L)} \right ) \,\,\,n\geq 1 }[/math]

where the eigenfunctions have been chosen so that their inner products equal the Kronecker delta.

Natural Frequencies

The equation in the time domain for a beam is

[math]\displaystyle{ m\partial_t^2 w + EI \partial_x^4 w = 0 }[/math]

so that, having obtained eigenvalues [math]\displaystyle{ \lambda_n }[/math], the natural frequency can be readily calculated :

[math]\displaystyle{ \omega_n = \lambda_n^2 \sqrt\frac{EI}{m} }[/math]