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 eigenfunctions which satisfy
[math]\displaystyle{ \partial_x^4 X_n = \lambda_n^4 X_n \,\,\, -L \leq x \leq L }[/math]
plus the edge conditions of zero bending moment and shear stress
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_{2n} x)}{\cos(\lambda_{2n} L)}+\frac{\cosh(\lambda_{2n} x)}{\cosh(\lambda_{2n} 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_{2n+1} x)}{\sin(\lambda_{2n+1} L)}+\frac{\sinh(\lambda_{2n+1} x)}{\sinh(\lambda_{2n+1} 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]
Matlab Code
A program to calculate the eigenvalues can be found here beam_ev.m
A program to calculate the eigenvectors can be found here beam_em.m
