Difference between revisions of "Eigenfunctions for a Uniform Free Beam"

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\frac{\partial^2}{\partial x^2} \frac{\partial\phi}{\partial z} = 0\mbox{ for } \;\;\;\; \mbox{ at } z = 0 \;\;\; x = \pm L.
 
\frac{\partial^2}{\partial x^2} \frac{\partial\phi}{\partial z} = 0\mbox{ for } \;\;\;\; \mbox{ at } z = 0 \;\;\; x = \pm L.
 
\end{matrix}</math></center>
 
\end{matrix}</math></center>
This solution is discussed further in [[Eigenfunctions for a Free Beam]].
 
  
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.
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General solution of the differential equation is :
 
 
General solution of the above stated equation is :
 
  
 
<center>
 
<center>
 
<math>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>
 
<math>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>
 
</center>
 
</center>
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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.
  
 
== Symmetric modes ==
 
== Symmetric modes ==

Revision as of 22:57, 6 November 2008

We can find a the eigenfunction which satisfy

[math]\displaystyle{ \partial_x^4 w_n = \lambda_n^4 w_n }[/math]

plus the edge conditions.

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

General solution of the differential equation 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.

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]

Having obtained eigenvalues [math]\displaystyle{ \lambda_n }[/math], natural requency can be readily calculated :

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

Symmetrical natural modes can be written in normalized form as :

[math]\displaystyle{ w_n(x) = \frac{1}{2}\left( \frac{\cos(\lambda_n x)}{\cos(\lambda_n l)}+\frac{\cosh(\lambda_n x)}{\cosh(\lambda_n l)} \right ) }[/math]

Skew-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]

Having obtained eigenvalues [math]\displaystyle{ \lambda_n }[/math], natural requency can be readily calculated :

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

Symmetrical natural modes can be written in normalized form as :

[math]\displaystyle{ w_n(x) = \frac{1}{2}\left( \frac{\cos(\lambda_n x)}{\cos(\lambda_n l)}+\frac{\cosh(\lambda_n x)}{\cosh(\lambda_n l)} \right ) }[/math]