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

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<center>
 
<center>
<math>C_1 = C_3 = 0 \Rightarrow w_n(x) = C_2 \cos(\lambda_n x) + C_4 \cosh(\lambda_n x)</math>
+
<math>C_2 = C_4 = 0 \Rightarrow w_n(x) = C_1 \sin(\lambda_n x) + C_3 \sinh(\lambda_n x)</math>
 
</center>
 
</center>
  
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<math>
 
<math>
 
\begin{bmatrix}
 
\begin{bmatrix}
- \cos(\lambda_n l)&\cosh(\lambda_n l)\\
+
- \sin(\lambda_n l)&\sinh(\lambda_n l)\\
\sin(\lambda_n l)&\sinh(\lambda_n l)\\  
+
-\cos(\lambda_n l)&\cosh(\lambda_n l)\\  
 
\end{bmatrix}
 
\end{bmatrix}
  
 
\begin{bmatrix}
 
\begin{bmatrix}
C_2\\
+
C_1\\
C_4\\  
+
C_3\\  
 
\end{bmatrix}
 
\end{bmatrix}
  
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For a nontrivial solution one gets:
 
For a nontrivial solution one gets:
 
<center>
 
<center>
<math>\tan(\lambda_n l)+\tanh(\lambda_n l)=0\,</math>
+
<math>-\tan(\lambda_n l)+\tanh(\lambda_n l)=0\,</math>
 
</center>
 
</center>
  
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Symmetrical natural modes can be written in normalized form as :
 
Symmetrical natural modes can be written in normalized form as :
 
<center>
 
<center>
<math>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>w_n(x) = \frac{1}{2}\left( \frac{\sin(\lambda_n x)}{\sin(\lambda_n l)}+\frac{\sinh(\lambda_n x)}{\sinh(\lambda_n l)} \right )
 
</math>
 
</math>
 
</center>
 
</center>

Revision as of 23:01, 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_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]

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{\sin(\lambda_n x)}{\sin(\lambda_n l)}+\frac{\sinh(\lambda_n x)}{\sinh(\lambda_n l)} \right ) }[/math]