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

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We can find a the eigenfunction which satisfy
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{{complete pages}}
 +
 
 +
== Introduction ==
 +
 
 +
We show here how to find the eigenfunction for a beam with free edge conditions.
 +
 
 +
== Equations ==
 +
 
 +
{{equations for a eigenfunction of a free beam}}
 +
 
 +
== Solution ==
 +
 
 +
General solution of the differential equation for <math>\lambda \neq 0</math> is
 +
<center>
 +
<math>X_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>
 +
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>\lambda = 0</math> ==
 +
 
 +
There are two modes for <math>\lambda = 0</math>  which are the two rigid body motions; they are given by
 +
{{rigid modes for an elastic plate}}
 +
 
 +
== Symmetric modes ==
 +
 
 +
<center>
 +
<math>C_1 = C_3 = 0 \Rightarrow w_n(x) = C_2 \cos(\lambda_n x) + C_4 \cosh(\lambda_n x)</math>
 +
</center>
 +
By imposing boundary conditions at <math>x = L</math> :
 +
<center>
 +
<math>
 +
\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>
 +
</center>
 +
 
 +
For a nontrivial solution one gets:
 
<center>
 
<center>
<math>\partial_x^4 w_n = \lambda_n^4 w_n</math>
+
<math>\tan(\lambda_n L)+\tanh(\lambda_n L)=0\,</math>
 
</center>
 
</center>
plus the edge conditions.
 
<center><math>\begin{matrix}
 
\frac{\partial^3}{\partial x^3} \frac{\partial\phi}{\partial z}= 0 \;\;\;\; \mbox{ at } z = 0 \;\;\; x = \pm L,
 
\end{matrix}</math></center>
 
<center><math>\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></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.
+
The first three roots are :
 +
 
 +
<center>
 +
<math>\lambda_0 L = 0, \lambda_2 L = 2.365, \lambda_4 L = 5.497\,</math>
 +
</center>
  
General solution of the above stated equation is:
 
  
 +
Symmetric natural modes can be written in normalized form as :
 
<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>X_{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>
 
</center>
 
</center>
 +
where the
 +
The symmetric modes have been normalised so that their inner products equal the Kronecker delta.
  
Symmetric modes
+
== Anti-symmetric modes ==
  
 
<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>
  
By imposing boundary conditions at <math>x = l</math> :
+
By imposing boundary conditions at <math>x = L</math> :
  
 
<center>
 
<center>
 
<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}
 
 
  =  
 
  =  
 
 
\begin{bmatrix}
 
\begin{bmatrix}
 
0\\
 
0\\
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</math>
 
</math>
 
</center>
 
</center>
 +
 +
For a nontrivial solution one gets:
 +
<center>
 +
<math>-\tan(\lambda_n L)+\tanh(\lambda_n L)=0\,</math>
 +
</center>
 +
 +
The first three roots are :
 +
 +
<center>
 +
<math>\lambda_1 L = 0, \lambda_3 L = 3.925, \lambda_5 L = 7.068\,</math>
 +
</center>
 +
 +
Anti-symmetric natural modes can be written in normalized form as :
 +
<center>
 +
<math>X_{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>
 +
</center>
 +
where the eigenfunctions have been chosen so that their inner products equal the Kronecker delta.
 +
 +
== Equations for a beam ==
 +
{{equations for a beam}}
 +
 +
== Solution for a uniform beam in [[Eigenfunctions for a Uniform Free Beam|eigenfunctions]] ==
 +
 +
{{solution for a uniform beam in eigenfunctions}}
 +
 +
 +
 +
== Matlab Code ==
 +
 +
A program to calculate the eigenvalues can be found here
 +
{{eigenvalues beam}}
 +
 +
A program to calculate the eigenvectors can be found here
 +
{{eigenvectors beam}}
 +
 +
[[Category:Complete Pages]]
 +
[[Category:Simple Linear Waves]]

Latest revision as of 16:12, 8 December 2009


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

[math]\displaystyle{ \begin{matrix} \partial_x^3 X_n= 0 \;\;\;\; \mbox{ at } z = 0 \;\;\; x = \pm L, \end{matrix} }[/math]
[math]\displaystyle{ \begin{matrix} \partial_x^2 X_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{ X_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{ X_0 = \frac{1}{\sqrt{2L}} }[/math]

and

[math]\displaystyle{ X_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{ X_{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{ X_{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.

Equations for a beam

There are various beam theories that can be used to describe the motion of the beam. The simplest theory is the Bernoulli-Euler Beam theory (other beam theories include the Timoshenko Beam theory and Reddy-Bickford Beam theory where shear deformation of higher order is considered). For a Bernoulli-Euler Beam, the equation of motion is given by the following

[math]\displaystyle{ \partial_x^2\left(\beta(x)\partial_x^2 \zeta\right) + \gamma(x) \partial_t^2 \zeta = p }[/math]

where [math]\displaystyle{ \beta(x) }[/math] is the non dimensionalised flexural rigidity, and [math]\displaystyle{ \gamma }[/math] is non-dimensionalised linear mass density function. Note that this equations simplifies if the plate has constant properties (and that [math]\displaystyle{ h }[/math] is the thickness of the plate, [math]\displaystyle{ p }[/math] is the pressure and [math]\displaystyle{ \zeta }[/math] is the plate vertical displacement) .

The edges of the plate can satisfy a range of boundary conditions. The natural boundary condition (i.e. free-edge boundary conditions).

[math]\displaystyle{ \partial_x^2 \zeta = 0, \,\,\partial_x^3 \zeta = 0 }[/math]

at the edges of the plate.

The problem is subject to the initial conditions

[math]\displaystyle{ \zeta(x,0)=f(x) \,\! }[/math]
[math]\displaystyle{ \partial_t \zeta(x,0)=g(x) }[/math]

Solution for a uniform beam in eigenfunctions

If the beam is uniform the equations can be written as

[math]\displaystyle{ \beta \frac{\partial^{4}\zeta}{\partial x^{4}} + \gamma \frac{\partial^{2}\zeta}{\partial t^{2}}=0 }[/math]

We can express the deflection as the series

[math]\displaystyle{ \zeta(x,t)=\sum_{n=0}^{\infty} A_n X_n(x) \cos(k_n t) + \sum_{n=2}^{\infty}B_n X_n(x) \frac{\sin(k_n t)}{k_n} }[/math]

where [math]\displaystyle{ X_n }[/math] are the Eigenfunctions for a Uniform Free Beam and [math]\displaystyle{ k_m = \lambda^2_n \sqrt{\beta/\gamma} }[/math] where [math]\displaystyle{ \lambda_n }[/math] are the eigenfunctions.

Then [math]\displaystyle{ A_n \,\! }[/math] and [math]\displaystyle{ B_n \,\! }[/math] can be found using orthogonality properties:

[math]\displaystyle{ A_n=\frac{\int_{-L}^{L}f(x)X_n(x)\mathrm{d}x}{\int_{-L}^{L}X_n(x)X_n(x)\mathrm{d}x} \,\! }[/math]
[math]\displaystyle{ B_n=\frac{\int_{-L}^{L}g(x)X_n(x)\mathrm{d}x}{\int_{-L}^{L}X_n(x)X_n(x)\mathrm{d}x} }[/math]

Note that we cannot give the plate an initial velocity that contains a rigid body motions which is why the sum starts at [math]\displaystyle{ n=2 }[/math] for time derivative.


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