# Waves on a Variable Beam

## Introduction

We consider here the equations for a non-uniform free beam. We begin by presenting the theory for a uniform beam.<br\>

An example of the motion for a non-uniform beam is demonstrated below:

## 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

$\partial_x^2\left(\beta(x)\partial_x^2 \zeta\right) + \gamma(x) \partial_t^2 \zeta = p$

where $\beta(x)$ is the non dimensionalised flexural rigidity, and $\gamma$ is non-dimensionalised linear mass density function. Note that this equations simplifies if the plate has constant properties (and that $h$ is the thickness of the plate, $p$ is the pressure and $\zeta$ 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).

$\partial_x^2 \zeta = 0, \,\,\partial_x^3 \zeta = 0$

at the edges of the plate.

The problem is subject to the initial conditions

$\zeta(x,0)=f(x) \,\!$
$\partial_t \zeta(x,0)=g(x)$

## Solution for a uniform beam in eigenfunctions with no pressure

If the beam is uniform the equations can be written as

$\beta \frac{\partial^{4}\zeta}{\partial x^{4}} + \gamma \frac{\partial^{2}\zeta}{\partial t^{2}}=0$

We can express the deflection as the series

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

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

Then $A_n \,\!$ and $B_n \,\!$ can be found using orthogonality properties:

$A_n=\frac{\int_{-L}^{L}f(x)X_n(x)\mathrm{d}x}{\int_{-L}^{L}X_n(x)X_n(x)\mathrm{d}x} \,\!$
$B_n=\frac{\int_{-L}^{L}g(x)X_n(x)\mathrm{d}x}{\int_{-L}^{L}X_n(x)X_n(x)\mathrm{d}x}$

Note that we cannot give the plate an initial velocity that contains a rigid body motions which is why the sum starts at $n=2$ for time derivative.

## Variational Techniques

As an alternative to solving the eigenvalue problem, we can equivalently minimise the Rayleigh Quotient (Linton and McIver 2001):

$R[\zeta]=\frac{\varepsilon[\zeta]}{H[\zeta]}$

where $\varepsilon[\zeta] \,\!$ is known as the energy functional, and $H[\zeta] \,\!$ is some constraint.

In the case of a uniform beam with zero transverse load, our energy functional is (Lanczos 1949):

$\varepsilon[\zeta]=\frac{\beta}{2}\int_{-L}^{L}\left(\frac{\partial^2 v}{\partial x^2} \right)^2 \mathrm{d}x$

If we choose

$H[\zeta]=\int_{-L}^{L}\gamma \zeta^2 \mathrm{d}x=1$

Then minimizing the Rayleigh Quotient can be expressed as a variational problem

min $R[\zeta]=\frac{\beta}{2}\int_{-L}^{L}\left(\frac{\partial^2 \zeta}{\partial x^2} \right)^2 \mathrm{d}x$

subject to $\int_{-L}^{L}\gamma \zeta^2 \mathrm{d}x=1$

which will in turn, also solve our eigenvalue problem.

## Rayleigh-Ritz method

We can essentially replace the variational problem of finding a $y(x) \,\!$ that extremises $J[y] \,\!$ to finding a set of constants $a_1, a_2, ..., a_N \,\!$ which extremise $J[a_1, a_2, ..., a_N] \,\!$, through using the Rayleigh-Ritz method.

We solve

$\frac{\partial}{\partial a_k}J[a_1, a_2, ..., a_N]=0 \,\!$

for all $k=1,2,...,N \,\!$.

In our example of non-uniform beams, the assumption is made that we are able to approximate the eigenfunctions for a non-uniform beam as a linear combination of the eigenfunctions for a linear beam:

$\widehat{X}_n=\sum_{i=1}^{N}a_{i}X_{i}(x) \,\!$

## Solution for a Non-uniform free beam in eigenfunctions

In the case of a non-uniform free beam, the Euler-Bernoulli beam equation becomes:

$\frac{\partial^2}{\partial x^2}\left( \beta(x) \frac{\partial^{2}\zeta}{\partial x^{2}}\right) +\gamma(x)\frac{\partial^{2}\zeta}{\partial t^{2}}=0 \,\!$

Using separation of variables, where $\zeta(x,t)=\widehat{X}(x)\widehat{T}(t) \,\!$, we obtain a corresponding eigenfunction problem (with eigenvalues denoted by $\mu_n=\widehat{k}_n^4 \,\!$):

$\frac{\partial^2}{\partial x^2}\left( \beta(x) \frac{\partial^{2} \widehat{X}_n}{\partial x^{2}}\right)=\gamma(x)\mu_n \widehat{X}_n \,\!$

which leads us to the following variational expression:

min $J[\widehat{X}_n]=\frac{1}{2}\int_{-L}^{L} \bigg\{\beta(x)(\widehat{X}_n^{''})^2 -\mu_n \gamma(x) \widehat{X}_n^2 \bigg\}\mathrm{d}x$

We can approximate this using Rayleigh-Ritz and obtain:

$J[\vec{a}]=\frac{1}{2}\int_{-L}^{L} \bigg\{\beta(x)[\sum_{i=1}^{N}a_{i}{X}_{i}^{''}]^2 -\mu_i \gamma(x) [\sum_{i=1}^{N}a_{i}{X}_{i}]^2 \bigg\}\mathrm{d}x \,\!$

Then extremising $J[\vec{a}] \,\!$ as follows

$\frac{\partial}{\partial a_k}J[\vec{a}]=0 \,\!$

for all $k=1,2,...,N \,\!$, allows us to obtain:

$\frac{\partial}{\partial a_k}J[\vec{a}]=\sum_{i=1}^{N} \left\{ \int_{-L}^{L}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x -\mu_i \int_{-L}^{L} \gamma(x){X}_{i}{X}_{k}\mathrm{d}x\right\}a_i=0 \,\!$

for all $k=1,2,...N \,\!$ (here $\mu_i \,\!$ denotes both the Lagrange multiplier and eigenvalues of a non-uniform beam)

## Converting to Matrix Form

If we define

$K_{ik}=\int_{-L}^{L}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!$
$M_{ik}=\int_{-L}^{L} \gamma(x){X}_{i}{X}_{k}\mathrm{d}x \,\!$

then for all $k=1,2,...N \,\!$:

$\frac{\partial}{\partial a_k}J[\mathbf{a}]=\sum_{i=1}^{N}(K_{ik}-\mu_i M_{ik})a_i=0 \,\!$

Let us create the matrix $K \,\!$, with elements $K_{ik} \,\!$, the matrix $M \,\!$, with elements $M_{ik} \,\!$, the sparse matrix $\Lambda \,\!$ with $\mu_i \,\!$ terms on the diagonal, and the vector $\mathbf{a} \,\!$. We can consequently express the above sum in the following way:

$(K-\Lambda M)\mathbf{a} = 0 \,\!$

So we can solve for $\mathbf{a} \,\!$ in any of the problems below to obtain coefficients $a_i \,\!$:

$(K-\Lambda M)\mathbf{a} = 0 \,\!$
$K\mathbf{a}=\Lambda M\mathbf{a} \,\!$
$(M^{-1}K)\mathbf{a}=\Lambda \mathbf{a} \,\!$

Rather than tediously evaluating an integral for each element of the matrices $K \,\!$ and $M \,\!$, we can break up the integral into subintervals with weights $w_h \,\!$ using Composite Simpson's Rule:

$\int_a^b f(x) dx \approx \frac{h}{3} \left[ f(x_0)+2 \sum_{j=1}^{n/2-1}f(x_{2j})+4\sum_{j=1}^{n/2}f(x_{2j-1})+f(x_n)\right] \,\!$
$\approx \frac{h}{3}\left[ f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+4f(x_{n-1})+f(x_n)\right] \,\!$

So we can express $K_{ik} \,\!$ as follows:

$K_{ik}=\int_{-L}^{L}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!$
$\approx \sum \beta(x_h){X}_{i}^{''}(x_h) {X}_{k}^{''}(x_h)w_h \,\!$
$\approx \vec{X}_{i}^{''} H \vec{X}_{k}^{''T} \,\!$

where

$\vec{X}_i^{''}=[X_i^{''}(x_1),X_i^{''}(x_2),...,X_i^{''}(x_h)] \,\!$

$H = \begin{bmatrix} \beta(x_1)w_1 & 0 & ... &0 \\ 0 & \beta(x_2)w_2 & ... &0 \\ \vdots & & \ddots&\vdots \\ 0 & ... & ...&\beta(x_h)w_h \end{bmatrix} \,\!$

and weights $w_h \,\!$ are defined as: $w_1=h/3, w_2=4h/3, ... , w_h=h/3 \,\!$.

We can extend this concept to the the full matrix $K \,\!$ if we form:

$K=X^{''}_{mat}H X^{''T}_{mat} \,\!$

where

$X^{''}_{mat} = \begin{bmatrix} {X}_{1}^{''}(x_1) & {X}_{1}^{''}(x_2) & ... & {X}^{''}_{1}(x_h) \\ {X}_{2}^{''}(x_1) & {X}_{2}^{''}(x_2) & ... & {X}^{''}_{2}(x_h) \\ \vdots & \vdots & & \vdots \\ {X}_{N}^{''}(x_1) & {X}_{N}^{''}(x_2) & ... & {X}^{''}_{N}(x_h) \end{bmatrix} \,\!$

For $M \,\!$ we use an identical approach:

$M=X_{mat}J X^{T}_{mat} \,\!$

where

$J = \begin{bmatrix} \gamma(x_1)w_1 & 0 & ... &0 \\ 0 & \gamma(x_2)w_2 & ... &0 \\ \vdots & & \ddots&\vdots \\ 0 & ... & ...&\gamma(x_h)w_h \end{bmatrix} \,\!$

$X_{mat} = \begin{bmatrix} {X}_{1}(x_1) & {X}_{1}(x_2) & ... & {X}_{1}(x_h) \\ {X}_{2}(x_1) & {X}_{2}(x_2) & ... & {X}_{2}(x_h) \\ \vdots & \vdots & & \vdots \\ {X}_{N}(x_1) & {X}_{N}(x_2) & ... & {X}_{N}(x_h) \end{bmatrix} \,\!$

## Evaluation

Using the MATLAB program eig.m, we can solve the eigenvalue problem

$K\vec{a}=\Lambda M\vec{a} \,\!$

using the either of the commands

[Ai_values,nonuniform_eigvals] = eig(M^(-1)*K)
[Ai_values,nonuniform_eigvals] = eig(K,M)

Where the diagonal of nonuniform_eigvals denotes the eigenvalues for the nonuniform beam ($\mu_n \,\!$), and each column of the matrix Ai_values represents the coefficients $a_1, a_2,... \,\!$ corresponding to $\widehat{X}_n \,\!$:

$\widehat{X}_n=\sum_{i=1}^{N}a_{i}X_{i}(x) \,\!$

We can obtain the matrix $\widehat{X}_{mat} \,\!$ by taking Xhat_mat= Ai_valuesTX_mat.

## Non-uniform beam revisited

We have already solved the eigenfunction problem. We now turn our attention to the time component arising from separation of variables:

$\widehat{T}_n^{''}+\mu_n\widehat{T}_n=0 \,\!$

which has solutions of the form

$\widehat{T}_n(t)=A_n \cos(\widehat{k}_n^2 t)+B_n \sin(\widehat{k}_n^2 t) \,\!$

Introducing the initial conditions

$v(x,0)=f(x) \,\!$
$\frac{\partial v(x,0)}{\partial t}=g(x)\,\!$

Then using the the first of these initial conditions we obtain:

$\sum_{n=0}^{\infty}A_n\widehat{X}_n(x)=f(x) \,\!$

multiplying by $m(x)\widehat{X}_n(x) \,\!$ and integrating allows us to obtain:

$A_n=\frac{\int_{-L}^{L}f(x)m(x)\widehat{X}_n(x)\mathrm{d}x}{\int_{-L}^{L}m(x)\widehat{X}_n(x)\widehat{X}_n(x)\mathrm{d}x} \,\!$

Then using the second initial condition, and applying the same technique yields

$B_n=\frac{1}{\widehat{k}_n^2}\frac{\int_{-L}^{L}g(x)m(x)\widehat{X}_n(x)\mathrm{d}x}{\int_{-L}^{L}m(x)\widehat{X}_n(x)\widehat{X}_n(x)\mathrm{d}x} \,\!$

Consequently,

$v(x,t)=A_0\widehat{X}_0(x)+A_1\widehat{X}_1(x)+\sum_{n=2}^{\infty} A_n\widehat{X}_n(x)\cos(\widehat{k}_n^2 t)+\sum_{n=2}^{\infty} B_n\widehat{X}_n(x) \sin(\widehat{k}_n^2 t) \,\!$

where $\widehat{X}_0(x) \,\!$ and $\widehat{X}_2(x) \,\!$ are the two rigid modes of the non-uniform beam. Note that $B_0 \,\!$ and $B_1 \,\!$ do not exist.

## Matlab Code

A program to solve for a variable beam can be found here variable_beam.m