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.

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]

Equations for a uniform beam

If the beam is uniform the equations can be written We model our beam with the Euler-Bernoulli Beam equation

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

where [math]\displaystyle{ D=EI\,\! }[/math] which is the elastic modulus multiplied by the moment of inertia and [math]\displaystyle{ m=\rho h\,\! }[/math] is the linear mass density.

If we use separation of variables:

[math]\displaystyle{ v(x,t)=X(x)T(t) \,\! }[/math]

After some manipulation we obtain

[math]\displaystyle{ \frac{D}{m}\frac{X^{''''}}{X}=-\frac{T^{''}}{T}=\overline{\omega}^{2}_{n} }[/math]

Note that this constant [math]\displaystyle{ \overline{\omega}^{2}_{n} \,\! }[/math] actually denotes the natural frequencies of the beam.

We now end up with two differential equations

[math]\displaystyle{ X_n^{''''}-k_n^4X_n=0 }[/math]

[math]\displaystyle{ T_n^{''}+\overline{\omega}^{2}_{n}T_n=0 }[/math]

where

[math]\displaystyle{ k_n^4=\frac{\overline{\omega}^{2}_{n} m}{D} }[/math]

In the case of a free beam, we have boundary conditions

[math]\displaystyle{ \frac{\partial^2 v}{\partial x^2}(\pm L,t)=0 }[/math]

[math]\displaystyle{ \frac{\partial^3 v}{\partial x^3}(\pm L,t)=0 }[/math]

For the case when [math]\displaystyle{ \lambda_n=k_n^4=0 \,\! }[/math] we obtain

[math]\displaystyle{ X_n(x)=Ax^3+Bx^2+Cx+D \,\! }[/math]

Using the boundary conditions above:

[math]\displaystyle{ X_n(x)=Cx+D \,\! }[/math]

From this we obtain the first two modes [math]\displaystyle{ X_0=1 \,\! }[/math] and [math]\displaystyle{ X_1=x \,\! }[/math]. These are known as the rigid modes (corresponding eigenvalues are zero). After normalising, they become

[math]\displaystyle{ X_0=\frac{1}{\sqrt{2L}} }[/math]

[math]\displaystyle{ X_1=\sqrt{\frac{3}{2L^3}}x }[/math]

For [math]\displaystyle{ \lambda_n \not=0 \,\! }[/math] the general solution is:

[math]\displaystyle{ X_n(x)=A \sin(k_n L)+B\cos(k_n L)+C\sinh(k_n L)+D\cosh(k_n L) \,\! }[/math]

We can split [math]\displaystyle{ X_n(x) \,\! }[/math] up into symmetric (even) and skew-symmetric (odd) modes. For symmetric modes, we set [math]\displaystyle{ A=C=0 \,\! }[/math], use the boundary conditions, normalise and obtain:

[math]\displaystyle{ X_{2n}(x)=\frac{1}{\sqrt{2L}}\left( \frac{\cos(k_n x)}{\cos(k_n L)}+\frac{\cosh(k_n x)}{\cosh(k_n L)}\right) }[/math]

where [math]\displaystyle{ k_n \,\! }[/math] are the roots of [math]\displaystyle{ \tan(k_n L)+\tanh(k_n L)=0 \,\! }[/math], which is the frequency equation for the symmetric modes of a free beam

For anti-symmetric modes, we set [math]\displaystyle{ B=D=0 \,\! }[/math], use the boundary conditions, normalise and obtain:

[math]\displaystyle{ X_{2n+1}(x)=\frac{1}{\sqrt{2L}}\left( \frac{\sin(k_n x)}{\sin(k_n L)}+\frac{\sinh(k_n x)}{\sinh(k_n L)}\right) }[/math]

where [math]\displaystyle{ k_n \,\! }[/math] are the roots of [math]\displaystyle{ \tan(k_n L)-\tanh(k_n L)=0 \,\! }[/math]. This equation is the frequency equation for anti-symmetric modes of a free beam.

For the time component we obtain:

[math]\displaystyle{ T(t)=A \cos (\overline{\omega}_{n}t)+ B \sin (\overline{\omega}_{n}t) }[/math]

We can express the deflection as the series

[math]\displaystyle{ v(x,t)=\sum_{n=0}^{\infty}v_n(x,t) }[/math]

where

[math]\displaystyle{ v_n(x,t)=X_n(x)[A_n \cos (\overline{\omega}_{n}t)+ B_n \sin (\overline{\omega}_{n}t)] }[/math]

which actually represents the characteristic vibration for each frequency (these are known as vibration modes).

If we introduce the initial conditions

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

[math]\displaystyle{ \frac{\partial v(x,0)}{\partial t}=g(x) }[/math]

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
- these modes drop off very quickly (ie [math]\displaystyle{ v_4 \,\! }[/math] oscillates about zero with negligible amplitude), so the higher order vibration modes can be ignored.
- As time progresses ([math]\displaystyle{ t \rightarrow \infty \,\! }[/math]), each mode will vibrate around the zero displacement line with frequency [math]\displaystyle{ \overline{\omega}_{n}\,\! }[/math].
- Having obtained eigenvalues [math]\displaystyle{ k_n \,\! }[/math], the natural frequencies can easily be obtained [math]\displaystyle{ \overline{\omega}_{n}=k_{n}^{2}\sqrt{\frac{D}{m}}\,\! }[/math].

Variational Techniques

As an alternative to solving the eigenvalue problem (\ref{first_eig_prob}), we can equivalently minimise the Rayleigh Quotient \cite{linton}:

[math]\displaystyle{ R[v]=\frac{\varepsilon[v]}{H[v]} }[/math]

where [math]\displaystyle{ \varepsilon[v] \,\! }[/math] is known as the energy functional, and [math]\displaystyle{ H[v] \,\! }[/math] is some constraint.

In the case of a uniform beam with zero transverse load, our energy functional is \cite{lanczos}:

[math]\displaystyle{ \varepsilon[v]=\frac{D}{2}\int_{-L}^{L}\left(\frac{\partial^2 v}{\partial x^2} \right)^2 \mathrm{d}x }[/math]

If we choose

[math]\displaystyle{ H[v]=\int_{-L}^{L}m v^2 \mathrm{d}x=1 }[/math]

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

min [math]\displaystyle{ R[v]=\frac{D}{2}\int_{-L}^{L}\left(\frac{\partial^2 v}{\partial x^2} \right)^2 \mathrm{d}x }[/math]

subject to [math]\displaystyle{ \int_{-L}^{L}m v^2 \mathrm{d}x=1 }[/math]

which will in turn, also solve our eigenvalue problem (\ref{first_eig_prob}).

Rayleigh-Ritz method

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

We solve

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

for all [math]\displaystyle{ k=1,2,...,N \,\! }[/math].

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:

[math]\displaystyle{ \widehat{X}_n=\sum_{i=1}^{N}a_{i}X_{i}(x) \,\! }[/math]

Non-uniform free beam

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

[math]\displaystyle{ \frac{\partial^2}{\partial x^2}\left( D(x) \frac{\partial^{2}v}{\partial x^{2}}\right) +m(x)\frac{\partial^{2}v}{\partial t^{2}}=0 \,\! }[/math]

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

[math]\displaystyle{ \frac{\partial^2}{\partial x^2}\left( D(x) \frac{\partial^{2} \widehat{X}_n}{\partial x^{2}}\right)=m(x)\mu_n \widehat{X}_n \,\! }[/math]

which leads us to the following variational expression:

min [math]\displaystyle{ J[\widehat{X}_n]=\frac{1}{2}\int_{-L}^{L} \bigg\{D(x)(\widehat{X}_n^{''})^2 -\mu_n m(x) \widehat{X}_n^2 \bigg\}dx }[/math]

We can approximate this using Rayleigh-Ritz and obtain:

[math]\displaystyle{ J[\vec{a}]=\frac{1}{2}\int_{-L}^{L} \bigg\{D(x)[\sum_{i=1}^{N}a_{i}{X}_{i}^{''}]^2 -\mu_i m(x) [\sum_{i=1}^{N}a_{i}{X}_{i}]^2 \bigg\}dx \,\! }[/math]

Then extremising [math]\displaystyle{ J[\vec{a}] \,\! }[/math] as follows

[math]\displaystyle{ \frac{\partial}{\partial a_k}J[\vec{a}]=0 \,\! }[/math]

for all [math]\displaystyle{ k=1,2,...,N \,\! }[/math], allows us to obtain:

[math]\displaystyle{ \frac{\partial}{\partial a_k}J[\vec{a}]=\sum_{i=1}^{N} \left\{ \int_{-L}^{L}D(x) {X}_{i}^{''} {X}_{k}^{''}dx -\mu_i \int_{-L}^{L} m(x){X}_{i}{X}_{k}dx\right\}a_i=0 \,\! }[/math]

for all [math]\displaystyle{ k=1,2,...N \,\! }[/math] (here [math]\displaystyle{ \mu_i \,\! }[/math] denotes both the Lagrange multiplier and eigenvalues of a non-uniform beam)

Converting to Matrix Form

If we define

[math]\displaystyle{ K_{ik}=\int_{-L}^{L}D(x) {X}_{i}^{''} {X}_{k}^{''}dx \,\! }[/math]

[math]\displaystyle{ M_{ik}=\int_{-L}^{L} m(x){X}_{i}{X}_{k}dx \,\! }[/math]

then for all [math]\displaystyle{ k=1,2,...N \,\! }[/math]:

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

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

[math]\displaystyle{ (K-\Lambda M)\vec{a}=\vec{0} \,\! }[/math]

So we can solve for [math]\displaystyle{ \vec{a} \,\! }[/math] in any of the problems below to obtain coefficients [math]\displaystyle{ a_i \,\! }[/math]:

[math]\displaystyle{ (K-\Lambda M)\vec{a}=\vec{0} \,\! }[/math]

[math]\displaystyle{ K\vec{a}=\Lambda M\vec{a} \,\! }[/math]

[math]\displaystyle{ (M^{-1}K)\vec{a}=\Lambda \vec{a} \,\! }[/math]

Quadrature

Rather than tediously evaluate an integral for each element of the matrices [math]\displaystyle{ K \,\! }[/math] and [math]\displaystyle{ M \,\! }[/math], we can break up the integral into subintervals with weights [math]\displaystyle{ w_h \,\! }[/math] using Composite Simpson's Rule:

[math]\displaystyle{ \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] \,\! }[/math]

[math]\displaystyle{ \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] \,\! }[/math]

So we can express [math]\displaystyle{ K_{ik} \,\! }[/math] as follows:

[math]\displaystyle{ K_{ik}=\int_{-L}^{L}D(x) {X}_{i}^{''} {X}_{k}^{''}dx \,\! }[/math]

[math]\displaystyle{ \approx \sum D(x_h){X}_{i}^{''}(x_h) {X}_{k}^{''}(x_h)w_h \,\! }[/math]

[math]\displaystyle{ \approx \vec{X}_{i}^{''} H \vec{X}_{k}^{''T} \,\! }[/math]

where

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

[math]\displaystyle{ H = \begin{bmatrix} D(x_1)w_1 & 0 & ... &0 \\ 0 & D(x_2)w_2 & ... &0 \\ \vdots & & \ddots&\vdots \\ 0 & ... & ...&D(x_h)w_h \end{bmatrix} \,\! }[/math]

and weights [math]\displaystyle{ w_h \,\! }[/math] are defined as: [math]\displaystyle{ w_1=h/3, w_2=4h/3, ... , w_h=h/3 \,\! }[/math].

We can extend this concept to the the full matrix [math]\displaystyle{ K \,\! }[/math] if we form:

[math]\displaystyle{ K=X^{''}_{mat}H X^{''T}_{mat} \,\! }[/math]

where

[math]\displaystyle{ 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} \,\! }[/math]

For [math]\displaystyle{ M \,\! }[/math] we use an identical approach:

[math]\displaystyle{ M=X_{mat}J X^{T}_{mat} \,\! }[/math]

where

[math]\displaystyle{ J = \begin{bmatrix} m(x_1)w_1 & 0 & ... &0 \\ 0 & m(x_2)w_2 & ... &0 \\ \vdots & & \ddots&\vdots \\ 0 & ... & ...&m(x_h)w_h \end{bmatrix} \,\! }[/math]

[math]\displaystyle{ 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} \,\! }[/math]

Evaluation

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

[math]\displaystyle{ K\vec{a}=\Lambda M\vec{a} \,\! }[/math]

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 ([math]\displaystyle{ \mu_n \,\! }[/math]), and each column of the matrix Ai_values represents the coefficients [math]\displaystyle{ a_1, a_2,... \,\! }[/math] corresponding to [math]\displaystyle{ \widehat{X}_n \,\! }[/math]:

[math]\displaystyle{ \widehat{X}_n=\sum_{i=1}^{N}a_{i}X_{i}(x) \,\! }[/math]

We can obtain the matrix [math]\displaystyle{ \widehat{X}_{mat} \,\! }[/math] by taking Xhat_mat= Ai_values^TX_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:

[math]\displaystyle{ \widehat{T}_n^{''}+\mu_n\widehat{T}_n=0 \,\! }[/math]

which has solutions of the form

[math]\displaystyle{ \widehat{T}_n(t)=A_n \cos(\widehat{k}_n^2 t)+B_n \sin(\widehat{k}_n^2 t) \,\! }[/math]

Introducing the initial conditions

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

[math]\displaystyle{ \frac{\partial v(x,0)}{\partial t}=g(x)\,\! }[/math]

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

[math]\displaystyle{ \sum_{n=0}^{\infty}A_n\widehat{X}_n(x)=f(x) \,\! }[/math]

multiplying by [math]\displaystyle{ m(x)\widehat{X}_n(x) \,\! }[/math] and integrating allows us to obtain:

[math]\displaystyle{ A_n=\frac{\int_{-L}^{L}f(x)m(x)\widehat{X}_n(x)dx}{\int_{-L}^{L}m(x)\widehat{X}_n(x)\widehat{X}_n(x)dx} \,\! }[/math]

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

[math]\displaystyle{ B_n=\frac{1}{\widehat{k}_n^2}\frac{\int_{-L}^{L}g(x)m(x)\widehat{X}_n(x)dx}{\int_{-L}^{L}m(x)\widehat{X}_n(x)\widehat{X}_n(x)dx} \,\! }[/math]

Consequently,

[math]\displaystyle{ 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) \,\! }[/math]

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

To Do List

MATLAB image of subplots
MATLAB code wikiwaves, em_beam_dd.m,ev_beam,em_beam needed
References for authors/sources, as well as for formulas inside a page (ie refer to formula 33)