Difference between revisions of "Waves on a Variable Beam"
Mike smith (talk | contribs) |
|||
(15 intermediate revisions by 3 users not shown) | |||
Line 8: | Line 8: | ||
{{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 for a Uniform Free Beam|eigenfunctions]] with no pressure == |
{{solution for a uniform beam in eigenfunctions}} | {{solution for a uniform beam in eigenfunctions}} | ||
Line 14: | Line 14: | ||
== Variational Techniques == | == Variational Techniques == | ||
As an alternative to solving the eigenvalue problem, we can equivalently minimise the [http://en.wikipedia.org/wiki/Rayleigh_quotient Rayleigh Quotient] ([[Linton and McIver 2001]]): | As an alternative to solving the eigenvalue problem, we can equivalently minimise the [http://en.wikipedia.org/wiki/Rayleigh_quotient Rayleigh Quotient] ([[Linton and McIver 2001]]): | ||
− | <center><math> R[ | + | <center><math> R[\zeta]=\frac{\varepsilon[\zeta]}{H[\zeta]} </math></center> |
− | where <math> \varepsilon[ | + | where <math> \varepsilon[\zeta] \,\!</math> is known as the energy functional, and <math> H[\zeta] \,\!</math> is some constraint. |
<br /><br /> | <br /><br /> | ||
In the case of a uniform beam with zero transverse load, our energy functional is ([[Lanczos 1949]]): | In the case of a uniform beam with zero transverse load, our energy functional is ([[Lanczos 1949]]): | ||
− | <center><math> \varepsilon[ | + | <center><math> \varepsilon[\zeta]=\frac{\beta}{2}\int_{-L}^{L}\left(\frac{\partial^2 v}{\partial x^2} \right)^2 \mathrm{d}x </math></center> |
If we choose | If we choose | ||
− | <center><math> H[ | + | <center><math> H[\zeta]=\int_{-L}^{L}\gamma \zeta^2 \mathrm{d}x=1 </math></center> |
Then minimizing the Rayleigh Quotient can be expressed as a variational problem <br /> | Then minimizing the Rayleigh Quotient can be expressed as a variational problem <br /> | ||
<center> | <center> | ||
− | min <math> R[ | + | min <math> R[\zeta]=\frac{\beta}{2}\int_{-L}^{L}\left(\frac{\partial^2 \zeta}{\partial x^2} \right)^2 \mathrm{d}x </math> |
− | subject to <math> \int_{-L}^{L} | + | subject to <math> \int_{-L}^{L}\gamma \zeta^2 \mathrm{d}x=1 </math></center> |
which will in turn, also solve our eigenvalue problem. | which will in turn, also solve our eigenvalue problem. | ||
Line 35: | Line 35: | ||
<center><math> \widehat{X}_n=\sum_{i=1}^{N}a_{i}X_{i}(x) \,\!</math></center> | <center><math> \widehat{X}_n=\sum_{i=1}^{N}a_{i}X_{i}(x) \,\!</math></center> | ||
− | == Non-uniform free beam == | + | == Solution for a Non-uniform free beam in eigenfunctions == |
In the case of a non-uniform free beam, the Euler-Bernoulli beam equation becomes: | In the case of a non-uniform free beam, the Euler-Bernoulli beam equation becomes: | ||
− | <center><math> \frac{\partial^2}{\partial x^2}\left( | + | <center><math> \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 \,\!</math></center> |
− | Using separation of variables, where <math> | + | Using separation of variables, where <math> \zeta(x,t)=\widehat{X}(x)\widehat{T}(t) \,\!</math>, we obtain a corresponding eigenfunction problem (with eigenvalues denoted by <math> \mu_n=\widehat{k}_n^4 \,\!</math>): |
− | <center><math> \frac{\partial^2}{\partial x^2}\left( | + | <center><math> \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 \,\!</math></center> |
which leads us to the following variational expression: | which leads us to the following variational expression: | ||
<center> | <center> | ||
− | min <math> J[\widehat{X}_n]=\frac{1}{2}\int_{-L}^{L} \bigg\{ | + | min <math> 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 </math> |
</center> | </center> | ||
We can approximate this using Rayleigh-Ritz and obtain: | We can approximate this using Rayleigh-Ritz and obtain: | ||
− | <center><math> J[\vec{a}]=\frac{1}{2}\int_{-L}^{L} \bigg\{ | + | <center><math> 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 \,\!</math></center> |
Then extremising <math> J[\vec{a}] \,\!</math> as follows | Then extremising <math> J[\vec{a}] \,\!</math> as follows | ||
<center><math> \frac{\partial}{\partial a_k}J[\vec{a}]=0 \,\!</math></center> | <center><math> \frac{\partial}{\partial a_k}J[\vec{a}]=0 \,\!</math></center> | ||
for all <math> k=1,2,...,N \,\!</math>, allows us to obtain: | for all <math> k=1,2,...,N \,\!</math>, allows us to obtain: | ||
− | <center><math> \frac{\partial}{\partial a_k}J[\vec{a}]=\sum_{i=1}^{N} \left\{ \int_{-L}^{L} | + | <center><math> \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 \,\!</math></center> |
− | for all <math> k=1,2,...N \,\!</math> (here <math> \mu_i \,\!</math> denotes both the Lagrange multiplier and eigenvalues of a non-uniform beam) | + | for all <math> k=1,2,...N \,\!</math> (here <math> \mu_i \,\!</math> denotes both the [http://en.wikipedia.org/wiki/Lagrange_multipliers Lagrange multiplier] and eigenvalues of a non-uniform beam) |
== Converting to Matrix Form == | == Converting to Matrix Form == | ||
If we define | If we define | ||
<center> | <center> | ||
− | :<math> K_{ik}=\int_{-L}^{L} | + | :<math> K_{ik}=\int_{-L}^{L}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!</math> |
− | :<math> M_{ik}=\int_{-L}^{L} | + | :<math> M_{ik}=\int_{-L}^{L} \gamma(x){X}_{i}{X}_{k}\mathrm{d}x \,\!</math> |
</center> | </center> | ||
then for all <math> k=1,2,...N \,\!</math>: | then for all <math> k=1,2,...N \,\!</math>: | ||
− | <center><math> \frac{\partial}{\partial a_k}J[\ | + | <center><math> \frac{\partial}{\partial a_k}J[\mathbf{a}]=\sum_{i=1}^{N}(K_{ik}-\mu_i M_{ik})a_i=0 \,\!</math></center> |
− | Let us create the matrix <math> K \,\!</math>, with elements <math> K_{ik} \,\!</math>, the matrix <math> M \,\!</math>, with elements <math> M_{ik} \,\!</math>, the sparse matrix <math> \Lambda \,\!</math> with <math> \mu_i \,\!</math> terms on the diagonal, and the vector <math> \ | + | Let us create the matrix <math> K \,\!</math>, with elements <math> K_{ik} \,\!</math>, the matrix <math> M \,\!</math>, with elements <math> M_{ik} \,\!</math>, the sparse matrix <math> \Lambda \,\!</math> with <math> \mu_i \,\!</math> terms on the diagonal, and the vector <math> \mathbf{a} \,\!</math>. We can consequently express the above sum in the following way: |
− | <center><math> (K-\Lambda M)\ | + | <center><math> (K-\Lambda M)\mathbf{a} = 0 \,\!</math></center> |
− | So we can solve for <math> \ | + | So we can solve for <math> \mathbf{a} \,\!</math> in any of the problems below to obtain coefficients <math> a_i \,\!</math>: |
<center> | <center> | ||
− | :<math> (K-\Lambda M)\ | + | :<math> (K-\Lambda M)\mathbf{a} = 0 \,\!</math> |
− | :<math> K\ | + | :<math> K\mathbf{a}=\Lambda M\mathbf{a} \,\!</math> |
− | :<math> (M^{-1}K)\ | + | :<math> (M^{-1}K)\mathbf{a}=\Lambda \mathbf{a} \,\!</math> |
</center> | </center> | ||
== Quadrature == | == Quadrature == | ||
− | Rather than tediously | + | Rather than tediously evaluating an integral for each element of the matrices <math> K \,\!</math> and <math> M \,\!</math>, we can break up the integral into subintervals with weights <math> w_h \,\!</math> using [http://en.wikipedia.org/wiki/Composite_simpson%27s_rule#Composite_Simpson.27s_rule Composite Simpson's Rule]: |
<center> | <center> | ||
:<math> \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> \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> | ||
Line 77: | Line 77: | ||
So we can express <math> K_{ik} \,\!</math> as follows: | So we can express <math> K_{ik} \,\!</math> as follows: | ||
<center> | <center> | ||
− | :<math> K_{ik}=\int_{-L}^{L} | + | :<math> K_{ik}=\int_{-L}^{L}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!</math> |
− | :<math> \approx \sum | + | :<math> \approx \sum \beta(x_h){X}_{i}^{''}(x_h) {X}_{k}^{''}(x_h)w_h \,\!</math> |
:<math> \approx \vec{X}_{i}^{''} H \vec{X}_{k}^{''T} \,\!</math> | :<math> \approx \vec{X}_{i}^{''} H \vec{X}_{k}^{''T} \,\!</math> | ||
</center> | </center> | ||
Line 85: | Line 85: | ||
<math> H = | <math> H = | ||
\begin{bmatrix} | \begin{bmatrix} | ||
− | + | \beta(x_1)w_1 & 0 & ... &0 \\ | |
− | 0 & | + | 0 & \beta(x_2)w_2 & ... &0 \\ |
\vdots & & \ddots&\vdots \\ | \vdots & & \ddots&\vdots \\ | ||
− | 0 & ... & ...& | + | 0 & ... & ...&\beta(x_h)w_h |
\end{bmatrix} | \end{bmatrix} | ||
\,\!</math> | \,\!</math> | ||
Line 112: | Line 112: | ||
<math> J = | <math> J = | ||
\begin{bmatrix} | \begin{bmatrix} | ||
− | + | \gamma(x_1)w_1 & 0 & ... &0 \\ | |
− | 0 & | + | 0 & \gamma(x_2)w_2 & ... &0 \\ |
\vdots & & \ddots&\vdots \\ | \vdots & & \ddots&\vdots \\ | ||
− | 0 & ... & ...& | + | 0 & ... & ...&\gamma(x_h)w_h |
\end{bmatrix} | \end{bmatrix} | ||
\,\!</math> | \,\!</math> | ||
Line 154: | Line 154: | ||
<center><math> \sum_{n=0}^{\infty}A_n\widehat{X}_n(x)=f(x) \,\!</math></center> | <center><math> \sum_{n=0}^{\infty}A_n\widehat{X}_n(x)=f(x) \,\!</math></center> | ||
multiplying by <math> m(x)\widehat{X}_n(x) \,\!</math> and integrating allows us to obtain: | multiplying by <math> m(x)\widehat{X}_n(x) \,\!</math> and integrating allows us to obtain: | ||
− | <center><math> A_n=\frac{\int_{-L}^{L}f(x)m(x)\widehat{X}_n(x) | + | <center><math> 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} \,\!</math></center> |
Then using the second initial condition, and applying the same technique yields | Then using the second initial condition, and applying the same technique yields | ||
− | <center><math> B_n=\frac{1}{\widehat{k}_n^2}\frac{\int_{-L}^{L}g(x)m(x)\widehat{X}_n(x) | + | <center><math> 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} \,\!</math></center> |
Consequently, | Consequently, | ||
<center><math> 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></center> | <center><math> 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></center> | ||
where <math> \widehat{X}_0(x) \,\!</math> and <math> \widehat{X}_2(x) \,\!</math> are the two rigid modes of the non-uniform beam. Note that <math> B_0 \,\!</math> and <math> B_1 \,\!</math> do not exist. | where <math> \widehat{X}_0(x) \,\!</math> and <math> \widehat{X}_2(x) \,\!</math> are the two rigid modes of the non-uniform beam. Note that <math> B_0 \,\!</math> and <math> B_1 \,\!</math> do not exist. | ||
− | |||
== Matlab Code == | == Matlab Code == | ||
Line 176: | Line 175: | ||
==References== | ==References== | ||
− | [[Lanczos 1949]] | + | [[Lanczos 1949]], |
− | [[Linton and McIver 2001]] | + | [[Linton and McIver 2001]], |
[[Rao 1986]] | [[Rao 1986]] | ||
[[Category:Complete Pages]] | [[Category:Complete Pages]] |
Latest revision as of 23:45, 2 July 2009
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
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).
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 with no pressure
If the beam is uniform the equations can be written as
We can express the deflection as the series
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.
Variational Techniques
As an alternative to solving the eigenvalue problem, we can equivalently minimise the Rayleigh Quotient (Linton and McIver 2001):
where [math]\displaystyle{ \varepsilon[\zeta] \,\! }[/math] is known as the energy functional, and [math]\displaystyle{ H[\zeta] \,\! }[/math] is some constraint.
In the case of a uniform beam with zero transverse load, our energy functional is (Lanczos 1949):
If we choose
Then minimizing the Rayleigh Quotient can be expressed as a variational problem
min [math]\displaystyle{ R[\zeta]=\frac{\beta}{2}\int_{-L}^{L}\left(\frac{\partial^2 \zeta}{\partial x^2} \right)^2 \mathrm{d}x }[/math]
subject to [math]\displaystyle{ \int_{-L}^{L}\gamma \zeta^2 \mathrm{d}x=1 }[/math]which will in turn, also solve our eigenvalue problem.
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
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:
Solution for a Non-uniform free beam in eigenfunctions
In the case of a non-uniform free beam, the Euler-Bernoulli beam equation becomes:
Using separation of variables, where [math]\displaystyle{ \zeta(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]):
which leads us to the following variational expression:
min [math]\displaystyle{ 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 }[/math]
We can approximate this using Rayleigh-Ritz and obtain:
Then extremising [math]\displaystyle{ J[\vec{a}] \,\! }[/math] as follows
for all [math]\displaystyle{ k=1,2,...,N \,\! }[/math], allows us to obtain:
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}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\! }[/math]
- [math]\displaystyle{ M_{ik}=\int_{-L}^{L} \gamma(x){X}_{i}{X}_{k}\mathrm{d}x \,\! }[/math]
then for all [math]\displaystyle{ k=1,2,...N \,\! }[/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{ \mathbf{a} \,\! }[/math]. We can consequently express the above sum in the following way:
So we can solve for [math]\displaystyle{ \mathbf{a} \,\! }[/math] in any of the problems below to obtain coefficients [math]\displaystyle{ a_i \,\! }[/math]:
- [math]\displaystyle{ (K-\Lambda M)\mathbf{a} = 0 \,\! }[/math]
- [math]\displaystyle{ K\mathbf{a}=\Lambda M\mathbf{a} \,\! }[/math]
- [math]\displaystyle{ (M^{-1}K)\mathbf{a}=\Lambda \mathbf{a} \,\! }[/math]
Quadrature
Rather than tediously evaluating 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}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\! }[/math]
- [math]\displaystyle{ \approx \sum \beta(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{ 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} \,\! }[/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:
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:
where
[math]\displaystyle{ 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} \,\! }[/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
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]:
We can obtain the matrix [math]\displaystyle{ \widehat{X}_{mat} \,\! }[/math] 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:
which has solutions of the form
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:
multiplying by [math]\displaystyle{ m(x)\widehat{X}_n(x) \,\! }[/math] and integrating allows us to obtain:
Then using the second initial condition, and applying the same technique yields
Consequently,
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.
Matlab Code
A program to solve for a variable beam can be found here variable_beam.m
Additional code
This program requires