Waves on a Variable Beam - Revision history

Mike smith: /* Quadrature */

2009-07-02T23:45:28Z

Quadrature

← Older revision		Revision as of 23:45, 2 July 2009
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	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}D(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!</math>		:<math> K_{ik}=\int_{-L}^{L}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!</math>
	:<math> \approx \sum D(x_h){X}_{i}^{''}(x_h) {X}_{k}^{''}(x_h)w_h \,\!</math>		:<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>
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	<math> H =		<math> H =
	\begin{bmatrix}		\begin{bmatrix}
	D(x_1)w_1 & 0 & ... &0 \\		\beta(x_1)w_1 & 0 & ... &0 \\
	0 & D(x_2)w_2 & ... &0 \\		0 & \beta(x_2)w_2 & ... &0 \\
	\vdots & & \ddots&\vdots \\		\vdots & & \ddots&\vdots \\
	0 & ... & ...&D(x_h)w_h		0 & ... & ...&\beta(x_h)w_h
	\end{bmatrix}		\end{bmatrix}
	\,\!</math>		\,\!</math>
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	<math> J =		<math> J =
	\begin{bmatrix}		\begin{bmatrix}
	m(x_1)w_1 & 0 & ... &0 \\		\gamma(x_1)w_1 & 0 & ... &0 \\
	0 & m(x_2)w_2 & ... &0 \\		0 & \gamma(x_2)w_2 & ... &0 \\
	\vdots & & \ddots&\vdots \\		\vdots & & \ddots&\vdots \\
	0 & ... & ...&m(x_h)w_h		0 & ... & ...&\gamma(x_h)w_h
	\end{bmatrix}		\end{bmatrix}
	\,\!</math>		\,\!</math>

Mike smith: /* Converting to Matrix Form */

2009-07-02T23:44:12Z

Converting to Matrix Form

← Older revision		Revision as of 23:44, 2 July 2009
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	If we define		If we define
	<center>		<center>
	:<math> K_{ik}=\int_{-L}^{L}D(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!</math>		:<math> K_{ik}=\int_{-L}^{L}\beta(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x \,\!</math>
	:<math> M_{ik}=\int_{-L}^{L} m(x){X}_{i}{X}_{k}\mathrm{d}x \,\!</math>		:<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>:

Mike smith: /* Solution for a Non-uniform free beam in eigenfunctions */

2009-07-02T23:43:52Z

Solution for a Non-uniform free beam in eigenfunctions

Mike smith: /* Variational Techniques */

2009-07-02T23:42:08Z

Variational Techniques

← Older revision		Revision as of 23:42, 2 July 2009
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	<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[\zeta]=\frac{D}{2}\int_{-L}^{L}\left(\frac{\partial^2 v}{\partial x^2} \right)^2 \mathrm{d}x </math></center>		<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[\zeta]=\int_{-L}^{L}m \zeta^2 \mathrm{d}x=1 </math></center>		<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[\zeta]=\frac{D}{2}\int_{-L}^{L}\left(\frac{\partial^2 \zeta}{\partial x^2} \right)^2 \mathrm{d}x </math>		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}m \zeta^2 \mathrm{d}x=1 </math></center>		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.

Meylan: /* Non-uniform free beam */

2009-04-09T10:43:43Z

Non-uniform free beam

← Older revision		Revision as of 10:43, 9 April 2009
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	<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( D(x) \frac{\partial^{2}\zeta}{\partial x^{2}}\right) +m(x)\frac{\partial^{2}\zeta}{\partial t^{2}}=0 \,\!</math></center>		<center><math> \frac{\partial^2}{\partial x^2}\left( D(x) \frac{\partial^{2}\zeta}{\partial x^{2}}\right) +m(x)\frac{\partial^{2}\zeta}{\partial t^{2}}=0 \,\!</math></center>

Meylan: /* Solution for a uniform beam in eigenfunctions */

2009-04-09T10:39:00Z

Solution for a uniform beam in eigenfunctions

← Older revision		Revision as of 10:39, 9 April 2009
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	{{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}}

Adi Kurniawan: /* References */

2009-04-09T06:59:47Z

References

← Older revision		Revision as of 06:59, 9 April 2009
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	==References==		==References==
	[[Lanczos 1949]]		[[Lanczos 1949]],
	[[Linton and McIver 2001]]		[[Linton and McIver 2001]],
	[[Rao 1986]]		[[Rao 1986]]

	[[Category:Complete Pages]]		[[Category:Complete Pages]]

Meylan: /* Converting to Matrix Form */

2009-04-08T10:57:44Z

Converting to Matrix Form

← Older revision		Revision as of 10:57, 8 April 2009
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	</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[\~~vec~~{a}]=\sum_{i=1}^{N}(K_{ik}-\mu_i M_{ik})a_i=0 \,\!</math></center>		<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> \~~vec~~{a} \,\!</math>. We can consequently express the above sum in the following way:		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)\~~vec~~{a}=~~\vec{~~0} \,\!</math></center>		<center><math> (K-\Lambda M)\mathbf{a} = 0 \,\!</math></center>
	So we can solve for <math> \~~vec~~{a} \,\!</math> in any of the problems below to obtain coefficients <math> a_i \,\!</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)\~~vec~~{a}=~~\vec{~~0} \,\!</math>		:<math> (K-\Lambda M)\mathbf{a} = 0 \,\!</math>
	:<math> K\~~vec~~{a}=\Lambda M\~~vec~~{a} \,\!</math>		:<math> K\mathbf{a}=\Lambda M\mathbf{a} \,\!</math>
	:<math> (M^{-1}K)\~~vec~~{a}=\Lambda \~~vec~~{a} \,\!</math>		:<math> (M^{-1}K)\mathbf{a}=\Lambda \mathbf{a} \,\!</math>
	</center>		</center>

Meylan: /* Non-uniform free beam */

2009-04-08T10:55:00Z

Non-uniform free beam

← Older revision		Revision as of 10:55, 8 April 2009
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	== Non-uniform free beam ==		== Non-uniform free beam ==
	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( D(x) \frac{\partial^{2}v}{\partial x^{2}}\right) +m(x)\frac{\partial^{2}v}{\partial t^{2}}=0 \,\!</math></center>		<center><math> \frac{\partial^2}{\partial x^2}\left( D(x) \frac{\partial^{2}\zeta}{\partial x^{2}}\right) +m(x)\frac{\partial^{2}\zeta}{\partial t^{2}}=0 \,\!</math></center>
	Using separation of variables, where <math> v(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>):		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( D(x) \frac{\partial^{2} \widehat{X}_n}{\partial x^{2}}\right)=m(x)\mu_n \widehat{X}_n \,\!</math></center>		<center><math> \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></center>
	which leads us to the following variational expression:		which leads us to the following variational expression:

Meylan: /* Variational Techniques */

2009-04-08T10:50:38Z

Variational Techniques

← Older revision		Revision as of 10:50, 8 April 2009
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	== 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[v]=\frac{\varepsilon[v]}{H[v]} </math></center>		<center><math> R[\zeta]=\frac{\varepsilon[\zeta]}{H[\zeta]} </math></center>
	where <math> \varepsilon[v] \,\!</math> is known as the energy functional, and <math> H[v] \,\!</math> is some constraint.		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[v]=\frac{D}{2}\int_{-L}^{L}\left(\frac{\partial^2 v}{\partial x^2} \right)^2 \mathrm{d}x </math></center>		<center><math> \varepsilon[\zeta]=\frac{D}{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[v]=\int_{-L}^{L}m v^2 \mathrm{d}x=1 </math></center>		<center><math> H[\zeta]=\int_{-L}^{L}m \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[v]=\frac{D}{2}\int_{-L}^{L}\left(\frac{\partial^2 v}{\partial x^2} \right)^2 \mathrm{d}x </math>		min <math> R[\zeta]=\frac{D}{2}\int_{-L}^{L}\left(\frac{\partial^2 \zeta}{\partial x^2} \right)^2 \mathrm{d}x </math>
	subject to <math> \int_{-L}^{L}m v^2 \mathrm{d}x=1 </math></center>		subject to <math> \int_{-L}^{L}m \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.

← Older revision		Revision as of 23:43, 2 July 2009
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	== Solution for a Non-uniform free beam in eigenfunctions ==		== 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( D(x) \frac{\partial^{2}\zeta}{\partial x^{2}}\right) +m(x)\frac{\partial^{2}\zeta}{\partial t^{2}}=0 \,\!</math></center>		<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> \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>):		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( D(x) \frac{\partial^{2} \widehat{X}_n}{\partial x^{2}}\right)=m(x)\mu_n \widehat{X}_n \,\!</math></center>		<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\{D(x)(\widehat{X}_n^{''})^2 -\mu_n m(x) \widehat{X}_n^2 \bigg\}\mathrm{d}x </math>		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\{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\}\mathrm{d}x \,\!</math></center>		<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}D(x) {X}_{i}^{''} {X}_{k}^{''}\mathrm{d}x -\mu_i \int_{-L}^{L} m(x){X}_{i}{X}_{k}\mathrm{d}x\right\}a_i=0 \,\!</math></center>		<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 [http://en.wikipedia.org/wiki/Lagrange_multipliers 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)