User:EGM4313.f14.Team1.Bishop/Report 3

=Report 3=

Problem Statement
Discuss the similarities and differences between the eigenpairs obtained from problem R2.5 and problem R2.6.

Problem Statement
Solve R2.6, but solve for the eigenpairs by setting the second coefficient in each eigenvector to -5, instead of the first coefficient to 1. Normalize the resulting eigenvectors with respect to to the mass matrix, and animate each eigenvector using desmos. Compare the animated eigenvectors in this problem to those in R2.6.

Problem Statement
generalized eigenvalue problem; continuation of R2.8.

use the same structure as in R2.8, with the same stiffness / mass coefficients as in the example in my lecture notes on p.53c-25 (i.e., sec.53c, p.53-21).

the parameters E,A,L for each of the 2 elastic bars are as defined in R2.8.

find the eigenpairs for this problem (set the 1st coefficient in each eigenvector to 1).

do the desmos animation of the 2 modes for both this problem.

discuss in detail (similarities and differences) the results for 3 cases: (1) the example in my lecture notes, (2) the structure in R3.4, and (3) the structure in the current problem.

the minimum is to observe the natural frequencies and the amplitudes of the modes, together with explanations for these observations (just like i did in class).

using the same initial conditions, and apply the same “case 1” forces (sec.53c, Eq.(1)-(2) p.53-22, h(t) = 1), as in the example in my lecture notes, find and solve the modal equations, satisfy the modal initial conditions, then combine the modal displacements to obtain the displacements for the MDOF system of this problem.

compare the displacements for 3 cases: (1) the example in my lecture notes, (2) the structure in R3.4, and (3) the structure in the current problem. plots and animations (if applicable, screenshots and web address) should be provided.

Solution
Use FBD from Problem 2.8 of Report 2.



Stiffness Coefficients(Spring Constants)
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$$
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k_1 = 14 - \sqrt{6}, \ k_2 = \sqrt{6}, \ k_3 = 3 - \sqrt{6} $$
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Mass Coefficients
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$$
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m_1 = 2, m_2 = \frac12 $$
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Find the Eigenvalues using the GEP (Generalized Eigenvalue Problem)


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$$
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\mathbf K \boldsymbol \phi = \beta \mathbf M \boldsymbol \phi $$
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Rewrite as:
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$$[ \mathbf K - \beta \mathbf M ] \boldsymbol \phi = \mathbf 0 $$
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To find non-trivial (non-zero) solutions for column matrix Φ, we must have


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$$ $$
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 * \mathbf K - \beta \mathbf M | = 0
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Expand Determinant
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$$ \begin{vmatrix} K_{11} - \beta m_1 & K_{12} \\ K_{21} & K_{22} - \beta m_2 \end{vmatrix} = 0 $$    (1)
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Find the Eigenvectors using the GEP
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$$ \mathbf K \boldsymbol \phi = \beta \mathbf M \boldsymbol \phi $$
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Expand the GEP


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$$ \begin{bmatrix} K_{11} & K_{12} \\ K_{21} & K_{22} \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} =
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\beta \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} $$
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Plug in values for constants and values obtained through the GEP for the Eigenvalues to Solve for Phi


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$$ ( \mathbf K - \beta_1 \mathbf M ) \boldsymbol \phi_1 = \mathbf 0 $$
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$$ \begin{bmatrix} K_{11} - \beta_1 * m_1 & K_{12} \\ K_{21} & K_{22} - \beta_1 * m_2 \end{bmatrix} \begin{Bmatrix} \phi_{11} \\
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\phi_{12} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$    (2)
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Simplify to Solve for $$ \phi_{12} $$. Remember $$ \phi_{11} $$ is given:


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$$ (K_{11} - \beta_1 * m_1)(\phi_{11}) + (K_{12})(\phi_{12}) = 0
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$$
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$$ \phi_{12} = \frac{-(K_{11} - \beta_1 * m_1)(\phi_{11})}{ (K_{12})}
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$$
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From R2.8, we know that the equation of motion is


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$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}   4 & -2 \\ -2 & 4  \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & \frac{91}{9} \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
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So, $$\mathbf M = \begin{bmatrix}  1.3 & 0 \\ 0 &  1.3 \end{bmatrix} $$ and $$ \mathbf K = \begin{bmatrix} 10 & -5 \\ -5 & \frac{91}{9} \end{bmatrix}$$.

Substituting into (1) gives


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$$ \begin{vmatrix} 10 - 1.3\beta & -5 \\ -5 & \frac{91}{9} - 1.3\beta \end{vmatrix} = 0 $$
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Which is equivalent to the quadratic equation


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$$ \frac{169\beta^2}{100} - \frac{2353\beta}{90} + \frac{685}{9} = 0 $$
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Solving for $$\beta$$ gives two solutions: the two eigenvalues.


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$$ \beta_1=\frac{-5(\sqrt{8101}-181)}{117} $$
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$$ \beta_2=\frac{5(\sqrt{8101}+181)}{117} $$


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Substituting $$\beta_1$$ into (2) gives


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$$ \begin{bmatrix} \frac{-1+\sqrt{8101}}{18} & -5 \\ -5 & \frac{1+\sqrt{8101}}{18} \end{bmatrix} \begin{Bmatrix} 1 \\ \phi_{12} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
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Solving for $$ \phi_{12}$$ gives


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$$ \frac{-1+\sqrt{8101}}{18} -5(\phi_{12}) = 0 $$
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$$ \phi_{12} = \frac{-1+\sqrt{8101}}{90} $$
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So the first eigenpair is $$\beta_1=\frac{-5(\sqrt{8101}-181)}{117}, \boldsymbol \phi_1=\begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix}$$.

Now we do the same with $$\beta_2$$ to get $$\boldsymbol \phi_2$$

Substituting $$\beta_2$$ into (2) gives


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$$ \begin{bmatrix} \frac{-1-\sqrt{8101}}{18} & -5 \\ -5 & \frac{1-\sqrt{8101}}{18} \end{bmatrix} \begin{Bmatrix} 1 \\ \phi_{22} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
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Solving for $$ \phi_{12}$$ gives


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$$ \frac{-1-\sqrt{8101}}{18} -5(\phi_{22}) = 0 $$
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$$ \phi_{22} = \frac{-1-\sqrt{8101}}{90} $$
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So the second eigenpair is $$\beta_2=\frac{5(\sqrt{8101}+181)}{117}, \boldsymbol \phi_2=\begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix}$$.

Now, to get the modes of the system, we need to normalize the eigenvectors. To normalize the eigenvectors, we need to find the modal mass of each eigenvector. The modal mass is calculated by the equation


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$$ \overline {m_i} = \boldsymbol \phi_i^T \mathbf M \boldsymbol \phi_i $$     (3)
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Where $$i$$ is $$1$$ or $$2$$, depending on which eigenvector is being normalized. Once the modal mass is calculated, the eigenvector is normalized by the equation


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$$ \overline {\boldsymbol \phi_i} = \frac{\boldsymbol \phi_i}{\sqrt{\overline{m_i}}} $$     (4)
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Starting by subsituting $$\boldsymbol \phi_1$$ into (3).


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$$ \overline {m_1} = \begin{bmatrix} 1 & \frac{-1+\sqrt{8101}}{90} \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix} $$
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$$ \overline {m_1} = \begin{bmatrix} 1 & \frac{-1+\sqrt{8101}}{90} \end{bmatrix} \begin{Bmatrix} 1.3 \\ \frac{13(\sqrt{8101}-1}{900} \end{Bmatrix} $$
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$$ \overline {m_1} = 1.3 + \frac{-13(\sqrt{8101}-4051)}{40500} = \frac{105313-13\sqrt{8101}}{40500} \approx 2.5714 $$
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So the first modal mass is $$\frac{105313-13\sqrt{8101}}{40500}$$. We calculate the second modal mass the same way.


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$$ \overline {m_2} = \begin{bmatrix} 1 & \frac{-1-\sqrt{8101}}{90} \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix} $$
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$$ \overline {m_2} = \begin{bmatrix} 1 & \frac{-1-\sqrt{8101}}{90} \end{bmatrix} \begin{Bmatrix} 1.3 \\ \frac{-13(\sqrt{8101}+1}{900} \end{Bmatrix} $$
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$$ \overline {m_2} = 1.3 + \frac{13(\sqrt{8101}+4051)}{40500} = \frac{105313+13\sqrt{8101}}{40500} \approx 2.6292 $$
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So the second modal mass is $$\frac{105313+13\sqrt{8101}}{40500}$$.

We can now normalize the eigenvectors by substituting each eigenvector and its respective modal mass into (4).


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$$ \overline {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313-13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \end{Bmatrix} \approx \begin{Bmatrix} 0.6236 \\ 0.6099 \end{Bmatrix} $$
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$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313+13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} \\ \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{Bmatrix} \approx \begin{Bmatrix} 0.6167 \\ -0.6236 \end{Bmatrix} $$
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To summarize,

Mode 1

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$$ \beta_1=\frac{-5(\sqrt{8101}-181)}{117} $$
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$$ {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313-13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \end{Bmatrix} $$
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Animation: https://www.desmos.com/calculator/ueb4vnare4

Mode 2

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$$ \beta_2=\frac{5(\sqrt{8101}+181)}{117} $$
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$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313+13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} \\ \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{Bmatrix} $$
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Animation: https://www.desmos.com/calculator/3mj7z91soy