User:EML4507.s13.team4ever.Bonner/Bonnerreport5p6

Problem R5.6
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Description
We are to solve Pb-53.6 p.53-13b over again using the modal superposition method.

Given:
We are given from Pb-53.6 that:

$$ m_1 = 3 $$,   $$m_2 = 2 $$

$$k_1 = 10 $$,   $$k_2 = 20 $$,    $$k_3 = 15 $$

$$c_1 = 1/2 $$,   $$c_2 = 1/4 $$,    $$ c_3 = 1/3 $$

$$F_1(t) = 0 $$,  $$F_2(t) = 0 $$

We also know from Pb-53.9 that:

$$ \lambda_1 = 4.765 $$

$$ \lambda_2 = 22.735 $$

$$ \phi_1 = \begin{bmatrix} 1.273\\ 1\\ \end{bmatrix} $$

$$ \phi_2 = \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} $$

$$ M= \begin{bmatrix} m_1 & 0\\ 0 & m_2\\ \end{bmatrix} $$ and $$ K= \begin{bmatrix} k_1 + k_2 & -k_2\\ -k_2 & k_2 + k_3\\ \end{bmatrix} $$ and $$ C= \begin{bmatrix} c_1 + c_2 & -c_2\\ -c_2 & c_2 + c_3\\ \end{bmatrix} $$

Therefore, we have:

$$ M= \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} $$ and $$ K= \begin{bmatrix} 30 & -20\\ -20 & 35\\ \end{bmatrix} $$ and $$ C= \begin{bmatrix} 3/4 & -1/4\\ -1/4 & 7/12\\ \end{bmatrix} $$

Solution:
From Pb-53.9, we already know the eigenvectors. To solve the problem using superposition, we know only need to find the modal coordinates of d(t) corresponding to the mode shapes. That is to say, we need to find z so as to solve the equation:

$$ d(t) = z_1 \phi_1 + z_2 \phi_2 $$

We can get this from:

$$ z^{''}_j + \bar{\phi}^T_i C \bar{\phi}_j z^{'}_j + (w_j)^2 z_j = \bar{\phi}^T_j F(t) $$

Since F(t) = 0, the above differential equation becomes homogeneous. However, we still need to find $$ \bar{\phi}_1 $$ and $$ \bar{\phi}_2 $$. These can be found from the equation:

$$ \bar{\phi}_i = \frac{\phi_i}{\sqrt{\phi^T_i M \phi_i}}   $$, therefore,

$$ \bar{\phi}_1 = \frac{ \begin{bmatrix} 1.273\\ 1\\ \end{bmatrix} }{\sqrt{\begin{bmatrix} 1.273 & 1\\ \end{bmatrix} \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} \begin{bmatrix} 1.273\\ 1\\ \end{bmatrix} }} = \frac{ \begin{bmatrix} 1.273\\ 1\\ \end{bmatrix} }{\sqrt{6.86 }} = \frac{ \begin{bmatrix} 1.273\\ 1\\ \end{bmatrix} }{2.619} = \begin{bmatrix} 0.485\\ 0.381\\ \end{bmatrix} $$

$$ \bar{\phi}_2 = \frac{ \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} }{\sqrt{\begin{bmatrix} -0.5234 & 1\\ \end{bmatrix} \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} }} = \frac{ \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} }{\sqrt{2.82 }} = \frac{ \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} }{1.68} = \begin{bmatrix} -0.311\\ 0.595\\ \end{bmatrix} $$

Remembering that $$ (w_j)^2 = \lambda_j $$, our first differential equation becomes:

$$ z^{''}_1 + \begin{bmatrix} 0.485 & 0.381\\ \end{bmatrix} \begin{bmatrix} 3/4 & -1/4\\ -1/4 & 7/12\\ \end{bmatrix} \begin{bmatrix} 0.485\\ 0.381\\ \end{bmatrix} z^{'}_1 + 4.765 z_1 = 0 $$

$$ z^{''}_1 + 0.1687 z^{'}_1 + 4.765 z_1 = 0 $$

Using the method of undetermined coefficients for solving differential equations, we arrive at:

$$ z_1 = A e^{-0.0843 t} cos(2.18 t) $$

We now need to find the modal coordinate initial condition using the following equaiton:

$$ z_i(0) = \bar{\phi}^T_i M d(0) $$

$$ z_1(0) = \begin{bmatrix} 0.485 & 0.381\\ \end{bmatrix} \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} \begin{bmatrix} -1\\ 2\\ \end{bmatrix} = 0.069 $$

We can easily find A from:

$$ z_1(0) = A (1) (1) = 0.069 $$

so

$$ z_1 = 0.069 e^{-0.0843 t} cos(2.18 t) $$

Our second differential equation is:

$$ z^{''}_2 + \begin{bmatrix} -0.311 & 0.595\\ \end{bmatrix} \begin{bmatrix} 3/4 & -1/4\\ -1/4 & 7/12\\ \end{bmatrix} \begin{bmatrix} -0.311\\ 0.595\\ \end{bmatrix} z^{'}_2 + 22.735 z_2 = 0 $$

$$ z^{''}_2 + 0.371 z^{'}_2 + 22.735 z_2 = 0 $$

Using the method of undetermined coefficients for solving differential equations once again, we obtain:

$$ z_2 = B e^{-0.185 t} cos(4.76 t) $$

We now need to find the modal coordinate initial condition using the following equaiton:

$$ z_i(0) = \bar{\phi}^T_i M d(0) $$

$$ z_2(0) = \begin{bmatrix} -0.311 & 0.595\\ \end{bmatrix} \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} \begin{bmatrix} -1\\ 2\\ \end{bmatrix} = 3.313 $$

We can easily find B from:

$$ z_2(0) = B (1) (1) = 3.313 $$

so

$$ z_2 = 3.313 e^{-0.0843 t} cos(2.18 t) $$

The solution using modal superposition can then be displayed as:

$$ d(t) = 0.069 e^{-0.0843 t} cos(2.18 t) \begin{bmatrix} 1.273\\ 1\\ \end{bmatrix} + 3.313 e^{-0.185 t} cos(4.76 t) \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} $$