Question 7.1: (a) Find the eigenvalues, and hence the natural frequencies,......
(a) Find the eigenvalues, and hence the natural frequencies, of the undamped 3-DOF system shown in Fig. 7.1, by solving the characteristic equation for its roots.
(b) Using the eigenvalues, find the eigenvectors by Gaussian elimination.
The numerical values of the masses m_{1}, m_{2} \mathrm{and} m_{3}, and the stiffnesses, k_{1}, k_{2} \mathrm{and} k_{3}, are as follows:
m_{1} = 2 kg; m_{2} = 1 kg; m_{3} = 2 kg;\\ k_{1} = 1000 N/m; k_{2} = 2000 N/m; k_{3} = 1000 N/m.
Part (a):
The stiffness matrix for the springs of this system was derived in Example 1.4. Using the method described in Section 6.1.2, the equations for the undamped unforced system are as follows:
\begin{bmatrix} m_{1} & 0 & 0 \\ 0 & m_{2} & 0 \\ 0 & 0 & m_{3} \end{bmatrix} \begin{Bmatrix} \ddot{z}_{1} \\ \ddot{z}_{2} \\ \ddot{z}_{3} \end{Bmatrix} + \begin{bmatrix} (k_{1}+k_{2}) & -k_{2} & 0 \\ -k_{2} & (k_{2}+k_{3}) & -k_{3} \\ 0 & -k_{3} & k_{3} \end{bmatrix} \begin{Bmatrix} z_{1} \\ z_{2} \\ z_{3} \end{Bmatrix} = 0 (A)
Substituting \left\{z\right\} = \left\{\overline{z} \right\} e^{\mathrm{i}\omega t} in the usual way, and inserting the numerical values gives
-\omega^{2} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix} + 10^{3} \begin{bmatrix} 3 & -2 & 0 \\ -2 & 3 & -1 \\ 0 & -1 & 1 \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix} = 0 (B)
Substituting \omega^{2} = \lambda and introducing the scaling \hat{\lambda} = 10^{-3} \lambda to simplify the matrices:
\left(\begin{bmatrix} 3 & -2 & 0 \\ -2 & 3 & -1 \\ 0 & -1 & 1 \end{bmatrix} – \hat{\lambda} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \right) \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix} = 0. (C)
The determinant D of Eq. (C) is now equated to 0:
D = \begin{bmatrix} (3 – 2\hat{\lambda}) & -2 & 0 \\ -2 & (3 – \hat{\lambda}) & -1 \\ 0 & -1 & (1 – 2\hat{\lambda}) \end{bmatrix} = 0 (D)
Using the standard rule for evaluating a determinant:
D = (3 – 2\hat{\lambda})\left[(3 – \hat{\lambda})(1 – 2\hat{\lambda}) – 1\right] + 2 \left[ (-2)(1 – 2\hat{\lambda}) \right] = 0 (E)
which when simplified gives
\hat{\lambda}^{3} – 5\hat{\lambda}^{2} + 4.25\hat{\lambda} – 0.5 = 0 (F)
Using a standard roots solution program, the scaled roots, \hat{\lambda}_{i}, as calculated, are:
\hat{\lambda}_{1} = 0.1401, \hat{\lambda}_{2} = 0.9017, \hat{\lambda}_{3} = 3.958However, since the roots, \hat{\lambda}_{i}, were defined as \hat{\lambda}_{i} = 10^{-3} \lambda_{i}, the eigenvalues of the physical problem, λ_{i}, are
λ_{1} = 140.1 = ω^{2}_{1}, λ_{2} = 901.7 = ω^{2}_{2}, λ_{3} = 3958 = ω^{2}_{3}The natural frequencies, ω_{i}, in rad/s are
ω_{1} = 11.83 rad/s; ω_{2} = 30.03 rad/s; ω_{2} = 62.91 rad/sPart (b):
From Eq. (C):
\begin{bmatrix} (3 – 2\hat{\lambda}) & -2 & 0 \\ -2 & (3 – \hat{\lambda}) & -1 \\ 0 & -1 & (1 – 2\hat{\lambda}) \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} (G)
The scaled roots \hat{λ}_{1},\hat{λ}_{2} \mathrm{and} \hat{λ}_{3} are now substituted, one at a time, into Eq. (G), and Gaussian elimination is used to find the corresponding eigenvectors. Substituting the first root, \hat{λ}_{1} = 0.1401 , into Eq. (G) gives
\begin{bmatrix} 2.7198 & -2 & 0 \\ -2 & 2.8599 & -1 \\ 0 & -1 & 0.7198 \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix}_{(1)} = \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} (H)
The vector has the subscript (1) to indicate that it is the eigenvector corresponding to the first eigenvalue.
Gaussian elimination is a standard method for solving any set of linear equations.
The general idea is to eliminate the terms in the matrix below the leading diagonal, then the values of \overline{z}_{1}, \overline{z}_{2} \mathrm{and} \overline{z}_{3} can be found by back-substitution.
In this case the term –2 in the first column can be removed by multiplying the first row by (–2/2.7198) = –0.7353 and subtracting it from the second row, giving
\begin{bmatrix} 2.7198 & -2 & 0 \\ 0 & 1.3893 & -1 \\ 0 & -1 & 0.7198 \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix}_{(1)} = \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} (I)
Without going further, from Eq. (I) we have the ratios:
\frac{\overline{z}_{2}}{\overline{z}_{3}} = 0.7198, \frac{\overline{z}_{1}}{\overline{z}_{2}} = 0.7353 \mathrm{and} \\ \frac{\overline{z}_{1}}{\overline{z}_{3}} = \frac{\overline{z}_{1}}{\overline{z}_{2}} \times \frac{\overline{z}_{2}}{\overline{z}_{3}} = (0.7353 \times 0.7198) = 0.5293If \overline{z}_{3} is now arbitrarily assigned the value 1, then the first eigenvector is
\begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix}_{(1)} = \begin{Bmatrix} 0.5293 \\ 0.7198 \\ 1.0000 \end{Bmatrix} (J)
The second eigenvector can be found in precisely the same way, by substituting the second scaled root, \hat{λ}_{2} = 0.9017 , into Eq. (G), and repeating the process of eliminating terms below the diagonal and so on. The whole process is then repeated for the third eigenvector.