Question 7.2: (a) Basing the calculation on the flexibility matrix, use th......

Part (a):
Although we intend to base the calculation on the flexibility matrix, [K]^{-1}, suppose for a moment that the homogeneous equations of motion, in terms of the stiffness matrix, [K], are

[M]\left\{\ddot{z}\right\} + [K]\left\{z\right\} = \left\{0\right\}                          (A)

Substituting \left\{z\right\} = \left\{\overline{z} \right\} e^{\mathrm{i}\omega t} in the usual way:

(- \omega^{2} [M] + [K])\left\{\overline{z} \right\} = 0                          (B)

Pre-multiplying by [K]^{-1},

– \omega^{2} [K]^{-1} [M] \left\{\overline{z} \right\} + [K]^{-1} [K] \left\{\overline{z} \right\} = 0                          (C)

leads to a standard form

[\underline{\underline{A}}] \left\{\overline{z} \right\} = \underline{\lambda} \left\{\overline{z} \right\}                          (D)

where

The flexibility matrix [K]^{-1} for the same system was derived in Example 1.4 as:

[K]^{-1} = \begin{bmatrix} 1/k_{1} & 1/k_{1} & 1/k_{1} \\ 1/k_{1} & (1/k_{1} + 1/k_{2}) & (1/k_{1} + 1/k_{2}) \\ 1/k_{1} & (1/k_{1} + 1/k_{2}) & (1/k_{1} + 1/k_{2} + 1/k_{3}) \end{bmatrix}                          (E)

Inserting the numerical values, the matrices [M] ,  [K]^{-1}   \mathrm{and}   [\underline{\underline{A}}] become

Equation (D), to be solved by matrix iteration, is then,

\begin{bmatrix} 2 & 1 & 2 \\ 2 & 1.5 & 3 \\ 2 & 1.5 & 5 \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix} = \hat{\lambda} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix}                         (F)

where

To find the first eigenvector \left\{\overline{z}\right\}_{1} and first scaled eigenvalue, \hat{\lambda}_{1}, a trial vector, \left\{\overline{z}\right\}, say,

\left\{\overline{z}\right\} = \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \\ \overline{z}_{3} \end{Bmatrix} = \begin{Bmatrix} 0.4 \\ 0.6 \\ 1.0 \end{Bmatrix}                       (G)

is inserted on the left side of Eq. (F). The equation is then evaluated with this assumption, normalizing the resulting vector by dividing by the largest element:

\begin{bmatrix} 2 & 1 & 2 \\ 2 & 1.5 & 3 \\ 2 & 1.5 & 5 \end{bmatrix}\begin{Bmatrix} 0.4 \\ 0.6 \\ 1 \end{Bmatrix} = \begin{Bmatrix} 3.4 \\ 4.7 \\ 6.7 \end{Bmatrix} = 6.7\begin{Bmatrix} 0.5074 \\ 0.7014 \\ 1 \end{Bmatrix}                      (H)

The process is repeated, using the normalized column on the right in place of the trial vector on the left. After four such repeats, we have

\begin{bmatrix} 2 & 1 & 2 \\ 2 & 1.5 & 3 \\ 2 & 1.5 & 5 \end{bmatrix}\begin{Bmatrix} 0.5292 \\ 0.7198 \\ 1 \end{Bmatrix} = \begin{Bmatrix} 3.7782 \\ 5.1381 \\ 7.1381 \end{Bmatrix} = 7.1381 \begin{Bmatrix} 0.5293 \\ 0.7198 \\ 1 \end{Bmatrix}                      (I)

The final vector on the right is now seen to agree with the first eigenvector from Example 7.1 by the Gaussian elimination method, to four significant figures.
The root \hat{\lambda} = 7.1381 = 10^{3} \underline{\lambda} = 10^{3} /\omega^{2} gives ω = 11.83 rad/s agreeing with the result of Example 7.1.

Part(b):
To find the second eigenvalue and eigenvector, the first eigenvector, already found, must be completely removed from the dynamic matrix [\underline{\underline{A}}]. This can be achieved by using the orthogonality relationship, Eq. (6.54):

\left\{\overline{z}\right\}^{T}_{j} [M] \left\{\overline{z}\right\}_{i} = 0                  (J)

At some point in the iteration procedure, the impure vector \left\{\overline{z}\right\} must consist of a linear combination of the three pure eigenvectors being sought

\left\{\overline{z} \right\} = C_{1} \left\{\overline{z} \right\}_{1} + C_{2} \left\{\overline{z} \right\}_{2} + C_{3} \left\{\overline{z} \right\}_{3}                       (K)

where C_{1},C_{2}   \mathrm{and}   C_{3} are constants. Pre-multiplying Eq. (K) by \left\{\overline{z}\right\}^{T}_{1} [M]:

\left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\} = C_{1} \left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\}_{1} + C_{2} \left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\}_{2} + C_{3} \left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\}_{3}                       (L)

Now due to the orthogonality relationship for eigenvectors, Eq. (J), the last two terms in Eq. (L) must be zero, and

\left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\} = C_{1} \left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\}_{1}                       (M)

To produce a vector \left\{\overline{z}\right\} free from \left\{\overline{z}\right\}_{1} we can now set C_{1} = 0 , and since clearly \left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\}_{1} ≠ 0 , it must be true that

\left\{\overline{z}\right\}^{T}_{1} [M] \left\{\overline{z}\right\} = 0                       (N)

or written out, using the notation:

The known values of s_{1}   \mathrm{and}   s_{2} can now be incorporated into a sweeping matrix, [S], where:

[S] = \begin{bmatrix} 0 & s_{1} & s_{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -0.6799 & -1.8892 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}                       (R)

This is incorporated into the original iteration equation, Eq. (D), as follows:

[\underline{\underline{A}}] [S] \left\{\overline{z} \right\} = \underline{\lambda} \left\{\overline{z} \right\}                       (S)

and Eq. (F) becomes

This is then solved for the second eigenvector and second eigenvalue, in precisely the same way as before, using an estimated vector to start the process.