Question 6.3: (a) Find the eigenvalues and eigenvectors of the undamped sy......

Part (a)
The equations of motion for this system, in terms of global coordinates z_{1}  \mathrm{and}  z_{2}, and with external forces F_{1}  \mathrm{and}  F_{2}, have already been derived as Eq. (6.10):

\begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix} \begin{Bmatrix} \ddot{z}_{1} \\ \ddot{z}_{2} \end{Bmatrix} + \begin{bmatrix} (k_{1}+k_{2}) & -k_{2} \\ -k_{2} & k_{2} \end{bmatrix} \begin{Bmatrix} {z}_{1} \\ {z}_{2} \end{Bmatrix} = \begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix}                         (A)

Inserting the given numerical values:

\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{Bmatrix} \ddot{z}_{1} \\ \ddot{z}_{2} \end{Bmatrix} + \begin{bmatrix} 20 & -10 \\ -10 & 10 \end{bmatrix} \begin{Bmatrix} {z}_{1} \\ {z}_{2} \end{Bmatrix} = \begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix}           (B)

Omitting the applied forces F_{1}  \mathrm{and}  F_{2} gives the homogeneous equations, from which the eigenvalues and eigenvectors can be found:

\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{Bmatrix} \ddot{z}_{1} \\ \ddot{z}_{2} \end{Bmatrix} + \begin{bmatrix} 20 & -10 \\ -10 & 10 \end{bmatrix} \begin{Bmatrix} {z}_{1} \\ {z}_{2} \end{Bmatrix} = 0            (C)

Arranging into the form of Eq. (6.45): ([K] – \lambda [M]) \left\{\overline{z} \right\} = 0, we have

\left(\begin{bmatrix} 20 & -10 \\ -10 & 10 \end{bmatrix} – \lambda \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \right) \begin{Bmatrix} \overline{z} _{1} \\ \overline{z} _{2} \end{Bmatrix} = 0                (D)

or

\begin{bmatrix} (20-\lambda) & -10 \\ -10 & (10-2\lambda) \end{bmatrix} \begin{Bmatrix} \overline{z} _{1} \\ \overline{z} _{2} \end{Bmatrix} = 0             (E)

The two eigenvalues, \lambda _{1}  \mathrm{and}  \lambda _{2}, are now found by writing the matrix in Eq. (E) as a determinant and equating it to zero:

\begin{bmatrix} (20-\lambda) & -10 \\ -10 & (10-2\lambda) \end{bmatrix} \begin{Bmatrix} \overline{z} _{1} \\ \overline{z} _{2} \end{Bmatrix} = 0,

or

(20 – λ)(10 – 2λ) – 100 = 0,

giving

λ² – 25λ + 50 = 0.

Solving this quadratic equation by the formula \lambda =\frac{25 ± \sqrt{25^{2} – 200}}{2} gives the lower eigenvalue, \lambda_{1} = 2.1922, and the higher eigenvalue, \lambda_{2} = 22.807.

The two natural frequencies are

ω_{1}=\sqrt{\lambda_{1}} = \sqrt{2.1922} = 1.480 rad/s, and

ω_{2}=\sqrt{\lambda_{2}} = \sqrt{22.807} = 4.775 rad/s

The two eigenvectors can now be found by substituting the eigenvalues, one at a time, into Eq. (E). Since the eigenvectors are arbitrary to an overall factor, we can only find the ratio (\overline{z}_{1}/\overline{z}_{2}) in each case. Either row of Eq. (E) can be used, giving

\left(\frac{\overline{z}_{1} }{\overline{z}_{2}} \right) = \left(\frac{10}{20 – \lambda} \right)  \mathrm{or}  \left(\frac{10 – 2\lambda}{10} \right)

Thus we have

for   \lambda = \lambda_{1} = 2.192,    \left(\frac{\overline{z}_{1} }{\overline{z}_{2}} \right)_{1} = 0.5615

for   \lambda = \lambda_{2} = 22.807,    \left(\frac{\overline{z}_{1} }{\overline{z}_{2}} \right)_{2} = -3.5615

The eigenvectors can now be called ‘normal modes’, and it is usual to write them as vectors:

\left\{\phi \right\} _{1} = \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix} _{1}                 \left\{\phi \right\} _{2} = \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix} _{2}              (G)

Using the first of the standard methods for normalizing the eigenvectors (making the largest absolute element in each column equal to unity), these are

\left\{\phi \right\} _{1} = \begin{Bmatrix} 0.5615 \\ 1 \end{Bmatrix}                 \left\{\phi \right\} _{2} = \begin{Bmatrix} 1 \\ -0.2807 \end{Bmatrix}              (H)

Part (b)
The procedure for transforming the equations of motion into normal coordinates is now as outlined in Section 6.2.3, which could be applied using any set of modes. What is special about a normal mode transformation, in contrast to an assumed mode transformation, however, is that the matrix [X] in Eq. (6.26):

{z} = [X] {q}

is now formed by using the eigenvectors of the original equations as its columns. The transformation matrix [X], in this case, is therefore:

[X] = \left[\left\{\phi \right\}_{1} \left\{\phi \right\}_{2} \right] = \begin{bmatrix} 0.5615 & 1 \\ 1 & -0.2807 \end{bmatrix}             (I)

From Eq. (6.31), the new mass matrix [\underline{M}], in modal coordinates, is given by:

[\underline{M}] = [X]^{T}[M][X]

or numerically:

[\underline{M}] = \begin{bmatrix} 0.5615 & 1 \\ 1 & -0.2807 \end{bmatrix}^{T} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 0.5615 & 1 \\ 1 & -0.2807 \end{bmatrix}=\begin{bmatrix} 2.315 & 0 \\ 0 & 1.157 \end{bmatrix}               (J)

From Eq. (6.33), the new stiffness matrix, [\underline{K}], in modal coordinates, is given by:

[\underline{K}] = [X]^{T}[K][X]

or numerically:

[\underline{K}] = \begin{bmatrix} 0.5615 & 1 \\ 1 & -0.2807 \end{bmatrix}^{T} \begin{bmatrix} 20 & -10 \\ -10 & 10 \end{bmatrix}\begin{bmatrix} 0.5615 & 1 \\ 1 & -0.2807 \end{bmatrix}=\begin{bmatrix} 5.075 & 0 \\ 0 & 26.40 \end{bmatrix}             (K)

The complete equations in normal mode coordinates are now given by substituting
Eqs (J) and (K) into Eq. (6.25):

[\underline{M}]\left\{\ddot{q} \right\} + [\underline{K}]\left\{q \right\} = \left\{Q\right\}

giving

\begin{bmatrix} 2.315 & 0 \\ 0 & 1.157 \end{bmatrix}\begin{Bmatrix} \ddot{q}_{1} \\ \ddot{q}_{2} \end{Bmatrix} + \begin{bmatrix} 5.075 & 0 \\ 0 & 26.40 \end{bmatrix} \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix} = \begin{Bmatrix} Q_{1} \\ Q_{2} \end{Bmatrix}                   (L)

It can be seen that the use of the eigenvectors as modes in the transformation has made the mass and stiffness matrices diagonal. Equation (L) now consists of two completely independent single-DOF equations:

\underline{m}_{11}\ddot{q}_{1} + \underline{k}_{11}q_{1} = Q_{1}             (M_{1})\\  \mathrm{and}   \\ \underline{m}_{22}\ddot{q}_{2} + \underline{k}_{22}q_{2} = Q_{2}             (M_{2})

where the numerical values of \underline{m}_{11}, \underline{m}_{22}, \underline{k}_{11}  \mathrm{and}  \underline{k}_{22} are clear from Eq. (L).

The quantities \underline{m}_{11}  \mathrm{and}  \underline{m}_{22} are known as the generalized (or modal) masses of modes 1 and 2, respectively, and similarly \underline{k}_{11}  \mathrm{and}  \underline{k}_{22} are the generalized (or modal) stiffnesses. The absolute numerical values of the generalized masses and stiffnesses have no significance, since they depend upon how the eigenvectors are scaled. However, \underline{k}_{11}/\underline{m}_{11} should always be equal to ω^{2}_{1}, and \underline{k}_{22}/\underline{m}_{22} should equal ω^{2}_{2} . It is easily verified that this is so in this example.

Part (c)
The other standard method for normalizing the eigenvectors is to scale them so that the new mass matrix, [\underline{M}], in modal coordinates, is equal to the unit matrix, [I].

[\underline{M}] = \begin{bmatrix} \underline{m}_{11} & 0 \\ 0 & \underline{m}_{22} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = [I]                (N)

The eigenvectors are then described as weighted normal or orthonormal, and the corresponding stiffness matrix, [\underline{K}], then becomes

[\underline{K}] = \begin{bmatrix} \underline{k}_{11} & 0 \\ 0 & \underline{k}_{22} \end{bmatrix} = \begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix} = \begin{bmatrix} ω^{2}_{1} & 0 \\ 0 & ω^{2}_{2} \end{bmatrix}                (O)

that is, a diagonal matrix of the squares of the natural frequencies. Scaling the eigenvectors so that the mass and stiffness matrices take these simple forms can be achieved in two ways.

Method(1):
If the diagonal mass and stiffness matrices in modal coordinates have already been evaluated, using some other scaling for the eigenvectors, as in Eqs (J) and (K), they can be re-scaled to orthonormal form by multiplying each eigenvector \left\{\phi \right\} _{i} by a factor \alpha _{i}, where

\alpha _{i} = \frac{1}{\sqrt{\underline{m}_{ii}}}            (P)

where \underline{m}_{ii} is the diagonal element of the new mass matrix (in modal coordinates) corresponding to eigenvector i. In this case, from Eq. (L):

\alpha _{1} = \frac{1}{\sqrt{2.315}}    \mathrm{and}    \alpha _{2} = \frac{1}{\sqrt{1.157}}

Multiplying the old vectors from Eq. (H) by these factors:

\left\{\phi \right\} ^{new}_{1}= \frac{1}{\sqrt{2.315}}\begin{Bmatrix} 0.5615 \\ 1 \end{Bmatrix} = \begin{Bmatrix} 0.3690 \\ 0.6572 \end{Bmatrix}                 (Q_{1})\\ \left\{\phi \right\} ^{new}_{2}= \frac{1}{\sqrt{1.15767}}\begin{Bmatrix} 1 \\ -0.2807 \end{Bmatrix} = \begin{Bmatrix} -0.9294 \\ 0.2609 \end{Bmatrix}               (Q_{2})

where Eqs (Q_{1})  \mathrm{and}  (Q_{2}) give the rescaled eigenvectors. The new transformation matrix, now based on orthonormal eigenvectors, is

[X]^{new} = \left[\left\{\phi \right\}_{1} \left\{\phi \right\}_{2} \right] = \begin{bmatrix} 0.3690 & -0.9294 \\ 0.6572 & 0.2609 \end{bmatrix}                (R)

It is easily verified that using this matrix in the transformation of the mass and stiffness matrices changes them to the forms of Eqs (N) and (O), respectively:

[\underline{M}] = [X]^{T}[M][X] = \begin{bmatrix} 0.3690 & -0.9294 \\ 0.6572 & 0.2609 \end{bmatrix}^{T}\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 0.3690 & -0.9294 \\ 0.6572 & 0.2609 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}              (S)

so [\underline{M}] becomes a unit matrix and

[\underline{K}] = [X]^{T}[K][X] = \begin{bmatrix} 0.3690 & -0.9294 \\ 0.6572 & 0.2609 \end{bmatrix}^{T} \begin{bmatrix} 20 & -10 \\ -10 & 10 \end{bmatrix}\begin{bmatrix} 0.3690 & -0.9294 \\ 0.6572 & 0.2609 \end{bmatrix} = \begin{bmatrix} 2.192 & 0 \\ 0 & 22.807 \end{bmatrix}             (T)

where the diagonal terms are now seen to be equal to the squares of the natural frequencies, ω^{2}_{1}  \mathrm{and}  ω^{2}_{2}.

Method (2):
The eigenvectors in orthonormal form can also be calculated directly from the original mass matrix, [M], and the eigenvectors in any form, by noting that in Eq. (P), the generalized mass, \underline{m}_{ii}, for mode i, in the modal (transformed) equations, can be calculated from

\underline{m}_{ii} = \left\{\underline{\phi } \right\}^{T}_{i}[M]\left\{\underline{\phi } \right\}_{i}               (U)

where \left\{\underline{\phi } \right\}_{i} represents eigenvector i in any arbitrary form, for example, in the nonstandard form:

\left\{\underline{\phi } \right\}_{1} = \begin{Bmatrix} 0.5615 \\ 1 \end{Bmatrix}  \mathrm{and}  \left\{\underline{\phi } \right\}_{2} = \begin{Bmatrix} -3.561 \\ 1 \end{Bmatrix}                (V_{1})(V_{2})

and [M] is the original mass matrix in global coordinates, in this example given as

[M] = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}

Equations (P) and (U) can be combined, giving the more usual expression for \alpha _{i}:

\alpha _{i} = \left(\frac{1}{\left\{\underline{\phi } \right\}^{T}_{i}[M]\left\{\underline{\phi } \right\}_{i}} \right)^{\frac{1}{2}}                (W)

Substituting the numerical values above into Eq. (W) gives \alpha _{1} = 0.6572 and \alpha _{2} = 0.2609. The orthonormal eigenvectors are then:

\left\{\phi \right\}_{1}=\alpha _{1} \left\{\underline{\phi } \right\}_{1}=0.6572 \begin{Bmatrix} 0.5615 \\ 1 \end{Bmatrix} = \begin{Bmatrix} 0.3690 \\ 0.6572 \end{Bmatrix}                     (X_{1}) \\ \left\{\phi \right\}_{2}=\alpha _{2} \left\{\underline{\phi } \right\}_{2}=0.260958 \begin{Bmatrix} -3.561 \\ 1 \end{Bmatrix} = \begin{Bmatrix} -0.9294 \\ 0.2609 \end{Bmatrix}                  (X_{2})

which are seen to be the same as the eigenvectors given in Eqs (Q_{1})  \mathrm{and}  (Q_{2}) by the first method.