Question 6.5: The simple structure shown in Fig. 6.5 consists of a beam, c......

Note: The numerical operations in this example were carried out using a larger number of significant figures than indicated.
Part (a):
The flexibility matrix for this problem was derived by the area-moment method in Example B1 of Appendix B:

[\alpha] = [K]^{-1} =\begin{bmatrix} \alpha _{11} & \alpha _{12} \\ \alpha _{21} & \alpha _{22} \end{bmatrix} =\frac{L^{3}}{EI} \begin{bmatrix} \frac{1}{24} & \frac{5}{48} \\ \frac{5}{48} & \frac{1}{3} \end{bmatrix}                     (A)

Part (b):
The equations of motion of the system in global coordinates can be written in the form of Eq. (6.76):

[M]\left\{\ddot{z} \right\} + [C]\left\{\dot{z} \right\} +[K]\left\{z\right\} = \left\{F\right\}                     (B)

To find the real eigenvalues and eigenvectors, [C] and {F} are omitted, leaving the undamped homogeneous equations:

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

With the usual substitution, valid for undamped systems: \left\{z\right\} = \left\{\overline{z}\right\}e^{\mathrm{i}\omega t}, Eq. (C) can be written as:

([K] – \omega^{2}[M])\left\{\overline{z}\right\} = 0                     (D)

In this example, the flexibility matrix, [K]^{-1}, is known, rather than the stiffness matrix, [K]. Therefore, Eq. (D) is pre-multiplied by [K]^{-1}, giving:

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

From Eq. (A) we already have

[K]^{-1} =\frac{L^{3}}{EI} \begin{bmatrix} \frac{1}{24} & \frac{5}{48} \\ \frac{5}{48} & \frac{1}{3} \end{bmatrix}

The mass matrix, [M], can be derived, as in Section 6.1.2, by noting that the masses, m_{1}  \mathrm{and}  m_{2}, produce inertia forces, m_{1}\ddot{z}_{1}  \mathrm{and}  m_{2}\ddot{z}_{2}, respectively, on the left side of the equation, so the mass matrix simply consists of the masses m_{1}  \mathrm{and}  m_{2} written as a diagonal matrix:

[M] = \begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix} =\begin{bmatrix} 10 & 0 \\ 0 &8 \end{bmatrix}                     (F)

Equation (E) then becomes

\left(\frac{L^{3}}{48EI} \begin{bmatrix}2 & 5 \\ 5 & 16 \end{bmatrix} \begin{bmatrix} 10 & 0 \\ 0 & 8 \end{bmatrix} – \frac{1}{\omega^{2}} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix} = 0                     (G)

which can be written as:

\left(\begin{bmatrix} 20 & 40 \\ 50 & 128 \end{bmatrix} – \Lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix} = 0                     (H)

where Λ is one of the two eigenvalues, defined here as:

\Lambda = \frac{48EI}{L^{3}\omega^{2}} = \frac{1.5 \times 10^{6}}{\omega^{2}}                     (I)

since EI = 2 \times 10^{6}, and L = 4, and ω is one of the two natural frequencies.

The two eigenvalues are given by the roots of the characteristic equation, obtained by multiplying out the following determinant, and equating to zero:

\begin{vmatrix} (20 – \Lambda) & 40 \\ 50 & (128 – \Lambda) \end{vmatrix} = (20 – \Lambda)(128 – \Lambda) – 2000 = 0 \\ \Lambda^{2} – 148\Lambda + 560 = 0                     (J)

The two roots, or eigenvalues, are

\Lambda = \frac{148 \pm \sqrt{148^{2} – 4(560)} }{2} = 74 \pm 70.114

that is Λ_{1} = 144.11,  \mathrm{and}  Λ_{2} = 3.885, in ascending order of natural frequency. From Eq. (I), the natural frequencies are ω_{1} = 102.02    rad/s   \mathrm{and}    ω_{2} = 621.30    rad/s.
The ratios (\overline{z}_{1}/\overline{z}_{2})_{1}  \mathrm{and}  (\overline{z}_{1}/\overline{z}_{2})_{2} can be found: by substituting each eigenvalue in turn into Eq. (H):

\left(\frac{\overline{z}_{1}}{\overline{z}_{2}} \right)_{1} = \frac{-40}{20 – \Lambda_{1}} = \frac{-40}{20 – 144.11} = 0.3222                    (K_{1}) \\ \left(\frac{\overline{z}_{1}}{\overline{z}_{2}} \right)_{2} = \frac{-40}{20 – \Lambda_{2}} = \frac{-40}{20 – 3.885} = -2.482                    (K_{2})

Temporarily setting \overline{z}_{2} to 1 in both cases, the two eigenvectors or mode shapes are

\left\{\underline{\phi}\right\}_{1} = \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix}_{1} = \begin{Bmatrix} 0.3222 \\ 1 \end{Bmatrix}                    (L_{1}) \\ \mathrm{and} \\ \left\{\underline{\phi}\right\}_{2} = \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix}_{2} = \begin{Bmatrix} -2.482 \\ 1 \end{Bmatrix}                    (L_{2})

These are sketched in Fig. 6.6. To rescale the eigenvectors, or mode shapes, above, so that they are orthonormal, each of them is multiplied by the scalar \alpha_{i}, given by the following expression, derived in Example 6.3 as Eq. (W):

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

where is \left\{\underline{\phi } \right\}_{i} = \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix}_{i} eigenvector i and [M] the mass matrix in global coordinates.

Thus for mode 1:

\alpha_{1} = \left(\begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix}_{1}^{T} \begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix}_{1}\right)^{-\frac{1}{2}} = \begin{Bmatrix} 0.3222 \\ 1 \end{Bmatrix}^{T} \begin{bmatrix} 10 & 0 \\ 0 & 8 \end{bmatrix} \begin{Bmatrix} 0.3222 \\ 1 \end{Bmatrix} = 0.3326                    (N_{1}) \\ \mathrm{and  for  mode  2:} \\ \alpha_{2} = \left(\begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix}_{2}^{T} \begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix} \begin{Bmatrix} \overline{z}_{1} \\ \overline{z}_{2} \end{Bmatrix}_{2}\right)^{-\frac{1}{2}} = \begin{Bmatrix} -2.482 \\ 1 \end{Bmatrix}^{T} \begin{bmatrix} 10 & 0 \\ 0 & 8 \end{bmatrix} \begin{Bmatrix} -2.482 \\ 1 \end{Bmatrix} = 0.1198                    (N_{2})

The orthonormal eigenvectors or mode shapes, \left\{\phi \right\}_{1}  \mathrm{and}  \left\{\phi \right\}_{2}, are

\left\{\phi \right\}_{1} = \alpha_{1} \left\{\underline{\phi} \right\}_{1} = 0.3326 \begin{Bmatrix} 0.3222 \\ 1 \end{Bmatrix} = \begin{Bmatrix} 0.1071 \\ 0.3326 \end{Bmatrix}                    (O_{1}) \\ \left\{\phi \right\}_{2} = \alpha_{2} \left\{\underline{\phi} \right\}_{2} = 0.1198 \begin{Bmatrix} -2.4822 \\ 1 \end{Bmatrix} = \begin{Bmatrix} -0.2975 \\ 0.1198 \end{Bmatrix}                    (O_{2})

It can be seen that the scaling applied to the original eigenvectors is immaterial, since the process of converting them to orthonormal form cancels out any overall scaling factor.
The modal matrix, [X], consists of the eigenvectors, in this case the orthonormal set, written as columns:

[X] = [\left\{\phi \right\}_{1} \left\{\phi \right\}_{2}] = \begin{bmatrix} 0.1071 & -0.2975 \\ 0.3326 & 0.1198 \end{bmatrix}                 (P)

The matrix [X] will later be used for two purposes:
(1) As the transformation matrix between the global coordinates, {z}, and the modal coordinates, {q}:

{z} = [X] {q}                                                     (Q)

(2) As the transformation matrix between the actual external forces, {F}, and the modal forces, {Q}:

\left\{Q\right\} = [X]^{T} \left\{F\right\}                                                 (R)

At this point it is as well to check that the transformation:

[\underline{M}] = [X]^{T} [M][X]                                                (S)

does, in fact, produce a unit matrix. Using the numerical values from Eqs (P) and (F):

[\underline{M}] = \begin{bmatrix} 0.1071 & -0.2975 \\ 0.3326 & 0.1198 \end{bmatrix}^{T}\begin{bmatrix} 10 & 0 \\ 0 & 8 \end{bmatrix} \begin{bmatrix} 0.1071 & -0.2975 \\ 0.3326 & 0.1198 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = [I]                           (T)

and this is clearly the case.
Part (c):
The essential relationships for forming the undamped equations of motion in modal coordinates, from the equations in global coordinates, were derived in Section 6.2.3. These apply whether the transformation is based on assumed (or arbitrary) modes, or, as in this case, normal modes. However, in the present case, the mass matrix and the stiffness matrix, in modal coordinates, will both be diagonal. Moreover, if the eigenvectors are scaled to orthonormal form, as here, the mass matrix in modal coordinates will become a unit matrix, and the corresponding stiffness matrix will consist of the squares of the undamped natural frequencies in rad/s on the leading diagonal.
The equations in global coordinates, Eq. (B), are now transformed to modal coordinates. The damping terms are initially omitted, and will be based on measured data in the modal equations. Eq. (B) then becomes

[M]\left\{\ddot{z}\right\} + [K]\left\{z\right\} = \left\{F\right\}                         (U)

Then the undamped equations of motion in normal mode coordinates will be

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

Now recalling that the modal matrix [X] was scaled to be orthonormal, i.e. to make [\underline{M}] a unit matrix, we already know that

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

Also, since the two equations represented by Eq. (V) are completely uncoupled, and we know the undamped natural frequencies, ω_{1}  \mathrm{and}  ω_{2}, it follows that

[\underline{K}] = \begin{bmatrix} \omega_{1}^{2} & 0 \\ 0 & \omega_{2}^{2} \end{bmatrix}

Therefore the undamped equations in normal coordinates are simply:

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} \ddot{q}_{1} \\ \ddot{q}_{2} \end{Bmatrix} + \begin{bmatrix} \omega_{1}^{2} & 0 \\ 0 & \omega_{2}^{2} \end{bmatrix} \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix} = \begin{Bmatrix} Q_{1} \\ Q_{2} \end{Bmatrix}                      (W)

The damping matrix can now be inserted. Since it is assumed that the damping does not couple the modes, the equations can continue to be treated as two completely separate single-DOF equations, and the damping term, \underline{c}_{ii}, for Mode i, is

\underline{c}_{ii} = 2\gamma_{i} \omega_{i} \underline{m}_{ii}                      (X)

where \gamma_{i} is the non-dimensional damping coefficient for Mode i, ω_{i} the normal mode undamped natural frequency for Mode i  \mathrm{and}  \underline{m}_{ii} the modal or generalized mass for Mode i, equal to unity, by definition, in this case.
The complete equations of motion are therefore:

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} \ddot{q}_{1} \\ \ddot{q}_{2} \end{Bmatrix} + \begin{bmatrix} 2\gamma_{1}\omega _{1} & 0 \\ 0 & 2\gamma_{2}\omega _{2} \end{bmatrix} \begin{Bmatrix} \dot{q}_{1} \\ \dot{q}_{2} \end{Bmatrix} + \begin{bmatrix} \omega_{1}^{2} & 0 \\ 0 & \omega_{2}^{2} \end{bmatrix} \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix} = \begin{Bmatrix} Q_{1} \\ Q_{2} \end{Bmatrix}                      (Y)

Numerically, since \omega_{1} = 102.02, \omega_{2} = 621.30   \mathrm{and}   \gamma_{1} = \gamma_{2} = 0.02: \\ 2\gamma_{1}\omega _{1} = 4.080,    2\gamma_{2}\omega _{2} = 24.85,    \omega_{1}^{2} = 10408,    \omega_{2}^{2} = 386020

In order to put the equations to practical use, we shall also require the relationship between the actual (global) displacements, {z}, and the modal displacements, {q}, which is

\left\{z\right\} =[X]\left\{q\right\}       \mathrm{or:}        \begin{Bmatrix} z_{1} \\ z_{2} \end{Bmatrix} = \begin{bmatrix} 0.1071 & -0.2975 \\ 0.3326 & 0.1198 \end{bmatrix} \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix}                      (Z)

and the relationship between the actual external loads, [F], and the modal forces, [Q], which is

\left\{Q\right\} = [X]^{T} \left\{F\right\}

or

\begin{Bmatrix} Q_{1} \\ Q_{2} \end{Bmatrix} = \begin{bmatrix} 0.1071 & -0.2975 \\ 0.3326 & 0.1198 \end{bmatrix}^{T} \begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix} = \begin{bmatrix} 0.1071 & 0.3326 \\ -0.2975 & 0.1198 \end{bmatrix}^{T} \begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix}                        (a)

Part (d):
Equation (Y) can now be treated as two uncoupled, single-DOF equations:

\ddot{q}_{1} + 2\gamma_{1} \omega_{1} \dot{q}_{1} + \omega^{2}_{1} q_{1} = Q_{1}                    (b_{1}) \\ \mathrm{and} \\ \ddot{q}_{2} + 2\gamma_{2} \omega_{2} \dot{q}_{2} + \omega^{2}_{2} q_{2} = Q_{2}                    (b_{2})

From Eq. (a), we have

Q_{1} = 0.1071F_{1} + 0.3326F_{2}                    (c_{1}) \\ \mathrm{and} \\ Q_{2} = -0.2975F_{1} + 0.1198F_{2}                    (c_{2})

The external applied force F_{1} is defined as a step force of 1000 N, and F_{2} is zero:

F_{1} = 1000 H(t),       F_{2} = 0

where H(t) is the unit step or Heaviside function.
Thus,

Q_{1} = (0.1071 × 1000 ) H(t)                    (d_{1})  \\ \mathrm{and} \\ Q_{2} = (-0.2975 × 1000 ) H(t)                    (d_{2})

From Eq. (B) of Example 3.4 in Section 3.1.3, the response of a single-DOF system to a step force of magnitude a is given as:

z = \frac{a}{m\omega^{2}_{n}} \left[1 – e^{-\gamma \omega_{n} t}\left(\cos \omega _{d}t + \frac{\gamma}{\sqrt{(1 – \gamma^{2})}}\sin \omega _{d}t \right) \right]                         (e)

where z is the displacement; m the mass; γ the viscous damping coefficient and ω_{n} the undamped natural frequency, all of a single-DOF system. Also, ω_{d} is the damped natural frequency given by \omega_{d} = \omega_{n} \sqrt{1 – \gamma^{2}} .

In Eq. (e), z is replaced by q_{1}  \mathrm{and}  q_{2}, in turn, and a is replaced by the magnitudes of the step modal forces, Q_{1}  \mathrm{and}   Q_{2} in turn, to give the responses of the two equations, Eqs (b_{1})  \mathrm{and}   (b_{2}):

q_{1} = \frac{0.1071\times 1000}{\omega^{2}_{1}} \left[1 – e^{-\gamma_{1} \omega_{1} t}\left(\cos \omega _{1d}t + \frac{\gamma_{1}}{\sqrt{(1 – \gamma^{2}_{1})}}\sin \omega _{1d}t \right) \right]                    (\mathrm{f_{1}}) \\ q_{2} = \frac{-0.2975\times 1000}{\omega^{2}_{2}} \left[1 – e^{-\gamma_{2} \omega_{2} t}\left(\cos \omega _{2d}t + \frac{\gamma_{2}}{\sqrt{(1 – \gamma^{2}_{2})}}\sin \omega _{2d}t \right) \right]                    (\mathrm{f_{2}})

Finally, the required displacement history, z_{1}, is given by Eq. (Z), which can be written, since z_{2} is not required in this case:

z_{1} = 0.1071 q_{1} – 0.2975 q_{2}                          (g)

where q_{1}  \mathrm{and}  q_{2} are given by Eqs (\mathrm{f_{1}})  \mathrm{and}   (\mathrm{f_{2}}). The numerical values are ω_{1} is the undamped natural frequency of Mode 1 = 102.02 rad/s; ω_{2} the undamped natural frequency of Mode 2 = 621.30 rad/s; γ_{1} the damping coefficient of Mode 1 = 0.02; γ_{2} the damping coefficient of Mode 2 = 0.02; ω_{1d} the damped natural frequency of Mode 1 = \omega_{1} \sqrt{1 – \gamma _{1}^{2}} = 102.02 \sqrt{1 – 0.02^{2}} = 101.99   \mathrm{and}    ω_{2d} the damped natural frequency of Mode 2 = \omega_{2} \sqrt{1 – \gamma _{2}^{2}} = 621.30 \sqrt{1 – 0.02^{2}} = 621.17. The time history of z_{1} is plotted in Fig. 6.7.