Question 6.4: Find the steady-state response of each mass in the following...
Find the steady-state response of each mass in the following two-degrees-offreedom system to a sinusoidal force of amplitude 10 N and frequency 6 Hz applied to the 1 kg mass.
Using the method presented in Section 6.2, the equations of motion, natural frequencies and modal vectors are obtained first.
\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{Bmatrix} \ddot{x}_{1} \\ \ddot{x}_{2} \end{Bmatrix} + \begin{bmatrix}2000 & – 1000 \\ – 1000 & 2000 \end{bmatrix}\begin{Bmatrix} x_{1} \\ x_{2} \end{Bmatrix} = \begin{Bmatrix} p_{1}(t) \\ 0 \end{Bmatrix}The eigenvalues are \omega _{1}^{2} = 634 s^{-2} and \omega _{2}^{2} = 2366 s^{-2} giving natural frequencies of 4.0 Hz and 7.7 Hz respectively.
The corresponding mode shape vectors are
\left\{X\right\}_{1} = \begin{Bmatrix} X_{11} \\ X_{21} \end{Bmatrix} = \begin{Bmatrix} 0.732 \\ 1.0 \end{Bmatrix} and \left\{X\right\}_{2} = \begin{Bmatrix} X_{12} \\ X_{22} \end{Bmatrix} = \begin{Bmatrix} -2.732 \\ 1.0 \end{Bmatrix}
These vectors are scaled by making X_{2} = 1.0 for each mode. Using equations (6.58) and (6.59), the modal mass and modal stiffness values are:
For r = s, \left\{X\right\}_{r}^{T} [M] \left\{X\right\}_{r} = M_{r} (6.58)
and \left\{X\right\}_{r}^{T} [K] \left\{X\right\}_{r} = K_{r} (6.59)
M_{1} = 2.54, M_{2} = 9.46, K_{1} = 1608 and K_{2} = 22 392.
Rescaling to unit modal mass using equation (6.58) gives the new modal matrix:
Using the new modal matrix, the reader is left to confirm that [\Phi ]^{T}[K][\Phi ]= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} and that [\Phi ]^{T}[K][\Phi ] = \begin{bmatrix} 634 & 0 \\ 0 & 2366 \end{bmatrix} As mentioned earlier, the new modal stiffnesses of K_{1} = 634 and K_{2} = 2366 are numerically equal to the squares of the corresponding natural frequencies.
From \begin{Bmatrix} f(t) \end{Bmatrix} =[\Phi ]^{T}\begin{Bmatrix} p(t) \end{Bmatrix}, the modal forces are
\begin{Bmatrix} f_{1}(t) \\ f_{2}(t) \end{Bmatrix} = \begin{bmatrix}0.460 & 0.628 \\ – 0.888 & 0.325 \end{bmatrix} \begin{Bmatrix} p_{1}(t) \\ 0 \end{Bmatrix} = \begin{Bmatrix} 0.460 \times p_{1}(t) \\ – 0.888 \times p_{1}(t) \end{Bmatrix}
The modal equation for mode 1 is
\ddot{q}_{1} + \omega _{1}^{2} q_{1} = f_{1}(t)or \ddot{q}_{1} + 634q_{1} = 0.460p_{1}(t) (6.66)
For the steady-state response, put p_{1} = 10e^{i\omega t} q_{1} and q_{1} = Q^{\star}_{1} e^{i\omega t} into (6.66) with an excitation frequency of 6 Hz to give:
Q_{1}^{\star } = \frac{4.60}{634 – (6 \times 2\pi )^{2}} = – 0.005 84Similarly for mode 2,
Q_{2}^{\star } = \frac{- 8.88}{2366 – (6 \times 2\pi )^{2}} = – 0.009 40From equation (6.65),
x_{j}(t) = \sum\limits_{r}{u_{jr}q_{r}(t)} (6.65)
\begin{Bmatrix} X_{1}^{\star} e^{i\omega t} \\ X_{2}^{\star} e^{i\omega t} \end{Bmatrix} = \begin{bmatrix} 0.460 & – 0.888 \\0.628 & 0.325\end{bmatrix} \begin{Bmatrix} Q_{1}^{\star}e^{i\omega t} \\ Q_{2}^{\star}e^{i\omega t} \end{Bmatrix}Hence, X_{1}^{\star} = 0.460 \times Q_{1}^{\star} – 0.888 \times Q_{2}^{\star} = 5.66 mm
and X_{2}^{\star} = 0.628 \times Q_{1}^{\star} + 0.325 \times Q_{2}^{\star} = – 6.72 mm
Notice that X_{1}^{\star} is positive, showing that mass 1 moves in phase with the force applied to it, while the negative sign for X_{2}^{\star} shows that mass 2 moves 180° out of phase with the applied force.
Repeating the exercise across a range of frequencies, the modal responses and the displacements of the two masses are as shown in Figures 6.69 and 6.70.
