Question 13.16: The two-degree-of-freedom system of Figure E13.16a is subjec...

The stiffness and mass matrices of the system are given by

\begin{aligned}&\mathbf{K}=\left[\begin{array}{rr}192 & -64 \\-64 & 64\end{array}\right] \\&\mathbf{M}=\left[\begin{array}{ll}2 & 0 \\0 & 1\end{array}\right]\end{aligned}

The frequencies and mode shapes of the system can be obtained by any of the standard methods discussed in Chapter 11. The calculated frequencies are

\begin{aligned}&\omega_{1}=4 \sqrt{2} \\&\omega_{2}=8 \sqrt{2}\end{aligned}

The corresponding mass-orthonormal mode shapes are

\begin{aligned}&\boldsymbol{\phi}_{1}=\frac{1}{\sqrt{6}}\left[\begin{array}{l}1 \\2\end{array}\right] \\&\boldsymbol{\phi}_{2}=\frac{1}{\sqrt{3}}\left[\begin{array}{r}1 \\-1\end{array}\right]\end{aligned}

The modal forces are given by

\begin{aligned}&p_{1}=\boldsymbol{\phi}_{1}^{T} g(t)\left[\begin{array}{l}1 \\\frac{1}{2}\end{array}\right]=\frac{2}{\sqrt{6}} g(t) \\&p_{2}=\boldsymbol{\phi}_{2}^{T} g(t)\left[\begin{array}{l}1 \\\frac{1}{2}\end{array}\right]=\frac{1}{2 \sqrt{3}} g(t)\end{aligned}

The modal equations are

\begin{aligned}\ddot{y}_{1}+32 y_{1} &=\frac{2}{\sqrt{6}} g(t) \\\ddot{y}_{2}+128 y_{2} &=\frac{1}{2 \sqrt{3}} g(t)\end{aligned}

For t<t_{1}, t_{1} being the duration of the rectangular pulse, the solutions of the two modal equations are given by

\begin{aligned}&y_{1}=\frac{p_{0}}{16 \sqrt{6}}(1-\cos 4 \sqrt{2} t) \\&y_{2}=\frac{p_{0}}{256 \sqrt{3}}(1-\cos 8 \sqrt{2} t)\end{aligned}

where p_{0}=10. The corresponding response in the physical coordinates is obtained from

For the free vibration era, t>t_{1}, the modal response can be obtained by finding the modal displacements and velocities at t=t_{1}=2 \mathrm{~s}. Thus

\begin{aligned}&y_{1}\left(t_{1}\right)=\frac{p_{0}}{16 \sqrt{6}}(1-\cos 8 \sqrt{2})=0.1753 \\&\dot{y}_{1}\left(t_{1}\right)=\frac{p_{0}}{4 \sqrt{3}} \sin 8 \sqrt{2}=-1.371\end{aligned}

and

\begin{aligned}&y_{2}\left(t_{1}\right)=\frac{p_{0}}{256 \sqrt{3}}(1-\cos 16 \sqrt{2})=0.0407 \\&\dot{y}_{2}\left(t_{1}\right)=\frac{\sqrt{2} p_{0}}{32 \sqrt{3}} \sin 16 \sqrt{2}=-0.1516\end{aligned}

The free vibration response is given by

Response in the physical coordinates is obtained by superposing the modal responses

\left[\begin{array}{l}u_{1} \\u_{2}\end{array}\right]=\left[\begin{array}{c}\frac{1}{\sqrt{6}} \\\frac{2}{\sqrt{6}}\end{array}\right] y_{1}+\left[\begin{array}{c}\frac{1}{\sqrt{3}} \\\frac{-1}{\sqrt{3}}\end{array}\right] y_{2}

The exact response of the system obtained as above for the entire range of t will be used for comparison with the frequency response obtained later.

The fundamentals mode period of the system is given by

T_{1}=\frac{2 \pi}{\omega_{1}}=1.11 \mathrm{~s}

The maximum response of the system will occur either within the duration of the rectangular pulse or in the free-vibration era after the force pulse has ceased to act. The maximum in the freevibration era will be attained within half-a-cycle of motion at a frequency of \omega_{1}. We therefore select a total time duration of 2.6 \mathrm{~s} to calculate the response; that time also becomes T_{0}, the period of the forcing function.

The augmented forcing function for the first mode is shown in Figure E13.16c. For analysis in the frequency domain, we first obtain the true periodic response to a periodic extension of the forcing function. This requires the evaluation of periodic impulse function given by

\bar{h}_{1}(t)=\frac{1}{2 M_{1} \omega_{1}}\left(\sin \omega_{1} t+\frac{\cos \omega_{1} t \sin \omega_{1} T_{0}}{1-\cos \omega_{1} T_{0}}\right)

Table E13.16a Modal frequency-domain analysis of two-degrees-of-freedom system.

where M_{1} is the modal mass of the first mode. Since the mode shape have been massorthonormalized, M_{1}=1.

The response in the first mode to a periodic extension of the forcing function is obtained by convolving the sampled forcing function of Figure E13.16c and \bar{b}_{1}(t). The sampling interval is chosen as 0.05 \mathrm{~s}. The response as obtained above is not equal to the transient response and must be corrected for the initial conditions. The displacement at t=0 in the periodic response is found to be 0.0213. The velocity at t=0 is calculated from Equation 9.85

\dot{\bar{u}}(0)=-\frac{4 \pi}{\left(T_0\right)^2} \sum_{n=0}^{N / 2} n \operatorname{Im}\{U(n \Delta \Omega)\} \qquad(9.85)

and works out to 1.001. The corrective responses are obtained from Equations 9.89 \mathrm{a} and 9.89 \mathrm{~b}