Question 18.4: Consider a five storey building (rigid floor beams and slabs......
Consider a five storey building (rigid floor beams and slabs) with lumped mass m at each floor, and same storeyed stiffness k. <F> = <0 0 –1 1 2). See Fig. 18.8.
Use the MATHEMATICA or MATLAB package. Assume k = 1 m = 1.
[m]=\left[\begin{array}{lllll}2 & & & & \\& 2 & & & \\& & 2 & & \\& & & 1 & \\& & & &1\end{array}\right]
|k|=\left[\begin{array}{ccccc}2 & -1 & & & \\-1 & 2 & -1 & & \\& -1 & 2 & -1 & \\& & -1 & 2 & -1 \\& & & -1 & 1\end{array}\right]
Using the MATLAB program we get five natural frequencies and five normalized mode shapes.
\omega_1=0.2439 ; \omega_2=0.6689 ; \omega_3=1.0 ; \omega_4=1.286 ; \omega_5=1.6850~ rad / s
Normal eigenvector
[\phi]=\left[\begin{array}{ccccc}-0.1585 & -0.3602 & -0.4472 & 0.3804 & 0.0193 \\-0.2982 & -0.3981 & -0 &-0.4976 & -0.0711 \\-0.4023 & -0.0797 & 0.4472 &0.2703 & 0.2423 \\-0.4586 & 0.31 & 0 . & 0.1441 &-0.8202 \\-0.4877 & 0.5610 & -0.4472 & -0.2204 &0.446\end{array}\right]
<F> = <0 0 –1 1 2>
\begin{aligned}& \Gamma_1=\phi_1^{ T } F=-1.0316 \\ \\ & \Gamma_2=\phi_2^{ T } F=1.5117 \\ \\ & \Gamma_3=\phi_3^{ T } F=-1.3416 \\ \\ & \Gamma_4=\phi_4^{ T } F=-0.5669\\ \\ & \Gamma_5=\phi_5^{ T } F=-0.1706 \end{aligned}
Check
\sum\limits_{i=1}^N~ \Gamma^{2}~i=\sum\limits_{i=1}^5 ~m_i
only when [F] = [m](I).
[F]=\sum~ \Gamma_n[m] \phi_n=[m][\phi][\Gamma]
[F]=\left[\begin{array}{ccccc} 0.3271 & -1.0891 & 1.2 & -0.4314 & -0.0066 \\ 0.6152 & -1.2036 & 0 & 0.5641 & 0.0243 \\ 0.8301 & -0.2410 & -1.2 & -0.3064 & -0.0827 \\ 0.4731 & 0.4686 & 0 & -0.0817 & 0.1399 \\ 0.5031 & 0.8481 & 0.6 & 0.1249 & -0.0761 \end{array}\right]
The modal expansion of the excitation vector is shown in Fig. 18.9.
