Question 18.12: Consider a five storey building (Example 18.1) (see Fig. 18.......
Consider a five storey building (Example 18.1) (see Fig. 18.32) whose properties are given. Calculate effective modal mass and height.
Using the MATHEMATICA package, the normalized eigenvector is given as
[\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]
The numerical analysis value is
< F >=<2 2 2 1 1>
The modal contribution of mass (see Fig. 18.33) is
\phi_1^{ T } F=\Gamma_1=-2.6644
\Gamma_2=-0.8051
\Gamma_3=-0.4472
\Gamma_4=0.23
\Gamma_5=0.0068
[F] = [m][φ][Γ]
The effective modal mass and modal height are shown in Fig. 18.34.
The ground acceleration \ddot{u}_g~(t) is defined by its numerical value as time instant equally spaced at ∆t. This time step is chosen small enough to define \ddot{u}_g~(t) and to determine accurately the response of the SDOF system:
\text { Base shear }=m\left(7.0926 A_1(t)+0.6471 A_2(t)+0.2 A_3(t)\right.
\left.+0.0528 A_4(t)+0.0003 A_5(t)\right)
\text { Base moment }=m h\left(7.0927 \times 3.079 A_1(t)+0.6471 \times(-1.563) A_2(t)\right.
+0.2 A_3(t)+0.0528 \times(-0.587) A_4(t)
\left.+0.0003 \times 7.67 A_5(t)\right)
