Question 18.5: For the five storey building example F(t) = Ff (t); = The sy......

                      f(t)=f_0 ~\sin \omega t ; \frac{\omega}{\omega_1}=\frac{T_1}{T}=0.75

                      D_n(t)=\frac{f_0}{\omega_n^2} R_{d n}~ \sin~ \omega t

                      R_{d n}=\frac{1}{1~-~\left(\omega / \omega_n\right)^2}

                      R_{d n} \text { may be }+ \text { ve or }- \text { ve }

                      R_n(t)=R^{s t}~ \bar{R}_n ~f_0~ R_{d n}~ \sin \omega t

                      R(t)=f_0 ~R^{s t}~\left(\bar{R}_n R_{d n}\right)~ \sin \omega t

Maximum value

                        R_0=f_0~ R^{s t}\left(\bar{R}_n~ R_{d n}\right)

                      V_b^{s t}=2 ; V_{b 0}=f_0~ V_b^{s t}~ \Sigma ~\bar{V}_{b n} ~R_{d n}

                       \frac{V_{b 0}}{f_0}=V_b^{s t}~ \sum \bar{V}_{b n}~ R_{d n}

For the five storey building {F} is resolved into model components as shown in Fig. 18.10. For example for mode 1 \omega_1=0.2439 .

                    R_{d 1}=\frac{1}{1~-~\left(\frac{\omega}{\omega_1}\right)^2}=\frac{1}{1~-~\left(\frac{0.1829}{0.2439}\right)^2}=2.284

                     \frac{V_{b 0}}{F_0}=\frac{5.6}{2}=2.8 \text { (See Table 18.2) }

If we include ‘p’ modes

                      \frac{V_{b 0}}{f_0}=1+\sum\limits_{n=1}^p \bar{V}_{b n}~ R_{d n}-\sum\limits_{n=1}^N \bar{V}_{b n}

The calculations are shown in Table 18.2.