Question 18.3: Solve Example 18.1 using a normalized vector Φ instead of ψ....

                      [m]=m\left[\begin{array}{lllll}1 & & & & \\& 1 & & & \\& & 1 & & \\& & & 1 & \\& & & & 1\end{array}\right] ;\{\psi\}=\frac{1}{5}\left\{\begin{array}{l}1 \\2 \\3 \\4 \\5\end{array}\right\};

                      M_1^2=\psi^{ T }~ m \psi=\frac{55~ m}{25} ;M_1=\sqrt{\frac{11~m}{5}}

                      \{\phi\}=\frac{1}{M_1}\{\psi\}=\frac{1}{\sqrt{55~m}}\left\{\begin{array}{l}1 \\2 \\3 \\4 \\5\end{array}\right\}

When a normalized vector is used, the participation factor is

                 \Gamma=\Sigma~ m_j \phi_j=\phi^{ T }~ m\{i\}=\frac{m}{\sqrt{55 ~m}}(1+2+3+4+5)=\frac{15~\sqrt{m}}{\sqrt{55}}

                u_j=\phi_j~ Z(t)

To determine Z(t) consider the dynamic equilibrium equation

                  \ddot{Z}(t)+\omega_n^2~ Z(t)=-\Gamma\ddot{u}_g

                  \omega_n^2=\phi^{ T } ~k \phi=\Sigma k_j\left(\phi_j-\phi_{j-1}\right)^2=\frac{k}{11~ m}

                  \omega_n=0.302 ~\sqrt{\frac{k}{m}} ; k=5530 \times 10^3 N / m ; m=45412~ kg

                  \omega_n=3.335~ rad / s ; T=\frac{2~ \pi}{\omega_n}=1.89 ~s

                  \frac{A}{g}=1.8 T^{-1} \times 0.25=0.238 ; A=0.238 \times 9.81=2.334~ m / s ^2

                  D=\frac{A}{\omega_n^2}=\frac{2.334}{3.335^2}=0.212

                  Z=\Gamma~ D=\frac{15~ \times~ 0.212 ~\sqrt{m}}{\sqrt{55}}

\{u\}=\{\phi\}~ Z=\frac{15 ~\times~ 0.212 ~\sqrt{m}}{55~ \sqrt{m}}\left\{\begin{array}{l}1 \\2 \\3 \\4 \\5\end{array}\right\}=\left\{\begin{array}{c}0.0572 \\0.1144 \\0.1716 \\0.228 \\0.286\end{array}\right\}

The results obtained are the same as the ones obtained in Example 18.2