Question 18.10: Figure 18.24 shows a two storey frame with flexural rigidity......

Mass matrix

                 [m]=m\left[\begin{array}{ll}1 & \\& 0.5\end{array}\right]

     The complete stiffness matrix to the corresponding to degrees of freedom is given by

k=\frac{E I}{h^3}\left[\begin{array}{cccccc}48 & -24 & 0 & 0 & -6 h & -6 h \\-24 & 24 & 6 h & 6 h & 6 h & 6 h \\0 & 6 h & 10 h^2 & 1 h^2 & 2 h^2 & 0 \\0 & 6 h & 1 h^2 & 10 h^2 & 0 & 2 h^2 \\-6 h & 6 h & 2 h^2 & 0 & 6 h^2 & 1 h^2 \\-6 h & 6 h & 0 & 2 h^2 & 1 h^2 & 6 h^2\end{array}\right]

Denoting

                 U^m=\left\{\begin{array}{l}u_1 \\u_2\end{array}\right\} ; U^s=\left\{\begin{array}{c}u_3 \\:\\u_i\end{array}\right\}\begin{gathered}U^m=\text { master degree of freedom } \\U^s=\text {slave degree of freedom }\end{gathered}

                 K^* m m=k_{m m}-k_{m s} ~k_{s s}^{-1} ~k_{s m}

                                =\frac{E I}{h^3}\left[\begin{array}{cc}37.15 & -15.12 \\-15.12 & 10.19\end{array}\right]

                          \omega_1=2.407~ \sqrt{\frac{E I}{m h^3}}

                          \omega_2=7.193 ~\sqrt{\frac{E I}{m h^3}}

\text { Eigenvector }=\left[\begin{array}{cc}0.482 & -1.037 \\1 & 1\end{array}\right]

          \text { Normalized eigenvector }=\frac{1}{\lceil{m}}\left[\begin{array}{cc}0.563 & -0.826 \\1.165 & 0.796\end{array}\right]

            \Gamma_1=\varphi_1^{ T }~ m(1)

                 =\frac{1}{\lceil m}<0.563\quad1.1687>\left[\begin{array}{ll}1 & \\& 0.5\end{array}\right]\left\{\begin{array}{l}1 \\1\end{array}\right\} m

                 =(0.563+0.584) \sqrt{m}

                 =1.147 \sqrt{m}

           \Gamma_2=\varphi_2^{ T }~ m(1)

                =<-0.826~~0 .796>\left[\begin{array}{ll}1 & \\& 0.5\end{array}\right]\left\{\begin{array}{l}1 \\1\end{array}\right\} \sqrt{m}

                =(-0.826+0.398) \sqrt{m}

                =-0.428 \sqrt{m}

(a) Floor displacement due to first mode

                U_1(t)=\left\{\begin{array}{l}u_1(t) \\u_2(t)\end{array}\right\}=1.147\left\{\begin{array}{l}0.563 \\1.168\end{array}\right\} D_1(t)

                          =\left\{\begin{array}{l}0.647 \\1.340\end{array}\right\} D_1(t)

The joint rotations can be obtained from

                U_s=-k_{s s}^{-1} ~k_{s m}~ U_m

                U_s=\frac{1}{h}\left[\begin{array}{cc}-0.164 & -0.411 \\-0.164 & -0.411 \\0.904 & -0.74 \\0.904 & -0.74\end{array}\right]\left\{\begin{array}{c}0.647 \\1.34\end{array}\right\} D_1(t)

                     =\frac{1}{h}\left[\begin{array}{c}-0.657 \\0.657 \\-0.407 \\-0.407\end{array}\right] D_1(t)

Floor displacements due to second mode

                U_2(t)=\left\{\begin{array}{l}u_1(t) \\u_2(t)\end{array}\right\}_2

                          =-0.425\left\{\begin{array}{c}-0.826\\0.796\end{array}\right\} D_2(t)

                         =\left\{\begin{array}{c}0.313\\-0.341\end{array}\right\} D_2(t)

The joint rotations for the structure

                \left(U_s\right)_2=\frac{1}{h}\left[\begin{array}{l}0.082 \\0.082 \\0.572\\0.572\end{array}\right] D_2(t)

Combining we get

                  U(t)=\left[\begin{array}{l}u_1(t) \\u_2(t)\end{array}\right]

                           =\left[\begin{array}{l}0.647 ~D_1(t)+0.753~ D_2(t) \\1.341~ D_1(t)-0.341 ~D_2(t)\end{array}\right]

                    U_s=\frac{1}{h}\left[\begin{array}{l}-0.657 ~D_1(t)+0.082 ~D_2(t) \\-0.657~ D_1(t)+0.082 ~D_2(t) \\-0.407~ D_1(t)+0.572~ D_2(t) \\-0.407 ~D_1(t)+0.582 ~D_2(t)\end{array}\right]

(b) The bending moment at the ends of the flexural elements are related to modal displacement as

                M_a=\frac{4 ~E I}{l} \theta_a+\frac{2~ E I}{l} \theta_b+\frac{6~ E I}{l^2} u_a-\frac{6 E I}{l^2} u_b

                M_b=\frac{2~ E I}{l} \theta_a+\frac{4 ~E I}{l} \theta_b+\frac{6 ~E I}{l^2} u_a-\frac{6~ E I}{l^2} u_b

Substituting the above quantities we get the moments at the ends ‘a’ and ‘b’ of the column (see Fig. 18.25)

                M_a=\frac{4 ~E I}{h} U_3+\frac{2~ E I}{h}(0)+\frac{6 ~E I}{h^2}\left(U_1\right)-\frac{6 ~E I}{h^2}(0)

                       =\frac{E I}{h^2}\left[1.254 D_1(t)+2.446 D_2(t)\right]

The bending moment at the right-hand is similar (see Fig. 18.26). Relating D_{n}(t) to A_{n}(t) as)

                D_1(t)=\frac{A_1(t)}{\omega_1^2}

                           =\frac{m h^3}{E I(2.407)^2}~ A_1(t)

                D_2(t)=\frac{A_2(t)}{\omega_2^2}

                           =\frac{m h^3}{E I(7.193)^2}~ A_2(t)

Solving we get the bending moment at the top of the column

                M_a=m h\left[0.216 ~A_1(t)+0.0403 ~A_2(t)\right]

Similarly bending moment at the right side of the beam

                M_b=\frac{2~ E I}{h} ~U_3+\frac{6~ E I}{h^2}~ U_1

                       =m h\left[0.443~ A_1(t)+0.0441 ~A_2(t)\right]

For the second floor beam L = 2h

                M_a=m h\left[-0.21~ A_1(t)+0.0332~ A_2(t)\right]

                 M_b=m h\left[-0.21 ~A_1(t)+0.0332~ A_2(t)\right]