Question 16.4: Design diagram of multistore frame is presented in Fig. 16.1......

The primary system is shown in Fig. 16.12b. For computation of unit reactions, we need to construct the bending moment diagrams due to unit displacements of the introduced constraints, and then for calculation of unit reactions consider the equilibrium condition for each crossbar.

Bending moment diagram caused by unit displacement of the constraint 1 is shown in Fig. 16.12c. Elastic curve is shown by dotted line. Since crossbars are absolutely rigid members, then joints cannot be rotated and each vertical member should be considered as fixed-fixed member. In this case, specified ordinates are 6 i / h  (Table A.4, Case 2). Now we need to show free-body diagram for each horizontal member using the closed sections (shown by dotted lines in Fig. 16.12 b ). Bending moments at cut sections are 6 i / h. Both moments may be equilibrated by two forces 12 \mathrm{i} / \mathrm{h}^2 . These forces should be transmitted on both crossbars. Positive unit reactions r_{11}, r_{21} , and r_{31}  are shown by dotted arrows.

Equilibrium condition for each crossbar leads to the following unit reactions

r_{11}=2 \frac{12 i}{h^2}=24 \frac{i}{h^2}, \quad r_{21}=-24 \frac{i}{h^2}, \quad r_{31}=0, \quad i=\frac{E I}{h}

Similarly, considering the second and third unit displacements, we get

\begin{aligned}& r_{12}=-24 \frac{i}{h^2}, \quad r_{22}=48 \frac{i}{h^2}, \quad r_{32}=-24 \frac{i}{h^2}, \\& r_{13}=0, \quad r_{23}=-24 \frac{i}{h^2}, \quad r_{33}=48 \frac{i}{h^2}\end{aligned}

Let r_0=24 i / h^2 . Equations (16.11) become

\begin{aligned}& \left(r_{11}-m_1 \omega^2\right) A_1+r_{12} A_2+\ldots+r_{1 n} A_n=0 \\& r_{21} A_1+\left(r_{22}-m_2 \omega^2\right) A_2+\ldots+r_{2 n} A_n=0 \\& \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot\left(r_{n n}-m_n \omega^2\right) A_n=0 \\+& r_{n 1} A_1+r_{n 2} A_2+\left(r_n\right.\end{aligned}                  (6.11)

\begin{aligned}& \left(r_0-m \omega^2\right) A_1-r_0 A_2+0 \cdot A_3=0 \\& -r_0 A_1+2\left(r_0-m \omega^2\right) A_2-r_0 A_3=0 \\& 0 \cdot A_1-r_0 A_2+2\left(r_0-m \omega^2\right) A_3=0\end{aligned}         (a)

The frequency equation is

D=\operatorname{det}\left[\begin{array}{ccc}r_0-m \omega^2 & -r_0 & 0 \\-r_0 & 2 r_0-2 m \omega^2 & -r_0 \\0 & -r_0 & 2 r_0-2 m \omega^2\end{array}\right]=0

If eigenvalue is denoted as \lambda=\frac{m \omega^2}{r_0}=\frac{m \omega^2 h^2}{24 i}, then the system (a) may be rewritten as

\begin{aligned}& (1-\lambda) A_1-A_2=0 \\& -A_1+2(1-\lambda) A_2-A_3=0 \\& 0-A_2+2(1-\lambda) A_3=0\end{aligned}        (b)

Eigenvalue equation is

\begin{aligned}& D=\operatorname{det}\left[\begin{array}{ccc}1-\lambda & -1 & 0 \\-1 & 2(1-\lambda) & -1 \\0 & -1 & 2(1-\lambda)\end{array}\right]=0 \\& 4(1-\lambda)^3-3(1-\lambda)=0 \quad \text { or } \quad(1-\lambda)\left[4(1-\lambda)^2-1\right]=0 \\&\end{aligned}

The eigenvalues in increasing order are

\lambda_1=1-\frac{\sqrt{3}}{2}, \quad \lambda_2=1, \quad \lambda_3=1+\frac{\sqrt{3}}{2}

Corresponding frequencies of free vibration (eigenfrequencies)

\omega_1^2=24\left(1-\frac{\sqrt{3}}{2}\right) \frac{E I}{m h^3}, \quad \omega_2^2=24 \frac{E I}{m h^3}, \quad \omega_3^2=24\left(1+\frac{\sqrt{3}}{2}\right) \frac{E I}{m h^3}

Mode shapes of vibration. Now we need to consider the system (b) for each calculated eigenvalue. If we denote \rho_2=A_2 / A_1  and \rho_3=A_3 / A_1 , then system (b) may be rewritten as

\begin{aligned}& (1-\lambda)-\rho_2=0 \\& -1+2(1-\lambda) \rho_2-\rho_3=0 \\& 0-\rho_2+2(1-\lambda) \rho_3=0\end{aligned}         (c)

This system should be solved with respect to \rho_2  and \rho_3  for each eigenvalue.

First mode \omega_1: \lambda_1=1-\frac{\sqrt{3}}{2} . Equations (c) become

\begin{aligned}& \left(1-\lambda_1\right)-\rho_2=0 \\& -1+2\left(1-\lambda_1\right) \rho_2-\rho_3=0 \\& -\rho_2+2\left(1-\lambda_1\right) \rho_3=0\end{aligned}

Solution of the first and second equations are \rho_2=\sqrt{3} / 2, \quad \rho_3=1 / 2 . The last equation is satisfied. Indeed

-\frac{\sqrt{3}}{2}+2\left[1-\left(1-\frac{\sqrt{3}}{2}\right)\right] \frac{1}{2}=0

The same procedure should be repeated for \lambda_2=1.0  and \lambda_3=1+\frac{\sqrt{3}}{2}.

The modal matrix \boldsymbol{\Phi}  is defined as

\boldsymbol{\Phi}=\left[\begin{array}{ccc}1 & 1 & 1 \\\sqrt{3} / 2 & 0 & -\sqrt{3} / 2 \\1 / 2 & -1 & 1 / 2\end{array}\right] 

First (principal) mode of vibration, which corresponds to smallest frequency \omega_1 , is presented in Fig. 16.12d.

Verification. Orthogonality conditions of the first and second modes of vibration should be written as follows:

\boldsymbol{\varphi}_1^T \mathbf{M} \boldsymbol{\varphi}_2=0

where \boldsymbol{\varphi}_1^T  is a transposed first column of the matrix \boldsymbol{\Phi} , and \mathbf{M}  is the mass matrix. In our case we have

   \boldsymbol{\varphi}_1^T \mathbf{M} \boldsymbol{\varphi}_2=\left\lfloor\begin{array}{lll} 1 & \sqrt{3} / 2 & 1 / 2 \end{array}\right] \cdot m\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \cdot\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\left\lfloor\begin{array}{lll} 1 & \sqrt{3} / 2 & 1 / 2 \end{array}\right] \cdot m\left[\begin{array}{c} 1 \\ 0 \\ -2 \end{array}\right]=1 \cdot 1+\frac{\sqrt{3}}{2} \cdot 0-\frac{1}{2} \cdot 2=0

Similarly, it is easy to check that \boldsymbol{\varphi}_1^T \mathbf{M} \boldsymbol{\varphi}_3=0 , and \boldsymbol{\varphi}_2^T \mathbf{M} \boldsymbol{\varphi}_3=0