Question 16.1: Design diagram of the frame is shown in Fig. 16.7a. Find eig......

The system has two degrees of freedom. Generalized coordinate are q_1  and q_2 . We need to apply unit forces in direction of q_1  and q_2 , and construct the bending moments diagram. Unit displacements are

  \begin{aligned} & \delta_{11}=\sum \int \frac{\bar{M}_1 \bar{M}_1}{E I} \mathrm{~d} x=\frac{\bar{M}_1 \times \bar{M}_1}{E I}=\frac{1}{2 E I} \cdot \frac{1}{2} \cdot 1 \cdot l \cdot l \cdot \frac{2}{3} \cdot 1 \cdot l+\frac{1}{E I} \cdot 1 \cdot l \cdot h \cdot 1 \cdot l=\frac{l^3}{6 E I}+\frac{l^2 h}{E I} ; \\ & \delta_{22}=\frac{\bar{M}_2 \times \bar{M}_2}{E I}=\frac{1}{E I} \cdot \frac{1}{2} \cdot 1 \cdot h \cdot h \cdot \frac{2}{3} 1 \cdot h=\frac{h^3}{3 E I} ; \\ & \delta_{12}=\delta_{21}=\frac{\bar{M}_1 \times \bar{M}_2}{E I}=\frac{1}{E I} \cdot \frac{1}{2} \cdot 1 \cdot h \cdot h \cdot 1 \cdot l=\frac{h^2 l}{2 E I} . \end{aligned}

Let h=2 l  and \delta_0=\frac{l^3}{6 E I} . In this case \delta_{11}=13 \delta_0 ; \delta_{22}=16 \delta_0 ; \delta_{12}=\delta_{21}=12 \delta_0 . Equation for calculation of amplitudes (14.4)

 P_{\lim }=\frac{6 M_y^{\mathrm{vert}}}{l}      (14.4)

\begin{aligned}& \left(13 \delta_0 m \omega^2-1\right) A_1+12 \delta_0 m \omega^2 A_2=0 \\& 12 \delta_0 m \omega^2 A_1+\left(16 \delta_0 m \omega^2-1\right) A_2=0\end{aligned}      ( a)

Let \lambda=\frac{1}{\delta_0 m \omega^2}=\frac{6 E I}{m \omega^2 l^3} . In this case equation (a) may be rewritten

\begin{aligned}& (13-\lambda) A_1+12 A_2=0 \\& 12 A_1+(16-\lambda) A_2=0\end{aligned}

Frequency equation becomes

D=\left[\begin{array}{cc}13-\lambda & 12 \\12 & 16-\lambda\end{array}\right]=(13-\lambda)(16-\lambda)-144=0

Roots in descending order are \lambda_1=26.593 ; \lambda_2=2.4066

Eigenfrequencies in increasing order are

\omega_1=\sqrt{\frac{6 E I}{\lambda_1 m l^3}}=0.4750 \sqrt{\frac{E I}{m l^3}}, \quad \omega_2=\sqrt{\frac{6 E I}{\lambda_2 m l^3}}=1.5789 \sqrt{\frac{E I}{m l^3}}

Control 1: According to (16.7) we have \frac{1}{\omega_1^2}+\frac{1}{\omega_2^2}=m_1 \delta_{11}+m_2 \delta_{22} . In our case

 \frac{1}{\omega_1^2}+\frac{1}{\omega_2^2}=m_1 \delta_{11}+m_2 \delta_{22}          (16.7)

\begin{aligned}& \frac{1}{\omega_1^2}+\frac{1}{\omega_2^2}=\frac{m l^3}{0.4750^2 E I}+\frac{m l^3}{1.5789^2 E I}=(4.4321+0.4011) \frac{m l^3}{E I}=4.8332 \frac{m l^3}{E I} \\& m_1 \delta_{11}+m_2 \delta_{22}=m \times 13 \delta_0+m \times 16 \delta_0=29 m \delta_0=29 m \frac{l^3}{6 E I}=4.8333 \frac{m l^3}{E I}\end{aligned}

Mode shape of vibration may be determined on the basis of equations (b).

For first mode \left(\lambda_1=26.593\right)  ratio of amplitudes are

\begin{aligned}& \frac{A_2}{A_1}=-\frac{13-\lambda}{12}=-\frac{13-26.593}{12}=1.1328 \\& \frac{A_2}{A_1}=-\frac{12}{16-\lambda}=-\frac{12}{16-26.593}=1.1328\end{aligned}

Assume that A_1=1 , so the first eigenvector \boldsymbol{\varphi}  becomes

\varphi=\left\lfloor\begin{array}{ll}\varphi_{11} & \varphi_{21}\end{array}\right\rfloor^T=\left\lfloor\begin{array}{ll}1 & 1.1328\end{array}\right\rfloor^T

For second mode \left(\lambda_2=2.4066\right)  ratio of amplitudes are

   \begin{aligned} & \frac{A_2}{A_1}=-\frac{13-2.4066}{12}=-0.8828 \\ & \frac{A_2}{A_1}=-\frac{12}{16-2.4066}=-0.8828 \end{aligned}

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

\boldsymbol{\Phi}=\left[\begin{array}{cc} 1 & 1 \\ 1.1328 & -0.8828 \end{array}\right]

Corresponding mode shapes of vibration are shown in Fig. 16.7b. It easy to check the scalar multiplication of two vectors \varphi_1^T M \varphi_2=0 , where M is diagonal mass matrix (Kiselev 1980). In our case we have

\left\lfloor\begin{array}{ll}1 & 1.1328\end{array}\right]\left[\begin{array}{cc}m & 0 \\0 & m\end{array}\right]\left[\begin{array}{c}1 \\-0.8828\end{array}\right]=(1-1.000036) m \cong 0

This result shows that both forms of vibrations are orthogonal. It means that the force which corresponds to the first mode does not produce a work on the displacements of the second mode, and vice versa.

Control 2: Relationships (16.8) for our case leads to the following results

 \left(\frac{A_2}{A_1}\right)_{\omega_1} \times\left(\frac{A_2}{A_1}\right)_{\omega_2}=-\frac{m_1}{m_2}        (16.8)

\left(\frac{A_2}{A_1}\right)_{\omega_1} \times\left(\frac{A_2}{A_1}\right)_{\omega_2}=1.1328(-0.8828)=-1.0 \quad \text { and } \quad-\frac{m_1}{m_2}=-1