Question 16.2: Design diagram of structure containing two hinged-end member......
Design diagram of structure containing two hinged-end members is shown in Fig. 16.8a. Modulus of elasticity E and area of cross section A are constant for both members; l and l \sqrt{3} are length of the members, \alpha=60^{\circ}, \beta=30^{\circ} . Find eigenvalues and modal matrix and present the mode shapes.
The structure has two degrees of freedom. The first and second unit states and corresponding internal forces S for each member are shown in Fig. 16.8b.
Equations (16.4a) for unknown amplitudes
\begin{array}{r}\left(m \delta_{11} \omega^2-1\right) A_1+m \delta_{12} \omega^2 A_2=0 \\m \delta_{21} \omega^2 A_1+\left(m \delta_{22} \omega^2-1\right) A_2=0\end{array} (b)
\begin{aligned}& \left(m_1 \delta_{11} \omega^2-1\right) A_1+m_2 \delta_{12} \omega^2 A_2=0 \\& m_1 \delta_{21} \omega^2 A_1+\left(m_2 \delta_{22} \omega^2-1\right) A_2=0 .\end{aligned} (16.4a)
Unit displacements
\begin{aligned}& \delta_{11}=\sum \int \frac{S_1 \cdot S_1}{E A} d s=\frac{1}{E A}\left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \cdot l+\frac{1}{2} \cdot \frac{1}{2} \cdot l \sqrt{3}\right)=\frac{l}{4 E A}(3+\sqrt{3}) \\& \delta_{22}=\sum \int \frac{S_2 \cdot S_2}{E A} d s=\frac{1}{E A}\left(\frac{1}{2} \cdot \frac{1}{2} \cdot l+\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \cdot l \sqrt{3}\right)=\frac{l}{4 E A}(1+3 \sqrt{3}) \\& \delta_{12}=\delta_{21}=\sum \int \frac{S_1 \cdot S_2}{E A} d s=\frac{1}{E A}\left(\frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot l-\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot l \sqrt{3}\right)=\frac{l}{4 E A}(\sqrt{3}-3)\end{aligned}
Let us denote \lambda=\frac{1}{m \delta_0 \omega^2}, \delta_0=\frac{l}{4 E A} , then
\delta_{11}=\delta_0(3+\sqrt{3}), \quad \delta_{22}=\delta_0(1+3 \sqrt{3}), \quad \delta_{12}=\delta_{21}=\delta_0(\sqrt{3}-3)
and equation (b) becomes
\begin{aligned}& (3+\sqrt{3}-\lambda) A_1+(\sqrt{3}-3) A_2=0 \\& (\sqrt{3}-3) A_1+(1+3 \sqrt{3}-\lambda) A_2=0\end{aligned} \quad \text { or } \quad \begin{array}{r}(4.7320-\lambda) A_1-1.2679 A_2=0 \\-1.2679 A_1+(6.1961-\lambda) A_2=0\end{array} (c)
Frequency equation
D=\left[\begin{array}{cc}4.7320-\lambda & -1.2679 \\-1.2679 & 6.1961-\lambda\end{array}\right]=0
Roots (in decreasing order) of frequency equation and corresponding (in decreasing order) eigenvalues are
\begin{aligned}& \lambda_1=6.9280 \rightarrow \omega_1^2=\frac{1}{\lambda_1 m \delta_0}=\frac{4 E A}{6.9280 \mathrm{ml}}=0.5774 \frac{E A}{m l} \\& \lambda_2=4.0002 \rightarrow \omega_2^2=\frac{1}{\lambda_2 m \delta_0}=0.9999 \frac{E A}{m l} \cong \frac{E A}{m l}\end{aligned}
Mode shape of vibration may be determined on the basis of equation (b). For first mode \left(\lambda_1=6.9280\right) the ratio of amplitudes is
\begin{aligned}& \frac{A_2}{A_1}=-\frac{3+\sqrt{3}-\lambda_1}{\sqrt{3}-3}=\frac{4.7320-6.9280}{1.2679}=-1.73=-\sqrt{3} \\& \frac{A_2}{A_1}=-\frac{\sqrt{3}-3}{1+3 \sqrt{3}-\lambda_1}=\frac{1.2679}{6.1961-6.9280}=-\sqrt{3}\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]^T=\left\lfloor\begin{array}{ll}1 & -\sqrt{3}\end{array}\right]^T
For second mode \left(\lambda_2=4.0002\right) the ratio of amplitudes are
\begin{aligned}& \frac{A_2}{A_1}=\frac{4.7320-4.0002}{1.2679}=0.577=\frac{1}{\sqrt{3}} \\& \frac{A_2}{A_1}=\frac{1.2679}{6.1961-4.0002}=0.577=\frac{1}{\sqrt{3}}\end{aligned}
The modal matrix \boldsymbol{\Phi} is then defined as
\boldsymbol{\Phi}=\left[\begin{array}{cc}1 & 1 \\-\sqrt{3} & 1 / \sqrt{3}\end{array}\right]
Corresponding mode shapes of vibration are shown in Fig. 16.8c. Orthogonality condition is (Kiselev 1980)
\left\lfloor\begin{array}{ll} 1 & -\sqrt{3} \end{array}\right\rfloor\left[\begin{array}{cc} m & 0 \\ 0 & m \end{array}\right]\left[\begin{array}{c} 1 \\ 1 / \sqrt{3} \end{array}\right]=0
Control
\begin{aligned} & \text { 1. } \frac{1}{\omega_1^2}+\frac{1}{\omega_2^2}=m_1 \delta_{11}+m_2 \delta_{22}: \\ & \frac{1}{\omega_1^2}+\frac{1}{\omega_2^2}=\frac{m l}{0.5774 E A}+\frac{m l}{0.9999 E A}=2.7320 \frac{\mathrm{ml}}{E A} \\ & m_1 \delta_{11}+m_2 \delta_{22}=m \frac{1}{4 E A}(3+\sqrt{3})+m \frac{1}{4 E A}(1+3 \sqrt{3})=2.7320 \frac{\mathrm{ml}}{E A} \end{aligned}
Control 2. \left(\frac{A_2}{A_1}\right)_{\omega_1} \times\left(\frac{A_2}{A_1}\right)_{\omega_2}=-\frac{m_1}{m_2}
\left(\frac{A_2}{A_1}\right)_{\omega_1} \times\left(\frac{A_2}{A_1}\right)_{\omega_2}=-\sqrt{3} \times \frac{1}{\sqrt{3}}=-1, \quad-\frac{m_1}{m_2}=-1