Question 16.3: The beam in Fig. 16.9a carries three equal concentrated mass......

The beam has three degrees of freedom. The bending moment diagrams caused by unit inertial forces are shown in Fig. 16.9b. Multiplication of corresponding bending moment diagrams leads to the following results for unit displacements

   \begin{aligned} \delta_{11} & =\int \frac{\bar{M}_1 \bar{M}_1}{E I} d x=\frac{1}{E I}\left(\frac{1}{2} \frac{l}{4} \frac{3 l}{16} \frac{2}{3} \frac{3 l}{16}+\frac{1}{2} \frac{3 l}{4} \frac{3 l}{16} \frac{2}{3} \frac{3 l}{16}\right)=\frac{9}{768} \frac{l^3}{E I} \\ \delta_{22} & =\int \frac{\bar{M}_2 \bar{M}_2}{E I} d x=\frac{16}{768} \frac{l^3}{E I}, \quad \delta_{33}=\int \frac{\bar{M}_3 \bar{M}_3}{E I} d x=\frac{9}{768} \frac{l^3}{E I} \\ \delta_{12} & =\delta_{21}=\int \frac{\bar{M}_1 \bar{M}_2}{E I} d x=\frac{11}{768} \frac{l^3}{E I}, \quad \delta_{13}=\delta_{31}=\int \frac{\bar{M}_1 \bar{M}_3}{E I} d x=\frac{7}{768} \frac{l^3}{E I} \\ \delta_{13} & =\delta_{31}=\int \frac{\bar{M}_1 \bar{M}_3}{E I} d x=\frac{7}{768} \frac{l^3}{E I} \\ \delta_{23} & =\delta_{32}=\delta_{12}=\delta_{21}=\frac{11}{768} \frac{l^3}{E I} \end{aligned}

Let \delta_0=\frac{l^3}{768 E I} . Matrix of unit displacements \mathbf{F}  (flexibility matrix) is

\mathbf{F}=\left[\delta_{i k}\right]=\left[\begin{array}{lll} \delta_{11} & \delta_{12} & \delta_{13} \\ \delta_{21} & \delta_{22} & \delta_{23} \\ \delta_{31} & \delta_{32} & \delta_{33} \end{array}\right]=\delta_0\left[\begin{array}{ccc} 9 & 11 & 7 \\ 11 & 16 & 11 \\ 7 & 11 & 9 \end{array}\right]=\delta_0 \mathbf{F}_0

Equations (16.4) with unknown amplitudes A_i  of mass m_i  are

 \begin{aligned} & \left(m_1 \delta_{11} \omega^2-1\right) A_1+m_2 \delta_{12} \omega^2 A_2+\ldots+m_n \delta_{1 n} \omega^2 A_n=0 \\ & m_1 \delta_{21} \omega^2 A_1+\left(m_2 \delta_{22} \omega^2-1\right) A_2+\ldots+m_n \delta_{2 n} \omega^2 A_n=0 \\ & m_1 \delta_{n 1} \omega^2 A_1+m_2 \delta_{n 2} \omega^2 A_2+\ldots+\left(m_n \delta_{n n} \omega^2-1\right) A_n=0 \end{aligned}       (16.4 )

   \begin{aligned} & \left(m_1 \delta_{11} \omega^2-1\right) A_1+m_2 \delta_{12} \omega^2 A_2+m_3 \delta_{13} \omega^2 A_3=0 \\ & m_1 \delta_{21} \omega^2 A_1+\left(m_2 \delta_{22} \omega^2-1\right) A_2+m_3 \delta_{23} \omega^2 A_3=0 \\ & m_1 \delta_{31} \omega^2 A_1+m_2 \delta_{32} \omega^2 A_2+\left(m_3 \delta_{33} \omega^2-1\right) A_3=0 \end{aligned}      (a)

In our case all masses m_i=m . Divide by m \delta_0 \omega^2  and denote \lambda=\frac{1}{m \delta_0 \omega^2} . Equations for amplitudes A_i

 \begin{aligned} & (9-\lambda) A_1+11 A_2+7 A_3=0 \\ & 11 A_1+(16-\lambda) A_2+11 A_3=0 \\ & 7 A_1+11 A_2+(9-\lambda) A_3=0 \end{aligned}      (b)

Frequency equation becomes

   D=\operatorname{det}\left[\begin{array}{ccc} 9-\lambda & 11 & 7 \\ 11 & 16-\lambda & 11 \\ 7 & 11 & 9-\lambda \end{array}\right]=0

Eigenvalues in descending order

 \begin{aligned} \lambda_1 & =31.5563, \\ \lambda_2 & =2.0, \\ \lambda_3 & =0.44365 \end{aligned}             (c)

Verification:

1. The sum of the eigenvalues is \lambda_1+\lambda_2+\lambda_3=31.5563+2.0+0.44365=34 ; on the other hand, the trace of the matrix \mathbf{F}_0 (i.e., the sum of diagonal members) is \operatorname{Tr}\left(\mathbf{F}_0\right)=9+16+9=34 .
2. The multiplication of the eigenvalues is \lambda_1 \cdot \lambda_2 \cdot \lambda_3=31.5563 \times 2.0 \times 0.44365=28 ; on the other hand, \operatorname{det} \mathbf{F}_0=28 . (Note that determinant of unit displacement matrix is strictly positive, i.e., \operatorname{det} \mathbf{F}>0 ).

Frequencies of the free vibration in increasing order

 \begin{aligned} \omega_1^2 & =\frac{1}{\lambda_1 m \delta_0}=\frac{768}{31.5563} \frac{E I}{m l^3}=24.337 \frac{E I}{m l^3} \rightarrow \omega_1=4.9333 \sqrt{\frac{E I}{m l^3}}, \\ \omega_2^2 & =\frac{1}{\lambda_2 m \delta_0}=\frac{768}{2.0} \frac{E I}{m l^3}=384 \frac{E I}{m l^3} \rightarrow \omega_2=19.5959 \sqrt{\frac{E I}{m l^3}}, \\ \omega_3^2 & =\frac{1}{\lambda_3 m \delta_0}=\frac{768}{0.44365} \frac{E I}{m l^3}=1731.09 \frac{E I}{m l^3} \rightarrow \omega_3=41.6064 \sqrt{\frac{E I}{m l^3}} \end{aligned}       (d)

For each  ith eigenvalue the set of equation (b) for calculation of amplitudes is

 \begin{aligned} & \left(9-\lambda_i\right) A_1+11 A_2+7 A_3=0 \\ & 11 A_1+\left(16-\lambda_i\right) A_2+11 A_3=0 \\ & 7 A_1+11 A_2+\left(9-\lambda_i\right) A_3=0 \end{aligned}     (e)

Equations (e) are divided by A_1 . Let \rho_2=\frac{A_2}{A_1}, \quad \rho_3=\frac{A_3}{A_1} . Equations for modes become

   \begin{aligned} & \left(9-\lambda_i\right)+11 \rho_{2 i}+7 \rho_{3 i}=0, \\ & 11+\left(16-\lambda_i\right) \rho_{2 i}+11 \rho_{3 i}=0, \\ & 7+11 \rho_{2 i}+\left(9-\lambda_i\right) \rho_{3 i}=0 \end{aligned}      (f)

Assuming A_1=1  we can calculate \rho_2  and \rho_3  for each calculated eigenvalue. For their calculation we can consider set of any two equations.

1.  Eigenvalue \lambda_1=31.5563

\begin{aligned}& (9-31.5563)+11 \rho_2+7 \rho_3=0 \\& 11+(16-31.5563) \rho_2+11 \rho_3=0\end{aligned}

Solution of these equations is \rho_2=1.4142, \rho_3=1.0 . Therefore the first (principal) mode is defined as y_{11}, \quad y_{21}=\sqrt{2} y_{11}, \quad y_{31}=y_{11} . If we assume that y_{11}=1 , then the eigenvector \boldsymbol{\varphi}_1  which corresponds to the frequency \omega_1  is \boldsymbol{\varphi}_1=\left\lfloor\begin{array}{lll}1.0 & \sqrt{2} & 1\end{array}\right]^T .

Corresponding mode shape of vibration is shown in Fig. 16.9c.

2. Eigenvalue \lambda_2=2.0 . In this case

\begin{aligned}& (9-2.0)+11 \rho_2+7 \rho_3=0 \\& 11+(16-2.0) \rho_2+11 \rho_3=0\end{aligned}

Solution of these equations is \rho_2=0.0, \quad \rho_3=-1.0 , and second eigenvector becomes \boldsymbol{\varphi}_2=\left\lfloor\begin{array}{lll}1.0 & 0.0 & -1\end{array}\right\rfloor^T ; this mode shape of vibration is shown in Fig. 16.9d.

3. Eigenvalue \lambda_3=0.44365 . In this case

\begin{aligned}& (9-0.44365)+11 \rho_2+7 \rho_3=0 \\& 11+(16-0.44365) \rho_2+11 \rho_3=0\end{aligned}

Solution of these equations is \rho_2=-1.4142, \quad \rho_3=1.0 , and eigenvector \boldsymbol{\varphi}_3=\left\lfloor\begin{array}{lll}1.0 & -\sqrt{2} & 1\end{array}\right]^T. Third mode shape of vibration is shown in Fig. 16.9 \mathrm{e} .

We can see that the number of the nodal points of the mode shape of vibration is one less than the number of the mode.

The modal matrix is defined as

   \boldsymbol{\Phi}=\left\lfloor\begin{array}{lll} \boldsymbol{\varphi}_1 & \boldsymbol{\varphi}_2 & \boldsymbol{\varphi}_3 \end{array}\right]=\left[\begin{array}{lll} \varphi_{11} & \varphi_{12} & \varphi_{13} \\ \varphi_{21} & \varphi_{22} & \varphi_{23} \\ \varphi_{31} & \varphi_{32} & \varphi_{33} \end{array}\right]=\left[\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{2} & 0.0 & -\sqrt{2} \\ 1 & -1 & 1 \end{array}\right],         (g)

where the i  th and k  th indexes at \varphi_{i k}  mean the number of mass and number of frequency (or form), respectively. The vectors \varphi_1  and \varphi_2  are orthogonal. Indeed

   \varphi_1^T \varphi_2=\left\lfloor\begin{array}{lll} 1.0 & \sqrt{2} & 1 \end{array}\right\rfloor\left[\begin{array}{c} 1 \\ 0.0 \\ -1 \end{array}\right]=1 \times 1+\sqrt{2} \times 0.0+1 \times(-1)=0

Similarly, it is easy to check the orthogonality conditions for other form of vibration: \varphi_1^T \varphi_3=\varphi_2^T \varphi_3=0 . Above procedure may be applied for analysis of free vibration of the arbitrary linear deformable structures. The detailed examples of free vibration of different types of uniform arches (three-hinged, hingeless, circular, parabolic, etc.) are presented in book (Karnovsky 2012).