Find two natrial frequencies of transverse vibration of a beam fixed at 60th axis. Use lumped mass matrix. Assume length of beam as 1 unit, FI = $10^6$ unit, $\zeta A = 10^6$ unit and Use consistent mass matrix.

a) Consistent Mass Matrix
$\hspace{2cm}$Element Matrix B given by,

Element matrix equation is given by
$\frac{EI}{h_e^3} \begin{bmatrix} 6&-3h_e&-6&-3h_e\\-3h_e&2h_e^2&3h_e&h_e^2\\-6&3h_e&6&3h_e\\-3h_e&h_e^2&3h_e&2h_e^2 \end{bmatrix} -\frac{w^2\,\delta\,A\,h_e}{420} \begin{bmatrix} 156&-22h_e&54&13h_e\\-22h_e&4h_e^2&-13h_e&-3h_e^2\\54&-13h_e&156&22h_e\\13h_e&-3h_e^2&22h_e&4h_e^2 \end{bmatrix} \begin{Bmatrix} V_1 \\ V_2 \\V_3\\V_4\end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2\\Q_3\\Q_4 \end{Bmatrix}$

Take $h_e=0.5\, and \,EI=\delta A=10^6$

$16 \begin{bmatrix} 6&-1.5&-6&-1.5\\-1.5&0.5&1.5&0.25\\-6&1.5&6&1.5\\-1.5&0.25&1.5&0.5 \end{bmatrix} -\frac{w^2}{840} \begin{bmatrix} 156&-11&54&6.5\\-11&1&-6.5&-7.5\\54&-6.5&156&11\\6.5&-7.5&11&1 \end{bmatrix} \begin{Bmatrix} V_1 \\ V_2 \\V_3\\V_4\end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2\\Q_3\\Q_4 \end{Bmatrix}$

$\therefore \begin{bmatrix} 96&-24&96&-24\\-24&8&24&4\\-96&24&96&24\\-24&4&24&8 \end{bmatrix} -{w^2} \begin{bmatrix} 0.186&-0.013&0.0645&0.0075\\-0.013&0.0012&-0.0075&-0.009 \\0.0645&-0.0075&0.186&0.013 \\0.0075&-0.009&0.013&0.0012 \end{bmatrix} \begin{Bmatrix} V_1 \\ V_2 \\V_3\\V_4\end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2\\Q_3\\Q_4 \end{Bmatrix}$

Global Matrix equation is given by,

$\therefore 10^6\begin{bmatrix} 96&-24&-96&-24&0&0\\-24&8&24&4&0&0\\-96&24&192&0&-96&24\\ -24&4&0&16&-24&4\\0&0&-96&24&96&24\\0&0&-24&4&24&8 \end{bmatrix} -{w^2} \begin{bmatrix} 0.186&-0.013&0.0645&0.0075&0&0\\-0.013&0.0012&-0.0075&-0.009 &0&0\\0.0645&-0.0075&0.372&0&0.0645&0.0075\\ 0.0075&-0.009&0&0.0024&-0.0075&-0.009\\0&0&0.0645&-0.0075&0.186&0.013 \\0&0&0.0075&-0.009&0.013&0.0012 \end{bmatrix} \begin{Bmatrix} V_1 \\ V_2 \\V_3\\V_4\\V_5\\V_6\end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2\\Q_3\\Q_4 \\Q_5\\Q_6\end{Bmatrix}$

Imposing global boundary condition.
$V_1=0,\,V_2=0,\,V_5=0,\,V_6=0$
For Balancing, $Q_3=0,\,Q_4=0,$ matrix reduces to,

$10^6 \begin{Bmatrix} \begin{bmatrix} 192 & 0 \\ 0 & 16 \end{bmatrix} -w^2 \begin{bmatrix} 0.372&0\\0&0.0024 \end{bmatrix} \end{Bmatrix} \begin{Bmatrix} V_3 \\ V_4 \end{Bmatrix} = \begin{Bmatrix} 0\\ 0\end{Bmatrix} $

$ i.e, \, \begin{bmatrix} 192-0.372w^2 & 0 \\ 0 & 16-0.0024w^2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix} $

For non-trival solution,
$\begin{vmatrix} 192-0.372\,w^2&0\\0&16-0.0024\,w^2 \end{vmatrix} =0$
$\therefore \,\, (192-0.372\,w^2)(16-0.0024\,w^2)=0 \hspace{0.3cm}=\gt\hspace{0.3cm} 0.00884w^4-6.4w^2+3072=0$
By Solving above equation we get,
$w^2=6722.91 \,\,and\,\, 516.91$
$\therefore\,\,w_1=82\,\,and\,\,w_2=22.74$

b) Lump Mass matrix
$\hspace{2cm}$Lumped mass Element matrix is given by
$\begin{bmatrix} M\end{bmatrix}^e=\delta\,Ah_e\begin{bmatrix} \frac{1}{2}&0&0&0\\0&(\frac{1}{78})h_e^2&0&0\\0&0&\frac{1}{2}&0\\0&0&0&(\frac{1}{78})h_e^2 \end{bmatrix}$

Now $\delta\,A=10^6\,\,\,h_e=0.5$
$\begin{bmatrix} M\end{bmatrix}^e=10^6\begin{bmatrix} 0.25&0&0&0\\0&0.0016&0&0\\0&0&0.25&0\\0&0&0&0.0016 \end{bmatrix}$

One assembly of two element global matrix will be,
$M=10^6\begin{bmatrix} 0.25&0&0&0&0&0\\0&0.0016&0&0&0&0\\0&0&0.5&0&0&0\\0&0&0&0.0032&0&0\\0&0&0&0&0.25&0\\0&0&0&0&0&0.0016 \end{bmatrix}$

Now global matrix equation will be,
$\therefore 10^6\begin{bmatrix} 96&-24&-96&-24&0&0\\-24&8&24&4&0&0\\-96&24&192&0&-96&24\\ -24&4&0&16&-24&4\\0&0&-96&24&96&24\\0&0&-24&4&24&8 \end{bmatrix} -{w^2} \begin{bmatrix} 0.25&0&0&0&0&0\\0&0.0016&0&0&0&0\\0&0&0.5&0&0&0\\0&0&0&0.0032&0&0\\0&0&0&0&0.25&0\\0&0&0&0&0&0.0016 \end{bmatrix} \begin{Bmatrix} V_1 \\ V_2 \\V_3\\V_4\\V_5\\V_6\end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2\\Q_3\\Q_4 \\Q_5\\Q_6\end{Bmatrix}$

Imposing boundary conditions,
$V_1=V_2=V_4=V_6=0,\,\, and \,\,\,Q_3=Q_4=0 \,\,for\, balancing\, matrix\, reduces \,to, $
$ i.e, \, \begin{bmatrix} 192-0.5w^2 & 0 \\ 0 & 16-0.0032w^2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix} $

For non-trival solution,
$\begin{vmatrix} 192-0.5\,w^2&0\\0&16-0.0032\,w^2 \end{vmatrix} =0$

$\therefore \,\, (192-0.5\,w^2)(16-0.0032\,w^2)=0 \hspace{0.3cm}=\gt\hspace{0.3cm} 0.0016w^4-8.164w^2+3072=0$
By Solving above equation we get,
$w^2=4693.42 \,\,and\,\, 409.084$
$\therefore\,\,w_1=68.5\,\,and\,\,w_2=20.22$