For a uniform cross section bar as shown in figure of length L = 1m made up of a material having $E = 2 \times 10^{11}N/m^2$ and $\delta = 7800 kg/m^3$. Estimate the natural frequencies of axial vibrations of the bar using both consistent and lumped mass matrices. Use a two element mesh. If the exact solution is given by the relation.

$w_i = \frac{i \pi}{2L}\sqrt{\frac{E}{\delta}}, \hspace{1cm}$ i = 1, 3, 5, ...... $\infty$

Compare your answers wih the comments, $A = 30 \times 10^{-6}m^2$

Lump Mass matrix is given by,
$\begin{bmatrix} M\end{bmatrix}^e=\frac{\delta A\, h_e}{2} \begin{bmatrix} 1&0\\0&1\end{bmatrix}$

Element matrix equation is given by
$\frac{AE}{h_e} \begin{bmatrix} 1&-1\\-1&1 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix}=\frac{w^2\,\delta\,A\,h_e}{2} \begin{bmatrix} 1&0\\0&1 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} $

Take $ E=2*10^{11} N/m^2,\,\,\,\delta=7800\,kg/m^3,\,\,h_e=0.5,$
Cancelling A on both sides

$\therefore \frac{2*10^{11}}{0.5} \begin{bmatrix} 1&-1\\-1&1 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} =w^2\,\frac{7800*0.5}{6} \begin{bmatrix} 1&0\\0&1 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} $

$10^8\begin{bmatrix} 4&-4\\-4&4 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} =w^2\begin{bmatrix} 1.95&0\\0&1.95 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} $

Global Matrix equation written as

$10^8 \begin{bmatrix} 4&-4&0\\-4&8&-4\\0&-4&4 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2\\U_3 \end{Bmatrix} =w^2\begin{bmatrix} 1.95&0&0\\0&3.9&0\\0&0&1.95 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2\\U_3 \end{Bmatrix} $

Imposing global boundary condition , $\,\,U_1=0$
Matrix reduces to

$10^8 \begin{bmatrix} 8 & -4 \\ -4 & 4 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} = w^2 \begin{bmatrix} 3.9 & 0 \\ 0 &1.95 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} $

$\begin{bmatrix} 8*10^8-3.9\,w^2&-4*10^8\,w^2\\-4*10^8\,w^2&4*10^8-1.95\,w^2 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} =0$

For non-trivial solution,
$\begin{vmatrix} 8*10^8-3.9\,w^2&-4*10^8\,w^2\\-4*10^8\,w^2&4*10^8-1.95\,w^2 \end{vmatrix} =0$

$\therefore \,\, (8*10^8-3.9\,w^2)(4*10^8-1.95\,w^2)-(-4*10^8)^2=0$

$\therefore 7.505w^4-31.2*10^8w^2+16*10^{16}=0$

By Solving above equation, we get
$\therefore w^2=355804832.9$ and $59918018.58$

$\therefore w=18862.790 \,rad/s$ and $w=7740.673 \,rad/s$

Exact Solution,
$wi=\frac{i\pi}{2L}\sqrt{\frac{E}{\delta}}$

For, i=1
$w_1=\frac{\pi}{2}\sqrt{\frac{2*10^{11}}{7800}}= 7957.2\, rad/sec$

For, i=2
$w_2=2\frac{\pi}{2}\sqrt{\frac{2*10^{11}}{7800}}= 23871.7\, rad/sec$