Find the natural frequency of axial vibration of a bar having cross sectional area as $30 \times 10^{-4}m^2$, 1m length with left end fixed. Take $E = 2 \times 10^{11} N/m^2$. Density of the material is $7800kg/m^3$. Take two linear elements.

Subject: Finite Element Analysis

Topic: Finite Element Formulation of Dynamics and Numerical Techniques

Difficulty: Medium

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}=w^2\,\delta\,A\,h_e \begin{bmatrix} 2&1\\1&2 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} $

Take $h_e=0.5$ & 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} 2&1\\1&2 \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} 1.3&0.65\\0.65&1.3 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} $

Element matrix B same for both the element
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} 1.3&0.65&0\\0.65&2.6&0.65\\0&0.65&1.3 \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_2 \\ U_3 \end{Bmatrix} = w^2 \begin{bmatrix} 2.6 & 0.65 \\ 0.65 &1.3 \end{bmatrix} \begin{Bmatrix} U_2 \\ U_3 \end{Bmatrix} $

$\begin{bmatrix} 8*10^8-2.6\,w^2&-4*10^8-0.65\,w^2\\-4*10^8-0.65\,w^2&4*10^8-1.35\,w^2 \end{bmatrix} \begin{Bmatrix} U_1 \\ U_2 \end{Bmatrix} =0$

For non-trivial solution,
$\begin{vmatrix} 8*10^8-2.6\,w^2&-4*10^8-0.65\,w^2\\-4*10^8-0.65\,w^2&4*10^8-1.35\,w^2 \end{vmatrix} =0$

$\therefore \,\, (8*10^8-2.6\,w^2)(4*10^8-1.35\,w^2)-(-4*10^8-0.65)^2=0$

$\therefore32*10^{16}-20.8*10^8w^2+3.38w^4-16*10^{16}-5.2*10^8w^2-0.4225w^4=0$

$\therefore 2.9575w^4-26*10^8w^2+16*10^{16}=0$

By Solving above equation, we get
$\therefore w^2=812539840.6$ and $66581038.5$

$\therefore w=28505.085 \,rad/s$ and $w=8159721 \,rad/s$