Find the natural frequency of axial vibrations of a bar of uniform cross section 20$mm^2$ and length Im. Take E = 2 x $10^5 N/mm^2$and $p=8000kg/m^3$. Take two linear elements.

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written 7.7 years ago by

teamques10 ★ 68k

• modified 7.7 years ago

$L = \frac{1}{2} = 0.5m$

Elemental matrix equation

$\frac{AE}{L} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix} = \frac{W^2 \delta A L}{6} \begin{bmatrix} \ 2 & 1 \\ \ 1 & 2 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}$

$\frac{E}{L} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix} = \frac{W^2 \delta A L}{6} \begin{bmatrix} \ 2 & 1 \\ \ 1 & 2 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}$

$\frac{2 \times 10^11}{0.5} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix} = \frac{w^28000 \times 0.5}{6} \begin{bmatrix} \ 2 & 1 \\ \ 1 & 2 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}$

$10^8 \begin{bmatrix} \ 4 & -4 \\ \ -4 & 4 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix} = w^2 \begin{bmatrix} \ 1.33 & 0.67 \\ \ 0.67 & 1.33 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}$

Global matrix equation

$10^8 \begin{bmatrix} \ 4 & -4 & 0 \\ \ -4 & 8 & -4 \\ \ 0 & -4 & 4 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \ u_3 \\ \end{Bmatrix} = w^2 \begin{bmatrix} \ 1.33 & 0.67 & 0 \\ \ 0.67 & 2.66 & 0.67 \\ \ 0 & 0.67 & 1.33 \\ \end{bmatrix} \begin{Bmatrix} \ u_1 \\ \ u_2 \\ \ u_3 \\ \end{Bmatrix}$

Boundary conditions $u_1 = 0$

$10^8 \begin{bmatrix} \ 8 & -4 \\ \ -4 & 4 \\ \end{bmatrix} \begin{Bmatrix} \ u_2 \\ \ u_3 \\ \end{Bmatrix} = w^2 \begin{bmatrix} \ 2.66 & 0.67 \\ \ 0.67 & 1.33 \\ \end{bmatrix} \begin{Bmatrix} \ u_2 \\ \ u_3 \\ \end{Bmatrix}$

$\begin{bmatrix} \ 8 \times 10^8 - 2.66w^2 & -4 \times 10^8 - 0.67w^2 \\ \ -4 \times 10^8 & 4 \times 10^8 - 1.33w^2 \\ \end{bmatrix} \begin{Bmatrix} \ u_2 \\ \ u_3 \\ \end{Bmatrix} = 0$

$\begin{bmatrix} \ 8 \times 10^8 - 2.66w^2 & -4 \times 10^8 - 0.67w^2 \\ \ -4 \times 10^8 & 4 \times 10^8 - 1.33w^2 \\ \end{bmatrix} = 0$

$( 8 \times 10^8 - 2.66w^2) (4 \times 10^8 - 1.33w^2) - (-4 \times 10^8 - 0.67w^2)^2 = 0\\ 3.1w^2 - 26.68 \times 10^8 w^2 + 16 \times 10^16 = 0\\ \therefore w^2 = 795787517.7 \hspace{0.2cm}\text {and} \hspace{0.2cm} 64857643 \\ \therefore w = 28209.7 rad/s \hspace{0.2cm} and \hspace{0.2cm} 8053 rad/s$

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