Consider a uniform cross section of bar of Length L made up of a material whose youngs modulus and density are given by E and S. Estimate the natural frequency of axial vibration of the bar using both consistent and lumped mass matrices.

Using one element for entire rod i.e. h$_e$ = l.

Using lumped mass matrix:

$$ \frac{AE}{h_e} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} = \frac{\omega^2 \zeta Ah_e}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} $$

Since h$_e$ = l

$$ \frac{AE}{l} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} = \omega^2 \zeta Al \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} $$

Applying boundary condition, u$_1$ = 0

$$ \frac{AE}{l} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} 0 \\ u_2 \end{Bmatrix} = \omega^2 \zeta Al \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{Bmatrix} 0 \\ u_2 \end{Bmatrix} $$

$ \therefore \frac{AE}{l} u_2 = \omega^2 \frac{\zeta Al}{2}u_2 \\ \omega^2 = \frac{2E}{\zeta l^2} \\ \therefore w = \sqrt{\frac{2E}{\zeta l^2}} = \frac{1.414}{l} \sqrt{\frac{E}{\zeta}}$

Using consistent mass matrix

$$ \frac{AE}{h_e} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} = \frac{w^2 \zeta Ah_e}{6} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} $$

But, h$_e$ = l

$$ \frac{AE}{l} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} = \frac{w^2 \zeta Al}{6} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix} $$

Applying boundary condition, u$_1$ = 0

$$ \frac{AE}{l} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} 0 \\ u_2 \end{Bmatrix} = \frac{\omega^2 \zeta Al}{6} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{Bmatrix} 0 \\ u_2 \end{Bmatrix} $$

$ \frac{AE}{l}u_2 = \frac{\omega^2 \zeta A l}{6} 2u_2 \\ \frac{E}{l} = \frac{\omega^2 \zeta l}{3} \\ \omega^2 = \frac{3E}{\zeta l^2} \\ \therefore \omega = \sqrt{\frac{3E}{\zeta}{l^2}} = \frac{1.732}{l} \sqrt{\frac{E}{\zeta}} $