Consider a uniform cross section bar of length L made up of a material whose Yong's modulus and density are given by E and $\rho$.Estimate the natural frequencies of axial vibration of the bar using both consistent and lumped mass matrices.

Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 10M

Year: Dec 2016

$L\rightarrow he \rightarrow $ length of element.

$E \rightarrow$Modular of elasticity.

$S\rightarrow $density of material

$A \rightarrow$ cross sectional area .

(i) Natural freoquencies $w_i$ using consistent mass matrix .

The element matrix is given by,

$\frac{AE}{he}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$ $\begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}= \frac{w^2 \rho Ahe}{6}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$ $\begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}$

$\begin{bmatrix} \ \frac{AE}{he} & -\frac{2w^2\rho Ahe}{6} & -\frac{AE}{he} & -\frac{w^2\rho Ahe}{6} \\ \ -\frac{AE}{he} & -\frac{w^2\rho Ahe}{6} & \frac{AE}{he} & -\frac{2w^2\rho Ahe}{6} \\ \end{bmatrix}$ $\begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}= 0$

for not trivial solution

$\hspace{0.6cm}\begin{vmatrix} \ \frac{AE}{he} & -\frac{2w^2\rho Ahe}{6} & -\frac{AE}{he} & -\frac{w^2\rho Ahe}{6} \\ \ -\frac{AE}{he} & -\frac{w^2\rho Ahe}{6} & \frac{AE}{he} & -\frac{2w^2\rho Ahe}{6} \\ \end{vmatrix}$ = 0

Put the values of A,E, he and $\rho$ and solve above determinant to calculate $w_1$ and $w_2 $

where, w - Natural freoquencies.

(ii) Natural freoquencies $w_i$ using lumped mass matrix :

lumped mass matrix is given by ,

$\hspace{2cm}\frac{\rho Ahe}{2}$ $\begin{bmatrix} \ 1 & 0 \\ \ 0 & 1 \\ \end{bmatrix}$

$\therefore$ Element matrix eq' becomes,

$\frac{E}{he}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$ $\begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}= \frac{w^2 \rho Ahe}{2}$ $\begin{bmatrix} \ 1 & 0 \\ \ 0 & 1 \\ \end{bmatrix}$ $\begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}$

$\hspace{0.6cm}\begin{vmatrix} \ \frac{E}{he} & -\frac{\rho he}{2}w^2 & -\frac{E}{he} \\ \ -\frac{E}{he} & -\frac{E}{he} & \frac{\rho he}{2}w^2 \\ \end{vmatrix}$ $\begin{Bmatrix} \ u_1 \\ \ u_2 \\ \end{Bmatrix}= 0$

for non trivial solution

$\hspace{0.6cm}\begin{vmatrix} \ \frac{E}{he} & -\frac{\rho he}{2}w^2 & -\frac{E}{he} \\ \ -\frac{E}{he} & -\frac{E}{he} & \frac{\rho he}{2}w^2 \\ \end{vmatrix}$ = 0

Put the values of E, $he$ and $\rho$ and solve above determinant to calculate $w_1$ and $w_2 $