A vibrometer having a natural frequency of 4 rad /s and damping ratio of 0.2 is attached to a structure that performs a harmonic motion. If the difference between the maximum and minimum recorded values is 8 mm, find the amplitude of motion of the vibrating structure when its frequency is 40 rad/s.

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

Marks: 10M

Year: may 2016

$x_1 = 1.5 \hspace{0.6cm} x_2 = 7 \hspace{0.6cm} x_3 = 4$

$y_1 = 2 \hspace{0.6cm} y_2 = 3.5 \hspace{0.6cm} y_3 = 7$

$\alpha_1 = x_2 y_3 -x_3 y_2=7 \times 7- 4 \times 3.5 =35$

$\alpha _2 = x+3 y_1 - x_1 y_3 =4 \times 2 -1 \times 5.7 = -2.5$

$\alpha_3 = x_1 y_2 - x_2 y_1 = 1.5\times3.5 - 7 \times2 = -8.75$

$\beta_1=y_2-y_3 = 3.5 -7 = -3.5$

$\beta = y_3 -y_1 = 7 -2 = 5$

$\beta_3 y_1-y_2 =2-3.5 = -1.5$

$r_1 =(x_2-x_3) = -(7-4)= -3$

$r_2= - (x_3-x_1)= -(4-1.5)= - 2.5$

$r_3 = -(x_1 - x_2) =-(1.5-7) = 5.5$

2A = $\begin{vmatrix} \ 1 & x_1 & y_1 \\ \ 1 & x_2 & y_2 \\ \ 1 & x_3 & y_3 \\ \end{vmatrix}$= $\begin{vmatrix} \ 1 & 1.5 & 2 \\ \ 1 & 7 & 3.5 \\ \ 1 & 4 & 7 \\ \end{vmatrix}$ = 23.75

N_1$=\frac{1}{2A}(\alpha_1 + \beta_1x+r_1y) = \frac{1}{23.75}(35-3.5x-3y)$

$\hspace{2cm}x = 3.85 ,y =4.8$

$\hspace{1cm}\therefore N_1 = \frac{1}{23.75}(35-3.5\times3.85-3\times4.8)= 0.3$

$\hspace{1.3cm}N_2 = \frac{1}{23.75}(-2.5+5\times3.85 - 2.5 \times 4.8) =0.2$

$\hspace{1.3cm}N_3 = \frac{1}{23.75}(-8.75-1.5\times3.85+5.5\times4.8) =0.5$

$\hspace{1cm}\therefore N_1+N_2+N_3 = 0.3+0.2+0.5=1$

$\hspace{5.6cm}$ Hence proved