One of the solution forms for free underdamped 1 d.o.F. vibration systems is given as : x(t) = $Ae^{-\epsilon m_et} sin (\omega_4t+\phi)$ where displacement amplitude A and phase angle $\phi$ are unknown constants Given initial disturbances in the form of displacement x$_0$ and v$_0$ , derive the expressions for the A and $\phi$

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

Marks: 5M

Year: may 2016

$x= 2.5, y= 2.5$

$\bar{x}= 2.5-2=0.5 ,\bar{y} = 2.5-2=0.5$

$\phi_1=\bigg(1-\frac{\bar{x}}{l}\bigg)\bigg(1-\frac{\bar{y}}{h}\bigg) = \bigg(1-\frac{0.5}{2}\bigg)\bigg(1-\frac{0.5}{1}\bigg)$

$=0.375 \hspace{5cm}L=2, h = 1$

$\phi_2=\frac{\bar{x}}{l}\bigg(1-\frac{\bar{y}}{h}\bigg)=\frac{0.5}{2} \bigg(1-\frac{0.5}{2}\bigg)= 0.125 $

$\phi_3 = \frac{\bar{x}\bar{y}}{Lh}=\frac{0.5\times0.5}{2\times1} = 0.125 $

$\phi_4\frac{\bar{y}}{h}\bigg(1-\frac{\bar{x}}{L}\bigg)=\frac{0.5}{2} \bigg(1-\frac{0.5}{2}\bigg) = 0.375$

$\phi_1+\phi_2+\phi_3+\phi_4 = 0.375+0.125+0.125+0.375 =1 $

$T(0.5 ,0.5)= T_1 \phi_1+T_2 \phi_2+T_3 \phi_3 +T_4 \phi_4$

$\hspace{1.9cm}=100 \times 0.375 +60 \times 0.125 + 50 \times 50 \times0.125 +90 \times 0.375$

$\hspace{1.9cm}=85^0 c$