A 10-kg mass is connected to a spring of stiffness 3.000 N/m and is released after giving an initial displacement of 100 mm .Assuming that the mass on horizontal dry surface.determine the position at which the mass comes to rest. Assume the coefficient of friction between the mass and the surface to be 0.12.

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

Marks: 10M

Year: may 2016

Apply principal of minimum , potential Energy,

$\pi$ = spring Energy + potential Energy .

$\hspace{0.3cm}=\frac{1}{2} k _1 (u_2-u_1)^2+ \frac{1}{2} k_2(u_5-u_2)^2+\frac{1}{2} k_3(u_3-u_2)^2$

$\hspace{1.3cm} +\frac {1}{2} k_4 (u_4-u_3)^![enter image description here][2]2-f_1 u_3 -f_2 u_5$

$\hspace{0.3cm}$But $u_1 = u_4 =0 $

$\pi = \frac{1}{2} k_1 (u_2)^2+\frac{1}{2} k_2 (u_5-u_2)^2+\frac{1}{2}k_3(u_3-u_2)^2$

$\hspace{1.3cm} = \frac{1}{2} k_4 .u_3^2 - f_1 u_3 - f_2 u_5$

$\hspace{1.3cm} \frac{\partial \pi}{\partial u _1}$

$\therefore \frac{\partial \pi}{\partial u_2} = k_1 u_2 +k_2 (u_5-u_2)(-1) + k_3 (u_3-u_2)(-1) = 0$

$\hspace{1.2cm} (k_1+k_2+k_3)u_2 - k_3 u_3 - k_5 u_5 = 0 $

$\therefore 550.u_2 - 150u_3 - 300 u_5 = 0 \hspace{2cm} ---(1)$

$\frac{\partial \pi}{\partial u_3} = 0+0+k_3 (u_3-u_2)(1) + k_4 u_3 - f_1 = 0 $

$\hspace{1cm}-k_3u_2 +(k_3 +k_4) u_3 = f_1 $

$ \hspace{1cm}-150u_2+350u_3 = 200 \times 10^3 \hspace{2cm}---(2)$

$\frac{\partial \pi}{\partial u_5} = 0+k_2(u_5-u_2) = f_2$

$\hspace{1cm}-k_2 u_2 + k_2 u_5 = f_2 $

$\hspace{1cm}--300u_2 + 300u_5 = 100 \times 10^3 \hspace{2cm}---(3)$

Eq' (1)(2) & (3) in matrix from,

$\begin{bmatrix} \ 550 & -150 & -300 \\ \ -150 & 350 & 0 \\ \ -300 & 0 & 300 \\ \end{bmatrix} \begin{bmatrix} \ u_2 \\ \ u_3 \\ \ u_5 \\ \end{bmatrix} = \begin{bmatrix} \ 0 \\ \ 200\times10^3 \\ \ 100 \times 10_3 \\ \end{bmatrix}$

$\therefore u_2 = 1000 mm$

$\hspace{0.3cm}u_3 = 1000 mm$

$\hspace{0.3cm} u_5 = 1333.33 mm.$