An air-conditioner weighs 200 kg. and is driven by a motor at 500 r .p m What is the required static deflection of an undamped isolator to achieve 80% isolation?

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written 7.7 years ago by

teamques10 ★ 68k

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using cosine Rule,

AB = $\sqrt{Ac^2 +Bc^2-2Ac Bc .cos 32 . 4}$

$\hspace{0.5cm}= 3.82$

using sine Rule,

$\hspace{0.5cm} \frac{sin 32.4}{3.82}= \frac{sin \alpha}{3}\frac{sin \beta}{6}$

$\alpha = 24.00^0 \hspace{0.8cm}$&$\hspace{0.8cm} \beta=57.31^0$

Ele No.	Nodes	L	A	0	c	s	cs	$c^2$	$s^2$	$\frac{AE}{L}$
1	1-3	6000	20	$24.88^0$	0.907	0.421	1.328	0.8226	0.177	666.67
2	2-3	3000	20	$180-57.31=122.7^0$	-0.54	0.841	-0.454	0.292	0.708	1333.3

Elemental stiffness matrix ,

$K = \frac{AE}{L}$ $\begin{bmatrix} \ c^2 & cs & -c^2 & -cs \\ \ cs & s^2 & -cs & -s^2 \\ \ -c^2 & -cs & c^2 & cs \\ \ -cs & -s^2 & cs & s^2 \\ \end{bmatrix} $ where $C \rightarrow cos \theta$ $S \rightarrow sin \theta$

$K ^1=\begin{bmatrix} \ 548.403 & 885.34 & -548.403 & -885.34 \\ \ 885.34 & 118.0 & -885.34 & -118.0 \\ \ -548.403 & -885.34 & 548.403 & 885.34 \\ \ -885.34 & -118.0 & 885.34 & 118.0 \\ \end{bmatrix} $

$K ^2=\begin{bmatrix} \ 389.33 & -605.33 & -389.33 & 605.33 \\ \ -605.33 & 944.00 & 605.33 & -944.0 \\ \ -389.33 & 605.33 & 389.33 & -605.33 \\ \ 605.33 & -944.0 & -605.33 & 944.0 \\ \end{bmatrix} $

Global stiffness matrix ,

$K ^2=\begin{bmatrix} \ 548.403 & 885.34 & 0 & 0 & -548.403 & -885.334 \\ \ 885.34 & 118.0 & 0 & 0 & -885.34 & -118.0 \\ \ 0 & 0 & 389.33 & -605.33 & -389.33 & 605.33 \\ \ 0 & 0 & -605.33 & 944.0 & 605.33 & -944.0 \\ \ -548.403 & -885.34 & -389.33 & 605.33 & 937.733 & 280.01 \\ \ -885.34 & -118.0 & 605.33 & -944.0 & 280.01 & 1062 \\ \end{bmatrix} $ $\begin{Bmatrix} \ x_1 \\ \ y_1 \\ \ x_2 \\ \ y_2 \\ \ x_3 \\ \ y_3 \\ \end{Bmatrix}$ $\begin{Bmatrix} \ fx_1 \\ \ fy_1 \\ \ fx_2 \\ \ fy_2 \\ \ fx_3 \\ \ fy_3 \\ \end{Bmatrix}$

B.Cs $x_1=0 ,y_1 = 0 ,x_2 = 0, y_2 = 0 'fyx_1=fy_1=fx_2=fy_2fx_3 =0 fy_3 = -5000 N$

$\begin{bmatrix} \ 937.733 & 280.01 \\ \ 280.01 & 1062 \\ \end{bmatrix} $ $\begin{Bmatrix} \ x_3 \\ \ y_3 \\ \end{Bmatrix} $ $\begin{bmatrix} \ 0 \\ \ -5000 \\ \end{bmatrix} $

$\hspace{1.3cm}\therefore x_3 = 1.526mm$

$\hspace{1.6cm}y_3 = -5.1104mm$

Stress

$\hspace{0.6cm} 6 = \frac{E}{L}[-c\hspace{0.2cm} -s\hspace{0.2cm} c \hspace{0.2cm} s]$ $\begin{Bmatrix} \ x_1 \\ \ y_1 \\ \ x_2 \\ \ y_2 \\ \end{Bmatrix} $

$\hspace{0.6cm} 6 _1= \frac{2 \times 10^5}{3000}[-0.907\hspace{0.2cm} --0.421\hspace{0.2cm} 0.907\hspace{0.2cm} 0.421]$ $\begin{Bmatrix} \ 0 \\ \ 0 \\ \ 1.526 \\ \ -5.1104 \\ \end{Bmatrix} $

$\hspace{0.9cm} =-25.58 N/mm^2=25.58 N/mm^2(compressive)$

$\hspace{0.6cm} 6 _2= \frac{2 \times 10^5}{3000}[0.54\hspace{0.2cm} --0.841\hspace{0.2cm} -0.54\hspace{0.2cm} 0.841]$ $\begin{Bmatrix} \ 0 \\ \ 0 \\ \ 1.526 \\ \ -5.1104 \\ \end{Bmatrix} $

$\hspace{0.9cm}= - 341.46 N/mm^2 = 341.46 N/mm^2(compressive)$

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