Numerical : 4

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Numerical : 4

written 4.9 years ago by

bhaveshkumarn.pasi • 500

dynamics of machinery

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written 4.9 years ago by

bhaveshkumarn.pasi • 500

A shaft carries four masses in parallel planes A,B,C and D. The mass at B and C are 18 kg and 12.5 kg respectively and each has a eccentricity of 60 mm.The mass at A and D have an eccentricity 80 mm. The angle between masses B and C is 100 $^\circ$ and that between mass B an A is 190 $^\circ$. The axial distance between plane A and B is 10 cm and that B and C is 20 cm. If the shaft is in complete dynamic balance determine:

The masses at A and D.
The distance between plane C and D.
The angular position of mass D.

Plane	Mass(m) kg	Eccentricity(r) m	cent.force$\div \omega^2$ (m.r)	Dist from plane A(L)	couple $\div \omega^2$ (m.r.L) $kg.m^2$
$A \ (R.P.)$	$m_A$	0.08	0.08 $\ M_A$	0	0
B	18	0.06	1.08	0.1	0.108
C	12.5	0.06	0.75	0.3	0.225
D	$M_D$	0.08	$0.08 \ M_D$	X	$0.08 \ M_D.x$

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The position of mass B assumed in horizontal direction?

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Couple Polygon : 1 cm = 0.05 kg$\cdot \ m^2$

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By measurement,

$0.08 \ MDx \ = \ Vector \ c'o' = 0.235 \ kg \cdot m^2$. Now draw vector OD parallel to o'c' to fix the direction of mass D.

Force Polygon : 1 cm = 0.25

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$0.08 \ M_A \ = \ Vector \ cd \ = \ 0.77 \ kg-m$

$\therefore{M_A} \ = \ 9.625 \ \mathrm{kg}$

$0.08 \ M_D \ = \ Vector \ do \ = \ 0.65 \ kgm$

$\therefore M_D \ = \ 8.125 \ kg$

$ \rightarrow \ 0.08 \ M_D \cdot x \ = \ 0.235$

$0.08 \ \times \ 8.125 \ \times \ x \ = \ 0.235$

$x \ = \ 0.3615 \mathrm \ {m} \ = \ 361.5 \ \mathrm{mm}$

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