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A shaft carries four rotating masses A,B,C,D in the order along its axis.

rA = 18cm, rB = 24cm, rC = 12cm and rD = 15cm, mB = 30kg, mC = 50kg and mD = 40kg.

The plane containing 'B' and 'C' are 30 cm apart, the angular spacing of planes containing 'C' and 'D' are 90 and 210 respectively. relative to 'B' measured in the same plane. If the shaft and masses are to be in completely dynamic balance,

Find:

  1. Mass and angular position of mass 'A'

  2. Position of planes 'A' and 'D'.

rA=0.18 m MA = ?
rB=0.24 m MB=30 kg
rC=0.12 m MC=50 kg
rD=0.15 m MD=40 kg

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Let, MA = Magnitude of mass A,

X = Distance between planes B and D,

Y = Distance between planes A and D

Plane Mass(m) kg Radius(r) m cent.force÷ω2 (mr) kg-m Dist from plane B(L)m couple ÷ω2 (mrL) kgm2
A mA 0.18 0.18 MA -Y -0.18 MAY
B(R.P.) 30 0.24 7.2 0 0
C 50 0.12 6 0.3 1.8
D 40 0.15 6 X 6X

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enter image description here

Force Polygon : 1cm = 1 kg - m

0.18 MA = 3.6 kg - m

MA = 20 kg

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  • The magnitude and angular position of mass Ma may be determined by drawing the force polygon to some suitable scale
  • Positions of planes A and D may be obtained by drawing couple polygon.

Couple Polygon :

  • From point c' and o' draw lines parallel to OD and OA respectively. such as they intersect at ' point d'

By measurement

6X = Vector c'd' = 2.3 kgm2

X=0.383 m enter image description here

We see from the couple polygon that the direction of vector c'd' is opposite to the direction of mass D. Therefore the planes of mass D is 0.383 n towards left of plane B and not towards right of plane B as already assumed.

  • Again by measurement,

    -0.18MAY = Vector o'd' = 3.6 kgm2

    -0.18×20×Y = 3.6, Y = -1m

  • The -ve sign indicates that plane A is not towards left of B as assumed but it is 1 m towards right of plane B.

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