written 5.8 years ago by |
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:
Mass and angular position of mass 'A'
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 |
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) kg−m2 |
---|---|---|---|---|---|
A | mA | 0.18 | 0.18 MA | -Y | -0.18 MA⋅Y |
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 |
Force Polygon : 1cm = 1 kg - m
0.18 MA = 3.6 kg - m
MA = 20 kg
- 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 kg−m2
∴X=0.383 m
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,
-0.18MA⋅Y = Vector o'd' = 3.6 kg−m2
-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.