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Balancing Rotating Masses
The balancing of rotating bodies is important to avoid the damaging effects of vibration.
Vibrations are noisy and uncomfortable. For example, when a car wheel is out of balance, the
ride can be quite unpleasant.
Consider below a single mass moving in a circular arc of radius with angular velocity
rad/s. The mass has a centripetal (centre seeking) acceleration given by
. By Newton
2 the centripetal force acting on the mass is
.
This force must be reacted at the centre of rotation, at the bearing. This reaction is called the
centrifugal force and is equal in magnitude and opposite in sense to the centripetal force.
The centrifugal force acting on the bearing is therefore given by
.
[You will experience the effect of a centrifugal force if you swing a mass on the end of a piece
of string around in a circle.]
This bearing force, for a given value of , is of constant magnitude but varying direction as it
sweeps around the bearing axis at angular velocity . The force is a source of bearing load,
vibration, noise, etc and constitutes an unbalanced force which increases with .
In order to eliminate or balance this bearing force, a second mass
may be added
diametrically opposite the original mass (via an extension of the rotating arm, for example) at
a radius such that
or
.
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Static Balance
In the above case where the two masses are diametrically opposed and
balanced condition is both statically and dynamically balanced.
, the
Static balance is achieved because the static moment of the masses about the bearing axis,
are equal. For the masses and shown, the anti-clockwise moment is
and the
clockwise moment is
.
Since
it follows that the above static moments balance. (A stationary shaft
carrying a system of masses that are statically balanced will have no tendency to rotate in its
bearings.)
Balancing of Co-Planar Masses
Diagram shows a system of co-planar masses rotating about a common centre with the same
angular velocity . The radii are , , etc and the masses are
,
, etc. Any out of
balance will have a detrimental effect on the bearings and will cause vibration, noise, etc.
Each mass has a centripetal force
acting on it. Reaction forces, acting at the bearing,
are centrifugal (equal and opposite to the centripetal). Therefore in general the system of coplanar concurrent forces could be replaced by a resultant out of balance force.
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System of Co-Planar Forces
Since these forces are vectors, a graphical solution is often the most convenient way of
determining the out of balance force. This is illustrated below. The balancing vector is the
one which closes the
polygon and its direction is as shown in the diagram. Obviously the
out of balance
will be in the opposite direction.
MR Polygon to scale
Note that an analytical solution may also be used and will be demonstrated using a tutorial
problem.
Co-Planar Balancing Tutorials
1. Two masses revolve together in the same plane at an angular distance 45o apart. The
first is a 3 kg mass at a radius of 225 mm, the second 5kg at 175 mm. Calculate the
out of balance force at 2 rev/s and the position of a 10 kg balancing mass required to
reduce this force to zero.
226 N; balance mass at 143 mm radius and 160 o 33' to 5 kg mass
2. Determine the resultant out of balance force at the centre of rotation ‘O’, when the
system shown below rotates at 10 rev/min and state its direction. What value of
balance weight would be required at 1 m radius and where should it be placed?
55.94 N; 33o clockwise from mass C
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System of Co-Planar Concurrent Masses Rotating at 10 rev/min
3. Four masses A, B, C, and D, rotate together in a plane about a common axis O. The
masses and radii of rotation are as follows: A, 2 kg, 0.6 m: B, 3 kg, 0.9 m; C, 4 kg, 1.2
m; D, 5kg, 1.5 m. The angles between the masses are : Angle AOB =30 o, Angle BOC =
60o, Angle COD = 120o. Find the resultant out of balance force ar 12 rev/s and the
radius of rotation and angular position of a 10 kg mass required for balance.
21.6 KN; 380 mm; 39o 7' to OA
Balancing of Multi-Planar Rotating Masses
If the masses of the system rotate in different planes, the centrifugal force in addition to
being out of balance, form couples which must be eliminated if dynamic equilibrium is to be
achieved. Such a typical system is shown below.
Multi-Planar Rotating System
The first step is to transfer each centrifugal force to a suitably chosen datum and them draw
and
polygons for force and couples respectively. The reasons for the
polygon is
explained below.
Suppose
is rotating as shown in the diagram below. The centrifugal force is
.
This also has a moment
about the bearing O, which tends to bend the shaft and is
continuously changing direction.
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Now imagine that masses
and
are attached to point O, such that the centrifugal forces
are not only equal and opposite, but equal to the centrifugal force at Q. The addition does
not affect the equilibrium of the shaft and results in a pure couple
combined with a
downward out of balance force at point O. See diagram (a) below. This force has, in effect,
been “transferred” from Q to O. This transformed system is shown in diagram (b).
(a)
(b)
This transfer of forces can be done for any number of masses and also for any reference
plane, thus converting the problem to a uniplanar balancing one. For complete dynamic
balance, force and couples must be balanced. This means drawing
and
polygons for
forces and couples respectively.
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X1 – X2 is the neutral axis of the shaft when deflected by the couple
. The
value (to
which the couple is proportional) may therefore be represented to a suitable scale by a
vector pq drawn from some point p on this axis in the direction it would be travelling by a
right hand screw. The
value due to the unbalance of each of a number of parts spaced
along the shaft may be represented in the reference plane in the same way so enabling a
second or couple polygon to be drawn in the plane.
To simplify the drawing and to remove the problem of remembering whether lags or
vice-versa, Dalby’s convention recommends turning the couple vector through 90 o to be in
line with the force vector. (Note that if a reference plane is between rotating masses, one is
taken positive, one negative, hence negative is drawn in the opposite direction.)
Worked Example
A rotor 150 mm long is unbalanced by a mass of 120 g at 240 mm radius at 50 mm from one
end and a mass of 90 g at 180 mm radius at 40 mm from the other end at 150o anti-clockwise
from the first mass. Determine the magnitude and position of balancing masses to be
attached to the ends of the rotor at 150 mm radius to give dynamic balance. A diagram of the
arrangement is given below.
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m (g)
r (mm)
mr
x (mm)
mrx
A
MA
150
150MA
0
0
B
120
240
28800
50
1440000
C
90
180
16200
110
1782000
D
MD
150
150MD
150
22500MD
Using this data
draw mrx polygon.
This value of MD = 40g can now be substituted into the mr column of table such that 150MD =
150 x 40 = 6000. Thus the mrx polygon can be drawn.
From the above, A = 15800 = 150MA, therefore MA = 105.3g.
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Multi-Planar Balancing Problems
1. A shaft has 4 disc A, B, C, and D along its length 100 mm apart. A mass of 0.8 kg is
placed on B at a radius of 20 mm. A mass of 2 kg is placed on C at a radius of 30 mm
and rotated 120o from the mass on B. Find the masses to be placed on A and D at a
radius of 25 mm that will produce total balance.
0.696 kg and 1.52 kg
2. The diagram below shows masses on two rotors in planes B and C. Determine the
masses to be added on rotors in planes A and D at radius 40 mm which will produce
static and dynamic balance.
1.9 kg at 177o; 2.2 kg at 141o
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