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. 1
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. 2
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 45 o 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; 33 o clockwise from mass C 3
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 = 60 o, Angle COD = 120 o. 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; 39 o 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. 4
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. 5
X 1 X 2 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 150 o 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. 6
m (g) r (mm) mr x (mm) mrx A M A 150 150M A 0 0 B 120 240 28800 50 1440000 C 90 180 16200 110 1782000 D M D 150 150M D 150 22500M D Using this data draw mrx polygon. This value of M D = 40g can now be substituted into the mr column of table such that 150M D = 150 x 40 = 6000. Thus the mrx polygon can be drawn. From the above, A = 15800 = 150M A, therefore M A = 105.3g. 7
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 120 o 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 177 o ; 2.2 kg at 141 o 8