Balancing of Rotating Masses

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1 Balancing of Rotating Masses 1

2 Balancing of Rotating Masses m Consider first a single mass m moving in a circular arc of radius r with an angular velocity rad/s. The mass has a centripetal (centre 2 seeking) acceleration given by a r. By Newton s second law the centripetal force acting 2 on the mass is F mr. Rotating Mass Centrifugal Force This force must be reacted at the centre of rotation, i.e. 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 2 on the bearing is therefore given by F mr. 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 M may be added diametrically opposite the original mass (via an extension of the rotating arm, for example) at a radius R such that MR 2 = mr 2 or MR = mr. 2

3 Static Balance In the above case where the two masses are diametrically opposed and MR = mr, the balanced condition is both statically and dynamically balanced. Static balance is achieved because the static moment of the masses about the bearing axis, are equal. For the masses M and m shown, the anti-clockwise moment is MgRCos and the clockwise moment is mgrcos. Static Balance Since MR = mr 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 r 1, r 2, etc and the masses are m 1, m 2, etc. Any out of balance will have a detrimental effect on the bearings and will cause vibration, noise, etc. m 4 m 1 r 4 r 2 r 1 Each mass has a centripetal force mr 2 acting on it. Reaction forces, acting at the bearing, are centrifugal (equal and opposite to the centripetal). Therefore, in general the system of co-planar concurrent forces could be replaced by a resultant out of balance force. r 3 R m 3 m 2 M Since these forces are vectors, a graphical approach is often the most convenient for determining the out of balance force. 3

4 m 2r 2 m 1r 1 MR (Balancing vector) m 4r 4 m 3r 3 If we let the resultant be MR 2, then the system could be balanced by adding mass M at a radius R in the correct direction. This is found graphically by plotting an mr polygon as shown above. Note that 2 can be ignored since it is common to all vectors. Worked Example 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? A 90kg D 45 kg 0.6m B 22.5kg 30 o 0.6m 60 o 45 o 45 o 0.3m The information to draw the mr polygon is best tabulated. C 45kg 0.6m System of co-planar concurrent masses rotating at 10 rev/min. m (kg) r (m) mr A B C D The mr polygon is then draw to scale. 4

5 B C mr polygon (NTS) D MR = 51 Required to balance the system. A 12 o The resultant out of balance force is in the opposite direction to that shown on the mr polygon. A Resultant out of balance force D MR = 51 = M x 1 Therefore, M = 51kg Out of balance force = MR 2 B 12 o 1m C 2 10 = 51 x 1 x 60 =55.94N 2 51kg 5

6 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 mr and mrx polygons for force and couples respectively. The reasons for the mrx polygon is explained below. Suppose m is rotating as shown in the diagram below. The centrifugal force is F = mrω 2. This also has a moment Fx about the bearing O, which tends to bend the shaft and is continuously changing direction. Now imagine that masses m 1 and m 2 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 Fx combined with a downward out of balance force F 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). 6

7 (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 mr and mrx polygons for forces and couples respectively. X 1 X 2 is the neutral axis of the shaft when deflected by the couple Fx. The mrx 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 mrx 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 C lags F 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.) 7

8 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. m (g) r (mm) mr x (mm) mrx A M A M A 0 0 B C D M D M D M 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 = Thus the mrx polygon can be drawn. 8

9 From the above, A = = 150M A, therefore M A = 105.3g. 9

10 Balancing of Rotating Masses Analytical Method A more modern and accurate approach to solving out-of-balance problems is by the use of an analytical methods, which is best illustrated by the use of a worked example. Again since all masses will be rotating with the same angular velocity we can continue to work in terms of mr and mrx values. To this end let s use the previous graphically worked example, which is given again below to see if a more accurate result can be achieved. Worked Example - Co-Planar 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? A 90kg 0.6m 45kg D 0.6m 60 o 45 o 30 o 45 o B 0.6m 0.3m C 22.5kg 45kg We begin by tabulating the information we have System of co-planar concurrent masses rotating at 10 rev/min. m(kg) r(m) rm A B C D E m E 1.0 m E x o 45 o 30 o 45 o We begin by summing the mr values in the x direction, taking those to the right as positive. + mr x = 54 cos cos cos cos 45 = kgm Next we sum the mr values in the y direction, taking those upward at positive 10

11 mr y + = 54 sin sin sin sin 45 = kgm For the resultant we have kgm kgm R mr θ R mr = = = m E 1 Therefore m E = kg Angle θ being given by tan θ = or θ = tan 1 ( ) Therefore θ = It follows that the out-of-balance force mrω 2 = ( 2π )2 = N A Resultant out of balance force D Original system with balancing mass added for complete dynamic balance. B o C 1m 50.57kg 11

12 Worked Example - Multi-Planar 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. m (g) r (mm) mr x (mm) mrx A M A M A 0 0 B C D M D M D M D o Net horizontal mrx ( +) = cos 60 = Net vertical mrx ( +) = sin 60 = Thus, we have R = = θ θ= tan = But = 22500M D 12

13 Therefore M D = kg say 40 kg Now = 6.6 So, it follows that M D = 40 k and will be placed at an angle of 6.6 o anticlockwise from zero datum. In the above table 150M D = = 6000, thus we have o Therefore, net horizontal mr ( +) = 6000 cos cos 60 = and net vertical mr ( +) = sin sin 60 = Therefore, we have θ R= = θ= tan = But 150M A = , therefore M A = g, say 104 g For complete balance we have: B (or from zero datum) 6.6 o D 60 O O C A 13

14 Tutorial Problems - Co-Planar 1. Determine the size and position of the mass required at 150 mm radius to balance the following co-planar system. 5 kg mass at 100 mm radius 10 kg mass at 75 mm radius, 90 o clockwise from the 5 kg mass 15 kg mass at 100 mm radius, 240 o anti-clockwise from 5 kg mass. Find also the size of each of two balancing weights which could be substituted for the single one already found if these are to be at 75 mm radius and positioned at 30 o and 150 o anticlockwise from 5 kg mass. 2. Two masses revolve together in the same plane at an angular distance of 45 o apart. The first is a 3 kg mass at a radius of 225 mm, the second 5 kg at 175 mm radius. Calculate the out-ofbalance force at 2 rev/s and the position of a 10 kg balance mass required to reduce this force to zero. (226 N; balance mass at 143 mm radius and 160 o 33 to 5 kg mass) 3. A casing is bolted to the face plate of a lathe. It is equivalent to 2 kg at 50 mm from the axis of rotation, another 1 kg at 75 mm radius and 4 kg at 25 mm radius. The angular positions are respectively, 0 o, 30 o, 75 o. Find the balance mass required at 150 mm radius to eliminate the outof-balance force. State the angular position of the balance mass. (1.56 kg; 215 o from 2 kg mass) 4. A turbine casing is placed on a rotating table mounted on a vertical axis. The casing is symmetrical except for projecting lug of mass 15 kg at a radius of 1.2 m and a cast pad of mass 25 kg at 0.9 m radius. The lug and the pad are positioned at right angles to one another. The casting is bolted down symmetrically with respect to the axis of rotation. Find the magnitude and position of the balancing mass required at a radius of 1.5 m. (19.2 kg at 141 o 41 to pad) 5. Two equal holes are drilled in a uniform circular disc at a radius of 400 mm from the axis. The mass of material removed is g. Calculate the resultant out-of-balance force if the holes are spaced at 90 o to each other and the speed of rotation is 1000 rev/min. Where should a mass be placed at a radius of 250 mm in order to balance the disc, and what should be its magnitude? (582 N; kg at 45 o to a drilled hole) 6. Three masses are bolted to a face plate as follows: 5 kg at 125 mm radius, 10 kg at 75 mm radius, and 7.5 kg at 100 mm radius. The masses must be arranged so that the face plate is in balance. Find the angular position of each the masses relative to the 5 kg mass. (Each at 114 o 36 to 5 kg mass) 7. 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, 5 kg, 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 at 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) 8. Four mass m 1, m 2, m 3, and m 4 are 200 kg, 300 kg, 240 kg, and 260 kg respectively. The corresponding radii of rotation are 0.2 m, 0.15 m, 0.25 m and 0.3 m respectively and the angles between successive masses are 45 o, 75 o, and 135 o. Determine the position and magnitude of the 14

15 balancing mass required, if its radius of rotation is 0.2 m. (Angles quoted are to be taken to be in the anti-clockwise direction.) (115 kg, 201 o anti-clockwise from the 200 kg mass) Multi-Planar 9. A shaft has 4 discs 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) 10. 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 ) 15

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[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.]

[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.] 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,