Balancing of Rotating Masses
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BALANCING OF ROTATING BODIES
SUMMARY In heavy industrial machines such as steam turbines, internal combustion engines and electric generators, unbalanced rotating bodies could cause vibration, which in turn could cause catastrophic failure. This chapter explains the importance of balancing rotating masses. It also explains both static and dynamic balance, i.e. balancing of coplanar and non-coplanar masses.
1. INTRODUCTION The balancing of rotating bodies is important to avoid vibrations. In heavy industrial machines such as steam turbines internal combustion engines and electric generators, vibration could cause catastrophic failure. Vibrations are noisy and uncomfortable and when a car wheel is out of balance, the ride is quite unpleasant. In the case of a simple wheel, balancing simply involves moving the centre of gravity to the centre of rotation but as we shall see, for longer and more complex bodies, there is more to it. For a body to be completely balanced it must have two things: static balance and dynamic balance. •
Static Balance (Single-plane balance). This occurs when the resultant of the centrifugal forces is equal to zero and the centre of gravity is on the axis of rotation.
•
Dynamic Balance (Two-plane balance). This occurs when there is no resulting turning moment along the axis.
2. STATIC BALANCE Despite its name, static balance does apply to things in motion. The unbalanced forces of concern are due to the accelerations of masses in the system. The requirement for static balance is simply that the sum of all forces on the moving system must be zero. Another name for static balance is single-plane balance, which means that the masses which are generating the inertia forces are in, or nearly in, the same plane. It is essentially a twodimensional problem. Some examples of common devices which meet this criterion, and thus can successfully be statically balanced, are: a single gear or pulley on a shaft, a bicycle or motorcycle tire and wheel, a thin flywheel, an airplane propeller, an individual turbine bladewheel (but not the entire turbine). The common denominator among these devices is that they are all short in the axial direction compared to the radial direction, and thus can be considered to exist in a single plane. An automobile tire and wheel is only marginally suited to static balancing as it is reasonably thick in the axial direction compared to its diameter. Despite this fact, auto tires are sometimes statically balanced. More often they are dynamically balanced. Figure l-a shows a link in the shape of a "vee", which is part of a linkage. We want to statically balance it. We can model this link dynamically as two point masses m1 and m2 concentrated at the local CGs of each "leg" of the link as shown in Figure l-b. These point masses each have a mass equal to that of the "leg" they replace and are supported on massless rods at the position (R1 or R2) of that leg's CG. We can solve for the required amount and location of a third "balance mass" to be added to the system at some location in order to satisfy the equilibrium.
Figure 1: Static Balancing
Assume that the system is rotating at some constant angular velocity ω. The accelerations of the masses will then be strictly centripetal (toward the centre), and the inertia forces will be centrifugal (away from the centre) as shown in Figure 1. Since the system is rotating, the figure shows a "freeze-frame" image of it. The position at which we "stop the action", for the
purpose of drawing the picture and doing the calculations, is both arbitrary and irrelevant to the computation. We will set up a coordinate system with its origin at the centre of rotation and resolve the inertial forces into components in that system. Writing the equilibrium equation for this system we get: 0 1
Note that the only forces acting on this system are the inertia forces. For balancing, it does not matter what external forces may be acting on the system. External forces cannot be balanced by making any changes to the system's internal geometry. Note that the ω2 terms cancel and equation (1-a) could be re-written as follows. 1
Breaking into x and y components: 1 The terms on the right sides are known. Then one can solve for the magnitude and direction of the product needed to balance the system.
1
1 is calculated from equation 1 , there is infinity of solutions After the product available. We can either select a value for and solve for the necessary radius at which it should be placed, or choose a desired radius and solve for the mass that should be placed there. Once a combination of and is chosen, it remains to design the physical counterweight. The chosen radius is the distance from the pivot to the CG of whatever shape we create for the counterweight mass. A possible shape for this counterweight is shown in Figure l-c. Its mass must be , distributed so as to place its CG at radius and at angle .
Example 1 (Static Balance) The system shown in Figure 1 has the following data: 1.2 1.135 at Ѳ 113.4 1.8 0.822 at Ѳ 48.8 Find the mass-radius product and its angular location needed to statically balance the system.
Solution: Ѳ
Ѳ 1.2 1.135 113.4 1.8 0.822 48.8
‐0.541 0.975
Ѳ 1.250 1.113
0.541 0.975 0.434 1.250 1.113 2.363 2.363 79.6 180 259.6 0.434 0.434 2.363 2.403 ·
This mass‐radius product of 2.403 · can be obtained with a variety of shapes appended to the assembly. Figure 1 shows a particular shape whose CG is at radius of 0.806 m at the required angle of 259.6 . The mass required for this counterweight design is then: 2.403 2.981 0.806 at a chosen CG radius of: 0.806
3. DYNAMIC BALANCE Dynamic balance is sometimes called two-plane balance. It requires that two criteria be met. The sum of the forces must be zero (static balance) plus the sum of the moments must also be zero. 0 2 0 These moments act in planes that include the axis of rotation of the assembly such as planes XZ and YZ in Figure 2. The moment's vector direction, or axis, is perpendicular to the assembly's axis of rotation. Any rotating object or assembly which is relatively long in the axial direction compared to the radial direction requires dynamic balancing for complete balance. It is possible for an object to be statically balanced but not be dynamically balanced. Consider the assembly in Figure 2. Two equal masses are at identical radii, 180o apart rotationally, but separated along the shaft length. A summation of -ma forces due to their rotation will be always zero. However, in the side view, their inertia forces form a couple which rotates with the masses about the shaft. This rocking couple causes a moment on the ground plane, alternately lifting and dropping the left and right ends of the shaft.
Some examples of devices which require dynamic balancing are: rollers, crankshafts, camshafts, axles, clusters of multiple gears, motor rotors, turbines, and propeller shafts. The common denominator among these devices is that their mass may be unevenly distributed both rotationally around their axis and also longitudinally along their axis.
Figure 2: Balanced Forces – Unbalanced Moments [1]
To correct dynamic imbalance requires either adding or removing the right amount of mass at the proper angular locations in two correction planes separated by some distance along the shaft. This will create the necessary counter forces to statically balance the system and also provide a counter couple to cancel the unbalanced moment. When an automobile tire and wheel is dynamically balanced, the two correction planes are the inner and outer edges of the wheel rim. Correction weights are added at the proper locations in each of these correction planes based on a measurement of the dynamic forces generated by the unbalanced, spinning wheel. It is always good practice to first statically balance all individual components that go into an assembly, if possible. This will reduce the amount of dynamic imbalance that must be corrected in the final assembly and also reduce the bending moment on the shaft. Consider the system of three lumped masses arranged around and along the shaft in Figure 3. Assume that, for some reason, they cannot be individually statically balanced within their own planes. We then create two correction planes labelled A and B. In this design example, the unbalanced masses m1, m2, m3 and their radii R1, R2, R3 are known along their angular locations t1, t2 and t3. We want to dynamically balance the system. A three-dimensional coordinate system is applied with the axis of rotation in the Z direction. Note that the system has again been stopped in an arbitrary freeze-frame position. Angular acceleration is assumed to be zero. The summation of forces is: 0 3 Dividing out the
and rearranging we get:
3 Breaking into x and y components: 3
Equations (3-c) have four unknowns in the form of products at plane A and products at plane B. To solve, we need the sum of the moments which we can take about a point in one of the correction planes such as point O. The moment arm (z-distance) of each force measured from plane A are labelled , , , in the figure; thus
Figure 3: Two‐plane Dynamic Balancing [1]
3 Dividing out the , breaking into x and y components and rearranging: The moment in the XZ plane (i.e., about the Y axis) is:
3
3 These can be solved for the products in x and y directions for correction plane B which can then be substituted into equation (3-c) to find the values needed in plane A. Equations (1d) and (1-e) can then be applied to each correction plane to find the angles at which the balance masses must be placed and the product needed in each plane. The physical
counterweights can then be deigned consistent with the constraints outlined in the section on and do not have to be the same value. static balance. Note that the radii
Example 2 (Dynamic Balance) The system shown in Figure 3 has the following data: 1.2 1.135 at Ѳ 113.4 1.8 0.822 at Ѳ 48.8 2.4 1.040 at Ѳ 251.4 The z-distances in metres from the plane A are: 0.854 1.701 2.396
3.097
Find the mass-radius products and their angular locations needed to dynamically balance the system using the correction planes A and B.
Solution: Ѳ Ѳ 1.2 1.135 0.854 113.4 1.8 0.822 1.701 48.8 2.4 1.040 2.396 251.4
Ѳ 1.250 1.113 ‐2.366
‐0.541 0.975 ‐0.796
‐0.462 1.660 ‐1.910
0.462
1.660 1.910 0.23 3.097 0.462 1.660 1.91 0.874 3.097 0.874 75.26 0.23 0.904 · 0.23 0.874
Solving equations (3‐c) for forces in x and y directions:
0.541
0.975
0.796
0.23
0.132
1.250
1.113
0.132
2.366
0.871 0.132 0.871
0.874
0.871
81.38 0.881
·
1.067 1.894 ‐5.668
These mass‐radius products can be obtained with a variety of shapes appended to the assembly in planes A and B. Many shapes are possible. As long as they provide the required mass‐radius products at the required angles in each correction plane, the system will dynamically balanced.
4. EXERCISES 4.1 A shaft carries four masses 200 kg, 300 kg, 240 kg and 260 kg respectively. The corresponding radii of rotation are 20 cm, 15 cm, 25 cm and 30 cm respectively and the angles between successive masses are 45o, 75o and 135o. Find the position and magnitude of the balance mass required, if its radius of rotation is 110 , 201 20 cm [2]. . 4.2 A shaft carries four masses A, B, C and D placed in parallel planes perpendicular to the shaft axis and in this order along the shaft. The masses of B and C are 36 kg and 25 kg respectively and both are assumed to be concentrated at a radius of 15 cm, while the masses A and D are both at radius of 20 cm. The angle between the radii of B and C is 100o and that between B and A is 190o, both angles are being measured in the same sense. The planes containing A and B are 25 cm apart and those containing B and C are 50 cm apart. If the shaft is to be in complete dynamic balance, determine [2]: a) The masses of A and D; b) The distance between the planes containing C and D, and c) The angular position of the mass D. . 19.5 , 16.5 252 , 13.2
5. CONCLUSION
Balancing of rotating masses in heavy industrial machines is very essential to reduce the unpleasant and dangerous vibration. Two balancing techniques have been introduced in this chapter, namely, static and dynamic balance. Two illustrative examples have been demonstrated in order to understand the two different techniques. Two exercises are left to the students to train themselves on solving balancing problems with final answers given to guide them.
6. REFERENCES [1] Norton, R.L. (1999): “DESIGN OF MACHINERY”, 2nd Edition, McGraw‐Hill, ISBN: 0‐07‐ 048395‐7, 1999. [2] Khurmi, R.S., Gupta, J.K. (1976): “THEORY OF MACHINES”, Eurasia Publishing House Ltd, 1976.
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