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Static & Dynamic Balance Experiment

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INTRODUCTION

Shafts which revolve at high speeds must be carefully balanced if they are not to be a source of vibration.  If the shaft is only just out of balance and. revolves slowly .the vibration may merely be a nuisance but catastrophic failure can occur at high speeds even if the imbalance is small.

For example if the front wheel of a car is slightly out of balance this may be felt as a vibration of the steering wheel.  However if the wheel is seriously out of balance, control of the car may be difficult and the wheel bearings and suspension will wear rapidly, especially if the frequency of vibration coincides with any of the natural frequencies of the system.  These problems can be avoided if a small mass is placed at a carefully determined point on the wheel rim.

It is even more important to ensure that the shaft and rotors of gas turbine engines are very accurately balanced, since they may rotate at speeds between 15,000 and 50,000 rev/min.  At such speeds even slight imbalance can cause vibration and rapid deterioration of the bearings leading to catastrophic failure of the engine.

It is not enough to place the balancing mass such that the shaft will remain in any stationary position, i.e. static balance. When the shaft rotates, periodic centrifugal forces may be developed which give rise to vibration.  The shaft has to be balanced both statically and dynamically.

Usually, shafts are balanced on a machine which tells the operator exactly where he should either place a balancing mass or remove material.  The apparatus requires the student to balance a shaft by calculation or by using a graphical technique, and then to assess the accuracy of his results by setting up and running a motor driven shaft.  The shaft is deliberately made out of balance by clamping four blocks to it, the student being required to find the positions of the third and/or fourth blocks necessary to statically and dynamically balance the shaft.

Aims

The goals of this experiment are:

  • To gain an understanding of the meaning of the terms static and dynamic balance.
  • To demonstrate the dynamic balancing of an unbalanced shaft using two added eccentric masses.

Theory

A shaft with masses mounted on it can be both statically and dynamically balanced.  If it is statically balanced, it will stay in any angular position without rotating.  If it is dynamically balanced, it can be rotated at any speed without vibration.  It will be shown that if a shaft is dynamically balanced it is automatically in static balance, but the reverse is not necessarily true.

Static Balance

Figure SDB1 shows a simple situation where two masses are mounted on a shaft.  If the shaft is to be statically balanced, the moment due to weight of mass (1) tending to rotate the shaft anti-clockwise must equal that of mass (2) trying to turn the shaft in the opposite direction.

Figure SDB 1 - Static Balance

 

Hence for static balance,

m1r1cosa1 = m2r2cosa2

 

The same principle holds if there are more than two masses mounted on the shaft, as shown in figure SDB2.

Figure SDB 2 - Static Balance of Three masses

 

The moments tending to turn the shaft due to the out of balance masses are:-

Mass

Moment

Direction

1

m1.g.r1cosa1

Anticlockwise

2

m2.g.r2cosa2

Clockwise

3

m3.g.r3cosa3

Clockwise

For static balance,

m1r1cosa1   =   m2r2cosa2 + m3r3cosa3

In general the values of m, r and a have to be chosen such that the shaft is in balance.  However, for this experiment the product W.r can be measured directly for each mass and only the angular positions have to be determined for static balance.

If the angular positions of two of the masses are fixed, the position of the third can be found either by trigonometry or by drawing.  The latter technique uses the idea that moments can be represented by vectors as shown in figure SDB3(a).  The moment vector has a length proportional to the product mr and is drawn parallel to the direction of the mass from the centre of rotation.

                

Figure SDB 3 - (a) MR Vectors from Centre to Mass       (b) Rearrangement to show Closed Polygon

For static balance the triangle of moments must close and the direction of the unknown moment is chosen accordingly.  If there are more than three masses, the moment figure is a closed polygon as shown in Figure SDB3(b).  The order in which the vectors are drawn does not matter, as indicated by the two examples on the figure.

If on drawing the closing vector, its direction is opposite to the assumed position of that mass, the position of the mass must be reversed for balance.

Dynamic Balance

The masses are subjected to centrifugal forces when the shaft is rotating.  Two conditions must be satisfied if the shaft is not to vibrate as it rotates:

a)     There must be no out of balance centrifugal force trying to deflect the shaft.

b)     There must be no out of balance moment or couple trying to twist the axis of the shaft.

If these conditions are not fulfilled, the shaft is not dynamically balanced.

 

Figure SDB 4 - Dynamic Out-of-Balance for a Two Mass System

 

Applying condition a) to the shaft shown in Figure SDB.4 gives:

F1 = F2

The centrifugal force is mrw2

Therefore:

m1r1w2   =  m2r2w2

Since the angular velocity, w, is common to both sides then for dynamic balance

m1r1   =  m2r2

This is the same result for the static balance of the shaft.  Therefore if a shaft is dynamically balanced it will also be statically balanced.

 

The second condition is satisfied by taking moments about some convenient datum such as one of the bearings.

Thus,

a1F1 = a2F2

For this simple case where m1 and m2 are diametrically opposite and F1 = F2 (condition a) then dynamic balance can only be achieved by having a1 = a2 which means that the two masses must be mounted at the same point on the shaft.

 

 

 

Unlike static balancing where the position of the masses along the shaft is not important, the dynamic twisting moments on the shaft have to be eliminated by placing the masses in carefully calculated positions.  If the shaft is statically balanced it does not follow that it is also dynamically balanced.

 

In order for static balance to be achieved the sum of the vectors representing the couple due to each rotor must form a closed polygon.  In the case where there are three rotors, the simplest arrangement to give balance is shown in

 

Figure SDB 5 - Dynamically Balanced Shaft with 3 Eccentric Masses

In this case, it is clear from the first requirement that:

F2 = 2 x F1

The second criterion then says that:

a2F2 = a1F1 + a3F1

2a2F1 = a1F1 + a3F1

a2 = (a1 + a3)/2

or that the eccentric mass in the middle has twice the m.r value of the two masses on either side and is equidistant from both masses.

The general case, where the eccentric masses differ on each rotor and the directions are not exactly opposite is shown in Figure SDB6. 

 

Figure SDB 6 - General Case for Three Out-of-Balance Masses

 

The method for balancing such shafts requires the addition of two extra eccentric masses to the system at locations chosen by the engineer.  These masses are determined in many ways.  One method will be outlined in the lectures, while this experiment shows a method that uses preset out of balance forces and determines the orientation and position of the masses relative to the eccentric masses already on the shaft.

 

Equipment:

The Static and Dynamic Balance Equipment consists of a shaft mounted on a plate isolated from the base by rubber bushes.  A motor, also attached to the plate, may be used to drive the shaft using a belt.  Four identical eccentric mass blocks are provided, together with four individual inserts which may be used to alter the imbalance on each block.  An extension shaft and pulley are stored on the base and used in conjunction with the string/buckets and ball bearings to determine the imbalance associated with each block.  Two hexagonal keys (Allen keys) are provided to clamp the blocks on the shaft and the inserts into the blocks.  Two guides, one on the pulley at the end of the shaft and one on the plate, are used to measure the relative angle and position along the shaft of each block.  At either side of the plate two clamps allow the plate to be locked to the base.  The plate should be clamped for the static parts of the experiment.  If the motor is in use then the clamps should be released and secured away from the plate by tightening the screws.

 

Figure SDB 7 - Eccentric Masses and Axial Positioning

 

Experimental Procedure:

The experiment is divided into three parts:

  1. Observation of the phenomenon
  2. Calculation of a solution to a posed problem
  3. Implementation of the solution

 

Demonstration of Static and Dynamic Balance

For this part of the experiment the four blocks will be used without the inserts.  This provides four identical eccentric masses for use in demonstration of static and dynamic balance.

Step 1 - Static Imbalance

Lock the plate to the base.  Attach one of the blocks securely to the shaft near the pulley.  The mark on the protractor should align with zero when the block is positioned against the guide.  Slide the guide clear of the block.  Rotate the shaft and observe the behaviour.  Record your observations.

Step 2 - Static Balance

Attach a second block near the opposite end of the shaft.  Align this block such that the protractor reads 180o.  Rotate the shaft and observe the behaviour.  Record this and compare with step 1.

Step 3 - Dynamic Imbalance

Release the plate and secure the clamps clear of the plate.  Attach the belt between the motor and the pulley on the end of the shaft.  Ensure that all loose components are removed from the equipment and then place the safety cover over the motor and shaft.  Switch on the motor controller and the motor.  Slowly increase the speed of the motor and observe the behaviour of the shaft and plate.  Record your observations.  Switch off both the motor & controller and allow the shaft to come to rest before removing cover.

Step 3 - Dynamic Balance

There are three steps to this part of the experiment

  1. Preparation of the imbalances
    Clamp the plate and remove the blocks from the shaft.  Insert the four circular imbalances into the blocks and clamp them securely.  Attach one of the blocks to the shaft with the protractor reading 0o.  Remove the drive belt from the motor and attach the pulley extension to the shaft so that the pulley overhangs the end of the bench.  Loop the string and buckets around the pulley so that there is no slip.  Add ball bearings to one bucket until the protractor reads 90o.  At this point the moment due to the bearings is equal to the eccentric moment of the block.  Record the number of bearings and repeat for the other three blocks.
  2. Calculation of the solution
    The problem for solution must be posed carefully if a satisfactory solution is to be found in the time available.  The relative axial position and angular orientation of the two largest eccentricities should be selected first and will represent the dynamic imbalance to be corrected.  An axial spacing of 100 mm and relative angle of 150o provide a reasonable starting point.  The main limitations are the total length of the shaft and the thickness of each block.  The solution is calculated graphically in two parts.  Initially, the static balance of the system is obtained by
    1. Draw vectors representing the two imbalances set above.  These have a length proportional to the number of ball bearings and a direction relative to the angular orientation.  The vector should go from the centre of the shaft along the block.  The drawing should like Figure SDB 3b with the m1r1 and m2r2 vectors only.
    2. On the graph, knowing the lengths of the other two imbalances, complete the four sided closed polygon.  Record the angular orientation of the two balancing vectors.

The axial position of the two balancing masses needs to be calculated next.  In this case, it is simplest to take the largest eccentricity as the reference axial location, eliminating it from this part of the calculation.

    1. On a new diagram draw a vector representing the axial turning moment of the second eccentricity.  The length is proportional to the number of ball bearings and the axial distance from the reference.
    2. Complete the closed triangle using the directions for the balancing eccentricities found in b.  From the scale calculate the axial distance associated with each eccentricity.

Assemble all the information into a table indicating the eccentricity (mr) of each block, its axial location (l) and its angular orientation (q).

Figure SDB 8 - Graphical Solution to Dynamic Balance

  1. Test the result
    Carefully attach the blocks to the shaft at the locations and orientations in the table.  Remove the pulley from the system and reattach the motor drive belt.  Release the platform clamps and secure them to the base.  Put safety cover in place and run motor.  Record whether dynamic balance has been achieved and if necessary revise calculations.

Reporting

Your report should give a detailed description of your observations and include all rough work and calculations.  The graphs should be clearly labelled and neat.

 

 

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Last modified: February 19, 2002