##### Document Text Contents

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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.

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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 two-

dimensional 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 blade-

wheel (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 . 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

Page 3

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

1

he terms on the right sides are known. Then one can solve for the magnitude and direction

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 2 terms

cancel and equation (1-a) could be re-written as follows.

Breaking into x and y components:

1

T

of the product needed to balance the system.

1

1

the product is calculated from equation 1 , there is infi ity of solutions

either

combination of and is chosen, it remains to design the physical counterweight.

,

After n

available. We can 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

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.

Page 4

Solution:

3. DYNAMIC BALANCE

ynamic balance is sometimes called two-plane balance. It requires that two criteria be met.

e z o ( ta nce) plus the sum of the moments must also be

D

T s tic bala

ze

nclude the axis of rotation of the assembly such as planes

otating 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

th assembly in

se arated along

the shaft length. A summation of -ma forces due to their rotation will be always zero.

eir inertia forces form a couple which rotates with the masses

he sum of the forces must b

ro.

er

hese moments act in planes that iT

XZ and YZ in Figure 2. The moment's vector direction, or axis, is perpendicular to the

assembly's axis of rotation.

Any r

object to be statically balanced but not be dynamically balanced. Consider e

Figure 2. Two equal masses are at identical radii, 180o apart rotationally, but p

However, in the side view, th

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.

Page 5

Some examples of devices which require dynamic balancing are: rollers, crankshafts,

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.

is always good practice to first statically balance all individual components that go into an

s t1, t2 and t3. We want to dynamically balance the system. A three-dimensional

oordinate system is applied with the axis of rotation in the Z direction. Note that the system

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.

It

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

location

c

has again been stopped in an arbitrary freeze-frame position. Angular acceleration is assumed

to be zero. The summation of forces is:

Dividing out the and rearranging we get:

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3

Breaking into x and y components:

3

Equations (3-c) have four unknowns in the form of products at plane A and products

t plane B. To solve, we need the sum of the moments which we can take about a point in one

n p 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:

a

of the correctio lanes such as point O.

3

3

hese can be solved for the products in x and y directions for correction plane B which

ane A. Equations (1-

lane to find the angles at which the

alance masses must be placed and the product needed in each plane. The physical

T

can then be substituted into equation (3-c) to find the values needed in pl

d) and (1-e) can then be applied to each correction p

b

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counterweights can then be deigned consistent with the constraints outlined in the section on

static balance. Note that the radii and do not have to be the same value.

Example 2

(Dynamic Balance)

The system shown in Figure 3 has the following data:

Find the mass-radius products and their angular locations needed to dynamically balance the

system using the correction planes A and B.

at

at

at

The z-distances in metres from the plane A are:

Solution:

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.

Page 2

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 two-

dimensional 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 blade-

wheel (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 . 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

Page 3

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

1

he terms on the right sides are known. Then one can solve for the magnitude and direction

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 2 terms

cancel and equation (1-a) could be re-written as follows.

Breaking into x and y components:

1

T

of the product needed to balance the system.

1

1

the product is calculated from equation 1 , there is infi ity of solutions

either

combination of and is chosen, it remains to design the physical counterweight.

,

After n

available. We can 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

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.

Page 4

Solution:

3. DYNAMIC BALANCE

ynamic balance is sometimes called two-plane balance. It requires that two criteria be met.

e z o ( ta nce) plus the sum of the moments must also be

D

T s tic bala

ze

nclude the axis of rotation of the assembly such as planes

otating 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

th assembly in

se arated along

the shaft length. A summation of -ma forces due to their rotation will be always zero.

eir inertia forces form a couple which rotates with the masses

he sum of the forces must b

ro.

er

hese moments act in planes that iT

XZ and YZ in Figure 2. The moment's vector direction, or axis, is perpendicular to the

assembly's axis of rotation.

Any r

object to be statically balanced but not be dynamically balanced. Consider e

Figure 2. Two equal masses are at identical radii, 180o apart rotationally, but p

However, in the side view, th

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.

Page 5

Some examples of devices which require dynamic balancing are: rollers, crankshafts,

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.

is always good practice to first statically balance all individual components that go into an

s t1, t2 and t3. We want to dynamically balance the system. A three-dimensional

oordinate system is applied with the axis of rotation in the Z direction. Note that the system

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.

It

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

location

c

has again been stopped in an arbitrary freeze-frame position. Angular acceleration is assumed

to be zero. The summation of forces is:

Dividing out the and rearranging we get:

Page 6

3

Breaking into x and y components:

3

Equations (3-c) have four unknowns in the form of products at plane A and products

t plane B. To solve, we need the sum of the moments which we can take about a point in one

n p 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:

a

of the correctio lanes such as point O.

3

3

hese can be solved for the products in x and y directions for correction plane B which

ane A. Equations (1-

lane to find the angles at which the

alance masses must be placed and the product needed in each plane. The physical

T

can then be substituted into equation (3-c) to find the values needed in pl

d) and (1-e) can then be applied to each correction p

b

Page 7

counterweights can then be deigned consistent with the constraints outlined in the section on

static balance. Note that the radii and do not have to be the same value.

Example 2

(Dynamic Balance)

The system shown in Figure 3 has the following data:

Find the mass-radius products and their angular locations needed to dynamically balance the

system using the correction planes A and B.

at

at

at

The z-distances in metres from the plane A are:

Solution: