DYNAMICS OF ROTATING SYSTEMS questions Mechanical Principles

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MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : DYNAMICS OF ROTATING SYSTEMS

LESSON 2 : THE BALANCE OF MASSES ROTATING

IN THE SAME PLANE

MP - 4 - 2

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

The dynamic forces which are produced by the rotating parts of machinery can

cause high alternating stresses in these components, leading to fatigue failure,

large cyclic loading on the connections between the components, e.g. bearings,

and are a major cause of unwanted vibration in machinery. The basic cause of

these dynamic, or inertia, forces is when the centre of mass of the component

does not coincide with the centreline of the shaft to which it is attached, i.e. its

axis of rotation. The component is said to be 'unbalanced' and the force

required to keep this mass rotating in circular motion is called the out-of-

balance, or the inertia, or the dynamic, or the centrifugal force.

All types of machinery which have rotating parts are susceptible to centrifugal

loading as manufacturing processes cannot ensure that the centre of mass and

the axis of rotation exactly coincide – some examples are engine flywheels,

brake discs, road wheels, gear wheels, rotors of electric motors, steam and gas

turbine rotors. Some components are asymmetrical by design, such as cams,

engine crankshafts, links and levers in high speed textile machinery. All of

these components need to be 'balanced' by deliberately introducing further

forces to oppose, or counterbalance, the inertia forces.

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________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

When you have completed this lesson you should be able to:

• identify the forces caused by unbalanced rotating components

• calculate the loads on the shaft and support bearings caused by

unbalanced components

• describe the function of counterbalancing in reducing or eliminating

the forces on the support bearings

• draw the vector diagram for the forces caused by the circular motion

of several masses rotating in the same plane

• use the force diagram to evaluate the required counterbalance size

and location.

________________________________________________________________________________________

STUDY ADVICE ________________________________________________________________________________________

You will need graph paper, ruler, protractor and set square to complete this

lesson.

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________________________________________________________________________________________

CENTRIFUGAL FORCE AND BEARING REACTIONS ________________________________________________________________________________________

Component parts of machinery that are attached to rotating shafts can cause

significant forces on the shaft when the centres of mass of the components do

not coincide with the axis of rotation.

FIG. 1

This system can be modelled by a 'point' mass, equal to that of the component,

rotating in a circular path about O with a radius r which is the same length as

OG [see FIGURE 2(a)]. This distance, by which the centre of mass of the

component is offset from the axis through O, is often called the eccentricity of

the mass centre. The mass is imagined to be connected to the shaft by a rod

having no mass.

Centre of mass

Component

Mass m kg Axis

of rotation

Support bearings

ω r

Rotating shaft

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FIG. 2

The point mass has an acceleration of r × ω2 in the radial direction directed towards the centre, FIGURE 2(b). This is due to the velocity of the mass

changing direction as the mass moves around the circle. This acceleration is

termed either (a) the radial or (b) the centripetal (meaning centre seeking)

component of acceleration, which has been introduced in an earlier lesson.

According to Newton’s 2nd law of motion, there must be a force applied to the

mass to cause that acceleration.

Where does it come from?

What is its magnitude?

In what direction is it?

What is the effect of the mass at G on the shaft?

You will need to recall Newton’s 3rd law of motion – 'to every action there is

an equal and opposite reaction'.

Circular path

ω rm

Shaft

Point mass at G of m kg

(a) (b)

aR = rω 2

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FIG. 3(a) and (b)

The acceleration can only be given to the mass by a force directed in the same

direction, and of magnitude = mass × acceleration = mrω2 N. This force is provided by the rod OG. The mass would prefer to travel in a straight line

(Newton’s 1st law), but the rod pulls the mass into the curved path, being in

tension as it does so – FIGURE 3(a). The rod applies an equal and opposite

force to the shaft, shown in FIGURE 3(b). An 'exploded' view of the system is

shown in Figure 3(c). The sketches of the separated parts with their forces are

called free body diagrams.

FIG. 3(c)

The point mass is subjected to a 'centre-seeking' force (a), of magnitude mrω2, by the rod, i.e. Newton's 2nd law.

(c)

mrω2

(a)

(b)

(c)

(d)

(a) (b)

mrω2 r

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The rod experiences an equal and opposite reaction force (b) – Newton's 3rd

law.

The force at the other end of the rod must be equal and opposite for the rod to

be in equilibrium (c) – Σ forces must be zero.

The shaft experiences an equal and opposite reaction to (c) of (d).

As this force is directed away from O it is termed the centrifugal force.

We can now represent the rotating system of FIGURE 1 in the following

manner.

FIG. 4

Can we derive an expression for the bearing reaction at A? We will need to

note that the shaft is in static equilibrium as it is not accelerating, and then

recall that the sum of the:

• x – components of all the forces on the shaft = 0

• y – components of all the forces on the shaft = 0

• moments of all the forces about a chosen point = 0

Bearing reaction

at A

Bearing reaction

at B

Axis of rotationB

a b

Centrifu gal force

on the sh aft

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FIG. 5

From FIGURE 5, take moments about B (clockwise moments being negative):

R a b mr b

R R mr b

A A

× +( ) ( ) × =

= +

– ω

Solving for bb

⎛ ⎝⎜

⎞ ⎠⎟

a b

Shaft

mrω2

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For the system shown in FIGURE 6, the following data is given:

mass of the body m = 0.8 kg

eccentricity OG = 5 cm

angular velocity ωω = 260 rad s–1

FIG. 6

Using the knowledge you have gained, calculate the following and choose the response

to the questions which is nearest your answer.

1. The centrifugal force on the shaft is:

(a) 270.4 kN

(b) 104 N

(d) 2.7 kN

Body

10 cm

14 cm

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2. The maximum force experienced by the bearing at A is:

(a) 158 kN

(b) 1.58 kN

(d) 2.7 kN

3. The maximum force experienced by the bearing at B is:

(a) 1.58 kN

(b) 2.7 kN

(d) 1.12 kN

4. If the eccentricity is halved and the speed of the shaft is doubled then the force on

the shaft is:

(a) the same

(b) doubled

(d) halved

...................................................................................................................................................

________________________________________________________________________________________

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Solution

1. (d) Force is mrω2 acting outwards from the shaft.

2. (b) Draw the free-body diagram

For equilibrium of the shaft, take moments about B so ΣMB = 0

3. (d) Now for equilibrium ΣF = 0

∴ ×( ) =

∴ =

2 7 1000

1 58

. – –

R R

A B

with kN downwards

thhen

RB = × ×

2 7 1000 1 58 1000

1 12

. – .

∴ × × × =

= ×

× =

24 100

2 7 1000 14 100

2 7 14 24

1000

– .

11577 3

1 58

kN�

24 cm

2.7 kN RB

14 cm B

As kg, cm and 260, thenm r F= = = = × ⎛ ⎝⎜

⎞ ⎠

0 8 5 0 8 5

100 . .ω ⎟⎟ ×

= =

260

2704

N 2.7 kN

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4. (b)

You can see that large forces can be generated by rapidly rotating components

with eccentric centres of mass. As can be seen in FIGURE 6, the 'out of

balance' force as it is often known causes bending in the shaft as well as

subjecting the bearings to a rotating or periodic force. The periodic force is a

major source of vibration in types of machinery that have rotating parts such as

car wheels, spin driers, engine crank shafts, turbine shafts and electric motor

rotors. One cause of the mass being off the axis of rotation is the asymmetry

of the rotating part (such as a cam), another being the inaccuracy in machining

large parts.

∴ New force is double the original force..

Force

New force

= × ( )

= ×

m r

2 2

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________________________________________________________________________________________

THE BALANCING OF A SINGLE ROTATING MASS ________________________________________________________________________________________

The most common method of eliminating the 'out of balance' force is to add

extra mass to the rotating part which provides its own centrifugal force of

equal magnitude but in the opposite direction to the out-of-balance force,

hence cancelling it out. This is known as balancing the rotating part. You may

have seen this process if you have taken your car to have its wheels balanced.

This extra mass is known as a counterbalance or counterweight. However, it

may be more convenient to remove mass from the unbalanced part, by drilling

holes for instance.

FIG. 7

In FIGURE 7, the rotating part is represented by its mass m, as a point mass,

rotating about the axis of the shaft O, with radius r. A counterbalance of mass

mc has also been attached to the shaft a distance rc from O.

Notice that it is attached on the other side of O to the mass of the part, and on

the same line which passes from m through O. This ensures that the

centrifugal force caused by the counterbalance has the same line of action as

that caused by m but of opposite sense, shown in FIGURE 8.

The counterbalance is attached at the same point on the shaft as the rod and so

rotates in the same plane.

Shaft

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It only remains to ensure that the magnitudes of the two forces are the same, so

that the sum of the forces acting on the shaft from the rotating parts is zero.

Therefore,

FIG. 8

We have a choice of fixing the value of either mc or rc and calculating the

other. The magnitude of ω does not affect our calculations. The rotating parts are then said to be in static balance.

Force due to mc

r Force due to m

m r m r O

m r m r

c c

ω ω2 2– =

=or

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Example 1

(a) FIGURE 9 shows a rotor which has a mass of 98 kg and is mounted, on a

shaft, eccentric to the axis of rotation by a distance OG of 2mm. What

mass would have to be placed at a radius of 11 cm from O to balance the

rotor?

FIG. 9

(b) What depth of hole would need to be drilled into the steel rotor for

balance, if the diameter of the hole is 3 cm, the centre of which is 8 cm

from the axis of rotation? Assume the rotor is long enough and that the

density of steel is 7800 kg m–3.

Counterbalance mc

Rotor O

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Solution

(a) To achieve a balance, the sum of the forces on the shaft from the rotating

parts must = 0

(b) First we must relate the mass of metal to be removed with the depth L.

Mass = density × volume where volume of the hole is that of a cylinder, i.e.

V D

= ×

= × ⎛ ⎝⎜

⎞ ⎠⎟

3 100

With cm

then V 4

7 07 10000

L .

∴ × ( ) × × × =

= =

OG 0

OG 2

1000 m,

m m r

r m

c c

ω ω2 2

11 100

–

aand kg

98 2

1000

∴ × ⎛ ⎝⎜

⎞ ⎠⎟

× ⎛ ⎝⎜

⎞ ⎠⎟

11 100

– 00

Thus 0.196 0.11 0

–

. .

= = 0 196 0 11

1 78

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FIG. 10

We can look upon this as a negative counterbalance, that is, a loss of

centrifugal force of mcrcω2, so it is located on the same side of the axis of rotation as the centre of mass of the part being balanced.

This is the amount of material to be removed by drilling.

∴ × ⎛ ⎝⎜

⎞ ⎠⎟

× ⎛ ⎝⎜

⎞ ⎠⎟

= as is n98 2

1000 8

100 0– m dc oow only 8 cm

∴ = ×

mc 98 2 1000

100 8

24 5 10

..45 kg

Axis through O8 cm

L 2

Side view

So mass of metal removed V

With 7800

then

= ×

mmass 7800

= ×

7 07 10000

5 51

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By adding (or subtracting) mass from the rotating part to balance it, this

has, in effect, moved the centre of mass of the combination onto the axis

of rotation and eliminated the load on the supporting bearings.

So 5.51

0.44 m

∴ =

2 45

2 45 5 51

44 cm

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________________________________________________________________________________________

BALANCING OF SEVERAL MASSES ROTATING IN THE SAME PLANE ________________________________________________________________________________________

If the rotating part is of complex shape and can be partitioned into an assembly

of simpler shapes, then we can represent the system as several masses rotating

in the same plane.

Example 2

Represent the component shown in FIGURE 11 by two point masses m1 and

m2 rotating at radii of r1 and r2 respectively.

The component is 1 cm thic k and has a density of 10 000 kg m –3

FIG. 11

10 cm 10 cm

60 cm

45 cm

1 100

kg cm –3 ⎛ ⎝⎜

⎞ ⎠⎟

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Solution

Partition the component into the two rectangles as shown in FIGURE 12.

FIG. 12

Its centre of mass is located in the middle of the rectangle, such that OG1 measures 50 cm. For area 2, the mass is 4.5 kg, located in the centre of A2 such that OG2 = 22.5 cm.

The angle between OG1 and OG2 is measured as 37° as the figure is drawn to

scale.

We will use this example to illustrate the balancing of several masses.

Area 1 10 50 500 cm

its volume 1 500 500 c

2= × =

∴ = × = mm

its mass 500 1

100 kg

∴ = × = 5

10A1G1

G2 2

2 .5

50 O

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THE POLYGON OF FORCES

Example 3

Determine the mass of a counterbalance for this L–shape if it is to be at a

radius of 30 cm.

Again, the sum of the centrifugal forces on the shaft must be zero. But as the

centrifugal forces are inclined to each other, we must sum the forces as

vectors. We can do this by drawing the polygon of forces, i.e. drawing the

representation of the rotating part again and calculating the magnitudes of the

centrifugal forces F1 and F2 as shown in FIGURE 13.

We are not given ω, but we do not actually need it. Neither do we need to convert cm into m, provided all lengths are in cm. So a reduced form of the

magnitude of force F1 is 250 kg cm.

Similarly, in reduced form, F m r2 2 2 2

4 5 22

= ×

. ..

101 25

2ω

= kg cm

F m r1 1 1 2

5 50

250

= × ×

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FIG. 13

FIGURE 13 shows the representation of the part plus the unknown

counterbalance mc. FIGURE 14 shows the configuration of the centrifugal

forces acting on the shaft, including the force due to the counterbalance which

is, as yet, unknown in magnitude and direction.

FIG. 14

For balance, the vector sum of these forces must be zero, which means the

resultant of all these forces = 0. This requires that the polygon of forces must

end up as a closed figure.

Shaft

mc rc

37°

F1 = 250

F2 = 101.25

Shaftrc

m2 r1

r2 θ

5 kg 50 cm 4.5 kg 22.5 cm 37°

m1 r1 m2 r2 θ

= = = = =

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Choosing a scale of 1 cm on the vector diagram FIGURE 15 to represent

50 kg cm reduced force, F1 is drawn at 37° to the vertical, starting from O, and

of length . Then F2 is drawn starting from the end of F1, of

The force due to the counterbalance (mcrc) must start from the end of F2 and

end at O, if the resultant force is to be zero. The length of this balancing force,

shown by a dashed line in FIGURE 15, comes to 6.8 cm, which represents

6.8 × 50 = 340 kg cm (= mc × rc) at an angle of 153° to the vertical or 153 + 37 = 190° from r1.

FIG. 15

γ 153°

The required mass is 340 30

kg at a radi= 11 33. uus of 30 cm.

length 101.25

cm in the vertical direction 50

2~ ..

250 50

cm 5 cm=

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Keep control of your calculations by putting the values in a table.

Item

m(kg)

4.5

r(cm)

22.5

θ°

F (kg cm)

250

101.25

The first column identifies which mass is being analysed. We have used C in the last cell to identify the counter balance.

The second column gives the magnitude of the mass and the last cell is unfilled as yet.

The eccentricity of each mass is entered in column 3

Column 4 contains the angle which radius line makes with some chosen datum.

The line of action of F1 in this case.

Column 5 contains the product in each row of column 2 with column 3, i.e. mr (the 'reduced' force F).

The highlighted cells are entered after the polygon of forces has been drawn and measured.

TABLE 1

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Fill in the missing values or symbols.

FIG. 16

The figure shows a small lever of mass 0.8 kg attached to a shaft at a radius of 3 cm.

(a) The reduced centrifugal force on the shaft F1 = = ––––––––––––––

A second lever is attached to the shaft, in the same plane as the first. This has a mass

of 0.4 kg at a radius of 5 cm from the axis of rotation and its centreline is 90° from the

centre line of lever 1.

(b) The reduced centrifugal force on the shaft due to lever 2

F2 = = ––––––––––––––

to find the required mass of the counterbalance which will have a centre of

mass rotating on a radius of 6 cm. Use the pole (starting point) of your

polygon of forces on the diagram opposite.

∴ m r Fc c c= =

at an angle = to the centre line oof lever 1

3 cm

Lever 1

Bearing BBearing A G1

5 cm

Side view

(i)

Lever 2

Lever 1

5 cm

Shaft

End view

(ii)

3 cm

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(d) If the parts had not been counterbalanced the centrifugal force on the shaft

would have been = –––––––––––––– N at an angular velocity of the shaft of

200 rad s–1.

(e) The maximum load of the bearings would have been = –––––––––––––– N

occurring on bearing ––––––––––––––.

+ POLE

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Solution

TABLE 2

(a) Reduced force for 1st lever = m1r1 = 0.8 × 3 = 2.4 kg cm

(b) Reduced force for 2nd lever = m2r2 = 0.4 × 5 = 2 kg cm at right angles to the first

• F1 will be represented by a line 2.4/0.4 = 6 cm in length drawn vertically

• F2 represented by a line 2/0.4 = 5 cm in length, at right-angles to F1 and drawn

from the end of F1. Thus:

at an angle of 220° from the centre line of the first mass.

F Fc cmeasured as 7.8 cm, the magnitude of iss

kg cm

at radius of 6 cm

= ×

∴

7 8 0 4

3 12

. .

r c

3 12 6

0 52

. /

. kg

Item

m(kg)

0.8

0.4

0.52

r(cm)

θ°

220

mr (kg cm)

2.4

2.0

3.12

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(d) The out of balance force, if no counterbalance is added, would have been F1 + F2 which is the same magnitude and direction as Fc but of opposite sense, i.e. 3.12 kg cm

at 40° to the centre line of the first lever.

(e)

= +

⎛ ⎝⎜

⎞ ⎠⎟

⎛

⎝⎜ ⎞

⎠⎟ m r

a b ω 2 shown previously

Maximum load is on bearing A and is RA = ×1248 55 8

780 ⎛ ⎝⎜

⎞ ⎠⎟

= N

∴ = The actual centrifugal force at raω 200 dd s

100 200 N

1248 N

–1

2= ×

3 12.

220°

F 2

F 1

F c

40°

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We can determine the mass required to be added to balance the centrifugal

forces caused by several masses rotating in the same plane using the same

technique as for the two masses. We only need to extend the number of rows

in the table and the number of forces in the polygon, which still has to close.

Carefully follow the steps in the next example and check each calculation and

each line in the polygon of forces.

Example 4

A rotating shaft carries four masses, m1 = 20 kg, m2 = 30 kg, m3 = 24 kg

and m4 = 26 kg. These rotate at radii of 9, 7, 10 and 12 cm and at angles,

measured from r1, of 45°, 75°, and 135° respectively. Determine the mass and

angular location of the mass to be added at a radius of 20 cm in order to

balance the shaft. All of the masses rotate in the same plane.

The solution is carried out on the following worksheet showing the use of the

table and the polygon of forces.

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Draw the configuration diagram with the radii of the masses at their given angular position.

Fill in the table with the given quantities of mass and radius. Calculate the 'force' m × r and fill in the last column. Only r for item A can be entered in the last row at this stage.

Choose a suitable scale and pole for the force diagram.

Draw the force diagram using the scaled values of each m × r drawn in sequence at their given angles.

Close the diagram by the line from point 4 back to the pole. This gives the force required to balance the system.

Measure the length and direction of this closing line and use the values to fill in the remaining cells in the table.

A O

135°

75°

45°

Item

m(kg) 20

r(cm) 9

θ° 0

135

255

mr(kg cm) 180

210

240

312

620

1 A

Scale: 1cm = 50 kg cm

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A rotating shaft carries the four masses, m1 = 20 kg, m2 = 30 kg, m3 = 24 kg and

m4 = 26 kg, rotating at radii r1 = 9 cm, r2 = 7 cm, r3 = 10 cm and r4 = 12 cm.

The angular displacement of r2 from r1 is 45°, r3 from r2 is 75° and r4 from r3 is 255°. Determine the mass to be added at a radius of 20 cm in order to balance the

shaft. All of the masses rotate in the same plane.

Carry out your solution by completing the table. Start, and finish, your polygon of

forces at pole O, shown on the following worksheet. A recommended scale for the

polygon is also given.

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Worksheet 2

Item

m(kg) 20

r(cm) 9

θ° 0

mr(kg cm)

3 2

Scale: 1cm = 50 kg cm

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Solution

Item

m(kg) 20

r(cm) 9

θ° 0

120

255

222

mr(kg cm) 180

210

240

312

100

3 2

Scale: 1cm = 50 kg cm

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. During the fabrication of a rotating component, its centre of mass

becomes offset from its centre of rotation by 5 mm. If the component has

a mass of 20 kg, at what radius should a mass of 150 gm be placed so that

the component will be balanced?

2. A rotating object is formed from two levers. One has a mass of 5 kg and

is 120 mm from the centre of rotation O, the other 6 kg at 200 mm from

O. The radii of the masses are at 90°. Where should a mass of 2 kg be

placed to balance the object?

3. A disc is mounted on a shaft between a pair of support bearings, spaced

230 mm apart, and is located 150 mm from the left-hand bearing. Three

masses are attached to the disc in the positions shown in the following

table.

1 2

0 m

200 mm

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Determine:

(a) the force exerted on each bearing when the shaft is rotating with an

angular velocity of 100 rad s–1

(b) the mass and angular position of a counterweight placed on the disc

at a radius of 150 mm which will eliminate the forces on the support

bearings.

MASS kg

1.5

RADIUS mm

100

125

87.5

ANGULAR POSITION

0°

60°

135°

measured counterclockwise from the radius of the first mass looking from the left hand end.

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

Counterweight must be placed on the line OG, at an angle of 180° from

OG, so that it causes a centrifugal force in the opposite direction to that

caused by the component. For equilibrium ΣF = 0.

O rB

Centrifugal force on the shaft at O

wher

= m r ω 2

ee kg

and is not given.

Use 're

= =

∴

5 1000

dduced' form of the force

kg m

= ×

m r

20 5

1000

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2. Label component masses as m1 at r1 and m2 at r2, where r1 and r2 are at

90°.

For equilibrium the sum of the centrifugal forces acting on the shaft must

be zero, i.e.

If FB is the centrifugal force caused by the counterweight, then using the

reduced form of the forces:

F F

= = × = ° = =

m r m r

1 1

2 2

5 120 600 kg mm at 90 from F2 66 200 1200 2

× = ° = = × °

kg mm at 0 at ? toF FB m r rB B B 2

F F F FB=∑ ∴ + + =0 the vector sum of 1 2 0

∴ × =

∴ ×

But kg

20 5

1000

150 1000

5 1

– m r

B B

0000 150 1000

1000 150

20 5 1000

2 3

0 6

–

= × ×

77 m.

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150

230

Support bearing A at left-hand side

Support bearing B

Shaft

Plane of rotation of the disc

3 2

135°

60°

Angular positions of the radii of the masses

1350 kg mm

θB = 207° FB F1

Scale: 1cm = 200 kg mm

Item

m(kg)

r(mm)

120

200

675

θ°

207

F(kg mm)

600

1200

1350

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(a) The sum of all of the forces acting on the shaft must be zero for

equilibrium, i.e. ΣF = 0 including the bearing reactions. The net, or resultant force acting on the shaft by the centrifugal force caused by the

three masses is represented by:

→ = =

→ =

OC kg mm kg m

and with rad s–

365 365

1000

100ω 11 OC represents a force of

N 365

1000 100 36502× =

Item

M(kg)

1.5

2.43

r(mm)

100

125

87.5

150

θ°

135

253

F(kg mm)

200

187.5

262.5

365

Scale 1cm = 50 kg mmC

O C

= 3

65 k

g m

Teesside University Open Learning (Engineering)

Now the shaft is loaded as shown:

where: RA is the force on the shaft by the bearing at A and RB is the force

on the shaft by the bearing at B. As the shaft in equilibrium then:

But the sum of the moments of those forces must also be zero, i.e.

ΣM = 0 taken about any point on the shaft.

Taking moments about A

3650 150 1000

230 1000

3650 150 230

238

× × =

= ×

– R

00 4

2380 4 3650 0 0

– – .

∴ + = =( )

∴ =

ΣF

11269 6. N

ΣF =

∴ + =

0– –R RA B 3650

Shaft

150 mm 230 mm

RBRA

3650 N

Teesside University Open Learning (Engineering)

(b) If another mass m c

is added to the disc at the given radius of 150 mm, so

that its centrifugal force is equal to 3650 N but in the opposite direction to

the resultant (represented by ), thus closing the polygon of forces,

then the rotating part or disc will exert no resulting force on the shaft and

thus eliminate the dynamic loads on the bearings.

→ ∴ =

∴ =

CO 365 kg mm

kg mm at an angm rc c 365 lle of 253 to but

∴ =

150

365 150

== 2 43. kg

→ CO

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

• A mass, eccentrically mounted on a rotating shaft will subject the shaft to

a force of magnitude mrω2 directed outwards from the shaft along the radius line joining the axis of rotation and the mass centre.

• The resulting loads on the support bearings can be evaluated by using the

conditions of static equilibrium, i.e. the sum of all of the forces on the

shaft is zero and the sum of the moments of those forces taken about one

of the bearings is zero.

• The effect of the centrifugal force on the bearings can be cancelled by

balancing the component. This can be achieved by either adding mass to

the components to create a force equal and opposite to the original out-of-

balance force, or by removing an equivalent amount of mass.

• The net, or resulting, out-of-balance force caused by several masses

rotating in the same plane can be found by the graphical summation of the

individual forces and closing the force polygon by the addition of the

centrifugal force from the counterbalance.

Teesside University Open Learning (Engineering)

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