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Balancing of Rotating Masses

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Balancing of Rotating Masses

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 .

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

seeking) acceleration given by 2ra . By

Newton’s second law the centripetal force acting

on the mass is 2mrF .

m

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

on the bearing is therefore given by 2mrF .

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 MR2 = mr2 or MR = mr.

Rotating Mass

Centrifugal Force

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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 r1, r2, etc and the masses are m1, m2, etc. Any out of balance will have a

detrimental effect on the bearings and will cause vibration, noise, etc.

Since these forces are vectors, a graphical approach is often the most convenient for determining the

out of balance force.

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.

R

M

m3

r3

r2

m2

r1

m1

m4

r4 Each mass has a centripetal force mr2 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.

Static Balance

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If we let the resultant be MR2, 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?

The information to draw the mr polygon is best tabulated.

m (kg) r (m) mr

A 90 0.6 54

B 22.5 0.6 13.5

C 45 0.3 13.5

D 45 0.6 27

The mr polygon is then draw to scale.

m1r1

m2r2

m3r3 m4r4

MR

(Balancing

vector)

A

B

C

D

45o

45o 30o

60o

0.6m

45kg

0.3m

0.6m

90kg

0.6m

22.5kg

System of co-planar concurrent

masses rotating at 10 rev/min.

45 kg

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A

B

C

D MR = 51

Required to balance the

system.

The resultant out of balance force is

in the opposite direction to that

shown on the mr polygon. 12o

mr polygon

(NTS)

A

B

C

D

1m

51kg

MR = 51 = M x 1

Therefore, M = 51kg

Out of balance force = MR2

= 51 x 1 x

2

60

102

=55.94N

Resultant out of

balance force

12o

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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 = 𝑚𝑟𝜔2. 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 𝑚1 and 𝑚2 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).

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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.

X1 – X2 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 90o 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.)

(a)

(b)

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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 150o 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 MA 150 150MA 0 0

B 120 240 28800 50 1440000

C 90 180 16200 110 1782000

D MD 150 150MD 150 22500MD

This value of MD = 40g can now be substituted into the mr column of table such that 150MD = 150 x

40 = 6000. Thus the mrx polygon can be drawn.

Using this data

draw mrx polygon.

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From the above, A = 15800 = 150MA, therefore MA = 105.3g.

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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?

We begin by tabulating the information we have

m(kg) r(m) rm

A 90 0.6 54

B 22.5 0.6 13.5

C 45 0.3 13.5

D 45 0.6 27

E mE 1.0 mE x 1

54

27

60o 45o

30o 45o

13.5

13.5

We begin by summing the mr values in the x direction, taking those to the right as positive.

→ +

∑ 𝑚𝑟𝑥 = −54 cos 60 − 13.5 cos 30 + 13.5 cos 45 + 27 cos 45 = −10.054 kgm

Next we sum the mr values in the y direction, taking those upward at positive

A

B

C

D

45o

45o 30o

60o

45kg

0.6m

45kg

0.3m

0.6m

90kg

0.6m

22.5kg

System of co-planar concurrent

masses rotating at 10 rev/min.

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∑ 𝑚𝑟𝑦 ↑ + = 54 sin 60 − 13.5 sin 30 − 13.5 sin 45 + 27 sin 45 = 49.56 kgm

For the resultant we have

10.054 kgm

49.56 kgm

Rmr

θ

Rmr = √10.0542 + 49.56 = 50.57 = mE×1

Therefore mE = 50.57 kg

Angle θ being given by tan θ =10.054

49.56 or θ = tan−1 (

10.054

49.56)

Therefore θ = 11.47°

It follows that the out-of-balance force mrω2 = 50.57×1× (2π×10

60)

2= 55.46 N

A

B

C

D

1m

50.57kg

Resultant out of

balance force

11.47o

Original system with balancing mass

added for complete dynamic balance.

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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 150o 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 MA 150 150MA 0 0

B 120 240 28800 50 1440000

C 90 180 16200 110 1782000

D MD 150 150MD 150 22500MD

1440000

60o

1782000

Net horizontal mrx (→+) = −1782000 cos 60 = −8.910×105

Net vertical mrx (↑+) = 1440000 − 1782000 sin 60 = −1.03257×105

Thus, we have

8.910×105

1.03257×105 𝑅 = √8910002 + 1032572 = 896963

θ θ= tan−1 891000

103257= 83.4°

But 896963 = 22500MD

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Therefore MD = 39.86 kg say 40 kg

Now 90 − 83.4 = 6.6°

So, it follows that MD = 40 k and will be placed at an angle of 6.6o anticlockwise from zero datum.

In the above table 150MD = 150×40 = 6000, thus we have

28800

6000

6.6o

600

16200

Therefore, net horizontal mr (→+) = 6000 cos 6.6 − 16200 cos 60 = −2139.76

and net vertical mr (↑+) = 28800 + 6000 sin 6.6 − 16200 sin 60 = 15460

Therefore, we have

θ

15460 R= √154602 + 2139.762 = 15607.4

2139.76 θ= tan−1 2139.76

15460= 7.88°

(or -82.12 from zero datum)

But 150MA = 15607.4, therefore MA = 104.05g, say 104 g

For complete balance we have:

B

D

6.6o

60O 82.12O

C A

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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, 90o clockwise from the 5 kg mass

15 kg mass at 100 mm radius, 240o 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 30o and 150o anti-

clockwise from 5 kg mass.

2. Two masses revolve together in the same plane at an angular distance of 45o 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-of-

balance 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 160o33’ 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, 0o, 30o, 75o. Find the balance mass required at 150 mm radius to eliminate the out-

of-balance force. State the angular position of the balance mass.

(1.56 kg; 215o 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 141o 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 187.5 g. Calculate the resultant out-of-balance force if the holes

are spaced at 90o 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; 0.211 kg at 45o 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 114o36’ 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 = 30o, angle BOC = 60o, angle COD = 120o.

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, 39o7’ to OA)

8. Four mass m1, m2, m3, and m4 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 45o, 75o, and 135o. Determine the position and magnitude of the

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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, 201o 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 120o

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 177o; 2.2 kg at 141o)

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