BALANCINGBy: E/C Sapid, Mark Gil

Static balance

(two masses in a plane)

Dynamic balance

(two masses

in a plane)

Method of

balancing Rotors

topics

Static Balance

The condition which exists in a body which has an absolutely even distribution of the weight mass around the axis of rotation

This occurs when when there is no resultant centrifugal force and the center of gravity is on the axis of rotation.

Static BalanceConsider a light arm pivoted freely at the fulcrum O and carrying masses m1, m2 at distances r1, r2 from O respectively. In general the arm will rotate about O and the system is said to be out of balance. For equilibrium there must be balance of moments about O. i.e.

m1g x r1 = m2g x r2

orm1r1 = m2r2

When in balance the arm may be set in any position and will remain at rest in that position. The weights are said to be in static balance and the center of gravity of the system is located at O.

Static balance (two masses in a

plane)Two masses in a plane

r1 r2

O

m1g m2g

Static balance (two masses in a

plane)

Dynamic Balance

The condition which exists in a rotating body when the axis about which it is forced to rotate, or to which reference is made, is parallel with a principal axis of inertia; no products of inertia about the center of gravity of the body exist in relation to the selected rotational axis.

This occurs when there is no resulting moment along the axis.

Dynamic Balance

Two masses in a plane

Dynamic balance (two masses in a

plane)

Now consider two light arms fixed to a shaft at bearing O and rotating with angular velocity ω. The arms are in the same plane and carry masses m1, m2 at radii r1, r2 respectively. Owing to the rotation each masses exerts an inertia force radially outward on the bearing O.

The force due to m1 is m1 ω2 r1 (OA in the force diagram)The force due to m2 is m2 ω2 r2 (AB in the force diagram) B

A O

Figure A

ω

m2

m1m1 ω2 r1

m2 ω2 r2

O

ω

m1m2

Dynamic balance (two masses in a

plane)

Figure C

Figure B

The resultant out of balance force on the bearing is given by OB in the force diagram.

when the dynamic load on the bearing is zero the rotating system is said to be in dynamic balance. The condition for no load at O is that two inertia force shall:

1. act along the same straight line but with opposite sense

2. Be equal in magnitude

the relative positions of the masses are as in the figure C shown earlier. The condition for equal inertia forces is:

m1 ω2 r1 = m2 ω2 r2

Thus, since ω2 is the same both massesm1r1 = m2r2

Two bodies in the same plane are in static balance when pivoted about a given axis they will be in dynamic balance at any speed when rotating about the same axis.

Dynamic balance (two masses in a

plane)

It was shown earlier that for a two mass system to be in static balance the mr product for each mass had to be the same. This is also the condition for the masses to be balanced when rotating and suggests a method for ensuring balance for rotating rotors such as turbine disks or car wheel assemblies.

Method of balancing

Rotors Method ofBalancing Rotors

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In practice it is usually possible to balance a rotor to an accuracy of 0.001 m-kg, i.e. the amount of residual unbalance is equivalent to a mass of 1 kg at 1mm radius. For a rotor of mass 10 tons this is equivalent to a displacement x of he center of gravity from the axis of rotation given by:

10 x 1000 x X = 0.0001 X = 10-7 m or 0.1 μm

The corresponding out-of-balance centrifugal force when running at 3600 rpm is

m1 ω2 r1 = 10 x 1000 x 2Л 3600 x 10-7 kg-m/s2 60

= 142 N

It is usual to limit the out-of-balance force to be greater than 1 percent of the rotor weight.

Method of balancing

Rotors

Sample Problem

?

Sample Problem

A shaft carries two rotating masses of 1.5 kg and 0.5 kg, attached at radii 0.6 m and 1.2 m, respectively, from the axis of the masses are shown in the figure below. Find the required angular position and radius of rotation r of a balance mass of 1 kg.

If no balance mass is used what is the out-of-balance force on the shaft bearing at 120 rpm?

B

AO

60o

1.2 m

0.6 m

o a

b

B

AO

156o 36’

Figure c

Figure b

Figure

Sample Problem

The mr values are 0.9 kg-m for A and 0.6 kg-m for B and these are represented by oa and ab, respectively, in the force polygon, figure b. The resultant out-of-balance mr value is given by ob in direction o to b. From a scale drawing:

ob = 1.31 kg-m

the equilibrant is equal and opposite to the out-of-balance force, and since the mr value for the balance mass is:

1 x r kg-m

Therefore, for balance

1 x r = 1.31 and r = 1.31 m

Thus the radius of rotation of the balance mass is 1.31 m

Solution:

Sample Problem The balance mass must be positioned so that its

inertia force is acting in direction b to o, i.e. at an angle of 156°36’ to the radius of mass A, as shown in figure c. If no balance mass is used

Out-of-balance force = ob x w2

= 1.31 x 2Л 120 kg-m/s2 60

= 207 N

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Thank youStatic balance is keeping balanced while still and dynamic balance is keeping balance while moving.