BALANCING

Static Unbalance

A disk-shaft system on rigid rails An unbalanced disk-shaft system mounted on bearings and rotated

For the unbalanced rotating system: Dynamic Bearing Reactions + Inertia Forces =0 -(FA + FB)i + mrG2i = 0

Both bearing reactions occur in the same plane and in the same direction. same direction. Balancing is performed on Balancing is performed on the plane of unbalance. After balancing no bearing reaction remains theoretically.

Dynamic Unbalance

A long rotor mounted in bearings at A and B

• If m1=m2 and r1=r2 the rotor is statically balanced but dynamically unbalanced.

Balancing may be performed on the planes of unbalances or any other two convenient planes. After balancing, no bearing reactions remain theoretically.

Both bearing reactions occur in the same plane but in opposite directions.

Unbalance Types

STATIC (Misalignment of axes)

DYNAMIC (Tilt of axes)

STATIC and DYNAMIC (GENERAL) (Called as DYNAMIC, in industry) (Both misalignment and tilt of axes)

Analysis of Static Unbalance

m1R12+ m2R2

2+ m3R32+ mcRc

2=0 F=0

• mcRc is the correction quantity. It is added to the plane of unbalances.

Analysis of Dynamic Unbalance Graphical Vectorial

Scalar

True moment polygon is 90 cw away

Industrial Balancing Machines

Balance Quality Grade Maximum Permissible Spesific Unbalance

MrG2=mr2

MrG=mr=u (Amont of Unbalance)

e=rG=mr/M=u/M (Spesific Unbalance) Empirically, e=constant G (Balance quality grade) (0.4-4000) For example: N=3000 rpm, G=6.3 give e=20m as max permissible eccentiricity. 1) As the rotor mass increases permissible unbalance also increases

(u=MrG ) 2) As the rotor speed increases permissible unbalance decreases

(e=constant )

For small masses and high speeds, higher quality balance grade is required.

Maximum Speed

Permissible spesific unbalance (u/M) OR Mass center’s eccentricity (e)

G. m. mr2

r

RA

rG

MrG2 M

G

Balancing A Single-Cylinder Engine

FC

mA

F = FA,B + FC

Eliminated

Primary component reduced 50% (Altough a new vertical component added)

Secondary component does not changed

Polar Diagram of Inertia Forces

Before adding mC:

After adding mC :

Imaginary-Mass Approach 1) Two imaginary rotary masses are

used for each reciprocating mass.

2) Each mass is equal to half the

equivalent reciprocating mass.

3) These imaginary masses rotate

about the crank center in opposite

directions and with equal angular

velocities.

4) The total inertia force in y-direction is

zero.

5) The inertia force in x-direction is

equal to the inertia force due to

reciprocating pisto mass.

A simple model to analyse

and recover the unbalance of

reciprocating piston mass.

Balance of Multicylinder Engines 2-Cylinder 2-Stroke Engine

First Harmonics:

Angles between imaginary masses : 360/2 = 180

S1 = 0

M1 0

2-Cylinder 2-Stroke Engine

Second Harmonics:

Angles between imaginary masses: 720/2 = 360

S2 0

M2 0

4-Cylinder 4-Stroke Engine

Firing Order: 1-2-4-3

First Harmonics:

Angles between imaginary masses : 720/4 = 180

S1 = 0

M1 = 0

4-Cylinder 4-Stroke Engine

Firing Order: 1-2-4-3

Second Harmonics:

Angles between imaginary masses : 1440/4 = 360

S2 0

M2 0

4-Cylinder 4-Stroke Engine

Firing Order: 1-2-3-4

First Harmonics:

Angles between imaginary masses : 720/4 = 180

S1 = 0

M1 0

4-Cylinder 4-Stroke Engine

Firing Order: 1-2-3-4

Second Harmonics:

Angles between imaginary masses : 1440/4 = 360

S2 0

M2 0

Balancing Linkages

The position of the mass center of whole linkage,

is tried to remain stationary.

Balancing Linkages (cont’d)

rs3

rs4

rs2

Vector loop-closure eq’n:

One of the vectors, for example may be drawn from

this eq’n: e.g.

Balancing Linkages (cont’d)

• This eq’n shows that the center of mass may be made stationary at the position, (1 and 3)

• if the following coefficients of the time-dependent terms vanish:

32 2

2 2 3 2 3 3

3

0jj r

m a e m r m a er

34 4

4 4 3 3

3

0jj r

m a e m a er

Simplification

3 3

3 3 3

j ja e r a e

32 2

2 2 3 3

3

0jj r

m a e m a er

These two

equations

must be

satisfied for

total force

balance

Balancing Linkages (cont’d)

• These equations yield the two sets of conditions:

• and

• and

• A study of these conditions show that the mass and its location can be specified in advance for any single link (3); and then full balance can be obtained by rearranging the masses of the other two links (2 and 4).

• Without disturbing 3rd link , counterweights are added to rotating links 2 and 4. In this procedure,

•

• relations must be satisfied.

44 4 3 3

3

rm a m a

r

2 3

4 3

0 0 0 * * *

i i i i i i i i im a m a m a

Unbalanced

linkage

Counterweight

22 2 3 3

3

rm a m a

r

Balancing Linkages (cont’d)

• Therefore gives,

• and

• If it is required to find ,

• condition must be satisfied.

* * 2 0 0 2 0 0 0( ) ( ) 2( )( ) cos(i i i i i i i i i i i im a m a m a m a m a

0 0 0* 1

0 0 0

sin sintan

cos cos

i i i i i ii

i i i i i i

m a m a

m a m a

*

im

* 0

i i im m m