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The 14th IFToMM World Congress, Taipei, Taiwan, October 25-30, 2015 DOI Number: 10.6567/IFToMM.14TH.WC.OS14.008

Stability Analysis of Self-Balancing System with Elastic Shaft

Tadeusz Majewski

Department of Industrial&Mechanical Engineering, Universidad de las Americas-Puebla, Puebla 72810, Mexico

e-mail: tadeusz.majewski@udlap.mx

Dariusz Szwedowicz

Centro Nacional de Investigacion y Desarrollo Tecnologico, Cuernavaca, Morelos, Mexico

e-mail: d.sz@cenidet.edu.mx

Abstract: This paper presents a method of balancing of a disk on

an elastic shaft. Free balls or rollers are used as compensating

elements for the disk´s unbalance. It is a self-balancing system.

With vibration of the disk, the balls change their positions with

respect to the disk which leads to a change in the system

unbalance. For some ranges of the disk speed, the balls are able to

compensate for unbalance. The vibratory forces which are

responsible for such behavior of the balls are defined. The paper

presents the way the vibratory forces depend on the system

parameters. Detailed analysis was done for the stability of the free

elements and it was shown when the balls can compensate for the

disk unbalance and eliminate the disk´s vibrations.

Keywords: Balancing, Vibrations, Stability

1 Introduction

The unbalance of the rotors is the most frequent source of

vibrations. Especially at a high speed of the rotor, the

dynamic forces and amplitude of vibrations can be large.

Many papers were published where this problem is

discussed [1].

The method of self-balancing of the rotors has been

known for some years [2-9]. In this method free elements

such as balls, rollers or sometime fluid are located inside

the rotor or in a special container fixed to the rotor and they

rotate together.

The problem of the rotors balancing with free elements is

more complex than the classical dynamic model of the

rotor because of the behavior of the rotors influence on the

behavior of the free elements and vice versa.

The free elements are no always able to compensate for the

unbalance, sometime they increase the system unbalance.

One of the more interesting applications of this method is

for hand-held equipment with rotating elements, e.g for

hand grinds or even optic discs [10, 11]. It can be a unique

method for balancing a machine with variable unbalance,

e. g. washing machines where the unbalance from

laundering may be very large and different for each

running or for centrifuges.

The author of this paper also studied the problem of

self-balancing and published some papers[12, 13].

2 Description of the system

The balancing system (Fig 1) consists of disk 1, drums 2

and 6, free balls 3 and 4, the elastic shaft 5 and its bearings

1 and 8 at the ends of the shaft. Sometimes the balls cannot

be placed on the same plane where there is unbalance and

in our model they are at a distance l2, l3. In each drum there

can be N≥2 balls with the same mass m and they change

their position around the circle at the radius R. The position

of the disk on the shaft is defined by l1+l2 and it is between

0.3l and 0.7l.

Figure 1 Scheme of the system

Figure 2 The angular coordinates of the disk

1 2 6

3 4

5

l1 l2 l3 l4

e C Z

X

Y

1 2 3 4 5 6 7 8

θ

Φ

ω

O

C

O

ωt

αi

Y

X x

y

Figure 3 Position of the ball inside the drum

3 Mathematical model of self-balancing

During rotation the position of the disk is defined by the

coordinates

Tyxo ],,,,[ 4444 q . (1)

The balls in the drum are at some distance from the disk

and the deformation of the shaft is different that the disk.

The displacement of the shaft in the planes 2-3 and 6-7

depends on the displacement of the disk and the distance

from the disk l2 or l3. This relation can be written as

N

iixQxdxcxxbx

1114114112 , (2)

N

jjyQydycyyby

1214214216 . (3)

The coefficients b11x, …, d21y depend on the length of each

segment of the shaft l1, l2, l3, l3.

The equations governing the behavior of the disk are as

follows

i j

jo QkQkQqKqccqM iogo 21)( (4)

The coefficient of the disk inertia, its damping and

gyroscopic are M, co, cg and the stiffness of the shaft is Ko.

Qo, Qi, Qj present the excitation from the rotor unbalance,

the balls in the left and right drums. The balls are located at

some distance from the disk and their influence on the disk

behavior is presented by the coefficient k1, k2.

The mass moment of the system inertia around the x axis

22211

211

22 )(sin NmcNmctMeII yyxdx . (5)

where Ixd is the mass moment of inertia of the disk.

The gyroscopic moment Mg around the axis x is

))(sin( 22211

211

22 NmcNmctMeJ

JM

yyzd

zgx

(6)

Some of the excitations from Eq. (4) have a form

tMeoxQ cos2 , (7)

)cos()( 2iiix tmRQ , (8)

22 )sin()( ltmRQ iii . (9)

The mass of the balls is very small and Eq. (4) may be

taken in a simpler form.

The resultant stiffness Ko of the shaft at the position of the

disk depends on the stiffness K(s) of all segments of the

shaft (s=1,2,3,4). The stiffness of one segment is a function

of its length. The coefficients K(s) are established from the

relations given for the bending of the shaft and they are

function of their length.

The relative motion of the balls with respect to the

reference frame which rotates with the disk is described by

the following equations

iicRityitxi /)]cos(2)sin(2[ , (10)

jjcRjtyjtxj /)]cos(6)sin(6[ . (11)

The behavior of the balls depends on the accelerations of

the shaft in the planes of the drum so the relations (2,3)

have to be used.

The balls change their position slowly with respect to the

disk and their excitations on the disk Qi, Qj can be taken as

harmonic.

4 Simulation of the behavior of the system

Some of the system parameters taken for the simulation

are: M=25 kg, Me=0÷0.05 kgm, EI=1650 Nm2 (stiffness of

the shaft), l=1.5 m, m=0.15 kg, and R=0.15 m.

The behavior of the system during self-balancing is

presented in Fig 4 - 5. By changing the position of the disk

on the shaft, the natural frequencies also change. The

vibrations of the drums with the balls depend on the disk´s

vibrations and the distance of the drum from the disk (Eqs.

2,3). It was observed that the method is effective only for a

rotor´s speed higher than the natural frequencies of the

disk. The diagrams in Fig 4 present the situation when the

disk is in the middle of the shaft, the balls are at a small

distance of 5 mm from the disk and its speed is ω=100

rad/s. The vibrations of the disk vanish, the final unbalance

of the system goes to zero when the balls go to their final

positions (Fig 5e).

0 50 100 150 200 250 300-10

0

10

Time

x,

y [

mm

]

Linear vibrations

0 50 100 150 200 250 300-5

0

5

Time

, [

min

]

Angular vibrations

0 50 100 150 200 250 300-500

0

500

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-400

-200

0

200

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-10

0

10

Time

x2 [

mm

]

Vibration x2

0 50 100 150 200 250 300-10

0

10

Time

x6 [

mm

]

Vibration x6

0 50 100 150 200 250 300-0.1

0

0.1

Time

DM

e [

kgm

]

Final unbalance

0 50 100 150 200 250 300-10

0

10

Time

x,

y [

mm

]

Linear vibrations

0 50 100 150 200 250 300-5

0

5

Time

, [

min

]

Angular vibrations

0 50 100 150 200 250 300-500

0

500

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-400

-200

0

200

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-10

0

10

Time

x2 [

mm

]

Vibration x2

0 50 100 150 200 250 300-10

0

10

Time

x6 [

mm

]

Vibration x6

0 50 100 150 200 250 300-0.1

0

0.1

Time

DM

e [

kgm

]

Final unbalance

0 50 100 150 200 250 300-10

0

10

Time

x,

y [

mm

]

Linear vibrations

0 50 100 150 200 250 300-5

0

5

Time

, [

min

]

Angular vibrations

0 50 100 150 200 250 300-500

0

500

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-400

-200

0

200

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-10

0

10

Time

x2 [

mm

]

Vibration x2

0 50 100 150 200 250 300-10

0

10

Time

x6 [

mm

]

Vibration x6

0 50 100 150 200 250 300-0.1

0

0.1

Time

DM

e [

kgm

]

Final unbalance

a)

b)

c)

d)

f)

Fig 4 Behavior of the system during balancing

a) linear disk vibrations, b) position of the balls in the

left drum, c) angular vibrations of the disk, d)

position of the balls in the right drum, e) final

unbalance DMe

The diagrams in Fig 5 are for the balls at the distance

l2=l3= 250 mm from the disk. The balls go to their final

positions which are different than for l2=l3= 5 mm (Fig

4). The balls do not eliminate the disk unbalance (Fig

5b) and there are the vibrations (Fig 5a). In this case the

balls compensate for only a part of the disk´s unbalance.

0 50 100 150 200 250 300-10

0

10

Time

x,

y [

mm

]

Linear vibrations

0 50 100 150 200 250 300-100

0

100

Time

, [

min

]

Angular vibrations

0 50 100 150 200 250 300-400

-200

0

200

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-400

-200

0

200

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-10

0

10

Time

x2 [

mm

]

Vibration x2

0 50 100 150 200 250 300-5

0

5

Time

x6 [

mm

]Vibration x6

0 50 100 150 200 250 300-0.2

0

0.2

Time

DM

e [

kgm

]

Final unbalance

0 50 100 150 200 250 300-10

0

10

Time

x,

y [

mm

]

Linear vibrations

0 50 100 150 200 250 300-100

0

100

Time

, [

min

]

Angular vibrations

0 50 100 150 200 250 300-400

-200

0

200

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-400

-200

0

200

Time

1,

2 [

deg]

Position of the balls

0 50 100 150 200 250 300-10

0

10

Time

x2 [

mm

]

Vibration x2

0 50 100 150 200 250 300-5

0

5

Time

x6 [

mm

]

Vibration x6

0 50 100 150 200 250 300-0.2

0

0.2

Time

DM

e [

kgm

]

Final unbalance

Fig 5 Behavior of the system during balancing

for l2=l3=250 mm

a) linear disk vibration, b) final unbalance

The diagrams from Fig 4, 5 are for a rotor´s speed higher

than the natural frequencies of the disk.

5 Vibratory forces

The right side of Eqs. (10,11) presents the inertial forces

(N/kg) which force the balls to change their position with

respect to the disk. As the excitations from the disk´s

unbalance and the balls are harmonic, then the vibrations

are also harmonic (1x). From the disk´s unbalance and the

excitation given by the ball ith or jth the vibrations can be

approximated as

tsotcot sincos)( aaoq , (12)

tsitciti sincos)( aaq . (13)

ttt sjcjj sincos)( aaq . (14)

The components aci , asi of the vibrations from ith ball

depend on the position of all balls in the left drum.

The inertial force which acts on the ball ith or jth are as

follows

)]cos(2)sin(2[ ityitxmví

F (15)

)]cos(6)sin(6[ jtyjtxmvjF . (16)

As the vibrations are harmonic then the vibratory forces

can be taken as the average during the period of vibration

T=2π/ω.

dt

T

ityitxT

miF 0

)]cos(2)sin(21

. (17)

Substituting Eqs. (2-3,12-14) into Eq. (17) one obtains the

average vibratory force as a sum of the vibratory forces

from each component of vibrations of the disk and the

forces generated by the balls

QyFQxFiFiFiyFixFiF (18)

Fix, Fiy, FiΦ, Fiθ are the components forces from the

vibrations; x(t), y(t), Φ(t), θ(t), respectively. Each

components of the vibratory force depends on the disk’s

vibration and the position of all balls

r

irl

ilidix FFFF . (19)

The component Fid is the result of disk vibration, the

components of the vibratory force Fil, Fir present the

interaction of the lth ball in the left drum and interaction of

rth ball in the right drum on the ball ith in the left drum.

For example

)]cos()sin([5.0 2lixsllixclil aamRF . (20)

When the vibratory forces are determined then the

positions of equilibrium αif, αjf of all the balls in both

drums are also defined.

0),...,1,,...,1( NrfrfNlflfiFiF (21)

0),...,1,,...,1( NrfrfNlflfjFjF (22)

The balls in their equilibrium positions may compensate

for a part or all of the disk´s unbalance. The Eqs. (21,22)

have some solutions for αif, αjf. Which of them the balls

will occupy and what is the resultant unbalance depend on

their stability in these positions.

6 Stability of the balls

There are some solutions to Eqs. (21, 22) and some of them

can be dynamically stable and others unstable. The

analysis of the stability of the balls should prove that they

are able to compensate or not the disk’s unbalance. As it

was shown earlier the balls may compensate for a part or

all of the unbalance (Fig 4 and 5). Lagrange and Dirichlet’s

theorem allows one to verify if the position of equilibrium

is stable. The system has the minimum potential energy at

these positions if the roots χ of Eq. (23) have negative real

parts [14].

0

3

3

2

3

1

3

...2

2

1

2

...2

1

1

1

FFF

FF

FF

(23)

The derivatives are calculated using Eqs. (18, 19).

a)

b)

r i

ir

l i

il

i

id

i

ix FFFF

(24)

l

ilF

l

ixF

r

irF

r

ixF

(25)

The result of the calculation is shown in Table 1 for

the disk at the middle of the shaft, balancing with one ball

in each drum and different distances of the balls from the

disk. The analysis is done for two speeds of the rotor; 100

rad/s which is higher than the natural frequencies of the

disk and 20 rad/s smaller than the natural frequencies.

When the speed is 100 rad/s and the distance of the balls

from the disk increases then their equilibrium positions

move away from the disk’s unbalance which results in an

increasing final unbalance.

For 20 rad/s the ball’s stable positions are close to the

disk’s unbalance and the balls increase the initial

unbalance of the disk.

ω

[rad/s]

l2=l3

[m]

α11f

[rad]

α21f

[rad]

χ1 χ2

100>

ωo

0.005 2.09 -2.09 -1.80 -0.598

100 0.01 2.09 -2.096 -1.799 -0.598

100 0.1 2.13 -2.14 -1.58 -0.6

100 0.2 2.28 -2.27 -0.92 -0.62

100 0.3 2.66 -2.64 -0.64 -0.046

20<ωo 0.005 -0.184 -0.189 -0.03 -0.01

Table 1 Roots of Eq. (17) for the disk at the position

l1+l2=l/2

The results in Table 2 are for the asymmetrical disk

position on the shaft l1+l2=2l/3.

ω

[rad/s]

l2=l3

[m]

α11f

[rad]

α21f

[rad]

λ1 λ2

100 0.005 4.19 -4.18 -2.02 -0.68

100 0.05 4.23 -4.13 -1.93 -0.66

100 0.1 4.18 -4.1 -1.65 -0.66

100 0.2 3.83 -3.98 -1.02 -0.45

100 0.3 3.17 -3.11 -1.02 -0.019

20< ωo 0.005 -0.44 -0.45 -0.029 -0.009

Table 1. Roots of Eq. (17) for the disk at the position

l1+l2=2l/3

By increasing the distance l2 or l3 (the distance of the balls

from the disk) the equilibrium positions of the balls

become stable but these positions are far from the positions

for which the unbalance would be compensated 100%.

Therefore the resultant unbalance increases as it is

presented in Fig 6 - the efficiency of the system decreases.

Fig 6 Resultant unbalance of the system vs. the distance of

the balls for l1+l2=l/2

Introducing a disturbance of the initial conditions of

the balls in Eqs. (5,10,11) one obtains the variational

equations of the system and they can also be used for

verification of the stability of the balls. The disturbance of

the balls is η11, η12, η21, η22 and their new positions are

α11=α11f+ η11, …, α22=α22f+ η22. (26)

The same for the disk q(t)=qo(t)+ η(t),

where η(t) are the variations of the disk´s position with

respect to its steady-state vibrations qo(t) when the balls

are at the positions α11f, α12f, α21f, α22f.

Fig 7 presents the behavior of the disk and the balls when

the disk is at the middle of the shaft, the balls are very close

to the disk, the speed is 100 rad/s and the displacement of

the balls from their initial position is 0.05 – 0.25 rad.

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

1,

2 [

mm

]

Linear vibrations

0 20 40 60 80 100 120 140 160 180 200-0.1

0

0.1

Time

3,

4 [

min

]

Angular vibrations

0 20 40 60 80 100 120 140 160 180 200-10

0

10

20

Time

11,

12 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-10

0

10

20

Time

21,

22 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

x2 [

mm

]

Vibration x2

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

x6 [

mm

]Vibration x6

0 20 40 60 80 100 120 140 160 180 2000.148

0.149

0.15

Time

DM

e [

kgm

]

Final unbalance

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

1,

2 [

mm

]

Linear vibrations

0 20 40 60 80 100 120 140 160 180 200-0.1

0

0.1

Time

3,

4 [

min

]

Angular vibrations

0 20 40 60 80 100 120 140 160 180 200-10

0

10

20

Time

11,

12 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-10

0

10

20

Time

21,

22 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

x2 [

mm

]

Vibration x2

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

x6 [

mm

]

Vibration x6

0 20 40 60 80 100 120 140 160 180 2000.148

0.149

0.15

Time

DM

e [

kgm

]

Final unbalance

Fig 7 Perturbation of the disk and the balls

(ω=100 rad/s, l2=l3= 5 mm, 0.5l)

The behavior of the system for asymmetrical positions of

the disk is presented in Fig 8.

[%]

l2 [m]

0 20 40 60 80 100 120 140 160 180 200-0.2

0

0.2

Time

1,

2 [

mm

]

Linear vibrations

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

3,

4 [

min

]

Angular vibrations

0 20 40 60 80 100 120 140 160 180 200-5

0

5

Time

11,

12 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-10

0

10

Time

21,

22 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-0.2

0

0.2

Time

x2 [

mm

]

Vibration x2

0 20 40 60 80 100 120 140 160 180 200-0.2

0

0.2

Time

x6 [

mm

]

Vibration x6

0 20 40 60 80 100 120 140 160 180 2000.1122

0.1123

0.1124

0.1125

Time

DM

e [

kgm

]

Final unbalance

0 20 40 60 80 100 120 140 160 180 200-0.2

0

0.2

Time

1,

2 [

mm

]

Linear vibrations

0 20 40 60 80 100 120 140 160 180 200-1

0

1

Time

3,

4 [

min

]

Angular vibrations

0 20 40 60 80 100 120 140 160 180 200-5

0

5

Time

11,

12 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-10

0

10

Time

21,

22 [

deg]

Position of the balls

0 20 40 60 80 100 120 140 160 180 200-0.2

0

0.2

Time

x2 [

mm

]

Vibration x2

0 20 40 60 80 100 120 140 160 180 200-0.2

0

0.2

Time

x6 [

mm

]

Vibration x6

0 20 40 60 80 100 120 140 160 180 2000.1122

0.1123

0.1124

0.1125

Time

DM

e [

kgm

]

Final unbalance

Fig 8 Perturbation of the disk and the balls

(ω=100 rad/s, l1+l2=2l/3)

The difference of the behavior of the balls in the left and

right drum can be seen. The period of the vibration of the

balls in the right drum is much greater than for the balls in

the left drum. It means that the margin of stability for the

balls in the right drum is smaller than for the balls in the

left drum. The vibrations of the right drum are smaller than

the left drum (l1+l2>l3+l4) so the vibratory forces in the

right drum are smaller and the margin of stability is also

smaller.

7 Conclusions

The paper has proven that the disk’s unbalance can be

completely compensated for if the balls are very close to

the disk. It was shown that with increasing the distance

between the balls and the disk the efficiency of the

balancing decreases. In what way the resultant unbalance

depends on the position of the balls is shown in Fig 6. The

efficiency of the method is smaller for the asymmetrical

position of the disk. The positions of the balls in which

they are able to compensate for a part or all of the disk’s

unbalance are stable for the speeds which are higher than

the disk´s natural frequencies. For low disk speed the balls

increase the unbalance of the system which limits its

application. The results of the paper maybe useful for

engineers who design rotors and want to use this method of

balancing. The vibratory forces are met in many dynamic

systems and they give some interesting phenomena.

Acknowledgements

This research is sponsored by the National Council of

Science and Technology (CONACYT), Mexico. (Grant No.

167877), VIBRATORY MECHANICS AND ITS

APPLICATION.

References 1. Muszynska A., Rotordynamics, CRC Taylor&Francis, 2005

2. Leblanc M., Automatic Balancer for Rotating Bodies. US

Patent 1,159,052, 1915

3. Thearle E. L., Means for dynamically balancing tools. USA

patent No 1,967,163, 1934

4. Thearle E. L., Automatic dynamic balancers. Machine

Design, No.9: 119–124, No.10: 103–106, No. 11: 149–153,

1950

5. Lee J., Van Moorhem W. K., Analytical and experimental

analysis of a self-compensating dynamic balancer in a

rotating mechanism. Transactions of ASME, 118, 468-475,

1996

6. L. Sperling, B. Ryzhik, Ch. Linz, H. Duckstein, Simulation

of two-plane automatic balancing of a rigid rotor.

Mathematics and Computers in Simulation, Vol. 58, Issues

4-6, 351-365, 2002

7. J. Chung, I. Jang, Dynamic response and stability analysis of

an automatic ball balancer for a flexible rotor, Journal of

Sound and Vibration 259 (1), 31–43, 2003

8. Rajalingham C., Bhat R. B., Complete balancing of a disk

mounted on a vertical cantilever shaft using a two ball

automatic balancer, Journal of Sound and Vibration Vol.

290 (1–2), 169–191, 2006

9. L Urbiola-Soto, M Lopez-Parra. Liquid self-balancing

device effects on flexible rotor stability. Shock and

Vibration 20 (1), 109-121, 2013

10. R. Rajalinghman, S. Rakheja. Whirl suppression in

hand-held power tool rotors using guided rolling balancers.

Journal of Sound and Vibration, 217(3) (1998), pp.453-466

11. W. Kim, J.Chung. Performance of automatic ball balancers

on optical disc drives. Journal of Mechanical Engineering

Science, 216, 2002, pp.1071-1080

12. Majewski T. Domagalski R., Meraz M., Automatic

Compensation of Dynamic Forces for the Rigid Rotor,

Journal of Theoretical and Applied Mechanics, Poland, No.

2: 379–403, 2007

13. Majewski T., Position error occurrence in self-balancers

used on rigid Rotors of rotating machinery. Mechanism and

Machine Theory 23 (1): 71 – 78, 1988

14. Blekhman I. I., Synchronization in Science and Technology,

ASME Press Translations, 1988, New York