balancing of rotating parts dom dom dom do m dom dom dom dom dom dom dom

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1. Unit 4

Balancing

Introduction to Balancing
‘Often an unbalance of forces is produced in rotary or reciprocating machinery duo to the inertia forces associated with the moving masses.
Balancing is the process of designing or modifying machinery so that the unbalance is reduced to an acceptable level and if possible is eliminated entirely.
particle or mass moving in a circular path experiences a centripetal acceleration and a force is required to produce it
“An equal and opposite force acting radially outwards acts on the axis of rotation and is known as centrifugal force.
‘This isa disturbing force on the axis of rotation, the magnitude of which is constant but the direction changes with the rotation of the mass,
In a revolving rotor, the centrifugal force remains balanced as long as the center of the mass of the rotor ies on the axis of the shaft.
When the conter of mass does not ie on the axis or there is an eccentricity, an unbalanced force is produced.

Need of Balancing

‘Tho high spood of onginos and other machines is a common phenomonon nou”
Itis, theroforo, vory ossontial that all the rotating and reciprocating parts should bo completely balancod as far as possible.

I these parts are not properly balanced, the dynamic forces are set up.

‘These forces not only increase the loads on bearings and stresses in the various members, but also produce unpleasant and even dangerous vibrations.

3. Considera distubing mass ml attached toa shalt rotating at w rad/s as show in ig

Letr be the radius of rotation ofthe mass my (Le. distance between the aris of rotation of the shatand the center of gravity ofthe mass my).

Wie know that the centrifugal force exerted by the mass my onthe

=m OF

This centrifugal force acts radially outwards and thus produces bending moment on the shat,

In order to counteract the effect ofthis force, a balancing mass (mg) may be atached in the same plane of rotation as that of disturbing mass (om) such that th centrifugal forces due to the two
masses are equal and opposite

Distt

Balancing mass

Let 1g be the raits of rotation ofthe mass mg (Le distance between the ais of rolation ofthe shaft and the center of gravity ofthe mas mp)

‘Therefore centrifugal force exerted by the mass my onthe ha

4. Centrifugal force due

F

may be split up
balancing mass (m,) is genera
The centri

tation of the shaft and the centre of

ravity of mass m, )

venient way. But the radius of n
10 reduce the balancing mass m;

I the mass and radius of

5. Equating both equations

MOT =m 0 M EM

In the previous arrangement for balancing gives rise to a couple which tends to rock the shaft in its bearings,

‘Theroforo, in order to put the system in complete balance, two balancing masses aro placed in two different planes, parallel to the plane of rotation of the disturbing
‘mass, in such a way that they satisty the following two conditions of equilibrium.
+. The net dynamic force acting on a shaft is zero. This requires that the line of action of three centrifugal forces must be same. In other words, the center of
line of masses of the system must lo on the axis ofthe rotation,
+ The net couple due to dynamic forces acting on a shaft is equal to zero. In other words, the algebraic sum of moments about any point on the plane must
bo zero.

Le.The system must be dynamic balance

ve a

6. Let these masses rotate about an axis throug

The magnitude and ofthe balancing mass may
+ Analytical Method

magnitude and direction of the bala

magnitud

to nd
“u

is the angle which the zosultant forces m

ed
The balancing force is equal to the resulta
+ Nox, find the magnitude of the balan

where.
m =Balancing ma

= Radius of rotatio

7. 2. Graphical Method

The magnitude and position of the balancing mass may also be obtained graphically as discussed below
+ First of all, draw the space diagram with the position of several masses, as shown in fig (a)
{Find out the centrifugal force (or product of mass and radius of rotation) exerted by each mass on the rotating shaft
+ Now, draw the vector diagram with the obtained centrifugal force, such that ab represents a centrifugal force exerted by the mass m, (or m, Xy)

magnitude and direction with suitable scale. Similarly draw be, cd and de to represent centrifugal forces of other masses mz, ms, and my

Resultant

A
“Foi

(a) Space diagram, (b) Vector diagram,
Now, as per polygon law of forces, the closing side ae represents the resultant force in magnitude and direction, as shown in fig.

‘The balancing force, is then, equal to the resultant force, but in opposite direction.
Now, find the magnitude of the balancing mass (m) at a given radius of rotation 1), such that,

moitr = Resultant Centrifugal for

mr = Resultant of mur, mar mar and mrs

8. ‘Balancing of Several Masses Rotating in Different Planes

ular position of the masses.

9. frene plane (R.P.). Th

may be obtained as discussed below :

Take om

Tabulate the data as shown in Table
from left 10 right

10. + A couple may be represented by a vector drawn perpendicular to the plane of the couple, The cc

and acts in he plane through O

uple C, introduced by transferring m to th reference p

and perpendicular tothe paper. The Vector representing this coup!

ras in the plane of the paper and perpendicular to O

mby OC, in Fig (). Similar, the vectors, OC,

OC, and OC, are drawn perpendicular 10 0

din the plane of paper

) Couple vectors turned

‘kwise through

11. crise trough aright angle forconverience of drewing as
r80C,,0C, and OC, are parallel an inthe same direction as

o
ando
while vector OC, is parallel to O

butin opposite direction. Hence th couple vectors ar drawn reall outwards fo the masses on one side ofthe reference plane and radially inward forthe masses onthe other side othe reference plane
as thon in () The vector do represents the balanced couple ouple Cy proportional te mye Ly Thereore
vector do
Mg

Cu = mu a hu = vector do’ or

la

Im this expresion, he cing mas min the plane M may be obtained and angle finclination 4 is may be measured tom fig (b)

“Cu
d
(e) Couple polygon. (f) Force polygon,
«ow dante oe polygons showing (Th eco eo nthe ction tom 010) represents the balanced fre Since he aan oes propano, ero,

Vector eo.

in he plane Lmay be obtained

12. Balancing of reciprocating masses

(Consider a horizontal reciprocating engine mechanism a shown in

fy Foren eu se vostngpare
nero du toesprca par,
= For on be ider le der val or normal fern sing be machen
Fore ang on he nat bearing or mas bering
agitado but oppo in deco, hero te balance each er

GF) ating along te ie of eciprocaion is lo cual and opposite

required tobe propery balan

vertical component of (Ley) ae equal and opposite and ths form ashaing couple of magnitude Kory, xX

From above, me ses tha De electa he in king force and a shaking cou re and a shaking couple vary in marido and direction during the en
every objectionable vib

eof balancing he reiprocatg masse i force and a shaking couple In mos ol the mechanins, we can res the shaking ore and à shaking couple by a
ng mae, bute ouai no raten 1 eliminate them completely. I other word, the reciprocatng ases ae ony partaly balanced

13. Primary and Secondary Unbalanced Forces of Reciprocating Masses

10 mechanism as shown in Fig.

= Maa tie mp

Length ot

ine of stoke PO,
Angular speed,
= Ratio of length ing rod to the crank radius = Lt

yen by De expression

14. exerted on he crak haft bearing (Le Fo) it equal and opposite nera force (F) This force is an unbalanced one and

marc ES |= ma root mal rx

The expression (m-o?-rcos8) is known as

00520

[nar | cated

Primary unbalanced force, Fp =m-0? -reos®
md secondary unbalanced force, = mg? «rx

(Characteristics of primary and secondary unbalanced forces

is maximum when 82 0 0 10" and dar force is maximum four times in one revlon of the rank The maximum secondary unbalance fore is

Fs

+ From above wo see tat secondary unbalanced foros 1/n times he maximum prinaryunbaa
+ Incaseolmoderatespeds he secondary unbalanced fre is small at may be neg
+ The unbalanced force due 1 ecir cion wl de othe evolving masses, the unbalanced force costar in magaitud bat are indirecto

15. Balancing of Single Cylinder Reciprocating Engine

‘The primary unbalanced force (marcos 9) may be considered as the component ofthe centrifugal force produced by a rotating mass m placed atthe crank radius r,as shown in Fig.

boos 8

7
marco 7

b

Bo’ bsind

‘The primary force acts from O 1 P along the line of stoke, Hence, balancing of primary foros is considered as equivalent to the balancing of mass m rotating atthe crank radins
‘This is balanced by having a mass B at a radius b placed diametrically opposite tothe crank pin
We know that centrifugal force due to mass B,
=B-w-b
and horizontal component of this force acting in opposite direction of primary force
= B-@ -bcos0
The primary force is balanced, if

B-@ bcos =m-w"-rcosO or

16. Alittle consideration will show, hat the primary force is completely balanced if .b = mır, but the centrifuga force produced due to the revolving mass B has also a vertical component
(Perpendicular t the line of stroke) of magnitude (B*bsin 9

‘This force remains unbalanced. The maximum value ofthis force is equal to Bab when i 90° and 270°, which is same as the maximum value ofthe primary force mur

1 above discussion, we see that inthe fist case, the primary unbalanced force acts along the line of stroke whereas in the second case, the unbalanced force acts along the

im value of the force remains same in both the cases. I is thus obvious, that the elect of the above method of balancing isto change the direction of the maximum
unbalanced force from the line of stroke tothe perpendicular of line of stoke.

Asa compromise, lta fraction ofthe reciprocating masses is balanced, such tha: mr = Bb

Therefore Unbalanced fre along he ine of stoke
=m-0* rc0s8-B- 0 -bc0s8

=m-@ -rcos8—c-m- a «r cos +: Bb=cmr
=(1-c)m-0* -rcos0

and unbalanced force along the erpendicalar othe ine of sroke

=B-o -bsin6=c-m-w -rsin8

Resultant unbalanced force at any instant

=y[t-om.* rend] + c-m-o"-rsin8 |

= m4 «raf(1=c] cos? 0+c sin?8

17. ‘Balancing o Primary Focus
"Tas following tr conditions must be ein arder to give the primary balance ol he eciprcatog parts fa ml liner engine:

2) The algebmic sm oh couple about an poinin he plane otro primary feroe must be qual ooo In the mods he primary couple pen mus los.
Tea, in cres a give he piman lanos fhe reciprocating pat olamullnder engine itis coven! o imagino e ecirocag mass o otero ei spe rakpis and o eat
the problem ase olor masse
tes
+ Fora wo-einder egin with cranks 180 condition I) may be sae, bt hi wilt an usblaroed cosple Tus he above method o primary balancing censo be applied is cas
2 Fu or hide engine with crank a 120 andihezeiprecaung meses per cylinder are same, the coin (1) wil be aed because horse ay be mpresented by
triangle Howe, by kg a lerne plane though one ofthe cnet center Ans, o couples vino pardo! exes wil remain and thee anna ah vectorial Hence he abore mind of balancing
fale nica ase
+ Forafour eyinder engine, similar song il ah that complete primary balance is possible
2 Reramulcoynder engine te primary forces may e completely balance by sulably ranging he cask ange, proide a! the number ol ras ae not le an our
* Tue clonng nde ofthe primary fre plygon gres the mamar unbulaned primary force atthe closing se othe primary couple palos ges the mami us alan primary couple
Balancing of Secondary Forces
When the connecting rod is ot to ong (een the cgi ofthe connecting rod considered) then the secondary dub fore dae tothe reciprocating mass aries

secondary ore,

oF

This expression may be written as

Fg =m.(20)?x—xeos28
m

‘Asin case of primary forces, he socordary ores may be considered tobe equivalen 1 ha component, par! othe ine of se, of he conga ros produced by an equal mass placed a ie imaginary
rankof length vn and rer ing atrio the speed of te al crn e 2) a sown Fi

18. Balancing of Secondary Forces of Multi-cylinder In-line Engines

Fg = mer

sal
This expression may be written as er

Imaginary crank
an

À
3

“Thu, in muli-oylinder inline engines, each imaginary secondary crank witha mass attached othe erankpin i inlined to th line of stroke at tice Ihe angle of he actual erank +
‘The values ofthe secondary forces and couples may be obtained by considering the revolving mass. This i done in the salar way as discussed for primary forces.

The following two conditions must be sisi in order to give a complete secondary balance of an engine
1 The algebraic sum of the secondary forces must be equal o ero In other word, the secondary force polygon must close, and
2.The algebraic sum of the couples about any point inthe plane of he secondary forces must be equal o 20ro In other words, the secondary couple polygon must lose.

Note: The closing sido ofthe secondary force polygon gives the maximum unbalanced secondary force and the closing side of the secondary couple polygon gives the maximum
unbalanced, ple