Balancing of rotating masses

• Balancing

– Rotating masses

• Single plane

• Different planes

– Reciprocating masses– Reciprocating masses

• Primary force and couple

• Secondary force and couple

1

Balancing of single rotating weight by a weight rotating

in the same plane

m

ω

r

θO

A weight W carried on a

weight less arm of length

r rotates with angular

velocity ω as shown. For

completely balancing the

b

θ

θ

O

Bbmr

bBrm

=∴

=22 ωω

completely balancing the

mass m, a weight B at

distance b is attached to

the same axle in a

diametrically opposite

direction.

2

Balancing a number of masses rotating in one plane by another weight

rotating in the same plane

• Consider the masses m1, m2,m3 revolving

the at radii r1,r2,r3 respectively in the

same plane. Then each mass produces a

centrifugal force acting radially outwards

from the axis of rotation.

• If F is the vector sum,

F= m1r1ω2+ m2r2ω2+ m3r3ω2

All the masses will be in balance if F=0.

r1r2

r

m1

m2

θ1

θ2

θ3

θAll the masses will be in balance if F=0.

If F is not zero, a counter weight of mass mc

at radius rc is introduced to balance the

masses

F= m1r1ω2+ m2r2ω2+ m3r3ω2 + mcrcω2=0

m1r1+ m2r2+ m3r3 + mcrc=0

Analytically, it can be solved by resolving the

forces into components along and

perpendicular to x axis.

r3

m3

mc

rc

θc

m1r1

m2r2

m3r3

mcrc

3

Analytical method

( ) ( )

+=

=+

=+

ccciii

ccciii

rmrmrm

rmrm

rmrm

θθ

θθ

θθ

cossin

0sinsin

0coscos

22( ) ( )

−

−=

+=

iii

iii

c

iiiiiicc

rm

rm

rmrmrm

θ

θθ

θθ

cos

sintan

cossin22

4

Balancing a number of masses rotating

in different planes

• It is proposed to reduce the case to the case

of several masses rotating in the same plane.

• In order to do so, the different masses rotating

in different planes will have to be transferred in different planes will have to be transferred

to one plane called as a reference plane.

5

Transfer of force from one plane to another

FF2=F

F1=F

A

B

a

Consider two planes A and B. Let F be the centrifugal force due to a rotating

mass m acting in the plane B. The equilibrium in the planes will not be

disturbed if two equal and opposite forces F1and F2each equal to and parallel

to F. then F and F1 constitute a couple of magnitude F.a and F2=F will be also

present. Thus the effect of transferring a rotating mass from one plane A to

another plane, B is to introduce in the plane B

an unbalanced equal force in the same direction and

an unbalanced couple whose magnitude is equal to the product of the force

and the distance between the two planes. 6

• Therefore, transfer of various forces in

different planes to a reference plane will result

in a number of unbalanced forces and

unbalanced couples in the reference plane.

Hence, to balance the system, the unbalanced

forces and couples are to be balanced. The forces and couples are to be balanced. The

conditions to be satisfied are

e planee referenc i from thce of masstandisa

arm

sses no. of man to n where i

rm

i

iii

ii

=

=

==

=

0

)(1

0

7

• couple acts in plane perpendicular to the plane in which the forces act. Its dire ction can be obtained by application of right hand screw rule. Thus the balancing couple will be at right angle to the forces producing it. However, in actual practice, the phase of couple diagram is changed through 90 degrees, thus enabling us to draw couples parallel to the forces.couples parallel to the forces.

• As a couple is produced by two forces, two balancing masses acting in two different planes will be required.

8

Masses in different planes

θ3

r3

3

R2

a

θ3

m

r3

3

R2

aR2

mR2

θR2

Couple polygon

m3r3a3

θR2

θ1

m1

θ2

m2

m3

r1

r2

R1

1

2

a1

a2

a3

aR2

mR1

θR1

θ1

m1

θ2

m2

m3

r1

r2

R1

1

2

a1

a2

a3

aR2

θR1

Force polygon

Couple polygon

9

• In order to balance the masses completely, introduce two masses mR1

and mR2 at radii rR1 and rR2 in the reference planes R1 and R2

respectively. Then the conditions for complete balance are

ygraphicallor ly analytical solved becan equations above The

)4........(0

)3......(0

)2.....(0

)1......(0

222

2

222

2

333

2

222

2

111

2211

2

22

2

11

2

33

2

22

2

11

=+

=+++

=++

=++++

RRRiii

RRR

RRRRii

RRRR

armarm

armarmarmarm

rmrmrm

rmrmrmrmrm

ωωωω

ωωωωω

( ) ( )

)8..(....................cos

sintan

)7........(cossin

(6) and (5) Eq. from

)6........(0sinsin

)5.......(0coscos

,components into forces theResolving

2

22

222

2222

2222

−

−=

+=

=+

=+

iiii

iiii

R

iiiiiiiiRRR

RRRRiiii

RRRRiiii

arm

arm

armarmarm

armarm

armarm

θ

θθ

θθ

θθ

θθ

10

( ) ( )

( )( )

)10..(....................coscos

sinsintan

)9...(coscossinsin

components its by taking solved becan Eq.(2)

Eq.(2),in (8) and (7) Eq. from values thegSubsitutin

222

222

1

2

222

2

22211

+−

+−=

+++=

RRRiii

RRRiii

R

RRRiiiRRRiiiRR

rmrm

rmrm

rmrmrmrmrm

θθ

θθθ

θθθθ

The Eq. (2) and (4) can be solved graphically by solving first The Eq. (2) and (4) can be solved graphically by solving first

Eq.(4) by drawing couple polygon to determine mR2rR2aR2 and

θR2 and then Eq.(2) through force polygon to determine

mR1rR1 and θR1 .

11

mmr

r

3.1501.0/03.15

)150sin60 x 5.00sin50 x 5.0()150cos60 x 5.00cos50 x 5.0( x 1.0 22

==

+++=

12

Balancing of reciprocating parts

• Acceleration of reciprocating parts in an

engine is given by

2coscos2

rts, ocating pathe recipre mass of If m is th

nrf

θθω

+=

13

force. ngacceleratisecondary thecalled is 2cos

and force ngacceleratiprimary thecalled is cos

2coscos

2

2

2

nmr

mr

nrmF

ss m isate the mato acceler required then force

θω

θω

θθω

+=

ceptengines ex, for all eater thance n is grsin

nmr

mr

1

force.secondary thecalled is 2cos

and forceprimary thecalled is cos

2

2

θω

θω

14

gected.nel is

ceondary forof balancing d engines, high spee

ceptengines ex, for all eater thance n is grsin

sec

1

Partial balancing of Primary force

• The primary force acts along the line of stroke as shown in the figure. The

primary force can be considered as a component of a centrifugal force due

to a mass m at the crank radius r. Thus, balancing of primary force is

A

O

m

θ

r

Bθ

to a mass m at the crank radius r. Thus, balancing of primary force is

equivalent to balancing of mass m.

• This can be done by attaching a mass B at a radius b in diametrically

opposite direction as shown such that Bb= mr

• By the above method, the primary force is completely balanced, but the

vertical component of the centrifugal force due to B = Bbω2 sinθ remains

and its maximum value, Bbω2 is again equal to the maximum magnitude

of the primary force, mrω2 .

• In other words, the effect of the above method of balancing is toun

change the direction of maximum unbalanced force from that along the

line of stroke to that perpendicular to it.15

• As a compromise, a fraction, c of the reciprocating parts is

balanced such that

Bb= cmr

As a result unbalanced force along the line of stroke = (1-

c)mrω2cosθ

Similarly, unbalanced force perpendicular to the line of stroke =

cmrω2sinθ

instantany at forceprimary unbalancedResultant ∴

16

( )[ ] [ ]

( )[ ] [ ]222

2222

sincos1

sincos1

instantany at forceprimary unbalancedResultant

θθω

θωθω

ccmr

cmrmrc

+−=

+−=

∴

The resultant unbalanced force will be minimum when c=1/2.

However, the unbalanced force along the line stroke is more

harmful than in a direction perpendicular to it. So, the common

practice is to balance two-third of the reciprocating parts.

Partial balancing of reciprocating parts of two cylinder locomotives

• The two cylinders of locomotive have their cranks at 90 degrees.

• On account of partial balancing of reciprocating parts, there is an

• 1. unbalanced force along the line of stroke and 2. unbalanced force perpendicular to the line of stroke.unbalanced force perpendicular to the line of stroke.

• The effect of unbalanced force along the line of stroke is to produce

– a)the variation of tractive force along the line of stroke

– b) the unbalanced couple

17

• The effect of unbalanced force perpendicular

to the line of stroke is to produce

– c) the variation of pressure on rails which results

in hammering action on the rails. The maximum

magnitude of the unbalanced force perpendicular

to the line of stroke is termed as hammer blow.to the line of stroke is termed as hammer blow.

18

• Consider a two cylinder engine with cranks at 90 degrees as shown.

m - mass of reciprocating parts per cylinder

c - fraction of reciprocating parts balanced

r - radius of crank

a - the distance between cylinder centre lines

1

2

12

19

1

a

( )

( ) ( )

( ) ( ) 22

2

2

)90cos(1cos1F stroke of line

along force tractiveunbalancedResultant

90cos1

2cylinder

cos1

1cylinder

of stroke of line thealong force unbalanced

:force tractiveofVariation a)

+−+−=

+−=

−=

θωθω

θω

θω

mrcmrc

mrc

mrc

20

( ) ( )

( )

oo

t

2

t

22

t

315or 135

1tan

0cossin

0

minimumor maximum be toFfor

)sin(cos1F

)90cos(1cos1F stroke of line

=

−=

=−−

=

−−=

+−+−=

θ

θ

θθ

θ

θθω

θωθω

d

dF

mrc

mrcmrc

t

( )

( )

force ein tractiv variationas to

referred are F of valuesminimum and Maximum

12

2

1

2

11 min. and max.

t

2

2

ω

ω

mrc

mr-cFt

−=

=

m

mm

21

b) Swaying couple

• Unbalanced forces along line of stroke for cylinders 1 and 2

constitute a couple. This couple is measured about line YY in

figure and its maximum magnitude is known as swaying couple

and occurs at θ=45o and 225o.

( ) ( ) ( )

( ) ( )sincos1

290cos1

2cos1,

2

22

amrc

amrc

amrcCCouple

θθω

θωθω

+−−=

+−+

−−=

22

( ) ( )

( ) ( )

( )2

1minmax

225 45

1tan

0cossin2

1

0 when minimumor maximum is couple the

sincos2

1

2

2

amrcorC

and

amrc

d

dC

d

dC

mrc

oo

ω

θ

θ

θθωθ

θ

θθω

−=

=

=

=+−−−=

=

+−−=

m

c) Hammer blow• The unbalanced force perpendicular to the line of

stroke due to balance mass B1 at radius b to balance reciprocating parts only= B1bω

2 sinθ

• The maximum magnitude of this force is known is hammer blow and occurs at θ=90o and 270o

• Hammer blow =B1bω2

• If P is the downwards pressure on rail due to dead load • If P is the downwards pressure on rail due to dead load and other loads, then net pressure =P± B1bω

2.

• If P- B1bω2 is negative, then the wheels will be lifted

from the rails. Therefore the limiting condition in order that the wheel does not lift from rails is P- B1bω

2 =0.

• The permissible value of angular speed obtained from the condition is ω=(P/B1b)1/2

23

Balancing of secondary force

r

tl

rmr

nmrFS

ωω

θω

2cos

2cos

2

2

2

=

=

24

tl

rm ωω 2cos .

4 . )2(

22

=

Balancing of In-line engines

Two cylinder in-line engines:

Refe

rence p

lane

θmra cosθ

Primary couple polygonθ

r

25

c a

Refe

rence p

lane

Consider a two cylinder vertical engine having equal reciprocating masses and

cranks 180o apart as shown. The crank angles are θ and 180+θ. Length of each

crank is r and the distance of cylinder center lines from the reference plane are

c and a. the reference plane is at the centre of left hand main bearing.

θ

Primary force polygon

Two cylinder in-line engines….

• Primary forces: From the primary force polygon diagram, it is completely balanced.

• Primary couples: From the primary couple polygon diagram, resultant primary couple is proportional to the vertical component of the resultant mra.resultant mra.

• Resultant primary couple= mrω2acosθ

• It is maximum at θ=0 and 180o and the maximum value=±mrω2a

26

• Secondary forces:

theof speed the twiceiscrank thisof speed The

4

2cos .4

. )2(

2

1

22 ωω

=

=

l

rrtaking

tl

rmFS

Two cylinder in-line engines….

27

ly.respective

2360 and 2 are anglescrank secondary the

,180 and are anglescrank actual when thesince

linein lie pinscrank Therefore, cranks. actual

theof speed the twiceiscrank thisof speed The

θθ

θθ

+

+

Two cylinder in-line engines….

• Resultant centrifugal force is proportional to 2mr1.

( )

l

rm

l

rm

22

oooo

22

2 valuemaximum

theandcrank actual theof revolutionper four times i.e.

270 and 180 ,90 ,0 occur when valuesmaximum The

2cos4

22 forcesecondary Resultant

ω

θ

θω

=

=

=

28

lm2 valuemaximum =

2θr1

a

2θ

mr1

mr1

Secondary

force polygon

2θ

mr1(a/2)

Secondary

couple polygon

• Secondary couples: From the secondary

couple polygon, secondary couples are

balanced.

• Resultant secondary couple=0.

Two cylinder in-line engines….

29

Four cylinder in-line engine

θ Re

fere

nce

pla

ne

r

30

aa a

2θr1

a aa

Primary forces are balanced

Primary couples are balanced

( ) θω 2cos4

24 forcesecondary Resultant 2

2

l

rm=

Secondary couples are balanced

Questions

31

Which one of the following can completely balance

several masses revolving in different planes on a

shaft?

(a)A single mass in one of the planes of the revolving

masses

32

masses

(b) A single mass in any one plane

(c)Two masses in any two planes

(d)Two equal masses in any two planes.

Ans. (c)

A rotor supported at A and B, carries two masses as

shown in the given figure. The rotor is

(a) dynamically balanced

(b) statically balanced

(c) statically and dynamically balanced

(d) not balanced.(d) not balanced.

33

Ans. (b)

• Masses B1, B2 and 9 kg are attached to a shaft

in parallel planes as shown in the figure. If the

shaft is rotating at 100 rpm, the mass B2 is

(a) 3 kg (b) 6 kg (c) 9 kg

(d) 27 kg

34

Solution:

9 x0.5=B2 x 1.5

B2=3 kg

Static balancing is satisfactory for low speed rotors but with increasing speeds, dynamic balancing becomes necessary. This is because, the

(a) Unbalanced couples are caused only at higher speeds

(b) Unbalanced forces are not dangerous at higher speeds

(c) Effects of unbalances are proportional to the square of the speed

(d) Effects of unbalances are directly proportional to the speed speed

Ans. (C)

35

• The balancing weights are introduced in

planes parallel to the plane of rotation of the

disturbing mass. To obtain complete dynamic

balance, the minimum number of balancing balance, the minimum number of balancing

weights to be introduced in different planes is

(a) 1 (b) 2 (c) 3 (d) 4

• Ans. b

36

A rigid rotor consists of a system of two masses located as shown in the given figure. The system is

(a) statically balanced

(b) dynamically balanced

(c) statically unbalanced

(d) both statically and dynamically unbalanced

Ans. a

centre of masses lie on the axis of rotation.

37

If the ratio of the length of connecting rod to the crank radius increases, then

(a) Primary unbalanced forces will increase

(b) Primary unbalanced forces will decrease

(c) Secondary unbalanced forces will increase (c) Secondary unbalanced forces will increase

(d) Secondary unbalanced forces will decrease

Ans. (d) Secondary force only involves ratio of length of connecting rod and crank radius.

If n increases, value of secondary force will decrease.

38

• A single cylinder, four-stroke I.C. engine rotating

at 900 rpm has a crank length of 50 mm and a

connecting rod length of 200 mm. If the effective

reciprocating mass of the engine is 1.2 kg, what is

the approximate magnitude of the maximum

'secondary force' created by the engine?

(a) 533 N (b) 666 N (c) 133 N (a) 533 N (b) 666 N (c) 133 N

(d) None of the above

Sol: Maximum secondary force= mrω2/n

39

• Consider the following statements:

An in-line four-cylinder four-stroke engine is

completely balanced for

1. primary forces 2. secondary forces 1. primary forces 2. secondary forces

3. primary couples 4. secondary couples

Of these statements:

(a) 1, 3 and 4 are correct (b) 1, 2 and 4 are correct

(b) (c) 1 and 3 are correct (d) 2 and 4 are correct

40