BALANCING OF RECIPROCATING MASSES - … · RECIPROCATING MASSES BALANCING OF RECIPROCATING MASSES...

VTU EDUSAT PROGRAMME-17

DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE

1

DYNAMICS OF MACHINES

Subject Code -10 ME 54

BALANCING

OF

RECIPROCATING MASSES

BALANCING

OF

RECIPROCATING MASSES

SLIDER CRANK MECHANISM:

PRIMARY AND SECONDARY ACCELERATING FORCE:

Acceleration of the reciprocating mass of a slider-crank mechanism is given by,

r

lnWhere

)(n

coscosr

pistonofonAcceleratiap

=

−−−−−−−−−

θ+θω=

=

122

And, the force required to accelerate the mass ‘m’ is

)(n

cosrmcosrm

n

coscosrmF

i

22

2

22

2

−−−−−−−−−θ

ω+θω=

θ+θω=

Notes Compiled by:

VIJAYAVITHAL BONGALE

ASSOCIATE PROFESSOR

DEPARTMENT OF MECHANICAL ENGINEERING

MALNAD COLLEGE OF ENGINEERING

HASSAN -573 202. KARNATAKA

Mobile:9448821954

E-mail:[email protected]

ADARSHA H G



2

The first term of the equation (2) , i.e. θω cosrm2

is called primary accelerating

force the second term n

cosrm

θω

22 is called the secondary accelerating force.

Maximum value of primary accelerating force is 2

ωrm

And Maximum value of secondary accelerating force is n

rm2

ω

Generally, ‘n’ value is much greater than one; the secondary force is small compared to

primary force and can be safely neglected for slow speed engines.

In Fig (a), the inertia force due to primary accelerating force is shown.

ADARSHA H G



3

In Fig (b), the forces acting on the engine frame due to inertia force are shown.

At ‘O’ the force exerted by the crankshaft on the main bearings has two components,

horizontal hF21 and vertical

vF21 .

hF21 is an horizontal force, which is an unbalanced shaking force.

vF

21 and vF

41 balance each other but form an unbalanced shaking couple.

The magnitude and direction of these unbalanced force and couple go on changing with

angle θ. The shaking force produces linear vibrations of the frame in horizontal direction,

whereas the shaking couple produces an oscillating vibration.

The shaking force hF21 is the only unbalanced force which may hamper the smooth

running of the engine and effort is made to balance the same.

However it is not at all possible to balance it completely and only some modifications can

be carried out.

BALANCING OF THE SHAKING FORCE:

Shaking force is being balanced by adding a rotating counter mass at radius ‘r’ directly

opposite the crank. This provides only a partial balance. This counter mass is in addition

to the mass used to balance the rotating unbalance due to the mass at the crank pin. This

is shown in figure (c).

ADARSHA H G



4

The horizontal component of the centrifugal force due to the balancing mass is

θω cosrm2

and this is in the line of stroke. This component neutralizes the unbalanced

reciprocating force. But the rotating mass also has a component θω sinrm2

perpendicular to the line of stroke which remains unbalanced. The unbalanced force is

zero at θ = 00 or 180

0 and maximum at the middle of the stroke i.e. θ = 90

0. The

magnitude or the maximum value of the unbalanced force remains the same i.e. equal to 2

ωrm . Thus instead of sliding to and fro on its mounting, the mechanism tends to

jump up and down.

To minimize the effect of the unbalance force a compromise is, usually made, is 3

2 of the

reciprocating mass is balanced or a value between 2

1 to

4

3.

If ‘c’ is the fraction of the reciprocating mass, then

θcosωrmcmassthebybalancedforceprimaryThe 2=

and

θcosωrmc)1(massthebyunbalancedforceprimaryThe 2−=

θsinωrmc

unbalancedremainswhichforcelcentrifugaofcomponentVertical2

=

In reciprocating engines, unbalance forces in the direction of the line of stroke are more

dangerous than the forces perpendicular to the line of stroke.

[ ] [ ]2222 θsinωrmcθcosωrc)m(1

instantanyatforceunbalancedResultant

+−=

The resultant unbalanced force is minimum when, 2

1=c

This method is just equivalent to as if a revolving mass at the crankpin is completely

balanced by providing a counter mass at the same radius diametrically opposite to the

crank. Thus if P

m is the mass at the crankpin and ‘c’ is the fraction of the reciprocating

mass ‘m’ to be balanced , the mass at the crankpin may be considered as P

mmc +

which is to be completely balanced.

ADARSHA H G



5

Problem 1:

A single –cylinder reciprocating engine has a reciprocating mass of 60 kg. The crank

rotates at 60 rpm and the stroke is 320 mm. The mass of the revolving parts at 160 mm

radius is 40 kg. If two-thirds of the reciprocating parts and the whole of the revolving

parts are to be balanced, determine the, (i) balance mass required at a radius of 350 mm

and (ii) unbalanced force when the crank has turned 500 from the top-dead centre.

Solution:

mm350r,3

2ckg,40m

mm320stroketheoflengthLrpm,60N

kg60partsingreciprocattheofmassm:Given

cP===

===

==

(i) Balance mass required at a radius of 350 mm

We have,

mm1602

320

2

Lr

rad/sπ260

60xπ2

60

Nπ2ω

===

===

kg804060x

3

2mmcM

MpincranktheatbalancedbetoMass

P=+=+=

=

and

kg 5736.350

160x80mi.e.

r

rMmthereforerMrm

c

c

ccc

==

==

(ii) Unbalanced force when the crank has turned 500 from the top-dead centre.

( )[ ] [ ]

( ) ( )

N209.9

50sin2πx0.16x60x3

250cos2πx0.16x60x

3

21

θsinωrmcθcosωrmc1

50θat force Unbalanced

2

02

2

02

2222

0

=

+

−=

+−=

=

ADARSHA H G



6

Problem 2:

The following data relate to a single cylinder reciprocating engine:

Mass of reciprocating parts = 40 kg

Mass of revolving parts = 30 kg at crank radius

Speed = 150 rpm, Stroke = 350 mm.

If 60 % of the reciprocating parts and all the revolving parts are to be balanced, determine

the,

(i) balance mass required at a radius of 320 mm and (ii) unbalanced force when the crank

has turned 450 from the top-dead centre.

Solution:

mm320r,%60c

mm350stroketheoflengthLrpm,150N ,kg30m

kg40partsingreciprocattheofmassm:Given

c

P

==

====

==

(i) Balance mass required at a radius of 350 mm

We have,

mm1752

350

2

Lr

rad/s15.760

150xπ2

60

Nπ2ω

===

===

kg543040x0.60mmcM

MpincranktheatbalancedbetoMass

P=+=+=

=

and

kg29.53320

175x54mi.e.

r

rMmthereforerMrm

c

c

ccc

==

==

(ii) Unbalanced force when the crank has turned 450 from the top-dead centre.

( )[ ] [ ]

( ) ( )[ ] ( )[ ]N880.7

45sin15.7x0.175x40x0.6045cos15.7x0.175x40x0.601

θsinωrmcθcosωrmc1

45θat force Unbalanced

2022

02

2222

0

=

+−=

+−=

=

ADARSHA H G



7

SECONDARY BALANCING:

Secondary acceleration force is equal to )(n

cosrm 1

22−−−−−−

θω

Its frequency is twice that of the primary force and the magnitude n

1 times the

magnitude of the primary force.

The secondary force is also equal to )(n

cos)(rm 2

4

22

2−−−−−−

θω

Consider, two cranks of an engine, one actual one and the other imaginary with the

following specifications.

Actual Imaginary

Angular velocity ω ω2

Length of crank r n

r

4

Mass at the crank pin m m

Thus, when the actual crank has turned through an angle tω=θ , the imaginary crank

would have turned an angle tω=θ 22

ADARSHA H G



8

Centrifugal force induced in the imaginary crank = ( )4n

2ωrm2

Component of this force along the line of stroke is = ( )

2θcos4n

2ωrm2

Thus the effect of the secondary force is equivalent to an imaginary crank of length n

r

4

rotating at double the angular velocity, i.e. twice of the engine speed. The imaginary

crank coincides with the actual at inner top-dead centre. At other times, it makes an angle

with the line of stroke equal to twice that of the engine crank.

The secondary couple about a reference plane is given by the multiplication of the

secondary force with the distance ‘ l ’ of the plane from the reference plane.

COMPLETE BALANCING OF RECIPROCATING PARTS

Conditions to be fulfilled:

1. Primary forces must balance i.e., primary force polygon is enclosed.

2. Primary couples must balance i.e., primary couple polygon is enclosed.

3. Secondary forces must balance i.e., secondary force polygon is enclosed.

4. Secondary couples must balance i.e., secondary couple polygon is enclosed.

Usually, it is not possible to satisfy all the above conditions fully for multi-cylinder

engine. Mostly some unbalanced force or couple would exist in the reciprocating engines.

BALANCING OF INLINE ENGINES:

An in-line engine is one wherein all the cylinders are arranged in a single line, one behind

the other. Many of the passenger cars such as Maruti 800, Zen, Santro, Honda-city,

Honda CR-V, Toyota corolla are the examples having four cinder in-line engines.

In a reciprocating engine, the reciprocating mass is transferred to the crankpin; the axial

component of the resulting centrifugal force parallel to the axis of the cylinder is the

primary unbalanced force.

Consider a shaft consisting of three equal cranks asymmetrically spaced. The crankpins

carry equivalent of three unequal reciprocating masses, then

ADARSHA H G



9

∑ −−−−−−−−−−−−−= (1)θcosωrmforcePrimary 2

∑ −−−−−−−−−−−−−= (2)θcoslωrmcouplePrimary 2

( )∑ −−−−−−−−−−−−−= (3)2θcos

n4

2ωrmforceSecondary

2

( )

∑

∑

−−−−−−−−−−−−−=

=

(4)2θcosln

ωrm

2θcosln4

2ωrmcoupleSecondary And

2

2

GRAPHICAL SOLUTION:

To solve the above equations graphically, first draw the ∑ θcosrm polygon (2

ω is

common to all forces). Then the axial component of the resultant forces )cosF(r

θ

multiplied by 2

ω provides the primary unbalanced force on the system at that moment.

This unbalanced force is zero when 0

90=θ and a maximum when0

0=θ .

ADARSHA H G



10

If the force polygon encloses, the resultant as well as the axial component will always be

zero and the system will be in primary balance.

Then,

00 == ∑∑ VPhPFandF

To find the secondary unbalance force, first find the positions of the imaginary secondary

cranks. Then transfer the reciprocating masses and multiply the same by ( )

n4

22

ω or

n

2ω

to get the secondary force.

In the same way primary and secondary couple ( m r l ) polygon can be drawn for

primary and secondary couples.

Case 1:

IN-LINE TWO-CYLINDER ENGINE

Two-cylinder engine, cranks are 1800 apart and have equal reciprocating masses.

ADARSHA H G



11

Taking a plane through the centre line as the reference plane,

[ ] 0θ)180(cosθcosωrmforcePrimary 2=++=

θcoslωrmθ)180(cos2

lθcos

2

lωrmcouplePrimary 22

=

+

−+=

Maximum values are lωrm 2 at 00 180and0θ=

[ ] θ2cosn

ωr2mθ)2360(cosθ2cos

n

ωrmforceSecondary

22

=++=

Maximum values are n

ωr2m 2

when 0000

0000

270and180,90,0θor

540and360,180,0θ2

=

=

ADARSHA H G



12

0)2θ360(cos2

l2θcos

2

l

n

ωrmcoupleSecondary

2

=

+

−+=

ANALYTICAL METHOD OF FINDING PRIMARY FORCES AND COUPLES

• First the positions of the cranks have to be taken in terms of 0θ

• The maximum values of these forces and couples vary instant to instant and are

equal to the values as given by the equivalent rotating masses at the crank pin.

If a particular position of the crank shaft is considered, the above expressions may not

give the maximum values.

For example, the maximum value of primary couple is lωrm 2 and this value is

obtained at crank positions 00 and 180

0. However, if the crank positions are assumed

at 900 and 270

0, the values obtained will be zero.

• If any particular position of the crank shaft is considered, then both X and Y

components of the force and couple can be taken to find the maximum values.

For example, if the crank positions considered as 1200 and 300

0, the primary couple

can be obtained as

( )

lωrm2

1

120180cos2

l120cos

2

lωrmcomponentX

2

0002

−=

+

−+=−

( )

lωrm2

3

120180sin2

l120sin

2

lωrmcomponentY

2

0002

=

+

−+=−

Therefore,

lωrm

lωrm2

3lωrm

2

1couplePrimary

2

2

2

2

2

=

+

−=

Case 2:

IN-LINE FOUR-CYLINDER FOUR-STROKE ENGINE

ADARSHA H G



13

This engine has tow outer as well as inner cranks (throws) in line. The inner throws are at

1800 to the outer throws. Thus the angular positions for the cranks are 0θ for the first,

00 θ180 + for the second, 00 θ180 + for the third and 0θ for the fourth.

ADARSHA H G



14

FINDING PRIMARY FORCES, PRIMARY COUPLES, SECONDARY FORCES

AND SECONDARY COUPLES:

Choose a plane passing through the middle bearing about which the arrangement is

symmetrical as the reference plane.

[ ]0

θcosθ)180(cosθ)180(cosθcosωrmforcePrimary 002

=

+++++=

0

θcos2

3lθ)180(cos

2

l

θ)180(cos2

lθcos

2

3l

ωrmcouplePrimary0

0

2

=

−++

−+

++

=

θ2cosn

ωr4m

θ2cosθ)2360(cos

θ)2360(cosθ2cos

n

ωrmforceSecondary

2

0

02

=

+++

++=

0000

0000

2

027and180,09,0θ

or054and360,018,02θat

n

ωrmvalueMaximum

=

=

=

0

2θcos2

3l)2θ360(cos

2

l

)2θ360(cos2

l2θcos

2

3l

n

ωrmcoupleSecondary

0

0

2

=

−++

−+

++

=

Thus the engine is not balanced in secondary forces.

ADARSHA H G



15

Problem 1:

A four-cylinder oil engine is in complete primary balance. The arrangement of the

reciprocating masses in different planes is as shown in figure. The stroke of each piston is

2 r mm. Determine the reciprocating mass of the cylinder 2 and the relative crank

position.

Solution:

r2

r2

2

Llengthcrank

kg480mkg,590m,?mkg,380m

:Given

4321

===

====

Plane Mass (m)

kg

Radius (r)

m

Cent.

Force/ω2

(m r )

kg m

Distance

from Ref

plane ‘2’

m

Couple/ ω2

( m r l )

kg m2

1 380 r 380 r -1.3 -494 r

2(RP) m2 r m2 r 0 0

3 590 r 590 r 2.8 1652 r

4 480 r 480 r 4.1 1968 r

ADARSHA H G



16

Analytical Method:

Choose plane 2 as the reference plane and0

30=θ .

.

Step 1:

Resolve the couples into their horizontal and vertical components and take their sums.

Sum of the horizontal components gives

)(coscos.,e.i

cosrcosrcosr

119681652494

0196801652494

41

4

0

1

−−−−−−−−−θ+=θ+

=θ++θ−

Sum of the vertical components gives

)(sinsin.,e.i

sinrsinrsinr

21968494

0196801652494

41

4

0

1

−−−−−−−−−θ=θ

=θ++θ−

Squaring and adding (1) and (2), we get

( ) ( ) ( )

( ) ( ) ( ) ( )

00

44

2

4

2

44

22

2

4

2

4

2

192.1or167.9θand0.978θcos

get,wesolvingOn

θsin1968θcos1968θcos1968x1652x21652494

θsin1968θcos19681652494

=−=

+++=

++=

.,e.i

Choosing one value, say 0

4167.9θ =

Dividing (2) by (1), we get

0

1

0

0

1

123.4θi.e.,

1.515272.28

412.53

)(167.9cos19681652

)(167.9sin1968θtan

=

−=−

+=

+=

Step 2:

Resolve the forces into their horizontal and vertical components and take their sums.


)(.cosmor

).(cosrcosrcosrm).(cosr

3588

0916748005904123380

22

00

22

0

−−−−−−−−−−−−−=θ

=++θ+

ADARSHA H G



17


)(.sinmor

).(sinrsinrsinrm).(sinr

49417

0916748005904123380

22

00

22

0

−−−−−−−−−−−−−−=θ

=++θ+


Anskg.m 14272

=


Ansr0

2

2

282θo

4.7288.5

417.9θtan

=

−=−

=

Graphical Method:

Step 1: Draw the couple diagram taking a suitable scale as shown.

ADARSHA H G



18

This diagram provides the relative direction of the masses 431mandm,m

.

Step 2: Now, draw the force polygon taking a suitable scale as shown.

This gives the direction and magnitude of mass m2.

The results are:

Anskgmorrrm

,,

427427

282123168

22

0

2

0

1

0

4

==

=θ=θ=θ

ADARSHA H G



19

Problem 2:

Each crank of a four- cylinder vertical engine is 225 mm. The reciprocating masses of the

first, second and fourth cranks are 100 kg, 120 kg and 100 kg and the planes of rotation

are 600 mm, 300 mm and 300 mm from the plane of rotation of the third crank.

Determine the mass of the reciprocating parts of the third cylinder and the relative

angular positions of the cranks if the engine is in complete primary balance.

Solution:

kg100mandkg120mkg,100m

mm225r

:Given

421===

=

Plane Mass (m)

kg

Radius (r)

m

Cent.

Force/ω2

(m r )

kg m

Distance

from Ref

plane ‘2’

m

Couple/ ω2

( m r l )

kg m2

1 100 0.225 22.5 -0.600 -13.5

2 120 0.225 27.0 -0.300 -8.1

3(RP) m3 0.225 0.225 m3 0 0

4 100 0.225 22.5 0.300 6.75

ADARSHA H G



20

Analytical Method:

Choose plane 3 as the reference plane and 0

10=θ .

Step 1:

Resolve the couples into their horizontal and vertical components and take their sums.


)(.cos.cos..,e.i

.cos.cos..,e.i

cos.cos.cos.

151375618

51375618

0756180513

42

42

42

0

−−−−−−−−−−θ=θ

+θ−=θ−

=θ+θ−−


)(sin.sin..,e.i

sin.sin.sin.

275618

0756180513

42

42

0

−−−−−−−−−θ=θ

=θ+θ−−


162.20365.61-182.2545.563θcos182.25i.e.,

182.25θcos182.25-45.563

182.25θcos182.25)θsinθ45.563(cos

θsin45.563182.25θcos182.25θcos45.56365.61

)θsin(6.7513.5)θcos(6.75(8.1)

4

4

44

2

4

2

4

2

44

2

2

4

2

4

2

=+=

+=

+−+=

++−=

+−=

Therefore, Ans27.13θand182.25

162.203θcos 0

44==


000

2

0

0

2

157.6718022.33-θi.e.,

1.5157.493

3.078

13.5-)(27.13cos6.75

)(27.13sin6.75θtan

=+=

−=−

==

Step 2:

Resolve the forces into their horizontal and vertical components and take their sums.


ADARSHA H G



21

(3)17.545θcosm0.225i.e.,

020.02θcosm0.22524.97522.5i.e.,

0)27.13(cos22.5θcosm0.225)(157.67cos27)0(cos22.5

33

33

0

33

00

−−−−−−−−−−−−−−=

=++−

=+++

And sum of the vertical components gives

(4)-20.518θsinm0.225i.e.,

010.26θsinm0.22510.258i.e.,

0)27.13(sin22.5θsinm0.225)(157.67sin27)0(sin22.5

33

33

0

33

00

−−−−−−−−−−−−−−=

=++

=+++


Anskg120kg119.98

0.225

20.518

0.225

17.545mi.e.,

20.518)(17.545)(m)(0.225

22

3

222

3

2

≈=

−+

−=

−+−=


Ans229.5θor

17.545-

20.518θtan

0

3

3

=

−=

ADARSHA H G



22

Problem 3:

The cranks of a four cylinder marine oil engine are at angular intervals of 900. The engine

speed is 70 rpm and the reciprocating mass per cylinder is 800 kg. The inner cranks are 1

m apart and are symmetrically arranged between outer cranks which are 2.6 m apart.

Each crank is 400 mm long.

Determine the firing order of the cylinders for the best balance of reciprocating masses

and also the magnitude of the unbalanced primary couple for that arrangement.

Analytical Solution:

17195(7.33)x0.4x800ωrm

s/rad7.3360

Nπ2ωm,0.4r,rpm70Nkg,800m

:Given

22==

=====

Note:

There are four cranks. They can be used in six different arrangements as shown. It

can be observed that in all the cases, primary forces are always balanced. Primary

couples in each case will be as under.

Taking 1 as the reference plane,

ADARSHA H G



23

( ) ( ) ( ) ( )

erchangedintarelandlcesin,onlymNCC

mN

...lllrmC

pp

p

4216

222

42

2

3

2

1

43761

43761

62808117195

==

=

−+−=−+−ω=

( ) ( ) ( ) ( )


mN

...lllrmC

pp

p

3225

222

32

2

4

2

2

47905

47905

81806217195

==

=

−+−=−+−ω=

( ) ( ) ( ) ( )


mN

...lllrmC

pp

p

3434

222

34

2

2

2

3

19448

19448

81628017195

==

=

−+−=−+−ω=

Thus the best arrangement is of 3rd

and 4th

. The firing orders are 1423 and 1324

respectively.

Unbalanced couple = 19448 N m.

Graphical solution:

ADARSHA H G



24

Case 3:

SIX – CYLINDER, FOUR –STROKE ENGINE

Crank positions for different cylinders for the firing order 142635 for clockwise rotation

of the crankshaft are, for

Since all the force and couple polygons close, it is inherently balanced engine for primary

and secondary forces and couples.

First 0

10=θ Second

0

2240=θ

654321

654321

rrrrrr

mmmmmm

And

=====

=====

Third 0

3120=θ Fourth

0

4120=θ

Fifth 0

5240=θ Sixth

0

60=θ

ADARSHA H G



25

Problem 1:

Each crank and the connecting rod of a six-cylinder four-stroke in-line engine are 60 mm

and 240 mm respectively. The pitch distances between the cylinder centre lines are 80

mm, 80 mm, 100 mm, 80 mm and 80 mm respectively. The reciprocating mass of each

cylinder is 1.4 kg. The engine speed is 1000 rpm. Determine the out-of-balance primary

and secondary forces and couples on the engine if the firing order be 142635. Take a

plane midway between the cylinders 3 and 4 as the reference plane.

Solution:

/srad72104.60

1000xπ2

60

Nπ2ωhave,We

rpm1000N

,kg1.4cylindereachofmassingreciprocatm

,mm240lengthrodconnectingl,mm60r

:Given

===

=

==

===

ADARSHA H G



26

Plane Mass (m) kg Radius (r)

m

Cent.

Force/ω2

(m r )

kg m

Distance

from Ref

plane ‘2’

m

Couple/ ω2

( m r l )

kg m2

1 1.4 0.06 0.084 0.21 0.01764

2 1.4 0.06 0.084 0.13 0.01092

3 1.4 0.06 0.084 0.05 0.0042

4 1.4 0.06 0.084 -0.05 -0.0042

5 1.4 0.06 0.084 -0.13 -0.01092

6 1.4 0.06 0.084 -0.21 -0.01764

Graphical Method:

Step 1:

Draw the primary force and primary couple polygons taking some convenient scales.

Note: For drawing these polygons take primary cranks position as the reference

NO UNBALANCED

PRIMARY FORCE

NO UNBALANCED

PRIMARY COUPLE

ADARSHA H G



27

Step 2:

Draw the secondary force and secondary couple polygons taking some convenient scales.

Note: For drawing these polygons take secondary cranks position as the reference

Problem 2:

The firing order of a six –cylinder vertical four-stroke in-line engine is 142635. The

piston stroke is 80 mm and length of each connecting rod is 180 mm. The pitch distances

between the cylinder centre lines are 80 mm, 80 mm, 120 mm, 80 mm and 80 mm

respectively. The reciprocating mass per cylinder is 1.2 kg and the engine speed is 2400

rpm. Determine the out-of-balance primary and secondary forces and couples on the

engine taking a plane midway between the cylinders 3 and 4 as the reference plane.

Solution:

/srad251.3360

2400xπ2

60

Nπ2ωhave,We

rpm2400N

,kg1.2cylindereachofmassingreciprocatm

,mm180lengthrodconnectingl,mm402

80

2

Lr

:Given

===

=

==

=====

NO UNBALANCED

SECONDARY FORCE

NO UNBALANCED

SECONDARY COUPLE

ADARSHA H G



28

Plane Mass (m) kg Radius (r)

m

Cent.

Force/ω2

(m r )

kg m

Distance

from Ref

plane ‘2’

m

Couple/ ω2

( m r l )

kg m2

1 1.2 0.04 0.048 0.22 0.01056

2 1.2 0.04 0.048 0.14 0.00672

3 1.2 0.04 0.048 0.06 0.00288

4 1.2 0.04 0.048 -0.06 -0.00288

5 1.2 0.04 0.048 -0.14 -0.00672

6 1.2 0.04 0.048 -0.22 -0.01056

Graphical Method:

Step 1:

Draw the primary force and primary couple polygons taking some convenient scales.

Note: For drawing these polygons take primary cranks position as the reference

Note: No primary unbalanced force or couple

ADARSHA H G



29

Step 2:

Draw the secondary force and secondary couple polygons taking some convenient scales.

Note: For drawing these polygons take secondary cranks position as the reference

Note: No secondary unbalanced force or couple

ADARSHA H G



30

Problem 3:

The stroke of each piston of a six-cylinder two-stroke inline engine is 320 mm and

the connecting rod is 800 mm long. The cylinder centre lines are spread at 500 mm.

The cranks are at 600 apart and the firing order is 145236. The reciprocating mass

per cylinder is 100 kg and the rotating parts are 50 kg per crank. Determine the out

of balance forces and couples about the mid plane if the engine rotates at 200 rpm.

Primary cranks position

Relative positions of Cranks in degrees

Firing

order θ1 θ2 θ3 θ4 θ5 θ6

142635 0 240 120 120 240 0

145236 0 180 240 60 120 300

Secondary cranks position

Relative positions of Cranks in degrees

Firing

order θ1 θ2 θ3 θ4 θ5 θ6

142635 0 120 240 240 120 0

145236 0 0 120 120 240 240

Calculation of primary forces and couples:

Total mass at the crank pin = 100 kg + 50 kg = 150 kg

Plane Mass (m)

kg

Radius (r)

m

Cent.

Force/ω2

(m r )

kg m

Distance

from Ref

plane

m

Couple/ ω2

( m r l )

kg m2

1 150 0.16 24 1.25 30

2 150 0.16 24 0.75 18

3 150 0.16 24 0.25 6

4 150 0.16 24 -0.25 -6

5 150 0.16 24 -0.75 -18

6 150 0.16 24 -1.25 -30

ADARSHA H G



31

ADARSHA H G



32

Calculation of secondary forces and couples:

Since rotating mass does not affect the secondary forces as they are only due to

second harmonics of the piston acceleration, the total mass at the crank is taken as

100 kg.

Plane Mass (m)

kg

Radius (r)

m

Cent.

Force/ω2

(m r )

kg m

Distance

from Ref

plane

m

Couple/ ω2

( m r l )

kg m2

1 100 0.16 16 1.25 20

2 100 0.16 16 0.75 12

3 100 0.16 16 0.25 4

4 100 0.16 16 -0.25 -4

5 100 0.16 16 -0.75 -12

6 100 0.16 16 -1.25 -20

ADARSHA H G



33

BALANCING OF V – ENGINE

Two Cylinder V-engine:

A common crank OA is operated by two connecting rods. The centre lines of the two –

cylinders are inclined at an angle α to the X-axis.

Let θ be the angle moved by the crank from the X-axis.

Determination of Primary force:

Primary force of 1 along line of stroke OB1 = )()(cosrm 12

−−−−−−α−θω

Primary force of 1 along X - axis = )(cos)(cosrm 22

−−−αα−θω

Primary force of 2 along line of stroke OB2 = )()(cosrm 32

−−−−−α+θω

Primary force of 2 along X-axis = )(cos)(cosrm 42

−−−αα+θω

[ ]

[ ]

(5)θcosαcosωrm2

αcosθcos2xαcosωrm

αsinθsinαcosθcosαsinθsinαcosθcosαcosωrm

)αθ(cos)αθ(cosαcosωrm

axis-X along force primary Total

22

2

2

2

−−−−−−−−−−−−=

=

−++=

++−=

ADARSHA H G



34

Similarly,

[ ]

[ ]

(6)θsinαsinωrm2

αsinθsin2xαsinωrm

αsinθsinαcosθcos(α)sinθsinαcosθ(cosαsinωrm

α)sinαθ(cos-αsin)αθ(cosωrm

axis-Z along force primary Total

22

2

2

2

−−−−−−−−−−−−=

=

−−+=

+−=

( ) ( )

( ) ( ) (7)θsinαsinθcosαcosωrm2

θsinαsinωrm2θcosαcosωrm2

force Primary Resultant

22222

222222

−−−−−+=

+=

and this resultant primary force will be at angle β with the X – axis, given by,

(8)θcosαcos

θsinαsinβtan

2

2

−−−−−−=

If0

902 =α , the resultant force will be equal to

( ) ( )

(9)ωrm

θsin45sinθcos45cosωrm22

2022022

−−−−−=

+

and (10)θtanθcos45cos

θsin45sinβtan

02

02

−−−−−−==

i.e., θβ = or it acts along the crank and therefore, can be completely balanced by a

mass at a suitable radius diametrically opposite to the crank, such that,

(11)----- rmrmrr

=

For a given value of α, the resultant force is maximum (Primary force), when

( ) ( )

( ) maximumisθsinαsinθcosαcos

or

maximumisθsinαsinθcosαcos

2424

2222

+

+

ADARSHA H G



35

Or

( )

0θcosθsin2xαsinθsinθcos2xαcos-i.e.,

0θsinαsinθcosαcosdθ

d

44

2424

=+

=+

[ ] 0αcos-αsinθ2sin i.e.,

0θ2sinxαsinθ2sinxαcos- i.e.,44

44

=

=+

As α is not zero, therefore for a given value of α , the resultant primary force is

maximum when .00θ=

Determination of Secondary force:

Secondary force of 1 along line of stroke OB1 is equal to

)()(cosn

rm12

2

−−−−−−α−θω

Secondary force of 1 along X - axis = )(cos)(cosn

rm22

2

−−−αα−θω

Secondary force of 2 along line of stroke OB2 =

)()(cosn

rm32

2

−−−−−α+θω

Primary force of 2 along X-axis = )(cos)(cosn

rm42

2

−−−αα+θω

Therefore,

[ ]

[ ]

(5)α2cosθ2cosαcosn

ωrm2

α2sinθ2sinα2cosθ2(cos)α2sinθ2sinα2cosθ2(cosαcosn

ωrm

)αθ(2cos)αθ(2cosαcosn

ωrm

axis-X along force secondary Total

2

2

2

−−−−−−−−−−−−=

−++=

++−=

ADARSHA H G



36

Similarly,

(6)α2sinθ2sinαsinn

ωrm2

axis-Z along force secondary Total2

−−−−−−−−−=

( ) ( ) (7)α2sinθ2sinαsinα2cosθ2cosαcosn

ωrm2

force Secondary Resultant

222

−−−−−+=

And (8)α2cosθ2cosαcos

α2sinθ2sinαsinβtan '

−−−−−−=

If 00 45αor90α2 == ,

Secondary force = )(sinn

rmsin

n

rm922

2

22 222

−−−−θω

=

θω

And (10)90βandβtan 0''−−−−−−=∞= i.e., the force acts along Z-

axis and is a harmonic force and special methods are needed to balance it.

Problem 1:

The cylinders of a twin V-engine are set at 600 angle with both pistons connected to a

single crank through their respective connecting rods. Each connecting rod is 600 mm

long and the crank radius is 120 mm. The total rotating mass is equivalent to 2 kg at the

crank radius and the reciprocating mass is 1.2 kg per piston. A balance mass is also fitted

opposite to the crank equivalent to 2.2 kg at a radius of 150 mm. Determine the

maximum and minimum values of the primary and secondary forces due to inertia of the

reciprocating and the rotating masses if the engine speed is 800 mm.

Solution:

5120

600

r

lnand/srad83.78

60

800xπ2

60

Nπ2ωhave,We

rpm800N

mm120radiuscrankr

mm600lengthrodconnectingl

mm150rkg,2.2massbalancingm

kg2 mass rotatingequivalentM

kg1.2pistoneachofmassingreciprocatm

:Given

CC

======

=

==

==

===

==

==

ADARSHA H G



37

Primary Force:

(1)θcosαcosωrm2axis-X along force primary Total 22

−−−−−−−−−−−−=

(2)θcosωrM

axisXalongmassrotatingtodue forcelCentrifuga2

−−−−−−−−−−−−=

−

(3)θcosωrm

axisXalongmassbalancingtodue forcelCentrifuga2

CC −−−−−−−−−−−−−=

−

Therefore total unbalance force along X –axis = (1) + (2) + (3)

That is

[ ]

( ) [ ]

( ) [ ] (4)-------Nθcos884.410.33-0.240.216θcos83.78

2.2x0.152x0.1230xcos2x1.2x0.12θcos83.78

rm-rMαcosrm2θcosω

θcosωrmθcosωrMθcosαcosωrm2

axis X along force UnbalanceTotal

2

022

CC

22

2

CC

222

=+=

−+=

+=

−+=

(5)θsαsωrm2axis-Z along force primary Total 22

−−−−−−−−−−−−= inin

ADARSHA H G



38

(6)θsωrM

axisZalongmassrotatingtodue forcelCentrifuga2

−−−−−−−−−−−−=

−

in

(7)θsωrm

axisZalongmassbalancingtodue forcelCentrifuga2

CC −−−−−−−−−−−−−=

−

in

Therefore total unbalance force along Z –axis = (5) + (6) + (7)

That is

[ ]

( ) [ ]

( ) [ ] (8)-------Nθsin126.34-0.33-0.240.072θsin83.78

2.2x0.152x0.1230xsin2x1.2x0.12θsin83.78

rm-rMαsinrm2θsinω

θsinωrmθsinωrMθsinαsinωrm2

axis -Z along force UnbalanceTotal

2

022

CC

22

2

CC

222

=+=

−+=

+=

−+=

( ) ( )

(9)15961.8θcos766219.25

θsin15961.8θcos782181.05

θsin126.34-θcos884.41


2

22

22

−−−−−+=

+=

+=

This is maximum, when 0

0=θ and minimum, when 0

90=θ

(11)N126.3415961.890cos766219.25

90θwhen i.e., force, Primary MinimumAnd

(10)N884.4115961.8766219.25

0θwhen i.e., force, Primary Maximum

02

0

0

−−−−−=+=

=

−−−−−=+=

=

Secondary force:

The rotating masses do not affect the secondary forces as they are only due to second

harmonics of the piston acceleration.

ADARSHA H G



39

( ) ( )

( )

( )

[ ] (12)2θθsin(0.18752θθcos(0.1875404.3

60sinθ2sin30sin

60cosθ2cos30cos

5

78)x0.12x(83.x1.22

α2sinθ2sinαsinα2cosθ2cosαcosn

ωrm2

force Secondary Resultant

22

200

2222

222

−−−−−+=

+=

+=

This is maximum, when 0

0=θ and minimum, when 0

180=θ

[ ]

[ ] (14)N175.07)018sin(0.1875)180cos(0.1875404.3

180θwhen i.e., force, secondary MinimumAnd

(13)N175.07)0sin(0.1875)0cos(0.1875404.3

0θwhen i.e., force, secondary Maximum

2020

0

2020

0

−−−−−=+=

=

−−−−−=−+=

=

BALANCING OF W, V-8 AND V-12 – ENGINES

BALANCING OF W ENGINE

In this engine three connecting rods are operated by a common crank.

( ) (1)1 αcos2θcosωrm

axis-X along force primary Total

22−−−−−−−−−−−−+=

(2)θsinαsinωrm2

zero)isaxisZalong3offorceprimarythe(since

engine,twinVtheinsasamebewillaxis-Z along force primary Total

22−−−−−−−−−−−−=

−

−

( ) ( )[ ] (3)θαsin2sin1α2cosθcosωrm


22222−−−−−++=

and this resultant primary force will be at angle β with the X – axis, given by,

ADARSHA H G



40

( )

(4)1α2cosθcos

θsinα2sinβtan

2

2

−−−−−−+

=

If 0

60=α ,

(5)ωrm

2

3


2−−−−−=

and

(6)θtanβtan −−−−−−==

i.e., θβ = or it acts along the crank and therefore, can be completely balanced by a

mass at a suitable radius diametrically opposite to the crank, such that,

(7)----- rmrmrr

=

(8)1α2cosαcosn

ωr2m2θcos

axis-X along force secondary Total

2

−−−−−−−−−−−−

+=

Total secondary force along Z –direction will be same as in the V-twin engine.

( ) ( )[ ] (9)2θsin2αsinα2sin12αcosα2cos2θcosn

ωrm

force secondary Resultant

222

−−−−−++=

( )(10)

1α2cosα2cosθ2cos

α2sinθ2sinα2sinβtan '

−−−−−−+

=

If 0

60=α ,

(11)2θcos2n

ωrm

axis-X along force Secondary

2

−−−−−−−−−−−−=

(12)2θsin2n

ωr3m

axis-Z along force Secondary

2

−−−−−−−−−−−−=

It is not possible to balance these forces simultaneously

ADARSHA H G



41

V-8 ENGINE

It consists of two banks of four cylinders each. The two banks are inclined to each other

in the shape of V. The analysis will depend on the arrangement of cylinders in each bank.

V-12 ENGINE

It consists of two banks of six cylinders each. The two banks are inclined to each other in

the shape of V. The analysis will depend on the arrangement of cylinders in each bank.

If the cranks of the six cylinders on one bank are arranged like the completely balanced

six cylinder, four stroke engine then, there is no unbalanced force or couple and thus the

engine is completely balanced.

BALANCING OF RADIAL ENGINES:

It is a multicylinder engine in which all the connecting rods are connected to a common

crank.

ADARSHA H G



42

Direct and reverse crank method of analysis:

In this all the forces exists in the same plane and hence no couple exist.

In a reciprocating engine the primary force is given by, θω cosrm2

which acts along

the line of stroke.

In direct and reverse crank method of analysis, a force identical to this force is generated

by two masses as follows.

1. A mass m/2, placed at the crank pin A and rotating at an angular velocity ω in the

counter clockwise direction.

2. A mass m/2, placed at the crank pin of an imaginary crank OA’ at the same angular

position as the real crank but in the opposite direction of the line of stroke. It is assumed

to rotate at an angular velocity ω in the clockwise direction (opposite).

3. While rotating, the two masses coincide only on the cylinder centre line.

The components of the centrifugal forces due to rotating masses along the line of stroke

are,

θω= cosrm

AatmasstoDue2

2

θω= cosrm

AatmasstoDue' 2

2

Thus, total force along the line of stroke = θω cosrm2

which is equal to the primary

force.

At any instant, the components of the centrifugal forces of these masses normal to the

line of stroke will be equal and opposite.

The crank rotating in the direction of engine rotation is known as the direct crank and

the imaginary crank rotating in the opposite direction is known as the reverse crank.

Now,

Secondary accelerating force is

θω=

θω=

θω

224

4

22

2

2

22

cos)(n

rm

n

cos)(rm

n

cosrm

ADARSHA H G



43

This force can also be generated by two masses in a similar way as follows.

1. A mass m/2, placed at the end of direct secondary crank of length n

r

4 at an angle 2θ

and rotating at an angular velocity 2ω in the counter clockwise direction.

2. A mass m/2, placed at the end of reverse secondary crank of length n

r

4 at an angle -2θ

and rotating at an angular velocity 2ω in the clockwise direction.

The components of the centrifugal forces due to rotating masses along the line of stroke

are,

2θcos2n

mrω2θcosω)(2

4n

r

2

mCatmasstoDue

2

2==

2θcos2n

mrω2θcosω)(2

4n

r

2

mC'atmasstoDue

2

2==

Thus, total force along the line of stroke =

2θcosn

mrω2θcosω)(2

4n

r

2

m 2

2=x2 which is equal to the secondary force.

References:

1. Theory of Machines by S.S.Rattan, Third Edition, Tata McGraw Hill Education

Private Limited.

2. Kinematics and Dynamics of Machinery by R. L. Norton, First Edition in SI units, Tata

McGraw Hill Education Private Limited.

ADARSHA H

G

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