balancing of reciprocating masses 1

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

=

+=

=

122

And, the force required to accelerate the mass m is

)(n

cosrmcosrm

n

coscosrmFi

22

2

22

2

+=

+=

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]



2

The first term of the equation (2) , i.e. cosrm 2 is called primary accelerating

force the second term n

cosrm

22 is called the secondary accelerating force.

Maximum value of primary accelerating force is 2rm

And Maximum value of secondary accelerating force is n

rm 2

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.



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.

vF21 and vF41 balance each other but form an unbalanced shaking couple.

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



4

The horizontal component of the centrifugal force due to the balancing mass is cosrm 2 and this is in the line of stroke. This component neutralizes the unbalanced

reciprocating force. But the rotating mass also has a component sinrm 2 perpendicular to the line of stroke which remains unbalanced. The unbalanced force is zero at = 00 or 1800 and maximum at the middle of the stroke i.e. = 900. The magnitude or the maximum value of the unbalanced force remains the same i.e. equal to

2rm . 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

32

of the

reciprocating mass is balanced or a value between 21

to 43

.

If c is the fraction of the reciprocating mass, then

cosrmcmassthebybalancedforceprimaryThe 2=

and

cosrmc)1(massthebyunbalancedforceprimaryThe 2=

sinrmc

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 sinrmccosrc)m(1instantanyatforceunbalancedResultant

+=

The resultant unbalanced force is minimum when, 21

=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 Pm 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 Pmmc + which is to be completely balanced.



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/spi260

60xpi2

60

Npi2

===

===

kg804060x3

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

50sin2pix0.16x60x3

250cos2pix0.16x60x

3

21

sinrmccosrmc1

50at force Unbalanced

2

02

2

02

2222

0

=

+

=

+=

=



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

150xpi2

60

Npi2

===

===

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

sinrmccosrmc1

45at force Unbalanced

202

202

2222

0

=

+=

+=

=



7

SECONDARY BALANCING:

Secondary acceleration force is equal to )(n

cosrm 122

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 24

22 2

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

Actual Imaginary Angular velocity 2

Length of crank r n

r

4

Mass at the crank pin m m

Thus, when the actual crank has turned through an angle t= , the imaginary crank would have turned an angle t= 22



8

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

2rm2

Component of this force along the line of stroke is = ( ) 2cos4n

2rm2

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



9

= (1)cosrmforcePrimary 2

= (2)coslrmcouplePrimary 2

( ) = (3)2cos

n4

2rmforceSecondary

2

( )

=

=

(4)2cosln

rm

2cosln4

2rmcoupleSecondary And

2

2

GRAPHICAL SOLUTION:

To solve the above equations graphically, first draw the cosrm polygon ( 2 is common to all forces). Then the axial component of the resultant forces )cosF(

r

multiplied by 2 provides the primary unbalanced force on the system at that moment. This unbalanced force is zero when 090= and a maximum when 00= .



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 == VPhP FandF

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 ( )

n42 2

or n

2

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.



11

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

[ ] 0)180(coscosrmforcePrimary 2 =++=

coslrm)180(cos2

lcos

2

lrmcouplePrimary 22 =

+

+=

Maximum values are lrm 2 at 00 180and0=

[ ] 2cosn

r2m)2360(cos2cos

n

rmforceSecondary

22

=++=

Maximum values are n

r2m 2 when

0000

0000

270and180,90,0or

540and360,180,02

=

=



12

0)2360(cos2

l2cos

2

l

n

rmcoupleSecondary

2

=

+

+=

ANALYTICAL METHOD OF FINDING PRIMARY FORCES AND COUPLES

First the positions of the cranks have to be taken in terms of 0 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 lrm 2 and this value is obtained at crank positions 00 and 1800. However, if the crank positions are assumed at 900 and 2700, 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 3000, the primary couple can be obtained as

( )lrm

2

1

120180cos2

l120cos

2

lrmcomponentX

2

0002

=

+

+=

( )lrm

2

3

120180sin2

l120sin

2

lrmcomponentY

2

0002

=

+

+=

Therefore,

lrm

lrm2

3lrm

2

1couplePrimary

2

2

2

2

2

=

+

=

Case 2: IN-LINE FOUR-CYLINDER FOUR-STROKE ENGINE



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 0 for the first,

00 180 + for the second, 00 180 + for the third and 0 for the fourth.



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

cos)180(cos)180(coscosrmforcePrimary 002

=

+++++=

0

cos2

3l)180(cos

2

l

)180(cos2

lcos

2

3l

rmcouplePrimary0

0

2

=

++

+

++

=

2cosn

r4m

2cos)2360(cos

)2360(cos2cos

n

rmforceSecondary

2

0

02

=

+++

++=

0000

0000

2

027and180,09,0

or054and360,018,02at

n

rmvalueMaximum

=

=

=

0

2cos2

3l)2360(cos

2

l

)2360(cos2

l2cos

2

3l

n

rmcoupleSecondary

0

0

2

=

++

+

++

=

Thus the engine is not balanced in secondary forces.



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/2

(m r ) kg m

Distance from Ref plane 2

m

Couple/ 2 ( 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



16

Analytical Method:

Choose plane 2 as the reference plane and 03 0= . .

Step 1: Resolve the couples into their horizontal and vertical components and take their sums.

Sum of the horizontal components gives

)(coscos.,e.icosrcosrcosr

1196816524940196801652494

41

40

1

+=+=++

Sum of the vertical components gives

)(sinsin.,e.isinrsinrsinr

219684940196801652494

41

40

1

==++

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

( ) ( ) ( )

( ) ( ) ( ) ( )00

44

2

4

2

44

22

2

4

2

4

2

192.1or167.9and0.978cos

get,wesolvingOn

sin1968cos1968cos1968x1652x21652494

sin1968cos19681652494

==

+++=

++=

.,e.i

Choosing one value, say 04167.9 =

Dividing (2) by (1), we get

0

1

0

0

1

123.4i.e.,

1.515272.28

412.53

)(167.9cos19681652

)(167.9sin1968tan

=

=

+=

+=

Step 2:

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


)(.cosmor).(cosrcosrcosrm).(cosr

35880916748005904123380

22

0022

0

==+++



17


)(.sinmor).(sinrsinrsinrm).(sinr

494170916748005904123380

22

0022

0

==+++


Anskg.m 14272 =

Dividing (4) by (3), we get Ansr 0

2

2

282o

4.7288.5

417.9tan

=

=

=

Graphical Method:

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



18

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

427427282123168

22

02

01

04

==

===



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===

=


m

Cent. Force/2

(m r ) kg m


m


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



20

Analytical Method:

Choose plane 3 as the reference plane and 01 0= . Step 1: Resolve the couples into their horizontal and vertical components and take their sums. Sum of the horizontal components gives

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

cos.cos.cos.

15137561851375618

0756180513

42

42

420

=+=

=+


)(sin.sin..,e.isin.sin.sin.

2756180756180513

42

420

==+


162.20365.61-182.2545.563cos182.25i.e.,

182.25cos182.25-45.563

182.25cos182.25)sin45.563(cos

sin45.563182.25cos182.25cos45.56365.61

)sin(6.7513.5)cos(6.75(8.1)

4

4

44

2

4

2

4

2

44

2

2

4

2

4

2

=+=

+=

++=

++=

+=

Therefore, Ans27.13and182.25

162.203cos 0

44==

Dividing (2) by (1), we get

000

2

0

0

2

157.6718022.33-i.e.,

1.5157.493

3.078

13.5-)(27.13cos6.75

)(27.13sin6.75tan

=+=

=

==

Step 2: Resolve the forces into their horizontal and vertical components and take their sums.




21

(3)17.545cosm0.225i.e.,

020.02cosm0.22524.97522.5i.e.,

0)27.13(cos22.5cosm0.225)(157.67cos27)0(cos22.5

33

33

0

33

00

=

=++

=+++

And sum of the vertical components gives

(4)-20.518sinm0.225i.e.,

010.26sinm0.22510.258i.e.,

0)27.13(sin22.5sinm0.225)(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

=

+

=

+=

Dividing (4) by (3), we get Ans229.5or

17.545-

20.518tan

0

3

3

=

=



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)x0.4x800rm

s/rad7.3360

Npi2m,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,



23

( ) ( ) ( ) ( )

erchangedintarelandlcesin,onlymNCCmN

...lllrmC

pp

p

4216

22242

23

21

4376143761

62808117195

==

=

+=+=

( ) ( ) ( ) ( )


...lllrmC

pp

p

3225

22232

24

22

4790547905

81806217195

==

=

+=+=

( ) ( ) ( ) ( )


...lllrmC

pp

p

3434

22234

22

23

1944819448

81628017195

==

=

+=+=

Thus the best arrangement is of 3rd and 4th. The firing orders are 1423 and 1324 respectively. Unbalanced couple = 19448 N m.

Graphical solution:



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 01 0= Second 02 240=

654321

654321

rrrrrr

mmmmmm

And

=====

=====

Third 03 120= Fourth 04 120= Fifth 05 240= Sixth 06 0=



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

1000xpi2

60

Npi2have,We

rpm1000N

,kg1.4cylindereachofmassingreciprocatm

,mm240lengthrodconnectingl,mm60r

:Given

===

=

==

===



26

Plane Mass (m) kg Radius (r) m

Cent. Force/2

(m r ) kg m


m


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



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

2400xpi2

60

Npi2have,We

rpm2400N

,kg1.2cylindereachofmassingreciprocatm

,mm180lengthrodconnectingl,mm402

80

2

Lr

:Given

===

=

==

=====

NO UNBALANCED SECONDARY FORCE

NO UNBALANCED SECONDARY COUPLE



28

Plane Mass (m) kg Radius (r) m

Cent. Force/2

(m r ) kg m


m


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



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



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 1 2 3 4 5 6

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 1 2 3 4 5 6

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


m

Cent. Force/2

(m r ) kg m

Distance from Ref

plane m


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



31



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.


m

Cent. Force/2

(m r ) kg m

Distance from Ref

plane m


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



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 to the X-axis. Let 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

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

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

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

[ ][ ]

(5)coscosrm2

coscos2xcosrm

sinsincoscossinsincoscoscosrm

)(cos)(coscosrm

axis-X along force primary Total

22

2

2

2

=

=

++=

++=



34

Similarly,

[ ][ ]

(6)sinsinrm2

sinsin2xsinrm

sinsincoscos()sinsincos(cossinrm

)sin(cos-sin)(cosrm

axis-Z along force primary Total

22

2

2

2

=

=

+=

+=

( ) ( )( ) ( ) (7)sinsincoscosrm2

sinsinrm2coscosrm2

force Primary Resultant

22222

222222

+=

+=

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

(8)coscos

sinsintan

2

2

=

If 0902 = , the resultant force will be equal to

( ) ( )(9)rm

sin45sincos45cosrm22

2022022

=

+

and (10)tancos45cos

sin45sintan

02

02

==

i.e., = 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 , the resultant force is maximum (Primary force), when

( ) ( )( ) maximumissinsincoscos

or

maximumissinsincoscos

2424

2222

+

+



35

Or

( )0cossin2xsinsincos2xcos-i.e.,

0sinsincoscosd

d

44

2424

=+

=+

[ ] 0cos-sin2sin i.e.,02sinxsin2sinxcos- i.e.,

44

44

=

=+

As is not zero, therefore for a given value of , the resultant primary force is maximum when .00=

Determination of Secondary force:

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

)()(cosn

rm 122

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

rm 222

Secondary force of 2 along line of stroke OB2 =

)()(cosn

rm 322

+

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

rm 422

+ Therefore,

[ ][ ]

(5)2cos2coscosn

rm2

2sin2sin2cos2(cos)2sin2sin2cos2(coscosn

rm

)(2cos)(2coscosn

rm

axis-X along force secondary Total

2

2

2

=

++=

++=



36

Similarly,

(6)2sin2sinsinn

rm2

axis-Z along force secondary Total2

=

( ) ( ) (7)2sin2sinsin2cos2coscosn

rm2

force Secondary Resultant

222

+=

And (8)2cos2coscos

2sin2sinsintan ' =

If 00 45or902 == ,

Secondary force = )(sinn

rmsinn

rm 922222 222

=

And (10)90andtan 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

800xpi2

60

Npi2have,We

rpm800N

mm120radiuscrankr

mm600lengthrodconnectingl

mm150rkg,2.2massbalancingm

kg2 mass rotatingequivalentM

kg1.2pistoneachofmassingreciprocatm

:Given

CC

======

=

==

==

===

==

==



37

Primary Force:

(1)coscosrm2axis-X along force primary Total 22 =

(2)cosrM

axisXalongmassrotatingtodue forcelCentrifuga2

=

(3)cosrm

axisXalongmassbalancingtodue forcelCentrifuga2

CC =

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

That is

[ ]( ) [ ]( ) [ ] (4)-------Ncos884.410.33-0.240.216cos83.78

2.2x0.152x0.1230xcos2x1.2x0.12cos83.78

rm-rMcosrm2cos

cosrmcosrMcoscosrm2

axis X along force UnbalanceTotal

2

022

CC

22

2

CC

222

=+=

+=

+=

+=

(5)ssrm2axis-Z along force primary Total 22 = inin



38

(6)srM

axisZalongmassrotatingtodue forcelCentrifuga2

=

in

(7)srm

axisZalongmassbalancingtodue forcelCentrifuga2

CC =

in

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

That is

[ ]( ) [ ]( ) [ ] (8)-------Nsin126.34-0.33-0.240.072sin83.78

2.2x0.152x0.1230xsin2x1.2x0.12sin83.78

rm-rMsinrm2sin

sinrmsinrMsinsinrm2

axis -Z along force UnbalanceTotal

2

022

CC

22

2

CC

222

=+=

+=

+=

+=

( ) ( )

(9)15961.8cos766219.25

sin15961.8cos782181.05

sin126.34-cos884.41


2

22

22

+=

+=

+=

This is maximum, when 00= and minimum, when 090=

(11)N126.3415961.890cos766219.25

90when i.e., force, Primary MinimumAnd

(10)N884.4115961.8766219.25

0when 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.



39

( ) ( )( )

( )[ ] (12)2sin(0.18752cos(0.1875404.3

60sin2sin30sin

60cos2cos30cos

5

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

2sin2sinsin2cos2coscosn

rm2

force Secondary Resultant

22

200

2222

222

+=

+=

+=

This is maximum, when 00= and minimum, when 0180=

[ ]

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

180when i.e., force, secondary MinimumAnd

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

0when 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 cos2cosrmaxis-X along force primary Total

22+=

(2)sinsinrm2

zero)isaxisZalong3offorceprimarythe(since

engine,twinVtheinsasamebewillaxis-Z along force primary Total

22=

( ) ( )[ ] (3)sin2sin12coscosrmforce Primary Resultant

22222++=

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



40

( ) (4)12coscossin2sin

tan2

2

+=

If 060= ,

(5)rm2

3


2=

and

(6)tantan ==

i.e., = 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)12coscosn

r2m2cos

axis-X along force secondary Total

2

+=

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

( ) ( )[ ] (9)2sin2sin2sin12cos2cos2cosn

rm

force secondary Resultant

222

++=

( ) (10)12cos2cos2cos2sin2sin2sin

tan ' +

=

If 060= ,

(11)2cos2n

rm

axis-X along force Secondary

2

=

(12)2sin2n

r3m

axis-Z along force Secondary

2

=

It is not possible to balance these forces simultaneously



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.



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, cosrm 2 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 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 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,

= cosrmAatmasstoDue 22

= cosrmAatmasstoDue ' 22

Thus, total force along the line of stroke = cosrm 2 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

=

=

224

4222

2

22

cos)(n

rm

n

cos)(rmn

cosrm



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 2

and rotating at an angular velocity 2 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 -2

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

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

2cos2n

mr2cos)(2

4n

r

2

mCatmasstoDue

2

2==

2cos2n

mr2cos)(2

4n

r

2

mC'atmasstoDue

2

2==

Thus, total force along the line of stroke =

2cosn

mr2cos)(2

4n

r

2

m 22=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.

balancing of reciprocating masses 1

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