Chapter 7 Balancing of Reciprocating Machines

We have already discussed how to obtain inertia forces of linkages (the four-bar and reciprocating

engine mechanisms). Such inertia forces cannot be avoided and we have to consider these forces in

designing of linkages. (These inertia forces increase stresses in members and loads in bearing). Inertia

force 2 , where is the angular velocity of links. For linkages with very high speed, inertia

forces (or loading) results in vibrations, noise emission and sometimes machine failure due to fatigue.

Inertia force of a reciprocating mass:

y

2A

l

3B

x

x

r+l

O2

r

Figure 1 A reciprocating mechanism

The displacement of the piston is given as

1/ 2

2

2

sincos 1x r l

n

= +

with n = l/r (1)

The velocity of the piston can be obtained as

0 .5

2 2

1 sin 2sin

2 1 sinx r

n n

= +

(2)

The acceleration of the piston can be obtained as

2 42

1.53 2 2

( 1)cos2 coscos

1 sin

nx r

n n

+ = +

(3)

In equation (1) the term 0.5

2 21 (sin ) / n is expanded bi-nominally to give a harmonic analysis. i.e

,x x and x can be represented as the cosine and sine terms of harmonics. The frequencies

representing integral multiples of the fundamental frequency are called harmonics, with the

fundamental frequency being the first harmonic. For example:

122

0 02 0n

fundamental harmonic second harmonic thn harmonic

A series of harmonic functions are known as the Fourier series. A harmonic function of any arbitrary

frequency is periodic i.e. it repeats itself at equal intervals of time /2=T , where T is the

period of excitation. If a function is periodic, it is not necessary that it will be harmonic.Let

tAx cos= is a harmonic function with period /2 or frequency . Any arbitrary periodic

functions can be represented by a convergent series of harmonic functions whose frequencies are

integral multiples of certain fundamental frequency ( )T/20 = .

Fourier Series:

1

0 02 1( ) ( cos sin )

n

r rrf t a a r t b r t

== + + (4)

with

/ 2

0

/ 2

2( )cos

T

r

T

a f t r t dtT

= and / 2

0

/ 2

2( )sin

T

r

T

b f t r tT

=

T/20 = ; 1,2,3,r =

where T is the period of function f(t).

Shaking force: Bearing loads due to inertia forces of rotating/reciprocating masses of mechanism

causes shaking of the mechanism due to its dynamic nature i.e. it changes with time. This bearing

loads due to inertia forces of mechanism are called shaking forces. Shaking force can be minimised by

balancing the inertia forces in opposition to each other such that little or no force is transmitted to

machine supports. The degree to which this shaking force is undesirable depends upon the frequency,

, of shaking force and the natural frequency, nf , of flexible members (such as shafts, bearings,

supports, etc.) through which the force is transmitted. For nf = we have undesirable resonance

condition.

Primary and secondary forces can be replaced by rotating masses as:

1. A mass bm at radius r at A and rotating with crank with the component in horizontal

direction will be same as by reciprocating mass bm of piston i.e. primary force.

123

2. A, mass bm at radius )4/( nr and rotating at twice the crank angular velocity . The

component in the horizontal direction will be same as that due to piston bm i.e. secondary

force.

The acceleration of piston can be obtained up to second harmonics as:

)2cos

(cos2

nrAp

+= (5)

where, n=l/r

F32

F32Sin

O2

A

mb2bsin

mb

2b

mb

mb2cos

F32 cos

primary force

=mbr2cos

Figure 2 A crack with a counter balance mass

0cos)sin(cos)sin( 122

32 =++ TbbmrF b

with cossincossin 223212 bmrFT b=

For a single cylinder engine the inertia force due to the reciprocating mass, 43 mmm BB += , will be

2 cos2cosBF m rn

= +

(6)

where Bm3 is one of the equivalent mass of connecting rod at piston and 4m is the mass of piston.

For F ve + the inertia force will be directed away from main bearing and for F ve the inertia

124

force towards the main bearing. Bearing load due to this inertia force will be in the same direction as

the inertia force.

In equation (6), various terms can be defined as follows:

First term:

cos2rmb (7)

is called primary force. It reaches its maximum value of 2rmb twice per revolution, when

0= and .

Second term:

nrmB

2cos2 (8)

is called secondary force. It reaches its maximum value of 2 /Bm r n four times per revolution of

crank when 0, / 2, and 3 / 2 = .

(i) Harmonic motion (Fundamental) is given as tAx cos= , where A is the amplitude and is the

frequency so that sinx A t = and 2 cosx A t = , where 2 /T = is the period and

t = .

The primary inertia force tAmF pi cos2+=

(ii) Second harmonic motion is given as tAx 2cos= , so that 2 sin 2x A t = and

24 cos2x A t = .

The secondary inertia force tAmtAmF si 2cos)4(2cos422 ==

Both the primary and secondary inertia forces represent a force, which is constant in direction but

varying in magnitude, and hence cannot be completely balanced by a rotating mass (or which inertia

force is varying in the direction but is constant in the magnitude).

Assumption: (i) The replacement of the connecting rod by two masses at its ends maintains

complete dynamic equivalence (ii) The crank is so designed that the combined centre of mass of

the crank and the apportioned connecting rod mass, 3m , at the crank pin lies on the axis of rotation,

and (iii) the engine speed is constant. For multi-cylinder engines, we shall apply these assumptions to

each constituent cylinder.

125

Bearing load: In reciprocating engine the bearing load due to the piston inertia can be obtained with

help of following free body diagram, which is self-explanatory.

Figure 3 (a) A reciprocating engine

From ABO2

sinxq =

xFT sin34=

qF .34= 2112 . TqF ==

21TT =

Figure 3(b) Free body diagram of the connecting rod

Figure 3 (c) Free body diagram of the piston

Figure 3 (d) Free body diagram of the crank

Figure 3 (e) Free body diagram of bed

Figure 3 (f) Equivalent system of Figure (e)

126

The full effect on the engine frame of the inertia of the reciprocating mass is equivalent to the force,

2 1(cos cos2 )m r n + , along the line of the stroke at O.

Figure 4 (a) (top) Bearing load due to rotating

mass (bottom) reaction force on the engine

frame

Figure 4 (b) (top) Centrifugal forces on disc (bottom)

Free body diagram of disc

From Figure 4, it can be seen that the frame will experience forces: (i) horizontal direction

cos2em and (ii) vertical direction sin2em . The primary inertia force can be balanced by a

counter mass bm at radius b such that b Bm b m r= .

Figure 5 A reciprocating machine with a counter balance mass

The component of inertia force bm along the line of stroke is cos2bmb . But a force

sin2bmb is then introduced in a direction (perpendicular) to line of stroke. In most practical

applications, the reciprocating mass is partially balanced by a rotating mass bm , reducing the inertia

force in the line of stroke and direction, such that

bmrcm bB =

127

where c is the fraction of reciprocating mass to be balanced and mB is the mass of the slider. The

unbalanced primary force along the line of action is given by

)coscos( 22 bmrm bB = or coscos.22 rcmrm Bb

or cos)1( 2rmc B (9)

The unbalance force in the vertical direction

sin2bmb = sin2rcmB (10)

Resultant unbalance force at any instant

= 22222 sincos)1( ccrmB + (11)

If the revolving and reciprocating (primary force) masses to be balanced simultaneously then,

rcmmrcmrmbm BABAb )( +=+= (12)

where Am is the mass of rotating components. The direction of resultant force at any instant is given

as

tan

1cos)1(tan

2

2

c

c

rmc

rcm

B

B

=

= (13)

(a)

(b)

(c) Equivalent of Figure (c)

Figure 6 Frame reaction in the horizontal direction

Resultant R will be rotating in opposite direction to crank (i.e. cw). For 1

2c = ; R is constant and

rotating with in cw direction. So primary force can be balanced as shown in Figure 7.

128

Figure 7

Conditions of the balance in a multi cylinder engine:

c2 c1

c3 c4

m2m3 m4

m1

stroke

12

3

4

2

3

m4

m3

m2m1

Figure 8 Multi-cylinder in-line engine

In Figure 8 1 , 2 and 3 are 4 are angular positions of the multi cylinder engine (Four engines in

Figure 8) and 2 , 3 and 4 are the phase difference of cranks 2, 3, 4 with respect to 1. The line of

stroke of each cylinder is assumed to in the horizontal plane through the crankshaft axis. The inertia

force of reciprocating masses will be situated in same horizontal plane. For each cylinder: r is the

crank radius, l is the connecting rod length, m is the total reciprocating mass (the piston and the

connecting rod) and i is the angle between cranks, which depends upon the firing order. Subscript i

represent the cylinder number (i = 1, 2, 3, 4).

Equilibrium of Primary forces:

Conditions for the complete equilibrium of primary forces are:

1. The algebraic sum of primary forces is zero and

2. The algebraic sum of their moments with respect to any of the transverse plane ( to

crankshaft) is equal to zero.

129

The resultant primary shaking force will be

nnnp rmrmrmrmF coscoscoscos2

32

3322

2212

11 ++++=

==

n

iiiirm

1

2 cos (14)

with ii += 1 ni = 3,2,1 (15)

where 1 is the variable and i are constants and 01 = . Substituting (14) into (15), we get

( ) += iiip rmF 12 cos (16)

or ===

n

iiii

n

iiii rmrm

11

2

11

2 sinsincos.cos

or = =

n

i

n

iiiiiii rmrm

1 11

21

2 )sin(sin)cos(cos (17)

The primary force, pF , will be maximum, when 01

=

pF i.e.

=

cos

sintan

11

mr

mr (18)

For pF to be zero for all values of 1 , then from eqn. (17), we get

0cos1

=

=

n

i

mr (19)

and

==

n

i

mr1

0sin (20)

Similarly, for moments, we can write

0cos1

=

=

n

i

mra (21)

and

0sin1

=

=

n

i

mra (22)

where a is the distance from the center line (line of stroke) of any cylinder to an arbitrarily selected

reference plane ( to the crank shaft axis). Thus, for complete balancing of primary forces and

moments, eqns. (19) to (22) should be satisfied.

130

Equilibrium of Secondary Forces: Conditions for complete equilibrium of the secondary forces

are:

1. The algebraic sum of the secondary forces is zero and

2. The algebraic sum of their moments with respect of any transverse plane is also zero.

Thus the resultant of secondary forces:

2

1

cos2n

s i i

i i

rF m r

i

=

with 1 i

i i

r

n l

=

(23)

or ( )2 11

cos2n

s i i i

i i

rF m r

l

= +

or 2 2

2 2

1 1cos2 cos2 sin 2 sin 2smr mr

Fl l

= (24)

(i) sF is maximum, when

1

0sF

=

which gives

2

1

1 2

sin 21tan

2cos2

mr

l

mr

l

=

(25)

(ii) For 0=sF i.e. criteria for the balancing of the secondary force is (from eqn. 24)

= 02cos2

l

mr (26)

and

= 02sin2

l

mr (27)

(iii) For balancing of moments, we have:

= 02sin.2

al

mr (28)

and

= 02cos.2

al

mr (29)

131

Thus for complete secondary balancing eqns. (26)-(29) has to be satisfied. Fundamental eqns. (19) to

(22) and eqns. (26) to (29) express completely the primary and secondary balances for reciprocating

engine with n-cylinders arranged in a line on one side of the crank shaft with the cylinder centre lies

all in the plane containing the crankshaft axis. Firing order affects the balance of multi cylinder

engine: For example Two-stroke engine: One cycle of operation in one revolution of crankshaft.

Interval between the crank= n/2 . Four- stroke engine: One cycle of operation in two revolutions of

crankshaft. Interval between the crank= n/4 , where n is numbers of cylinder in the multi-cylinder

engine.

(A) Crank arrangement for two-stroke engine:

(1) The crank arrangement can be so chosen to make the cylinder firing not in a sequential order

but distribute this on either side of the crankshaft with respect to its middle plane.

(2) Once crank arrangement is fixed there may be several choice on firing order itself (see

specially four- stroke engine).

Cranks are rotating in the cw direction. Firing will start say from engine 1. Next cylinder, which is

2/ in phase in ccw, will be ready for firing (because of completion of compression stroke). Next

firing can be done in engine 4 then in 2 and finally in 3. Hence, the firing order will be: 1-4-2-3.

Another version can be 1-3-2-4. For given crank arrangement (see in Figure 9 for 4, 5 and 6 cylinders)

for two- stroke engine only unique firing order is possible i.e. for 4 cylinder: 1-4-2-3, 5 cylinder: 1-5-

2-4-3 and 6 cylinder: 1-6-2-5-3-4.

1

2

3490

Firing order 1-4-2-3

1 2 3 4

(a) Four cylinder engine (interval=2

4 2

= )

132

1

5

24

372

1 2 3 4 5

Firing order 1-5-2-4-3

(b) Five cylinder engine: interval=2

725

o =

1 2 3 4 5 61

4

3

5

2

6

60

Firing order 1-6-2-5-3-4

(c) Six cylinder engine: interval=2

606

o =

Figure 9 Two-stroke engine

(B) Four stroke engine: For a given crank arrangement there may be several firing order possible.

Refer to Figure 10. For 4- cylinder the firing order could be: 1-4-2-3 or 1-3-2-4; for 5-cylinder the

firing order could be: 1-5-3-4-2 (only one) and for 6- cylinder the firing order could be: 1-6-2-5-3-4,

1-6-4-5-3-4 or 1-3-4-5-6-2.

1 2

4 3

1 2 3 4

Firing order 1-4-2-3

(a) Four cylinder engine: internal 4

4

= =

133

1

4

32

3

1 2 3 4 5

Firing order 1-5-2-4-3

(b) Five cylinder: interval 4

1445

= =

4

53

2

144

1-0

5-144

2-288

4-432 72

3-576 216

1 5

63

4

2

Firing order 1-6-2-5-3-4

1 2 3 4 5 6

(c) Six cylinder: interval4

1206

= =

Figure 10 Four-stroke engine

Balancing of Radial Engines (Direct and reverse crank method)

B

A

O2

Figure 11(a) Slider-crack mechanism

A

A

O

Direct crank

Reverse crank

Fig. 1(a)

Figure 11(b) Primary direct and reverse crank

Reverse crank OA is an image of direct crank OA. Direct crank and reverse crank rotates in opposite

direction. Let rm is the reciprocating mass at B, then primary force will be cos2rmr . This is

2

rm

2

rm

134

equivalent to the component of the centrifugal force produced by mass rm at A. Let total

reciprocating mass rm is divided into two parts i.e. 2/rm for each in the direct crank and in the

reverse crank as shown in Fig. 11(a). The centrifugal force acting on primary direct and reverse crank

is 2

2

rm r . Total component in the direction of line of stroke 22 cos2

rm r =

2 cosrm r = . The

component in vertical direction cancel each other i.e. 2 sin2

rm r .

The secondary force is ( )22 cos24

r

rm

n

, where rm is replaced by the mass of 2/rm each

placed at c and c as shown in Figure 11(c) with 4r

oc ocn

= = =

crank radius of secondary crank.

Secondary crank oc rotates at 2 rad/sec . Secondary crank co rotates at 2 rad/sec .

c

o

2

2

reversecrank

c

directcrank

Figure 11(b) Secondary direct & reverse crank

Three cylinders radial engine: Line of stroke is 120. All three lines of stroke are in one plane.

Shaking forces from reciprocating parts of each cylinder will be acting along the line of strokes in a

single plane. Let rm be the mass of reciprocating part of each cylinder.

Cylinder no. Direction of rotation Crank angle Direction of rotation Crank angle

1 ccw 0 cw 0

2 ccw 240 cw 120

3 ccw 120 cw 240

Primary unbalance: mass3

2

rm= ; crank radius = r; speed= cw; 232

rp

mF r= . It can be balanced

by a balancing mass bm at radius b i.e. 2

3 rb

mbm = and placed at opposite to the crank.

135

1

Line of stroke

connecting rod

for cylinder 1

connecting rod for cylinder 3

connectingrod forcylinder2

120

120120

32lineofstroke

Figure 12(a) Three cylinders radial engine

1,2,3 1

3 2

Direct primarycranks

Reverse primarycranks

Figure 12(b)

1

3 2

1,2,3

Direct Secodary cranks(Balanced by itself)

Reverse secondarycranks

22

Figure 12(c)

Secondary unbalance:mass2

3 rm= , radius n

r

4= ,

r

ln = and speed 2 ccw= + . Hence,

( )23 22 4

rs

m rF

n =

; by gear train we have to obtain secondary balancing.

rm

rm rm

2

rm

2

rm

23 r

m

2

rm

2

rm

23 r

m

2

rm

2

rm

136

Tutorial Problems:

(1) A single cylinder horizontal oil engine has a crank 190.5 mm long and a connecting rod 838.2 mm

long. The revolving parts are equivalent to 489.3 N at crank radius and the piston and gudgeon pin

weigh 400.32 N. The connecting rod weighs 511.52 N and its center of gravity is 266.7 mm from the

crank pin centre. Revolving balance weights are introduced at a radius of 215.9 mm on extensions of

the crank webs in order to balance all the revolving parts and one-half of the reciprocating parts. Find

the magnitude of the total balance weight and neglecting the obliquity of the connecting rod.

( ) . . / 0i e l r , the nature and magnitude of the residual unbalanced force on the engine. RPM=300 cw.

(Answer: 987.456 N, 5390.976 N and revolving at 300 rpm ccw).

(2) A four cylinder Marin oil engine has the cranks arranged at angular interval of 90. The inner

cranks are 1.22 m apart and are placed symmetrically between the out cranks, which are 3.05 m apart.

Each crank is 457.2 mm long, the engine runs at 90 rpm; and the weight of the reciprocating parts for

each cylinder is 8006.4 N. In which order should the cranks be arranged for the best balance of the

reciprocating masses, and what will then be the magnitude of the unbalanced primary couple?

(Answer: 1,4,2,3;42.82 kN-m)

(3) In a three-cylinder radial engine all three connecting rods act on a single crank. The cylinder

center lines are set at 120 , the weight of the reciprocating parts per line is 22.24 N, the crank length

is 76.2 mm, the connecting rod length is 279.4 mm and the rpm are 1800 cw. Determine with regard

to the inertia of the reciprocating parts (a) the balance weight to be fitted at 101.6 mm radius to give

primary balance (b) the nature and magnitude of the secondary unbalanced force (c) whether the

fourth and sixth order forces are balanced or unbalanced. (Answer: (a) 25 N (b) Constant magnitude

2508.67 N, 3600 rpm ccw (c) th4 harmonic is unbalanced, th6 harmonic is balanced).

Primary Balance of Multi-cylinder In-line Engines:

The conditions in order to give primary balance of the reciprocating masses are:

= 0cos. 2 rmrec (i.e. Algebraic sum of primary forces shall be zero) and

= 0cos. 2. ramrec (i.e. Algebraic sum of the (primary force) couple about any point in the plane of force shall also be zero).

where a is the distance of the plane of rotation of the crank from a parallel reference plane. Above

two equations should satisfy for the angular position of the crankshaft relative to the dead center

(reference line). The primary force due to the reciprocating mass is identical with the component

parallel to the line of stroke of the centrifugal force produced by an equal mass attached to and

137

revolving with the crank pin. Let the reciprocating masses are attached to crank pin at A, B, C and D

are revolving with crank as shown in Figure 13(a). The force polygon (centrifugal forces) may be

drawn. (i.e. oa, ab, bc and cd). Primary forces of the individual cylinder are equal to the component of

oa, ab, bc and cd along the line of stroke (say PQ). Now ef, fg ,gh and he are primary forces and since

algebraic sum of these are equal to zero (for this configuration), the engine is balanced for this

configuration. But suppose cranks turns clockwise through an angle , then the effect will be same as

crank shaft is fixed and line of stroke is rotating ccw by angle. Let us say SP is now new line of

stroke. Now 1, , and kl lm mn nk are primary forces and these algebraic sum is not zero, but it is equal

to 1kk (primary force is not balanced). The primary force will be only balanced when 1k coincides

with k or point d with o in force polygon. i.e. the centrifugal force polygon is a closed one.

A B

CD

O

Four cylinder engineOA,OB,OC,OD are four cranks

Figure 13(a) Four cylinder engine

b

a

oS

lm k k

1 np

e

h

fc

d

Qtoline ofstroke

Force polygon of the centrifugalforces produced when revolvingmasses respectively equal toreciprocating masses at crank pin

Figure 13(b) Force polygon

In similar way the primary couple can only be balanced if the couple polygon for the corresponding

centrifugal forces is closed. Hence, if a system of reciprocating masses is to be primary balance, the

system of revolving masses, which is obtained by substituting an equal revolving mass at the crank

pin for each reciprocating mass, must be balanced. The problem of the primary balance of

reciprocating masses may therefore be solved by using the method for revolving masses.

Secondary Balance of Multi-cylinder In-line Engine: The conditions for the complete secondary

balance of an engine are

( )22 cos2 04

rec

rm

n =

and

138

( )22 cos2 04

rec

rm a

n =

for all angular positions of the crank shaft relative to the line of stroke. As for the primary balance,

these conditions can only be satisfied if the force and couple polygons are closed for the

corresponding system of revolving masses. If to each imaginary secondary crank (angular

velocity= 2 and crank length ( )/ 4r n ) a revolving mass is attached, equal to the corresponding

reciprocating mass, then the system thus obtained must be completely balanced.

Example 1: A shaft carries four masses in parallel planes A, B, C and D in this order, along it. The

masses at B and C are 18 kg and 12.5 kg, respectively, and each has an eccentricity of 6 cm. The

masses at A and D have an eccentricity of 8 cm. The angle between the masses at B and C is 1000,

and that between the masses at B and A is 1900 (both angles measured in the same direction). The

axial distance between the planes A and B is 10 cm. And between B and C is 20 cm. If the shaft is in

complete dynamic balance, determine (i) the masses at and D (ii) the distance between the planes C

and D, and (iii) the angular position of the mass at D.

Solution:

B (Reference)

ld 1000

D

10cm 20cm CDl 1900 0

A B C D

A Figure 14 Multi-cylinder engine

Given data are: 18 6 ; 12.5 6 ; 8 B B C C A Dm kg r cm m kg r cm r r cm= = = = = =

Aim is to obtain: ? ?; ? ? (with respect to B)A D CD Dm m l = = = =

Condition of dynamics balancing are

cos 0; sin 0; cos 0; sin 0mr mr mrl mrl = = = =

Taking reference as the crank B for the phase measurement (i.e. B = 0) and for the moment the

cylinder A as reference and are as follows

139

cos cos 0; sin sin 0

cos cos cos 0; sin sin sin 0

B B B C C C C d d d C C C C d d d d

A A A B B C C C d d d A A A C C C d d d

m r l m r l m r m r l m r l

m r m r m r m r n m r m r m r

+ + = + =

+ + + = + + =

Hence,

1080 390.7 8 cos 0; 2215.82 8 sin 0

7.88 108 13 8 cos 0; 1.39 73.86 8 sin 0

d d d d d d

A d d A d d

m l m l

m m m m

+ = + =

+ + = + + =

or 8 cos 689.3 (1)

8 sin 2215.82

d d d

d d d

m l

m l

=

= (2)

7.88 8 cos 95 (3)

1.39 8 sin 73.86 (4)

A d d

A d d

m m

m m

+ =

+ =

Equations (1) and (2) will give tan 3.2146d = 252.72 so in third quadrant.d = So that

90=ddlm (5)

Equations (3) and (4) gives

7.88 2.3763 95; 1.39 7.639 73.86

9.67 kg; 7.91 kg; 36.66 cm; 6.66 cm

A d A d

A d d cd

m m m m

m m l l

= =

= = = =

Using Tabular method calculation are much easier. Take moments about plane A and plane D with

phase reference as crank B. Choosing plane D as first reference as advantage that mD and D will not

appear in two moment equation and mA and lCD can be calculated from this two equation directly. So

choice of reference plane should be such that maximum number of unknown can be eliminated (at

least two). Second reference plane can be chosen as A to eliminate mA from equation.

Example 2: By direct and reverse cranks method, consider the W-engine in which there are three

rows of cylinders. An engine of this type has four cylinders in each row and the crankshaft is of the

normal flat type with one connecting rod from each row coupled directly to each crankpin. The

middle row of cylinders is vertical and the other two rows are inclined at 600 to the vertical. The

weight of the reciprocating parts is 26.69 N per cylinder; the cranks are 7.6 cm long. The connecting

140

rods 27.9 cm long and the rpm 2000 cw. Find the maximum and minimum values of the secondary

disturbing force on the engine (Figure 15).

Figure-15 W-engine Solution:

26.69 N; 7.6 cm; 3.67; 2.72 kg; 27.9 cm; 2000 rpm 209.43 rad/secrec recl

w r n m l cr

= = = = = = = =

2

1 3

600 600

Figure-16 W-engine

Secondary cracks

2

(m/2) 2 (m/2)

600 600

(a) Direct (b) reverse

2 2 (m/2) (m/2) Resultant Unbalance

mmm

=

+ 5.02

22

424

)2(2

22 r

m

n

r

m=

+

n

m

n

rm

22

4)2( =

m/2

2 2

141

Total 4 rows: 2

4 9882.2 Nm r

n= Total 4 rows:

2

2 4941.1 Nm r

n=

(i). At present position Resultant is minimum:

2

2

Minimum unbalance 24941.1 N (2 )2 4

m r

n

= = +

;

0 0 0

2 2 2 20, 90 ; 180 , 270 = = = =

(ii). For 0 0 0 0 02 1 345 (or 60 45 or 300 45 ) = = + = + the resultant is maximum.

2

m

m/2

2

maximum unbalance 23

(2 ) 14823.3 N2 4

m r

n= =

0 0 0 0

4 4 4 445 , 135 ; 225 , 315 = = = =

Exercise Problems:

1. A four-cylinder marine oil engine has the cranks arranged at angular intervals of 900. The inner

crank are 1.2 m apart and are placed symmetrically between the outer cranks, which are 3.0m apart.

Each crank is 45.7 cm long, the engine runs at 90 rpm, and the weight of the reciprocating parts for

each cylinder is 8006 N. In which order should the cranks be arranged for the best balance of the

reciprocating masses, and what will then be the magnitude of the unbalanced primary couple?

2. The reciprocating masses for three cylinders of a four-crank engine weight 3047, 5078 and 8125

kgs and the centerlines of the se cylinders are 3.66 m, 2.59 m. and 1.07 m. respectively from that of

the fourth cylinder. Find the fourth reciprocating mass and the angles between the cranks so that they

may be mutually balanced for primary forces and couples. If the cranks are each 0.61 m long, the

connecting distributing rods 2.75 m long and the rpm 60, find the maximum value of the secondary

distributing force and crank positions at which it occurs.

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