Download chapter 7 balancing of linkages

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
αω 2 , where ω 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
A
2
ω
l
3
r
φ
O2
B
x
x
r+l
Figure 1 A reciprocating mechanism
The displacement of the piston is given as
1/ 2
 sin 2 θ 
x = r cosθ + l 1 − 2 
n 

with
n = l/r
(1)
The velocity of the piston can be obtained as

s in 2 θ
 1 
x = − ω r  s in θ + 

0 .5

 2 n  1 − n − 2 s in 2 θ 







(2)
The acceleration of the piston can be obtained as


( n 2 − 1)cos 2θ + cos 4 θ 
x = −ω 2 r  cosθ +
1.5


n3 1 − n −2 sin 2 θ 


In equation (1) the term 1 − (sin 2 θ ) / n 2 
(3)
0.5
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:
ω0
2ω
fundamental harmonic
nω
…
0
0
n th harmonic
second harmonic
A series of harmonic functions are known as the Fourier series. A harmonic function of any arbitrary
frequency ω is periodic i.e. it repeats itself at equal intervals of time T = 2π / ω , where T is the
period of excitation. If a function is periodic, it is not necessary that it will be harmonic.Let
x = A cos ωt is a harmonic function with period 2π / ω or frequency ω . Any arbitrary periodic
functions can be represented by a convergent series of harmonic functions whose frequencies are
integral multiples of certain fundamental frequency ω 0 (= 2π / T ) .
Fourier Series:
f (t ) = 12 a0 + ∑ r =1 (ar cos rωt + br sin rω0t )
n
(4)
with
T /2
ar =
2
f (t ) cos rω0 t dt
T −T∫/ 2
ω 0 = 2π / T ;
T /2
and
br =
2
f (t )sin rω0 t
T −T∫/ 2
r = 1, 2,3, ⋅⋅⋅
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,
ω , of shaking force and the natural frequency, ωnf , of flexible members (such as shafts, bearings,
supports, etc.) through which the force is transmitted. For ω = ωnf we have undesirable resonance
condition.
Primary and secondary forces can be replaced by rotating masses as:
1. A mass mb at radius r at A and rotating with crank with ω the component in horizontal
direction will be same as by reciprocating mass mb of piston i.e. primary force.
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2. A, mass mb at radius ( r / 4n) and rotating at twice the crank angular velocity ω . The
component in the horizontal direction will be same as that due to piston mb i.e. secondary
force.
The acceleration of piston can be obtained up to second harmonics as:
A p = −ω 2 r (cos θ +
where,
cos 2θ
)
n
(5)
n=l/r
primary force
F32 cosφ =mbrω 2cosθ
A
φ
F32
F32Sin φ
θ
θ
mbω2cosφ
mb
O2
mbω2 bsin φ
mbω2 b
Figure 2 A crack with a counter balance mass
( F32 sin φ ) r cos θ + (mb ω 2 b sin φ )b cos θ + T12 = 0
with T12 = rF32 sin φ cos θ − mb ω 2 b 2 sin φ cos θ
For a single cylinder engine the inertia force due to the reciprocating mass, m B = m3 B + m4 , will be
cos 2θ 

F = mBω 2 r  cos θ +

n 

(6)
where m3 B is one of the equivalent mass of connecting rod at piston and m 4 is the mass of piston.
For F → + ve the inertia force will be directed away from main bearing and for F → −ve the inertia
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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:
mb rω 2 cos θ
(7)
is called primary force. It reaches its maximum value of mb rω 2 twice per revolution, when
θ = 0 and π .
Second term:
mBω 2 r
cos 2θ
n
(8)
is called secondary force. It reaches its maximum value of mBω 2 r / n four times per revolution of
crank when θ = 0, π / 2, π and 3π / 2 .
(i) Harmonic motion (Fundamental) is given as x = A cos ω t , where A is the amplitude and ω is the
x = − ω 2 A cos ω t , where T = 2π / ω is the period and
frequency so that x = −ω A sin ω t and θ =ωt .
The primary inertia force Fi p = + mω 2 A cos ω t
(ii) Second harmonic motion is given as
x = A cos 2ω t , so that x = −2ω A sin 2ω t and
x = −4ω 2 A cos 2ω t .
The secondary inertia force Fi s = 4mω 2 A cos 2ω t = (4m)ω 2 A cos 2ω t
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, m3 , 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.
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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.
From
O2 AB
q = x sin φ
T = F34 sin φ x
= F34 . q = F12 . q = T21
T = T21
Figure 3 (a) A reciprocating engine
Figure 3 (c) Free body diagram of the piston
Figure 3(b) Free body diagram of the connecting rod
Figure 3 (d) Free body diagram of the crank
Figure 3 (f) Equivalent system of Figure (e)
Figure 3 (e) Free body diagram of bed
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The full effect on the engine frame of the inertia of the reciprocating mass is equivalent to the force,
mω 2 r (cos θ + n −1 cos 2θ ) , 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
mω 2 e cos θ and (ii) vertical direction mω 2 e sin θ . The primary inertia force can be balanced by a
counter mass mb at radius b such that mb b = mB r .
Figure 5 A reciprocating machine with a counter balance mass
The component of inertia force mb along the line of stroke is mb bω 2 cos θ . But a force
mb bω 2 sin θ 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 mb , reducing the inertia
force in the line of stroke and ⊥ direction, such that
cmB r = mbb
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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
= (mB r cosθω 2 − mbbω 2 cosθ ) or
mb .rω 2 cosθ − cmB rω 2 cosθ
or (1 − c)mB rω 2 cos θ
(9)
The unbalance force in the vertical direction
mbbω 2 sin θ = cmB rω 2 sin θ
(10)
Resultant unbalance force at any instant
= mB rω 2 (1 − c) 2 cos 2 θ + c 2 sin 2 θ
(11)
If the revolving and reciprocating (primary force) masses to be balanced simultaneously then,
mbb = mAr + cmB r = (mA+ cmB )r
(12)
where mA is the mass of rotating components. The direction of resultant force at any instant is given
as
tan φ =
cmB rω 2θ
c
tan θ
=
2
(1 − c)mB rω cosθ 1 − c
(b)
(a)
(13)
(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 c = 12 ; R is constant and
rotating with ω in cw direction. So primary force can be balanced as shown in Figure 7.
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Figure 7
Conditions of the balance in a multi cylinder engine:
c1
c2
2
2
3
m2
m3 m4
m1
1
3
m3
m2
m1
4
c3
stroke
c4
m4
Figure 8 Multi-cylinder in-line engine
In Figure 8 θ1 , θ 2 and θ 3 are θ 4 are angular positions of the multi cylinder engine (Four engines in
Figure 8) and φ 2 , φ3 and φ 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 φi 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.
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The resultant primary shaking force will be
F p = m1r1ω 2 cos θ1 + m2 r2ω 2 cos θ 2 + m3 r3ω 2 cos θ 3 + ⋅ ⋅ ⋅ ⋅ + mn rnω 2 cos θ n
n
= ∑ mi ri ω 2 cos θ i
(14)
i =1
with θ i = θ1 + φi
i = 1,2,3 ⋅ ⋅ ⋅ n
(15)
where θ1 is the variable and φi are constants and φ1 = 0 . Substituting (14) into (15), we get
F p = ∑ mi ri ω 2 cos(θ1 + φi )
or
or
(16)
n
n
i =1
i =1
= ∑ mi ri ω 2 cos θ1 . cos φi − ∑ mi ri ω 2 sin θ1 sin φi
n
n
i =1
i =1
ω 2 cos θ1 ( ∑ mi ri cos φi ) − ω 2 sin θ1 ( ∑ mi ri sin φi )
The primary force, F p , will be maximum, when
δF p
δθ1
(17)
= 0 i.e.
 ∑ mr sin φ 

 ∑ mr cos φ 
θ1 = tan −1 −
(18)
For F p to be zero for all values of θ1 , then from eqn. (17), we get
n

 ∑ mr cos φ  = 0
 i =1

(19)
 n mr sin φ = 0 
∑

 i =1

(20)
and
Similarly, for moments, we can write
n

 ∑ mra cos φ  = 0
 i =1

(21)
 n mra sin φ  = 0
∑

 i =1

(22)
and
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.
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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:
n
r
Fs = ∑ mi riω 2   cos 2θ
 i i
i −1
with
 1 ri 
 = 
 ni li 
(23)
n
r
or Fs = ∑ mi riω 2   cos 2 (θ1 + φi )
 l i
i −1
or
 mr 2

 mr 2

Fs = ω 2 cos 2θ1  ∑
cos 2φ  − ω 2 sin 2θ1  ∑
sin 2φ 
l
l




(i) Fs is maximum, when
(24)
∂Fs
= 0 which gives
∂θ1


mr 2
sin 2φ 
∑

1
l
θ1 = tan −1  −

2
mr
2

cos 2φ 
 ∑ l

(25)
(ii) For Fs = 0 i.e. criteria for the balancing of the secondary force is (from eqn. 24)
∑
mr 2
cos 2φ = 0
l
(26)
∑
mr 2
sin 2φ = 0
l
(27)
and
(iii) For balancing of moments, we have:
∑
mr 2
.a sin 2φ = 0
l
(28)
and
∑
mr 2
.a cos 2φ = 0
l
(29)
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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= 2π / n . Four- stroke engine: One cycle of operation in two revolutions of
crankshaft. Interval between the crank= 4π / n , 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
π / 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: 1-52-4-3 and 6 cylinder: 1-6-2-5-3-4.
1
1
2
3
4
°
4
90
3
2
Firing order 1-4-2-3
(a) Four cylinder engine (interval=
131
2π π
= )
4
2
5
2
1
1
°
3
4
5
3
72
2
4
Firing order 1-5-2-4-3
(b) Five cylinder engine: interval=
1
1
6
2
3
2π
= 72o
5
4
5
6
4
60°
3
2
5
Firing order 1-6-2-5-3-4
(c) Six cylinder engine: interval=
2π
= 60o
6
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 =
4π
=π
4
132
4
144°
1
1
2
3
4
5
5
3
4
2
3
1-0
3
5-144
2
2-288
Firing order 1-5-2-4-3
(b) Five cylinder: interval =
1
1 5
2
3
4-432 ≡ 72°
4π
= 144
5
4
3-576 ≡ 216°
5
6
4
6
3
2
Firing order 1-6-2-5-3-4
(c) Six cylinder: interval =
4π
= 120
6
Figure 10 Four-stroke engine
Balancing of Radial Engines (Direct and reverse crank method)
mr
2
Direct crank
A
A
mr
2
B
O
A
Reverse crank
Fig. 1(a)
O2
Figure 11(b) Primary direct and reverse crank
Figure 11(a) Slider-crack mechanism
Reverse crank OA′ is an image of direct crank OA. Direct crank and reverse crank rotates in opposite
direction. Let mr is the reciprocating mass at B, then primary force will be mr ω 2 r cos θ . This is
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equivalent to the component of the centrifugal force produced by mass mr at A. Let total
reciprocating mass m r is divided into two parts i.e. m r / 2 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
mr 2
m 
ω r . Total component in the direction of line of stroke = 2  r  ω 2 r cos θ = mrω 2 r cosθ . The
2
 2 
component in vertical direction cancel each other i.e.
mr 2
ω r sin θ .
2
2 r 
The secondary force is mr ( 2ω )   cos 2θ , where mr is replaced by the mass of mr / 2 each
 4n 
 4r 
placed at c and c ′ as shown in Figure 11(c) with oc = oc′ =   = crank radius of secondary crank.
 n 
Secondary crank oc rotates at 2ω rad/sec . Secondary crank oc ′ rotates at −2ω rad/sec .
c
direct
crank
2
o
2
c
reverse
crank
Figure 11(b) Secondary direct & reverse crank
Three cylinders radial engine: Line of stroke is 120º. 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 m r be the mass of reciprocating part of each cylinder.
Cylinder no.
Direction of rotation
Crank angle
Direction of rotation
Crank angle
1
ccw
0º
cw
0º
2
ccw
240º
cw
120º
3
ccw
120º
cw
240º
Primary unbalance: mass =
3mr
3m
; crank radius = r; speed= ω cw; Fp = r ω 2 r . It can be balanced
2
2
by a balancing mass mb at radius b i.e. mb b =
3mr
and placed at opposite to the crank.
2
134
mr
1
Line of stroke
connecting rod
for cylinder 1
120
120
120°
connecting
rod for
cylinder2
mr
connecting rod
for cylinder 3
line
of
stroke
2
3
m
r
Figure 12(a) Three cylinders radial engine
1,2,3
 mr 


 2 
 mr 
3×

2 

1
 mr 

 2 
2
3
 mr 


 2 
Direct primary
cranks
Reverse primary
cranks
Figure 12(b)
1  mr 
m 
3×  r 
 2 


 2 
 mr

 2


3
2
 mr 


2 2 
Direct Secodary cranks
(Balanced by itself)
1,2,3
2
Reverse secondary
cranks
Figure 12(c)
Secondary unbalance:mass =
3m r
r
l
, radius =
, n=
4n
r
2
and speed
2 r 
 3m 
Fs =  r  ( 2ω )   ; by gear train we have to obtain secondary balancing.
 4n 
 2 
135
= +2ω ccw . Hence,
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.
( i.e. l / r 〉〉 0 ) , 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 90º. 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 120º , 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) 4 th harmonic is unbalanced, 6 th harmonic is balanced).
Primary Balance of Multi-cylinder In-line Engines:
The conditions in order to give primary balance of the reciprocating masses are:
2
∑ mrec .ω r cosθ = 0 (i.e. Algebraic sum of primary forces shall be zero)
and
2
∑ mrec. .ω ra cosθ = 0 (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
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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 γ , then the effect will be same as
crank shaft is fixed and line of stroke is rotating ccw by γ angle. Let us say SP is now new line of
stroke. Now kl , lm, mn and nk1 are primary forces and these algebraic sum is not zero, but it is equal
to kk1 (primary force is not balanced). The primary force will be only balanced when k1 coincides
with k or point d with o in force polygon. i.e. the centrifugal force polygon is a closed one.
Q
b
a
A
S
D
O
f
h
c
B
C
o
l
mk
to
line of
stroke
e
d
γ
k1
n
p
Force polygon of the centrifugal
forces produced when revolving
masses respectively equal to
reciprocating masses at crank pin
Four cylinder engine
OA,OB,OC,OD are four cranks
Figure 13(b) Force polygon
Figure 13(a) Four cylinder engine
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
∑ m ( 2ω )
rec
2
 r 
  cos 2θ = 0
 4n 
and
137
∑ m ( 2ω )
rec
2
 r 
 a cos 2θ = 0
 4n 
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= 2ω and crank length ( r / 4n ) ) 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)
1000
ld
D
10cm
A
20cm
B
0
lCD
C
190
φ0
D
A
Figure 14 Multi-cylinder engine
Given data are: mB = 18 kg
Aim is to obtain: mA = ?
rB = 6 cm;
mD = ?; lCD = ?
mC = 12.5 kg
rC = 6 cm;
rA = rD = 8 cm
θ D = ? (with respect to B)
Condition of dynamics balancing are
∑ mr cos φ = 0;
∑ mr sin φ = 0;
∑ mrl cos φ = 0;
∑ mrl sin φ = 0
Taking reference as the crank B for the phase measurement (i.e. φB = 0) and for the moment the
cylinder A as reference and are as follows
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mB rB lB + mC rC lC cos φC + md rd cos φd = 0;
mC rC lC sin φC + md rd ld sin φd = 0
m A rA cos φ A + mB rB + mC rC cos φC + md rd cos nφd = 0;
mA rA sin φ A + mC rC sin φC + md rd sin φd = 0
Hence,
1080 − 390.7 + 8md ld cos φd = 0;
2215.82 + 8md ld sin φd = 0
− 7.88mA + 108 − 13 + 8md cos φd = 0;
or
− 1.39m A + 73.86 + 8md sin φd = 0
8md ld cos φd = −689.3
(1)
8md ld sin φd = −2215.82
(2)
− 7.88mA + 8md cos φd = −95
(3)
− 1.39 mA + 8md sin φd = −73.86
(4)
Equations (1) and (2) will give tanφd = 3.2146 ∴ φd = 252.72 so in third quadrant. So that
md ld = 90
(5)
Equations (3) and (4) gives
−7.88mA − 2.3763md = −95;
∴ m A = 9.67 kg;
− 1.39mA − 7.639md = −73.86
md = 7.91 kg;
ld = 36.66 cm;
lcd = 6.66 cm
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 φD 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
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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:
wrec = 26.69 N; r = 7.6 cm; n =
l
= 3.67; mrec = 2.72 kg; l = 27.9 cm; ω = 2000 rpm cω = 209.43 rad/sec
r
2
1
3
600
600
ω
Figure-16 W-engine
Secondary cracks
2
(m/2)
(m/2)
2
600 600
(a) Direct
(b) reverse
2ω
2ω
(m/2)
(m/2)
Resultant Unbalance
m

m
+ 2  × 0. 5 = m
2
2


m
r
 r  m
+ ( 2ω) 2   = ω 2
2
4
 4n  2
2
 r  mω
m(2ω) 2   =
n
 4n 
m/2
2ω
2ω
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Total 4 rows:
4
mω2 r
= 9882.2 N
n
Total 4 rows:
2
mω2 r
= 4941.1 N
n
(i). At present position Resultant is minimum:
2ω
2ω
m
 r 
+ (2ω)2   ;
2
 4n 
0
0
0
↑ θ2 = 0, θ2 = 90 ↓; ↑ θ2 = 180 , θ2 = 270 ↓
Minimum unbalance = 4941.1 N ↑=
(ii). For θ2 = 450 (or θ1 = 600 + 450 or θ3 = 3000 + 450 ) the resultant is maximum.
2ω
m
m/2
2ω
3m
r
(2ω)2
= 14823.3 N
2
4n
↑ θ4 = 450 , θ4 = 1350 ↓; ↑ θ4 = 2250 , θ4 = 3150 ↓
maximum unbalance =
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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