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 αω , 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 ω 2 A l 3 B x x r+l O 2 r φ Figure 1 A reciprocating mechanism The displacement of the piston is given as 1/2 2 2 sin cos 1 x r l n θ θ = + - with n = l/r (1) The velocity of the piston can be obtained as 0.5 2 2 1 sin 2 sin 2 1 sin x r n n θ ω θ θ - =- + - (2) The acceleration of the piston can be obtained as 2 4 2 1.5 3 2 2 ( 1) cos 2 cos cos 1 sin n x r n n θ θ ω θ θ - - + =- + - (3) In equation (1) the term 0.5 2 2 1 (sin )/ n θ - is expanded bi-nominally to give a harmonic analysis. i.e , xx 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:
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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 2αω , 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
ω 2A
l
3B
x
x
r+l
O2
r
φ
Figure 1 A reciprocating mechanism
The displacement of the piston is given as
1/ 2
2
2
sincos 1x r l
n
θθ
= + −
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
θω θ
θ−
= − + −
� (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
θ θω θ
θ−
− + = − + −
�� (3)
In equation (1) the term 0.5
2 21 (sin ) / nθ − 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:
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0ω 02ω … 0ωn
fundamental harmonic second harmonic thn 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 ωπ /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 ωcos= is a harmonic function with period ωπ /2 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 ( )T/20 πω = .
Fourier Series:
1
0 02 1( ) ( cos sin )
n
r rrf t a a r t b r tω ω
== + +∑ (4)
with
/ 2
0
/ 2
2( )cos
T
r
T
a f t r t dtT
ω−
= ∫ and / 2
0
/ 2
2( )sin
T
r
T
b f t r tT
ω−
= ∫
T/20 πω = ; 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,
ω , of shaking force and the natural frequency, nfω , of flexible members (such as shafts, bearings,
supports, etc.) through which the force is transmitted. For nfω ω= 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 ω the component in horizontal
direction will be same as by reciprocating mass bm of piston i.e. primary force.
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2. A, mass bm at radius )4/( nr and rotating at twice the crank angular velocity ω . 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
θθω +−= (5)
where, n=l/r
F32
F32Sin φ
O2
A
mbω2bsin φ
φ
mb
ω2b
mb
mbω2cosφ
θ
θ
F32 cosφ
primary force
=mbrω2cosθ
Figure 2 A crack with a counter balance mass
0cos)sin(cos)sin( 122
32 =++ TbbmrF b θφωθφ
with θφωθφ 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 r
n
θω θ = +
(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
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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:
θω cos2rmb (7)
is called primary force. It reaches its maximum value of 2ωrmb twice per revolution, when
0=θ and π .
Second term:
nrmB
θω
2cos2 (8)
is called secondary force. It reaches its maximum value of 2 /Bm r nω four times per revolution of
crank when 0, / 2, and 3 / 2θ π π π= .
(i) Harmonic motion (Fundamental) is given as tAx ωcos= , where A is the amplitude and ω is the
frequency so that sinx A tω ω= −� and 2 cosx A tω ω= −�� , where 2 /T π ω= is the period and
tθ ω= .
The primary inertia force tAmF pi ωω cos2+=
(ii) Second harmonic motion is given as tAx ω2cos= , so that 2 sin 2x A tω ω= −� and
24 cos2x A tω ω= −�� .
The secondary inertia force tAmtAmF si ωωωω 2cos)4(2cos4 22 ==
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.
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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.
Figure 3 (a) A reciprocating engine
From ABO2
φsinxq =
xFT φsin34=
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)
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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 nω θ θ−+ , 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
θω cos2em and (ii) vertical direction θω 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 θω cos2bmb . But a force
θω 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 =
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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