balancing of reciprocating masses - notes
TRANSCRIPT
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VTU EDUSAT PROGRAMME-17
DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE1
DYNAMICS OF MACHINES
Subject Code -10 ME 54
BALANCING
OF
RECIPROCATING MASSES
BALANCINGOF
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
coscosrmF
i
22
2
22
2
+=
+=
Notes Compiled by:
VIJAYAVITHAL BONGALEASSOCIATE PROFESSOR
DEPARTMENT OF MECHANICAL ENGINEERING
MALNAD COLLEGE OF ENGINEERINGHASSAN -573 202. KARNATAKA
Mobile:9448821954
E-mail:[email protected]
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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,
horizontalhF21 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 forcehF21 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 additionto the mass used to balance the rotating unbalance due to the mass at the crank pin. This
is shown in figure (c).
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DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE5
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 revolvingparts 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 500from 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/s260
60x2
60
N2
===
===
kg804060x3
2mmcM
MpincranktheatbalancedbetoMass
P =+=+=
=
and
kg5736.350
160x80mi.e.
rrMmthereforerMrm
c
c
ccc
==
==
(ii) Unbalanced force when the crank has turned 500from the top-dead centre.
( )[ ] [ ]
( ) ( )
N209.9
50sin2x0.16x60x3
250cos2x0.16x60x
3
21
sinrmccosrmc1
50atforceUnbalanced
202
202
2222
0
=
+
=
+=
=
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DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE6
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 radiusSpeed = 150 rpm, Stroke = 350 mm.
If 60 % of the reciprocating parts and all the revolving parts are to be balanced, determinethe,
(i) balance mass required at a radius of 320 mm and (ii) unbalanced force when the crank
has turned 450from 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
150x2
60
N2
===
===
kg543040x0.60mmcMMpincranktheatbalancedbetoMass
P =+=+=
=
and
kg29.53320
175x54mi.e.
r
rMmthereforerMrm
c
c
ccc
==
==
(ii) Unbalanced force when the crank has turned 450from the top-dead centre.
( )[ ] [ ]
( ) ( )[ ] ( )[ N880.7
45sin15.7x0.175x40x0.6045cos15.7x0.175x40x0.601
sinrmccosrmc145atforceUnbalanced
022
02
2222
0
=
+=
+=
=
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SECONDARY BALANCING:
Secondary acceleration force is equal to )(n
cosrm 1
22
Its frequency is twice that of the primary force and the magnituden
1 times the
magnitude of the primary force.
The secondary force is also equal to )(n
cos)(rm 2
4
22
2
Consider, two cranks of an engine, one actual one and the other imaginary with thefollowing specifications.
Actual ImaginaryAngular 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
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DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE9
= (1)cosrmforcePrimary 2
= (2)coslrmcouplePrimary 2
( ) = (3)2cos
n4
2rmforceSecondary
2
( )
=
=
(4)2cosln
rm
2cosln4
2rmcoupleSecondaryAnd
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 by2
provides the primary unbalanced force on the system at that moment.
This unbalanced force is zero when090= and a maximum when 00= .
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If the force polygon encloses, the resultant as well as the axial component will always be
zero and the system will be inprimary 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( )
n4
22
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 1800apart and have equal reciprocating masses.
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Taking a plane through the centre lineas the reference plane,
[ ] 0)180(coscosrmforcePrimary 2 =++=
coslrm)180(cos2
lcos
2
lrmcouplePrimary 22 =
+
+=
Maximum values are lrm 2 at 00 180and0=
[ ] 2cosnr2m
)2360(cos2cosn
rmforceSecondary
22
=++=
Maximum values aren
r2m 2 when0000
0000
270and180,90,0or
540and360,180,02
=
=
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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 areequal 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 00and 1800. However, if the crank positions are assumed
at 900and 2700, the values obtained will be zero.
If any particular position of the crank shaft is considered, then both X and Ycomponents of the force and couple can be taken to find the maximum values.
For example, if the crank positions considered as 1200and 300
0, the primary couple
can be obtained as
( )
lrm2
1
120180cos2
l120cos
2
lrmcomponentX
2
0002
=
+
+=
( )
lrm2
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
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FINDING PRIMARY FORCES, PRIMARY COUPLES, SECONDARY FORCESAND 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
rmcouplePrimary 0
0
2
=
++
+
++
=
2cosn
r4m
2cos)2360(cos
)2360(cos2cos
n
rmforceSecondary
2
0
02
=
+++
++=
0000
0000
2
027and180,09,0
or054and360,018,02at
nrmvalueMaximum
=
=
=
0
2cos
2
3l)2360(cos
2
l
)2360(cos2
l2cos
2
3l
n
rmcoupleSecondary
0
0
2
=
++
+
++
=
Thus the engine is not balanced in secondary forces.
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DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE15
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
===
====
PlaneMass (m)
kg
Radius (r)
m
Cent.
Force/2
(m r )
kg m
Distance
from Ref
plane 2
m
Couple/ 2
( m r l )
kg m
2
1 380 r 380 r -1.3 -494 r
2(RP) m2 r m2r 0 0
3 590 r 590 r 2.8 1652 r
4 480 r 480 r 4.1 1968 r
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Analytical Method:
Choose plane 2 as the reference plane and0
30= .
.
Step 1:
Resolve the couples into their horizontal and vertical components and take their sums.
Sum of the horizontal components gives
)(coscos.,e.i
cosrcosrcosr
119681652494
0196801652494
41
4
0
1
+=+
=++
Sum of the vertical components gives
)(sinsin.,e.i
sinrsinrsinr
21968494
0196801652494
41
4
0
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 04 167.9 =
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.
Sum of the horizontal components gives
)(.cosmor
).(cosrcosrcosrm).(cosr
3588
0916748005904123380
22
00
22
0
=
=+++
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Sum of the vertical components gives
)(.sinmor
).(sinrsinrsinrm).(sinr
49417
0916748005904123380
22
00
22
0
=
=+++
Squaring and adding (3) and (4), we get
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.
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This diagram provides the relative direction of the masses 431mandm,m
.
Step 2:Now, draw the force polygon taking a suitable scale as shown.
This gives the direction and magnitude of mass m2.
The results are:
Anskgmorrrm
,,
427427
282123168
22
0
2
0
1
0
4
==
===
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DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE19
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 ===
=
PlaneMass (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
1 100 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
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Analytical Method:
Choose plane 3 as the reference plane and0
10= .
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.
151375618
51375618
0756180513
42
42
42
0
=
+=
=+
Sum of the vertical components gives
)(sin.sin..,e.i
sin.sin.sin.
275618
0756180513
42
42
0
=
=+
Squaring and adding (1) and (2), we get
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.
Sum of the horizontal components gives
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(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
=
=++
=+++
Squaring and adding (3) and (4), we get
Anskg120kg119.98
0.225
20.518
0.225
17.545
mi.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
=
=
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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
N2m,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,
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( ) ( ) ( ) ( )
erchangedintarelandlcesin,onlymNCC
mN
...lllrmC
pp
p
4216
222
42
2
3
2
1
43761
43761
62808117195
==
=
+=+=
( ) ( ) ( ) ( )
erchangedintarelandlcesin,onlymNCC
mN
...lllrmC
pp
p
3225
222
32
2
4
2
2
47905
47905
81806217195
==
=
+=+=
( ) ( ) ( ) ( )
erchangedintarelandlcesin,onlymNCC
mN
...lllrmC
pp
p
3434
222
34
2
2
2
3
19448
19448
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:
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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.
First0
10= Second
0
2240=
654321
654321
rrrrrr
mmmmmm
And
=====
=====
Third0
3120= Fourth
0
4120=
Fifth0
5240= Sixth 0
60=
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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 eachcylinder 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
1000x2
60
N2have,We
rpm1000N
,kg1.4cylindereachofmassingreciprocatm,mm240lengthrodconnectingl,mm60r
:Given
===
=
==
===
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Plane Mass (m) kgRadius (r)
m
Cent.
Force/2
(m r )kg m
Distance
from Ref
plane 2m
Couple/ 2
( m r l )
kg m2
1 1.4 0.06 0.084 0.21 0.01764
2 1.4 0.06 0.084 0.13 0.010923 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
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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 mmrespectively. 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
2400x2
60
N2have,We
rpm2400N
,kg1.2cylindereachofmassingreciprocatm
,mm180lengthrodconnectingl,mm402
80
2
Lr
:Given
===
=
==
=====
NO UNBALANCED
SECONDARY FORCE
NO UNBALANCED
SECONDARY COUPLE
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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
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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 600apart 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 outof 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
order1 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
order1 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
PlaneMass (m)
kg
Radius (r)
m
Cent.
Force/2
(m r )
kg m
Distance
from Ref
plane
m
Couple/ 2
( m r l )
kg m2
1 150 0.16 24 1.25 30
2 150 0.16 24 0.75 183 150 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
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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.
PlaneMass (m)
kg
Radius (r)
m
Cent.
Force/2
(m r )
kg m
Distance
from Ref
plane
m
Couple/ 2
( m r l )
kg m2
1 100 0.16 16 1.25 20
2 100 0.16 16 0.75 12
3 100 0.16 16 0.25 4
4 100 0.16 16 -0.25 -4
5 100 0.16 16 -0.75 -12
6 100 0.16 16 -1.25 -20
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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-XalongforceprimaryTotal
22
2
2
2
=
=
++=
++=
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DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE34
Similarly,
[ ]
[ ]
(6)sinsinrm2
sinsin2xsinrm
sinsincoscos()sinsincos(cossinrm
)sin(cos-sin)(cosrm
axis-ZalongforceprimaryTotal
22
2
2
2
=
=
+=
+=
( ) ( )
( ) ( ) (7)sinsincoscosrm2
sinsinrm2coscosrm2
forcePrimaryResultant
22222
222222
+=
+=
and this resultant primary force will be at angle with the X axis, given by,
(8)coscos
sinsintan
2
2
=
If0902 = , the resultant force will be equal to
( ) ( )
(9)rm
sin45sincos45cosrm22
2022022
=
+
and (10)tancos45cossin45sintan 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)-----rmrm rr =
For a given value of , the resultant force is maximum (Primary force), when
( ) ( )
( ) maximumissinsincoscosor
maximumissinsincoscos
2424
2222
+
+
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Or
( )
0cossin2xsinsincos2xcos-i.e.,
0sinsincoscosd
d
44
2424
=+
=+
[ ] 0cos-sin2sini.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 OB1is equal to
)()(cosn
rm12
2
Secondary force of 1 along X - axis = )(cos)(cosn
rm22
2
Secondary force of 2 along line of stroke OB2=
)()(cosn
rm32
2
+
Primary force of 2 along X-axis = )(cos)(cosn
rm42
2
+
Therefore,
[ ]
[ ]
(5)2cos2coscosn
rm2
2sin2sin2cos2(cos)2sin2sin2cos2(coscosn
rm
)(2cos)(2coscos
n
rm
axis-XalongforcesecondaryTotal
2
2
2
=
++=
++=
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Similarly,
(6)2sin2sinsinn
rm2
axis-ZalongforcesecondaryTotal2
=
( ) ( ) (7)2sin2sinsin2cos2coscosn
rm2
forceSecondaryResultant
222
+=
And (8)2cos2coscos
2sin2sinsintan ' =
If 00 45or902 == ,
Secondary force = )(sinn
rmsin
n
rm922
2
22 222
=
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 fittedopposite 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
800x2
60
N2have,We
rpm800N
mm120radiuscrankr
mm600lengthrodconnectingl
mm150rkg,2.2massbalancingmkg2massrotatingequivalentM
kg1.2pistoneachofmassingreciprocatm
:Given
CC
======
=
==
==
=====
==
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Primary Force:
(1)coscosrm2axis-XalongforceprimaryTotal 22 =
(2)cosrM
axisXalongmassrotatingtodueforcelCentrifuga2
=
(3)cosrm
axisXalongmassbalancingtodueforcelCentrifuga2
CC =
Therefore total unbalance force along X axis = (1) + (2) + (3)
That is
[ ]( ) [ ]
( ) [ ] (4)-------Ncos884.410.33-0.240.216cos83.78
2.2x0.152x0.1230xcos2x1.2x0.12cos83.78rm-rMcosrm2cos
cosrmcosrMcoscosrm2
axisXalongforceUnbalanceTotal
2
022CC
22
2
CC
222
=+=
+=
+=
+=
(5)ssrm2axis-ZalongforceprimaryTotal 22 = inin
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(6)srM
axisZalongmassrotatingtodueforcelCentrifuga2
=
in
(7)srm
axisZalongmassbalancingtodueforcelCentrifuga2
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-ZalongforceUnbalanceTotal
2
022
CC
22
2
CC
222
=+=
+=
+=
+=
( ) ( )
(9)15961.8cos766219.25
sin15961.8cos782181.05
sin126.34-cos884.41
forcePrimaryResultant
2
22
22
+=
+=
+=
This is maximum, when00= and minimum, when 090=
(11)N126.3415961.890cos766219.25
90wheni.e.,force,PrimaryMinimumAnd
(10)N884.4115961.8766219.25
0wheni.e.,force,PrimaryMaximum
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.
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( ) ( )
( )
( )[ ] (12)2sin(0.18752cos(0.1875404.3
60sin2sin30sin
60cos2cos30cos
5
78)x0.12x(83.x1.22
2sin2sinsin2cos2coscosn
rm2
forceSecondaryResultant
22
200
2222
222
+=
+
=
+=
This is maximum, when00= and minimum, when
0180=
[ ]
[ ] (14)N175.07)018sin(0.1875)180cos(0.1875404.3
180wheni.e.,force,secondaryMinimumAnd
(13)N175.07)0sin(0.1875)0cos(0.1875404.3
0wheni.e.,force,secondaryMaximum
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)1cos2cosrm
axis-XalongforceprimaryTotal
22+=
(2)sinsinrm2
zero)isaxisZalong3offorceprimarythe(since
engine,twinVtheinsasamebewillaxis-ZalongforceprimaryTotal
22=
( ) ( )[ ] (3)sin2sin12coscosrmforcePrimaryResultant
22222++=
and this resultant primary force will be at angle with the X axis, given by,
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DYNAMICS OF MACHINES VIJAYAVITHAL BONGALE40
( )
(4)12coscos
sin2sintan
2
2
+
=
If060= ,
(5)rm2
3
forcePrimaryResultant
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)-----rmrm rr =
(8)12coscosn
r2m2cos
axis-XalongforcesecondaryTotal
2
+=
Total secondary force along Z direction will be same as in the V-twin engine.
( ) ( )[ ] (9)2sin2sin2sin12cos2cos2cosn
rm
forcesecondaryResultant
222
++=
( ) (10)
12cos2cos2cos
2sin2sin2sintan '
+=
If0
60= ,
(11)2cos2n
rm
axis-XalongforceSecondary
2
=
(12)2sin2n
r3m
axis-ZalongforceSecondary
2
=
It is not possible to balance these forces simultaneously
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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, cosrm2
which acts alongthe 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 assumedto 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 strokeare,
= cosrm
AatmasstoDue2
2
= cosrm
AatmasstoDue' 2
2
Thus, total force along the line of stroke = cosrm2
which is equal to the primary
force.
At any instant, the components of the centrifugal forces of these masses normal to the
line of stroke will be equal and opposite.The crank rotating in the direction of engine rotation is known as the direct crankand
the imaginary crank rotating in the opposite direction is known as the reverse crank.
Now,
Secondary accelerating force is
=
=
224
4
22
2
2
22
cos)(n
rm
n
cos)(rm
n
cosrm
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