Unit 5 Balancing of RECIPROCATING MASSES
Unit 5 Balancing of RECIPROCATING MASSES
Unit 5 Balancing of RECIPROCATING MASSES
DYNAMICS OF MACHINES
Subject Code -10 ME 54
BALANCING
OF
RECIPROCATING MASSES
Notes Compiled by:
VIJAYAVITHAL BONGALE
ASSOCIATE PROFESSOR
DEPARTMENT OF MECHANICAL ENGINEERING
BALANCING
MALNAD COLLEGE OF ENGINEERING
HASSAN -573OF
202. KARNATAKA
RECIPROCATING MASSES
Mobile:9448821954
E-mail:vvb@mcehassan.ac.in
a p = Accelerati on of piston
cos 2 θ
= r ω 2 cos θ + − − − − − − − − − (1)
n
l
Where n =
r
cos 2 θ
Fi = m r ω2 cos θ +
n
cos 2 θ
= m r ω2 cos θ + m r ω2 − − − − − − − − − ( 2)
n
In Fig (a), the inertia force due to primary accelerating force is shown.
At ‘O’ the force exerted by the crankshaft on the main bearings has two components,
horizontal F21h v
and vertical F21 .
F21v and F41v balance each other but form an unbalanced shaking couple.
The magnitude and direction of these unbalanced force and couple go on changing with
angle θ. The shaking force produces linear vibrations of the frame in horizontal direction,
whereas the shaking couple produces an oscillating vibration.
h
The shaking force F21 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.
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).
The horizontal component of the centrifugal force due to the balancing mass is
m r ω2 cos θ and this is in the line of stroke. This component neutralizes the unbalanced
2
reciprocating force. But the rotating mass also has a component m r ω sin θ
perpendicular to the line of stroke which remains unbalanced. The unbalanced force is
zero at θ = 00 or 1800 and maximum at the middle of the stroke i.e. θ = 900. The
magnitude or the maximum value of the unbalanced force remains the same i.e. equal to
m r ω2 . Thus instead of sliding to and fro on its mounting, the mechanism tends to
jump up and down.
2
To minimize the effect of the unbalance force a compromise is, usually made, is of the
3
1 3
reciprocating mass is balanced or a value between to .
2 4
If ‘c’ is the fraction of the reciprocating mass, then
and
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.
1
The resultant unbalanced force is minimum when, c =
2
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 m P 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 c m + m P
which is to be completely balanced.
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:
Given : m = mass of the reciprocat ing parts = 60 kg
N = 60 rpm, L = length of the stroke = 320 mm
2
mP = 40 kg, c = , rc = 350 mm
3
2 πN 2 π x 60
ω= = = 2 π rad/s
We have, 60 60
L 320
r= = = 160 mm
2 2
Mr
mc rc = M r therefore mc =
and rc
80 x 160
i.e. mc = = 36. 57 kg
350
(ii) Unbalanced force when the crank has turned 500 from the top-dead centre.
Unbalanced force at θ = 50 0
= [(1 − c )mr ω 2
cos θ] + [c mr ω2 sinθ]
2 2
2 2
2 2 2 2
= 1 − x 60 x 0.16 x (2π) cos 50 0 + x 60 x 0.16 x(2π) sin50 0
3 3
= 209.9 N
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:
2 πN 2 π x 150
ω= = = 15.7 rad/s
We have, 60 60
L 350
r= = = 175 mm
2 2
Mr
mc rc = M r therefore mc =
and rc
54 x 175
i.e. mc = = 29.53 kg
320
(ii) Unbalanced force when the crank has turned 450 from the top-dead centre.
Unbalanced force at θ = 45 0
= [(1 − c )mr ω 2
cos θ] + [c mr ω2 sin θ]
2 2
= 880.7 N
SECONDARY BALANCING:
2 cos 2θ
Secondary acceleration force is equal to m r ω − − − − − −(1)
n
Its frequency is twice that of the primary force and the magnitude 1 times the
n
magnitude of the primary force.
2 cos 2θ
The secondary force is also equal to m r ( 2ω) − − − − − −( 2)
4n
Consider, two cranks of an engine, one actual one and the other imaginary with the
following specifications.
Actual Imaginary
Angular velocity ω 2ω
r
Length of crank r
4n
Mass at the crank pin m m
Thus, when the actual crank has turned through an angle θ = ω t , the imaginary crank
would have turned an angle 2θ = 2 ω t
2
m r (2ω )
Centrifugal force induced in the imaginary crank =
4n
2
m r (2ω )
Component of this force along the line of stroke is = cos 2θ
4n
r
Thus the effect of the secondary force is equivalent to an imaginary crank of length
4n
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.
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.
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
Primary force = ∑ m rω 2
cos θ − − − − − − − − − − − − − (1)
Primary couple = ∑ m rω 2
l cos θ − − − − − − − − − − − − − (2)
2
(2ω)
Secondary force = ∑ mr
4n
cos 2θ − − − − − − − − − − − − − (3)
2
(2ω)
And Secondary couple = ∑ mr
4n
l cos 2θ
ω2
= ∑mr n
l cos 2θ − − − − − − − − − − − − − (4)
GRAPHICAL SOLUTION:
2
To solve the above equations graphically, first draw the ∑ m r cos θ polygon ( ω is
common to all forces). Then the axial component of the resultant forces ( Fr cos θ )
2
multiplied by ω provides the primary unbalanced force on the system at that moment.
0 0
This unbalanced force is zero when θ = 90 and a maximum when θ = 0 .
∑F Ph
= 0 and ∑F PV
=0
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
(2ω) 2
or
ω2
4n n
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.
l l
Primary couple = m r ω2 cos θ + − cos (180 + θ) = m r ω2 l cos θ
2 2
m r ω2 2
m r ω2 l l
Secondary couple = cos 2θ + − cos ( 360 + 2θ ) = 0
n 2 2
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 m r ω2 l 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
l
X − component = m r ω 2 cos 120 0 l
+ − cos (180 0
+ 120 0
)
2 2
1
= − mrω2 l
2
l l
Y − component = mr ω 2 sin120 0 + − sin (180 0 + 120 0 )
2 2
3
= m r ω2 l
2
2
3
2
1 2
Therefore, Primary couple = − 2 mr ω l + 2 mr ω l
2
2
= mr ω l
Case 2:
IN-LINE FOUR-CYLINDER FOUR-STROKE ENGINE
Choose a plane passing through the middle bearing about which the arrangement is
symmetrical as the reference plane.
3l l 0
2 cos θ + 2 cos (180 + θ)
Primary couple = m r ω2
+ − l cos (180 0 + θ) + − 3l cos θ
2
2
=0
m r ω2
Maximum value =
n
at 2θ = 0 0 ,18 0 0 , 360 0 and 54 0 0 or
θ = 0 0 , 9 0 0 ,180 0 and 27 0 0
3l l
cos 2θ + cos (360 0 + 2θ )
2
m rω 2 2
Secondary couple = =0
n l 3l
+ − cos (360 + 2θ ) + − cos 2θ
2
0
2
Problem 1:
Solution:
Given :
m1 = 380 kg, m2 = ? , m3 = 590 kg, m4 = 480 kg
L 2r
crank length = = =r
2 2
Cent. Distance
Couple/ ω2
Mass (m) Radius (r) Force/ω2 from Ref
Plane (mrl)
kg m (m r ) plane ‘2’
kg m2
kg m m
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
Analytical Method:
0
Choose plane 2 as the reference plane and θ 3 = 0 .
.
Step 1:
Resolve the couples into their horizontal and vertical components and take their sums.
Step 2:
Resolve the forces into their horizontal and vertical components and take their sums.
m 2 = 427.1 kg Ans
− 417.9
Dividing (4) by (3), we get tan θ2 = = − 4.72
88.5
or θ 2 = 282 0 Ans
Graphical Method:
Step 2: Now, draw the force polygon taking a suitable scale as shown.
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:
Given :
r = 225 mm
m1 = 100 kg, m2 = 120 kg and m4 = 100 kg
Cent. Distance
Couple/ ω2
Mass (m) Radius (r) Force/ω2 from Ref
Plane (mrl)
kg m (m r ) plane ‘2’
kg m2
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
Analytical Method:
0
Choose plane 3 as the reference plane and θ1 = 0 .
Step 1:
Resolve the couples into their horizontal and vertical components and take their sums.
Sum of the horizontal components gives
Step 2:
Resolve the forces into their horizontal and vertical components and take their sums.
− 20.518
Dividing (4) by (3), we get tan θ 3 =
- 17.545
or θ 3 = 229.5 0 Ans
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:
Given :
2 πN
m = 800 kg, N = 70 rpm , r = 0.4 m, ω = = 7.33 rad / s
60
m r ω 2 = 800 x 0.4 x (7.33) 2 = 17195
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.
C p1 = m r ω 2 (− l ) + (l
3
2
2 − l 4 ) = 17195
2
(− 1.8) + (0.8 − 2.6)
2 2
= 43761 N m
C p 6 = C p1 = 43761 N m only , sin ce l 2 and l 4 are int erchanged
C p 2 = m r ω2 (− l ) + (l
4
2
2 − l 3 ) = 17195
2
(− 2.6) + (0.8 −1.8)
2 2
= 47905 N m
C p 5 = C p 2 = 47905 N m only , sin ce l 2 and l 3 are int erchanged
C p 3 = m r ω2 (− l ) + (l
2
2
4 − l 3 ) = 17195
2
(− 0.8) + (2.6 −1.8)
2 2
= 19448 N m
C p 4 = C p 3 = 19448 N m only , sin ce l 4 and l 3 are int erchanged
Thus the best arrangement is of 3rd and 4th. The firing orders are 1423 and 1324
respectively.
Unbalanced couple = 19448 N m.
Graphical solution:
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
First θ1 = 0
0
Second θ 2 = 240
0 And
0 0 m1 = m2 = m3 = m4 = m5 = m6
Third θ 3 =120 Fourth θ 4 =120
r1 = r2 = r3 = r4 = r5 = r6
0 0
Fifth θ 5 = 240 Sixth θ 6 = 0
Since all the force and couple polygons close, it is inherently balanced engine for primary
and secondary forces and couples.
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:
Given :
r = 60 mm , l = connecting rod length = 240 mm ,
m = reciprocat ing mass of each cylinder =1.4 kg ,
N = 1000 rpm
2 πN 2 π x 1000
We have, ω = = = 104. 72 rad /s
60 60
Cent. Distance
Couple/ ω2
Radius (r) Force/ω2 from Ref
Plane Mass (m) kg (mrl)
m (m r ) plane ‘2’
kg m2
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
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
NO UNBALANCED
SECONDARY FORCE
NO UNBALANCED
SECONDARY COUPLE
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:
Given :
L 80
r= = = 40 mm , l = connecting rod length = 180 mm ,
2 2
m = reciprocat ing mass of each cylinder = 1.2 kg ,
N = 2400 rpm
2 π N 2 π x 2400
We have, ω = = = 251.33 rad /s
60 60
Cent. Distance
Couple/ ω2
Radius (r) Force/ω2 from Ref
Plane Mass (m) kg (mrl)
m (m r ) plane ‘2’
kg m2
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
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 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.
Cent. Distance
Couple/ ω2
Mass (m) Radius (r) Force/ω2 from Ref
Plane (mrl)
kg m (m r ) plane
kg m2
kg m 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
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.
Cent. Distance
Couple/ ω2
Mass (m) Radius (r) Force/ω2 from Ref
Plane (mrl)
kg m (m r ) plane
kg m2
kg m 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
BALANCING OF V – ENGINE
A common crank OA is operated by two connecting rods. The centre lines of the two –
cylinders are inclined at an angle α to the X-axis.
Let θ be the angle moved by the crank from the X-axis.
2
Primary force of 1 along X - axis = m r ω cos ( θ − α ) cos α − − − ( 2)
2
Primary force of 2 along line of stroke OB2 = m r ω cos ( θ + α ) − − − − − (3)
2
Primary force of 2 along X-axis = m r ω cos ( θ + α ) cos α − − − ( 4)
= 2 mr ω 2 (cos 2
α cos θ) + (sin2 α sin θ) − − − − − (7)
2 2
and this resultant primary force will be at angle β with the X – axis, given by,
sin2 α sin θ
tanβ = − − − − − −(8)
cos 2 α cos θ
0
If 2α = 90 , the resultant force will be equal to
2 2
2 mr ω 2 (cos 2
45 0 cos θ) + (sin2 45 0 sin θ)
= mr ω 2 − − − − − (9)
i.e., β = θ 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,
mr rr = mr - - - - - (11)
For a given value of α, the resultant force is maximum (Primary force), when
(cos 2
α cos θ) + (sin2 α sin θ)
2 2
is maximum
or
(cos 4
α cos 2 θ + sin 4 α sin2 θ ) is maximum
Or
d
(cos 4 α cos 2 θ + sin4 α sin2 θ ) = 0
dθ
i.e., - cos 4 α x 2 cos θ sin θ + sin 4 α x 2 sin θ cos θ = 0
As α is not zero, therefore for a given value of α , the resultant primary force is
maximum when θ = 0 0 .
m r ω2
cos 2 ( θ − α ) − − − − − −(1)
n
m r ω2
Secondary force of 1 along X - axis = cos 2 ( θ − α ) cos α − − − ( 2)
n
Secondary force of 2 along line of stroke OB2 =
m r ω2
cos 2 ( θ + α ) − − − − − (3)
n
m r ω2
Primary force of 2 along X-axis = cos 2 ( θ + α ) cos α − − − ( 4)
n
Therefore,
If 2 α = 90 0 or α = 45 0 ,
2
2 m r ω2 sin 2 θ m r ω2
Secondary force = = 2 sin 2 θ − − − −(9)
n 2 n
And tan β ' = ∞ and β ' = 90 0 − − − − − −(10) 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:
Given :
m = reciprocat ing mass of each piston = 1.2 kg
M = equivalent rotating mass = 2 kg
m C = balancing mass = 2.2 kg, rC = 150 mm
l = connecting rod length = 600 mm
r = crank radius = 120 mm
N = 800 rpm
2 π N 2 π x 800 l 600
We have, ω = = = 83.78 rad /s and n = = =5
60 60 r 120
Primary Force:
That is
Total Unbalance force along X axis
= 2 mr ω2 cos 2 α cos θ + Mr ω2 cos θ − mC rC ω2 cos θ
= ω2 cos θ [2 mr cos 2 α + Mr - mC rC ]
= (83.78 ) cos θ [2x1.2x0.12xcos 30 0 + 2x0.12 − 2.2x0.15]
2 2
That is
0 0
This is maximum, when θ = 0 and minimum, when θ = 90
Secondary force:
The rotating masses do not affect the secondary forces as they are only due to second
harmonics of the piston acceleration.
=
2 x1.2 x0.12x(83. 78)2 (cos 30 cos 2 θ cos 60 )
2 2 2
0 0
This is maximum, when θ = 0 and minimum, when θ =180
BALANCING OF W ENGINE
Total primary force along Z - axis will be same a s in the V − twin engine,
(since the primary force of 3 along Z − axis is zero)
= 2 m r ω 2 sin2 α sin θ − − − − − − − − − − − −(2)
and this resultant primary force will be at angle β with the X – axis, given by,
2sin2 α sin θ
tan β = − − − − − −(4)
cos θ(2cos 2 α + 1)
0
If α = 60 ,
Resultant Primary force
3
and
= m r ω 2 − − − − − (5)
2
i.e., β = θ 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,
mr rr = mr - - - - - (7)
2m r ω 2
= cos 2θ cos α cos 2 α + 1 − − − − − − − − − − − −(8)
n
Total secondary force along Z –direction will be same as in the V-twin engine.
0
If α = 60 ,
mr ω2
= cos 2θ − − − − − − − − − − − −(11)
2n
3m r ω 2
= sin 2θ − − − − − − − − − − − −(12)
2n
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.
It is a multicylinder engine in which all the connecting rods are connected to a common
crank.
In this all the forces exists in the same plane and hence no couple exist.
2
In a reciprocating engine the primary force is given by, m r ω cos θ 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 ω 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 ω 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,
m 2
Due to mass at A = r ω cos θ
2
m 2
Due to mass at A ' = r ω cos θ
2
2
Thus, total force along the line of stroke = m r ω cos θ 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,
r
1. A mass m/2, placed at the end of direct secondary crank of length at an angle 2θ
4n
and rotating at an angular velocity 2ω in the counter clockwise direction.
r
2. A mass m/2, placed at the end of reverse secondary crank of length at an angle -2θ
4n
and rotating at an angular velocity 2ω in the clockwise direction.
The components of the centrifugal forces due to rotating masses along the line of stroke
are,
m r mrω2
Due to mass at C = (2 ω)2 cos 2θ = cos 2θ
2 4n 2n
m r mrω2
Due to mass at C' = (2 ω)2 cos 2θ = cos 2θ
2 4n 2n
m r 2 mrω 2
2x (2 ω) cos 2θ = cos 2θ which is equal to the secondary force.
2 4n n
References: