This document discusses various types of machine balancing. It begins by defining static and dynamic balancing. Static balancing deals with balancing forces when a machine is at rest, while dynamic balancing deals with balancing forces during motion. It then discusses balancing of single and multiple rotating masses, as well as reciprocating masses. Methods for analytically and graphically balancing multiple masses are provided. The document also covers balancing of engines with different cylinder configurations, including inline, V-shaped, radial, and locomotive engines. Partial balancing techniques are discussed for reducing unbalanced forces in locomotives.

1. THEORY OF MACHINE
BALANCING
By: KIRAN J. BHINGARE
Guided By: Prof. K. R. Madavi

2. CONTENTS
• Static Balancing
• Dynamic Balancing
• Balancing of Several Masses in Different Planes
• Force Balancing of Linkages
• Balancing of Reciprocating Masses
• Balancing of Locomotives
• Effect of Partial Balancing in Locomotives
• Balancing of Inline Engine
• Balancing of V,W,V-8 and V-12 Engines
• Balancing of Radial Engines

4. THEORY OF MACHINE
• The Machine receives energy in some available form and uses it
to do some particular type of work.

5. STATIC
• It deals with the study the forces acting on machine in rest.
DYNAMICS
• It deals with the study the forces acting on various parts of a
machine

6. KINEMATICS
• It is the study of motion, quite apart from the forces which
produce that motion.
• It is the study of position, displacement, rotation, speed, velocity
and acceleration.
KINETICS
• It is the study of inertia force which arise due to combined
effect

7. BALANCING
• Balancing is defined as the process of designing (or modifying) a
machine in which unbalanced forces is minimum.
• Balancing is the process of attempting to improve the mass
distribution of a body so that it rotates in its bearings without
unbalanced centrifugal forces.

8. OBJECTIVES OF BALANCING
1. That the center of gravity of the system remains stationery
during a complete revolution of the crank shaft and
2. That the couples involved in acceleration of the different moving
parts balance each other.

9. STATIC & DYNAMIC UNBALANCE

10. STATIC BALANCING
• Static balancing is a balance of forces due to action of gravity.
• A body is said to be in static balance when its center of gravity is
in the axis of rotation.

11. DYNAMIC BALANCING
• Dynamic balance is a balance due to the action of inertia forces.
• A body is said to be in dynamic balance when the resultant
moments or couples, which involved in the acceleration of
different moving parts is equal to zero.
• The conditions of dynamic balance are met, the conditions of
static balance are also met.

12. BALANCING OF ROTATING MASSES
• The process of providing the second mass in order to counteract
the effect of the centrifugal force of the first mass, is called
balancing of rotating masses.

13. Balancing of a Single Rotating Mass By a
Single Mass Rotating in the Same Plane
• We know that the centrifugal force
exerted by the mass m1 on the shaft,
FC1= m1.r1.w2
• Centrifugal force due to mass m2,
FC2 = m2.r2.w2
• Equating equations,
m1.w2.r1 = m2.w2.r2
m1.r1 = m2.r2

14. Balancing of a Single Rotating Mass By Two
Masses Rotating in Different Planes
• This masses satisfy the following two conditions of equilibrium.
1. The net dynamic force acting on the shaft is equal to zero. This is
the condition for static balancing.
2. The net couple due to the dynamic forces acting on the shaft is
equal to zero.
• The conditions (1) and (2) together give dynamic balancing.
• The following two possibilities may arise while attaching the two
balancing masses :
1. The plane of the disturbing mass may be in between the planes of
the two balancing masses, and
2. The plane of the disturbing mass may lie on the left or right of the
two planes containing the balancing masses.

15. Balancing of a Single Rotating Mass By Two
Masses Rotating in Different Planes
The plane of the disturbing mass may be in
between the planes of the two balancing
masses.
• Centrifugal force exerted by the mass
m, m1 and m2 are resp.,
FC= m.r.w2
FC1= m1.r1.w2
FC2= m2.r2.w2
• Since the net force acting on the shaft
must be equal to zero,
FC= FC1+FC2
m.r= m1.r1 + m2.r2

16. Balancing of a Single Rotating Mass By Two
Masses Rotating in Different Planes
The plane of the disturbing mass may be in between the planes of
the two balancing masses.
• Taking moment at P,
FC2.l = FC.l1
m2.r2.l = m.r.l

17. Balancing of a Single Rotating Mass By Two
Masses Rotating in Different Planes
When the plane of the disturbing mass
lies on one end of the planes of the
balancing masses.
• the following conditions must be
satisfied in order to balance the
system, i.e.
FC + FC2 = FC1
m.r + m2.r2 = m1.r1
• Taking moment at P, to find the
balancing force in the plane L
FC1.l = FC.l2
m1.r1.l = m.r.l2

18. Balancing of a Single Rotating Mass By Two
Masses Rotating in Different Planes
• Taking moments about Q, to find the balancing force in the plane
M,
FC2.l = FC.l1
m2.r2.l = m.r.l1

19. Balancing of Several Masses Rotating in the
Same Plane
The magnitude and position of
the balancing mass may be
found out by,
A. Analytical Method
B. Graphical Method

20. Balancing of Several Masses Rotating in the
Same Plane
A. Analytical Method
1. Find out centrifugal forces
2. Resolve the centrifugal forces horizontally and vertically and
find their sums
ΣH = m1 × r1 cosθ1 + m2 × r2 cosθ2 + …………
ΣV = m1 × r1 sinθ1 + m2 × r2 sinθ2 + …………
3. Magnitude of the resultant centrifugal force,
FC = ΣH 2 + (ΣV)2
4. If θ is the angle, which the resultant force makes with the
horizontal, then
tanθ = ΣV / ΣH

21. Balancing of Several Masses Rotating in the
Same Plane
5. The balancing force is then equal to the resultant force, but
in opposite direction.
6. Now find out the magnitude of the balancing mass, such that
FC = m× r
Where, m = Balancing mass, and
r = Its radius of rotation.

22. Balancing of Several Masses Rotating in the
Same Plane
B. Graphical Method
1. First of all, draw the space diagram with the positions of the several masses.
2. Find out the centrifugal force exerted by each mass on the rotating shaft.
3. Now draw the vector diagram with the obtained centrifugal forces, such that
ab represents the centrifugal force exerted by the mass m1 (or m1.r1) in
magnitude and direction to some suitable scale. Similarly, draw bc, cd and de.
4. Now, as per polygon law of forces, the closing side ae represents the resultant
force in magnitude and direction.
5. The balancing force is, then, equal to the resultant force, but in opposite
direction.
6. Now find out the magnitude of the balancing mass (m) at a given radius of
rotation (r), such that
m × w2 × r= Resultant centrifugal force
m.r = Resultant of m1.r1, m2.r2, m3.r3 and m4.r4

23. Balancing of Several Masses Rotating in
Different Planes
• The following two conditions must be satisfied :
1. The forces in the reference plane must balance, i.e. the
resultant force must be zero.
2. The couples about the reference plane must balance, i.e. the
resultant couple must be zero.

24. Balancing of Several Masses Rotating in
Different Planes

25. Balancing of Several Masses Rotating in
Different Planes
Plane
(1)
Mass (m)
(2)
Radius(
r)
(3)
Cent.force
w2
(m.r)
(4)
Distance
from
Plane L
(l) (5)
Couple ¸
w2
(m.r.l)
(6)
1 m1 r1 m1.r1
–l1
– m1.r1.l1
L(R.P.) mL rL mL.rL 0 0
2 m2 r2 m2.r2
l2
m2.r2.l2
3 M3 r3 m3.r3
l3
m3.r3.l3
M Mm rm mm.rm lM mm.rm.lm
4 m4 r4 m4.r4 l4 m4.r4.l4
The magnitude of the balancing masses mL and mM in planes L and M may be obtained as
discussed below :
• Take one of the planes, say L as the reference plane (R.P.). The distances of all the other
planes to the left of the reference plane may be regarded as negative, and those to the right
as positive.

26. BALANCING OF RECIPROCATING MASSES

27. • Thus the purpose of balancing the reciprocating masses is to
eliminate the shaking force and a shaking couple.
• In most of the mechanisms, we can reduce the shaking force and
a shaking couple by adding appropriate balancing mass, but it is
usually not practical to eliminate them completely.
• In other words, the reciprocating masses are only partially
balanced.
BALANCING OF RECIPROCATING MASSES

28. BALANCING OF RECIPROCATING MASSES

29. BALANCING OF RECIPROCATING MASSES
• Let, FR = Force required to accelerate the reciprocating parts,
FI = Inertia force due to reciprocating parts,
FN = Force on the sides of the cylinder walls or normal force acting
on the cross-head guides, and
FB = Force acting on the crankshaft bearing or main bearing.
• Since FR and FI are equal in magnitude but opposite in direction,
therefore they balance each other.
• The horizontal component of FB (i.e. FBH) acting along the line of
reciprocation is also equal and opposite to FI.
• This force FBH = FU is an unbalanced force or shaking force and
required to be properly balanced.

30. Primary and Secondary Unbalanced Forces of
Reciprocating Masses
• The acceleration of the reciprocating parts is approximately given
by the expression,
ar =r.w2 (cosθ +
cos2θ
𝑛
)
• Hence, Inertia force due to reciprocating parts or force required to
accelerate the reciprocating parts,
FI=FR= mass × acceleration = m.r.w2 (cosθ +
cos2θ
𝑛
)
• Unbalanced force,
FU= m.r.w2 (cosθ +
cos2θ
𝑛
= m.r.w2.cosθ + m.r.w2 cos2θ
𝑛
= FP + FS

31. BALANCING OF LOCOMOTIVES

32. Partial Balancing of Locomotives
• The locomotives, usually, have two cylinders with cranks placed at
right angles to each other in order to have uniformity in turning
moment diagram.
1. Inside cylinder locomotives: In the inside cylinder locomotives,
the two cylinders are placed in between the planes of two
driving wheels as shown in Fig.(a)
2. Outside cylinder locomotives: in the outside cylinder
locomotives, the two cylinders are placed outside the driving
wheels, one on each side of the driving wheel, as shown in
Fig.(b)

33. Partial Balancing of Locomotives

34. Effect of Partial Balancing of Reciprocating
Parts of Two Cylinder Locomotives
• Due to this partial balancing of the reciprocating parts, there is an
unbalanced primary force along the line of stroke
• also an unbalanced primary force perpendicular to the line of
stroke.
1. Variation in tractive force along the line of stroke
2. Swaying couple.

35. Variation of Tractive Force
• The resultant unbalanced force due to the two cylinders, along
the line of stroke, is known as tractive force.
Variation of tractive force.

36. Variation of Tractive Force
• Let m = Mass of the reciprocating parts per cylinder, and
• c = Fraction of the reciprocating parts to be balanced. We know
that unbalanced force along the line of stroke for cylinder 1
• = (1 – c)m.w2 .r cosq
• Similarly, unbalanced force along the line of stroke for cylinder 2,
= (1 - c)m.w2 × r cos(90° + q)
As per definition, the tractive force, FT = Resultant unbalanced
force along the line of stroke
= (1 - c)m.w2 .r cos q+ (1 - c)m.w2 .r cos(90° + q)
= (1 - c)m.w2 .r(cosq -sin q)

37. Variation of Tractive Force
• The tractive force is maximum or minimum when (cos q – sin q ) is
maximum or mini- mum. For (cos q – sin q ) to be maximum or
minimum,
d/d q (cos q- sin q) = 0 or
-sin q - cos q = 0 or
-sin q = cos q
tan q= -1 or q= 135° or 315°
Thus, the tractive force is maximum or minimum when q = 135° or
315°.
Maximum and minimum value of the tractive force or the variation
in tractive force
= ±(1 - c) m.w2 .r(cos135° - sin135°) = ± 2 (1 - c) m.w2 .r
= ±(1 - c).w2 .r
= ±(1 - c) m.w2 .r(cos135° - sin135°) = ± 2 (1 - c) m.w2 .r

38. Swaying Couple
Swaying couple.

39. Swaying Couple
• Let a = Distance between the centre lines of the two cylinders.
• = (1 - c)m.w2 .r cos q* a/2 (1 - c)m.w2 .r cos (90° + q) a/2
• = (1 - c)m.w2 .r * a/2(cos q + sin q)
• The swaying couple is maximum or minimum when (cosq+ sin q) is
maximum or minimum. For (cosq+ sin q) to be maximum or
minimum,
d /d q (cos q+ sin q) = 0
Or -sin q + cos q = 0
or
-sin q = - cos q
tan q = 1 or q= 45° or 225°
-

40. Swaying Couple
• Thus, the swaying couple is maximum or minimum when q = 45° or
225°.
• Maximum and minimum value of the swaying couple
=  (1 - c)m.w2 .r  a/2 (cos 45 + sin 45) =  a/ root(2) (1 - c)m.w2 .r
=
.

41. Balancing of Primary Forces of Multi-
cylinder In-line Engines
• The multi-cylinder engines with the cylinder center lines in the
same plane and on the same side of the center line of the
crankshaft, are known as In-line engines.
• The following two conditions must be satisfied:
1. The algebraic sum of the primary forces must be equal to zero.
2. The algebraic sum of the couples about any point in the plane of
the primary forces must be equal to zero.

42. Balancing of Primary Forces of Multi-
cylinder In-line Engines
• ‘For a multi-cylinder engine,
• the primary forces may be completely balanced by suitably
• arranging the crank angles, provided that the number of cranks
are not less than four’

43. Balancing of Secondary Forces of Multi-
cylinder In-line Engines
Secondary force.

44. Balancing of Secondary Forces of Multi-
cylinder In-line Engines
• When the connecting rod is not too long (i.e. when the obliquity of the
connecting rod is considered), then the secondary disturbing force due
to the reciprocating mass arises.
• We have discussed in Art. 22.2, that the secondary force,
𝐹𝑠 = 𝑚. 𝑤2. 𝑟 × cos
2𝜃
𝑛
• This expression may be written as,
•
𝐹𝑠 = 𝑚(2𝑤)2× r/4n × cos 2𝜃

45. Balancing of Secondary Forces of Multi-
cylinder In-line Engines
• The algebraic sum of the secondary forces must be equal to zero.
In other words,
• the secondary force polygon must close,
• The algebraic sum of the couples about any point in the plane of
the secondary forces must be equal to zero.
• In other words, the secondary couple polygon must close.

46. Balancing of V-engines
Balancing of V-engines.

47. Balancing of V-engines
• Consider a symmetrical two cylinder V-engine as shown in Fig.
• The common crank OC is driven by two connecting rods PC and QC.
• The lines of stroke OP and OQ are inclined to the vertical OY, at an
angle a as shown in Fig 22.33.
• Let m = Mass of reciprocating parts per cylinder,
• l = Length of connecting rod,
• r = Radius of crank,
• n = Ratio of length of connecting rod to crank radius = l / r
• q = Inclination of crank to the vertical at any instant,
• w = Angular velocity of crank.

48. Balancing of V-engines
• We know that inertia force due to reciprocating parts of cylinder
1, along the line of stroke
= 𝑚. 𝑤. 𝑟2[cos ∝ −𝜃 + cos
2 ∝−𝜃
𝑛
]
• and the inertia force due to reciprocating parts of cylinder 2,
along the line of stroke
= 𝑚. 𝑤. 𝑟2[cos ∝ −𝜃 + cos
2 ∝−𝜃
𝑛
]

49. Balancing of Radial Engine
Balancing of Radial Engine
• A radial engine is one in which all the cylinders are arranged
circumferentially as shown in Fig.
• These engines were quite popularly used in aircrafts during World
War II.

50. Balancing of Radial Engine
• we proceed with the assumption that all cylinders are identical
and the cylinders are spaced at uniform interval (
2𝜋
𝑛
)around the
circumference. Thus the crank angle for an cylinder will be given
by: 𝜃𝑖 = 𝜃1 + (𝑖 − 1)(
2𝜋
𝑛
)
• Inertia force due to reciprocating parts of the cylinder is given
by: 𝐼. 𝐹 = 𝑚. 𝑟2 [cos 𝜃𝑖 + 𝑟/𝑙 cos 2𝜃𝑖]
• Resolving this along global Cartesian X-Y axes, we have,
• 𝐼. 𝐹 = 𝑚𝑟2 [cos 𝜃𝑖 + 𝑟/𝑙 cos 2𝜃𝑖] cos[
𝑖−1 2𝜋
𝑛
]
• 𝐼. 𝐹 = 𝑚𝑟2 cos 𝜃𝑖 +
𝑟
𝑙 cos 2𝜃𝑖
sin[
𝑖−1 2𝜋
𝑛
]

51. THANK YOU
• When summed up for all the cylinders, we get the resultant inertia
forces for the whole radial engine.