Unit: V
Balancing and Vibration
Chapter: 12
Balancing of Rotating Masses
Learning Objectives
While reading and after studying this chapter, you will be able to:
• Understand the importance of balancing in machines.
• Distinguish static balance with dynamic balance using appropriate examples.
• State conditions for the complete balancing of rotating masses.
• Describe the balancing of several rotating masses in the same plane.
• Explain how the different masses rotating in different planes are balanced.
SYLLABUS
Balancing single rotating mass - Balancing of several coplanar rotating masses - Static and
dynamic balancing - Balancing of several masses rotating in different planes.
INTRODUCTION
• A machine has many moving parts. Some machines have rotating motion and some of them
have reciprocating motion. If these moving parts are not in complete balance, inertia forces
are set up which may cause excessive noise, vibration, wear and tear of the system.
• Thus, balancing is the process of designing or modifying machinery so that the unbalance
is reduced to an acceptable level and if possible is eliminated entirely.
• When a particle or mass moving in a circular path, it experiences a centripetal force acting
radially inwards. An equal and opposite force acting radially outwards on the axis of rotation
and is known as centrifugal force. This is a disturbing force, and its magnitude remains
constant but the direction changes with the rotation of the mass.
The centrifugal disturbing force, F =mω r
C
2
where
m = Mass of the rotating component in kg,
ω = Angular velocity of the component in rad/s = 2πN/60,
N = Speed of the component in rpm., and
r = Distance of C.G. of mass from the axis of rotation in metres.
• In fact, due to modern trend of development of high speed machines and vehicles, the problem of
balancing becomes greater importance for mechanical engineers. This type of unbalance problem is very
common in steam turbine rotors, engine crankshafts, rotary compressors, centrifugal pumps, etc.
1. Types of Balancing
Study of the problem of balancing can be divided into two classes:
1. Balancing of rotating masses, and
2. Balancing of reciprocating masses.
In this chapter the balancing of rotating parts of machines are considered.
BALANCING OF SINGLE ROTATING MASS
Consider a mass of m attached to shaft rotating at o rad/s, as shown in Fig.12.1(a). Let r be
the radius of rotation (i.e., distance between the axis of rotation of the shaft and the C.G. of
the mass m) of the mass 'm'.
We know that the centrifugal force (i.e., disturbing force) FC = m ω2 r, producing out-of-
balance effect acting radially outwards on the shaft. This out-of-balance force can be
balanced in any one of the following two ways.
1. By Introducing Single Revolving Mass in the Same Plane
The disturbing mass m is balanced by introducing a counter mass or balancing mass m at B
radius of rotation r diametrically opposite to m in the same plane, rotating with same angular
B
velocity ω rad/s, as shown in Fig.12.1(b).
We know that, Disturbing force, FC1 = m ω2 r The value of rB may be kept larger to reduce the
and Balancing force, FC2 = mB ω2 rB value of balancing mass mB.
But for balancing, FC1 = FC2 or m ω2 r = mB ω2 Note
rB
The product mB rB or m r is very often called
as the mass moment.
2. By Introducing Two Revolving Masses in Different Planes
Sometimes it is not possible to introduce balancing mass in the same plane in which
disturbing mass m is placed. In that case two masses can be placed in different planes.
If the balancing mass and disturbing mass lie in different planes, disturbing mass cannot be
balanced by a single mass as there will be a couple left unbalanced. In such cases, at least
two balancing masses are required for complete balancing and the three masses are arranged
in such a way that the resultant force and couple on the shaft are zero, as shown in Fig.12.2.
Let m = Mass of the disturbing body acting in plane A,
m and m = Masses of the two balancing bodies acting in plane B and C respectively,
B1 B2
r, r and r = Distance of C.G of m, m and m from the axis of rotation respectively,
B1 B2 B1 B2
l = Distance between the planes A and B,
1
l = Distance between the planes A and C, and
2
l = l + l = Distance between the planes B and C.
1 2
We know that the masses m, m and m experience the centrifugal forces as F , F , and F
B1 B2 C C1 C2
which are given as
(i) For balancing of the system, the centrifugal force of the disturbing mass must be equal to
the sum of centrifugal forces of the balancing masses.
(ii) We also know that, for complete balance of the system, the sum of moments must be
zero.
Taking moments about C, we get
Using equations (i) and (ii), we can find the balancing masses m and m .
B1 B2
BALANCING OF SEVERAL MASSES ROTATING IN THE SAME
PLANE (BALANCING OF SEVERAL COPLANAR ROTATING
MASSES)
If several masses are rigidly attached to a shaft at different radii in one plane perpendicular to
the shaft and the shaft is made to rotate, each mass will set up out-of-balance force on the
shaft. In this case, complete balance can be obtained by placing only one balancing mass in
the same plane.
Illustration: Consider any number of masses (say, three masses) of magnitude m , m and m
1 2 3
at distances of r , r and r from the axis of the rotating shaft. Their relative positions are
1 2 3
indicated by angles θ , θ and θ as shown in Fig.12.3(a).
1 2 3
The magnitude and position of the balancing mass can be obtained analytically or
graphically as discussed below.
1. Analytical Method
Step 1: First of all, find out the centrifugal force (or mass moments i.e., the product of the
mass and its radius of rotation) exerted by each mass on the rotating shaft.
• Since all the masses are connected to the shaft, all will have the same angular velocity o,
we need not to calculate the actual magnitude of centrifugal force, therefore F = mr.
C
Step 2: Resolve the centrifugal forces horizontally and vertically. Then find their sums Σ F H
and Σ Ε V
Step 3: Find the magnitude of the resultant centrifugal force (F ).
CR
Step 4: The balancing force is equal to the resultant force, but in opposite direction. Now
find out the magnitude of the balancing mass, such that
where m = Balancing mass in kg, and
B
r = Its radius of rotation in metres.
B
Step 5: The angle (θ ) made by the resultant force F with the horizontal is given by
R CR
We know that the balancing mass is placed in the opposite direction to the resultant force
direction. Therefore, the angle (θ ) made by the balancing mass with the horizontal is given
B
by
2. Graphical Method
Step 1: First, draw the space diagram with the position of the several masses, as shown in
Fig.12.3(a).
Step 2: Find out the centrifugal force (ie., the product of the mass and its radius of rotation)
exerted by each mass on the rotating shaft.
Step 3: Now draw the force polygon with the obtained centrifugal forces by polygon law of
forces, such that vector oa represents the centrifugal force exerted by the mass m (or m r )
1 1 1
in magnitude and direction to some suitable scale.
Similarly, draw vectors ab and bc to represent centrifugal forces of other masses m and m 2 3
(or m r and m r ).
2 2 3 3
Step 4: According to polygon law of forces, the closing side vector oc (from o to c and not
from c to o) represents the resultant force in magnitude and direction, as shown in
Fig.12.3(b). The balancing force is equal to the resultant force, but in opposite direction.
Now find out the magnitude of the balancing mass må at a given radius of rotation r , such B
that
m r Resultant centrifugal force = Vector oc.
B B
• It may be noted that space diagrams need not to be drawn to the scale. Since the diagram is
only to show the angle of inclination of several masses, therefore the angle of inclinations
must be marked accurately.
Step 5: To obtain the position of balancing mass in space diagram, draw a line parallel to the
resultant force but in opposite direction. Now measure the angle of inclination of this line in
the same sense from the horizontal line OX, which gives the angle of inclination of the
balancing mass (θ ). B
Note
If the force polygon closes in loop, then the system is called the balanced system.
Example 12.1
A rigid-rotor has all its unbalance in one plane and can be considered to consist of three
masses m = 5 kg, m = 3 kg at an angle 165° counter clockwise from mɲ, and m = 8 kg at
1 2 3
angle 85° clockwise from my. The radii r = 20 cm, r = 8 cm, r = 14 cm. Determine the
1 2 3
balancing mass required at a radius of 10 cm. Specify the location of this mass with
respect to m . 1
Given data:
m = 5 kg; m = 3 kg; m = 8 kg; r = 20 cm = 0.2 m; r = 8 cm = 0.08 m; r = 14 cm = 0.14 m;
1 2 3 1 2 3
r = 10 cm = 0.1 m.
B
Solution:
This problem can be solved by both analytical and graphical methods, as below.
Method 1: Analytical Method
Step 1: To find the centrifugal forces
We know that the magnitude of centrifugal force is proportional to the product of mass and
its radius of rotation, therefore
Step 2: To resolve the centrifugal forces horizontally and vertically.
Resolving the centrifugal force horizontally, we get
Here the angles θ , θ and θ should be measured in the same direction i.e., either in clockwise
1 2 3
direction or in counter clockwise direction. Assuming the mass m lies horizontally, we get θ
1 1
= 0°. Given that m is at an angle 165° counter clockwise from m , therefore θ = 165°. But m
2 1 2 3
is at an angle of 85° measured clockwise from m , therefore θ = 360° - 85° = 275°.
1 3
Resolving the centrifugal forces vertically, we get
Step 3: To find the magnitude of the resultant centrifugal force
Resultant centrifugal force is given by
Step 4: To find the magnitude of balancing mass
We know that the balancing mass m r must be equal to resultant centrifugal force by
B B
magnitude.
Step 5: To find angle made by the balancing mass
We know that the angle made by resultant centrifugal force with the horizontal line,
Method 2: Graphical Method
Step 1: The space/configuration diagram with the position of the given three masses can be
drawn as shown in Fig. 12.4(a).
Step 2: The centrifugal force exerted by each mass are found as
Step 3: Now the force polygon i.e., vector diagram can be drawn to some suitable scale (say,
1 cm = 0.2 kg.m), as shown in Fig. 12.4(b).
[First draw the vector oa that represents m r = 1 kg.m parallel to OA. Then from a draw
1 1
vector ab that represents m r = 0.24 kg.m parallel to OB; from b draw vector bc that
2 2
represents m r = 1.12 kg.m parallel to OC.]
3 3
Step 4: We know that the closing side of the force polygon (vector co) represents the
resultant force by magnitude and direction.
By measurement from force polygon, we get
F = vector co = 1.4 kg.m
CR
Step 5: Angle of inclination of balancing mass is determined by drawing a line parallel to
vector co in the space diagram, as shown in Fig.12.4(a) (in dotted lines). Now 0B can be
measured from space diagram as θ = 130° from m measured in counter clockwise
B 1
direction Ans.
STATIC VS DYNAMIC BALANCING
1. Static Balancing
• A system of rotating masses is said to be in static balance if the combined mass centre of
the system lies on the axis of rotation.
• Condition for static balancing: The net dynamic force acting on the shaft is equal to
zero. In other words, the centre of the masses of the system must lie on the axis of rotation.
2. Dynamic Balancing
• When several masses rotate in different planes, in addition to the out-of-balance centrifugal
forces, the couple also produced. A system of rotating masses is in dynamic balance when
there does not exist any resultant centrifugal forces as well as resultant couple._
• Conditions for dynamic balancing:
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.
In other words, the algebraic sum of moments about any point in the plane must be zero.
