This document discusses balancing of rotating masses. It defines balancing as reducing unbalance to an acceptable level or eliminating it entirely. There are two types of balancing: static and dynamic balancing of rotating masses, and balancing of reciprocating masses. Balancing of rotating masses involves attaching a mass to counteract the centrifugal force from an initial disturbing mass. Several cases are discussed, like balancing a single mass with one or two other masses in the same or different planes. Analytical and graphical methods to determine the required balancing masses and their positions are also presented.

1. ME8594 - DYNAMICS OF MACHINES
UNIT-II-BALANCING
(BALANCING OF ROTATING MASSES)
By,
Dr.S.SURESH,
Assistant Professor,
Department of Mechanical Engineering
Jayalakshmi Institute of Technology.

2. WHAT IS MEANT BY BALANCING
 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.

3. Different types of balancing
1. Balancing of rotating masses
I. Static balancing
II. Dynamic balancing
2. Balancing of reciprocating masses.

4. BALANCING OF ROTATING MASSES
Certain mass (m1) is attached to a rotating shaft
It exerts some centrifugal force
Bend the shaft and to produce vibrations
In order to prevent the effect of centrifugal force
Another mass (m2) is attached to the opposite
side of the shaft
centrifugal force of both the masses are made to
be equal and opposite.
rotating masses are balanced

5. The following cases are important from
the subject point of view:
1. Balancing of a single rotating mass by a
single mass rotating in the same plane.
2. Balancing of a single rotating mass by two
masses rotating in different planes.
BALANCING OF ROTATING MASSES

6. 3. Balancing of different
masses rotating in the
same plane.
4. Balancing of different
masses rotating in
different planes.

7. Let r1 = radius of rotation of the mass m1
m1 = a disturbing mass attached to a shaft rotating at
ω rad/s.
Centrifugal force exerted by the mass m1 =
1) Balancing of a single rotating mass by a single mass
rotating in the same plane.
m2 = balancing mass
For complete balance = FC1 = FC2
CASE 1

8. 3. BALANCING OF SEVERAL MASSES ROTATING IN THE SAME PLANE
m1, m2, m3 and m4 = Magnitude of masses (Kg)
r1, r2, r3 and r4 = Radius of rotation (m)
θ1, θ2, θ3 and θ4 = angles of these masses with the horizontal line OX
ω = Constant angular velocity (rad/s).
CASE 3

9. The magnitude and position of the balancing mass may be found out
analytically or graphically.
1. Analytical method
(i) Find out the centrifugal force
Fc = m . R
(ii) Resolve the centrifugal forces horizontally and vertically and
find their sums, i.e. ∑ H and ∑V
∑ H = m1 r1 cos θ1 + m2 r2 cos θ2 + m3 r3 cos θ3
∑V = m1 r1 sin θ1 + m2 r2 sin θ2 + m3 r3 sin θ3
(iii) Magnitude of the resultant centrifugal force

10. (iv) If θ is the angle, which the resultant force make
with the horizontal
tan θ = ∑V / ∑H
(v) The balancing force is then equal to the resultant
force, but in opposite direction.
(vi) Now find out the magnitude of the balancing
mass, such that

11. Example:
1. Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and
260 kg respectively. The corresponding radii of rotation are 0.2 m,
0.15 m, 0.25 m and 0.3 m respectively and the angles between
successive masses are 45°, 75° and 135°. Find the position and
magnitude of the balance mass required, if its radius of rotation is
0.2 m.
Given Data:
m1= 200 kg ; m2= 300 kg ; m3= 240 kg ; m4= 260 kg ;
r1= 0.2 m ;r2= 0.15 m ; r3= 0.25 m ; r4= 0.3 m ;
θ1 = 0° ;
θ2 = 45° ;
θ3 = 45° + 75° = 120° ;
θ4 = 45° + 75°+ 135° = 255° ;
r = 0.2 m
To find:
Magnitude and position of the balance mass (m & θ)

12. SOLUTION:
 Magnitude of centrifugal forces
1. Analytical method
 Resolving m1.r1, m2.r2, m3.r3 and m4.r4 horizontally,
 Resolving m1.r1, m2.r2, m3.r3 and m4.r4 vertically,
= m1 r1 cos θ1 + m2 r2 cos θ2 + m3 r3 cos θ3
= m1 r1 sin θ1 + m2 r2 sin θ2 + m3 r3 sin θ3

13.  Resultant centrifugal force
=
We know that ...
R = m.r = 23.2
 Direction of Resultant force
 Angle of the balancing mass from the horizontal mass of 200 kg
m = 116 kg Ans.
θ = 201.48° Ans.

15.  Graphical Method
1. Draw the space diagram showing the positions of all the given masses
as shown in Fig.
Space diagram
2. Centrifugal force of each mass

16. 3. Draw the vector diagram with the above values, to some suitable scale
Vector Diagram
Space Diagram
Scale: 1 cm: 10 kg.m
Measurement from Vector diagram
Find that
Balancing force is equal to resultant force but
opposite in direction
ae = Resultant Force = 23 kg-m

17. 4. BALANCING OF DIFFERENT MASSES ROTATING
IN DIFFERENT PLANES
 In order to have a complete balance of the several revolving masses 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.
CASE 4

18. m 1, m 2, m 3 & m 4 = Revolving in planes 1, 2, 3 and 4
m L & m M = Balancing masses in the plane M and N
Rp = Reference Plane
Position of planes of the masses Angular position of the masses
CASE 4. BALANCING OF DIFFERENT MASSES ROTATING IN DIFFERENT PLANES

19. EXAMPLE 1
A shaft carries four masses A, B, C and D of magnitude 200 kg, 300 kg, 400 kg and 200 kg
respectively and revolving at radii 80 mm, 70 mm, 60 mm and 80 mm in planes measured from A at
300 mm, 400 mm and 700 mm. The angles between the cranks measured anticlockwise are A to B
45°, B to C 70° and C to D 120°. The balancing masses are to be placed in planes X and Y. The
distance between the planes A and X is 100 mm, between X and Y is 400 mm and between Y and D is
200 mm. If the balancing masses revolve at a radius of 100 mm, find their magnitudes and angular
positions.
CASE 4- Balancing of different masses rotating in different planes
Given data: To find
m A = 200 kg , rA = 80 mm (i) Magnitude of Balancing masses (mX, mY)
m B = 300 kg, rB = 70 mm (ii) Angular positions of Balancing masses
m C = 400 kg, rC = 60 mm (θX , θY)
m D = 200 kg, rD = 80 mm

20. Solution:
Plane Mass (m)
(Kg)
Radius (r)
(m)
Cent. Force
(Fc=m.r)
(kg.m)
Distance from
RP (l) (m)
Couple
(m.r.l)
(Kg.m2)
A
X (RP)
B
C
Y
D
200
m
X
300
400
m
Y
200
0.08
0.1
0.07
0.06
0.1
0.08
16
0.1 m
X
21
24
0.1 m
Y
16
-0.1
0
0.2
0.3
0.4
0.6
-1.6
0
4.2
7.2
0.04 m
Y
9.6

21. • Couple Polygon
Scale:
1 cm = 1 kg.m 2
By measurement from couple polygon
0.04 m
Y
= Vector d
|
o
|
= 7.3 Kg-m
2
Angular position of mY

22.  Force Polygon
Scale:
1 cm = 5 kg.m
By measurement from force polygon Angular position of mX