The document explains the concepts of static and dynamic balance in rotating systems with two masses in a plane. Static balance is achieved when the moments about a pivot are equal, while dynamic balance occurs when the inertia forces act along the same line and are equal in magnitude. Methods for balancing rotors, such as turbine disks or car wheels, are discussed along with a sample problem illustrating the application of these principles.

1. BALANCING
By: E/C Sapid, Mark Gil

2. topics
Dynamic
balance
(two
masses in
a plane)

3. Static Balance
The condition which exists in a body which has an
absolutely even distribution of the weight mass around the
axis of rotation
This occurs when when there is no resultant
centrifugal force and the center of gravity is on the axis of
rotation.

4. Two masses
Static
balance
(two masses
in a plane) Static Balance
in a plane
Consider a light arm pivoted freely at the fulcrum O and
carrying masses m1, m2 at distances r1, r2 from O
respectively. In general the arm will rotate about O and the
system is said to be out of balance. For equilibrium there
must be balance of moments about O. i.e.
m1g x r1 = m2g x r2
or
m1r1 = m2r2
When in balance the arm may be set in any position and will
remain at rest in that position. The weights are said to be in
static balance and the center of gravity of the system is
located at O.

5. Static
balance
(two masses
in a plane)

6. Dynamic Balance
The condition which exists in a rotating body when
the axis about which it is forced to rotate, or to which
reference is made, is parallel with a principal axis of inertia;
no products of inertia about the center of gravity of the body
exist in relation to the selected rotational axis.
This occurs when there is no resulting moment
along the axis.

7. Dynamic
balance (two
masses in a
Two masses
Dynamic Balance
in a plane
plane)
Now consider two light arms fixed to a shaft at bearing O
and rotating with angular velocity ω. The arms are in the
same plane and carry masses m1, m2 at radii r1, r2
respectively. Owing to the rotation each masses exerts an
inertia force radially outward on the bearing O.
The force due to m1 is m1 ω2 r1 (OA in the force diagram)
The force due to m2 is m2 ω2 r2 (AB in the force diagram)

8. Dynamic
balance (two
masses in a
plane)

9. Dynamic
balance (two The resultant out of balance force on the bearing is
masses in a given by OB in the force diagram.
plane)
when the dynamic load on the bearing is zero the
rotating system is said to be in dynamic balance. The
condition for no load at O is that two inertia force shall:
1. act along the same straight line but with opposite
sense
2. Be equal in magnitude
the relative positions of the masses are as in the
figure C shown earlier. The condition for equal inertia
forces is:
m1 ω2 r1 = m2 ω2 r2
Thus, since ω2 is the same both masses
m1r1 = m2r2
Two bodies in the same plane are in static balance when
pivoted about a given axis they will be in dynamic balance
at any speed when rotating about the same axis.

10. Method of
balancing Balancing
Method of
Rotors
Rotors
It was shown earlier that for a two mass system to be in
static balance the mr product for each mass had to be the
same. This is also the condition for the masses to be
balanced when rotating and suggests a method for ensuring
balance for rotating rotors such as turbine disks or car
wheel assemblies.

11. Method of
balancing In practice it is usually possible to balance a rotor to an
Rotors accuracy of 0.001 m-kg, i.e. the amount of residual
unbalance is equivalent to a mass of 1 kg at 1mm radius.
For a rotor of mass 10 tons this is equivalent to a
displacement x of he center of gravity from the axis of
rotation given by:
10 x 1000 x X = 0.0001
X = 10-7 m or 0.1 μm
The corresponding out-of-balance centrifugal force when
running at 3600 rpm is
m1 ω2 r1 = 10 x 1000 x 2Л 3600 x 10-7 kg-m/s2
60
= 142 N
It is usual to limit the out-of-balance force to be greater than
1 percent of the rotor weight.

12. Sample
Problem

13. Sample
A shaft carries two rotating masses of 1.5 kg and
Problem
0.5 kg, attached at radii 0.6 m and 1.2 m, respectively, from
the axis of the masses are shown in the figure below. Find
the required angular position and radius of rotation r of a
balance mass of 1 kg.
If no balance mass is used what is the out-of-
balance force on the shaft bearing at 120 rpm?

14. Sample
Problem Solution:
The mr values are 0.9 kg-m for A and 0.6 kg-m for B and
these are represented by oa and ab, respectively, in the
force polygon, figure b. The resultant out-of-balance mr
value is given by ob in direction o to b. From a scale
drawing:
ob = 1.31 kg-m
the equilibrant is equal and opposite to the out-of-balance
force, and since the mr value for the balance mass is:
1 x r kg-m
Therefore, for balance
1 x r = 1.31 and r = 1.31 m
Thus the radius of rotation of the balance mass is 1.31 m

15. Sample
Problem The balance mass must be positioned so that its inertia
force is acting in direction b to o, i.e. at an angle of 156°36’
to the radius of mass A, as shown in figure c. If no balance
mass is used
Out-of-balance force = ob x w2
= 1.31 x 2Л 120 kg-m/s2
60
= 207 N

16. Thank you
Static balance is keeping balanced while still and dynamic balance is keeping balance while moving .