Balancing of rotating members

Balancing is a process of adjusting the

mass distribution in a rotor which has
unbalance
 Unbalance is caused by the disturbing force
due to offset of center of mass from the axis
of rotation.
 Centrifugal force is generated due to radial
acceleration of unbalanced mass
component.
 Balancing is the correction of this phenomena
by the removal or addition of mass
 rotating machines, reciprocating machines,

and vehicles

 Mass balancing is necessary for

quiet operation,

high speeds ,

long bearing life,

operator comfort,
 Benefits of balancing
 Increase quality of operation.

 Minimize vibration.

 Minimize audible and signal noises.

 Minimize structural fatigue stresses.

 Minimize operator annoyance and fatigue.

 Increase bearing life.

 Minimize power loss.

Rotating Unbalance occurs due to

● The shape of the rotor is unsymmetrical

● Eccentricity exists due to a machining error

● The material is not uniform (Castings)

● A deformation due to a distortion

An eccentricity exists due to a gap of fitting
An eccentricity exists in the inner ring of rolling bearing
Non-uniformity exists in either keys or key seats
Non-uniformity exists in the mass of flange
. Types of Unbalance

Static Unbalance

Dynamic Unbalance
 STATIC BALANCING
(SINGLE PLANE BALANCING)
 Single plane balancing

Adequate for rotors which are short in length,

such as pulleys and fans
F=mr
2

m
 Magnitude of
r unbalance
O2
Vibration
Elasticity of the bearing
occurs
Balancing of several masses revolving in the same plane using a Single
balancing mass

m3r3 2 y
m3  2
m2r2 
m2
2
1
3 m1 x
bearing m1r1 2

m4 mb
m4r4 2
Graphical method of determination magnitude and Angular position of the
balancing mass

m4r4 2

m3r3 2
b

m b r b 2
O m2r2 2
m1r1 2

Force vector polygon

Determination of magnitude and Angular position of the
balancing mass

m1r1 2 cos 1+ m2r2 2 cos  2

+ m3r3 2cos  3+ m4r4 2 cos  4
= mb cos b
m1r1 2 sin 1+ m2r2 2 sin  2
+ m3r3 2sin  3+ m4r4 2 sin  4
= mb sin b
magnitude ‘m b’ and position ‘b’ can be determined
by solving the above two equations.
A circular disc mounted on a shaft carries three
attached masses of 4 kg, 3 kg and 2.5 kg at radial
distances of 75 mm, 85 mm and 50 mm and at the
angular positions of 45o, 135o and 240o respectively.

The angular positions are measured ccw from the

reference line along x-axis.

Determine the amount of the counter-mass at a radial

distance of 75 mm required for the static balance
Four masses 200 kg, 300 kg, 240 kg and 260 kg at
corresponding radii of rotation 0.2 m, 0.15 m, 0.25
m and 0.3 m and angle between successive
masses are 45o, 75o and 135o.

Find the position and magnitude of the balance

mass required, if its radius of rotation is 0.2 m.
 Dynamic Balancing

When several masses rotates in different planes,

the centrifugal force, in addition to being out of
balance, also forms couple
 Dynamic or "Dual-Plane" balancing

Dynamic balancing is required for components

such as shafts and multi-rotor assemblies.
Dynamic or "Dual-Plane" balancing

Statically balanced
mr
2
but dynamically unbalanced

r
r
Brg A Brg B
l
mr
2

Load on each support Brg

due to unbalance = (m r  l)/ L
2
Several masses revolving in different planes
Apply dynamic couple on the rotating shaft

Dynamic unbalance
Balancing of several masses rotating in different planes

A B C D

Fc
End view Fb


Fd
Fa

L M
Plane Mass Radius Force / 2, Dist. From Couple / 2
M r M r =F , ref plane Mrl =C
( kg) (cm) (kg. cm) l , (cm) (kg cm 2)
A Ma ra Mara -la -Mara la

L Ml rl Ml rl 0 0
(Ref.plane)

B Mb rb Mbrb lb Mbrb lb
C Mc rc Mcrc lc Mcrc lc
M Mm rm Mmrm d Mmrmd

D Md rd Mdrd ld Mdrdld
 A B C D
Fm la
Fc
lb
lc
Fb
ld
Fa

Fl d
Fd
L, M
End view
Ref plane side view of the planes
 Fc
Fm =?
Cc
Fd
Fb
Fc
Fa Cb Fm
Cd Ca
Fl=Ml rl Fb

F l =? Cm=Mmrmd Fa

Fd Couple polygon force polygon

From couple polygon, by measurement, Cm = Mm X r m X d

From force polygon, by measurement, Fl = Ml X rl
For a multi plane unbalances as shown in Figure obtain the
balancing required
For dynamic balance the sum of forces and moments about any plane
should be zero
Tabular calculation for the moment about the plane L

0o
Tabular calculation for the moment about the plane R
 A shaft carries four masses in parallel planes A, B, C & D.
 The masses at B & C are 18 kg & 12.5 kg respectively and
each has an eccentricity of 6 cm.
 The masses at A & D have an eccentricity of 8 cm.
 The angle between the masses at B & C is 100o and that
between B & A is 190o both angles measured in the
same sense.
 The axial dist. between planes A & B is 10 cm and that
between B & C is 20 cm. If the shaft is complete dynamic
balance,

Determine,
1 masses at A & D
2. Distance between plane C &D
3. The angular position of the mass at D
 18 kg
A B C D
10 cm  =100o  =190 o

20 cm
12.5 kg

ld End view
Ma

Plane Mass Radius Force / 2, Dist. Couple / 2

M r Mr, From ref Mrl
kg cm kg. cm plane kg cm 2
l , cm
A Ma=? 8 8 Ma 0 0
B 18 6 108 10 1080
C 12.5 6 75 30 2250
D Md=? 8 8 Md ld=? 8 Md ld
 A B C D 18 kg

10 cm  =100o  =190 o

20 cm
12.5 kg
Md
ld
Ma

Couple polygon force polygon

75
2,250
108 8 Md =63.5 kg. cm
1080
d= 203o
8 Ma = 78 kg .cm
8 Md ld= 2312 kg cm 2
O
O
From the couple polygon,

By measurement, 8 Md ld= 2,312 kg cm 2

M = 2312 / 8 = 289 kg cm
d ld
d= 203o
From force polygon,

By measurement, 8 Md = 63.5 kg cm

8 Ma = 78.0 kg cm

Md = 7.94 kg

Ma = 9.75 kg

ld = 289 /7.94 = 36.4 cm

 Shaft with
rotors

Bearing 1 Bearing 2

Unbalanced force on the bearing –rotor system

Balancing of Reciprocating Engines
• Moving elements: Piston, crank, connecting
rod

• Vibrations due to
Periodic variations of gas pressure in cylinder
Inertia forces associated with moving parts
• Unbalanced Forces Due to Fluctuations in Gas Pressure

• Expanding gas in cylinder exerts force F on piston

• F can be resolved into F/cos Φ and F tan Φ.

• F/cos Φ induces torque Mt which rotates the crank shaft

• Figure shows equilibrium forces on stationary parts of

the engine  F 
M t   r cos
 cos 
• Total resultant force =0
Unbalanced Forces Due to Fluctuations in Gas Pressure
• Unbalanced Forces Due to Fluctuations
in Gas Pressure

r cos
• Resultant torque MQ=FhtanΦ where h
sin 

Fr cos
MQ 
cos
Torque induced at crank shaft is felt at engine
support.
• Unbalanced Forces Due to Inertia of Moving Parts

Acceleration of Piston

xP  r  l  r cos  l cos
 r  l  r cost  l 1  sin  2

But l sin   r sin   r sin t

r2
Hence cos  1  2 sin 2 t
l
Substituti ng into xP ,
2
r
xP  r  l  r cost  l 1  2 sin t
2

l
• Unbalanced Forces Due to Inertia of Moving Parts
Acceleration of Piston
r 1 
In general,  and 1   1
l 4 2
2
Hence xP  r 1  cost   sin 2 t
r
2l
 r   r 
or xP  r 1    r  cost  cos 2t 
 2l   4l 
 r 
x P  r  sin t  sin 2t 
 2l 
 r 
xP  r  cost  cos 2t 
2

 l 
Unbalanced Forces Due to Inertia of Moving Parts
Acceleration of the Crankpin
Vertical and horizontal displacements of crankpin C:

xC  OA  AB  l  r 1  cost 
yC  CB  r sin t
xC  r sin t
y C  r cost
xC  r cost
2

yC  r sin t

2
Unbalanced Forces Due to Inertia of Moving Parts
Acceleration of the Crankpin

Fx  m p xp  mc xc
r
 m p  mc r cost  m p
2 2
2
cos 2t
l
Primary part Secondary part

Fy  m p xp  mc xc
 mc r sin t
2

Primary part
Balancing of Reciprocating Engines

mc can be made zero by counterbalancing the crank. Hence Fy

can be reduced to zero.

mp always positive hence vertical unbalanced force Fx always

exists.

Thus single cylinder engine is inherently unbalanced.

In multi-cylinder engine, we can balance inertia forces by proper

arrangement of cranks
Balancing of Reciprocating Engine
Balancing of Reciprocating Engines

For force balance,

N
Fx total   Fx i  0
i 1

F    Fy i  0
N

y total
i 1

r 2 2
F   m  m  r cost     m 
x i p c i
2
i p i cos2t  2 i 
l
F   m  r sint   
y i c i
2
i

Assume (mp)i = mp and (mc)i = mc for i=1,2,…,N

Balancing of Reciprocating Engines

Assume (mp)i = mp and (mc)i = mc for i=1,2,…,N

When t=0, conditions for total force balance:

N N N

 cos 
i 1
i  0,  cos 2
i 1
i  0,  sin 
i 1
i 0

Moments about the z and x-axes:

M z   Fx i li  0, M z   Fy i li
N N

i 2 i 2

Necessary conditions for balancing of moments:

N N N

 l cos 
i 2
i i  0,  l cos 2
i 2
i i  0,  l sin 
i 2
i i 0
Balancing of Reciprocating Engines

Thus we can arrange the cylinders of a

multicylinder reciprocating engine to satisfy Eq

The engine will be completely balanced against

the inertia forces and moments.
In a two cylinder V engine the center lines of cylinder are
inclined at an angle of 45o on either side of the central
vertical line.

The mass of the reciprocating parts per cylinder is 0.5 kg

crank radius of 35 mm and connecting root length as 130
mm.

Show that the vertical force on this engine due to

secondary unbalance is absent.

Also show that if suitable mass is attached to the crank at

appropriate distance, primary inertia force can be totally
balanced.

Find the max value of secondary inertia force on this

engine for rpm = 3000
Primary Force Analysis
Total resultant primary force

Primary force can be completely balanced by placing

suitable mass at suitable radius diametrically opposite to
crank and having magnitude
 Secondary Force Analysis

Vertical force due to secondary unbalance is absent