Short Notes: Balancing of Reciprocating and Rotating Masses | Short Notes for Mechanical Engineering PDF Download

 Page 1


Balancing of Reciprocating and Rotating Masses 
Balancing
• Balancing Is defined as the process of designing a machine in which 
unbalance force is minimum. The rotating and reciprocating parts of a high 
speed engine if are not properly balanced, the dynamic forces will be setup.
Balancing of rotating Masses
• If the centre of mass of rotating machines does not lie on the axis of rotation, 
the inertia force is given by Fi = m u2 e
where, m = mass of the machine 
c o = angular speed of the machine
e = eccentricity i.e., the distance from the centre of mass to the axis of 
rotation.
Internal and external Balancing
• Let a shaft carry an unbalanced mass mi with the centre of mass located at 
distance e from the axis of rotation. The shaft can be completely balancing by 
adding a mass m- 1 at a distance e -i from the axis of rotation diametrically 
opposite to m so that,
mu)2e = m^u)2e ^
(where, = speed of rotation of the shaft)
me = m^e^
In this case R A = 0, Rs = 0
Page 2


Balancing of Reciprocating and Rotating Masses 
Balancing
• Balancing Is defined as the process of designing a machine in which 
unbalance force is minimum. The rotating and reciprocating parts of a high 
speed engine if are not properly balanced, the dynamic forces will be setup.
Balancing of rotating Masses
• If the centre of mass of rotating machines does not lie on the axis of rotation, 
the inertia force is given by Fi = m u2 e
where, m = mass of the machine 
c o = angular speed of the machine
e = eccentricity i.e., the distance from the centre of mass to the axis of 
rotation.
Internal and external Balancing
• Let a shaft carry an unbalanced mass mi with the centre of mass located at 
distance e from the axis of rotation. The shaft can be completely balancing by 
adding a mass m- 1 at a distance e -i from the axis of rotation diametrically 
opposite to m so that,
mu)2e = m^u)2e ^
(where, = speed of rotation of the shaft)
me = m^e^
In this case R A = 0, Rs = 0
UIUUIU
e
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Balancing diagram
The shaft will also be free from dynamic bending stress. This situation is referred 
to as internal balancing.
The dynamic reactions R a and R b can be reduced to zero, by adding two balancing 
masses m y and m2 at distances ei and e2 respectively from the shaft in the same 
axial plane as m but on the opposite side of the axis of rotation.
me = m7 e 7 + m2e2 
mea = m-ie-iai + m2e2a
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Describe balancing diagram
• This situation is referred as external balancing where the entire length of 
the shaft is not from dynamic bending stresses.
Static Balancing
• If a shaft carries a number of unbalanced masses such that the centre of 
mass of the system is said to be statically balanced.
• Statically balancing is defined as below
+ m2 r; +m.r3 + m (rc = 0 
T « r + m crc = 0 
Ewrcos 9 + m c re cos9 = 0 
5Z w : rsin 9+ m e r( sin 9 = 0
tan0f =
— T^mrsin 9
— 2 Z m r cos 9
Static balancing system
Dynamic Balancing
Page 3


Balancing of Reciprocating and Rotating Masses 
Balancing
• Balancing Is defined as the process of designing a machine in which 
unbalance force is minimum. The rotating and reciprocating parts of a high 
speed engine if are not properly balanced, the dynamic forces will be setup.
Balancing of rotating Masses
• If the centre of mass of rotating machines does not lie on the axis of rotation, 
the inertia force is given by Fi = m u2 e
where, m = mass of the machine 
c o = angular speed of the machine
e = eccentricity i.e., the distance from the centre of mass to the axis of 
rotation.
Internal and external Balancing
• Let a shaft carry an unbalanced mass mi with the centre of mass located at 
distance e from the axis of rotation. The shaft can be completely balancing by 
adding a mass m- 1 at a distance e -i from the axis of rotation diametrically 
opposite to m so that,
mu)2e = m^u)2e ^
(where, = speed of rotation of the shaft)
me = m^e^
In this case R A = 0, Rs = 0
UIUUIU
e
//////////
•
vvvwww
A
® 1 A
B
m 1
Balancing diagram
The shaft will also be free from dynamic bending stress. This situation is referred 
to as internal balancing.
The dynamic reactions R a and R b can be reduced to zero, by adding two balancing 
masses m y and m2 at distances ei and e2 respectively from the shaft in the same 
axial plane as m but on the opposite side of the axis of rotation.
me = m7 e 7 + m2e2 
mea = m-ie-iai + m2e2a
11 a '
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uuiuut e
vw vvvfa ;
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Describe balancing diagram
• This situation is referred as external balancing where the entire length of 
the shaft is not from dynamic bending stresses.
Static Balancing
• If a shaft carries a number of unbalanced masses such that the centre of 
mass of the system is said to be statically balanced.
• Statically balancing is defined as below
+ m2 r; +m.r3 + m (rc = 0 
T « r + m crc = 0 
Ewrcos 9 + m c re cos9 = 0 
5Z w : rsin 9+ m e r( sin 9 = 0
tan0f =
— T^mrsin 9
— 2 Z m r cos 9
Static balancing system
Dynamic Balancing
• A system of rotating masses in dynamic balance when there does not exist 
any resultant centrifugal force as well as resultant couple.
mx rx -I- m,r, 
mlrll1= m2 r2 l2 
Y. w r + m c rc le = 0
Dynamic balancing system
Balancing of Several Masses in Different Planes
• Let there be a rotor revolving with a uniform angular velocity c j, mi, m2 and m3 
are the masses attached to the rotor at radii r-1 , r2 and r3 respectively. The 
masses m-1 , m2 and m3 rotate in planes 1, 2, 3 respectively. In order to 
complete balancing, Let introduce two counter masses
in .
and
m<
at radii
and
respectively.
Balancing in different planes with masses 
Balancing Forces
According to figure, the balancing forces are defined as
Page 4


Balancing of Reciprocating and Rotating Masses 
Balancing
• Balancing Is defined as the process of designing a machine in which 
unbalance force is minimum. The rotating and reciprocating parts of a high 
speed engine if are not properly balanced, the dynamic forces will be setup.
Balancing of rotating Masses
• If the centre of mass of rotating machines does not lie on the axis of rotation, 
the inertia force is given by Fi = m u2 e
where, m = mass of the machine 
c o = angular speed of the machine
e = eccentricity i.e., the distance from the centre of mass to the axis of 
rotation.
Internal and external Balancing
• Let a shaft carry an unbalanced mass mi with the centre of mass located at 
distance e from the axis of rotation. The shaft can be completely balancing by 
adding a mass m- 1 at a distance e -i from the axis of rotation diametrically 
opposite to m so that,
mu)2e = m^u)2e ^
(where, = speed of rotation of the shaft)
me = m^e^
In this case R A = 0, Rs = 0
UIUUIU
e
//////////
•
vvvwww
A
® 1 A
B
m 1
Balancing diagram
The shaft will also be free from dynamic bending stress. This situation is referred 
to as internal balancing.
The dynamic reactions R a and R b can be reduced to zero, by adding two balancing 
masses m y and m2 at distances ei and e2 respectively from the shaft in the same 
axial plane as m but on the opposite side of the axis of rotation.
me = m7 e 7 + m2e2 
mea = m-ie-iai + m2e2a
11 a '
K
uuiuut e
vw vvvfa ;
© 1 ®21
: m y m2C
1111111111
t ,
wwww
®2
Describe balancing diagram
• This situation is referred as external balancing where the entire length of 
the shaft is not from dynamic bending stresses.
Static Balancing
• If a shaft carries a number of unbalanced masses such that the centre of 
mass of the system is said to be statically balanced.
• Statically balancing is defined as below
+ m2 r; +m.r3 + m (rc = 0 
T « r + m crc = 0 
Ewrcos 9 + m c re cos9 = 0 
5Z w : rsin 9+ m e r( sin 9 = 0
tan0f =
— T^mrsin 9
— 2 Z m r cos 9
Static balancing system
Dynamic Balancing
• A system of rotating masses in dynamic balance when there does not exist 
any resultant centrifugal force as well as resultant couple.
mx rx -I- m,r, 
mlrll1= m2 r2 l2 
Y. w r + m c rc le = 0
Dynamic balancing system
Balancing of Several Masses in Different Planes
• Let there be a rotor revolving with a uniform angular velocity c j, mi, m2 and m3 
are the masses attached to the rotor at radii r-1 , r2 and r3 respectively. The 
masses m-1 , m2 and m3 rotate in planes 1, 2, 3 respectively. In order to 
complete balancing, Let introduce two counter masses
in .
and
m<
at radii
and
respectively.
Balancing in different planes with masses 
Balancing Forces
According to figure, the balancing forces are defined as
E » ir + m t r + m c r =0
Balancing Couples
• Let be
placed in reference plane and 
be /2 from reference plane
Y,mri cos9 + m c r ( /f cos0, = 0
E»J rl sin 9 + m t r c l c sin 9 , = 0
. E m W sin#
tan#c = -¦ = --------------
1 22mrl cos#
— (E mrl s'm O+rn, r cos9, )
tan 9 C = -----—------------------= —¦ ----- - 1 -
— (Ew r'cos^+ni r cos 9( )
• By complex method,
m lr lllL9l+ m 2r 1l1Z92 + m 3r }l 3Z .9} + m t r c J tiZ .0 t =0 
J2mrlZ.9+ m ; r ( l{ Z0f = 0
Y.mrZ.6 + E '» t r e <6 — 0 
Balancing of Reciprocating Mass
• This diagram shows the condition of the balancing of reciprocating mass
and
m raf 
cos 0
m ' 
2 m
Primary force
t
mria2 sin 0
Balancing of reciprocating mass
Force required to accelerate mass is
Page 5


Balancing of Reciprocating and Rotating Masses 
Balancing
• Balancing Is defined as the process of designing a machine in which 
unbalance force is minimum. The rotating and reciprocating parts of a high 
speed engine if are not properly balanced, the dynamic forces will be setup.
Balancing of rotating Masses
• If the centre of mass of rotating machines does not lie on the axis of rotation, 
the inertia force is given by Fi = m u2 e
where, m = mass of the machine 
c o = angular speed of the machine
e = eccentricity i.e., the distance from the centre of mass to the axis of 
rotation.
Internal and external Balancing
• Let a shaft carry an unbalanced mass mi with the centre of mass located at 
distance e from the axis of rotation. The shaft can be completely balancing by 
adding a mass m- 1 at a distance e -i from the axis of rotation diametrically 
opposite to m so that,
mu)2e = m^u)2e ^
(where, = speed of rotation of the shaft)
me = m^e^
In this case R A = 0, Rs = 0
UIUUIU
e
//////////
•
vvvwww
A
® 1 A
B
m 1
Balancing diagram
The shaft will also be free from dynamic bending stress. This situation is referred 
to as internal balancing.
The dynamic reactions R a and R b can be reduced to zero, by adding two balancing 
masses m y and m2 at distances ei and e2 respectively from the shaft in the same 
axial plane as m but on the opposite side of the axis of rotation.
me = m7 e 7 + m2e2 
mea = m-ie-iai + m2e2a
11 a '
K
uuiuut e
vw vvvfa ;
© 1 ®21
: m y m2C
1111111111
t ,
wwww
®2
Describe balancing diagram
• This situation is referred as external balancing where the entire length of 
the shaft is not from dynamic bending stresses.
Static Balancing
• If a shaft carries a number of unbalanced masses such that the centre of 
mass of the system is said to be statically balanced.
• Statically balancing is defined as below
+ m2 r; +m.r3 + m (rc = 0 
T « r + m crc = 0 
Ewrcos 9 + m c re cos9 = 0 
5Z w : rsin 9+ m e r( sin 9 = 0
tan0f =
— T^mrsin 9
— 2 Z m r cos 9
Static balancing system
Dynamic Balancing
• A system of rotating masses in dynamic balance when there does not exist 
any resultant centrifugal force as well as resultant couple.
mx rx -I- m,r, 
mlrll1= m2 r2 l2 
Y. w r + m c rc le = 0
Dynamic balancing system
Balancing of Several Masses in Different Planes
• Let there be a rotor revolving with a uniform angular velocity c j, mi, m2 and m3 
are the masses attached to the rotor at radii r-1 , r2 and r3 respectively. The 
masses m-1 , m2 and m3 rotate in planes 1, 2, 3 respectively. In order to 
complete balancing, Let introduce two counter masses
in .
and
m<
at radii
and
respectively.
Balancing in different planes with masses 
Balancing Forces
According to figure, the balancing forces are defined as
E » ir + m t r + m c r =0
Balancing Couples
• Let be
placed in reference plane and 
be /2 from reference plane
Y,mri cos9 + m c r ( /f cos0, = 0
E»J rl sin 9 + m t r c l c sin 9 , = 0
. E m W sin#
tan#c = -¦ = --------------
1 22mrl cos#
— (E mrl s'm O+rn, r cos9, )
tan 9 C = -----—------------------= —¦ ----- - 1 -
— (Ew r'cos^+ni r cos 9( )
• By complex method,
m lr lllL9l+ m 2r 1l1Z92 + m 3r }l 3Z .9} + m t r c J tiZ .0 t =0 
J2mrlZ.9+ m ; r ( l{ Z0f = 0
Y.mrZ.6 + E '» t r e <6 — 0 
Balancing of Reciprocating Mass
• This diagram shows the condition of the balancing of reciprocating mass
and
m raf 
cos 0
m ' 
2 m
Primary force
t
mria2 sin 0
Balancing of reciprocating mass
Force required to accelerate mass is
E - 2 /\ 2 C O S d
F = m r a) cos a - m r o -------
----------- v ----------- - ft
Pr pr.an acctltranKg poet -------------v----------- *
StzoKdan occtltram g poet
Primary Balancing of Reciprocating Mass
• If c is the fraction of the reciprocating mass, primary force balanced by the 
mass
= cmr u)2 cos 9
• Primary force unbalanced by the mass
= (l-c)mr u)2 cos 9
• Vertical component of centrifugal force which remains unbalanced
= cmr u)2 sin Q
• Resultant unbalanced force at any instant
=J [(l -c ) m r o' cos d)\ +(cm r a 1 sin d)1
The resultant unbalanced force is minimum when
1
c= -
" J
Secondary Balancing of Reciprocating Mass
• The secondary balancing of reciprocating mass is defined as the given below.
• Secondary force
> cos 2d 
= m ra~--------
n
mr( 2a)1
cos 2d 
4 n
• Primary forces must balance i.e., primary force polygon is enclosed.
• Primary couples must balance i.e., primary couple polygon is enclosed.
• Secondary forces must balance i.e., secondary force polygon is enclosed.
• Secondary couples must balance i.e., secondary couple polygon is enclosed.
Balancing of Inline Engine
• The in-line engine is a multi-cylinder engine in which the line of stroke of all 
reciprocating parts are placed parallel to each other. The net force acting on 
inline is equal to zero. For this, the line of action of the forces must be same, 
or in other words, the centre of mass of the system lie on the line.
• Primary force = mr w2[cos 9 + cos(18O°+0)]=O
• Primary couple = mr c j2I cos 9
Maximum value at 9 = 0° and 180