MAURBHANJ SCHOOL OF ENGINEERING�BARIPADA : 757107 , ODISHA�

Branch : Mechanical Engineering

Semester : 4^th Sem

Subject : THEORY OF MACHINE

Topic : BALANCING OF MACHINE PARTS

Faculty : Er. Malabika Nayak

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Balancing

"is the process of attempting to improve the mass distribution

of a body so that it rotates in its bearings without unbalanced centrifugal forces”

WHAT IS BALANCING OF ROTATING MEMBERS?�

Balancing means a process of restoring a rotor which has unbalance to a balanced state by adjusting the mass distribution of the rotor about its axis of rotation

• Mass balancing is routine for rotating machines,some reciprocating machines and vehicles
• Mass balancing is necessary for quiet operation, high speeds , long bearing life, operator comfort, controls free of malfunctioning, or a "quality" feel

Rotating components for balancing

Unbalanced force on the bearing –rotor system

Bearing 1

Bearing 2

Shaft with

rotors

• Unbalance is caused by the displacement of the mass centerline from the axis of rotation.
• Centrifugal force of "heavy" point of a rotor exceeds the centrifugal force exerted by the light side of the rotor and pulls the entire rotor in the direction of the heavy point.
• Balancing is the correction of this phenomena by the removal or addition of mass

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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 a rotor which has unbalance

causes the following problems.�

• The whole machine vibrates.

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• Noise occurs due to vibration of

the whole machine.

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• Abrasion of bearings may shorten

the life of the machine.��

NEED FOR BALANCING

Rotating Unbalance occurs due to the following reasons.

� ● The shape of the rotor is unsymmetrical.�� ● Un symmetrical exists due to a

machining error. � ● The material is not uniform, especially in

Castings.

● A deformation exists 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

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• Unbalance due to

unequal distribution

of masses�

• Unbalance due to

unequal distance of masses

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. Types of Unbalance Static Unbalance Dynamic Unbalance�

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STATIC BALANCING

(SINGLE PLANE BALANCING)

Single plane balancing

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Adequate for rotors which are short in length,

such as pulleys and fans

O₂

m

r

ω

F = m r ω²

Magnitude of

unbalance

Vibration

occurs

Elasticity of the bearing

θ₁

θ₂

θ₃

m₄r₄ ω²

m₁r₁ ω²

m₂r₂ ω²

m₃r₃ ω²

Balancing of several masses

revolving in the same plane using a

Single balancing mass

bearing

m _b

m₁r₁ ω²

m₂r₂ ω²

m₃r₃ ω²

m₄r₄ ω²

m _b r _b ω²

θ_b

Graphical method of determination

magnitude and

Angular position of the balancing mass

Force vector polygon

O

m₁r₁ ω² cos θ₁+ m₂r₂ ω² cos θ ₂

+ m₃r₃ ω²cos θ ₃+ m₄r₄ ω² cos θ ₄

= m_b cos θ_b

m₁r₁ ω² sin θ₁+ m₂r₂ ω² sin θ ₂

+ m₃r₃ ω²sin θ ₃+ m₄r₄ ω² sin θ ₄

= m_b sin θ_b

magnitude ‘m _b’ and position ‘θ_b’ can be determined

by solving the above two equations.

Determination of magnitude and

Angular position of the balancing mass

Dynamic or "Dual-Plane" balancing

Dynamic balancing is required for components such as shafts and multi-rotor assemblies.

m r ω²

m r ω²

l

Brg A

Brg B

Statically balanced

but dynamically unbalanced

Load on each support Brg

due to unbalance = (m r ω² l)/ L

r

r

Dynamic or "Dual-Plane" balancing

On an arbitrary plane C

Several masses revolving in different planes

Apply dynamic couple on the rotating shaft

Dynamic unbalance

ω

Balancing of several masses rotating in different planes

F _a

F _b

F _c

F _d

A

B

C

D

L

M

End view

ω

l_a

l_b

l_c

l_d

d

A

B

C

D

L,

Ref plane

M

F_c

F_b

F_a

F_d

F_m

F _l

End view

side view of the planes

F_c

F_b

F_a

F_d

F_m =?

F _l =?

Couple polygon

force polygon

C_a

C_c

C_d

C_b

C_m=M_mr_md

F_a

F_b

F_c

F_d

F_m

F_l=M_l r_l

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From couple polygon, by measurement, C_m = M_m X r _m X d

From force polygon, by measurement, F_l = M_l X r_l

A shaft carries four masses in parallel planes A,B,C,&D in this order. 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 100 ^o and that between B & A is 190^o both angles measured in the same sense. The axial dist. between planes A & B is 10cm 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

Example :

A

B

C

D

10 cm

20 cm

l _d

18 kg

12.5 kg

M _a

θ =190 ^o

θ =100^o

End view

A

B

C

D

10 cm

20 cm

l _d

18 kg

12.5 kg

M _a

θ =190 ^o

θ =100^o

O

1080

2,250

8 M_d l_d= 2312 kg cm ²

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O

108

75

8 M_d =63.5 kg. cm

8 M_a = 78 kg .cm

Couple polygon

force polygon

θ_d= 203^o

M _d

From the couple polygon,

By measurement, 8 M_d l_d= 2,312 kg cm ²

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^∴ M_d l_d = 2312 / 8 = 289 kg cm

θ_d= 203^o

From force polygon,

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By measurement, 8 M_d = 63.5 kg cm

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8 M_a = 78.0 kg cm

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M_d = 7.94 kg

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M_a = 9.75 kg

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l_d = 289 /7.94 = 36.4 cm

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