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Machine Balancing

Here are the key steps to solve this problem: 1) Draw FBDs for rotors A and B showing centrifugal forces due to eccentricity 2) Resolve forces into x and y components 3) Sum forces in x and y directions and set equal to bearing forces 4) Solve for maximum speed such that bearing forces don't exceed 20 kN 5) Determine which bearing will fail first based on calculated forces

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Ruben D-ewallarz
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0% found this document useful (0 votes)
33 views19 pages

Machine Balancing

Here are the key steps to solve this problem: 1) Draw FBDs for rotors A and B showing centrifugal forces due to eccentricity 2) Resolve forces into x and y components 3) Sum forces in x and y directions and set equal to bearing forces 4) Solve for maximum speed such that bearing forces don't exceed 20 kN 5) Determine which bearing will fail first based on calculated forces

Uploaded by

Ruben D-ewallarz
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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BALANCING

Rotor
CLO3
Analyse and solve
problems in
balancing of rotating
and reciprocating
machineries
Balancing
a) Rotor
Balancing
a) Rotor
Balancing
a) Rotor
Balancing

 Balancing of rotating bodies is


important to avoid vibration
• Heavy industrial machines: Vibration cause
catastrophic failures

 Vibration
• Noisy and uncomfortable
• In car: unpleasant ride
Balancing

 Balanced system: 2 conditions


 Static balance
• No resulting centrifugal force and the
centre of gravity is on the axis of rotation.
• Σ 𝑚𝑟 = 0

 Dynamic balance
• No resulting couple along the axis.
• Σ 𝑚𝑟𝑑 = 0
Centrifugal Force
Case Study 1
 A vehicle of mass 𝑚 is taking a turn
with constant velocity, 𝑣
 Dynamic Analysis
𝐹 = 𝑚𝑎𝑛 𝑣 constant, 𝑎𝑡 = 0

[FBD] = [Kinetics Diagram]


𝑣2
Centrifugal Force, 𝐹 = 𝑚 = 𝑚𝜔2 𝑟
𝜌
Centrifugal Force
Case Study 2
 Consider a lump mass 𝑚 rotating with
a constant angular velocity, 𝜔
 Dynamic Analysis
𝐹 = 𝑚𝑎𝑛 𝑣 constant, 𝑎𝑡 = 0

[FBD] = [Kinetics Diagram]


𝑣2
Centrifugal Force, 𝐹 = 𝑚 = 𝑚𝜔2 𝑟
𝑟
Balancing 1 Plane
 For an unbalanced disc, the centre of gravity 𝐺 will
always rest below the centre of rotation when the
disc eventually stopped turning
 A balanced disc will stop at a random position

 How to balance?
• Balancing Force = Centrifugal Force
𝑚𝑏 𝜔2 𝑟𝑏 = 𝑚𝜔2 𝑟
𝑚𝑏 𝑟𝑏 = 𝑚𝑟 since 𝜔 is the same
𝑟𝑏
𝑟 Resultant unbalance + balancing force = 0
𝑚𝑟 = 0 vectorially
Unbalance Types
 Eccentricity
• 𝐺 of the disc located at an eccentricity 𝑒
• Effective mass: the mass of the disc, 𝑀𝑒
 Unbalance Mass
• Rotating mass that causes the shaft to be
unbalance
• Effective mass: the unbalance mass, 𝑚
 Addition or Removal of Mass
• The shaft is initially balanced
• Addition or removal of mass causes unbalance
• Effective mass: the added or removed mass, ±𝑚
Multi Plane
Balancing
 Unbalance mass located at different planes
 Couples tends to rock the shaft
 Two balancing masses are placed at two
different planes
 Must satisfy two conditions:
 Static balance
• Σ 𝑚𝑟 = 0
 Dynamic balance
• Σ 𝑚𝑟𝑑 = 0
Types of Problems
 Type 1: to balance the shaft by
adding correction masses at two
different planes

 Type 2: to determine the forces


acting on the bearings.
To Balance Shaft
 Solution Strategy
1. Identify the type of unbalance
2. Include all unbalance planes and the
two correction planes in the table
3. Ignore all bearings and planes already
in balanced condition
4. The reference plane is usually chosen
where there is an unknown quantity
Force on Bearings
 Solution Strategy
1. Identify the type of unbalance
2. Include all unbalance planes and the
two bearings in the table
3. Ignore all planes already in balanced
condition
4. The reference plane is chosen where
there is an unknown quantity, usually
it is one of the bearings
Question 1
The figure shows a completely balanced system comprising 6 discs,
𝐴, 𝐵, 𝐶, 𝐷, 𝐸, and 𝐹 of mass 10, 8, 6, 7, 9 and 5 kg respectively.
Some modifications are observed at discs 𝐴, 𝐶, 𝐷 and 𝐸 as follow:
Disc 𝐴 addition of 0.3 kg at radius 150 mm and 𝜃 = 0o
Disc 𝐶 removal of 0.35 kg at radius 100 mm and 𝜃 = 270o
Disc 𝐷 addition of 0.2 kg at radius 200 mm and 𝜃 = 180o
Disc 𝐸 removal of 0.3 kg at radius 100 mm and 𝜃 = 90o
If the shaft is rotating at a constant 1000 rpm, determine the
dynamics force acting on bearings 𝐿 and 𝑅.
𝐴 𝐵 𝐶 𝐷 𝐸 𝐹
𝐿 𝑅 90o

180o 0o

5 cm 5 cm 10 cm 5 cm 5 cm 5 cm 10 cm 270o
Solution Method
Plane 𝒎 𝒓 𝒎𝒓 𝒅 𝒎𝒓𝒅 𝜽 Direction

 Solution Method
• Polygon 𝑚𝑟𝑑 and 𝑚𝑟
• Resolutions method, 𝑚𝑟𝑑 = 0 and 𝑚𝑟 = 0
Question 2
The figure shows a shaft carrying 5 discs, 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 each of
mass 20 kg. Initially the centre of mass is at the axis of rotation.
Some modifications are made on the discs as follow:
Disc 𝐵 removal of 0.3 kg at radius 0.2 m and 𝜃 = 90o
Disc 𝐶 addition of 0.2 kg at radius 0.2 m and 𝜃 = 180o
Disc 𝐷 addition of 0.1 kg at radius 0.2 m and 𝜃 = 90o
For complete balance, determine the masses (magnitude and
direction) to be added to discs 𝐴 and 𝐸, both at radius 0.1 m.
𝐴 𝐵 𝐶 𝐷 𝐸

𝑋 𝑌 90o

180o 0o

270o
0.1 m 0.1 m 0.2 m 0.2 m 0.1 m 0.1 m
Question 4
The figure shows a shaft carrying a 40 kg mass motor 𝐶 and two
rotors 𝐴 and 𝐵 of mass 10 kg each. The shaft and motor are initially
balanced and supported by two bearings 𝐿 and 𝑅. If rotors 𝐴 and 𝐵
have a radius off eccentricity of 2 cm at 180o and 1 cm at 90o
respectively, determine the maximum operating speed of the shaft
if the strength of each bearing is 20 kN. State which bearing will fail
first. Neglect the static force on bearings.
𝐴 𝐶 𝐵

𝐿 𝑅 90o
𝐶. 𝐺
180o 0o

270o
20 cm 10 cm 30 cm 10 cm 20 cm

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