Chapter3 Overall

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Mechanics of Machines

BDA2033Lecture #05

University Tun Hussein Onn Malaysia (UTHM),

Faculty of Mechanical and Manufacturing,

Department of Mechanics


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Todays Objectives:

Students will be able to:

1. explain static and dynamic unbalance

2. solve problems relate to the static and dynamic balancing of

rotating machinery

Learning topics:

Introduction

Application

Balancing of rigid

rotor

CHAPTERS : BALANCING


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APPLICATION

A tire balancer machine

Wheel-balancing machines are use

to provide accurate balance. There

are two general designs : one type

spins the wheel it is off the vehicle

and the other spins the wheel while

one the vehicle. They both use the

vibrations that occur when the

wheel is rotating to locate any

unbalance in the wheel and tyre.

They check both static and dynamic

balance.


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APPLICATION (Cont.d)

There are 2 types of balance (and unbalance) for tire :

static and dynamic.

centrifugal force acting on a

balanced and unbalanced wheeleffect of static and dynamic

unbalance on a wheel

= mr2


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APPLICATION (Cont.d)

A wheel and tire in static balance, free to rotate on their axle, will remain in anyposition to which it is turned.

If out of balance, the heavy spot which causes the unbalance will rotate the wheel

until it is at the bottom.

This static unbalance can be balanced by fitting weights to the wheel rim directly

opposite the heavy spot.

Static unbalance

0CF (Shaking force)


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BALANCING OF RIGID ROTORS

F = mr2

= r

=

(a) A mass that is moving in a circular path undergoes angular acceleration, and

there is a dynamic force, referred to as centrifugal force, associated with the

acceleration. The centrifugal force is exerted by the mass on the rod, and is

transmitted to the bearing.

(b) The shaft is subject to centrifugal force because the center of mass of the

rotor does not lie on the shaft centerline

Introduction


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BALANCING OF RIGID ROTORS (Cont.d)

Static unbalance

Static unbalance caused by an eccentric mass on a rotating shaft

0CF


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Dynamic unbalance

Dynamic unbalance due to eccentric masses at multiple axial location on a

rotating shaft

0

0

C

C

M

F


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Static balancing

Figure 2

(a) a static unbalance can be

eliminated by the addition of a single

counterweight mc at the proper radial

distance rc and angle c.(b) graphical determination of the

counterweight size and location

Figure beside shows a rigid rotor

setup. The rotor is assumed to be

rotating with constant angular

velocity .



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0332211 CCrmrmrmrm

233

222

211 rmrmrmFtotal

022332

222

11 ccrmrmrmrmF

0coscoscoscos 333222111 CCCrmrmrmrm

0sinsinsinsin 333222111 CCCrmrmrmrm

Sum of a centrifugal force produces by the original masses :

To balance the rotor, sum of the vector forces must be zero :

The quantity of can be factored out :

The equation above can be solved mathematically by dividing it into y dan z

component :



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21

2333222111

2333222111

sinsin

coscoscos

rmainrmrm

rmrmrmrmCC

333222111

333222111

coscoscos

sinsinsinarctan

rmrmrm

rmrmrm

C

Solving for mCrC and C we have :



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EXAMPLE 1 (Static Balancing of a Rotor)

The rotor of the figure shown rotates with a speed of 40 rpm has the following

properties :

700.123m2

00.195m1

angle()r (m)m (kg)mass

Determine the amount and location of the

counterweight required for static balance.

Then calculate the centrifugal force generate by the counterweight.

Answer :

mcr

c= 11.25 kgm

c = 197.48

Fc = 19.7 N

70m1

m2r2

r1


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EXAMPLE 2 (Static Balancing of a Rotor)

The rotor of the figure shown has the following properties :

20035010m3

15030023m2

6025034m1

angle()r (mm)m (kg)mass



Answer :

mcr

c=9.3 kg.mm

c = 226

M4 15 170


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EXERCISE 1 (Static Balancing of a Rotor)


602m3

802m2

803m1

r (mm)m (kg)mass



Answer :

mcr

c= 227.8 kgmm

c = 297.0

1= 60

2= 150

3= 225


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Dynamic balancing

Figure 3(a) In general, dynamic balancing

requires the use of two counter-

weights. Shown are counter-

balances placed in arbitrarily

selected planes at axial position

P and Q.(b) Graphical determination of

counterweight 2.

(c) Graphical determination of

counterweight 1.

Figure beside shows a rigid rotor

setup. The rotor is assumed to be

rotating with constant angular

velocity .



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For static balance, the sum of all the centrifugal forces must be zero :

02211332211 cccc rmrmrmrmrm

0

2

22

2

11

2

33

2

22

2

11

ccccrmrmrmrmrmF


Additional condition for dynamic balance is that the sum of the moments of the

centrifugal forces about any arbitrary point on the rotor must be zero :

Here we will take moments about point P.

022

2232

3322

2212

11 ccc srmsrmsrmsrm


0222333222111 ccc srmsrmsrmsrm

0coscoscoscos 2222333322221111 cccc srmsrmsrmsrm

0sinsinsinsin 2222333322221111 cccc srmsrmsrmsrm

The equation above can be solved mathematically by dividing it into y dan z

component :



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21

2222333222111

2222333222111

11sinsinsin

coscoscoscos

ccc

ccc

cc

rmrmainrmrm

rmrmrmrmrm

222333222111

2223332221111

coscoscoscos

sinsinsinsinarctan

ccc

ccc

crmrmrmrm

rmrmrmrm

Solving for mC2rC2 and C2 we have :

21

2333322221111

2333322221111

222

sinsinsin

coscoscos1

rsmrsmrsm

rsmrsmrsm

srm

c

cc

333322221111

3333222211112

coscoscos

sinsinsinarctan

rsmrsmrsm

rsmrsmrsm

c

Next solving for mC1rC1 and C1 we have :



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Example 2EXAMPLE 1( Dynamic Balancing of a Rotor)


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EXAMPLE ( Dynamic Balancing of a Rotor)


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EXAMPLE 2 ( Dynamic Balancing of a Rotor)


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EXAMPLE 2( Dynamic Balancing of a Rotor)


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IN CLASS TUTORIAL (Dynamic Balancing of a Rotor)

170602m3

300902m2

60803m1

angle()r (mm)m (kg)mass


Determine the amount and location of the counterweight in plane A and B required

for complete balance.

Answer :

The total axial length is 1000 mm between

bearings. Counterweight are to be placed

in planes that are 100 mm from each bearing.

The axial distances are then

l1 = 200 mm, l2 = 500 mm, l3 = 700 mm,

and lR = 800 mm

mRrR = 32.2 kgmm

R = 57.7

mLrL = 147.9 kgmm

L = 222.5


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BALANCING OF RECIPROCATING MASSESBALANCING OF RECIPROCATING MASSES

Internal

engine


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BALANCING OF RECIPROCATING MASSESBALANCING OF RECIPROCATING MASSES


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PARTIAL BALANCING OF PRIMARY INERTIA FORCEPARTIAL BALANCING OF PRIMARY INERTIA FORCE


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PARTIAL BALANCING OF PRIMARY INERTIA FORCEPARTIAL BALANCING OF PRIMARY INERTIA FORCE


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BALANCING OF MULTIPLE CYLINDER ENGINEBALANCING OF MULTIPLE CYLINDER ENGINE INLINE ENGINEINLINE ENGINE


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B.Balancing for secondary inertia forceon your own


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