lect balancing

7/23/2019 Lect Balancing

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ME 3507: Theory of Machines

Balancing

Dr. Faraz Junejo


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Ojec!i"es

Balance simple rotating objects

and pin jointed fourbar linkages


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#n!ro$uc!ion

Shaking forces and shaking torques

(Chapter 11) will lead to vibrations

Shaking forces and shaking torques areproduced by unbalanced rotating

membersnbalanced rotating mass used in

machinery where vibration is required! e"#vibration function in cell phone

$or common machinery! vibration is not adesired feature (leads to noise, faster wearand tear) and can be removed throughbalancing


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%layStation & 'ual Shock controller


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%ur&oses of BalancingMaking machine elements center of mass or

gravity coincide with their rotating centers

The center of mass of some machine may notcoincide with their rotating centers, the reasons

are: The asymmetry of the structure Uneven distribution of materials Errors in machining , casting and forging

Improper boring y keys y assembly


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'en!rifugal Force! particle made to travel along a circular

path generates a centrifugal force directedoutward along the radial line form thecenter of rotation to the particle

! s the particle rotates about the centerpoint! so does the centrifugal force


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'en!rifugal Force(con!$.)

! 'en!rifugal force is an inertia force and is

actually the body*+s reaction to an e"ternallyapplied force

! $or circular motion the e*!ernal force is knownas centripetal force ,he centripetal force acts on

the particle in a radially inward direction

! ,hey both have the same magnitude but di-er inthe direction of action


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'en!rifugal Force (con!$.)

! Similarly! a rotor with mass center slightly

displaced form the a"is of rotation will generatecentrifugal force

! ,his is the force associated with static unbalance

! ,he shaft supports counteract the forces ofunbalance *./ the e"ternally applied centripetalforce

! 0t should be noted here that the quantity m. r is knownas unbalanceand that centrifugal force is the product of

unbalance and angular velocity squared hileunbalance force (Fcentrifugal) increases rapidly with speed!

the unbalance quantity itself (m r) does not change atall


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nalance: 1u,,ary

rotor is unbalanced if its centre of gravity(S) or principal a"is of inertia does notcoincide with the a"is of rotation

'uring rotation the centre of gravity thenmoves on a circular pathand the rotor e"erts

centrifugal forces and moments on its frame

s system is rotating atsome constant angularvelocity ,he accelerationsof the masses will then bestrictly cen!ri&e!al (towardthe center) ! and the inertiaforces will be centrifugal(away from the center)


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nalance: 1u,,ary (con!$.)

0n addition to mass! the factors

governing the centrifugal force of a bodyare the radius of the circular path andthe angular velocity2speed

F = m r 2

s the centrifugal force increases with

the square of the speed! good balancing

is e"tremely important! particularly for

high.speed machines


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1!a!ic Balancing

pplies to things in motion even though

the word 3static+ is used

lso known as single2&lane alancing

pplied to rotating massesthat are in! or

nearly in the same plane (&' problem)

4"# Bicycle wheel! gear! grinding wheels!

fan propeller


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'o,,on fea!ure# ll objects are short inthe a"ial direction compared to the radialdirection,he a"ial dimension Bis smallercompared to the diameter D( usually B2'56& )

7assesalmost lie on a single plane8equirement for static balance# Sum of all

forces on system must be 9ero! :$ ; 6


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E*a,&le

! 0f the system is a simple disc then

static balance is all that is needed

! Consider a thin disc or wheel onwhich the centre of gravity is not the

same as the centre of rotation


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! 0f the centre of gravity is distance r fromthe centre of rotation then when it spins at

rad2s! centrifugal force is produced! ,his has a formula '.F.+ M /r where M

is the mass of the disc

! ,his is the out of balance force 0n order tocancel it out an equal and opposite force isneeded ,his is simply done by adding amass M2at a radius r2as shown ,he two

forces must have the same magnitudes


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! %lacing a suitable mass at a suitableradius moves the centre of gravity to thecentre of rotation

! ,his balance holds true at all speeds downto 9ero hence it is balanced so long as theproducts of M and r are equal and

opposite


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$igure shows a link in the shape of a vee

which is part of a linkage e want tostatically balance it

e can model this link dynamically as twopoint masses m1 and m2 concentrated at

the local !s of each


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,hese point masses eachhave a mass equal to thatof the s C?e can solve for the

required amount andlocation of a third


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e will set up a coordinate system with its origin atthe center of rotationand resolve the inertial forcesinto components in that system riting vector

equation ($ @ ma ; 6) for this system we get#


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Aote that the only forces acting on this system

are the inertia forces$or balancing! it does not matter what e"ternal

forces may be acting on the system

"#ternal forces cannot be balanced by makingany changes to the system>s internal geometry

Aote that the /!er,s is cancelle$! b2c forbalancing as discussed earlier! it also does not

matter how fast the system is rotating! onlythat it is rotating (will determine themagnitudes of these forces! but we will maketheir sum to be 9ero anyway for balancing)

e,e,er 44


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1!a!ic Balancing (con!$.)


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1!a!ic Balancing (con!$.)


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nce a combination of ,and

is chosen! it remains to

design the physical

counterweight ,he chosen ra$ius is the

distance from the pivot to theC? of whatever shape we createfor the counterweight mass

ur simple dynamic model!used to calculate the ,

&ro$uc!! assumed a point massand a massless rod,hese idealdevices do not e"ist

possible shape for thiscounterweightis shown in$igure 1&.1c 0ts mass must bemb! distributed so as to place its

C? at radius$bat angle b


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E*a,&le: 6i"en :,he system shown in the

right Dgure has the followingdata#

m1;1&kg 81;11EFm

GH11EIJ

m&;1Kkg 8&;6K&&m GHIKKJ

L;I6rad2sec

Fin$:,he mass.radius product

and its angular location needed tostatically balance the system%lease note desired value for$b%&'(&)m

E l 6 ( !$ )


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(1) esol"e !he &osi!ion "ec!ors in!o *8 y

co,&onen!s:

6+6.635,9663.;8 6*+20.;56,8

6y+6.0;/,

/+0.


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(E) 0f a value for $b%&'(&)misdesired! the mass required for this

counterweight design is#

mb+(/.;0/g2,)C(0.,)

+/.


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E*ercise: 6

! ,hree masses ! B and C are placed on a

balanced disc as shown at radii of 1&6mm! 166 mm and K6 mm respectively ,hemasses are 1 kg! 6F kg and 6N kgrespectively $ind the Ith mass which

should be added at a radius of M6 mm inorder to statically balance the system

ns?er:The massrequired is 2'* kg at

2&('+o

counterclockwise ofve #-a#is'


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1!a!ic BalanceC1ingle %laneBalancesually the counterweightmb are placed atas large a rotating radius as is practicable to

minimi9e the amount of the added mass

n alternative approach could be by

removing mass in the direction of imbalance(eg drilling a hole as the open circlemb)!rather than by adding counterweights to it

ny number of masses in a disk-like rotor

can be balanced by adding a single mass orremoving a mass at an appropriate position

,his is also called single2&lane alance.


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Dyna,ic Balancing

lso known as two-plane balancing

pplied to rotating objects that arerelatively longer in the a"ial directioncompared to the radial direction

4"# Car tire! squirrel cage fan!turbine rotor

l


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E*a,&le

! Consider & masses statically balanced as

shown but acting at di-erent places alongthe a"is

! 0t is clear that even with static balance!centrifugal force will produce a turning momentabout the centre of gravity for the system

$or s!a!ic alance

Mr+ MBrB


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!0n this case! the problem is solved byadding equal and opposite forces atthe two points as shown

E*a,&le (con!$.)

Consider the assembly in $igure ,wo equal


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Consider the assembly in $igure ,wo equal

masses are at identical radii! 1K66apart

rotationally! but separated along the shaft length


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summation of inertial forces due to theirrotation will be always 9ero (ie staticallybalanced) Oowever! in the side view! their

inertia forces form acouplewhich rotateswith the masses about the shaft

,he couple will act on the frame and tend

to produce rotation vibration of the frame,hecri!erionfor the balancing in such

cases is#

Both the vector sum of all inertia forces andthe vector sum of all moments of inertia

forces about any point must be 9ero! ie

PFi;6 and PMi;6


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Dyna,ic Balancing (con!$.)

euire,en!s for $yna,ic alancing#

@ Sum of all forces must be 9ero! :$ ; 6@ Sum of all moments must be 9ero! :7; 6

8equires addition of balancing weights in

two correction planes separated by some

distance along the shaft !o &ro"i$e a

coun!er cou&le to cancel the unbalanced

moment

E*a,&le: Dyna,ic


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E*a,&le: Dyna,icalancing

hen an automobile tire and wheel is

dynamically balanced! the two correction

planes are the inner and outer edges of the

wheel rim

Correction weights are added at the proper

locations in each of these correction planes

based on a measurement of the dynamic

forces generated by the unbalanced! spinning

wheel


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Dyna,ic alancing: 1u,,ary (con!$.)

C id th t f th d


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Consider the system of three masses arrangedaround and along the shaft in $igure 1&.E

e then create two correction planes labeledAand

B 0n this design e"ample! the unbalanced massesm1! m&! mEand their radii 81! 8&! 8Eare known

along with their angular locations 1! &! and E

E' di t t i li d ith th


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E' coordinate system is applied with thea#is of rotationin the Q direction

Aote that the system has again beenstopped in an arbitrary free9e.frameposition ngular acceleration is assumed tobe 9ero ,he summation of forces is#


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4 ti 1& I b


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4quation 1&I c becomes


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Dyna,ic Balancing (con!$.)

d th i l l !i i d t


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and their angular loca!ions required todynamically balance the system using thecorrection planes . and /


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E*a,&le: / (1u,,ary)

Balancing inages


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Balancing inages

Balancingforces

the C7 of the objectto coincide with thecenter of rotation

C7 of objectbecomes stationaryas a result of this

Static balancing isin e-ect making theC7 of the objectstationary

Balancing inages (con!$ )


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Balancing inages (con!$.)

Balancingforces

the C7 of the objectto coincide with thecenter of rotation

C7 of objectbecomes stationaryas a result of this

Static balancing isin e-ect making theM of the ob0ectstationary

Balancing inages (con!$ )


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Balancing inages (con!$.)

e will employ static balancing on a linkage!

because mass of fourbar linkage is distributed ina single plane

C7 of entire linkage will moveas linkage moves


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inage Balancing 1!ra!egy

$orce C7 of entire linkage! Mtto be

s!a!ionary'o this by adding & balancing weights! mA

and mB! to links & and I (because they are

in pure rotation)


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Balancing of ; ar linage

! ,otal mss of the system is given by

,! + ,/ = ,3= ,;! Aow the total mass moment about the

origin & must be equal to the sum of

mass moments due to individual links! ie

PMO2

; mt8t; m&8&R mE8ER mI8I

here! 8&! 8Eand 8Iare individual C7 oflinks located in global system centered at&using position vectors

o a o n age


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o a o n age#.r.t. $2

Gec!ors e&resen!ing in eng!hs


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Gec!ors e&resen!ing in eng!hs


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4"pression for /


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4"pression for /

4"pression for 3


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4"pression for ;


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4"pression for ;


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4"pression for B


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4"pression for B

4"pansion of inkage M equation


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4"pansion of inkage M equation


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oca!ion an$ ,ass of alancing ?eigh!s


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g g

Tand Ugives orientation of 8and 8Brelative &

and I respectively

ny combination of mand 8are acceptable solong as their product is equal to mA&A(same with

mB&B)

E*ercise: 6


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$igure shows an unbalanced fourbar linkage

Balance this linkage using two balancingweights of m; mB; 166gm %lease note!

all dimensions are in mm! unless statedotherwise

,he distance and angular orientation of thesecond balancing weight mB! relative to link

I has been determined for you,herevalues are shown in Dgure

'etermine the distance8and angular

orientation of the Hrs! alancing ?eigh!,! relative to link &

E*ercise: 6

E*ercise: 6


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E*ercise: 6


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lect balancing

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