jefcott model with ofset disc

8/18/2019 Jefcott Model With Ofset Disc

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2.5 A Jefcott Rotor Model with an Ofset Disc

Figure 2.20(a) show a more general case of the Jecott rotor when the rigid disc is

placed with some oset from the mid-span. With a and b locate the position of the disc

in a shaft of length l. The spin speed of the shaft is considered as constant. For such

rotors apart from two transverse displacements of the center of disc i.e. x and y the

tilting of disc a!out the x and y -a"is i.e. φ x andφ y also occurs# and it ma$es the rotor

s%stem as a four &'Fs. For the present anal%sis the rotar% inertia of the disc is

considered however the eect of the g%roscopic moment has !een neglected. n Fig.

2.20(!) points C and G represent the geometrical center and the center of gravit% of

disc respectivel%. The angle Φ represent the phase !etween the force and the

response.

Jecott rotor with an oset disc


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From Figure 2.20(!) we can have the following relations for the eccentricit%

(2.)

where e" and e% are components of the eccentricit% e in the x and y -directions

respectivel% (in fact these components of eccentricit% are in the plane of disc that is

inclined).

From Figure 2.20(c) e*uations of motion of the disc in the y - and φ x directions can !e

written as

(2.+)

and

(2.,)

where m is the disc mass Id is the diametral mass moment of inertia a!out the x -

a"is f y is the reaction force and M yz is the reaction moment. t should !e noted that the

moment is ta$en a!out the point . From a!ove e*uations it can !e o!served that

e*uations are non-linearl% coupled with the angular (titling) component of

displacement φ x .


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(2.)

and

(2.+0)

where Id is the diametral mass moment of inertia a!out the y -a"is f x is the reaction force

and M zx is the reaction moment. /*uations (2.) and (2.+0) are also non-linearl%

coupled with the angular component of displacement φ y . owever two transverse

planes (i.e. y - z and z - x ) motions are not coupled and that will allow two-plane motion to

anal%1e independent of each other i.e. set of e*uations (2.+) and (2.,) and

e*uations (2.) and (2.+0) can !e solved independent of each other.

n!alance forces can !e simpli3ed (i.e. !% lineari1ation) with the assumption of small

angular displacement (i.e. cosφ x 4 cosφ y 5 6) and e*uations (2.+) and (2.) can !e

simpli3ed as

(2.+6)


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and

(2.+2)

7ow e*uations (2.+6) (2.,) (2.+2) and (2.+0) are assem!led as

(2.+8)

which can !e written in matri" notation as

(2.+9)

With

where :M; represents the mass matri"


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shaft displacements at the disc location with the help of in@uence coeAcients as

(Timoshen$o and Boung 6,)

(2.+C)

where represent the displacement at ith station due to a unit force at jth station

$eeping all other forces to 1ero. t should !e noted that the displacement and force

terms are used as general sense so that displacement can !e a linear or an angular

displacement whereas the force can !e a force or a moment. The coupling of the force

and the displacement in two orthogonal planes has not !een considered !ecause of the

s%mmetr% of the shaft. /*uation (2.+C) can !e written in a matri" form as

(2.+)

with

where EI is the !eam @e"ure length parameters a and b are de3ned in Figure 2.26(a)

with . From the simple !eam de@ection theor% we can get these in@uence coeAcients

(Timoshen$o and Boung 6,). /*uation (2.+) can !e written as

(2.++)

where k ij is the stiness coeAcient and de3ned as force at ith station due to a unit

displacement at jth station $eeping all other displacements to 1ero. Dimilarl% since theshaft is s%mmetric a!out its rotation a"is we can o!tain

(2.+,)

/*uations (2.++) and (2.+,) can !e com!ined in matri" form as


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(2.+)

with

7oting e*uation (2.+) the nonlinear reaction force vector ta$es the following form

(2.,0)

E!ove e*uation contains product of the linear and angular displacements which ma$es

the s%stem e*uations as nonlinear. The present anal%sis considers onl% linear s%stems

so contri!utions from these nonlinear terms can !e ignored with the assumption of

small displacements. 'n su!stituting reactions forces and moments from e*uation

(2.+) into e*uations of motion i.e. e*uation (2.+9) we get

(2.,6)

with

2.C.6? Calculation of natural freuencie!? For o!taining natural fre*uencies of the

s%stem the determinant of the d%namic stiness matri" :; 4 (:G; - "2:H;) should !e

e*uated to 1ero and solved for " which gives four natural fre*uencies of the rotor

s%stem. t should !e noted that since two orthogonal plane motions are uncoupled (i.e.


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corresponding to y and φ x and x andφ y ). hence e*uations of motion of each plane could

!e solved independentl% This would ma$e the si1e of : # ; matri" to half. t will !e

illustrated through e"amples su!se*uentl%. Hore general method !ased on the eigen

value pro!lem will !e discussed in su!se*uent sections.

2.C.2? $nbalance forced re!%on!e? The un!alance forcing with fre*uenc% " can !e

written as

(2.,2)

where


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the phase of response rather than the time histor%. The present method gives the

response in fre*uenc% domain. When the damping term is also present the a!ove

un!alance response procedure can easil% handle additional damping term and the

d%namic stiness will ta$e the following form

(2.,+)

where :C; is the damping matri". t should !e noted that : # ; is now a comple" matri"

and !% the numerical simulation critical speeds can !e o!tained !% noticing pea$s of

responses while var%ing the spin speed of the shaft. The procedure for o!taining

damped natural fre*uencies will !e discussed su!se*uentl%. The anal%sis of the present

section is e*uall% valid for other !oundar% conditions. The onl% change would !e the

e"pressions of in@uence coeAcients corresponding to new !oundar% conditions (e.g.

cantilever 3"ed-3"ed free-free overhang etc.).

2..C.8? 'earin( reaction force!? Iearings are in the present stud% assumed to transmit

onl% forces and not moments. Forces transmitted through !earings are those which are

related to the de@ection of the shaft as shown in Figure 2.22 on the y - z plane.

'n ta$ing moments a!out ends (left) and K (right) of the shaft we have

(2.,,)

and

(2.,)


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Lonsidering a single plane ( y - z ) motion and neglecting the rotational displacement Φ x the natural fre*uenc% can !e o!tained as (refer section 2.C.6)

which gives

Example 2.6? '!tain transverse natural fre*uencies of an oset Jecott rotor s%stem asshown in Figure 2.2C. Ta$e the mass of the disc m 4 60 $g the diametral mass momentof inertia Id 4 0.02 $g-m2 and the disc is placed at 0.2C m from the right support. Theshaft has the diameter of 60 mm and total length of the span is 6 m. The shaft is

assumed to !e massless. se the in@uence coeAcient method. Ta$e shaft BoungNsmodulus E 4 2.6 M 6066 7>m2. 7eglect the g%roscopic eect and ta$e one plane motiononl%.

)olution* n@uence coeAcients for a linear and angular diaplacements ( y φ)correspoding to a force (f ) and a moment (M) acting at the disc are de3ned as

For the present pro!lem onl% single plane motion is considered. For free vi!ration frome*uation (2.,6) we get


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t can !e seen that there is a small dierence in the fundamental natural fre*uenc% dueto pure translation motion (2.C rad>s) with that of the fundamental natural fre*uenc%of the coupled s%stem (2.9 rad>s) and a large dierence in the natural fre*uenc% forthe pure tilting motion (6,, rad>s) with the second natural fre*uenc% of the coupleds%stem (20 rad>s).

(iii) &or the +exible !haft and ri(id bearin(! (Hethod 2)? 7ow the in@uence coeAcientmethod is used. Iearing forces are given as

where the reaction forces from the disc can !e e"pressed as

with

where

&isplacement vectors are related with the un!alance force as

with


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n view of a!ove e*uations !earing reaction forces can !e written as

From a!ove e*uations we have

and

which is same as !% previous method. t would !e interesting to var% the spin speed andplot the !earing forces with it. t should !e noted since the disc is at the mid-spanhence there is no contri!ution of the diametral mass moment of inertia on to !earingreactions. f there had !een couple un!alance then the diametral mass moment of inertia would have aected !earing reactions. Es an e"ercise ta$e the disc location fromthe left support a 4 0.8 l and o!tain !earing !earings for the same.

Example 2.8. Find the transverse natural fre*uenc% of a rotor s%stem as shown inFigure 2.2. Lonsider the shaft as massless and is made of steel with 2.6(60)66 7>m2 of the BoungNs modulus and +,00 $g>m8 of the mass densit%. The disc has 60 $g of themass. The shaft is simpl% supported at ends.

)olution* Lonsidering onl% the linear displacement 3rst we will o!tain the stiness (or

the in@uence coeAcient ) for Figure 2.80 using the energ% method. 'n ta$ing theforce and moment !alances we have


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which gives reaction forces as

Iending moments are o!tained at various segments of the shaft to get the strainenerg% of the s%stem. 'n ta$ing the moment !alance in the free !od% diagram asshown in Figure 2.86 of a shaft segment for 0.0 O x O 0. we get

(a)

'n ta$ing the moment !alance in the free !od% diagram as shown in Figure 2.82 of theshaft segment for 0. O x O 6.0 we get


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(!)

The strain energ% is e"pressed as

The linear displacement is e"pressed as

'n su!stituting !ending moment e"pression from e*uations (a) and (!) we get

The stiness is given as

where

which gives the natural fre*uenc% as

t should !e noted that the tilting motion of the disc has not considered. For the coupledlinear and angular motions natural fre*uencies of the s%stem can !e o!tained as ane"ercise !% o!taining corresponding in@uence coeAcients.

Example 2.9 '!tain the !ending natural fre*uenc% for the s%nchronous motion of arotor as shown in Figure 2.88. The rotor is assumed to !e 3"ed supported at one end.

Ta$e mass of the disc m 4 6 $g. The shaft is assumed to !e massless and its length anddiameter are 0.2 m and 0.06 m respectivel%. Ta$e shaft BoungNs modulus E 4 2.6 M6066 7>m2.


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)olution? et us assume for simplicit% that there is no coupling !etween the linear andangular motions. Lonsidering onl% the linear displacement the transverse stiness forthis case would !e

(a)

with

(!)

ence the natural fre*uenc% would !e

2.6 Alternatie !a" o# $%ppression o# &ritical $peeds

n the present section an interesting phenomenon will !e dealt in which a critical speedwill !e shown to !e eliminated !% suita!l% choosing s%stem parameters. For thispurpose the Jecott rotor model with a disc oset has !een chosen. 7ow for a detailedin depth anal%sis a closed form e"pression for the response is o!tained !% de3ningfollowing comple" displacements

(2.9)

/*uations of motion (2.,6) can !e written as

(2.C)


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and

(2.)

with . et the solution !e

where md is the mass of the disc and Id is the diametral mass moment of inertia. et thesolution !e

(2.+)

where R and Φr

are the translational and rotational whirl amplitudesrespectivel%# Φr and Φφare the phase of the translational and rotational whirl amplitudesrespectivel% (these are all real *uantities)# so that

(2.,)

'n su!stituting e*uations (2.+) and (2.,) into e*uations of motion (2.C)-(2.) weget

(2.)

(2.600)

/*uation (2.600) can !e e"pressed as

(2.606)

'n su!stituting e*uation (2.606) into e*uation (2.) we get

(2.602)


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'n e*uating the real and imaginar% parts of !oth sides of e*uation (2.602) we get

(2.608)

and

(2.609)

From e*uation (2.609) we get

(2.60C)

which means there will not !e an% phase dierence !etween the force and theresponse. 'n su!stituting phase information in e*uation (2.609) we get

(2.60)

which is the whirl amplitude and the condition of resonance can !e o!tained !%e*uating the denominator of e*uation (2.60) to 1ero

(2.60+)

where "cr represents the critical speed. I% de3ning

(2.60,)

/*uation (2.60+) can !e written as

(2.60)

The solution of the a!ove pol%nomial can !e e"pressed as


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or

(2.660)

which gives critical speeds of the rotor s%stem (the outer most ne(ati,e !i(n has nomeaning since fre*uenc% can not !e negative). ence for the case when the rotor is notmounted at the mid-span there are two critical speeds due to coupling of the linear andangular displacements. The a!ove solution (i.e. e*uation (2.660)) can !e more criticall%anal%sed as follows. t can !e seen that terms inside the 3rst s*uare root is alwa%s

positive i.e. since it can !e rearranged as

(2.666)

t can !e seen that the a!ove condition !e alwa%s true since all individualterms "r "P "rφ and"Φr are the real *uantit%. owever if the following condition is validfor terms inside the 3rst s*uare root

(2.662)

then it gives two real critical speeds since e*uation (2.609) gives two real roots.owever if the following condition prevails

(2.668)

then it gives onl% one real critical speed since the other root will !e comple". Figures2.89 (a) and (!) give these two cases respectivel%. t can !e seen that for the 3rst casetwo distinct pea$s corresponds to two critical speeds. For the second case onl% one

critical speed is o!served and since s%stem parameters chosen are dierent hence thisvalue is dierent as compared to the previous case. owever there is anti-resonancewith ver% low amplitude of vi!rations. The following data is ta$en for the simulation? thedisc mass 4 6 $g the un!alance mass eccentricit% 4 0.0006 m the diametral massmoment of inertia 4 0.08 $gm2 k 66 4 6000 7>m k 22 4 7>m k 624 600 7>m and k 26 4 0.C7>m. For the disc at the center of the shaft span we have k 62 4 k 26 4 0 so /*n.(2.669)!ecomes


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(2.669)

which is same as discussed in the previous section for the Jecott rotor. The response is

shown in Figure 2.89(c). t can !e o!served that it has onl% one critical speed whichma% not coincide with the critical speeds o!tained !% e*uation (2.660) in Figures 2.89(a)and (!). owever there will !e another critical speed corresponding to angulardisplacement and it is illustrated su!se*uentl%.


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'n su!stituting e*uation (2.60) into e*uation (2.606) we get


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(2.66C)

'n e*uating imaginar% parts of e*uation (2.66C) we get

(2.66)

which means there will not !e an% phase dierence !etween the rotationaldisplacement and the force also also since there is no damping in the s%stem. 'nsu!stituting phase information in e*uation (2.66C) we get

(2.66+)

which is the whirl amplitude of angular displacement and the condition of resonancecan !e o!tained !% e*uating the denominator of e*uation (2.66+) to 1ero which is sameas in e*uations (2.60) and (2.660) for the linear displacement. For the disc at thecenter of the shaft span we have $62 4 $26 4 0 e*uation !ecomes

(2.66,)

which gives critical speeds as

(2.66)

which is the case when the disc is at the center of the shaft span and the linear andangular displacements are uncoupled.

For the single plane motion from e*uation (2.6) we have

with

(2.620)


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The !earing force amplitude and phase can !e o!tained from e*uation (2.620). Iearing

reaction forces will have similar trend in the variation with spin speed as that of theresponse since it has the same denominator Q as that of the response. t can !eshown from e*uation (2.620) that forces transmitted through !earings have also ama"imum at s%stem critical speeds. These forces are d%namic forces and aresuperimposed on an% stead% loads which ma% !e present for e"ample due to gravit%loading. n real s%stems which are designed to operate a!ove their critical speeds themachine would normall% !e run through the critical speed ver% *uic$l% so that ver%large vi!rations and forces associated with the resonance do not have suAcient time to!uild up. Dame is true during the run-down where some form of !ra$ing ma% !eemplo%ed. f the s%stem is to run at the critical speed and vi!rations are allowed to !uildup then either the shaft will fracture and a catastrophic failure will result or there ma%!e suAcient damping in the s%stem to simpl% limit the vi!ration and force amplitudes tosome ver% large (however tolera!le) value

Concludin( Remark!? The present chapter e"plains various simple rotor models in use todescri!e some of the important rotor !ehaviour especiall% natural fre*uencies andcritical speeds (i.e. the shaft spin at which the amplitude of rotor is ma"imum). Iasicterminologies generall% used to descri!e the rotor d%namic characteristics areintroduced. For a single-&'F s%stem the natural fre*uenc% and hence the critical speeddecrease !% small amount due to damping. owever in the Jecott rotor model it isshown that critical speed increases slightl% due the increase in damping in the s%stem.Epart from the amplitude of the rotor vi!rations it is shown that the phase !etween theforce and the response is also important parameters to understand the rotor !ehaviourespeciall% at the critical speeds where it changes of the order of 6,0R. The damping isshown to !e an important factor in suppressing the rotor vi!rations at the resonance. t

is shown that the Jecott rotor is ver% a !asic model to understand several importantphenomena of the rotor s%stem. owever several other phenomena also emanate fromsupports and for this the !asic understanding support d%namics is ver% important. Themotivation of the ne"t chapter would !e to 3nd out d%namic parameters of the rollingelement and h%drod%namic !earings and seals in isolation to the shaft. This will help inunderstanding some of the insta!ilities which occurs due to support d%namics.

Exercise 'ro(lems

Exercise 2.)? For a single degree of freedom damped rotor s%stem o!tain an

e"pression for the fre*uenc% ratio for which damped response amplitude!ecomes ma"imum (i.e. location of the critical speed). Dhow that it is alwa%s more than

the undamped natural fre*uenc% of the s%stem. What is the ma"imum feasi!le value of damping ratio for under-damped s%stem is possi!le.

:int? &ierential the denominator of the un!alance response (- >e) e"pression with

respect to the fre*uenc% ratio and e*uate is to 1ero. Enswer? ;


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Exercise 2.2? et us de3ne a new fre*uenc% ratio in terms of the damped natural

fre*uenc% i.e. with . '!tain an e"pression for the amplituderatio (- >e) and the phase Φ in terms of the new fre*uenc% ratio de3ned. Slot theamplitude ratio and the phase versus the new fre*uenc% ratio and discuss the results.

'!tain an e"pression for the fre*uenc% ratio ( ) for which damped responseamplitude !ecomes ma"imum. What is the ma"imum feasi!le value of damping ratio forunder-damped s%stem is possi!le.

: .n!/er ? #

for is a comple" *uantit%. The ma"imum feasi!le value of damping ratio

for under-damped s%stem will remain the same ;.

Exercise 2.*? '!tain transverse critical speeds of an overhung rotor s%stem as shownin Figure /2.8. Ta$e the mass of the disc m 4 60 $g the diametral mass moment of inertia Id 4 0.02 $g-m2. The shaft diameter is 60 mm and total length of the span is 0.2m. The shaft is assumed to !e massless and its BoungNs modulus E 4 2.6 M 6066 7>m2.7eglect the g%roscopic eect and ta$e one plane motion onl%.

n@uence coeAcients are given as

: .n!/er ? With the diametral mass moment of inertia eect two natural fre*uencies will

e"ist? 4 C.CC rad>s and 4 699.62 rad>s. f the linear and angular motion is

uncoupled then 4 C.C rad>s and 4 +6.,0 rad>s. n case diametralmass moment of inertia is 1ero and no coupling !etween the linear and angular

motion 4 .6 rad>s;.

Exercise 2.+, '!tain the transverse critical speed of a rotor s%stem as shown in Figure/2.9. Ta$e the mass of the disc m 4 C $g and the diametral mass moment of inertia Id 4 0.02 $g-m2. Ta$e shaft length a 4 0.8 m and b 4 0.+ m. The diameter of theshaft is 60 mm. 7eglect the g%roscopic eect.


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For the present case in@uence coeAcients are given as

and .

:Enswer? 4 62.69 rad>s and 4 660.29 rad>s. With negligi!le diametral mass

moment of inertia 4 C., rad>s;

Exercise 2.5? '!tain the !earing reaction forces and moments of an overhung rotor at

rotor speeds of (i) 0.C (ii) 0.C( ) and (iii) 6.C # where and arethe 3rst and second !ending natural fre*uencies respectivel%. Ta$e the mass of thedisc m 4 60 $g the diametral mass moment of inertia Id 4 0.02 $g-m2. The disc has aresidual un!alance of 2C g-cm. The shaft diameter is 60 mm and the total length of thespan is 0.C m. The shaft is assumed to !e massless and its BoungNs modulus E 4 2.6 M6066 7>m2. Ta$e one plane motion onl%.

n@uence coeAcients are given as .

: .n!/er ? 4 6C.0 rad>s and 4 208.+ rad>s.# (i) KE 4 6.9C, M 60 7 HE 4-6.600, 7m (ii) KE4 -8.288 M 6062 7 HE 4 2.9290 M 6062 7m (iii) KE 4 -2.662C M 609 7HE 4 6.C,86 M 6097m;.

Exercise 2.6? Find transverse natural fre*uencies of an overhung rotor s%stem asshown in Figure /2.. Lonsider the shaft as massless and is made of steel with the

BoungNs modulus of 2.6(60)66 7>m2. E disc is mounted at the free end of the shaft withthe mass of 60 $g and the diametral mass moment of inertia of 0.09 $g-m2. n thediagram all dimensions are in cm.


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: .n!/er ? For the pure translator% motion? 6200.+ rad>s and for pure rotar% motion?C6. rad>s. For anal%sis of com!ined translato% motion refer Lhapter ,;

Exercise 2.-? (a) While the Jecott rotor is whirling with the help of the center of gravit% the center spinning of the disc and the !earing a"is draw their relativepositions in an a"ial plane when the rotor is (i) !elow the critical speed (ii) at critical

speed and (iii) a!ove the critical speed. (!) &e3ne following terms? natural fre*uenc%and critical speed of a rotor# s%nchronous and as%nchronous whirls.

Exercise 2.8? n a design stage of a rotor-!earing s%stem it has !een found that its oneof the critical speed is ver% close to the 3"ed operating speed of the rotor. ist what arethe design modi3cations a designer can do to overcome this pro!lem.

Exercise 2.9? E cantilever shaft of 6 m length ( l) and 80 mm diameter (d) has a C $gmass (m) attached at its free end with negligi!l% small diametral mass moment of inertia. The shaft has a through hole parallel to the shaft a"is of diameter 8 mm (di) which is verticall% !elow the shaft center with the distance !etween the centers of theshaft and the hole as mm (e). Lonsider no cross coupling in two orthogonal directions

as well as !etween the linear and angular displacements# and o!tain the transversenatural fre*uencies of the shaft s%stem in two principal planes. Lonsider the shaft asmassless and BoungNs modulus E 4 2.6 M 6066 7>m2.

:int? Find the e*uivalent stiness of the shaft in two principal directions and then

o!tain natural fre*uencies? and

and

4 +0.CC rad>s 4 +0.+, rad>s;.

jefcott model with ofset disc

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