AD/SSCfrech.No.89

AcceleratorDevelopmentDepartment

BROOKHAVEN NATIONAL LABORATORY

AssociatedUniversities,Inc.

Upton. New York 11973

SSC Technical Note No. 89368-16RHIC-MD-107

SSCL-N-742

INSTABILITIES OF BELLOWS: DEPENDENCEONINTERNAL PRESSURE,END SUPPORTS,AND INTERACTIONS

IN ACCELERATOR MAGNET SYSTEMS

R.P. Shun andM.L Rehak

November20, 1990

Tableof Contents

for

Instabilitiesof Bellows: DependenceonInternal Pressure,End Supports,andInteractions

in AcceleratorMagnet Systems

Stimmaiy andConclusions 1

I. Introduction 6

II. Basic Derivation for Bellows underPressure 8

ifia. Axial and LateralSpring Constants,and Convolution WallStressesandDeflectionsunder Bellows Compressionor Tension 16

11Th. DeflectionsandStressesof Bellows ConvolutionsunderInternal Pressure 28

IV. Spring-SupportedBellows Model 34

V. Magnet End Misalignmentand LateralSpring Constants 40

VI. Bellows Interactionsin MagnetSystems 45

VIIa. NumericalResults:Spring Constants,Wall Stressesand Deflections 53

VIIb. NumericalResults:Spring-SupportedBellows Model 59

VIIc. NumericalResults:Bellows Interactionsin MagnetSystems 68

VIII. Numerical Method 74

References 76

Appendix: MACSYMA Code for Magnet Bellows Assembly 126

Appendix: Fortran Codefor MagnetBellows Assembly 128

Appendix: File Namesof Additional Codes 130

Instabilities ofBellows: DependenceonInternal Pressure,End Supports, and Interactions

in AcceleratorMagnet Systems

R.P. Shutt and ML. Rehak

Summaryand Conclusions

In this paperwe aredealingwith internallypressurizedbellows which areto be used

extensively for SSC and RHIC magnet interconnections. In Section II we show how an

internally pressurizedbellowswhoseendsarenot free to move but aresupportedmore or

less rigidly can becomeunstableat somewell-determinedcritical pressurep1, similarly to

a bar that is compressedaxially, resulting in Euler-typebuckling.

Interconnectionbellows must not only accommodatehelium coolant flow and

pressures,but alsomust be axially precompressedwhen warm andextendedwhen cold in

order to allow for the thermal shrinkageof superconductingmagnetsduring cooldown.

From manufacturersone can obtain pressureratings,spring constants,diameters,length,

convolutiongeometry,cycling fatigue propertiesandmanufacturinginaccuracies.Making

use of our requirementsfor the mentionedparameters,we give in Sections IIIa,b an

approximatetheory for bellows, having assumedthat, for our purposes,radial widths of

convolutionsaresmall comparedto bellowsdiameters.This givesusa possibility to specify

bellows detailsthat will satisfr our requirementsfor spring constants,compressionsand

extensions,andpressuresincluding margins. We canthustestwhetherourrequirementswill

enableus to obtain reasonablyavailablebellows. Specifically, we will be able to calculate

convolutionwall stressesand motions without collapsing convolutionsunder operating

conditionsand beyond,when instability occurs. Basedon the results,we can write our

specificationsfor manufacturersto consider. Large numbers of small or large, and

expensive,bellows will be needed,and they mustnot fail sinceneedfor replacementwill

meana major interruptionof acceleratoroperation.

1

SectionsIV to VI arevalid for any size of bellows, sincethe derivationsgiven there

requireknowledgeof only easilyobtainablebellowsparameters.In SectionIV bellows ends

areassumedto be supportedby two kinds of springs,namelyspringsactingperpendicularly

to the bellows axisandothersactingtorsionally,providing a torquewhen the bellows ends’

angle is varied. If the torsion springs were not provided, the endswould be considered

"hinged". This "spring-support"modelhasprovidedmuch insight into thevariousconditions

underwhich a bellows can becomeunstable. The importanceof properbellows support

becomes apparent, in addition to selection of proper bellows design parameters.

Manufacturing inaccuracies resulting in prebent shapesof purchasedbellows, and

installation inaccuraciesresulting in lateral or angularoffsets of bellows ends are also

investigated. Instabilitiesare due to the mentionedEuler-typebuckling and end support

spring properties. End-to-endoffsets and initial bellows prebendsbefore installation

reinforcetheseinstabilitiesand increasebellows distortionsandstresseseven at operating

pressures.

In Section V we addresspossiblemisalignrnentsof adjacentmagnet ends due to

bellows and installationinaccuracies. For instance,assumingthat magnetendsare well-

aligned before bellows installation, prebendsor offsets of bellows ends can disturb the

alignment,especially if the bellows’ lateral spring constantis large or if the magnetend

stiffness is small. The latter could, in principle, be increasedby moving magnetsupports

as closeto the endsaspossible. Interconnectioninstallationdifficulties would thenhave to

be considered. Nevertheless,acceleratorperformance is quite sensitive to magnet

misalignments,especiallyconcerningquadrupoles,and thereforeaffects requirementsfor

bellows behaviorunderoperatingconditions.

In SectionVI, we havestudieda model consistingof threemagnetsinterconnected

by two bellows. Eachmagnetcanbe supportedby up to five supportswhoselateralstiffness

is taken into accountin addition to the magnetstiffness. Vertical magnetsupportstiffness

is larger thanlateral. A resultingmagnetendstiffnesscanthen becalculatedandusedfor

comparisonwith thespring-supportmodeldiscussedin SectionIV. Someof the instabilities

already found in the spring-supportmodel can be split into two peaks here, due to

2

asymmetriesof componentsor bellows inaccuracies. In particular, however, interactions

betweenbellows areof interesthere;what is the effectof a bucklingbellows on bellows in

neighboring interconnections? To answer this question,we have also introduced the

possibility of large forcesactingat bellows endsin a direction perpendicularto the magnet

axis. This would modelthe effectof a grosslydistorted,but still pressurized,bellowson the

magnetsystem. Sincethe effect is transmittedthroughsupportedmagnets,the numberand

stiffnessof magnetsupportswill thusalsoplay an important role.

Calculationswere performedusing a symbolic algebra manipulationcode called

MACSYMA SectionVIII which hasbeenextremelyvaluablein solving themanyalgebraic

equationsfollowing from the previousSections. The program provided matrix inversions

and thus algebraicsolutionsor matrix coefficientswhich couldbe enteredinto a numerical

FORTRAN program. The algebraicsolutionsfor the spring-supportmodel allowed us to

understandmany of the features that can cause bellows collapse or rupture. The

FORTRAN programprovided the large matrix inversionneededfor Section VI, and of

course,all the numericaloutput data MACSYMA and FORTRAN program codes are

describedin the Appendices.

In Section VII we discussnumericalresultsas well assomealgebraicexpressions

including limiting valuesfor somequantitiessuchas springvalues. Examplesare givenfor

SSC aswell asfor R}IIC. Additional detailsconcerningeachacceleratorwill be presented

in separateMagnetDivision Notes.

Our calculationsconfirmthatbellowswall stressesgenerallywill haveto be very high

LB x io psi if onewants to obtain bellows that are usablein applicationsof the kind

requiredhere. Therefore,in many casesthe material will be allowed to yield somewhat

during every operatingcycle axial deflection,bending,pressurizing,cool-down/warm-up.

Material fatigue must then be taken into account, which usually is included in

manufacturers’specifications. For our purpose,we would speci’ about2000 cycles.

In the spring-supportmodelwe find that aprebentshapeof a bellows we usesine,

cosineandparabolicshapescanstimulatetheinstabilitiesthatmayalgebraicallyappearas

indeterminate,even for ideally straight shapes. Our bellows are "designed"for the first

3

instability to occurat alowestpressureof 450psi. Another"peak" will thenappearat 1800

psi, andmore atstill higherpressures.We aremostly concernedthat the stressesor strainsin the bellows walls will not exceedspecifications. An indication is the maximum axial

compressionor elongation,uniform for a whole bellows or localizedsomewherealongthe

bellows convolution. We thereforepresentmaximumconvolutionelongationsfound in the

calculations,which are comparedwith manufacturers’specifications. As the pressureis

raised in a bellows wall, elongationsincreasegraduallyand can reach largevalues at our

critical pressureof 450 psi. To us the value reachedat our maximumoperatingpressure

of 300 psi is particularly important. Thus, after a bellows is designed for the critical

pressure,we mustalsomeetthe criterion not to exceedallowable elongationat operating

pressure,which is determinedby the rateof rise of the elongationvs. pressurefunction.

The rise rate is very muchaffectedby the endsupport of a bellows. The two ends

may be forced into laterallyor angularlyoffsetpositionsduring installation. Furthermore,

equivalent lateral and torsional spring propertiesof the support structureplay a very

important role. These spring propertiesproduce additional peaks, usually at higher

pressure,but also below 450 psi if, especially, the torsional support is not stiff enough.

Torsionalstiffnessashigh as5 x il5 lb inch/radianis requiredfor someof the largebellows

andmagnetsconsideredhere,independentof lateral stiffnessor pressure.Fortunately,we

can easily providevalues > IC?.

Due to the possibleend offsets and due to "support peaks"the rise rate of the

bellows elongation increases:support peaks can be placedabove 450 psi by sufficient

torsional stiffnessof endsupports. However,the presenceof the supportpeaksincreases

the rise rate of elongationvs. pressure. The presenceof pre-bent shapesadditionally

increasesthe rise rate.

The springsupportmodel findingsrecurin themagnetassemblymodelSectionVI,

exceptthat now, asalreadymentionedabove,becauseof assemblyasymmetriesanddue to

the presenceof two bellows, somepeaksfound previouslywill be split into doublepeaks.

The magnet assemblymodel allows us to study the effects of one bellows on a

neighboringone. Any such effects would, of course,also dependon the rigidity of the

4

magnetsupports Using Fermilabas well as BNL supports,we find that interactionsare

very small in the pressureregion 450 psi of interestto us.

Onecansimulatea"catastrophe"at onebellowsby applyingoneor more largeforces

there laterally and look at the distortion of the neighboringone. We show that such a

disturbancein one bellows affectsthe otherone little at or below operatingpressure.

Qurgeneralconclusionis that,with sufficientattentionto bellowsdesignandsupport

detail also including the magnetsupportsif large bellows are used,practically all sizes,

largeor small,of bellows can be usedfor interconnectionsbetweenmagnets.However,we

recommendstrongly that any of the bellows that are to be used in the large quantities

requiredfor the acceleratorsshouldstill be testedvery carefully in test set-upsthatsimulate

conditionsto which the bellows will be exposed.

5

I. Introduction

Bellows representone of the most commonly used componentsin. engineering

structures.They areusedto containvacuum,or gasesor liquids underpressure.They can

providefor accommodationof differencesin thermalexpansionbetweendifferentstructures.

Bellows can be bent, stretched,or compressedin various ways when used to connect

adjoining but not well-alignedassemblies.Theycan be obtainedin a multitudeof shapes,

sizes,andmaterials. They may consist of single or multiple layers.

Many applications of bellows merely require design of adjoining components,

resultingspecificationsfor connectingbellows, andmakingproperselectionsfrom catalogues

available from a large number of manufacturers. In other cases,special, not readily

availablebellows may haveto be manufactured.

As vital componentsin much of moderntechnology,bellows must be as carefully

consideredin designas otherparts in an assembly. They must satisfy all requirements,

including a sufficient margin. They must not be overstressedin any direction, mustnot be

damagedin transit,duringinstallationor operation.They mustnot leakgasesor fluids, and

their endsmust be properly supported. Material fatigue must be taken into account if

cycling is intended.

When axially compressedor internally pressurized,bellows can becomeunstable,

leading to gross distortion or complete failure. Ef several bellows are contained in an

assembly,failure modesmight interact. Of course,externalpressurecan also"buckle" a

bellows, but only similarly to a plain cylinder.

In particleacceleratorsusedfor high energyphysicsresearch,bellowshavebeenused

successfullyfor a long time to connectbeamtubespassingthroughmagnetswhosemagnetic

field, interacting with acceleratedelectrically charged particles, provides the required

circular pathsfor the particles. In recenttimes, acceleratormagnetshavebeenbuilt to be

superconductingTevatronatFermilab,HERA atDESY. In thedesignstagein the U.S.A.

are the SuperconductingSuper Collider SSC in Texas and the Relativistic Heavy Ion

Collider RHIC at BrookhavenNationalLaboratory. RHIC will containmanyhundreds

of superconductingmagnetsand SSC about 10,000.

6

For superconductingmagnets,one needsmany bellows for connectionof varioushelium cooling transfer lines in additionto beamtube connectingbellows.

Large bellows are also consideredas a very feasible and strong possibility for

connectingthe largetubularshellsthatsupportthemagnetiron yokesandsuperconducting

coils andcontainsupercriticalhelium for magnetcooling. In principle, every magnetcould

be self-contained,with endplatesclosing off theendsof the shells. Theseendplateswould

then be connectedby much smaller bellows for the beam tube connectionsand other

bellows to passcooling fluid, electricbus bars,and instrumentationconnections. However,

the spaceavailable in the magnetinterconnectionregion may be quite limited, and to

reduceit further by introductionof thementionedendplates,transitions,extrabellows,etc.,

would seemnot justified unlessit turnsout thatdirectmagnetconnectionwith largebellows

is not feasible.

It shouldbe mentionedthat, with large interconnectionbellows full use is made of

availablevolume to limit helium pressureincreasewhena magnet"quenches",when much

of the magneticfield energy is transferredto the helium coolant in a short time. In this

eventahelium reservoir,suchasthe interconnectionvolume, alsoservesto reducepressure

by reductionof temperaturedue to partial mixing and helium compressibility.

Assuming that a large bellows has been designed properly concerningstresses,

support, availability, etc., we intend to presentin this report, 1 a suitable theoretical

treatmentof bellows properties,2 a spring-supportedbellows model, in order to develop

necessarydesignfeaturesfor bellows andendsupportsso that instabilitieswill not occurin

the bellows pressureoperatingregion, including some margin, 3 a model consisting of

three superconductingacceleratormagnetsconnectedby two large bellows, in order to

ascertainthat supportrequirementsare satisfied,and in order to study interactioneffects

betweenthe two bellows. Reliability of bellows for our applicationwill be stressed.

The calculationshave been performed with computer programsanalytically and

numerically,whicheverapproachseemedmostsuitablefor aparticularphaseof this work.

The programsare equallyadaptableto RHIC and SSC magnetconfigurations,such

as dipole-quadrupole-dipoleDOD, ODO, DDD, etc.

7

II. Basic Derivationfor Bellows underPressure

Here we wish to derive an analytic expressionwhich will take into account the

possibly destabiliring effect of internal pressureon a bellows. External pressurehas a

stabilizing effect up to a limit. The bellows will be assumedto havean arbitrarily bent

shape before installation, due to manufacturinginaccuracies. For simplicity, the bent

bellows axis is to remain in a plane.

Theendsof thebellows are to be quite rigidly supportedbut areto be opentowards

adjoining pipes. Figure 1 showsan elementof a bellows with averagediameterD, local

axial bend-radiusp, and length As = pAy. Pressurep producesa force df,, = p dA on

elementdA of the bellowswall, whenintegratedover convolutions,asfor a plain cylinder.

Local forcesdueto pressureinside convolutionswill be treatedbelow. With dA = ½D

dØAs,,,, where As,, = length alongwall at angle *. andexpressing

- p + 71PAz

we obtain for the componentof df0, in a direction that is parallel to the bendingplane,

r_P9c0s111P+ Dcos$f$A

Integrationfrom = 0 to 2w resultsin

Sf- itD42p5X

for the total force due to p on the length element As. Integral of componentsof d

perpendicularto the bendingplane is zero. Note that this force dependsonly on the

bendingangle Ay = As/p. Using Cartesiancoordinates,

8

y

Figure 1.

x

9

P 1 +

- 1 + y’Zh12AxI,

__

‘ X

4 1+y’2

We will be interestedonly in small bellowsdeflections.Large deflectionsarenot admissible

becausewall bendingstressesbecomelargeand, also the bellows may then interfere with

various structurescontainedin it. Thereforesimply

Sf ItD2P 115x 14

x,y planeto coincidewith bendingplanewhich nowactsonly in the y-direction,neglecting

small componentsalongx. The negativesign is usedheresincewe shall assumethat df>

O when y" is C 0.

We can now setup the equationfor bendingof a bellows underpressure. In Fig. 2,

thecentralaxisof a bellows is shownin an arbitrarily bent planeshapesupportedby forces

± F,,, ± F, and momentsM1 and Mci, where

+ - F0 + F- f I-i - 0 2

for equilibrium. Here, the bellows is assumedto be straight, not preshapedbefore

installation.

* Bendingdeflectiony at x can be found by solving the differential equation

- . - F&oY-r, + + px-x’df

10

Ryc

Mc

yo

0 > X

2df=rcDpy"dx/4

Mc2

y’O

- xc

Figure2.

-Fyo

11

where E, I are "equivalent"valuesfor elasticmodulusandmomentof inertia for a bellowsrepresentedby a cylinder asdiscussedbelow. Accordingto eq. 1

-- itDp fx-x’y"x’dx’

y’x’x-x’ + yx’ i:

yx-yO-y’Ox

*Therefore

- jFJYO-y-F,.r + Me,

where

F - F + 4a1 10 4

F, - F...nD2PYIo 4b

F,, is the axial bellowsendsupportforce before pressureis introduced,for instancedue to

axial precompression.F, gives lateral support at x = 0. yO and y’ 0 are given as

boundaryconditionsor can be determinedasa part of the solutionof eq. 4. Assume,for

instance,that we give asboundaryconditionsat x = 0

y-yO- y’O

12

andat x =

y - yQ-

Solutionof eq. 4 requiresdeterminationoftwo integrationconstants.Furthermorewe must

find M0 andF. Thus the four boundaryconditionssuffice. Moment M can be found

from eq. 2, if required.

Equation4 is an Euler-typeequationdescribingthe effect of an axial end force F,

on a bar. For a sufficiently largeendforce bucklingcanoccur. Thus, for a bellows internal

pressurep canresultin buckling "squirming", if thebellowsis supportedat the endsaswas

assumed.

So far we have neglectedthe effect of shearforces on bending of the bellows.

Axially, the bellowsactslike aspringwhosespring constantdependson variousparameters.

To be called "K", it will be derivedbelow. Laterally, without bending,the bellows must act

approximatelylike a cylinder of diameterD with the bellows convolutionwall thicknesst,

but with a muchsmallerequivalentelastic modulusthan that of a cylinder.

Consideringbendingof a bar whoseendsare fixed andwhich is loadedaxially by

forces ± F, a critical force existswherethe bar becomesunstable. A simple expression,

due to Euler gives eq. 5,

-

5

where E1., I, t are elastic modulus, moment of inertia of cross section, and length,

respectively. Ends are assumedto be fixed. Equation5 doesnot take shearforces into

* account. An expressiontaking deflections due to both bending and shear forces into

accounthasbeengiven by Timoshenkot1l. The critical force is now expressedby

13

- 116x2 E,I,112_i

A,G,6

Cfl A,G, 2

whereA,. and O arecross sectionandshearmodulusof the bar.

Concerningpure shear,we have to treat the bellows as a cylinder whose average

crosssectionis A = ,rDt.

0‘ 21+v

whereE5 will be the elastic modulusof the bellows material. If v 0.25 = Poisson’sratio,

Gr 0.4 E5.

TheproductEl concernsthe axialbehaviorof the bellows. Particularly,E must be

an "equivalent"numberreferringto theaxial bellowsspringconstantK. The spring constant

also called "stiffness"of abar would be

- Elastic modulus x crosssectionlength

For a spring, or bellows, onecan define then an equivalentelasticmodulus

A6a

if K is given. A = A, t = U. Therefore,in eq. 6 we will setE = E. Ir will have an

average value

- - tD+.4_D_.4_

n36b

if t C C j Therefore

14

- 6c

It follows that

2itKD O.4nDtE,P - 1+ -l

O.4aE, 2

Typical averagevalues for bellows consideredin this report will be

K = 3000 lbs./inch

D 13.5" or less

U 10" or less

t 0.03"

E =3xlO7psi

which resultsin

16x2 El 2xKD

___

- -

_____

- O.071c1AG, O.4il6,

Thereforeit follows from eq. 6 that

Pen-

____

- p 6d

for our case;deflectionsdue to shear aresmall here,thus justifring the simple approach

takento derive eq. 3. The reasonis that here E,. C C 0,.. If D were considerably larger or* U smaller,eq. 6 would haveto be used.

15

lila. Axial and Lateral Spring Constants,and Convolution Wall Stressesand

DefiectionsunderBellows Compressionor Tension.

Bellowsmanufacturersprovide informationon bellowsparametersin theircatalogues

andupon inquiries. However,sometimesparametersfor a particularapplication,such as

ours, are not readily available. Also, interconnectinglarge numbersof superconducting

magnetsrequiresas much attentionto andqualityof design,and reliability of components,

as the magnetsthemselves;repair of a failed interconnectioncan require nearly as much

time as replacing a faulty magnet. In order to judge details of a bellows under

consideration,such as spring constants,stressesand deflections under pressure,etc.,

derivationsaregivenin thepresentsectionenablingus to correlateobtainableparameters.

Sincein our applicationsthe radial bellows convolutionwidth d will be muchsmallerthan

diameterD, it is admissibleTimoshenkot11,et al. to reducethe problemfrom threeto two

dimensions: We considera "corrugatedsheet"of width irD. The resultsagreenumerically

well with someformulas given in the literature.

One-halfof a suitableshapefor a convolution is drawn in Fig. 3. We define

d = convolutionwidth

r1 = "valley" radiusbetweenconvolutions

t = wall thickness

r = r1 + t/2 = averageconvolution radius

b = straight sectionlength in convolution

F = force appliedat bellows ends

M = momentrequired to satisfy boundaryconditions

D = averagediameterof bellows

wiaOi,2 = radial deflections, to be >0 when directed toward center of

circle formedby r

v1381,2 = azimuthaldeflection, to be CO when directedtoward l,2 = 0

yx = deflectionof straight section

Nb = numberof convolutions in bellows

16

F

Figure3.

Mc

17

Sincewe have assumedthat d cc D, we expectsymmetry around point P at the

centerof the straight section when no pressureis applied in the bellows. We will be

interestedin the axial elongation of the bellows upon application of force F and in

accompanyingmaximumstresses,expectedto occurat = r/2. Becauseof the indicated

symmetry, the elongationof one convolutionshouldbe 4yx = b/2 andof the bellows:

7

The axial spring constantis then

8

Without symmetry one would have to write A = 2Nb yb + v,ir/2. v, was

defined, is indicatedin Fig. 3, andwill be treatedbelow.

For equilibrium we have

2M - Fb+2r

Accordingto Timoshenko21,for a bent curved bar of radius r

,, dw1 w1 M019

which is also valid for a sectionof a cylinder if

1 _v2

wherev = Poisson’sratio and Eb’ the elasticmodulusof the material.

18

12

From Fig. 3 we obtain

__

- - w+__Fr1-sthO-M 101 E/fr 1 C

The positive sign for the secondterm is due to F and M actingat 9 = w/2, not at

= 0. Solution for eq. 10:

- A1cosO1+B1s1n01_!!Q1cosO1_sinO1+d1 11

if

1 E/b

- -M -Fr-?-.EJ,,

For the convolutionstraight section:

- -J....Fx-M-Fr 12

y - Ax+B-_.L FLM -FrL- 13E,/ 6 ‘ ‘2

Boundaryconditions:

19

eiE.: w-012 1

6-0: y--w1/ /w1-y

For the resulting deflectionat x =

-

_____

2

y - b/6+br+3br2+ivr3 13a

This is the deflectionin the axial direction for one-quarterof a convolution. For the axial

spring constantwe obtain then

K- lvEbt3rD 14

3Q1

since Nb = t/4r if the bellows straight sectionsareparallel to eachother which they are

not necessarilyseebelow.

The maximumtension/compressionstressin the convolutionwall is

M01 MO,ta_61 -

_____

- 2!b

and the maximumfor Cm occursat 9 = w/2, leading to

20

2Etrr+bt2A "150m21

where A was definedas the total axial deflectionof the bellows eq.7.

One can show that our simplecalculationis valid by estimatingthe hoop stresson

the bellows that would ensuewithout approximation. For this purposewe will calculate

= 0, for which Timoshenkoand Geregive

- o 16

Therefore

vi -

G1 is found from the boundarycondition

@: v-Q12 1

v1O is then easily calculatedand by about this amountthe outer convolution diameter

would be compressedand the inner diameterexpandedif no approximationhad been

used. The resultinghoop stresswould amountto only 2 to 3% of the calculatedbending

stress. We concludethat for our bellows d C c 0 the above-givencalculationis a good

approximation. More generalcalculations,including d not C C D, often approximatethe

shapeof the convolutionsby alternatingflat sheetsandshort cylinders. For our present

purposewe prefer the procedureusedabove. See, however, also A. Laupa and N.A.

Wellt81.

According to manufacturers,the convolutionstraight sectionsarenot quite parallel,

meaningthat Nb is not quite = L/4r. Figure 4 shows the actualshapeof a convolution.

The "pitch" of a bellows is thus given by

21

A

I bc

Jflgure 4.

22

A-4r+ThP 17

from which angle fi can be calculatedif A, r or r, and t, and b are given. Radial

convolutionwidth d is

d-2r1-P÷b+t 18

From eqs.17 and 18:

- d-t-2r 1+ 4rA-4r 19

4r d-t-2r2

- d-t-2r1-J3 20

From manufacturers’data,we obtain

- 0.094 reid 540

- 0.340"

for somerelevantbellows.

This value for p we shall use for our calculations. For different bellowsparameters

$ may alsobe different but can be calculatedfrom eq. 19. If then r1, t andb aregiven, we

can determineA, d, and Nb = t/A.

We cannow modify our original procedure,rememberingthat fi is small. First, eq.

5 gave the critical buckling force F at the ends of a bar which are fixed. For our

calculationswe will not be allowed to assumethatendsarerigidly fixed; thebellows will be

welded to magnet ends which are somewhatflexible, laterally as well as rotationally.Thereforewe will encounterbuckling modeswhere

F -

a

23

Making useof eq. 6c,

2 rnF - 21

a

Thus, in ordernot to exceedthe critical statewe must demandthat

K 8tF/r2D2 21a

At installation,at room temperature,bellows will be precompressedby an amount

A. Magnetsandbellows aredesignedfor a maximumoperatingpressurep. Adding some

margin to p we obtainPcr Thenwe can set

pit2, -n+ a 22

Cr4

from which follows, with eq. 21a,

F- AIl_ 8A1 23

1 n2D2

We cannow chosecritical pressurePa for a bellows anduse the resultingF to determine

the bellows parameters.

Summarizing,we shall now assumethat for a bellows

valley radius r1

inner diameterD1

radialconvolutionwidth d

angle $ of convolutionstraight sections

bellows length t

maximumpressurepa = p + margin

24

precompressionA

aregiven. Thenwe can calculate

averagediameterD = + d

- P D241... sueq. 23

Assumeguessa likely value for wall thicknesst and find

r = r1 + t/2

straight sectionlength

= d-t-2r 1-fl eq.20

1 - .kLbc2r+3bcrl+itrs eq.13a

pitch length A = 4r + 2 d-t-2r1-$p eq. 17

numberof convolutions:

24A

improvedvalue for t:

-6NryF, 25

iv Eb

where

- E/i_v2

25

The obtainedvalue for t can now be substitutedfor the first-guessedvalue for iterationconvergencewill occurafter very few iterations.

The maximumwall stressis, substituting t eq.25 into eq. 15,

- E "bC 6rF ‘326

it y I 2 N6

For axial spring constantK we use eq. 21a23:

K’ - 2tPa

it22 SAl

For a lateral spring constant,both bellows endsfixed to be parallel to bellows axis:

K - i.s122 27

which can be provedeasilyby anappropriatebendingcalculation,or by applying resultsfor

the spring-supportmodel to be consideredbelow seeSectionV.

Finally, a simple calculation,consideringthe total integratedlength of the bellows

convolutions,gives for the averagehoop stressdue to pressurePa

2/v 282t1 .- d-4+-- I

2

Note that ak increaseswhen the magnetsystemis cooled down, due to an increaseof

bellows length L, without adding to the numberof convolutions,Nb.

26

Resultsof calculationsfor different possibleparameterswill be given below, after

stability calculations have indicatedwhat values are required for K, K, a,. Special

attentionwill have to be paid to availability of wall thicknesses,convolutionshapes,etc.

27

hUb. Defiectionsand StressesofBellows ConvolutionsunderInternal Pressure

In Sectionlila. we havediscussedeffectson bellowsdue to forcesappliedat the ends

of the bellows. Here we will considereffects of internalpressure. Total defiectionsand

stressescanbe obtainedby superposition.FiguresSaandSb illustratethe problem,showing

half a convolution. The small effect of angle p is neglectedhere. We assumethat the

bellowsendsaresupportedagainstaxial motion. This meansthat at # = S ir/2 no axial

motion is allowed:

.!i !a1_o21,.2

Pressurep is supportedat 9 = = ir/2 by forces F1, F2 and0, andsinceat 8, =

dw/dO, = dw,/dU, = 0 dueto requiredsymmetry,momentsM1 andM, arealsorequired.

For equilibrium:

F1÷F2-pd-t - 0 29

a -29a

_MJ+M3+F1d_t+2Gr_Pd_t1_t_2Pr2- 0 30

In this Sectionall quantitiesF1, F2, M1, M2, 0 refer only to a one-inch length of the

bellows circumference.Referring to Fig. 5b, the momentdue to pressureforce element

pr1deç acting at 9 is, using componentsof the force element,

28

MciFt

FigureSit.

FigureSb.

29

- -pr,dWsineçrcose1-r,cosec+coseçrpine’1-nine1/--pr1rsin&1-01dei

Integrating from S to üç, total momentat 8:

MeOi - -prr1-sinO1 31

1Following eqs. 9, 10, and 12, we can now set up the bendingequationsfzere I&-_.J:

____

- -w1÷- -pr1rl -sinO1+F1r1-sinO1+Grcose1-Me,E616

or

d2w1

____

- -w +

d6_____F1-pr4rl-sinO1+Grcos61--M, 32

-

- 2+r4,r_j!J_Fir+x_Gr+MciY"

Eb4

x+r,2 r

2÷_!._pr4l -coso2-F1l+r1+sinO22dO2

-Gs2-cosO2+M, 34

÷prQ+ninO2+,2-cosO2

+tj +F51fl02

In additionwe have

35-WI

30

362

asgiven in eq. 16.

Boundaryequationsare

‘C

-

thy

rdO1- 0

- 036a

x-0

W2-y/ /62-0 w2-y

62_i!. w2-O

V2 - 0

Solutions for eqs.32 and34 are

- Al,zcosOi3+BiasinStffL6icos6iasin6i.237

+

if one haswritten eqs.32 and 34 in the form

31

d’w2 -

dO12 -

Integratingeqs.35, 36 resultsin

- A14sin013-B13cos012_i!e13sin613+2cos6132 38

C1.2j_013c05613-sinOi3+duOi.2+Eia

Equation33 leadsto

1 4 L/ 3 lax ox ay -

12 6 2

We will have to determineintegrationconstantsA1,2, B12, E12, C1, D1, and alsoF1, F,, 0,

M1, M2, a total of 13 valuesfor which we haveequilibrium conditions29, 29a,30, and the

10 given boundaryconditions. Substitutingequilibrium condition 29a into the boundary

conditionsresultsin 10 linear equationswhich are solved numericallyasdescribedbelow.

M2 and F2 can be found directly from eqs.29, 30. Having determinedw1, w2, y, one can

determinestressesfrom

Ma--

where1, = 2/t. For M useeither sideof eqs.32, 33, 34. For instance,

d2w Eta,4 14 -

-

_____

ox -

All resultswill be discussedbelow. Besidesmaximumstresses,the maximumvalueof y will

32

be of particularinterest;if y is too large, neighboringstraight sectionsmay contacteach

other,in which casethe propertiesof the bellows will changedrastically.

33

IV. Spring-SupportedBellowsModel

In order to understandandexplain the behaviorof bellows usedfor interconnectingacceleratormagnets,it hasbeenuseful to treat a model consistingof a bellowswhoseends

aresupportedby springs. Thesespringsare to model the effectson the magnetendswhich

could be deflected laterally by forces acting perpendicularly to the magnet axis, and

rotationallyby moments.Both forcesandmomentscan producelateraldeflectionsaswell

as rotationsof the bellows ends.

For the spring supportedbellows model we will use two straight springs at the

bellowsends,with springconstantsk1 and k2 lbs./inch,andtwo torsionspringswith spring

constantsk and 1S2 lbs. inch/radian.

Manufactureditems can adhereto ideal shapesonly within tolerances. Thus, we

cannotassumethat bellows can be ideally straight but will be obtainedpre-bentto some

shapethat could be expressedby aFourier series. For our purposeit is not necessaryto

carry the higher harmonicsbut it is sufficient to carry two of the lowest ones. The prebern

shapeshall be

yo - d1sin.5!+d1_cos.! 41

with d1, d2 the amplitudes,t the bellows length. y0 = 0 at x = 0 and t. The lowestmode

alsorepresentativeof apossibleoffsetmoded3 cosirx/t will causebucklingof thebellows

at the lowestpressure.The highermodecos2irx/t would causebuckling only at4 times

the lowestpressurebut it significantly affects the deflectionsof the bellows,and therefore

stresses,evenat much lower pressures,in the bellows operatingregion.

In SectionII it wasshownthat theeffectof internalpressureon an arbitrarilyshaped

bellowscanbe representedby an axial forcethatmay causebucklinganda lateralforce that

dependson the pressureandthe angle y’O that thebellows axis subtendsat x = 0 see

eq. 4a,b.

Figure6 shows the axis of a bellows, which departsfrom straightnessby a function

y0 such as given in eq. 41. In addition, due to small fabricationerrors of the bellows

34

Fwy’l+y’Ol

l±y’Ol

-Pa

y’O+y’OO

+ y’OoMel

61

02

Pa

Figure6.

‘01

1

x 1<2

x=O x=I

35

mountings, the endsare to be offset laterally by length c. At the ends, the bellows ismountedon laterally actingsprings,k1 and k2, androtationof the ends is limited by torsionsprings k, 1S2 The bellows is to be axially compressedby length A and internallypressurizedto pressurep. The bellowsendsarethenexposedto equivalentaxial end forceseqs.4a

TI:- ± 42

S d

Lateral supportis to be given by forces F1 and F2. Deflection y is to be measuredfrom the unstressed,prebentshapey0x of the bellows. If

y’O+yO a Q and /e+y - 0

we have F1 = F2. if the end slopesarenot zero, then, similarly to eq. 4b, F1, F, become

F1 -F -FfrO +y’O

F2-F2+FY:c+Y’c

as the total lateralforceson the bellows ends,due to the spring supportsandnon-zeroend

slopes. [EW- 1W For equilibrium:

2ad 43- F4Y’O-Y’C+ I

making use of eq. 41. Signsin eq. 43 areconsistentwith Fig. 6.

MomentsM1, M2 arealso required for endsupport. For equilibrium:

.fMgl_Fa81_6l4F1_F4/O+ i - o 44

Bendingdeflectionsy at; due to the appliedforces,arenow found from

-45

36

F1 is the force that is due to compressionor extensionof supportspring k1. Therefore

F

similarly

F2

For boundaryconditions,M1 andMa determinethe endslopesdue to the bendingforces:

y’O - .._L. 46aka

My19 - 46b

ka

and for the total end deflections:

FyO - +6 - +.L 46c

y62_L 46d

Boundaryconditions46a to 46d, togetherwith equilibriumconditions43 and44 will be used

to determineyx. The six conditionssuffice to find F1, F2, M1, M2, and two integration

constantsfor eq. 45 whose solution is

y - Acoscax+ Bsjnwx+ c1sin!! + c2cos?-!E+ Dx+G

Substitutioninto eq. 45 resultsin

- F4/Et" 47a

andexpressionsfor C1, C2, D1, and0, while A and B arethe integrationconstantsthatneed

to be addressed.Of special interestare

37

C1 - d192 ca92 4Th

- wc2ø12

which ate due to the prebentshapeof the bellows seeeq. 41. We can thereforeexpectbuckling squirmingof the bellows when

- iv fds0

and

- 2ic d.O

or when

Pg-

t;:2 48

F -

_____

49

in agreementwith Euler-typebuckling.

We must not exceedallowed extensionor compressionof the bellows, to be called

AEma.

Maximum bellows stressa due to bendingexcludingstressdue to local pressureon

convolutionwall seeSection11Th, which mustbe added.

i

__

Z Z 2nt

Strain is

aE

Thus

33

"DQa

-

____

and, including precompressionA:

- +IY"ID

which must be checkedfor all cases.

In addition to the buckling modeseqs.48, 49 due to bellows prebending,thereare

importantadditional onesdue to the endsupportsk1, ic, k, k, which will be considered

in detail below, when the procedurefor solution of the boundarycondition matrix and

numericalresultsarediscussed.

39

V. MagnetEndMisalignmentand Lateral Spring Constants

The acceleratormagnets must be aligned within strict tolerances. If bellowsmountingsaremisaligned,making it necessaryto distort the bellows laterally, the magnetendsmay be misalignedin an oppositedirection,dependingon lateral bellows andmagnet

endstiffnesses.This occursmostly duringbellows installation,when it may be necessaryto

force the bellows to ashapewith offset ends.

Referto Fig. 7. Let k, be the lateral spring constantof the magnetends and2d

their relativemisalignment. Then± F = k d aretheforceson the bellows ends. Moments

M aregiven by 2M Ft for equilibrium. Then the bendingequationis simply

a -

El E1 2

At

x-O: y-C-dy/-oy-d

- 0

Oneobtains

F- l2-2dEI 51

and for the total misalignmentof the magnetends

2d- -

____

k_9 k_92 521+

24E1 I. 3KD

If k, -. , d would haveto - 0, and, from eq. 51

f.. 1ZEI15MD1gC

as was givenfor the lateral bellows springconstantK in eq.27 and is provedhere. Thus

40

F

x0 x1

Figure7.

HN

t.

k x

Fd

km

41

also

2d-52a

2K,

It is seen here eq. 52 that, in order to keep magnetend misalignmentsmall, ç mustremainsmall, lateralmagnetendstiffness magnetsupportandshell stiffnessesshouldbelarge, andK should not be large. SinceIC - K and, aswill be seen,buckling of bellowsoccurs at pressuresthat are proportional to K, the latter should only be as large asnecessary.The "slendernessratio" for the bellowsD/t occurringin eq. 52 is determinedby the requiredgeometryof the interconnection. For bellows dimensionsconsideredformagnetinterconnectionswe may have2d c/l.5. If magnetendmisalignment2 d is to be

1 mm, ç must be c 1.5 mm 0.06". Magnet support stiffness could be increasedbymoving magnet supportscloser to the bellows ends. However, this would most likelyinterfere with assemblyof the interconnections.

Bellows mounting misalignmentscan occur in random directions. The magnetsupportsare considerablystiffer vertically than horizontallyso that 2 d would be smallervertically for a given ç. It must be rememberedthat c must also remain small becauselateral bellows stiffnessK times ç is the force neededto adjust the bellows before welding

to the magnetends. For ç z 0.06", Kc could be 500 lbs. This lateral force must be

provided in additionto the force neededto precompressthe bellows axially, namely K.,

which, for instance,for K = 3000 lbs,/inchand A = 0.5" becomes1500 lbs. Necessaryjigswill have to be provided, and the bellows endmountingdesignedto minimize ç.

Somehavesuggestedanalternateway to mount the bellows, so that the latter would

nothaveto bedistortedlaterally. Figure 8 indicatesthearrangement,by usingring sections,

cut from cylindersat properangles. The largeaxial compressionforce must, of course,still

be provided. It is not clear that this method,requiringadditionalwelding, would result in

a substantialdecreaseof ç. When internalpressureis applied,momentsare producedon

bellows andmagnetendsdue to the asymmetricalmounting.

42

QXISIfri

/ II

magnetwedge-cutring cnserts ends

I/

/ I

/I

nisoJignnentFigure8.

totaL C

43

A possibility to eliminateany offset of the bellows would be to butt-weldone end of

it to a narrow flange welded to one end of the magnetshell. This possibility has been

testedsuccessfully.

44

VI. BellowsInteractionsin MagnetSystems.

In Fig. 9 we show schematicallythreemagnetsand two bellows interconnectingthe

magnets. Supportsfor the magnetsareshownat locationsx1 to x19. Supportforcesare F1

to F19. The wholesystemis assumedto bepreshapedbeforeapplyingforcesaccordingto

afunctiony0x. Deflectionsyx 1 a 19 areto be measured,startingat y0x, so that

the total deviationfrom abscissax is yx + y0x. Concerningy0x, dipolesarecircularly

curved,quadrupolespackagedtogetherwith sextupoleandcorrectormagnetsarestraight,

but muchshorterthan dipoles,and interconnectionsarestraight. y0x shall representan

averagecurve of the system. For 1 n 18, forces Fa will be expressedby

- k60 53

if the k, are the supportspring constants,andhaving called 6,, - y,,x. For n = 19,

F19 - k1y18x19k19619 53a

Note that forces F act only approximatelyin the y-direction as shown in Fig. 9. For the

presentproblemsmall deflectionsthis approximationis adequate.

Two bellows arelocatedbetween;andx7, andx13 andx14. At theselocationsk6713,14

= 0 so that thereareno magnetsupportforces eq.53. However,one can apply forces

F671314for the purposeof analyzingeffectsof "bellows catastrophes"on adjacentmagnets

and other bellows. Such a "catastrophe"may be collapseof a bellows due to buckling,

resultingin grossdistortionof the bellowsandthereforepossibly large laterallyacting forces

due to the internalpressure. When a bellows buckles,it will not necessarilyrerleasethe

pressurein it but rather deform the convolutions until they are either compressedtill

contactor "straightened"to becomea smooth surface.

In our calculationswe assumethatlargediameterbellowsareemployed. Therefore

forces on bellows and"adjacentcomponentscan becomelarge. On smaller bellows the

forcesof concernto us are,of course,alsosmaller,decreasingproportionallyto the average

bellows cross section.

In Fig. 9, the bellowsendsareshownto be offset laterally from the magnetends,at

;.7,1s,14 by c671314, and by angle 16.7,13.14. While adjacentmagnetsmust be alignedwithin

45

Fe £7 F14

F4 FS

Fe

"2

I‘.0

P’S PlO Fit F12

.p13

14

F.

Pie Pby

Iyo1

.0

magnet..-

bMlon

.‘..

bellow. magnetxZ x3 ‘4 *5 x6 xl ze *0 *10 xli zlZ a 3 *14 xIS *16 xl? ito

&

small tolerances,the offsets can be due to inaccuraciesof the bellows end mountingsor

welding procedures.

It hasbeenshownin SectionII, eqs.4a,4b, that the effectsof internalpressurep on

arbitrarily shapedor distortedcylindersorbellowscan becalculatedby merelyapplyingend

forces ±irD2p/4 alongthe, howeverdeflected,axis of thepressurizedcontainer. D, again,

is the liter diameterof a cylinder or averagediameterof a bellows. Force componentsin

directionsx and y, if yx representsthe bellows axis, areat x = 0

FO-irD2p 1-y’O2z IrD2P -F,1,

if y’O is small, and

F,O - "1’y"O - F_y’O

Replacingzero by length t gives correspondingexpressionsat x = t. If additional forces

act, we obtain eqs. 4a, 4b. Adding bellows precompressionforce AK to Fr we define

n2

F -AK’.p0

-AK+F 54a -a

which is to act approximatelyin thex-direction. F,j’ is to act in they-direction.FzF0

and F,., - F,,,,yx1 areshownatx = x1, y = y1 in Fig. 9. Actually bellows precompression

force AK actsat axially fixed locationx3, but for simplicity we shall locate it at x = x1, which

is unimportantfor this calculation. In orderalso to analyzetheeffectof unequalpressures

at magnet ends, as could be encounteredduring asymmetricalmagnet quenches,it is

assumed,andshown at x x10 that force -F1, actingat x10, providesequilibrium together

with the relevant force in the y-direction for F1 at x = x1. From here on force

irD2p- AK

+

b- AK + F_b is to act, having replaced pressure Pa by Pb’ At

47

x - x19, -F z -Fb must then act for equilibrium, and Ffr, -

As mentionedabove, thereareno supportsat x = ;,7,13,14where k71314 = 0. Y&7.13.14

indicate deflections at the bellows ends. This leaves magnetsupports at the remaining

locations. Eachof the threemaznetsis shownwith five supports. The actual numberof

supportsfor a given magnetcan be adjustedby setting the k, # 0 at the supportlocations.

Elsewherek = 0. Magnetsupportscan thus be specifiedby enteringknown valuesfor

the Ic. We can then analyze SSC dipoles D five supports for magnet to cryostat,

quadrupoles0 three supports, RHIC dipoles three supports, quadrupoles two

supports.ThusanycombinationsuchasDDD, DQD,etc.,canbe studied. The SSCdipole

cryostatis to haveonly two supportsto the floor. This will considerablydecreasethesystem

stiffnessprovided by the five rnagnet-to-ciyostatsupports.

We proceedto derive relevantrelationsfor analysis. Refer to Fig. 9. Thesagittas

of our system,comparedto cord length x19 - x1 is found to be very small. Therefore we

can approximatethe arc by a parabolawhich is given by

4sy0x - ___xx19-x

x19

having calledx1 = 0. 4s0/4 is the curvatureof the systemwhich we shall,for our purpose,

merely cafl approximatelyequal to the dipole curvature0 - 4s/L2, ifs = dipole sagitta, t

= length. Therefore

y0x - *rx19 -x 54a

Referringto Fig. 9, we can nowwrite down differentialequationsfor bendingof the

magnetandbellows in the 18 regions; - ;. 1 S n 5 18. For simplebending,where

length heightor width of abar or, as hasbeen shown, for bellows,

I,

-

___

Examples: for region x2 -

48

- F061 -y4x -y1 - F1x-x1 + Fjy1’x + 0x19x 55

Fö- -! seeeq. 53

k1

F-a - wDp4/4

t elastic modulusof magnetshell including Poissonratio

= moment of inertiaof magnetcross section combination of shell andyoke

laminations.

For region x7 - ;: bellows:

-M6-F061 Y0xY4-E Fx-x

+ + 756

-F4 fr6sin [.!AE] + d6 i_cos{.!]}J

73 - ¶fyx3 takesinto accounta torque exertedby the centersupportof the magnet

causedby distortionof the shell resulting in angley at x3.

d and d6 determinethe amount of prebendof the bellows before installation see

SectionIV, eq. 41. Similarly, for d5, d13 for bellows prebendin region x14 - x13.

All otherdM, d©1, = 0.

= E seeeq. 6a

= I seeeq. 6b

A generalequation,valid in all regions,can be written:

49

y,’- [F [s -y0x + y0x10 -y _dmsin{!j!J - [ []JJ 57

- S F,x-x + Fa8i -y0x10- w +F,oxr1

In this equationit hasbeenassumedthat the slopesyx arecØx19 slopesatx1 = 0 and

x19 dueto dipolecurvature.v shallbe thesum of torquesthatmaybe exertedby supports,

suchas 73

for n-1,2, v-03n9, v-73

10n16, V0-’73+710

175n18, v-y3+10+y17

For 710, 717 replace y3’x3 by yx10 ,y,<x17 , respectively. Furthermore,

1Wforlsns9:Fa-F,-Fa F,_F,,4s

i0niS: F -Fbi, -F

Equation 57 represents 18 second order differential equations, requiring

determinationof 36 integrationconstants.In addition,valuesof thesupportforcesF must

be determined.Theseforces are, of course,known to be equal to zero where a support

springconstantIc = 0, such as at ;7,13,14 unlessF714# 0. For the SSCmagnets,there

are five supports for dipoles and three for quadrupoles,for RElIC, three and two,

respectively. Therefore, at most 15 additional quantitiesmust be found, for a total of 51.

There aretwo equilibriumconditions for the system;1 S forces = 0, 2 5 moments

= 0.

S forces:

E F 58

50

MAX BELLOWS LOCAL ELONGATION1C0--4

SSG dddzet6’O.OOOin

-° a zet7=0.000inq -

- zetl30.000 inzetl4O.000 in

o et60.000 ina

2 - et7=0.000inet130.000 inet14’O.OOOin

I - ds6=0.000in4s130.000inf’O= 0.0 lbs13: 0.0 lbs

- K 2759.2lbs/in

N00 -

-4

4-

DLI

r I-4- *414j 4

o 100 200 300 400 bOO 600 700 500 000Prespsi

$2$DUAY:CROSSUM.BUC>FOROOI.DAT;9t FIGURE D2

Table 9! Results: node location, stiffness, force at p.,

node xi in ki lbs/in Fi lbs

1 0.00 25000.00 87.702 135.69 25000.00 35.17

3 271.38 25000.00 22.61

4 407.07 25000.00 24.105 542.76 25000.00 27.57

6 606.38 0.00 0.00

7 615.68 0.00 0.00

8 679.30 25000.00 27.419 814.99 25000.00 24.9010 960.68 25000.00 25.73

11 1086.37 25000.00 24.8912 1222.06 25000.00 27.42

13 1285.68 0.00 0.0014 1294.98 0.00 0.0015 1358.60 25000.00 27.57

16 1494.29 25000.00 24.10

17 1629.98 25000.00 22.6218 1765.67 25000.00 35.1619 1901.36 25000.00 81.70

FIGURE D28

101

plot file: 2DUA7:[ROSSTJM.BUCIF0ROO1.DAT;185machine: SSCmode: ddd

Table 1: Input: bellows design parameters

rim Thin din j3 un pupil flu Ebpsi

0.060 14.300 0.500 0.094 9.300 450.000 1.000 3.297E+07

Table 2: Results: bellows properties

Nb rmn Din Am tin bcin Klbs/in Kilba/in un426 0.072 14.800 0.356 0.025 0.342 2759.212 1.048E+04 3.1SIE+01

Table 3: Results: maximum bellows stressesat

p,

opsi tp.,1p51 Cpp5t Y,na’fl

1.523E+05 2.778E+04 6.640E+04 6.782E+04 J 2.967E+04 1.472E-03

Table 4: Input: bellows misalignments

6 in 7 in 13 in 14 in q6 i7 q13 ‘I4

0.030 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

d46 in d113 in d6 in 443 in

0.020 0.000 0.020 0.000

Table 6: Input: dipole properties

I sin Loin °"o D,,,in tin 1mM4 Empsi

0.200 670. 63.6 j 10.5 0.188 3.000E+07

Table 7: Input: quadrupoleproperties

L,in oehqin Din tin j Iqin4 Eqpsi

210. 52.5 10.5 0.188 3.000E+07

Table 8: Results: bellows maximum elongation at

ppsi dl6in dll3in

3.000E+02 5.000E+00 1.002E-I-00

FIGURE D3A

103

SSC dqdzet6=0.030inzetl=0.000inzett3=0.000inzetl4=O.000 inet6=0.000inetl=0.000 inet130000 inetj4=0.000 inds6=0.020in4s13=O.000in

f6=-12500.0 lbs13= 0.0 lbsK 2759.2lbs/in

MAX BELLOWS LOCAL ELONGATION

Pros psi

FIGURE D4

Table 9: Results: node location,stiffness, force at

node xi in km lbs/in Fi lbs

1 0. 25000. 96.2 136. 25000. 119.3 271. 25000. -580.

4 407. 25000. -1808.

5 543. 25000. 8384.

6 606. 0. -12500.7 616. 0. 12500.

8 668. 25000. -8502.

9 694. 0. 0.10 721. 0. 0.11 747. 0. 0.12 773. 25000. 2067.

13 826. 0. 0.14 835. 0. 0.15 899. 25000. 731.

16 1034. 25000. -193.17 1170. 25000. -30.18 1306. 25000. 45.

19 1441. 25000. 69.

FIGURE D4B

107

FIRST BELLOWS DEFLECTION AT 300 psi

I....

I........

I... .4....:. .. .4.... I,...

o 4SSCddd

zet60.030 inA

.;iu zet7O.000in0.

- zetl3O.000 in0

zetl4O.000 inet60.000 inetlO.000in

- et130.000inetl4r0.000 inds60.020in

4s130.000 in- f6 0.0 lbs

Af13r 0.0 lbs

K 2759.2 lbs/in0N0 -

0

AC

YC

A

N

00

605 607 606 609 610 611 612 513 614 615 656 617IC in

$2$D0A7:<ROSSUM.BUC>FOROOI.DAT;90 FIG URE D6

SSC ddqzet6O.000inzetlnO.000inzetl3O.000 inzetl4--.030 inet60.000 inet?11000inet130.000in

ett4-0.000 indsô=0.000inds130.020in6: 0.0 lbsf13= 0.0 lbsK 2759.2lbs/in

MAX BELLOWSLOCAL ELONGATION

.9-4

500

Prespsi

FIGURE D8$29JIJA%<ROSSUM.DUC>F0ft0Ol.DST;95

plot file: 2DUA7:[ROSSUM.BUCFOR00I.DAT;91machine: RHIC mode: dqd

Table 1: Input: bellows design parameters

"jin D4in din $ tin pcrpsi fin Ebpsi0.060 7.630 0.500 0.094 6.000 450.000 0.500 3.297E+07

Table 2: Results: bellows properties

Nb Tin Din Am tin bcin KIbs/in Klinlbs un4

17 0.012 8.130 0.352 0.024 0.346 1784.526 4.9l5E+03 5.137E+00

Table 3: Results: maximumbellows stressesat p.,

0msa P5 U psi 01 J5i Cral Psi 0p. Y.

1.066E+05 1.534E+04 6.869E+04 7.013E+04 3.071E+04 1.548E-03

Table 4: Input: bellows misalignments

6 in 7 in 13 in 14 in q6 q7 q13 q14

0.030 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

d*6’in d,13 in d56 in 413 in0.020 0.000 0.020 0.000

Table 6: Input: dipole properties

sin Lpin ovhin Dmin t,,in twin4 Empsi2.00 394. 55.4 10.5 0.188 111. 3.000E+07

Table 7: Input: quadrupole properties

L,in ovhqin D,in tin Iqin4 Eqpsi I170. 48.0 10.5 0.188 I I 3.000E+07

Table 8: Results: bellows maximumelongation at p.,

p.,,psi dl6in dll3in

3.000E+02 I liE-al I 5.139E-F

FIGURE DR1A 113

RI-tIC dqdzet6O.000inzet7’.0.000inzetl3rO.000 inzet.14=0.000inet60.000inet.70.000inet130.000inet140.000inds60.000in4s130.000in6: 0.0 lbs13: 0.0 lbsK 1784.5 lbs/in

MAX BELLOWSLOCAL ELONGATION

.9-4

‘0

‘I

0 100 200 300 400 500 500 too oooProspsi

900

$2$DUA7:cROSSUM.BUC>POROOI.DAT;95 FIGURE DR2

Table 9: Results: node location, stiffness, force at p.,

node xi in ki lbs/in Fi lbs

1 0.00 25000.00 402.98

2 70.80 0.10 0.00

3 141.60 25000.00 276.46

4 212.40 0.10 0.00

5 283.20 25000.00 205.65

6 338.60 0.00 0.00

7 344.60 0.00 0.00

8 392.60 25000.00 149.29

9 411.10 0.10 0.00

10 429.60 0.10 0.00

11 448.10 0.10 0.00

12 466.60 25000.00 149.27

13 514.60 0.00 0.00

14 520.60 0.00 0.00

15 576.00 25000.00 205.64

16 646.80 0.10 0.00

17 717.60 25000.00 276.49

18 788.40 0.10 0.00

19 869.20 25000.00 402.96

FIGURE DR2B

117

plot file: 2DUAT:[H.OSSUM.BUCJFOROOI.DAT;12machine: RHIC mode: dqd

Table 1: Input: bellows design parameters

rein Djin din 0 lin pcrpsi fin Ebpsi

0.060 7.630 0.500 0.094 5.000 450.000 0.500 3.297E+07

Table 2: Results: bellows properties

Nb rin Din Am tin bcin Kibs/ja Kllbs/in tin417 0.072 8.130 0.352 0.024 0.346 1784.526 4.915E+03 5.137E+00

Table 3: Results: maximum bellows stressesat

Urn.. pSl e, psi psi U.3 psi a y,4. in

i.066E-i-05 1.534E÷04 6.869E+04 7.013E+04 3.071E+04 1.548E-03

Table 4: Input: bellows misalignments

6 in 7 in 13 in 14 in q6 q7 ‘$13 i$!4

0.030 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

d,6 in d,13 in 46 in 413 in0.020 0.000 0.020 0.000

Table 6: Input: dipole properties

s in LD in orhin D in tm in lm in4 Em psi

2.00 394. 55.4 10.5 0.188 111. 3.000E+07

Table 7: Input: quadrupole properties

L, in ovhq in D5 in 1, in Iq in4 Eq psi170. 48.0 10.5 0.188 111. 3.000E+07

Table 8: Results: bellows maximumelongationat

I dl6in dll3in

3.000E+02 2.165E+00 6.724E.01

FIGURE DR3A119

ERIC dpizet6O.030inzetlO.000 inzeU3O.000inzetl4zr0.000inet60.000 inet70.000inet130.000inet14z0.000inds60.020indsl3=0.000In

8 0.0 lbsf13 0.0 lbsK 1784.5 lbs/in

ASSEMBLYDEFLECTIONSAT 303.5psi

.9

NJ JUn

$2$DUA7:c9OSStJjj.5JCFcJgOOl.DA1’;91 FIGURE DR4

MAX BELLOWSLOCAL ELONGATION

C- .1....’. * .1 I... .1....:. * .1

RHtC qdqzetfi0.030 in

LSGIND zet7O.000ina. an+.SLS

0* zetl3zO.000in

zetl4=-.030 inet60.000 inet7-0.000in

- etl3zO.000inS

et140.000inds6=O.020 in

ds130.02Oin

DLI 6 0.0 lbsS

£ 113: 0.0 lbs

K 1784.5 lbs/ina

N. aa

o too 200 300 400 500 600 700 800 900Free psi

$2$DIJA7:<ROSSUII.BUC>VOR0O1.flAT;102 FIGURE DR6

MAX BELLOWSLOCAL ELONGATION.1.,. *

RUEC dqdzetôO.030inzetlO.000 in

+.fl*s

__

- zetl3O.000 in

a zetl4-.030 inet60.000 inetlO.000 inet130.000inetl4-0.000 in

a dsôO.020 in0 dslSO.020 inm.

- f6 0.0 lbso fia 0.0 lbs

£* ilLS -a

a aS

a

0 100 200 300 400 500 600 700 500 900Pros psi

$2$DUA7:<ROSSUM.SUC>FOItOOt.DAT; 504 FIG URE DRS

A, B of eq.64. Functionpfor opensa tile to which thefortran expressionswill be written,

then calls dfor described above, andalsocreatesfortranexpressionsfor C, D, Es, Ec and

G which are neededin addition to & B in eq.64 to obtain y.

127

The valuesfor omegaappearingin thesolutionsof the differentialequationsare then

defined. They are neededin the subroutineCOEF which are the MACSYMA generated

coefficientscalled Qi,j, the vector of right-handsides is W. Q and W are copied into

arraysA and B andinvertedusingthenumericalroutinesLUDCMP andLUBKSB from [6].

Upon inversionof the matrix, array B containsthe solutionsXA, XB, F which areneeded

to obtain the explicit expressionsfor deflections,slopesandcurvaturesy, y’, y". Additional

coefficientsE, 0, D, which are neededfor theseexpressionsaregeneratedby MACSYMA

are given by subroutineIND. Finally the maximumlocal bellows elongationDL can be

computed.

SubroutinePLOT producesfour types of output in addition to a table of values.

Option 1 plots maximumlocal bellowselongationversuspressure,therearetwo curves,one

for eachbellows. Option 2 plots support deflections for each support versuspressure.Option 3 givesthe deflectedshapeof the magnet-bellowsassemblyat pcr andpop. Option

4 providesthe deflectedshapeof the bellows at per andpop. Thesearewritten in the plotfile FOROO1.DAT. The Table containingall the variousdesignparametersandsummaryof the output is in the file ECHO.TEX which must be further processedwith the TEXprogram.

129

c24 dispfunbcl,bc2,bc3,bc4,bc5,bc6,eq;

iii n2e24 bcl ft + f2 - Pa -

ktikt2

ft f2 iiie25 bc2 iii + ii2 - fa + fi - Pa N-- 1

ki k2 kti

:1e26 bc3 cypo - --

ktl

.2e27 bc4fl::cyplfl----

kt2

fte28 bc5 expandratsiipcyO- z -

ki

f2e29 bc6 cy1 - -

k2

ft ftp].fayx+yO---+flx------x+u1

ki ktie30 eq ::expandyp

ei

d30. done

c31 disptunsol,ptor;

e3l solo sol partsolveleqQ,iwik, 1

e32 pfor writetileO’workcoef.for",

detrig : trigsiipdenoirbspartsol,1,

detriqdetsi:p : expand--------------, fortran{deterp substis, detsiw,

lktk2ktlkt2

FORk TURU 6 W tortranOsubstls,s3k, fortrancd substls,xd,131

fortrancq = substOls, xq, closefileQ

FIGURE E4d32 done

ci ‘derivative of y w.r.t. x’Sdisptunyp,pab,bp,hep;

c14e14 ypi, j := paM!, j + hpi, j + hepi, j

e15 paM!, j := - o.i xai sinoii xj + oii xbi cosoI! xj

e16 bpi, j zci + 2 zdi xj

e17 hepi, j := IF I = 6 TEEN xinit x6

ELSE IF I = 13 TEEN tinit x13, zesi p11 cospil xi - xinit

+ zec1 - 2 p11 sin2 pu xj - xinitfl

d17 done

c18 "list of un]atowns"$dispfwiu;

c19e19 ui := IF I <= 18 TEEN xai ELSE IF 1 <:36 TEEN xbi - 18

ELSE IF! <= 41 THENti -36 ELSE IF j<: 46 THENfi -34

ELSE IF < 51 TEEN fi - 32

d19 done

FIGURE E6

133

c50 "torsional rigidity of supportsat centerof the 3 .agnets"$dispfungas;

c5le51 gaLl:: IF i = 3 TEEN gai3 ELSE IF j: 10 TEEN gatio

ELSE IF I = 17 THEN gatl7 ELSE 0

dSI done

c52 "product LI for eachof the 3 Mgnets"$dispfunaIl;

c53e53 apJ.j =1? <= STEEN at ELSE IF I = 6 THEN a].

ELSE IF i <= 12 TEEN aq ELSE IF u : 13 THEN al ELSE a.

d53 done

c54 "axial force"$dispfunfab;

c55e55 fabi := IF i <: 9 THEN fa ELSE fb

d55 done

c56 "useful functions of au, fab"$dispfuna,b;

c57fabi

e57 aiaul1

suufj, j, 1, 1e58 bi := -

auli

dSS done

c59 "function expressingthe curvatureof the magnet and derivatives"$disptunfO,fOp,fopp;

c60eGo fOi :: - p xI xi - x19

eGi fOpi :2 - 2 xi - xi9 p

e62 foppj := - 2 p

d62 done 135FIGURE ES

c28 "boundary condition at the supportsrelating force to displaceuent"$dispfunceq;

c29fi

e29 ceqi :: IF i <: 5 THEN expandyi, 1 -

sM U

fj + 2ELSE Ui <: lQTEENexpandyi+2,1+2 ----------

ski + 2

fi+4ELSE IFI <=l4TEENexpandyi+4,1+4

ski + 4

fj + 4ELSEIFi:lsrllENexpandyi+3,i+4-------___fl

ski + 4

d29 done

c30 "boundary condition for continuity of deflection"$dispfuneq;

c31e31 egi :: -abi, i + 1 + abi + 1,1 + 1

+expandratsiup-hi,1+1 +hi+l, 1+1 -bei, 1+1

+ hei + 1, I + 1 - zeti + 1

d31 done

c32 "boundary condition for continuity of slope"$dispfunpeq;

c33e33 peqi :2 - pabi, I + 1 + pabi + 1, i + 1

+ expandratsiup-bpi, I + 1 + bpi + 1, 1 + 1 - bepi, I + 1

+ hepi + 1, 1 + 1 - et1 + 1

d33 done

c34 "nouent equation’$dispfunmou;

c35flO f19

e35 uou := uoi : expandfb - - thU9 - fOlOfl

sklO sk19

f1 fIO+ fa N- - ------ - fOIO -

skl sk1O 1372

iou + suigaui - fu x19 - xi, 1, 1, 19 + fwa p x 19

FIGURE MO

TECHNICAL NOTE DISTRIBUTION

Building 902

Blake, A. Building 911Cottinghani,J. 0.Garber,M. AD Library, R. CohenGhosh,A.. P. Hughes1 plus one for ea.authorGreene,AGupta.,R. Building 460Herrera,J.Kahn, S. Samios,N. P.Kelly, E.Morgan, 0. Building 510Muratore, J.Ozaki, S. Banks, M.Prodell, ARehak,M. For SSC PanersRohrer, E.P.Sampson,W. FNALShutt, R.Thompson,P. Mantsch,P.Wanderer,P. Strait, J.Willen, E.

sscLBuilding 1005S

Bush,T.Brown, D. Coombes,R.Claus,J. Ooodzeit, C.Courant,E. Kelly, V.Dell, F. Tompkins, J.Gavigan,M. Library CopyLee, S.Y.Milutinovic, J.Parzen,0.Rhoades-Brown, M. Taylor, C.Ruggiero, S.Sondericker, J.Tepekian, S. S&P in Masnet Division for all MDWolf,L papers

S&P in Accelerator Physics Division forall Cogenic Papersand J. Briggs.R. Daradi H. Hildebrand. W. Koitmer

S moments:

Fb610-y0x19+y,r10-619-EFz19-x59

irD2p 2+ 73+710+717 +

In addition,we need49 boundaryconditionsto solvefor thementioned51 forcesand

integration constants:as defined,

y,,x,, -6,,- ns 18, ii *6,7,13,14 60

This resultsin 18 - 4 = 14 equations. One additional equationmust be

-

61

Continuity y to y+1 "nodes" equationsare

y,1x_1-c,1 -y,,x41 62

n,1 y,,’x,,41 63

Equations62, 63 arevalid when lsn17, thereforeresult in 2 x 17 = 34 more equations,

for a total of 14 + 1 + 34 = 49, asneeded. Only S’6,71314 n7114 will be *0 for the system

drawn in Fig. 9.

Note that, in Fig. 9, c7,14n7,14 aredrawn so that they must be enteredasnegative

values;for instance,y41 - ç7 -y6x7; since in Fig. 9 y7x7 cy6x,, ç7 must be <0: for

abruptincreasesof yx or y’x with; c or n > 0 and for decreasesc or ‘i < 0.

The solutionfor equation57 is

y,,x -A,,cosø,x + B,,sino,ft + Cflt + D,z2It 2 64

+ Ecncos[_!Ex_xnJ +

51

0,, - F,, /E,j,,1/2

D,, -

-1

Em dm

{32{

- iiiEs,,

-..d4421} fi..[jJ

- -F0610+y0r10 + F0 6 + Fx + t’,, -2øEJ,,

The are to be determined, together with the F, by solving eqs.58 to 63.

52

Vila NumericalResults:Spring Constants,Wall Stressesand Deflections

The stepby stepbellowsdesignproceduresummarizedin lila was appliedto obtain

the following numericalresults. The input parametersthat one is free to chooseare given

in Table 1, the rem2iningbellows designparametersresultingfrom the precedingchoices

are shownin Table2, stressesa, h at operatingpressurep0, areshownin Table3. The

ratio D17d mustbe largefor thetheory to apply. Thefirst threecasescorrespondto bellows

whose diameteris smaller than the shell of the magnetin order to reduce forces on the

magnetsupports. In the last two cases,where the bellows diameteris largerthanthat of

the shell, it is assumedthat the supportshavebeendesignedto carry the resultingtoads.

When the bellows is internallypressurizedthe deflectionsandstressesarecomputed

following equationsin hUb. The systemof 10 boundaryconditions and three equilibrium

conditionseq.36a is solved using techniquesdiscussedin the sectionon the Numerical

Method. The maximumstressesin thetwo circularsectionsof a convolution.a, aN2, and

in the straight section a correspondingto eq. 40 appearin Table 3 along with themaximumdeflectionof the straight section,denotedy. Plots of wall deflectionswI, w2

in polar coordinatesare shown in Fig. al for the first case of the table, corresponding

stresses°r" appearin Fig. a2. These must be added to 0mfl to obtain the total

maximumstress.The total stressis largebut admissiblefor fatigue to the manufacturers

for a numberof about2000 cycles.

The wall deflectiony is shown in Fig. a3 and the stress appearsin Fig. a4. It is

verified thaty, is less than r1 so that the convolutionwalls do not touchwhen the bellows

is pressurizedTable 3.

53

Table 1: Input: bellows designparameters

r1in Din din $ là papal Un Ebpsi0.060 7.630 0.500 0.094 6.000 450.000 0.500 3.297E+070.060 7.630 0.500 0.094 7.000 450.000 0.500 3.297E+07

0.060 7.630 0.500 0.094 8.000 450.000 0.500 3.297E+070.060 11.800 0.500 0.094 9.300 450.000 0.500 3.297E±070.060 14.300 0.700 0.094 9.300 450.000 1.000 3.297E+07

Table 2: Results:bellow, properties

Nb tin Din Am tin bcin XiS/in KItS/in tin417 0.072 8.130 0.352 0.024 0.346 1784.526 4.915E+03 5.137E+0019 0.073 8.130 0.356 0.026 0.342 2095.286 4.240E+03 5.516E+0022 0.074 8.130 0.361 0.029 0.337 2410.052 3.734E+03 6.049E+O026 0.073 12.300 0.356 0.027 0.342 2732.325 7.169E+03 1.943E+0123 0.075 15.000 0.401 0.031 0.533 2756.610 1.076E+04 4.080E+01

Table 3: Results:matmumbellows stressesat

e,, psi Ok Psi 01 PSI 0’n2 P e,1 Vm.s in1.066E+05 t.534E+04 6.869E+04 7.013E+04 3.071E+04 1.548E.03

1 .037E+05 1.484E+04 5.942E+04 6.077E+04 2.649E+04 1 .246E-03

9.964E+04 1.345E+04 4.924E+04 5.047E+04 2.185E+04 9.405E.04

8.083E+04 2.157E+04 5.741E+04 5.873E+04 2.557E+04 1.184E-03

1.181E+05 1.883E+04 7.946E+04 8.090E+04 3.719E+04 2.558E-03

54

preès’urized bellowsdeflection:wI,w2 vs theta

*

I * I *

+ + pop 300.0 psi-q

. ri=O.060 in+- St

Di= 7.6 inC-

d=0.500inC +4 + + beto.094

++ 16.000 in+++

+ + pen 450SpsiCIi +++ * Eb 3.3E+O7psi+++

++

.8 +.+

Nb=17

nO.072 inra A

* 4A bc 0.346 in

o LA K= 1784.5 lbs/ina-

* 46o11. 44

44

C.

aC.

0.0 0.2 0.4 04 1.0 14 1.6

theta

$2$DUA?:CROSSUWDUC>FOR001.DLT;SO FIGURE Al

0.0 0.2 0.5

theta1.6

pop= 300.0 psiri=0.060 in

Di 7.6 in

d=0.500 in

betO.0941=6.000inpen 450.0 psiEb= 3.3E+07psiNb=17

nO.072 intO.024 inbc 0.346 in

K 1784.5lbs/in

CC0CC ‘zags.

As I45I

pressurizedbellowsstresses:sigpwl,sigpw2vs theta

+4+

++++++

+++++++++

++++++++++++

+++

++

+++

Io

I-‘0

N:0N

A A A AA A

A

00

IA A4

A A A A A A 44A

0000

70000C

‘-‘I

A4

A4444

AA4 1

4AAAA

0.4 0.6 1.0 £2 1.4

$2flUA7;<ROSSUILEUC>POROO1.DAT;60 FIGURE A2

pop 300.0psi

ri-0.060 in

Di 7.6 ind0500 inbetO.0941=6.000inpen 450.0 psi

Eb= 3.3E+07psiNb=17

nO.072 intzo.024 Inbe 0.346 in

K 1784.5lbs/in

LIGIND

pressurizedbellowsdeflection:y vs x

CCso.4

4AAA444

A

*1.4

AA

A

A

A

A

A

A

A

a.4

A

A

A

A

A

A

A

CC-I-i Oj

.4

A

A

A

A

A

A

0C0.4

A

A

A

A

A

C

A

A

A

A

A

A

A

A

A

N 0.00

A

0.04

A

0.12

I in

0.16

A

0.32 0.36

$2$DUAI:<ROSSUILDUC>FORO01.DLT;60 FIGURE A3

‘pressurizedbellowsstresses:sigpxvs x00 * I I I * IC-0N, I -A1

pop 300.0 psiA

* 1 riO.060 in0A0 A Di 7.6in0-0

AA do.500 in

A betO.O94A

A 1=8.000inA

C pen 450.0 psiA A

Eb= 3.3E+07 psiA A

Nb=17-4 a A0 nO.072in

0. Ap.C A t0.024 inA A

A bc 0.346inAA A K 1784.5 lbWin

0 A A0C -0 6 A

V A AA A

A AA A

A A0 A A

AC A

AAAAAAAA0aI,

0CC

,_0.0.0 0’De* O!IE - 020 * 024 026x in

$2$DUA7:<HOSSUM.UUC>POROOI.DAT;S0 FIG URE A4

VIIb. NumericalResults:Spring--Supportedbellowsmodel

In order to obtain the bellows deflection yx W, eq.47,one must solve a system

of 6 equationsformed by the six boundaryconditions eqsA3, 44, 46a to 46d for the 6

unknowns,A, B, Mci, Mc2, Fl, F2. Seethe NumericalMethod sectionfor details.

The solution reveals that A and B have a common denominatorwhich is the

determinant of the matrix shown in Fig. ci along with the surface determinant=0 as a

function of pressure and stifftiesses. When the determinant approacheszero, y becomes

unboundedor indeterminate andthe bellows is unstable.The determinantis, asexpected,

a symmetricalexpression with respect to left and right since the indices i and 2 can be

interchangedwithout alteringits value. it is a function of Fa both directly and through 0,

eq. 47a and of kl, k2, k, k. With this expressionone can find the critical buckling load

of the bellows dueto the supportsystemonly, that is thevalue of Fa or pressurep which

will reduce the determinant to zero for given stiffnesses.For simplicity, the realistic

assumptionsthat Ic11 k2 andki = k2 areusedfrom now on. The precompressionis A =

0.

sssflasssssssss sSc

Fig. bi shows a plot of the maximum bellows elongationdefined in eq.SO as a

function of pressure.The bellows was designedto havea critical load at 450 psi due to an

assumedinitial sine shapeprebend. Accordingto the last casein the preceedingthree

tables,this determinesan axial stiffnessof K=2760lb/in for a bellows of innerdiameterD,

= 14.3 in and length93 in, thesupportshave assumedstiffnessesof k11 = 3.e7 lb in/radian

andid = 7e3 lb/in. There is a peakat 600 psi andoneat 1875 psi. The expectedpeak,for

which thebellowswasdesignedat450 psi, is absentbecausey eq.47 is indeterminatezero

appearsboth in the numeratorand denominator. Neverthelessthe bellows should be

designed as having a critical load at that value since deviations could lead to large

deflections. Seide [3] and Haringx [4,5] discussand presentexperimentaldata on the

stabilityof internallypressurizedbellows. Haringx showsthat abellows with clampedends

59

canfail at relatively low valuesfor pressure. Keepingin mind that therecan be a peakat

= ir, the spring modelwill be usedhereonly to find the buckling load due to support

details.

We show next how the analysisof the determinantalone allows one to predict the

critical load due to the springsupportsonly. Thedeterminantis solved first for the variable

Id, thesolution is plotted for severalvaluesof Ic11 as a functionof p in Fig. b2. Figuree2

showsthe equationgiving the valueof Ii which will makethe determinantgo to zero for

variousvalues of l. Asymptotic expressionswhere l goes to zero or infinity appearing

in Fig. e3 areplotted ascurves 1 and7 in Fig. b2. A horizontalline drawn at ki =7e3 lb/in

intersectsthe curveof solutionswith k11 = 3e7 lb in/radianwhen p is about610 psi. Curves

aretruncatedat 20000 lb/in. With k11=0, p is only somewhatlarger,at about650 psi.

An equivalentway to predict the buckling load is to solve the determinantfor the

variablek asa function of ki andp. Thereare two solutionssincethe determinantis a

quadraticfunctionof Ic11. Thefirst one, denotedk12 in Fig. b3, showsthat a horizontalline

drawnat 3e7 lb in/radianwould intersectthedeterminant=0 curve for Il = 7e3 lb/in at 610

psi. The secondroot, denoted‘Sib in Fig. b4 showsthat for all valuesof ki the determinant

is equal to zero at 1875 psi when Ic11 =3e7 lb in/radian. We have thus explainedthe origin

of the two peaks in Fig. bi: they can be traced to a given combination of support

stiffnesses. Figures.b2,b3, b4 can be usedto ensurethat peaks due to supports remain

always abovethe critical load for which thebellowsis designed.Expressionsfor ki, kja, ‘Sib

which causethedeterminantto be equal to zero are shownin Fig. e2. Limits when w = 0

are given in Fig. e3 for the caseswhereki=0 andki infinite.

We now give an examplewhere the supportpeak occurs below the ot = if peak.

Figureb4 predicts thata peakcould be obtainedat 425 psi if ki =7000 lb/in andk = 1.e4

lb in/radian, this is verified in Fig. b5 which showsthe425 psi peak. In additionthereare

thepeaksat 650 psi andat 1830 psi shownin Fig. b2 and/orb3.

Table 4 lists peak locations accordingto their origin for two values of k11. The

influenceof themountingoffset c is to acceleratethe rate of rise to the peaks,but not to

causethem. It is a multiplying factor of the numeratorof y and doesnot appearin the

60

1detertnant fa U-- +

k2

1 21- 0 sin1 0 +

fa fa

1 1fa --- + --- o cosil o

kt2ktl

1 1+ - 1 --- sin1 0 - I

}ct2ktl

fa 1 sin1 o

}ctlkt2

1 0 sin1

+

o

fw1.

+ 2 cos1

FIGURE El

- fw sin;J. 0

ktlkt2

0 cos1o

1.

k2kt2

0

+ *-----

klktl0 1 - cos1o

a - 1

DETERMINANT-C

0a

SURFACE

‘a

0

t

1.,

61

2 2

kl : - ktl - a fw a sin la- a wo COS la

2 2+fafw-2faocoslai+kt22faktlo sintlo

2 2 3

+ a fw - 2 a a cas1 a - a fw 0 + 2 a fw - 2 a sin1 a

2 2 2/kt2 ktl 0 sin 1 o - Ia sin1 a +a cos Cia -2° cas1o +a

- a sin]. a + a 1 a cos1a + ktl fa 1 a casia - a sin1 a

2

+ a 1 sin1 a

ktla= U Ic. 2falocoslo -2tasinio

2+ a fw -2a 0 cos1 a - atwo

11+faJcJ.lcl 2 sinlo -Ia + -+-- a fwcoslo -1 +2a

Ic2kl

2+ Ii a N - 2 Ia a cos1a - a two

2 2Jk2ki21a sinlo +4ocosio-4o-2fao sinlo

2- 2 fa Ii a sin1 a

ktlb:.k2k12asinlo-2falocoslo

2+ 2 a - a 1w a cos1 o + afwo

1 1+falcik2 2 sinia -Ia + -+-- ofwcoslo -1 +2a

k2k1

2+ IC. 2 a - a fw a cosl a + a 1w a

/k2 2:a sinla +4bcos1a-4a -2faasinuo62

- 2 a Ic]. a sin1. a

FIGURE E2

limit ki, ktiØ

2 I. - 2 Ia

limit ki, ktiinfinity

2 Ia o winO a

I a sinl a + 2 casI a - 2

limit ktia, k10, and with asø

1w - I. casI a - fw * 1. 9w I

a sinl a ‘ 2

J;m;t ktla, k1nflnity, and with *sø2

Ia I tasl a - I. I 3D K

Iasinlo+2co.I,-2 4

limit ktib, ki.a, and with a2

Iato,lo.fa Dl

sinlo 4

Hmit ktlb, kiwinfinity, and with ø2

2 Ia sinI a - I. I a cosl a - fa I a D K

2 4I a sinI * 2 a cosl a - 2 a

ktta as a function ot Iii. wfl*n C%p/I22 2 22

%p D K 4 kI I * 9 ?w I - %p D Kktla

2S4 ki I

FIGURE E3

63

denominator.The influenceof c can be seenby comparingflgs.b6 and b7 which differ only

in that c = 0.03 in Fig. b6 and zero in Fig. b7. The first supportpeak is no longer visiblewhen c=0 but it could be large if c were increased.

Table 4: Pressurepsi valuesat peaks

‘S 3e7 lb in/radian 1e4 lb in/radian

cot = w peak 450 450

cot = 2w peak 1800 1800

ki, ktia peak 610 650, 1830

‘Sib peak 1875 425

The effect of the sine shapeof the bellows can be seenin Fig. b7 wheredl = 0.02;

thesecondsupportpeakis largerthan in Fig. bi. This effect is further accentuatedin Fig.

b8 when d2=0.02where there is also a peakat cot =2ir at 4 x 450 = 1800 psi which here

is maskedby the second support peak. In addition to c and dl, the presenceof d2

contributesto the rise of the curve evenat operatingpressure.Thecorrespondingbellows

deflectedshapeis shownin Fig. b9.

We determinenextconditionsfor supportstiffnessessuchthat support peaksremain

alwaysabovethe cot = ,r peakfor any pressure.A conservativevaluefor ‘S is given in Fig.

e3 by the limit of when ki is infinite and w=0. This value is 3D2K/4 for any value of

kl, eveninfinity. A less conservativevalue can be obtainedby using the relation between

‘Sia and Ii given in Fig. e3 for cot = n. The root 1cib is zero at cot = ir for all valuesof

Ii. Figure b3 can be used to verir that no peak will occur below cot = it if these

guidelinesare followed. Figure b4 indicatesthat the 3D2K/4 value for ‘S is three times

higher than neededin that figure to avoid low peaks. The torsional stiffnessfor an SSC

dipole with two support postsis estimatedin the next sectionat L5e7 lb in/radianwhich

is much higher thanthe 4.6e5lb in/radianobtainedfrom the proposedlimit Is =3D2K/4.

64

sassassass R.I-IIC

Figure bri shows a plot of the maximumbellows elongationdefined in eq.5O asa

function of pressure.The bellows wasdesignedto have a critical load at 450 psi due to an

assumedinitial sine shapeprebend.According to the first casein the preceedingthree

tables this determinesan axial stiffnessof K 1785 lb/in for a bellows of inner diameter

= 7.63.in and length 6 in. thesupportshaveassumedstiffnessesof ‘S = 3.e7 lb in/radian

andkl =7e3 lb in. Thereis a peakat 800 psi and one at 1900psi. The expectedpeak, for

which the bellowswasdesignedat 450psi, is absentbecausey eq.47 is indeterminatezero

appearsboth in the numeratorand denominator.

Neverthelessthe bellows shouldbe designedas having a critical load at that value

since deviationscould lead to large deflections. Seide [3] and Haringc [4,5] discussand

presentexpethnentaldataon thestability of internally pressurizedbellows. Haringx shows

thata bellowswith clampedendscan fail at relatively low valuesfor pressure. Keepingin

mind that therecan be a peakat cot = it, the spring modelwill be usedhere only to find

the buckling load due to supportdetails.

We show next how the analysisof the determinantalone allows one to predict the

critical load dueto thespring supportsonly. The determinantis solvedfirst for the variable

kl, the solution is plotted for severalvaluesof ls asa function of p in Fig. br2. Figuree2

showsthe equationgiving the value of ki which will makethe determinantgo to zero for

variousvaluesof ‘Se. Asymptoticexpressionswhere k11 goesto zero or infinity appearing

in Fig. e3 areplotted as curves1 and7 in Fig. br2. A horizontalline drawnat kl =7e3 lb/in

intersectsthe curve of solutions with ‘S =3e7 lb in/radianwhenp is about 800 psi. With

= 0, p is 875 psi.

An equivalentway to predict the buckling load is to solve the determinant for the

variable ‘S asa function of kl and p. Thereare two solutionssincethe determinantis a

quadraticfunctionof k. The first one,denotedk in Fig. br3, showsthat a horizontalline

drawnat 3e7lb in/radian would intersectthedeterminant=0curve for kl =7e3 lb/in at 800

psi. The second root, denoted ‘Sib in Fig. br4 shows that for all values of kl the

determinantis equal to zero at 1900 psiwhen lç13e7lb in/radian.We havethusexplained

65

the origin of the two peaksin Fig. brl: theycanbe tracedto agiven combinationof supportstiffnesses. Figuresbr2, br3, br4 can be usedto ensurethat peaksdue to supportsremainalways abovethe critical load for which the bellows is designed.

Expressionsfor ki, ‘Sib which causethe determinantto be equal to zero areshown in Fig. ci Limits when w=0 aregiven in Fig. e3 for the caseswherekl=0 andklinfinite. . We now give an examplewherethe supportpeakoccursbelow the cot = it peak.Figurebr4 predictsthata peakcould be obtainedat 340 psi if kl = 7000 lb/in and ‘S = l.e4lb in/radian, this is verified in Fig. br5. In addition thereare the peaksat 875 psi and at1700 psi shownin Fig. br2 and/orbr3. Table5 listspeaklocationsaccordingto their originfor two valuesof ‘Si. The influenceof the mountingoffset c is to acceleratethe rateof riseto the peaks,but not to causethem. It is a multiplying factor of the numeratorof y anddoesnot appearin the denominator.The influenceof c canbeseenby comparingFigs. bróand br7 which differ only in that c=0.03 in Fig. br6 andzero in Fig. br7. The first support

peakis no longer visible when c = 0 but it could be large if c were increased.

The effect of the sineshapeof the bellows can be seenin Fig. br7 wheredl = 0.02;thesecondsupportpeakis larger thanin Fig. brl. This effect is further accentuatedin Fig.

br8 when d2=0.02where thereis also a peakat cot = 2w at 4 x 450=1800psi which here

is maskedby the secondsupport peak. In addition to c and dl, the presenceof d2contributes considerably to the rise of the curve even at operating pressure. The

correspondingbellows deflectedshapeis shownin Fig. br9.

Finally Fig. brlO shows the maximum elongationfor a bellows of large diameter,

D= 11.8 in and93 in in length for the sameparametervalues as in Fig. br8.

Table5: Pressurepsi values at peaks

‘S 3e7 lb in/radian 1e4 lb in/radian

wt=irpeak 450 450

cot = 2w peak 1800 1800

kl, ‘Sia peak 800 875, 1700

‘Sib peak 1900 340

66

We determinenextconditionsfor supportstiffnessessuchthatsupportpeaksremain

alwaysabove thecot = w peakfor anypressure.A conservativevaluefor lcd is given in Fig.

e3 by the limit of Isia when kl is infinite and co=0. This value is 3D2K/4 for any value of

kl, eveninfinity. A lessconservativevaluecanbe obtainedby therelationbetween‘Sia and

ki given in Fig. e3 for cot = ,r. The root ‘Sib is zeroat cot = w for all valuesof kl. Figure

br3 canbe usedto verify that no peak will occur below cot = it if theseguidelinesare

followed. Figurebr4 indicatesthat the ‘S 3D2K/4 value for ‘S is three times higher than

neededin that figure to avoid low peaks. The torsional stiffnessfor RHIC dipoles is

estimatedin the next sectionat 3e7 lb in/radianwhich is much higher than the 88500 lb

in/radianobtainedfrom the proposedlimit Ic11 = 3D2K/4.

67

VIIc NumericalResults:Bellows interactionsin MagnetSystems

MacnetequivalentStiffness

In order to relateresultsfrom the magnetbellows systemto the spring-supported

model it is desirableto haveanapproximatevaluefor the equivalentlateralspring constantand torsionalstiffnessof the magnetends.

The magnetis modeledasan axially compressedbeamwith five supportsto which

a force or momentis applied at one end. This is a classical statistically indeterminate

problemsolvedby applying the equationsderivedin VI for the magnet- bellowsassembly

to one magnetonly.

Oneof the major differencesof concernto us in this problem betweenRHIC and

SSC magnetsand alsobetweendipoles andquadrupoleslies in the locationand number

ofsupportswhich is adjustedby settingthe relevantsupportstiffnessesto zero in the general

modelwith five supports. Entriesin Table 6 show similar stiffnesseswith the exceptionof

the last case;SSC dipoles are mountedon five supportsinside the cryostatwhich itself is

mountedon two supportsto theground.The cryostatsareinterconnectedby bellows.The

actualstiffnessof the magnetcryostatassemblyis betweenthe two extremecasesof five

supportsand two supports. Tests can be made to measurethe stiffness of the magnet

cryostat assembly.The lateral stiffness k, and torsional stiffness ‘S are practically linear

functionsof theaxial force equivalentto thepressure,but their increaseis sosmall that they

canbe consideredconstant.

Table6: Magnet stiffness

Magnet Ic lb/in ‘SIb in/radian

RHIC quadrupole 6600 3.0 e7

RHIC dipole 7500 3.2e7

SSCdipole 5 support 6300 3.0 e7

SSCdipole 2 support 530 1.5e7

68

*ssaSS$**$

Magnet bellows assembly

The first example of a magnet bellows assembly is for a dipole-dipole-dipole

combinationwith SSCdimensions. The bellows inner diameteris 14.3 in, its length is 9.3

in, andthe axial stiffnesswhich will producea first peakat 450 psi is K=2760 lb/in, the

precompressionA =1 in, the safetyfactor over the operatingpressureof 300 psi is thus 1.5.

In Fig. dl the maximum local elongation of the two bellows is plotted versuspressure,the

initial offset of the first bellows is c6 =0.03 in, it has an initially bent shapewith ds6= 0.02

anddc6=O.02 in. The uppercurve madeof trianglescorrespondsto the first bellows, the

lower curve madeof crossescorrespondsto the secondbellows.Maximum elongationsare

denotedin the legenddlaanddlb for the first and secondbellows respectively,all values

are truncatedat 5 in for betterdisplayof details.

Figure dla summarizes the parametervalues for the bellows, mounting offsets,

magnetparameters,shows the stressesand the maximum local elongationsat operating

pressure.Figured lb indicateswherethesupportsarelocated,what their stiffnessesareand

what forces areexerted on them, bellows are located at nodes6, 7, 13, 14.

Peaks

There is a small peak in the curve for the first bellows at 450 psi which requiresa

fine meshto be detected.It revealsthe indeterminacyat cot = it discussedin the spring-

supportedmodel. The rate of rise to this cot = ir peak is controlled by c, dsó and dcó

which arealsoequal to 0.02 in. Thereis a rising trend in the curve as it proceedstoward

the cot = 2w peakat 1800 psi.

The curve correspondingto the first bellows which hasthe mountingoffset shows

a large rise to the first supportpeakat 600 psi. The secondbellows hasa smallerpeakat

the samelocationbut doesnot seemto rise below that. This indicatesvery little, if any,

interactionbetweenbellows. Going back to Fig. b2, a peak at about 600 psi for large ‘Swould occur at ki =6000 lb/in. This numberagreeswith the 6300 lb/in computedabove

in the sectionon magnetequivalentstiffness.

69

Mounting and shaDeof bellows

Figure d2 shows the ideal casewhere thereareno mountingoffsetsand where thebellows are almoststraight. Accompanyingtablesareshownin flgs.d2a,d2b. Comparison

with Fig. dl shows that as with the spring-supported model, c7114, 7j314 and ds6,l3,

dc6,13 increasethe rateof rise to the support peak. The effect of an initial angle t issimilar to the effect of c, for instancereplacingc6=0.03by p6=0.004would leave Fig. dl

unchanged.If both bellows were mounted in the samemannerand had the same initial

shapes,then the lower curveof Fig. dl would coincide with the upper one. In the spring-

supported model Fig. b6 there was no visible support peak when c=0, but there is one

here becausethe assemblywasassumedto have an initial average parabolic shape due to

the magnets’sagitta.

Bellows interaction

Figure d3 correspondsto the casewhere lateral forces f7 =-f6 = 12500 p/p if

p p and f7 = -f6 =12500lbs if p p,, areapplied to the first bellows accordingto

Fig. 9 in order to simulate failure in a dipole-dipole-dipoleassembly.The upper bellows

curve hasexceededthe5 in cutoff limit while the lower bellows curveremainsflat denoting

no interaction between adjacentbellows at operating pressure. A more pronounced

interactioneffect is visible in Fig. d4 for the dipole-quadrupole-dipoleassemblywherethe

second bellows showsa maximumelongationof 0.5 in abovethe initial precompressionof

1 in. Complete failure of one bellows doesnot cause adjacent bellows to fail.

Deflectedshapeof assemblyand bellows

Figure d4 shows the deflectedshape of the assemblycorrespondingto Fig. dl at

operating pressure.In order to better see the effect of the internal pressure on the system,

only thedeparturey of the assemblyfrom theinitial shapehas beenplotted. The total final

shapeis the superposition of the initial parabolic shape, the sine andcosinebellows shapes

if they apply, and y. The Table in Fig. dib can be used to identify bellows and dipole

locations. The 0.03 in offset marks the end of the first dipole andthe beginningof the first

bellows. The offset can also be seen in the bellows deflectedshapein Fig. dS, and in Fig.

d6.

70

Additional cases

A possiblealternateconfigurationis shownin Fig. d6 with adipole-quadrupole-dipole

assembly.Both bellows aremountedwith offsetsandprebentshapesin orderto maintain

symmetrywith respectto the centerof the assemblyand, asexpected,both curvescoincide.

The dipole-dipole-quadrupoleconfiguration where the second bellows only is

mountedwith theusualoffsetand bentshapeis shownin Fig. d7. The secondbellowscurve

is higher than the first one becauseit is the one with the offset.

""""" RI-BC e.sss*s$s$

Magnetbellows assembly

The first example of a magnet bellows assemblyis for a dipole-quadrupole-dipole

combinationwith RHIC dimensions. The bellows inner diameteris 7.63 in, its length is 6

in, and the axial stiffness which will produce a first peakat 450 psi is K= 1785 lb/in, the

precompressionA =0.5 in, the safetyfactor over the operatingpressureof 300 psi is thus 1.5.

In Fig. dri the maximum local elongation of the two bellows is plotted versus pressure, the

initial offsetof the first bellows is c6=0.03 in, it hasan initially bent shapewith ds6=0.02

and dcó= 0.02 in. The uppercurve made of trianglescorrespondsto the first bellows, the

lower curve made of crossescorrespondsto the secondbellows. Maximum elongationsare

denotedin the legenddla and dlb for the first and secondbellows respectively, all values

are truncatedat 5 in for betterdisplay of details.

Figure dna summarizesthe parameter values for the bellows, mounting offsets,

magnetparameters,shows the stressesand the maximum local elongationsat operating

pressure. Figure drib indicates where the supports are located, what their stiffnessesare

andwhat forces are exerted on them, bellows are located at nodes6, 7, 13, 14.

Peaks

There is a small peak in the curve for the first bellows at 450 psi which requires a

fine mesh to be detected. It reveals the indeterminacy at cot = r discussedin the spring-

supportedmodel. The rate of rise to this cot = it peak is controlledby c, dsó and dc6

71

which arealso equal to 0.02 in. Thereis aSing trend in the curveasit proceedstowardthe cot 2r peakat 1800 psi.

The curvecorrespondingto the first bellows which hasthe mountingoffset showsa largesupportpeakat 700 psi. The secondbellows hasa smallerpeakat the samelocationbut doesnot seemto rise below that. This indicatesvery little, if any, interactionbetweenbellows., Going back to Fig. br2, a peak at about700 psi for large k would occur atkl=5000 lb/in.

Mounting and shapeof bellows

Figure dr2 shows the ideal casewherethereareno mountingoffsets and wherethe

bellowsarealmoststraight. Accompanyingtablesare shownin figs.dr2a,dr2b. Comparison

with Fig. dni shows that as with the spring-supportedmodel, c671314, p6,7,13,14 and dsô,13,

dc6,13 increasethe rate of rise to the support peak. The effect of the initial angle v is

similar to the effect of c, for instancereplacingc6 = 0.03 by "i6 =0.004 would leaveFig. dr 1unchanged. If both bellows were mountedin the samemannerand had the sameinitial

shapes,then the lower curveof Fig. dnl would coincidewith theupperone. In the spring-

supportedmodel Fig. bro therewas no visible supportpeakwhen c=0, but there is one

herebecausethe assemblywas assumedto have an initial averageparabolicshapedue to

the magnets’sagitta.

Bellows interaction

Figure dr3 correspondsto the casewherelateral forces f7 -f6 =3000 p/p0 if p

p andf7 =-f6 =3000 lbs if p p are appliedto the first bellowsaccordingto Fig.

9 in orderto simulatefailure in a dipole-quadrupole-dipoleassembly.The interactioneffect

on the secondbellows is small since the secondbellows showsa maximumelongationof

0.17 in above the initial 0.5 in precompression.Completefailure of one bellows doesnot

causeadjacentbellows to fail.

Deflectedshapeof assemblyand bellows

Figure dr4 shows the deflectedshapeof the assemblycorrespondingto Fig. dri at

operatingpressure.In order to betterseethe effectof the internalpressureon the system,

only thedeparturey of the assemblyfrom the initial shapehasbeenplotted. The total final

- 72

shapeis the superpositionof the initial parabolicshape,the sineandcosinebellows shapes

if they apply, and y. The Table in Fig. drib can be usedto identify bellows and dipole

locations.The 0.03 in offset marks the endof the first dipole and the beginningof the first

bellows. The offset canalso be seenin the bellows deflectedshapein Fig. dr4, and in Fig.

dr5.

Additional cases

A possiblealternateconfiguration is shown in Fig. dn6 with a quadrupole-dipole

quadrupoleassembly. Both bellows aremountedwith identical offsetsand prebentshapes

in order to maintain symmetrywith respectto the centerof the assemblyand as expected

both curves coincide.

Double peaks

In Fig. dr7 bellowswith diameterD1 =11.8 and length 1 = 9.3 in havebeenusedwith

a dipole-quadrupole-dipoleconfiguration. The first bellows only is mountedwith an offset

c6=.03 in and ds6=dc6=0.02in. This choice of bellows parametersaccentuatesthe fact

that in general thereare two peaks due to supportswhere therewas one in the spring

supportmodel. Closeobservationof Fig. dri revealsthat therearereally two peakswhich

are slightly offset at 700 psi. Indeedmountingonly onebellows with an offset introduces

an asymmetryin the problem. In Fig. dr7 the first bellowsbecomesunstablefirst at 630 psi,

thenthesecondbellows becomesunstableat 760 psi. A peakin onecurve always coincides

with a lesserpeak in the other revealingsome interaction betweenbellows when they

becomeunstabledue to the supportsystem. When the asymmetryis removedby letting

c14=-c6 andusing bellows with identical prebentshapesthe two peaksnow mergeat 740

psi in Fig. dr8.""""" SSC and RHIC """""

The two types of instabilitieswhich can occur, the Euler-typeof buckling and that

due to thesupportsystemhave beenidentified. The supportpeakor peaksshouldalways

be above the cot = it peak, this is controlled mostly by ensuringa large value for the

torsional stiffnessof thebellows supports. Interactionbetweenbellows affectsonly support

peakswhich arekept well above the 450 psi cot = it.

73

["IlL NumericalMethodMost of the algebrarequiredfor this problem was performedusing the symbolic

manipulationcode MACSYMA. This code has proved to be of invaluable help bothnumerically and analytically. Some of MACSYMA’s capabilities and limitations are

illustrated in the following four cases.

Convolutionwall stressesand deflections

The numerical part of the internally pressurizedbellows problem rests on theinversion of a 1OX1O matrix which is performed analytically here using MACSYMA.

Eq.36ain ifib describesthe linearsystemof 10 equationsandunknownswhich aresolved.Thesevery long expressionsare translatedinto FORTRAN and written into a file forfurther processinggraphics,parametricstudy using conventionalfortran code. In this

example where results are so lengthy, the availability of an analytical solution offers

marginal improvementover a numericalmatrix inversion.

Snring supportedbellows model

Thenumericalpartof this problemis reducedto finding the six unknownsA, B, Mci,

Mc2, Fi, F2 using the six boundaryconditionseqs.43,44, 46a to 46d.

Resultsare now of manageablesize and can lend themselvesto analyticalinvestigationin

addition to the usual procedureof translation into FORTRAN. The list of boundary

conditionsand eq.47 appearin Fig. e4, function solo solves the system,function pfor

opens a file in which it writes the appropriate FORTRAN statements. pfor. also

rearrangesthedeterminantof thesystemto producethe expressionalreadyseenin Fig. e1.

In the sectionon NumericalResults it was shown how this determinantwas solved for ki

andk andhow this allowedus to predict critical loads. In this respectMACSYMA was

critical to the comprehensionof the problem.

Bellows interactionsin magnetsystems

The numericalpart of the problemin chapterVI is reducedto solving a systemof

51 equationseqs.58to 63 and 51 unknownswhich is written using matrix notation as:

XQ] = W. Inverting analyticallythe matrix with MACSYMA was not attempteddue to its

size,insteadMACSYMA wasused to find the coefficientsof [01 which is almostimpossible

74

to do by hand due to the massiveamount of algebra involved. Given the list of 51

unknownsand 51 equations,MACSYMA gathersall the coefficientsof the unknownsand

collects the terms forming W. The result is translatedinto fortran and electronically

transferred to a numerical computer code where the matrix inversion is performed

numerically. A discussionof the codeis includedin the appendix.

Magnetlateralspring constant

The lateralspring stiffnessof onemagnetis obtainedby slightly modifring the code

for the previousproblemsincedeflection,boundaryconditionsandequilibrium equations

are thesame. Equationsdescribingthe first magnetonly arekept: the numberof supports

is reducedfrom 19 to 6 and functionsdescribingbellows areomitted. A force or moment

is appliedat the endof the magnetand the correspondingdeflectionor slope is computed,

thusgiving lateralstiffness andtorsional stiffnesses.This lastexampleshowshow thesame

MACSYMA codecan be adaptedto two differentproblemswith similar equations.Oneof

the drawbacksof MACSYMA is that resultsarepresentedin a form which is oftennot the

most concise. A considerableamountof effort must be exertedto reduceexpressionsto

their mostadvantageousrepresentation.

REFERENCES

Existing work on stability of internally pressurizedbellows

The most recent reference,ret [3], contains an extensive historical review and

discussionof work relatedto internally pressurizedbellows. It investigatesthe stability of

a cantileveredbellows with a movableendwhich is permittedonly to rotateabouta fixed

point on the longitudinal axis of the beam. This case is not applicable to the present

problem.

The other two references,[4], [5] show that buckling can occur in internally

pressurizedbellowsat arelatively low pressureevenwhentheendsareclampedratherthan

simply supported. The bellows stiffness is computedbasedon corrugationsof rectangular

shape.

Roark [7] containsadditional referencesthan [3-5] and a discussionof internally

pressurizedbellows.

75

References

[1] 5. TimoshenkoandJ. Gere,Theoryof ElasticStability, Mc Craw-Hill, 2nd edition,p.143

[2] 5. TimoshenkoandJ. Gere,Theoryof ElasticStability, Mc Craw-Hill, 2nd edition,p.279

[3] J. Seide, ‘The effect of Pressureon the Bending Characteristicsof an ActuatorSystem,Journalof Applied Mechanics," Sept. 1960,pp429-437.

[4] J.A. Haringx, "Instability of Thin Walled Cylinders Subjectedto Internal Pressure,"Philips ResearchReports,vol. 7,1952, pp.112-118.

[5] J.A. Haringx,"Instability of BellowsSubjectedto InternalPressure",Philips ResearchReports,vol. 7, 1952, pp.189-196.

[6] W. Press,B. Flannery, S. Teukoisky,W. Vetterling, NumericalRecipes,CambridgeUniversity Press.

[711 W. Young, "Roark’s Formulasfor StressandStrain", 6th ed., MacCraw-Hill, 1989.

[8] A. Laupa and NA. Wet "Analysis of U-shapedExpansion Joints", Journal ofApplied Mechanics,March 1962, p.112.

76

Sprinçj-supportedBellowsmodel:rnaxelongvs pres- .1. * .1...

SscK 2760.0in

LIQUID D= 15.0 inAI

£ A zet 0.03 inktl= 3.OE+07 Ibsinkt2 3.OE-f 07 Ibsin

k1 7.OE+03 lbs/Ink2 7.OE+03 lbs/rn41=0.000 ind20.000 in

A I

A

A Aa

A£

0-----

o zoo 400 500 000 1000 1200 1400 1600 1500 2000 2200press psi

$Z$DUA7<RQSSUU.DUC>FOROOI.DAT;51 NC URE B 1

0

000Xxg,

0 XaU’0

0 LIV

a

200

k11

0

0

k,1

K= 2760.Olbs/rn

0= t5.0 inI ktl=0

2 1±1= I.OE+043 ktl= 2.0}H-054 ktl= 3.OE+05

5 ktl= 1.OE+0ft6 ktl= 1.OE-t-O77 ktlinf

Spring Supportedmodel:ki vs Pres

LICINIA4-‘-30*4VSs-I

4yN4

VN

9* 4

ssc

V.

9* 49U4

*‘

0

VC

x

00

‘C

0

I

0

0‘C

*

x

‘C

= 3d,

N

1e4,

V

= 7e30

0000N

Iv

= 7e3

‘C

Pres psi1400 1600 1500 2000

$211UA7:<ROSSLJM.BUC>yo8001 .DAT;?’? FICURE B2

Spring Supportedmodel:ktla vs Pres

Ssc.lC 2760.Olbs/inD 15.0 in1 kl=0

2 kl= 1.OE+03

3 kl= 3.OE+03

4 kl= 5.OE+03

5 kl= 7,OE+036 kt= 9.OE+037 klinf

U

I

Prospsi

$2$DUA7:CIIOSSUII.SUC>POROOI.DAT;102 FICURE B3

Spring Supportedmodel:ktlb vs PresC00 * .1... .1 * I. * I... I.

N* LEVIS SSC

A10 +*2 K 2760.Olbs/mno x.3o

D 15.0 ino *.I uo 4-1 kl=0

00o 2 kl= 1.OE+03

1 3 kl=o 4 k1 5.OE+0320- 5 kl= 7.OE+0300N 6 kl= 9.OE+03

do

‘

lAm k116 = KD2/4klinf

&j-too,O

I--

-- I*0

II

I. U0 U00 U0

k.i1e4, ki=7e3

00o k41 = 3e7, k1 = 74

00

0 200 400 500 800 1000 1200 1400 1500 1500 2000Prospsi

$2$DUA7:cROSSUtBUC>FOR001.DAT;77 FIGURE 114

Spring-supportedBellowsmodel:maxelongvs pres

55K= 2760.0in

D 15.0 inzet 0.03 inktl 1.OE+04 ibsinktE 1.OE+04 Ibsink1 7.OE+03 lbsfln

k2 7.OE+03 lbs/ind10.020in42=0.020 in

p

Co pros psi

$2$DUAT:<ROSSUM.BUC>FOROOI.DAT;75 FIG URE B5

Spring-supportedBellowsmodel:maxelongvs pres

85K 2760.0in

0= 15.0 inzet 0.00 inktl= 3.OE+07Lbsin

kt2= 3.OE-i-07 Ibsink1 7.OE+03 lbs/in

k2 7.OE+03 lbs/ind10.020ind20.000in

pros psi

$2$DUA7;CROSSUM.BUC>F0R00I.DAT;S6 FIGURE BO

Spring-supportedBellowsmodel:maxelongvs pres* * *..l__ -._ * * I . - - I

A

A

A

A

A

A

A

A

AA

A££

A

8scK= 2760.0inD= 15.0 inzet 0.03 inktl 3.OE+07 ibsin

kt2 3.OE+07 Ibsin

k1 7.OE+03 lbs/Ink2 7.OE+03 lbs/in41=0.020 in

420.000 in

I-’,

0 200 400 500 500 1000

pros1200 1400

psi1500 1500 2000

L1CU*A I

0-

C

4-

pI-sn

‘0

N

-4.

A

£

A

AA

AA

A

2200

$2$DUA7:cROSSUILBUC>FOROOI.DAT;53 FIGURE B7

Spring-supportedBellowsmodel:max elongvs pres- * * I * I I. I. * .1 .1 -

- ssc- K 2760.0in

IJOiND D 15.0 inA1

C- A - zet 0.03 inA 1

A ktl= 3.OE-f07 ibsinA

kt2 3.OE+07 ibsink1 7.OE+03 lbs/In

A k2 7.OE+03 Lbs/in41=0.020 ind20.020 in

A IA I

N- A -A

£6

0 200 400 500 000 1000 1200 1400 1600 1500 2000 2200pros psi

FIGURE 88

AA

AA

$2$DUA?;-cItOSSUM.BUC>VOROOI.DAT;S?

Spring-supportedBellowsmodel:yvs x at 300 psi.1. *. I.. I I. * I - .1. - .1. *

a. sSc-

A K 2760*Oin15.0 in

0 A-Io zet 0.03 inA- ktl 3.OE+07 Ibsin

- Akt2 3.OE+07lbsin

- k1 7.OE+03 lbs/In

k2 7.OE-i-03 lbs/in

0 d10.020 in

I A - d20.020in-9 - I

A

z in

$2$DUA7:CROSSUM.BUC>FOROOI.DAT;S7 FIG URE 89

Spring-supportedBellowsmodel:max elongvs pres- I - I * I I I. * I * I * I. I -

- RHIC- K 1800.0 in

- D 8.1 inAI -

A - - zet 0.03 in- ktI 3.OE+07 ibsin- kt2= 3.OE÷07 lbsin

- - k1 7.OE+03 lbs/InA - k2= 7.0E+03 Lbs/in

- - 41=0.000in- d2=0.000in

A IH

N- -A

A

AA

-4.

LAA A

I

A

I -o 200 400 500 500 1000 1200 1400 1600 1800 2000 2200pros psi

$2$DUA7I<ROSSUWDUC>FOR00i.DAT;52 FIGURE BRI

- Spring Supportedmodel:ki vs Pres

RHICK 1800.OLbs/inD 8.1 in

1 ktl=02 ktl= 1.OE-I-043 ktl= 2.OE+054 ktl= 3.OE+055 ktl= 1.OE-i-06

6 ktl= 1.OE+077 ktlinf

0000N

0000N

00

0000-4

C0

8I 0

0000-4

00

0000N

Co Prespsi

$2$DUAI:CROSSUM.8UC>FOROOI.0AT78 FIGURE BR2

- Spring Supportedmodel:ktla vs Pres

RHICK 1600.Olbs/inD 8.1 in

1 kl=02 kI= 1.OE+03

3 kl= 3.OEtO34 kl= 5,OE+03

5 kl= 7.OE+03

6 kl= 9.OE+037 klinf

U

I

Pros psi

$2$DUAYXROSSUILDUC>YOR001.DAT;149 FIGURE BR3

00000N-4

0000-00-4

- Spring Supportedmodel:ktlb vs Pres

SCSIA14-Ixml04V-SU-I‘4-,

000.0

000C0-0

a0

lAm k,16 = KD2/4k1 -.o010

RHICK 1800.Olbs/in0= &lin

1 kl=02 kl= LOE+033 kl= 3.OE-i-034 kl= 5.OE+035 kt= 7.OE+036 kl= 9.0Et037 klinf

a UatIrdII...,

*I% -U

UI

U -U

k11 = 1e4, Sci = 7t3 k11 = 3d, k1 = 74 -

/

400 500 600 1000 1200 1400 1600 1500 2000Prospsi

0-.0

00

I0

Co‘0

200

$2$DUA7:<ROSSUILBIJC>f05001.Dfl;IO5 FIGURE 804

RHIC

K= 1600.0 inD filmzet 0.03 inktl= LOE+04 ibsinkt2 1.OE+04 lbsin

k1 70E+03lbs/Ink2 7.OE+03 lbs/in

dl=0.020 ind2=0.020in

Sprint-supported Bellowsmodel:max elongvs pres

LIQUIDA I

C - a A

I,AA

A

A A

A

A

-9C

A

A

AA

AA

A

A

A

A

A

A

A

AA

A

AA

N

A

AA

AA

AA

A

AA

A

AAA

,1A

A

AA

A

0 200 400 500 800 1000 1200pros

I400psi

1600

$2$DUA?<ROSSUM.BUC>FOR001.DAT?9 FIGURE BR5

Spring-supportedBellowsmodel:max elongvs pres

RHIC

K 1800M in0= 8.1 inzet 000 inktl= 3.OE+07Ibsin

kt2 3.OE+07 Ibsink1’ 7.OE+03 lbs/Ink2 7.OE-i-03 lbs/in

d10.020 ind20.000 in

.8

‘0 pros psi

$2$DUA7:<ROSSUM.BUC>FOROOI,DA1;56 FIG URE BRa

Spring-supportedBellowsmodel:maxelongvs pres

RHIC

lC 1600.0 inD 6.linzet 0.03 inkt1 3.OE+07 ibsinktE 3.OE+07lbsin

k1 7.OE+03 lbs/Ink2 7.OE-s-03 lbs/in41=0.020in420.000 in

-9C

lOGO 1200pros psi

$2$DUA%<ROSSUM.DUC>F01100l.DAT;64 FIGURE BR?

A

A

Spring-supportedBellowsmodel:maxelongvs pres

-I ..I. I I. - I - I I. I -

RH1C

K= 1800.0inD 8.lin

A1 -0- A

_________

- zetfl.O3tnA 1

AA kt1 3.OE+07lbsinA

A kt2 3.OE+07Ibsin

k1 7.OE+03 lbs/In- k2 7ME+03 lbs/in

d10.020 in42=0.020in

.9 IA

AA

NA A -

A

0 200 400 - 600 800 1000 1200 1400 1600 1500 2000 2200- pros . - psi

FIGURE BRS$2$DUA7:<ROSSIJIL.BUC>FOR001.DAT;SS

Spring-supportedBellowsmodel:yvs x at 300psi

- £ I - I I I I

A RHIC

A K 1600.0 inD= 6.1 in

AI -a A zet 003 in

ktl 30E+07 lbsinkt2 3ME+07 Ibsin

* A

- ki 7.OE+03 lbs/In

k2 7ME+03 lbs/InA d10.020 in

I - 42=0.020 inC I

A

A I

I

AC-

80-0

100 A

0

0 1 3 4 5 7x in

$2$DUA7:cROSSU$.SUC>FOR00I.DAT;88 FIGURE BRO

Spring-supportedBellowsmodel:max elongvs pres-- ---I I I I - I. I .1 I I -

- ERIC

K.= 2732.0inD= 12.3 in

AIA

A 1- zet 0.03 in

A ktl 3.OE+07ibsina kt2= 3.OE-i-07 lbsin

A kl= 7.OE+03 lbs/In- k2= 7.OE+03 lbs/in

41=0.020in42=0.020 in

.9 I1

A

IA I

N AA

00 200 400 500 800 1000 1200 1400 1600 1500 2000 2200

pros psi

AA

AA

$2$DUA7<ROSSUM.SUC>FOROOI.DLT;59 FIGURE BR1 0

- MAX BELLOWSLOCAL ELONGATIONq

____________________________________________________________________

I I I I. I I I

SSC dddzet6=0M30 in

0

icoum A - zet7=0000inA1U4-IL’

zetl3O.000inA

C . zetl4=0.000 in

- et6=O.000in- e17=0.000in

0

- etl3=0.000 in- etl4=0.000 in

q - A ds60.020inN - -

£9 4s130.000in

f6 0.0 lbs0

- f13 0.0 lbsA

K 2759.2lbs/InA

C.1- -

AtA

A AqN A A DISk

A%A A

.4.

9?tIiII,uuIuIIIIu1IyuuIuIuIuIyIuIuIuILIIIlIIlIIIIIIur+I+fflInhIIIIIIIIIIuIII

OLD0-L IlIIIII

0 100 200 300 400 500 500 700 500 000Pros psi

FIGURE DlS2$DUA7:CROSSUM.BUC>VOlt00l.DAT;20

plot file: 2DUA7;[ROSStJM.BUCFOROOi.DAT;90machine: SSC mode: ddd

Table 1: Input: bellows design parameters

r in Dj in din 0 un pcr psi Un Ebpsi

0.060 14.300 0.500 0M94 L300 450.000 1.000 3.297E+07

Table 2: Results: bellows properties

Nb tin Din Am tin bcin KIbs/in Kllbs/in ha4

26 0.072 14.800 0.356 0.025 0,342 2759.212 L048E-+04 3.151E+01 ITable 3: Results: maximum bellows stresses at P.p

0,n. PI Cp 51 9pst pSI °,.1 P" 0p. lena W

1.523E+05 2.778E+04 6.640E+04 6.782E+04 2.967E+04 1.472E-03

Table 4: Input: bellows misalignments

6 in 7 in 13 in 14 in ‘i5 r7 ii3 ‘114

0.030 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

d16in d13in &6in d1l3in0.020 0.000 0.020 0.000

Table 0: Input: dipole properties

s in LD in ovhj, in D in t,, in 1mM4 Em psi0.200 670. 63.6 10.5 0.188 111. &000E+07

Table 7: Input: quadrupole properties

I L,in I I D,in t,in I hqin4 I Eqpsi II 210. 52.5 I 10.5 I 0.188 I I 3.000E+07

Table 8: Results: bellow, maximum elongation at p.,

I *1 I dl6in I dll3in II 3.000E+02 1.262E+00 I 1.000E+oo I

FIGURE D1A 97

Table 9: Results:node location,stiffness, force at

node xi in ki lbs/in Fi lbs1 0.00 25000.00 87.54

2 135.69 25000.00 35.69

3 271.38 25000.00 24.72

4 407.07 25000.00 17.59

5 542.76 25000.00 .12.71

6 606.38 0-00 0.00

7 615.68 0.00 0.00

8 679.30 25000.00 115.289 814.99 25000.00 -13.68

10 950.68 25000.00 18.37

11 1086.37 25000.00 26.95

12 1222.06 25000.00 27.80

13 1285.68 0.00 0.00

14 1294.98 0.00 0.00

15 1358.60 25000.00 27.45

16 1494.29 25000.00 24.13

17 1629.98 25000.00 22.63

18 1765.67 25000.00 -35.16

19 1901.36 25000.00 87.70

FIGURE D1B

98

- MAX BELLOWSLOCAL ELONGATION1

£ I * I .1. I.. I. * I..,. I. * I

-.SSC* 444

zetGO.000inzetlO.000 in

0,- -.4 zotl3O.000 in

zetl4O.000 in

o etBO.000 in0

2 etlO.000 inet130.000in

- et140.000in- ds&0.000 in

.9 4s130.000inf6 0.0 lbs

I - f13 0.0 lbs

- K 2759.2 lbs/in

N00--4

-DLI

C

0 100 200 300 400 500 500 700 800 000Pros psi

$2$DJj7:CROSSIJM.BUC>FOROOI.flAT;92 FIGURE D2

plot file: 2DUA7:[ROSSUM3UCJFOROOI.DAT;92machine: SSC mode: ddd

Table 1: Input: bellows design parameters-

r,in Thin din 0 un pcrpsi Un Ebp,i

0.060 14.300 0.500 0.094 9.300 450.000 1.000 3297E+07

Table 2: Results: bellows properties

Nb tin Din Am tin bcin Klbs/in Kllbs/in 11,026 0.072 14.800 0.356 0.025 0.342 2759.212 1.048E+04 3J51E+01J

Table 3: Result,: maximum bellows stressesat

0"P1 I oPsi I o1psi 0’,.2P51 I ,. I y.nain

1.523E+05 2.778E+04 &640E+04 6.782E+04 2.967E+04 1.472E.-03

Table 4: Input: bellow, misalignment.

6 in 7 in 13 in 14 in q6 $7 n13 nl4

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

d,6 in d,13 in d6 in d113 in0.000 0.000 0.000 0.000

Table 8: Input: dipole properties

sin Lv in °1’D D D in t in 1mM4 Em psi

0.200 670. 63.6 10.5 0.188 111. 3.000E+07

Table 7: Input: quadrupole properties

L, in ovhqin D, in t, in Iqin4 Eqpsi

210. 52.5 10.5 0.188 111. 3.000E+07

Table 8: Results: bellows maximumelongation at p.,

p..,psi dl6in dll3in

I 3.000E+02 L000E.-e-00 1.000E+00 J

FIGURE D2A 100

Table 9: Results: nodelocation, stiffness,force at p,

node xi in ki lbs/in Fi lbs1 0.00 25000.00 87.70

2 135.69 25000.00 35.173 271.38 25000.00 22.614 407.07 25000.00 24.10

5 542.76 25000.00 27.576 606.38 0.00 0.00

7 615.68 0.00 0.00

8 679.30 25000.00 27.41

9 814.99 25000.00 24.9010 950.68 25000.00 25.7311 1086.37 25000.00 24.89

12 1222.06 25000.00 27.4213 1285.68 0.00 0.0014 1294.98 0.00 0.0015 1358.60 25000.00 27.5716 1494.29 25000.00 24.10

17 1629S8 25000.00 22.6218 1765.67 25000.00 35.1619 1901.36 25000.00 87.70

FIGURE D2B

101

- MAX BELLOWSLOCAL ELONGATION- I.. * I I.. .1. .1.. .1. I.

4.SSC ddd

- zetOO.030 in0

LISIN

______________________________________

- zeL70.000 in- A.aM A DLA4-813- zetl3O.000inA

q A - zetl4O000 inA et60.000 in

A et7=O.000in04 A

- et130.000inA

et140.000inA

q A - ds80.020in*1

9 A 4s130.000inA

t6-12500.0 lbsA

A + - f13= 0.0 lbsA K 2759.2 lbs/in

A

C ASI. -

A

A +

A

q Aor

A

A +A

+AA -+

A +A +

+ DLB0

‘-IIlItlIIJIIJIIIUIIiIIiIIJIYIIJIUIIJIU.w._ -.1-- t’i.ii£00 200 300 400 500 000 700 500 900

Li Prospsi

$2$DUA7:<ROSSUM.UUC>POR001.DAT;185 FIG URE D3

plot file: 2DUA7:[ROSSUM.BUCJFOROOI.DAT;l85machine: SSC mode: ddd

Table 1: Input: bellow, design parameters

rin Thin din /3 tin pcrpsi in Ebpsi0.080 14.300 0.500 0.094 9.300 450.000 1.000 3.297E+07 I

Table 22 Results: bellows properties

Nb rin Din Am tin bcin Kibs/in Kllbi/in 110

26 0.072 14.800 0.356 0.025 0.342 2759.212 1.048E+04 3.l5lE+0l ITable 3: Results: maximum bellows stressesat p.,

Cm.. Psi ek psi t.i psi en pSi tip. i,.. UI

1.523E+05 2.778E+04 6.840E+04 6.782E+04 2.967E+04 1.472&03

Table 4: Input: bellows misalignment.

6 in 7 in 13 in 14 in q6 ii7 iia ‘l40.030 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

46 in 413 in 46 in 413 in0.020 0.000 0.020 0.000

Table 8: Input: dipole properties

s in LD in ovh in Den in in 1w in4 Em psi0.200 670. 63.6 10.5 0.188 111. 3.000E+U7

Table 7: Input: quadrupole properties

L, in ovhqin D, in 1, in Iq in4 Eq psi

210. 52.5 10.5 0.188 111. 3.000E+07

Table 8: Results: bellows maximum elongation at p.,

ni I dl6in I dll3in II 3.000E+02 I 5.000E+00 1.002E+00 I

FIGURE D3A103

Table 9: Results: node location,stiffness,force at p.,

node xi in ki lbs/in Fi lbs1 0. 25000. 112.

2 136. 25000. 114.3 271. 25000. -549.4 407. 25000. -1681.5 543. 25000. 7975.

6 606. 0. -12500.7 616. 0. 12500.

8 679. 25000. -7870.

9 815. 25000. 1685.

10 951. 25000. 594.11 1086. 25000. -51.

12 1222. 25000. -3.

13 1286. 0. 0.14 1295. 0. 0.15 1359. 25000. 33.

16 1494. 25000. 23.

17 1630. 25000. 23.18 1766. 25000. 35.

19 1901. 25000. 84.

FIGURE D3B

104

SSC dqdzetfiO.030in

zet70.000inzetl30.000 in

zett4O.000in

et60000 in

et70.000inet130.000in

et140.000in

ds60.020in4s130.000inf6=-12500.Olbs

13 0.0 lbs

K 2759.2lbs/in

MAX BELLOWSLOCAL ELONGATION

a-I-4

0 ioa 200 300 400 00 000 700 500 900

Pres psi

$2$DUA7<ROSSUM.BUC>FOROOI.DAT;l0 FIGURE D4

plot file: 2DUAT:[ROSSUM.BUCJFORO0I .DAT;i0

machine: SSC mode: dqd

Table 1: Input: bellows design parameters

rjin D4in din liii pcr psi Un Ebpsi

0.060 14.300__j_0.500 0.094 9.300 450.000 1.000 3.297E÷07

Table 2: Results: bellows properties

Nb

26

rin Din Am

0.072 14.800J__0.356__

f_tin0.025

bcin KIbs/in Kllbs/is un4

0.342 2759.212 1.048E+04 3.151E+01

Table 3: Results: maximum bellows stressesat p.,

Om.PSi tihP5l I OI5 ti3p5i 0_p. Y,naLfl

I 1.523E+05 I 2.778E+04 I 6.640E-i.04 I 6.782E+04 I 2.967E+04 j 1.472E-03 ITable 4: Input: bellows misalignment.

6 in 7 in 13 in 14 in ,6 q7 t13 71140.030 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

46 in 413 in 46 in 413 in0.020 0000 0.020 0.000

Table 6: Input: dipole properties

I sin I LDin ovh0in I Dmin t,in I 1mM4 I Empsi II 0.200 670. I 63.6 10.5 I 0.188 I I 3.000E+07 I

Table 7: Input: quadrupole properties

I L,in I ot’hqin I D,in I t,in I IqiO I Eqpsi

I 210. 52.5 I 10.5 I 0.188 iii. I 3.000E+07 ITable 8: Results: bellows maximumelongationat

p.psi dl6in dll3in

I 3.000E+02 I &000E+00 1.502E+00

FIGURE D4A106

Table 9: Results: node location, stiffness, force at p.,

node xi in ki lbs/in Fi lbs1 0. 25000. 96.

2 136. 25000. 119.

3 271. 25000. -580.

4 407. 25000. -1806.

5 543. 25000. 8384.

6 608. 0. -12500.

7 616. 0. 12500.

8 668. 25000. -8502.

9 694. 0. 0.10 121. 0. 0.

11 747. 0. 0.

12 773. 25000. 2067.

13 826. 0. 0.

14 835. 0. 0.

15 899. 25000. 731.

16 1034. 25000. -193.17 1170. 25000. -30.

18 1306. 25000. 45.

19 1441. 25000. 69.

FIGURE D4B

107

ASSEMBLYDEFLECTIONSAT 303.5psi.1.. I. .1 .1...’.. - * .1... 2

0-

- SSCddd

a zet6O.030in

‘loINs zet7r=OMOOin

8 2 - zetl3O.000inzetl4O000 in

Aet60.000 in

et7O.000Iii

A etl3O.000 inet140000indsOO.O2Oun

Ads130.000int8 0.0 lbs

A f13 0.0 lbs- K 2759.2lbs/in

rprr’

DIPOLF DIPU’E

__________________

rse ow; araows

0

0 200 400 500 800 1000 1200 1400 *800 1800 2000Xiii

FIGURE D5$21DU17<ROSSUM.BUCF0R00l.0AT;90

FIRSTBELLOWSDEFLECTIONAT 300 psi

0 £- SSCddd

zet6O.030 inA

ALIOsNI zetl=0.000in

AnY

zett3o.000inA zetl40.000 in

et60.000 in

m et70.000inA

et130.000in-

- et140.000inds60.020in

9 ds130.000in- f6 0.0 lbs

A 13- 0.0 lbsKz 2759.2 lbs/in

CDl0- -C

A

C

Y -0

A -

00

606 60? 606 609 110 611 6*2 613 614 615 616 6*?Xiii

$2$DUA7:CROSSUM.UUC>FOR00*.DAT;90 FIGURE D6

SSC dqdzetBO.030inzetY0.000 in

zetl3O.000inzetL4-.030 inetOO.000 inetThO.000 in

et130.000 inet140.000 inds60.020inc1s130.020 inf6 0.0 lbsf13r 0.0 LbsK 2759.2lbs/in

MAX BELLOWSLOCAL ELONGATION

.9

400 500Prospat

FIGURE Dl$2$DUA?:<ROSSUM.BUC>FOROOI.DAT;9?

MAX BELLOWSLOCAL ELONGATION.1. -.1. * .1... * .1. * .1.. .1 * .1.

C -

- SSC ddqzet6O.000in

C4+

- zetlO.000 inzetl3=0.000in

zetl4-.030 in

- etG=0.000inet?0.000in

0

- etl3-O.000 in- etl4=O000in

‘a - ds@0.000in55_ +

9 ds130.020inf6 0.0 lbs

+- f13 0.0 lbs

K 2759.2 lbs/in+

CDl. +

+ +

a + + Dlii+

-**rt +0 100 200 300 400 600 600 700 eoo coo

Pros psi

$2flUfl:cROsSUM.BVC>FOfto0i.DAT;96 FIG URE D8

MAX BELLOWSLOCAL ELONGATION* .1 * - I * * *

RHIC dqdzetaz0.030in

sans zet70.000in+-JI*a -

a zetl3=0.000inzetl4zrOMOO inet60.000inetl=0.000 in

etl3=0.000in

etl40.000 in

4s60.020in

.9 ds130.000in- f6 0.0 lbs

A f13 0.0 lbs+ K 1784.5 lbs/in

Dl- -4’

DIA

0 - 100 - 200 30 400 500 - - 600 * 700 - 800 MOOProspsi

$2$0UA7:<ROSSUI&.DUC>VOR00j.DST;91 FIGURE DR1

plot file: 2DUA7:[ROSSUM.BUCPOROO1.DAT;91machine: RSIC mode: dqd

Table 1: Input: bellows design parameters

is-in Thin din j fi un pcrpsi Un Ebpsi

0.060 7.630 ro.500 0.094 6.000 450.000 0.500 L297E+07 ITable 2: Results: bellows properties

Nb tin Din zUn tin bcin Klbs/in Klinlbs fin’

17 0.072 8.130 0.352 0.024 0.346 1784.526 4.915E+03 5.137E+00 ITable 3: Results: maximumbellows stressesat p

0m5. p51 0_h psi °t psi °n psi 0_. y,, in

1.066E+05 1.534E+04 6.869E+04 tOl3E+04 3.071E+04 1.548E-03

Table 4: Input: bellows misalignments

6 in 7 in 13 in 14 in li6 $7 i$13 n140.030 0-000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

d56in .413 in 46 in 413 in

0.020 0.000 0.020 0.000

Table 6: Input: dipole properties

1 sin LDIn ovhoin D,,,in t_in 1mM’ Empsi

I 2.00 ‘- I "-‘i I 10.5 o*iss I i- I 3.000E+07

Table 7: Input: quadrupole properties

I L,in oehqin I D,in I t,in Iqin’ I Eqpsi I170. 48M 10.5 0.188 I I 3.000E+07 I

Table 8: Results: bellows maximum elongation at p.,

p., psi dlO in dll3 in3.000E+02 7.511E-01 5.139E-01

FIGURE DR1A 113

Table 9: Results: node location, stiffness, force as p.,

node xi in 11 l/in Fi lbs1. 0.00 25004.00 404.582 70.80 0.10 0.00

- 3 141.60 25000.00 280.094 212.40 0.10 0.005 283.20 25000.00 168.42

- 6 338.60 0.00 0.007 344.60 0.00 0.00

- 8 392.00 25000.00 213-019 411.10 0.10 0.00UI 429.60 0.10 0-00

11 448.10 0.10 0.0012 466.60 25000.00 121.71.13 514.60 0-00 0.00

14 520.60 0.00 0.0015 576.00 25000.00 199.7416 646.80 0.10 0.00

17 717.60 25000.00 278.0318 788.40 0.10 0.0019 859.20 25000.00 403.15

FIGURE DRiB

114

RElIC dqdzet6O.000 inzet7’O.OOO inzetl3O.000 inzetl4=0.000in

etöO.000inet?’=O*OOO inet130.000inetI40.000 inds60.000inds130.000inf6= 0.0 lbsf13= 0.0 lbsK 1784.5lbs/in

MAX BELLOWSLOCAL ELONGATION

.9

‘-‘I

400 600

Prospsi

$2$DUA7:cRQssUJgBUC>Foft001.Dfl;93 FIGURE DR2

plot file: 2DUAT:ROSSUM.BUCJFOROOI.DAT;93machine: R.HIC mode: dqd

Table 1: Input: bellows design parameters

r in D in din p [ tin pcr psi Un Ebpsi0.060 1.630 0.500 0.094 J 6.000 450.000 0.500 3.297E+07 I

Table 2: Results: bellows properties

Nb rin Din Am tin bcin Klbs/ia Kllbs/in17 0.072 8.130 0.352 0.024 0.346 1784.526 4.915E+03 5.137E+00

Table 3: Results:maximum bellows stressesat p.,

tv.’0. psi e, psi e,,, psi o-,,2 psi C,,. y. in1.066E+05 1.534E+04 6.869E+04 7.013E+04 3.OTIE+04 1.548E-03

Table 4: Input: bellows misalignments

6 in 7 in 13 in 14 in ‘16 r17 ‘p13 rj140.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Table 5: Input: bellows initial shape

46 in 413 in 46 in 413 in

0.000 0.000 0.000 0.000

Table 6: Input: dipole properties

5 ID Lp in ovhp in D in tm in 1mm’ Em psi

2.00 394. 55.4 10.5 0.188 111. 3.000E-i-01

Table 7: Input: qnadrupoleproperties

L,in ovhqin D,in t,in Iqin’ Eqpsi

170. 48.0 105 0.188 111. 3.000E+07

Table 8: Results: bellows maximumelongationat p.,

p, psi

3.000E+02

dl6 in

5.051E-01

dlI3 in

5.052E-01

FIGURE DR2A116

Table 0: Results:node location, stiffness, force at p.,

node 11 in ki lbs/in Fi lbs

1 0.00 25000.00 402.982 70.80 0.10 Q00

3 141.60 25000.00 276.464 212.40 0.10 0.005 283.20 25000.00 206.65

6 338.60 0.00 0.00

7 344.60 0.00 0.00

8 392.60 25000.00 149.29

9 411.10 0.10 0.0010 429.60 0.10 0.00

11 448.10 0.10 0.0012 466.60 25000.00 149.2713 514.60 0.00 0.0014 520.60 0.00 0.0015 576.00 25000.00 205.6416 646.80 0.10 0.0017 717.60 25000.00 276.4918 788.40 0.10 0.0019 859.20 25000.00 402.96

FIGURE DR.2B

117

MAX BELLOWSLOCAL ELONGATION

I * .1....’. .1...:..C.

RHIC dqdzet6=O.030inzet7=0.000in

AS ma+5113

____

- zetl3zO.000flU

zetl4=0.000in

/ et6-0.000inetl=0.000 in

aA

- etl3=0.000 in

J etl4=0.000in

4f ds60.020 indsl3=0.000 in

‘3- DLB t6 -3000.0 lbs

* _00 f13 0.0 lbs

/K 1784.5 lbs/in

-.

0-.--.-

100 200 300 400 500 600 700 800 900Prespsi

$2$DUA7;<ROSSIJM.BUC>FOR0Ot.DAt12 FIGURE DR3

plot file: 2DUA7:[ROSSUM.BUC]FOR001.DAT;12machine: B.HIC mode: dqd

Table 1: Input: bellows design parameters

rin Thin din $ f lin pcrpsi Un Ebpsi0.060 7.630 0.500 0.094 6.000 450.000 0.500 3.297E+07

Table 2: Results: bellows properties

Nb rin Din Am tin bein KIbs/ja Kilbu/in Em4

17 0.072 8.130 0.352 0.024 0.346 1784.526 t915E+03 5.l37E+O0

Table 3: Results: maximum bellows stressesat p.,,

0ms* psi ei psi c.,1 psi psi I.. in

1.066E+05 1.534E+04 6.869E.-s-04 7.013E+04 3.071E+04 1.548E-03

Table 4: Input: bellows misaliguments

I 6in 7in I o3iin I 14mn I 6 I ‘;U I ‘X4 II.03o I ooo I 0.000 I 0.000 I 0.000 I 0.000 o.ooo o.ooo

Table 5: Input: bellows initial shape

46 in 413 in 461n d13 in0.020 0.000 0.020 0.000

Table 6: Input: dipole properties

I sin I LDin I ouhin D,,in I ‘mIt’ j 1mM’ I Empsi II 2.00 I I "- I 10.5 I 0.188 I I 3.000E+07

Table 7: Input: quadrupole properties

I Lqin ovhQin D,in I iin Iqmn’ I Eqpsi

I 170. I 48.0 I 10.5 0.188 I 111- 3.000E+07

Table 8: Results: bellows maximumelongationat

f p.,psi dl6in I dll3in

I 3.000E+02 I 2.165E+00 I 6.724E-01 I

FIGURE DR3A119

Table 0: Results: node location, stiffness, force at p.,

node xi in ki lbs/in Fi lbs1 0. 60000. 447.

2 71. 0. 0.

3 142. 60000. -725.4 212. 0. 0.

5 283. 60000. 3457.6 339. 0. -3000.

7 345. 0. 3000.

8 393. 60000. -3541.

9 411. 0. 0.10 430. 0. 0.

11 448. 0. 0.

12 467. 60000. 1465.

13 515. 0. 0.

14 521. 0. 0.

15 576. 60000. 337.16 647. 0. 0.17 718. 60000. 233.

18 188. 0. 0.

19 859. 60000. 395.

FIGURE DR3B

120

RHIC dqdzetfi0.030 in

zet7O.000 in

zetl3O.000inzetl4O.000 in

et60.000inet70.000in

et130.000inet140.000indsOO.020in4s130.000Inf6 0.0 lbsfia 0.0 lbsK 1784.5 lbs/in

ASSEMBLYDEFLECTIONSAT 303.5 psi

.9

L INn

$2$DUA7:cROSSUM.BUC>FOROOI.DAT;91 FIGURE DR4

RHIC dqdzet6tO.030in

zet7O.000inzetl3-O.000inzetl4O.000 in

et60.000inet7=0.000inetl3=0.000inet140.000inds60.020 in4s130.000inf6 0.0 lbs

f13 0.0 lbsIC 1784.5lbs/in

FIRSTBELLOWSDEFLECTIONAT 300psi

-M

tJ JI in

$2$DUA7:<ROSSIJM.DUC>VOROOI.DAT;01 FIGURE fiRs

MAX BELLOWSLOCAL ELONGATION.1...... * .1.... 1....... * .1....

* RHIC qdqzetfiO.030 in

zet70.000in+5612

10 . zetl3O.000in

zetl4-=-.030inet60.000inet70.000 in

et130.000 inet140.000inds60.020 in

a ds130.020inC’ - ff6 0.0 lbs

a a f13= 0.0 lbsa K= 1784.5 lbs/in

aa

aa

0 tOO 200 300 400 000 600 700 800 900Pres psi

$2$DUA7:<ROSSUM.BUC>FOROOI.Dkt;102 FIGURE DR6

MAX BELLOWSLOCAL ELONGATION.1... I * .1..... * .1. * I -

ERIC dqdzet60M30 in

LIQUID zet70.000ina. 5tH+ .613

a - zetl3O.000inzet14r0.000in

AetOO.000in.

+et70.000in

- et130.000 inet140.000 in

A ds60.020inds130.000in

a + - t6= 0.0 lbs

+ t13= 0.0 lbsl@ 2732.3 lbs/in

AA

ACa -

Af ADUA

A A

Al- +A

A tow

- __#t+ ++

fllffiq99m?itIIIIIIII,IIIIIlIIIIIIIII1I11I7 11111111 W.

o -0 100 200 300 400 500 600 700 000 900

LN Prespsi

$2$DUA7:dtOSSUM.SUC>FOROOI.DAT;106 FIGURE Dli?

MAX BELLOWSLOCAL ELONGATION.1 I. * .1 * .1. .1

RUIC dqd

zet6O.030inzet7rO000 in

+. DLIC. . zetl3O.000 in

a zetI4-.030 inet60.000inetlO.000 in

aet130.000inet140.000in

a ds60.020 in

ads130.020in

at6= 0.0 lbs

a f13 0.0 lbsa V.- 9’7q9 1 1K J

a DLI Ar

aa a

aa

0 100 200 300 400 500 600 700 800 900Prespsi

$2$0UA7<ROSSUM.SUC>FORo01.OAT;104 FIGURE DR8

AppenditMACSYIvL4codefor magnetbellowsassembly

There are 19 supportsand 10 nodesbetweensupports. On Fig. eS the deflected

shapeof the bellowsis definedasa function of i numberof the spanbetweentwo supports

and j nodenumberon this interval. For instanceif thereare 10 nodesbetweensupports,

yl8,1O denotesthe deflectionof the last nodeof the eighteenthspanbetweensupports18and 19. y is definedin termsof threefunctions: one function containingthe unknownsxa,

xb correspondingto A, B in eq. 64, one function which is a polynomial expressionof x

denotedh, and one function of trigonometric expressionsof x. y is split into threeexpressionsto reducethe numberof operationsthat MACSYMA has to perform, this will

be shown later. Similar functions are defined for y’ in Fig. e6. One could have useda

commandto differentiatey eachtime that y’ is called but in order to minimize the number

of operationsthat MACSYMA has to perform,explicit expressionsfor y’ areprovided. On

Fig. e7 the functionswhich are called in the polynomial part of y aredefined. ZO, ZC, ZD

correspondto 0, C, D, in eq.64.

FigureeS showsvarious functions, the initial shapeof the whole systemdenotedby

yO in the text is calledherefO, all its derivativesarealso explicitly providedto increasethe

speedof executionof the program. On Fig. e9, one finds the bellows offsetsc and r, and

E5, E of eq.64 . Fig. elO showstheboundaryconditionsand the momentequation. In eq.62

two expressionsof y for adjacentintervalsbut at the samex locationaresubtracted.Since

the trigonometricfunctionsdo not cancel,operationsareperformedonly on thepolynomial

part of y functionh, thussavingunnecessaryoperationsandjustifying the initial definition

of y in termsof threefunctions.

On Fig. eli one finds the functionbci calling all the boundaryconditionsstarting

with eq.58 which expressesequilibrium of forces. Function dfor first calls a boundary

condition equationbci, then isolateswith the commandcoeff all the coefficientsof the

unknowns ui which it denotesby qi,j in the fortran expressioncreatedfor further

processing.Thesecoefficients,multiplied by the correspondingunknowns,aresubtracted

from bci to obtainthe remainderwi. wi is the ith componentof vector W in the final

matrix equationX [0] =W which will be invertednumericallyto obtainvector X containing

126

A, B of eq.64. Functionpfor opensa file to which the fortranexpressionswill be written,

then calls dfor describedabove,andalsocreatesfortranexpressionsfor C, D, Es, Ec and

O which are neededin addition to A, B in eq.64 to obtain y.

127

AppendicFortran codefor magnetbellowsassembly

The numericalmethodconsistsof a simple matrix inversion. The systemof three

magnetsand two bellows is modeledby a grid of 19 points. Bellows are locatedbetweennodes 6 and 7 and betweennodes 13 and 14. Although there is a total of 19 nodescorrespondingto the assemblyof three magnetswith 5 supportseach,one can set the

supportstiffness to zero and achievethe desiredcombinationof dipoles and quadrupoles

with any number lessthan 16 of supports. The programtreatsRHIC and SSC magnets

andall applicablecombinationsof dipoles and quadrupoles.

Themain programstartsby defining the sizeof the matrix to be inverted,the bellows

canhavebent shapesof sineor cosine functions,ds6 anddcó refer to the sine andcosine

componentsfor the first bellows while the index 13 refers to the secondbellows.

The values f6, f7, f 13, f 14 of array f define the external forces which are

appliedat the bellows nodesused to study interactionbetweenbellows.

All other values of f are the supportforces on the posts. The critical pressurepcr is the

pressureat which the systemwill first becomeunstable,by settinga safetyfactor onegets

an operatingpressurepop. The bellows is then designedto fail at this critical pressure.

The options for machinetype and magnetsequencearegiven next.

SubroutineGEOdefinesnodelocation andmaterialpropertiesusing the previously

selectedoptions. SubroutineAX specifiesall the bellows designdetailsto build a bellows

whose axial stiffness is such that it will fail at the previouslydefined critical pressure.

SubroutineINT appliesonly if one wantsto know what the stressesin the bellows are once

they are internally pressurized. INT calls COEFINT where MACSYMA generated

expressionsfor the constantsof integrationsare given. It alsocalls PLOTINT andENCINT

for plotting the deflectedshapesandstresses.

MAT gives the material properties. In order to clearly seethe peakat the critical

pressure,the mesh is refmed betweenpressurespin2 to pfin2 which are centeredaround

pzo=pcr. Alternatively one could have a uniformly fine mesh of icm points by setting

mu =1. The overall meshstartsat pin andendsat pfin. SubroutineFORCE simply setsup

the array presof pressureswith variablemeshdescribedabove.

128

The valuesfor omegaappearingin thesolutionsof thedifferentialequationsare then

defined. They are neededin the subroutineCOEF which arethe MACSYMA generated

coefficientscalled Qi,j, the vector of right-handsides is W. Q and W are copied into

arraysA and B andinvertedusingthe numericalroutinesLUDCMP andLUBKSB from [6].

Upon inversionof the matrix, array B containsthe solutionsXA, XB, F which are needed

to obtain the explicit expressionsfor deflections,slopesandcurvaturesy, y’, y". Additional

coefficientsE, 0, D, which areneededfor theseexpressionsaregenerated by MACSYMA

are given by subroutineIND. Finally the maximum local bellows elongationDL can be

computed.

SubroutinePLOT producesfour types of output in addition to a table of values.

Option 1 plots maximumlocalbellowselongationversuspressure,thereare two curves,one

for eachbellows. Option 2 plots support deflections for eachsupport versus pressure.

Option 3 gives the deflectedshapeof the magnet-bellowsassemblyat pcr andpop. Option

4 providesthe deflectedshapeof the bellows at pcr andpop. Thesearewritten in the plot

file FOROOI.DAT. The Tablecontainingall the variousdesignparametersandsummary

of the output is in the file ECHO.TEX which must be further processedwith the TEX

program.

129

AppenditFile namesof additional Codes

The main program is in BELFOR, the commonblock is in BELCOM.FOR, output

routines are in the file BELOUT.FOR, material and geometric properties are in

BELGEO.FOR,bellows designparametersandoutputare in BELAX.FOR, the numerical

algorithm for matrix inversion are in BELNUM.FOR, the MACSYMA code

BELLOWS.MAC generatesthecoefficientswhich arein BELCOEF.FOR.BEL.COM is the

commandfile which compilesthe main programand links all the othersubroutines. The

bellows designparametersarecomputedin themain programfor onebellows,theyarealso

available for a numberof casesin BELDES.FORwhich includesBELDESCOM.FOR.

The spring-supportmodeldeflectedshapeis computedin SPRING.FORwhich calls

the subroutinegeneratedby the MACSYMA codeSPRING2.MAC,SPRINGCOEF2.FOR

and includesSPRINGCOM.FOR as the commonblock. This routine also computesthe

three roots of the determinant=0 equationwhich are createdby the MACSYMA code

DEtMAC.

130

c24 dispfunbcl,bc2,bc3,bc4,bcs,bc6,eq;

1 .2e24 bcl fi + 12- 1w --- -

ktlkt2

1112 iie25 bc2 :=.1+.2_fa

klk2 ktl

.1e26 bc3 cypo -

ktl

.2e27 bc4 cypl -

kt2

fte28 bcs expandratsi.pcyO- z -

‘U

12e29 bc6 cyl - -

k2

ft fyi].fayx+yO---+flx------x+il

‘U ktte30 eq expandyp

ei

d30 done

c31 dispfunsol,pfor;

en solo sot partsolveleqfl, tunIc, 1

e32 pfor writefile"workcoef.for",

detrig : triqsiipdenoirbspartsol,1,

detrigdetsi.p expaiid-- , fortrandeteri substls,detsiip,

lklk2ktlkt2

Folk lIEU 6 DO fortransubstls,s3k, fortrancd= substJ.s,xdfl,131

fortrancq substts,xgfl, closefilefl

FIGURE E4d32 done

CS "deflectedshapey, i is the interval and j the node nuiberi

c6 dispfuny;

e6 yi, j abi, j + hi, j + hefl, j

d6 done

c7 "ab contains the trigonoaetric functions of x"$dispfunab;

c81e8 abOi, j := xai coso.i xj + xbi sinoai xj

d8 done

c9 ii is a polyno.ial expressionof x"$dispfunb;

dO2

elO bi, j zci xj + zdi x j + zqi

dlO done

dil "he containsthe trigonoietric expressionsdue to the prebentsbape9dispfunhe;

d12e12 hei, j := IF i = 6 THEIC xinit x6

ELSE IF I = 1 TEE! xinit x13, zesi sinpil xj - xinit

+ zeci cos2p11 xj - xinit

dl2 done FIGUREES

132

d13 "derivative of y w.r.t. x’$disptunyp,pab,hp,hep;

c14e14 ypi, j := pabi, j + hpi, j + hepi, j

elS pabi, j :: - o.i xai sinoii xj + o.i xbi cosori xj

e16 hpi, j :: zci + 2 zdi xj

e17 hepi, j := tF 1 6 THEN xinit x6

ELSE IF 1:13 THEN xinit x13, zesi pit cospil xj - xinit

+ zeci - 2 pit sin2 p11 xj - dnit

d17 done

c18 "list of unimowns’$dispfunu;

d19e19 ui : IF i :18 THEN xaj ELSE IF i 36 THEN xbi -18

ELSEIFI a 41THENfi -36 UE IF1 <:46Tfi-34

ELSE IF i <= 51 THEN fi - 32

d19 done

FIGURE E6

133

c69 "functions called by functions h and hp"$dispfunzg,g,zc,zd;

c70gi - 2 zdi

e70zgi ::xgc:--_--_________,IFi=6THENxgc-dc6ai

ELSE IF i = 13 THEN xgc - dcl3 ELSE xgc

fa f1e71 gi :: IF I = 1 THEN gin

sk1

110ELSE IF i = 10 THEN gin ft ------ + 1010

sk10

11 110 gin + suLga.j + 1j xj, j, 1, i+ Ia ----- - ------ - 1010, ----------------------------------

sk1 skOlO aili

bi p x19 fwae72 zci := ---- - p x19 +

ai fabi

e73 zdi :: p

d73 doneFIGURE E7

134

d50 "torsional rigidity of supportsat centerof the 3 .agnetsidispfunga.;

cSle51 ga.i := IF I = 3 THEN ga.3 ELSE IF *= 10 TEEN gaulO

ELSE IF 1=17 THEN gail7 ELSE 0

dSl done

c52 "product El for each of the 3 .agnets"$dispfuna.l;

c53e53 a.li := IF I <= 5 THEN a* ELSE IF 6 THEN at

ELSE IF 1<: 12 THEN aq ELSE IF j= 13 THEN al ELSE a.

d53 done

cM "axial force"$dispfunfab;

cSSe55 fabi := 1? 1 a 9 THEN fa ELSE lb

d55 done

c56 "useful functions of ail, fab"$dispfuna,b;

c57fabi

e57 ai :=aili

suifj, j, 1, 1e58 bi := -

aili

dSS done

c59 "function expressingthe curvatureof the .agnet andderivatives"$dispfunfO,fOp,fOpp;

cGOeGO fOi := - p xi xi - x19

e61 fOpi := - 2 xi - x19 p

e62 foppj := - 2 p

d62 done 135FIGURE ES

c63 "bellows touting lateral and angular.isalignients"$dispfunzet,et;

c64e64 zeti = IF j = 6 TEEN *zet6 ELSE IF j = 7 TEEN zet7

ELSE IF i= 13 TEEN zetl3 ELSE IF 1= 14 THEN zetl4 ELSE 0

eGS eti := IF = 6 THEN et6 ELSE IF i = 7 TED et7

ELSE IF j = 13 THEN etl3 ELSE IF j = 14 THEN etl4 ELSE 0

d65 done

c66 "terE due to initial cosine and sine shapeof the bellows"$dispfunzec,zes;

c67ai

dc62

2 pue67 zeci := IF = 6 THEN -

ai1-- -

22 pit

aiddl3 -------

22 pil

ELSEIF1:13THEN---------------ELSEOai

1 -

22 pil

aids6 ----

2pit

e68 zesi :=IF1=6THD-ai

1-----2

pu

aidsl3 -

2pit

ELSEIF1=I3TEEN--------ELSEO

1-

136pit

FIGURE E9

c28 "boundary condition at the supports relating force to displace.ent"Sdispfunceq;

c29f i

e29 ceqI := IF I a 5 TEEN expandyi, i -

ski i

fi+2ELSE IF I <= 10 TEEN expandyI + 2,1 + 2-

ski + 2

f1+4ELSE IFi al4THENexpandyi +4, i +4--

ski + 4

fI + 4ELSEIFI=lSTHEIiexpandyi+3,i+4---------

skU + 4

d29 done

c30 "boundary condition for continuity of deflection"$dispfuneq;

c31e31 egi := - abi, I + 1 + abi + 1,1 + 1

+expandratsiip-hI,i+fl+hi+1,i+l-bei,i+l

+ hei + 1, 1 + 1 - zeti + 1

d31 done

c32 "boundary condition for continuity of slope9dlspfunpeq;

c33e33 peqi :=-pabi, i +1 +pabi +1,1+1

+ expandratsiip-bpi, i + 1 + hpi + 1, I + 1 - bepi, I + 1

+ hepi + 1, i. + 1 - eti + 1

d33 done

c34 "uo.ent equation"$dispfun.o.;

c35flO f19

e35 .o. := moi : expandfb - - fO19 - 1010

sk1O sk19

f1 1104 Ia -- - ------ - 1010 -

skt sk1O 1372

10. + suigaii - fu x19 - xI, 1, 1, 19 + fwa p x 19

FIGURE ElO

c38 "list of all the boundary conditionsstarting with equilibriut of forces"Sdispfuzubc;

c39e39 bci :=IFI= lTHENsuifj, j, 1,19 - fwa+fwb x19 p

ELSE IFi al8THENeqi- 1 ELSE IFi <=3STHENpeqI- 18

ELSE IF i a 50 TEEN ceqi - 35 ELSE oifl

d39 done

c40 "gather all the coefficients of the unbowns"$dispfundfor;

c41e4l dfori := s:0, be: bci, FORjTHRU5I

DO zs : coeffbe,uj, IFzs SoTEENfortranqi,j is,

s : s + is uj, sube : ratsiips - be,

IF sube I 0 THEN fortranwi = sub.

d41 done

c42 "write coefficients in fortran to file"$dlspfunpfor;

c43e43 ptor :: writefile"belcoefs.for", FOR I TEED 51 DO dfori,

FOR i THRU 18 DO .fortranxci = zci, fortranxdi = zdi,

fortranxgi = zgi, lortranxecu = zeci, fortranxesi = zesi,

closefile"belcoef5.for’

d43 done FIGURE Eli

138

TECHNICAL NOTE DISTRIBUTION

Buildina 902

Blake, A. Building 911Cottingham,J. 3.Gaiter, M. AD Library, R. CohenOhosh, A. P. Hughes 1 plus one for ea.authorGreene,A.Gupta,R. Building 460Herrera,J.Kahn, S. Samios,N. P.Kelly, E.Morgan,G. Building 510Muratore,J.Ozaki, S. Banks, Ni.Prodell,A.Rehak M. For SSCPapersRohrer, E.P.Sampson,W. FNALShun,R.Thompson,P. Mantsch,P.Wanderer,P. Strait, J.Willen, E.

SnBuildin2 bOSS

Bush,T.Brown, D. Coombes,R.Claus,J. Goodzeit,C.Courant,E. Kelly, V.Dell, F. Tompkins,J.Gavigan, M. Library CopyLee, S.Y.Milutinovic, J. LDI.IParzen,G.

* Rhoades-Brown,M. Taylor, C.Ruggiero,S.Sondericker,J.Tepekian,S. S&P in Magnet Division for all M.DWolf, L papers

S&P in AcceleratorPhysics Division forall CrvoenicPapersandJ. Brias.R. Da2radL H. Hildebrand.W. Kolimer