4 LinearlyActingExternalandInternal DrumBrakes Linearlyactingdrumbrakesarethosefittedwithshoeswhich,whenacti- vated,approachthedrumbymovingparalleltoaradiusthroughthecen- teroftheshoe.TypicallinearlyactingdrumbrakesareillustratedinFigures 1–3. Analysisoflinearlyactingbrakesincludesthoseinwhichthecentrally pivotedshoesareattachedtopivotedlevers,asinFigure1.Includingbrakes ofthisdesignwithinthecategoryoflinearlyactingbrakesisjustifiedifthey aredesignedsothattheappliedforcesontheshoesandliningsactalongthe radiioftheshaftsthattheygripwhenthebrakesareapplied.Brakesofthis typemayacteitheruponbrakedrumsordirectlyuponrotatingshaftsandare suitableforuseinheavy-dutyapplications,suchasfoundinminingand constructionequipmentandinmaterials-handlingmachinery. Internallinearlyactingdrumbrakes,suchasusedontrucksinEurope, maybedesignedasinFigure2.Eitherpneumaticorhydrauliccylindersor cams may be used to force the shoes outwardly against the drum. The cylin- ders or cams also serve as anchors to prevent rotation and react against the vehicle frame. The springs shown are to retract the shoes when the brake is released. A collection of segmented brake pads (backing plate plus lining) along the entire circumference of the drum may be arranged as in Figure 3 to move outwardly against a drum, as in Figure 3a, or inwardly against a drum, as in Figure3b.Thebrakepads,orshoes,arethemselvesconstrainedagainst Copyright © 2004 Marcel Dekker, Inc. rotationbyanchorpinsthatfitintoshortradialslotsbetweentheshoesand areattachedtotherimofthecircularframe,asshowninFigure3a.Brake actuationisaccomplishedbyusingairtoexpandthenormallyflatelasto- meric-fabricannulartubeshowninthatfigurebetweenthebrakepadsandthe circularframe.Whendesignedtomoveinwardlyagainstadrum,asinFigure 3b,thebrakeliningisrivetedtoadifferentlycontouredbackingplatewhich hasshouldersateachendtoresistatwistingtorqueandwhichisfittedwitha centralslotthatacceptstheanchorpintotheouterframeateachsideofthe shoe.Thisradialslotallowsthepadtomoverapidlyinwardbutprevents tangentialmotion. F IGURE 1Linearlyacting,centrallypivotedshoebrake.(CourtesyoftheHindon Corp., Charleston, SC) Chapter 468 Copyright © 2004 Marcel Dekker, Inc. I.BRAKINGTORQUEANDMOMENTSFORCENTRALLY PIVOTEDEXTERNALSHOES Tocalculatethetorque,wemustfirstfindanexpressionfortheliningpres- sure.GuidedbythegeometryshowninFigure4,weseethattheliningpres- surewillbegivenby p¼kDcosuð1-1Þ intermsoftheliningdeformationDiftheshoeanddrumareassumedtobe absolutelyrigid.Maximumpressureoccurswhenu g 0,sothatp max =kD. Thusequation(1-1)becomes p¼p max cosuð1-2Þ andtheincrementaltangentialfrictionforceisgivenby dF¼Ap max cosurwduð1-3Þ F IGURE 2Linearlyacting,twin-shoe,internaldrumbrakewithpneumaticactiva- tion (Girling Twinstop). (Reprinted with permission. n1977 Society of Automotive Engineers, Inc.) Linearly Acting External and Internal Drum Brakes 69 Copyright © 2004 Marcel Dekker, Inc. sothebrakingtorquebecomes T¼Ap max r 2 w Z u 2 u 1 cosudu¼Ap max r 2 wðsinu 2 Àsinu 1 Þð1-4Þ IndesignsdifferentfromthoseshowninFigure1itmayprove convenienttohavetheshoepivotedaboutapointataradialdistanceRon theaxisofsymmetry,suchaspointAinFigure4.ThemomentM p duetothe pressureontheliningiszeroaboutpointAbecauseofthesymmetryofthe shoeaboutthispoint.NosuchsymmetryexistsforthefrictionmomentM f , however,sofromtheincrementalmomentduetofriction dM f ¼ðAp max rwcosuduÞðRcosuÀrÞ F IGURE 3Rimbrakeswithpneumaticactivation(alsousedasrimclutches). (Courtesy of Eaton Corp., Airflex Division, Cleveland, Ohio.) Chapter 470 Copyright © 2004 Marcel Dekker, Inc. it follows that M f ¼ Ap max rw Z u 2 u 1 R cos 2 u À r cos u ÀÁ du ¼ Ap max rw R f 0 2 þ 1 4 ðsin 2u 2 À sin 2u 1 Þ !& ð1-5Þ À rðsin u 2 À sin u 1 Þ ' where f 0 = u 2 À u 1 . The expression in equation (1-5) may be simplified by observing that the symmetry of the shoe about A requires that À u 1 ¼ u 2 ¼ B 0 2 ð1-6Þ where B 0 is the angle subtended by the lining. Substitution of these values into equation (1-5) leads to M f ¼ Ap max rw R 2 ðf 0 þ sin f 0 ÞÀ2r sin f 0 2 ! ð1-5aÞ F IGURE 3 Continued. Linearly Acting External and Internal Drum Brakes 71 Copyright © 2004 Marcel Dekker, Inc. whichsuggeststhatmomentM f willvanishiftheshoeispivotedat R¼r 4sinðf 0 =2Þ f 0 þsinf 0 ð1-7Þ UponplottingR/rweobtainFigure5,whereintheratioincreases smoothlyfrom1.0atf 0 =0to1.273atf 0 =krad.=180j.Thisclearly indicatesthatitisimpossibletofindapivotpointforwhichM f =0foran internallinearlyactingshoe.Thisconclusionis,ofcourse,unaffectedbythe F IGURE 4Geometryusedfortheanalysisofalinearlyactingexternalshoe. Chapter 472 Copyright © 2004 Marcel Dekker, Inc. signreversalfoundintheexpression(RcosuÀr)whenequation(1-5a)is appliedtoaninternalshoe.Thesignreversalsimplychangesthedirectionof rotationimpliedbyapositivevalueofM f ,aswasdiscussedinanearlier section. Thenearlyhorizontalportionfrom0jtoabout30jimpliesthatfor externalshoeswhichsubtendananglelessthan30j,anychangesinthelength oftheshoethatdonotincreasethesubtendedanglebeyond30jwillhavea negligibleeffectontheR/rratio.Thiscorrelateswiththeshort-shoesegments usedinthebrakesshowninFigure3.Moreover,thevalueofM f causedbya deviationfromtheR/rratiothatyieldsazerovalueofM f willbesmalliff 0 remainssmall.Inparticular,iftheRvaluethatyieldsM f =0isreplacedby R+yRinequation(1-5a),themomentduetofrictionwillincreasetoonly Ap max rw yR 2 ðf 0 þsinf 0 Þ whichissmallenoughtobeeasilyresistedbytheshouldersshownontheshoes inFigure3. ActivationforceF s andtangentialforceF t onasymmetricallyplaced pivotaregivenbytherelations F s ¼2p max rw Z f 0 =2 0 cos 2 udu¼ 1 2 p max rwðf 0 þsinf 0 Þð1-8Þ and F t ¼2Ap max rw Z f 0 =2 0 cos 2 udu¼AF s ð1-9Þ F IGURE 5VariationofR/rwithanglef 0 . Linearly Acting External and Internal Drum Brakes 73 Copyright © 2004 Marcel Dekker, Inc. Letusdefinetheefficiencyofabrakeastheratioofthetorqueprovided bythebraketothetorquethatcouldbehadbyapplyingtheforcedirectlyto thedrum,orshaft.Accordingtothisdefinition,theefficiencybecomes T F s r ¼ Ap max r 2 wðsinu 2 Àsinu 1 Þ ð1=2Þp max r 2 wðf 0 þsinf 0 Þ ¼2A sinu 2 Àsinu 1 f 0 þsinf 0 ð1-10Þ Uponsubstitutingforu 1 andu 2 inequation(1-8)andrecallingequations(1-6) and(1-7)wefindthat T F s r ¼ 4Asinðf 0 =2Þ f 0 þsinf 0 ¼A R r ð1-11Þ wheretheright-handsidehasalreadybeenplottedinFigure5.Fromthat figurewefindthatalthoughmaximumefficiencymaybeachievedonlyifeach shoeandliningextendoverhalfofthedrum,orshaft,relativelylittleefficiency islostiftheliningextendsoveronly160jinsteadof180j.This,togetherwith thenearimpossibilityofmaintaininggoodcontactbetweentheliningand thedrumneartheendsofashoesubtending180jatthecenterofthedrum, accountsfortheangulardimensionsofthebrakeliningsshowninFigure1. Finally,itfollowsfromequation(1-11)thatiftheshoeissymmetrically pivotedandifequation(1-7)holds,theappliedtorqueisgivenby T¼ARF s ð1-12Þ II.BRAKINGTORQUEANDMOMENTSFORSYMMETRICALLY SUPPORTEDINTERNALSHOES Pressurepandbrakingtorqueareagaingivenbyequations(1-2)and(1-4), respectively,foraninternalshoemovedagainstarotatingdrumalongaline paralleltoitsaxisofsymmetry,lineOB,inFigure6.Inthefollowinganalysis it may be more descriptive to measure the angle along the shoe from the end rather than from the middle because the activation forces are now applied at the ends. Denote this angle by f. Since the expression for the torque is unaffected by this choice of angle, substitution of equation (1-6) into equation (1-4) shows it can be given by T ¼ 2Ap max r 2 w sin f 0 2 ð2-1Þ The pressure distribution described by equation (1-2) may be rewritten in terms of f according to p ¼ p max cosðf À aÞð2-2Þ Chapter 474 Copyright © 2004 Marcel Dekker, Inc. Let the shoe be restrained at A to prevent it from rotating with the drum, with the restraint moving with the shoe. This may be accomplished using guide pins and/or plates which may also serve as anchors to transfer braking torque from the shoes to the appropriate structure. The moment M p about A due to pressure p is given by integration of dM p ¼ðprw dfÞR sin f ¼ p max Rrw cosðf À aÞ sin f d f ð2-3Þ to obtain M p ¼ p max wRr Z f 2 f 1 ðcos a cos f þ sin a sin fÞ sin f df ð2-4Þ which may be integrated directly to give M p ¼ p max 4 wRr 2 cos a ðsin 2 f 2 À sin 2 f 1 Þ Â ð2-5Þ þ sin að2f 0 À sin 2f 2 þ sin 2f 1 Þ Let a represent the central angle from the R vector to the middle of the shoe (i.e., from R to the transverse plane of symmetry through radius OB in Figure 6), so that f 1 ¼ a À f 0 2 f 2 ¼ a þ f 0 2 ð2-6Þ F IGURE 6 Geometry used in the analysis of a linearly acting internal shoe drum brake. Linearly Acting External and Internal Drum Brakes 75 Copyright © 2004 Marcel Dekker, Inc. Aftersubstitutionforf 1 andf 2 fromequations(2-6)andusingcommon trigonometricidentities,M p maybewrittenas M p ¼ p max 2 rRwðf 0 þsinf 0 Þsinað2-7Þ Similarly,themomentaboutAduetofrictionmaybefoundfrom dM f ¼ðaprwdfÞðrÀRcosfÞð2-8Þ whichwiththeaidofequation(2-2)leadstotheintegral M f ¼p max rwA Z f 2 f 1 ðcosacosfþsinasinfÞðrÀRcosfÞdfð2-9Þ which,uponintegration,produces M f ¼ A 4 rwp max 4rðsinf 2 Àsinf 1 Þcosa½ þ4rðcosf 1 Àcosf 2 ÞsinaÀRð2f 0 þsin2f 2 ð2-10Þ Àsin2f 1 ÞcosaÀ2Rðsin 2 f 2 f sin 2 f 1 Þsina Substitutionforf 1 andf 2 fromequations(2-6)intoequation(2-10)anduse ofcommontrigonometricidentitiesenablesequation(2-10)tobesimplied toread M f ¼ p max 2 Arw4rsin f 0 2 ÀRðf 0 þsinf 0 Þcosa ! ð2-11Þ AccordingtothegeometryshowninFigure6,apositiveM p corre- sponds to clockwise rotation of the shoe about point A and positive M f corresponds to counterclockwise rotation when the drum rotation is from the opposite end of the shoe toward point A. Reversing the direction of drum rotation will not affect the implied direction of shoe rotation due to M p but will reverse the direction by a positive M f ; that is, positive M f will then imply clockwise rotation about point A. This last observation is of academic interest only, however, if the shoes are supported at each end, because in that case each shoe will tend to pivot about the end toward which the drum rotates, regardless of the direction of rotation of the drum. For symmetrically supported symmetric shoes it follows that the shoes will be free of self-locking if M p À M f > 0 ð2-12Þ Chapter 476 Copyright © 2004 Marcel Dekker, Inc. [...]... Acting External and Internal Drum Brakes 77 Substitution for Mp and Mf from equations (2-7) and (2-11) into equation (2-12) yields h pmax r f0 Rrw ðf0 þ sin f0 Þsin a À 4A sin Mp À M f ¼ R 2 2 ð2-13Þ i þ Aðf0 þ sin f0 Þcos a > 0 Condition (2-12) is, according to equation (2-13), equivalent to the condition r f0 < ðf0 þ sin f0 Þðsin a þ A cos aÞ sin 4A ð2-14Þ 2 R for internal, linearly acting brakes Consequently,... to ensure that the brake will not become self-locking when it is applied III DESIGN EXAMPLES Example 4.1 Design an external, linearly acting, twin-shoe brake to provide a braking torque of 2700 N-m when acting on a flywheel hub 260 mm in diameter The lining material to be used here has a design maximum pressure of 3.41 MPa and A = 0.41 Since the torque on either an external or an internal shoe is given... the drum, or 29.74 mm from the drum surface Example 4.2 Design an internal, linearly acting, twin-shoe drum brake to provide a braking torque of 413,000 in - 1b acting on a drum whose maximum inside diameter may be 26.0 in The lining material to be used has a maximum design pressure of 450 psi and a friction coefficient of 0.50 or greater over the design temperature range Substitution into the expression... 6:990 in: ð2:2689 þ sin 130B Þðsin 70B þ 0:5 cos 70B Þ Linearly Acting External and Internal Drum Brakes 79 Equal forces that must be applied at points A and C in Figure 6 to achieve the 450 psi maximum pressure may be found from equation (1-8) after replacing Fs with 2Fs, where Fs in equation (3-1) represents the force at A and at C Thus Fs ¼ pmax rwðf0 þ sinðf 0 ÞÞ 4    450   13ð3Þ 130 þ sin 130... 90% of the maximum theoretical torque (i.e., for f0 = 180j) may be obtained from f0 = 128.3j, that 95% may be had from f0 = 143.6j, and 98% may be had from f0 = 157.0j It we select f0 = 145j for each shoe, assume that each shoe will supply half of the design braking torque, and solve equation (2-1) for w, we find that T w¼ 2Apmax r2 sinðf0 =2Þ ¼ 1350  103 0:82ð3:41Þð130Þ2 sin 72:5B ¼ 29:95 mm ! 30 mm... Acting External and Internal Drum Brakes h1 k Mf Mp p pmax R r T w a D u A f Length available for the shoe structure equivalent spring constant for lining material (mtÀ2) moment due to friction (ml2tÀ2) moment due to pressure (ml2tÀ2) lining pressure (mlÀ1tÀ2) maximum lining pressure (mlÀ1tÀ2) radius to effective pivot point from the drum center(l) drum radius (l) braking torque (ml2tÀ2) shoe and lining... p Space available for such a cylinder may be found from the geometry in Figure 6 by finding the distance from a plane P through the center of the drum and perpendicular to the u = 0 line The available distance will be twice this value Angle h between R and plane P may be found from 2h ¼ 180j À f0 À 2f1 ð3-2Þ so that the distance h0 from point A to the corresponding point on the opposite shoe becomes... points A, or C, and the drum surface as a function of radius R; Curve 2: Height h0 available for a hydraulic cylinder as a function of R All dimensions in inches which for R = 10 inches yields     10  cos 20 13 sin acos À 3:4202 ¼ 5:56299 13 180 which may suggest that selecting a larger value for R would give more space for the hydraulic cylinders that force the shoes against the drum and would also... for the hydraulic cylinders that force the shoes against the drum and would also require less material in each shoe Plotting h0 and h1 for other values of R results in the curves shown in Figure 7 Distances are measured along chords that pass through points A on opposing shoes and through points C on opposing shoes IV NOTATION Fs Ft h0 Copyright © 2004 Marcel Dekker, Inc Force in the transverse plane... 609 N Copyright © 2004 Marcel Dekker, Inc 78 Chapter 4 So if the brake is to be pneumatically activated, as shown in Figure 1, the pressure and diaphragm diameter are related according to Fs ¼ kr2 pdia Upon solving this relation for the diaphragm pressure pdia, and using an active diaphragm diameter of 250 mm, we find that the line pressure to the diaphragm must be 4.20 atm Finally, if the shoes are . pivotedshoesareattachedtopivotedlevers,asinFigure1.Includingbrakes ofthisdesignwithinthecategoryoflinearlyactingbrakesisjustifiedifthey aredesignedsothattheappliedforcesontheshoesandliningsactalongthe. suitableforuseinheavy-dutyapplications,suchasfoundinminingand constructionequipmentandinmaterials-handlingmachinery. Internallinearlyactingdrumbrakes,suchasusedontrucksinEurope, maybedesignedasinFigure2.Eitherpneumaticorhydrauliccylindersor