TORSION OF CIRCULAR SHAFTS AND TUBES材料强度教材Strength of Materials

TORSIONTORSION OFOF CIRCULARCIRCULAR SHAFTSSHAFTS ANDAND TUBESTUBES材料强度教材材料强度教材StrengthStrength ofof MaterialsMaterials 16Torsion of circular shafts and th in wal ledtubes16 1Introduction InChapter3we introducedthe conceptsof shearing stress andshearing strain these havean importantapplication intorsion problems Such problemsarise inshafts transmitting heavy torques in eentricallyloaded beams in aircraftwings andfuselages and many other instances These problemsare veryplex ingeneral and atth s elementarystage we can gono furtherthan studyinguniform torsion of circular shafts thin walled tubes and thin walled opensections 16 2Torsion of a thin circular tubeThe simplesttorsion problemis that of the twisting of a uniform thincircular tube the tubeshown in Figure16 1is ofthickness f and the mean radius of the wall isr L is the length of the tube Shearing stresses T areapplied around the circumference of the tube ateach end and inopposite directions Figure16 1Torsion of a thin walled circulartube If thestressesTare uniformaround theboundary the total torque Tateach end of the tube is T 25rrt Tr 2xr2t T 16 1 Thus the shearing stressaround theCircumference due to anapplied torque Tis 16 2 T T 2nr2t368Torsion of circular shafts and thin walled tubesWe considernext thestrains causedby theseshearing stresses We notefirstly thatplementary shearing stresses areset upin the wall parallel to the longitudinal axis of the tube If6s is a smalllength of the circumferencethen anelement of the wallABCD Figure16 1 is in a state of pure shearing stress If theremote end of the tube is assumed notto twist then thelongitudinal element ABCD isdistorted into the parallelogramABC D Figure16 1 the angle of shearingstrain beingT Y 16 3 G if the material is elastic and has a shearing or rigidity modulus G But if8is the angle of twist of the nearend of the tube we have yL re 16 4 Hence 16 5 e YL this isgiven by WL and from equation 16 5 this is equal toT 16 6 0 L Gr16 3Torsion of solid circular shafts Thetorsion of a thmcirculartube isarelatively simpleproblem asthe shearing stress may be assumedconstant throughout the wall thickness The caseof a solid circular shaft ismore plexbecause the shearing stresses are variableover the cross section of the shaft The solid circular shaft of Figure16 2hasa length Land radiusa in the cross section Figure16 2Torsion of a solid circular shaft Torsion ofsolid circular shafts369When equal and opposite torques Tare applied ateach endabout a longitudinal axiswe assumethat0 thetwistingis uniformalong the shaft that is all normal cross sections the same distanceapart sufferequal relativerotation ii cross sections remainplane during twisting and iii rahi remain straight duringtwisting If8is the relative angle of twist of thetwo ends of thesha elemental tube ofhckness6r and at radiusr is ey L then the shearingstrain yof an 16 7 If the material is elastic and hasa shearing or rigidity modulus G Section3 4 then thecircumferential shearing stress on this elemental tube is The thicknessof the elementaltube is6r so the total torque onthis tube is 21cr6r rr 2nr2r6r The total torque on the shaft is then T oa27cr2rdr Onsubstituting for T from equation 16 8 we haveT 2n loa r3dr 16 8 16 9 Now 16 10 na2271La r3dr This is the polar second moment of area of the cross section about an axisthrough the centre and isusually denotedby J Then equation 16 9 may bewritten GJB L T 16 11 370We maybine equations 16 8 and 16 11 in the form Torsion of circular shafts and thin walled tubesGO I T J rL 16 12 We seefrom equation 16 8 that T increases linearlywith r from zeroat thecentre of the shaftto GaB L at the circumference Along anyradius ofthe cross section the shearing stresses arenormal to the radius and in the planeof the cross section Figure16 3 16 4Torsion of a hollow circular shaftIt frequentlyarises that a torque is transmittedby a hollow circular shaft Suppose a and u2are the internal andexternal radii respectively of sucha shaft Figure16 4 We makethe samegeneral assumptionsas in the torsion of a solid circular shaft If r is the shearing stress at radiusr thetotal torque on the shaft is T fa 2 r d r 16 13 Figure16 3Variation of shearing stressesFigure16 4Cross section of a hollowover the cross section forelastic torsion of a solid circularbar circular shaft If weassume as before that radii remainstraight duringtwisting and that the materialiselastic we haveGA LT Then equation 16 13 bees 16 14 T T 25rr3dr GJB Lwhere Torsion of a hollowcircular shaft371J J o 2xr3dr 16 15 Here J is the polar second moment of areaor more generally the torsion constant of the cross section aboutan axisthrough thecentre J has the valueJ y2xr3dr E a a 2 16 16 Thus for bothhollow and solid shafts we have the relationshipT T G9J rL Problem16 1What torque applied to a hollowcircular shaftof25cm outside diameter and17 5cm insidediameter willproduce a maximum shearing stress of75MN m2in the material Cambridge So IutionWe haver 12 5cm rz 8 75cm Thenx J 0 125 4 0 0875 4 0 292x m42If the shearing stress is limited to75MN m the torque is Problem16 2A ship s propellershaft hasexternal andinternal diametersof25cm and15cm What powercan betransmitted at110rev minute with amaximum shearing stress of75MN m2 and whatwill thenbe thetwist indegrees of a10m length of the shaft G 80GN m2 Cambridge Torsion of circular shaftsand thin walled tubes372Solution Inthis caserl 0 125m r2 0 075m I 10m xJ 0 125 4 0 075 4 0 335x m42Then At110rev min thepower generatedis Theangle oftwist is 0 075radians 4 3 e TL 201x io3 IO GJ 80 x io9 0 335x10 3 Problem16 3A solidcircular shaftof25cm diameteris to be replacedby a hollow shaft the ratioof the external tointernal lametersbeing2to1 Find thesize of the hollowshaft if the maximum shearing stress is to be the same asfor thesolid shaft What percentageeconomy in mass willth s changeeffect Cambridge Solution Letr bethe insideralus of the newshafi then 2r the outside radiusof the new shaft7t Jfor thenew shaft 16r4 r4 7 57 27t Jfor theold shaft x 0 125 4 0 384x m42Torsion of a hollowcircular shaft373If Tis the appliedtorque the maximum shearing stressfor theold shaft is T 0 125 0 384x and that for thenewone isIf theseare equal T 0 125 T 2r 0 384x7 5xr4 Then r3 0 261x m3or r 0 640m Hencetheinternal diameter willbe0 128m and theexternal diameter0 256m area ofnew cross section 0 128 2 0 064 o 785area ofold cross section 0 125 2 Thus the savinginmassis about21 Problem16 4A shp s propellershaft transmits7 5x lo6W at240rev min The shafthas aninternal diameter of15cm Calculate theminimum permissibleexternaldiameterifthe shearing stress in the shaft istobelimitedto150MN m Cam bridge Solution If Tis the torque on the shaft then ThusT 298kNm374If dlis the outsidediameter of the shaft then Torsion ofcircular shaftsand thin walled tubes1 J d 0 1504 m432If the shearing stress is limitedto150MN m2 then Td 150 x lo6w Thus Td 300 x106 J Onsubstituting for J andT 298x103 d 300 x lo6 d 0 1504 This gives L 4 3 L I0 1500 150 0On solvingthls bytrial and error we getd 1 54 0 150 0 231m ord 23 1cm16 5Principal stresses in atwisted shaftIt isimportant to appreciate thatuniform torsion ofcircular shafts of the form discussedin Section16 3 involves noshearing betweenconcentric elemental tubes of the shaft Shearing stressesT our in a cross section of the shaft and plementaryshearing stressesparallelto thelongitudinal axis Figure16 5 Figure16 5Principal stressesin the outer surface of atwisted circular shaft Torsion bined with thrustor tension375An elementABCD in the surface of the shaft isin astateof pureshear The principalplane makesangles of45 with theaxisof the shaft therefore and the principal stresses are T If theelementABCDis square then the principal planesare ACand BD The direct stress onAC ispressive andof magnituder the directstress onBD istensile andof thesame magnitude Principal planes such asAC cutthe surface of the shaft in a helix for abrittle material weak in tension we shouldexpect breakdownin atorsion testto ourby tensile fracture alongplanessuchas BD The failure of atwisted bar of abrittle materialis shown in Figure16 6 Figure16 6Failure intorsion ofa circularbar ofbrittle castiron showing atendency to tensilefractureacross ahelix on the surfaceof thespecimen The torsionalfailureof ductile materialsours when the shearing stresses attain the yield stressof the material The greatestshearing stressesina circularshaftourinacross section and along the length of the shaft A circularbar ofa ductilematerial usuallyfails bybreaking offover anormal cross section as shown in Figure16 7 Figure16 7Failure of torsionofacircularbarofductile castiron showing ashearing failureover anormalcross section of the bar 16 6Torsion binedwith thrustor tensionWhen acircularshaft is subjected to longitudinalthrust or tension as wellas twisting the direct stresses due to thelongitudinal loadmust bebinedwith the shearing stresses due to torsionin orderto evaluatethe principal stressesin the shaft Suppose theshaftisaxially loadedintensionso that there isa longitudinaldirectstress T at all pointsof theshaft Figure16 8Shearing anddirectstressesdue tobined torsionand tension 376Torsion ofcircular shaftsand thin walled tubesIfTis the shearing stressat any point then weare interestedin theprincipal stressesof thesystem shownin Figure16 8 for thissystem theprincipalstresses from equations 5 12 havethe values LOf L T222and the maximum shearing stress from equation 5 14 is 16 17 16 18 Problem16 5A steel shaft 20cm externaldiameter and7 5cm sternal is subjected to atwisting moment of30kNm andathrust of50kN Find the shearing stress due to the torquealone and the percentageincrease whenthe thrustis takeninto aount RhJC Solution Forthis case we haverl 0 100m r20 0375m22A r r 0 0270m2The pressivestressisNow J 2 r r 0 00247m42The shearing stressdueto torquealone is The maximumshearing stressdueto the binedloading is Torsion binedwith thrustor tension3771 L d 4t i 1 53MN m2 c 2Problem16 6A thinsteel tube of2 5cm diameter and0 16cm thicknesshas anaxial pullof10kN and anaxial torque of23 5Nm appliedto it Find themagnitude anddirection of theprincipalstresses at any point Cambridge Solution Itwdl beeasier and sufficientlyaurate to neglectthe variationin the shearing stressfrom the inside to theoutsideof the tube Let T the meanshearing stressduetotorsion r themean radius 0 0109mt thehckness 0 016m then the moment of thetotal resistanceto shear 2d t 1 19x10 6 TNm If this isequal to23 5Nm then T 19 75MN m2The areaof thecross section isapproximately27trt 0 1098x lO 3m2Hence the tensile stressiso 10 x lo3 91 1m m20 1098x lO 3378The principalstressesare Torsion ofcircular shaftsand thin walled tubes1ri f 91 1f99 3 MN mZ22Then0 4 1MN m2 r2 95 2MN mZ thepositive signdenoting tension The planesacross which they actmake angles8and 8 n 2 withtheaxis where39 5 0 434tm2e 2r tJ91 1giving8 11 75 16 7Strain energy of elastic torsion InSection16 3we foundthat the torque twist relationshipfor acircularshafthasthe form T GJBLThis shows that the angle oftwist 8 of oneend relativetothe other increases linearlywith T If oneendof theshaftis assumedtobefEed then the work donein twisting theotherend throughan angle8is thearea underthe T B relationship Figure16 9 This workis conservedin theshaft asstrain energy which hasthe value12u Te 16 19 Figure16 9Linear torque twist relationshipand strain energyofelastictorsion Plastic torsionofacircularshaft379On usingequation 16 11 we mayeliminate either8orT a dwe haveu behaviour iselastic up toashearing stress T theshearing modulus beingG Beyond thelimit ofproportionality shearingproceeds ata constantstressT Thls behaviouris nearlytrue of mild steelwith awell defined yieldpoint If weare dealingwith asolidcircularshaft then after the onsetof plasticityin thesurface fibrestheshearing stresses varyradially in theformshowninFigure16 11 The materialwithin aradius bis stillelastic thematerialbeyond aradius bis plasticand iseverywhere stressedtotheyieldstress sY Figure16 10Idealized shearingstress strain Figure16 11Elastic plastic torsionofasolid curveofmildsteel circularshaft The torquesustained by the elastiore isq Ji Ty b3T b2 16 22 380where subscripts1refer tothe elastiore The torquesustained by theouter plastic zoneis Torsion ofcircular shaftsand thin walled tubes2112 11r dT T a b 3 16 23 The totaltorqueon theshaftis Theangle oftwist of the elastiore is 16 24 16 25 where Lis the length of theshaft We assumethat theouterplasticregion suffersthesameangle oftwist this istantamount toassuming thatradiiremainstraightduringplastic torsionof theshaft Equation 16 25 gives Then thetorquebees Atthe onsetof plasticityThen for anyother conditionoftorsion 16 26 16 27 16 28 16 29 which givesor T 16 30 Plastic torsionofacircularshaft381and equation 16 27 bees T the formsof thesestress strain curvesare similartotensileand pressivestress strain curves as showninFigure16 13 In theelastic rangeofamaterial C ywhere Gistheshearingmodulus ofthematerial Section3 4 Figure16 13Forms ofshearingstress strain curvesfor mildsteel andfor aluminiumlight alloys It isimportant toappreciate that theshearingstress strain curvecannot bedirectly deducedfrom atorsion testofasolidcircularbar although thelimit ofproportionality canbe estimatedreasonably aurately 16 9Torsion of thin tubesof noncircular cross section In general theproblem ofthetorsionofashaftof non circular cross section isa plexone intheparticular casewhentheshaftisahollowthin tube wecandevelop however a simpletheory givingresults thatare sufficientlyaurate forengineering purposes Consider a thin walled closedtube ofuniform sectionthroughout itslength The thicknessofthe wall at any pointist Figure16 14 although thismay varyat pointsaround the circumference ofthetube Suppose torquesTareappliedto eachend sothatthetube twists aboutalongitudinalaxisCz We assumethatthetorqueTisdistributed over the endofthetube intheformofshearing stresses whichare paralleltothe tangent tothewallatany point Figure16 14 and thatthe ends ofthetube arefree fromaxial restraint If theshearingstressatanypoint ofthe circumferenceisT then anequal plementaryshearingstressis setup alongthelength ofthetube Consider theequilibrium ofthe section ABCD ofthewall iftheshearingstressT atanypointis uniformthroughoutthewallthickness then theshearing forcetransmitted over the edgeBC isT tperunit length Torsion ofthin tubesofnon circularcross section383Figure16 14Torsion ofa thin walled tube of any cross section For longitudinalequilibrium ofABCD wemust havethat Tt onBC isequalandoppositeto rt onAD but the sectionABCDis anarbitrary one and wemust havethat rt is constantfor allparts ofthetube Suppose hs constantvalue of rtisTt q 16 35 The symbol q iscalledtheshearflow it hasthe unitsofaload perunit lengthofthecircumferenceofthetube Suppose wemeasure a distance sroundthetube fromsome point0on thecircumference Figure16 14 The forceacting alongthetangentto anelement oflength6sinthecross section isrt6s Suppose risthelengthofthe perpendicularfrom thecentre oftwist Conto thetangent Then themomentofthe force t6saboutC is10 6sThe totaltorqueonthecross section ofthetube is thereforeT rtrak 16 36 Q QQwhere theintegration iscarried outoverthewhole ofthecircumference But Ttis constant and equal to q for all valuesofs Then T rt rds q rds 16 37 Now rds istwice thearea A enclosed by thecentreline ofthewallofthetube andsoT 2Aq 16 38 The shearingstressatanypointis then384Torsion ofcircularshaftsandthin walled tubes 16 39 To find theangleoftwistofthetubeweconsider thestrain energy stored inLe tube and equatethis tothework done bythe torquesT intwistingthetube When amaterialissubjectedtoshearing stressesrthestrain energystored perunit volumeof materialis fromequation 3 9 t2G where Gistheshearing or rigidity modulusofthematerial In thetube theshearing stressesare varyingaround thecircumference butnot alongthelengthofthetube Thenthestrainenergystored inalongitudinalelement oflengthL width6sandthickness tis LtGs Thetotal strainenergystoredinthetubeistherefore Ltcis f 16 40 where theintegration iscarried outoverthewhole circumferenceofthetube But rtisconstant andequaltoq and wemay write 16 41 If theendsofthetubetwist relativeto eachother by an angle8 thentheworkdonebythetorques Tis12w m Onequating Uand W we haveq2L cise G TJ tBut fromequation 16 38 we haveT4 2A 16 42 16 43 16 44 Torsion ofa flatrectangular strip385Then equation 16 43 may bewritten e T 1 16 45 2A GFor atubeofdor mhckness t 0 E 16 46 4A2G twhere Sisthetotal circumferenceofthetube Equation 16 45 canbewritten intheformTL e GJ where4A2ds J rJisthetorsionconstantfor thesection for circularcross sections Jisequaltothe polarsecond momentof area but thisis nottrue ingeneral 16 10Torsion ofa flatrectangular stripA longflat stripof rectangularcross section hasa breadthb thickness t and lengthL For uniform torsion aboutthe centroidofthecross section the stripmaybetreated asa setof concentricthin hollowtubes all twisted bythesame amount Figure16 15Torsion ofa thinstrip 386Torsion ofcircularshaftsandthin walled tubesConsider suchan elementaltube whichis rectangularin shapethe longer sides beingadistanceIf6Tis thetorque carriedby thiselementaltubethentheshearingstress inthelonger sidesof yfrom thecentral axisofthe strip the duchessofthetubeis6y Figure16 15 thetubeis6T 16 47 T 4bY SYwhere bisassumedvery much greater than t This relationshipgives 16 48 dT 4byr4For theangleoftwistoftheelementaltubewe have fromequation 16 46 2bLST16b y G6yo 16 49 where Listhelengthofthe strip This givesthe fiutherrelationship0 16 50 dT 8by G 4L Onparing equations 16 48 and 16 50 wehaveT 2yG 16 51 This showsthattheshearingstressT vaneslinearly throughoutthe thicknessofthestrip having amaximum valueinthesurfaceofT Gt 16 52 An importantfeature isthattheshearingstresses 5act paralleltothelongersideb ofthestrip andthattheir directionsreverse overthe thicknessofthestrip This approximatesolution givesan inexactpicture oftheshearingstresses nea