(48)   

   

CHAPTER3:TORSION

Introduction: In this chapter structural members andmachine parts that are in torsionwill be considered. Morespecifically, you will analyze the stresses and strains inmembers of circular cross section subjected to twistingmoments, or torques, T and T'. Members in torsion areencountered in many engineering applications. The mostcommonapplicationisprovidedbytransmissionshafts,whichareusedtotransmitpowerfromonepointtoanother.Theseshaftscanbesolidorhollow.

AnalysisofStressandStrain

Now Consider a shaft ABsubjected at Aand B to equal andoppositetorquesTandT',wepassasectionperpendiculartotheaxisoftheshaftthroughsomearbitrarypointCasshown.Basedonthefree‐bodydiagramoftheportionBCoftheshaftandequilibriumwehave:

→

Also note that shear cannot take place in one plane only.Considertheverysmallelementofshaftshown.Weknowthatthetorqueappliedtotheshaftproducesshearingstressesτonthe faces perpendicular to the axis of the shaft. But theconditions of equilibrium require the existence of equalstressesonthefacesformedbythetwoplanescontainingtheaxisoftheshaft.Suchshearingstressesoccurintorsioncanbedemonstratedbyconsideringa“shaft”madeofseparateslatspinnedatbothendstodisksasshownbelow.Ifmarkingshavebeen painted on two adjoining slats, it is observed that theslatsslidewithrespecttoeachotherwhenequalandoppositetorquesareappliedtotheendsoftheshaft.Whileslidingwillnot actually takeplace in a shaftmadeof a homogeneous and cohesivematerial, the tendency for sliding willexist, showing that stresses occur onlongitudinalplanesaswellasonplanesperpendiculartotheaxisoftheshaft.

 

(49)   

   

Nowconsideracircularshaftthatisattachedtoafixedsupportatone end. If a torqueT is applied to the other end, the shaftwilltwist,withitsfreeendrotatingthroughanangleφcalledtheangleoftwist.Whenacircularshaftissubjectedtotorsion,everycrosssectionremainsplaneandundistorted.Inotherwords,whilethevarious cross sections along the shaft rotate through differentamounts,eachcrosssectionrotatesasasolidrigidslab(thisisnotthecaseforashaftwithsquarecrosssectionasshown).We will now determine the distribution of shearing strains in acircular shaft of length L and radius c that has been twistedthroughanangleφ.Detachingfromtheshaftacylinderofradiusr,we consider the small square element formed by two adjacentcirclesandtwoadjacentstraightlinestracedonthesurfaceofthecylinderbeforeany load isapplied.As theshaft issubjectedtoatorsional load, the element deforms into a rhombus. Recall thattheshearingstrainγinagivenelementismeasuredbythechangeintheanglesformedbythesidesofthatelement.Sincethecirclesdefining twoof the sidesof the element consideredhere remainunchanged, the shearing strain γ must be equal to the anglebetweenlinesABandA'B:

→ → →

→

→ →

:2

, :2

→ →

HowcanwemeasureGbyatorsiontest?

(50)   

   

Shaftwithintermediatetorque 

/ / / 0 

→  

ShaftwithContinuouslyVaryingLoadsorDimensions

 

NormalStressinTorsion

Uptothispoint,ouranalysisofstressesinashafthasbeenlimitedtoshearingstresses.Thisisduetothefactthattheelementwehad selectedwasoriented in suchaway thatitsfaceswereeitherparallelorperpendiculartotheaxisofthe shaft. We know from earlier discussions that normalstresses, shearing stresses, or a combination of bothmaybe found under the same loading condition, dependingupontheorientationof theelementthathasbeenchosen.Considerthestressesandresultingforcesonfacesthatareat45°totheaxisoftheshaft(noshearingforceactsalongDC):

2 cos 45 ° √2√2

√2

FailuremodelinDuctileandBrittleMaterialsDuctile materials generally fail in shear. Therefore, whensubjectedtotorsion,aspecimenmadeofaductilematerialbreaksalongaplaneperpendiculartoitslongitudinalaxis.Ontheotherhand,brittlematerialsareweaker intensionthaninshear.Thus,whensubjectedtotorsion,aspecimenmade of a brittle material tends to break along surfacesthatareperpendiculartothedirectioninwhichtensionismaximum,i.e.,alongsurfacesforminga45ᴼanglewiththelongitudinalaxisofthespecimen.

 

 

(51)   

   

 

Example1:Knowingthata10mmdiameterholeisdrilledthroughAD,determine(a)theshaft inwhichthemaximumshearingstressoccurs,(b)themagnitudeofthatstress.

 

TAB

TBC

TCD

PartAB:

↶ 0 → 90 0 → 

90  

90000 10

2 10 5

61.1  

PartBC:

↶ 0 → 270 90 0

→ 

180  

180000 10

2 10 5

122.2  

 

PartCD:

↶ 0 → 90 270

110 0 → 

290  

290000 10

2 10 5

196.9 →  

(52)   

   

 

Example2: The aluminium rodAB(G= 27GPa) is bonded to the brass rodBD(G=39GPa). Knowing that portion CDof the brassrod is hollow and has an inner diameter of40mm,determinetheangleoftwistatA.

 

Statics:

↷ 0 → 800 0 → 

800  

↷ 0 → 800 1600 0

→ 2400  

↷ 0 → 800 1600 0

→ 2400  

 

/ / /  

 

0 

 

800000 400

2 18 27000

2400000 375

2 30 39000

2400000 250

2 30 20 39000

0.072 0.018 0.015 0.1056.02° 

 

TCD

TBC

TAB

(53)   

   

 

Example 3: The shaft (G = 80 GPa) has adiameter of 14 mm, determine the angle oftwistatB.

↷ 0 → 150 0

→ 150  

↷ 0 → 150 280 0

→ 130  

↷ 0 → 150 280

40 0 

→ 170  

/ / /  

 

0 

 

150000 400

2 7 80000

130000 300

2 7 80000

170000 500

2 7 80000 

0.2 0.13 0.28 0.2112.1° 12.1° ↷ 

 

TBC

TCD

TDA

(54)   

   

 

 

Example 5: For the shaft and loadingshown(G=75GPa,d=80mm,L=800mm)determinetheangleoftwistatB.

 

Example4: For the given shaft and loadingshowthat:

2 3

→

2,

→

2

2 3

→

1 1

2 40 750005000 /

5000 /

2 40 75000

5000 / 800 /2

2 40 75000

0.0053 0.3°

 

(55)   

   

 

Example 6: The designof the gear‐and‐shaftsystem shown requiresthat steel shafts of thesame diameter be usedfor bothABandCD. It isfurtherrequiredthatτmax≤ 60 MPa and that theangle φD through whichendDof shaftCDrotatesnotexceed1.5°.Knowingthat G = 77 GPa,determine the requireddiameteroftheshafts.

 

0 → 0 0 → 0 

→10040

1000 2500  

Shearstressremainingsmallerthan60MPa:

MaximumshearstressoccursinshaftABasTAB>TCD

60 →

2500000

260 → 29.82 → 59.64  

AngleoftwistatDremainingsmallerthan1.5°:

/ →1.5180

 

1.5180

→1.5180

1000000 600

2 77000 

0.026184961

:  

→ /  

→2500000 400

2 77000→

8268→

10040

8268 20670 

0.026184961

→ 0.026184961 20670

→ 31.45 → 62.9  

 

(56)   

   

 

Example 7: Two solid steel shafts (G = 77.2 GPa) areconnected to a couplingdiskBand to fixed supports atAandC.Fortheloadingshown,determine(a)thereactionateachsupport,(b)themaximumshearingstressinshaftAB,(c)themaximumshearingstressinshaftBC.

0 → 1.4

Thesystemisstaticallyindeterminate.Compatibilityequation:

0 → / / 0

0 0

0 →

0 → 1.4

250

2 19

1400000 200

2 250

→ 294938 295 → 1400 295 1105 → 295 → 1.4 295 1400

1105

1105000 25

2 2545  

 

295000 19

2 1927.4  

 

TA

TC

A

C

TBC

TAB

TC

TC

(57)   

   

: 4 1 →

Compatibilityequation:

→ →2 36 27 27000 2 27 77000

→

1.32 2

1.72 2.27

Maximumshearstressinthesteelcore:

2.27 10 27

2 27 73.42

MaximumshearstressintheAluminiumjacket:

1.72 10 36

2 36 27 34.3

TheangleoftwistatA:

1.72 10 2500

2 36 27 27 0000.088 5.05°

 

2.5m

Example 8: A torque ofmagnitudeT= 4 kNm is appliedat end Aof the composite shaftshown. Knowing that themodulusofrigidity is77GPa forthe steel and 27 GPa for thealuminum, determine (a) themaximum shearing stress in thesteel core, (b) the maximumshearing stress in the aluminumjacket,(c)theangleoftwistatA.

0MPa

73.4MPa

34.3MPa

25.7MPa

(58)   

   

TBR1:shaftsaremadeofA‐36steel(G=75GPa).Eachhasadiameterof25mmand they are connectedusing the gearsfixedtotheirends.Theirotherendsareattached to fixed supports at A and B.They are also supported by journalbearings at C and D, which allow freerotationoftheshaftsalongtheiraxes.Ifatorqueof500Nmisappliedtothegearat Eas shown, determine the reactionsatAandBaswellastheangleoftwistatE(1390).Answer: TB = 222.22 Nm, TA=55.6 Nm,φE=1.66°

FromStatics(equilibrium)oftorques:

: 0 → 500 0 1  

: 0 → 0 2  

Wehavetwoequationsthreeunknowns(TA,TB,andF),soweneedacompatibilityequation:

→ / / →

0 = 0

→ 100 =50 →

→ 4 3

, , 55.6 , 222.22

/ /

55600 1500

2 12.5 750000.029

1.66°

 

 

(59)   

   

TBR2:shafts(1)and(2)haveadiameterof20mmandshaft(3)has a diameter of 25 mm. Thesupports allow free rotation ofthe shafts along their axes.Determine the maximal shearstress in shaft (1). Also, findrotation of gears C and E.AssumethatG=80GPaandL=400mm(1391).

 

 

(60)   

   

TBR 3: Find the maximalshear stress in shafts (1) and(3)aswellasrotationofgearsCandE(1392).

 

0 ,  0

→5442

1.286 → 460

1.286 → 1.286 460 1  

→ : 2 : CompatibilityEquations:

→ / / / /

→

54 352

42 252

→ 2.988 2

107.6 , 321.6

321.6 10 352

2352

38.2 ,107.6 10 25

2

2252

35.1

/ /107.6 10 400

2 252 28000

0.04 2.3°

/ / / /

460000 200

2 352 28000

321.6 10 400

2352 28000

0.0535 3.06° 

(61)   

   

TBR4:For the gear system shown findmaximalT0 so thatmaximal shear stress in the thin‐walledshaftABremainssmallerthan40MPa.BasedonthecalculatedT0determinerotationsofNandMaswellasmaximalstressinshoftCD(G=60GPa)(1393).

400 0② 

200 0② 

→ 2 1 ① 

: 400 200 → 2 ② → 24  

→1000

2 45 600002

12004 3600 60000

4 605

⑧ 

0.000155 0.002222 → 14.31 2 ① 

1 2 → 0.03375 0.483 ② 

2→ 40

0.033752 3600 5

→ 42663978 42.6 ④ 

0.483 42663978 45

2 45144 ②

1000

2 45 60000

0.0533 3.05°,2

1.52°① 

Y

Y

x

(c) view Y-Y

90 mm

(b) view X-X

1.0 m

B

Hollow Shaft

(a)

N (rN = 200 mm)

M (rM = 400 mm)

A X

D

1.2 m

Solid Shaft

X

To

t = 5 mm

C

60 mm

60 mm

(62)   

   

DesignofTransmissionShafts

StressConcentrationinCircularShaft

 

The principal specifications to be met inthedesignofa transmissionshaftarethepowerto be transmitted and the speedofrotation of the shaft. The role of thedesigner is to select thematerial and thedimensions of the cross section of theshaft, so that the maximum shearingstressallowableinthematerialwillnotbeexceeded when the shaft is transmittingtherequiredpoweratthespecifiedspeed.

Todeterminethetorqueexertedontheshaft,werecallfromelementarydynamicsthatthepowerPassociatedwiththerotationofarigidbodysubjectedtoatorqueTis:

(Watt)=(N.m)(rad/sec)

2 wherefisfrequencyofrotationanditsunitis1/secorHz

→ 2 →2

∴ →

 

/  

(63)   

   

 

Example9: The stepped shaft shownmusttransmit 40 kW at a speed of 720 rpm.Determinetheminimumradiusrofthefilletifanallowablestressof36MPaisnot tobeexceeded.

40 , 720

→40000

720 260

530.52

2

36

530.52 10 452

2452

1.21 → 1.21 2 

→ → ≅ 0.25 → 0.25 0.25 45 10.8  

 

(64)   

   

TorsionofNoncircularMembers

 

An important feature of the torsionaldeformation of noncircular prismatic bars isthewrapingofthecrosssections.Thetheoryofelasticitymaybeusedtorelatethetorqueapplied to such noncircular prismaticmembers to the resulting stress distributionandangleoftwist.

For thin‐walledopen‐sectionmembers ofuniform thickness (as those shown above) the sameformulation can be used to determine maximal stress and angle of twist. Maximal stress isapproximatelythesameoverthelong sidesurfaces exceptinthevicinityoftheshortsides.

 

 

131 0.63 5 

x

L

 

(65)   

   

 

Example 10: Segments AB and BC ofthe shaft have circular and squarecrosssections,respectively.Theshaftismade from A‐36 steel (G = 75 GPa)withanallowableshearstressof75 andanangleof twistatendAwhich is not allowed to exceed 0.02rad.DeterminethemaximumallowabletorqueT that canbeappliedat endA.TheshaftisfixedatC.

MaximumshearstressinshaftAB:

→ 75 30

2 30 → 3180860  

→ 3.18  

 

MaximumshearstressinshaftBC:

12 , 90 → 1 → 1 0.208 2 0.1406 

→1

2 → 750.208 90 902 2

 

→ 11372400 11.37  

 

MaximumangleoftwistatA:

/ / → 0.022

3 0 

→ 0.02 600

2 30 75000

6000.1406 90 90 75000

 

→ 2795311 → 2.79  

(66)   

   

 

Example11:A3‐m‐longsteelanglehasanL203×152×12.7crosssection.FromTablebelowwefind that the thicknessof thesection is12.7mmandthatitsareais4350mm2.Knowingthatτall=50MPa and thatG=77.2GPa, and ignoring theeffectofstressconcentrations,determine(a)thelargest torque T that can be applied, (b) thecorrespondingangleoftwist.

 

→ 50  

12.7 → 4350 12.7 → 342.52  

→342.5212.7

26.97 5 → 131 0.63

131 0.63

12.7342.52

0.325 

→ 50 → 50 0.325 342.52 12.7

→ 899242  

 

899242 30000.325 342.52 12.7 77200

0.1532 8.78° 

 

(67)   

   

 

Example 12: A 2.4‐m‐long steel member (G = 77 GPa) has aW200×46.1crosssection.KnowingthatT=560Nmdetermine(a)maximum shearing stress along lines a‐a and b‐b as well as theangleoftwist.

 

FromAppendixCfortheflanges:b=11mmanda=203mmsowehavea/b=18.45>5

131 0.63

131 0.63

11203

0.322

2 560  

FromAppendixCfortheweb:b=7.2mmanda=d‐2tf=203‐2(11)=181mm,a/b=25.14>5

131 0.63

131 0.63

7.2181

0.325

Thesystemisstaticallyindeterminatesoweneedacompatibilityequation: →

→ 3.96 → 248.61 62.78  

248.61 10 0.322 203 11

31.43  

62.78 10 0.325 181 7.2

20.58  

0.089 5.1° 

 

(68)   

   

 

Example13:Ahollowtubewithradial finsissubjectedtoatorqueT=2kNm.Findthetorque transmitted to the fins and themaximumshearstress.

2 8 1 → Compatibilityequation:

386

5

131 0.63

131 0.63

638

0.300

→2 41 35

0.3 38 6→

845.3 2 2.34 , 1981.25

Thefinscarrylessthan1%ofthetorque.

1981.25 10 41

2 41 3539

2.34 100.3 38 65.71

→ 39

(69)   

   

Thin‐walledHollowShafts(noncircularclosedsection)

As indicated the determination of stresses in noncircularmembersgenerallyrequirestheuseofadvancedmathematicalmethods. In the caseof thin‐walledhollownoncircular shafts(such as light‐weighted frameworks in aircrafts andspacecraft),however,agoodapproximationofthedistributionof stresses in the shaft can be obtained by a simplecomputation. Consider a hollowprismatic1 cylindrical (crosssection does not vary along the length of the member)memberofnoncircularclosed2sectionsubjectedtoatorsionalloading.While the thickness tof thewallmayvarywithinatransversesection3, itwillbeassumedthat itremainssmallcompared to theotherdimensions4 of the member. As t issmall we can assume that shear stress remains constantthroughwall thickness5. We now detach from the membertheportionofwallABboundedbytwotransverseplanesatadistance ∆x from each other, and by two longitudinal planesperpendiculartothewall.Consideringequilibrium:

→ ∆ ∆ → 

qiscalledshearflow.

We now detach a small element from the wall portion AB.Sincetheupperandlowerfacesofthiselementarepartofthefreesurfaceofthehollowmember,thestressesonthesefacesare equal to zero. It follows that the stress componentsindicated on the other faces by dashed arrows are also zero,while those representedbysolidarrowsareequal.Thus,theshearing stressatanypointofa transverse sectionof thehollowmemberisparalleltothewallsurface6.

WewillnowderivearelationbetweenthetorqueTappliedtoahollowmemberandtheshearflowqinitswall:

2

2 2 →2

→

This isaverageshearstressas it isbasedonassumptionthatshear stress does not vary across wall thickness. Maximalshearstressoccurswheretisminimal.

 

Oisanarbitrarypoint

(70)   

   

=theaverageshearstressactingoveraparticularthicknessofthetube.Wecallitaveragestressasweassumedthatshearstressremainsconstantthroughwallthickness.T=theresultantinternaltorqueatthecrosssectiont=thethicknessofthetubewhere istobedetermined=themeanareaenclosedwithintheboundaryofthecenterlineofthetube’sthickness.

AngleofTwistTheangleoftwistofathin‐walledtubeoflengthLcanbedeterminedusingenergymethods(wewillseeitinStrengthofMaterialsII).IfthematerialbehavesinalinearelasticmannerandGistheshearmodulus,thenthisangle( ),giveninradians,canbeexpressedas:

Heretheintegrationmustbeperformedaroundtheentireboundaryofthetube’scross‐sectionalarea(seetheexamplebelow).Iftremainsconstantthroughthesectionthenwecanwrite:

whereSisthelengthofthecenterline.

 

Example 14: A torque T = 5kNm is applied to a hollowshaft having the cross sectionshown.Determinetheshearingstress at points a and b. FindangleoftwistifL=2mandG=77GPa. 125 2 5

75 2 37935

2 5000000

2 7935 652.5

2 5000000

2 7935 1031.5

4 5 10 2000

4 7935 770002

125 106

275 610

0.0269 1.5°

(71)   

   

 

 

TBR5(1390):Ashafthasthecrosssectionshown and is made of 2014‐T6 aluminiumalloy(G=27GPa)havinganallowableshearstress ofτall=125MPa. If the angle of twistper meter length is not allowed to exceed0.03 rad, determine the requiredminimumwall thickness twhen the shaft is subjectedtoatorqueofT=15kNm.

75tan 30°

1502 2

75 18578.51

2 → 12515000000

2 18578.51 2

→ 3.22

4

Example 15: A 90‐Nm torque is applied to ahollow shaft having the cross section shown.Determinetheshearingstressatpointsaandb. 2 39 13 13 13

439 2377.6

2 90000

2 2377.6 44.73

2 90000

2 2377.6 29.46

4 4 2524

2132

2 39 /42

→4 →

0.031000

15 000 000

4 18578.51 2 27 000

2 75sin30° 75

→ 7.18

(72)   

   

r1r2

Example 16: Calculateshear stress at mid‐thicknessand theangleoftwist from elastic torsionformula and once fromthin‐walledtheory.

Elastictorsionformula(stress):

2

2 

→  

Thin‐walledformula(stress):

2 2 2

2  

Iftheshaftisthin‐walledwehave: :

2

2 24 2

→  

Forverythin‐walledshaftsthethin‐walledformulagivestheexactsolutionandbecomesequaltotheelasticformula

Elastictorsionformula(angleoftwist):

2

2 2

4 4 2

2 2 4

Iftheshaftisthin‐walledwehave: :

22

42

→

(73)   

   

 

Example 17: Equal torques areappliedtothin‐walledtubesofthesame length L, same thickness t,andsameradiusc.Oneofthetubeshasbeenslit lengthwiseasshown.Determine(a)theratioτb/τaofthemaximum shearing stresses in thetubes, (b) the ratio φb/φa of theanglesoftwistofthetubes.

2 2

0.333 2

0.333 22

→0.333

3

 

c

4 42

2 

0.333 23

2→

32

2 3 

Example 18: A circular tube (1) and a squaretube (2) are constructed of the same materialand subjected to the same torque. Both tubeshave the same length, samewall thickness, andsamecross‐sectionalarea.Whataretheratiosoftheirshearstressesandanglesoftwist?2 4 → 2

2 2

2 2 2 /42

2 2 40.79

 

4 42

2,

4 44 8

 

2 8 160.62

Theseresultsshowthatthecirculartubenotonlyhasa21%lowershearstressthandoesthesquaretubebutalsoagreaterstiffnessagainstrotation.