534 CHAPTER 18 Combined Open and Closed Section Beams
Thetorsionalrigidityofthecompletesectionisthen
GJ= 5000 × 107 + 6 × 107 = 5006 × 107 Nmm^2
Inallunrestrainedtorsionproblems,thetorqueisrelatedtotherateoftwistbytheexpression
T=GJ
dθ
dz
Theangleoftwistperunitlengthisthereforegivenby
dθ
dz
=
T
GJ
=
10 × 106
5006 × 107
=0.0002rad/mm
SubstitutingforTinEq.(17.1)fromEq.(17.4),weobtaintheshearflowintheclosedsection.Thus,
qcl=
(GJ)cl
2 A
dθ
dz
=
5000 × 107
2 × 20000
×0.0002
fromwhich
qcl=250N/mm
Themaximumshearstressintheclosedsectionisthen250/1.5=166.7N/mm^2.
Intheopenportionofthesection,themaximumshearstressisobtaineddirectlyfromEq.(17.10)
andis
τmax,op= 25000 × 2 ×0.0002=10N/mm^2
Itcanbeseenfromtheabovethatintermsofstrengthandstiffness,theclosedportionofthewingsection
dominates.Thisdominancemaybeusedtodeterminethewarpingdistribution.Havingfirstfoundthe
positionofthecenteroftwist(theshearcenter),thewarpingoftheclosedportioniscalculatedusing
the method described in Section 17.1. The warping in the walls 13 and 34 is then determined using
Eq.(17.19),inwhichtheoriginforthesweptareaARistakenatthepoint1andthevalueofwarping
isthatpreviouslycalculatedfortheclosedportionat1.
Problems..............................................................................................
P.18.1 ThebeamsectionofExample18.1(seeFig.18.2)issubjectedtoabendingmomentinaverticalplaneof
20kNm.Calculatethemaximumdirectstressinthecrosssectionofthebeam.
Ans. 172.5N/mm^2.
P.18.2 AwingboxhasthecrosssectionshowndiagrammaticallyinFig.P.18.2andsupportsashearloadof100
kNinitsverticalplaneofsymmetry.Calculatetheshearstressatthemidpointoftheweb36ifthethicknessofall
wallsis2mm.
Ans. 89.7N/mm^2.