15.2 Unsymmetrical Bending 435
15.2.2 Resolution of Bending Moments
A bending momentMapplied in any longitudinal plane parallel to thezaxis may be resolved into
componentsMxandMybythenormalrulesofvectors.However,avisualappreciationofthesituation
isoftenhelpful.ReferringtoFig.15.11,weseethatabendingmomentMinaplaneatanangleθto
Oxmayhavecomponentsofdifferingsigndependingonthesizeofθ.Inbothcases,forthesenseof
Mshown
Mx=Msinθ
My=Mcosθ
which give, forθ<π/2,MxandMypositive (Fig. 15.11(a)) and forθ>π/2,Mxpositive andMy
negative(Fig.15.11(b)).
15.2.3 Direct Stress Distribution due to Bending
ConsiderabeamhavingthearbitrarycrosssectionshowninFig.15.12(a).Thebeamsupportsbending
momentsMxandMyandbendsaboutsomeaxisinitscrosssectionwhichisthereforeanaxisofzero
stressoraneutralaxis(NA).LetussupposethattheoriginofaxescoincideswiththecentroidCofthe
crosssectionandthattheneutralaxisisadistancepfromC.Thedirectstressσzonanelementofarea
δAatapoint(x,y)andadistanceξfromtheneutralaxisis,fromthethirdofEq.(1.42)
σz=Eεz (15.13)
If the beam is bent to a radius of curvatureρabout the neutral axis at this particular section then,
sinceplanesectionsareassumedtoremainplaneafterbending,andbyacomparisonwithsymmetrical
bendingtheory
εz=
ξ
ρ
Fig.15.11
Resolution of bending moments: (a)θ< 90 ◦and (b)θ> 90 ◦.