15.1 Symmetrical Bending 425
Fig.15.2
Symmetrical section beams.
Fig.15.3
Beam subjected to a pure sagging bending moment.
15.1.2 Direct Stress Distribution
Consideralengthofbeam(Fig.15.4(a))thatissubjectedtoapure,saggingbendingmoment,M,applied
inaverticalplane;thebeamcrosssectionhasaverticalaxisofsymmetryasshowninFig.15.4(b).The
bendingmomentwillcausethelengthofbeamtobendinasimilarmannertothatshowninFig.15.3(a)
sothataneutralplanewillexistwhichis,asyet,unknowndistancesy 1 andy 2 fromthetopandbottom
ofthebeam,respectively.CoordinatesofallpointsinthebeamarereferredtoaxesOxyz,inwhichthe
originOliesintheneutralplaneofthebeam.Weshallnowinvestigatethebehaviorofanelemental
length,δz,ofthebeamformedbyparallelsectionsMINandPGQ(Fig.15.4(a))andalsothefiberST
of cross-sectional areaδAa distanceyabove the neutral plane. Clearly, before bending takes place
MP=IG=ST=NQ=δz.
ThebendingmomentMcausesthelengthofbeamtobendaboutacenterofcurvatureCasshownin
Fig.15.5(a).Sincetheelementissmallinlengthandapuremomentisapplied,wecantakethecurved
shapeofthebeamtobecircularwitharadiusofcurvatureRmeasuredtotheneutralplane.Thisisa
usefulreferencepoint,since,aswehaveseen,strainsandstressesarezerointheneutralplane.
ThepreviouslyparallelplanesectionsMINandPGQremainplaneaswehavedemonstratedbutare
nowinclinedatanangleδθtoeachother.ThelengthMPisnowshorterthanδzasisST,whileNQis
longer;IG,beingintheneutralplane,isstilloflengthδz.SincethefiberSThaschangedinlength,it