15.1 Symmetrical Bending 429Fig.15.6
Direct stress distribution in beam of Example 15.1.
The cross section of the beam is doubly symmetrical so that the centroid, C, of the section, and
thereforetheoriginofaxes,coincideswiththemidpointoftheweb.Furthermore,thebendingmoment
isappliedtothebeamsectioninaverticalplanesothatthexaxisbecomestheneutralaxisofthebeam
section;therefore,weneedtocalculatethesecondmomentofarea,Ixx,aboutthisaxis.
Ixx=200 × 3003
12
−
175 × 2603
12
=193.7× 106 mm^4 (seeSection15.4)FromEq.(15.9),thedistributionofdirectstress,σz,isgivenby
σz=−100 × 106
193.7× 106
y=−0.52y (i)Thedirectstress,therefore,varieslinearlythroughthedepthofthesectionfromavalue
−0.52×(+ 150 )=−78N/mm^2 (compression)atthetopofthebeamto
−0.52×(− 150 )=+78N/mm^2 (tension)atthebottomasshowninFig.15.5(b).