15.1 Symmetrical Bending 429
Fig.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).