430 CHAPTER 15 Bending of Open and Closed, Thin-Walled Beams
Example 15.2
NowdeterminethedistributionofdirectstressinthebeamofExample15.1ifthebendingmomentis
appliedinahorizontalplaneandinaclockwisesenseaboutCywhenviewedinthedirectionyC.
Inthiscase,thebeamwillbendabouttheverticalyaxiswhichthereforebecomestheneutralaxis
ofthesection.Thus,Eq.(15.9)becomes
σz=
M
Iyy
x,(i)
whereIyyisthesecondmomentofareaofthebeamsectionabouttheyaxis.AgainfromSection15.4,
Iyy= 2 ×
20 × 2003
12
+
260 × 253
12
=27.0× 106 mm^4
Hence,substitutingforMandIyyinEq.(i)
σz=
100 × 106
27.0× 106
x=3.7x
We have not specified a sign convention for bending moments applied in a horizontal plane.
However, a physical appreciation of the problem shows that the left-hand edges of the beam are
in compression, while the right-hand edges are in tension. Again the distribution is linear and
varies from 3.7×(−100)=−370N/mm^2 (compression) at the left-hand edges of each flange to
3.7×(+100)=+370N/mm^2 (tension)attheright-handedges.
WenotethatthemaximumstressesinthisexampleareverymuchgreaterthanthoseinExample15.1.
Thisisduetothefactthatthebulkofthematerialinthebeamsectionisconcentratedintheregionof
theneutralaxiswherethestressesarelow.TheuseofanI-sectioninthismannerwouldthereforebe
structurallyinefficient.
Example 15.3
ThebeamsectionofExample15.1issubjectedtoabendingmomentof100kNmappliedinaplane
paralleltothelongitudinalaxisofthebeambutinclinedat30◦totheleftofvertical.Thesenseofthe
bendingmomentisclockwisewhenviewedfromtheleft-handedgeofthebeamsection.Determinethe
distributionofdirectstress.
The bending moment is first resolved into two components,Mxin a vertical plane andMyin a
horizontalplane.Equation(15.9)maythenbewrittenintwoforms
σz=
Mx
Ixx
y σz=
My
Iyy
x (i)
Theseparatedistributionscanthenbedeterminedandsuperimposed.Amoredirectmethodistocombine
thetwoequations(i)togivethetotaldirectstressatanypoint(x,y)inthesection.Thus,
σz=
Mx
Ixx
y+
My
Iyy
x (ii)