538 CHAPTER 19 Structural Idealization
Fig.19.2
Idealization of a wing section.
dueto,say,bendingofthewingwouldbesmall.Furthermore,thedifferencebetweenthedistancesof
thestringercentroidsandtheadjacentskinfromthewingsectionaxisissmall.Itwouldbereasonable
toassume,therefore,thatthedirectstressisconstantoverthestringercrosssections.Wecouldtherefore
replacethestringersandsparflangesbyconcentrationsofarea,knownasbooms,overwhichthedirect
stressisconstantandwhicharelocatedalongthemidlineoftheskin,asshowninFig.19.2.Inwing
andfuselagesectionsofthetypeshowninFig.19.1,thestringersandsparflangescarrymostofthe
directstresses,whiletheskinismainlyeffectiveinresistingshearstresses,althoughitalsocarriessome
of the direct stresses. The idealization shown in Fig. 19.2 may therefore be taken a stage further by
assumingthatalldirectstressesarecarriedbythebooms,whiletheskiniseffectiveonlyinshear.The
directstress-carryingcapacityoftheskinmaybeallowedforbyincreasingeachboomareabyanarea
equivalent to the direct stress-carrying capacity of the adjacent skin panels. The calculation of these
equivalent areas will generally depend on an initial assumption as to the form of the distribution of
directstressinaboom/skinpanel.
19.2 IdealizationofaPanel...............................................................................
SupposethatwewishtoidealizethepanelofFig.19.3(a)intoacombinationofdirectstress-carrying
boomsandshear-stress-only-carryingskin,asshowninFig.19.3(b).InFig.19.3(a),thedirectstress-
carryingthicknesstDoftheskinisequaltoitsactualthicknesst,whileinFig.19.3(b),tD=0.Suppose
alsothatthedirectstressdistributionintheactualpanelvarieslinearlyfromanunknownvalueσ 1 to
an unknown valueσ 2. Clearly the analysis should predict the extremes of stressσ 1 andσ 2 , although
thedistributionofdirectstressisobviouslylost.Sincetheloadingproducingthedirectstressesinthe
actual and idealized panels must be the same, we can equate moments to obtain expressions for the
boomareasB 1 andB 2 .Thus,takingmomentsabouttheright-handedgeofeachpanel,
σ 2 tD
b^2
2
+
1
2
(σ 1 −σ 2 )tDb
2
3
b=σ 1 B 1 b
fromwhich
B 1 =
tDb
6
(
2 +
σ 2
σ 1