Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

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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

)

(19.1)
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