608 CHAPTER 22 Wings
Theresultantoftheshearflowsq 2 andq 3 mustbeequivalenttotheappliedtorque.Hence,formoments
aboutthecenterofsymmetryatanysectioninbay③andusingEq.(19.10),
200 × 800 q 2 + 200 × 800 q 3 = 10 × 106 Nmm
or
q 2 +q 3 =62.5 (ii)
SolvingEqs.(i)and(ii),weobtain
q 2 =46.9N/mm q 3 =15.6N/mm
ComparisonwiththeresultsofEq.(17.1)showsthattheshearflowsareincreasedbyafactorof1.5in
theupperandlowerskinpanelsanddecreasedbyafactorof0.5inthesparwebs.
Theflangeloadsareinequilibriumwiththeresultantsoftheshearflowsintheadjacentskinpanels
andsparwebs.Thus,forexample,inthetopflangeofthefrontspar,
P(st.4500)= 0
P(st.3000)= 1500 q 2 − 1500 q 3 =46875N(compression)
P(st.2250)= 1500 q 2 − 1500 q 3 − 750 q 1 = 0
Theloadsalongtheremainderoftheflangefollowfromantisymmetry,givingthedistributionshown
in Fig. 22.19. The load distribution in the bottom flange of the rear spar will be identical to that
shown in Fig. 22.19, while the distributions in the bottom flange of the front spar and the top flange
oftherearsparwillbereversed.Wenotethattheflangeloadsarezeroatthebuilt-inendofthewing
(station0).Generally,however,additionalstressesareinducedbythewarpingrestraintatthebuilt-in
end(see[Ref.1]).Theloadsonthewingribsoneitherinboardoroutboardendofthecutoutarefound
byconsideringtheshearflowsintheskinpanelsandsparwebsimmediatelyinboardandoutboardof
therib.Thus,fortheribatstation3000,weobtaintheshearflowdistributionshowninFig.22.20.
In Example 22.6, we implicitly assumed in the analysis that the local effects of the cutout were
completelydissipatedwithinthelengthoftheadjoiningbayswhichwereequalinlengthtothecutout
bay.ThevalidityofthisassumptionreliesonSt.Venant’sprinciple(Section2.4).Itmaygenerallybe
Fig.22.19
Distribution of load in the top flange of the front spar of the wing of Example 22.6.