Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

22.8 Cutouts in Wings 611

SolvingEqs.(i)and(ii)gives

q′ 32 =117.6N/mm q′ 14 =53.1N/mm

The final shear flows in bay②are found by superimposingq′ 12 ,q′ 32 ,andq′ 14 on the shear flows in
Fig.22.22,givingthedistributionshowninFig.22.24.Alternatively,theseshearflowscouldhavebeen
founddirectlybyconsideringtheequilibriumofthecutoutbayundertheactionoftheappliedshear
loads.
The correction shear flows in bay②(Fig. 22.23) will also modify the shear flow distributions in
bays①and③.ThecorrectionshearflowstobeappliedtothoseshowninFig.22.22forbay③(thosein
bay①willbeidentical)maybefoundbydeterminingtheflangeloadscorrespondingtothecorrection
shearflowsinbay②.
Itcanbeseenfromthemagnitudesanddirectionsofthesecorrectionshearflows(Fig.22.23)thatat
anysectioninbay②,theloadsintheupperandlowerflangesofthefrontsparareequalinmagnitude
butoppositeindirection,similarfortherearspar.Thus,thecorrectionshearflowsinbay②produce
anidenticalsystemofflangeloadstothatshowninFig.22.17forthecutoutbaysinthewingstructure
ofExample22.6.Itfollowsthatthesecorrectionshearflowsproducedifferentialbendingofthefront
andrearsparsinbay②andthatthesparbendingmomentsandhencetheflangeloadsarezeroatthe
midbaypoints.Therefore,atstation3000,theflangeloadsare

P 1 =(75.9+53.1)× 500 =64500N(compression)

P 4 =64500N(tension)

P 2 =(75.9+117.6)× 500 =96750N(tension)

P 3 =96750N(tension)

These flange loads produce correction shear flowsq′′ 21 ,q′′ 43 ,q′′ 23 ,andq′′ 41 in the skin panels and spar
websofbay③,asshowninFig.22.25.Thus,forequilibriumofflange1,

1000 q′′ 41 + 1000 q′′ 21 =64500N (iii)

andforequilibriumofflange2,

1000 q′′ 21 + 1000 q′′ 23 =96750N (iv)

Fig.22.24

Final shear flows (N/mm) in the cutout bay of the wing box of Example 22.7.