Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

22.8 Cutouts in Wings 607

Midwaybetweenstations1500and3000apointofcontraflexureoccursinthefrontandrearsparsso
thatatthispointthebendingmomentiszero.Hence,

200 P= 12500 ×750Nmm

sothat

P=46875N

Alternatively,Pmaybefoundbyconsideringtheequilibriumofeitherofthesparflanges.Thus,

2 P= 1500 q 1 = 1500 ×62.5N

whence

P=46875N

TheflangeloadsParereactedbyloadsintheflangesofbays①and③.Theseflangeloadsaretransmitted
totheadjacentsparwebsandskinpanelsasshowninFig.22.18forbay③andmodifytheshearflow
distributiongivenbyEq.(17.1).Forequilibriumofflange1,

1500 q 2 − 1500 q 3 =P=46875N

or

q 2 −q 3 =31.3 (i)

Fig.22.18

Loads on bay③of the wing of Example 22.6.