Page 420 - Aircraft Stuctures for Engineering Student

10.3 Wings 401

Fig. 10.35 Final shear flow system in Cell II.

Fig. 10.35. Since the cell does not twist then, from Eq. (9.42)
dB 1 ds
-=- q-=o
dz 2AIIGf11 t
or

Hence, from Fig. 10.35 and taking anticlockwise torques as positive
ds
fII qb 7 - qI1611 + qIbI,II f ~III~III.11 = 0
giving

(10.37)

The first term on the right-hand side of Eq. (10.37) represents the proportion of the
'open section' shear flow qb which acts as a constant shear flow around the cell to
cancel out the twist due to qb; the second and third terms counteract the twist due
to qr and qllI. Rewriting Eq. (10.37)
411 = QiI + C1,rrqr + CI1I:lIqIII (10.38)
in which CIqII is the correction carry over factor from cell I to cell I1 and CIIIqII is the
correction carry over factor from cell 111 to cell 11.
As a first approximation in the solution we neglect the effect of the shear flows in
adjacent cells so that
411 = 41
Similarly
41 = rlf? qIII = 4h
Substituting these values in Eq. (10.38) we have
(10.39)

or
411 = 41 + 41 (10.40)