Page 408 - Aircraft Stuctures for Engineering Student

10.3 Wings 389

I I I‘
U’

Fig. 10.24 Moment equilibrium of Rth cell.

Substituting for qR in terms of the ‘open section’ shear flow qb and the redundant
shear flow qs‘s;O,R, we have

Mq,R = fR %PO dS + qs,O,R fRPO dS
or

Mq,R = fR qbP0 ds f 2ARqs,0:R

The sum of the moments from the individual cells is equivalent to the moment of the
externally applied loads about the same point. Thus, for the wing section of Fig. 10.22

(10.29)

If the moment centre is chosen to coincide with the point of intersection of the lines of
action of S, and S,, Eq. (10.29) becomes

(10.30)

Example 10.8
The wing section of Example 10.6 (Fig. 10.17) carries a vertically upward shear load
of 86.8 kN in the plane of the web 572. The section has been idealized such that the
booms resist all the direct stresses while the walls are effective only in shear. If the
shear modulus of all walls is 27 600 N/m2 except for the wall 78 for which it is
three times this value, calculate the shear flow distribution in the section and the
rate of twist. Additional data are given below.

Wall Length (mm) Thickness (mm) Cell area (mm’)
~~
12, 56 1023 1.22 AI = 265000
23 1274 1.63 AI1 = 213 000
34 2200 2.03 A111 = 413 000
483 400 2.64
572 460 2.64
61 330 1.63
78 1270 1.22