Page 389 - Aircraft Stuctures for Engineering Student

370 Stress analysis of aircraft components

From column @

From column @

From column @

From Eqs (10.15)
Sx,w = 0, Sy,w = 100 - 33.4 = 66.6 kN

The shear flow distribution in the walls of the beam is now found using the method
described in Section 9.9. Since, for this beam, Zxy = 0 and Sx = Sx,w = 0, Eq. (9.80)
reduces to

We now ‘cut’ one of the walls, say 16. The resulting ‘open section’ shear flow is
given by
66.6 x IO3
qb = - BrYr
5.4 x 108 r=l

or

Thus
qb.16 =
qb.12 = 0 - 1.23 X X 900 X 300 = -33.2N/mm

qb,23 = -33.2 - 1.23 x x 1200 x 300 = -77.5N/mm
qb,34 = -77.5 - 1.23 x x 900 x 300 = -110.7N/mm
qb.45 = -77.5 N/mm (from symmetry)
qb,56 = -33.2 N/mm (from symmetry)

giving the distribution shown in Fig. 10.7. Taking moments about the centre of
symmetry we have, from Eq. (10.16)
-100 x lo3 x 600 = 2 x 33.2 x 600 x 300 + 2 x 77.5 x 600 x 300

+ 110.7 x 600 x 600 + 2 x 1200 x 600qs,0