Page 409 - Aircraft Stuctures for Engineering Student

390 Stress analysis of aircraft components

Choosing GREF as 27 600 N/mm2 then, from Eq. (10.27)

3 x 27600
t;, = x 1.22 = 3.66mm
27 600
Hence
1270
67, = - - 347
-
3.66
Also

612 = 656 = 840, 623 = 783, 634 = 1083, 63, = 57, 684 = 95, 687 = 347,
627 = 68, 675 106, 616 = 202

We now ‘cut’ the top skin panels in each cell and calculate the ‘open section’ shear
flows using Eq. (9.75) which, since the wing section is idealized, singly symmetrical
(as far as the direct stress carrying area is concerned) and is subjected to a vertical
shear load only, reduces to

where, from Example 10.6, Ixx = 809 x 106mm4. Thus, from Eq. (i)
86.8 x IO3 n
qb = - B,~, = - 1.07 x 10-~ B,~, (ii)
809 x lo6 r=l
r=l
Since qb = 0 at each ‘cut’, then qb = 0 for the skin panels 12,23 and 34. The remaining
qb shear flows are now calculated using Eq. (ii). Note that the order of the numerals in
the subscript of qb indicates the direction of movement from boom to boom.

qb:27 = -1.07 x x 3880 x 230 = -95.5N/mm
qb,J6 = -1.07 x x 2580 x 165 = -45.5N/mm

qb$5 -45.5 - 1.07 x io-4 x 2580 x (-165) = 0

qb.57 = .-1.07 X X 3880 X (-230) = 95.5N/=
qb;38 = -1.07 X X 3230 X 200 = -b9.0N/mm
qb.48 = - 1.07 X lop4 X 3230 x (-200) = 69.0 N/mm

Therefore, as ,,83 = qb.48 (or qb,72 = qb,57), &7g = 0. The distribution of the shear
flows is show n Fig. 10.25. The values of 6 and qb are now substituted in Eq. (10.28)
for each cell in turn.
For cell I
_- 1 [qs,o:l( 1083 + 95 + 57) - 57qS,o,11 + 69 x 95 + 69 x 571 (iii)
d0
dz - 2 x 265 OOOGREF