Page 358 - Aircraft Stuctures for Engineering Student

9.9 Effect of idealization 339

Eq. (9.80) reduces to

in which
[y+y = 2(200 x 302 + 250 x 100’ + 400 x 100’ + 100 x 502) = 13.86 x lo6 mm4
Equation (i) then becomes
10 x io3
“ = - 13.86 x lo6 r=l Bryr + qs,o

i.e.
n
qs = -7.22 x 10-~ B,~, + (ii)
r=l
‘Cutting’ the beam section in the wall 23 (any wall may be chosen) and calculating the
‘basic’ shear flow distribution qb from the hst term on the right-hand side of Eq. (ii)
we have

qb:23 =
qb,34 = -7.22 X 10-4(400 X 100) = -28.9N/mm

qb,4j = -28.9 - 7.22 x io-4(ioo x 50) = -32.5N/m.m
qb,56 = qb:34 = -28.9 N/mm (by symmetry)
qb.67 = qb;23 = (by
qb.21 = -7.22 X 10-4(250 X 100) = -18.1N/mm

qb.18 = -18.1 - 7.22 x 10-4(200 x 30) = -22.4N/mm

N/mm (by
qb:87 = qb:21 = -18a1
Taking moments about the intersection of the line of action of the shear load and the
horizontal axis of symmetry and referring to the results of Eqs (9.76) and (9.77) we
have, from Eq. (9.38)

0 = [qb,81 X 60 x 480 + 2qbT12(240 x 100 + 70 x 240) + 2qb,23 x 240 x 100
-2qb.43 x 120 x 100 - qb;54 x 100 x 1201 + 2 x 972OOqs,0
Substituting the above values of qb in this equation gives
qs,o = -5.4N/=

the negative sign indicating that qs,0 acts in a clockwise sense.
In any wall the final shear flow is given by qs = qb + qs,o so that
q21 = -18.1 + 5.4 = -12.7N/~n.m = q87

q23 = -5.4N/mm = q67