Page 325 - Aircraft Stuctures for Engineering Student

306 Open and closed, thin-walled beams

3
1 9a 6a

Fig. 9.26 Closed section beam of Example 9.6.

Eq. (9.35) simplifies to
S
t
q --2
s - p Yd.S+%,O
Ixx 0
in which
I,, = 2 [ 1; t ( $SI>’ dsl + 1: f ( As2>’ ds2]

Evaluating this expression gives I,, = 1 152a3t.
The basic shear flow distribution qb is obtained from the first term in Eq. (i). Thus,
for the wall 41

In the wall 12

which gives

(iii)
The qb distributions in the walls 23 and 34 follow from symmetry. Hence from
Eq. (9.48)

giving
”
qs,o = -
(58.7;)
1 1 52a3