Page 330 - Aircraft Stuctures for Engineering Student

9.5 Torsion of closed section beams 31 1

where

In Eq. (i)
(: t)
wo=O, S=2 -+- and A=ab

From 0 to 1,0 < s1 < b/2 and

Note that Sos and Aos are both positive.

Substitution for So, and Aos from Eq. (ii) in Eq. (i) shows that the warping distribu-
tion in the wall 01, wol, is linear. Also

which gives
IV] =-(----) a (iii)
T
b
8abG tb
The remainder of the warping distribution may be deduced from symmetry and the
fact that the warping must be zero at points where the axes of symmetry and the
walls of the cross-section intersect. It follows that
w2 = -w1 = -w3 = w4
giving the distribution shown in Fig. 9.31. Note that the warping distribution will take
the form shown in Fig. 9.31 as long as Tis positive and b/tb > a/t,. If either of these
conditions is reversed w1 and w3 will become negative and H:., and w4 positive. In the
case when b/tb = a/t, the warping is zero at all points in the cross-section.

Fig. 9.31 Warping distribution in the rectangular section beam of Example 9.7.