Page 319 - Aircraft Stuctures for Engineering Student

300 Open and closed, thin-walled beams

On the bottom flange, y = -h/2 so that from Eq. (ii) we have
6S,
(iii)
q12 = h2(1 + 6b/h) s1

Equating the clockwise moments of the internal shears about the mid-point of the web
to the clockwise moment of the applied shear load about the same point gives

or, by substitution from Eq. (iii)

from which
3b2
Is = h(1 + 6h/h)
In the case of an unsymmetrical section, the coordinates (Js, qs) of the shear centre
referred to some convenient point in the cross-section would be obtained by first
determining Es in a similar manner to that of Example 9.5 and then finding qs by
applying a shear load S, through the shear centre. In both cases the choice of a
web/flange junction as a moment centre reduces the amount of computation.

hear of closed section beams

The solution for a shear loaded closed section beam follows a similar pattern to that
described in Section 9.3 for an open section beam but with two important differences.
First, the shear loads may be applied through points in the cross-section other than
the shear centre so that torsional as well as shear effects are included. This is possible
since, as we shall see, shear stresses produced by torsion in closed section beams have
exactly the same form as shear stresses produced by shear, unlike shear stresses due to
shear and torsion in open section beams. Secondly, it is generally not possible to
choose an origin for s at which the value of shear flow is known. Consider the
closed section beam of arbitrary section shown in Fig. 9.23. The shear loads S, and
S,. are applied through any point in the cross-section and, in general, cause direct
bending stresses and shear flows which are related by the equilibrium equation
(9.22). We assume that hoop stresses and body forces are absent. Thus
do;
dq
-+r-=o
as az
From this point the analysis is identical to that for a shear loaded open section beam
until we reach the stage of integrating Eq. (9.33), namely