Page 316 - Aircraft Stuctures for Engineering Student

9.3 Shear of open section beams 297

The origin for our system of reference axes coincides with the centroid of the
section at the mid-point of the web. From antisymmetry we also deduce by inspection
that the shear centre occupies the same position. Since S, is applied through the shear
centre then no torsion exists and the shear flow distribution is given by Eq. (9.34) in
which S, = 0, i.e.

or

SY
qs = IxxI,, - I$ (Ix, txds - I,, tY ds)

The second moments of area of the section have previously been determined in
Example 9.3 and are

Substituting these values in Eq. (i) we obtain
s,
qs = - (10.32~ - 6.84~) ds (ii)
h3 10
On the bottom flange 12, y = -h/2 and x = -h/2 + sl, where 0 < s1 < h/2. Therefore

giving

(iii)

Hence at 1 (sl = 0), q1 = 0 and at 2 (sl = h/2), q2 = 0.42SJh. Further examination
of Eq. (iii) shows that the shear flow distribution on the bottom flange is parabolic
with a change of sign (Le. direction) at s1 = 0.336h. For values of s1 < 0.336h, q12
is negative and therefore in the opposite direction to sl.
In the web 23, y = -h/2 + s2, where 0 < s2 < h and x = 0. Thus

We note in Eq. (iv) that the shear flow is not zero when s2 = 0 but equal to the value
obtained by inserting s1 = h/2 in Eq. (iii), i.e. q2 = 0.42Sy/h. Integration of Eq. (iv)
yields
S
q23 = (0.42h2 + 3.42h.Y~ - 3.424)

This distribution is symmetrical about Cx with a maximum value at s2 = h/2(y = 0)
and the shear flow is positive at all points in the web.
The shear flow distribution in the upper flange may be deduced from antisymmetry
so that the complete distribution is of the form shown in Fig. 9.20.