Page 314 - Aircraft Stuctures for Engineering Student

9.3 Shear of open section beams 295

and
- p - - XR sin +- + yR cos +- (9.29)
dv,
de
de
de
=
dz dz dz dz
Also from Eq. (9.27)
de du + -sm +
dv .
3 = p- + -cos (9.30)
dz dz dz dz
Comparing the coefficients of Eqs (9.29) and (9.30) we see that
dvldz duldz
XR = -- (9.31)
dO/dz I YR =- dQ/dz

The open section beam of arbitrary section shown in Fig. 9.18 supports shear loads S,
and Sy such that there is no twisting of the beam cross-section. For this condition to
be valid the shear loads must both pass through a particular point in the cross-section
known as the shear centre (see also Section 11.5).
Since there are no hoop stresses in the beam the shear flows and direct stresses
acting on an element of the beam wall are related by Eq. (9.22), i.e.
aq do,
-+t-=o
as dz
We assume that the direct stresses are obtained with sufficient accuracy from basic
bending theory so that from Eq. (9.6)
-- - [(aM,/az)Ixx - (awaz)r.Xyl + [(aMx/wryy - (dMy/wx,l
acz
dZ IxxI,, - I:, Ixxryy - I&

't

Fig. 9.18 Shear loading of open section beam.