Page 318 - Aircraft Stuctures for Engineering Student

9.3 Shear of open section beams 299

t
3 4
t A h

2

11
S C X
h
2
2u’

Fig. 9.22 Determination of shear centre position of channel section of Example 9.5

The shear centre S lies on the horizontal axis of symmetry at some distance &, say,
from the web. If we apply an arbitrary shear load Sy through the shear centre then the
shear flow distribution is given by Eq. (9.34) and the moment about any point in the
cross-section produced by these shear flows is equivalent to the moment of the applied
shear load. Sy appears on both sides of the resulting equation and may therefore be
eliminated to leave &.
For the channel section, Cx is an axis of symmetry so that Ixy = 0. Also S, = 0 and
therefore Eq. (9.34) simplifies to

where
‘:1
I xx- -2bt (;y +-=- ;;( I+- ?)
-

Substituting for I,, in Eq. (i) we have

(ii)
-IZSy
‘”h3(1+6b/h) Pds o
The amount of computation involved may be reduced by giving some thought to
the requirements of the problem. In this case we are asked to find the position of
the shear centre only, not a complete shear flow distribution. From symmetry it is
clear that the moments of the resultant shears on the top and bottom flanges about
the mid-point of the web are numerically equal and act in the same rotational
sense. Furthermore, the moment of the web shear about the same point is zero. We
deduce that it is only necessary to obtain the shear flow distribution on either the
top or bottom flange for a solution. Alternatively, choosing a web/flange junction
as a moment centre leads to the same conclusion.