Page 300 - Aircraft Stuctures for Engineering Student

9.1 Bending of open and closed section beams 281

In the case where the beam cross-section has either (or both) Cx or Cy as an axis of
symmetry the product second moment of area Ixy is zero and Cxy are principal axes.
Equation (9.7) then reduces to

Further, if either M,, or M, is zero then

(9.9)
Equations (9.8) and (9.9) are those derived for the bending of beams having at least a
singly symmetrical cross-section. It may also be noted that in Eqs (9.9) a= = 0 when,
for the first equation, y = 0 and for the second equation when x = 0. Therefore, in
symmetrical bending theory the x axis becomes the neutral axis when M,, = 0 and
the y axis becomes the neutral axis when Mx = 0. Thus we see that the position of
the neutral axis depends on the form of the applied loading as well as the geometrical
properties of the cross-section.
There exists, in any unsymmetrical cross-section, a centroidal set of axes for which
the product second moment of area is zero. These axes are then principal axes and the
direct stress distribution referred to these axes takes the simplified form of Eqs (9.8) or
(9.9). It would therefore appear that the amount of computation can be reduced if
these axes are used. This is not the case, however, unless the principal axes are obvious
from inspection since the calculation of the position of the principal axes, the princi-
pal sectional properties and the coordinates of points at which the stresses are to be
determined consumes a greater amount of time than direct use of Eqs (9.6) or (9.7) for
an arbitrary, but convenient set of centroidal axes.

9.1.4 Position of the neutral axis

The neutral axis always passes through the centroid of area of a beam’s cross-section
but its inclination a (see Fig. 9.4(b)) to the x axis depends on the form of the applied
loading and the geometrical properties of the beam’s cross-section.
At all points on the neutral axis the direct stress is zero. Therefore, from Eq. (9.6)

where xh:A and JJ~A are the coordinates of any point on the neutral axis. Hence
JJNA - - MJxx - M.Jx.v
-
-
XNA M.rIy.v - M/.xy
or, referring to Fig. 9.4(b) and noting that when a is positive x~A and yNA are of
opposite sign
My?rx - MxIxy
tana = (9.10)
MJyy - MJxy