Page 304 - Aircraft Stuctures for Engineering Student

9.1 Bending of open and closed section beams 285

Differentiating Eqs (9.14) twice with respect to z and then substituting for C from
Eq. (9.13) we obtain

(9.15)

In the derivation of Eq. (9.6) we see that

(9.16)

Substituting in Eqs (9.16) for sinalp and cosalp from Eqs (9.15) and writing
ut' = d2u/d3, v" = d2v/d3 we have

(9.17)

It is instructive to rearrange Eqs (9.17) as follows
{ zi} = -E[ 2 t] (see derivation of Eq. (9.6)) (9.18)
:I}
{

i.e.

(9.19)

The first of Eqs (9.19) shows that M, produces curvatures, that is deflections, in
both the xz and yz planes even though M - 0; similarly for My when Mx = 0.
y:
Thus, for example, an unsymmetrical beam will deflect both vertically and horizon-
tally even though the loading is entirely in a vertical plane. Similarly, vertical and
horizontal components of deflection in an unsymmetrical beam are produced by
horizontal loads.
For a beam having either Cx or Cy (or both) as an axis of symmetry, IXy = 0 and
Eqs (9.17) reduce to

(9.20)

which are the equations of symmetrical bending theory.

Example 9.2
Determine the horizontal and vertical components of the tip deflection of the canti-
lever shown in Fig. 9.8. The second moments of area of its unsymmetrical section
are L, I,T LY.
and

From Eqs (9.17)