Page 302 - Aircraft Stuctures for Engineering Student

9.1 Bending of open and closed section beams 283

Since M, = 1500 N m and My = 0 we have, from Eq. (9.7)
0, = 1.5~ - 0.39~

in which the units are N and mm.
By inspection of Eq. (i) we see that a, will be a maximum at F where x = -8 mm,
y = -66.4mm. Thus

mzz:max = -96 N/mm2 (compressive)
In some cases the maximum value cannot be obtained by inspection so that values of
CJ- at several points must be calculated.

9.1.5 Load intensity, shear force and bending moment
relationships, general case

Consider an element of length Sz of a beam of unsymmetrical cross-section subjected
to shear forces, bending moments and a distributed load of varying intensity, all in
the yz plane as shown in Fig. 9.6. The forces and moments are positive in accordance
with the sign convention previously adopted. Over the length of the element we
may assume that the intensity of the distributed load is constant. Therefore, for
equilibrium of the element in the y direction

+
(.. + SSZ) wy sz - sy = 0

from which
8%
w,, = --
dZ
Taking moments about A we have
(S#
(Mx+TSz) - (S,+%Sz)Sz- w,2- M,=O

Y

Fig. 9.6 Equilibrium of beam element supporting a general force system in the yz plane.