Page 344 - Aircraft Stuctures for Engineering Student

9.7 Analysis of combined open and closed sections 325

The analysis of the open part of the beam section is now complete since the shear
flow distribution in the walls 67 and 78 follows from symmetry. To determine the
shear flow distribution in the closed part of the section we must use the method
described in Section 9.4 in which the line of action of the shear load is known.
Thus we ‘cut’ the closed part of the section at some convenient point, obtain the qb
or ‘open section’ shear flows for the complete section and then take moments as in
Eqs (9.37) or (9.38). However, in this case, we may use the symmetry of the section
and loading to deduce that the final value of shear flow must be zero at the mid-
4r
points of the walls 36 and 45, i.e. qs = qs,o = 0 at these points. Hence
403 = -69.0 x 10- 2 x 75&3

and q3 = - 104 N/mm in the wall 03. It follows that for equilibrium of shear flows at 3,
4r
q3, in the wall 34, must be equal to -138.5 - 104 = -242.5N/mm. Hence
q34 = -69.0 x 10- 2(75 - ~4) ds4 - 242.5

which gives
+
q34 = -1.04~~ 69.0 x - 242.5 (vi)
Examination of Eq. (vi) shows that q34 has a maximum value of -281.7 N/mm at
s4 = 75mm; also q4 = -172.5N/mm. Finally, the distribution of shear flow in the
wall 94 is given by

q94 = -69.0 x lop4 2(-125) ds5
Lj
i.e.
q94 = 1.73s5 (vii)
The complete distribution is shown in Fig. 9.43.

9.7.3 Torsion

Generally, in the torsion of composite sections, the closed portion is dominant since
its torsional stiffness is far greater than that of the attached open section portion
which may therefore be frequently ignored in the calculation of torsional stiffness;
shear stresses should, however, be checked in this part of the section.

Example 9.10
Find the angle of twist per unit length in the wing whose cross-section is shown in
Fig. 9.44 when it is subjected to a torque of 10kNm. Find also the maximum
shear stress in the section. G = 25 000 N/mm2.
Wall 12 (outer) = 900mm. Nose cell area = 20 OOOmm2.