Page 431 - Aircraft Stuctures for Engineering Student

41 2 Stress analysis of aircraft components

Thus, substituting for q1 and q2 obtained from Eq. (10.17) and noting that
S,,,2 = S,,l - W, we have

in which qs,o is calculated using Eq. (9.37) where the shear load is W and
-w
qb = BrYr
Ixx r=l
The method of determining the shear flow distribution applied to the periphery of a
fuselage frame is identical to the method of solution (or the alternative method) of
Example 10.5.
Having determined the shear flow distribution around the periphery of the frame,
the frame itself may be analysed for distributions of bending moment, shear force and
normal force, as described in Section 4.7.

10.4.2 Wing ribs

Wing ribs perform similar functions to those performed by fuselage frames. They
maintain the shape of the wing section, assist in transmitting external loads to the
wing skin and reduce the column length of the stringers. Their geometry, however,
is usually different in that they are frequently of unsymmetrical shape and possess
webs which are continuous except for lightness holes and openings for control runs.
Wing ribs are subjected to loading systems which are similar to those applied to
fuselage frames. External loads applied in the plane of the rib produce a change in
shear force in the wing across the rib; this induces reaction shear flows around its
periphery. These are calculated using the methods described in Chapter 9 and in
Section 10.3. To illustrate the method of rib analysis we shall use the example of a
three-flange wing section in which, as we noted in Section 10.3, the shear flow
distribution is statically determinate.

Example 10.14
Calculate the shear flows in the web panels and the axial loads in the flanges of the
wing rib shown in Fig. 10.48. Assume that the web of the rib is effective only in
shear while the resistance of the wing to bending moments is provided entirely by
the three flanges 1,2 and 3.

Since the wing bending moments are resisted entirely by the flanges 1,2 and 3, the
shear flows developed in the wing skin are constant between the flanges. Using the
method described in Section 10.3 for a three-flange wing section we have, resolving
forces horizontally
600q12 - 600q23 = 12 000 N (9
Resolving vertically

3ooq31 - 3ooq23 15 000 N