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Chapter 23

Q. 23.7

A wing box has the skin panel on its undersurface removed between stations 2000 and 3000 and carries lift and drag loads which are constant between stations 1000 and 4000 as shown in Fig. 23.21(a). Determine the shear flows in the skin panels and spar webs and also the loads in the wing ribs at the inboard and outboard ends of the cut-out bay.

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Step-by-Step

Verified Solution

Assume that all bending moments are resisted by the spar flanges while the skin panels and spar webs are effective only in shear.
The simplest approach is first to determine the shear flows in the skin panels and spar webs as though the wing box were continuous and then to apply an equal and opposite shear flow to that calculated around the edges of the cut-out. The shear flows in the wing box without the cut-out will be the same in each bay and are calculated using the method described in Section 20.3 and illustrated in Example 20.4. This gives the shear flow distribution shown in Fig. 23.22.

We now consider bay ② and apply a shear flow of 75.9 N/mm in the wall 34 in the opposite sense to that shown in Fig. 23.22. This reduces the shear flow in the wall 34 to zero and, in effect, restores the cut-out to bay ②. The shear flows in the remaining walls of the cut-out bay will no longer be equivalent to the externally applied shear loads so that corrections are required. Consider the cut-out bay (Fig. 23.23) with the shear flow of 75.9 N/mm applied in the opposite sense to that shown in Fig. 23.22. The correction shear flows q_{12}^{\prime}, q_{32}^{\prime} \text { and } q_{14}^{\prime} may be found using statics. Thus, resolving forces horizontally we have

800 q_{12}^{\prime}=800 \times 75.9 \mathrm{~N}
whence
q_{12}^{\prime}=75.9 \mathrm{~N} / \mathrm{mm}

Resolving forces vertically
200 q_{32}^{\prime}=50 q_{12}^{\prime}-50 \times 75.9-300 q_{14}^{\prime}=0 (i)
and taking moments about O in Fig. 23.21(b) we obtain
2 \times 52000 q_{12}^{\prime}-2 \times 40000 q_{32}^{\prime}+2 \times 52000 \times 75.9-2 \times 60000 q_{14}^{\prime}=0 (ii)
Solving Eqs (i) and (ii) gives

q_{32}^{\prime}=117.6 \mathrm{~N} / \mathrm{mm} \quad q_{14}^{\prime}=53.1 \mathrm{~N} / \mathrm{mm}

The final shear flows in bay ② are found by superimposing q_{12}^{\prime}, q_{32}^{\prime} \text { and } q_{14}^{\prime} on the shear flows in Fig. 23.22, giving the distribution shown in Fig. 23.24. Alternatively, these shear flows could have been found directly by considering the equilibrium of the cut-out bay under the action of the applied shear loads.

The correction shear flows in bay ② (Fig. 23.23) will also modify the shear flow distributions in bays ① and ③. The correction shear flows to be applied to those shown in Fig. 23.22 for bay ③ (those in bay ① will be identical) may be found by determining the flange loads corresponding to the correction shear flows in bay ②.

It can be seen from the magnitudes and directions of these correction shear flows (Fig. 23.23) that at any section in bay ② the loads in the upper and lower flanges of the front spar are equal in magnitude but opposite in direction; similarly for the rear spar.
Thus, the correction shear flows in bay ② produce an identical system of flange loads to that shown in Fig. 23.17 for the cut-out bays in the wing structure of Example 23.6.

It follows that these correction shear flows produce differential bending of the front and rear spars in bay ② and that the spar bending moments and hence the flange loads are zero at the mid-bay points. Therefore, at station 3000 the flange loads are

\begin{aligned}&P_1=(75.9+53.1) \times 500=64500 \mathrm{~N} \text { (compression) } \\&P_4=64500 \mathrm{~N} \text { (tension) } \\&P_2=(75.9+117.6) \times 500=96750 \mathrm{~N} \text { (tension) } \\&P_3=96750 \mathrm{~N} \text { (tension) }\end{aligned}

These flange loads produce correction shear flows q_{21}^{\prime \prime}, q_{43}^{\prime \prime}, q_{23}^{\prime \prime} \text { and } q_{41}^{\prime \prime} in the skin panels and spar webs of bay ③ as shown in Fig. 23.25. Thus for equilibrium of flange 1

1000 q_{41}^{\prime \prime}+1000 q_{21}^{\prime \prime}=64500 \mathrm{~N}  (iii)

and for equilibrium of flange 2

1000 q_{21}^{\prime \prime}+1000 q_{23}^{\prime \prime}=96750 \mathrm{~N}  (iv)

For equilibrium in the chordwise direction at any section in bay ③

800 q_{21}^{\prime \prime}=800 q_{43}^{\prime \prime}

or
q_{21}^{\prime \prime}=q_{43}^{\prime \prime}(v)
Finally, for vertical equilibrium at any section in bay ③

300 q_{41}^{\prime \prime}+50 q_{43}^{\prime \prime}+50 q_{21}^{\prime \prime}-200 q_{23}^{\prime \prime}=0  (vi)

Simultaneous solution of Eqs (iii)–(vi) gives

q_{21}^{\prime \prime}=q_{43}^{\prime \prime}=38.0 \mathrm{~N} / \mathrm{mm} \quad q_{23}^{\prime \prime}=58.8 \mathrm{~N} / \mathrm{mm} \quad q_{41}^{\prime \prime}=26.6 \mathrm{~N} / \mathrm{mm}

Superimposing these correction shear flows on those shown in Fig. 23.22 gives the final shear flow distribution in bay ③ as shown in Fig. 23.26. The rib loads at stations 2000 and 3000 are found as before by adding algebraically the shear flows in the skin panels and spar webs on each side of the rib. Thus, at station 3000 we obtain the shear flows acting around the periphery of the rib as shown in Fig. 23.27. The shear flows applied to the rib at the inboard end of the cut-out bay will be equal in magnitude but opposite in direction.

Note that in this example only the shear loads on the wing box between stations 1000 and 4000 are given. We cannot therefore determine the final values of the loads in the spar flanges since we do not know the values of the bending moments at these positions caused by loads acting on other parts of the wing.

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