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## Q. 24.2

Calculate the shear flows in the web panels and the axial loads in the flanges of the wing rib shown in Fig. 24.9. 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.

## Verified Solution

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 23.1 for a three-flange wing section we have, resolving forces horizontally
$600 q_{12}-600 q_{23}=12000 \mathrm{~N}$ (i)
Resolving vertically
$300 q_{31}-300 q_{23}=15000 \mathrm{~N}$ (ii)
$2(50000+95000) q_{23}+2 \times 95000 q_{12}=-15000 \times 300 \mathrm{~N} \mathrm{~mm}$ (iii)
Solution of Eqs (i)–(iii) gives
$q_{12}=13.0 \mathrm{~N} / \mathrm{mm} \quad q_{23}=-7.0 \mathrm{~N} / \mathrm{mm} \quad q_{31}=43.0 \mathrm{~N} / \mathrm{mm}$
Consider now the nose portion of the rib shown in Fig. 24.10 and suppose that the shear flow in the web immediately to the left of the stiffener 24 is $q_1$. The total vertical shear force $S_{y, 1}$ at this section is given by
$S_{y, 1}$ = 7.0 × 300 = 2100 N
The horizontal components of the rib flange loads resist the bending moment at this section. Thus

$P_{x, 4}=P_{x, 2}=\frac{2 \times 50000 \times 7.0}{300}=2333.3 \mathrm{~N}$

The corresponding vertical components are then
$P_{y, 2}=P_{y, 4}=2333.3 \tan 15^{\circ}=625.2 \mathrm{~N}$
Thus the shear force carried by the web is 2100 − 2 × 625.2 = 849.6 N. Hence

$q_1=\frac{849.6}{300}=2.8 \mathrm{~N} / \mathrm{mm}$

The axial loads in the rib flanges at this section are given by
$P_2=P_4=\left(2333.3^2+625.2^2\right)^{1 / 2}=2415.6 \mathrm{~N}$
The rib flange loads and web panel shear flows, at a vertical section immediately to the left of the intermediate web stiffener 56, are found by considering the free body diagram shown in Fig. 24.11. At this section the rib flanges have zero slope so that the flange loads $P_5 \text { and } P_6$ are obtained directly from the value of bending moment at this section. Thus

$P_5 = P_6$ = 2[(50 000 + 46 000) × 7.0 − 49 000 × 13.0]/320 = 218.8 N

The shear force at this section is resisted solely by the web. Hence
320$q_2$ = 7.0 × 300 + 7.0 × 10 − 13.0 × 10 = 2040 N

so that
$q_2$ = 6.4 N/mm
The shear flow in the rib immediately to the right of stiffener 56 is found most simply by considering the vertical equilibrium of stiffener 56 as shown in Fig. 24.12. Thus
320$q_3$ = 6.4 × 320 + 15 000
which gives
$q_3$ = 53.3 N/mm

Finally, we shall consider the rib flange loads and the web shear flow at a section
immediately forward of stiffener 31. From Fig. 24.13, in which we take moments about the point 3

$M_3=2[(50000+95000) \times 7.0-95000 \times 13.0]+15000 \times 300=4.06 \times 10^6 \mathrm{~N} \mathrm{~mm}$

The horizontal components of the flange loads at this section are then

$P_{x, 1}=P_{x, 3}=\frac{4.06 \times 10^6}{300}=13533.3 \mathrm{~N}$

and the vertical components are
$P_{y, 1}=P_{y, 3}=3626.2 \mathrm{~N}$
Hence

$P_1=P_3=\sqrt{13533.3^2+3626.2^2}=14010.7 \mathrm{~N}$

The total shear force at this section is 15 000 + 300 × 7.0 = 17 100 N. Therefore, the shear force resisted by the web is 17 100 − 2 × 3626.2 = 9847.6 N so that the shear flow $q_3$ in the web at this section is

$q_3=\frac{9847.6}{300}=32.8 \mathrm{~N} / \mathrm{mm}$