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Chapter 24

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.

Screenshot 2022-10-11 130734

Step-by-Step

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)
Taking moments about flange 3
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
320q_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
320q_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}
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