Part of a wing section is in the form of the two-cell box shown in Fig. 20.4(a) in which the vertical spars are connected to the wing skin through angle sections all having a crosssectional area of 300 mm². Idealize the section into an arrangement of direct stress carrying booms and shear stress

Question

Part of a wing section is in the form of the two-cell box shown in Fig. 20.4(a) in which the vertical spars are connected to the wing skin through angle sections all having a crosssectional area of 300 mm². Idealize the section into an arrangement of direct stress carrying booms and shear stress only carrying panels suitable for resisting bending moments in a vertical plane. Position the booms at the spar/skin junctions.

Accepted Answer

The idealized section is shown in Fig. 20.4(b) in which, from symmetry, [latex]B_1 = B_6, B_2 = B_5, B_3 = B_4[/latex]. Since the section is required to resist bending moments in a vertical plane the direct stress at any point in the actual wing section is directly proportional to its distance from the horizontal axis of symmetry. Further, the distribution of direct stress in all the panels will be linear so that either of Eqs (20.1) or (20.2) may be used. We note that, in addition to contributions from adjacent panels, the boom areas include the existing spar flanges. Hence

[latex]B_1=\frac{t_{\mathrm{D}} b}{6}\left(2+\frac{\sigma_2}{\sigma_1} ight)[/latex] (20.1)

[latex]B_2=\frac{t_{\mathrm{D}} b}{6}\left(2+\frac{\sigma_1}{\sigma_2} ight)[/latex] (20.2)

[latex]B_1=300+\frac{3.0 imes 400}{6}\left(2+\frac{\sigma_6}{\sigma_1} ight)+\frac{2.0 imes 600}{6}\left(2+\frac{\sigma_2}{\sigma_1} ight)[/latex]

or

[latex]B_1=300+\frac{3.0 imes 400}{6}(2-1)+\frac{2.0 imes 600}{6}\left(2+\frac{150}{200} ight)[/latex]

which gives
[latex]B_1\left(=B_6 ight)=1050 \mathrm{~mm}^2[/latex]
Also

[latex]B_2=2 imes 300+\frac{2.0 imes 600}{6}\left(2+\frac{\sigma_1}{\sigma_2} ight)+\frac{2.5 imes 300}{6}\left(2+\frac{\sigma_5}{\sigma_2} ight)+\frac{1.5 imes 600}{6}\left(2+\frac{\sigma_3}{\sigma_2} ight)[/latex]

i.e.

[latex]B_2=2 imes 300+\frac{2.0 imes 600}{6}\left(2+\frac{200}{150} ight)+\frac{2.5 imes 300}{6}(2-1)+\frac{1.5 imes 600}{6}\left(2+\frac{100}{150} ight)[/latex]

from which
[latex]B_2\left(=B_5 ight)=1791.7 \mathrm{~mm}^2[/latex]
Finally

[latex]B_3=300+\frac{1.5 imes 600}{6}\left(2+\frac{\sigma_2}{\sigma_3} ight)+\frac{2.0 imes 200}{6}\left(2+\frac{\sigma_4}{\sigma_3} ight)[/latex]

i.e.

[latex]B_3=300+\frac{1.5 imes 600}{6}\left(2+\frac{150}{100} ight)+\frac{2.0 imes 200}{6}(2-1)[/latex]

so that
[latex]B_3\left(=B_4 ight)=891.7 \mathrm{~mm}^2[/latex]

Question 20.1: Part of a wing section is in the form of the two-cell box sh...

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