A singly symmetric wing section consists of two closed cells and one open cell (see Fig. P.23.7). The webs 25, 34 and the walls 12, 56 are straight, while all other walls are curved. All walls of the section are assumed to be effective in carrying shear stresses only, direct stresses being carried

Question

A singly symmetric wing section consists of two closed cells and one open cell (see Fig. P.23.7). The webs 25, 34 and the walls 12, 56 are straight, while all other walls are curved. All walls of the section are assumed to be effective in carrying shear stresses only, direct stresses being carried by booms 1–6. Calculate the distance xS of the shear centre S aft of the web 34. The shear modulus G is the same for all walls.

[latex]\begin{array}{llllllc}\hline ext { Wall } & ext { Length }(\mathrm{mm}) & ext { Thickness }(\mathrm{mm}) & ext { Boom } & ext { Area }\left(\mathrm{mm}^2 ight) & ext { Cell } & ext { Area }\left(\mathrm{mm}^2 ight) \\hline 12,56 & 510 & 0.559 & 1,6 & 645 & ext { I } & 93000 \23,45 & 765 & 0.915 & 2,5 & 1290 & ext { II } & 258000 \34^{\circ} & 1015 & 0.559 & 3,4 & 1935 & & \34^{\mathrm{i}} & 304 & 2.030 & & & & \25 & 304 & 1.625 & & & & \\hline\end{array}[/latex]

Accepted Answer

Referring to Fig. P.23.7 the horizontal x axis is an axis of symmetry so that [latex]I_{xy}[/latex] = 0 and the shear centre lies on this axis. Further, applying an arbitrary shear load, [latex]S_y[/latex], through the shear centre then [latex]S_x[/latex] = 0 and Eq. (20.6) simplifies to

[latex]\begin{aligned}q_s=&-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)\end{aligned}[/latex] (20.6)

[latex]q_{\mathrm{b}}=-\frac{S_y}{I_{x x}} \sum_{r=1}^n B_r y_r[/latex] (i)

in which
[latex]I_{x x}=2 imes 645 imes 102^2+2 imes 1290 imes 152^2+2 imes 1935 imes 150^2=162.4 imes 10^6 \mathrm{~mm}^4[/latex]
Eq. (i) then becomes

[latex]q_{\mathrm{b}}=-6.16 imes 10^{-9} S_y \sum_{r=1}^n B_r y_r[/latex] (ii)

‘Cut’ the walls 34° and 23. Then, from Eq. (ii)

[latex]\begin{aligned}&q_{\mathrm{b}, 34^{\circ}}=q_{\mathrm{b}, 23}=0=q_{\mathrm{b}, 45} \quad( ext { from symmetry) } \&q_{\mathrm{b}, 43^{\mathrm{i}}}=-6.16 imes 10^{-9} S_y imes 1935 imes(-152)=1.81 imes 10^{-3} S_y \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 65}=-6.16 imes 10^{-9} S_y imes 645 imes(-102)=0.41 imes 10^{-3} S_y \mathrm{~N} / \mathrm{mm}=q_{\mathrm{b}, 21} \&\quad ext { (from symmetry) } \&q_{\mathrm{b}, 52}=0.41 imes 10^{-3} S_y-6.16 imes 10^{-9} S_y imes 1290 imes(-152)=1.62 imes 10^{-3} S_y \mathrm{~N} / \mathrm{mm}\end{aligned}[/latex]

Since the shear load, [latex]S_y[/latex], is applied through the shear centre of the section the rate of twist, dθ/dz, is zero. Thus, for Cell I, Eq. (23.10) reduces to

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 A_R G}\left(-q_{s, 0, R-1} \delta_{R-1, R}+q_{s, 0, R} \delta_R-q_{s, 0, R+1} \delta_{R+1, R}+\oint_R q_{\mathrm{b}} \frac{\mathrm{d} s}{t} ight)[/latex] (23.10)

[latex]0=q_{s, 0, \mathrm{I}}\left(\delta_{34^{\circ}}+\delta_{34^{\mathrm{i}}} ight)-q_{s, 0, \mathrm{II}} \delta_{34^{\mathrm{i}}}+q_{\mathrm{b}, 43^{\mathrm{i}}} \delta_{34^{\mathrm{i}}}[/latex] (iii)

and for Cell II

[latex]0=-q_{s, 0, \mathrm{I}} \delta_{43^{\mathrm{i}}}+q_{s, 0,\mathrm{II}}\left(\delta_{23}+\delta_{34^{\mathrm{i}}}+\delta_{45}+\delta_{52} ight)+q_{\mathrm{b}, 52} \delta_{52}-q_{\mathrm{b}, 43^{\mathrm{i}}} \delta_{43^{\mathrm{i}}}[/latex] (iv)

in which

[latex]\begin{aligned}\delta_{34^{\circ}} &=1015 / 0.559=1815.7 \quad \delta_{34^{\mathrm{i}}}=304 / 2.030=149.8 \\delta_{23} &=\delta_{45}=765 / 0.915=836.1 \\delta_{25} &=304 / 1.625=187.1\end{aligned}[/latex]

Thus Eq. (iii) becomes

[latex]1965.5 q_{s, 0, \mathrm{I}}-149.8 q_{s, 0, \mathrm{II}}+0.271 S_y=0[/latex]

or

[latex]q_{s, 0, \mathrm{I}}-0.076 q_{s, 0, \mathrm{II}}+0.138 imes 10^{-3} S_y=0[/latex] (v)

and Eq. (iv) becomes
[latex]-149.8 q_{s, 0, \mathrm{I}}+2009.1 q_{s, 0, \mathrm{II}}+319.64 imes 10^{-4} S_y=0[/latex]
or

[latex]q_{s, 0, \mathrm{I}}-13.411 q_{s, 0, \mathrm{II}}-0.213 imes 10^{-3} S_y=0[/latex] (vi)

Subtracting Eq. (vi) from (v)

[latex]13.335 q_{s, 0, \mathrm{II}}+0.351 imes 10^{-3} S_y=0[/latex]

whence
[latex]q_{s, 0, \mathrm{II}}=-0.026 imes 10^{-3} S_y[/latex]
Then from Eq. (vi)
[latex]q_{s .0 . \mathrm{I}}=-0.139 imes 10^{-3} S_y[/latex]
Now taking moments about the mid-point of the web 43

[latex]\begin{aligned}S_y x_s=&-2 q_{\mathrm{b}, 21}(508 imes 152+50 imes 762)+q_{\mathrm{b}, 52} imes 304 imes 762+2 imes 258000 q_{s, 0, \mathrm{II}} \&+2 imes 93000 q_{s, 0, \mathrm{I}}\end{aligned}[/latex]

from which
[latex]x_s[/latex] = 241.4 mm

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