The idealized cross-section of a two-cell thin-walled wing box is shown in Fig. P.23.5. If the wing box supports a load of 44 500 N acting along the web 25, calculate the shear flow distribution. The shear modulus G is the same for all walls of the wing box.

Question

[latex]\begin{array}{llllc}\hline ext { Wall } & ext { Length }(\mathrm{mm}) & ext { Thickness }(\mathrm{mm}) & ext { Boom } & ext { Area }\left(\mathrm{mm}^2 ight) \\hline 16 & 254 & 1.625 & 1,6 & 1290 \25 & 406 & 2.032 & 2,5 & 1936 \34 & 202 & 1.220 & 3,4 & 645 \12,56 & 647 & 0.915 & & \23,45 & 775 & 0.559 & & \ ext { Cell areas: } A_{\mathrm{I}}=232000 \mathrm{~mm}^2, A_{\mathrm{II}}=258000 \mathrm{~mm}^2 & & \\hline\end{array}[/latex]

Accepted Answer

In Eq. (23.10) the [latex]q_b[/latex] shear flow distribution is given by Eq. (20.6) in which, since the x axis is an axis of symmetry (Fig. S.23.5), [latex]I_{xy}[/latex] = 0; also [latex]S_x[/latex] = 0. Thus

[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]\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]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 1290 imes 127^2+2 imes 1936 imes 203^2+2 imes 645 imes 101^2=214.3 imes 10^6 \mathrm{~mm}^4[/latex]

Then Eq. (i) becomes

[latex]q_{\mathrm{b}}=-\frac{44500}{214.3 imes 10^6} \sum_{r=1}^n B_r y_r=-2.08 imes 10^{-4} \sum_{r=1}^n B_r y_r[/latex]

‘Cut’ the walls 65 and 54. Then

[latex]\begin{aligned}&q_{\mathrm{b}, 65}=q_{\mathrm{b}, 54}=0 \&q_{\mathrm{b}, 61}=-2.08 imes 10^{-4} imes 1290 imes 127=-32.8 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 12}=q_{\mathrm{b}, 23}=0 \quad( ext { from symmetry }) \&q_{\mathrm{b}, 25}=-2.08 imes 10^{-4} imes 1936(-203)=81.7 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 34}=-2.08 imes 10^{-4} imes 645(-101)=13.6 \mathrm{~N} / \mathrm{mm}\end{aligned}[/latex]

From Eq. (23.10) for Cell I

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{J}} G}\left[q_{s, 0, \mathrm{I}}\left(\delta_{56}+\delta_{61}+\delta_{12}+\delta_{24} ight)-q_{s, 0, \mathrm{I}} \delta_{25}+q_{\mathrm{b}, 25} \delta_{25}+q_{\mathrm{b}, 61} \delta_{61} ight][/latex] (ii)

For Cell II

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{II}} G}\left[-q_{s, 0, \mathrm{I}} \delta_{25}+q_{s, 0,\mathrm{II}}\left(\delta_{45}+\delta_{52}+\delta_{23}+\delta_{34} ight)+q_{\mathrm{b}, 34} \delta_{34}+q_{\mathrm{b}, 52} \delta_{25} ight][/latex] (iii)

in which

[latex]\begin{aligned}&\delta_{56}=\delta_{12}=647 / 0.915=707.1 \quad \delta_{45}=\delta_{23}=775 / 0.559=1386.4 \&\delta_{61}=254 / 1.625=156.3 \quad \delta_{52}=406 / 2.032=199.8 \quad \delta_{34}=202 / 1.220=165.6\end{aligned}[/latex]

Substituting these values in Eqs (ii) and (iii)

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 232000 G}\left(1770.3 q_{s, 0, \mathrm{I}}-199.8 q_{s, 0, \mathrm{II}}+11197.0 ight)[/latex] (iv)

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 258000 G}\left(-199.8 q_{s, 0, \mathrm{I}}+3138.2 q_{s, 0, \mathrm{II}}-14071.5 ight)[/latex] (v)

Also, taking moments about the mid-point of the web 25 and from Eq. (23.11) (or Eq. (23.12))

[latex]S_x \eta_0-S_y \xi_0=\sum_{R=1}^N M_{q, R}=\sum_{R=1}^N \oint_R q_{\mathrm{b}} p_0 \mathrm{~d} s+\sum_{R=1}^N 2 A_R q_{s, 0, R}[/latex] (23.11)

[latex]0=\sum_{R=1}^N \oint_R q_{\mathrm{b}} p_0 \mathrm{~d} s+\sum_{R=1}^N 2 A_R q_{s, 0, R}[/latex] (23.12)

[latex]0=13.6 imes 202 imes 763-32.8 imes 254 imes 635+2 A_{\mathrm{I}} q_{s, 0, \mathrm{I}}+2 A_{\mathrm{II}} q_{s, 0, \mathrm{II}}[/latex] (vi)

Equating Eqs (iv) and (v)
[latex]q_{s, 0, \mathrm{I}}-1.55 q_{s, 0, \mathrm{II}}+12.23=0[/latex] (vii)
From Eq. (vi)
[latex]q_{s, 0, \mathrm{I}}+1.11 q_{s, 0, \mathrm{II}}-6.88=0[/latex] (viii)
Subtracting Eq. (viii) from (vii) gives
[latex]q_{s, 0, \mathrm{II}}=7.2 \mathrm{~N} / \mathrm{mm}[/latex]

Then, from Eq. (vii)
[latex]q_{s, 0, \mathrm{I}}=-1.1 \mathrm{~N} / \mathrm{mm}[/latex]

Thus

[latex]\begin{aligned}&q_{16}=32.8+1.1=33.9 \mathrm{~N} / \mathrm{mm} \quad q_{65}=q_{21}=1.1 \mathrm{~N} / \mathrm{mm} \&q_{45}=q_{23}=7.2 \mathrm{~N} / \mathrm{mm} \quad q_{34}=13.6+7.2=20.8 \mathrm{~N} / \mathrm{mm} \&q_{25}=81.7-1.1-7.2=73.4 \mathrm{~N} / \mathrm{mm}\end{aligned}[/latex]

Question p.23.5: The idealized cross-section of a two-cell thin-walled wing b...

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