A portion of a tapered, three-cell wing has singly symmetrical idealized cross-sections 1000 mm apart as shown in Fig. P.23.8. A bending moment Mx = 1800 N m and a shear load Sy = 12 000 N in the plane of the web 52 are applied at the larger cross-section. Calculate the forces in the booms and the

Question

A portion of a tapered, three-cell wing has singly symmetrical idealized cross-sections 1000 mm apart as shown in Fig. P.23.8. A bending moment [latex]M_x[/latex] = 1800 N m and a shear load [latex]S_y[/latex] = 12 000 N in the plane of the web 52 are applied at the larger cross-section. Calculate the forces in the booms and the shear flow distribution at this cross-section. The modulus G is constant throughout. Section dimensions at the larger cross-section are given below.

[latex]\begin{array}{lrlllll} 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 & 600 & 1.0 & 1,6 & 600 & ext { I } & 100000 \23,45 & 800 & 1.0 & 2,5 & 800 & ext { II } & 260000 \34^{\circ} & 1200 & 0.6 & 3,4 & 800 & ext { III } & 180000 \34^{\mathrm{i}} & 320 & 2.0 & & & & \25 & 320 & 2.0 & & & & \16 & 210 & 1.5 & & & & \\hline\end{array}[/latex]

Accepted Answer

The direct stresses in the booms are given by the first of Eqs (16.21) in which, referring to Fig. P.23.8, at the larger cross-section[latex]\sigma_z=\frac{M_x}{I_{x x}} y \quad \text { or } \quad \sigma_z=\frac{M_y}{I_{y y}} x[/latex] (16.21)

[latex]I_{x x}=2 \times 600 \times 105^2+4 \times 800 \times 160^2=95.2 \times 10^6 \mathrm{~mm}^4[/latex]

Then, from Eq. (21.8)

[latex]P_{z, r}=\sigma_{z, r} B_r[/latex] (21.8)

[latex]P_{z, r}=\sigma_{z, r} B_r=\frac{M_x B_r}{I_{x x}} y_r[/latex]

or

[latex]P_{z, r}=\frac{1800 \times 10^3}{95.2 \times 10^6} B_r y_r=1.89 \times 10^{-2} B_r y_r[/latex] (i)

The components of boom load in the y and x directions (see Fig. 21.4(a) for the axis system) are found using Eqs. (21.9) and (21.10). Then, choosing the intersection of the web 52 and the horizontal axis of symmetry (the x axis) as the moment centre

[latex]P_{y, r}=P_{z, r} \frac{\delta y_r}{\delta z}[/latex] (21.9)

[latex]P_{x, r}=P_{z, r} \frac{\delta x_r}{\delta z}[/latex] (21.10)

Table S.23.8

[latex]\begin{array}{lrrrrrrrrrr}\hline \text { Boom } & P_{z, r}(\mathrm{~N}) & \delta y_r /\delta z & \delta x_r / \delta z & P_{y, r}(\mathrm{~N}) & P_{x, r}(\mathrm{~N}) & P_r(\mathrm{~N}) & \begin{array}{l}\eta_r \(\mathrm{~mm})\end{array} & \begin{array}{l}\xi_r \(\mathrm{~mm})\end{array} & \begin{array}{l}P_{y, r} \xi_r \(\mathrm{Nmm})\end{array} & \begin{array}{l}P_{x, r} \eta_r \(\mathrm{Nmm})\end{array} \\hline 1 & 1190.7 & 0.045 & -0.12 & 53.6 & -142.9 & 1200.4 & 590 & 105 & 31624 & -15004.5 \2 & 2419.2 & 0.060 & 0 & 145.2 & 0 & 2423.6 & 0 & 160 & 0 & 0 \3 & 2419.2 & 0.060 & 0.18 & 145.2 & 435.5 & 2462.4 & 790 & 160 & -114708 & 69680 \4 & -2419.2 & -0.060 & 0.18 & 145.2 & -435.5 & -2462.4 & 790 & 160 & -114708 & 69680 \5 & -2419.2 & -0.060 & 0 & 145.2 & 0 & -2423.6 & 0 & 160 & 0 & 0 \6 & -1190.7 & -0.045 & -0.12 & 53.6 & 142.9 & -1200.4 & 590 & 105 & 31624 & -15004.5 \\hline\end{array}[/latex]

and defining the boom positions in relation to the moment centre as in Fig. 21.5 the moments corresponding to the boom loads are calculated in Table S.23.8. In Table S.23.8 anticlockwise moments about the moment centre are positive, clockwise negative. Also

[latex]\begin{aligned}\sum_{r=1}^n P_{x, r} &=0 \\sum_{r=1}^n P_{y, r} &=688.0 \mathrm{~N} \\sum_{r=1}^n P_{y, r} \xi_r&=-166168 \mathrm{~N} \mathrm{~mm} \\sum_{r=1}^n P_{x, r} \eta_r &=109351 \mathrm{~N} \mathrm{~mm}\end{aligned}[/latex]

The shear load resisted by the shear stresses in the webs and panels is then
[latex]S_y[/latex] = 12 000 − 688 = 11 312 N
‘Cut’ the walls 12, 23 and 34° in the larger cross-section. Then, from Eq. (20.6) and noting that [latex]I_{xy}[/latex] = 0

[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}\right)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r\right) \&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r\right)\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.e.

[latex]q_{\mathrm{b}}=-\frac{11312}{95.2 \times 10^6} \sum_{r=1}^n B_r y_r=-1.188 \times 10^{-4} \sum_{r=1}^n B_r y_r[/latex]

Thus

[latex]\begin{aligned}&q_{\mathrm{b}, 12}=q_{\mathrm{b}, 23}=q_{\mathrm{b}, 34^{\circ}}=q_{\mathrm{b}, 45}=q_{\mathrm{b}, 56}=0 \&q_{\mathrm{b}, 61}=-1.188 \times 10^{-4} \times 600 \times(-105)=7.48 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 52}=-1.188 \times 10^{-4} \times 800 \times(-160)=15.21 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 43^{\mathrm{i}}}=-1.188 \times 10^{-4} \times 800 \times(-160)=15.21 \mathrm{~N} / \mathrm{mm}\end{aligned}[/latex]

From Eq. (23.10) for Cell I

[latex]\frac{\mathrm{d} \theta}{\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}\right)[/latex] (23.10)

[latex]\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{I}} G}\left[q_{s, 0, \mathrm{I}}\left(\delta_{34^{\circ}}+\delta_{34^{\mathrm{i}}}\right)-q_{s, 0, \mathrm{II}} \delta_{34^{\mathrm{i}}}+q_{\mathrm{b}, 43^i} \delta_{43^{\mathrm{i}}}\right][/latex] (ii)

For Cell II

[latex]\begin{aligned}\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{II}} G} & {\left[-q_{s, 0, \mathrm{I}} \delta_{34^{\mathrm{i}}}+q_{s, 0, \mathrm{II}}\left(\delta_{23}+\delta_{34^{\mathrm{i}}}+\delta_{45}+\delta_{52}\right)-q_{s, 0, \mathrm{III}} \delta_{52}\right.} \&\left.+q_{\mathrm{b}, 52} \delta_{52}-q_{\mathrm{b}, 43^{\mathrm{i}}} \delta_{43^{\mathrm{i}}}\right]\end{aligned}[/latex] (iii)

For Cell III

[latex]\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{III}} G}\left[-q_{s, 0, \mathrm{II}} \delta_{52}+q_{s, 0, \mathrm{III}}\left(\delta_{12}+\delta_{25}+\delta_{56}+\delta_{61}\right)+q_{\mathrm{b}, 61} \delta_{61}-q_{\mathrm{b}, 52}\delta_{52}\right] \text { (iv) }[/latex]

in which

[latex]\begin{aligned}\delta_{12} &=\delta_{56}=600 / 1.0=600\quad \delta_{23}=\delta_{45}=800 / 1.0=800 \\delta_{34^{\circ}} &=1200 / 0.6=2000 \quad\delta_{34^{\mathrm{i}}}=320 / 2.0=160 \quad\delta_{52}=320 / 2.0=160 \\delta_{61} &=210 / 1.5=140\end{aligned}[/latex]

Substituting these values in Eqs (ii)–(iv)

[latex]\begin{aligned}\frac{\mathrm{d} \theta}{\mathrm{d} z} &=\frac{1}{2 \times 100000 G}\left(2160 q_{s, 0, \mathrm{I}}-160 q_{s, 0, \mathrm{II}}+2433.6\right) (v) \\frac{\mathrm{d} \theta}{\mathrm{d} z} &=\frac{1}{2 \times 260000 G}\left(-160 q_{s, 0, \mathrm{I}}+1920 q_{s, 0, \mathrm{II}}-160 q_{s, 0, \mathrm{III}}\right) (vi)\\frac{\mathrm{d} \theta}{\mathrm{d} z} &=\frac{1}{2 \times 180000 G}\left(-160 q_{s, 0,\mathrm{II}}+1500 q_{s, 0, \mathrm{III}}-1384.8\right) (vii)\end{aligned}[/latex]

Also, taking moments about the mid-point of web 52, i.e. the moment centre (see Eq. (23.13))

[latex]S_x \eta_0-S_y \xi_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}-\sum_{r=1}^m P_{x, r} \eta_r+\sum_{r=1}^m P_{y, r} \xi_r[/latex] (23.13)

[latex]\begin{aligned}0=& q_{\mathrm{b}, 61} \times 210 \times 590-q_{\mathrm{b}, 43^{\mathrm{i}}} \times 320 \times 790+2 A_{\mathrm{I}} q_{s, 0, \mathrm{I}}+2 A_{\mathrm{II}} q_{s, 0, \mathrm{II}} \&+2 A_{\mathrm{III}} q_{s, 0, \mathrm{III}}+\sum_{r=1}^n P_{x, r} \eta_r+\sum_{r=1}^n P_{y, r} \xi_r\end{aligned}[/latex] (viii)

Substituting the appropriate values in Eq. (viii) and simplifying gives
[latex]q_{s, 0, \mathrm{I}}+2.6 q_{s, 0, \mathrm{II}}+1.8 q_{s, 0, \mathrm{III}}-14.88=0[/latex] (ix)

Equating Eqs (v) and (vi)
[latex]q_{s, 0, \mathrm{I}}-0.404 q_{s, 0, \mathrm{II}}+0.028 q_{s, 0, \mathrm{III}}+1.095=0[/latex] (x)

Equating Eqs (v) and (vii)
[latex]q_{s, 0, \mathrm{I}}-0.033 q_{s, 0, \mathrm{II}}-0.386 q_{s, 0, \mathrm{III}}+1.483=0[/latex] (xi)
Now subtracting Eq. (x) from (ix)
[latex]q_{s, 0, \mathrm{II}}+0.590 q_{s, 0, \mathrm{III}}-5.318=0[/latex] (xii)
and subtracting Eq. (xi) from (ix)
[latex]q_{s, 0, \mathrm{II}}+0.830 q_{s, 0, \mathrm{III}}-6.215=0[/latex] (xiii)
Finally, subtracting Eq. (xiii) from (xii) gives

[latex]q_{s, 0, \mathrm{III}}=3.74 \mathrm{~N} / \mathrm{mm}[/latex]
Then, from Eq. (xiii)
[latex]q_{s, 0, \mathrm{II}}=3.11 \mathrm{~N} / \mathrm{mm}[/latex]
and from Eq. (ix)
[latex]q_{s, 0, \mathrm{I}}=0.06 \mathrm{~N} / \mathrm{mm}[/latex]
The complete shear flow distribution is then

[latex]\begin{aligned}q_{12} &=q_{56}=3.74 \mathrm{~N} / \mathrm{mm} \quad q_{32}=q_{45}=3.11 \mathrm{~N} / \mathrm{mm} \q_{34^{\circ}} &=0.06 \mathrm{~N} / \mathrm{mm} \quad q_{43^i}=12.16 \mathrm{~N} / \mathrm{mm} \q_{52} &=14.58 \mathrm{~N} / \mathrm{mm} \quad q_{61}=11.22 \mathrm{~N} / \mathrm{mm}\end{aligned}[/latex]

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