The central cell of a wing has the idealized section shown in Fig. P.23.1. If the lift and drag loads on the wing produce bending moments of −120 000 N m and −30 000 N m, respectively at the section shown, calculate the direct stresses in the booms. Neglect axial constraint effects and assume that

Question

The central cell of a wing has the idealized section shown in Fig. P.23.1. If the lift and drag loads on the wing produce bending moments of −120 000 N m and −30 000 N m, respectively at the section shown, calculate the direct stresses in the booms. Neglect axial constraint effects and assume that the lift and drag vectors are in vertical and horizontal planes.

Boom areas:

$\begin{aligned}&B_1=B_4=B_5=B_8=1000 \mathrm{~mm}^2 \\&B_2=B_3=B_6=B_7=600 \mathrm{~mm}^2\end{aligned}$

Boom areas:

[latex]\begin{aligned}&B_1=B_4=B_5=B_8=1000 \mathrm{~mm}^2 \&B_2=B_3=B_6=B_7=600 \mathrm{~mm}^2\end{aligned}[/latex]

Accepted Answer

The beam section is unsymmetrical and [latex]M_x=-120000 \mathrm{Nm}, \quad M_y=-30000 \mathrm{Nm}[/latex] . Therefore, the direct stresses in the booms are given by Eq. (16.18), i.e.

[latex]\sigma_z=\left(\frac{M_y I_{x x}-M_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight) x+\left(\frac{M_x I_{y y}-M_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight) y[/latex] (i)

n Fig. S.23.1 [latex]\bar{x}[/latex] = 600 mm by inspection. Also, taking moments of area about the line of the bottom booms
(4 × 1000 + 4 × 600)[latex]\bar{y}[/latex] = 1000 × 240 + 1000 × 180 + 600 × 220 + 600 × 200
from which
[latex]\bar{y}[/latex] = 105 mm
Then

[latex]\begin{aligned}I_{x x}=& 2 imes 1000 imes 105^2+2 imes 600 imes 105^2+1000 imes 135^2+1000 imes 75^2+600 imes 115^2 \&+600 imes 95^2=72.5 imes 10^6 \mathrm{~mm}^4 \I_{y y}=& 4 imes 1000 imes 600^2+4 imes 600 imes 200^2=1536.0 imes 10^6 \mathrm{~mm}^4 \I_{x y}=& 1000[(-600)(135)+(600)(75)]+600[(-200)(115)+(200)(95)] \=&-38.4 imes 10^6 \mathrm{~mm}^4\end{aligned}[/latex]

Table S.23.1

[latex]\begin{array}{llllllllr}\hline ext { Boom } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\hline x(\mathrm{~mm}) & -600 & -200 & 200 & 600 & 600 & 200 & -200 & -600 \y(\mathrm{~mm}) & 135 & 115 & 95 & 75 & -105 & -105 & -105 & -105 \\sigma_z\left(\mathrm{~N} / \mathrm{mm}^2 ight) & -190.7& -181.7 & -172.8 & -163.8 & 140.0 & 164.8 & 189.6 & 214.4 \\hline\end{array}[/latex]

Note that the sum of the contributions of booms 5, 6, 7 and 8 to [latex]I_{xy}[/latex] is zero. Substituting for [latex]M_x, M_y, I_{x x}[/latex], etc. in Eq. (i) gives
[latex]\sigma_z=-0.062 x-1.688 y[/latex] (ii)
The solution is completed in Table S.23.1.

Question p.23.1: The central cell of a wing has the idealized section shown i...

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