Chapter 23
Q. p.23.1
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}
Step-by-Step
Verified Solution
The beam section is unsymmetrical and M_x=-120000 \mathrm{Nm}, \quad M_y=-30000 \mathrm{Nm} . Therefore, the direct stresses in the booms are given by Eq. (16.18), i.e.
\sigma_z=\left(\frac{M_y I_{x x}-M_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right) x+\left(\frac{M_x I_{y y}-M_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right) y (i)
n Fig. S.23.1 \bar{x} = 600 mm by inspection. Also, taking moments of area about the line of the bottom booms
(4 × 1000 + 4 × 600)\bar{y} = 1000 × 240 + 1000 × 180 + 600 × 220 + 600 × 200
from which
\bar{y} = 105 mm
Then
Table S.23.1
Note that the sum of the contributions of booms 5, 6, 7 and 8 to I_{xy} is zero. Substituting for M_x, M_y, I_{x x}, etc. in Eq. (i) gives
\sigma_z=-0.062 x-1.688 y (ii)
The solution is completed in Table S.23.1.
