Question

A two-cell wing box has wall thickness 1.5 mm and shear modulus \(G = 27\ \text{GPa}\). A torque of 12 kNm is applied. Determine the shear stress in wall AD (round off to one decimal place). 

Hint:

For multi-cell torsion, use compatibility of twist: \(\frac{q_1}{A_1} = \frac{q_2}{A_2}\). Shear stress follows \(\tau = q/t\).

Answer 1

Correct Answer: 2.1

Convert dimensions:
Cell heights = \(300\ \text{mm} = 0.3\ \text{m}\)
Widths = \(200\ \text{mm} = 0.2\ \text{m}\), \(300\ \text{mm} = 0.3\ \text{m}\)
Thickness: \(t = 1.5\ \text{mm} = 0.0015\ \text{m}\) Areas of the two cells:
\[ A_1 = 0.3 \times 0.2 = 0.06\ \text{m}^2, A_2 = 0.3 \times 0.3 = 0.09\ \text{m}^2 \] Total applied torque:
\[ T = 12\ \text{kNm} = 12000\ \text{Nm} \] For multi-cell torsion:
\[ T = 2A_1 q_1 + 2A_2 q_2 \] Compatibility (same twist per unit length):
\[ \frac{q_1}{A_1} = \frac{q_2}{A_2} $\Rightarrow$ q_2 = q_1\left(\frac{A_2}{A_1}\right) = 1.5 q_1 \] Substitute into torque equation:
\[ T = 2A_1 q_1 + 2A_2 (1.5q_1) \] \[ T = 2(0.06)q_1 + 2(0.09)(1.5q_1) \] \[ T = 0.12 q_1 + 0.27 q_1 = 0.39 q_1 \] Thus:
\[ q_1 = \frac{12000}{0.39} = 30769\ \text{N/m} \] Shear flow in wall AD is \(q_1\). Shear stress:
\[ \tau = \frac{q_1}{t} = \frac{30769}{0.0015} = 2.05\times10^7\ \text{Pa} \] \[ \tau = 2.05\ \text{MPa} \approx 2.1\ \text{MPa} \]