A thin-walled closed single-cell torque box of an aircraft wing has an enclosed cross-sectional area $A_m = 0.50\text{ m}^2$ and a uniform wall skin thickness $t = 2.0\text{ mm}$. If the box is subjected to a pure twisting torque $T = 30.0\text{ kN}\cdot\text{m}$, what is the resulting shear stress $\tau$ in the skin (in MPa)?

Answer Key:

15.0

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Governing Bredt-Batho Theory:** For thin-walled closed single-cell tubes subjected to torque, shear flow $q$ is constant along the perimeter. 2. **Stage 2: Shear Flow Formula:** * $T = 2 A_m q \implies q = \frac{T}{2 A_m}$. * Where $A_m$ is the area enclosed by the median line of the wall cross-section. 3. **Stage 3: Calculation of Shear Flow:** * $T = 30.0\text{ kN}\cdot\text{m} = 30,000\text{ N}\cdot\text{m}$. * $A_m = 0.50\text{ m}^2$. * $q = \frac{30,000}{2 \times 0.50} = 30,000\text{ N/m}$. 4. **Stage 4: Shear Stress in the Skin:** * $\tau = \frac{q}{t} = \frac{30,000\text{ N/m}}{0.002\text{ m}} = 15,000,000\text{ Pa} = 15.0\text{ MPa}$. 5. **Stage 5: Official Reference:** T.H.G. Megson, *Aircraft Structures for Engineering Students*, Chapter 18.

← Prev Question Next Question →