A single-cell thin-walled closed box beam has cross-sectional enclosed area $A_m = 0.40\text{ m}^2$ and uniform wall thickness $t = 2.0\text{ mm}$. If an applied torque $T = 8000\text{ N}\cdot\text{m}$ acts on the beam, what is the shear stress $\tau$ in the wall in MPa?

Answer Key:

5.0

### Solution: Bredt-Batho formula for single-cell tube: $q = \frac{T}{2 A_m} = \frac{8000}{2 \times 0.40} = \frac{8000}{0.80} = 10000\text{ N/m}$. Shear stress $\tau = \frac{q}{t} = \frac{10000\text{ N/m}}{0.002\text{ m}} = 5 \times 10^6\text{ Pa} = 5.0\text{ MPa}$. Correct: 5.0.

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