A thin circular cylindrical fuselage section of radius $R = 1.5\text{ m}$ and wall thickness $t = 1.5\text{ mm}$ is pressurized to an internal cabin differential pressure $p = 60\text{ kPa}$. The longitudinal tensile stress $\sigma_L$ (in $\text{MPa}$) in the fuselage skin is:

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### 1. Strategy & Theoretical Principles From membrane theory for thin-walled pressurized cylindrical shells: - Hoop (circumferential) stress: $\sigma_h = \frac{p R}{t}$ - Longitudinal (axial) stress: $\sigma_L = \frac{p R}{2t} = \frac{1}{2} \sigma_h$ ### 2. Analytical Formulation & Calculation Given: - Pressure $p = 60\text{ kPa} = 60 \times 10^3\text{ Pa}$ - Radius $R = 1.5\text{ m}$ - Thickness $t = 1.5\text{ mm} = 1.5 \times 10^{-3}\text{ m}$ $$\sigma_L = \frac{p R}{2 t} = \frac{(60 \times 10^3) \times 1.5}{2 \times (1.5 \times 10^{-3})} = \frac{90 \times 10^3}{3.0 \times 10^{-3}} = 30 \times 10^6\text{ Pa} = 30.0\text{ MPa}$$ ### 3. Distractor Forensics - Candidate calculating hoop stress gets $\frac{pR}{t} = 60.0\text{ MPa}$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Longitudinal stress is exactly half of the hoop stress: $\sigma_L = \frac{1}{2} \sigma_h$. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 1.

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