Discussion
A fuselage of \( 2 \) m radius and \( 1.6 \) mm skin has \( I = \pi r^3 t \). Its second moment of area is nearest to
Answer: B
Correct answer: 0.040 m^4
For a thin circular shell,
\[ I = \pi r^3 t = \pi \times 2^3 \times 0.0016 = \pi \times 8 \times 0.0016 \approx 0.040 \ \mathrm{m}^4 \]The thickness must be converted to metres before substituting - using 1.6 directly would give an answer a thousand times too large. The cube on the radius is why fuselage stiffness rises so steeply with diameter.
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