Discussion
A fuselage skin of \( 1.2 \) mm thickness and \( 1.8 \) m radius is to be limited to \( 90 \) MPa hoop stress. The maximum cabin differential is
Answer: D
Rearranging \( \sigma = pr/t \) gives \( p = \sigma t/r = 90 \times 10^6 \times 0.0012/1.8 = 60\,000 \) Pa. A differential of 60 kPa is representative of a transport cabin at cruise altitude, so this skin is about right for the duty.
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