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## Because it is not stated in

Because it is not stated in the problem, let us assume that the 48 kips is an axial load.

SolutionInner radius, $r = 12.6 ~ \text{inches}$

Outer radius, $R = 12.6 + 0.5 = 13.1 ~ \text{inches}$

Calculate the cross-sectional area of the tube:

$A = \pi (R^2 - r^2)$

Calculate the normal stress.

$\sigma = \dfrac{P}{A}$ where $P = 48^k$

Your answer will be in the unit ksi (kilopound per square inch).