A hollow circular tube of inner radius 12.6 inches and a thickness
of 0.5 inches is subjected to a load of 48 kips. Determine the normal
stress.
strength of materials: normal stress of hollow circular tube
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Because it is not stated in the problem, let us assume that the 48 kips is an axial load.
Solution
Inner 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).