Question

A wooden barrel is made by using semi-circular hoops that are spaced every s 12 in. as shown in the fieure. The barrel is subjected internal pressure of p 1 psi. Assume the barrel has a cylindrical shape with an inside diameter of d 25 in, thickness of t 0.5 n, and height of h 42 in. The hoops are holding the staves (wood pieces) together to resist againat the internal pressure. Hoops are from steel [E-29000 ksi] and have a thickness of a-0.0625 in. and width of b -1 in. Determine the tensile stress in the steel hoops psi

Answer 1

Equilibrium for the steel hoop: from the FBD sigma F_x = 0 2P - internal pressure * diameter of barrel * S = 0 2P - 1 * 25 * S = 0 Where S is spacing between every two steel hoop. S = 12 inch therefore, P = 25 * 12/2 = 150kip Hoop stress for the steel sigma_1 = sigma_allow = P/A cross sectional area of hoop A = 2 * pi * R * t * b A = 2 pi * 12.5 * 0.0625 * 1 A = 4.90 inch^2 sigma_allow = 150/4.90 = 30.61 ksi