Oxygen gas at a pressure of 20 MPa is stored in a thin cylinder of thickness 2.5 mm and a mean diameter of 50 mm. The longitudinal stress in the cylinder is

Concept:By thin-walled cylinder, we mean that the thickness ‘t' is very much smaller than the radius Ri and we may quantify this by stating that the ratio t / Ri of the thickness of radius should be less than 0.1.In a thin shell circumferential STRESS is \({\sigma _c} = \frac{{Pd}}{{2t}}\;\) andLongitudinal stress will be half of the circumferential stress i.E. \({\sigma _l} = \frac{{Pd}}{{4t}}\).where P is the pressure, t is the thickness of cylinder, d is the diameter of the cylinderCalculation:Given:p = 20 MPa, t = 2.5 mm and d = 50 mm.Longitudinal stress\({\sigma _l} = \frac{{Pd}}{{4t}}\)\({\sigma _l} = \frac{{20\times 50}}{{4\times 2.5}} =100~MPa\)Hoop stress\({\sigma _h} = \frac{{Pd}}{{2t}} = {\sigma _1}\)Hoop strain\({\epsilon_h} = \frac{{Pd}}{{4tE}}\left( {2 - \MU } \right)\)Longitudinal stress\({\sigma _L} = \frac{{Pd}}{{4t}} = {\sigma _2} = \frac{{{\sigma _1}}}{2}\)Longitudinal strain\({\epsilon_L} = \frac{{Pd}}{{4tE}}\left( {1 - 2\mu } \right) = \frac{{{\sigma _2}}}{E}\left( {1 - 2\mu } \right)\)