A thin-walled cylinder of radius r and thickness t is open at both ends, and fits snugly between two rigid walls under ambient conditions, as shown in the figure.The material of the cylinder has Young’s modulus E. Poisson’s ratio v, and coefficient of thermal expansion α. What is the minimum rise in temperature ΔT of the cylinder (assume uniform cylinder temperature with no buckling of the cylinder) required to prevent gas leakage if the cylinder has to store the gas at an internal pressure of p above the atmosphere?

Concept:When a thin-walled CYLINDER is subjected to internal fluid pressure then there are TWO stresses ACTING upon the cylinderHoop stress in the circumferential DIRECTION \({\sigma _h} = \frac{{pr}}{t}\)and longitudinal stress \({\sigma _l} = \frac{{pr}}{{2t}}\)Where p is the internal pressure of the fluid in the cylinder above atmospheric pressure, r is the RADIUS of the cylinder and t is the thickness of the cylinder.If the cylinder is open from the sides then longitudinal stress is zero.Strain due to these stress\({\epsilon_h} = \frac{1}{E}\left( {{\sigma _h} - \nu {\sigma _l}} \right)\)\({\epsilon_l} = \frac{1}{E}\left( {{\sigma _l} - \nu {\sigma _h}} \right)\)ν is Poisson’s ratio.Thermal strain due to temperature rise is given byϵth = αΔTα is the coefficient of thermal expansion, ΔT is the change in temperatureCalculation:It is given that both the ends of the cylinder are open therefore the longitudinal stress acting will be equal to zero. The longitudinal stress, σlong = 0The strain in the longitudinal direction will be due to the hoop stress only.Now the temperature of the cylinder is increased by ΔT, then in order to avoid the leakage, the sum of strains due to increased temperature and the strain due to hoop stress (in the longitudinal direction) should be equal to zero.Therefore,∴ ϵth + ϵl = 0 \({\epsilon_{th}} = \alpha \left( {{\rm{\Delta }}T} \right)\) \({\epsilon_l} = \frac{1}{E}\left( {0 - \nu \frac{{pr}}{{t}}\;} \right)\)\(\therefore \alpha \left( {{\rm{\Delta }}T} \right) + \left( {\frac{{ - vpr}}{{Et\;}}\;} \right) = 0\;\;\;\)\({\rm{\Delta }}T = \frac{{vpr}}{{atE}}\)