A spherical steel pressure vessel 400 mm in diameter with a wall thickness of 20 mm, is coated with a brittle layer that cracks when strain exceeds100 x 10-7. What internal pressure will cause the layer to develop cracks? (E = 200 GPa, µ = 0.3)

Concept:The longitudinal and hoop stress in the thin spherical vessel is given by,\({\varepsilon _H} = {\varepsilon _L}=\varepsilon = \frac{{PD}}{{4tE}}\left( {1 - \mu } \right)\)CalculationGivend = 400 mm, t = 20 mm, E = 200 GPa, μ = 0.3Maximum strain, εmax = 100 × 10-7When strain in the SPHERE exceeds the maximum allowable strain, There will be a FAILURE.\( {\varepsilon } = \frac{{pd}}{{4tE}}\left( {1 - \mu } \right)\)\(100 \times {10^{ - 7}} = \frac{{P \times 400}}{{4 \times 20 \times 200 \times {{10}^3}}}\left( {1 - 0.3} \right)\)P = 0.57 MPaDescriptionCylindrical vesselSpherical VesselLongitudinal stress\({\sigma _l} = \frac{{pd}}{{4T}}\)\({\sigma _l} = \frac{{pd}}{{4t}}\)Hoop stress\({\sigma _h} = \frac{{pd}}{{2t}}\)\({\sigma _h} = \frac{{pd}}{{4t}}\)Longitudinal strain\({\varepsilon _l} = \frac{{pd}}{{4tE}}\left( {1 - 2\mu } \right)\)\({\varepsilon _l} = \frac{{pd}}{{4tE}}\left( {1 - \mu } \right)\)Hoop strain\({\varepsilon _h} = \frac{{pd}}{{4tE}}\left( {2 - \mu } \right)\)\({\varepsilon _h} = \frac{{pd}}{{4tE}}\left( {1 - \mu } \right)\)Volumetric strain\({\varepsilon _v} = \frac{{pd}}{{4tE}}\left( {5 - 4\mu } \right)\)\({\varepsilon _v} = \frac{{3pd}}{{4tE}}\left( {1 - \mu } \right)\)