**Introduction:**

The wall thickness of the cylindrical and hemispherical. portion is different. While the internal diameter of both the portions is assumed to be equal.

**Description:**

Let us now consider the vessel with hemispherical ends.

Let the cylindrical vassal is subjected to an internal pressure p.

**For the Cylindrical Portion**

**\**

**For The Hemispherical Ends:**

Because of the symmetry of the sphere the stresses set up owing to internal pressure will be two mutually

perpendicular hoops or circumferential stresses of equal values. Again the radial stresses are neglected in

comparison to the hoop stresses as with this cylinder having thickness to diametre less than1:20.

Consider the equilibrium of the half – sphere

Force on half-sphere owing to internal pressure = pressure x projected Area =** p. πd ^{2}/4**

Fig – shown the (by way of dotted lines) the tendency, for the cylindrical portion and the spherical ends to expand by a different amount under the action of internal pressure. So owing to difference in stress, the two portions (i.e. cylindrical and spherical ends) expand by a different amount. This incompatibly of deformations causes a local bending and sheering stresses in the neighborhood of the joint. Since there must be physical continuity between the ends and the cylindrical portion, for this reason, properly curved ends must be used for pressure vessels.

Thus equating the two strains in order that there shall be no distortion of the junction

But for general steel works ν = 0.3, therefore, the thickness ratios becomes

i.e. the thickness of the cylinder walls must be approximately 2.4 times that of the hemispheroid ends for no distortion of the junction to occur.