LECTURE 17
Let us now consider the vessel with hemispherical ends. The wall thickness of the cylindrical and hemispherical portion is different. While the internal diameter of both the portions is assumed to be equal Let the cylindrical vassal is subjected to an internal pressure p.
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
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
0.7/1.7 or
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.
(A) The stresses set up in the walls of a thin cylinder owing to an internal pressure p are : (i) Circumferential or loop stress
(ii) Longitudinal or axial stress
Where d is the internal diametre and t is the wall thickness of the cylinder. then Longitudinal strain Hoop stain _{L} ] (B) Change of internal volume of cylinder under pressure (C) Fro thin spheres circumferential or loop stress
Consider a thin ring or cylinder as shown in Fig below subjected to a radial internal pressure p caused by the centrifugal effect of its own mass when rotating. The centrifugal effect on a unit length of the circumference is p = m w
Here the radial pressure ‘p' is acting per unit length and is caused by the centrifugal effect if its own mass when rotating. Thus considering the equilibrium of half the ring shown in the figure, 2F = p x 2r (assuming unit length), as 2r is the projected area F = pr Where F is the hoop tension set up owing to rotation. The cylinder wall is assumed to be so thin that the centrifugal effect can be assumed constant across the wall thickness. F = mass x acceleration = This tension is transmitted through the complete circumference and therefore is resisted by the complete cross – sectional area. hoop stress = F/A = Where A is the cross – sectional area of the ring. Now with unit length assumed m/A is the mass of the material per unit volume, i.e. the density r . hoop stress =
r . w ^{2} . r^{2} |