Thin CylindersThin cylinders are those whose diameter is more than 20 times thethickness of the shell.The following assumptions are made in order to derive the expressions for the stresses in thin cylinders :

(a)The stresses are uniformly distributed through the thickness of the wall. (b) The ends of the cylindrical shall are not supported from sides. (c) The weight of the cylinder and that of the fluid contained inside are not taken into account. (d) The atmosphere pressure is taken as the reference pressure.

Longitudinal Stress (l):

The internal stress is called longitudinal stress, because it acts parallel to the longitudinal axes of cylinder, indicating the direction in which it is acting and its nature will be tensile.

Where p = internal pressure of fluid D = diameter of cylinder t = thickness of cylinder

Circumferential Stresses or Hoop stresses (h):

The stresses acting in the circumferential direction are called hoop stresses or circumferential stresses or tangential stresses and these will be tensile in nature.

It can be stated at this stage that the internal stresses in a thin cylindrical shell are acting in the longitudinal and circumferential directions and hence they are named as longitudinal stress and hoop stress respectively, and both are tensile in nature.Where p = internal pressure of fluid D = diameter of cylinder t = thickness of cylinder

Example: 1 A cylindrical boiler is 2.5 m in diameter and 20 mm in thickness and it carries steam at a pressure of 1.0 N/mm2. Find the stresses in the shell.

Solution: Diameter of the shell, d = 2.5 m = 2500 mm. Thickness of the shell, t = 20 mm. Internal pressure, p = 1.0 N/mm2.Longitudinal Stress:Circumferential Stress:

Example: 2 A thin cylindrical vessel of 2 m diameter and 4 m length contains a particular gas at a pressure of 1.65 N/mm2. If the permissible tensile stress of the material of the shell is 150 N/mm2, find the minimum thickness required. Solution:In a thin cylindrical shell, circumferential stress will be higher, since it is double that of longitudinal stress. Hence, maximum stress is reached in the circumferential direction. If thickness required is t, thenThus, minimum thickness required is 11 mm.

Thick CylindersThick cylinders whose wall thickness is large such that they can no longer be considered as thin i.e. the diameter-thickness ratio is less than 20. The following assumptions are made in order to derive the expressions for the stresses in thick cylinders :Assumptions (a) The material of the cylinder is homogeneous and isotropic. (b) Plane section perpendicular to the longitudinal axis of the cylinder remain plane even after the application of the internal pressure. This implies that the longitudinal strain is same at all points of the cylinder. (c) All fibres of the material are free to expand or contract independently without being confined by the adjacent fibres.

Consider a thick walled cylinder with open ends as shown in fig. It is loaded by internal pressure Pi and external pressure Po as shown in fig. It has inner radius a and outer radius b.

The tangential stress in this shell is t , the radial stress on the inner surface is r and that on the outer surface is r+dr, where dr is the increment due to the variation of pressure across the cylinder wall.

For equilibrium, taking summation of forces The product dr dr being neglected because it is very small compared to the other quantities, this reduces toAs longitudinal strain z is constant for all fibers, so applying Hooks law for triaxial stress, we have(1)

Since z, E, z and are all constants, it follows that r+t is a constant throughout the cross section. Let this constant be 2A, so that(2)Adding equ. (1) and equ. (2)Separating the variables, we obtain

Integration gives(3)Substitute this value in equ (2), gives(4)

The values of the constants A and B are determined by substituting in equ. (3) the known values of r at the inner and outer surfaces of the cylinder. These values areThus we obtainSolving above equations simultaneously will give the values of A and B.From (a)(a)(b)

Put this in (b)Put the value of B in (a) we can get value of A(5)(6)

Substituting the values of A and B in equation (3) and (4) gives the following general expressions for r and t at any point:(7)These are known as Lams equations.

Example 3: The internal and external diameters of a thick hollow cylinder are 80 mm and 120 mm respectively. It is subjected to an external pressure of 40 N/mm2 and an internal pressure of 120 N/mm2. Calculate the circumferential stress at the external and internal surfaces and determine the radial and circumferential stresses at the mean radius. Solution: Given data, a = 80/2 = 40 mm b = 120/2 = 60 mm po = 40 N/mm2 pi = 120 N/mm2Circumferential stress at external surface (at r = 60 mm)= 88 N/mm2

Circumferential stress at internal surface (at r = 40 mm)= 168 N/mm2Radial & Circumferential stress at mean radius (at r = 50 mm)= 68 N/mm2= 116 N/mm2