( BDA 30303 ) CHAPTER V THICK CYLINDER SOLID MECHANICS II ( BDA 30303 ) CHAPTER V THICK CYLINDER

Chapter V Thick Cylinders Lame equation for Tangential or hoop stress Radial stress, Graft method, Composite cylinder.

Introduction Thick cylinders are designed to withstand high internal pressure about 40 to 60 MPa A wall is considered thick if it is 1/10 the cylinder radius or more. Presence of radial stress ( no negligence) Variable hoop stress Complexity involved in contrast to thin cylinders which has Low pressure Negligible radial stress and Constant hoop stress

Lame’ Theory The problem of determining the tangential stress and the radial stress at any point on a thick walled cylinder, in terms of applied pressure and the dimensions, was first solved by the French elastician Gabriel Lame in 1833. His analysis is commonly known as Lame’s theory Hence, in order to address the complexity involved in the thick cylinders, Lame’s theory is used Assumptions made in lame’s theory (i) the material is homogenous and isotropic (ii) Plane transverse sections remain plane under the action of internal pressure. (iii) The material is stressed within elastic limit as per Hook’s Law (iv) All the fibres of the material are stressed independently without being constrained by the adjacent fibres

Lame’s Theory (Contd.) Consider a thick cylinder ( as shown in Figure ). The stresses acting on an element of unit length at radius r as shown in the figure. The radial stress increasing from r to r+ rdr over the element thickness dr (all stresses are assumed tensile)

Lame’s Theory (Contd.) For radial equibrium of the element:

LAME’s theory The force balance for equilibrium in the radial direction:

Lame’s Theory (Contd.)

Lame’s Theory (Contd.) Lame’s Equation

Lame’s Theory (Contd.) Boundary conditions for thick-walled cylinder ri ro Po Note: pressure is compression Pi

Lame’s Theory (Contd.) Constants A and B can be determined from the boundary conditions r=ri and r=ro

Lame’s Theory (Contd.) Then Substituting the constants

Lame’s Theory (Contd.) Now consider the cross-section of a thick cylinder with closed ends subjected to an internal pressure Pi and external pressure Po. For horizontal equilibrium:-

Lame’s Theory (Contd.) Maximum Shear Stress Since H is normally tensile, whilst R is compressive and both exceed L in magnitude:-

Special Case Case 1: A cylinder subjected to internal pressure. Pi=P and Po=0 These equations show that R is always a compressive stress and H is a tensile stress.

Special Case (Contd.) Case 2: A cylinder subjected to external pressure. Pi=0 and Po=P

Problem 1 A thick cylindrical shell with inner radius 10 cm and outer radius 16 cm is subjected to an internal pressure of 70 MPa. Find the maximum and minimum hoop stresses Given a =10 cm and b = 16 cm the hoop stress at r=ri=a=10 cm (Maximum at the inner radius) is Similarly the hoop stress at r=ro =16 cm is

Problem 2 Calculate the thickness of metal necessary for a cylindrical shell of internal radius 160mm to withstand an internal pressure of 25 MPa, if maximum permissible tensile stress is 125 MPa Given Pi= P=25 MPa and Maximum hoop stress , i.e., H at r = ri is 125 MPa. Considering Po as zero ( no external pressure) , using the formula for hoop stress at r=ri So the thickness is b-a = 180-160=20 mm

Compound (Composite ) Cylinder A compound cylinder is made by press- fitting one or more jackets around an inner cylinder. The purpose is that the outer cylinder puts pressure on the outside of the inner cylinder. This means that the bore is put into compression In a compound cylinder, the outer cylinder is having inner diameter smaller than outer diameter of inner cylinder The inner cylinder is shrink fit to outer cylinder by heating and cooling On cooling the contact pressure is developed at the junction of two cylinders

Compound Cylinder (Contd.) It introduces compressive tangential stress at the inner cylinder and Tensile stresses at outer cylinder. Inner cylinder subjected to external pressure and outer cylinder subjected to internal pressure

Compound Cylinder (Contd.) The method of solution for compound cylinders constructed from similar materials is to break the problem down into three separate effects: Shrinkage pressure only on the inside cylinder Shrinkage pressure only on the outside cylinder Internal pressure only on the complete cylinder

Compound Cylinder (Contd.) For each of the resulting load conditions there are two known values of radial stress which enable the Lame’s constant to be determined in each case.

Compound Cylinder (Contd.) Consider the thick compound cylinder as shown in Figure:- ri rf r0 Outer cylinder R1 = Inner radius of inner cylinder Rc = Outer radius of inner cylinder (common) R2 = Outer radius of outer cylinder P = Radial pressure at the junction of the two cylinder Inner cylinder

Compound Cylinder (Contd.)

Compound Cylinder (Contd.) Values of Constants A1, A2,B1 and B2 can be found if the radial pressure is known Hoop stress can also be found by using relative expressions If the fluid pressure is admitted inside the compound shell, it will be resisted by both the shells The resultant shell will be the algebraic some of initial stresses and Those due to fluid pressure

Problem 3 A compound cylinder is made by shrinking a tube of 160 mm internal diameter and 20 mm thick over another tube of 160 mm external diameter and 20 mm thick. The radial pressure at the common surface, after shrinking is 80 kgf/cm2. Find the final stress setup across the section when the compound cylinder is subjected to an internal fluid pressure of 600 kgf/cm2 Solution:

Solution (Contd.) Applying Lame’s equation for inner cylinder without fluid pressure

Solution (Contd.)

Solution (Contd..) Now, apply the Lame’s equation for the inner cylinder only after the fluid under pressure of 600 kg/cm2 admitted.

Solution (Contd..)

Difference in Radius due to Shrinking The radial pressure is same for inner radius of outer tube and outer radius of inner tube Difference in radius is given as Tensile strain at the shell is Increase in inner radius Decrease in outer radius Subscript 2 =outer tube Subscript 1 =inner tube

Problem 4 A steel cylinder of 300 mm external diameter is to be shrunk to another steel cylinder of 150 mm internal diameter. After shrinking the diameter at the junction is 250 mm and radial pressure at the juction is 28 N/mm2. Find the original difference in radii at the junction. Take E=2x105 N/mm2 [Answer: 0.134 mm]