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THICK CYLINDERS Course Instructor: Dr. Sri Kalyana Rama J Introduction A cylinder is considered to be ‘thick’ if the ratio of the inner diameter to the thickness of the walls is < 20 If the cylinder is long in comparison with its diameter, the longitudinal stress is assumed to be uniform across the thickness of the cylinder wall σc = circumferential stress varies with radius σr = radial stress varies with radius σL = Longitudinal stress is constant Hydraulic or pneumatic actuators Stresses in Thick Cylinders Stress element Wall thickness Longitudinal stress l (closed ends) t Small element Tangential stress θ Hoop stress r Radial stress Lame’s Theory and Expression Assumptions made in lame’s theory The material is homogeneous and isotropic Plane transverse sections remain plane under the action of internal pressure. The material is stressed within elastic limit as per Hooke’s Law All the fibres of the material are stressed independently without being constrained by the adjacent fibres Lame’s Equations a and b are Lame’s constants evaluated from known internal and external pressure values and radius Special Cases of Compound Cylinders Applications ❖ pi = 0 (Internal pressure) ▪ Submarine ▪ Vacuum chamber ▪ Shrink fit ▪ Buried pipe ❖ po = 0 (External pressure) ▪ Gun barrel ▪ Liquid- or gas-carrying pipe ▪ Hydraulic cylinder ▪ Gas storage tank Special Cases r2 r P2 CASE-I External pressure = p2 Internal pressure = p1 CASE-II External pressure = 0 Internal pressure = p1 CASE-III External pressure = p2 Internal pressure = 0 𝜎𝑟 = 𝑃1 𝑎𝑡 𝑟 = 𝑟1 𝜎𝑅 = 𝑃2 𝑎𝑡 𝑟 = 𝑟2 𝜎𝑟 = 𝑃1 𝑎𝑡 𝑟 = 𝑟1 𝜎𝑅 = 𝑃2 = 0 𝑎𝑡 𝑟 = 𝑟2 𝜎𝑟 = 𝑃1 = 0 𝑎𝑡 𝑟 = 𝑟1 𝜎𝑅 = 𝑃2 𝑎𝑡 𝑟 = 𝑟2 1 P1 𝜎𝑟 = 𝑃1 𝑎𝑡 𝑟 = 𝑟1 𝜎𝑅 = 𝑃2 𝑎𝑡 𝑟 = 𝑟2 𝜎𝑅 = 𝑃2 = 0 Longitudinal and Shear stresses Source: Strength of Materials by RK Rajput Numerical Problem A thick walled closed end cylinder is made of an Aluminum alloy with E = 72 Gpa and poisson’s ratio = 0.33, has inside diameter of 200mm and outside diameter of 800mm. The cylinder is subjected to internal fluid pressure of 150 Mpa. Determine the principal stresses and maximum shear stress at a point on the inside surface of the cylinder. Also determine the increase in inside diameter due to fluid pressue. i ii Source: Strength of Materials by RK Rajput Earlier I have covered till this and the same is available in your notes Compound Cylinders or Composite Cylinders A compound cylinder is made by press-fitting 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 Cylinders or Composite Cylinders • 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 Press or shrink fits http://www-mdp.eng.cam.ac.uk/web/library/enginfo/textbooks_dvd_only/DAN/cylinders/compound/compound.html http://ent.uthm.edu.my/ Compound Cylinders or Composite Cylinders Ideal Case Worst case for tight fit Worst case for loose fit https://www.engineersedge.com/calculators/machine-design/press-fit/press-fit-equations.html Demonstration Compound Cylinders or Composite Cylinders The method of solution for compound cylinders constructed from similar materials is to break the problem down into three separate effects: a) b) c) Shrinkage pressure only on the inside cylinder Shrinkage pressure only on the outside cylinder Internal pressure only on the complete cylinder http://ent.uthm.edu.my/ Compound Cylinders or Composite Cylinders Source: Strength of Materials by RK Rajput Compound Cylinders or Composite Cylinders Source: Strength of Materials by RK Rajput A compound cylinder is made by shrinking one tube to another is subjected to an internal pressure of 90 MN/m2.. Before the fluid is admitted, the internal and external diameters of the compound cylinder are of 180mm and 300mm and the diameter at the junction is 240mm. If after shrinking on, the radial pressure at the common surface is 12 MN/m2, determine the final stresses developed in the compound cylinder. Numerical Problem Source: Strength of Materials by RK Rajput Source: Strength of Materials by RK Rajput Final circumferential stresses at different points are tabulated below From step (a) From step (b) Source: Strength of Materials by RK Rajput Necessary difference of radii for shrinkage Source: Strength of Materials by RK Rajput From the previous example, calculate the necessary difference in diameter of the two tubes at the common surface before shrinking on, so as to produce a radial pressure of 12 MN/m2. Also calculate the minimum temperature to which the outer tube should be heated before it can be slipped on Source: Strength of Materials by RK Rajput Design of Thick Cylindrical Shell using Failure Theories Source: Strength of Materials by RK Rajput Design of Thick Cylindrical Shell using Failure Theories Source: Strength of Materials by RK Rajput Design of Thick Cylindrical Shell using Failure Theories Source: Strength of Materials by RK Rajput Numerical Problem A thick cylindrical shell is of 300mm internal diameter and has to withstand an internal pressure of 30 MN/m2. If the permissible tensile stress of the shell is 150 MN/m2 and poisson’s ratio is 0.3, calculate the thickness of the metal necessary for the cylinder using all the 1st four failure theories. Neglect longitudinal direct stress. Source: Strength of Materials by RK Rajput