Uploaded by Sri Kalyana Rama Jyosyula

# Thick Cylinders-Theory, Derivations and Numerical Problems

```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 &lt; 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
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
Lame’s Theory and Expression
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
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
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