The Thick-Walled cylinder

1. Namas Chandra
Advanced Mechanics of Materials Chapter 11-1
EGM 5653
The Thick Walled CylinderThe Thick Walled Cylinder
CE 527
Duaa FADIL
Submitted by:
Submitted to:
Asst.Prof.Dr. Nildem TAYŞİ
Advanced Mechanics of Materials

2. Namas Chandra
Advanced Mechanics of Materials Chapter 11-2
EGM 5653CE 527
Presentation OverviewPresentation Overview

Introduction
Basic Relations
Closed End Cylinders
Open Cylinder
Stress Components and Radial Displacement for
Constant Temperature
Stresses and Deformations in Hollow cylinder
Advanced Mechanics of Materials

3. Namas Chandra
Advanced Mechanics of Materials Chapter 11-3
EGM 5653
Introduction
 This chapter deals with the basic
relations for axisymmetric
deformation of a thick walled cylinder
In most applications cylinder wall
thickness is constant, and is subjected
to uniform internal pressure p1,
uniform external pressure p2, and a
temperature change ΔT
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Advanced Mechanics of Materials

4. Namas Chandra
Advanced Mechanics of Materials Chapter 11-4
EGM 5653
Introduction contd.

Solutions are derived for open cylinders or for cylinders with
negligible “end cap” effects.

The solutions are axisymmetrical- function of only radial
coordinate r

Thick walled cylinders are used in industry as pressure vessels,
pipes, gun tubes etc..
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Advanced Mechanics of Materials

5. Namas Chandra
Advanced Mechanics of Materials Chapter 11-5
EGM 5653
Basic Relations
Equations of equilibrium derived neglecting the body force
( )rr
rr rr
d d
r or r
dr dr
θθ θθ
σ
σ σ σ σ= − =
Strain- Displacement Relations and Compatibility Condition
, ,rr zz
u u w
r r z
θθε ε ε
∂ ∂
= = =
∂ ∂
Three relations for extensional strain are
where u= u(r,z) and w= w(r,z) are
displacement components in the r and z
directions
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Advanced Mechanics of Materials

6. Namas Chandra
Advanced Mechanics of Materials Chapter 11-6
EGM 5653
Basic Relations Contd.
At sections far from the end shear
stress components = 0 and we
assume
εzz= constant. Therefore, by
eliminating u = u(r)
( )
rr rr
d d r
r or
dr dr
θθ θθ
θθ
ε ε
ε ε ε= − =
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Advanced Mechanics of Materials

7. Namas Chandra
Advanced Mechanics of Materials Chapter 11-7
EGM 5653
Basic Relations Contd.
Stress Strain Temperature Relations
The Cylinder material is assumed to be Isotropic and linearly elastic
The Stress- Strain temperature relations are:
[ ]
[ ]
[ ]
1
( )
1
( )
1
( ) constant
rr rr ZZ
rr ZZ
zz zz rr
T
E
T
E
T
E
θθ
θθ θθ
θθ
ε σ ν σ σ α
ε σ ν σ σ α
ε σ ν σ σ α
= − + + ∆
= − + + ∆
= − + + ∆ =
Where, E is Modulus of Elasticity
ν is Poisson’s ratio
α is the Coefficient of linear thermal expansion
ΔT is the change in the temp. from the uniform reference temp.
CE 527
Advanced Mechanics of Materials

8. Namas Chandra
Advanced Mechanics of Materials Chapter 11-8
EGM 5653
Closed End Cylinders
Stress Components at sections far from the ends
.
The expressions for the stress components σrr,σθθ, σzz for a cylinder
with closed ends and subjected to internal pressure p1, external
pressure p2, axial load P and temperature change ΔT.
From the equation of equilibrium, the strain compatibility equation and
the stress- strain temperature relations we get the differential
expression, 0
1
rr
d E T
dr
θθ
α
σ σ
ν
∆ 
+ + = ÷
− 
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Advanced Mechanics of Materials

9. Namas Chandra
Advanced Mechanics of Materials Chapter 11-9
EGM 5653
Closed End Cylinders
Stress Components at sections far from the ends contd.
Eliminating the stress component σθθ and integrating, we get
2
2
12 2 2
1
(1 )
r
rr
a
CE a
Trdr C
r r r
α
σ
ν
 
= − ∆ + − + ÷
−  
∫
Using this in the previous expression and evaluating σθθ, we get
2
2
12 2 2
1
(1 ) 1
r
a
CE E T a
Trdr C
r r r
θθ
α α
σ
ν ν
 ∆
= − ∆ − + + − ÷
− −  
∫
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Advanced Mechanics of Materials

10. Namas Chandra
Advanced Mechanics of Materials Chapter 11-10
EGM 5653
Closed End Cylinders
Stress Components at sections far from the ends contd.
The effects of temperature are self- equilibrating. The expression for εzz
at section far away from the closed ends of the cylinder can be written
in the form
( )
2 2
1 2closed end 2 2 2 2 2 2
1 2 2
( )
( ) ( ) ( )
b
zz
a
P
p a p b Trdr
E b a b a E b a
ν α
ε
π
−
= − + + ∆
− − − ∫
( )
2 2
1 2
closed end 2 2 2 2 2 2
2
( ) 1 (1 )( )
b
zz
a
p a p b P E T E
Trdr
b a b a b a
α α
σ
π ν ν
− ∆
= + − + ∆
− − − − − ∫
The expression for σzz at section far away from the ends of the cylinder can
be written in the form
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Advanced Mechanics of Materials

11. Namas Chandra
Advanced Mechanics of Materials Chapter 11-11
EGM 5653
Open Cylinder
 No axial loads applied on its ends.
The equilibrium equation of an axial portion of the
cylinder is:
2 0
b
zz
a
r dzπ σ =∫
2 2
2 1
(open end) 2 2 2 2
2 ( ) 2
( ) ( )
b
zz
a
p b p a
Trdr
b a E b a
ν α
ε
−
= + ∆
− − ∫
The expression for εzz and σzz may be written
in the form
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Advanced Mechanics of Materials

12. Namas Chandra
Advanced Mechanics of Materials Chapter 11-12
EGM 5653
Stress Components and Radial Displacement
for Constant Temperature
.
For a closed cylinder (with end caps) in the absence of temperature
change ΔT = 0 the stress components are obtained as
2 2 2 2
1 2
1 22 2 2 2 2
( )
( )
rr
p a p b a b
p p
b a r b a
σ
−
= − −
− −
2 2 2 2
1 2
1 22 2 2 2 2
( )
( )
p a p b a b
p p
b a r b a
θθσ
−
= + −
− −
2 2
1 2
2 2 2 2
constant
( )
zz
p a p b P
b a b a
σ
π
−
= + =
− −
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Advanced Mechanics of Materials

13. Namas Chandra
Advanced Mechanics of Materials Chapter 11-13
EGM 5653
Radial Displacement for an Closed Cylinder
The radial displacement u for a point in a thick wall closed cylinder may
be written as
2 2
2 2
(closed end) 1 2 1 22 2 2
(1 )
(1 2 )( ) ( )
( )
r a b P
u p a p b p p
E b a r
ν
ν ν
π
 +
= − − + − − 
−  
Radial Displacement for an Open Cylinder
2 2
2 2
(open end) 1 2 1 22 2 2
(1 )
(1 )( ) ( )
( )
r a b
u p a p b p p
E b a r
ν
ν
 +
= − − + − 
−  
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Advanced Mechanics of Materials

14. Namas Chandra
Advanced Mechanics of Materials Chapter 11-14
EGM 5653
Example 1
Stresses and Deformations in Hollow cylinder
A thick walled closed-end cylinder is made of an aluminum alloy,
E = 72 GPa, ν = 0.33,Inner Dia. = 200mm, Outer Dia. = 800 mm,
Internal Pressure = 150 MPa.
Determine the Principal stresses, Maximum shear stress at the inner
radius (r= a = 100 mm), and the increase in the inside diameter caused
by the internal pressure
Solution:
2p 0and r a= =
2 2
1 12 2
150rr
a b
p p MPa
b a
σ
−
= = − = −
−
2 2 2 2
1 2 2 2 2
100 400
150 170
400 100
a b
p MPa
b a
θθσ
+ +
= = =
− −
2 2
1 2 2 2 2
100
150 10
400 100
zz
a
p MPa
b a
σ = = =
− −
The Principal stresses for the conditions that
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Advanced Mechanics of Materials

15. Namas Chandra
Advanced Mechanics of Materials Chapter 11-15
EGM 5653
Example 1
Stresses and Deformations in Hollow cylinder
contd.
max min
max
170 ( 150)
160
2 2
MPa
σ σ
τ
− − −
= = =
2 0and r ap P= = =
The maximum shear stress is given by the equation
The increase in the inner diameter caused by the internal pressure
is equal to twice the radial displacement for the conditions
( )
2 21
( ) 2 2
2 2
2 2
(1 2 ) (1 )
( )
150(100)
1 0.66 100 (1 0.33)400
72,000(400 100 )
0.3003
r a
p a
u a b
E b a
mm
ν ν=
 = − + + −
 = − + + −
=
The increase in the internal pressure caused by the internal pressure
is 0.6006 mm
CE 527
Advanced Mechanics of Materials

16. Namas Chandra
Advanced Mechanics of Materials Chapter 11-16
EGM 5653
Advanced Mechanics of Materials
CE 527
QuestionsQuestions

17. Namas Chandra
Advanced Mechanics of Materials Chapter 11-17
EGM 5653
Advanced Mechanics of Materials
CE 527
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