elastic stress
TRANSCRIPT
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-1
EGM 5653
CHAPTER 11The Thick Walled Cylinder
EGM 5653Advanced Mechanics of Materials
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-2
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 subjectedto uniform internal pressure p1,
uniform external pressure p2, and a
temperature change ΔT
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-3
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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Namas ChandraAdvanced Mechanics of Materials Chapter 11-4
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11.1 Basic Relations
Equations of equilibrium derived neglecting the body force
( )rr rr rr
d d r or r
dr dr
Strain- Displacement Relations and Compatibility Condition
, ,rr zz
u u w
r r z
Three relations for extensional strain are
where u= u(r,z) and w= w(r,z) aredisplacement components in the r and z
directions
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-5
EGM 5653
11.1 Basic Relations Contd.
At sections far from the end shear
stress components = 0 and we
assume
εzz = constant. Therefore, byeliminating u = u(r)
( )rr rr
d d r r or
dr dr
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-6
EGM 5653
11.1 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
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.
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-7
EGM 5653
11.2 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
1rr
d E T
dr
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-8
EGM 5653
11.2 Closed End Cylinders Stress Components at sections far from the ends contd.
Eliminating the stress component σθθ and integrating, we get
2
212 2 2
1(1 )
r
rr
a
C E aTrdr C
r r r
Using this in the previous expression and evaluating σθθ, we get
2
212 2 2
1(1 ) 1
r
a
C E E T aTrdr C
r r r
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-9
EGM 5653
11.2 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
2 2
1 2closed end 2 2 2 2 2 22
( ) 1 (1 )( )
b
zz
a
p a p b P E T E Trdr b a b a b a
The expression for σzz at section far away from the ends of the cylinder can
be written in the form
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-10
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 dz
2 2
2 1(open end) 2 2 2 2
2 ( ) 2
( ) ( )
b
zz
a
p b p aTrdr
b a E b a
The expression for εzz
and σzz
may be written
in the form
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-11
EGM 5653
11.3 Stress Components and Radial Displacementfor 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
2 2 2 2
1 21 22 2 2 2 2
( )( )
p a p b a b p p
b a r b a
2 2
1 2
2 2 2 2 constant( ) zz
p a p b P
b a b a
EGM 5653
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-12
EGM 5653
11.3.2 Radial Displacement for an Closed Cylinder
The radial displacement u for a point in a thick wall closed cylinder may
be written as
2 22 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
11.3.3 Radial Displacement for an Open Cylinder
2 22 2
(open end) 1 2 1 22 2 2
(1 )(1 )( ) ( )
( )
r a bu p a p b p p
E b a r
EGM 5653
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-13
EGM 5653
Example 11.2 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:2 p 0 and r a
2 2
1 12 2150rr
a b p p MPa
b a
2 2 2 2
1 2 2 2 2
100 400
150 170400 100
a b
p MPab a
2 2
1 2 2 2 2
100150 10
400 100 zz
a p MPa
b a
The Principal stresses for the conditions that
EGM 5653
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-14
EGM 5653
Example 11.2 Stresses and Deformations in Hollow cylinder contd.
max minmax
170 ( 150)160
2 2 MPa
2 0and r a p 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 au a b
E b a
mm
The increase in the internal pressure caused by the internal pressure
is 0.6006 mm
EGM 5653
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-15
EGM 5653
11.4 Criteria of Failure
Recap
Maximum Principal stress criterion – Design of Brittle isotropic materials
– If the principal stress of largest magnitude is the tensile stress
Maximum shear stress or the Octahedral shear- stress criterion
– Design of Ductile isotropic materials
11.4.1 Failure of Brittle Materials
Maximum principal stress = Ultimate tensile Strength σu
Condition for Failure
At sections far removed from the ends
Maximum Principal stress = Circumferential stress σθθ(r=a)
or axial stress σzz
EGM 5653
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-16
EGM 5653
11.4.2 Failure of Ductile Materials
.
General Yielding Failure
Yielding at sections other than the points of stressconcentration
Thick walled cylinders occasionally subjected to static
loads or peak loads
Member has yielded over a considerable region as with
fully plastic loads
Fatigue failure
Subjected to repeated pressurizations (loading and unloading)
Found predominantly around the region of stress
concentration Maximum shear stress or maximum octahedral shear stress to
be determined at the regions of stress concentration
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-17
EGM 5653
11.4.3 Material Response Data for Design
General Yielding
Property : Yield stressCriteria : Maximum shear stress or Octahedral shear stress
Fatigue Failure
Property : Fatigue strength
Criteria : Maximum shear stress and Octahedral shear stress in conjunction
Values obtained from tests of either a tension specimen or hollow thin
walled cylinder in torsion
The hollow thin walled cylinder specimen values led to more accurate
prediction of thick walled cylinders than the tension specimen
The critical state of stress is usually at the inner wall , for a pressure
loading it is only pure shear in addition to hydrostatic state of stress
Values from tensile test specimen and hollow thin-wall tube
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-18
11.4.3 Material Response Data for Design contd.
Since for most materials the hydrostatic stress does not affect
yielding yielding is caused by the pure shear
Hence maximum shear- stress criterion and octahedral shear-stress
criterion predict with errors of <1% in comparison with 15.5 %
11.4.4 Ideal Residual Stress Distributions for
Composite Open Cylinders
Stress distributions in a closed cylinder at initiation of yielding (b=2a)
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-19
11.4.4 Ideal Residual Stress Distributions forComposite Open Cylinders contd.
Stress distributions in composite cylinder made of brittle material that fails
at inner radius of both cylinders simultaneously
(a) Residual stress distributions
(b) Total stress distributions
EGM 5653
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-20
11.4.4 Ideal Residual Stress Distributions forComposite Open Cylinders contd.
Stress distributions in composite cylinder made of ductile material
that fails at inner radius of both cylinders simultaneously
(a) Residual stress distributions
(b) Total Stress distributions
EGM 5653
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-21
Example 11.5Yield of a Composite Thick Wall Cylinder
Problem: Consider a Composite cylinder,
Inner cylinder: Inner Radii a =10 mm and outer radii ci=25.072 mm
Outer cylinder: Inner Radii co =25 mm and outer radii b= 50 mm
Ductile steel ( E=200 GPa and ν=0.29)
Determine minimum yield stress for a factor of safety SF=1.75Solution:
It is necessary to consider the initiation of yielding for theinside of both the cylinders.σzz =0 for both the cylinders
At the inside of the inner cylinder, the radial and circumferential
stresses for a pressure (SF)p1 are,(1.75)(300) 525
(1.75)(325) 450.2 118.6
rr MPa
MPa
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Namas ChandraAdvanced Mechanics of Materials Chapter 11-22
Example 11.5Yield of a Composite Thick Wall Cylinder contd.
2 2 21(118.6 525) (525) (118.6) 593.3
2Y MPa
At the inside of the outer cylinder, the radial and circumferential
stresses for a pressure (SF)p1 are,
(1.75)(37.5) 189.1 245.7
(1.75)(62.5) 315.1 424.5
rr MPa
MPa
2 2 21(424.5 254.7) (254.7) (424.5)
2
594.3 593.3
Y
MPa MPa
In an ideal design, the required yield stress will be the same for the
inner and the outer cylinders.