4.1 thick walled cylinders
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Stress distribution in thick walled cylinder under pressure
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
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Equilibrium of element in cylindrical coordinates (Plane stress)
Consider equilibrium of element in radial ( r ) direction:
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
1 1 1sin , co s 12 2 2
For small angle approximation
Also neglecting second and higher order terms and dividing by r r z
Rearranging
(eq 1)1 0r r r Rr r r
+ + + =
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Consider equilibrium of element in tangential direction:
Using small angle approximation and neglecting second and higher order termsand dividing by r r z
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
Due to symmetry:
Body symmetrical about z axis
stress depends on r only and
and shear
0
=
0r =
(eq 2)
(eq 2) vanishes
1 2 0r r r r r
+ + + =
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Strain-displacement of an element in cylindrical coordinates (Plane
strain)
change of length of AD =
displacement of D to D in r -direction =
total strain in r -direction therefore =
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
displacement of B to B tangentially =
displacement of B to B in radial direction ( r + u ) =
total strain in tangential-direction therefore =
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shear strain
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
Due to symmetry:
Body symmetrical about z axis
No tangential displacement v
and shear strain = 0
The strain equationabove reduce to
r ur
= 1 v u
r r
= +1
r v u v
r r r
= +
r u d ur d r
= =
ur =
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(eq 2) vanishes and (eq 1) becomes:
0r r d
d r r + = 0 z
d
d z
=
and z is constant
Again due to symmetry: Body symmetrical about z axis stress depends on r only and and shear
0
=
0r =
(eq 3)
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
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r
=
differentiatesubstitute
In z direction:
(eq 4)
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
Comparing (Eq 5) and (Eq 6)
r r
d
r d r
=
(eq 6)
Substitute (Eq 5) into (Eq 4) and from (eq 3)
0 zd
d r
=
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( ) 0ro
r
r i
d d d r + =
Integrate (Eq 6) from inner to outer radius:
From (eq3)
Multiply by r 2
(eq 7)
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
Which is equivalent to
Integrating2r
B A
r
=
Rearranging
(eq 8)
2
B A
r = +
From (eq7) (eq 9)
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Radial and hoop stresses: Internal and external pressure
,i r iat r r p = =
Boundary conditions:
,o r oat r r p = =
From (eq 8)
B eliminatin A or B:
Negative sign forcompressive stress
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
Substitute into (eq 8) and (eq 9)
2 222 2 21 1 1
1o i
r i or r p p k
k r r =
2 2
22 2 21 1 11
o ii or r p p k k r r
= + +
o
i
r k
r =
Or
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Internal pressure only:maximum hoop and redial stress occur at inner radius
0o p =
2 2
2 2 21i i or
o i
p r r
r r r
=
or
ir
2 2
2 2 21i i o
o i
p r r
r r r
= +
2 2
2 2i o
io i
r r pr r
+
2
2 2
2 oi
o i
r p
r r
r
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
External pressure only:maximum hoop occurs at inner radiusmaximum redial stress occur at outer radius
0i p =or
ir 2 2
2 2i o
oo i
r r p r r
+
2
2 2
2 oo
o i
r p
r r
2 2
2 2 21o o ir
o i
p r r r r r
=
2 2
2 2 21o o i
o i
p r r
r r r
= +
r
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Axial stress: Internal and external pressure
( )2 2 2 2 0 z o i i i o or r p r p r + =2 2
2 2
i i o o
zo i
p r p r
r r
=
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
Cylinder free to change in lengthunder internal pressure only
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Maximum shear stress in the cylinder:
z r > >Since
Yielding in the cylinder:
2
2
2:
1i
i Y r k p
at r r Tresca criterionk
= = =
2 1k
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
2
2i Y p
k
=
Von Misses stress in the cylinder:
( ) ( ) ( )2 22 22r r z z Y
+ + =
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Shrink fit cylinder assembly
Methods of containing high pressure
highest stress lower than in single cylinder
cylinders fitted by interference fitouter cylinder heated and inner cylinder cooled. When under ambient temperature inner cylindersubject to hoop compression and outer cylinder subject to hoop tension
under internal pressure the resultant stress = stress du to shrink fit + stress due to internal pressure
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
Autofrettage:
Wire wound cylinder:
Internal pressure applied until cylinder yields
Plastic deformation occurs for a portion of thickness at inner diameter
Residual stress occurs when pressure removed compressive hoop stress
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Stresses due to Shrink fit only:
1. at interference of two cylinders radial stresses equal
To solve for A, B, C and D need 4 known parameters:
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
( ) 20i n n e r r i
i
Br r A
r = = = 2. At inner radius of inner cylinder
3. At outer radius of outer cylinder ( ) 20ou te r r oo
Dr r C r
= = =
4. Need to know interference pressure ( ) 2i n n e r r o m m
m
Br r r p A
r = = = =
( ) ( )2 2in n er o u ter r o m r i mm m
B Dr r r A C r r r r r
= = = = = = =
( ) 2o u te r r i m m
m
Dr r r p C
r
= = = =
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displacement of inner cylinder inwards
displacement of outer cylinder outwards
Interference due to shrink fit:
i n n e r i n n e r in n er in n er r
in n er in n er E E
= o u te r o u t e r
o u te r o u ter r o u te r o u ter E E
=
( )in n e r o u ter o u te r in n er mu u r = + =
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
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Example 2
A bronze bush of 25 mm wall thickness is to be shrunk onto a steel cylinder 20 mm thick having 100 mm outer diameter.If an interface pressure of 69 MN/m 2 is required, determine the interface between bush and shaft.Steel E = 207 GN/m 2, v = 0.28; bronze E = 100 GN/m 2, v = 0.29.
Inner cylinder:
( ) 203 0
i nner r i
Br r A = = =
( ) 26 9 5 0i n n e r r o m
Br r r M P a A = = = =
49 .7 0 3 1 1 0 ; 1 0 7 .8 1 B A= =
Outer cylinder (bronze bush):
( ) 205 0
ou te r r o
Dr r C = = =
( ) 26 9 5 0o u te r r i m
Dr r r M P a C = = = =
35 5 .2 ; 3 1 0 .5 1 0C D= =
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
( ) 2 1 4 6 . 6 25 0i nner
m
Br r A M P a
= = + =
4( ) 6.1167 10inner inner
inner inner r o m inner inner
at r r r E E
= = = =
( ) 2 1 7 9 . 45 0ou te r
m
Dr r C M P a
= = + =
( ) 0 .0 0 2ou te r ou te r
o u te r o u ter r i m o u te r ou te r
a t r r r E E
= = = =
( ) 0 . 1 2 9 9in n e r o u te r o u te r in n e r mu u r m m = + = =
at r = r m = 50 mm : at r = r m = 50 mm :
at r = r m = 50 mm :
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Example 3
A vessel is to be used for internal pressure up to 207 MN/m 2. It consists of two hollow steel cylinderswhich are shrunk one on the other. The inner tube has an inner diameter of 200 mm and a nominalexternal diameter of 300 mm, while the outer tube is 300 mm nominal and 400 mm for the inner andouter diameters respectively. The interference at the surface of the two cylinders is 0.1 mm.Determine the radial and circumferential stresses at the bores and outside surfaces.The axial stress is to be neglected. E = 207 MN/m 2. Compare the stress distribution with thatfor a single steel cylinder, having the same overall dimensions, subject to the same internal pressure
Inner cylinder:
( ) 20 1 0 0i n n e r r i
Br r A = = =
Due to shrink fit only:
( ) 20 2 0 0ou te r r o
Dr r C = = =
B D
Outer cylinder:
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
( )0.1 150outer inner outer inner
inner outer outer inner outer inner r r m outer outer inner inner
u u r v v E E E E
= = + = = +
2100 2200
1 3 8o u te r in n e r =
: in n e r o u te r m r r a t r = 2 2: 1 5 0 1 5 0m
B Da t r A C =
: in n e r o u te r m r r a t r =
54 .0 2 5 1 0 ; 4 0 .2 5 B M P a A M P a= =
62 8 .7 5 ; 1 .1 5 1 0C M P a D M P a= =
0.1 150outer inner
outer inner E E =
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,( ) 2207 100ir r r F E = = =
Single cylinder under internal pressure only:
,( ) 20
200or r r
F E = = =
2200
F E =
66 9 ; 2 .7 6 1 0 E M P a F M P a= =
Adding coefficients for resultant stresses:
Inner cylinder:5
2 2 2
2 3.5 75 1 02 8 . 7 4i nner r
B F A E
r r r
= + =
6
2 2 2
3 .9 1 1 09 7 . 7 5o u t e r r
D F C E
r r r
= + =
Outer cylinder:
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition100 120 140 160 180 200-100
-80
-60
-40
-20
0
20
40
60
80r n s resses
r
100 120 140 160 180 200-300
-200
-100
0
100
200
300
400nge cy n er un er n erna pressure
100 120 140 160 180 200-300
-200
-100
0
100
200
300esu an s resses
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Stresses due to Autofrettage :
Considering only ideal elastic-plastic material. Maximum shear stress (Tresca) theory of yielding will beadopted
Plastic stress starts at bore and progresses to radius a
2 2a o p a r
plastic zone first
cylinder
elastic zone second cylindera
Consider 2 cylinders with interface at r=a
for outer cylinder in an elastic zone,starting yielding at inner radius:
2 2a o p a r
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
2 2 2r
or a a
= elastic - plastic
zone at r=a
2 2 2
or a a
=
Maximum shear stress in the second cylinder:
z r > >Since
22 2 2
a o Y
o
p r r a
= =
( )2 222Y
a oo
P r ar
= (eq 10)
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plastic zone firstcylinder
elastic zone second cylinder
elastic - plastic
zone at r=a
a
0r r d
d r r + = (eq 3)
using (eq 10):
;i r ia t r r p = =
(eq 11)
( )2 222Y
a oo
P r ar
=
Internal pressure to cause yielding up to r=a :
To obtain expression for inner cylinder in plastic zone
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
or
r Y d
d r r
=
; r aa t r a p = = =
substitute for C
(eq 12)
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Residual stress after Plastic deformation:
or
ir 2 2
m ax2 2 21
i or
o i
p r r r r r
=
2 2m ax2 2 2
1i o
o i
p r r
r r r
= +
To obtain stresses on off loading calculate r and with p i = p max (pressure to cause yielding but in reverse direction)
NOTE: m axi p p ve= = +
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
100 120 140 160 180 200-150
-100
-50
0
50
100
150
200Single cylinder under plastic stresses
r
100 120 140 160 180 200-250
-200
-150
-100
-50
0
50
100
150 Stress after off-loading
r
100 120 140 160 180 200-140
-120
-100
-80
-60
-40
-20
0
20
40
60 Residual stresses
r
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100 120 140 160 180 200-150
-100
-50
0
50
100
150
200Single cylinder under plastic stresses
r
100 120 140 160 180 200-300
-200
-100
0
100
200Stress after off-loading
r
100 120 140 160 180 200-150
-100
-50
0
50Residual stresses
r
Autofrettaged cylinder under internal pressure:
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
100 120 140 160 180 200-200
-100
0
100
200
300
400
Single cylinder under internal pressure
100 120 140 160 180 200-200
-100
0
100
200
300
Autofrattaged cylinder under int pressure
r
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Thin-Walled Pressure VesselsThin wall refers to a vessel having an inner-radiusto-wall-thickness ratio of 10 or more.
For cylindrical vessels under normal loading, thereare normal stresses in the circumferential or hoop direction and in the lon itudinal or axial direction .
( )10 / t r
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
directionallongitudininstressnormal 2
directionhoopinstressnormal
2
1
t pr t
pr
=
=
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Thin-Walled Pressure VesselsThe following is a summary of loadings that can beapplied onto a member:a) Normal Forceb) Shear Forcec Bendin Moment
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
d) Torsional Momente) Thin-Walled Pressure Vesselsf) Superposition
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Example 8.3The tank has an inner radius of 600 mm and a thickness of 12 mm. It is filled to thetop with water having a specific weight of w = 10 kNm 3. If it is made of steel havinga specific weight of st = 78 kNm 3, determine the state of stress at point A. The tankis open at the top.
Solution:The weight of the tank is
Mechanics of Material 7Mechanics of Material 7 thth EditionEdition
( ) kN56.311000600
1000612
78 = == st st st V W
The pressure on the tank at level A is ( )( ) kPa10110 === z p wFor circumferential and longitudinal stress, we have
( )( )
( ) ( )[ ] (Ans) kPa9.7756.3
(Ans) kPa50010
210006002
10006122
1000121000600
1
=
==
===
st
st
AW
t pr
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