4.1 Thick Walled Cylinders

7/30/2019 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


2/27

Equilibrium of element in cylindrical coordinates (Plane stress)

Consider equilibrium of element in radial ( r ) direction:


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

+ + + =


3/27

Consider equilibrium of element in tangential direction:

Using small angle approximation and neglecting second and higher order termsand dividing by r r z


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

+ + + =


4/27

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 =


displacement of B to B tangentially =

displacement of B to B in radial direction ( r + u ) =

total strain in tangential-direction therefore =


5/27

shear strain


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 =


6/27

(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)



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r

=

differentiatesubstitute

In z direction:

(eq 4)


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

=


8/27

( ) 0ro

r

r i

d d d r + =

Integrate (Eq 6) from inner to outer radius:

From (eq3)

Multiply by r 2

(eq 7)


Which is equivalent to

Integrating2r

B A

r

=

Rearranging

(eq 8)

2

B A

r = +

From (eq7) (eq 9)


9/27

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


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


10/27

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


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


11/27


12/27

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

=


Cylinder free to change in lengthunder internal pressure only


13/27

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


2

2i Y p

k

=

Von Misses stress in the cylinder:

( ) ( ) ( )2 22 22r r z z Y

+ + =


14/27

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


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


15/27

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:


( ) 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

= = = =


16/27

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 = + =



17/27

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= =


( ) 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 :


18/27

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:


( )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 =


19/27

,( ) 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


20/27

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


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)


21/27

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


or

r Y d

d r r

=

; r aa t r a p = = =

substitute for C

(eq 12)


22/27


23/27

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= = +


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


24/27

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:


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


25/27

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


directionallongitudininstressnormal 2

directionhoopinstressnormal

2

1

t pr t

pr

=

=


26/27

Thin-Walled Pressure VesselsThe following is a summary of loadings that can beapplied onto a member:a) Normal Forceb) Shear Forcec Bendin Moment


d) Torsional Momente) Thin-Walled Pressure Vesselsf) Superposition


27/27

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


( ) 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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4.1 Thick Walled Cylinders