thick wall cylinder

7/28/2019 Thick Wall Cylinder

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EM 424: Exact solution for thick wall pressure vessel

Pressure loading of a thick walled cylinder is a relatively simple example where we can

demonstrate explicitly how we can satisfy all the governing equations and boundaryconditions to arrive at a complete solution of a stress analysis problem. Figures 1(a), (b)

show the geometry and loading of the problem.

Fig 1(a) Stresses on the cylinder surfaces (the z-axis is

outwards from this planar view)

Fig. 1 (b) Stresses on the surfaces of the cylinder and the pressure loading.

rr

rr r

r

zr

z

ri

re

AB

C

rrr

z

rr

rz

rzzz

zzzr

zrpi

pe

AB

CD

L


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It is assumed that the internal pressure, pi , and the external pressure, pe , are uniform

along the entire inner and outer surfaces as shown in Fig. 1(b) and that all the othersurfaces of the cylinder are unloaded. In this case the boundary conditions for this

problem are:

for all 0 2 ,0 z L

on the inner and outer surfaces we have

( )

( ) ( )

( )

( ) ( )

, ,

, , , , 0

, ,

, , , , 0

i

i i

e

e e

rr ir r

r rzr r r r

rr er r

r rzr r r r

r z p

r z r z

r z p

r z r z

=

= =

=

= =

=

= =

=

= =

(1)

and for all 0 2 , i er r r on the cylinder ends

( ) ( ) ( )

( ) ( ) ( )

0 0 0, , , , , , 0

, , , , , , 0

zz zr zz z z

zz zr zz L z L z L

r z r z r z

r z r z r z

= = =

= = =

= = =

= = =(2)

Given that the pressure loading is all radially directed, it is reasonable to assume that the

only stresses developed in the cylinder are the normal stresses rr , and to assume

that these stresses are only a function of r, i.e.

( ) ( ),0

rr rr

r rz z zz

r r

= =

= = = =(3)

In this case the boundary conditions of Eq. (2) are satisfied identically for the cylinder

ends and the only boundary conditions in Eq. (1) that are not satisfied identically are

( )

( )

i

e

rr ir r

rr er r

r p

r p

=

=

=

= (4)

Under these assumptions, the equations of equilibrium, which in cylindrical coordinates

are


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10

2 10

10

rr rrr rzr

r r z

zzr zr zz z

fr r r z

fr r r z

fr r r z

+ + + + =

+ + + + =

+ + + + =

(5)

reduce to only one non-trivial equation

0rrrr

r r

+ =

(6)

The stress state given by Eq. (3) is one of plane stress so that the 3-D stress-strain

relations for a homogeneous, isotropic elastic solid reduce to

( )

( )

( )

1

1

0

0

0

rr rr

rr

zz rr

rr

zz

rzrz

E

E

E

G

G

G

=

=

= +

= =

= =

= =

(7)

or, inverting these relations, we have the usual plane stress equations

( )

( )

( )

2

2

1

1

rr rr

rr

zz rr

E

E

E

= +

= +

= +

(8)

To guarantee that we automatically satisfy compatibility, we will work directly with thedisplacements as the fundamental unknowns for this problem. Because of the symmetry

we expect that there will be no displacement component , u

, in the cylinder and that the

other displacements will also be independent of. Also, because the pressure loading isuniform in z, we expect that the radial displacement is only a function of r, i.e.


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( )r ru u r= . In this case the strains in the cylinder, which are given in cylindrical

coordinates as

1, ,

1 1

2

1

2

1 1

2

r r zrr zz

rr

z rrz

zz

uu u u

r r r z

u uu

r r r

u u

r z

uu

r z

= = + =

= +

= +

= +

(9)

reduce to

, ,

0

r r zrr zz

r z rz

du u du

dr r dz

= = =

= = =

(10)

Placing the strain-displacement relations in Eq.(10) into the stress-strain relations of Eq.(8) gives

( )

2

2

1

1

r rrr

r r

zzz rr

E du u

dr r

E u du

r dr

du

dz E

= +

= +

= = +

(11)

so that when the stresses of Eq. (11) are substituted into the equilibrium equation (Eq.

(6)), we find

2

2 2

10r r r

d u du u

dr r dr r + = (12)

It is easy to solve this equation for the displacement since we can rewrite it equivalently

in the form

( )1

0rd d

rudr r dr

=

(13)


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Integrating once on r gives

( ) 1rd

ru C r dr

= (14)

where 1C is a constant of integration. Integrating Eq.(14) once more then yields

1 2 21

2r

C C Cu r C r

r r

= + = + (15)

where 1 1 2/ 2,C C C= are both constants of integration. If the displacement expression in

Eq. (15) is placed into the stress-strain relations of Eq. (11), one finds

21 2

21 2

1

1 1

1 1

2

1

rr

zz

E E CC

rE E C

Cr

C constant

=

+= +

+

= =

(16)

If we define two new constants, A, B as

1 2,1 1

E EA C B C

= =

+(17)

then we can rewrite Eq.(16) as

2

2

2

rr

zzz

BA

r

BA

r

du Aconstant

dz E

=

= +

= = =

(18)

At this point our solution for the stresses (Eq. (18)) and the displacement (Eq. (15))

satisfy equilibrium, compatibility, and the stress-strain relations and are given in terms oftwo unknown constants ( C1 and C2 or A and B). To find these constants we must satisfy

the boundary conditions (Eq. (4)) which , together with Eq.(18) yield


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2

2

i

i

e

e

BA p

r

BA p

r

=

=

(19)

whose solution is

( )2 22 22 2 2 2

,i e i ei i e e

e i e i

r r p pr p r pA B

r r r r

= =

(20)

which gives the stresses and displacements as:

( )

2 2 2 2

2 2 2 2 2 2

2 2 2 2

2 2 2 2 2 2

2 2 2 2

2 2 2 2

2 2

1 1

1 1

2 2

1

i i e e e irr

e i e i

i i e e e i

e i e i

i i e e i i e ezzz z

e i e i

i i er

p r r p r r

r r r r r r

p r r p r r

r r r r r r

r p r p r p r pduu z C

dz E r r E r r

r p r pu

E

=

= + +

= = = +

=

( ) ( )2 2

2 2 2 2

1 1e i i ee

e i e i

r r p pr

r r E r r r

+ +

(21)

where C is an arbitrary constant displacement (translation) in the z-direction.

The solution of Eq. (21) is an exact solution to our pressurized cylinder problem

since it satisfies all the governing equations and boundary conditions. One can show thatfor a pressurized cylinder subject to an internal pressure only, the highest stresses occur

on the inner wall where we have

2 2

2 2 2

2 2

2 2 2

1

1

i

i

i i err ir r

e i i

i i e

r re i i

p r rp

r r r

p r r

r r r

=

=

= =

= +

(22)

For a thin wall cylinder these stresses become


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( )

2 2 2 2 2

2 2 21

2 2

2

i

i

rr ir r

i i e i e i i e i

r re i i e i e i

i e i i m

p

p r r p r r p r r t

r r r t r r t r r

p r r p r

t t

=

=

=

+ += + = = + + +

+

=

(23)

in terms of the thickness e it r r= and the mean radius ( ) / 2m e ir r r= + . This agrees with

the result for the thin wall cylindrical pressure vessel hoop stress obtained in Strength ofMaterials (where the small radial stress component is normally ignored).

thick wall cylinder

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