Handout Notes for CylindersAA

8/13/2019 Handout Notes for CylindersAA

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Stress Analysis for Thick-walled Cylinders

Handout Notes for MEC53 Mechanics of Deformable BodiesFall 2011

1. Introduction

Applications of pressurized vessels/tanks with thick walls are popularly seen in many

industries. Examples include engine cylinders, boilers, nuclear reactor cores, petroleum

pipelines, and weapons such as gun barrels, etc. The previously learned thin-walled

cylinder problem gives simple stress solutions that, however, are not applicable to thick

walled containers. You may recall that in deriving the thin-wall tank theory, we assumed

uniform stress across the wall thickness. Yet as the thicker the wall is, the more the

assumption deviates from the reality. Up to certain thickness to diameter ratio, the thin-wall

based solutions are not valid any more. In general, to apply thin-wall theory to the

cylinders with mean-diameter to thickness ratio less than 10 can cause a relative error instress solution greater than 5%. The ratio of 10 is considered a threshold limit for the

thin-wall solution to stay valid. For pressurized cylinders with wall thicker than that, one

usually applies the exact solutions that are obtained based on the theory of linear-elasticity.

In mechanics of materials, closed-form solutions, such as those we have learned to treat

beams, shafts and other simple structure members, are developed with certain simplifications

and assumptions to avoid mathematic complexity. The analysis of thick-walled cylinders

follows an approach of elasticity, which gives the exact solutions. Elasticity is a primary

subject insolid mechanicand it treats a linear-elastically deformed structure problem in way

of solving a set of partial differential equations. Yet to solve these equations under certain

boundary conditions could be mathematically challenging. As the result, majority of the

practical problems have no such closed form analytical solutions available to use. Instead,

experimental or numerical approaches are needed for solving most practical problems. The

methods of strain gauges, photo-elasticity, finite-difference and finite-element are among

many such methods.

The topic discussed below serves a brief exposure to how the theory of elasticity is

followed to obtain a theoretical solution for a simple linear-elastic problem. Upon

mathematical derivation for a tank under static loading, we further extend the approach to

treat a rotating wheel/disk problem. Furthermore, since many such pressure vessels are

subject to high temperature and temperature variation, the concept of thermal stress and the

analysis of such stress in a pressure tank will be discussed. Finally, the shrink-fit

technology is introduced and the stresses in a shrink-fit made composite tank are analyzed.


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2. Analysis for equilibrium of a small element

The analysis as follows is valid for the cylinders that are axis-symmetric in both

geometry and loading. Along longitudinal axis of an un-capped cylinder, all cross sections

are subjected to identical load conditions so that a simple two-dimensional analysis is valid

for whole cylinder. Due to center symmetry, a polar coordinate system is used for the

convenience of the analysis, and the displacement of any point in the cross section isone-dimensional along the radial direction. Also, the tank has zero shear stress randthe

normal stresses r and are both functions of r, i.e., independent of.

With the above analysis, we now consider a differential unit element of the cross-section

as shown in Fig.1 (a) and (b). The force equilibrium along r-direction of the element is

considered and a relationship is established as follows:

0))(( drdrdddrrd

rrr or

0 drdrrdrr

.

This latter expression reduces to a D. E. (differential equation) as

rdr

drr

(1)

Fig1 (a) Fig1 (b)

The analysis of a thick walled cylinder therefore becomes simply finding the solution for that

differential equation (Eqn. (1)). Since both r and are unknown functions of r, the

solution of equation (1) is explored taking advantage of the inter-relation between the stress

components.

3. Analysis for geometrical relationship for the deformed element

Referring to Fig. 2, ABCD is an original element before it is deformed and ABCD is its

deformed counterpart as the cylinder is pressurized. The radial displacement uas shown in

the figure is resulted from the cylinder expansion upon being pressurized. Side AB of the

element moves to AB by an amount u, while Side CD has a displacement of duu .

r

d

r r

drr+dr


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Based on the definition for normal strains, strains along the radial direction and the

tangential strain are derived as follows:

BD

BDBBDDBD

BD

BDDBr

''''

or dr

du

dr

druduudrr

(2)

andr

u

rd

rddur

AB

ABBA

)('' (3)

4. Stress-strain relationshipHookes law

The 2D linear elastic stress-strain relationship (Hookes law) in polar terms is expressed as:

)(1

rr

E

)(1

r

E

.

Alternatively, the inversed strain-stress relationship is expressions as

E

E

r

r

r

)1

(

)1

(

2

2

in whichEand are elastic constants.

5. Differential Equation for radial displacement u(r)

Applying the strain-displacement relationship in Eqn. (2) and (3) into Eqn. (4) and (5),

the stress-displacement relationships are obtained to give r and expressed in the same

displacement term u(r)as follows:

)(1

)(

1

2

2

dr

du

r

uE

r

u

dr

duEr

Substitute (6) into the equilibrium equation (1) to obtain

dr

du

r

u

r

u

r

u

dr

du

rr

u

dr

du

rdr

udE

2222

2

2

1

10

(4)

(5)

(6)

dr

Br

d

C

AC

u

BD

D

u +du

A

Fig2


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By further reduction, it yields the following D.E. for the displacement uas a function of r:

01

22

2

r

u

dr

du

rdr

ud

6. General solution for the differential equation

Equation (7) is a field equation (namely the equation valid within the entire domain

specified by the tanks cross-section) for the differential equation problem. To solve it, Eqn.

(7) is rewritten into 0)()( r

u

dr

d

dr

du

dr

dand is then integrated twice with respect to r. The

operation yields the general displacement solution for the cylinder as

u= C1r+ C2/r (8)

where both C1and C2are constants to be determined by the B.C.(boundary condition) that the cylinder is

subjected to. Substituting Eqn. (8) into (6), the stress solution is obtained as follows

2

2

1

2

2

1

''

''

r

CC

r

CC

r

(9)

in which C1and C

2are constants, different from C1and C2but can also be determined simply by applying

B.C.of the problem.

7. Particular solution for cylinders with specific B.C.

For a problem as specified in Fig. 3, in which the internal pressure Pi, the external

pressure Po, the inner radius Ri and the outer radius Ro are all given, the B.C. are

mathematically specified asiir

PR )( and 00 )( PRr . Imposing theB.C. in Eqn. (9)

it comes to

2

21 /'' ii RCCP (10)

and2

0210 /'' RCCP (11)

Solving Eqn. (10) and (11) simultaneously, it obtains

)/()('

)/()('

22

0

22

002

22

0

2

00

2

1

iii

iii

RRRRPPC

RRRPRPC

(7)

Ri

Pi

P0

Fig. 3

R0


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Substituting C1 and C

2 back in Eqn.(9), the stress solution for a thick-walled cylinder is

obtainedas follows:

222

0

22

00

22

0

2

00

2

222

0

22

00

22

0

2

00

2

1)(

1)(

rRR

RRPP

RR

RPRP

rRR

RRPP

RR

RPRP

i

ii

i

ii

i

ii

i

ii

r

Recalling Eqn. (3) and applying the Hookes law to the strain term, it yields the displacement

uin terms of both stress components as:

)(r

E

rru

(14)

Substituting (12) and (13) into (14), uis solved as

222

0

2200

22

0

200

2 1)()1(1/

rRR

RRPP

RR

RPRPEru

i

ii

i

ii (15)

Eqn. (12), (13) and (15) constitute a set of particular solutions for an internally and

externally pressurized, axis-symmetric and thick-walled cylinder. Given the cylinders

inner and outer diameters and the materials elastic constants, the radial displacement and the

normal stress components at an arbitrary point in a cross section are but the functions of the

position r of the point (independent of ). With the equations, the stresses are readily

obtainable if the inner and external pressures are given. It is noted that both the tangential

displacement and the shear stress vanish owing to the axial symmetry of the problem.

8. Application examples

(a) High pressure vessels

In this case, the condition Pi >> P0holds. Considering P0 is close to zero, Eqn. (12)

and (13) can be simplified to

2

2

0

22

0

2

2

2

0

22

0

2

222

22

0

22

0

2

1

11

r

R

RR

RP

r

R

RR

RP

rRR

RRP

RR

RP

i

ii

i

ii

i

ii

i

ii

r

Noting that rRoor equivalently, 01 2

2

0 r

R, r0 can be proved from Eqn. (16). This

indicates that the radial stress is always compressive in an internally pressured cylinder.

(12)

(13)

(16)

(17)


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Also, since ,1)1(2

2

0 r

Rit can be proved that > 0 from Eqn. (17). Eqn (17) indicates

the normal stress in tangential direction (similar to the hoop stress in thin-walled cases) is

always tensile. In particular, at r = Ro (the outer surface), r = 0 and 220

22

i

ii

RR

RP

are

obtained. And for r=Ri(in the inner surface), 220

2

0

2)(

i

ii

ir

RR

RRPandP

.

(b) Containers under high hydrostatic pressure or vacuum containers

Contrarily to (a), in practical applications such as cylindrical tanks sit in seabed or high

vacuum containers, Po>> Pi so that Piis negligible. Eqn. (12) and (13) can be simplified

to become

2

2

22

0

2

00

222

0

22

00

22

0

2

00

2

2

22

0

2

00222

0

22

0022

0

2

00

11

11

r

R

RR

RP

rRR

RRP

RR

RP

r

R

RR

RP

rRR

RRP

RR

RP

i

ii

i

i

i

ii

i

i

r

Noting that r>Ri, the following expressions hold true

0,0)1(

0,0)1(

2

2

2

2

r

R

r

R

i

r

i

In particular, at r=Ri, r = 0, and 220

2

002

iRR

RP

.

And at r=R0,

22

0

22

00

2

0

2

22

0

2

00

02

0

2

22

0

2

00 )()1(,)1(i

ii

i

i

i

r

RR

RRP

R

R

RR

RPP

R

R

RR

RP

For pressure vessels, not only is always tensile, its magnitude is also greater than that

of r, which indicates that the stress along tangential direction is a main concern. For case

(b), both r and are compressive and is greater in magnitude. The buckling failure can

be an issue for vacuum chamber when the wall thickness is not very thick. The

schematics in Fig. 4 show the distribution patterns for rand across the wall thickness for

both case (a) and (b). Note that in both cases the highest tangential stress (in magnitude)

is always located in the inner surface.

(18)

(19)

(compressive)

(compressive)


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Case (a) Case (b)

Fig.4

P0

Pi=0

r

Pi

r

P0=0

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Handout Notes for CylindersAA