Thermal stress analysis of pressure vessels with cylindrical skirt ...

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1964

Thermal stress analysis of pressure vessels

with cylindrical skirt supports.

McManus, James P.

http://hdl.handle.net/10945/13284

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U. S. Naval Postgraduate School

Monte ...-, California

V7

THERMAL STRESS ANALYSIS OF PRESSURE VESSELS

WITH CYLINDRICAL SKIRT SUPPORTS

James P u McManns

THERMAL STRESS ANALYSIS OF PRESSURE VESSELS

WITH CYLINDRICAL SKIRT SUPPORTS

by

James P. McManus

Lieutenant , United States Naval Re -

Submitted in partial fulfillment ofthe requirements for the degree of

MASTER OF SCIENCEIN

MECHANICAL ENGINEERING

United States Naval Postgraduate SchMonterey, California

19 6 4

(vies A-fccmv£ Jg&

Library

W. S. Naval Pomtfirndnu** Sch«»f u u i

Monterey., ,„ fornia

^1!. - ^auuaf bch.ol

Monterey, California

THERMAL STRESS ANALYSIS OF PRESSURE VESSELS

WITH CYLINDRICAL SKIRT SUPPORTS

by

James Po McManus

This work is accepted as fulfilling

the thesis requirements for the degree of

MASTER OF SCIENCE

IN

MECHANICAL ENGINEERING

from the

United States Naval Postgraduate School

ABSTRACT

A method is described for determination of stresses

near the circumferential junction of shell , head, and skirt

of a pressure vessel subjected to mechanical and thermal

loadings » Considerable latitude is permitted for variation

in geometrical, mechanical, and loading characteristics;

however, it is presumed that there is complete symmetry

about the axis of the vessel. A digital computer program

is appended to facilitate solution of the problem..

ii

-

ACKNOWLEDGEMENT

The writer extends his thanks to Dr. John E u Brock

,

the faculty advisor for this thesis* Br. Brock 9 s helpful

advice and criticism have made working on the thesis a very

rewarding experience

„

iii

TABLE OF CONTENTS

Section Title Page

1 D Introduction 1

2. Theory 7

3

.

Method 16

4. Digital Computer Program 34

5. Conclusion 3$

6» Bibliography 40

7. Appendix A - Instructions for Use of Program AlSkirt

& e Appendix B - Listing of Program Skirt and BlLogical Flow Chart

9o Appendix C - Sample Problems CI

IV

1. Introduction

The chemical and petroleum industries have constructed

many pressure vessels with vertical axes and cylindrical

skirt supports,, Some of these vessels contain fluids at

temperatures as high as 1000°F with resultant high heat

flows through the skirt support .jjO* These heat flowssand

the thermal stresses resulting, give rise to the "Hot

Skirt Problem" which is here generalized to that of deter-

mining the stress distribution near the junction of shells

head, and supporting skirt of a vertical pressure vessels

resulting from any combination of the following loadings

;

lo Changes in temperature proceeding axially along

any of these elements (shell, head, skirt

)

e

2. Changes in temperature (assumed to vary linearly)

through the thickness of an element

„

3» Internal pressure in the shell and head but not

in the skirt

.

4. An additional axial loading on any or all of the

elements. (Could result from the dead weight of

one of the elements.)

5. A moment applied to the junction of the three

elements by agencies external to the elements.

6. A radially outward distributed loading applied to

the junction by agencies external to the elements.

^Numbers in brackets refer to bibliography,page 40.

Throughout it Is assumed that the three elements are

joined on a common circumference, that they and their load-

ings are axially symmetric , that the material of the

elements remains elastic , and that the structure always

remains stable and all deformations are T

' small". It is

also assumed that the walls of the elements are "thin"o

The theory of thin axisymmetric cylinders (Section 2)

shows that all except axial loadings applied a sufficient

distance from the end of the cylinder (i.eo, junction end),

influence that end very slightly • Axial loadings however

are transmitted without change . Also one may consider

such shell configurations as cones, ellipsoids, etc, as if

they were cylinders for purposes of calculating their

reactions close to the junction end due to influences at

that end, providing such shells are reasonably "deep" o^

Only a finite length of the shell need be considered since

any nonaxial loading has negligible effect upon the junction

end if it is remote enough . This nonaxial loading may

therefore be taken to be zero, which suggests a finite

length with the remote end free.

Thus the present solution Involves three finite

cylinders, joined together at their common junction ends-

To be as general as possible, this solution allows the

material properties of the shells to vary axially and even

*See footnote, page 127, Ref* 3 Part II u

the radius and thickness to vary "slowly" , The variance

of radius and thickness must be kept within reasonable

limits to assure the desired accuracy in the solutionss

since terms resulting from axial derivatives of these

quantities are not included,,

Each cylinder is dealt with in turn,, performing the

following sequence of calculations:

1. The moment, shear, slope (belling), and radial

deflection are found and recorded for each of

n (a preselected integer) points along the

disconnected finite cylinder, resulting from the

application of Loads 1 through L as previously

listed

o

2. The same quantities are calculated and recordeds

resulting from a unit moment (unit moment per unit

circumference) applied to the junction end of the

cylinder.

3. The same quantities are calculated and recorded,

resulting from a unit shear (unit radial load per

unit circumference) applied to the junction end

of the cylinder

o

Equations of statics may be written relating the end

moments and shears to the externally applied loads numbers

5 and 6. Using the junction end slopes and radial

deformations found in Calculations 19 2, and 3 9

equations

of slope and displacement continuity may be written

relating again the end moments and shears „ These relations

result in six linear equations with six unknowns (the

three end moments and three end shears) which may be solved

for the junction end moment and shear on each cylinder

«

When the end conditions are known for an element , the

deflection, moment , and shear along the length of the

element may be calculated using the results of calculations

1, 2, and 3, which have been retained „ The stresses are

then calculated using the deflection and moment

Two methods are presented for handling internal

pressure „ The first considers pressure loading at the

outset as an input into Calculation lo This is permissible

if the local geometry is essentially that of a cylinder

If, however, either the shell or head is not cylindrical

in shape at the junction, all loads but pressure are

included in Calculation 1. The effects of pressure are

computed separately for shell and head using membrane

theory. The resultant end deflections due to pressure are

included in the continuity equations and the membrane

pressure stresses are added to the stresses calculated from

all other loadings (including the bending stress caused

by dissimiliar deflections due to pressure

)

A finite difference iterative procedure is used to

solve the fourth order differential equation presented in

Section 3 and is the "heart" of the solution „ Essentially

an assumption of the radial deformation of the cylinder is

made and an integration is performed to find a new shape

„

To assist convergence, an averaging process is used but for

some shells convergence may still be difficult to attain

The original goal was to develop a method which

could be easily performed on a desk calculator in a

reasonable length of time Due to the difficulty in

attaining convergence this goal has only been partially met

If one had a problem with cylinders of constant parameters

so that the numerical solution need only be performed for

Calculation 1, with an analytical solution available for

Calculations 2 and 3, a hand solution might be feasible To

solve more complicated problems a digital computer program

has been written in Fortran language and is listed in

Appendix B„ This program allows one to consider shells and

loadings, the solutions for which would take a prohibitive

amount of time by hando

Three cylinders were assumed to be present in developing

the method and the computer program „ However if one wishes

to consider a vessel of but two abutting elements, one

may choose a very small length for the unwanted cylinder

,

thereby effectively eliminating its influence

„

Some previous work has been published on a hot skirt

problem. Weil and Murphy |l| have developed a solution

adapting the equations of the analytical solution of a beam

on an elastic foundation to a cylinder,, However these

equations require a constant "beam" or constant cross

section cylinder This restricts the problem to a constant

axial thermal gradient, and constant temperature difference

across the wall thickness <> These restrictions are severe

and limit greatly the usefulness of their solution

6

2. Theory-

Consider a small element cut from a cylinder with

dimensions dx, ad , and t in the x, y sand z directions a

The cylinder is of radius a, and has a wall thickness t u

Figure 1 illustrates the element,

V+dV

axis

Figure 1„ Element Cut from Cylinder

The following symbols will be used in this section:

a - radius of cylinder

D = Et*/l2{l-/i2 ) r longitudinal stiffness of a unit width of

element, (also called plate flexural rigidity )

d s infinitesimal central angle

E s Young T s Modulus

o

e„,v and e__ - centerline extensional strain in x and z

directions respectively due to all but thermal

causes.

ebx anc* ebz s bending strain on inner wall surface in x

and z directions respectively due to all

but thermal causes; and positive as indicated

in equations (5) and (6).

e v and e „ - extensional strain in x and z directionsx z

respectively,

Mx b moment (as illustrated) per unit circumference

„

Mz s moment (as illustrated) per unit axial length

„

P - ap/2 + P s axial force per unit circumference

P s axial force per unit circumference from all causes but

pressure

o

p = internal pressure

.

s ; radial deflection. (Positive outward .

)

t z wall thickness.

T - temperature at centerline of wall.

4 T - inner wall temperature minus outer wall temperature.

V - radial force per unit circumference. (Positive as shown.)

x, y, and z s cartesian coordinates as shown in Figure 1.

o< s coefficient of linear thermal expansion,,

V s curvature in x, y plane.

8

px s radius of curvature in x, y plane

,

(Tx and CTZ a axial and circumferential tensile stresses

respectively,,

jx - Poisson*s ratio

For axial equilibrium:

ft/2(1) P = J C^dy

For equilibrium in the y direction:

ft/2padSdx + ad6V - ade(V+dV) + di9dx i CT dy

'-t/2 z

This reduces to:

(2) E -P-T J-t/F**7 '

Taking moments about axis o=o:

dxad0(Mx + Vdx) * (pad^dx)™ = (M* + dMx )ad6 +

(dddx/^V.dy) fe^ ~t/2 2

which reduces to:

dMY(3)

dx

Also from elementary statics

ft/2 ft/2(4a, b) Mx « - / CTxydy and Mz - - i <Tydy

/-t/2 '-t/2 z

9

Assuming the longitudinal strain is linear through

the wall:

(5) ex - emx - ^ebx -*Z4T * XTt t

Assuming circumferential strain is linear through

the wall:

< 6 > ez * emz - &ehz -«ZAT! XTt t

These assumptions are reasonable if t^a c Otherwise

the distribution may be nonlinear Thus^ for example , the

stress distribution due to internal pressure in a thick

cylinder is hyperbolic (Lame/ Solution)

.

Figure 2 shows the cross section of a piece cut from

the cylinder wall of width ad£ and length dx Q The strains

are greatly magnified for illustrative purposes » From

similar sectors J

Xxdx s dx/p x 3- (ebx + <*AT/2)dx/(t/2)

.

10

Outside surface of wall

dxle^-HxT) (dx(ebx+<*4T/2))

Figure 2 Cross Section of Cylinder Wall

Presuming that M^<1, Xx ^r<L_s Thereforedx dx'

(7)^s = (2e +^T)/t.dx' bx

(£) e z= s/(a+y)^rs/a (Here again we assume t^aj

Comparing (6) and (£), and equating coefficients of like

powers of y:

(9) ebz = --cAT/2

(10) emz - s/a -ocT„

The fundamental stress-strain relations are:

11

(id <rx- [E/(i-/t

2)J [(vAsi - f^bx^bzU

(12) (TzS [E/d-^jJ [(e^t/te^) - 2Z(ebzVy ebx )J

Therefore:

't/2

U3) PSA/I Cr*dT " ^t/(l^

2J [(e^/^e^)]

(U) g = P " a Xt/ir«dy = P" Et/a)/(1^2

iI[(emz^ en>xi]

(15) Mx = -J^f^ = [Et2/6(l-^2 !] O'bx^/^ta'J

(16) Mz

='J^.Jfm**'

' [^Ml-A^lJ f< ebz+/«e

bx )7

Combining (7), (9), and (15), we get:

(17) S2« £ 4(1^ T

dx2 D ^

Eliminating e^ from (13) and (14)°

mai dv - n *p Et6mz

Substituting (10):

dV /<P . Ettt T Ets(19) cE a p " a"

+ ~~a~ " ^2~ °

From (15) and (9):

(20) ebx =6(1°/^ )Mx + 4J*A1

,

Et2 2

12

and from (13) and (10)

(21) e B ^M2) P " M *^"T^ Et

Summarizing:

Ets12

(22b) ^k s Vdx

(22c)d̂x 2

If it is assumed that T=P= AT-09and that a, E 5/# ?

and

t are constants, equations (22) reduce to?

(23) D^-3-^s -Ets

dx^ a2

The solution to (23) assuming a semi-infinite cylinder.

is:

(24) s a o V-j^cos/Jx +/5Mxl (cos^x - sin^x)2|d u

where V^ and M^ are the end loadings , and /S s

y 4a 2D

d 2sMx then equals D— ,

dx2

After M^. and s have been determined either by numerical

solution of (22) or by using (24), the stresses in the

13

and from (13) and (10):

(21) emv a <M2) P " M *^°mx Et a

Summarizing:

(22a) cBT ' P + —a— a a2

(22b) ^x = Vdx

(22c) ^ s ^ + UL^fl^I .

dx2 D t

If it is assumed that T^P- 4T-09and that a

sE 9/# s

and

t are constants, equations (22) reduce to:

(23) D^-3-^s -Ets

dx^ a2

The solution to (23) assuming a semi-infinite cylinder.

is:

/?x

(24) s s6

3 Vjcos/sx f/fiM^cos^x sin^x)

where V^ and M^ are the end loadings , and fi Et

4a 2D

^2„Mx then equals D

,d2s

dx2

After M^ and s have been determined either by numerical

solution of (22) or by using (24)> the stresses in the

13

cylinder may be determined with equations (9) , (10) , (20)

,

(21), (11), and (12) . The results are:

or . p 6Mxv x inner = — + —^

(25) 6MC"x

pouter =

tX2

t

°~z

Esinner = -r E*T +/^r

crzEs

outer = ~ MPE"T + —

- E*4T2

+6/Mx

t2

+E*JT

2—

6/MX2

(26)

A single equivalent stress is frequently used as a

basis of assessing the severity of loading on a structure

in which the state of stress is not a simple uniaxial

stress. The equivalent stress employed here is- that of

the distortion energy theory (equivalently, of the octa-

hedral shearing stress theory) . If the principal stresse;

are (T~ , (T and zero, as they are at the outside of thex z

cylinders, the equivalent stress is:

<rA \n2

(TCP , G2

v d = V x - vx v z + v zd = V x - vx z + z

On the interior surface, if a pressure p is acting,

third principal stress is =p rather than zero However,

the same formula is used since (T and (f are at least onex z

or two orders of magnitude greater than p c Thus:

14

(27) o".«.

= \(cP"

- (trcr)-

" 7^"d outer \ x outer x z outer z

and:

(2d) CT. . = VE3

. - (T(D. +(T2

d inner ^ x inner x z inner z

15

3 . Method

Figure 3 shows the basic three-cylinder problem that

is to be solved, with the numbering convention used, and

the sign conventions for the externally applied moment,

shear force, radial deflection, and slope when the cylinders

are considered jointly.

Positive radial deflection for all cylinders

Positive slope for all cylinders

Cylinder 1

Cylinder 2

+x _axis.

* V" Positive shear for all/^ oo , . .

cylinders

o

M Positive moment for allcylinders o

Cylinder 3

ijQ axi a _^hx

Figure 3. Positions of Three Cylindersand Sign Conventions

16

The basic procedure to be followed in the solution is

to solve equations (22) numerically or (23) analytically

to find the reaction of each cylinder to (1) the given

loadings, (2) a unit junction end moment*, and (3) a unit

junction end shear* when this cylinder is "free" (io6c,

disconnected from the junction) „ The three cylinders are

then conceived to be rejoined, and by using the above in-

formation along with the equations of statics and continuity.,

the reactions (i.e., the actual end loadings) of the three

cylinders are determined

»

The analytic solution of equation (23) is given in

equation (24) . However in the numerical solution of (22)

we choose a finite length for the cylinder . The only load

applied to the remote end is a moment equal to that caused

by the radial thermal gradient. From equation (24)

the radial deflection at x = $//3 is about 0.7% of the

radial deflection at x = for any given loadings applied

at x = Oo Thus, if L, the length of the finite cylinder,

is set equal to 5/& there is only a small error in the

radial deflection at the junction end due to neglecting

the loading applied at x = 5/^6 by the portion of the cylinder

^Throughout this thesis "shear" means the radiallyoutward directed load per unit circumferential length and"moment" means moment per unit circumferential (or axial)length.

17

beyond this point . However an error results if there are

nonlinearities close to the remote endo See "Discussion

of Results," Appendix C»

12 I3_

i

station numbers

\L 15 y, i n-4,1 n-3 b=g I

cylinder wall

axis

Mxn

Figure 4, Finite Cylinder Divided into n-1Increments Showing Loads at Junctionand Remote EndSo

The following iterative procedure was developed by

John E Brock, Professor of Mechanical Engineering, Uo So

Naval Postgraduate School, Monterey, Calif „ , and modified

by the writer to assist convergence

„

Figure 4 shows a cylinder divided into n-1 equal

increments with stations 1 through n, (With a modification

of the integration method used, the requirement of equal

increments could be withdrawn*,) At each station, values

IS

of all the parameters of the cylinder and all loadings are

assumed known

„

Equations (22) may be restated as follows:

dV EtAT jU? Ets+ —T-- /C

T" -—V

(29)

dx P a a I^

dxdx

M = (Vdxx ^

2 M

ds fd2 s

,

Let M* = M + D(H/0«4TX t

(30) m* - /Vdx + D(1+/0*4T' t

(3D d2 s _ M*. 2 Ddx

Let s = A + Bx + f (x) where A and B are constants and

f(x) is normalized in such a way that f(0) - f (L) - o

Also let Q = M. o Then from (29), (30), and (31):dx

Q = p + Et£T _£P . Et rA + Bx + f(x)]3. a. ~.£~ •

(32) = Q - AS| - BEt| . Etf(x)

a a a

19

where Q = p + 1^1 _/^a a

V = JQdx = Vj^ + f Qdx - a/ I| dx - BM^xdx-J

l|f(x)dx

Let % = ka

(33) V = V - kf kdx - B/kxdx - / kf(x)dx

where V = V, + / Qdx1 so

Next

M* = M +J Vdx - A^kdxdx - B^Jkxdxdx -C fkf(x)6x6x

so that

D(H*)«4T+ t

'A/A /J(/X • X/'A

(34) M* = M* - kjofkdxdx - Bj£^kxdxdx -^ ^kf (x)dxdx

where M* = M , + ( Vdx + D(lyfo<4Txl /» t

An initial assumption is made for f (x) ; conveniently,

one may take f (x)=0 e Therefore the only unknowns in

equations (33) and (34) are A and B»

It is also assumed that at x = L (the remote end of the

cylinder) , M* = V = o This is a reasonable assumption

because as x approaches infinity, with a constant radial

thermal gradient, the axial curvature of the cylinder wall

approaches zero<> L has been chosen long enough so that the

conditions at x - L approach those at infinity which allows

M* = V = Oo The two equations and two unknowns are:n n n

20

(35) - Vn

- AI kdx - B^ kxdx - jQ kf(x)dx

(36) - Mjj - A^> X kdxdx - B ^ ^ kxdxdx - ^ ^kf(x)dx

Solving for A and B;

[vn - /kf ( x) dx] [^jkxdxdxj - [/kxdx] fe~jfikf ( x) dxdx.

{^kdxJtC^kxdxdxJ -JZkxdxJft^kdxdxJ

Kkdx][j£-i^V(x)dxdx]-[v^^

(37) A

(36) B =DHK kxdxdxJ -ft

kxdxjp^W

Substituting these values into gives

(3D d2 s _ M*

dx

Let

D

(39) g(x) =[*<£*

'* dx

(40) f-^x) =^g(x)dx

dx =j ^tt dx

It will be noted that g(x) / ~ and f-i (x) / s becausedx *

the initial conditions have not been included „ We normalize

f-, (x) to obtain

(4D f (x) = f_(x) - fl^ L)x2 1 L

and it may be seen that f?(0) = f

?(L) = 0,

If f2 ( x ) is used for *"(*) directly in equation (32) to

perform the next iteration, the process usually diverges

rapidly» To force convergence, the following averaging

21

procedure is usecL

1. Let numbers m and C . .chosen as described later,mm*

be specifiedo

2 Let d be the value of x for which f2(x) is maximal

„

3o Determine a number C according to the rule:

If f(d) - 0, C - Cmin ;

If |f(d)/f2(d)| >1, C = Cm

.

n ;

If 0<|f(d)/f2(d)|<l, C equals the larger

of Cmin

or |f(d)/f2(d)| m

.

4« Obtain a new function f«(x) by the formula:

(42) f3(x) = Cf

2(x) + (l-C)f(x)o

fo(x) is the function to be substituted into the equation

(32) for f(x) in the next iteration The process has

converged when f2( x ) ^^f (x) •

The fraction lltjjL I is used solely to reduce thefTTdTl

number of iterations required for convergence „ It may be

seen that when convergence has almost been attained, the

fraction C will become progressively larger resulting in

a higher percentage of fp(x) being used in the next

iteration^,

The number m must be kept to a minimum and Cs

kept

to a naximum to reduce the number of iterations to a

minimum o It was found by trial and error that a value of

22

$ or 9 for m and „06 to o 0$ for C . generally insured

convergence „ One problem required an m of 15 for reasons

unknown to the writer „ Generally if divergence is exper-

ienced, increasing m and decreasing Cm^n seems to eliminate

the divergence

o

When convergence has been attained , M , s, and &| are

calculated as follows:

(43) mx = M*- D(1+^T

dq fi(L)(44) 45 = B + g(x)

dx " &x ~' L

(45) s = A +Bx + f2(x)

Using the sign conventions established in Figure 3 and

with the understanding that "loadings" stand for the pressure,

thermal loadings, and axial loading in each cylinder, let;

£ ,,1 o'^al ~ Ju110^ 011 end deflections of the disconnected

Cylinders 1,2, and 3 respectively due to their

loadings

.

Oiid? ' 3=

Juncti°n enc* deflections of Cylinders 1,2, and

3 respectively caused by joining the cylinders

at the junction,. The total deflection of

Cylinder 1 would then bep t.+ 1 °

<P , & 2 , 4> o = junction end slopes (subscripts have the same

1 » 2 * 3 meanings as those for o )

.

^ol'^o2 ,Mo3

=Juncti°n en(* moments of Cylinders 1,2, and 3

respectively,,

23

V 2>V 2}V -d junction end shear loadings of Cylinders i,2s

and 3 respectively u

Y0O and M00 = externally applied shear and moment at

junction

•

Ay.^ = slope at junction end of Cylinder 1 produced

by application of unit shear to junction end.

A £ = deflection at junction end of Cylinder 1

produced by application of unit shear to

junction end

A . = slope at junction end of Cylinder 1 produced

by application of unit moment to junction end

A ~ = deflection at junction end of Cylinder 1

produced by application of unit moment to

junction end.

B and C denote similar coefficients for Cylinders 2 and 3

respectively, with the same subscripts and definitions as

for A.

Note that <pol> d>o2

, Av0 , A^, B^, B

m<p, Cm ^

9and Cm $

are the negatives of the end conditions calculated by the

above iterative process, or analytical process , due to the

difference in sign conventions between Figures 3 and 4°

Figure 5 shows Cylinders 1,2, and 3, situated as in Figure

4 but with sign conventions as in Figure 3 S and as used in

these equations

o

24

4.V, Positive Shear Load

M-, Positive Moment

3t

CYLINDERS 1 AND 2

PositiveSlope

V

PositiveDeflection axi;

V, Positive Shear Load/^ 1

"M-, Positive Moment

CYLINDER 3

PositiveSlope

-iPositiveDeflection axis

Figure 5„ View of Cylinders 1,2, and 3, ShowingSign Conventions Used in Equations (46),(47), and (4$)

»

25

(46)

From statics, compatibility, and other considerations

ol o2 03 oo

ol o2 03 oo

*1 +*ol

= S2

+4)2

*1 +4l

=*3

+^o3

1*1+

*ol " ^2 +*o2

*1+

'ol " ^3+

*o3

Vol\*> + MolAm4 =AV .A^ + M ,A . -£01 vg ol m£ 1

02 v# o2 m<i> 2

V ?B

,

r, + M ~Bmf -'L

<c v© o<c m£ x

Vo3

Cv*

+ Mo3

Cm* =*3

Wv* + Mo3

Cmi " S

Solution of equations (46) yield:

p2Qi - P1Q2Mo2 - N

XP2

- PA= NXQ2 =

N2QX03 N

XP2

I N2PX

(47)M = M - M - M01 oo o2 o3

vo3

= (Vi^da - jxmo2

-kiMo3 )

Vo2

= ^/D1)(HrE

1Vo3-F^ - G^

ol oo o2 o3

where,

26

US)

Dl=

Do -

D, -

D, -

E-

Eo =

E, = A

Av* + Bv<*

K4>

\s + BvS

\sAy

*

Av*+ cY d>

E, = A 'vi

F-, -m 9 m a>

F2 = Am*

F3 = Am6

+ Bm6

F, = A

G, = A

m6

Go =

G, =

G, =

mo> m<I>

Vj

Amr + CmC

H. = d>ol

ol

ol

I, =

I« =

*3

ol

D2E1

D3E2

D4E3

c£ + V A + M A02 00 v* 00 m4>

£> + V A + M A ^03 00 v# 00 md>

S+VA + M A02 00 vS 00 raS

$ + V A + M A03 00 v£ 00 m $

D1E2

D2E3

D3E4

27

(43)

Jl

— D F2 1

- D F12J2

= D F3 2

- D F2*3

J3

= D F„43" D

3F4

Kl

= D G2 1

" D1G2

K2

S3 D G3 2

~ D2G3

S= D

4G3

" D3°4

Ll

= D2H1

~ D1H2

L2= D

3H2

" D2H3

L3

= D4H3

- D H.j 4

h = Vl - J1J2

N2= J

3J2 - V3

Pl

= J2K1 " I

1K2

P2

= I3K2

- I2K3

«1= J

2L1 " I

1L2

Q2

=s J3L2-I

2L3

Let ]VL , NL and M, represent the axial moments in

Cylinders 1,2, and 3 respectively. Mu> ML2>

and My

represent the axial moments in Cylinders 1,2, and 3 caused

by the thermal, pressure and axial loadings . M ,, M «,

and M ~ represent the axial moments in Cylinders 1,2, and

3 respectively caused by application of a unit shear load

to their junction ends. MLi> ^W?' an(* ^ml represent the

axial moments in Cylinders 1,2, and 3 respectively caused

by application of a unit moment to their junction ends

23

The subscripts for V and s have comparable meanings

„

Then:

Mlx - ML1 + Vol*Vl + MoAlK2x - ML2 + V 2*V2

+ M^K^

M3x = ML3 + Vo3Mv3 " Mo3^3Vlx = VL1

+ VolVvl+ MolVml

(49) V2x - VL2+ Vo2Vv2

+ Mo2Vm2

V3x - VL3+ Vo3 Vv3 - M

o3Vm3

slx= SL1

+ Vol svl+ Molsml

L22x= SL2

+ Vo2 sv2

+ Mo2 sm2

3x " SL3+ V

o3 Sv3Mo3

sm3

Substituting the results of (49) into (25), (26\ (27), and

(2#) gives the desired stresses along the eyiinderso

In the above development it was assumed that the head

and shell (Cylinders 2 and 3) were in fact cylinders and

that their pressure stresses could be computed accordingly

Call this method Option 1 If one or both are not cylinders

but some other shell of revolution (ellipsoid, cone, etc ) ,

an error will result in the stress computations by consid-

ering them cylinders,, (See problems 10 and 20, Appendix Cj

One way to reduce this error is to use the above methods

to obtain the reactions resulting from all loadings but

pressure, and then superpose the separately calculated

reactions due to pressure <> Call this method Option 2

29

Skirt

Shell

AXIS

Figure 6. Shell Head, and Skirt Showing Rl and R2o

Consider the shell and head shown in Figure 6 In

this particular case the head is Cylinder 2 and the vessel

shell is Cylinder 3 but the following development will

allow a noncylindrical element for either the shell, head s

or both. The vessel shell is a cylinder, the head is some

other shell of revolution » R is the radius of the vessel

at the junction, R-, is the radius of curvature of the

meridian of the head at the junction, and R?

is the radius

of curvature of the line in the head perpendicular to the

Rmeridian at the junction,, From Figure 6, % = CosO, where

is formed by the axis and the tangent to the meridian at

the junction,.

30

From membrane theory: [3]

(50) °\+

02 = £Rl

R2

t

where CT and CT are the meridional and circumferential (or

hoop) stresses respectively, p is the inner gage pressure,

and t is the wall thickness » At the junction the axial

component of V^ must equal ££ to satisfy static

s

There-

fore:

^ P' 1 2tCos0 2tR 2t :

From ( 50 ) :

Let £ ! be the radial deflection due to pressure*

(52) £' » fR rITb . P_„^2lo LE 2j U t 2tR 2tJ

There is also an inward radial force at the junction,

V* due to the component of CC perpendicular to the axis*

(53) V£Q=-0^tSina= -P^ -

*l2

The suggested procedure for computing the effects of

pressure in Option 2 is first to calculate the inputs to

31

equations (48) setting internal pressure equal to zero but

considering all other loadings and external loadings. £ 9

o2

and <C* are then calculated and added to £ ^ and S *° V 9

w o3 o2 o3 00

is calculated as the sum of the contributions of both

Cylinders 2 and 3 and is added to V a Using equations

(54)

(49) s (25) » and (26) , the longitudinal and circum-

ferential stresses are computed,, However 5, it will be noted

that these stresses contain only those pressure stresses

caused by the dissimilar deflections and consequent bending

of Cylinders 2 and 3° The direct membrane pressure stresses

must still be added as follows

:

G~x outer " °x outer (^om (25)) + ^1

(Tx inner = °lc inner ^ from (25) >+ °"l

^z outer " °z outer (from (26)) +0"

2

C*z inner r °z inner Cfrom (26))+Q~

2

These values may then be substituted into (2?) and (28) to

obtain the equivalent stresses.-

Note that with Option 1 the effects of internal press-

ure are calculated in the numerical solution of the differ-

ential equations and therefore any variations in the

parameters of either cylinder will be taken into account

These variations will cause moments and shears in the

cylinder itself before consideration is given to its junction

32

with the others o With Option 2, the only moments and

shears due to pressure that are calculated are those caused

by the dissimilar deflections at the junction end of each

cylinder o Either option treats all three shells as cylinders

when calculating the effects of other than pressure loadings

„

For this reason either option will err in treating axial

force in a shell with other than zero meridianal curvature

since, among other things, the cylindrical equations do not

consider the radial component of CTx Q

33

4o Digital Computer Program

At first the goal of this thesis was to develop a

method in which a complicated hot skirt problem could be

solved on a desk calculator in no more than a dayo This

goal was not met due to the number of iterations necessary

for convergence

o

To make the method more useful , a digital computer

program which closely parallels the method outlined in the

previous section was written in the CDC 1604 Fortran 60

language o There are some small deviations in sequence of

operations mainly due to the fact that the program progress=

ively grew as it became more apparent that a manual solution

was becoming impractical for any problem in which the

parameters of the cylinders varied axially A standard

library program is used to solve equations (46)0

The program also provides for longitudinal slots in a

portion of a cylinder (usually the skirt) „ Cheng and Weil

[2J have indicated that slotting the skirt at the junction

substantially reduces the stresses in the cylinderso

The slots are considered narrow enough so that the

loss of metal in them may be neglected » The slotted section

is considered to be a series of separate cantilever plates

with no interaction between adjacent plates and ? conse-

quently » no CT~ along their slotted lengths, where (T is

34

the stress at the midplane in the z direction,. There is

of course some Gl_ at the junction end of a slotted skirt

due to the strain of the junction itself and therefore

this is not an exact representation of the actual situation,,

However provision is made in the program to start and end

the slot at any distance from the junction one might choose

,

and a way might be found to simulate the junction conditions

by starting the slot at some predictable distance from the

junction This slotted shell capability was provided

mainly so that others may have a useful tool for system-

atically investigating the effects of such slots

The following alterations are made to the equations

in the program at those stations where the cylinder is

slotted:

L Q of equation (32) is set equal to zero since all

terms except p are functions of interactions between

adjacent longitudinal segments of the cylindero If

the cylinder were slotted it could carry no internal

pressure

o

2o For the same reasons all the terms except the

6Mx^/^2 term are set equal to zero in equations

(26).

The program uses a midordinate integration scheme

which, as programmed, wastes about half as much memory space

35

in the computor as it employs . It was originally used

because it is systematic and very convenient for hand

calculations. If limitations should thereby be imposed

upon the utility of the program, more compact integrating

methods could be used.

The program was written so that only the minimum

number of input cards necessary to describe the problem

is required • For example if only the first and last

stations of a cylinder are described, the program uses the

cylinder parameters of the first station for the remaining

intermediate stations and makes a linear interpolation of

the thermal and axial load inputs. The internal pressure

is assumed constant. See Appendix A for a detailed

explanation.

Three output options are available to the user:

1. Only the maximum equivalent stress (based on the

distortion energy theory) and its location is given

for each cylinder.

2. The inner and outer longitudinal, circumferential

,

and equivalent stresses are given for each station

of each cylinder.

3 . The inner and outer longitudinal and circumferential

stresses and the moment and shear are given for

each station of each cylinder.

36

With each option ail of the input data is repeated, and a

short explanation of the output is printed «,

The program takes about one minute to compile and

each problem requires from one half minute to one minute

It took about seven minutes to compile and solve the

eleven problems in Appendix Co The program uses only the

normal Fortran functions and language as described by

McCrackenOy and therefore should be easily adaptable to

any large digital computer having a Fortran system

J 7

5o Conclusion

Even though the method was developed specifically to

solve the "Hot Skirt" problem, the reader ie reminded that

with suitable manipulations of the input, many other

problems may be solved

„

The only restrictions are the following:;

1 Elastic behavior

o

2o Axial symmetry

„

3" Maximum of three shells arranged as in Figure 3°

4o Constant internal pressure in Cylinders 2 and 3 onlyo

5o "Thin" wallo

6o Linear temperature variation in radial direction

7o Configurations of shells such that their reactions

to an end load may be approximated by regarding

them as cylinders of the same radius, wall thickness

,

and material o (Thus, treating a very shallow

elliptical head as such probably would result in

appreciable error „)

Some useful extensions of the development would be:

1c, A complete revision of the theory so as to include

a radius of curvature in the xz plane u This would

eliminate the restrictions imposed by the use of

the cylindrical equations <>

2o An alteration to allow inelastic behavior „ This

33

would probably take the form of a series of

repetive stress calculations, increasing the

loading until yielding begins, and then a different

set of calculations until the full loading is

applied

.

3. A modification to accommodate "thick" shells such

as are found in reactor pressure vessels

.

4° A modification to allow radially variable material

parameters such as would be found in a reactor

pressure vessel with internal cladding*

5- Modification to include the moments due to axial

forces acting through offsets in radii or through

radial deflections

.

39

BIBLIOGRAPHY

Numbers 1 through 5 cited in text or Appendices

lo Weil, NoAo and Murphy, J D Jo, "Design and Analysis ofWelded Pressure-Vessel Skirt Supports/' Journal ofEngineering for Industry , Volo #2, No u 1, February I960,

2o Cheng, DoHo and Weil, NoAc, "The Junction Problem ofSolid-Slotted Cylindrical Shells/ 1 Journal of AppliedMechanics , Volo 27, No, 2, June 1960 o

3o Timoshenko, S«, Strength of Materials > Parts I and II 9

Do Van Nostrand Company, Inc , Princeton, New Jersey,March 1956 Q

4o McCracken, DoDo, A Guide to Fortran Programming , JohnWiley & Sons, Inc., New York , NoYo , 1961

o

5o Bergman, D J o, "Temperature Gradients in Skirt Supports

of Hot Vessels/ 7 ASME - Paper 62-PET-41 for meetingSeptember 23-26, 1962

6o Bijlaard, PoPo, and Dohrmann, R„J,"Thermal-Stress

Analysis of Irregular Shapes," Journal of Engineeringfor Industry , Volo 33 , Noo 4, November 1961

7c Weil, NoAo, and Rapasky, F Q So, "Experience with Vesselsof Delayed-Coking Units," Proceedings of API, Divisionof Refining , Volo 38, No„ III, 1958V

40

APPENDIX A

Instructions for use of Program Skirt

These pages are intended to serve both as Appendix A

of the thesis

"Thermal Stress Analysis of Pressure Vessels

with Cylindrical Skirt Supports"

by James P. McManus, May 1964

,

and as an independent document describing the employment

of Program Skirt to solve certain stress-analytical

problems for pressure vessels The theory upon which

Program Skirt is based is developed in detail in the above

mentioned thesis. Those interested in studying the larger

document should make inquiry from Head Librarian: \J So

Naval Postgraduate School, Monterey, California

„

Given three axisymmetric shells joined together and

numbered as shown in Figure (Al) c These shells may be

subjected to the following axisymmetric loads:

lo Changes in temperature proceeding axially along any

or ail of the shells

.

2o Changes in temperature (assumed to vary linearily)

through the thickness of a shell

3° Internal pressure in Shells 2 and 3 but not in Shell

lo

4° An additional axial loading on any or all of the

Al

shells o (Could be the dead weight of one of the

shells and its eontentSo)

5. A moment applied to the junction of the three shellj

by agencies external to the shells

o

60 A radially outward distributed loading applied to

the junction by agencies external to the shells

„

*f Positive radial deflection for all cylinders

Positive slope for all cylinders

Cylinder 1kv~„ Positive shear for allA OO

e:

M Positive moment forall cylinders

o

+x

Cylinder 2

_axis_

Cylinder 3

aio axis _±x

Figure Al, Positions of Three Cylinders andSign Conventions

Program Skirt solves the problem using simple numerical

techniques o The user must choose a finite length for each

shell such that loads applied at the remote end of the shell

have no appreciable effect on the junction end, except that

A2

axial loads are fully transmitted. Suggested values for

the lengths are given in the remarks on Card 2o Each

finite shell is then divided into n~l equal increments with

the dividing stations numbered from 1 at the junction to n

at the remote endo Data for each station such as shell

parameters and loadings is either read into the computers,

or created by the computer from data given about adjacent

stationso

Two options are available for computing the effects of

internal pressure Either option may be used if the shells

are uniform and are cylindrical or have small meridional

curvature near the junction „ Option 1 is to be preferred

if the shells are cylindrical or have small meridional

curvature, but there are variations in thickness.* material

properties, etc Option 2 is to be preferred if there are

no variations of thickness or material properties but the

meridional curvature at the junction is not small „ Neither

option is capable of obtaining accurate results if the

shell is not relatively deep c

The following assumptions have been made in the program

L The three shells act as elastic cylinders „ The

shells may have "slowly variable Tr radii, thickness,

and material properties, but are still treated as

cylinders from station to station

A3

2c Temperature varies linearly through the wall thick-

116 bb b

3* Shear and curvature are negligible at the remote

endo If there are nonlinearities near the remote

end such as variation in the shell parameters , non-

linear axial thermal gradient , or a changing radial

gradient or axial load, the values of stress

obtained near the remote end will be inaccurate

*

Description of Input Cards

The input consists of all the parameters necessary to

describe the problem plus some control parameters to direct

the method of solution and the format of the output * In

the descriptions below the items given from left to right

are: Column Numbers, Format, Parameter , Remarks* When a

parameter has an TT I T? Format Field, its last digit must be

in the last column of its field*

Card Type I (One card only)

1-10 110 Problem number May be 1 through 999* If

it is zero or blank, the

program stops*

11-20 110 Output option 1, 2, or 3

number 1 Only maximum equivalent

energy stresses* and loca-

^Equivalent on basis of distortion energy theory*

A4

21-30 FlOoO Externally

applied moment

to junction

31=40 FlOoO Externally

applied shear

to junction

41=50 FlOoO Pressure

51-60 110 Code for hand-

ling pressure

tion given for each

cylinder

o

2 Longitudinal s circumfer-

ential and equivalent

stresses given for each

station of each cylinder,

3 Longitudinal and circum-

ferential stresses, and

moment and shear given

for each station of each

Positive if causing com-

pression on outer surface

of Cylinders 1 and 2* (Mo=

ment per unit circumference )

Positive if directed radially

outward o (Shear load per

unit circumferenceo

)

Gage pressure in Cylinders

2 and 3 °

If 1, Option 1 is usedo (See

above o) If 2, Option 2 is

usedo

A5

61-70 F10.0 Minimum

fraction

1- 5 15

6-10 15

11-15 15

16-20 15

The minimum value of a frac-

tion C . used in themmaveraging routine for assis-

ting convergence o Recommend

from .06 to .08, If left

blank, .06 substituted.

Values less than .06 might

be necessary if divergence

is experienced.

Card Type 2 (Three cards required)

Cylinder number 1 , 2 , or 3- Must be in

sequence.

Maximum of 100,Number of

increments

Interval of

printout for

input data

Exponent for

forcing

convergence

Example: If 2, every other

station reported; if 5, every

fifth station reported, etc.

Suggest 8 to 12 „ If diver-

gence is experienced, use a

higher value. The higher

the exponent , the greater

the computer time required

11 substituted if left blank

„

A6

21=30 FlOoO Minimum x

31-40 F10.0 Maximum x

41^50 FlOoO Convergence

criterion

51-60 F1CL0 Value of x at

beginning of

slots

61=70 F10.0 Value of x at

end of slots

1= 5 15

Card Type 3

Station number

See below u * x at junction

endc

See beloWo* x at remote

endo

Suggested values from 01 to

o05o This is an approximate

indication of the relative

error in the re suit s°

Smaller values require

longer computer time

For use with longitudinally-

slotted sylinderso

For use with longitudinally

slotted cylinderso

(Two or more cards foreach cylinder)

1 through n, where n is the

number of increments plus one

*The program allows x, the axial dimension in inches,to be other than zero at the junction if the user so desiresIt is recommended that the length of the cylinder be approx-imately six times the square root of the radius times thewall thickness

s or 6Vato

A?

6-20 Fl5°0 Young *s Modulus

at this station

21-35 F15o0 Mean radius of

cylinder

at this station

36-50 F15o0 Poisson's ratio at

this station

51-65 F15.0 Wall thickness at

this station

66-30 F15o0 Thermal coefficient

of expansion at

this station

Note: There must be Cards 3 for the first and last stations

.

The first station is number 1 and the last is n c If

there is no card for an intermediate station, the

values for the previous station are substituted

o

Therefore to change a value, simply feed in a card

for the appropriate station c The cards submitted

must be in numerical sequence by station number „ The

minimum number of Cards 3 for a cylinder is two, the

maximum number is n Q

Card Type 4 (Two or more cards foreach cylinder)

1-10 110 Station number

AS

11-20 FICO Mean temperature

at this station

21-30 FlOoO Radial tempera-

ture difference

31-40 FlOoO Axial load

Temperature at inside wall

minus temperature at outside

wallo

Total axial load imposed

upon cylinder due to all

causes except pressure

«

Positive if causing tension,,

Note that this is not load

per unit circumference , but

total loado

Note: There must be Cards 4 for the first and last stations „

If no card is submitted for a series of stations , a

linear interpolation of the temperatures and load is

made between stations that have cards . The cards

must be in sequence . Same minimum and maximum number

of cards as with Card 3

«

Card Type 5 (No cards or two cards)(Used only when Option 2 is used for pressure stresses,

1-10 F10 o Radius of meridian at junction „ (R-, in Figure

A2J Infinity substituted if blank

11-20 FlOoO Radius of circumference at junction divided by

the cosine of the angle between the tangent to

A9

the meridian and the axis (R2 in Figure A2j

Radius of junction substituted if blank.

Note: There must be Cards 5 after the last Card 4 for

Cylinders 2 and 3 whenever the pressure Option 2 is

usedo

- ___ J3kirt

Shell

Rl AXIS

Figure A2„ Shelly Head, and Skirt o ShowingR, and R?c

Following is an example of the sequence of data cards

for the input of one problem to the program

„

A10

Pressure stresscalculated byOption 1

Pre,cal<Opt:

ssure stress:ulated byIon 2

Card 1 Card 12

3

2

3

3

43

4Cylinder 1

4 42

3

2

3

3 3

4 4 Cylinder 2

4no card

45

2

3

2

3

3 3

4 4 Cylinder 3

4no card

45

-Denotes the series of Cards 3 and 4 necessary to describe

the problem,,

If the problem involves fewer than three cylinders, the

effects of the cylinders not to be considered may be elimina-

ted by setting their length equal to a very small number,,

The program will still give results for these shortened

cylinders but they might be inaccurate due to round off

All

error and should be ignored

„

The minimum fraction (columns 61-70, Card 1) and

exponent for forcing convergence (columns 16-20, Card 2)

are values used in an averaging process to assist in

obtaining convergence of the numerical solution of the

fourth order differential equation of a cylinder „ These

values are named Cm^n and m respectively in the main body

of the the si So

The convergence criterion (columns 41-50, Card 2) is

multiplied by the maximum radial deflection of the cylinder

and this value is compared with the change in radial de-

flection at each station for the last two successive

iterations „ If this change for every station is less than

the maximum deflection times the convergence criterion,

convergence is deemed to have been attainedo

The program is stopped by placing a blank card in

place of Card lo

A12

APPENDIX B

Listing of Program Skirt and Logical Flow Chart

List of Variables in Main Body of Program

Name inProgram

AA

AAA

AINC

AL

AMBAR

AMO

AMOO

AV

AX

B

Counterpartin Thesis

Numerous

M*

M.

A

V

M

oo

- £*Vdx

ol' o2'

o2M_ OJ M

ol'

Numerous

o3

Description

Inputs to continuity andstatics equationso (Equa-tions (46) in thesis,reduced to 6X6 arrayo)

Instruction code for inte-gration subroutine TTZigzag TT

o

if no divergence exper-ienced in problem; 1 ifdivergence experiencedo

Number of increments

„

Coefficient of linearthermal expansion «,

Junction end moment plusthe moment necessary tocompensate for that imposedby the radial thermalgradient*

Moment at junction endo

Externally applied momentto junction

Deflection at junction endo

The end shear and momentfor each cylinder

Inputs to continuity andstatics equationso (Equa=tions (46) in thesis,reduced to 6X6 arrayo)

Bl

BM

BM1

BM2

BM3

BV

CURVE

CX

CXI

CX2

c

CI

C2

E

EI

ERROR

FM

M, M

ML

*

Mm

B

X*

(k)(x)

E

D

Moment in cylinder wall perunit circumference

o

Moment in cylinder wallcaused by thermal, pressure

.

and axial loadings

„

Moment in cylinder wallcaused by applying a unitshear to the junction endu

Moment in cylinder wallcaused by applying a unitmoment to the junction end

Slope of a straight linedrawn between radial deflec-tions plotted for junctionand remote endSo

Axial curvature of cylinderwall,

cx = (c)U)

'0

Jo

LCXdX

LCXldX

Pressure necessary to in-crease the radius of thecylinder one inch»

/oLCdX

•XCldX

min

Young ? s Modulus

.

Axial stiffness of a unitwidth segment of thecylinder wallo

Convergence criterionfractiono

Minimum fraction in convexgence assistance routine

B2

FRAC

FX

FXC

FXC1

FXC 2

FIX

F2X

GX

H

ICYL

IEXP

IIF

INC

INC2

IND

IPROB

IWILL

JJ

Kl

C

f(x)

(k)f(x)

fl<x >

f2(x)

g(x)

t

ra

n-1

Option 1 or 2

Fraction used in conver-gence assistance routine.

Deviation of radial deflec-tion from a linear function,

FXC = (FX)(C)

JXFXCdX

yoXFXCldX

/oLGXdX

fo

Normalized F1X

X(BM/D)dx

Thickness of cylinder wall»

Cylinder number

o

Exponent in convergenceassistance routine

Printout option

Number of equal incrementsinto which the cylinder isdivided*,

(2) (INC) + 1

Interval of printout ofinput data Q

Problem number

»

Pressure stress calculationoptlono

Code for type of calculationpresently in progress

.

First station in a sequencein which the loadings werenot read into the computer.

B3

K2

K3

L

LL

NAA

PSI

PSI1

R

Rl

R2

SIG1

SIG2

SL

SLOT

SLOT1

ST

P

P

a

R:

R.

°i

^2

dsdx

STCI

*EAT2

z inner

First station in a sequencein which the loadings wereread into the computer

o

(K2 - Kl + 2)/2

Station number where maxi-mum F2X occurs

u

Station number where maxi-mum deflection occursc

Number of Iterations

„

Internal pressure.

Internal pressure

„

Radius of cylinder

Meridional radius of curva-ture of at junction,,

Radius of curvature of lineperpendicular to meridianat junction

o

Longitudinal membrane stressdue to pressure

c

Circumferential membranestress due to pressure a

Axial slope of cylinder wall

Value of X at end of slotSo

Value of X at beginning ofslots.

Stress at surface of wall ofan infinite cylinder sub-jected to a radial tempera-ture gradient

o

Circumferential stress oninner wall surface

„

B4

STCO

STLI

STLO

TL

TLINC

TP

TPI

TR

TRINC

UM

V

VBAR

VBAR1

VO

VOO

VI

z outer

^x inner

cnx outer

AT

MV

Voo

Circumferential stress onon outer wall surface

„

Longitudinal stress oninner wall surface

Longitudinal stress onouter wall surface

.

Mean temperature

Increment of linear inter-polation for TL 3

Number of 53 word blocksneeded to store one para-meter of cylinder,.

Total number of 53 wordblocks necessary to storeon tape the intermediatevalues of one problem

Difference between innerand outer wall temperature

.

Increment of linear inter-polation for TRo

Poisson ? s ratio

.

Radially outward directedload per unit circumference

Constant shear imposed atjunction end.

/ xVBARdX

Shear at junction endo

Externally applied shearforce to junction

Shear in cylinder wallcaused by thermal, axial,and pressure loadings

„

B5

V2

V3

WT

WTINC

W

Wl

W2

X

XDEL

XMAX

XMIN

Y

Yl

v

Y2

V,m

(P*)(2rra)

a~

Shear in cylinder wallcaused by applying a unitshear to the junction end.

Shear in cylinder wallcaused by applying a unitmoment to the junction end,

Total axial load caused byall influences other thanpressure

o

Increment of linear inter-polation for WTo

A+Bx+f(x)) Distributed axial loadsapplied to cylinder includ-ing simulated thermal loads

n

s

v

/XWdX'0

/XWldX

Axial dimension

o

Length of interval betweenstationso

Value of X at remote end

Value of X at junction

„

Radial deflection.,

Radial deflection of thecylinder wall caused bythermal, pressure, and axialloadings a

Radial deflection of thecylinder wall caused byapplying a unit shear tothe junction endo

B6

Y3 s Radial deflection of thecylinder wall caused by-

applying a unit moment tothe junction end,

The following variables are those used in Subroutine Zigzags

Variable Description

A Instruction code for integration

B Interval length <,

D Integrand

.

E Integral

.

M (2) (number of increments) + 1

The following variables are those used in Subroutine Answer

„

A Equivalent stress on outerwall surface

.

B Equivalent stress on innerwall surface

„

BM Moment

.

C Outer longitudinal stress

„

D Inner longitudinal stress.

E Outer circumferential stress.

F Maximum equivalent stress.

G Inner circumferential stress.

KK Denotes whether maximum equivalentstress is on inner or outer surfaceof wall.

KKK Station number of location ofmaximum equivalent stress.

L Cylinder number.

B7

M Number of increments

.

N Printout code.

NN Problem number.

V Shear.

X Longitudinal dimension.

The symbols used in the following logical flow chart are

from the "Flowcharting Template," Form X20-3020, of Interna-

tional Business Machines Corporation, and are described below.

The numbers in the locator (circle) and connector (inverted

house) correspond to the sequence numbers at the extreme

right of the program listing.

Use

The beginning of end of a program.

The location of the first statementof the following operation in theprogram.

Trapezoid Input/Output Any function of an input/output device

Connector

Symbol

Oval Terminal

Circle Locator

InvertedHouse

Diamond Decision

Rectangle Processing

An entry from, or an exit to, anotherpart of the program flowchart.

A decision is made for routing todifferent branches of the program.

A group of instruction performinga processing function.

Subroutine GAUSS2 is not charted or listed because it is

a standard library program for solution of simultaneous

linear equations. The subroutine was written by C. B. Bailey

and is listed in "F2 UTEX LINEQN," CO OP Manual for CDC 1604,

December, I960.

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ENTER INPUTSTO CONT. ANDSTAT. EQNS.WITH LOADSAPPLIED TOCYLINDER 1

SAME INPUTS BUTLOADS APPLIEDTO CYLINDER 2

SAM5AME INPUTS BUT-LOADS APPLIEDTO CYLINDER 3

SA?ffi INPUTS :

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APPENDIX C

Sample Problems

Elementary problems were set up to show the capabili-

ties of the program c No pretense is made that the values

used are representative loadings c The problems are des-

cribed belowo No significance should be attached to the

numbering system of the problems, A summary of the equi-

valent stress in each cylinder of each problem is given in

Table 1 and sample printouts of problems 20,21,22j,&Q Sland

90 along with the input cards for these problems, are given

after page C6o

Problem 10 - Three identical cylinders, 16 inches longs, 50

inch radius , o 5 inch wall thickness , Young *s Modulus

3X10' psi, Poisson"s Ratio 0o3» thermal coefficient of

expansion 7 ^ 7X10-6 inches/inch °F, number of increments

25 o 100 psig internal pressure in Cylinders 2 and 3; no

thermal or mechnical loadings other than pressure „ The

method of solution used was Option 1 (See Appendix A).

Printout option 3*

Problem 20 - Same as problem 10 except Option 2 used for

calculation In cylinder 2, R^ = 50 inches, R2- 50

inches ; ioe c, hemispherical heacL In Cylinder 3* % i s

infinite, R^ 50 inches

«

Problem 21 - Same as problem 20 but printout option 2 is

CI

used.

Problem 22 - Same as problem 21 but printout option 1

Problem 30 = Same as problem 20 with following thermal

loading added,

Cylinder 1 Medial wall temperature = T = 300 -

400(l-e^ 025x ) »F U

Cylinders 2 and 3 T = 300 +• 200(l~e= o025x

) °F.

Problem 40 - Same as problem 30 with the following thermal

loading added

:

Cylinder 1 Axial temperature difference -^T = for all x,

Cylinders 2 and 3 4T = (25) (T^Tj)

where T^ is the median temperature at station i and Tt is

the median temperature at the junction

Problem 50 - Same as problem 40 with an axial load added;

Cylinder 1 Axial load = P* = =5000 - 47<,5x Lb.

Cylinder 2 P* = 3500-47. 5x Lb.

Cylinder 3 P* - -1500+47* 5x Lbo

Problem 60 - Same as problem 50 but the thickness of

Cylinder., 3 is progressively reduced.Thickness of Cylinder 3

Oo0fx^2c56 o5 Inches

2 o 56^x^5 o 76 .45 Inches

5 * 76^x^3 o 96 .40 Inches

3. 96^x<L2 .16 .35 Inches

12ol6^xil6.0 o30 Inches

C2

Problem 70 - Same as problem 60 but Cylinder 1 (skirt) is

slotted from x = to x = 10 inchest

Problem BO - Same as problem 70 but externally applied

moment and shear are added c

External moment = M = 500 Midkkoo in

External shear = V = 500 J£oo in

Problem 90 - Same as problem 20 but length of cylinder 1 :

reduced to 10=

' inches, effectively eliminating it from

the problem

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Discussion of Results

Since the sample problems are somewhat complex and are

used mainly to display the capabilities of the program., the

purpose of the following discussion is mainly to point out

those results which the writer felt most interesting and

to offer what seem to be plausible explanations of these

re suit So

Problems 20 and 90 have no loadings but pressure and

differ only in that 20 has a skirt and 90 does not In

comparing the stresses in Table 1 of Cylinders 2 and 3> it

will be noted that the addition of the skirt did not

grossly affect the stresses or their locations The

stresses did increase somewhat due to the moment exerted

by the skirt upon the junction, but this increase was

partially balanced by the skirt's constraint on the outward

movement of the junction* This may be seen by comparing

the circumferential stresses of Cylinders 2 and 3 of the

two problems . The results for Cylinder 1 of problem 90,

should be neglected

Note that the application of an external moment and

shear on problem SO considerably changes the equivalent

stress at the remote end of Cylinder 3 from what it was in

problem 70 e.

Since the length of the cylinder was supposedly

chosen such that a force applied at the junction should not

C5

affect the remote end, this change in stress might be

puzzlingo However the junction end moment in Cylinder 3

of problem 70 was -88 in~lb* and in problem 80 was =559

in~lb., an increase of over 500% o Also the junction end

shear increases from 59 lb/in to 471 lb/in, an increase of

about 700%o Since, due to the sign conventions for Cylinder

3, (See Figure 5 of text) a moment and shear of opposite

signs have a cumulative effect on the moment , this 14

percent change in stress at the end is not unreasonable

From the printout of problem 80 it will be noted that

at station 25 of Cylinder 3 the stresses seem to change

radically from those of the previous station The probable

reason for this is that in this particular cylinder a large

nonlinearity is imposed upon the whole length of the

cylinder by the decrease in wall thickness. Such a non-

linearity will create internal moments and shears in the

wallc However we are still forcing the shear and curvature

to be zero at the end (Station 26) and therefore a large

compensation occurs near the endo For accurate results

near this point, a longer cylinder must be used with more

intervals

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