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Theses and Dissertations 1. Thesis and Dissertation Collection, all items
1964
Thermal stress analysis of pressure vessels
with cylindrical skirt supports.
McManus, James P.
http://hdl.handle.net/10945/13284
Downloaded from NPS Archive: Calhoun
Library
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.
BS
PROGRAM SKIRT
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= 1
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 :
UNIT SHEARAPPLIED TOCYLINDER 1
SAME INPUTSUNIT SHEARAPLIED TO3YLINDER 2
it!
BUT
AME INPUTS BUTIT SHEAR
PPLIED TOYLINDER 3
SAME INPUTS BUIONIT MOMENTAPPLIED TOCYLINDER 1
AME INPUTS BUI
IT MOMENTAPPLIED TOCYLINDER 2
SAME INPUTS BUTUNIT MOMENTAPPLIED TOCYLINDER 3
CORRECT Y,AND BM„CALCULATE ST,
w r o.
WRITE TAPE1 ST,E.
2910
Bll
SKIRT (continued)
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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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C4
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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