Dynamics Technology, Inc.
Any opinfGfls,findingsfconclusiOnsor recommendations expressed in thispublication are those of the author(s)and. do not necessarily reflect the viewsof the National Science foundation.
DT-7814-2
EARTHQUAKE RESPONSE OF SEA-BASED
STORAGE TANKS BY AHYBRID ELEMENT METHOD-
THEORY AND COMPUTER ANALYSIS
By: SHIH-CHI LEE
MARCH 1981
PREPARED FOR: EARTHQUAKE HAZARD MITIGATION PROGRAMDIVISION OF PROBLEM-FoCUSED RESEARCH
ApPLICATIONSNATIONAL SCIENCE FOUNDATION
REPIlODUC EO BYNAnONAl TECHNICALINFORMATION SERVICE
O.S. DEPARTMENT OF COMMERCESPRltlllflElD, VA 22161
DYNAMICS TECHNOLOGY) INC.22939 HAWTHORNE BLVD,) SUITE 200TORRANCE) CALI FORNIA 90505(213) 373-0666
This report has undergone an extensive internal review
before publication, both for technical and non-technical
content, by the Project Manager, the President, and an
independent internal review committee.
Project Manager:
President:
Internal Review:
, i
Ii
FOREWORD
This work was supported by the Earthquake Hazard Mitigation Program ofthe Nati onal Sci ence Foundati on, Washi ngton, D.C. under grant PFR 7919949, which is a continuation of PFR 78-09866.
This study addresses the problem of dynamic response of submergedstorage tanks subject to earthquake excitations, in an attempt to formulate a general evaluation procedure using the hybrid-finite elementmethod and to synthesize a comprehensive and predictive computer codefor engineering applications. This technical report presents the formulation and encoding of the research findings.
The author would like to acknowledge the stimulating discussions andinvaluable technical assistance provided by Professor C.C. Tung of North
CarolinA State University, Dr. C. Y. Liaw of EG &G, Professor P. Liu ofCornell University, Mr. K. C. Chang and Mr. B. P. Richman throughout thecourse of this investigation.
. "
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ABSTRACT
An effective method for the linear analysis of dynamic response of sub
merged underwater oi 1 storage tanks to 1oadi ngs of earthquake exci ta
tions is presented. The tank is axisymmetric in shape, has a flexible
wall/roof. A general hybrid-finite element solution procedure has been
formulated, wherein the tank structure, the interior fluids, as well as
the near field of the exterior water region are discretized into a
toroi da1 mesh network. The tank di sp1acement is expressed as a super
position of the first few modes of the structure's free vibration.
Contribution from the hydrodynbamic interaction to the coupled motion is
obtained by solving the Laplace equation with the appropriate boundary
conditions, which includes a matching to the exterior far-field pressure
(analyti c) representati on to simpl Hy the computati onal process. The
effects of fluids surrounding and inside the tank are studied. It is
demonstrated that these effects are, in general, significant on the tank
earthquake response analysis •
.. A comprehensive and predictive computer program for use in such tank
response analysis is developed for design engineering applications.
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TABLE OF CONTENTS
PAGE
FOREYJORD. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • il;
ABSTRACT. • • • • • • • • • • • • • • • • •.• • • • • • • • • • • • • • •.• • • • • • • • • • • • • • • • • • • • • • • • • • • ..i v
TABLE OF CONTENTS ••••••••••••••••••••••••••••••••••••••••••••••••••• .:i;I:IitV
1.
2.
3.
4.
5.
INTRODUCTION ....•. ·..•.•...•.••.•.......••.•••••.•.•••••.•.••..••
RELATED PAST WORK •••••••••••••••••••••••••••••••••••••••••••••••
FORfvlJLATION •••••••••••••••••••••••••••••••••••••••••••••••••••••
NUMERICAL RESULTS/VALIDATION ••••••••••••••••••••••••••••••••••••
SUMMARY AND FUTURE WORK •••••••••••••••••••••••••••••••••••••••••
1
3
5
35
41
REFERENCES ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
FIGURES •••.••••••••••••.•••••••••••••••••••••••••••••••••••••••••.••
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APPENDIX A:
APPENDIX B:
ERST User's Guide ..•••••••..•••.••.•••••.••..••••••••••
Program Listing .
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Dynamics Technology, Inc.DT-7814-2
INTRODUCTION
In recent years, exploration and production of oil offshore have
increased tremendously. Many of the new fields are far into the ocean
and the adjacent land area is often desolate and uninhabited, rendering
the conventional method of using pipelines and onshore tenninals more
and more costly. For countries in scarcity of land space (such as
Japan), util ization of the sea may become necessary for the purpose of
10ng-tenn petroleum storage. The alternative of storing crude at the
producing sites to be shipped via tankers to harbors is thus becoming
more and more economical, both in terms of capital investment and in
operating cost.
The first submerged underwater oil storage tank was placed in service in
December of 1969. Referred to as the Khazzan Dubai #1, the tank is
located in 150 feet of water, 60 mi 1es off the shore of the Truci a1
Coast in the Arabian Gulf (Chamberin, 1970). It weighs 15,000 ton and
holds 500,000 barrels of oil. It has the appearance (.Figures 1 and 2)
of an inverted funnel with a 270-foot-diameter base and a roof which is
a portion of a 180-foot-radius hemisphere. A conical transition con
nects the roof to a 30-foot-diameter shaft that extends above the ocean
surface. The "bottomless" tank rests on the ocean floor and operates on
the water displacement principle; it is always filled with either water
or oil or a combination of the two. Filling is accomplished by placing
oil through the shaft, the additional weight of the oil on the water
creates a pressure imbal ance whi cn forces the water out of the tank
through openings in the wall. Deep-well pumps are used for discharging
oil. As oil is withdrawn, inflow of water takes place, replacing the
removed oil.
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New concepts and designs of oil storage tanks have since been developedfor all areas of the world. Some of the tanks rest on ocean floors andare either completely sUbmerged or partially submerged with part of thetank protruding above water; other designs take the form of floating,bottomless tanks, moored to the ocean floors.
The primary forcing function these storage tanks must be designed to
resi st is that due to waves in the heavi est storm; and for bottomsupported tanks located in earthquake-prone areas, seismic-inducedhydrodynamic forces must, of course, also be considered. Because of theexorbitant cost incurred in the construction of these huge structures,
and the environmental hazards associated with the failure of such structures, an accurate evaluation of the hydrodynamic forces is vital. Inorder to predict the response of an underwater tank to waves and earthquakes, the development of a reliable computation method is a pressing
need for the construction of storage tanks in seismic areas. This present study is specifically aimed at the earthquake-induced response fora completely sUbmerged tank filled with oil and water.
Earthquake analysis of such cantilever structures requires special considerations which do not arise in land-based structures; any procedurefor analysis must recognize the additional dynamic forces and modifications in the dynamic properties caused by the surrounding water and thefluids inside. If the tank is perfectly rigid, the motions of thefluids inside and outside the structure may be treated independently.However, to accommodate for the more stable structures made of flexibleresilient materials, extra care needs to be exercised in studying theirdynamic behavior, by virtue of the fact that the structure and fluidmoti ons are coupl ed. The effect of the tank structural deformati on onthe dynamic response is the emphasis of this investigation.
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Dynamics Technology, Inc.DT-7814-2
2. RELATED PAST WORK
There have been few works that directly address the problem intended in
the present study. The ones bearing the closest relationship appear tobe that of Takayama (1976) and Helou (1981). Takayama treated transientwaves inside a vibrated oil storage; the tank is a rigid rectangular orcylindrical structure, submerged undersea. However, no attention wasgiven to the effect of surrounding water, and no numerical results arefurnished. Helou extended the problem wherein the tank wall is flexible. However, he was only concerned with cylindrical tanks whereanalytic solution can be found; no numerical scheme was developed for
general applications.
A wealth of research papers do exist which provide valuable sources ofpertinent information, most having to do with the hydrodynamic pressure
distribution of structures under earthquake excitations. For land-basedtanks of simple geometries under the assumption of iriviscid, compressible or incompressible fluid, and irrotational motion of small amplitudes, many sol uti on procedures have been formul ated. Fi rst, Jacobsen(1949) evaluated the dynamic mass effect of fluid inside a cylindricaltank and outside a cylindrical pier, when the base experiences an impulsive seismic load. Then, Housner (1956) set the foundation of generalearthquake-proof design analysis by introducing a simple approximationmethod which avoids partial differential equations and infinite series.Thereafter. many works appear which deal with the deformation of tankstructure. Notable among them are Baron and Skal ak (1962). Arya,Thakkar and Goyal (1972), and Yang (1976). who use the Rayl eigh-Ritzmethod; and Edwards (1969) and Shaaban and Nash (1975). who employ thefinite-element method. In all cases, fluid is treated as a continuumand appropriate shell theories are selected for the development(Sanders, 1959; Flugge, 1960; Basu and Gould, 1975; Ghosh and Wilson,1975). In Yang's work, the nature of the impul sive and convectiveeffects is carefully identified, and he based his dynamic analysis on
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uynamlcs lecnnology, lnc.DT-7814-2
assumed mode shapes of the tanks free vibration. Wu et ale (1975)
developed a computer program to calculate the natural frequencies of the
f1 ui d-tank system. Earthquake analysis of the resonant osci 11 ati on
(sloshing) phenomena in elastic shells can be found in Chester (1968),
Faltinsen (1974), Aslam, Godden and Scalise (1979) and ~i, Foda and
Tong (1979).
So far we have been quoting only works concerning ground tanks. The
problem associated with motion of water surrounding sUbmerged tanks is
more difficult to solve. For objects of simple geometries, attempts
were made usi ng the SChwi nger vari ati ona1 techni que (Bl ack and ~i,
1970), the Galerkin method (Garrett, 1971), and the integral equation
method (Garrison and Seetharama Rao, 1971). Tung (1979) also pursued
the problem of sUbmerged bodies subject to harmonic ground excitation,
using a semi-analytical method to obtain the hydrodynamic forces and
confirmed the insignificance of the gravity effect, as long as the exci
tation frequencies are moderately high. For objects of more complicated
shapes, numerical methods must be employed. The finite-element method,
known for its versatility, was used by Chakrabarti and Chopra (1972) and
Liaw and Chopra (1973) in studies of seismic response of gravity dams
and intake towers. This approach is further enhanced by the adoption of
an analytic super-element, thereby reducing the mesh requirement in the
far field (ordinarily it is required that the outer truncation boundary
must be far enough away from the longest waves). The so-called "hybri d ll
finite-element method, which combines judiciously finite-element solu
tion for fluid motion near the object and an analytic representation for
the far field, has been proven to be highly efficient. Among the pion
eers are Berkhoff (1972), Bai and Yeung (1974), Chen and ~i (1974) and
Vue, Chen and ~i (1976).
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Dynamlcs lecnnology, lnc.DT-7814-2
3. FORMULATION
Al though the subject of sei smi c response of sUbmerged storage tanks is
relatively new and unexplored, as we have pointed out in the previous
chapter, much of the pertinent analysis tools have been developed. It
is the purpose of this research to utilize these tools in formulating a
general hybri d-fi ni te el ement sol uti on procedure, for the hydrodynamic
response of flexible underwater storage tanks subject to earthquake
motions; and to synthesize this procedure into a comprehensive predic
tive computer code which can be used for engineering applications.
The tank structure in question is of general axisymmetric shape, has a
flexible wall and/or roof, and is rigidly attached to the ocean floor.
Thi s assumpti on that the support foundati on does not move rel ative to
the ground reduces the scope somewhat, since the effect of marine soil
structure i nteracti on can have si gnifi cant consequence on the hydro
dynamic analysis. However, it is expected that this variation can be
accommodated by our hybrid-finite element procedure, and will be dealt
with in a future study. The tank will be completely filled with oil
and/or water and sealed (this last restriction can be lifted by simply
changing the input format). Finally, in the following presentation, we
simplify matters by ignoring irregular bottom topography and depletion
of i nteri or compartments. Thei r presence can be handl ed strai ghtfor
wardly by carefully discretizing these components into finite elements.
The equation of motion of the tank structure can be written in terms of
the structure discretization as
.. .[M]{x} + [C]{x} + [K]{x} = - {F} (3.0)
where [M], [C] and [K] are the mass, damping and stiffness matrices of
the system, respectively. {~}, {x} and {x} are the acceleration, velo-
city and displacement vector of the structure relative to its base,
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Dynamics Technology, Inc.DT-7814-2
*{F} is the load vector of the external forces, which includes the
hydrodynamic pressure {Ps} due to the presence of fluids inside and out
side of the system. For nodal circles on the inner shell interfacing
with the interior oil and water, {Ps} is sought from the equation of
motion governing the dynamic interaction between the shell and the
interior fluids; the same goes for {Ps} at nodal circles on the outer
tank surface. Thus, before attacking (3.0), we need to solve two bound
ary val ue prob1 em suitably formul ated based on the two appropri ate
Laplace equations. The flexibility of the tank shell is entered into
both radi al boundary condi tions inf1 uenced by the ground acce1 eration,..consequently {Ps} intertwines with {x}.
Both boundary val ue prob1 em are sol ved usi ng the vari ati anal princi pl e
of finite element theory. In the "far awai' exterior region, a matching
of analytic representation of {Ps} is invoked (this region is to have
been rid of all geometrical irregularities). Once {Ps} is obtained (in.. ..
terms of {x} and the earthquake ground acceleration {fh}, (3.0) can be
solved by transformation into modal coordinates wherein the displacement
is expressed in terms of the fi rst few modes of the tank force vibra
ti on. The resul tant 1i near second-order differenti a1 equati on in the
generalized displacement amplitude can be solved by the ordinary step
by-step integration schemes.
* Vectors enclosed in braces { } are associated with the appropriate(interior fluids domain, structure, or exterior water region discretizations) nodal coordinates. Thus, if NO is the number ofnodal-circles in the oil domain, {P } is the column vector of nodalpressure distribution of dimension ~JO, ordered in the global nodalnumber sequence. However, in the case of structure discretization,the vector is represented by the (r,z,e) coordinates at each nodalcircle, making {P} a vector of dimension 3N, if N is the number ofnodal-circles in the structure assemblage. The symbol "-+" will bereserved for the ordi nary three-di mens,; ana1 conti nuum vectors.
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Dynamics Technology, Inc.DT-7814-2
3.1 Dynamic Behavior of Fluids
3.1.1 Interior Fluids
A. Problem Formulation
Consi der a axi symmetri c tank sUbmerged underwater whi ch is fill ed wi th
oil on top of water (more generally, two fluids of densities Po and Pw'where pw > po) with respective heights of ho and hw as shown in Figure 3
(ho depends on the radial distance from the axis). Choose the coordinate system as depicted in Figure 3 with the axis of tank symmetry being
the z-axis and the direction of earthquake ground motion being the(positive) x-axis, and the origin at the center of oil-water interface.We will also need the usual (r,z,e) cylindrical coordinate system.
-l-Let v(x,y,z;t) denote the velocity vector of the fluid particle at(x,y,z;t). Then, based on the usual assumptions of inviscid and incompressible fluids, irrotationa1 motion and small amplitude waves, thelinearized momentum equations of fluid motion read
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oVo _ 1""M- - - V'PPo 0
(3.1)
We use the suffi xes 0 and w to represent quanti ti es perta i ni ng to oi 1and water, respectively.
Take the curl of (3.1) and apply the continuity equations
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Dynamics Technology, Inc.DT-7814-2
? a'i/ • v =0
? a'i/ • v =w
we obtain
'i/2p = ao
(3.2)
(3.3)
'i/2p = aw
~
Let fh(t) denote the horizontal ground acceleration induced by an earth-
quake, th~n since the translatory velocity in the x-direction should be•equal to f h , we have
CPo ? ?
[(fh + ~) • n] across aVowfuI- - Po
aPw? ?
(3.4)
an-- - Pw [(fh + ~) • n] across oVww
In (3A), ~ = n· 'ii, and nis the unit outward normal to the tank
boundary. x is the accelerati~n of the tank structure relative to the
ground. The presence of the x term is di ctated by the fl exibil i ty of
the tank shell. Now if the tank had a roof whi ch is ri gi d, we woul d
have
along aVot •
However, to allow for the general case where the II roof ll is also flexible
(and more likely, inseparable from the wall as in the case of a half
dome tank), we will use the same boundary condition as in (3.4). Conse
quently, we include cVot as part of cVoW" The assumption of a rigid
floor support implies that
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Dynamics Technology, Inc.DT-7814-2
(3.5)
Now, if we let C(x,Y,z;t) stand for the displacement of the oil-waterinterface from its equilibrium position, the (linearized) dynamic inter
facial condition affirms the continuity of the total pressure such that
along z = 0 • (3.6)
And the (linearized) kinematic condition maintains fluid particles on
interface to stay on the interface (continuity of vertical velocity) so
that
along z = l; • (3.7)
Under the small amplitude assumption, only minor error is incurred by
evaluating equation (3.7) along z = O. If we assume that the gravitational effects are small compared to the forced oscillation so that
Equation (3.6) is then reduced to
(3.8)
along z = 0 , (3.9)
so that sloshing waves at the oil and water interface are ignored.
Here, Land T are the characteristic scales of length and time, respectively. This assumption can be justified because earthquake excitationsare generally of high frequencies. In our validation of the computer
results (§ 4), we included such surface wave effects and found that the
calculated response was practically frequency independent for moderately
high frequency values.
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UJII\oA"11VJ 1~"1111VIV':JJ'
DT-7814-2
B. Solution via Variational Principle
Since the tanks under consideration are of general shape except foraxisymmetry, no closed form analytic solution is readily available. Wenow present as a prelude to the finite element solution scheme, a variational functional in Po and Pw whose stationarity is equivalent to thesystem of equations of §A being satisfied. It is well-known that ingeneral, the less restrictive Galerkin method produces results identicalto those one would obtain from any variational principle. We choose touse the variational method to illuminate the process, demonstrating thematrix assembly along the way. The Galerkin method will be employed inthe exterior problem to simplify our presentation.
Consider the functional
+1_1_ (VP ) 2 dV + 1 pw w wV 2pw oV
W ww
1 1 oP- (P -p ) ~ dA.2 W 0 uZ 1
oV. Po1
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~ ~
[( fn+ ~) • n] d,\w
(3.10)
UYIIClllll~::> I~~IIIIUIU!:JY, J.II~.
DT-7814-2
where we have used (VP) 2 to denote V'P • V'P. Cl early if Po and Pwsatisfy the system of equations of §A, then of(Po'Pw) = O. We now showthat, conversely, if of(Po'Pw} = 0 then the system of §A is solved. Forsimplicity, we will drop all the differential symbol in the integralsprovided that there is no chance of confusion.
Now,
1 1 ooP 1 1 OPw_ "7l:"""" (p -P ) __w - ~ _ 6PLP
Ww 0 oz LP
Woz w
oV i oV i
1 oP+ 1 ~ oP
2pw OZ 0oV.
1
1- Vp • VoP +Pw w w L. ww
1 1 MPo 1- "7l:"""" (p - p ) - - -.r:-
LPO
W 0 oz LPOOV
i
+ 1~L
1
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(3.11)
DT-7814-2
From Green's identity, we have
(3.12)
oP0 oPo. ~ .Note that along oV i we have art = - az- S1nce n 15 pointing downward.
oPw oPwAnd from ail =az- we deduce
Ivw
1- VP • voP =Pw w w 1 1 oPw
--- oPp woV w OZ
wb
L oPw oP + 1 _1 _oP_w oPPw OZ w P on w
oV www
(3.13)
Substituting (3.12) and (3.13) into (3.14) we obtain
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i 1 (lOPw 1 oP0)+ . "2" P az - p ~ 6P0
oV. w 01
1 1 (1 oPw 1 oP0)+ "7} - -- - - -- 6P. c. P QZ P OZ Wov. W 0
1
1 1 o6Pw~ (P -P ) --c.Pw W 0 OZ
oV.1
J1.- V<P oP + 1 1 apw-~6PwPw W W PwV oVwbw
L 1.-epw+ * * ·in)+ Pw [(fn+ x) 6Pw • (3.14)
Pw onww
If we observe now that each 6-differential quantity can be variedindependently, for 6F(Po 'Pw} to be zero, each of the integrals in (3.14)must be identically zero. We thus have the system of equations of §A,and our statement is verified.
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c. Finite-Element Approximation
By util izing the axisymmetry of the tank, the interior fluid domain
composed of Vo and Vw may be discretized into a finite-element system
comprising of toroidal elements with quadrilateral (vertical) cross
section (cylindrical elements at the axis of symmetry). If we use the
cylindrical coordinates (r,z,e) to define the global coordinate system,
and the natural coordinates (s,t,e) within each element as the local
coordinate system (see Figure 4, -1 ~ s ~ 1, -1 ~ t ~ 1) we can write
the coordinate transformation on element e as
r = ~ e e {Ne}T{re},L..J Nk(s,t) r k =k=l
(3.15)
where
N~ = (1-s)(1-t)/4
Ne = (1+s)(1+t)/43
Ne = (1+s)(1-t)/42
Ne = (1-s)(1+t)/44
(3.16)
are the bilinear interpolation functions, and (r~, z~, e), k = 1, ••• 4
are the global coordinates for the four nodes of the quadrilateral
cross-section. Within each element e, the hydrodynamic pressure is then
expressed as
(3.17)
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From §B. the system of equations of §A can be solved by imposing
of(Po'Pw) = O. We now do this at the element level, assembling theelement stiffness matrices and solve the unknowns {pe}.
We first rewrite (3.10) as
+ f 1 (V'P)2 dV2pw w wVw
I 1 oPw- 7"P Pw""5Z dAi
oV. w1
1 1 oPw+ -.r::- P -~- dA.c.p 0 vZ 1oV . W
1
1 1 oPo- "2P Pw~ dAi
oV. 01 -15-
I I
1
I I
2
I I
- 3
I I
4
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and then evaluate each of these integrals numerically based on thefinite-element scheme in the following sections. Due to the similaritybetween each Ik and Ik, only the details of Ik will be presented.
Cl. Integral 11
If we assume that the domain Vo is divided into Eo number of elements,we can approximate 11 by
Denote the integral in the above formula by [K eJ, then sinceVo
1dVoe =121tI rdrdz de ,V e 0 A eo 0
Aoe being the area of the finite-element (vertical) cross-section, and
since
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o __ cosS ~ _ si n8 0ax - or rae'(3.18)
o _ s; nS ~ + cose ~oy - - or r 08
We can rewrite [K e] asVo
tl e 2~o (H~e}TU~e}Cos2e +HnTg~e}cos2e +{Ner{Ne}s~~2e)rdrdZde.a Ao
Thus the (;,j)-th entry of [K e] is. V
o
'Jt J~ONi e aNjeoN; e oNj e[K ]'0 =. -- -- + ---- +
V e , J ~ e or or . OZ OZo 'Ao
NoeNoe)
, J rdrdz, i ,j=1,2,3 ,4.r 2
To facilitate the evaluation of [K ] it is convenient to transforme i j'Va
the integral into the local coordinates of e. Thus,
1 1
Iedrdz = 1 f IJ Idsdt ,A -1 -1o
where IJI is the determinant of the Jacobian
(
e
( ~~ ~~) f ~s1 e T ~~e[J] = = l {N } r e
or oz 0 3
ot at ot r4e
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The last relation is obtained from (3.16). Now (K ] .. can be evaluaV e 1J
ted over element e using the natural coordinates. 0
We can write 11 in terms of matrices and vectors relating to all nodesif we extend the {poe}·s to {Po} by placing zeros at the appropriatecomponents. Recall that the components of {Po} is ordered according tothe global nodal-circle number sequencing. Then
{3.19}
where [KV ] is the global matrix whose (i ,j)-th entry iso
L [K e]Vo e(i)e(j)
The sum is taken over all el ements e where nodes i and j belong tosimultaneously (e(i) denotes the local node number of i). Sinceeach [K e] is symmetric, so is [KV ].
Vo 0
C2. Integral 12
Assuming that Vo has Eow elements in contact with aVow' we can approximate 12 by
(3.20)
Under the structural finite-element discretization, we introduce thenodal acceleration vector {X} (cf. Figure 5):
.. T .. .. .. .. .
{X} = (Xlr,Xlz,X1e'~r'x2z,X2e'···· ,XNr'xNZ'XNe)·
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(3.21)
Dynamics Technology, Inc.DT-7814-2
And if for each ~, the local angl e between the normal ~n and the XY-
plane, we defi ne the Nx3N transformation matrix [G] by
cos~, si n~, 0 0 0 0 0 0 0 0 0
0 0 0 cos 13, si n~, 0 0 0 0 0 0
[G](~} = 0 0 0 0 0 0 cos~, si n~, 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 cos~, si n~, 0
(3.22)
We can then write
(3.23)
along the A~w surface. Where we have used the same bilinearinterpolation functions of (3.16), now acting on the inner fluidstructure interfacial nodes.
Now we can write
(3.24)
where z~ and z~ delimit the vertical extent of element e on Vow' and the
global matrix [0] is assembled from the element matrices
e=l, ••• ,Eow
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(3.25)
and
Dynamics Technology, Inc.DT-7814-2
..(1,0,-1,1,0,-1, ••• ,1,0,-1) f h (3.26)
{L} is the vector which translates ground motion to loads on the nodal-
ci rcl es.
C3. Integral 13
EiI = L:3 e=1
cos 2e dA. e1
Ei=L:
e=l
where r/ and rue delimit the radial extent of element e. Also note
that Ne and its derivatives are to be evaluated at z = 0.
Again, we can write 13 in global form as
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(3.27)
Ei=L:
e=1
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C4. Integral 14
ioN e T
_1_ ({N e}T {p e}{_w_ l {p en cos 2e dA. ee 2p 0 0 oz (w 1
ov. W1
e
lr u oN e T
{p e}T 1t {N e} {_W_}o e 2Pw 0 oz
rJ!.
Note that we have added subscri pts 0 and w to Ne here because thi sintegral involves both the upper and the lower layers. In terms of thelocal coordinates, it is remarked that for {No
e }, we have t = -1 andoN e T
for { o~ } we have t = 1.
In the global form we have
(3.28)
D. Total Global Matrix
"Summarizing and treating {Po} and {Pw} as unknowns, the functional(3.10) becomes
F({Po}'{Pw}) = {po}T[KV ]{Po} + {pw}T[KV ]{Pw}o w
+ {po}T[O] ({fh}+{~}) + {pw}T[W] ({fh}~{~})
(3.29)
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For F to be stationary, we require
of 0---oP ok
~= 0oPwk
k = 1, ••• ,NO
k = 1, .•• ,NW. (3.30)
Since Po = Pw at the interface, we have a set of NO + NW - (number ofinterfacial nodes) equations, which gives
and
(3.31)
If we define {Pow} to be the vector of interior nodal pressure distributions, ordered according to the global nodal number sequence (thus norepetition on interface), we can combine the matrix equations of (3.31)
to assemble
(3.32)
To solve for the inner pressure distribution (at least in the case of a
rigid tank, when {x} = 0), we can solve (3.32) by Gaussian eliminationroutines serving this purpose are readily available. However, as {x} is
still unknown in the general flexible case, (3.32) will be used as inputto the general structural equation of motion to solve for the earthquake
response later.
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3.1.2 Exterior Water Region
A. Problem Formulation
Consider the same axisymmetric tank of §3.1.1 now sUbmerged in water ofdepth H as depicted in Figure 3. We use the same coordinate system as
before, extended to the exteri or water regi on, i ntroduci ng only the
additional boundary surfaces.
Again, under the assumptions of water being inviscid and incompressible,
undergoing motion which is irrotational, and that wave amplitudes aresmall, the governing field equation is
v2p = 0W
in V (3.33)
where Pw is the water hydrodynami c pressure. $i nce water is the only
fluid of concern here, we will henceforth omit the subscript w.
The rigid floor support implies the homogeneous boundary condition at
the sea bottom since we only consider horizontal ground motion:
oP - 0oz - z = -hw (3.34)
Along the laternal surface of the tank, we have as before
* 7:oP - [( f ) ~n]1frI - - Pw h+x • across $i ' (3.35)
7:X enters because of the flexibility of the tank shell.
For the free surface boundary condition, generally one combines the
dynamic condition P = pwg(rrz) with the kinematic condition
02.... = 1 oP . 02p oP" - - - to obtal n - = - 9 - at z = o.ot2 Pw oz ot 2 aZ
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UI-/O.L't-~
However, as demonstrated by Liaw and Chopra (1973), the surface waveshave negligible effects on the hydrodynamic response of the fluid-tanksystem except at very low frequenci es. Si nce most of the earthquakeexcitati on energy is contai ned in hi gher frequency components, we cansafely assume that the gravity effects are of little consequence. Consequently, we use
P = 0 at z = H-hw (3.36)
as the free surface boundary condition.
Finally, since we are dealing with an infinite domain, in order to havea bounded solution, we further require that
P ~ 0 as r ~ ex> • (3.37)
These condi ti ons are duly adjusted, if necessary, to account for any
bottom irregularities.
B. Solution via Galerkin1s Method/Hybrid Element Approximation
Since the exterior water region is an infinite domain, the size of thefininte-element discretization is an important issue. A straightforwardapplication of the finite-element method, even with a domain truncationadjusted to the convergence rate, may be potenti ally cumbersome andcostly, due to the conflicting requirements that the size of theelements be a fraction of the shortest wavelength, and that the outertruncation boundary be far enough away from the longest waves. Instead,we adopt the hybrid approach of using the available analytic solutionfor P at a few wavelengths away from the tank, as soon as most of the
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significant geometrical irregularities are passed. We introduce afictitious (cylindrical) surface Sf here and use the analytic P to gobackward to match the finite-element solution.
Again using a finite-element discretization of the exterior regionbetween Si and Sf composing of ring-shaped elements having quadrilateral(vertical) cross-sections, we have, as in (3.17) of §3.1.1
4pe = ~ N ecose • p
ke ,
~1 k
(3.38)
{Ne} is as defined in (3.16), but note that it now acts on the exteriornodal-circles. Let us denote Nkecose, the finite-element interpolationfunctions, by Tk
e• The Galerkin criterion requires that
j( ~k(V2P)dV =0 • k =1,2,3,4.
From Green's theorem we have
(3.39)k = 1,2,3,4,-Iv (VTk 'VP) dV + L\ ~~ dS = 0 ,
where S is the union of all the surface of the discretized domain.
Now
Iv(V\. vP)dV
4 f (OT e oT e oT e oT e 1 oT e oT e)LL: k.R. k.R. k.R. ee= Ve .R.=1 V C5r~ + 15Z ---az- + -;2 5e 5"e P.R. dV
e
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The similarity between the last expression under the intergral signs and
that of [K e] in Section Cl of §3.1.1 is quite apparent. By letting kVo
run from 1 through 4, we can assemble the element matrix [HVe], and itis then a trivial matter to assemble this into the global form of
(3.41)
Recall that {P} is a vector of dimension NE, NE being the number of
exterior water nodal-circles.
The second term of (3.39) can be expressed as
(3.42)
By (3.34) and (3.36), the first and third terms on the right-hand-side
of (3.42) vanish. Since P = 0 on SSt the corresponding rows and columnsin the matrix [HV] also vanish. Th-e solution procedure is thereforesimpl ified by removing all equations associated with nodes at the surface. From (3.35) we have
If we use the same {Nse } as defined in (3.16) of §3.1.1 C, acting now on
the exterior water-tank interfacial nodes, based on (3.23), we can
rewrite21t Z e
r \ ~~ dS i=2:11 u - pw\e{Nse}T[GJ({fh}+{~}) case rdzdeJS
ise a Zj.e
-PwTtl" Nke {NSe}T[GJ( {fh}+{~})dZ
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(3.43)
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If we let k run from 1 through 4, we can then assemble the followingglobal form
(3.44)
To evaluate the last term of (3.42) we need to know the pressure distri
bution on Sf' which is obtained from the analytic representation for Poutside of Sf'
CD
PE = L: (lmK1(kmr) cos km(Z+hw) cose •m=l
(3.45)
where k = (2m-l) 1t, and K1 is the modifi ed Bessel function of them 2H
second kind of order 1. <lm is to be determined. Consequently, along Sfwe have
(3.46)
where rf is the radial coordinate of Sf.leave with ~ terms, we can write (3.45) as
MoP - L- - - d exon m m 'm=l
with
If we truncate the series to
(3.47)
Therefore,
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(3.48)
Now the <Xm's are determined (in terms of p) from the continuity require
ment of pressure across the fictitious surface Sf:
(3.49)
Using dm as the weighting function, we get
[ dmPdS f =1dlEdS f (3.50)
Sf Sf
4Substituting the expression pe = L Tkepke and the expression (3.45)
k=lfor PE at Sf (truncated to Mterms) into (3.50), we have
4 M
L L q e p e = 2: 0:. a . •k=l e mk k j=l J mJ
Sf
where
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(3.51)
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=
o if m '" j
(3.52)
Substituting (3.51) into (3.48) we get
(3.53)
Define the Mx4 element matrix [Qe] as having qm~ as the (m,k)-th entry,and the MxM diagonal matrix [A]-l of entries l/amm , we can fonn the
global equivalent
(3.54)
Consequently, we can write (3.39) in the equivalent global form
(3.55)
As in §3.1.1 D, (3.55) will be input to the equation of motion to solve
for the earth quake response in the following section.
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3.2 Dynamic Behavior of Fluid-Structure System
The equation of motion of the tank structure under consideration can be
written in terms of our finite-element discretization as
.. .[M]{x} + [C]{x} + [K]{x} = - {E} - {PS}' (3.56)
with {x} the vector of nodal displacements relative to the tank support,expressed in the nodal r, z and e components. [M] is the mass matrix
(3.57)
with Mk the mass of the tank materialwhich is lumped at the k-th node.stiffness matrices.
between neighboring nodal-circles,[C] and [K] are the dampi ng and
(3.58)
Finally, {Ps} is the vector of nodal loads associated with the hydro
dynami c pressures. As these pressures act only on the inner and outersurfaces of the tank, the elements in {Ps} corresponding to non-interfaci al nodes are zero. Indeed, {Ps} has many zero entri es, si nce thepressures act in the direction of the surface normal, thus all ecomponents (circumferential) vanish, and if a section of the interface(i nner or outer) is cyl; ndri cal, the correspond; ng z-components al sovanish.
Since the materials used in the construction of the tank is assumed tobe flexible, the structural deformation entails the coupling of the freevibration with the hydrodynamic interaction. Generally based on thealgorithm selected, the coupling of this sort can be categorized into aweak and a strong one. In weak coupling, the fluid pressure is firstused to "drive" the structure into a new shape, and a new pressure field
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in turn is evaluated using this new configuration. Strong couplingavoi ds these cycl es, forces acti n9 on the structural nodes and on thefluids are determined simultaneously. We use the latter.
If we now stri p off all non-i nterfaci al nodal components from {Pow}((3.32) of §3.1.1) and {P} ((3.55) of §3.1.2), rewrite them in terms ofthe r,z,e components, and extend them to {Ps}; at the same time pickingout the accompanying terms from [K INT] and [HV] + [Q]T[Q] to reassemblethe coefficient matrix for {Ps}' we can combine (3.32) and (3.55) into
(3.59)
Here [F] also incorporates the contributions from [OW] and [BJ.
Assuming that inversion of [HJ can be done efficiently, we can then substitute (3.59) into (3.56) to get
([MJ - [Hr1[FJ){~} + [c]{x} + [KJ{x} = - ([MJ - [Hr1[FJ){fh }
(3.60)
This equation has the standard form of a second-order linear ordinarydifferential equation, which can be solved straightforwardly by a numberof conventional time-integration schemes. However, it is quite obviousthat inverting [HJ should not be recommended. The alternative is to
utilize the nodal superposition method commonly used in structural.. .analysis. Therein the structural response x,x and x are expressed bythe eigenvectors (mode shapes {~}) of the undamped structural vibrations(without fluids). The {~}IS are obtained from the following eigenvalue
problem
(3.61)
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where Wj denotes the j-th eigenvalue (natural vibration frequency) of
the structure. It should be pointed out here that. in general. the mode
shapes and natural frequencies of our entire coupled fluid-tank system
are di fferent from the {~j} and Wj here. For the enti re system. no
precise physical meanings can be imparted to {$j} and wj ' except
that {$j} is a set of linearly independent vectors out of which struc-
, tural response can be composed of. For that matter. just about any set
of linear independent vectors can be used in this approach. were it not
for the advantage of our particular set {4>} that it diagonizes the
matrices [M]. [C] and [K]:
{4>j}T[MHq>j} = [M;]
{$.}T[C]{q>.} = [C~] =J J J
*21;;. w. [M.]J J J
2 *w. [M.]J J
(3.62)
(3.63)
* * *[MjJ. [Cj ] and [K j ] are diagonal matrices referred to as the general-
ized mass. generalized damping and generalized stiffness matrices. I;;j
is the j-th mode damping ratio (assumed to be small).
Using {$j}' any arbitrary displacment {x} can be expressed as a linear
combination of them:
J
{x} = L {q>.} y .•j=l J J
The above expansi on is exact if J is equal to the total number of
degrees of freedom, 3N, of the structure fi nite-el ement system because
the {~j}IS form a basis of a space of dimension 3N. Usually. for earth
quake type of excitati ons, the responses can be approximated by the
first few modes fairly well.
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Dynamics Technology, Inc.DT-7814-2
Now (3.59) can be written as
J
[H]{PS
} = - [F]{fh
} - L: [F]{4>.} V.•j=l J J
If we rewrite {P s } as
J
{Ps} = ~ {PSj} Yj{ J=l
where {P Sj } is the solution of
and {P so } the solution of
[H]{PSO
} = - [F]{L} •
(3.64)
(3.65)
(3.66)
(3.67)
Note that (3.66) and (3.67) can be solved without inverting [H]. And
since in the numerical process, the solution is gotten in a piecemeal
fashion, one need not worry about the possible singular behavior of [H]
during its construction.
{P so} can be vi ewed as the pressure response due to the ri gi d body
motion of the tank, and {P sj } is the pressure response due to the j-th
mode of tank free vibration. Substituting the expressions for {x} and
{P s } into (3.56), multiplying on the left by {4>j}T and using the ortho
gonality property of the mode shapes, we obtain
* T .. * • *([M.] + {4>.} {P .})Y. + [C.]Y. + [K.]Y.J J SJ J J J J J
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j = 1, •.• ,J (3.68)
Dynamics Technology, Inc.DT-7814-2
The term {<I>j}T{PSj } can be interpreted as the modal added mass matrix,
and {<pj}T{pso} the generalized hydrodynamic force due to tank rigid-bodymotion. To solve (3.68), commonly used time-integration can be applied.
It shoul d be poi nted out that the added-mass matri x in (3.68) is notdi agonal. Consequently, the system is coupl ed. Thi s system caul d, ofcourse, be transformed into an uncoupled set by usingtlie mode shapes ofthe coupled fluid-tank system, which are eigenvectors of
We choose to solve (3.68) more straightforwardly.
The use of normal modes of the structural free vibration to reduce thenumber of unknown coordi nates may be vi ewed as an appl i cati on of theRitz method.
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Dynamics Technology, Inc.OT-7814-2
4. NUMERICAL RESULTS/VALIDATION
A computer program named ERST, for Earthquake Response of Sea-Based
Storage Tanks, has been developed to implement the formulation of §3 to
evaluate the elasto-hydrodynamic response of axisymmetric storage tankssubmerged underwater induced by earthquake ground moti ons. A detailed
discussion of ERST is given in Appendix A and the full program listing
is furnished in Appendix B. In this section, we document some results
obtained from the computer codes which were used to validate the
program.
The simplest test is to see whether we can reproduce the well-known
series solutions for cases where the tank structure is a rigid circularcylinder submerged underwater. Due to rigidity, the inner fluid motion
and the outer water motion are uncoupled from the tank vibration. Consequently, we can test the interior response and the exterior response
separately.
.Si nce many of the avail abl e resul ts are obtai ned with gravity effectincluded, we modified our program accordingly (we used the variationalprinciple in the new coding to aid in our validation; the results werecompared to be within 2% of ERST which uses the Galerldn scheme). We
first compared the work of Tung (1979) studying (exterior) hydrodynamic
forces on submerged cylindrical tanks under ground excitation. Con
sidering a tank of relative dimensions H: (H-hw) = 2 and R: H = 1, underthe assumed earthquake ground acceleration of e- iwt with w= 10 rad/sec,
we found excellent agreement with the (analytic) data presented inFigure 4 of Tung's paper. We reproduced the relevant portion of the
curves in our Figure 6. The pressures are evaluated on the tank wall at
e = 00•
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(4.2)
Dynamics Technology, Inc.DT-7814-2
For the pressure response due to the interior fluid motion, we comparedwith the results of Takayama (1976) and Helou (1981). Again, the tank
is a hollow circular cylinder with a flat top. Here, R=10 ft, ho=hw=5ft, po=0.86 and Pw=1 in g/cm3• He selected a finite-element idealization of 20 elements distributed symmetrically relative to the oil-waterinterface. In Figure 7, the x-z plane section of the assemblage ispresented. From the sources quoted above, we have, for the analyti crepresentation hydrodynamic pressure of oil and water,
ex>
p = -p R cose e-iwt_po L }- G~ cosh km(Z-ho) cose e- iwt ,o 0 m=l m
(4.1)
- < z < ho '
. t ex> 1 sinh k h _i wtP = -p R cose e-1 w+p.2: Gil m 0 cosh km{z+hw) cose ew w w m=l ~ m sinh. kmhw
-h < z < 0 ,w
if the ~stem undergoes a ground velocity (n.B., not acceleration) e- iwt;
In the above formula,
kmCmJ1(kmR)
K -M w2m m
and
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Dynamics Technology, Inc.DT-7814-2
(4.3)
(4.4)
(4.5)
km for m= 1,2, •.• are the roots of Ji(kmR} = O.
Pressures for a range of frequencies are calculated using the aboveformulae and our computer program. As can be seen from the results presented in Figures 8 through 11, the comparisons are quite satisfactory.Pressure distribution normalized by the excitation frequency are plottedagainst the nodes at the inner tank wall (€FOO). The figures are for w
= 0.5 rad/sec, 1.1 rad/sec, 6 rad/sec and 10 rad/sec. Notice that sincegravi ty effects are consi dered here, Po does not equal to Pw exactly.Also, note that the curves are for hydrodynamic pressures only, onewould need to add on the hydrostatic pressures to extrapolate thelocation of equal pressures (where z = C).
Close exami nati on of Fi gures 8 through 11 reveals that, whil e there islittle change of the pressure curves from w = 0.5 rad/sec to w = 1.1
rad/sec, there is significant (trend-reversal) difference from there onto w = 6 rad/sec. Eventually, the curves "stabil ized" and becomepractically frequency independent (this is confirmed from calculation ofa number of frequencies ranging from 3.7 rad/sec, 4.8 rad/sec to 60rad/sec). Now, with the gravitational effects included, one couldexpect the fluid-structure system to exhibit sloshing phenomena at thenatural frequencies ~ where
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Dynaml cs Tecnno lOgy, Inc.DT-7814-2
(4.6)
For the data used in our test, these frequencies are
Since the density difference between oil and water is small (0.14 g/cm3)
the sloshing amplitudes will be small. Within the frequency spectrum of
importance due to earthquake excitations, only high modes of sloshing
waves are expected. These are short-length waves of minor importance.
Since the pressure distribution is independent of frequencies in this
spectrum, this reinforces our belief that as far as our system configu
ration is concerned, the gravitational effects can be safely ignored, as
we did in our analysis.
To ascertain that our 20-element finite-element scheme has converged
enough to be believed, we increased the assemblage to 100 elements,
whose layout is basically the same as that of the 20-element case,
except that 10 columns are used. We present the results in Tables 1 and
2 (for w = 1.1 rad/sec and 10 rad/sec). As one can see, although the
results are certainly more accurate for the finer mesh, the results of
the 20-element discretization is very respectable, and from the
computational stand-point, by far more cost-efficient.
Although we know of no analytical results for the response problem of an
axisymmetric tank with inclined wall, we made several calculations using
a 20-element discretization on a slant tank similar to the one we used
above, except that an inclination (from vertical axis of 0°, 15° and.
30°) is imposed on the si de wall. R = 10 feet at the tank base.
Figures 12 and 13 show that pressure distribution along the side wall
(e = 0°) for w = 1 rad/sec and w = 6 rad/sec, respectively. For
~ = 15°, it is calculated that resonance occurs at w '" 2.5, 4.5, ••• (see
Figure 14). Again, for earthquake bound frequencies, the sloshing
phenomena will be negligible.
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Dynamics Technology, Inc.DT-7814-2
ERST has been implemented on the CDC Cyber 176 Computer of the Uni ted
Computing Systems at Dallas, Texas, under the NOS/BE operating system.
For the simple rigid circular cylindrical structure with a 20-element
symmetri c fi ni te-el ement di scretizati on as described above, undergoing
harmonic ground acceleration, a typical run requires less than 2 seconds
CPU time.
Our last series of tests involve a simple flexible (circular) cylindri
cal tank under a ground acceleration of sinwt, for various values of
w. The results are presented in Figures 15 through 17. Here R = 10 m,
ho = hw = 5m and H = 20 m. A uniformly positi oned 20-el ement fi ni te
element system is selected for the interior fluid domain, which is then
extended naturally to form a 45-element mesh for the exterior water and
a 12-element one for the tank structure. In Figure 15, w = 1 Hz, the
hydrodynamic pressure di stributi on for the fi rst two modes of response
is pl ottedfor t = .25 sec and .75 sec (1/4 and 3/4 cycl es after the
initial excitation). In Figures 16 and 17, w = 10 Hz, and we show
response of the fi rst mode as well as the fi rst two modes for t = .025. .
sec and .075 sec. Notice that the interior pressure forces are approx-
imately 1.5 to 2 times in o'fPlue to that of the exterior pressure. Also
notice that in Figure 15, ozo is close to zero even though the II roof ll is
not assumed to be rigid; but in Figures 16 and 17, larger displacements
step in and change the pressure distribution now that frequency is
bigger. The tank's natural vibration frequencies are ~ ::< 16.7 Hz,
U2 ::< 33.7 Hz.
We were unable to make a comparison of our results to that of Helou's
(1981) presented in his Figure 4.3 for, unfortunately, his results areoP
incorrect as evi denced by the fact that ozw ;/; 0 at the ocean floor,
violating the rigid boundary condition there.
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Dynamics Technology, Inc.OT-7814-2
Table 1. Pressure distribution at wall (9 = 0°)under different finite element discretizations.
P/w at Wall
z 20-element 100-element Analytical
5.0 -8.438 -8.432 -8.4312.0 -8.410 -8.404 -8.4031.0 -8.386 -8.381 -8.3810.66 -8.376 -8.372 -8.3720.33 -8.366 -8.361 -8.3610 -8.354 -8.349 -8.349
0 -10.286 -10.292 -10.292-0.33 -10.272 -10.278 -10.276-0.66 -10.260 -10.265 -10.266-1.0 -10.248 -10.254 -10.255-2.0 -10.221 -10.228 -10.228-5.0 -10.188 -10.195 -10.196
w = 1.1 rad/sec
Table 2. Pressure distribution at wall (9 = 0°)under different finite element discretizations.
P/w at Wall
z 20-element 100-element Analytical
5.0 -8.999 -8.996 -8.9962.0 -9.088 -9.078 -9.0751.0 -9.168 -9.159 -9.1530.66 -9.203 -9.198 -9.2010.33 -9.244 -9.247 -9.2500 -9.293 -9.322 -9.342
0 -9.194 -9.162 -9.137-0.33 -9.251 -9.247 -9.241-0.66 -9.298 -9.304 -9.309-1.0 -9.339 -9.350 -9.357-2.0 -9.432 -9.444 -9.446-5.0 -9.536 -9.539 -9.539
w = 10 rad/sec
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Dynamics Technology, Inc.DT-7814-2
5. SUMMARY AND FUTURES WORK
A rational and effective method to assess the effects of fluid-structureinteraction, on the dynamic response of submerged underwater oil storagetanks under earthquake excitations, has been developed. Our approach isthe hybrid-finite element method, discretizing the tank structure, theinterior fluids, as well as the near field of the exterior water regioninto a ring-shaped mesh network. Solution for the hydrodynamic pressuredistribution is substituted into the structural equation of motion inmodal coordinates to obtain the system displacement information. A comprehensive and predictive computer code is developed for design engineering applications. The program is described in the Appendices. Ourprogram has ben validated against known analytical solutions and shownto be effective an<t accurate with the added flexibility for arbitraryaXisymmetric tank shapes.
Duri ng thi s invest; gati on, we al so reconfi rmed the fact that gravi tyeffects can be safely neglected in evaluating the hydrodynamic pressure
induced at the wall of a submerged tank, such as the one under study.It is also observed that due to the unappreci ab1e difference of Po and
Pw' changes in tne ratio ho/hw above do not significantly influence thefluid-structure response.
From (3.68), it can be seen that the hydrodynamic interaction caused bythe fluids inside and outside the tank contributes to the structuralequation of motion in the forms of additional terms which can be viewedas added mass and added exci tati ons. The added i nerti a of waterincreases the natural period of the tank free oscillation, and decreases
(depends on the deflected shape) the modal damping ratios. However, thepresence of water does not influence the stiffness matrix [K].
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Dynamics Technology, Inc.DT-7814-2
Finally, the incorporation of analyltic representation of Pw in the
hybrid approach proves to be a useful cost-saver especially for tanks of
odd shape and wi th i rregul ar bottom topography. It isantici pated that
this super-element should play an even bigger role in the treatment of
sail-structure interactions.
The computer program developed in this study can be used, with appro
priate modifications, for the following studies:
1. Tanks of open bottoms.
2. Tanks which are floating or moored.
3. Tanks which are non-axisymmetric.
4. Effects of non-rigid tank support and marine soil-structureinteraction.
Understanding the nonlinear behavior of the kind of systems we have been
i nvesti gati ng shaul d, of course, be the ultimate study goal, but the
numerical details it involves are very complicated. An in-depth analy
si s may al so be needed to rel ax our assumptions that verti cal ground
motion and fluid incompressibility can be neglected. For the former,
other modes of shell vibration, such as the breathing mode may need to
be considered. Compressibility may become important for high values of
excitation frequencies (for large-scale earthquakes). Indeed, with
sound speed in water in the vicinity of 4720 ft/sec such high frequency
excitations give rise to sound waves of lengths comparable to the tank
dimension and the water depth. For squatty tanks, the compressibility
effect may not be ignored.
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Dynamics Technology, Inc.DT-7814-2
REFERENCES
1. Arya, A.S., Thakkar, S.K. and Goyal, A.C., "Vibr>ation Ana1-ysis ofThin CyUnd:Y'iea1- Container>s~" Journal of the Engineering MechanicsDivision, Proceedings of the ASCE, Vol. 97, No. EM2, April 1971.
2. Aslam, M., Godden, \!J.G. and Scalise, D.T., "Ear>thquake SLoshing inAnnu1-ar> and CyUndr>iea1- Tanks~" Journal of the Engineering Division, Proceedings of the ASCE, Vol. 105, No. EM3, June 1979.
3. Bai, K.J. and Yeung, R., 'Wumer>iea1- S01-utions of Fr>ee-SUr>faee F1-0~
'Pr>ob1-ems~" Proceedi ngs of the 10th Symposi um on Naval Hydrodynamics, Cambridge, Mass., 1974.
4. Baron, M.L. and Skalak, R., "Fr>ee Vibr>ation of F1-uid-Fi1-1-ed Cy1-ind'Y'iea1- SheUs~" Journal of the Engineering Mechanics Division,Proceedings of the ASCE, Vol. 88, No. EM3, June 1962.
5. Basu, P.K. and Gould, P.L., "SHORE-II~ SheU of Revo1-ution FiniteE1-ement Pr>ogr>am - Statie Case~ Theor'etiea1- Manua1-~" StructuralDivision, Department of Civil Engineering, Washington University,St. Louis, Missouri, March 1975.
6. Berkhoff, J. C. W., "Computation of Combined Refr>aetion-Diffr>aetion~"
Proceedings of the 13th Coastal Engineering Conference, Vancouver,1972.
7. Black, J.L. and Mei, C.C., "Seatter>ing and Radiation of Water>Waves~ " Ral ph M. Parsons Laboratory for Water Resources and Hydrodynami cs Report No. 121, Massachusetts Insti tute of Technology,Cambridge, Mass., April 1970.
8. Chak rabarti, P. and, Chopra, A. K., "Ear>thquake Response of Gr>avityDams Ine1-uding Reser>voir> Inter>aetion Effeets~ " EarthquakeEngi neeri ng Research Center Report No. EERC 72-6, Uni versity ofCalifornia, Berkeley, Calif., December 1972.
9. Chamberl in, R. S. , "Khazzan Dubai 1: Design~ Constr>uetion andIn8taUation~" Proceedings of the Second Annual Offshore Technology Conference, Paper No. OTC 1192, Houston, Texas, April 1970.
10. Chen, H.S. and r~i, C.C., '~8ei1-1-ations and Wave For>ees in an Offshor>e Har>bor> (AppUeations of Hybr>id Finite E1-ement Method toWater>-Wave Seatter>ingJ ~" Ral ph M. Parsons Laboratory for WaterResources and Hydrodynamics Report No. 19, Massachusetts Instituteof Technology, Cambridge, Mass., August 1974.
-43-
uynamlcs lechnology, Inc.DT-7814-2
REFERENCES (Continued)
H. Chester, W., "Resonant OsaiUation of Watero Waves,," Proceedi ngs ofRoyal Society of London, Vol. 306, 1968.
12. Edwards, N.W., ,~ Prooaeduroe foro the Dynamia Analysis of Thin-WalledCylindroiaal Liquid . Sf;oroage Tanks Subjeated to Lateroal GrooundMotions,," Ph.D. Dissertation, University of Michigan, Ann Arbor,Mich., 1969.
13. Faltinsen, O.M., "A Nonlinearo Theoroy of sLoshing in ReatangularoTanks,," Journal of Ship Research, Vol. 18, No.4, 1974.
14. F1Ugge, W., Stresses in Shells, Springer-Verlag, Berlin, 1960.
15. Garrett, C.J.R., "Wave Foroaes on a Ciroaularo Doak,," Journal of FluidMechanics, i§.., Part 1, 1971.
16. Garrison, C.J. and Seetharama Rao, V., "Interoaation of Waves withSubmeroged Objeats,," Journal of Waterways, Harbors and CoastalEngineering Division, Proceedings of· the ASCE, May 1971.
17 • Ghosh, S. and Wi 1son, E. L., "Dynamia Sf;roess Analysis ofAxisymmetroia Sf;rouaturoes Under' Arobitroaroy Loading,," Earthquake EngineeringResearch Center Report No. EERC 69-10, University of Cal iforni a,Berkeley, California, September 1969. Revised by Lin, C-J.,September 1975. '
18. He lou, A. H., "Seismia Analysis of Submeroged Under'1JJatero Oil Sf;oroageTanks,," Ph.D. Dissertation, North Carolina State University,Raleigh, North Carolina, 1981.
19. Housner, G.W., "Dynamia Proessuro(:Js on Aaaelemted Fluid Container's,,"Bulletin of the Seismological Society of Jlmerica, Vol. 47, No.1,January 1957.
20. Jacobsen, L.S., "Impulsive Hydroodynamias of Fluid Inside Cylin-:droiaal Tanks and of a Fluid Suroroounding a Cylindroiaal Peiro,,"Bulletin of the Seismological Society of Jlmerica, Vol. 39, July1949.
21. Li aw, C. Y. and Chopra, A.K., "Earothquake Response ofAxisyrrunetroiaT~er' S!;r'uatur'es Sur'roounded by Water',," Earthquake EngineeringResearch Center Report No. EERC 73-25, University of California,Berkeley, California, October 1973.
-44-
Dynamics Technology, Inc.DT-7814-2
REFERENCES (Continued)
22. Mei, C.C., Foda, M.A. and Tong, P., "Exa~t and Hybpid-Etement SOlu.. tion fop the Vibmtion of Thin Elastic Stpuctupes Seated on the
Sea Floop~" Applied Ocean Research, Vol. 1, No.2, April 1979.
23. Sanders, J.L., Jr., "An Imppoved Fipst-Apppoximation Theory fopThin SheUs~" National Aeronautical and Space Administration Technical Report No. R-24, NASA, Washington, D.C., 1959.
24. Shaaban, S.H. and Nash, W.A., "Response of an Empty CyUndpica'tGpound-suppopted Liquid stomge Tank to Base Excitation~ "Engineering Research Institute Report, Department of CivilEngineering, University of Massachusetts, Amherst, Mass., 1975.
25. Takayama , T., "Theory of Tr>ansient Fluid Waves in a VibpatedStopage Tank~" The Port and Harbor Research Institute, Vol. 15,No.2, Ministry of Transportation, Nagase, Yokosuka, Japan, June1976.
26. Tu ng , C. C• , "Hydpodynamic Fopces on SUbmepged Vepticat CipcuZapCyUndpicaZ Tanks Undep Gpound Excitation~ 1/ Appl i ed OceanResearch, Vol. 1, No.2, April 1979.
27. Wu, C.I., Monzakis, T., Nash, \~.A. and Col one" , J.M., "NatupaZFpequencies of CyZindpicaZ Liquid Stopage Containeps~" EngineeringResearch Institute, University of Massachusetts, Amherst, Mass.,1975.
28. Yang, J. Y., "Dynamic Behaviop of FZuid-Tank Sy8tems~" Ph. D. Di ssertation, Rice University, Houston, Texas, March 1976.
29. Yue, D.K.P., Chen, H.S. and Mei, C.C., I~ Hybpid Element Method fopCaZcuZating Thpee-DimensionaZ Watep WaVe Scatteping~" Ral ph M.Parsons Laboratory for Water Resources and Hydrodynami cs ReportNo. 215, Massachusetts Institute of Technology, Cambridge, Mass.,August 1976.
-45-
Fi gure 1. Secti 0 T'n hrough Kh 'azzan Dubai 1
-46-
f---- _ ..__.- ~
t----..- -- _. ---- . I--\l--l...
rt
!t50.:0·t• . _ .•
------~
Figure 2. Major Dimensions of Khazzan Dubai 1
-47-
(actuallyfarther away)
H
v
Figure 3. CoordinateSystem for AxisymmetricSubmerged Storage Tank
x
--- - -- --- -- - -- - - - ----
---------- ----
I
I
II
'-5I fI,I-,n,
1--__- II,I
I,I,I
III
--n
-48-
2
Global Coordinate System
5
2
t
I._+_.I
H,-I,B)
z
(1,-1, B)L------__._ r
Local Coordinate System
Figure 4. Finite Element Coordinate System
-49-
X~cos8
xnn
x~ cas8
g::~:
.//
I
/ zh\\\
'\.""' '-
nTH. NODAL
CIRCLE
FINITE ELEMENTIDEALIZATION
GROUND DISPLACEMENT RELATIVE DISPLACEMENTSOF NODAL-CIRCLE n
Figure 5. Axisymmetric Structure Subjected to Horizontal Ground Motion
-50-
H
h +zw
~
1.0.;----~-~.
, , , ,\
\\
\
\\
\
\\
\\
\\\,
\\\,I\
\III
II
II\II
I
i
2p/pR (m/s )
0.2 0.3 0.4 0.5
Figure 6. Pressures on wall of cylindrical tank e =0°.
-- H/(H-h ) = 2; ----- H/(H-hw) = 5, R/H = 1'vI
-51-
3'
/!- --C_
1/1 !1/3 ~f
~t
I1 6
..
2 7
.~ 84 I 9t:; ! 10
11 ! 1612 1711 18
II 14 19
15 20
I~~----------~'-O(:----------~l'15/ 5/
~Interface
Figure 7. Interior Fluid Domain Finite Element Discretization
-52-
w =0.5 rad/sec.
m FEM
- Analytical
Z FEM Analytical
5.0 -16.639 -16.6372.0 -16.631 -16.629
15' 1.0 -16.623 -16.623
Po 0.66 -16.619 -16.6190.33 -16.616 -16.616
_I0 -16.612 -16.612
Pw51
Z FEM Analytical
0 -19.491 -19.491I ·1I- -0.33 -19.485 -19.485
10 1-0.66 -19.481 -19.481-1.0 -19.478 -19.478-2.0 -19.470 -19.470-5.0 -19.460 -19.460
Z
5.0
Figure 8.
3.0
2.0
1.0
r--j----.........,.:l-.-. P/ w
-1.0
-2.0
-3.0
-4.0
r-5.0
Pressure Distribution at Inner Wall (8 =0°) of aSubmerged Cylindrical Tank
-53-
lb-·secft2
j ~,ii1~ytical.,
Z FEM
5.0 -16.373 i -]',6.359
2.0 -16.319 -It.3051.0 -16.272 i -1£.2620.66 -16.253 ~ 16.2450.33 -16.233 I 1.6.224 !
0 -16.210 J -16.200 i
iZ FEM Analytical II
0 -19.959 -1'9.970-0.33 -19.932 19.939,-0.66 -19.908 , 19.920-1.0 -19.885 ; -19.899-2.0 -19.833 -19.846-5.0 I -19.769 -19.784z
.. [10'
T5'
__-T-r-T"--r-.,....,-....-.--r-
P-r-W-;--]--r-..,..-r-r-';"" -1 5 '
w =1.1 rad/sec.
lSI FEM- Analytical
..;.20
-16.3 -16.1
5.0
4.0
3.0
2.0
1.0
-1.0
-2.0
-3.0
lbP/w -.secft2
-4.0
-5.0
Figure 9. Pressure Distribution at Inner vIall (e ==0°) of aSubmerged Cylindrical Tank
-54-
T5't------p-w------I 1"5'
w =6.0 rad/sec.
11 FEM- Analytical
R
iI
-17.5
z
Z FEM Analyti cal
5.0 -17.531 -17.5262.0 -17.743 -17.7311.0 -17.943 -17.9560.66 -18.0300.33 -18.133 ,
0 -18.261 -19.374
Z FEM Analytical
0 -17.574 -17.440-0.33 -17.723-0.66 -17.842-1.0 -17.945 -17.927-2.0 -18.176 -18.191-5.0 -18.422 -18.430
-18.5
Figure 10. Pressure Distribution at Inner l·~al1 (e =0°) of aSubmerged Cylindrical Tank
-55-
t. FEM Analytical
5.0 -17.462 -17.456
15'
2.0 -17.634 -17.609Po 1.0 -17.789 -17.760
0.66 -17.857 -17 .853
Pw 15'
0.33 -17.937 -17 .9490 -18.032 -18.127
Z FEM Analytical
10 I ~ 0 -17.753 -17.729-0.33 -17.951 . -17.931-0.66 -18.042 -18.063-1.0 -18.121 -18.156
w = 10.0 radjsec-2.0 -18.302 -18.333
Z-5.0 -18.504 -18.509
• FEM- Analytical
Figure 11. Pressure Distribution at Inner Wall (8 =0°) of aSubmerged Cylindrical Tank
-56-
z
B =Inclination Angle
w = 1.0 rad/sec.
-1.0
-5.0
1.0
5.0
.i--------P/w Jb_. secft2
-7.8
\ !\
11 /1 '
illI ,,I / /iiI..
Fi gure 12. Pressure Di stri bution at Inner ~'Ja11 (e = 0°) of aSubmerged Inclined Tank
-57-
z
-5.0
8 =0° 8 = 15° 8 = 30°
Y5.0
8 = Incl i nati on Angl e
\ \ w=6.0 rad/sec.
J 1.0;
lb
-19.41 -7.8P/w -oseeft-2
-1.0
Figure 13. Pressure Distribution at Inner to/all (8 =0°) of aSubmerged Inclined Tank
-58-
10.
9.8.
7.
Incl
inat
ion
Ang
le=
15°
5.
6.
w(r
ad/s
ec)
4.3
.2.
1.
o
11;;1
X10
2
, U"1
\.0 I
10.
20.
ft.
Fig
ure
14.
Oil
-Wat
erIn
terf
ace
Dis
plac
emen
t-F
requ
ency
Cur
veof
aSu
bmer
ged
Incl
ined
Tank
w=
2rr
rad/
sec
Mod
esth
roug
h2
----
----
Pre
ssur
eon
Inne
rW
all
----
Pre
ssur
eon
Ext
erio
rW
all
t=
.75
sec
15.m
"" ",
,
1\
\\ \
1\
\ \ \ \
1\ \ ,\
\ \ \
IO
..5
I.:]
1.5
\.
L.1
,./
//
//
I/
/ /I I
I I I f , I I Jt
=.7
5se
cI I , I I I , I I I
t=
.25
sec
I I I I I I
-2.
I-1
.5-1
.:-t
II
j;
\T
P(k
sf)
....
1t
III
I I , I \ \ I It=
25se
cI
•
I \ , I \ , I I I I ,
I O"l o t
Figu
re15
.P
ress
ure
Dis
trib
utio
nat
Tank
Wal
l(s
=0°
)o
fa
Subm
erge
dF
lexi
ble
Tank
w=
20n
rad/
sec
Mod
esth
roug
h1
----
----
-P
ress
ure
onIn
ner
Wal
l----
Pre
ssur
eon
Ext
erio
rW
all
2.
t=
.075
sec
I I I I I I I I It
=.0
25se
cI I I I I \
"-"-, ,
", ," \ \ \ \ \ \ \ \ \ \ \ \ \ \
.5\1
.o.rm
I I I I I J , I I I It
=.0
75se
cI I I I I
"/
//
/I
II
I I I I J I' I I J I I J-1
./-.
5-1
.5
t=
.02
5se
c
-2.
I O'l ...... ,
Fig
ure
16.
Pre
ssur
eD
istr
ibut
ion
atTa
nkW
all
(8=
0°)
of
aSU
bmer
ged
Fle
xibl
eTa
nk
w=
2011"
radj
sec
Mod
esth
rou
gh
2
----
----
-P
ress
ure
onIn
ner
Wal
l
----
Pre
ssu
reon
Ex
teri
or
Wall
At
-.5
2.
t=
.075
sec
\
\ \ \ \ \ \ \ \ I I I
1.I
\1.5~
-f1
_--:
:---
:::--
.t-'ll
-I
y-,,-
P(k
sf)
I f I I I I I
t=
.025
sec; I I I f I I I I I
...."-
""-""-
"-.....
'\
" \
.5_
5.m
o. -5.
I I I 1 I· I ( I I )t=
=.0
75se
c, I I I r· I I I J
,/
t'/
/
II
I I r I , , I r I ( I ,-1
.5-1
.I
II
I-2
.--f
t:=
.02
5se
c
f-I
• (1) N I
Fig
ure
17.
Pre
ssu
reD
istr
ibu
tio
nat
Tan
k~Jall
(e=
0°)
of
aS
ubm
erge
dF
lex
ible
Tan
k
r C'I
W ,
LAYO
UT
Figu
re18
.ER
STPr
ogra
mS
truc
ture
[~~ST
I
-{111
z
K
L
I J
~-------------------------J--r
Figure 19. Finite Element Nodal Points Ordering Scheme
-64-
z
r
[]E
xter
ior
Wat
erN
odal
-Cir
cle
()S
tru
ctu
ral
Nod
al-C
ircl
e
Li
Inte
rio
rF
luid
sN
odal
-Cir
cles
o~
y----
----
-crr
-I I I
II
I
-----
----
----0
----
-----
----
[~}
-----
----
----
----
-0-·
----
-----
_.---
---~
J II
II
I
I2
I3
II
1Q
P7"
l~
l7"'
<rn
r-:
.:_
_._._
_.__..~:
_._._._.
_:
:.:~.It:.:.::J.~
I en (J1 I
:::
.-:
:
Fig
ure
20.
Fin
ite
Ele
men
tA
ssem
blag
esSh
own
atC
ross
-Sec
tion
e=
0°.
(Not
eth
atex
teri
or
wat
erno
dese
quen
cest
arts
atst
ruct
ure
inte
rfac
e.)
Dynamics Technology, Inc.DT-7814-2
Appendix A. USER 1 S GUIDE TO CO/VPUTER PROGRAM IlERST"
The computer program named ERST, for Earthquake Response of Sea-Based
Storage Tanks, is developed and synthesized to evaluate the elasto
hydrodynamic response of axisymmetric storage tanks sUbmerged underwater
induced by earthquake ground motions.
The tank structure is ri gi dly attached to the ocean floor, has a fl ex
ible wall. The tank has a vertical axis of symmetry, is submerged
underwater (structures protrudi ng out of ocean surface can al so be
handled with minimum modifications) and filled with two different layers
of fluids. The whole system is discretized into three assemblages of
toroidal finite elements for: the interior fluids domain, the tank
shell, and the near field exterior water region.
Nodal displacements, stresses, as well as hydrodynamic pressures are
output as the response to the input (horizontal) ground acceleration.
Both the dynamic and the static responses are eval uated. A si gnificant
portion of the code pertaining to the dynamic behavior of exterior
water-structure interaction is based on the EATSW program written by
Dr. C. Y. Liaw for earthquake analysis of axisymmetric intake tower
structures surrounded by water, purchased through the NISEE/Computer
Applications Program, and modified to interface with the developed
i nteri or fl ui ds-structure i nteracti on program through the ki nd assi s
tance of Dr. Liaw.
Figure 18 gives the general code structure with accompanying descrip
tions of the functions of the subroutines and the tape files employed.
Note that a plotting scheme is built in based on the available sub
routi nes GRAFl1, PLOT, OPLT, MULPLT (i n subrouti nes LAYOUT, EIGEN and
TSTEP) interfaced to a Calcomp plotter. Users are advised to modify
this portion to suit onels own environment.
-66-
ERST
LAYOUT
ELEMENT -
TOTAL
STATIC
EIGENEXTSOLEXT
INTBNDWTH
SOLINT
TSTEP
RESPON
TAPElTAPE2
TAPE3TAPE4TAPE6TAPE7TAPE8TAPE9
Dynamics Technology, Inc.DT-7814-2
Driver program
Determines structure characteristics and finite-elementconfigurationsDetermines element stress strain, mass and stiffnessmatrices
Assembles global mass and stiffness matrices; also evaluates hydrostatic contribution for static analysisComputes static displacements and stress for staticanalysisDetermines structural free vibration modes and frequencies
Prepares for exterior water finite-element system set-upComputes general i zed force vector and mass matri x contributed from exterior water and superelementPrepares for interior fluids finite-element system set-upDetermines interior fluid element matrix bandwidthComputes generalized forces vector and mass matrix contributed from interior fluidsSolves for modal equation of motion by time-step integration
Calculates dynamic response by modal superpositions
Stores element mass and stiffness matrices informationStores compacted mass matrix as well as x-displacementresponse informationStores element static response informationStores total stress informationStores plot informationStores Y-displacmeent response informationStores eigenmodes and eigenvalues informationStores hydrodynamic pressure as well as interpolatedearthquake data information
-67-
Dynamics Technology, Inc.DT-7814-2
In the following, we describe the data deck structure to be used forexecuting the program.
1. TITLE CARD (8AUH
II. STRUCTURE INFORMATION (NAMELIST:TANK)
NUMNP: number of structural nodal-circles
NUMEL: number of structural elements
NUMMAT: number of different structural materials
NfvODE: number of modes of vibration included. If NMODE=0and IGRAV=0, only static responses are evaluated.
WLO: level of water surrounding the structure (in feet)
WLI: level of water in the interior of a hollow structure(i n feet)
NPO: number of structural nodal-circles on the exteriorsurface affected by the surrounding water
NPI: number of structural nodal-circles on the interiorsurface affected by interior fluids
IGRAV=0, perform static analysis only
=1, perform dynamic analysis only
=2, perform static as well as dynamic analysis
READFRQ=.FALSE., compute and write to TAPES frequencies andmode shapes of the structure without the surrounding water
READFRQ=.TRUE., skip calculation of frequencies and modeshapes; instead read from TAPE8
PLOTANK=.TRUE., plot structure elements
PLOTfvOO=.TRUE., plot mode shapes
-68-
Dynamics Technology, Inc.DT-7814-2
I II. MATER-I AL PROPERTY INFORMATION
One set of cards must be suppl i ed for each different material used instructure.
Card 1 (I5,5X,All)
Columns 1 - 5: Material sequence number
11 - 21: Either 'ISOTROPIC' or 'ORTHOTROPIC' to indicatematerial property
Card 21 (3FI0.0): Properties of Isotropic Material
Columns 1 - 10: Modulus of elasticity (in ksf - kips per square foot)
Properties of Orthotropic Materi a1
En· modul us of el asti ci ty (in ksf)
Es • modulus of elasticity (in ksf)
Et • modulus of el asti ci ty (in ksf)
vns ' Poisson's ratio
Vnt' Poisson's ratio
Vst· Poisson's ratio
Properties of Orthotropic Material(continued from Card 20)
11 - 20:
21 - 30:
31 - 40:
41 - 50:
51 - 60:
Card 3 (5F10.0):
Columns 1 - HI:
11 - 20: Poisson's ratio
21 - 30: Mass density (in kip-sec2/ft4)
Card 20 (6F10.0):
Columns 1 - 10:
11 - 20:
21 - 30:
31 - 40:
Gns ' shear modulus (in ksf)
Gnt , shear modulus (in ksf)
Gst • shear modulus (in ksf)
Mass density (in kip-sec2/ft4)
41 - 50: Angl e ~ in degrees measured counter-cl ockwi se fromthe r-axis to the n-axis
The n-s axes are the principal axes for the orthotropic material. and tis the tangential direction of the axisymmetric coordinates.
-69-
Dynamics Technology, Inc.DT-7814-2
IV. STRUCTURAL NODAL-CIRCLE CARDS (I5,F5.~,2F10.~,2I5,2F1~.0)
Columns 1 - 5: Nodal-circle number
8 - 1~: Boundary condition code"1" in column 8 if r-displacement is restrained
"1" in column 9 if z-displacement is restrained"1" in column 10 if 9-displacment is restrained
11 - 20: r-ordi nate (i n feet)
21 - 30: z-ordinate (in feet)
31 - 40: Used for layer generation, otherwise leave blank.
Nodal-circle cards must be in numerical sequence. If cards are omittedand Columns 31 - 60 are left blank, the omitted nodal-points are generated along a straight line between the defined nodal-points. (See Note1); or if Col umns 31 - 60 are not bl ank, they are generated in 1ayers(see Note 2).
Note 1: Straight line generation
If the (L-1) cards for nodal-circles N+1, N+2, ••• ,N+L-1 are omitted andColumns 31 - 60 of the card for nodal-circle N are left blank, theomitted nodal-circles are generated at equal intervals on the straightline joining nodes Nand (N+L).
Note 2: Layer generation
Layer generation may be used after two rows of nodal-circles are CO!11pl etely defi ned. If, on the card for node N, the foll O\'Ii ng data isspecified:
Columns 31 - 35: t()D: module, m (> 0)
36 - 40: NLIM: limit of generation (> N)
41 - 50 : FACX: ampl ification factor f r (default f =1)r
51 - 60: FACY: amplification factor f z (default fz=1)
the r-z coordinates of points N+1,N+2, ••• ,NLIM are generated by the formulas:
-70-
Dynamics Technology, Inc.DT-7814-2
for k = N+1, ••• ,NLI M. If NLI M = NUr1NP, no more nodal-cards areneeded. If NLIM < NUMNP, the card for circle (NLIM+1) must follow.
The boundary condition code for generated nodal-circles is set equal tozero, i.e., these nodal-circles are unrestrained in all r, z and e coordinates.
v. STRUCTURAL ELEMENT CARDS (615)
Columns 1 - 5: Element number
6 - 10: Nodal-Point I
11 - 15: Nodal-Point J
16 - 20: Nodal-Point K
21 - 25: Nodal-Point L
The maximum difference "b"between these numbers is anindication of the bandwidthof the Stiffness Matrix. "b"may be minimized by a judiciousnumbering of nodal points.
26 - 30: Material identification
Nodal-point numbers I, J, K and L must be in sequence in a counterclockwise direction around the element (cf. Figure 19). Element cardsmust be in element number sequence. If element cards are omitted, theprogram automatically generates the omitted information by incrementingby one the preceding I, J, K and L. The material identification for thegenerated card is set equal to the corresponding value on the lastcard. The last element card must always be supplied. Triangularelements are also permissible; they are identified by repeating the lastnodal number (i.e., I, J, K, K).
-71-
Dynamics Technology, Inc.DT-7814-2
VI. WATER PRESSURE CARDS
Card Set 1 (1615): This set of cards are to be omitted, if NPO = 0.
Cplumns 1 - 5: Nodal-circles of the structure affected by the
6 - 10: surrounding water; n1,n2, ••• ,nk = NPO, starting
etc. from roof center.
Card Set 2 (1615): This set of cards are to be omitted, if NPI = O.
Columns 1 - 5: Nodal-circles of the structure affected by the
6 - 10: interior fluids; m1,m2, .•• ,mL' L = NPI, starting
etc. from inner roof center.
VII. EXTERIOR FLUID CARDS
Card 1. FLUID DOMAIN DISCRETIZATION (NAMELIST: EXTFLD)
NUMPEX: number of fluid nodal-circles
NUELEX: number of fluid elements
NUINTEX: number of nodal-circles on the fluid-structure interface in the finite element idealization of the fluid
NUFSF: number of nodal circles on the free surface in thefinite element idealization of the fluid and at thez-axis.
NUCOF: number of coefficients of super-element pressurerepresentation.
-72-
Dynamics Technology, Inc.DT-7814-2
Card 2. FLUID NODAL-CIRCLES (2I5,2Fl~.0,2I5,2F10.~)
Nodal-circles in the finite element idealization of the fluid domainmust be numbered in such a way, that the first NUINTEX nodal-circles arelocated on the structure-fluid interface (cf. Figure 20). The numberingmust start at the top as shown there. In the finite element idealization for the fluid domain, nodal-circles on the interface must be provided to coincide with the nodal-circle in the structural idealization.Additional nodal-circles can be included in the idealization of thefluid domain, as shown in Figure 20.
One card for each nodal-circle containing the folloWing information mustbe provided.
Columns 1 - 5: Fluid nodal-circle number
6 - 1~: ICOOE: Boundary condition code for fluid nodalcircle. If the fluid nodal-circle is on theinterface and coincides with a structuralnodal-circle, ICOOE = the number of the coincident structural nodal circle.
If the fluid nodal-circle is on the interfacebut does not coincide with a structuralnodal-circle, then ICOOE = -1.
If the fluid nodal-circle is on the free surface of the fluid domain, ICOOE = -2.
For all other nodal-circles, leave Columns26-30 blank. Special attention must be givento fluid nodal-circle number 1; it must coincide with the roof center of the tank.
11 - 20: r-ordinate (in feet)
21 - 30: z-ordinate (in feet)
31 - 60: Used for layer generation; otherwise leave blank.
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Dynamics Technology, Inc.DT-7814-2
Nodal-circle cards must be in numerical sequence starting from one. Ifcards are omitted and Columns 31-60 are left blank, the omitted nodalpoints are generated along a straight ine between the defined nodalcircles (see IV, Note 1); or if columns 31-60 are used, they are generated in layers (see IV, Note 2). The boundary condition code (ICOOE) ofa generated nodal circle is set equal to the value of ICODE on the lastcard.
Card 3. FLUID ELEMENTS (915)
Columns 1 - 5 Element number
6 - 10: Nodal-Circle I
11 - 15: Nodal-Circle J
16 - 20 : Nodal-Circle K
21 - 25: Nodal-Circle L
Fluid Element Surface Code (=2, hybrid surface)
26 - 30: surface IJ
31 - 35: surface JK
36 - 40: surface KL
41 - 45: surface LI
Nodal-circle numbers I, -J, K and L must be in sequence in a counterclockwise direction around the element (cf. Figure 19). Element cardsmust be in element number sequence. If element cards are omitted, theprogram automati ca lly generates the omitted i nformati on by i ncrementi ngby one the preceding I, J, K and L. The last element card must alwaysbe supplied. Triangular elements are also permissible; they areidentified by repeating the last nodal number (i.e., I, J, K, K)
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Dynamics Technology, Inc.DT-7814-2
VIII. INTERIOR FLUID CARDS
Card 1. FLUID DOMAIN DISCRETIZATION (NAMELIST: INTFLD)
NUNPIN: number of fluid nodal-circles
NUELIN: number of fluid elements
ROIL: specific gravity of oil
NUINTIN: number of elements at oil/water interface
NUELOIL: number of oil elements; the oil elements must benumbered from 1 to NUELOIL
2. FLUID NODAL-CIRCLES (215,2FI0.0,215,2FI0.0)
One card for each nodal-circle containing the following information mustbe provided.
Columns 1 - 5: Fluid nodal-circle number
6 - 10: ICODE: Boundary condition code for fluid nodalcircle. If the fluid nodal-circle is on thewall and coincides with a structural nodalcircle, ICODE = the number of the coincidentstructural nodal-circle.
If the fluid nodal-circle is both on the walland on the interface and coi nci des with astructural nodal-circle, then ICODE = - (thenumber of the coincident structural nodalcircle).
11 - 20: r-ordinate (in feet)
21 - 30: z-ordinate (in feet)
31 - 60: Used for layer generation; otherwise leave blank.
Nodal-circle cards must be in numerical sequence starting from one. Ifcards are omitted and Columns 31-60 are left blank, the omitted nodalpoints are generated along a straight line between the defined nodalcircles (see IV, Note 1), or if Columns 31-60 are used, they aregenerated in layers (see IV, Note 2). The boundary condition code(ICODE) of a generated nodal-circle is set equal to zero.
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Dynamics Technology, Inc.DT-7814-2
Card 3. FLUID ELEMENTS (SIS)
Columns 1 - 5: Element number
6 - 1~: Nodal-circle I
11 - 15: Nodal-circle J
16 - 2~: Nodal-circle K
21 - 25: Nodal-circle L
Nodal-circle numbers I, J, K and L must be in sequence in a counterclockwise direction around the element (cf. Figure 19). Element cardsmust be in element number sequence. If element cards are omitted, theprogram automatically generates the omitted information by incrementingby one the preceding I, J, K and L. The last element card must alwaysbe supplied. Triangular elements are also permissible; they areidentified by repeating the last number (i.e., I, J, K, K).
Card 4. OIL/WATER INTERFACIAL ELEMENT (1615)
List of NUINTIN element numbers on the oil-water interface.
IX. RESPONSE CONTROL CARDS (NAMELIST: RESPONS)
NGRD:=l, if only one component of ground motion, al~ng e = ~9,
is to be considered
=2, if two components of ground motion, along e = ~o ande = 9~o, are to be included.
NT: number of integration steps in time
DT: time interval in step-by-step integration
NXFH: number of ordinates describing time history of goundmotion component 1 along e = ~o.
NYFH: number of ordinates describing time history of groundmotion component 2 along e = 9~o.
PLOTALL=.FALSE., ground motion not plotted.
=.TRUE., plot ground motion.
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Dynamics Technology, Inc.DT-7814-2
X. DAMPING RATIO CARDS (7F10.0)
Columns 1 - 10: Damping ratio for first mode of vibration of thetank.
11 - 20: Damping ratio for second mode of vibration of thetank.
21 - 30: Damping ratio for third mode of vibration of thetank.
etc.
XI. GROUND ACCELERATION CARDS
(i) FIRST COtvPONENT - Along = ~o
Card 1 TITLE (8A10)
Card 2 ACCELERATION (6(F6.3,F6.4),8X)
NXFH time-acceleration pairs describing the time-history of the component of ground acce1erati on along e = ~o are to be spec ifi ed on thesecards, with six pairs per card. Time must be expressed in seconds andaccelerations as multiples of g, the acceleration due to gravity.
(i i) SECOND COtvPONENT - Along e :: 90°
Card 1 TITLE (8A10)
Card 2 ACCELERATION (6(F6.3,F6.4),8X)
NYFH time-acceleration pairs describing the time-history of the component of ground acceleration along e :: 90 0 are to be specified on thesecards, with six pairs per card. Time must be expressed in seconds andaccelerations in multiples of g.
-77-
Dynamics TechnOlogy, Inc.DT-7814-2
XII. OUTPUT INFORMATION CARDS
Card 1. OUTPUT CONTROL (NAMELIST: PRINTIT)
NPRINT:
NNODE:
NNEL:
NANGLE:
Print interval. Nodal-circle displacement amplitudesand element stress amplitudes are written on TAPE3every NPRINT time-intervals. If printed output oftime history response is required (i .e., NNODE ! (3and/or NNEL f 0), nodal-circle displacements and/orelement stresses are also printed every NPRINT timeinterval s.
Total number of nodal-circles at which time-historyof displacements is to be printed.
Total number of elements at which time-history ofstresses is to be printed.
Total number of different directions around the circumference of a nodal-circle or axisymmetric elementat which displacement and stresses are to be printed.
Card 2. ANGLE SELECTION (8F10.0)
List of NANGLE values of angles (in degrees) describing directions alongthe circumference at which displacements and stresses are to beprinted. These cards are to be omitted in NANGLE =0°.
Card 3. NODAL CIRCLE SELECTION (1615)
List of NNODE nodal circle numbers at which displacements are to beprinted. These cards are to be omitted if NNODE = 0.
Card 4. ELEMENT SELECTION (1615)
Li st of NNEL el ement numbers for which stresses are to be pri nted.These cards are to be omitted if NNEL = 0.
-78-
Dynamics Technology, Inc.DT-7814-2
OUTPUT
The following is printed by the program (note that some of these may besuppressed according to the options provided in Cards II and XII).
1. First set of input data:options, etc.
structural and material properties,
2. Hydrostatic loads: i.e., equivalent nodal-circle loads due tohydrostatic pressure of the surrounding water and fluids inside thetank. Nodal-ci rcl e di spl acements and el ement stresses for stati cloads. Added-mass due to fluids inside the tank.
3. Frequencies and mode shapes of the structure, including the effectof i nteri or fl uids but not the effect of water su rroundi ng thestructure.
4. Second set of input data: geometric data for the surrounding andinterior fluids, structural nodal-points affected by hydrodynamici nteracti on.
5. The generalized mass matrix and the generalized force vector,including hydrodynamic effects.
6. Thi rd set of input data: response data i ncl udi ng number of timesteps, time increment, and modal damping ratios, earthquake acceleration data, control data for output of time history of response.
7. Displacements of selected nodal circles (see card group XII),stresses in selected elements along selected angles at instants oftime, determi ned by the pri nt interval NPRINT. The di spl acementsand stresses printed include the static values at the beginning ofthe earthquake motion.
8. The peak values of displacement amplitudes of each nodal-circle andamplitudes of stress in each element and with time at which theyoccur during the earthquake. These peak values exclude the staticvalues.
-79-
Dynaml cs Techno! 09Y, Inc.OT-7814-2
9. The following quantities are written on TAPE3:
Log.ical Record 1: NUMNP, NUMEL, NMOOE, OT, NT, NPRINT, IGRAV, NGRD
Starting with Record 2, two records are written for every NPRINTtime-intervals for each ground motion component using the followingtwo statements:
WRITE (3) X
WRITE (3) STRESS
X and STRESS are one-dimensional arrays, dimensioned properly sothat X(3*1-2), X(3*I-1) and X(3*I) are the r, z and e-components,respectively, of the displacement amplitude at nodal-circle I,
1=1, ••• ,Nl.!M'JP. Similarly, STRESS (6*N-5), STRESS (6*N-4)' STRESS( 6*N-3)'STRESS(6*N-2), STRESS{6*N-1) and STRESS{6*N) are the amplitudes ofthe six components, (jrr, (jzz, (jee, (jrz, (jre and ;e in the elementnumber N, N=1,2, ••• ,NUMEL.
If IGRAV = I, the dynami c responses start wi th the fi rst set of Xand STRESS. However, if IGRAV = 2, the first set of X and STRESS isthe static response; the dynamic response starts with the secondset.
STORAGE REQUIREMENTS
The cord storage requi rements of the program are separated into fi xedand variable parts with the fixed part consisting of instructions, nonsubscripted variables, and those arrays which do not depend on the sizeof the individual problem. The variable part is stored in Array A,which appears in the blank COMMON statement.
The bl ank COMMON storage requi rements of the program can be changeddepending on the size of the problem to be solved. This is done byusing the RFL job control statement to increase the program size:
RFL(MMAX)
MMAX (in octal) is the total memory words requested. N=MMAX-45000 isthe available memory size for blank COMMON storage.
-80-
-.., .. _.... ~- . __ .... _.-;;;J.,J,
OT-7814-2
The value for N must exceed each of the following:
(1) 3*NUMNP*(4 + MBANO) + 11*NUMEL + 11*NUMMAT + NBC + NPO + NPI
(2) 3*NUMNP*(8 + MBANO) + 5*NUMEL + 11*NUMMAT + NBC + NMODE*(NMODE + 1)
(3) 5*NUMNP + (2 + 3*NUMNP + NMODE + 3*NUINTEX)*NMODE + (3 + NEBAND +NMODE)*NUMPEX + 8*NUELEX + NUINTEX + NUFSF + (1 + NEBANO +NM:lDE)*NUCOF
(4) 5*NUMNP + (2 + 3*NUMNP + NMOOE)*NMODE + 7*NUELIN + (4 + 2*NMODE +NIBANO)*NUNPIN
(5) 5*NUMNP + (408 + 3*NUMNP + 2*NMOOE)*Nr~OE + 2*NXFH + 2*NYFH + NT
(6) 852 + (22 + 4*NMOOE)*NUMNP + 805*NMOOE + 42*NUMEL + NM:lOE + NNEL +NANGLE + 2*NT
where:
NUMNP: number of nodal circles in the structural idealization
NUMEL: number of elements in the structural idealization
NUMMAT: number of different structural materials
NBC: number of structural displacement constraints in thestructural idealization
NPO: number of nodal circles on the exterior surface of thestructure affected by the surrounding water
NP I : number of nodal ci rc1es on the i nteri or su rface of thestructure affected by the fluids in the tank interior
MBANO = 3*( MB + 1)
MB = max MB i , i = 1, NUMELi
MBi :
NMODE:
NUMPEX:
NUELEX:
di fference between the 1argest and small est structualnodal-circle numbers for structural element i
number of modes of vibration included
number of nodal circles in the exterior water idealization
number of elements in the exterior water idealization
-81-
UYIIClIIII \,;:) I ~\';lHlU I U!:lY, HI\,;.
OT-7814-2
NUINTEX: number of fluid nodal-circles on the exterior structurefluid interface
NUFSF: number of exterior fluid nodal-circles on the freesurface of the fluid
NEBANO = NEB + 1
NEB = max NEBi , i = 1, NUELEXi
NEBi :
NUNPIN:
NUELIN:
difference between the largest and smallest nodal circlenumbers for exterior fluid element i
number of nodal circles in the interior fluid idealization
number of elements in the interior fluid idealization
NIBANO = NIB + 1
NIB = max NIB., i = 1, NUELINi 1
NIBi : difference between the largest and smallest nodal circlenumbers for interior fluid element
NXFH: number of ordinates describing time-history of firstground motion component, along e = 0°
NYFH: number of ordi nates describi ng time-hi story of secondground motion component, along e = 90°
NT: number of integration time steps
NNODE total number of structural nodal-ci rcl es at whi ch timehistory of displacements is to be printed
NNEL: total number of structural elements for which timehistory of stresses is to be printed
NANGLE: total number of di fferent 1ocati ons around the circumference at which di spl acements and stresses are to beprinted
If only frequencies and mode shapes are desired along with staticanalysis, it suffices to check (1) and (2) above.
-82-
"'J11_,UI_.., 1__ .lll_.""jJ'
DT-7814-2
The computer time required for solution depends on a number offactors. The more important ones are the number of i ntegrati on timesteps (NT), the number of structural nodal ci rcl es (NUMNP), the bandwidth (MBAND) of the structural stiffness matrix (the nodal circlesshould be numbered in a manner which minimizes the bandwidth), thenumber of structural elements (NUMEL), the number of modes of vibrationto be included (NMODE), interval for printing and writing (NPRINT), andthe number of nodal-circles (NUNPEX, NUNPIN) and elements (NUELEX,NUELIN) in the idealization of the fluid domain.
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Dynamics Technology, Inc.DT-7814-2
APPENDIX B. PROGRAM IlERSTIl LISTINGS
-84-
PRO
GRA
ME
RST
76
/17
6O
PT=1
STA
TIC
FT
N4.
8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
TH
ISPR
OG
RAM
SHO
ULD
BE
CO
MPI
LED
USI
NG
THE
FTN
"ST
AT
IC"
OPT
ION
WH
EREB
YM
TOP
CAN
BE
AD
JUST
EDD
YN
AM
ICA
LLY
AT
RUN
TIM
EV
IATH
ECM
OR~FL
PAR
AM
ETER
ONTH
EJO
BC
AR
D.
~ I
5
10
15 20
25
30
PRO
GRA
ME
RS
T(I
NP
UT
.OU
TP
UT
.TA
PE
1.T
AP
E2.
TA
PE
3.T
AP
E4.
TA
PE
6=O
UT
PU
T.
1T
AP
E7.
TA
PE
B,T
AP
E9)
C C**
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
*C
EAR
THQ
UA
KE
RE
SPO
NSE
OF
SEA
-BA
SED
STO
RA
GE
TAN
KS
c***
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
C C C C CC
OM
MO
N/F
CO
NST
/PI,
ROW
.W
LOC
OM
MO
N/C
NT
RL
/NU
MN
P,N
AN
GL
E,M
BA
ND
,NB
C,N
F,N
PO.N
PI,T
RA
CE
.RE
AD
FRQ
DIM
EN
SIO
NH
ED
(8)
COM
MON
A(2
00
00
)LO
GIC
AL
REA
DFR
G,P
LOTA
NK
.PLO
TMO
DLO
GIC
AL
PLO
TAC
CN
AM
ELIS
TIT
AN
KI
NU
MN
P,N
UM
EL.
NU
MM
AT,
NM
OD
E.W
LO
,WL
I.N
PO,
NP
I.IG
RA
V,
1R
EAD
FRG
,PLO
TAN
K,P
LOTM
OD
NA
MEL
IST
IEX
TF
LD
IN
UN
PEX
,N
UEL
EX,
NU
INT
EX
.NU
FSF,
NU
CO
FN
AM
ELIS
T/I
NT
FL
D/
NU
NPI
N.N
UE
LIN
,N
UIN
TIN
,NU
EL
OIL
.RO
ILN
AM
ELIS
T/R
ES
PO
NS
IN
GR
D,N
T,D
T.N
XFH
,NY
FH,P
LO
TA
CC
NA
MEL
IST
IPR
INT
ITI
NPR
INT
.NN
OD
E.N
NE
L.N
AN
GL
ED
ATA
PL
OT
AN
K/.F
.I,
PL
OT
MO
D/.
F.I.
PL
OT
AC
C/.
F.I,
RE
AD
FR
Q/.
F./
PI=
3.
14
15
92
65
RO
W=
0.0
62
5/3
2.2
C CG
ETTH
EA
VA
ILA
BLE
BLA
NK
COM
MON
MEM
ORY
SIZ
EC
CA
LLPF
L<
MTO
P)
MST
OR
=M
TOP
-LO
CF
(A(!
»+
!
35
C C CRE
AD
AND
PRIN
TO
FC
ON
TRO
LIN
FOR
MA
TIO
N
REA
D1
00
0.
HED
REA
DTA
NKPR
INT
TAN
KPR
INT
20
00
.H
ED
.NU
MN
P,N
UM
EL
.NU
MM
AT
,NM
OD
E.W
LO
.WL
I.NPO
,NPI
40
45 50 55
C CRE
AD
DA
TAAN
DD
ETER
MIN
EEL
EMEN
TPR
OPE
RT
IES
CN
2=I+
NU
MN
PN
3=N
2+N
UM
NP
N4=
N3+
5*N
UM
EL
N5=
N4+
NU
Mt1
AT
*9
N6=
N5+
NU
MM
AT
N8=
N6+
NU
MM
AT
NEG
==3*
NU
MN
PN
9==N
S+N
EQC
ALL
MSC
HEC
K(M
STO
R,-N
9+N
UM
NP.
"LA
YO
UT
")
CA
LLL
AY
OU
T(A
(1),
A(N
2),
A(N
3).
A(N
4).
A(N
5).
A(N
6).
A(N
S).
A(N
9).
1N
UM
MA
T.N
UM
EL,P
LOTA
NK
lN
F=O
IF(
IGR
AV
.E
G.
1l
NF=
11
20
CA
LLE
LE
ME
NT
(A(I
).A
(N2
).A
(N3
).A
(N4
),A
(N5
).A
(N6
).A
(N8
l.N
F,N
UM
EL
)c C
FORM
TOTA
LM
ASS
AN
DS
TIF
FN
ES
SM
ATR
ICES
PRO
GR
AM
ER
ST
c
76
/17
6O
PT
=l
ST
AT
ICFT
N4
.8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
2
N9=
N8+
NB
C6
0N
10=
N9+
NE
GN
l1=
Nl0
+N
EG
N12
=N
11+
(NE
G*M
BA
ND
)N
13=
N12
+N
PO
CA
LLM
SCH
EC
K(M
STO
R,
N1
3+
NP
!'"T
OT
AL
")6
5C
ALL
TO
TA
L(A
(1),
A(N
2),
A(N
S),
A(N
9),
A(N
l0),
A(N
l1),
A(N
12
),A
(N1
3),
WL
O1
,WL
I,N
PO
,NE
G,M
BA
ND
,NU
ME
L,
IGR
AV
)c C
SOL
VE
FOR
ST
AT
ICLO
AD
SC
70
NU
M=6
*NU
MEL
IF(N
F.
NE
.O)
GO
TO1
50
N14
=N
13+
NP
IC
ALL
MSC
HE
CK
(MST
OR
,N
14+N
UM
,"S
TA
TIC
")C
ALL
ST
AT
IC(A
(N8
),A
(N9
),A
(N1
0),
A(N
11
),A
(N1
4),
NE
G,M
BA
ND
75
l,NU
M,N
UM
EL
)1
50
IF(N
F.
EG.
0)
GO
TO2
00
~8
0
85
C CM
OD
ESH
APE
SA
ND
NA
TUR
AL
FRE
QU
EN
CIE
SC
N12
=N11
+NM
OD
EN
13=
N12
+(N
EG
*MB
AN
D)
N14
=N
13+
NE
QN
15=
N14
+N
EG
N16
=N
15+
NE
QN
17=
N16
+N
EG
CA
llM
SCH
EC
K(M
STO
R,N
17+
NH
OD
E**
2,"E
IGE
N")
CA
LLE
IGE
N(A
(NB
),A
(N9
),A
(Nl0
),A
(Nl1
),A
(N1
2),
A(N
13
),A
(Nl
14
),A
(N1
5),
A(N
16
),A
(N1
7),
NE
G,N
HO
DE
,MB
AN
D,A
(lJ,
A(N
2J,
PlO
TH
OD
)
90
95
10
0
10
5
11
0
C C C C C C
Sal
VE
FOR
HY
DR
OD
YN
AM
ICR
ESP
ON
SE(I
NC
OM
PR
ES
SIB
LE
FL
UID
)
N13
=N
12+
(NE
Q*N
MO
DE
)N
TE
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NU
EL
ININ
TE
GE
RR
EFS
211
48
14
91
50
151
15
31
55
16
22
46
36
NU
EL
OIL
INT
EG
ER
RE
FS2
11
53
16
22
46
31
NU
FSF
ItH
EG
ER
RE
FS2
01
19
12
51
32
13
72
46
30
NU
INT
EX
INT
EG
ER
RE
FS2
01
19
12
012
11
26
13
72
46
35
NU
INT
ININ
TE
GE
RR
EFS
21
15
3
PRO
GR
AM
ER
ST7
6/1
76
OP
T=
lS
TA
TIC
FTN
4.8
+4
98
03
/16
/81
16.
39
.3
9PA
GE
7
VA
RIA
BL
ES
SNT
YPE
RE
LO
CA
TIO
N2
46
66
NUM
INT
EG
ER
RE
FS7
37
42
10
22
32
26
22
82
30
23
22
33
24
3D
EFI
NE
D7
02
46
21
NU
MEL
INT
EG
ER
RE
FS1
83
84
451
55
65
70
74
24
32
46
22
NU
MM
AT
INT
EG
ER
RE
FS1
83
84
54
64
751
0N
UM
NP
INT
EG
ER
CN
TRL
RE
FS1
31
83
84
24
34
85
0.2
34
23
52
43
24
62
6N
UN
PEX
INT
EG
ER
RE
FS2
01
19
12
21
23
12
613
11
37
24
63
3N
UN
PIN
INT
EG
ER
RE
FS21
14
51
46
14
71
53
15
51
57
15
81
59
16
01
62
24
64
3N
XFH
INT
EG
ER
RE
FS2
21
77
17
81
89
DE
FIN
ED
19
52
46
44
NY
FHIN
TE
GE
RR
EFS
22
19
52
46
62
N10
INT
EG
ER
RE
FS61
65
74
87
93
12
21
26
13
71
47
15
31
62
17
51
89
21
62
43
DE
FIN
ED
60
12
11
46
17
42
15
24
66
3N
11IN
TE
GE
RR
EFS
62
65
74
80
87
96
12
31
26
13
71
48
15
31
55
16
21
77
18
92
17
24
3D
EFI
NE
D61
12
21
47
17
52
16
24
66
4N
12IN
TE
GE
RR
EFS
63
65
81
87
92
12
41
26
13
71
49
15
31
55
15
71
62
17
81
89
21
82
43
DE
FIN
ED
62
80
12
31
48
17
72
17
24
66
5N
13IN
TE
GE
RR
EFS
64
65
72
82
87
95
12
51
26
13
21
37
15
01
53
15
81
62
17
9
-Q1
89
21
92
43
DE
FIN
ED
63
819
21
24
14
91
57
17
82
18
24
66
7N
14IN
TE
GE
RR
EFS
73
74
83
87
13
31
37
15
11
53
15
51
59
16
21
80
18
92
20
24
3D
EFI
NE
D7
28
21
32
15
01
58
17
92
19
24
67
0N
15IN
TE
GE
RR
EFS
84
87
13
41
37
15
21
60
16
218
11
89
22
12
43
DE
FIN
ED
83
13
315
11
59
18
02
20
24
67
1N
16IN
TE
GE
RR
EFS
85
87
13
51
37
161
18
21
89
22
22
43
DE
FIN
ED
84
13
41
60
181
22
12
46
72
N17
INT
EG
ER
RE
FS8
68
71
36
18
31
89
22
32
43
DE
FIN
ED
85
13
51
82
22
22
47
04
N18
INT
EG
ER
RE
FS1
84
18
92
24
24
3D
EFI
NE
D1
83
22
32
47
05
N19
INT
EG
ER
RE
FS1
85
18
92
25
24
3D
EFI
NE
D1
84
22
42
46
52
N2
INT
EG
ER
RE
FS4
351
55
65
87
20
52
09
24
3D
EFI
NE
D4
22
04
24
70
6N
20IN
TE
GE
RR
EFS
18
61
89
22
62
43
DE
FIN
ED
18
52
25
24
70
7N
21IN
TE
GE
RR
EFS
18
71
89
22
72
43
DE
FIN
ED
18
62
26
24
71
0N
22
INT
EG
ER
RE
FS1
88
22
82
43
DE
FIN
ED
18
72
27
24
71
3N
23IN
TE
GE
RR
EFS
22
92
43
DE
FIN
ED
22
82
47
14
N24
ItH
EG
ER
RE
FS2
30
24
3D
EFI
NE
D2
29
24
71
5N
25IN
TE
GE
RR
EFS
23
12
43
DE
FIN
ED
23
02
47
16
N26
INT
EG
ER
RE
FS2
32
24
3D
EFI
NE
D2
31
24
71
7N
27It
HE
GE
RR
EFS
23
32
43
DE
FIN
ED
23
22
47
20
N28
INT
EG
ER
RE
FS2
34
24
3D
EFI
NE
D2
33
24
72
1N
29IN
TE
GE
RR
EFS
23
52
43
DE
FIN
ED
23
42
46
53
N3
ItH
EG
ER
RE
FS4
451
55
99
10
51
37
20
62
43
DE
FIN
ED
43
20
52
47
22
N30
INT
EG
ER
RE
FS2
36
24
3D
EFI
NE
D2
35
24
72
3N
31IN
TE
GE
R'
RE
FS2
37
24
3D
EFI
NE
D2
36
24
72
4N
32IN
TE
GE
R.
RE
FS2
38
24
3D
EFI
NE
D2
37
24
72
5N
33IN
TE
GE
RR
EFS
23
92
43
DE
FIN
ED
23
82
47
26
N34
INT
EG
ER
RE
FS2
40
24
3D
EFI
NE
D2
39
PRO
GR
AM
ER
ST7
6/1
76
OP
T=
lST
AT
ICFT
N4
.8+
49
80
3/1
6/8
116
.3
9.
39
PAG
E9
INL
INE
FUN
CT
ION
ST
YPE
AR
GS
DEF
LIN
ER
EFE
RE
NC
ES
LOC
FIN
TE
GE
R1
INT
RIN
31
NA
ME
LIS
TS
EX
TFL
DIN
TFL
DP
RIN
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RE
SPO
NS
TAN
K
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LIN
E2
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32
2 18
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FER
EN
CE
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13
11
41
43
14
42
01
20
21
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16
93
63
7
~ lw.J
STA
TEM
ENT
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BE
LS
22
70
61
20
22
76
71
50
23
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79
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37
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00
23
55
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00
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44
51
20
00
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24
51
62
02
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33
20
21
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24
55
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04
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45
57
20
50
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61
03
00
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1T
COM
MO
NB
LOC
KS
LEN
GTH
FCO
NST
3C
NTR
L9
//
20
00
0
DEF
LIN
ER
EFE
RE
NC
ES
55
25
27
671
17
71
96
20
01
92
24
97
62
53
24
92
50
25
43
52
55
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26
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70
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617
12
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11
52
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92
74
25
3
ST
AT
IST
ICS
PRO
GR
Al1
LEN
GTH
BU
FFE
RLE
NG
THSC
MLA
BEL
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MM
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THSC
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LAN
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20
00
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MU
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30
03
82
17
62
81
4B
47
04
0B
15
39
92
02 12
20
00
0
SUB
RO
UTI
NE
MSC
HEC
K7
6/1
76
OPT
=l
STA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E
5
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KM
SCH
ECK
SUB
RO
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MSC
HEC
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MA
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RET
UR
NPR
INT
5.M
ESSA
GE.
MN
EED
-MM
AX
5FO
RMA
T(1
1*
---
NO
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GH
MEM
ORY
FOR
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RO
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0,
11
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ED
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9.*
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ORD
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GRA
MS
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P"N
OT
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UG
HM
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RY"
END
SYM
BO
LIC
REF
EREN
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MAP
(R=
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ENTR
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REF
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CES
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23
VA
RIA
BL
ES
SNTY
PER
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EN
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TFM
TW
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TEM
ENT
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ER
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15
Ff1T
54
ST
AT
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ICS
PRO
GRA
MLE
NG
TH41
133
352
0001
3SC
MU
SED
4 3 3
DE
FIN
ED 4 4
2D
EFI
NE
DD
EFI
NE
D2 2
SUB
RO
UT
INE
LAY
OU
T7
6/1
76
OP
T=
lST
AT
ICFT
N4
.8+
49
80
3/1
6/8
116
.39
.3
9PA
GE
5
10
SUB
RO
UTI
NE
LA
YO
UT
(R,
Z,IX
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O,W
AN
G,N
EB
C,C
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E,N
UM
MA
T,N
UM
EL
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DT
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C C**
****
****
****
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****
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****
****
****
****
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GEO
MET
RYAN
DM
ATE
RIA
LPR
OPE
RT
IES
OF
STR
UC
TUR
EA
RERE
AD
IN.
CN
OD
AL
POIN
TS
CAN
BE
GEN
ERA
TED
AT
EQU
AL
INTE
RV
AL
ORIN
LAV
ERS.
C**
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
*C
LOG
ICA
LPL
OTT
DIt1
EN
SIO
NR
(1),
Z(1
),NE
13C
(1),
COD
E(1
),IX
(NU
MEL
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),E
(9,
1),
ROC1
),W
ANG
C1
1),
NP
TS
(10
0),
IFIG
(10
0),
RP
LO
T(S
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0),
ZP
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00
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OM
MO
N/C
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NG
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AT
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15
20
25
30
C C CRE
AD
AND
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TO
FM
ATE
RIA
LPR
OPE
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IES
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10
M=l
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MM
AT
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D1
00
1,
MTY
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MA
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MA
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09
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D1
01
0,
EE
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U,R
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TY
PE
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INT
20
06
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TY
PE,E
E,E
EU
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YPE
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AN
G(M
TYPE
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02
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3)
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ND
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LE.
6)
E(
I,M
TY
PE)=
EE
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(I.
GT.
6)
E(I
,MT
YP
E)=
EE
/(2
.*(1
.+E
EU
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02
CO
NTI
NU
EGO
TO1
10
10
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AD
10
11
,(E
CI,
MT
YPE
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RO
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TY
PE),
WA
NG
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PE)
PRIN
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00
7,
MTY
PE,
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YP
E),
I=1,
9),R
OC
MT
YPE
),W
AN
GC
MT
YPE
)1
10
CO
NTI
NU
E
3S 40
45
50
55
C CRE
AD
AND
PRIN
TN
OD
AL
POIN
TD
ATA
CPR
INT
20
04
L"'O
N13C
=O1
20
REA
D1
00
2,
N,C
OD
EC
N),
R(N
),Z
CN
),M
OD
,NL
IM,F
AC
R,F
AC
ZL
l=L
+l
IF(N
-L1
)1
90
,17
0,1
30
13
0IF
CL
.LE
.O)
GOTO
18
0D
IV=
N-L
DR
=(R
(N)-
RC
L»
/DIV
DZ
=(Z
CN
)-Z
CL
»/D
IVM
=N-l
DO1
40
K=L
1,M
CO
DE
OO
=O.O
R(K
)=R
CK
-l)+
DR
140
Z(K)~Z(K-1)+DZ
GOTO
17
01
50
Lt"
'N+
lIr
(FA
CR
.LE
.O.O
lFA
CR
ul.
0IF
(FA
CZ
.LE
.O
.0
)F
AC
Z=
1.0
PRIN
T2
10
0,
MO
D,N
LIM
,FA
CR
,FA
CZ
16
0N
=N+l
Nl=
N-M
OD
N2=
NI-
MO
DIF
(Nl.
LE
.0.O
R.N
2.L
E.
0)
GOTO
20
0
SUB
RO
UT
INE
LAY
OU
T7
6/1
76
OPT
:;1ST
AT
ICFT
N4
.8+
49
80
3/1
6/8
116
.39
.3
9PA
GE
2
.-()
0',
60 65
70
75 80
85
C C C
CO
DE
(N)=
O.O
R(N
)=R
(Nl)
+F
AC
R*
(R(N
l)-R
(N2
»Z
(N)=
Z(N
l)+
FA
CZ
*(Z
(Nl)
-ZIN
2»
IF(N
.L
T.
NL
IM)
GOTO
160
MO
D=O
17
0L=
NPR
INT
20
02
,(K
,CO
DE
O\)
,R(K
).,Z
<lO
,K=
L1
,N)
CO
D=C
OD
E(N
)IF
(CO
O.E
Q.O
.O)
GOTO
17
50
=1
00
DO1
72
J:;
I,3
IF(C
OD
.LT
.D
)GO
TO1
72
NI3
C=N
BC
+lN
EB
C(N
BC
)=3*
N-3
+J
CO
D=C
OD
-D1
72
D=
D/I
0.
17
5IF
(MO
D.G
T.
0)
GOTO
15
0IF
(N-N
UM
NP)
12
0,2
20
,21
01
80
PRIN
T2
13
0ST
OP
19
0PR
INT
21
40
,N
STO
P2
00
PRIN
T2
15
0ST
OP
21
0PR
INT
21
60
,N
,NU
MN
PST
OP
22
0C
ON
TIN
UE
REA
DAN
DPR
INT
OF
ELEM
ENT
PRO
PER
TIE
S
90
95
10
0
10
5
11
0
PRIN
T20
01N
=OM
BAN
D=O
23
0RE
AD
10
03
,M
,(I
X(M
,I)
,1=
1,5
)
24
0N
=N+l
IF(M
.EO
.N
)GO
TO2
60
DO2
50
1=
1,4
25
01X
(N,
l)=
lX(N
-l,
1)+
1IX
(N,
5)=
IX(N
-l,
5)
26
0PR
INT
20
03
,N
,<
IX(N
,I)
,1
=1
,5)
IFIN
.EG
.N
UM
EL)
GOTO
27
0IF
(N.
EG.
M)
GOTO
23
0GO
TO2
40
27
0IF
<.N
OT.
PLO
TT
)GO
TO2
80
DO2
78
N=1
,NU
ME
LN
PT
S(N
)=5
IF1G
(N}=
2DO
27
5lo
ot.
4R
PLO
T(
1.
N)::
:R(
IX(N
,I»
27
5Z
PL
OT
(!,N
):;Z
(lX
(N,l
»R
PL
OT
(5,N
)=R
PL
OT
(I,N
)2
78
ZP
LO
T(5
,N):
;ZP
LO
T(1
,N)
CA
LLO
PLT
CA
LLM
UL
PLT
(RPL
OT
,Z
PLO
T,N
PTS,
NU
ME
L,
1,5
,1
00
,"R
-AX
1S
",6
,1
"Z-A
XIS
",6,
9,IF
IG)
CA
LLPL
OT
IO.
,0.
,99
9)
CD
ETER
MIN
EBA
ND
WID
TH
SUB
RO
UT
INE
LAY
OU
T7
6/1
76
OP
T=
lS
TA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E3
R-O
RD
Z-O
RD
II)
,I3
,lS
HIS
OT
RO
PIC
II=
,F1
6.6
/6H
RHO
=,F
16
.6
,
MA
TE
RIA
L*n
LK
OR
TH
OT
RO
PIC
IIE
T=
,FI6
.6,6
HK
/SF
T.
IIN
US
T=
,F1
6.6
1G
ST
=,F
I6.6
1
J
28
0M
B=O
DO2
90
N=
l,N
UM
EL
DO
29
01
=1
,4DO
29
0J=
I,4
MM
=IA
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11
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ll=
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1.0
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C(2
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CC
2.3
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(4,4
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E(7
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CN
F.E
G.
0)
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TO7
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C5
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CO
NT
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00
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DO
DO
15
01
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CN
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l,3
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0.1
40
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(I.J
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(I.J
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(I.N
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C(I
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C(t~,
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70
10
0
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0
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40
15
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10
20
40
25
35
30
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50
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00
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29
57
79
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G)
S2
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S*
SS
C2=
CS
*CS
SC
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C2
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52
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4)=
SC
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0(2
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-SC
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1.
00
(4.1
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(4,2
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(4,1
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BR
OU
TIN
ESS
LAW
76
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6O
PT
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ST
AT
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N4
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49
80
3/1
6/8
11
6.3
9.3
9PA
GE
2
60
65
70
75
D(4
,4)=
C2
-S2
D(5
,5
)=1
.0
D(6
,6
)=1
.0
DO
28
7JJ=
1,6
Dl=
C(1
,1)*
D(
1.
JJ)+
C(
1,
2)*
D(2
,JJ)+
C(1
,3
)*D
(3,
JJ)
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C(2
,1
)*D
(1
,JJ)
+C(2
,2
)*D
(2,
JJ)
+C(2
,3
)*D
(3,JJ)
D3
=C
(3.1
)*D
(l,J
J)+
C(3
,2)*
D(2
,JJ)+
C(3
,3)*
D(3
,JJ)
D4
=C
(4,4
)*D
(4,J
J)D
5=
C(5
,5)*
D(5
,JJ)
D6
=C
(6,6
)*D
(6,J
J)D
O2
87
II=
JJ,6
CC(I
I,
JJ)"
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1,
II
)*D
1+
D(2
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)*D
2+
D(3
,I
I>*
D3
+D
(4,II
)*D
4+
D(5
,I1
)*D
5+
D1
(6,
II
)*D
62
87
CC
(JJ,
II)"
'CC
(II,
JJ)
DO
30
01
=1
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DO
30
0J=
l,6
30
0C
(I,J
)=C
C(I
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50
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5
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10
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16
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19
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23
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67
68
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98
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313
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63
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10
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29
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62
15
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30
36
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NO
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51
40
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17
30
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72
74
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NO
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25
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74
313
INST
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K
ST
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LE
NG
TH
4061
32
62
5200
013
SCM
USE
D
PAG
E3
2*
36
2*
37
SUB
RO
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ST
RA
IN7
6/1
76
OP
T=
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TA
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4.S
+4
98
03
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16
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PAG
E
~ C'\
5
10
15
20
25
30
35
40
45
50
55
SUB
RO
UT
INE
ST
RA
IN(S
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R.
ZZ
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C.N
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CD
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B(6
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S(6
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T(6
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R(6
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Z(6
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ON
RR
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H(6
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ION
II(6
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L(b
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16
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L/9
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11
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2.
17
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50
1=
1.
lOS
50
8(1
)=0
.0sr
1=
1.0
-6S
P=
1.
O+
ST
M=1
.O
-TT
P=
1.0
+T
H(
1)=
SM
*TM
/4.
H(2
)=S
P*
TM
/4.
H(3
)=S
P*
TP
/4.
H(4
)=S
M*
TP
/4.
H(5
)=(1
.0-S
*S
)H
(6
)=
(1
.O
-T*T
>H
S(
1)=
-T1
'1/4
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S(2
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S(1
)H
S(3
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P/4
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S(4
)=-H
S(3
)H
S(5
)=-2
.0*
SH
S(b
)=O
.H
T(
1)=
-SM
/4.
HT
(2)=
-SP
/4.
HT
(3)=
-HT
(2)
HT
(4)=
-HT
<1
)H
T(5
)=0
.H
T(6
)=-2
.0*
TP
ZT
=H
T(1
)*Z
Z(1
)+H
T(2
)*Z
Z(2
)+H
T(3
)*Z
Z(3
)+H
T(4
)*Z
Z(4
)P
ZS
=H
S(1
)*Z
Z(1
)+H
S(2
)*Z
Z(2
)+H
S(3
)*Z
Z(3
)+H
S(4
)*Z
Z(4
)P
RS
=H
S(l
l*R
R(1
)+H
S(2
l*R
R(2
)+H
S(3
)*R
R(3
l+H
S(4
)*R
R(4
lP
RT
=H
T(1
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R(1
)+H
T(2
)*R
R(2
)+H
T(3
)*R
R(3
)+H
T(4
)*R
R(4
)X
J=P
RS
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ZT
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T*
PZ
SP
SR
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ZT
/XJ
PT
R=
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S/X
JP
SZ
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RT
/XJ
PT
Z=
PR
S/X
JD
O1
00
1=
1.
6H
R(I
)=P
SR
*H
S(I
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TR
*H
T(I
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00
HZ
(I)=
PS
Z*
HS
(I)+
PT
Z*
HT
(I)
R=
H(1
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R(1
)+H
(2)*
RR
(2)+
H(3
)*R
R(3
)+H
(4)*
RR
(4)
DO2
00
K=
1.6
I=II
(K)
J=>
JJ(K
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L(K
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(1.
I)=H
R(K
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(2.J
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Z(K
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(3.1
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(K)/
RB
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(3,
I)8
(4.
I)=
HZ
CK
)8
(4.
J)
=HR
00
SUB
RO
UT
INE
ST
RA
IN7
6/1
76
OP
T=
lS
TA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E2
60
13<5
,I
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(3,
L)
0(5
,L
)=8
(1,
1>
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,I)
0(6
,J)=
B(5
,1>
20
0B
(6,L
)=B
(2,J
)F
AC
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J*R
RET
UR
NEN
D
SYM
BO
LIC
RE
FER
EN
CE
MA
P(R
=2
)
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YP
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TS
DE
FL
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RE
FER
EN
CE
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ST
RA
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63
VA
RIA
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ES
SNT
YPE
RE
LO
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N0
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RA
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EF
S7
55
58
2*
59
60
61
DE
FIN
ED
11
25
25
35
45
55
65
75
85
96
061
0FA
CR
EA
LF
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EFI
NE
D1
62
0H
REA
LA
RR
AY
F.
P.
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FS
84
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75
4D
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D1
17
18
19
20
212
2
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42
HR
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LA
RR
AY
RE
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25
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NE
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52
26
HS
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LA
RR
AY
RE
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72
42
64
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64
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74
54
6'I
DE
FIN
ED
23
24
25
26
27
28
23
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TR
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RR
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RE
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73
13
24
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54
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54
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93
03
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23
33
42
50
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RE
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75
35
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04
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3*
45
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52
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56
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114
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56
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10
22
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50
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25
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12
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03
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35
40
45
50
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15
01
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15
0E
LM
AS
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O1
70
J=
l,1
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(J)=
O.O
DO
16
01
=1
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Bn
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O.
01
60
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J)
=0
.0
DO
17
01
=1
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17
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=(R
R(1
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R(2
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R(3
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R(4
»/4
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(1)+
ZZ
(2)+
ZZ
(3)+
ZZ
(4»
/4.0
DO
50
01
1=
1,2
DO
50
0JJ=
1,2
CA
LL
ST
RA
IN(S
S(I
I),S
S(J
J),R
R,
ZZ
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C,N
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(N.
EG
.0
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AC
=2
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AC
DO
40
0J=
1.
18
01
=(0
(1,
1)*
B(1
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0(1
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B(2
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0(1
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B(3
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0(1
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AC
02
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(2,
1)*
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0(2
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1)
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8(3
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5)
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1,
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(3,
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00
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06
40
0S
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S(I,
J)
IF(N
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O.
0)
GO
TO4
70
DO
45
01
=1
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50
EL
MA
SS
(I)=
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MA
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C*
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HO
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NE
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OTO
50
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Z*
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(B)=
P(B
)+B
Z*
H(4
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00
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NT
INU
EIF
(N.N
E.
0)
GO
TO5
10
DO
50
51
=1
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85
05
IF(S
(I.I
l.E
G.O
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S(I.I
)=
l,O
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ST
RE
SS
DIS
PL
AC
EM
EN
TM
AT
RIX
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RO
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MA
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6/1
76
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03
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B(K
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TE
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EG
RE
ES
OF
FREE
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MC
DO
55
0N
N=
I.6
L=
18
-NN
70
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L+
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50
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O5
40
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54
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B(J
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75
DO
55
0J=
I.L
55
0S
(I.J
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(I,J
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(K.J
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E.O
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OTO
57
0D
O5
60
1=
9.
12
56
0S
(1
.1
)=0
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0C C
RE
LO
CA
TE
ST
IFF
NE
SS
MA
TR
IXA
ND
LOA
DV
ECTO
R'-
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.....5
70
DO
58
01
=1
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t:lQ
(I
)=P
(I)
85
DO
58
0J=
I.1
25
80
QK
(I.J
)=S
(I.J
)D
O5
90
J=1
.6
DO
59
0K
=1.
12
59
0Q
C(J
.K)=
B(J
.K)
90
RET
UR
NEN
D
SYM
BO
LIC
RE
FER
EN
CE
MA
P(R
=2
)
EN
TR
YP
OIN
TS
DE
FL
INE
RE
FER
EN
CE
S3
t1A
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90
VA
RIA
BL
ES
SNT
YPE
RE
LO
CA
TIO
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20
BR
EA
LA
RR
AY
RE
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82
46
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76
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96
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16
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36
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56
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76
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06
42
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48
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EF
INE
D1
76
26
47
46
74
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EA
LA
RR
AY
RE
FS
85
96
4D
EF
INE
D1
65
11
BZ
RE
AL
RE
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48
49
50
51D
EF
INE
D4
75
15
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EA
LR
EF
S7
47
6D
EF
INE
D7
20
DR
EA
LA
RR
AY
F.
P.
RE
FS
86
*2
76
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96
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16
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36
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56
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76
4D
EF
INE
D1
50
3D
lR
EA
LR
EF
S4
0D
EF
INE
D2
75
04
02
RE
AL
RE
FS
40
DE
FIN
ED
29
50
5D
3R
EA
LR
EF
S4
0D
EF
INE
D3
15
06
D4
RE
AL
RE
FS
40
DE
FIN
ED
33
SUB
RO
UT
INE
MA
STIF
76
/17
6o
PT
=l
ST
AT
ICFT
N4.
8+
49
80
3/1
6/8
11
6.3
9:3
9PA
GE
3
VA
RIA
BL
ES
SNT
YPE
RE
LO
CA
TIO
N5
07
05
REA
LR
EFS
40
DE
FIN
ED
35
51
0D
6R
EAL
RE
FS4
0D
EFI
NE
D3
70
ELM
ASS
REA
LA
RRA
YF
.P.
RE
FS7
45
DE
FIN
ED
11
24
55
02
FAC
REA
LR
EFS
24
25
27
29
313
33
53
74
54
75
9D
EFI
NE
D2
54
75
FZR
EAL
RE
FS4
7D
EFI
NE
D1
01
57
6H
REA
LA
RRA
Y.R
EFS
82
44
54
84
95
051
59
47
6I
INT
EG
ER
RE
FS1
21
61
71
98
*4
02
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23
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54
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56
23
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47
22
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42
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62
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92
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42
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6D
EFI
NE
D11
15
18
39
44
54
60
717
88
35
00
II
INT
EG
ER
RE
FS2
4D
EFI
NE
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24
77
JIN
TE
GE
RR
EFS
14
161
71
96
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76
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96
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16
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36
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56
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73
92
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02
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26
23
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43
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43
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62
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62
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9D
EFI
NE
D1
32
661
73
75
85
87
50
1JJ
INT
EG
ER
RE
FS2
4D
EFI
NE
D2
35
12
KIN
TE
GE
RR
EFS
2*
64
3*
72
74
76
2*
89
DE
FIN
ED
63
70
88
51
4L
INT
EG
ER
RE
FS7
071
75
DE
FIN
ED
69
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INT
EG
ER
F.P.
RE
FS2
42
54
34
65
35
97
7D
EFI
NE
D1
51
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INT
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ER
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FS6
9D
EFI
NE
D6
8.....
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55
4P
REA
LA
RR
AY
RE
FS8
48
49
50
518
4.... ....,
DE
FIN
ED
14
48
49
50
510
GR
EAL
ARR
AY
F.
P.R
EFS
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EFI
NE
D1
84
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REA
LA
RR
AY
F.P.
RE
FS7
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FIN
ED
18
90
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EAL
AR
RA
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RE
FS7
DE
FIN
ED
18
60
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REA
LF
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RE
FS1
04
5D
EFI
NE
D1
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REA
LF.
P.D
EFI
NE
D1
20
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REA
LA
RR
AY
F.
P.R
EFS
74
*2
02
45
9D
EFI
NE
D1
10
50
SR
EAL
AR
RA
YR
EFS
84
04
25
52
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22
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68
6D
EFI
NE
D1
94
04
25
57
67
95
16
SS
REA
LA
RR
AY
RE
FS7
2*
24
DE
FIN
ED
90
Zf1
REA
LF
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DE
FIN
ED
121
0ZZ
REA
LA
RRA
YF
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RE
FS7
4*
21
24
59
DE
FIN
ED
EX
TE
RN
AL
ST
YPE
AR
GS
RE
FER
EN
CE
SST
RA
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24
59
STA
TEM
ENT
LA
BE
LS
DEF
LIN
ER
EFE
RE
NC
ES
01
50
12
110
16
01
71
50
17
01
91
31
80
40
04
22
63
90
45
04
54
42
35
47
04
64
32
51
50
05
22
22
34
60
50
55
55
42
70
51
05
95
30
53
06
46
061
63
05
40
74
73
05
50
76
68
717
50
56
07
97
84
05
57
08
37
70
58
08
68
38
5
SUB
RO
UT
INE
WR
ITE
I7
6/1
76
OP
T=
1S
TA
TIC
FTN
4.
8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
' ......
lI.i
5
10
15
20
SUB
RO
UT
INE
WR
ITE
1(A
,LA
,N,N
UM
EL
)C C
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
**C
WR
ITE
MA
TRIX
AO
NT
AP
EI
AS
AN
AR
RA
YC
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
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DIM
EN
SIO
NA
(LA
)C
OM
MO
N/B
UF
/B(2
830l
LB
=2
83
0IF
(N.
NE.
1)
GO
TO1
00
REW
IND
1M
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30
SMA
SSR
EAL
ARR
AY
F.P.
RE
FS1
2D
EFI
NE
D1
14S
5R
EAL
ARR
AY
LS4A
RG
RE
FS10
31
0S
TR
ES
REA
LA
RRA
YF.
P.R
EFS
12
50
DE
FIN
ED
13
67
TRA
CE
REA
LC
NTR
LR
EFS
91
24
XCR
EAL
LS4A
RG
RE
FS10
43
0X
SR
EAL
ARR
AY
F.P
.R
EFS
12
192
031
47
3*
48
49
DE
FIN
ED
14
71
25
YCR
EAL
LS4A
RG
RE
FS1
04
3
FIL
EN
AM
ESM
ODE
OU
TPU
TFM
TW
RIT
ES4
04
34
548
TA
PEI
UN
FI1T
REA
DS
14M
OT
ION
1522
SUB
RO
UT
INE
ST
AT
IC7
6/1
76
OP
T=
lS
TA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E3
FIL
EN
AM
ESM
OD
ET
AP
E3
UN
FMT
WR
ITE
S4
95
0M
OTI
ON
13
EX
TE
RN
AL
ST
YPE
AR
GS
RE
FER
EN
CE
SB
AN
SOL
61
92
0R
EA
Dl
42
4
STA
TE
ME
NT
LA
BE
LS
DE
FL
INE
RE
FER
EN
CE
S0
50
18
16
01
80
31
27
29
01
90
33
32
01
95
37
35
02
70
INA
CT
IVE
39
38
12
72
80
42
2*
38
03
00
44
23
03
60
47
46
27
22
00
0FM
T51
40
30
62
00
1F/
'"lT
54
43
31
12
00
9FM
T5
54
53
26
20
10
FMT
58
48
LO
OPS
LA
BE
LIN
DEX
FRO
M-T
OLE
NG
THP
RO
PE
RT
IES
26
50
K1
61
84B
INST
AC
K5
53
00
N2
34
464
BEX
TR
EFS
NO
TIN
NE
R6
41
80
I2
73
117
BN
OT
INN
ER
73
18
0J
29
31
5BIN
STA
CK
10
61
90
J3
23
33B
INST
ACJ
.{.....
12
01
95
I3
53
72B
INST
AC
K~
14
63
60
I4
64
73B
INST
AC
KC
'l1
54
N4
84
812
BEX
TR
EF
S
COM
MO
NB
LO
CK
SLE
NG
THC
NT
RL
8L
S4A
RG
28
3
ST
AT
IST
ICS
PRO
GR
AM
LEN
GTH
36
5B
24
5SC
ML
AB
EL
ED
COM
MO
NLE
NG
TH4
43
B2
91
52
00
0B
SCM
USE
D
SU
BR
OU
TIN
EB
AN
SOL
76
/17
6O
PT
=l
ST
AT
ICFT
N4
.8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
25
18
0.....
20
0N 'i
30
30
0
35
35
04
00
405
10
15
20
45
C C C C C C C C C
SU
BR
OU
TIN
EB
AN
SO
L(N
N,M
M,N
DIM
,A,B
,KK
)
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
LIN
EA
RE
GU
AT
ION
SOL
VE
RFO
RSY
MM
ET
RIC
BA
ND
EDM
AT
RIC
ES
KK
=0
TR
IAN
GU
LA
RIZ
ES
BA
ND
MA
TR
IXA
KK
=1
RE
DU
CE
SA
ND
BA
CK
SU
BS
TIT
UT
ES
VEC
TOR
BK
K=
2B
AC
KS
UB
ST
ITU
TE
SV
ECTO
RB
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
**
DIM
EN
SIO
NA
(ND
IM,
1),
B(1
)
NR
=N
N-
1IF
(KK
-l)
10
0,3
00
,40
01
00
DO
20
0N
=1
,NR
M=
N-
1IF
(A(N
.1
).E
G.
o.)
A(N
,1
)=
1.
OE
-16
PIV
OT
=A
(N,l
)M
R=
MIN
D(M
M,N
N-M
)D
O2
00
L=
2,M
RC
=A
(N,
Ll/
PIV
OT
IF(C
.EG
.O
.)
GO
TO2
00
I=
M+
LJ
=0
DO
18
0K
=L
,MR
J=
J+
1A
(I,J
)=
A(I
,J)
-C
*A
(N,K
)A
(N,
L)
=C
CO
NT
INU
EIF
(A(N
N,
1).
EG
.O
.)
A(N
N,1
)=
1.O
E-1
6G
OTO
50
0D
O3
50
N=
1,N
RM
=N
-1
MR
=M
INO
(MM
,NN
-M)
C=
B(N
)B
(N)
=C
IA(N
,1
)D
O3
50
L=
2,M
RI
=M
+L
B(I
)=
B(I
I-
A(N
,L)*
CB
(NN
)=
B(N
Nl/
A(N
N,
1)
DO
45
0K
=2
,NN
M=
NN
-K
N=
M+
1M
R=
MIN
D(M
M,K
)D
O4
50
L=
2,M
RI
=M
+L
45
0B
(N)
=B
(N)
-A
(N,L
l*B
(I)
50
0R
ET
UR
NEN
D
SYM
BO
LIC
RE
FER
EN
CE
MA
P(R
=2
)
SUB
RO
UT
INE
BA
NSO
l7
6/1
76
OP
T=
lS
TA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E2
ENTR
YP
OIN
TS
DE
FL
INE
RE
FER
EN
CE
S3
BA
NS
Ol
14
7
VA
RIA
BL
ES
SNT
YPE
RE
LO
CA
TIO
N0
AR
EAL
AR
RA
YF
.P
.R
EFS
10
16
17
20
2*
26
29
35
38
39
46
DE
FIN
ED
11
62
62
72
90
BR
EAL
AR
RA
YF
.P.
RE
FS1
03
43
83
92
*4
6D
EFI
NE
D1
35
38
39
46
16
7C
REA
LR
EF
S2
12
62
73
53
8D
EFI
NE
D2
03
41
70
III
HE
GE
RR
EFS
2*
26
2*
38
46
DE
FIN
ED
22
37
45
17
1.J
INT
EG
ER
RE
FS2
52
*2
6D
EFI
NE
D2
32
51
72
KIN
TE
GE
RR
EF
S2
641
43
DE
FW
ED
24
40
0K
KIN
TE
GE
RF
.P
.R
EF
S1
3D
EFI
NE
D1
16
6L
INT
EG
ER
RE
FS
20
22
24
27
37
38
45
46
DE
FIN
ED
19
36
44
16
3M
INT
EG
ER
RE
FS1
82
23
33
74
24
5D
EFI
NE
D1
53
24
10
Mt1
INT
EG
ER
F.
P.
RE
FS
18
33
43
DE
FIN
ED
11
65
MR
IIH
EG
ER
RE
FS
19
24
36
44
DE
FIN
ED
18
33
43
16
2N
INT
EG
ER
RE
FS
15
2*
16
17
20
26
27
32
34
2*
35
38
3*
46
DE
FIN
ED
14
31
42
0N
DIl1
INT
EG
ER
F.
P.R
EF
S1
0D
EF
INE
D1
0NN
INT
EG
ER
F.
P.R
EF
S1
21
82
*2
93
33
*3
94
041
'.D
EFI
NE
D1
~1
61
NRI~JTEGER
RE
FS1
43
1D
EFI
NE
D1
2C'
Q1
64
PIV
OT
RE
AL
RE
FS2
0D
EF
INE
D1
7
INL
IN
EFU
NC
TIO
NS
TY
PEA
RG
SD
EFL
INE
RE
FER
EN
CE
SM
INO
INT
EG
ER
0IN
TR
IN1
83
84
3
STA
TE
ME
NT
LA
BE
LS
OE
FL
INE
RE
FER
EN
CE
S0
10
0IN
AC
TIV
E1
41
30
18
02
62
46
12
00
28
14
19
21
73
30
03
11
30
35
03
83
13
61
23
40
03
91
30
45
04
64
04
41
56
50
04
73
0
LO
OPS
lAB
EL
IND
EX
FRO
M-T
OLE
NG
THP
RO
PE
RT
IES
15
20
0N
14
28
51
8N
OT
INN
ER
31
20
0L
19
28
33
8N
OT
INN
ER
50
18
0K
24
26
58
INST
AC
K7
43
50
N3
13
82
78
NO
TIN
NE
R1
13
35
0L
36
38
58
INST
AC
K13
14
50
K4
04
62
58
NO
TIN
NE
R1
46
45
0L
44
46
58
INST
AC
K
ST
AT
IST
ICS
PRO
GR
AM
LEN
GTH
22
18
14
55
20
00
8SC
l1U
SED
SU
BR
OU
TIN
EB
NE
IGN
76
/17
6O
PT
=l
ST
AT
ICFT
N4
.8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
SUB
RO
UT
INE
BN
EIG
N(N
N,M
M,N
MO
,NB
C,E
V,N
EB
C,S
MA
SS
,A,B
,V,R
,CN
,SN
,TR
AC
E)
CD
IMEN
SIO
NEV
(1
),N
EBC
(1
),S
t1A
SS(
1),
A(N
N,
1),
B(
1),
R(
1),
SN(
1),
CN
(1
),1
V(
1)
RE
DU
CE
TOC
LA
SS
ICA
LE
IGE
NV
AL
UE
PRO
BLE
MA
*X
~
5
10
15
20
25
30
35
40
45
C C C C C C C C C
INIT
IAL
IZE
TO
L=
1.O
E-1
2IF
LA
G=
0N
LOO
P=
5N
SMA
X=
50
PS
HIF
T=
O.
Mt1
1=
Mt"i
+1
NW=
NN
*MM
NE
IG=
0N
R=
NNN
NR
=N
N-
1R
EW
IND
2
DO
12
0I
=1
,N
NX
=S
MA
SS
(I)
IF(X
.G
T.0
.)G
OTO
11
0P
RIN
T1
2,
I1
2FO
RM
AT
(30H
ON
EG
.O
RZ
ER
OM
ASS
,E
QU
AT
ION
IFL
AG
=1
GO
TO1
20
11
0S
MA
SS
(I)
=1.
/SG
RT
(X)
12
0C
ON
TIN
UE
IF(I
FL
AG
.N
E.0
)S
TO
PD
O1
30
I=
1,N
NL
=I
-1
MR
=M
INO
(MM
,NN
-I+
l)D
O1
30
J=
1,M
RK
=L
+J
13
0A
(I,J
)=
A(I
,J)*
SM
AS
S(I
)*S
MA
SS
(K)
IMPO
SEB
OU
ND
AR
YC
ON
DIT
ION
SO
NA
IF(N
BC
.L
E.
0)
GO
TO1
50
DO
14
0N
=1,
NB
CI
=N
EBC
(N)·
A(I
,1
)=
10
0.
*TR
AC
E1
40
CO
NT
INU
E
15
)
E*X
50 55
C CC
OM
PAC
TM
AT
RIX
AIN
TO
Al-
DA
RR
AY
VC
15
0D
O1
60
J=
2,M
ML
=N
N*
(J-1
)M
=N
N-
J+
1D
O1
60
I=
1,
MK
=L
+I
16
0V
(K)
=A
(I,J
)W
RIT
E(2
)(V
(I
),I=
1,
NW
)C C
COM
PUTE
SMA
LLES
TEI
GEN
VA
LUE
AND
ASS
OCI
ATE
EIG
ENV
ECTO
ROF
A
SUB
RO
UT
INE
BN
EIG
N7
6/1
76
OP
T=
lST
AT
ICFT
N4.
8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
2
......
VJ~
60
65
70
75
80
85 90
95
10
0
10
5
11
0
CBY
INV
ER
SEIT
ER
AT
ION
C1
65
NE
IG=
NE
IG+
1E
l=
O.S
HIF
T=
O.
NS
=0
KK
T=
2C
ALL
BA
NS
OL
(NR
,MM
,NN
,V,B
,O)
DO1
70
I=
NR
,NN
17
0B
(Il
=O
.DO
180
I=
1,N
R1
80
B(I)
=1.
IF(N
BC
.LE
.0
)GO
TO2
00
PO1
90
N=
1,N
BC
I=
NE
BC
(N)
19
0B
(I)
=0.
20
0N
S=
NS
+1
CA
LLB
AN
SOL
(NR
,MM
,NN
,V,B
,KK
T)
KK
T=
1E
=O
.DO
22
0I
=1,
NR
IF(A
IlS
(B(I
ll.
GT.
AB
S(E
llE
=B
(I)
22
0C
ON
TIN
UE
E=
1.
IEE
PS=
(E-E
l)/E
*1
00
.DO
23
0I
=1,
NR
23
0B
(I)
=B
(I)*
EE1
=E
IF(A
BS
(EP
S).
GT.
1..
AN
D.
NS
.LT
.1
5)
GOTO
20
0N
L=
NL
OO
P'-
32
50
DO2
60
I=
1,N
R2
60
R(I
)=
B(I
)N
S=
NS
+1
CA
LLB
AN
SOL
(NR
,MM
,NN
,V,B
,1
)E
=O
.DO
30
0I
=1,
NR
IF(A
BS
(B(I
».G
T.A
BS
(E»
E=
B(I
)3
00
CO
NTI
NU
EDM
AX=
O.
SUI1
D=
O.
DO3
20
I=
1,N
RB
(I)
..B
(I
)IE
D=
AB
S(B
(I)-
R(I
»SU
MD
..SU
MD
+D
**2
IF(D
.GT
.D
MA
X)
DMAX
=D
32
0C
ON
TIN
UE
IF(D
MA
X.L
E.
TO
l.O
R.N
S.G
E.
NSM
AX
)GO
TO4
00
NL..
NL+
1IF
(NL
.Li.
NlO
OP)
GOTO
25
0R
EWIN
D2
REA
D(2
)(V
(I),
I=l,
NW
)NL
=0
X=
o.Y
=O
.DO
34
0I
=1,
NR
X=
X+
B(I
)*R
(I)
34
0y
=y
+B
(I)*
8(I
)
SUB
RO
UT
INE
BN
EIG
N7
6/1
76
OP
T=
1S
TA
TIC
FTN
4.
8+
49
80
3/1
6/8
11
6.
39
.3
9PA
GE
3
......~ "-
11
5
12
0
12
5
13
0
13
5
14
0
14
5
15
0
15
5
16
0
16
5
17
0
SH
IFT
=S
HIF
T+
AM
AX
1(1
.-4
.*S
UM
D,0
.9)*
X/(
Y*
E)
DO3
50
I=
1,N
R3
50
V(I
)=
V(I
)-
SH
IFT
CA
LLB
AN
SO
L(N
R,M
M,N
N,V
,B,O
)G
OTO
25
04
00
X=
o.y
=O
.D
O4
20
I=
1,
NR
X=
X+
B(I
)*R
(I)
42
0Y
=Y
+B
(I)*
B(I
)S
HIF
T=
SH
IFT
+X
/(Y
*E
)E
V(N
EIG
)=
SH
IFT
+P
SH
IFT
SH
IFT
=S
HIF
T-
TOL
P5
HIF
T=
EV
(NE
IG)
-TO
LY
=S
GR
T(Y
)D
O4
30
I=
1,N
N4
30
R(I
)=:
B(I
)/Y
IF(N
EIG
.G
E.N
MO
)G
OTO
65
0c C
DE
FLA
TE
BA
ND
MA
TRIX
CR
EWIN
D2
REA
D(2
)(V
(I),
I=l,
NW
)D
O4
50
NX=:
1,N
RFB
=R
(NX
)IF
(FB
.t·J
E.0
.)G
OTO
48
04
50
CO
NT
INU
E4
80
DO
50
0I
=1
,NR
L=
NW+
IV
(I)
=V
(I)
-S
HIF
T5
00
veL>
=o.
NR
S=
NR-
1N
R1=:
NR+
1G
l=
R(1
)**
25
2=
O.
C=
1.
DO
60
0I
=1
,NR
SK
=1
+1
G=
Ql
+R
(K)*
*2
IF(I
.LT
.NX
)G
OTO
55
05
2=
Gl/
0C
=R
(K)/
SO
RT
(O)
IF(F
B.L
T.0
.)C
=-C
55
0S
=S
GR
T(5
2)
C2
=C
*CS
N(I
)=
SC
N(I
l=C
01..
Gl"'N
N+
IA
ll=
V(I
)A
22=
V(K
)A
12=:
V(l)
X=
2.*A
12*S
*CV
(I)
=A
ll*
C2
+A
22
*S
2-
XV
(K)
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03
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SS
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P.
RE
FS8
12
91
DE
FIN
ED
17
27
50
RR
EAL
AR
RA
YF
.P.
RE
FS
82
*2
63
03
*4
18
8D
EF
INE
D1
213
041
0R
OIL
RE
AL
F.
P.
RE
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5D
EF
INE
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RE
AL
AR
RA
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EFS
82
*2
73
13
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28
8D
EFI
NE
D1
21
31
42
FIL
EN
AM
ESM
OD
EIN
PUT
FMT
RE
AD
S2
15
9f6
OU
TPU
TFM
TW
RIT
ES
36
49
515
35
58
48
58
68
88
99
1
STA
TE
ME
NT
LA
BE
LS
DE
FL
INE
RE
FER
EN
CE
S11
20
21
48
03
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AC
TIV
E2
42
30
40
31
29
55
50
33
47
67
60
37
44
SUB
RO
UT
INE
INT
76
/17
6O
PT
=l
ST
AT
ICFT
N4
.8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
4
STA
TE
ME
NT
LA
BE
LS
DE
FL
INE
RE
FER
EN
CE
S1
12
70
46
23
32
12
08
04
92
41
23
90
512
31
26
10
05
34
013
11
10
55
48
13
41
20
57
48
13
51
30
59
66
15
21
40
60
67
01
50
63
62
16
61
60
64
611
72
18
071
65
02
74
73
71
02
80
75
74
03
00
80
77
03
10
88
87
03
20
91
90
41
44
00
FMT
92
85
42
26
00
FMT
93
86
43
17
00
FMT
94
88
4·34
80
0FM
T9
58
94
46
90
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T9
791
45
11
00
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1T9
88
44
60
10
02
FMT
99
21
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64
10
03
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10
05
9~
46
61
02
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lT1
01
76
-t:.
47
02
10
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02
36
50
12
13
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04
49
50
62
14
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05
515
15
21
50
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10
65
35
23
21
60
FMT
10
75
5
LO
OPS
LA
BE
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DE
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GTH
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ER
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S4
74
0K
29
31
413
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K1
40
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95
911
13EX
TR
EFS
16
31
50
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26
321
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CK
17
72
74
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17
321
3IN
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CK
20
62
80
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47
521
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CK
22
33
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78
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CK
23
53
10
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D8
78
813
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TR
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52
32
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90
91
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RE
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TIN
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55
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19
111
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TR
EF
S
ST
AT
IST
ICS
PRO
GR
AM
LEN
GTH
5651
33
73
5200
013
SCM
USE
D
SUB
RO
UTI
NE
BNDW
TH7
6/1
76
OP
T=
lST
AT
ICFT
N4.
8+
49
80
3/1
6/8
11
6.3
9.3
9PA
GE
1
1010
C C
15C C
20
20
.......~ \7
-,2
5
30
30
5
35 40 45 50
C C C
SUB
RO
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NE
BN
DW
TH
IR.Z
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UM
EL.N
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IME
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(NU
MN
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ME
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EL
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TE
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ME
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ICA
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INT
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=0DO
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=l.N
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EL
DO10
1=
1.3
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I+1
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IN.4
IDIF
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BS
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l)-I
XIN
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DIF
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AX
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AX
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=ID
IFC
ON
TIN
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KFO
RIN
CR
EASE
INBA
ND
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THD
UE
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TER
ELEM
ENT
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MB
INA
TIO
NS
OF
NO
DE
POIN
TS
NEA
RIN
TER
FAC
E
1-11
=0N
2=0
N1=
N1+
1IF
(N1.
GT.
NU
ME
l)GO
TO4
0IF
(.N
OT
.IN
TE
IN1»
GOTO
20
IF(O
IlIN
1»
11=
1IF
(OIL
IN1»
J1=
2IF
(OIL
IN1»
12=
4IF
<.N
OT.
OIL
<N
ll)
11=
4IF
(.N
OT.
OIl
IN1
»Jl=
3IF
I.N
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<N
1»12
=1
R1 =
RI
IXIN
1,
11)
)Z
1=Z
IIX
IN1,
11»
N2=
N2+
1IF
IN2.
GT
.NU
ME
l)PR
INT
10
0IF
(N2.
GT
.NU
ME
l)ST
OP
R2=
RI
IXIN
2,12
»Z
2=
Z(I
XIN
2.I
2»
IF(R
l.N
E.R
2.O
R.Z
l.N
E.Z
2)GO
TO3
0
"11A
TCH
ING
"EL
EMEN
TA
LON
GIN
TE
RFA
CE
FOU
ND
IDIF
1=IA
BS
(IX
(N1.
11)-
IX(N
2.
1»ID
IF2=
IAB
SI
IXIN
1.1
1)-
IX
(N2.
2)
)ID
IF3=
IAB
SI
IXIN
1,11
)-IX
(N2
.3»
IDIF
4=IA
BSII
X(N
1.11
)-IX
IN2.
4)
)ID
IF5=
IAB
SI
IXIN
1,J1
)-IX
IN2
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IDIF
6=
IAB
SC
IX(N
l,Jl
)-IX
(N2
.2»
IDIF
7=
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S(I
XIN
1.J
l)-I
XIN
2,3
»ID
IF8
=IA
BS
IIX
INl,
Jl)-
IXIN
2.4
»ID
IF=
MA
XO
(ID
IF1,
IDIF
2.
IDIF
3.
IDIF
4.
IDIF
5.
IDIF
6.
IDIF
?ID
IFS
)IF
(ID
IF.
GT.
MA
XD
IF)
MA
XD
IF=I
DIF
N2=
0GO
TO2
04
0C
ON
TIN
UE
NB
AN
D=2
*MA
XD
IF+l
10
0FO
RMA
T1*
ELEM
ENT
NO
TFO
UN
D*)
RETU
RNEN
D
SUB
RO
UT
INE
BND
WTH
76
/17
6O
PT
=l
ST
AT
ICFT
N4.
8+
49
80
3/1
6/8
116
.3
9.
39
PAG
E2
SYM
BO
LIC
RE
FER
EN
CE
MA
P(R
=2
)
ENTR
YP
OIN
TS
DE
FL
INE
RE
FER
EN
CE
S3
BND
WTH
15
3
VA
RIA
BL
ES
SNT
YPE
RE
LO
CA
TIO
N2
27
IIN
TE
GE
RR
EF
S7
9D
EF
INE
D6
23
2IO
IFIN
TE
GE
RR
EF
S2
*1
02
*4
7D
EFI
NE
D9
46
24
4IO
IFl
INT
EG
ER
RE
FS4
6D
EFI
NE
D3
82
45
IOIF
2IN
TE
GE
RR
EF
S4
6D
EF
INE
D3
92
46
IOIF
3IN
TE
GE
RR
EF
S4
6D
EFI
NE
D4
02
47
IOIF
4IN
TE
GE
RR
EF
S4
6D
EFI
NE
D41
25
0IO
IF5
INT
EG
ER
RE
FS
46
DE
FIN
ED
42
25
1ID
IF6
INT
EG
ER
RE
FS
46
DE
FIN
ED
43
25
2IO
IF7
INT
EG
ER
RE
FS
46
DE
FIN
ED
44
25
3ID
IF8
INT
EG
ER
RE
FS
46
DE
FIN
ED
45
0IN
TE
LO
GIC
AL
AR
RA
YF.
P.
RE
FS
23
20
DE
FIN
ED
0IX
I~HEGER
AR
RA
YF.
P.
RE
FS
22
*9
27
28
32
33
2*
38
2*
39
2*
40
2*
41
2*
42
2*
43
2*
44
2*
45
DE
FIN
ED
12
35
I1
INT
EG
ER
RE
FS
27
28
38
39
40
41D
IOFI
NED
21
.,2
42
37
12
INT
EG
ER
RE
FS
32
33
DE
FIN
ED
23
26
.......
23
1J
INT
EG
ER
RE
FS
9D
EF
INE
D8
q..
23
0Jt
1IN
INT
EG
ER
RE
FS
8D
EF
INE
D7
'"2
36
Jl
INT
EG
ER
RE
FS
42
43
44
45
DE
FIN
ED
22
25
22
5M
AX
DIF
INT
EG
ER
RE
FS
10
47
51D
EFI
NE
D4
10
47
22
6N
INT
EG
ER
RE
FS
2*
9D
EF
INE
D5
0N
BAN
OIN
TE
GE
RF
.P
.D
EF
INE
D1
510
NU
MEL
INT
EG
ER
F.P
.R
EF
S3
*2
51
93
03
1D
EF
INE
D1
0N
UM
NO
DIN
TE
GE
RF.
P.
RE
FS
2*
2D
EF
INE
D1
23
3N
lIN
TE
GE
RR
EF
S1
81
92
02
12
22
32
42
52
62
72
83
83
94
0-4
14
24
34
44
5D
EFI
NE
D1
61
82
34
N2
I~HEGER
RE
FS
29
30
31
32
33
38
39
40
414
24
34
4·
45
DE
FIN
ED
17
29
48
0O
ILL
OG
ICA
LA
RR
AY
F.
P.
RE
FS
2...
..3
21
22
23
24
25
26
DE
FIN
ED
10
RR
EAL
AR
RA
YF
.P.
RE
FS
22
73
2D
EFI
NE
D2
40
Rl
REA
LR
EF
S3
4D
EFI
NE
D2
72
42
R2
REA
LR
EF
S3
4D
EFI
NE
D3
20
ZR
EAL
AR
RA
YF.
P.
RE
FS
22
83
3D
EFI
NE
D1
24
1Z
lR
EA
LR
EF
S3
4D
EFI
NE
D2
82
43
Z2
RE
AL
RE
FS
34
DE
FIN
ED
33
FIL
EN
AM
ESt1
0DE
OU
TPU
TFM
TW
RIT
ES
30
INL
INE
.FU
NC
TIO
NS
TY
PEA
RG
SD
EF
LIN
ER
EFE
RE
NC
ES
lAB
SIN
TE
GE
R1
INT
RIN
93
83
94
041
42
43
44
45
MA
XO
INT
EG
ER
0IN
TR
IN4
6
....... "" ""-.I
SUB
RO
UT
INE
BND
WTH
76
/17
6O
PT
=l
ST
AT
ICFT
N4.
8+
49
80
3/1
6/8
11
6.
39
.3
9PA
GE
3
STA
TE
ME
NT
LA
BE
LS
DE
FL
INE
RE
FER
EN
CE
S0
10
115
68
42
20
18
20
49
10
63
02
93
42
07
40
50
19
22
01
00
FMT
52
30
LO
OPS
LA
BE
LIN
DEX
FRO
M-T
OLE
NG
THP
RO
PE
RT
IES
12
10
N5
1127
BN
OT
INN
ER
13
10
I6
1123
BN
OT
INN
ER
25
10
..J8
116B
.INST
AC
K
ST
AT
IST
ICS
PRO
GR
AM
LEN
GTH
26
0B
17
65
20
00
BSC
MU
SED
SUB
RO
UT
INE
SOL
INT
76
/17
6O
PT
=l
STA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E
20
..... ...., ~2
5
52
30
60
65
35 40
68
70
5 10
15 45 50
55
SUB
RO
UTI
NE
SO
LIN
T(A
,BI,
BII
.R,Z
,IC
OD
E.IX
.D,H
M,H
B,H
L,
1R
01,N
UM
EL
,NB
AN
D,N
P,N
EG
,NM
O,N
UE
LO
IL)
CD
IME
NSI
ON
A(N
EG
,1
).B
I(N
MO
),B
II
(NM
O,N
MO
),IX
(NU
MEL
,4
).R
(NP
),Z
(NP
),1
ICO
DE
(NP
),H
M(N
P,N
BA
ND
),H
B(N
P).
HL
(NP
.NM
O),
D(N
MO
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DIM
EN
SIO
NR
R(4
).Z
Z(4
),H
H(4
,4
),H
(4,
4),
58
(2).
IJ(3
)CO
MM
ONIF
CO
NS
TI
PI,
RO
W,H
SD
ATA
SS1
-.5
77
35
02
69
18
96
3,
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77
35
02
69
18
96
31
DO3
01=
1.N
PHB
(I)
=0
.DO
22
M=l
,NM
OD
(M,
1)=
0.
22
HL
(1.
M)=
O.
DO3
0J=
1.
NBA
ND
30
HM
(1
.J)
=O
.DO
10
0N
=l.N
UM
EL
IJ(l
)=O
IJ(2
)=0
IJ(3
)=0
11-\=
0RO
=RO
WIF
(N.L
E.N
UE
LO
IL)
RO
=R01
*RO
WDO
60
1=
1,4
II=
IX(N
,I)
IF(I
CO
DE
(II)
.EG
.0
)GO
TO52
11-\=
11-\+
1IJ
(II-
\)=
IIR
R(I
)=R
(II)
ZZ
(I)=
Z(I
IlDO
65
1=
1,4
DO6
5J=
1,4
HH
(1
.J)
=O
.DO
70
11
=1
,2DO
70
JJ=
1,2
CA
LLF
OR
MH
(8S
(II)
.SS
(JJ)
,RR
.ZZ
,H)
DO6
81=
1.4
DO6
8J=
1.
4H
H(I
,J)=
HH
(I,J
)+H
(I,J
)H
H(J
,I
)=H
H(
1.J)
CO
NTI
NU
EDO
80
1=1.
4I
I=IX
(N,
I)DO
80
J=1
,4JJ
=IX
(N,J
)-II
+l
IF(J
J.L
T.
1)
GOTO
80
HM
(II,
JJ)=
HM
CII
.JJ)
+H
H(I
,J)/
RO
80
CO
NTI
NU
EIF
(IK
.LT
.2)
GOTO
10
0N
N=
II-\
-lDO
85
I=l,
NN
II=
IJ(I
)JJ
=IJ
CI+
l)Il
=IA
BS
(IC
OD
E(I
I»J1
=IA
BS
CIC
OD
E(J
J»R
IJ=
-R(I
I)+
RC
JJ)
ZIJ
=Z
(II)
-Z(J
J)8I
J=S
GR
TC
RIJ
**
2+
ZIJ
**
2)*
SIG
N(
1.
,Z
IJ)
SUB
RO
UT
INE
SOL
INT
76
/17
6O
PT=1
STA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E2
<l'-.. --('
\
60
65
70
75
80 85
90
95
10
0
10
5
CS
N=
-ZIJ
/SIJ
S5
N=
-RIJ
/SIJ
C51~=A13S(C
SN)
IFIZ
IJ.
GT
.O.)
SS
N=
-S5N
ZIJ
=A
135(
ZIJ
)DO
85
K=1
>2
Al=
l1.
-65
IK)
)/2
.A
4=I1
.+
6SIK
))1
2.
RF
=R
lIl)
*A
l+R
CJJ
)*A
4R
RR
=R
F*S
IJ*C
SN
/2.
HB
III)
=H
13
(II)
+R
RR
*A
lH
l3(J
J)=
H13
(JJ)
+R
RR
*A4
DO8
511
=1,
NMO
TI=
AI3
*II
-2,M
)*C
SN
+A
(3*
Il-l
,M)*
S6
NT
J=A
(3*
JI-2
,M)*
CS
N+
A(3
*Jl
-l,M
)*S
SN
AF
=IT
I*A
l+T
J*A
4)*
SIJ
/2.
*RF
HL
III,
M)=
HL
III,
M)+
AF
*A
lH
LIJ
J,M
)=H
LIJ
J,M
)+A
F*
A4
DIM
,I
I)=
DIM
,I
I)+
AF
*Al
DIM
,JJ)
=D
IM,J
J)+
AF
*A
48
5C
ON
TIN
UE
10
0C
ON
TIN
UE
CA
LLB
AN
SL1(
R,N
P,N
BA
ND
,HM
.HB
.1
)C
ALL
13A
NSL
IIR
,N
P.N
BA
ND
,HM
,HB
.2
)DO
11
011
=1.
NMO
11
0C
ALL
BA
NS
LII
R,N
P,N
BA
ND
,HM
,HL
(1.M
),2)
NW
ALL
=ODO
13
0I=
I.N
P1
30
IF(I
CO
DE
(I).
NE
.O)
NW
ALL
=NW
ALL
+lW
RIT
E(9
)N
WA
LLPR
INT
10
20
DO2
00
I=l,
NP
IF(I
CO
DE
(I).
EG
.O)
GOTO
20
0W
RIT
E(9
)IA
BS
(IC
OD
E(I
»,H
B(I
),(H
UI.
M),
M=
I,N
MO
)P
RIN
T1
02
5,
LH
B(I
),(H
UI,
M),
M=
1,N
MO
)DO
15
0M
=1,
NMO
BI
<l1)
=BI
(M)+
D(M
,I
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CD
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AS
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AS
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ZZ
(2)+
AS
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AS
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(4)
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AS
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AT
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AT
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RR
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AT
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RR
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A(3
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JD
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SR
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T(I
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TR
50
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AS
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PS
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AT
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PT
ZDO
10
01
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DO
10
0J=
I,4
H(I
,J)=
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AR
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AZ
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AZ
(J»
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J/R
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A(J
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H(J
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=H
(I,J
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ETU
RN
END
SYM
BO
LIC
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65
70
75
80
85
90
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5
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40
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00
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03
60
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40
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00
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20
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TE
P7
6/1
76
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T=
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4.8
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98
03
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16
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PAG
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INT
20
12
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12
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10
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20
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81
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20
22
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ON
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UE
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20
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=0.
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GR
AV
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TO2
4
SUB
RO
UT
INE
RES
PON
76
/17
6O
PT
=l
STA
TIC
FTN
4.8
+4
98
03
/16
/81
16.
39
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9PA
GE
2
C CRE
AD
FRO
MT
APE
3TH
EST
AT
ICR
ESP
ON
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60
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INT
20
08
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MN
P,N
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EL,
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T,
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25
DO2
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28
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81
=1
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M(
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MN
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ON
TIN
UE
75
MN
=OREI~IND
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EL
CA
LLR
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UM
EL
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0t1
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1N+l
DO31
J=1
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2
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MM
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N)=
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(J)
~DO
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1,6
SS
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,J.M
N)=
SS
(I,J
)8
531
CO
NTI
NU
EIF
(N.
EQ
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ME
L)
GOTO
32
IF<t
1N.L
T.
10
)GO
TO4
03
2W
RIT
E(4
)LM
MW
RIT
E(4
)SS
M9
0M
N=O
40
CO
NTI
NU
EDO
50
I=I,
NE
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1AX
(I)=
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(I
)=0
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550
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X(
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)=0.
60
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AX
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10
0R
EIH
ND
9DO
90
1=1,
NUM
NPH
B(I
)=O
.DO
90
11N=
1.NM
O9
0H
L(I,
MN
)::;:O
,10
5DO
100
1=
1,2
READ
(9)
NWAL
LIF
(NW
ALL
.L
E.
0)
GOTO
10
0DO
95
J=
l,N
WA
LL9
5RE
AD
(9)
N.H
B(N
).(H
L(N
.MN
).M
N=
I.N
MO
)1
10
10
0C
ON
TIN
UE
REA
D(9
)FH
XIF
(NG
RD
.EQ
.2)
REA
D(9
)FH
YC C
REA
DFR
OM
TA
PE2
AND
TA
PE7
THE
MO
DA
LR
ESP
ON
SES
SUB
RO
UT
INE
RES
PON
76
/17
6O
PT=1
STA
TIC
FTN
4.8
+4
98
03
/16
/81
16
.39
.39
PAG
E3
12
0
20
0
12
5
13
02
05
13
52
10
11
5C
REI
HN
D2
REI
HN
D7
IBLO
CK
=OTS
=O.
IT=
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M=O
IBL
OC
K=I
BL
OC
K+1
IF(I
BL
OC
K.
EG.
NB
LOC
K)
NB=
MBU
FRE
AD
(2)
(XB
UF
(J),
J=1
,NB
)RE
AD
(2)
(AX
BU
F(J
),J=
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B)
IF(N
GR
D.
EG.
2)
REA
D(7
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BU
F(J
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B)
IF(N
GR
D.
EG
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REA
D(7
)(A
YB
UF
(J),
J=1
,NB
)DO
21
01=
1.NM
OM
=M+1
IF(N
GR
D.E
G.2
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M(I
)=Y
BU
F(M
)IF
(NG
RD
.EG
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YM
(I)=
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BU
F(M
)A
XM
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AX
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F(M
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M(I
)=X
BU
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OD
E.N
E.O
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NEL
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E.O
)PR
INT
20
50
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"-1
40
~ ~
14
5
22
0
15
0
15
5
16
0
16
5
17
0
C CC
OM
PUTE
DIS
PLA
CE
ME
NT
CIF
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OD
E.N
E.0
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INT
20
51
DO2
50
I=l,N
UM
NP
SUM
1=0.
SUt1
2=0.
SUt1
3=0.
DO2
20
MN
=l,N
MO
SUM
1=SU
M1+
XM
(MN
)*A
(3*I
-2,M
N)
SUM
2=SU
M2+
XM
(MN
)*A
(3*I
-1,M
N)
SUM
3=SU
M3+
XM
(MN
)*A
(3*I
,MN
)X
(3*
I-2
)=S
UM
1*
12
.X
(3*
I-1
)=S
UM
2*
12
.X
(3*
I)=
SUM
3*12
.DO
23
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1,3
II=
3*
I-3
+J
XA
BS=
AB
S(X
(II
))
XX
=AB
S(X
MA
X(I
I»IF
(XX
.GE
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AB
S)GO
TO2
30
XM
AX
(II)
=X
(II)
TX
(II)
=T
S2
30
CO
NTI
NU
EIF
(NG
RD
.EG
.1
)GO
TO2
34
SUM
1=0.
Sut
12=
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t13=
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18
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19
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50
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24
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26
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28
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T1
6.4
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CL
EN
O.
=0
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16
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RA
RY
.DT
CP
Lo
T.
16
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0.
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6.
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XIT
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6.
40
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DA
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6.4
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TP
UT
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01
6.4
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9.M
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76
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D)
16
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40
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0.
10
20
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4.6
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16
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00
16
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.EJ
END
OF
JOB
.A
N