Dynamics Technology, Inc.

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.

. "

11/

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.

f

IY

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 •••.••••••••••••.•••••••••••••••••••••••••••••••••••••••••.••

43

46

APPENDIX A:

APPENDIX B:

ERST User's Guide ..•••••••..•••.••.•••••.••..••••••••••

Program Listing .

v

66

84

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.

-1-


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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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

-3-

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).

-4-

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,

-5-


*{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.

-6-


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

-l-

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

-7-


? 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

-8-


(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

-10-

~ ~

[( 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

-11-

(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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DT-7814-2

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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DT-7814-2

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

DT-7814-2

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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DT-7814-2

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

-17-

DT-7814-2

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)·

-18-

(3.21)


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

-19-

(3.25)

and


..(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

-20-

(3.27)

Ei=L:

e=1


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)

-21-


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.

-22-


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

-23-

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

-24-

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

-25-


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

-26-

(3.43)


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,

-27-


(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

-28-

(3.51)


=

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.

-29-

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

-30-


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)

-31-

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.

-32-


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

-33-

j = 1, •.• ,J (3.68)


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.

-34-

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•

-35-

(4.2)


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

-36-


(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

-37-

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.

-38-


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.

-39-

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

-40-


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].

-41-


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.

-42-


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-


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.)


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


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-


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-


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-


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-


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-


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-


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)


31 - 60: Used for layer generation; otherwise leave blank.

-73-


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)

-74-


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)


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.

-75-


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.

-76-


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-


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.

-83-


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

MP

=N

l0-1

WR

ITE

(1)

(A(I

),I=

N9,

NT

EM

P)

NT

EM

P=

N13

-1W

RIT

E(1

)(A

(!),

I=N

l1,N

TE

MP

)B

AC

KSP

AC

E1

BA

CK

SPA

CE

1N4=N3+r~EG

N5=

N4+

NM

OD

EN

6=N

5+(N

EG

*NM

OD

E)

N7=

N6+

NM

OD

EN

8=N

7+(N

MO

DE

*NM

OD

E)

NTE

I1p

..N

4-1

REA

D(1

)(A

(I"

I=N

3,N

TEM

P)N

TE

I1P=

N6-

1R

EA

D(

1)

(A(I)

,I=

N4

,N

TE

MP)

BA

CK

SPA

CE

1B

AC

KSP

AC

E1

WA

TER

TR

EA

TE

DA

SA

FIN

ITE

-EL

EM

EN

TSY

STE

M

REA

DE

XT

FLD

PR

INT

EX

TFL

D

115

PRO

GRA

MER

ST7

6/1

76

OP

T=

lST

AT

IC

PRIN

T2

04

0

FTN

4.8

+4

98

03

/16

/81

16.

39.

39

PAG

E3

O(J

"""'"

12

0

125

130

135

140

145

15

0

155

160

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EA

LA

RR

AY

F.

P.R

EF

S8

25

DE

FIN

ED

11

51

ZZR

EA

LA

RR

AY

RE

FS

12

26

DE

FIN

ED

25

~E

XT

ER

NA

LS

TY

PEA

RG

SR

EF

ER

EN

CE

S11

AS

TIF

112

6~

SSLA~~

42

1W

RIT

EI

43

4

STA

TE

ME

NT

LA

BE

LS

DE

FL

INE

RE

FER

EN

CE

S0

30

51

81

73

43

07

22

16

03

10

25

22

03

20

32

27

29

04

00

34

14

LO

OPS

LA

BE

LIN

DE

XFR

OM

-TO

LE

NG

TH

PR

OP

ER

TIE

S1

24

00

N1

43

471

BEX

TR

EF

SN

OT

INN

ER

23

30

5M

t11

71

83B

INST

AC

K4

23

10

II

22

25

56

INST

AC

K5

33

20

I2

73

221

BN

OT

INN

ER

64

32

0J

29

32

4BIN

STA

CK

C0I

1MO

NB

LO

CK

SL

EN

GT

HL

S4A

RG

28

3

ST

AT

IST

ICS

PRO

GR

AM

LE

NG

TH

20:3

813

1G

CML

AB

EL

ED

COM

MO

NL

EN

GT

H4

33

32

83

52

00

0B

SCM

USE

D

SUB

RO

UT

INE

SSLA

W7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.

8+

49

80

3/1

6/8

11

6.3

9.3

9PA

GE

SUB

RO

UT

INE

SS

LA

WC

EE

,BE

TA

,C,N

F)

5

C C*

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

CS

TR

ES

SS

TR

AIN

LAW

INN

-S-T

SYST

EM

C*

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

C

RO

TA

TE

MA

TE

RIA

LP

RO

PE

RT

IES

TOR

-Z-T

HSY

STE

M

DIM

EN

SIO

NE

E(9

),C

C6

,6),

DC

6.6

),C

C(6

,6)

DO

65

11

=1

.6D

O6

51-

\1-\=

1,6

C(

II.

1-\1

-\)=

0.

65

0(

II"'

,K)

=0

.C

(1.1

)=1

.0/E

E(1

)C

(1

.2

)=-E

E(4

)*C

(1,

1)

C(

1.3

)=

-EE

C5)*

CC1

,1

)C

(2,

ll=

CC

1,

2)

C(2

.2)=

1.0

/EE

(2)

C(2

,3)=

-EE

(6)*

C(2

,2)

C(3

,1l=

C(l

,3)

C(3

.2)=

CC

2.3

)C

(3.3

)=1

.0/E

E(3

)C

(4,4

)=E

E(7

)IF

CN

F.E

G.

0)

GO

TO7

0C

C5

.5)=

EE

(8)

C(6

.6)=

EE

(9)

CO

NT

INU

ED

O2

00

N=

l.3

DD

=C

(N,N

)D

O1

00

J=

l.3

C(N

.J)=

-C(N

,J)/

DO

DO

15

01

=1

,3IF

CN

-I)

11

0.1

50

,11

0D

O1

40

J=

l,3

IF(N

-J)

12

0.1

40

.12

0C

(I.J

)=C

(I.J

)+C

(I.N

)*C

(N,J

)C

ON

TIN

UE

C(I

.N)=

C(I

,N)/

DD

C(t~,

N)=

1.

0/0

0C

ON

TIN

UE

70

10

0

11

0

12

01

40

15

0

20

0C C C

15

10

20

40

25

35

30

........

I;;:) \...1

45

50

55

IF(B

ET

A.

EG

.0

.0)

GO

TO5

00

AN

G=

BE

TA

/57.

29

57

79

5S

S=

SIN

(AN

G)

CS

=C

OS

(AN

G)

S2

=S

S*

SS

C2=

CS

*CS

SC

=S

S*C

SO

(1

.1)=

C2

O(1

.2)=

52

O(

1.

4)=

SC

0(2

.1)=

S2

0(2

.2)=

C2

0(2

,4)=

-SC

0(3

.3)=

1.

00

(4.1

)=-2

.*S

C0

(4,2

)=-0

(4,1

)

SU

BR

OU

TIN

ESS

LAW

76

/17

6O

PT

=l

ST

AT

ICFT

N4

.8+

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)

D2=

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)"

'D(

1,

II

)*D

1+

D(2

,II

)*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

,6

DO

30

0J=

l,6

30

0C

(I,J

)=C

C(I

,J)

50

0R

ET

UR

NEN

D

'.SY

MB

OL

ICR

EFE

RE

NC

EM

AP

(R"'

2)

<:> i=:

EN

TR

YP

OIN

TS

DE

FL

INE

RE

FE

RE

NC

ES

3S

SL

AlJ

17

5

VA

RIA

BL

ES

SNT

YPE

RE

LO

CA

TIO

N2

45

AN

GR

EA

LR

EF

S4

44

5D

EF

INE

D4

30

BE

TA

RE

AL

F.

P.

RE

FS

42

43

DE

FIN

ED

10

CR

EA

LA

RR

AY

F.

P.

RE

FS

71

31

41

51

71

81

92

72

93

*3

43

63

*6

23

*6

33

*6

46

56

66

7D

EF

INE

D1

10

12

13

14

15

16

17

18

19

20

21

23

24

29

34

36

37

74

32

6C

CR

EA

LA

RR

AY

RE

FS

77

17

4D

EF

INE

D6

97

12

47

CS

RE

AL

RE

FS

2*

47

,..,4

8D

EF

INE

D4

52

51

C2

RE

AL

RE

FS

49

53

58

DE

FIN

ED

47

26

2D

RE

AL

AR

RA

YR

EF

S7

57

3*

62

3*

63

3*

64

65

66

67

6*

69

DE

FIN

ED

114

95

051

52

53

54

55

56

57

58

59

60

24

2D

DR

EA

LR

EF

S2

93

63

7D

EF

INE

D2

72

54

Dl

RE

AL

.R

EF

S6

9D

EF

INE

D6

22

55

D2

RE

AL

RE

FS

69

DE

FIN

ED

63

25

6D

3R

EA

LR

EF

S6

9D

EF

INE

D6

42

57

D4

REAL

R[;F

SG9

DEF

INii:

Ot.J

:J~~60

D5

RttA

LR

tJ'S

69

DE

:F'tN

ED

66

26

1D

6R

tAL

~EFS

69

DE

FIN

ED

67

0E

ER

EA

LA

RR

AY

F.

P.

RE

FS

71

21

31

41

61

72

02

12

32

4D

EF

INE

D1

24

4I

INT

EG

ER

RE

FS

31

3*

34

2*

36

2*

74

DE

FIN

ED

30

72

23

7II

INT

EG

ER

RE

FS

10

117

*6

92

*7

1D

EF

INE

D8

68

24

3J

INT

EG

ER

RE

FS

2*

29

33

3*

34

2*

74

DE

FIN

ED

28

32

73

25

3JJ

INT

EG

ER

RE

FS

3*

62

3*

63

3*

64

65

66

67

68

SUB

RO

UT

INE

SSLA

W7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.39

.39

VA

RIA

BL

ES

SNT

YPE

RE

LO

GA

TIO

N6

92

*7

1D

EF

INE

D6

12

40

KKIN

TE

GE

RR

EF

S1

011

DE

FIN

ED

92

41

NIN

TE

GE

RR

EF

S2

*2

72

*2

931

33

2*

34

DE

FIN

ED

26

0N

FIN

TE

GE

RF

.P

.R

EF

S2

2D

EF

INE

D1

25

2SC

REA

LR

EF

S51

54

56

DE

FIN

ED

48

24

65

5R

EA

LR

EF

52

*4

64

8D

EF

INE

D4

42

50

52

RE

AL

RE

FS

50

52

58

DEFIt~ED

46

EX

TE

RN

AL

ST

YPE

AR

GS

RE

FER

EN

CE

SC

OS

REA

L1

LIB

RA

RY

45

SIN

RE

AL

1L

IBR

AR

Y4

4

STA

TE

ME

NT

LA

BE

LS

DE

FL

INE

RE

FER

EN

CE

S0

65

11

89

44

70

25

22

01

00

29

28

01

10

INA

CT

IVE

32

2*

31

01

20

INA

CT

IVE

34

2*

33

10

11

40

35

32

33

10

41

50

36

30

31

02

00

38

26

02

87

71

61

68

"-0

30

07

47

27

3

<:>2

33

50

07

54

2It

;L

OO

PSF

Rot

'l-T

OP

RO

PE

RT

IES

LA

BE

LIN

DE

XLE

NG

TH7

65

II8

1114

·13N

OT

INN

ER

15

65

KK

911

313

INST

AC

K4

52

00

N2

63

852

13N

OT

INN

ER

54

10

0J

28

29

4BIN

STA

CK

62

15

0I

30

36

3013

NO

TIN

NE

R7

51

40

J3

23

561

3IN

STA

CK

14

22

87

JJ

61

71

5413

NO

TIN

NE

R1

77

28

7II

68

7113

13O

PT2

17

30

0I

72

74

14B

NO

TIN

NE

R2

25

30

0J

73

74

313

INST

AC

K

ST

AT

IST

ICS

PRO

GR

At1

LE

NG

TH

4061

32

62

5200

013

SCM

USE

D

PAG

E3

2*

36

2*

37

SUB

RO

UT

INE

ST

RA

IN7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.S

+4

98

03

/16

/81

16

.39

.39

PAG

E

~ C'\

5

10

15

20

25

30

35

40

45

50

55

SUB

RO

UT

INE

ST

RA

IN(S

.T.R

R.

ZZ

.H.B

,FA

C.N

)C C

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

*C

FOR

MS

TR

AIN

DIS

PLA

CE

ME

NT

MA

TR

IXc*

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

CD

IME

NS

ION

B(6

.lS

),H

S(6

).H

T(6

).H

R(6

).H

Z(6

)D

IME

NSI

ON

RR

(4).

ZZ

(4).

H(6

)D

IME

NS

ION

II(6

),JJ(b

).L

L(b

)D

ATA

11

/1.2

.3.4

.13

.14

/.JJ/5

.6

.7

.S

.1

5.

16

/.L

L/9

.1

0.

11

.1

2.

17

.1

8/

DO

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

.H

S(2

)=-H

S(1

)H

S(3

)=T

P/4

.H

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

)*R

R(1

)+H

T(2

)*R

R(2

)+H

T(3

)*R

R(3

)+H

T(4

)*R

R(4

)X

J=P

RS

*P

ZT

-PR

T*

PZ

SP

SR

=P

ZT

/XJ

PT

R=

-PZ

S/X

JP

SZ

=-P

RT

/XJ

PT

Z=

PR

S/X

JD

O1

00

1=

1.

6H

R(I

)=P

SR

*H

S(I

)+P

TR

*H

T(I

)1

00

HZ

(I)=

PS

Z*

HS

(I)+

PT

Z*

HT

(I)

R=

H(1

)*R

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

)L

=L

L(K

)8

(1.

I)=H

R(K

)8

(2.J

)=H

Z(K

)8

(3.1

)=H

(K)/

RB

(3,L

)=N

*B

(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

)=-B

(3,

L)

0(5

,L

)=8

(1,

1>

-8(3

,I)

0(6

,J)=

B(5

,1>

20

0B

(6,L

)=B

(2,J

)F

AC

=X

J*R

RET

UR

NEN

D

SYM

BO

LIC

RE

FER

EN

CE

MA

P(R

=2

)

ENTR

YP

OIN

TS

DE

FL

INE

RE

FER

EN

CE

S3

ST

RA

IN1

63

VA

RIA

BL

ES

SNT

YPE

RE

LO

CA

TIO

N0

8R

EAL

AR

RA

YF

.P

.R

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

.P

.D

EFI

NE

D1

62

0H

REA

LA

RR

AY

F.

P.

RE

FS

84

*4

75

4D

EFI

NE

D1

17

18

19

20

212

2

~2

42

HR

REA

LA

RR

AY

RE

FS

'75

25

7D

EFI

NE

D4

52

26

HS

REA

LA

RR

AY

RE

FS

72

42

64

*3

64

*3

74

54

6'I

DE

FIN

ED

23

24

25

26

27

28

23

4H

TR

EA

LA

RR

AY

RE

FS

73

13

24

*3

54

*3

84

54

6D

EFI

NE

D2

93

03

13

23

33

42

50

HZ

REA

LA

RR

AY

RE

FS

75

35

6D

EF

INE

D4

62

04

IIN

TE

GE

RR

EFS

12

3*

45

3*

46

52

54

55

56

58

2*

59

60

DE

FIN

ED

114

44

92

56

II

INT

EG

ER

AR

RA

YR

EF

S9

49

DE

FIN

ED

10

22

4J

INT

EG

ER

RE

FS

53

57

60

61D

EFI

NE

D5

02

64

JJ

INT

EG

ER

AR

RA

YR

EF

S9

50

DE

FIN

ED

10

22

3K

INT

EG

ER

RE

FS

49

50

515

25

35

45

65

7D

EFI

NE

D4

82

25

LIN

TE

GE

RR

EF

S5

55

85

961

DE

FIN

ED

512

72

LL

INT

EG

ER

AR

RA

YR

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S9

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FIN

ED

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0N

INT

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ER

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P.

RE

FS

55

DE

FIN

ED

12

13

PR

SR

EAL

RE

FS

39

43

DE

FIN

ED

37

21

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39

42

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FIN

ED

38

21

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RE

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5J;

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NED

40

22

0P

SZ

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LR

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S4

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NE

D4

22

17

PTR

REA

LR

EF

S4

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12

21

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LR

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NE

D4

32

12

PZ

SR

EAL

RE

FS

39

41D

EFI

NE

D3

62

11

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REA

L.R

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:::J9

40D

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,IED

35

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,...)

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62

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lNIW

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AR

RA

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02

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06

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40

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S1

51

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D2

07

TMR

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RE

FS

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I1

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NE

D1

52

10

TP

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20

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FIN

ED

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RO

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ST

RA

IN7

6/1

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T=

lS

TA

TIC

VA

RIA

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YPE

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ND

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RA

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TE

ME

NT

LA

BE

LS

DEF

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ER

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NC

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05

01

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10

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IES

12

50

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10

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44

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lOB

INST

AC

K1

44

20

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612

58

OPT

ST

AT

IST

ICS

PRO

GR

AM

LEN

GTH

30

38

19

55

20

00

BSC

MU

SED

39 8

FTN

4.8

+4

98

4*

35

4*

36

03

/16

/81

16

.39

.39

DE

FIN

ED

PAG

E3

SU

BR

OU

TIN

EM

AS

TIF

76

/17

6O

PT

=l

ST

AT

ICF

TN

4.8

+4

98

03

/16

/81

16

.39

.39

PAG

E1

" ~ -Q

5

10

15

20

25

30

35

40

45

50

55

SU

BR

OU

TIN

EM

AS

TIF

(RR

,Z

Z,R

M.

ZM

,EL

MA

SS

,O,G

K,G

C,G

,RH

O,N

)C C

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

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**

**

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CFO

RM

EL

EM

EN

TM

ASS

MA

TR

IXM

AN

DE

LE

ME

NT

ST

IFF

NE

SS

MA

TR

IXK

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*C

DIM

EN

SIO

NS

S(2

),R

R(4

),Z

Z(4

).E

LM

AS

S(4

),G

K(

12

,12

).G

C(6

,1

2>

'Q(

12

)D

rt1E

NS

ION

B(6

,1

8),

88

(6,

18

),S

(18

.18

),D

(6.6

),P

(18

),H

(6)

DA

TAS

S/-

O.

57

73

50

26

91

89

63

,0.

57

73

50

26

91

89

63

/F

Z=

-RH

O*

32

.2

DO

15

01

=1

,4

15

0E

LM

AS

S(I

)=O

.OD

O1

70

J=

l,1

8P

(J)=

O.O

DO

16

01

=1

,6B

Bn

.J)=

O.

01

60

B(I.

J)

=0

.0

DO

17

01

=1

,18

17

0S

(I.J

)=O

.0

RM

=(R

R(1

)+R

R(2

)+R

R(3

)+R

R(4

»/4

.0Z

M=

(ZZ

(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

,H,B

,FA

C,N

)IF

(N.

EG

.0

)F

AC

=2

.*F

AC

DO

40

0J=

1.

18

01

=(0

(1,

1)*

B(1

,J)+

0(1

,2)*

B(2

,J)+

0(1

,3)*

B(3

,J)+

0(1

,4)*

B(4

,J)+

0(1

,5)

1*

B(5

,J)+

0(1

,6)*

B(6

.J»

*F

AC

02

=(0

(2,

1)*

B(I

,J)+

0(2

,2)*

B(2

,J)+

0(2

,3)*

B(3

,J)+

0(2

,4)*

B(4

,J)+

0(2

,5)

1*

B(5

,J)+

0(2

,6)*

B(6

,J»

*F

AC

03

=<0

(3,

1)

*B(1

,J)+

0(3

,2

)*8

(2,

J)+

0(3

,3

)*

8(3

,J)

+O

(3,

'4)*

B(4

,,,

)+0

(3.

5)

1*

B(5

,J)+

0(3

,6)*

8(6

,J»

*F

AC

04

=<

0(4

.1)*

8(

1,

,,)1

+0

(4,

2H

I8(2

,..

)+0

(4/3

)*8

(3,

..)+

D(4

,4

)*B

(4,

,,)+

0(4

.5)

1*

B(S

,J)+

0(4

,6)*

8(6

.J»

*F

AC

05

=(0

(5.

1)*

B(I

,..)

+D

(5,2

)*B

(2,J

)+0

(5.3

)*B

(3,J

)+D

(5.4

)*B

(4,.

.)+

0(5

,5)

1*

B(5

,..)

+0

(5,6

)*8

(6,J

»*

FA

C0

6=

(0(6

.1

)*B

(1,.

.)+

D(6

,2)*

B(2

,..)

+D

(6.3

)*B

(3,J

)+D

(6,4

)*B

(4,J

)+D

(6,5

)1

*B

(5,J

)+0

(6,6

)*B

(6,

J»*

FA

CD

O4

00

I=J,

18

S(

I.J)

=S

(L

J)+

B(1

,I

>*

Ol+

B(2

,I

)*0

2+

B(3

,I

)*D

3+

8(4

,I

)*D

4+

B(5

,I

)*0

5+

B(

16

,1)*

06

40

0S

(J,

I)=

S(I,

J)

IF(N

.E

O.

0)

GO

TO4

70

DO

45

01

=1

,44

50

EL

MA

SS

(I)=

EL

MA

SS

(I)+

FA

C*

H(I

)*R

HO

47

0IF

(N.

NE

.0

)G

OTO

50

0B

Z=

FZ

*FA

CP

(S)=

P(S

)+B

Z*

H(l

)P

(6)=

P(6

)+B

Z*

H(2

)P

(7)=

P(7

)+B

Z*

H(3

)P

(B)=

P(B

)+B

Z*

H(4

)5

00

CO

NT

INU

EIF

(N.N

E.

0)

GO

TO5

10

DO

50

51

=1

,1

85

05

IF(S

(I.I

l.E

G.O

.O)

S(I.I

)=

l,O

C CFO

RM

ST

RE

SS

DIS

PL

AC

EM

EN

TM

AT

RIX

SUB

RO

UT

INE

MA

ST

IF7

6/1

76

OP

T=

1S

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.3

9.

39

PAG

E2

C5

10

CA

LLS

TR

AIN

(O.

0,0

.O

.RR

,Z

Z,H

.BB

.FA

C.N

)6

0DO

53

01

=1

,6D

O5

30

J=1

.1

8B

(1.

J)=

O.

0D

O5

30

1\=

1.6

53

0B

(I.J

)=B

(I.J

)+D

(I.K

)*B

B(K

,J)

65

C CE

LIM

INA

TE

EXTR

AD

EG

RE

ES

OF

FREE

DO

MC

DO

55

0N

N=

I.6

L=

18

-NN

70

K=

L+

lD

O5

50

1=

1,

LC

=S

(I.

K)

ISO

\.K

)D

O5

40

J=

I.6

54

0B

(J.

I)=

B(J

.I)-C*B(~K)

75

DO

55

0J=

I.L

55

0S

(I.J

)=S

(I,J

)-C

*S

(K.J

)IF

(N.N

E.O

)G

OTO

57

0D

O5

60

1=

9.

12

56

0S

(1

.1

)=0

.8

0C C

RE

LO

CA

TE

ST

IFF

NE

SS

MA

TR

IXA

ND

LOA

DV

ECTO

R'-

.C

.....5

70

DO

58

01

=1

.12

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

STli=

"1

90

VA

RIA

BL

ES

SNT

YPE

RE

LO

CA

TIO

N5

20

BR

EA

LA

RR

AY

RE

FS

82

46

*2

76

*2

96

*3

16

*3

36

*3

56

*3

76

*4

06

42

*7

48

9D

EF

INE

D1

76

26

47

46

74

BBR

EA

LA

RR

AY

RE

FS

85

96

4D

EF

INE

D1

65

11

BZ

RE

AL

RE

FS

48

49

50

51D

EF

INE

D4

75

15

CR

EA

LR

EF

S7

47

6D

EF

INE

D7

20

DR

EA

LA

RR

AY

F.

P.

RE

FS

86

*2

76

*2

96

*3

16

*3

36

*3

56

*3

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

*4

23

*4

54

*5

56

23

*6

47

22

*7

42

*7

62

*7

92

*8

42

*8

6D

EFI

NE

D11

15

18

39

44

54

60

717

88

35

00

II

INT

EG

ER

RE

FS2

4D

EFI

NE

D2

24

77

JIN

TE

GE

RR

EFS

14

161

71

96

*2

76

*2

96

*3

16

*3

36

*3

56

*3

73

92

*4

02

*4

26

23

*6

43

*7

43

*7

62

*8

62

*8

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

0N

INT

EG

ER

F.P.

RE

FS2

42

54

34

65

35

97

7D

EFI

NE

D1

51

3NN

INT

EG

ER

RE

FS6

9D

EFI

NE

D6

8.....

.1

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

7D

EFI

NE

D1

84

0OC

REA

LA

RR

AY

F.P.

RE

FS7

DE

FIN

ED

18

90

OKR

EAL

AR

RA

YF

.P.

RE

FS7

DE

FIN

ED

18

60

RHO

REA

LF

.P.

RE

FS1

04

5D

EFI

NE

D1

0RM

REA

LF.

P.D

EFI

NE

D1

20

0RR

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

*7

22

*7

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

.P.

DE

FIN

ED

121

0ZZ

REA

LA

RRA

YF

.P.

RE

FS7

4*

21

24

59

DE

FIN

ED

EX

TE

RN

AL

ST

YPE

AR

GS

RE

FER

EN

CE

SST

RA

IN8

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

****

****

****

****

****

****

****

****

****

****

****

****

****

****

****

**C

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

=O1

00

MM

=M+L

ADO

20

01

=1

.LA

II=

I+M

20

0B

(II)

=A

(I)

M=M

MIF

(N.

EO.

NU

MEL

)G

OTO

30

0IF

«M

+L

A).

LE

.L

B)

GO

TO4

00

30

0W

RIT

E(1

lB

M=O

40

0R

ETU

RN

END

SYM

BO

LIC

RE

FER

EN

CE

MA

P(R

=2

l

EN

TR

YP

OIN

TS

DE

FL

INE

RE

FER

EN

CE

S3

WR

ITE

I1

22

VA

RIA

BL

ES

SNT

YPE

RE

LO

CA

TIO

N0

AR

EA

LA

RR

AY

F.P

.R

EF

S7

16

DE

FIN

ED

10

BR

EA

LA

RR

AY

BU

FR

EF

S8

20

DE

FIN

ED

16

52

IIN

TE

GE

RR

EF

S1

51

6D

EF

INE

D1

45

3I

IIN

TE

GE

RR

EF

S1

6D

EF

INE

D1

50

LAIN

TE

GE

RF

.P

.R

EF

S7

13

14

19

DE

FIN

ED

47

LBIN

TE

GE

RR

EF

S1

9D

EF

INE

D9

50

MIN

TE

GE

RR

EF

S1

31

51

9D

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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)

=:A

22*C

2+

Al1

*S

2+

XV

eL)

=A

12

*(C

2-S

2)

+(A

11

-A2

2)*

S*

CMR

=M

INO

(I,M

M)

SUB

RO

UT

INE

BN

EIG

N7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.39

.39

PAG

E4

....... la.I~

17

5

18

0

18

5

19

0

19

5

20

0

20

5

21

0

21

5

22

0

22

5

IF(M

R.

LE

.1

)GO

TO5

70

L1

=I

DO

56

0J

=2,

MR

L1

=L

l+

NN

RL

2=

Ll

+NN

Al

=V

(U)

A2

=V

(L2

)V

(Ll)

=A

l*C

-A

2*S

56

0V

(L2

)=

A2*

C+

Al*

S5

70

MR

=M

INO

(MM

1,N

R1

-I)

IF(M

R.

LT

.3

)G

OTO

60

0L

2=

KD

O5

80

J=

3,M

RL

2=

L2

+N

NL

l=

L2

+N

NR

Ai

'"V

(U)

A2

=V

(L2

)V

(Ll)

=A

l*C

-A

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58

0V

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00

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NT

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DE

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TE

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TO

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(I),

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CN

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IF(N

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65

DO

62

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NB

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16

20

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NT

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16

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OF

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ND

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70

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1)

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00

LL

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00

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N-

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d;)

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N)

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M+

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00

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DO

80

0J

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K.N

MO

Al

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<I,

J)

A2

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(K,J

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(I,J

l=

Al*

CN

(Il

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2*

SN

(I)

SUB

RO

UT

INE

BN

EIG

N7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.39

.39

PAG

E5

80

0A

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I=

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>-

Al*

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(II

23

0D

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30

K=

1,

Ll

M=

LL

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DO

83

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ME

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11

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2DO

82

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NN

TEM

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A(I

tJ)

24

0A

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A(I

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l)8

20

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l1=

TEM

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30

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00

DO

92

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28

22

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24

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47

12

37

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18

91

90

22

82

29

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ED

17

71

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22

61

23

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22

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12

33

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81

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17

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57

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99

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41

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76

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38

49

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16

71

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91

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91

90

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01

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15

70

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98

22

82

29

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FIN

ED

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121

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31

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16

91

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ED

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12

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ED

10

01

21

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10

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02

12

11

ER

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FS

79

81

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82

84

85

94

99

11

51

25

DE

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ED

77

79

81

92

94

12

12

EP

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FS

86

DE

FIN

ED

82

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AL

AR

RA

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EF

S3

12

82

33

23

4D

EF

INE

D1

12

62

36

23

7

SUB

RO

UT

INE

BN

EIG

N7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.39

.39

PAG

E6

VA

RIA

l3L

ES

SNT

YPE

RE

LO

CA

TIO

N1

20

5E

lR

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RE

FS

82

23

52

36

DE

FIN

ED

61

85

23

31

24

3E

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35

23

7D

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34

12

21

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LR

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S1

40

15

7D

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D1

39

11

75

IIN

TE

GE

RR

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23

25

29

33

34

3*

37

44

53

54

55

67

69

73

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79

2*

84

2*

89

2*

94

2*

99

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10

01

08

2*

11

32

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14

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11

72

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23

2*

12

42

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31

13

71

43

2*

14

41

52

15

41

60

161

26

31

64

16

817

11

73

18

11

97

3*

19

82

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21

13

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19

22

42

26

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22

82

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29

23

92

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24

12

44

24

52

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FIN

ED

22

32

43

52

55

66

68

72

78

83

88

93

98

10

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11

61

22

13

01

37

14

215

11

97

19

82

02

21

02

19

22

32

38

24

31

16

4IF

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GIN

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31

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ED

92

71

20

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36

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50

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26

22

72

28

22

92

33

23

42

36

23

72

39

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24

02

41

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7D

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54

91

74

18

42

25

23

22

46

12

02

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GE

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37

54

15

31

56

16

51

69

18

32

27

22

92

31

DE

FIN

ED

36

53

15

22

24

23

01

24

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ER

RE

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25

DE

FIN

ED

22

11

21

0K

KT

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ER

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75

DE

FIN

ED

64

76

11

77

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GE

RR

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36

53

14

51

66

17

02

23

DE

FIN

ED

33

50

14

31

63

22

2

t1

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INT

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ER

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14

23

02

31

DE

FIN

ED

21

3

'f:.

12

35

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ER

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17

51

76

17

71

79

18

71

89

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FIN

ED

17

31

75

18

61

23

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GE

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17

81

80

18

51

86

18

81

90

DE

FIN

ED

17

61

83

18

51

20

4M

INT

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ER

RE

FS

52

21

62

19

22

12

32

DE

FIN

ED

512

15

23

10

Mt1

INT

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ER

F.P.

RE

FS1

31

43

44

96

57

59

11

18

17

1D

EFI

NE

D1

11

70

MM

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TE

GE

RR

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S1

81

DE

FIN

ED

13

12

00

MR

INT

EG

ER

RE

FS

35

17

21

74

18

21

84

DE

FIN

ED

34

17

11

81

12

03

NIN

TE

GE

RR

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S4

37

22

02

20

32

15

DE

FIN

ED

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214-

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27

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20

02

01

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FIN

ED

10

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TE

GE

RA

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F.

P.R

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34

37

22

02

DE

FIN

ED

12

03

11

72

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TE

GE

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60

12

61

28

13

2D

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56

01

21

3N

LIN

TE

GE

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10

51

06

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71

05

10

91

16

5N

LOO

PIN

TE

GE

RR

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87

10

6D

EFI

NE

D1

00

NMO

INT

EG

ER

F.

P.R

EF

S1

32

21

12

12

21

32

15

22

52

46

DE

FIN

ED

10

NN

INT

EG

ER

F.

P.R

EFS

31

41

61

72

23

23

45

051

65

66

75

91

11

81

30

16

31

76

18

51

98

21

02

16

21

92

38

24

3D

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D1

11

74

NN

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TE

GE

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17

51

86

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71

17

3NR

INT

EG

ER

RE

FS6

56

66

87

57

88

38

891

93

98

11

21

16

11

81

22

13

81

42

14

61

47

19

92

23

DE

FIN

ED

16

19

92

17

SUB

RO

UT

INE

BN

EIG

N7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.39

.39

PAG

E7

VA

RIA

BL

ES

SNT

YPE

RE

LO

CA

TIO

N1

22

2N

RS

INT

EG

ER

RE

FS

15

12

17

22

2D

EF

INE

D1

46

21

61

22

3N

R1

INT

EG

ER

RE

FS

18

1D

EFI

NE

D1

47

,1

20

7N

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TE

GE

RR

EF

S7

48

69

01

04

DE

FIN

ED

63

74

90

11

66

NSM

AX

INT

EG

ER

RE

FS1

04

DE

FIN

ED

111

17

1NW

HH

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ER

RE

FS

55

10

81

37

14

31

97

DE

FIN

ED

14

12

20

NXIt

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GE

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13

91

54

20

3D

EFI

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38

11

67

PS

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FS

12

6D

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21

28

1:;:2

270

REA

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55

15

61

62

DE

FIN

ED

15

31

22

401

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53

15

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48

16

20

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RA

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31

00

11

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23

13

91

48

15

31

56

19

82

11

DE

FIN

ED

18

913

11

23

0S

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60

16

71

70

17

91

80

18

91

90

DE

FIN

ED

15

81

20

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11

51

17

12

51

26

12

71

44

DE

FIN

ED

62

11

51

25

12

70

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24

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98

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88

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13

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44

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16

91

70

17

91

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18

91

90

11

76

XR

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24

29

11

31

15

12

31

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81

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24

52

47

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11

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17

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14

11

51

24

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51

29

13

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11

41

21

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FIL

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55

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18

10

71

36

19

6

EX

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RN

AL

ST

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FER

EN

CE

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AN

SOL

66

57

59

11

18

SOR

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29

12

91

56

15

8

INL

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CT

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NS

TY

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LIN

ER

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NC

ES

AB

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2*

79

86

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94

10

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17

118

1

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51

10

29

24

40

12

03

02

22

80

13

03

73

23

50

14

04

54

2

SUB

RO

UT

INE

BN

EIG

N7

6/1

76

OP

T=

lS

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.39

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PAG

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67

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69

68

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73

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07

47

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60

22

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23

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48

32

52

25

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06

11

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26

08

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30

09

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30

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03

40

11

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12

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71

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11

11

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12

31

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71

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66

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G,

l),H

M(N

EG

F,

l),H

B(l

),H

L(N

EG

F,

1),

1D

(NM

O,

1),

BI

(1),

BI

I(N

MO

,1

),IJ

FS

(1),

IJB

C(1

),R(l

),

Z(1

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(NU

FEL

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OM

MO

N/C

NTR

L/N

UM

NP,

NA

NG

LE,

MB

AN

D,N

BC

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NPP

O,

NP

PI,

TRA

CE

CO

MM

ON

/FC

ON

ST/P

I,RO

W,H

SD

ATA

SS

I-0

.57

73

50

26

91

89

63

,0.5

77

35

02

69

18

96

31

15C

PRIN

T2

00

6,

(IJB

C(1

),I=

l,N

UIN

TF

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INT

20

08

,(

IJF

S(I)

,1=

1,N

UFS

F)C C

GEN

ERA

TEBO

UN

DA

RYD

ISPL

AC

EM

EN

TS

20

CDO

97

K=

l,N

UIN

TF

II=

IJB

C(K

)IF

(K.

EG

.NU

INT

F)

GOTO

92

KJI

,=K

+l

--2S

91JJ

=IJ

BC

(KK

)

-F-

IF(J

J.N

E.O

)GO

TO9

2OQ

KK

=KK

+lGO

TO91

92

DO9

6M

=l,N

MO

30

IF(I

I.E

G.O

)GO

TO9

4DO

93

1=1,

3IU

=3

*K

-I+

l9

3D

(M,K

I)=

A(3

*II

-I+

l,M

)GO

TO9

63

59

4S

=(R

(K-l

)-R

(K»

**

2+

(Z(K

-l)-

Z(K

»*

*2

S=

SG

RT

(S)

Sl=

(R(K

-l)-

R(K

K»

**

2+

(Z(K

-l)-

Z(K

K»

**

2S

l=S

QR

T<

Sl

)R

S=

S/S

l4

0DO

95

1=

1,3

KI=

3*

K-I

+l

D(M

,KI)

=D

(M,K

I-3

)+(A

(3*

JJ-I

+1

,M)-

D(M

,KI-

3»

*R

S9

5C

ON

TIN

UE

96

CO

NTI

NU

E4

59

7C

ON

TIN

UE

C CFO

RMM

ATR

IXH

C

so 55

DO1

00

1=

1,

NEG

FDO

10

0J=

l,N

FB

AN

D1

00

Ht1

(I,

J)=

O.

NU

CO

F=N

EGF-

NU

FNP

NU

F1=

t~UFNP+

1B

=0

.5*

PI/

HS

DO6

00

N=

l,NU

FE

LDO

15

01=

1,4

II=

IX(N

,1)

8UB

RO

UT

INE

SOL

EX

T7

6/1

76

OP

T=

1S

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.3

9.

39

PAG

E2

70

25

03

00

75

32

0

80

......

35

0

-c.8

5

-..Q

37

03

71

90

37

2

37

3

95

37

5

60

65

10

0

10

5

11

0

RR<

I)=

R<

II

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50

ZZ

<l)

=Z

CII

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20

01

=1

.4D

O2

00

J=

I.4

20

0H

HC

LJ)

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00

11

=1

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O3

00

JJ=

1.2

CA

LL

FO

RM

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JJ).

RR

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Z.H

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50

1=

1,4

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25

0J=

I.4

HH

<I.

J)=

HH

CI.

J)+

HC

I.J)

HH

CJ,

I)=

HH

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J)

CO

NT

INU

ED

O3

20

1=

1.

4II

=IX

CN

,I)

DO

32

0J=

I.4

JJ=

IXC

N.J

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+l

IFC

JJ.

LT

.1

)G

OTO

32

0H

MC

II,J

J)=

HM

CII

,JJ)

+H

H<

I.J)

CO

NT

INU

EIF

CN

UC

OF.

LE

.0

)G

OTO

60

0D

O3

50

1=

5.8

11

=1

-4IF

CIX

CN

.I).

EG

.2)

GO

TO3

70

CO

NT

INU

EG

OTO

60

0G

OTO

C3

71

.3

71

,3

72

.3

73

).II

I1=

IXC

N.

II)

I2=

IXC

N.I

l+1

)G

OTO

37

51

1=

IXC

N.4

)1

2=

IX(N

,3)

GO

TO3

75

Il=

IXC

N.1

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2=

IXC

N.4

)S

IJ=

AB

S(Z

(Il)

-Z<

I2»

BR

=B

*R

Cll

)D

O3

80

K=

I.2

A2

=C

l.0

-88

(K»

/2.0

A3

=(1

.0+

88

(K»

/2.0

ZIJ

=Z

(11

)*A

2+

ZC

12

)*A

3D

O3

80

M=

I.N

UC

OF

Jl=

NU

F1

+M

-II

J2=

NU

F1

+M

-12

E1

G=

C2

.0*

M-1

.)*

BR

CA

LL

BE

SN

KS

CE

1G

.3.F

K)

HK

=0

.5*

SIJ

*E

1G

*<

FK

(1)+

FK

<3

»*

CO

S<

B*

ZIJ

)H

M<

Il.J

l)=

HM

<Il

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+H

K*

A2

HM

CI2

.J2

)=H

MC

12

.J2

)+H

K*

A3

38

0C

ON

TIN

UE

M=O

DO

40

0I=

NU

F1.

NE

GF

M=M

+lE

IG=

C2

.0*

M-l

.0)*

BR

CA

LL

BE

SN

K8

CE

IG.3

.FK

)H

M(

II1

)=

EIG

*H

S/4

.0

*F

K(2

)*<

FK

(1

)+F

K(3

))

SUB

RO

UT

INE

SOLE

XT

76

/17

6O

PT

=l

STA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.39

.39

PAG

E3

11

54

00

CO

NTI

NU

E6

00

CO

NTI

NU

EC C

FRE

E-S

UR

FAC

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DA

RYC

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ION

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20

DO6

50

1=1.

NU

FSF

II=

IJF

S(I

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(II,

1)=

1.

0IF

(II.

EG.

1)

GOTO

64

0K

M=M

INO

(II,

NFB

AN

D)

125

DO6

30

K=2

,KM

IK=

II-K

+l

63

0H

M(I

K,K

1=

0.0

64

0DO

65

0J=

2.N

FB

AN

D6

50

HM

(II,

J)=

O.O

13

0C C

FORM

MA

TR

ICE

SDO

ANO

OJ

CDO

70

01=

1,N

EQF

H13

(I)=

O.

13

5DO

70

0J=

1.

NMO

70

0H

L<I.

J1=

0.

NN

=NU

1NT

F-1

DO7

50

1=1,

NNJ=

I+l

......

14

0R

IJ=

-R(I

)+R

(J)

~Z

IJ=

Z(I

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(J)

Q"

SIJ

=S

OR

T(R

IJ*

*2

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IJ*

*2

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SN

=Z

IJ/S

IJS

SN

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IJ/S

IJ1

45

DO7

50

K=1

.2

Al=

(1.0

-SS

(K»

/2.0

A4

=(1

.0+

SS

(K')

/2.0

RF

=(R

(Il*

A1+

R(J

1*A

41*R

OW

HB

(I)=

HB

(I)+

RF

*A

1*

ZIJ

/21

50

HB

(J)=

HB

(J)+

RF

*A

4*

ZIJ

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75

0M

=1.

Nt1

0T

I=0

(M,3

*I-

21

*C

SN

+D

(M,3

*1

-11

*S

SN

TJ=

D(M

,3*

J-2

1*

CS

N+

D(M

,3*

J-1

)*S

SN

AF

=T

I*A

1+T

J*A

41

55

HL

(I,M

'=H

L(I

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RF

*A

1*

SIJ

*A

F/2

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L(J

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HL

(J,M

)+R

F*

A4

*S

IJ*

AF

/2.0

75

0C

ON

TIN

UE

C CIN

ITIA

LIZ

E13

1A

ND

131

I1

60

CDO

77

0M

=l,N

MO

[JI

(1);

,;(}

.b

d7

1b

N=

i.N

HtI

77

0B

II(M

,N

'=O

.1

65

CA

LLB

AN

SL1(

R.N

EG

F,N

F13A

ND

,HM

.HB

.1

)C

ALL

BA

NSL

1(R

,NE

OF

,N

FBA

NP,

HM

,HB

,2

)0

07

80

l=l.

NU

INT

FDO

78

0M

=l,N

MO

78

0D

(M,l

)=H

L<

l,M

)/R

OW

17

0DO

79

0t1

=1.

NMO

79

0C

ALL

BA

NS

L1(

R,N

EG

F,N

FB

AN

D,H

M,H

L(1

,M),

2)

SUB

RO

UTI

NE

SOLE

XT

76

/17

6O

PT=l

STA

TIC

FTN

4.8

+4

98

03

/16

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16

.39

.39

PAG

E4

' \t\ ......

175

18

0

18

5

19

0

19

5

20

0

20

5

21

0

21

5

22

0

C CR

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FRE

E-S

UR

FAC

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UN

DA

RYC

ON

DIT

ION

CDO

80

0I=

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II=

IJF

SII

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GT.

NU

INT

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00

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(II)

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79

9M

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MO

79

9H

UIL

M)=

O.

80

0C

ON

TIN

UE

PRIN

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01C

ALL

PRIN

TA

IHB

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GF.

1)

PRIN

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00

2.

CA

LLPR

INT

AIH

L,N

EG

F.N

MO

)N

WA

LL=O

DO8

05

J=l,

NU

INT

FIF

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CIJ

).G

T.O

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WA

LL=N

WA

LL+1

80

5C

ON

TIN

UE

WR

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(9)

NW

ALL

DO8

10

J=l,

NU

INT

FIF

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ITE

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IJB

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BIJ

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=1.N

MO

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81

0M

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NMO

BII

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BII

M)+

DIM

,,)*

HB

IJ)

DO8

10

N=M

,NM

OB

IIIM

.N)=

BII

IM.N

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IM.J

)*H

LI,

).N

)8

10

BII

IN,M

)=B

II(M

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PRIN

T2

01

0PR

INT

20

50

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I<

I),

1=1.

NM

Dl

PR

WT

20

12

PRIN

T2

05

0.

«([

HI(

LJ)

,I=

l,N

MO

),J=

1.N

MO

lDO

85

01=

1.NM

O8

50

BlI

n.

1)=

1.

+B

II(I

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DO8

70

I=I,

NM

O-

DO8

70

M=l

,NU

MNP

K=3

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ER

M=-

A(K

,Il

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AS

S(K

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IK-2

,I)

*SM

AS

SIK

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87

0B

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III)

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ER

MPR

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20

03

CA

LLP

RIN

TA

IBI,

NM

O,l

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INT

20

04

CA

LLPR

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AIB

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MO

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O)

2001

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OF

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20

02

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AT

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20

03

FORM

AT

11

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ERA

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00

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11

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00

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20

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15

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RN

END

SUB

RO

UT

INE

SOL

EX

T7

6/1

76

OPT

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TA

TIC

FTN

4.8

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98

03

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16

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PAG

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BO

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3

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43

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54

14

23

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81

49

15

41

55

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14

61

40

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91

06

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14

07

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98

14

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54

15

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16

17

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99

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87

90

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RO

UTI

NE

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**

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*C

DIM

EN

SIO

NF

I(5

0),

FK

(50

)IF

(X-2

.)

2,3

,32

T=.

5*X

10

T=

T*T

FK

(1)=

(C(C

L0

00

00

74

0*

T+

.00

01

07

50

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02

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69

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1.

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06

97

56

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27

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77

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56

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44

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15

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33

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20

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19

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25

33

14

14

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PC

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3.K

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)+F

K(N

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RE

TU

RN

END

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BO

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EN

CE

MA

P(R

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INE

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17

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18

27

28

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DE

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111

13

17

18

22

24

27

28

31

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MA

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TE

GE

RF

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9D

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INT

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RE

FS

30

3*

31

DE

FIN

ED

29

16

0T

RE

AL

RE

FS

2*

10

6*

11

6*

13

--1

66

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26

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43

1D

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70

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98

03

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39

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LIC

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FER

EN

CE

MAP

(R=

2)

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YP

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6O

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lST

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86

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66

80

81

83

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00

RF

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73

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62

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71

72

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59

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70

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54

91

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7

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98

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39

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40

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98

03

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PAG

E

:;

10 15

SUB

RO

UTI

NE

PR

INT

A(A

,NR

,NC

)C c*

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

CPR

INT

MA

TRIX

A.C

**

**

**

**

**

**

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**

**

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**

**

**

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**

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**

**

**

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**

**

**

**

*C

DIM

ENSI

ON

A(N

R,

1)

DO10

0J=

1.N

C.8

JH=

J+7

IF(J

H-N

C)

75

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50

JH=N

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1000

,(N

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JH)

DO10

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1,NR

10

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IDO

l,I.

(A(

I,K

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JH)

10

00

FORM

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14

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72

5200

013

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USE

D

SUB

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UT

INE

FOR

MH

76

/17

6O

PT

=l

ST

AT

ICFT

N4.

8+

49

80

3/1

6/8

11

6.3

9.3

9PA

GE

- -....j -{::,

5

10 15

20

25

30

35

40

SUB

RO

UT

INE

FO

RM

H(S

,T,R

R,

ZZ

,H)

c C*

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

**

CFO

RM

FL

UID

EL

EM

EN

TM

AT

RIX

HC*********************~*****************************************

CD

IME

NSI

ON

RR

(4),

ZZ

(4),

H(4

,4),

A(4

),A

S(4

),A

T(4

),A

R(4

),A

Z(4

)SM

=1.

-SS

P=

1.

+ST

M=1

.-T

TP

=1.

+T

A(

1)=

SM

*TM

/4.

A(2

)=

SP

*H

l/4

.A

(3)=

SP

*T

P/4

.A

(4)=

St1

*T

P/4

.A

S(

1)=

-TM

/4.

AS

(2)=

-AS

(1)

AS

(3)=

TP

/4.

AS

(4)=

-AS

(3)

AT

(1

)=-S

M/4

.A

T(2

)=-S

P/4

.A

T(3

)=-A

T(2

)A

T(4

)=-A

T(1

)P

ZT

=A

T(1

)*Z

Z(1

)+A

T(2

)*Z

Z{

2)+

AT

(3)*

ZZ

(3)+

AT

(4)*

ZZ

(4)

PZ

S=

AS

(1)*

ZZ

(1)+

AS

(2)*

ZZ

(2)+

AS

(3)*

ZZ

(3)+

AS

(4)*

ZZ

(4)

PR

S=

AS

(1)*

RR

(1)+

AS

(2)*

RR

(2)+

AS

(3)*

RR

(3)+

AS

(4)*

RR

(4)

PR

T=

AT

(1)*

RR

(1)+

AT

(2)*

RR

(2)+

AT

(3)*

RR

(3)+

AT

(4)*

RR

(4)

XJ=

PR

S*

PZ

T-P

RT

*P

ZS

R=

A(1

)*R

R(1

)+A

(2)*

RR

(2)+

A(3

)*R

R(3

)+A

(4)*

RR

(4)

FA

C=

XJ*

RP

SR

=P

ZT

/XJ

PT

R=

-PZ

S/X

JP

SZ

=-P

RT

/XJ

PT

Z=

PR

S/X

JD

O5

01

=1

,4A

R(I

)=A

S(I

)*P

SR

+A

T(I

)*P

TR

50

AZ

(I)=

AS

(I)*

PS

Z+

AT

(I)*

PT

ZDO

10

01

=1

,4

DO

10

0J=

I,4

H(I

,J)=

FA

C*

(AR

(I)*

AR

(J)+

AZ

(I)*

AZ

(J»

+X

J/R

*A

(I)*

A(J

)1

00

H(J

,I)

=H

(I,J

)R

ETU

RN

END

SYM

BO

LIC

RE

FER

EN

CE

MA

P(R

=2

)

EN

TR

YP

OIN

TS

DE

FL

INE

RE

FER

EN

CE

S3

FOR

r'lH

14

2

VA

RIA

BL

ES

SNT

YPE

RE

LO

CA

TIO

N2

04

AR

EA

LA

RR

AY

RE

FS

74

*2

92

*4

0D

EF

INE

D1

21

31

41

56'

;R

EA

LA

RR

AY

RE

FS

72

*4

0D

EF

INE

D3

6.

SUB

RO

UT

INE

FOR

MH

76

/17

6O

PT

=l

ST

AT

ICFT

N4.

8+

49

80

3/1

6/8

11

6.3

9.3

9PA

GE

2

VA

RIA

BL

ES

SNT

YPE

RE

LO

CA

TIO

N2

10

AS

REA

LA

RR

AY

RE

FS7

17

19

4*

25

4*

26

36

37

DE

FIN

ED

16

17

18

192

14

AT

REA

LA

RR

AY

RE

FS7

22

23

4*

24

4*

27

36

37

DE

FIN

ED

20

212

22

32

24

AZR

EAL

AR

RA

YR

EF

S7

2*

40

DE

FIN

ED

37

17

5FA

CR

EAL

RE

FS4

0D

EFI

NE

D3

00

HR

EAL

AR

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NO

DE

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OTO

23

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35

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DE

23

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23

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25

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MPU

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RE

SS

UR

EC

23

7C

ON

TIN

UE

SU

MX

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FH

X(I

T)*

HB

(I)

SUI1

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0.IF

(NG

RD

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G.

2)

SU

MY

P=

FH

Y(I

T)*

HB

(I)

DO

23

8M

N=

l.NM

OS

UM

XP

=S

UM

XP

+A

XM

(MN

)*H

L(I

,MN

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(NG

RD

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)S

UM

YP

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UM

YP

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YM

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38

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NT

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EC C

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MPU

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ND

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INT

FOR

DIF

FE

RE

NT

AN

GL

ES

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40

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l,N

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GL

ET

HE

TA

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G(J

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IS

UM

l=X

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I-2

)*C

OS

(TH

ET

A)+

Y(3

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2)*

SIN

(TH

ET

A)+

XS

T(3

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2)

SUM

2=X

(3*

I-l)

*C

OS

(TH

ET

A)+

Y(3

*I-

l)*

SIN

(TH

ET

A)+

XS

Y(3

*I-

l)SU

M3=

X(3

*I)

*S

IN(T

HE

TA

)+Y

(3*

I)*

CO

S(T

HE

TA

)+X

ST

(3*

I)SU

M4=

SU

MX

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OS

(TH

ET

A)+

SU

MY

P*S

IN(T

HE

TA

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40

PR

INT

20

60

,I,

SU

Ml,

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2.SU

M3,

SUM

4,X

AN

G(J

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50

CO

NT

INU

E

21

0

21

5

22

0

22

5

C C C C C C

WR

ITE

ON

TA

PE

3D

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LA

CE

ME

NT

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FA

LL

NO

DA

LC

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LE

S

WR

ITE

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NT

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SS

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EL

.NE

.0

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(4)

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39

0M

N=1

.1

0N

=N

+l

IF(N

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ME

L)

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TO4

00

DO

35

01

=1

,6

SIG

X(I

)=O

.S

IGY

(I

)=0

.D

O3

50

J=

1.

12

.)J=

LM

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,MN

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(NG

RD

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.1

)G

OTO

35

0S

IGY

(I)=

SIG

Y(I

)+S

SM

(I,J

,MN

)*Y

(J.)

SUB

RO

UT

INE

RE

SPO

N7

6/1

76

OP

T=

1S

TA

TIC

FTN

4.8

+4

98

03

/16

/81

16

.3

9.

39

PAG

E5

" ~ \l)

23

0

23

5

24

0

24

5

25

0

25

5

26

0

26

5

27

0

27

5

28

0

28

5

35

0S

IGX

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=S

IGX

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J,M

N)*

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JJ)

DO3

60

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OTO

36

0S

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87

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J)3

60

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J)=

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57

87

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IS

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ES

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S)

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(II)

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ES

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(II)

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NT

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OTO

36

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36

61

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TT

MA

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AB

S)

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66

ST

TM

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(II)

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ES

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{II

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66

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NT

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67

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NT

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EL

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GO

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90

IF(N

UM

EL.

EG

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EL

)G

OTO

36

9D

O3

68

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l.N

NE

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EL

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90

36

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70

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LE

TH

ET

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ET

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UM

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ET

A)+

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RE

ST

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HE

TA

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UM

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OS

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ET

A)+

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RE

ST

{NN

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SIG

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HE

TA

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SIG

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OS

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ET

A)+

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RE

ST

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HE

TA

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UM

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SIG

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HE

TA

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TR

ES

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N-l

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HE

TA

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UM

6=

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HE

TA

)+S

TR

ES

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N)+

SIG

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)*C

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(TH

ET

A)

37

0P

RIN

T2

11

0,

N.S

UM

l.S

UM

2,S

UM

3,S

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5,S

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90

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NT

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EIF

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LT

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UM

EL)

GO

TO3

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C CW

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ES

OF

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LE

LE

ME

NT

SC

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ITE

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RE

SY

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S+

DT

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INT

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M+

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OTO

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20

55

00

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K)

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50

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00

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22

00

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NG

LE

SUB

RO

UT

INE

RE

SPO

N7

6/1

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T=

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98

03

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END

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JOB

.A

N

Dynamics Technology, Inc.

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