Embed Size (px)
Citation preview
I
SHAPE OPTIMIZATION OF ARCH DAMS
by
DEPREZ JOSEPH LEON
Ingenieur Civil des Constructions, Tniversite
de ieege (1964)
Submitted in partial fulfillment
of the requirements for the degree of
Master of Science
at the
assachussetts Institute of Technology
(M.S. September 1968)
Signature of AuthorDepartment of Civil Engine ering, June 24, 1968'
Certified byThesis Supervisor
Accepted byChairmanDepartmental Committee on Graduate Students
Archiveg
MAY 1 3 19704LI&RARI.
MITLibrariesDocument Services
Room 14-055177 Massachusetts AvenueCambridge, MA 02139Ph: 617.253.2800Email: [email protected]://Iibraries.mit.edu/docs
DISCLAIMER OF QUALITY
Due to the condition of the original material, there are unavoidableflaws in this reproduction. We have made every effort possible toprovide you with the best copy available. If you are dissatisfied withthis product and find it unusable, please contact Document Services assoon as possible.
Thank you.
Due to the tightness of the binding, some text runs intothe gutter.
II
ABSTRACT
SHAPE OPTINIZATTON OF ARCH DAMS
by
DEPREZ JOSEPH lEON
Submitted to the Department of Civil Engineering on June 24,1968 in partial fulfillment of the requirements for thedegree of Master of Science.
The following work intends to set up a method ofrational determination of arch dams shapes and optimaluse of the concrete.
The method is based on the theory of membranes incartesian coordinates, in which the equilibrium equation,from a design point of view, can be considered as an equationin the shape Z, provided the stress function F is alreadydetermined. The basic hypothesis for the determination ofF is to write the equality of axial stress resultants inthe horizontal and vertical directions, which yields anhyperbolic partial differential equation in F. Then, tvwoelliptic differential equations (equilibrium and compati-bility) with their boundary conditions are solved for theshape Z and the normal displacement .
The whole problem is solved by use of finite diffe-rence methods. Because of computer storage limitation, amethod of block relaxation must be used for the twoelliptic equations, but it has not been possible to achieveconvergence. However, a/simplified problem with easierboundary conditions yields the conclusion that the basichypoth.-esis for F is not good enough to provide usableshapes for arch dams. The corresponding stress resultantsare too small and require high curvatures; moreover, theyyield values such that the problem is no longer mathema-tically well-defined.
From all these results, the fina] conclusion is thatthe problem ought to be studied on a different basis,wherethe determination of F should take into account the equi-librium equation in the direction normal to the membrane.
Thesis Supervisor: Z.P.Elias
Assistant Professor, Department of Civil EngineeringTitle:
ITI
ACKNOLEFDGIVENT
I express my acknowledgments to my Thesis Advisor,
Professor Z.M.Elias, whose very constructive and helpful
suggestions through this whole work greatly contributed
to its final achievement.
IV
CONTV1.NTS
PageT. Title page
II. Abstract II
III. Acknowledgment III
IV. Table of contents IV
V. Body of text 1
F o reword. Introduction 1
Chapter I. Membrane theory in cartesiancoordinates 3
1. Differential equation of equilibrium 32. Compatibility equation 83. Boundary conditions 94. Principal stresses 12
Chapter II. Method of determination of themembrane shape 16
1. Introduction 162. Determination of the stress function F 163. Determination of the thickness h 204. Determination of the shape Z 215. Summary of the iterative method 266. Remarks 26
Chapter ITTI. Computation of derivatives bymeans of finite differences 29
1. Introduction 292. Determination of Zx 29
3. Determination of Z, T 34
4. Determination of Z,, 35
5. Determination of Z,39
6. Determination of Zxy 41
7. Computation of Z, 46
8. Computation of Z, Z, Z, 49
9. Conclusion 51
Chapter IV. Determination of the stress resul tants 52
1. Determination of the fictive surface outsidethe domain of the dam 52
2. Characteristic lines of the hyperbolic equatiomfor F 53
V.
3. Computation of Jx and Jy 544. Determination of F by the method of characteristics 55
Chapter V. Determrrination of Z and A 59
1. Determination of Z 592. Determination of ,\ 62
Chapter VI. Programming considerations 64
1. Storage of arrays 642. Purpose of subroutines 64
Chapter VII. Results and conclusions 66
1. Boundary curve. Dimensions 662. General problem 663. Simplified problem 684. Determination of F for a given shape 705. Final conclusion 72
Figures 73
Appendix I. Determination of Z for the generalproblem: Newton-Raphson method 83
Appendix II. Shape for fixed boundary curve f5
Bibl iography 1 39
1.
Foreword. Introduction.
In the last ten years, the methods of analysis
of arch and more precisely shell dams have become more
and more accurate due mainly to the development of
hkigh-speed computers and subsequent methods which other-
wise would have been unusable (finite differences, fi-
nite elements, dynamic relaxa+ion, complete adjustement
methods ).
However, these methods apply only to structures
whose shape and thickness (variable of each point )
are already chosen primarily, on a more or less empi-
rical basis: comparison with previous structures, use
of approximating formules. In this point of view, the
above methods of analysis appear like an accurate veri-
fication of a structure whose design has been done very
roughly.
In a recent past, under the impulsion of Mr. Sera-
fim, the National Civil Engineering Laboratory in lisbon
developed experiments in order to get the best shape for
arch dams. The basic idea was to load rubber membranes
with thicknesses proportional to the anticipated thick-
ness in the actual dam, whithir boundaries similar
to the valley slopes. The membranes would be loaded in
the opposite direction then the dams, that is, upwards
with loads proportional to the weight and from the down-
stream side with water pressure. Such loads would infla-
te the membrane and give it a form of equilibrium in
tension. By changing the distribution of thicknesses
and the density of the loads, forms having a convenient
angle of insertion in the valley were found.(I)
However, this method was purel> experimental and nee-
ded already that the distribution of thicknesses be known.
2.
The intent of the following work is to present
an analytical method that could result in finding the
shape and thedistribution of thicknesses such as to
achieve an optimum in-plane state of stress in the con-
crete.
Based on a paper by Prof. Z. M. Elias (II), the
method is, until now, limited to membrane shapes.
Since, however, it is recognized that the membrane sta-
te of stress is very close to the actual state of stress
in the main part of arch dams, the membrane shape de-
termined as explained later could be used as a good
initial shape for an accurate analysis, provided the
thicknesses are increased near the edges in order to
resist the moments and shears that will surely develop
in this edge zone (generally very limited in space).
Chapter I*. Membrane Theory in Cartesian
Coordinates.
This chapter defines the system of axes, the sta-
tical quantities, the strains and displacements dealt
with in the equations describing the behavior of the
shell.
The letter symbols are chosen in conformity with
those used in the already mentioned reference (II);
all the formulas stated in this chapter have been de-
veloped in this same paper.
1. Differential equations of equilibrium
Let us chose:- the cartesian system of axes
x, y, z (fig.1.) , the x-y plane being vertical and
the y axis horizontal, lying in the z plane containing
the crest of the dam.
We shall study a symmetric shape of valley,
such that we adopt the x-z plane as plane of symmetry,
for the sake of convenience.
The shape of the dam will be described by the
two parameters x and y, through the equation Z= Z(x,y).
The thickness will vary from point to point according
to the law: h=h(x,y).
Defining i, i, k unit vectors along the x,y,z
axis respectively, the position vector of a point of
the membrane is given by
r = x i +y +zk (1-1)
We chose a local reference frame, two base vec-
tors of which are tangent to the parametric lines and
the other normal to the middle surface. Such a frame
4.
will include:
r = i + Zqx k
, = j + Z,y k
r = ryx ~y=-z~i - Z,,j + k
(1-2)
(1-3)
(1-4)
A comma indicates differentiation with regard
the variables following it.
We let t , T2 , n denote the unit vectors
to
of
this local reference frame,
r, = t2
rqv = 2 T2
n
with
21
S22
= r, . r,
= *
'12 = x * 'y.
1 + Zx
= 1 + Z
+ z .z,vx y
=1 2 + Z2Px py
212
The angle 0 between the coordinate lines is such
412 = 1 2 cosO
o( =c'2 sine
writ ing
(1-5)
(1-6)
(1-7)
(1-8)
(1-9)
(1-10)
(1-11)
that
(1-12)
(1-13)
_I 2 0(21 2
5.
In this membrane structure, the internal stress
components reduce to Nil, N22, N12 and such that
we can write
N 1 =Nil U
N2 = N2 1 t1
(fig.2)
+ N1 2 t2
+ N2 2 2
Considering next
NY per unit length of
N1 o 2dy = N dy
N2 o(dx = N7Nd2 1 n y
and defining
dx
the stress resultants N and
the x-y plane, such that (fig.2):
(1-16)
(1-17)
(1-16')
(1 -17')
N = N r + N rx xx ' y '
NY = N vXrlyNy='x yy -7y
we get:
orxY , 0(2 1
Nxx
N 11
2 c<i(1-18)
Similarly, it
Nxy
N12
c<2=
N = N 2 1yx 1 01
NN =< 22yy 1 (2
can be established
= N12N 12
N21
(1-14)
(1-15)
that
(1-19)
(1-20)
(1-21)
6.
The equilibrium equations of the small element
ABCD , whose projection on the x-y plane is abcd ,can
be written
(Ti 12) + (N2 41 1 1 f )x(c2 F2 )1 = 0 or
with (1-16), (1-17), (1-5), (1-6) and (1-7) :
Nxp + N , + pin = 0 or
N'px + N , + p =0 (1-22)
where j is the surface load per unit area of the
middle surface of the membrane.
In the case of arch dams, the main effect, for
usual heights, is due to the hydrostatic pressure.
This effect only has been considered in the following
study. Actually, the dead weight cannot be neglected,
but stresses due to it develop during the building and are
very dependent on the mode of construction (separate
plots). They can hardly be computed accurately using
the homogeneous shell theory. They will be taken into
account by chosing a reduced allowable stress in the
concrete.
Hence, we can write,if 6 represents the specific
weight of water, and if the dam is completely filled
(worst loading condition) :
p = -lxn
Equation (1-22) becomes:
N +N5 , -+ xNn = 0 (1-23)
Taking the scalar product of (1-23) by i, jand rn yields:
zN + N + xo 'xxx yx,y C
Nxyx
= 0
z+ N + Y x y 0
yy9 y Ok
N Z, + N Z, + N Z,xx xx xy yx yx 'xy
The monent equilibrium in membrane theory yields
N 1 2 = N2 1Hence N
yx(see (1-19) and (1-20) ),
Finally, equations (24) to (26) can be written:
N + N +xx,x xy,y
N + N9Sxy. x yyy
N Z,xx xx
Sx Z =0'x
+6 x Z9y
+ 2 N Zxy 'xy
=0
+ N Zyy 'yy
If we define
J x=0
Jy = IX Zy(XX)dX0
it can be seen that equations
satisfied by letting :
xx Fyy-J x
N = F - Jyy xx y
N = N =- Fxy yx 'xy
(1-28) and (1-29)
(1-33
-34)
(1-35)
Finally, substituting in
tions reduce to :
(1-30), the equilibrium equa-
7.
(1 -24)
(1-25)
=2rd
(1-26)
+ Nyy yy
(1-28)
(1-29)
(1-30)= x,2
(1-31)
(1-32)
are
,
X z\ Z ,) d X
8.
,yy Z,r - 2F, Z, + Z, = x+ Z, JX+ Z, J
(1 -31 )
2. Compatibility equation
Defining the complementary strain energy in terms
of the components N , N , N , and using a varia-
tional formulation, the resulting Euler equation can be
written:
Z, yy yy + Z9y 2Z y Xy xE2x'yy 'yy 'XX 'x xy xxtyy xylxy
+ E = 0 (1-32)
Where E , E and E are strain quantities related
to N , N , and N by the formulos :
1 (Exx -o(Eh 1 0/ 1N + o(2 Nyy + 2c(12N -(1+9) o(2 N
(1-33)
y2 y +201Nl - (1+'X 2 Nyy =Eh 2(Nxx+ 2
4) l2xy
(1 -34)
xy Eh 12 2 + 2 N+ x 2 yy +2v(1 2NxY] + (1+,)(2 Nxy
(1-35)
E being Young's modulus, V Poisson's ratio and >, aLagrange multiplier Which can be shown to be identical
to the component of the displacement vector along the
z axis.
9.
3. Boundary conditions
- As to the boundary conditions, we must distinguish
between the crest of the dam, and the abutments, to
which correspond different conditions.
a) Crest
The top edge of the dam is usually built monoli-
thically with a road which joins together both sides of
the valley. This road acts structurally as a curved edge
beam whose main stiffness is in the horizontal plane.
Hence, the membrane boundary conditions along the
top of the dam should be edge beam conditions, expres-
sing the compatibility of the in-plane displacements of
the membrane and the corresponding displacements of the
edge beam.
However, practically we can consider the top of
the dam as a free edge, for the following reasons:
a. Because of stability requirements, the upper part
of the dam must have a non negligible minimum thick-
ness, such that the stiffness of the edge beam vs.
the edge zone will remain fairly small. We can
conclude that the effect of the edge beam will affect
only in a small amount the total behavior of the
membrane.
b. A precast road, simply laid on the crest without
rigid connexion to it, can always be planned, such
as to achieve the membrane free edge conditions (no
compatibility for the horizontal tangential displa-
cement ), while still stiffening the shell in its
normal direction.
For these reasons, we can consider the top edge
free of shear forces and submitted to the in-plane com-
ponent of the dead weight of the road. For the sake of
convenience, if 0(1 and c2 remain close to 1 (see 1-18)
we can write:
10.
N = p (1-36)
N = 0 (1-37)xy
where p is the constant avera.e weight of the road per
unit length.
Remark : It must be noticed that, in a membrane analysis
point of view, two stress boundary conditions as equations
(1-36) and (1-37) cannot be imposed simultaneously on
the same edge. In this case, one of these equations
must be replaced by a displacement condition (or a free
edge condition, taking into account inextensional bending),
as it is derived later (equation (1-37) is then replaced
by equation (2-13) ). However, it will also be shown
that, in a design point of view, both stress boundary
conditions cn nonetheless be used for the determination
of F, without contradicting the above membrane theory
limitation (see pages 23 and 24).
b) Abutments
Along the sides of the valley, the boundary con-
ditions should express the equality of the in-plane
displacements of the membrane with the corresponding
displacements of the foundation. Although the valley
slopes undergo some deformation which, if the rock was
an elastic body, could be taken into account by means
of a potential. function, we shall restrict ourselves
to a rigid foundation.
In this case, the Euler boundary conditions have
been established (II) as
E + X( z, 5 + Z,, ) = 0 (1-38)
ssle + X ( szn 'n - 2 LEns +j\(Zns -'zs2s ]s
- s (.nn +XZ, ) = 0 (1-39)
10 a.
where
s, E and Enn are strains defined with regard to
tangential and normal directions to the boundary, and are
related to E , E and E by :
E = k C 2os2 + 2 E cos+ sir{ + Eyy sin 2J
E x sin2{ -2 E xy cos{> sinT +TE ssCos2
(1-40)
(1-41)
En= E ( cos 2 i 2 ) ( - sn + E - 6 , ) sin} Cos 1- ) ;
11.
is the angle between the x axis and the normal to
the projected boundary curve, measured positively from
the x- axis (fig.3 );
( ), and ( ),n are the derivatives with regard to
the tangential and the normal directions to the bounda-
ry curve of any quantity ( )Using transformation of variables theorems, we
can find :
( )s = - sin § (
( ),n = cos i( ), x + sin
), + cos J ( ),y
( ),y
This leads to :
Z9ss = ( -sin§ Z, + cost Z'y ),S
= -cosZ, . ,s -sink . Z'y . 4,s -sink (-sin4 Z,n
+ cos .Z, ) +cos ( -sinZyx + cos .Z, ) or
Z, s= -_fs . Z9n + (sin 2 + z, + cos 2+ Z, -2si4cosiZ, )(1-45)
Similarly, we can establish:
Z ns = C>,s . Z '5 + [sin< cos4 ( Z, yy
+ (cos 2{ _sin2c)) Z, j
Z'nn = cos 2 Z, + sin 2 1 Zyy + 2sini cosi Z,(1-47)
Substituting in (1-38) and (1-39), we can express the
boundary conditions in terms of quantities defined with
regard to the x and y axes . We find, after simplifica-
tions:
(1-43)
(1-44)
-Zxx
(1-46)
.
12.
(6 +, Xz, ) sin2 § -2(E +X Z, ) sin cos I +
E + X Z, ) cos 2 = ' (1-48)
-sin 2 cos i., + sinl(1+cos 2 §) -2sin3 §.<i
-2cos 3 t.G + cos§ (1+sin 2j) Y - sin~cos2
+, [XX (cos 2 - 2sin 2 ) +E (sin 2 - 2c s2 §)
+ 6 sin4 cosEy +X -sin2 cosj Z, + sin+( 2cos2
- sin 24) Z, + cost ( 2sin 2 1) - cos2 < ) Z, - sincos2( Z,
+,s zx ( cos 2 - 2sin 2 ) + Z, ( sin 2 ( - 2 cs 2
+ 6 sins Cs Z, +A,\4. - s) cos Z
-2 sin3 Z, + cos ( 1 + sin2)) Zyy
+ , sin I9 ( 1+ cos 2 ) ZO - 2 cos 34 Zq.
- sin cos 21 z,,] = 0 (1-49)
4. Principal Stresses.
Although the statical quantities dealt with in the
above formulas are N , N and N , they do not
represent the actual stress resultants in the membrane.
Besides, the stress resultants N 1 , 2 2 , N12and N21 are defined with regard to non-orthogonal direc-
tions and cannot be used directly to determine the prin-cipal stresses.
Let us consider (fig.4a ) a small parallelogram
ABCD whose projection on the x-y plane is the small rec-
tangle abcd of lengths dx and dy , with its stress
resultants Ni ,vectors i and
N22
T 2*
, N12 and N 21 , referred to unit
At the same point, we consider a small
submitted to stress resultants N 1' ,
(Fig.4b) referred to unit vectors
rectangle
1N22 ,N1and
We have obviously:
t2 =i 2
= t, sin + Tv cos&
writing the equality of force
N1 1 T + N12 T2
vectors on
= N t + N
CD
(1-50) and
(1-51) ;
, we have:
, orusing2 (1-50)
and (1-51)
N11 sine ti
This yields
+ ( N11 cosO + N12= Ni ti
I 2 I+ N 1 2 t 2
1 1 = Ni1 sin 0
N12 = N11 cos9 + N12
Writing now the equilibrium
ADF (fig. 5), we have :
|I TI. (N22 2 + N2 1 T1 ) + jFI I. (N 11
- N2 1 ti ) = 0
of the small triangle
+ N1 2 T2)
, or, since
and
13.
ABEF
and N21
2ti
(1-52)
(1-53)
+ TF (-N 2 I t
I1Kf I I sin 0
tg'
14.
(IT = I1II Cos
( N21 sing + N 1 1
, using again
sinO cosO ) tt
(1-50) and (1-51) we get
+ ( N 22 + N2 1cos 0 :
+N 11 cos 2O + N12
which yields :
N21 = N21 + N 111
N22 = sin N22
cosO )t= N21
Cos 0 = N 12
+ ( N12 + N21 ) cosO + N11
Calling Ce
in terms of
the principalt t
N 1- ,0 N22
stresses, we obtain easily,
, and N12
h-ce = N1 2+ N2 2
Using (1-52) to (1-55)
Ni sin2 0+ N22
, we have
+ 2 N12sin)
cosO + N 11
N 11 sin 2 - N 2 2 - 2N 1 2 cosO - Ni1 cos 2o ) 2
s 2sin
+ 4 ( N1 2 +
Using formulas (1-18)
N11 cos 0)2
to (1-21), (1-12) and (1-13)
get:
sin0 + N22sinO t
(1-54)
cos 2 o(1-55)
1
1 -N2 2
+4 N122
h Ge
(1-55')
+- 1
cos 0
, we
i I
2
2h, Ge. sine = N 0 N 2 + 2N a12
xy 0/1 -c ~2
2 2~ 12)
1 2
C<(1
-Nyy 1-2N 1 2XY C c1 < 2
I,
1< 2a 1C ~2 ) C 2 2
multiplying both sides
2h. e.( = N -2xx 1
+ \(Nxx
2 2
2
by
+ N 2
-Nyy
1 C2 yields
+ 2N 12
2
2
- 2Nxy (12
N +12xx 22
or, using (1-11)
+ N 2 + 2N C12
+ 4N2 o( - 2NXX Nxy 1 2yy ( Co 2
+ 4 o120 2 1 22
15.
x
+ 4 Nxy
2
+ N
+42 2)
2h. e.c =
tf N
N 0(xx. 1
+ N ( 2012 )
(1-56)
I2
Nxy Nxx N N
16.
Chapter II. Method of determination of
the Membrane Shape.
I. Introduction
In the previous chapter, we have developed all
the equations which determine the behavior of the membra-
ne.
For an analysis of a given membrane, we would
solve the system of partial differential equations (1- 31)
and (1-32) , with boundary conditions (1-36) , (2-.13)
(1-48) and (1-49) for the unknowns F and X . These
linear equations would lead to the determination of the
state of stress at each point of the membrane.
However, it should be noted that we could as well
chose as unknowns Z and \ , provided the stress functionbe given.
We would thus determine a shape for which the
given stress resultants will be in equilibrium with the
external load, and which will take normal displacements
such that the displacement boundary conditions and the
compatibility equation will be satisfied.
In this case, we need to solve a system of two
non-linear partial differential equations with non-
linear boundary conditions.
II. ]Determination of the stress function F.
In an optimization point of view, the stress func-
tion F should be determined such as to achieve, in some
sense, the "best" use of the concrete in the dam.
Of course, this could be done in many ways. Non
considered orientations could be, for instance, to use
17.
the methods of linear programming to minimize the total
volume of the dam. Another approach could use least
squares methods to minimize the difference between the
two in-plane principal stresses.
It has been decided to determine the function F
in a more precise way, by considering a purely struc-
tural point of view.
If we notice that one principal stress (normal
to the downstream face of the dam) is always equal to
zero, we can conclude that the concrete will be best
used if the in-plane normal stresses are the same in
any direction.
Indeed, in this case, provided that
a) We can consider concrete as an isotropic material;
b) The thickness is such that this unique in-plane
principal stress is equal to the rupture stress of
the concrete in compression, all. the Mohr's circles
corresponding to faces normal to the downstream
face of the dam will be tangent to the intrinsic
surface of concrete, while all other Mohr's circles
will be within that surface (IV, page 487).
If the in-plane principal stresses were different,
there would be only one circle tangent to the intrinsic
curve, in which case concrete would not be so
thoroughly used.
The equality of the principal stresses in both
directions can be described by one condition, expressing
in terms of F (by use of equations (1-33') to (1-35') )that the quantity under the square root sign in formula
(1-55') is equal. to zero. As a sum of two squares, this
condition is equivalent to the two following equations:
N11 N22 (2-1)
N1 2 =N 0 (2-2)
with the notation used in paragraph 4 of chapter I.
We can express the above condition in terms of
18.
the stress function F and obtain in this way two partial
differential equations for F; they are usually not
compatible . We must conclude to the impossibility of
achieving the above state of stress and must therefore,
either eliminate one of the above conditions, or chose
an intermediate between them.
If we notice now that the shear stress is equal
to zero along the y-axis (by symmetry), we can conclude
that, provided it does not increase too quickly when
going away from this centerline, the above conditions
will be almost realized in the main part of the dam,
if we impose everywhere the condition (2-1).
Of course, the assumption that N12 = N2 1 will remain
small will have to be checked from theshape which will
obtain finally; more precisely, in order to avoid ten-
sion, we will have to impose IN 1 '<IN I (see Mohr's
circle, fig.6).
Using (1-52) and (1-55), condition (2-1) yields:
N11 sina =s ( N22 + 2N12 cosO + N11 cos2e) or
2 in 22 112N = 2N + 12 + N 12
1 1 - 2 . 2 22 1 2 .1 11 2
1 2 1 2
or, multiplying by a and using (1-18) to (1-21) :
1(2 2\ 2 2 <2N 2 1' =1 K N + 2N c2 1xxC42 12), cl 1 yy xy 12 1 2
or
N 2 2 = 4 N + 2N C1 2xx (c< */-2 2 yy xy 12 2
In terms of F, we get:
19
(F - J ) .2 _2 (FJ *4 -2F C4 < (2,yy x 12 ,xx y 2 ,xy 12 2
or
4 2 2 2 2,xx 2 ,yy 12) ,xy 12 2 y 2
- J (o2 ~ 2) (2-3)
This partial differential equation for F is always
hyperbolic, since its discriminant is
AC B2 _4 2 2) 12 2 )2
_c44 /2<~2
Well-posed boundary conditions will be two "initial"
boundary conditions defined by equations (1-36) and (1-37)
expressed in terms of F :
F = p + Jx(x=0) (2-4)
F = 0 (2-5)
Quite generally, the domain of dependance of any
point of the dam (formed by the two dharacteristic lines
passing through this point) will extend beyond the pro-
jected surface of the membrane. This will require to
define a fictive surface continuing the actual surface
beyond the projected boundary curve.
Besides, if we notice that the compatibility condition
involves second derivatives of the stress resultants
(through same derivatives of the strains), we would like
F to be continuous up to the fourth derivative. This
20.
needs that the coefficients of equation (2-3) be con-
tinuous up to the second derivative, or the shape func-
tion of the membrane be continuous up to the third.de-
rivative.
This increased difficulty in the definition of
the fictive shape, as well as the need to define a non-
regular mesh in the finite-difference solution of the
differential equation (due to the varying slopes of the
characteristic lines) have led to consider, in first
approximation, an easier equation for F ,i.e
N = N orxx yy'
F - F = J - J (2-7),,xx ,yy y
with the same initial conditions.
It should be noticed that equations (2-3) and (2-7)
are the same for shallow membranes (Z and Z <<1),and do not differ very much as long as the slopes do not
become too large (Z and Z Z 0.5 , for instance),
which is achieved in practical shapes for arch dams.
Hence, unless the final shape differs strongly from prac-
tical shapes, both equations should give similar results,
with easier requirements for the last one.
III. Determination of the thickness h.
In chapter IV , we shall see in a more precise way
how to determine JX, Jy, F, and the stress resultants
Nxx , Nyy N by means of finite-difference methods.
Once the distribution of stress resultants is
computed, we can find the thickmess by prescribing the
maximum compressive stress to be equal to a given fraction
21.
of the ultimate strength of concrete in compression.
Indeed, because of the conditions imposed on the
stress function (see paragraph 11-2), we can expect
that no tension will develop in the membrane and that
rupture depends only on the compressive strength of
concrete.
In order to take into account the inextensional
bending and the edge-zone states of stress, the allowa-
ble stress in concrete will be limited to a small frac-
tion of the total strength Rbr b
Ri.e. n br with 5 - n '4 10 , for instance.
Using equation (1-56) with the minus sign yields:
h n N 2 + N 2 + 2N - N2 42Rbo [1xx 1 yy 2 xy 12 xx 1
br
+ N2 o44 + 4N 2 C2 <2 - 2N N 2 _ 2yy 2 xy 1 2 xx yy 12)
+ 4 o 02 N N + 4 o 02 N N (2-8)12 1 xy xx 12 2 xy yy
IV. Determination of the shape Z.
As it was seen in chapter I, the determination of
the shape will require the resolution of the two non-
linear partial differential equations (1-31) and (1-32)
(equilibrium and compatibility equations).
Generally, we can expect both equations to be of
the elliptic type. Indeed,
a) for equation (1-31), the discriminant is AC -B2
N N - N2xx yy xy
22.
it is positive if INTT remains smaler than IN Il = INyyI
which we expect according to the way in which the stress
function has been determined (see paragraph TI-2).
Anyway, this discriminant will certainly be positive
in the main part of the domain of integration .
b) For equation (1-32) considered as an equation in Xthe discriminant is AC - B2 = Z Z, - z 2
xx 'yy 'xy
For a shallow membrane, this quantity is nothing but the
Gaussian curvature of the surface. Even if the membrane
is not shallow,it can be shown that it is proportional
to the Gaussian curvature and thus, is positive for
elliptic surfaces such as usual arch dam.: shapes. Hence,
if the obtained shape is not far from usual dams,
equation (1-32) will be of the elliptic type.
In conclusion, we can expect the problem to be
completely of the elliptic type; it will be well, defined
if two boundary conditions are imposed 7long the whole
boundary curve (one for each equation). Along the slopes
of the valley, equations (1-48) and (1-49) have already been
defined. We need two boundary conditions for Z and Xalong the crest.
An easy condition for Z will be obtained by imposing
the shape of the crest : Z = Z(x=o) (2-9).
This shape could be obtained from existing dams. We
could also define the slope Z, x at x=O , but that con-
dition could lead to shapes differing more from actual
structures.
To be consistent with the membrane theory, we
can hardly prescribe a condition for A which is not
23.
related to membrane boundary conditions. For instance,
a condition X= 0 at x = 0 , taking into account the
presence of a stiff horizontal edge beam, would not
be compatible with the membrane assumptions.
Two approaches are possible:
a) Expressing that the crest is a. free edge, we would
use the boundary conditions derived by Reissner (V),
involving the stress couples and transverse shears re-
lated to inextensional bending. This would lead to two
conditions involving Z and X , along the top of the dam.In this case, of course, Z would not be fixed at the
crest.
However, this approach does not consider the pre-
sence of the edge beam of the top, and allows non negli-
gible inextensional bending to occur.
We prefer to use the other alternative:
b) We prescribe a condition which is compatible with
the presence of the edge beam and involves negligible
inextensional bending by imposing that the tangential
horizontal displacement be equal to zero along the top
of the dam.
The mathematical expression of this condition
will be derived from results developed in reference
III, p.15.
We have:
yy = U2,y - Z'yy uz (2-10)
where u and u2 are defined by
u = u i + U 2 j+ uz rn , u being the displacement
vector. If we write also :
u = u rx + u2 r, + un rn , we have the relation
24.
( ref. III, p.10 ).
U u + 042 'U2 12 U1 + C 2 U
Hence (2-10) can be written:
=(1 u1 +o( 2 u ) ,, - Z, ui (2-11)
The above condition is similar to u2 = 0 (ng'component
of displacement along r, ) .Since it has been seen that uz is identical to X [(II)],condition (2-11) yields:
S = (ei u~) - Z, (2-12)yy 12 1 '9y - 9yy (-2
Finally, if we notice that 2= 0 at y = 0 and
Ui = 0 at the junction of the crest with the valley
slopes (because of the boundary conditions along the
abutments), we can expect o*12 u to remain small and
have negligible derivative, We get finally the boun-
dary condition for X :
- E y(2-13)'yy
It should be noticed that this condition is not con-
tradictory with the discussion developed in paragraph
I-3 . In fact, the final shape and its normal displa-
cement X will satisfy both N = ( and u2 = 0 at
the top of the dam. We can even conclude that this
approach eliminates the choice between a completely
free or a fully restrained top edge, by satisfying si-
multeneously conditions relative to these alternatives.
Now, because of the complexity of the equations
25.
at hand, an analytic approach to the solution is un-
thinkable. We must devise a numerical solution and the
way in which the problem is formulated suggests ob-
viously the use of finite-difference techniques.
However. we must notice that there are two un-
knowns at each point, Clearly, for a mesh dense enough
to provide a significant accuracy, this multiplies the
number of equations by two and the size of tle matrix
of coefficients by four. Besides, the non-linearity
of the equations forbides the-direct solution of the
total system of equations.
For these two reasons, we must determine a linea-
rized method which divides the problem. A classical
approach would be to use a Newton-Raphson method.
The corresponding formulas are developed in appendix I
However, if .e remark that h does not appear in the dif-
ferential equation of equilibrium (1-31), we can set up
a method of iteration where the only change from one step
of iteration to the following one affects the boundary
conditions. In first approach, it has been decided to
use this method, whose prograrmation wns more direct,
hoping that the sensitiveness to the boundary condi-
tions would be small enough to guarantee convergence.
Assuming that a starting value of Xis known
along the abutments, we can solve the differential equa-
tion of equilibrium (1-31) for Z, with boundary condi-
tions (1-48) and (2-0). Using tis shape and formula
(2-8), we can determine the new distribution of thickness,compute the strains, and solve now the differential equa-
tion of compatibility (1-32) with boundary conditions(1-40)
and (2-13).
The cycle can now begin again until convergence
within a given accuracy is obtained. It is easily seen
that both problems are now linearized.
26.
V. Summary of the iterative method.
The former developments can be summed up in the
following general scheme of resolution.
a) Start from an arbitrary surface Without any know-
ledge of the bolution, a good starting shape can be
close to actual dam shapes.
b) From this shape, compute J and J ..x y
c) Integrate the hyperbolic equation (2-7), with boun-
dary conditions (2-4) and (2;-5).
d) From F(x,y), compute N , N , and N .xy
e) Determine the thickness from formula (2-8).
f) Compute the strains E E and E by formu-xx ' yy xylas (1-33) to (1-35).
g) Determine initial values of X on the valley slopes,1sing condition (1-48) with the starting shape and the
above strains.
h) Solve equation (1-31) with boundaryconditions
(1-48) and (2-9) for Z.
i) Determine a new distribution of thickness and of
strains, without changing stress resultants.
j) Solve equation (1-32) with boundary conditions
(1-49) and (2-13) for X.1) Come back to step h) until two successive shapes
do not differ from a given tolerance.
m) If steps h) to 1) yield a converging result, come
back to step b), and begin again the same proce-
dure, skipping steps e), f), and g) and using the
last values of X as new starting value.n) Do the above cycle until the whole procedure con-
verges towards a final shape.
VI. Remarks.
a) It should be noted thatsindependantly of assump-
tions relative to the convergence of the iterations
(which depends very much upon the choice of a star-
ting shape), the applicability of the method rests
upon two important assumptions, i.e.:
I. The stress function will be such that IN xy (4N xxi
in order for the problem to be mathematically well-
defined.
2. The final shape will not be too different from actual
shapes. More precisely, its curvatures should re-
main fairly small.
In fact, it is impossible to foresee whether these
conditions will bezsatisfied. It is one purpose of this
work to determine if the problem of shape optimization
can be solved on the previous basis and lead to a prac-
tical solution.
b) When integrating the equilibrium equation (1-31)
with boundary conditions (1-48) And (2-9), we see that
(1-48) involves only second derivatives of Z.
If we have found a solution Z of this problem, it
is obvious that any other function Z + ax , with a
arbitrary, will also satisfy the equations. We can spe-
cify the solution by imposing the value of Z at one
point of the boundary curve. For convenience, we shall
chose the bottom point of the dam, at y = 0 . Of
course, at this point, theboundary condition (1-48)
will not be exactly satisfied, since this equation is
replaced here by Z = Zb, where Zb is the prescribed
value of Z at the bottom.
c) We have to define a criterion in order to deter-
mine when the iteration can be stopped. This will be
done in the following way.
27.
28.
Let Z1 (xy) represent the value of Z at step
(i), and Zi+1 (xy) the same value at step (i+1).
We compute
22S = E(Z) and
R zi+1 - Z I where the summation is done
for all points of the mesh where the value of Z is
Computed. The iteration will be stopped if R/S be-
comes smaller than a chosen convergence tolerance.
d) An easier problem could be to determine a shape
which satisfies the differential equation of equili-
brium and where the stress resultants are such that
Ny x= Nyy , but where the boundary curve is prescribed.
We would only have to solve equation (1-31) with
Z prescribed all around the boundary curve. We should
still use an iteration method since each integration of
(1-31) determines a new shape to which different stress
resultants correspond.
After convergence, we could solve the compati-
bility equation for X , with one of boundary conditions(1-48) or (1-49). Obviously, one of these equations
would not be satisfied, but the corresponding shape
Z and displacement > could be used as starting values
for the general problem, with better chances of conver-
gence.
29.
Chapter III. Computation of derivatives
by means of finite differences.
1. Introduction
Since all the partial differetial equations of
the problem will be integrated by means of finite dif-
ference methods, we must define their coefficients at
the nodes of the mesh where the finite difference equa-
tions will be written. Besides, with the shape conti-
nuously varying, an analytical computation of these de-
rivatives is not possible. We have to determine them
by means of Lagrange interpolation polynomials, using
values of the interpolated function at the mesh nodes.
The formulas developed in this chapter will be
used in general subroutines called from the main pro-
gram to compute all the quantities appearing in the par-
tial differential equations.
2. Determination of Z, .
We shall have to develop differet formulas accor-
ding to the position of the point where the derivative
is computed. In order to be consistent, we shall se-
lect formulas with same order of truncation errors,
although some extrapolation formulas may have a higher
order of accuracy.
Besides, because of programming considerations,
it has been decided to use boundary points only at the
intersection of the boundary curve and mesh lines para-
llel to the y-axis. This requires a somewhat different
approach for the determination of derivatives of boun-
dary points, compared to generally used formulas.
30,
a) Interior node (fig.7)
Using the well-known central difference formula,
we write:
(Z, ) = (Zi+1, - Zi-1,j) / (2h) + O(h2 )(3-1)
where h is the mesh size in the x-direction.
This same formula applies for points on the
x-axis.
b) Nodes on or next to the boundary
Two cases will be considered, according to the
situations depicted in figures 8 and 9.
v4) Case _of _figure -8.
The following formulas will be used when
yi+1,b ii,j with the notation of figure 8.
Using a Lagrange interpolation formula through
points (i,j-2) , (i,j-1) and A, we can write:
1'.-n z~Z,x(ij) 1+n Zx(ij-2)
+ (+n) x(A) +
2(1-n)~ n Z,x _j)
0(k 3 ) (3-2)
where k is the mesh size in the y-direction.
Similarly, we can write:
(1+m) (1+m-n)Zlx(ivb) - 1+n Z~x(ijj-) (2+m)(1+m) 'X
+ n(1+n) Z~(A)
+ (2+m)(1+m-n') Z' . + 0(k 3 )n 'x( jj-1 )(3-3)
Ztx(i,j-2) and Z 'x(i,j-1) have already been computed from
formula(3-1); Zx(A) will be computed
31.
from
Z(i+lb) - z (B) ) + O(h2) , where
Z(B) is determined by a T-agrange interpolation formu-
la through points (i-1,j-1) , (i-1,j) and (i-1,b).
This yields:
Z( = - 41[Z (i+1,b)'x(A) 2h (+,)
+ Z . j)n(n-1-e)
e
Since all Z, are determined
of O(h 2), formulas (3-2) and
order of error.
M3 Case of _f iuE2_9.
- Z(il1
(n-1 )-(n-1 -e-)1 +e
n(n+1),b) e (1+e)
+ O(h2)
(3-4)with a truncation error
(3-3) will have this same
In this case,
formula (3-1). Pro
Z,x(i j) is already computed by
ceeding as before, we write:
Z,x(ijb)m(m-n1+n z,x(i, j-1)
(1+m)( -n)-n Zx,)
+ + M + 0(k 3 )
where Z,x(A) is given by:
A) - 1 +Zx (A) = h Z(i+1
.(1+n) (n-e),a) e - z (1+n)n
(i-1,b) (1+e) e J(3-6)
Z,x(A)12h
Z,x(A) (3-5)
,b) - Z(i-1,j-1)n(n-e)1 +e
+ 7il + O(h2 )
- Z(i-19j-l)
Here again, formula
Note: In formulas ((3-5) is of order O(h2 ).
3-6) and (3-4), it could happen
that e be higher than 1
case, we set e=1
(see fig.10). In this
and replace
Z (i-1,b) by Z(i-1,j+1) in form
The quantities m and n are
equal to 1.
c) Nodes on the
ulas (3-4) and (3-6).
always smaller or
crest (fig.11)
Let us determine the interpolation polynomial
passing through points
We have:
+
(X2(x-x2 3) -4)
-6h 3
(X-X 1) (x-x ) (X-x )
2h3
(1,j), (2,j), (3,j), and (4,j).
Z(1 tj)
(x-x 1 ) (x-x 2 ) (x-x 4)
-2h3
(x-x) (x-x 2) (x-x7)
6h3
Differentiating and replacing x by xi
+ 0(h)
we get:
,j) = - 11Z( - 18Z(2 ,j)
+ 0(h 3 ) .
+ 9Z( 3 ,) -2Z
(3-7)
At the junction of the crest and the valley
(fig.12), we use a Lagrange extrapolation formula with
the values of Z, determined above:
32.
-r
+
33.
Z 'x(1,k.) = 3Zk(lk-1) - 3 Zix(1,k-2) + Z'x(1,k-3)
+ 0(k) (3-8)
Here, the truncation error is of order O(h3 ) because
the derivatives are derived from extrapolation formu-
las whose accuracy is smaller.
d) Special nodes.
The above formulas can be used at a majority of
nodes, for afbitrary boundary curves. However, depen-
ding on the particular curve at hand, it can happen
that the previous formulas are not usable (when the
width of the dam bebomes small, at the bottom of the
valley).
For the shape of the valley considered later, we have
found the situation drawn in fig.13.
We can. easily get a formula similar to (3-7):
S1- F 2Z 1) - 18Z(Sl1Zlx(s,1) 6h -2Z ,1 + 9Z(s-2,1 - 89s11
+ 11Z(s,1 + 0(h3 ) (3-9)
with similar formulas for Zx(S,2 ) and Z, .
Finally, using a horizontal extrapolation, we
get:
Z~x ZY, m(1+m) -Zf
Z,x(sb) x(s 1 ) 2 - Z'x(s,2).m.(2+m)
+ Z (2+m)(1+m) + 0(k3 )(3-10)
The order of the error is again 0(h3 ), for the s.ame
reason as for the top nodes.
3. Determination of z,y
a) Interior mode (fig.14)
We have the usual formula:
y(i,j) - 2k Z(i, j+1) - Z j 1 + O(k2 )
(3-11)
b) Nodes on the y-axis
Because of symmetry, we have, on the y-axis:
z y(i,1) =0
c) Boundary nodes (fig.15)
Writing the interpolation polynomi-al through
points (i,j-1), (i,j) and (ib), we have:
z ) =~.1) =Y~ b)
(1+m)k2 (ij-
+ 2Z)(y-y )m(1+m)k2 (ib)
Deriving and writing su
we get:
m-1+ ZM (i -Z i, j)
1+ m+ Z(ib) I
1 mkE 1+m
( ~-1 -) (Y~7b9t - 2.mk
+ 0(k3)
ccessively y=yj
1-m-m
+ O(k2 )
Z(i )
(3-12)
1+m
34.
z
(3-12)
Z(i,j)
1= kg-
Z 9y(ijb) Z
+ 1 +2mm + 0(k2) (3-13)m (1+m) (ib) +
Remark: If the boundary node is one node of the rec-
tangular regular mesh, formula (3-13) can be
used with m=1.
However, in this case, a formula similar to
(3-9) could be programmed more easily and was
therefore used, with a reduced truncation error.
4. Determination of Z,
It should be noted first that second derivatives
of Z and other quantities appear only in the elliptic
partial differential equations of equilibrium and com-
patibility, with their boundary conditions. Hence,
these derivatives ought to be computed only at the nodes
where the finite difference representation of these
equations is used. For this reason, they have not been
computed on the top edge of the dam where the boundary
conditions do not involve other second derivatives than
Z, y, which is known analytically from the prescribed
value of Z at x=0.
a) Interior node (fig.7)
We have the well-known central difference formula:
Z- 'Z(i, 9j) - 2Z + Z(i-1,j) + O(h2
(3-14)
b) Boundary points
As for Z, , we must distinguish the situations
sketched in figures 8 and 9.
Using an horizontal extrapolation formula through
35.
points (i,j-2)
Z xx(i ,b)
, (i, j-1)
= Z'xx(i, j-2 )
+ Z'xx(ij-1)
+ Z xx(A)
and A , we have:
(1+m) (1+m-n)1+n
(2+m) (1+m-n)-n
(2+m) (1+m)(1+n) n
+ 0(k 3 )
Writing now Z'xx(A) =[Z(i+1,b) - 2Z(A) + Z(B)
and expressing Z(A) and Z
polation through points
and (i-1,j-1)
the final
Z xx(i,b)
((B) byi, j-1)
1
h2
an horizontal inter-
, (i,9j) , (ijb) ,
, (i-1,j) , (i-1,b) respectively, we get
formula:
h2
(+M) ( 1+M-11)1+n
.(2+m) (1+m-n)-n
{[ i+1, j-2)
+ [z(+1 i
+ Z(i+1
j-1 )
2Z(i, j-2)
- 2Z . .(i,j-1)
- 2 (Zb) ~(Z (i, j-1)
+ i-1, j -2)
t Z . .j-
(n-1) _(n-1 -m)+m
+ Z~j n(n-1 -m)+ Z, j) -M
+ Zb (n-1) n(ib) (1+m) m
+(Z(i-1(n-1) (n-I-e)
,j-1) 1+e (i-1, j)n (n-1ie)
-e
n (n-1(i-1,b) e (1+e)
)1)j
(2+m) (1+m)n (1+n)
36.
))
+ O(h2)
(3-15)
The truncation error has the order of the error
, since all other interpolating formu-
las are of higher order.
Using an horizontal interpolation
can now write:
formula, we
j)=-Z~xx(ij 2 )m + 2Z
2 +m 'xx(i,m
j-1) +in
+ Z,xx(ib) (2+m)(1+m)(
where the order of the truncation error is still
same as in (3-15).
Case ngeat (fig, s
Using exactly the same procedure as above,
Z x, xx(i,b) = Z,m(m-n)1 +n + Z'xx(ij)
(I+m) (m-n)-n
m(1+M)'xx(A) n(1+n) + 0(k3 )
in termes of Z:
Z xx(i,b)
m(m-n)1+n
+[Z Z(i+1
1fZ(
+ Z(i+1 ,j)
( ),b) -2 ( Z
i+1 , j-1)
- 2Z
- 2Z
(i, j)
n(n-m)1+m
(ij-1 )
+ (ij)
( i-i, j -1
(1+m) (m-n)-n
(1+n)-(n-m)-M
(1 +n)n+ Z(i~b) (1+m)m
)) Z n(n-e)
(-1,j-1 ) 1+e + Z(i 1 I)
of each Z,
Zxx(i,
16)
the
we get:
or,
37.
(1+n)(n-e)-e
+ 0 (h2)
After simplification,
Z xx(ib)
+ Z(il b) n (1 +n)e (1+e)
(1+m)m(1+n) n
we get:
= 1Zi+1 j-1) (i-1, j-1 )I
+ EZ(i+1
+ Z(i+1,
+) (i-1,)]
b) - 2Z(i
(1+m)(m-n)-n
j-1 )b(b) I-("(i-1,
S(1+n) (n-e)j) -e
S +m (1 +n )n
Remark: When we have
the previous
we set e=1
c) Special points
+ 0 (h 2 ) (3-17)
the situation depicted in fig.10,
formulascan be used, provided
and replaceZ(i-1,b)
by j+1).
The above formulas are not applicable in the last
and last but one rows of the mesh (see fig, 13).
With the notation of fig.13, we can write the
Lagrange interpolation polynomial passing through points
(s-3,2) , (s-2,2) , (s-1,2) and (s,2)
(x,2) =-6h3 ^ (s-3,2)
38.
m(m-n)1 +n
+ Z (i
n(n-e)1+e
+ Z ( i (1+n)n,b) (1+e)e
:
(1~ s - s -1- S)2h3
(x s-3) (x-x s-2) (x-xs)
-2h 3Z (s-1
(xs-3 s-2) (X-x 1 )
6h3
+
+
Deriving twice and setting x=x.
Z xx (s,2) *2 FZ(s-3,2) + 4Z(s-2,2)
+ 0 (h 2 )
We write a similar formula for Zxx(s 3 ) and Zxx(s+1,1)
Finally, using horizontal extrapolation through points
(s,1) , (s,2) and (s,3), we compute:
Z xx(sb) 'xx(s , 1)
+ Z'xx(s,3 )
- Z2
(2+m) (1+m)2
xx (s, 2)
+ 0(k3)
Here again, the total error is of
(truncation error on each term.)
5. Determination of Z,yy .
a) Interior node
(3-19)2order h
(fig. 14)
We use the central difference formula:
12 I'Z(ij-1)
- 2Z (ij)+ Z(i,j+l)
(3-20),-
39/
,2)
+ 0(h4 )+
, we get
+ 2Z(s,2)] (3-18)
Z , y ~ i, j) + O(k 2)
- 5Z (S-1,2)
b) Boundary points (fig.15)
Let us write the Lagrange interpolation polyno-
mial passing through points (i, j-2)and (ib). We have:
(y-y 2_) (y-y (y-y)
-2 ( 2+m)k 0Z (i, j-2)
(y-y
(1+m) k0
-2mk3
(Y-y i-2) 7-Y1_) (Y-y1)
m(2+m) (1 +m)k3Z(ib)
Deriving
we get:
y' Iyy(it j)
twice and setting
1
k2 SI'2+m
successively
Z(i j-2)
y=y. and y yb
4-2m Z1+m (ij-1)
- -m Zrn (i, +
) +6
+ m(2+m)(+n
(3-21)
and
1k2 U1 +2m
2+m
3+2m Zm~ (it j)
Z(i, j-2)
+ 6
+ 4Z(ijl)
Z (ib)] + O(k2 )
(3-22)
40.
Z(i,y)
+
+
+ 0 (k )
Z(ib) + O(k2 )
Z yy(i, b)
, ( i,. j -1 ) , ( i, j )
J-2) (Y~y) (Y~7Y)
41.
c) Nodes on the x-axis
Using symmetry considerations
we have:
2
kZ(i,2) - Z (i)7j
and formula
+ 0 (Ic2 )
d) Special node
The only point where the above formulas are not
applicable is the node at the bottom of the dam (fig.13)
Using an extrapolation formula similar to formu-
la (3-8), we have:
Zlyy(s+1,1) = Ztyy(s-2,1) S3Z, yy(s,1)
+ 0 (h3 )such that the truncation error is still
(3-24)
of order K2
6. Determination of Zxy
a) Interior node (fig.16)
We have the well-kn own formula:
Z(i+l , j+1 ) - Z (i+1, j-1 ) - Z(i-1,j+1)
+ 0 (h 2) + 0 (k2 )
Boundary node
c) Case y(i1 Lb y .
(i,j) we write:
(3-20),
(3-23)
Zxy( i ) 4hk
+ Z
b)
(3-25)
(fig.9)
- 3Z 1yy(s-1,1 )
9 j-1 ) I
For point
zxy(i, j)1
2h- Zy(i-19i)j + 0 (h2 )
Using now a formula similar to (3-12), we can write:
Z, xy( i )1 j
=2hk1
b) n (n+1) - Z1-n
(i+1,j) n
(i+1 , j-1 )n
n+1 - Z(i-11
,b) e(e+1)
+ 0 (h2 ) + 0 (k )
For points
write:
(i,b), using a horizontal extrapolation,
Ztxy(i,b) = Z,(1+m)m
2 - Z,xy(i j-1)
+ Z .xy(ij) (2+m) (1+m)- + c (k3),where Z xy(ij-2)
z(i,j-1) and where Z, x
and (3-26).
are given in terms
The total truncation
of Z by formulas (3-25)
error is still of
order L0 (h2 ) + 0 (k2
This yields:
14hkZ xy(ib) i+1 , j-1) - 7 i+1 , j-3) i-1 , j-1)
+ Z -.lj.. (1+m)m+ (i-1,9 j-3)1 2 j)- Z(i+1, j)
(2+m) + Z(i+1,b) n (n+1-) (?i+1, j) n
42.
- Z (i-1i j)1-ee
(3-26)
we
(2+m)m
~ Z (i-1,- (i+1,t j -2)
j-2) m+ Z (i-1
Z 'y(i+19j)
(i-1 , -1) -e+1
n,j-1) 1+n
,j-1) -1+e+ Z
- Z(i-1,b)1
e(1 +e)
(2+m) (1+m)}
(1-e)(i-1, j) e
+ 0 (h2) + 0 (k2 )
Remark: If we have the situation
(3-27)of figure 10, we use
the same formulas as before, with e=1 , replacing
(i-1,b) by Z(i-1,j+l).
Case y(i+1 L 2J' L1L (fig. 8)
Z, xy( iwith e=1 and
-1) will be determined from formula
Z(i-1,b) replaced by Z(i-lj+1)
Now, using a vertical interpolation for Z,points (i-2,j) (i-1,j) and (i,j) , we have:
(3-26)
through
Zy (xvj)-(XXi- 1) (x-xi)
2h2
(X-xi-2)(X-Xj)
h 2
Z,y(i-2, j)
z'y(i-1 , j )
(x~x i-2) (x-Ai- 1)
2h 2z'y(i,j) + 0 (h3 )
Deriving and setting x=xi, we
Zxy(I, j) 2h Z,y(i-2, j) + 3Zy(ij)]
+ 0 (h2 )
Expressing Z,y in terms
(3-28-a)
of Z , we get finally:
z
get:
43.
- Z (i+1
- 4Z, y~
Z, xy(i j)
+ 1d (1 +d)
4~ e(1+e)
2kh ~
Z
Z
(i-2,b)
(i-1 ,b)
d1 Z1 +d- (i-2, j-1
Z+ e1 +e
3m1+m Z
(i-1, j-1 )
(i, j-1 )
1d ,d ( - 9j
+ 4(1~ee
3 (1-m)-m Z
+ '3MO +iM)Z (ib) + 0 (h2 ) + 0 (k2) (3-28-b)
This formula can always be adapted to different situa-
tions by setting, if need be, d or e = 1 and repla-
cing Z(i-2,b)
Finally,
and/or Z(i-1,b)
using a horizontal
by Z(i-2,j+1)
trapolation, w
and/or Z(i-1, j+1)
e write:
Z xy(i, b) 'xy(i, j-2)(1+m)m - Z, xy(itj1)
(2+m) (1+m)2 + 0 (k3)
(3-29)
where Z~xy(ij-2), Z, xy(ij-1) and Z, j are given
in terms of Z
respectively.
o(h 2) + O(k2 ).
by formulas (3-25),(3-26) and (3-28)
The truncation error is still of order
c) Nodes on the x-axis
By symmetry, we have
ZXY(il) = 0
(2+m)m
xy(i, j)
44.
45.
d) Special points (fig.13)
Writing a vertical extrapolation, we have :
Ztxy(s,2) = - Ztxy(s-2,2) + 2Z xy(s-1,2)
In terms
Z,xy(s,2 )
of Z ,we cam write:
- 4hk 2Z (S9) - 2Z )
(s-2,1) - s-1
(j2) + 0 (k2)
Using a similar formula for the
'xy(s,3) 414h k
- m1+m 2)
next column, we get:
- 4(1-m)m
+ 4m (17+T)
- Z(s-1,4)
+ 0 (h2 )
Finally,
Z(sb) - 2Z(s-2 , 4 )
(s-1 ,2) (s-3,4) - Z (s-3,2)j
+ 0 (k2 )
an horizontal extrapolation yields
Z xy(s,b)= - m(2+m) Z xy(s, 2) + (2+m) (1 +i)
2
+ 0 (k3)
+ O(h 2 )
- 2Z (s-2,3)
+ 2Z
+ 0
3) + Z_(s-3,)
(3-30)
+ 2Z(s-2,2)
(3-31)
Z'xy(s,3)
Z (0
(3-32)
- Z (s-3,1 )+ Z (s-1, 11)
46.
This formula can easily be expressed in terms
of Z with formulas (3-30) and (3-31). The total
truncation error is of order O(h ) + O(k )
6, Computation of Z,Xxx
The third derivatives of Z appear only in the
second boundary equation. Hence, they ought to be com-
puted only at nodes on the boundary curve.
a) Case y(i+1,b) Y(ij) (fig.8)
Using horizontal extrapolation, we have:
Z (1+m) (1+m-n)'xxx(ib) 2 Ztxxx(i,j- 2 ) 1+n
(2+m) (1+m-nl+ Zxx i9 _)_
+ Z x (2+m) (1+m) + 0 (k3)xxx(A) n(1+n)
Using for Zxxx(A) a formula similar to (3-4),
we have, in terms of Z,x :
Z2xxx(i,b) = -{Z9xx(i+1,j-2) - xx(i-1,j-2)
(1+m) (1+m-n) + Z1+n +Z'xx(i+1,j-1) Z'xx(i-1,j-1)
(2+m) (1+m-n) (n-1)(n-1-e)-n + Zgxx(i+) x1, 1+e
+ Zxx (ij)
(2+m) (1 +m) +nl(1+n 5-
n(n-1 -e)e
0 (h2)
- Z,xx(i-1,b)
+ 0 (k2 )
n(n-1)e (1 +e)
(3-33)
b) Case Y(i+1,b)> (f ig.9)
Using the same kind of extrapolation through
points (i,j-1). (i,j) and(A) we have:
Z xxx(i,b)
+LZ,xx(i+l
+ Z xx(i+1
(1+n)(n-e)e
1h Z,xx(i+1
,j) - 'xx(i-1,
,b) - Z'xx(i'1,
Z,~ 'xx(i-1l
, j-1 ) - Z,xx(i-l
(1+m)(m-n)-n
n(n-e)j-1) 1 +e
-1+n n,b) (1+e)e I
+ Zxx(il
m (1+m)n (1+n)
+ 0 (h2 ) + 0 (k2)
c) Special points (fig.13)
Using a vertical extrapolation formula, we have:
Zlxxx(s,,1) = Zxxx(s-3,1) - 3Z,xxx(s-2 ,1) + 3Z,xxx(s-1,1)
+ 0 (h3 )
and similar formulas for Zxxx(s,2 ) and Zxxx(s3)
47
m (M-n)1 +n
j )
(3-34)
(iI j )
c,,,) Node (s, j
48.
Using now a horizontal extrapolation formula,
we have finally, in terms of Z, :
Z xxx(s,b)
+ Z
2h 'xx(s-2
,xx(s-3, 1)
'xx(s-2,2)
+ 3Z,xx(sl)
- Z'xx(s-4,2)
,1) - Z'xx(s-4,1
- 3Z,xx(s-2,1)
- 3Z,xx(s- 1 , 2 )
-3Zxx(s-1, 1)
2
+ 3Z,xx(s-3,2 )
+ 3Z,xx(s,2 ) - 3Z,xx(s-2,
2 )
- Zxx(s-4,3)
- 3Z'xx(s-2 ,3)j
(1+m)2
+ 3Z,xx(s-3,3)
+0
The final error is still of order
Node r(s+ u 1.
Using a formula similar to
(k3) + 0 (h)
0(h2 )
(5-9),'
Z,xxx(s+1,1) 6h -2Zxx(s-2,1 )
+ 9Z,xx(s-1,1)
,xx (s, ) + 11Z,xx(s+1,1)] + 0(h3)
such that the total error is still
(3-36)0 (h2)
m(m+2) +IZ xx(s-2,3)
+ NZ,
(3-35)
we get:
- 18Z
Z qxx(S-193)
of order
49.
)Node on the second row (fig,_18)
Since Z9XX is not determined on the first row
(top edge), the above formulas are not applicable.
Using a formula similar to (3-28-a), we write, with
the notation of fig.18:
z 1 -- NZx(1p + 4Z~x(p. 2xxx(1,p-2)p-2) xx(2,p-2)
- Z xx(3,p- 2 ) + 0 (h )
and similar formulas for ZXXX(lP-i) and Z, (1, )
We can now compute Z,xxx(1,b) by an extrapolation for-
mula similar to the one used in formula (3-35)
Z,xxx(i~b) == Zxxx(I,p-2 ) m(+m) - Z,XXX(Pl) m(2+m)
+ Z XXX(lp) (2+m)2 (1+m) + 0(k5 ) (3-37)
such that the total error is still of order 0 (h2)
7. Computation of Z, ; Z ; Z, .xx xayy yyy
Here again these derivatives must be computed
only at nodes on the boundary curve. We compute them
by a derivation with respect to y of the quantities
Z, Z7,y and Z~ yy respedtively.
a) Ordinary point
Let f denote one of these last functions.
Using fig.15 and writing the Lagrange interpolation
50.
polynomial through points
and (ib),we get:
-2(2+m)h 3
( y-y j-2) ~7yj Y-b)
(1 +m)h3
y-y~j -2) ~7 Y J -1) Y-b
- 2mh 3
m(y-y-2) (~-2+1)
m(1+m) (2+m)h3
Deriving and setting
'y(i,b) hm 1+m)2 (2+m)
f(i, j-2)
+ 0 (k)
, we get finally:
f(i, j-2)m(2+m)
+ 1+m
(1+m) (2+m)2m
+ 0 (h 3 )
3m2 + 6m + 2f(i, j) + m(1+m) (2+m) f(ib)
(3-38)
Z ,Thsi' formula can be used at any point with f z Z,or Z, yy.
b) Special point (fig.13)
At point (s+1,1) , we cannot use the above formu-
We have, from symmetry considerations:
Sxxy - (Zxy),
and
Zyyy (Z, Y)
f(i,y)
4
la.
z =0
( i, j-2 ) ) (i, j-1 ) / ( i, j)
(Y-yj-1) (y -YJ) (Y-yb)
51.
Now, we can write
Z xyy(s+1,1) = Z~xyy(s- 2 ,1) - 3Z,xyy(s-1,1) + 3Zxyy(s,l)
+ 0 (h2)
In terms of Z, we have, using the antisymmetry of
Zx :
Zxyy(s+1,1) =1 Z,xy(s-2,2) - 3Z,xy(s-1,2) + 3Zxy(s,2)
+ 0 (k2 ) (3-39)
8. Conclusion
All the previous formulas determine, within a
maximum truncation error of 0 (h 2) or 0 (k 2) (usually
we write h=k ), all the derivatiVes which appear in the
formulas describing the problem. The next two chapters
will deal with the numerical integration of these equa-
tions, which becomes now possible.
52.
Chapter IV. Determination of the stress
resultants.
1. Determination of the fictive surface outside the
---------- domain of the dam.--------------------
As it was outlined in Chapter II, we must extend
the actual surface of the dam outside the boundary
curve in order to determine the solution of the diffe-
rential equation (2-7) at any point of the dam. We
also know that this fictive surface should be conti-
nuous with the actual one up to the second derivatives,
along the boundary curve.
Since the only quantities depending on Z ir (2-7)
are J and J , involving Z, and Z,Y we can write,
developing in Taylor's series at the boundary point,
along lines x = x. :
Zfx = Z'x(i,b) + (y-yb) Ztxy(i,b) (4-1)
Z,y 'y(ib) + (Y-y) Z'yy(ipb) (4-2)
These functions will be continuous with actual
slopes up to the first derivatives along the boundary
curve.
Had. equation (2-3) been solved, we would have
added a term
2 2Z . and Z . to
2 'xyy(i,b) 2 'yyy(i,b)
equation (4-1) and (4-2) respectively to meat the same
continuity requirements.
2. Characteristic lines of the hyperbolic differen-
--------- tial equation for F.------------------
If we solve equation (2-7), which is the non-
homogeneous wave equation, we know that the characte-
ristic lines are straight lines inclined at 450 on
the x and y axes. Hence, using a square mesh (h=k),
all mesh nodes will be on characteristic lines, provi-
ding an easy integration of equation (2-7) by the
method of characteristics.
Besides, by drawing from one point of the x-axis
two characteristic lines such that the total projected
area of the dam be inside these lines and the line
x=O (fig, 19), we can easily determine the domain where
the fictive shape must be defined, J and J comvuted
and equation (2-7) integrated.
Remark: If we had solved equation (2-3) for F, the
characteristic lines would have the directions defined
by
2 212 2 d+22d<1 dyx x
which yields:
y 12+d ~ 2X2
The slopes varying from point to point, equation
(2-3) should be solved by a more complicate method such
as Hartree's hybrid method (VI, page 445)
Besides, for stability requirements, the mesh
can non longer be square, but such that
412± _Xk( _2 h
C 2
53.
54.
Sincd we do not know a priori the final shape,hthis would need to adjust the mesh ratio during
the iteration process, which would clearly complicate
the problem wery much. This is the main reason why
we adopted (2-7) rather than (2-3) in an elementary
approach.
3. Computation of J and J .x--y
a) Determination of J
).eriving (1-31) with respect to K, we get:
J = '6 x Z, (4-3)
which is an ordinary differential equation in x which
can be solved for each value y=y. of y.
The initial condition can be chosen arbitrarily.
For convenience, we take
J (x=O) = 0 (4-4)
This equation can be solved easily using a second-
order Runge-Kutta method, with the general equation:
J . =I + --- h Wxz, .x(n+l,j) = x(n,j) 2 (n,j)
+ (6xZ, ) (n+1, j) (4-5),
for all j and for values of x within the triangular
domain determined in paragraph (4-2).
Remark: The above approach is derived from
reference VII. Of course, it yields formulas quite
similar to usual integration formulas.(Newton-Cotes
formulas).
b) Determination of J
Deriving (1-32) with respect to y , we get:
Jyy
= ' xz,y (4-6)
which is also an ordinary differential equation in y
which can be solved for each value x = x of x.
Here, from symmetry considerations, we decide to
chose
Jy (y=0)
Using the
= 0 (4-7)
same Runge-Kutta method, we have:
= + - -k (i-n) +y(i,n) 2 1 ' y ,i, n) +
'y (in+1))
valid for all i and for values of y within
(4-8)
the same
triangular domain as before.
4. Determination of F by the method of characteristics.
Introduction
Let us look at the equation
a u, + b u, + c u = f
This equation, together with expressions of differen-
tials of uf and u, can be written in matrix form:
b c u, f y
dy 0 uxy d (u, )
dx dy u,7 v d(u, )
(4-9)
a)
r
a
dx
0
55.
56.
Investigating under which condition the deriva-
tives ux , u, and u, are uniquely determined
at a point P by values of ux and u,y on a curve 'yields the condition:
a
dx
0
a(dy) 2
b
dy
dx
c
0
dy, or
- b dx dy + c(dx) 2 0 (4-10) -
Writing the left hand side equal to zero will yield the
characteristic directions and
If we remain on these characteristic lines, the condi-
tions for (4-9)
a
dx
0
f
d (u, )d(u, )
ady d(u,x) - f
to be compatible can be written:
c
0
dy
= 0 , which gives:
dx dy + c dx d(uY)= 0 or
a -- d(u, ) + c d(u, y) - f dy = 0 (4-11)
This equation is valid along each characteristic line,
b) Integration of equation (2-7)
With ( d =1, (H (3 =-1 ,
a = 1 , b =.0, c = -1 f = J - J' yx
u 3F
equations (4-11) can be written:
and
d(F. ) - d(F, )
-d(F, ) - d(F,Sy)
= (Jy - J )dy
= (J y- J )dy
With the notation of fig.20,
characteristic lines, we wil
on lines a<
on lines
and integrating along the
1 have the equations:
,j - (F i,j-1- ( y )i+1
+ (F, ).
42k y
+ (F )
, + (y
i,j+1 - (F )i+
- X) ~i~j-1~ (4-12)
, j + (F9, )ij+
- k
Solving for (
(FX)i+1,j = 1
+ (Fy)i,j+1
- J )i+1 , + ((y - J)i, j+1)
and (F )i+1
2L(F, it j-1
+ kL 2
+(J - J)i, + 1j I
j
(4-13)
yields:
and
y i+1, j
+ (F, ) _
1-4-
2 -,(F ) ij-1
+F9, )i,j+1
+ (F,)i, j+
y~ x i, j+1
- (F, y )
~ y x i,
(4-15)
+ 9,)ij+1
+ (Jy - J)
(4-14)
j-1
57.
- J X)i+1
(Jy - J X)i+1, j
58.
c) Initial conditions
It is easily verified that conditions (2-4) and
(2-5) can be expressed in terms of F, and F, by:
FX = 0
F1y = py (4-16)
on x =0
It should be noted that this condition is compul-
sory on the part of line x = 0 belonging to the actual
dam. Other initial cona..ditions could be chosen on the
part of this line belonging to the fictive surface,
provided FI and F,y remain continuous up the third
derivative, in order not to introduce discontinuities
in the stress resultants and their derivatives up to
the second order,
d) Computation of stress resultants
Equations (4-14) and (4-15), together with ini-
tial conditions (4-16), allow to determine F, and F,yat each point of the triangular domain of integration
Using formulas (1-33) to (1-35) and computing
numerically the first derivatives of F, and F, by
formulas similar to (3-1) and (3-11), the stress resul-
tants can easily be determined at each mesh node.
Chapter V. Determination of Z and \
1. Determination of Z
It has been shown in chapter II that Z was de-
termined through integration of equation (1-31) with
boundary conditions (1-48) and (2-9).
Taking as unknowns the values of Z at each in-
terior node and at the boundary nodes (not on the top
edge where the value of Z is prescribed initially), and
writing at each point the finite difference expression
of the differential equation in terms of Z (using
formulas derived in Chapter III)., we can determine as
many equations as unknowns.
Solving this system of linear equations will
yield the solution for Z at each node.
To illustrate the procedure of filling the matrix
of coefficients row by row, we shall consider an inte-
rior node and a boundary node.
a) Interior node.
Introducing equations (3-14) , (3-20) and (3-25)
into (1-31), the ecuation at point (i,j) will involve
the following coefficients:
Unknown Coefficients:
Z. .N /y 2hk(i-1,j-1) N xy 2hk
Z(i+-11) xy
59.
Z (i+1, j)
Z(i-1, j+1 )
Z (i, j+1 )
Z(i+1 , j+1)
Constant term
All other coefficients will
b)
be equal to
Boundary node.
Let us consider the boundary node
tion of fig.9. Introducing formulas
and (3-27) into equation (1-48) will
wing coefficients
in the situa-
(3-17)
involve the
, ( 3-22)f ollo-
Coefficient
-2 Xsin{ Cos{ imZ(i , j3)
Z (i1 j-23)
il, -3)
(1+m) / 8hk
0
+2 \sin cos{ m
+2 X sin cos > M
(1+m) / 8hk
(2+m) / 4hk
Z (i, j-2) - X cos 2 { 1+ 2m /
- 2\ sin4 cos4 m(2+m) / 4hkZ(i+1, j-2)
Z(i-1 j-1 )>sin2
h2
\sink cosi+ 2hk
m(m-n) + m(1+m) (n-e )1+n + 1+n) (l+e)
m(1+m) e (1+m) (2+m)2 1 +e
60.
-2N / h2 - 2N /
N / h2
-N / 2hk
yy /
N xy/ 2hk
zero.
Unkn owm
Ix oe2
61.
4 \ cos 2 j/ k2
sin2 1 m (-n
h 2 1+n
sin §cosi m(1+m)2hk 2
_ Xsin 2 %
h 2 1
sinf cosi2hk
n(1+m) (2+m)]1 +n J
(1+m)(m-n) _ m(1+m)(n-e)n en
[m(2+m) + (1+m)(2+m)(1-ee
2 3+2mkcos 2
Ssin2
h2
[-m(2+m) -
Xsin2 k
h 2
(1+m) (M-n)n
Xsini Cos2hk
(1 -n) (2+m) (1 +mn
m (1 +m)e(1+e)
(1+m) (2+m)e +e)
2 sin2 % B
h 2-
+XCos 29
k 2
nX sin 2 nh2 r
sin i cos2hk
+\sin! cosf2hk
(1 +m)1 +n)
6m(2+m)
(1+m)1 +n)
(1 +m) (2+m)n(n+1
Constant term - Exx(i,b)sin21 - Eyy(ib)cos2
All other coefficientsc) Other cases
+ 2 Exy(i,b)sin§ cos
are equal to zero.
Any other situation can be handled in a similar way,
using formulas established in Chapter ITT.
j-1 )
(i+1, j)Z
Z (i-1 ,b)
Z(ib)
1b)
62.
d) Remarks
1. For equations written at points on row x=h,the contribution of points on row x=O must be taken
into account in the independant term. For a general
node (1,j), this additional contribution will be:
N F1 NConstant term: k - Zj1) + ZN(o,+1) - Z
Similar contributions can be determined for points
near or on the boundary curve.
2. As it was seen in chapter II, the equation
written at point (i+1,1) in fig.13 (bottom point)
will be
Z(i+11) =Zb
Although this equation could have been elimina-
ted in a way similar to what was done in the previous
remark, it has been left as above, setting the coeffi-
cient of Z(i+1 1) equal to 1 and acll other coefficients
equal to zero, while the constant term was set equal
to ZbeThe reason for this is that this equation was
introduced only after the completion of the program.
e) Method of resolution.
Although the matrix is an obvious band matrix,
it is not ascertained that a Gauss-Seidel iterative
method will converge since nothing insures that the
matrix is positive definite.
A'dire't method of elimination, exposed in (VIi)
has been used without any difficulty.
2. Determination of Xg
63.
In a quite similar way, we can set up the matrix
of coefficients and the vector of independant terms
which represent the finite-difference solution of equa-
tion (1-32) with boundary conditions (1-49) and (2-13)
All the expressions for X,. and Xy on the
boundary were established in chapter III, and can be
applied easily here.
It should be noted that, formally, the differential
equation (1-32) for is the same as the differential
equation for Z (1-31). This allows to use the same
subroutine to fill the matrix rows relative to interior
nodes, with different parameters.
64.
Chapter VI. Programming considerations.
1. Storage of arrays
Rather than storing all quantities defined at
each node in rectangular two-dimensional arrays, which
a) are computationnally time-consuming
b) would require more memories than strictly neces-
sary, since the projected surface of the dam is
not a rectangle,
only one-dimensional ariwys have been used throughout
the program.
An additional matrix INT has been defined such
that the position of any quantity Z(i,j) in the matrix
Z was referred to as Z(INT(I)+J), while the quantity
Z(itb) was Z(INT(I+1)).
It should be noted that, in order to reduce round-
off error in the formulas of chapter III, the pt-ogram
modifies the boundary coordinate y(i,b) when
y(ib) - y.j) 0.05 k (with the notation of fig.8)
It sets y(ib) Y(ij) , but still considers two
different memories for this double point to which cor-
respond now equal values of the function.
For functions not defined on the top edge such
that Zx 7..., another auxiliary matrix INS has
been defined, such that Zxx(it) is referred to as
ZXX(INS(I-1)+J) and Z,xx(ib) is ZXX(INS(I)).
2. Purpose of subroutines.
Many subroutines have been used to perforn all
65.
the sequential operations needed by the method. They
will not be detailed, but their general purpose is
outlined here;
COXX : computes second derivatives with respect to
x.
COXY : computes second cross derivatives.
COYY : computes second derivatives with respect to
y.DERIV : computes first derivatives with respect to
x and to y
COXXXB : computes Z, yy on the boundary
COYB : computes Z, , Z,,yy, Z,Yyy on the boundary
PARINT : computes J and Jx y
HART :determines F, and F,y by te _method of
characteristics.
FORCES : computes N xJ N yy and N y from F, and
F~y
THICK : computes the thickness at each point
FILI1 :fills the matrix of coefficients for Z and \at interior points
FIRSYS :completes the matrix of coefficients for Z
and calls SYSTEM
SECSYS :does the same work for X.
SYSTEM : solves a system of linear equations.
PRI :prints internal one-dimensional ar-ras as
a two-dimensional arnj.
All these subroutines are simple applications of
the formulas exDosed previously.
66.
Chapter VII. Results and conclusions
1. Boundary curve. Dimensions.
The particular valley shape which was considered
is a parabolic one, with the boundary curve described
by the equation:
y2 = - + 1(7-1)
where W is the width at the top and H is the height of
the dam.
Allthe quantities have been expressed in the meter-
kilogram system of urits. We have chosen W = H = 100m.
As other needed parameters, we have chosen:
R br = 50 Kg/cm 2 = 500,000 Kg/M 2 and - = 1
A square mesh size of 5m, was used, such that
the total number of nodes considered in the elliptic
equations was 157 (fig.21).
The fictive shape needed to determine F was
a triangle whose horizontal and vertical side were
equal to 120m. (fig.22)
2. General problem.
The success of the method outlined in chapter 2
clearly depends on the convergence of the iteration,
which is itself related to the starting assumed shape
for the membrane.
Initially, starting shapes similar to actual
arch dams were used, such that :
Z(x=O,y=O) e Z(xo0;=) These shapes were descri-
bed by equations such that
Z = 8.56 ( \A71 -1) - 0.63x - 15 3 - 0.012y2 1+
(7-1)
None of these shapes has given a converging
sbheme ; however, the observation of the partial results
has shown that J and Jy remained too small for such
shapes, thenfore influencing very little the solution F
of equation (2-7), and providing very small values of
stress resultants.
In order to increase J and Jy, other initial
shapes were used, for which
Z(x=O,'=100) - Z(x=,y=) = 50m.
Even in this case, no convergence was obtained; Howe-
ver, from all the partial results that were got, three
important conclusions could be drawn:
1) The successive shapes obtained through the itera-
tions had a slope Zy at x=O which was continuously
increasing and seemed to tend to Z, =00
2) All iterated shapes had very great curvatures Zyand Z, yy. This is easily explained when looking at
the stress resultants which were fairly small, such
that the differential equilibrium of a small element
required great curvatures. Besides, these stress
resultants were not changing very much from one
iteration to the next, even when the shape was chan-
ging very much.
3) Contrarily to the assumption done in chapter II, after
the first step of the iteration, INxy I has beenfound higher that IN XXI = 1T ( in some parts of the
xx yy
67.
68.
dam next to the boundary.
3) Simplified problem
Rather than spending too much time to realize
the convergence of a problem that could be mathemati-
cally not well defined and yield. non practicable shapes
it has been decided to solve an easier similar problem
and investigate whether the above conclusions were
inherent to the method or were a consequence of the
diverging scheme. This problem consisted in finding
the shape and the distribution of thickness such that
N = N and C = R , but for which the boun-xx yy max brdary curve was prescribed. This needed the resolution
of equation (1-31') for Z, with the boundary condition
Z = Zbound (7-2)
The displacement X could then be found by inte-gration of the compatibility equation (1-32) with one
of the boundary equations (1-48) or (1-49). Of course
the not-used of these boundary conditions would be trans-
gressed. We have used (1-48) which corresponds to a
zero extensional strain of the projected boundary curve,
thus allowing changes of curvatures of this curve(II).
After many trials, it has been possible to de-
termine a converging solution for Z by changing at
each step the solution at step i-1 by 20 percent of
the difference between the obtained solution Zg and Z_i-1
Z. =Z. + 0.20 ( Zt- Z.) (7-3)
The final results are given in Appendix 2.
They confirm all the conclusions drawn in the pre-
vious paragraph, i.e. that
69.
1) the obtained shape has much higher curvatures to
provide a practically usable dam shape;
2) The slope Z, at x=O is almost horizontal;9x3) Ny being still higher than Nx , the problem
is not mathematically well-posed.
Besides, this shape presents curvatures of diffe-
rent signs at the bottom, which means that the final
surface is'NVntirely elliptic.
Hence, it seems that we have now enough results
to conclude that the assumtion on F , i.e. equation
(2-7), done in chapter II, is not able to lead usable
results for arch dam shapes.
In fact, the assumption N = N ignores thexxyy
difference in the structural behavior of vertical ele-
ments (cantilevers) and horizontal elements (arches);
this difference, chiefly at the top of the dam, forbides
N N to be realized, unless very high slopes Z,are obtained in this region. These slopes explain
further the high curvatures in lower parts of the dam,
as well as the change of sign of Z,
Another a proach should be used for the deter-
mination of Z , probably based on some methods of linearprogramming and optimization. This is beyond the scope
of this work and was not examined.
Remarks:
1) In order to decrease Ny vs.. INXXI , we have stu-
died the dbove problem by introducing shear stresses
on the top edge, acting in the direction opposite to the
shear stresses derived from (2-7), expecting these boun-
dary stresses to propagate throughout the dam. This hope
was not fulfilled and the same features as before appea-
red in the results.
In the same point of view, we tried to modify the
fictive shape defined outside the domain of the dam in
70.
order to change J Y Here again, this artifice did not
bring any valuable results.
4) Determination of F for a given shape
In order to determine a condition for F, more
actual than equation (2-7) , the inverse problem of fin-
ding the stress resultants corresponding to a given
shape has been tried. This problem is quite similar to
the -general problem, and the method of solution is exac-
tly the same, with the difference that equation (1-31)
and its boundary 'condition (1-48) are now solved for F
instead of Z. In this case, they are written:
Zyy Fyxx - 2 Zxy Fyxy + Z, y Fyy = xx + z J
+ xx x (7-5)
with the boundary condition:
(F J ) 2 CV - (1+v) sin 2 + 2 cos2
- 2 >12 in2 cos + (Fy - J ) b sin2
+ 2 - (1+v)<( cos 2 ) - 2 #12 , sincos
+ FS- 2 c1 2 sin 2 - 2 o12 22xy 12 1 nr 12 2 Cos2
71.
L2 g 1 2 - (1+v)o sin Cos Eh.<X(Z, sin2
+ Z, cos 2j - 2Z, sin cos (7-6)
The iteration scheme consisted first in assuming
a given stress function, deduce the strains, solve (1-32)
with (1-49) for X , solve then (7-5) and (7-6) for F ,
Vzsing in (7-6) the values of X determined previously,and then begin again the iteration if the convergence
is not achieved within the precision of the conver-
gence factor. On the top edge, the boundary conditions
are
F = (77)2for F and
= - Z(7-8)Z~yy for ,XIexpressing thiat we have a
diaphragm-type support.
Here again, although many artifices were used to
provide convergence, no result was obtained.
Even a method which works for very sensitive pro-
blems did not yield any result:
assuming Xinitial = 0 , and
Finitial [py2 (1+0.05 x2) - 10x2(25 +1.25x)
+ xy2 (50 + 3.75x) + 150.10 6x j /2
(stress fu]ction yielding a distribution of stress re-
sultants similar to physically assumed actual values)
and then changing at each step\ and F by 10 percent of
thefewly determined values and the previous ones, seemed
to converge in the first steps, but began soon to diver-
ge again.
72.
5. Final conclusion
All the results show that the problem of the shape
optimization rests upon an assumption which is not valuable,
i.e. equation (2-7) for F . This problem should be approached
in a different manner, based on other hypotheses for
the determination of F. More precisely, the only factors
influencing mainly the stress function F are Jx, Jy and
the road weight p acting on the top edge, i.e. quantities
which have only secondary effects (such as p) or are
generally negligible for membranes whose curvatures
and slopes are fairly small (such as Jx and J y). Hence,
it is quite normal that the derived stress resultants
remained too small and required high curvatures for diffe-
rential equilibrium to be satisfied. A more accurate
approach should be to set up a condition for P by taking
into account the equation which predominantly influences
the behav'ior of the membrane, i.e. this normal equilibrium
equation. A more sophisticated approach,which probably
represents the best the behavior of the shell, consists
in minimizing the strain energy due to bending vs. the
strain energy due to stretching. This approach ought
to consider inextensinal bending and would certainly
yield much more difficult equations to solve.
As to the questionsof finding a solution for the
general problem of obtention of a.shape Z,or of determi-
nation of F for a given shape, all the equations have been
verified by slide rule computation with the printed
results, and residuals were found to be approximately
equal to zero, showing the accuracy of the program.
The obtention of a solution depends on having an initial
stress function close enough to the final one, as well
as determining an iteration scheme susceptible to
72 a.
yield convergence. However, rather than using in first
place this iterative method for a small mesh size,
starting from initial values whose accuracy is completely
unknown, it would be better to apply direct methods of
solution for a coarser mesh, determine initial values
fairly close to the final ones and start then the itera-
tive method for the rfined mesh, from this improved
initial solution. At this point, the convergence rate of
the iteration could certainly be increased very much by
a Newton-Raphson method linearizing the non-linear
problem.
c tJ
N IoD 4
zy
-fig. x
Nyxc
C4 u
x
73
Ixxa6 1
74
Ix
-- Pi. 3--
2:1) A
F
ft
IE
N'"
-Pitg. 4-ca --
I.,A
F
NU
i. -
t, v9
--Pis. 4-b-
~~Na
Nz2 t
F N21t,F 5D
Nitl
I4N z
-. c.6-
Th
G6
IN, I -)NL' I
4
76
0Ii~i.-ID"
al* I 4
e'i-I-I I'_____________ -
A;
nk
L~I.
rn k
I . I, -1- 4 - I I
4j.-z. LJ.--I I I I
LIj43 I. ij -z
eD(
ImI
A
4il
9-
i-.I6 J'
lh
1~
i-ij -3 -'
i j-3
-f'Pt. 7
Lj-.2
LK
"*uj j-zI I
'LJ-3
-i I- K-)
JL
- pip'.
I +',i -1 i+tjl
I .Ij.z
- fig.
&.ij-. 2 I -I~J-II - I - I - - £
i~i -a I tj Ii - 2 -wI
i4.&O.2 i -t 1
hk
'il
/
//LIe b
L nk
10 am-
-~ I
77
ItKI-3 1, K-i
3','
41j
i4tri-I
- fcg.
F 0
Pig
78
S-3,1 S-3Z . S-3,S.-3,
I- IsjZ S--,3 A
S-1 x, __ ,_
-fig.15
6
1-i,
41j-1
o.J
I -Ij41
"+i,1
-rn-I K 1~
16
S
IP-2
2,.1
ip-'
_,__-_ _ _3,p- 3,P
79
'5lp l>b
r -I-
i - i - i
-- f'ir,. Ia--
4
I/+iij
(f~C)
.f- . aO-
09'
.
mj-'I
I
'p 21~
50rn.
3 ~1 5 7 I
123 as I7V la8 2.q 30
34 35 36 37 38 39 4_ __ '.
44 45- 46 47 48 4 13S o S _52
54 £5 6 7 58 59 Go 61 6
11 6 4 (.7 68 69 70 71 _ 72
74 75 ?6 177 178_
85 86 187
92 93. 94 95 96
I lo 104 10s5
lii ~ ~Il9 j52.o j12.a
IZM: 125 .
13i 1 Z 133
137 138 133
143 1414161
P18 149
I&3 a~q 05
7
1126 -_I 7 28
13'l '3~'
'73 So __
88 B3
2.
81 e.
) 51
37 9% 33 e
ILL- '2.3
13o
I -3
140 1 4 2.
147.
-- PC z. -
&
3 _11 1q i f. 17 118 19
t
I
20
it
E00
83 84
Io?1o0
~Io~
A--
o
41
'3
I
, ,
62aq
U-
lic lilt I IIZ 1113
Is0i
82.
12o 'vi
ixy
K4
Appendix I
Determination of. Z for the general problem: Newton-
--------------- Raphson method. --------------------
The Newton-Raphson method consists here in wri-
and
Ni-1 i-1
ieplacing into equation
equation:
(1-31) yields the iteration
Nxx(i-1) (Z. - SZ1 i-1 ,vxx yy(i-1 ) (Zi -EZ.i)i 1-1 yy
+ 2 Nxy(i.. )
or, assuming Z
N xx(i-1
x-1
z.1,x X
(Zi i-1 ) + Z2i- 1xy
=sZi-2
yy(i-1) iyy + xy(i-1)
+ N x(i)SZ(i-2 ),xx yy(i-1) Z(i-2),yy
+ 2 N zxy(i-1 ) (i-2),xy
Equation (1-32) for can be treated in a similar way
and yields:
( i-i ) , y x. -2 z (i- ) ,xy i~j xy
yy(i-1),xx + 2 xy(i-1),xy
+ Z2i- 1
Z.1,xy
83 .
- 9Zi-1ting Z _g 1= Z i
Z (i1)x -X ,y
xx(i-1 ),7y
+ Z .X . +S.O.(i-l),xx (i-2),yy .(i-1),yy (i-2.),xx
- 2Z(i- 1 ) ,xy (-2).,xy
Boundary conditions (1-48) and (1-49) can be treated in
a similar way and the iteration scheme described in
chapter II could be applied to the transfrrmed equations
above, with their modified boundary conditions.
For a converging solution, the rate of convergence.
should be greater.
This method was not used because it was set up
only at the end and needed too important modifications
in the program in order to b'e finished in time.
84.
'VEL C MOD 0 MAIN DATE = 68147 22/46/24
DIMENSION X(21),Y(25),Z(175),ZX(175),ZY(175),YB(21), INT(22),I PO(22 )RJX(320)r, RJY 32 )CrFX( 320O),pFY( 320
2) ,RNX(?175) ,RNY(175) ,RNXY(175) ,TH(175 ),AUX(24649) ,VEC( 157),PHI( 21),3~IN(271PHI D (2 1EP X(175Tv, EPY (175 ) ,EPXY ( 175 ),EXB( 2)i,EPXY B(-21),I, I4S(157) ,ZXYB (21), ZYYB(21) ,ZXX(157) ,ZXY(157) ,ZYY(157), EPXDX( 175),5E PX DY ( 175)~,iPYiX (175) ,PYDPYt 1!5),EPXYDX i7 5) -EPXYDY (175), 5Y ( 21f6ZXXX(20),ZX)XY(20),ZXYY(2C),ZYYY(20),SEC(175b),ZPR(175),EPYXX(157),7EPXYY(157) ,EPXYXV(157)
COMMON FXFYEUVALENCE (U()E()
SH(R,P)=8.56*((P+1.)**0.5-1.)-0.23*P-0.004*(l.+P/100.)*R**2-15.*P-* 3/iC. t *6PINT(A,BCDE)=(-2.*A+9.*B-18.*C+11.*D)/(6.*E)P LAG ( A ,BvC ,D) =(3.*A-4.* B+C)/(2.*D)READ(5,IC) H,W,M,GAPSIGPO
I GFORMA T(E 14.7,1XE14.7,1Xi12,1X,E14.7.,-X,-E14.7,l-X,-E14.7)INDCDE L X=HMDELY=DELXMP=M+I
DO 12 I=1,MP12 Y(I)=DELY*(I-1)
MP=M+1A=- W**2 / (4 .*H)B=W** 214.INTMi) =0DO 13 I=1,MPYB(I)=(A*X(I)+B)**0.5INT(I+1)=INI(I)+YB(I)/DELY+2.05JR=INT( I+1 1- NTIIF(A 13 StYB (I)-Y(JR-))-0. 05*DELY) 16 16,13
16 YB(I.J=Y(JR-1)13 CONTINUE
00 14_I=1,MPJR=INT(I+i)-INT(I)-1DO 15_K=1,JRJS=INT(I )+K
15 Z(JS)=SH( Y(K),X(I))J S= IN T (1+1)
14 Z(JS)=SH( Bfl),vXH))INS(1)=0DO 31_1=1PML=0JX=INT(I+2)IF(Z(JX)-Z(JX-1)) 32,33,32
- - - --- - - - - - - - - - - - - -
86
V --L Q -OD -MAIN INDATE 6814722/46/24
33 L=1 _32 INS(1+1)=IN S(I )+INT( I+2)-INT ( I+1)-L31 CONTINUE
DO 30 1=1,0PHIII)=1.57C796-ATAN(H/(8.*YB(L)))
30 PHID(I)=8.*H**2/(H**2+64.4YB(I)**2)**1.5PHI(MP)=C.PHID(MP) 8. /HWRITE(6,4l) (PHI (1)_PHIJ( I,_I=1MP)
41 FORMAU('i',//(2(3X,E14.7)WRTE(617
17 FORMAT(1'f ,64X,'Z (M.)'/63X,'++++++//)CALL PRI (Z,I NTMP,0)CALL COXX(ZLXXINTINSMDELX,Y,YB)CALL COXV (ZXYtINTINSMDELXYYB)CALL COYY(Z,Z YYtNTINS ,MDELXYYB)WRI TE 6, icC)
100 FORMA(x ,'zxx'/I>3'+ ++CALL PRI(ZXX,INSM,0)WRITE(6,2C)
20 FOR MATI'',64 X ,'lXY 'A3/6 3X +++++ ++//)CALL PRI(ZXY ,INSM,0)WR ITE 16,121)--
21 FORMAT({'',64X,'ZYY',/63X,'+++++++//)CALLPRI_(ZYYyI NS1M,0)CALL CIXXXB ( ZXXZXXXINSMDELX ,Y ,YB)CALL_ CUYB (Z X, ZXXY INS ,., DE ELX ,Y, Y B )CALL COYB(ZXYZXYYINS,M,DELXYYB,-1)CALL COYBfZYYtZY 'YINS_,,DELY ,,YB--f)WRITE ( 6,4c) (ZXX.((I )-,ZXXY (H ) ?ZXYY (I) ,zyyy =IM)
IS=,6*M/5+1-7 CALLDER IV( Z .ZXiZY,.DELXY8iNTaMPY DELY,0.)
WR ITE (6,18)
CALL PRI(ZXINTMP,0)
19 FORMAT(l't63Xr'LY (M/M)',/62Xt,'+++++++++//)CALLPRLZ NAPAO--------------JP=INT( 2)ZXYi(1) =PINTZX ( JP-4),ZX (JP-3) ZX ( JP-2,ZX ( J P-1) DEL X)------ZYYB(l)=PINT(ZY(JP-4) ,LY(JP-3) ,ZY(JP-2),ZY( JP-1) DELXDU_ 36 1=2_.,M P _ _ _ _ _ _ _ _
JP=INS(I)- - Z X Y B ( I- ) -= Z X Y P ) ----36 ZYYB(I)=!ZYY (JP)
PALLPARINT JXRJYIPOINTX, X GA DELX Y ZXYB ZY 0ELYZYYBS-IYBI
87
EVELC, MOD C MAIN -- DATE 68147 22/46/24
CALL HART( I POIS ,MP ,RJX ,RJYDELXPY)CALL FORCES( IP ,1 TRNXRNYRNXYOELXDELYpRJXRJYpMMP)WRITE (6,24)
24 FORMAT( >161X,'NXX (KG/M)'/60X,....+++++++/CALL PRI (RNX ,INT ,MP,1)WRITE (6,25)
25 FORMAT (I',61X , 'NYY (KG/M) '/60X,'++++++++++++// )...CALL PRI (RNYINTMP,1)WRI TE (6,26)
26 FORMAT( Ill51 XNXiY~ I(KG/)/6~,+~++++++++++/CALL PRI (RNXYINT ,MP ,1)CALL THICK(RNX ,RNY,RNXYZXZYSIGTH TPWRITE(6,27)
27 FORMAT( '11',6CX,'1HICKNESS (M)' ,/59X,'+++++++++++++++//)CALL PRI(THINTMP,1)
53 CALL FIRSYS(RN~,RNYRT\XYAUXVECDELX X ZXZYGAzINTM,INS,YtYB, EPXEPYEPXYEPXB ,EPYBPEPXYB, IND, PHItHtDISZS EC)WRITE (6,17)CALL PRI(ZIINTMPO)U=Z(1)**2R=ZSEC ( 1)**2Sq=( LSEC (1I)-,Z (1) ) 4'*2MS=INT(MP)+1DO 50 1=2,M-'-U=U+Z(I_) **2R=R+Z SEC (I) **2
50 S=S+(ZSEC(lI)-Z(I))**2R=R**0.5U=U**C.5SS* C.5T=S/RWRITE(6,61) SRL , T
6 FORMATI4(3XE14. 7)-//)IF(S/R-0.05) 54,54,51
F IND=IND9(IF(IND-3) JC,70,90
7C MT=INT(MP)+2DO 71 I=1,MJ
11 Z(J )=Z SEC (I )O.40*Z SECGO TO 91
90 IF(IND-6) 9202,9392 MT=INTIMP)+2
DO 94 I=1,M_94 ZI)=ZSEC(I)+0.80*Z (I)-ZSEC(I))
GO TO '91 - - - - - - -- - -
93 IFI ND-IC)S1, 91,56L 91WR ITE(6t17)
CALL PRI(ZINTMP,0)
EVEL 0, MOD C MAIN DATE =68147 22/46/24---- ------------------ ------------ 2_2- --......... 4.
CALL COXX(ZZXXINTINSM,0ELXY ,YB) -- _------
CALL COVY(LZYYINT,INSMDELX,YYB)CALLCOXY(2 INT SMD ELXY YB)WRITE (6,100)CALL PRI (ZXXINSM0)WRITE ( 6,21)CALL PRI (ZYYINS,M,O)WRITE (6,2C)
PRI (ZX. tINS ,M,0)GU TO 57
5-6 WRIIE(.6,5E)58F0 RMA 1I/ I CONVE RGENCE HAS BEEN OBTAIND' ) -
JS=INT(MP+1)WRITE(7,1C1) (Z( ),I I 1 i1 JS)
M FORMAT( 5(E15. 8 1 X) IGO TO 59
54 WRITE(6,6C)60 F0RMA T / END- OF'1 T ERLTfI LST SU RFCE IS FINAL SHAPE')
JS=INT(MP+1)WRITE( 7, 101) (Z(I),I 1 JS)DO 35 l=1,MPJX=INTII fliJV=INT( 1+1)-iDO 35 K=JXJVAP=1.+ZX(K)**2 .......
AS=1.+ZY(K )**2A T=ZX (K)* ZY (K)AL=(AP*A-'-A1**2) **0.5A Z=RNX ( K) *A P+RNY (K) *A S+2. *RNX Y( K) *ATEPX(K)=(AZ*AP-(1.+PO) *AL**2*RNY(K) )/ (TH(K)*AL)EPY(K)=(AZ*AS-(1.+PO)*AL*42*R'NX(IK ) ) /-( T H ( K )* AL)EPXY(K)=(AZ*AT+(1..+PU)*AL**2*RNXY(K))/(TH(K)*AL)
5 CONTINUEJX=INT(2)-i
37 DISHI )=EPY(I)/0.024DIS(JX+1)=DIS(JX)CALL BOUND(EPX,EPXBINTYYBZDELX)CALL BOUD(EP PYBI NT,,YB ZOELXCALL BOUND(EPXYEPXYBI NT,YYBZDELX)WRITE (6,42)
A 2 FORMA T(I'63XY'EPX'v /62 X i+-.-CALL PRI(EPXINIMP,0)WRITE (6,43)
43 FORMAT('l63XEPY,/62x,++++/CALL PRI (EPYINTMP,0)W I TE (6,44)
44 FORMAT('1',63X,'EPXY',/62X ++++++//)
---------------
------------
EVEL C, MOD 0 MAI N DATE 68147 22/46/24
CALL PRI(EPXYINTMP,0) -CALL DERIV(EPXEPXDXEPXDYDELXYBINTMPY, DELYO.)C0AL DRIQ_(PYEPYDX,EPYDY,DErLX,YE3,[NTMPY, _ DELY,O .CALL DERIV(EP) YEPXYDXEPXYDYDELXYBINTMPY DELY, 1.)CALL COXX{EPY,EPYXXINTINSMDELXYYB)CALL COYY EPXEPXYY~, ITINNS ,MDELXYYB)CALL COXY(EPXYEPXYXY, INTINS,MDELXY,YB)
52 CALL SECS YSEPX EPYEPXY,DIS , ZPHI ,PHIDAUX ,VEC, DELX ,X, INT, INSM,1YYBZXXZY'YZXYEPXDXEPXDYEPYDXEPYDYEPXYDXEPXYDYZXXXZXXY,2ZXYYtZYYYEPYX,~EPXYY~,EPXYXY)WRITE (6,38)
3 8 FOMAT ( 1~6+X,~E*L..A1,~/63( +++++ ++CALL PRI (DISINTMP, O)JS=INT (MP+1)WRITE(7,1C1) (DIS(I)tI=1,JS)
5' ALLEXITEND
-- . .- -- - - - --.e i.i .m m = m 4, m a s a.. .i~ a ,11- . . ta-- .-. -.ie e. e = ,.9 .. . . 4 . .~a .,., i.. .... , .1. .p. . . . , , -.
SU BROUTI INE 0ER IV Z2ZYE 6 LMjjD__,__DIMENSION L(175),ZX(175) ,ZY(175),YB(21),INT(22),Y(25)
__ _ L L A 1 =A- 3q _*C - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -PlNT(A,B,CDE)=(-2.*A+9.*F3-18.*C+1l.*D')/(6.*E)
DO 15 I=1,M~SJR= IN T ( I+1)- 2JP= INT (I )+2
Q Q R J B . - - - - - - - - - -- - - - - - - - - - - - -- - - - - - - - - - - -10 ZY(K)=(L(K+1)-Z(K-1) )/(2.*-tCELY)
IF (Z (JR+2)-Z(JR+1 ) j 6jj ~ h-- - - - -- - - - -- - - -- - - - -- - - - -- - - -
RLA=(YE( I)-Y(JQ))_LIE L YZY(JR+1)=(Z(JR+2)/(RLA4'(l.-+RLA))-Z{JR)*RLA/q1.+RLA)-Z(JR+1)*(1.-RL
-- -- - -- - -- - -- - -- - -- - -- - -- - -- - -- - -- - -- - -- -ZY(JR+2)=(Z(JR+2)*(1.+2.*RLA)/(RLA*(1.+RLA))-Z(JR+1)*(l.+RLA)/RLA1+Z(JR)*RLA/(1.+RL A)LDELX-- - - - - - - - - - - - - - - - - - - - - - - - - - -GO TO 13
7 ZYfJR+1 )=PINT(7fJR-2)2ZfJR-l).Z(JR)j7,1R+1)JPFLX)ZY( JR+2) =ZY(JR+l)
L3-LYLZ11EmllLJPL!QLILLELX-------------------------------------------------------------15 CONTINUE
&Elu1Ii~~±L--------------------------------------------------------------------JR=INT(PP-2)+1
JQINTIMP -1)+lJP=JINT (IP)+I.
Z Y j p+ 1 -ZY ( jp)
DO030 K=2,MSIR=INI(K+l )-TNT( K-1IF(YBM$-Y(JR)) 50,51,50
LJ&aJ&s--------------------------------------------------------------------------------0 DO 31 I=1,JR
~J~aN~lK1+L-------------------------------------------------------------------------JP= INT (K-1)+ IJ1O= INT iK-+1I )+ IIF(YEC(+l)-Y(I)) 32v3203
* - ----------------------------------------------GO TO 31
RMA=(YB(K)-Y(I) )/DELX-REA= (YB (K- ) -Y (I I ) DFLXA=(1.-PNA)/( l.+RNA)
C=2./(RNA*(1.+RNA))R- I--(l .- +REA-) +Z( JPR)tRNAA ( RNA-__
11.-REA)/REA-Z(JP+1)*RNA*(RNA-1.)/(REA*(l.+REA)))/(2.*OELX)
---------------------------------------------------------------------------------------
---------------------------------------------------------------- 91 -----------------
------------- --------- D1132J3-L ------------
ZX(JS)=A*A=tl.+RMA)*(I.+RMA-RNA)/(l.+RNA)
2 .+RV A L"#z (I .+RMt-RN A) - /RNA ----------------------------C=(2.+RPiA)"-'tl.+RMA)/(RNA*(I.+RNA))
ZX( JS+1 )=A* ZMAS-2 ) +B31 CONTINLE
IF(YB(9+1)-Y(JR) ) 30,3CO55 RNA=(YB(K+I)-Y(JR) )/DELX
RMA= (YBIF(YB(X-1)-Y(JR)-DELX) 36737,37
36 REA=(YB ( K- l)-Y( JR)GO TO 38
37 REA=.l.38 A=RMA*(RPA-RNA)/(I.+RNA)
&UU8-M=RK, AURR A -------------------------------------------------------C=RMA*(l.+RMA)I(RNA*(I.+RNA))
--- JS=INT (K)+JR ----------------------------------------------------------------JP= INT (K-I)+JR
JQ= IAT (K+ U+JRAUX=(Z(JQ-+l)-Z(JP-1)'RNA*(RNA- REA)/(l.+REA)+Z(JP)*(I.+RNA)*(RNA-RE
---------------------------ZX(JS+1)=A*ZX(JS-I)+B*ZX(JS)+C*AUX
-j-F-cy-a- t-K i-=Y-(-j-R±-LLi-ao-*-5?-,-ac ------------------------------------------------------52 ZX(JS+2)=ZX(JS+l)
0 - COA-LLN L FJP=INT(PP-4)+l
-----------------------------------------------------------------JR=INT(PP-2)+l
-------------------------------------------------------------------J'T=INT(MP)+l
S)=Pl IJZ(JP)--- JQ),Z( R)j7jJS),P-[)El/-X(JS+I)=PINTtZ(JP41),Z(JQ+I),Z(JR+l)tZ(JS+I)vDELX)
-------------------------------JA=INT(t)P)-INT(MF-I)-l
X -------------------------------------------------------ZX(JS+3)=ZX(JSI*RLA*(l.+RLA)12.-ZX(JS+I)*(2.+RLA)*RLA+ZX(JS+2)*(2.
I+Rl A)*(I.+RLA-U-2,ZX(JT)=PINT(Z(JQ)tZ(JR),Z(JS)tZ(JT)tDELX)
JR=INT(2)-2
JP=INT(2)+l-XQ= LM 3 ) +1
JS=INT(4)+I
ZX(JR+I)=PI(ZX(JR-2)vZX(JR-l)tZX(JR))
RETURN
-------------------
--------------------------------------- ----------------------------------------------
9z
EVEL 1, MOD 1 COXX DATE 68157 21/31/59
SUBROUTINE COXX(AAXXINTINSM,DELX,YYB)DIMENSION A(175) ,AXX(157) ,INT(22) ,INS(21),(25),YB(21)PI (At ,B ,O ,D , S) =( -A+ 4. *B-5.*C+2 .*D)/SS=DELX** 2M S=M- 1DO 1C I=1,MSJX=INS(I)+1JV=INS(1+1)-i00 11 K=JXJVJA=K-INS[I)IFI Y(JA)-YB( I+2)) 12,12,11
12 JS=INT(I)+JAJT=INT(I+1)+JAJZ=INT (I+2)+JAAXX(K )=(A (JS)-2,*A (JT)+A(JZ)) /S
11 CONTINUE10 CONTINUE
MS=M-2DO 13 I=1,MSJV=INS(I+1)JA=INS(I+11-INS(I)-iRMA=( YB ( I+1)-Y(JA) ) /DELXJS=INT(I )+JA-2JT=INT( I+1)+JA-2JZ=INT( I+2)+JA-2IF(Y(JA)-YB(1+2)) 14,15,15
15 REA=( YB'(I)-Y(JA)) /DELXRNA=(YB( I+2)-Y(JA-1))/DELXB=( 1.+RMA) *(1.+RMA-RNA )/(S* (1.+R NA )C=(2.+RMA) *(1.+RMA-RNA)/ (S*RNA)D=12.+RMA)*(1.+RMA) /(S*RNA*(1.+RNA)AXX(JV)=A(JS)*B+A(JS+1)*(-C+D*(RNA-1.)*(RNA-1.-REA)/(1.+REA))+A(JS1+2)*D*RNA*(1.+REA-RNA)/REA+A(JS+3)*D*RNA*(RNA-1.)/(REA*(1.+REA ))-A2(JT)*2.*B+A(JT+1)*2.*(C-D*(RNA-1. )*(RNA-1.-RMA)/(1.+RMA)I+A(JT+2)*3D*2.*RNA*(RNA-1.-RMA)/RMA+A(JT+3)*2.*D*RNA*(1.-RNA)/(RM A*( 1.+RMA))4+A(JZ)*B-A(JZ+1) *C+A(JZ+2)*DGO TO 13
14 RNA=(YB (I+2)-Y(JA) ) /DELXIF(YB(I)-Y(JA)-DELX) 16,16,17
16 REA=(YB(I)-Y(JA))/DELXGO TO 18
17 REA=1.18 B=lMA*(RMA-RNA)/((1.+RNA)*S)
C=( 1.+RMA)*(RNA-RMA) /(RNA*S)D=(1.+RMA)*RMA/(RNA*(.+RNA)*S)AXX(JV)=A(JS+1)*(B+D*RNA*(RNA-REA)/(1.+REAH)+A(JS+2)*(C-D*(1.+RNA)
1*(RNA-REA) /REA) +A( JS+3)*D*RNA* I.+RNA)/( REA*(1.+REA) )-A(JT+3 )*2./S2+A (JZ+1) *B+A(JZ+2) *C+A( JZ+3) *D
93
EVEL 1, MOD 1 COXX DATE .68157 21/31/59
13 CONTINUEJV=INS(M)JA=JV-INS(M-1)-1RMA=(YB (M)-Y(JA)) /DELXB=RMA* 11.+RMA ) /2.C=-( 2.+RMA )*RMAD=( 2.+RM A).* (1.+RMA )/2.JS=INT(M-3)JT = INT ( M- 21JZ=INT(M- 1)JW=IN T(M)J X =IN T (M+1)AXX("JV-2)=PI(A(JS+2),A(JT+2),A(JZ+2),A(JW+2),S)AXX(JV-1)=PI(A(JS+3) ,A(JT+3) ,tA(JZ+3) ,A(J4+3),S)AXX(JV)=D*AXX(JV-)+C*AXX(JV-2)+B*AXX(JV-3)JV=INS(M)+1AXX(JV)=PI(A(JT+1),A(JZ+1),A(JW+i1),A(JX+1),S)DO 20 I=1,MSJA=INS( 1+1)-IN S( I)-lRMA=(YB(I+1)-Y(JA)) /DELXJV=INS( I+1)-iIF(Y(JA)-YB(L+2)) 20,20,21
21 AXX(JV)=-RMA*AXX(JV-2)/(2.+RMA)+2.*RMA*AXX(JV-1)/(1.+RMA)+2.*AXX1(JV+1)/1(2.+RMA)*(1.+RMA))
20 CONTINUERETURNEND.
EVEL 1, MOD 1 COXY DATE= 68157 21/31/59
SUBROUTINE COXY(A ,A XYINT ,INS ,M,DELXYYB)DIMENSION A(175) ,AXY(157) ,INT (22) ,I NS(21), Y(25), YB( 21P 1 (4B, ,D) =-D*A /(I.+ D) - (1.-D) *B/D+C/ (Dlw,-1.+D)S=DELX**2MS=M- 2Do 10 I=1,MSJV=IN S( I)+1AXY(JV)=0.JV=JV+1JX=INS(I+1)-1DO 11 K=JVJXJA=K- INS( I)IF(Y(JA+1)-YB (I+2)) 12 ,12 ,13
12 JS=INTII+2)+JAJT=INTLI )+JAAXY(K)=(A(JS+1)-A(JS-1)-A(JT+1)+A(JT-1))/(4.*SIGO TO 11
13 [F(Y(JA)-YB(I+2)) 14,11,1114 RL A=(YB(I+2)-Y(JA ) /DELX
IF(YB(I)-Y(JA)-DELX) 16,17,1716 REA=(YB(I)-Y(JA))/DELX
GO TO 1817 REA=1.18 J S=IN T( I+2) +JA
J T=INT ( I) +JAAXY( K) =( A( J S+1) /(R LA*(1.+RLA) )-A (JS)*(1. -RL A) /RL A-A( JS-1 )*RLA/ (1.+RLA)-A (JT+1) /( RE A*(1.+REA) )+A (JT )*(.-REA) fREA+A(,JT -1)*REA/( 1.+EA2))/(2.*S)
11 CONTINUE10 CONTINUE
DO 20 1=1,MSJV=IN.S 1+ 1)JA =JV-INS( I )-1RMA=(YB(I+1)-Y(JA)) /DELXB=RMA* ( 1.+RMA) ( 8. *S)C=-RMA*( 2.+RMA ) /(4.*S)D=(2.+RMA)*(1.+RMA)/(8.*S)IF(Y(JA)-YB(I+2)) 21,22,22
22 IF(Y3(I-1)-Y(JA)-DELX) 23 ,23 ,2423 RDA=(YB (I-1)-Y(JA)) /DELX
GO TO 2524 RDA=1.25 REA=(YB(I)-Y(JA))/DELX
RNA=( YB( 1+2)-YIJA-1) )/DELXJZ=I.NT( I-1)+JA+ 1JS=INT (I )+JA+1JT=INT( I+1)+JA+1J W=IN T( I+2 )+JA
EVEL 1, MOD 1 COXY DATE 68157 21/31/59
A XY (JV-1)=P.(1A( JZ-2) , A(JZ-1)PA (JZ) ,RD A )-4.*P(A( J S -2),A(JS-1),IA (JS ) ,REA )+3.*PI( A(JT-2) p A(JT-1)PA(JTI),RMA))/(2.*S)AAXY(J V) =A XY ( JV-3) *RMA* (I.+ RMA) /2.-AXY(JV-2) *12 .+RMA )*RMA+AXY( JV-1I
1*(2.+RMA )*( I.+RMA)/2.GO TO 20
21 RNA=(YB 1+2)-Y(JA))/DELXIF(Y3(I)-Y(JA)-DELX) 26,25,27
26 REA=(YB (I )-Y(JA) ) /DELXGO TO 28
27 REA=l.28 JS=INTI) )+JA
JW=INT(I+2)+JA-A XY ( JV)=B*(A (JW-1) -A( JW-3) -A ( JS-1+A (JS -3) )+C*( At JAI -A( J W-2)- A( J S)1+A (JS- 2) )+D*(-2.*RNA*A (JW-1) /(1.+RNA) -2.*(1.-RNA )*AtJ W ) /RN A+2.*A( J2W+1)/(RNA*(1.+RNA))+2.*REA*A(JS-1)/(1.+REA)+2.*(1.-REA,*A(JS)/REA3-2.*AIJS+1)/(REA*(1.+REA)))
20 CONTINUEJV=INS(M)JA=INS(M)-INS(M-1) -1RMA=(Y3 (M)-Y(JA) ) /DELXB=RMA* (2.+R MA)C=( 2.*RMA )*( 1.+RMA) /2.JZ=INT(M-'3)+1JS-=INT( M-2)+1JT=INT( M-1) +1JW=INT(M)+1AXY(JV- 3)=0.AXYIJV-2)=(2.*(A(JW+2)-A(JW)-A(JS+2+A(JS)-A(JT+2 +A(JT)+A(JZ+211-A(JZ) ) /(4.* S)AXY(JV-1)=(4.*(-RMA*A(JW+1)/(1.+R MA)-(1.-RMA)*A(J 4+2)/RA-A+A(JW+3)/
1( MA* (1.+R MA ) -2. *A ( J S+3)+2.A ( JS+1 ) -A JT + 3) +AM J T+11+ A( JZ+3-A ( J Z2+1))/(4.*S)AXY(JV)=-B*AXY( JV-2)+C*AXY(JV-1)JV=IN S(M)+1AXY(JV)=0.RETURNEND
------------ U11 --------------- -O-II3-2J3-1 -------------
SUBROUJINE CQYY(AAY ,-I-NT.INS.:.nE X-Y-ya)DIMENSION A(175)iAYY(157)tINT(22)vINSt2l)vY(25)tYB(21)
------------------------------------------------------------------------MS=M-l
-DO- C- 5 ----------------------------------------------------------------------JV=INSII)+lJX=INT(I+I)+lAYY(JV)=(A(JX+I)-A(JX))*2./S
----------------------------------------------------------------------JA = I N S ( I + I I N S ( I
--- LE-LKUAJ-- YULLItIII-1-1.,12A.1 ------------------------------------------------------12 JV=INS(1+1)-l
QO TC 1311 JV=INS(I+I)-2
------------------------------------------------------JS=INT(1+1)+INS(1+1)-INS(l)-l
--------------I-A(JS)"'(3.-RLA)/RLA+A(jS+1)*6./(PLA-"(l.+RLA)*(2.+RLA)))/S
.13 DO 1 1 1(=JXIJVJS=INT(1+1)+K-INS(l)
---------------------------------------------RMA=(YB(I+1)-Y(JA-1))/DELX
Sill -----------------------------------------------------JT=lNS(I+,l)
IMA)/RMA+A(JS)*6./(RMA*(2.+RMA)))/SLOL-LLLLIMLE ------------------------------------------------------------------------
JT=INS(P)+l--- dXaJ-bLT-i.L=2-L± L --------------------------------------------------------------------
JV=INT(M-1)+l.17=INIfV)+IAYY(JT)=(-A(JX)+A(JX+I)+3.*A(JV)-3.*A(JV+1)-',.*A(JZ)+3.*Atjz+1))*z
------------------------------------------------------------------------------RETURN
---LN ------------------------------------------------------------------------------
-----------------------------------------------------------------------------------
------------------------------------------------------ 7 ----------------------------
-----------------------------------------------------
-----------------------------------------------------------------------------------
57
L 0r MOD 0 COYB DATE = 68130 15/52/13
SUBROUTI NE COYB(AAYINSM,DELX,YYB,L)DIMENSION A.(157),AY(20),INS(21),Y(25),YB(21)iS=M-1DO 1. I=1,MSJA=INS(I+)-I NS( 1)-l
NJ f I )+JAJR =IN S ( 1+1)IF(YB(I+2)-Y(JA)) 11,11,12
2 RMA=( YB( I+1)-Y(JA) ) /DELX
GO TO 131 RMA=(YB (I+l)-Y(JA-1) ) /DELXIF(I-MS) 14,15,14
4 T=A (JP-3)JP=JP-1GOTO 13
5 T=A(JP-1)*LJP=JP-1
3 AY(I)=(-T*RMA*(1.+RMA)/(2.*(2.+RMA))+A(JP-1)*(2.+RMA)*RMA/(l.+RMA)1-A(JP)*(2.+RMA)*(1.+RMA)/(2.*RMA)+A(JR)*(3.*RMA**2+6.*RMA+2.)/(RMA2*( 1.+RMA)*(2.+RMA) )J/DELX
o CONTINUEJP=INS(M-3)+1JR=I NS (M-2)+IJQ=INS(M-1 )+1 __
P=(A(JP+1)-A(JP+1)*L)/(2.*DELX)Q=(A(JR+l)-A(JR+1)*L)/(2.*DELX)R=(A (JQ+1 )-A (JQ+I)*L)/ (2.*DELX)A Y (M =P-3.*Q+3. *R
END
98
L 0, MOD 0 COXXXB DATE = 68130 15/52/13
SUBROUTINE COXXXB(AAX,INSMDELXYYB)DIMENSION A(157),AX(20),INS(21),Y(25),YB(21)PLA (A ,B, C ,D )=(-1.5*A+2.*8-0 .5*C )/DPINT(A ,B,C ,D,E) =(-2.*A+9..*B-18. *C+11 .*D)/(6.*E)P1 (A ,B ,C ) =A-3.*B+3.*CPOL (A ,B ,C ,D,E)=A*D*(D-E)/(1.+ E)-B*(1.+D)*(D-E)/E+C*D*( 1.+D)/( E(1.
1+E)) _ _ _ _ _ _ _ _
MS=M-300 10 I=2,M-r
JP=INS(1-1)+JAJQ=INSI1 )+JAJT=INS( 1)JR=IN Sf I+1)+JARMA=(YB(_+1-Y(JA )/DELXREA=( YB (I) -Y (JA) /DEL XIFIYB(I +2)-Y(JA)) 14,14,15
4R NAYfB f(+ 2)-Y(JA-1) ) /DELXD=4 1.+RMA)*(1.+RMA-RNA) / (1.+RNA)B=-(2.+RMA)*(1.+RMA-RNA)/RNAC=f2.+RMA)*(1.+RMA)/(RNA*(1.+RNA))P=(A(JR-2)-A(JP-2))/(2.*DELX)Q=IA (JR-)-A(JP-1) /2.*DELX)R=(A(JR)-A(JP-1)*(RNA-1.)*(NA-J++AI*RNA*(RNA-.-IREA)/REA-A(JP+1)*RNA*(RNA-1.)/(REA*(1.+REA)))/(2.*DELX)AX(I1) =D*P+B*Q+C*RGO TO 10
5 RNA=( YB(12-(A)DLR=(A (JR+1)-A (JP-1) *RNA*( RNA-REA) / (1.+REA )+A(J P)*1.+RN A) *(RNA-R EA)REAA(J~T )*(I.+RNA)*RNA/(REA* (1.+REA)))/(2,*E)IF(YB(I+3)-Y(JA)) 16,17,17
6 D=(1.+RMA)*(RMA-RNA)/(2.+RNA)3=(2.+RMA)*(RMA-RNA) /(I.+RNA)C=(2Z.+RMA) *(1.+RMA)/MI1.+RNA)*N(2.+RA-P=(A(JR-2)-A(JP-2))/(2. *DELX)Q=(AIJR-1)-A(JP-1)) /(2.*DELX)AX ( I ) =D*P-B*Q+C*RGO TO 10
7D=RMA*(RMA-RNA)/(1.+RNA)B=( 1.+RMA)*(RMA-RNA) /RNAC=RMA*(1.+RMA) /( RNA*(.+RNA)P=(A(JR-1)-A(JP-1))/(2.*DELX)Q=(AJRI-A(JP) )/(2.*DELX)AX( I) =D*P-B*Q+C*R
o CONTINUEJA=I NSfM-1) -INS( M-2 )-1JP=INS(M-5) +JAJQ=INS(M-4J)+JA
L -0,1 -MD C - - - -- - -COXXXB - - - -- - -DAT E- =-68 130 15/5 2/1-3
JR=INS(M-3)+JAJ S =I NS( M- 2) +J A
RMA(YB(M-1.)-Y(J A)) /DEL XP-POI)ApJA- I JP )P,A (JP+I 11RMAL.)
REA =(YB (M-3) -Y UA) )/DELX - - - --- ~ -
Q=POL (A (JQ-1),tA(JQ),YA(JQ+2)RfAE)REA=(YB(M-2)--Y(JA))/DELX ___________
R=POL (A (JR-1) p A( JR) , A(JR+1 ) PRMAREA)*AX(M-2)-=PINT(P,Q,R,A(JS+i) DELX)
JAP = IN SI ) S(M - 1 )A-- - - - - - - - - - - - -- - - - - - - - - - - -JQ=IN\S(11-')+JAJR=INS(II-3)+JA___________JS=INS( M-2)+JAJT=INS(M-1)+JAB=IAI JR-2)-A (JP-2 )/ (2**DELX)
* -C=(A(JS-2)-A(JQ-2) ) /!2.*DELX) .---------- ---
D=(A(JT-2)-A{JR-2) )/(2.*DELX)P=PI (BC D)B = (A (JR- 1) -A (JP-1) (2 .*DEL X)
*C=A (JS-1)-A(JQ-1) (2. *DELX) --
D=(A(T-I)-(JR-1))/ (2.*DELX-~Q=PI(BtC ,D) -------
BAJ R) -A (J)/2*EC =( A(J S)-A (JQ))/(2. *DL X)RMA= (YB (M)-Y(JA) ) /DELXD=I-A (JT-2)*RMA/(2.+RMA)+A(JT-1 )*2.*RMA1 (1.+RMA)+A( JT+i)*2,/(i.IRMA)*(2.+RMA))-A.(JR))/(2.*'%DE LX)
KRKA=(YB()(A-)/EXAM M-1) =POL( PQ,RRMA,1.)JP=INS IM-3?+1JQ=INS(M-2)+1JR=IN 5(1-1)+1J S =1N-S(M)+ 1AX(M)=PNT(A(fP)A(JQ)A(JAjS),oDELX)JA=I N S (2) -1 NS (I ) -1___JP=INS( 1)+JA
-JQ=INS(2)+JA
-JR= IN S(3)+JA-- P=PLA(A(JP-2),A(JQ-2),A(JR-2),DE LX)
RLA=IYB(4)-Y(JA) ) /DELXB=-A(JR-2)*"RLA/(2.+RLA)+A(JR-)*2.cRLA/(,+RLA)+A(JR+.j*2,,((I,..
IRLA )4(2.+RLA))
RMA=(Y8(2)-Y(JA-1))/DELX -------- - - --
----------------------------------------------------------------------
loo
EL 1 MCD 0 PARINT DATE 68118 06/08/07
SUBROUTINEPARINT(MRJXt,RJY, IPOINT,ZXX,GADELXYZXYBZY,DELY,1ZYYBISYB)DIMENSIONRJX(320),RJY(32C),IPO(22),INT(22),ZX(175),X(21),Y(25).
1ZY( 175) , ZXYB (21) ,ZYYB ( 21) ,YB(21)DO 24 1=1.IS
24 RJX( 1)=0.IPO(1)=0DO 22 I=1,MJP=Is-I+1IPO(1+1)=IPO(I)+JP
-JL = I NT(U 1+1')-iJS=INT(1+2)-INT(I+1)-iDO 23 K=1 ,JSJT= INT( I+1)+KJW=1P0 ( 1+1)+K(JV=INT(I)+KJX=IPO( )1+K
23 RJX(JW)=RJX(JX)+(ZX(JV)*X(I)+ZX(JT)*X(I+1))*GA*DELX/2.JS=JS+1IQ=INT(I+2)+INT(I)-2*INT(1+1)
IF(IQ) 26,25,2526_DO ._Z-7.4S, J -----.------
JW=IPO tI+1)+K
JT=INT(I)+K
27 RJX(JW)=RJX(JX)+(ZX(JT)*X(I)+iZX(JV)+(YIK)-YB(1+1))*ZXYB(I+1))*X( I1+-1)_ *GA*DELX/2.
25 JQ=JQ+1DO__2J _J=J.P ___ ______
JW=IPO(1+1)+K__JX ~JPJJIIJ+ K_JT=INT(I+1)
28 RJX(JW)=RJX(JX)+((ZX(JT)+(Y(K)-YE(I))*ZXYB(I)) *X(I)+(ZX(JV)+(Y(K)i1-LU+1A*XiB~i±+1U)X A ±+1)GA*DEL X/? --.- ________
22 CONTINUE
DO 30 1=1,MPJ=IT 1±I1)-INI( ) -i
JX=IPO(I)+1RJY(JX)=O.DO 29 K=2,JS~JX=IPO(U±+K-- -- *.-.- -
JV=INT(I )+K29_RJYJX )=8J-(JX-)+(ZY.(JV-1)+ZY(JV))*GA*X(I)*DELY/2.-
JS=JS+1
--Q---- -- P. R N- - -AT E 681.18-61-0810-0- - ---
JP= pLS- L +_00 31 K=JS,JPJX = IE ( ii) + KJV= INT ( I+1)-1IE t K-.LAO ,AL 4 . .A... ---
41 RJY(JX)=RJY(JX-I)+(ZY(JV)+ZY(JV+I)+ZYYB( I)*(Y(K)-YB(I)))*GA*X(I)*
GO TO 314Q.LY J4l=R J X-_URA YZY(JV± 1-2 ZYY l *1YKiA-- ---1---YBA*2--
1*Xf( I)*DELY/2.31_COtUIE30 CONTINUE
RETURN__ED.
*- "mn. .. ii. ,=..e @.1.1 edme e -= ,- e .@=11 m-la e we eem e msir me.m ~ se = == ,am sim-,- -- --------------- ----- ----m * unne es. e ehum ,mse ui mm ~ a~a s.a - m~ i _ o -
102.----------------------------- 7 ------------------------------------------------------
1EL_0vMOD_ 0 ------ HART ---------- DATE = 68093 11/55/42 -----------
SUBROUTINE HART( IPOII-SgMPPRJXgRJYtDELX-iPY)DIMENSION FX(320),FY(320),.IPO(22)vRJX(320),RJY(320)?Y(25)COMMON FXjFY
10 _k illsFX(K)=O -- -------------------------------------------------------------------------
10 FY(K)=P*Y(K)DO 11 1=2,MPJV=IPO(I)+.lJX=IPO(1-1)+lFY(JV)=O.FX(JV)= (RJY(JX+1)-RJX(JX+I)+RJY(JV)-RJX(JV) U EULZ,,-tr-ljx+l)JT=IPO(1+1)JV=JV+lDO 12 K=JVPJT -------------------------------------------------- ----------------JX=IPO(1-1)+K-IPO(I)
FY(K)=(DELX*(RJY(JX+I)-RJY(JX-I)-RJX(JX+I)+RJX(JX-1))+2.*(FX(JX+I) -------------I-FX(JX-1)+FY(JX-1)+FY(JX+I)))/4.
12 FX(K)=(--DELX*(RJY(JX+I)+RJY(JX-I)-RJX(JX+I)-RJX(JX-I)+2.*(RJY(K)-RIJX(K)))+2.*(FX(JX+1)+FX(JX-I)+FY(JX+I)-FY(JX-1)))/4.
,11--CONTINUE --------------------------------------------------------------------------RETURN
.___END -------------------------------------------------------------------------
---------------------------------------------------------------------
------------------------------------------------------------------------------------
---------------
---------------------------------------------------------------------------------
iEL 0,MOD0 FORCES
SUBROUTINE FORCES( IPOINT,RNXRNY,RNXYDELX,DELYRJXRJY,M,M
PLAG(ABCDE)=(-2.*A+9.*-18.*C-+11.*D)/(6.*E)DIMENSION FX(20 )FY(320),IPO(22)tINT(22),RNX(175)175),RJX(320),RJY(320)COMMON FXtFYDO 10 I=1,MPJX INTTI+2JV=INT(I+1)JW=IPO(I)+RNX(JX-1)=FY(JW+1)/DELY-RJX(JW)RNXY (J ) =0.IF(I-MP) 12,10,10
L2 DO 11 K=JX,JVJW=IPO(I)+K-INT(I)RNX(K)=(FY(JW+)-FY(JW-1))/(2.*DELY)-R JX(JW)
L1 RNXY(K)=-(FX(JW+1)-FX(JW-1))/(2.*DELY)LO CONTINUE
JX=INT( 2)
JV=IPO(2)+I
JT=IPO(4)+I13 RNY I)P LAG FXJT FX(J W)FX(JV),FX(I),DELX)-RJY(
DO 14 I=2,MJX=INT(I )+IJV=INT( 1+l)DO 15 K=JX,JVJZ=IPO (I )+K-INT( I)
-I= I P 10(I -1 +IK INT I.JW=IPO(I+1)+K-INT(I)
15RNY(K)=(FX(JW)-FX(JT))/(2.*DELX)-RJY(JZ)[4 CONTINUE
JZ= IPO( MP)+1JX=I NT(TMP) +1J T I P0(MP-I)+ 1JW=IPO(MP-2)+1
1V=I(M-P-.3)+1RNY(JX)=PLAG(FX(JV),FX(JW),FX(JT),FX(JZ),DELX)-RJYRE TURNEN D
,RNY( 175) PRNXY(1
I)
(JZ)
DATE = 68094 19/57/45
104
1, MOD 1 FILL DATE = 68168 03/59/19
SUBROUTINE FILL(AUXVECRNXRNYRNXYNAINTS,MINSZ,YYB,DELXW)"DI MENS ION AUX 124649T"VECT157T, RNX (175)tRNY( 175) ,RNXY(1INS(21),Z(175),Y(25),W(175),YB(21)
DO 10 11,246490 AUX(I)=0.
DO 11 I=1,NA1 VEC(I)=0.
JX= I NT (2 2+T~--AUX(1)=-(RNX(JX)+RNY(JX))*2./SIJ=1I+NS2 *NA ---AUX( IJ)=RNX(JX)/SAUX(158)=RNYfJXT*2./SVEC I1)=-RNX( JX)*WUl)/SM=M-1 --
DO 12 I=2,MSJX=INS(I)+1IJ=JX+(JX-1)*NAJV= INTII+1)'+IAUX(IJ)=-(RNX(JV)+RNY(JV))*2./SIJ=IJ +NA...AUXI IJ)=2.*RNY(JV)/SIJ=JX+INS(I-1)*NAAUX( IJ)=RNX(JV)/S1J=JXINS(It*NA--........
2 AUX(IJ)=RNX(JV)/SMS=M-2 ~DO 13 I=1,MSJX=1NS (If2JA=INT(1+2)IF I ZIJA)YZLJA 20,21720
l JV= INS (I+1)-1GO TO 22
0 JV=INS(I+1)-22 -DO _i4 JX,~JV ----- . .-- .-..-.-- . - . . . .----.-.. ..---..--..--. -
IJ=K+(K-1)*NAJT=INT1I+1)+K-lNSAUX(IJ)=-(RNX(JT)+RNY(JT) )*2./SIJ=IJ+AAUX( IJ)=RNY(JT)/SIJ=IJ-2*NA.AUX(IJ)=RNY(JT)/S
6 VEC(K)=L-RNX(JT)*W(K)*2.+RNXY(JT)*(W(K+1)-W(K-1)) )/(2.*S)GO TO4 0-- ---- ----
5 IJ=K+(K+ INS( I-1) -INS ( I)-2) *NAAUXIIJ)=RNXY(JT)/(2.*S)IJ=I J+NAAUX( IJ)=RNX(JT) /S
'Os
L 1, MOD FILL DATE = 68168 03/59/19
IJ=IJ+AAUX ( IJ)---RNXYJTV/12.*S
0 JA=K-INS(I)+1IJ=K+(K+INS(I+1)-INS(I)-2)*NAIF(Y(JA)-Y8(I+2)) 18,18,19
8 AUX IJ) =-RNXY(JT) /2.*SIJ=IJ+NAAUX-(J)=RNXTJTT7-IJ=IJ+NAAUX(IJ)=RNXYJT/(2.*SV -GO TO 14
9 RL A= (YB(B-+2)- Y CJA-1I))-/DEuX-AUX(IJ)=-RNXY(JT)*RLA/(S*(1.+RLA))IJ=I iA+K - - - --
AUX(IJ)=(RNX(JT)-RNXY(JT)*(1.-RLA)/RLA)/S
AUX(IJ)=RNXY(JT)/(RLA*(I.+RLA)*S)4 CONTINUE -~3 CONTINUEMS=M-2 -- - - - -- - - ~
00 25 I=1,MSJA=.INT(I+2)-IFlZ(JA)-Z(JA-1)) 26,25,26
6 Jv=INSTI-1)+rJA=INS(I+1)-iNSlI)-lJR= INT{ITliT+JA ------
IF(Y(JA)-YB(1+2)) 27,28,288 RMA=(YBtI+1T-YTJATT/DEEXRNA=(YB( I+2)-Y(JA-1) )/DELXRE A=C(YB(IY-Y(J~fl/DELX -IF(YB(I-l)-Y(JA)-DELX) 500,501,501
0 RDA=(YB(I 4)-YTJAy)/oELXGO TO 502
1 RDA=T. - -
2 JT=INS(I-2)+JA-1IJ=JV+(JT-1T*NA -
AUX(IJ)=-RNXY(JR)*RDA/(U1.+RDA)*S)IJ=IJ+tNA -- --
AUX(IJ)=-RNXY(JR)*tl.-RDA)/(RDA*S)IJ= I J+NA +NAUX(IJ)=RNXY(JR)/(RDA*(1.+RDA)*S)JT=INS(I-1)+JA 2 --
IJ=JV+(JT-1)*NAAUX(IJ)TRNX(JR)*T RMA/(2.+RMAT+2.*(1. RMA-RNA)/( 2.
1)/SIJ=IJ+NAAUXIIJ)=RNXY(JR)*4.*REA/((1.+REA)*S)+RNX(JR)*(-2.*(
I(I1.+RMA)*RNA)TRNA .j*(RNA -l.-REA)*2./((1.+REA*(
-RMA)*(1.+RNA0)
1 .+RMA-RNA)/1.+RNA)*RNA)
L 1, MOD 1 FILL DATE 68168 03/59/19
2+2.*RMA/(i.+RMA))/S
AUX( IJ)=RNXY(JR)*4.*(1.-REA)/(REA*S)+RNX(JR)*2.*(REA+l.-RNA)/( REA
IJ=IJ+t\AAUX(IJ)=_RNXY(JR)*4./1REA*(1.+REA)*S)+RNXIJR)*2.*(RNA-I./(REA* -.-
It 1.+REA)*S*(1.+RNA)JT=INS VI)+JA-2IJ=JV+(JT-1)*NAAU X IJ)=RNX IJR)*(2~.*RMA/C(2.+RMA)-4.*(1.+RMA-RNA)/((RNA)*T2.+1RMA)))/S-RNY(JR)*(I.-RMA)/(2.+RA)*S)I J= IJ+NA-AUX[IJ)=-RNXY(JR)*3.*RMA/t(1.+RMA)*S)+RNY(JR)*(4.-2.*RMA)/((1.+
1RMA)*SJ+RNX(JR)T*(-4*RMA/(1e+RMA)+4.*(1.+RMA-RNA)/((1.+RMAT*RNA)24.*(RNA-1.)*(RNA-1.-RMA)/IRNA*(1.+RNA)*(1.+RMA)))/SIJ=1J+NAAUX(IJ)=-RNXY(JR)*3.*(1.-RMA)/(RMA*S)-RNY(JR)*(3.-RMA)/(RMA*S)L+RNX(JR)*4.*1RNA-1.-RMA')(RMA*(1.+RNA)*S)-IJ=IJ+NAAUX ( IJ)=RNXY (TJR) *3./(RMA*TI*+RMA)*S)RNY (JR)T*6.. (RMA*(T. RMA)1L2.+RMA)*S)+RNX(JR)*4.*(1.-RNA)/(RMA*( .+RMA)*(1.+RNA)*S)JT=INS(+1)+JA2IJ=JV+(JT-1)*NAAUX'( I J=RN Xf( JR) *"(-RMA/ (2 -RA T2.*11~--RMA-RNX)7(T(2-.VR MA) * ,+RNAT-...........
1) /S
AUX(IJ)=RNX(JR)*(2.*RMA/11.+RMA)-2.*(1.+RMA-RNA)/(RNA*I1.+RMA)))/SIJ=IJ+NA_ - - -- -
AUX(IJ)=RNX(JR)*2./(RNA*(1.+RNA)*S)GO TO~25
7 IJ=JV+(JV-3)*NA
JX=INT ( 1+2)-iAUX.J...Y.....R.A..-.T...-....-.
IJ=IJ+NAAUX(IJ)=RNYIJX) *14.-2.4*RLA)/((1.T4RLA )*S)-IJ=IJ+NAAUX(IJ)=-( 2.*RNXJX)-RNY( JX)*(3.-RLA)/RLA)/SIJ=IJ+NAAUXLTJ)=6.*RNY(JX)/IRLA* (1.+RLA )-* 12.+RLA)*S)RLA=(YB(I+2)-Y(JA))/DELXIJ=JV+(INS(I1+2)-3)*NA..-AUX(IJ)=-RNXY(JX)*RLA/U(1.+RLA)*S)
AUX(IJ)=(RNX(JX)-RNXY(JX)*(l.-RLA)/RLA)/SIJ=IJ+NA - -
AUX(IJ)=RNXY(JX)/(RLA*(i.+RLA)*S)I F (11 '141-,41,42
101
EL 1, MOD 1 FILL DATE = 68168 03/59/19
41 VE C (JV)= (-RNX (JX )*W (JV)*2.+RNXY ( JX) *( W(J V+1) -W (JV-1GO-T0-25 -
42 IJ=JV+(INS CI-1)+JA-2)*NAIF(YIJA+I)-YB(I)")W29,29,30
29 AUXIIJ)=RNXY(JX)/(2.*S)IJ=IJ+NAAUX(IJ)=RNX(JX)/SI J= IJ+NA ~~AUX(IJ)=-RNXY(JX)/(2.*S)GD To ~25-
30 RLA=tYB(l)-YIJA))/DELXAUX ( IJ)-=RNXY TJXT*RLA/ -(I1.+RL A )*S)IJ=IJ+NAAUX(IJ])(RNX(JX)+RNXYJXT*T1.-RI~AT/RLA)/IJ=IJ+NAAUX(IJ)=-RNXY-JXI/(RLA*u7;+RLA)*S
25 CONTINUEJV=INS(M)-2 - -- ~~IJ=JV+INS(M-2)*NAJX=iNTfMJ+2 *,*
AUX(IJ)=(RNXY(JX)+RNX(JX))/SIJ=IJ+2*NA ---AUX(IJ)=-RNXY( JX) /S
- IJ= JVINS(M-1T*NA -
AUX(IJ)=(-2.*RNX(JX)+RNY(JX)-RNXY(JX)) /S
AUX( IJ)=-2.*RNY( JX)/SIJ = Ij4-NA-AUX(IJ)=(RNXY(JX)+RNY(JX))/SIJ=J V +NS(M)*N A......- ..AUX(IJ)=RNX(JX)/SRMA =(Y B(M)-Y(3)/D[ELXJR=INT(M)+3JV=INS(M-1 -JT=INS(M-3)+2AJ=JV+CJT-1) *NAAUX( IJ)=-RNXY(JR)/(2.*S)IJ= IJ+NAAUX(IJ)=-RNX(JR)/SIJ=iJ+fNAAUX(IJ)=RNXY(JR)/(2.*S)JT= INS (M-2T+2IJ=JV+(JT-1)*NAAUX(IJ)=RNXYT3RT7SIJ=IJ+NAAUX(IJ)=RNXLJR)*4./SIJ=IJ+NAAUX ( IJ =-RNXY( jR ) /S
) /) )/2.*S)
J08
L 1, MOD l FILL DATE = 68168 03/59/19
JT=INS(M-1)+2IJ=JV+(JT-ITYNA ---
AUX(IJ)=RNXYIJR)/(2.*S)-IJ=IJ+NA
AUX(IJ)=-RNX(JR)*5./SIJ=1J+NA-AUX( IJ)=-RNXY(JR)/(2.*S)IJ=JV+INS(M*NAAUX(IJ)=RNY(JR)*(RMA-1.)/(t2.+RMA)*S)IJ=IJNAAUX(IJ)=RNY(JR)*(4.-2.*RMA)/f(1.+RMA)*S)-RNXY(JR)*2.*RMA/((1.+RMA)1*S)IJ=IJ+1'AAUX(IJ )=-RNYTJR)*T3.-RMA)/(RMA*S) -RNXY(JR)*2.*(l.-RMA)/(RMA*S)
1+RNX(JR)*2./SIJ=I J+NAAUX(IJ)=RNY(JR)*6./(RMA*(1.+RMA)*(2.+RMA)*S)+RNXY(JR)*2./(RMA*(l.
1+RMA)*S)RETURNEND
-----------
03
-- OD ~~~--~~~ -FIRSYS DATE 68167--~~ -1-3/51 ---
SUBROU-T-iNEFI-R--YStZ-X-X-j-ZYY i -XYAUXiVEC-,DELX X, ZXZYGATlIUNTM,1INS,YYB,INDPfPHIHDIS, ZSEC, RJXRJYTH, ZXS, ZYS, ZXYSPO, STRRJXB,~2RJYBTLL~-.----- -
GIMENSIGN ZXXfl57)tZYY1 157), ZXY(157),AUX(24649), VEC(157), X(21),ZX1t-7tr~t-7trt15ttN(2)-----NS (21 t -f-Y 25-Y-Y 8(-21I-4
2PHI(21),-30--S-1-75t-"S-L-1--5-7-)-f-ZLSEC1-75- )-ZXS-17 5)ZY S-4175-)-ZX Y S(1-75-)-CDE-20),
4FJH(20),HIJ(20),STR(175),RJXBL21), RJYB(21),RJX(175),RJYL175),TH
MP=M+
NA=157--- Att--CHANGEt(ZXTiZXS-i-NT-i-NS-iM-f -. t---- -
CALL CHANGE(ZYYZYStINT#INS,M,1.)- CALLtCHiANGE(Z YZX Y-STINT-NS--hr-T --.----------------------
DO 90 I=1,1--JX-INTi+2------------------------------- -
BA=1.+ZX( JXI**2B&8-1+ZYJXt**2BC=ZX (JX)*ZY(JX)BDAtBA+BBi-.**O5- - ---A=SINIPHI(1+1) )**28=COS-VPH IVItY*42 --------------------- - - - --- -
C=2.*SIN(PHI(1+1))*CGS(PHIII+l))- ~CaEtli tBA * -- t.~+PU *BD1*AV8 B**2*B=8C*B B*AC
FJH(I )=BA**2*A+ IBA*BB-(1.+PO)*BD)*B-BC*BA*C90~W17JI~ =i~2*BC48A*A=2.*BC*B B*Bt ~2~.4BC**2-(1. +PO )*BD)* C72
CALL FILLIAUXVECZYSZXSZXYSNAINTSMINSZYYBDELXSTR)--=Is.-1---------.---------- ----
CO 60 I=1,NS
JA=INStI +1)-INS( 1)-i-~iV=i-NS +T)-1------------- ---------- -------
DO 61 K=JX,JV
RLA=1.+ZXIJT)**2+ZY (JT )**261 VEC1~K-TVEtTfGA*RA *X1+1 XX K) *R JX I JT )+2 YY C K) 1*RJY U JT )
60 CONTINUE
DO 62 I=1,MS
JP=INT I+2)-- A=i.TZX-t-Pt**2+Z Y (-JP)**2 -__ _ ____________
62 VEC(JV)=VEC(JV)-TH(JP)*A*DISIJP)*IZXXJV)*SINIPHI(1+1))**2+ZYY(JV)-- I4CO SIPHl-ii-)-**2-2*ZX Y iJV )*S I N (PHI (I +1) )*COS PHI-( ) I 1I)+R-JYB------ --
2(I+1)*CDEII)+RJXB( I+1)*FJHII)3S=M2-- - - --- --
00 32 I=1, MS
JZ=INT( 1+21
RLA=CDEI 1)
34 IJ=JV+( INSI I)-4) *NA
-REA=tY Bt11-YJA ) D ELX
-RNA=A(Y B-Iil-2-)-YIJ-A--- ) DEL XAUXIJ)=(1.*RMA)*1.+RMA-RNA)*RLA/ (I 1.+RNA)*S)
AUXIJ)=C-(2.RMA)*(1.+RMA-RNA)/RNAI(RNA-1,)*(RNA-l.-REA)*2.-RMA)I-, -RMA/fIRNA*t1--- +RNA-)-*-(i.+REA )-)) *RL A/ S--- -__________ _ _ _ _ _ _ _
IJ=IJ+NA
I J=J+NA
IJ=JV+IINSC 1+1 )-4)*NA
IJ=1J+NA
1+RMA)/LRNA*( 1.+RNA) )) *RLAIS
AUX(IJ)=2**RNA-.-RMA)*(2.RMA)*(.+RA)*RLA/(RMA*(1.+RNA)*S)
AUX(IJP=-2,*RNA-1.)*(2.+RMA)*RLA/(RMA*t1.+RNA)*S)
AUXIJ)=11.+RMA).*i1.+RMA-RNA)*RLA/ (11.+RNA)*S)
AUX IJ =-f2,+Rt4A)*l 1.+RMA-RNA) *RLA/IRNA*S)
AUX( IJ)=1 2,+RMA)*( 1.+RMA)*RLA/IRNA*( 1.+RNA)*S)GTV2~--------------------------------
33 RM~A=(YBiIl+1)-YIJA))/DELX
IF4YB(L )-Y(JA)-DELX) 35,35,36 __
3-REAVYBt j-Y"IJAX)IIO/Df X -----.--..-.----...-.-- -. .
IFI1-1)-38#391,383 9-VE CtJV.)-; R4A*-RM A-RNk)t+RNA-)-+lRNA--R EA)*(I 1.+RMA)*RMA/( (.1--+REA)+
1 *11+RNA)))*STR(JA-)+(1.+RMA)*RNA-R4A)/RNA(REA-RNA)*RMA*(1l.+-2--RN)71 RNA*REA)-)-ST-RI-J-A-)4- 1. -- +RM A )4*RMA/ (-(-10e- +REA-) *RE A) TRJf1)*L.....-3A/S+VECI JV)
38 IJ=JV+IINSII)-3)*NA
36 REA=1 .
37 AUXI IJ)=(RMA*(RMA-RNA)/41.,+RNA)+(RNA-REA)*RMA*(1.+RMA)/( l1.+REA)*(
I J=I J+NA
-- AU-X(Li-J-)=(---+RMA-)*N-RMA) /RNA+ (REARNA)*RMA* 1-0- !RA*N))*-----
LRLA/S
AUX(IJ)=tl.+RMA)*RMA*RLA/(REA*(.+REA)*S)
AUX( IJ)=-2.*RLAIS
AUXIJ)=RMA*(iRMA-RNA)*RtAI ( (1+RNA)*S)
AUX IIJ )=(1.+RM.A)*IRI4A- RNA J*RL A/ (-RNA*S)
f -1A-
AUX( IJ)= ( 1.+RMAJ *RMA*RLA/L (1.+RNA)*RNA*S)
32-CQNT-[('dE------------------------------------- -----------------------------
JA=JV-INS(M4-1)-1-Rt-A=COETM-t:)
IJ=JV+l INSI M-4)+JA-2) *NA
AUX( IJ)=RMA*12.+RMA)#RLA/S
I J=I J+ KA
V=AUX( Li)
AUXLI J)=-4**T
AUXL IJ)=-4,P*V
AUXLIJ) =RMhA* (1.+RMA)*RLA/ (2.*S)
IJ=IJ+NA--AU'X CUJ-P)-5,e*1 ------------- ----- - - ---- - -
IJ=IJ+NA,-A U XtI-J Tw5-,*4V - _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _
IJ=JV+( INS-1 M-1)+JA-3)*NA
IAU=I J NA - - - - ~ - - -- - - - - ---------- - - -
I J=I J+NA
-AUX (IJ-);-2 , V -_ _ _ _ - _ _ _ _ _ _ _ _ _ _ _ _
IJ=JV+t INS (N )+JA-3)*NA
MS=M-1
JZ=INTI 1+2)
ELIFMOUA------------FISYS --- DATE--= -68167 --- - 1 / 1/l
RLA=FJ-( 1)
RI4A=( YB[ 1+1 )-YIJA) I /DELX
AUX( IJ)=AUX(IJ)-(1.+2.*RMA)*RLA/( (2.+RMA)*S)
AUXi IJ)=AUXIIIJ)+4.*RLA/S-I-J=J+NA-- - --- - - - ----------------- --- ---
AUX(IJ)=AUX(IJ)-13+2.*RMA)*RLA/RMA*S)
AUX( IJ)=AUX(IJ)+6,*RLA/{RMA*(2.+RMAP*S)4-7-C 0NT-IiNUE ____ __
JV=I NS(M)+1
AUX(IJ) =AUX( IJ)+l.
CO 48 1=19M.S
-J-V-NS 11 1
R LA =H I J I) 4. *S J
50 IFIYB(1-1)-Y(fJA)-DrELX) 51,51,52
IJ=JV*(INSL I-1)-3)*NA
52 RDA=1.
53 REA=-(YBtl)-Y(JA))/DELX
RNA=(YB(H+2)-Y(JA-1DIELX
IJ=IJ+NA
IJ=IJ+NA-- U~X11ljT= Auxri-J)+2.+RMA-J* (1.-+RMA) *RLA/ (R DA*- (1. O-+R DA)) -
IJ=JV+( INSI 1)-5)*NA-- 7UXVIII--=AUX (-IJ-)+ (-IY +RMA-)*RMA*RLA-/2-.------ .. l_. -__. __-_._ - I--.------
IJ=IJ+t'A-- AUXt-l V.-=AUXtV1J)---42.;+RMA)-*RMA*RL-A - -. . . .---- ----------
IJ= IJ+NA-~AtXttIJt-=AUXf-J)-4;(R-MA2;---4-.*-RE A* (t2 .*+R MA) /(1-6 +R EA) )-*RL A*(1 RA---__
IJ=IJ+NA--- AUXtP1J-=AUXI-1J-i+LRMA+4.*( 1.-R-EA)*fi 1.+RM'A) /REA) *RLA*(2.+RMA) - ..
IJ=IJ+NAtAUUJ-)=-AtJX(-J)-* 2-+RMA)*(1- RMA)*RLA/(RE A*(1 + R EA)IJ=JVNl INS 11+1 )-3 ) *NA
EL~vMOB4-----------IRSY -~ -- DATE--=--68 L67------- 1T3-5t-------------
---- A-UXtIJ)-AUXt--J)-3-*RMA*(2+RMA)*RL-A -- - _ ________
IJ= IJ +NA~-AUX1---J-)=AUXtiJ-)--3-** ( --- RM A )*(1.+RM A) * 2. +RM'A J *RLAIRMA.-
IJ= 4+ NA--- AUX-I-IJJ- "- -UXtl-J-I+3.46*2+RMA)*RLARMA ----
I J=J V+ A I NS 1+2 )-4) *NA-AJXI--J-=-AUX---J'-RMA*-1--+ RM-A)-*RL-A/ 2. _______
IJ=IJ+hA
IJ=IJ+NA
14=1 J+NA-- AU-X-L--J-)-AUX-(J-1-2--*RMA-*1-2;-+RMA) *RLA/ ( -RN-A* U-+RNAf-)--
GO TO 48
T=RMA*i.+RM'A)*RLA/2*
-WA= (2.*RMA)*t 1.+RMA)*RLA/2.
1F(YBII)-Y(JA)-DELX) 54v54,55
fF4I1U 56,57,56
57- V EC4V-=VECt-J-V-)- t-iR-MA*1-tI-.+RMA--) *4 SIR (--JA-3)---ST R-1-J - i2-R M A*(20+--IRMA)*(STR(JA-2)-STR(JA))+(2,+RMA)*(1.+RMA)*IREA*STR(JA-1)/(l.+REA
2itR- -T RtJAt/ R E -S R (J A 1) -14CR E-A* I, + R E W) I IiR LCO TO 58
GO TO059
IJ=JV+(I11S4 I -6) *NA
59- U=IJTAW~T T ___
IJ=IJ+NA
IJ=IJ+NA
IJ=[J+NA
58 IJ=JV+IINStI+2)-5)*NA
IJ=IJ+t'A
AUXt1J)-lAUX1J-)+V-WA*2.0**t 1-RNA)- /RNA----...---I J=I J+NA
EL 1, t4OO--------FIRSYS ----- DATE =68167 ~ - 17/31/51 --- --
--- AUX-1-J.AVX/--+A*2-/ ( RNA* ( j.+R NA))- -_
48 CONTINUEJZ=i NTLH+1l-JV=INSIM)JA=JV -NS-IM-1i---RLA=HIJ( M-1)/14.*S)
-fMA1-Y-BM-)-Y4Jk)t/DEL-x- --- - - __ _ -_________
T=-RMA*(2.+RMA)*RLAV(i2+R MAI*1.+ RMA) *RL A/-2.-- -- -IJ=JV+IINS ( M-4)+JA-3)*NA
tX( t-1J=AUf~ti)-1- --IJ=IJ+NAAUXtP-J+=AUXt-JJ-VIJ=IJ+NA
IJ=IJ+NA
I J=I J+ NA
AUXJI-=AUXtIJ)+ - --- -IJ=JV+t INS (M-3)+JA-3)*NA-AUX-=AUX --+2--------IJ=IJ+NAAUXtIJ=AUX (i-J4+2.*V--------------------------- - ------
I J=I J+ NAAUXVI-JT=~AUXiJ-2.~4 - - - - ----
IJ=IJ+NA-~AUXtiTWA=flrj)--z.*v
IJ=JV*+ INS (M-2)+JA-3)*NA~AUX~t J=AUXTrJ-J+T----- -- --- --
IJ=IJ+NA
AUXThJAAUX1TTV- - - _ --IJ=IJ+NA
IJ=JV+ INS(M-1)RJA--3)*NA
AUXLTTj)=AU- t21rn-------
IJ=IJ+NA
IJ=J+( -1)*157
20 READ(8) (AUX(K),K=IJJT)-~CALSYSTEM(1571-AUX-R-VM+TSOL- NA-- LLL)-
IF(IND) 112,113,112
~EV1j DL ~ FISYSDATE-=68 167 - --------3/-r- --
1J=1+L I-i)*157
21. WRITEM8 (ALX(K),K=JJTJ
GO ZSECI=STRII)
JX=INS(I)+1
CO 64 K=JXJV
64 STR(JT)=SQLIK)-JA=J V-J-X-+- -
IFIY(JA)-YB(I+1)iJ 63v65,63
STIT*1 )=STRIJT)
RETURN
E- -- - - - - - ---- -- - - _ _ _ _ _
-Ei-OA--------E -Y DAT 68 167--- - - - 17/---31/---5t --- ---
--- SUBROU-lNE-SEC-S-YS-iEPX-i-EPVs EPXYip D IS, oZ,-P H 4PH ID AU XjV EC, -DEtA-X-iN T-1INS,MtYVYBZXXIZYYtZXYEPXDXEPXDYEPYDX,EPYOY ,EPXVDXEPXYDV, ZXXX,
-- 2-X-X-YZXYY,-ZY-YY-,EPYXX,#-EPXYVVEPXYXYvO[AtKKKitIND)DIMENSION EPX(l15) ,EPY(175),EPXVIL75),DIS(175),PZ(175),PHI(21),PHID
--------- AUX-L24-6't9-)-VEC-( 157-) ,XI 21), NTL 22) ,I -NS(-21) pY2),B2),X(1
271'ZYYLL57),ZXY(I57),WXX(175),WYY(175),WXY1175)tEPYXXIL57),EPXYY(-3-1-5-7-h--E PX-YXYt-1-5-7--E PXDX (-175-)-,E P XD Y i175-)-- E-PYD X-(-l7-5-)-EPY DYI 15-)-4EPXYDX(175),EPXYOY4175),LX XX(20),ZXXY(201,LXYY(20)tZYYYL2O),PSOL(
- 5-15-7- h-OVDA-175)----00 10 I1=1,rM
JA=INT( 1+1 )*K-INS I)J
UYIY J A)= ZYY (K)
10 CONTINUE
S=DEL X**2-- -- CALL-- ILL-1-tX-,'VECWYY-iWXX,-W-XV, NA,I NT-,-SiM ,I-N-SiZiY-YB-DE-LX-,D-1IS)---------
JP=INS( I)*1
13 VECLK)=VEC(K)-EPXYYIK)+2.*EPXYXY(K)-EPYXXIK)
00 50 I=1,MS___
JP=INSI I)+1
A=SIN IPH I ( 1+1))
AC=A** 2
JR=INT(I-+1)4+INS( 1+1 )-INS(I)
1YD)X(JRh+2.*BC*B*EPXYDYCJR)-B*(l.+AC)*EPVOX(JR)+A*BC*EPYDY(JR)-PHID
IJ=J Q+1+JQ*NA-AUX- (-I-JJ-=AU X-4-1-J1-i-A C*B *ZX-XX-U I A* Q .*BCm-AC--) *Z-XXYI [)+B*t24A Bt
IZXYYII)-A*BC*ZYYYII)+PHIDtI+1)*IZXXJQ+IU)*IBC-2.*AC)+ZYY(JQ+1)*
RLA=I-AC*B*ZXX(JQ+I)-2.*AC*A*ZXY(JQII)+B*(1.+AC)*ZYY(JQ+1).)/(2.*
RNA-- Y B( I+ 1)-V (1JA) V/DEL X
117
VEV~i 0D~I-----------SECYS --- DATE 68167 1 / 15~
IF(FI-MS) 60,61t60
14 RNA~tYII+2)-YIJA-1))/DELX
IJ=JQ+1+tJS-3)*NA--- AUX-1-J=A----J-.RLA-* 1-1-+RMA) * (4-.+RMA-RNA) /i 1 * +RNA--)---.------
lj=lJ+NA-- AUX(J)-=-AUX--J -+RL-A*12. +RM A) * I (1.+R MA-RNA) I e.-*RM-A)* 1'- RNA)*URNA-----------
11 -REA) I ( 1.+REAJ*L1, +RNA) /RNA
AUXLIJP=AUX(IJJ+RLA*(2.,+RMA)*L11+RMA)*IRNA-l.-REA)/.((l.+RNA)*REA)
A4UXLIJ)=AUX(1J)-8LA*12..RMA)*(1..+RMA)*(RNA-1.)/{L1.+RNA)*REA*(1.+
IJ=JQ+1.(JQ4JA-2)*NA--AUX{--J)=AUXi-J)+RLA*(1-.+RMAJ*(l.7+RMA-RNA)/ (1.+R-NA-) ---
IJ=lJ+NA
1J~lj+NAAUX 4 I -)--AUX-IJ )RL-A*t2.fRMAI-*4 1.- +R M Ail1I RNA*- 110+ R NA-)--GO TO 20
IF(YBiI)-YtJA)J'DELX) 16916,17
GO TO 18
18 IFI-11 19,21,19g
11.+RMA)*IRNA-REA) / (41+RNA)* (1.+REA)) )+DIS(JQ)*I (1 .+RMA)*(RMA-RNA)
3*11.e-REAf)-- - -- - -------- ---- -- .---------- - - - - - - -------- - - - - - ---
19 IJ=JQ+Ii(JS-2)*NA---AUX-41IJ)=--AUX-t1J-QRLA-*RMA*14-,RMA-RNA)+4-.+RMA-)*(RNA -REA)/(1.+REA))/111.+RNA)
AUX4IJj=A1X(IJ)+RLA*1.+RMA)*4(RMA-RNA)+RMA*(RNA-REA)/REA)/RNA
AUX(IJ)=AUXL IJ)-RLA*(1.+RMA)*RMA/ 14 .+REA)*REA)
AUXI IJ )=AUX4 U )*RLA*RMA*4 RMA-RNA)/14 .+RNA)
AUXI JJ=AUX4IJ)-RLA*(1.+RMA)*(RMA-RNA)/RNA
AUX IJ)A=AUX(IIJ) +RlA*(1+RMA)*RMA/ (RNA* (1 *+RNA))
61 IJ=JQ+1+1NSIt4-4)*NA
Il8
VE~jMQA SC-SDATE---68L67---7 ----/ -1---
V=-RLA*RMA*( 2. RMA)/3,
AUX(IJ)=AUX(IJ)-2.*T
AUX UIJ)=AUXI IJ)-2.*V
AUX I J )=AUX I IJ )-2.-*WA
AUX( IJI=AUXIJ)+9.,*T
AUX( I JJ =AUXL U )+9. *V
AUXtII=AUX(IJId9.*WA
AUX4 IJ)AUX I J)- 18 .*T
AUXIIJI=AUX(IJ)-18.*V___-_____.-_ ___-
AUX(L IJ) =AUX i[IJ1- 18. *WA
AUXI I J)=AUX ILJ 1+I1.*T
AUXI IJJ=AUXI IJ)+11.*V
AUX( IJ)=AUX(1J)+II..*WA
1DELX)
I J=JQ+1+( JS-3)*NA
IJ=L J+NA
AU=IJ)NAUI)+A2.RA 2+M)/I*+M)
IJ=IJ+NA-- AUXtU1JJ-AUX(ILJ )-+RtA*f2y+1--.RMA*2 +MAJ/-RMA+4- TM*fl+M----
iRMA))
JQ=-INS(M ).I
IJ=JQ+1NS{ M-3)*NA
IJ=JQ*I NSI M-2)*NA
I J=JQ+ INS (M)*NA
19
VE--- -MSD--~-~------SEC SYS DATE =68167 7/3/51--
AAUXt1-AUXttJt*Iid+t-*R-A-ZXYY ( M) +PH ID (M+1)* ( ZXX (JQ) -2 .*Z-YYt-JQ )t)JR=INT( M+1)+1
---- VECJQ)=VEC-JQ-)+EPXYDY J Ri*2.-EPYD X (JR) -P HID (M+1)* (EPX (JR) -2. *EPYl(JR))-IFH-ND )-14 . l 10 -
110 REWIND 8
IJ=1+(I-1)*157
120 READ(8) (AUX(K),K=IJJT)1-CAL-.SYST-EM-157i-AUXVEC-SOL ,N AKKK)
IFfIND) 112,113,112113-00 421-L41, 57 - -- ___-__ __
IJ=1+ (1-1)*157
121 WRITEL8) LAUX(K)PK=IJJT)
100 DIA(I)=DIS(I)DO-30-I1,9M--JX=INStI)+13V=INStI~el- ---- - ------
DG 31 K=JXJV- 1T=1NT IT4+KT)tN NST-~ - - -
31 DISIJT)=SOL{K)
IF(Y(JA)-YB(1+1)) 30,35,3035-J11rNT-(~+t)INSAI*14-INS (I)----
DISiJT+1)=DIS(JT)3--ONTINUE---- --------------
RETURN- END.---
120
EVEL 1, MOD 1 SYSTEM DATE = 68164 01/12/27
SUBROUTINE SYSTE H( NA ,C ,X ,NALLL)DTMEN5TON-A24-649TC-(57) ,D (157) T I POS (157) , X (157)DOUBLE PRECISION DGL~ TO T,400 F)~.LUL-
5 DO 65 I=lN
00 68 J=2,NITJT+ J- 1)*NA- _ -_
IF(ABS(Q)-ABS(AIJ)) 67,68,68~ET T-=~AT1~JT68 CONTINUE
IJ=I+(J- ])*NA69 -AT TJT =A1T -JT7Q -~ - -- ~ ~ ~~- ~- ----- - - ---- -
C(I)=C I)/Q-5CN1TTUE----------~----~~-----~------
DO 10 K=1,N-- -IPn's(icrs- -~----~ ~-~~~- ~ ~--~----
10 D(K)=A(K)ALL MA XRk TN 1,JMAXFP J -____
A(1)=UMAX~TPOSTTT=TP~- ~
D(IP)=D(l)
15 A I I )=D )/A(1)U0 10OfI2,NfDO 20 J=1,N
20 D(J)=A(IA)JAUX~=I T ~~ - - ----.-----.-. ~...-DO 50 K=1,JAUX
IA=K+( I-1)*NA
D(IP ) = (K)JXP=K+1~- - -
DO 50_KL=JAUXP,NIAL KL + (K ~i NA
50 D(KL )=D(KL)-AI IAL)*A (IA)
28 II.=+(I-1)*NA
IPOS(L )=IP
Ix=l+1IFT-D6NT03T,3i0, -
31 DO 60 KM=I X,N- IAM=KM+TI-1)T*NA.-
12
-VEL 1, MOD 1 SYSTEtM DATE = 68164 01/12/27
60 A(IAM)=D(KM)/AIII)r0 0 C NTTiE- - ---- ~ _____
PR=A ( 1)~D0781 K=2, N
781 PR=PR*A(158*K-157)'WRTTET6782T PR -- -
782 FORMATI /' DETERMINANT ,E14.7//)400UO CVT0~T~=TiN ~ ~- -_______
110 D ( I )=G I)
DO 200 l=1,NM-IPETPSTT~C( I ) =D( IP)UTiTPJDT T H - -___-___-_
IJK=I+1-DO- 200 L JK---NJI=J+ (I-I )*NA2OA i~TJ=DtJ~FAT JI) *C(I- -
NN=N+(N-1)*NAXTNT-DTNT7/ATNNT .---- -------- __ _______
NAUX=N-1ITU -50 Tl NAiJ.......-...-. -- -.- - -- - ----
IZ=N-I
D1 1Z)=C(IZ)~ 0U-S F-J -- ~ ZF N~ -----.--------- - _ _ _ _ _ _ _
I Z J=I Z+ ( J-1) *N A
I Z I=I Z+( I L-1 )* NA5XTZ CT1T1T AT - 2 - -I--
LLL =2ENTUNE ND
-------------- --- -------- ---------- --- ----- ---- -- ---- --------- 1-2-24 --- ----- ----- ---- -- --
VU- -QiL-IAQCI -Q-- -JAAX A R -------------- DA-T.E-=--
5 BROU11NE MAXAR(X;la-,-LE-,-U-V-AX-,.-LS..)-.---DIMENSION X( 157)
----------------------DUMAX=X(IB)
------------------------------ -----------------------IF(IB-IE) 15920i2C
15 IBP=IB41DU 10 I=IBPIE
11 DUMAX=X(l)
10 CONTINLE20 UMAX=DUPAX
RETURN
---------------------------------------------------------------------------------------
------------------------------------------------------------------------------------
-------------------------- -------------------------------------------------------------
----------------------------------------------------------------------------------------
------------------------------------------------------------------------------------
123'I
VEL 0, MOD 0 PRI DATE = 68094 19/57/45
SUBROUTINE PRI(ZDIMNSN5ION Z (175)DO 10 I=1,MPJR= INT ( I )+1JP=INT( I+1) -L RWRITE(6,11) (Z(K
11 FORMAT(8(2X,E14.WRITE(6,T12)
12 FORMAT(/)10 CONTINUE
RETURNEND
,INTMP,LR), INT ( 22)
),K=JRJP)7))
-- -- ----- ----- ----- ----- ------.. . .- ~ ~---------- ----------- fa
S~~i~VFU1CTP ?R-Ny-t N)( Y I,7 X Z Y', SI ,THNMffD IMENSI ON RNX( 1*75) ,RNY( 175) ,RNXY(175) ,ZY(175) ,7Yf 175)IT Ht 175), TNT(
On 10toI?1MP
JV=INT( 1+1)
A = I+7 X fK )4?
C=ZX ( K BZY( K.
20.5)
12 THfK)=1.
10 CONTINUERE-TURNFND
----------------------------------------------------------- ------------------
--- aOLN 0 ------------- DAT-E-=- -0-1132-13-1 ------------
DIMENSION A(175)vB(21)tINT(22),Y(25),YB(21),Z(175)
-------------------------JV=INT(1+1)
IF(Z(-JV)-Z tV--7-- C, , -------------------
12 B(l )=A(JV-1)GO TO 10
I I JX=INT ( 1+1)- INT( I )--l
IF(I-2C) 15,16715
-15-T=A(J-V-4)GO TG 13
16 T = A j-J)L- Z-)13 B(I)=-RLA*(I.+RLA)*(2.+RLA)*T/6.4RLA,,(l.-+RLA),-",(3.+RLA)*A(JV-3)/2.
-------------21)/6.
10 ----------20 CONTINUE
REIURNEND
---------------------------------------------------------
--------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------
-----------------------------------------------------------------------------------
126
NXX (KG/M)
_-_S--_ 1-5 .o as
-0.5000COCE 05 -C.5CCOCOE 05 -C.5000COOE 05 -0.5000000E-0.5000000E C5 -0.5CC0000E 05 -C.5000COGE 05
-0.1084623E 06 -0.1082E95E C6 -0.1077589E 06 -0.1068309E-0.9038844E 05 -0.7971075E C5
-0.2176222E 06 -C.2172361E CE -0.2160516E 06 -0.2139838E401~0 7 21E 0 014 C0 6 6 SE C6
05 -0.5000000E 05
06 -0.
06 -0.
-0.3138727E 06 -0.31':7173E CE -0.3122491E EC6 -0.3101526E 06-O.2746082E 06 -C.21C6632E 06
-0.4031135E 06 -0.4028526E C6 -G.4C21004E 06 -0.4009811E 06-0.4053463E 06
-O.A8b6SC7E CC -OE87961E C6-0.4E92797E 06 -0.4907841L 06-0-.5604650E C6
1054307E 06
2108724E 06
-0.3070964E 06
-0.3998298E 06
-0.4949331E C6
-C.5000000E 05 _-0.5O00O01:0 05 -0.50000JOE 05
-. 1034290E 06 -0.1006026E 06 -0.9654231E 05
0.2064395E 06 -0.2001586E 06 -0.1905716E 06
-0.3029595E 06 -0.29739e6E C6 -0.2890037E 06
-.. 3994681E 06 -C.4G26879E 06 -0.4105166E 06
-0.5068807E 06 -0.5362512E 06 -0.5726286E 06~
-0.57213351E C6 -. 5730152E C6 -0.5761844E 06-0.7271980E 06
-0.65,48857E 06 -0.6573830E C6 -C.6663238E 06-0.8966697E C6
-0.7399536E 06 -0.7459577 C6 -0.7691414E C6
-0.83A14 2 E C6 -0.E500966E 06 -0.9066259E C6
-0.9599125E 06 -0.5561399E C(6-0.1104737E 07
-0.1161959 E0C7 -0 .1220 596E C 7 -.1379344 E 0 7
-0.1494356E 07 -0.156053iE C7 -0.1734781E 07
-0.1984796E 07 -0.2037122E C -C.2160628E
-0.262C457E C7 -C.262865IL C- -C.2645816E
-0.3327256E C7 -0.32E6917E C7 -0.3185091E:
-. 40187401G 07 -0.39514SCE C7 -0.3157617E
C7
07
G7
C7
-0.5836514E 06
-O .6886206 E 06
-0.8261836E 06
-0.1019225E 07
-0.1278012E 07
-0.6018801E 06 -C.6448746E 06 -0.7161663E 06 -0.7752569E 06
-0.7404896E 06 -0.8331339E 06 -0.9459202E_06 -0.1008835E 07
-0.9319824E- 06 -0.1077350E 07 -0.1212641E 07 -. 1242087E 07
-0.1183908E 07 -0.1364518E 07 -0.1476427E 07 -0.1434161E 07
-0.1485399E 07 -0.1656433E 07 -C._17C3875E 07 -0.1599515E 07
-0.1595642E 07 -0.1800493E
-0.1941347E 07 -0.2094966E
-0.2289813E07 -0.2361182E
-0.2650561E 07
-0.3042202E 07
-0.3475182E 07
-0.2606837E
07 -C.1917517E 07 -0. 1E96686E
07 0.2138398E 07 -0.2052996
07 -0.2323245E 07
07
07
07 -0.2485901E 07
-0.2858204E 07 -0.2646636E 07
-0.3149619 E 07
s -0.464679CE C7 -C.L,562113E C7 -0.4318214E 07 -0.3939030E 07
90 -C.5177582E 07 -C.56525E C-I -0.4816998EC7 -0.4392179E 07
95-0.5596464E:07 -C.5503234iE C? -0.5221934: C7
too -0.58940831: C7 _
jI
127
-.- (KG/M)
0 __ _ __ _ __ __ ___4..So .So__ __ _
-0.4361843E C5 -C.4 64C6CE C -C.437C940E (5 -0.4383616E 05-0 .4T 5 -C .572E (5 C.43,7E C5
-0.1211331E C6 -C.12100L4E C6 -0.1205992E C6 -0.1199029E 06-0.1092341E C6 -0.1043978E C6
-0.2143107E C6 -C.2140371E C6 -0.2130154E 06 -0.2112441E 06'b- . 18-2 J 06 -C.15701.54E C6
-0.312C344E 06 -0.3116772E C6 -C.3105964E C6 -0.3087692E 06
-0.401968CE C6 -G.4C17641E C6 -0.4ClU981E C6 -0.4004393L 06-0.4 197213E 06
-C.4876337E 06 -C.4E78162E C6 -C.4885531E 06 -0.4905541E 06-C.5803588E C6
-0.57C8706: C-.r718553E C6 -C.5753639E-0.7482812E 06
-0.6532821E C 6-G.6559166E EL -C.6653131E-0.9112860 EC6
-C.728013lE 06 -0.7441S9CE C6 -0.7679907E
-0.8326700E Cd -0.84E2589E C6 -C.9055781E
-0.9582256E C 6-C.9z741-7E C6 -0.1104308E
-0.1161552E 07 -C.1220525E C 1-C.1380361E
C0 14.6292E 07 C-156283SE C -C.1738188E
-0.1989d83E 07 -C.204255SE C7 -C.2167091E
-C.2628689E C7 -C.2637393E C? -C.2655487E
-0.3338609E C7 -0.32,8553E C7 -. 3197696E
-0.40322521: 07 -G. -96531 EE C7 -C.3772538E
-0.4661675E C-1
-0.5193332E C-
15 -0.5612424E (7
00 -C.538713EC7
-C.4577355E C7 -C.4334680E
06 -0.5835142E 06
C6
06
G,7
-0.6885527 E 06
0.8263191E 06
-0.1019835E 07
-0.1279527E 07
-0.44C4500E 05 -0.4438620E 05 -0.4497643E 05 -0.4610286E 05
-0.1188694E C6 -0.1174315E 06 -r.1154855E C6 -0.1128687E 06
-0.2086099EC6 -0.2049356E_06 -0. 1999392E 06 -C.1930r37E 06
-0.3061802E 06 -0.3028786E 06 -0.2990545E 06 -0.2957612E 06
-0.3999342E Co -C.4008010E C6 -C.4r69188E 06 -0.4237051E 06
-0.4956227E C6
-0.6030738E 06
-0. 7423476E 06
-0.9347214E 06
-0.1187649E C7
-0.1490224E C7
-0.1598665E 07 -0.1806627E C7
07 -0.1946647E
07
07
07
C7
07
-0.2298034E
-0.2662080 L
07
07
07
-0.3056910E 07
-0.3492439E 07
-0.3956273E
-C.5102651E 7 4-C.4E34279E 7C -0.4393985E
-0.21C3227E C7
-C.5094504E 06 -0.5437581E C6 -0.5929407E 06
-0.6 4 9C206E 06 -0.7279660E 06 -0.80C9566E06
-0.8392517E 06 C.96111 04E 06 -0.1C30991E 07
-0.1085229E C7 -C.1228808E 07 -0.1256419E 07
-0.1373632E 07 -0.1493803E 07 -0.1442892E 07
-0.1666392E C7 -C.1722247E 07 -C.1589018E 07
-C.1928555E 07 -0.198964E C7
-C.2-152730E C7 -0.2055991E 07
-0.2372607E C7 -G.2339240E 07
-0.2622267E 07 -C.2498190E 07
-0.2876845E C7 -0.2643428E 07
-0.316243.4E 07
07
07
-C.5511130E 07 -C.5198413E C7
I
NXY (KG/M)
Zo 2.5
0.0 -G c -C.G - C.
-- c
0.0 ____ .93446E8E C3 C.1898875E 04 G.2929000E 04G.12C4180E C5 C.15Z0197E C5
0.0 C. 3 S C76CCE C4 C.7955598E C4 0.1231440E 050 .5652780E CS C.37155CE C5
0. 0 C.91039EE C4 C.1900370E 05 0.2955720E 050.1592C52E c6 .228001E CC
0.0 0.1158G1CE C5 C.3600280E C5 0.5635240E 050.3457472E 06
0 .0 C.2S_27720E CS C. 6C23 880E0.6262707E 06
0.0 C.4s535C4CE C5 C.9397356E0.1UCC6ICE 07
0.0 C.67533C0E C5 C.1415652EC . 143 l 5 CE. C 7
0.0 C.SS%62CE C5 C.2134120E
0.0 C.1492192E C6 C.3314112E
0.235864CE CC .582704E
0.0 C .33124E E6 C.779596T7
0.0 C.5482I44E CC 0.1C92814E
0.0 C.72i3C-56E C6 0.1392164E
0.0 G. EE5E37E C6 0.1604400E
0.0 0.EE44S6CE C6 0.1691558E
0.0 C.A7C8GCE C6 G.1646092E
0.76547E4E C6 C.1 502195E
050.9515356-E 05
C5 0.15C5792E 06
06
06
06
C 6
C 7
07
C7
c 7
C 7
C.6.5887.6E C6 C.1303692E C7
-0.0 C. 1ECS6E (6 0.1005267E G7
loo 0.0
0.2334952E 06
0.3666847E 06
0.5746047E 06
0.8664256E 06
0.1228563E 07
C.1614024E 07
0.1963659E 07
0.2226374E
C.2357689E
0.2338075E
0.2176595E
0. 19C9081E
07
07
07
07
07
-0.0
0.4074925E 04
0.1723020E 05
-C.0
0.5
-- .2
-C.0 -0.C
5415348E 04 C.7083324E 04 0.9288523E 04
2310560EC5 C. 3065720E 05 0. 4111670E 05
0.4168660E 05 0.5663420E 05 C. 767825CE 05
0.8035119E_05 0.1112080E 06 0.1568472E 06
O.1379100 E 06-
0.2252632E I06
0.3653492ECL
0.5830864E 06
0.8887024E 06
0.127C713E C7
0.1686128E C7
0.2C84880E 07
0.2434347E 07
0.27C3609E 07
0.2860499E 07
0.2877113E 07
0 .1976356E 06 C. 293C827E 06
0.1C80317E 06
C.2328127E 06
0.4387224E 06
0.3385904E 06 0._5097224E 06 0.7371540E 06
0.5589436E 06 0.817944CE 06 0.1121958E 07
0.8694192E 06 0.12Q8448E C7 0.1551747E_07
C.1259496E 07 0.1640587E C7 C.1979217E 07
C.1687764E_ 07 0.2067619E C7 0.2386134E 07
C.2106579E 07
0.2486601E 07
0.2461062F
0.2902361E
07
07
C.2805504E 07
C.3C19715E C7
0.3211641E 07
ii0, 0-4oS-45 Is
0.0
129
THICKNESS (M)
0-4. S-4S to-. 5~0.933127CE CC C.'2~7026E CO C.~1921.18E CC0.93327GE CC C.S217026E CC C.3192118E 00
0.1037485E Cl C. I33463E Cl C.1C21229E C1-C. 73G927bE cc C .6-154834E cc
++++ +++++++
is 2.o
C .9009496 E 00
0.1000232E Il
0.9550372E 00 .'94902CE CC C.9546736E 00 0.9549991E 00114960E Cl C.129453E Cl
0.9260101E CC C.917562E CC 0.9594769E CC 0.99(9460E 00b.191>4814 8 C 0.1C3114E C
0.98_49643E 00 0.1C02492E Cl C.1051190b 01. 0.1126798E 010.2871340E Cl
0.l07Gc85E Cl C.1114941E Cl 0.1208912E 01 0.1333394E 010.371380'ik 01
0.1.151824E Cl C.127389E Cl 0.412990E Cl 0.158(655k 01.4257765E 01
00-1316298E 01 0. 64665E C C.1649172E0.4552734E 01
C.1481156E Cl C.167913CE CI C.1936715E
C.16c'58! Cl C.19 'i9S46E EC C.2347747E
0.19S7836E Cl 0.236561_6E Cl 0.2961653E
C.249262EC C1 C. 321934E Cl C.3808996E
0.3291852E 01 0.364934E Cl
0.443634k 01 0.5173759L Cl
0.5889995E 01 C.6524156E Cl
01
C I
01
C.4866326E CI
. 6C36824E
C. 7224116E
01
CI
0. 19329E
0.2319489.E
0.2913187E
0.3633828E
0.4616328E
,^.5622042E
0.6651014E
0.7719538E
0.8736135E 00
0.9695555
0
C.8350218E 00 0.7816107E CC 0.7C75562E 00
E 00 C.9279504E 00 0.8741138E CC
.9573680E 00
0.104C327E 01
0.1233075E Cl
0.1499914E I
0 135147E_ C
01 0.2281904ECi
01 0.2877179E Cl
01 0.3605944E CI
01 0.4427397E 0
01 0.5289078E 01
01 0.6187483E Cl
01 0.7160890E01
01 0.8184296E 01
_0.9651114E 00 C.%856083E CO
0.1116776E 01 0. 1240529E 01
0.1386966E 01 0.1638825E 01
0.1751290E Cl C.2184347E 01
C.223720E ] _
0.2857534E 01
0.3568897E O.
C.4319S50E Cl
0.5076491E 01
0.5877186E 01
0.6791150E CI
.2861360E Cl
0.8C77548E 00
C.1C343+-8E 01
0.1466335E 01-
D.2106415E 01
0.2898602E 01.
0.3708883E 01
0.3594948E01 0.4470872E 01
0.4328644E 01
.5038566k 01
5789806E
C. 67053 83E
0. 7750128E
01
01
01
0.5163542F 0 1
0.5948563E 01
0.6827621E O1
0.7809747k 01
0.8817041E Cl
'.7435316k Cl C.7871976E Cl
O.E881802E Cl C.S131e2GE Cl
C.8381701E Cl
C.9'448075E 01
0.8764175E 01
0.9712172E 01
0.9163499E 01
0.9990223k E-
C.9684966E 01
0.1G14584E C2 0.1023174E r2 0.1 36132E 02 0.1047742E
0.1118611E 02 0.111368CE C2 C.1106788E 02 0.1099312E 02
0.1198921E 02 C.11E241 E C2 C.1148997E 02
-G-.12.57548E_0
02
54
0-4o _-45 IO-~o I5
0.0 -C.99 90E-C -0.3S99997E 00 -0.8999993E-O.6399995E Cl -C. 99994E CI -G.99991E 01 -0.9999991E
0.372343s- 02 C.37C2058E C2 C.3637137E 02 0.3526118E0.1900563E 02 C.1063E C2 C.1280770E 01
.4858427E C2 C.4b836180E C2 C.4768527E 02 G.4652539E0.2882715E 02 C.1791144E C2 G.7615294E Cl
0.5484148E 02 0.5i61CUSE (2 C.5390916E C2 0.5270444EO.308E0 .T72EC2 0.1 2404 37E 02
0.589C105E C2 0.5865968E 02 C.5792482E 02 C.5666156E0.3551958E C2 C.163468CE C2
0.6159074E 02 C.6133577E (2 C.6C55884E 02 0.5922099E0.3503253E 02 C.1972823E C2
O.632535iE C2 C.62_8i66E C2 C.6215193E 02 0.6071953E0.3163919E 02 C.2269502E C2
0001
-0.1599998E Cl -0.2499997E 01 -0.3599997E 01 -0.4899996E 01
02 0.3364146E 02 0.3142E30E 02 0.2847615E 02 C.24516
02 0.4482582E C2 0.4248787E 02 0.3933665E 02 0.35038
02 0.5093501E C2 0.4849156E 02 C.4517453E 02 0.4C571
02 0.5480205E_02 0.5222466E 02 C.4870088E 02 0.43721
02 0.5724620E C2 0.5449631E 02 C.5070340E 02 0.45174
02 0.5859572E 02 0.5561569E 02 0.5141969E 02 G.44867
)98E 02
804E
1-09E
197E
491E 02
787E 02
C.64G6078E C2 C. -6";76746E C2 0.6287068E 020.2669272E 02 G.2533186E C2
0.641C390E 02 C.6378311EC2 0.6279994E 020.2769072E 02
0.634332 0C2 C.630773_6E_ C2 C.6198589E 020.298086 2 E C2
0.620856GE 02 0.6168857E C2 C.6C47440E --0.20.3169563E 02
.6C12088 EC2 C.59684t8E C2 0.5835144E C2
0.5766728L U2 C.57211E C2 C.5577322E 02
0.5492949E 02 0.54-4620E C2 C.5296498E 02
L.5213'i9&E C2 L.5e165C97E C2 C.5C174L9E.02
0.9i'96815E C2 C.49C0216E C2 C.4759494E 02
0.6131615E 02
0.6 108809 E
3.6008188E
0.5835498E
02
02
02
0.5601743E 02
C.5326880E 02
0.5038443E 02
0.4764713E 02
0.4524216E 02
L
80
S
02
02
02
02
0.5899570EC2 0.5569231E 02 C.5083875E 02 C.4269987E 02
0.5851205E C2 0.5475923E 02 C.49CC9C6E C2 0.3933850F 02
0.5718845E C2 0.5287524E 02 0.4620654E 02 0.3569394E 32
0.5510150E (2 0.5021794E 02
0 .5 242400E E C2. 0. 471378 8E 02-
0.4292397E 02
0. 39 79102E
0.4946132E 02 C.4412521E 02 C.3734944E
3.4657617E 02 0.4155989E 02 0.3613733E
0.4406636E 02 C.396C826E 02 0.3722284E
0.4202324E C2 0.3811119E 02 C.3811119E
0.4036536E 02
02
C2
02
02
C2-
0.3256554E 02
C.333765DE 02
0.3485574E 02
0.3880000E 02
0.3928529E C2
0.3956215E 02
S '-----*------- ~ - *1~0.15864E3E C2 C.5905757E C3 0.59810
-ENI UT IEATNLAS SUR5 F1NAL SHAPE
23E 03
V
C .2686333E-C 1
15o
02
02
02
C.L7C324tE C2 0. 660219E 02 C.4531523E C2 0.4320403E
0.4485252E 02 C.A4466S6E C2 0.4332C56E C2 0.41453C6E
0.4290126E 02 0.42 5566 (2 C.4154744E C2 0.3988997E
0.ill2646E C2 C.4CE2629E C2 0.3992531E 02 0.3962489E
G.39L6690E G2 0 .96690E (2
Z.(M.)
----------
0-4. _..4_ _.,____
E*LAMBDA (KGIM)++++++++++++4++ 4-
0.3263484E C7 0.3228229E C7 C.3122151E 07 0.2944469EO717935 06-.39i926-6 EC6 6-C.2206767E C7 -0.2206767E
-C .15C1688E C8 -C.15C4734E C8 -0.1505655E± C8 -0.1504392E-0.1404618E C8 -C.1233699E L8-0.8713174E 07
-C.2641592E C8 -C.266013CE C8 -0.2664357E G8 -0.2652674EUU.2976tE 08 -. 5232761 C -. 109165 8
-0.34714,66E C8 -C.34L09C6E C8 -C.3466643E (8 -0.3420578E-0.2128806E± 08 -0.1118266E C8 -0.8112317E (7
-C.4C8C672E 08 -0.4C72082E CE -C.4009846E C8
0707
0.2693199E 07
08 -0.1500488E 08
08
08
-C.3893875E 08-0.249A466E C8 -C.2E12261E Ce
-0.4481942E 08 -C.44A35;63E C8 -C.4305682E 08 -0.4119438E 08-0.2429064E 08 -0.1072343E C
-0.4727270E EC8 -G.4620053E C8 -C.4433902E C8 -0.4197038E 08-0.2069955E C8 -C.1017416E CE
-0.4876915E C8 -0.4706133E CE -C.4481050E C8 -0.4201171E 08-0.1885350E 08 -0.1794685E C8
-0.4983768E CE -C.4762856E C8 -C.4500408E 08 -0.4171296E 08-0.1307456E 08
-0.509137'9E C8 -C.48229C6E CC -0.4519547E C8 -0.4139195E 08-0.1713107E 08
-0.5231931E CE -(..4E94174E C8 -0.45637501E C8 -0.4152899E 08-0.2310608E 08
-0.54061701E C8 -0.4976896E CC -0.4663262E C8 -0.4288035E 08
-0.5542578E± C8 -C.5C57E51E CE -C.4845666E 08 -0.4635826E 08
-0.5472918E 08 -0.5097970E CE -C.5128011E (,8 -0.5212286E 08
-0.49961841E C8 -C.506C358E CE -0.5489232E C8 -0.5761376E 08
-0.4083691E CE -C.492413E CE -C.5850480E 08 -0.5752091E 08
-0.3031594E 08 -0.4480165E CE -0.6000845E C8 -0.4879622E 08
-0.2244565E_08 -C.3 7423CE C8 -0.5528914E C8 -0.3771893E 08
-0.1852696E 08 -C.2E76832E C8 -G.4171216 C8 -0.3181042E 08
-0.1668402E C8 -C.1894416E C -C.2176354E 08 -0.2207994E1 08
-0.1436338E± C8-C.1I 3633 EE CC
-0.2620800E 08
-0.3332718E 08
-0.3724096E 08
-0.3883008E 08
-0.3907776E C8
-0.3850237E 08
-0.3727766-E-08
-0.3541691E C8
-0.3278235E C8
-0.2903698E C6
'5
-0.3150880E 08 -0.2666488EC8 -0.2578914E
-0.2740550E 08 -0.2318811E C8 -0.2467376E
-0.216C254E 08 -0.1967786E 08 -0.2280238E
-C.142 1107E
-0.239257,O0E (,8-0.6762424E
-0.1844237E C8 -- 0.1751477E
08 -0.1691830E
C7 -0.1237685E
07 -0.2719733E
08
08
08
08 -0.2198810E 08
C8 -0.22011'21 E08
08
-0.1645243E C8 -0.1268709E 07 -0.2351293E 08
0.1927466E C8 -0.2933171E 08 --0. 2933171E C8
-0.3157518E C8
-0.2885427E 08
-0.2631264E C8
-0.2720123E 083S.
100
0.2364629E 07 0.1950086E 07 0.1428127E 07
-C.1492998E 08 -0.1479738E 08 -0.1455252E 08
-0.2560853E± C8 -C.2459382E 08 -0.2294475E 08
-0.3191387E 08 -C.2981347E 08 -0.2673205E 08
-0.3500266E 08 -0.3222739E C8 -0.2887442E 08
-0.3599064E 08 -0.3270312E 08 -0.2892509E 08
-0.3562576E 08 -0.3166922E 08 -0.2771178E 08 _
-0.3417138E 08 -0.2%61922E 08 -0.2723992E 08
Grt of z
Cuts throupIh horizont-L. pLaine-s
Scale:
~c~o ,".
K1
W-Vwmff_ - a f I PO u-r
'0 3 i
*140
1
2o
.so
Graphs of Z
Cuts throujk ver tic0 plLanes
So 30 ZoI I I _____________________ ____________________
4P 0V.
4.
(30
cot
IyScale : Soo
zf" C' 7o Go 1o
'C0
"II
to
20
30
~4
.50
Go
70
ft0So
90
loo
X ( rn)
........ ....
" 7f 2A 3 - 39 i i 30E i
Vz k N N pt ftI
14
0
i
Soo
"a
Soo
-I -
-0
Ss thfr*tA5 vc-rht l pnes
J,
S.- .. . ... . . ... ... . I -
10 r.
CLts throu9 k verettc.o tonesj =o JA3 m.
Scale
I I
Mae
Soo
ix
Cuts hrou 5 vertcnes
40 m.
tx
la3ro-ris of N.,, for diferent
vdues J
SooeR4
Fl
lN
N
N -
N
Zo
30-
So
6o
7o
8o/
NK
I
Kv
N.
N
/
loo
K
7I
)
t:(M)
iv i - qc i
i i i IV i i -4. i %
i -OF
qii;;;;< 44
-7-
f
I
Da.agrbams of No.N , eoc different
vaLues of
10' K3/i
77,
7-7--,
~a.
//
<79,
i i i i
i t v
121b
l o
//I
I/
//
3o
4o
(0
70
90
to
Ix (,h)
0
4.
b-,
A
14
29 i 20EFlip- -- IF I-r
Z- I - Az -- Gm0AF ir---lt-
IA4
Od-I I
Bibliography
I. "New shapes for arch dams", Laginha Serafim, Civil
Engineering ASCE, November 1966.
II. "Cartesian Formulation of membrane theory", Z.N .Elias
.Journal of the Engineering Mechanics Division, Pro-
ceedings ASCE, vol.93, I1" EM2, April 1967.
III. "Thin shell theory in Cartesian Coordinates", MIT
Technical Report, Civil Engineering Department,
R67-45, September 1967. Z.M.Elias.
IV. "Resistance des Materiaux, tome II", C.Massonnet,
Sciences et Lettres, Liege.
V. "Variational Considerations for elastic Beams and
Shells", E. Reissner, Journal of the Engineering
Mechanics Division, Proceedings ASCE, vol.90,
N EMI, February 1962.
VI. "Non linear partial differential equations in Engi-
neering", W,F,Ames, Academic Press, New-York 1965
VII. "A first course in Numerical Analysis", A. Ralston,
Mc.Graw Hill Book company.
VIII."Engineering Analysis", S.H. Crandall, Mc.G-aw Hill
Book company, 1965
IX. "Theory of Arch Dams", J.R. Rydzewski, Pergamon Press,1965
146.
X. "Effect of Foundation Movements on the Stresses
in Arch Dams", J.L. Serafim, P.J Pahl, Z.M. Elias,
MIT Technical Report, Civil Engineering Departe-
ment, R65-57, November 1965.
![Company [1970] - Trumpet 3](https://img.pdfslide.us/doc/110x75/55cf8530550346484b8ba49a/company-1970-trumpet-3.jpg)
