I
OF
OPTIMIZATION
SHAPE
ARCH
DAMS
by
JOSEPH
DEPREZ
LEON
Constructions, Tniversite
des
Civil
de
ieege (1964)
Ingenieur
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 Author
Department of Civil Engine ering, June 24, 1968'
Certified by
Thesis Supervisor
Accepted by
ChairmanDepartmental Committee on Graduate Students
Archiveg
MAY 1 3 1970
4
LI&RARI.
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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 the
degree of Master of Science.
The following work intends to set up a method of
rational determination of arch dams shapes and optimal
use of the concrete.
The method is based on the theory of membranes in
cartesian coordinates, in which the equilibrium equation,
from a design point of view, can be considered as an equation
in the shape Z, provided the stress function F is already
determined. The basic hypothesis for the determination of
F is to write the equality of axial stress resultants in
the horizontal and vertical directions, which yields an
hyperbolic partial differential equation in F. Then, tvwo
elliptic differential equations (equilibrium and compatibility) with their boundary conditions are solved for the
shape Z and the normal displacement .
The whole problem is solved by use of finite difference methods. Because of computer storage limitation, a
method of block relaxation must be used for the two
elliptic equations, but it has not been possible to achieve
convergence. However, a/simplified problem with easier
boundary conditions yields the conclusion that the basic
good enough to provide usable
hypoth.-esis for F is not
shapes for arch dams. The corresponding stress resultants
are too small and require high curvatures; moreover, they
yield values such that the problem is no longer mathematically well-defined.
From all these results, the fina] conclusion is that
the problem ought to be studied on a different basis,where
the determination of F should take into account the equilibrium equation in the direction normal to the membrane.
Thesis Supervisor:
Title:
Assistant Professor,
Z.P.Elias
Department of Civil Engineering
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
Page
T.
II.
III.
IV.
V.
Title page
II
III
IV
Abstract
Acknowledgment
Table of contents
Body of text
1
F o reword. Introduction
Chapter I. Membrane theory in cartesian
coordinates
1.
2.
3.
4.
Differential equation of equilibrium
Compatibility equation
Boundary conditions
Principal stresses
Chapter II. Method of determination of the
membrane shape
1.
2.
3.
4.
5.
6.
Introduction
Determination of the stress function F
Determination of the thickness h
Determination of the shape Z
Summary of the iterative method
Remarks
Chapter ITTI.
1.
Computation of derivatives by
means of finite differences
3
3
8
9
12
16
16
16
20
21
26
26
29
29
Introduction
29
2. Determination of Zx
3.
Determination of Z,
4.
Determination of Z,,
34
T
35
5. Determination of Z,39
6. Determination of Zxy
7. Computation of Z,
8. Computation of Z,
1
41
46
Z,
Z,
9. Conclusion
Chapter IV. Determination of the stress resul tants
1. Determination of the fictive surface outside
the domain of the dam
2. Characteristic lines of the hyperbolic equatiom
for F
49
51
52
52
53
V.
3. Computation of Jx and Jy
4. Determination of F by the method of characteristics
Chapter V.
Determrrination of Z and A
1. Determination of Z
2. Determination of ,\
Chapter VI. Programming considerations
1. Storage of arrays
2. Purpose of subroutines
Chapter VII. Results and conclusions
1.
2.
3.
4.
5.
Boundary curve. Dimensions
General problem
Simplified problem
Determination of F for a given shape
Final conclusion
54
55
59
59
62
64
64
64
66
66
66
68
70
72
Figures
73
Appendix I. Determination of Z for the general
problem: 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 otherwise would have been unusable (finite differences, finite 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 empirical basis: comparison with previous structures, use
of approximating formules. In this point of view, the
above methods of analysis appear like an accurate verification of a structure whose design has been done very
roughly.
In a recent past, under the impulsion of Mr. Serafim, 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 thickness 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 downstream side with water pressure. Such loads would inflate 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 needed 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 concrete.
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 state of stress is very close to the actual state of stress
in the main part of arch dams, the membrane shape determined 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 statical 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 developed 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).
i, i, k unit vectors along the x,y,z
Defining
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 vectors 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
(1-2)
= j + Z,y k
(1-3)
=
,
r
= ryx
~y
=-z~i
- Z,,j + k
(1-4)
A comma indicates differentiation with regard to
the variables following it.
We let t , T2 , n denote the unit vectors of
this local reference frame, writ ing
r,
=
t2
rqv =
2 T2
(1-5)
(1-6)
(1-7)
n
with
2 = r,
1
S2
2
'12
.
r,
= 1
*
=
(1-8)
1 + Z x
+ Z
(1-9)
z + .z,
vx
x *
=
'y.
2 + Z2
Px
py
=1
The angle
0
(1-10)
y
_I 2 0(2
1
2
2
12
(1-11)
between the coordinate lines is such
that
412 = 1 2 cosO
o( =c'2
sine
(1-12)
(1-13)
5.
In this membrane structure, the internal stress
such that
components reduce to Nil, N 2 2 , N 1 2 and
(fig.2)
we can write
N 1 =Nil
U
+ N1 2
t2
N2 = N2 1
t1 + N 2 2
2
(1-14)
(1-15)
Considering next the stress resultants N
NY per unit length of the x-y plane, such that
and
(fig.2):
dy
(1-16)
N2 o(dx = N7 dx
Nd2 1
n y
(1-17)
N1 o 2 dy = N
and defining
+ N
r
Nx = N
xx
'
NYyNy='x
= N vXrl
r
y
(1-16')
'
(1-17')
yy
-7y
we get:
xY ,
0(2
or
1
N 11
Nxx
(1-18)
2 c<i
Similarly, it can be established that
N1
=
N
xy
N
yx
=
2
c<2
N21
1
01
=
N1 2
N12
(1-19)
N21
(1-20)
N
N
yy
=<
1
22
(2
(1-21)
6.
The equilibrium equations of the small element
ABCD , whose projection on the x-y plane is abcd ,can
be written
with (1-16),
+ N ,
Nxp
1
+ (N2 41
(Ti 12)
N'px + N ,
(1-5),
(1-17),
+ pin
+
p
= 0
1f
or
)x(c2 F 2 )1 = 0
(1-6) and (1-7) :
or
(1-22)
=0
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
j
where
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
They can hardly be computed accurately using
plots).
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:
,
-+ xNn = 0
N
+N5
and
Taking the scalar product of
rn yields:
(1-23)
(1-23)
by i,
j
7.
N
N
N
+ N
xxx
+ N
xyx
+
yx,y
z
xo
'
C
z
+ Yx
yy9 y
Ok
= 0
(1 -24)
0
(1-25)
y
xx Z,xx + Nxy Z,yx + Nyx Z,'xy + Nyy
yy
=2rd
(1-26)
The monent equilibrium in membrane theory yields
N1 2 = N
,
21
Hence
Finally,
N
+ N
xx,x
yx
equations (24)
+
xy,y
N
to (26)
can be written:
Sx Z 'x =0
(1-28)
=0
(1-29)
+6 x Z9y
NSxy. x + N9
yyy
(see (1-19) and (1-20) ),
N xx Z, xx + 2 N xy Z'xy
+ N
yy
Z
'yy
=
x,2
(1-30)
If we define
J x=
dX
(1-31)
IX Zy(XX)dX
(1-32)
X z\
Z
,)
0
Jy =
0
it can be seen that equations (1-28)
and (1-29)
are
satisfied by letting :
xx
N
N
yy
xy
Fyy
= F
xx
-J x
- J
== N
yx
(1-33
-34)
y
F
'xy
Finally, substituting in (1-30),
tions reduce to :
(1-35)
the equilibrium equa-
8.
,yy
- 2F,
Z,r
Z,
Z,
= x+
Z,
+
JX+ Z,
(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,
x'yy
yy
yy + Z9y
'yy
+E
2Z
'XX
y
Xyxy
'x
xxtyy
= 0
Where
E
to N
, N
xylxy
(1-32)
, E
and E
,
1 (E
xx -o(Eh
xE2
1
and
N
are strain quantities related
by the formulos :
+ o(2 Nyy + 2c(12N
0/ 1N
-(1+9) o(2 N
(1-33)
y
y2
yy =Eh
xy
Eh
2(Nxx+
12
2
2
+
x+
2
2 Nyy
+201Nl
l2xy
+2v(1 2 NxY]
-
(1+'X
2
N
4)
(1-34)
+ (1+,)(2 Nxy
(1-35)
E being Young's modulus, V Poisson's ratio and >, a
Lagrange multiplier Which can be shown to be identical
to the component of the displacement vector along the
z
axis.
J
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 monolithically 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, expressing 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 thickness, 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 displacement ), 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 component 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 conditions 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
+
ssle
+ X (
szn
- s (.nn
+XZ,
Z,,
)
=
0
'n
)
=
0
(1-38)
- 2
LEns
+j\(Zns
-'zs2s
(1-39)
]s
10 a.
where
and Enn are strains defined with regard to
E
s,
tangential and normal directions to the boundary, and are
by :
and E
, E
related to E
k
Cos22
E ss
Ex
sin2{ -2
En=
E
( cos
E
=
cos+ sir{
+ 2 E
2
E xy cos{> sinT
i
2
) +( E
+ Eyy sin 2J
Cos2
+T
(1-40)
(1-41)
- 6 , ) sn
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 );
),n are the derivatives with regard to
), and (
(
the tangential and the normal directions to the boundary curve of any quantity ( ) .
Using transformation of variables theorems, we
can find :
(
)s
= - sin § (
(
),n
= cos i(
),
+ cos J (
), x + sin
(1-43)
),y
( ),y
(1-44)
This leads to :
Z9ss = ( -sin§ Z,
= -cosZ,
+ cos
.Z,
Z, s=
-_fs
)
+ cost Z'y ),S
.
,s
+cos
-sink . Z'y .
4,s
( -sinZyx + cos
. Z9n + (sin 2 + z,
-sink (-sin4 Z,n
)
.Z,
+ cos 2+ Z,
or
-2si4cosiZ,
(1-45)
Similarly, we can establish:
Z ns = C>,s . Z '5
+ [sin<
+ (cos 2{ _sin 2c))
Z'nn = cos 2
Z,
cos4 ( Z, yy
-Zxx
Z, j
(1-46)
+ sin 2 1 Zyy + 2sini cosi Z,
(1-47)
(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 simplificaSubstituting in
tions:
)
12.
(6
+,
Xz, ) sin2 § -2(E
E
+ X
-sin 2
-2cos 3
+,
Z,
cos
) cos 2
+
6 sin4 cosEy
-
sin 24) Z,
-2 sin3
2
Cs
Z,
+X
Z,
+A,\4.
+ cos
,
sin I9 ( 1+
-
sin
cos 21 z,,]
-
(sin2
-
2c s 2 §)
+ sin+( 2cos 2
cos 2 < ) Z,
2sin 2 ) + Z,
-
sin~cos 2
-
cosj Z,
-sin2
( cos 2
+
4.
-2sin3 §.<i
Y
- 2sin 2 ) +E
+ cost ( 2sin 2 1)
zx
+ 6 sins
(1-48)
'
+ cos§ (1+sin 2 j)
[XX (cos
cos I +
sin
+ sinl(1+cos 2 §)
i.,
t.G
+,s
=
)
+X Z,
- s)
-
sincos 2 ( Z,
( sin 2 ( - 2 cs
cos
2
Z
( 1 + sin2)) Zyy
cos 2 ) ZO
=
- 2 cos 3 4 Zq.
0
(1-49)
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
, 2 2 , N1 2
1
and N21 are defined with regard to non-orthogonal directions and cannot be used directly to determine the principal stresses.
Let us consider (fig.4a ) a small parallelogram
ABCD whose projection on the x-y plane is the small rectangle
abcd
of lengths
dx
and
dy , with its stress
13.
resultants
Ni
, N22 , N12 and N 2 1
i
and T 2*
vectors
,
referred to unit
At the same point, we consider a small rectangle
ABEF
submitted to stress resultants
N21 (Fig.4b)
and
N 1' ,
referred to unit vectors
N22 ,N1 1 2
tg' and ti
We have obviously:
t 2 =i
=
(1-50)
2
t, sin
(1-51)
+ Tv cos&
writing the equality of force vectors on
N1 1 T
+ N1 2 T
2
N
=
t
+ N
CD
,
and
;
we have:
, orusing (1-50)
2
and (1-51)
N1 1 sine
ti
+
(
=
N11 cosO + N 1 2
Ni
I2 I
ti + N1 2 t 2
This yields
1
1 = Ni sin 0
1
(1-52)
(1-53)
N1 2 = N 11 cos9 + N
12
Writing now the equilibrium of the small triangle
ADF (fig. 5), we have :
+
|I TI.
(N22
TF
(-N
2
I
I sin 0
I1Kf
2 + N2 1 T1 )
It
- N2 1 ti )
+ jFI I.
=
and
0
(N 11
+ N1
, or, since
2
T2
)
14.
= I1II Cos
(IT
( N21
+N
11
,
(1-50) and (1-51)
using again
we get
sing + N1 1 sinO cosO ) tt
+ ( N 22 + N
cos 0 :
21
cos 2 O + N 12 cosO )t= N21 sin0
sinO
+ N2
2
t
which yields :
N2 1 = N 2 1
0
+ N 11 Cos
1
N 2 2 = sin
N2
2
=
(1-54)
N1 2
+ ( N 1 2 + N 2 1 ) cosO + N
11
cos 2
o
(1-55)
Calling Ce
in terms of
h-ce =
N1
the principal stresses, we obtain easily,
t
t
N 1 - ,0 N22 , and N 1 2
2+ N2 2
i
1
1 -N
+4 N 1
2
22
2
I
(1-55')
Using
(1-52)
Ni
h Ge
+- 1
to (1-55)
, we have
sin2 0+ N 2 + 2 N12
2
2 sin)
N1 1
sin 2
- N22 -
2N 1 2 cosO - Ni1
cos 2 o )2
sins2
+ 4 ( N1 2 + N11
Using formulas (1-18)
get:
cosO + N
cos 0
11
cos 0)2
to (1-21),
(1-12)
and (1-13)
, we
15.
2h, Ge.
sine =
0
N
N
2 2
~ 12)
2
C<(1
+ 4
multiplying both sides by
e.( = N
-2
1
xx
2h.
+ N
12
Cc1 < 2
I
I,
)
2
C 2
1 C2 yields
+ 2N
2
12
2
2
+
1
-2N XY
a12
xy 0 /1 -c ~2
2
1< 2
a1C~2
Nxy + N
+ 2N
2
-N
yy
x
1
2
2
-N
\(Nxx
yy
- 2Nxy (12
2
2
+4
N
xx
+12
2
2
)
2
2
or, using (1-11)
2h.
tf
e.c
=
N
+ 4 o120 2
Nxx.0(1
+ N
(
Nxy Nxx
+ N
2
+ 2N
+ 4N2xy o( 1
122
C12
2NXX N
2yy
N
N
( Co 2
(1-56)
2
012
)
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 membrane.
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 function
be 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 nonlinear boundary conditions.
II.
]Determination of the stress function F.
In an optimization point of view, the stress function 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 structural 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)
b)
We can consider concrete as an isotropic material;
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
N1 2
=N
(2-1)
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
=
N 2 1 will remain
small will have to be checked from theshape which will
obtain finally; more precisely, in order to avoid ten-
'<IN
sion, we will have to impose IN 1
I
(see Mohr's
circle, fig.6).
Using (1-52)
N11
11
(
sina =s
2
N
and (1-55),
- 2
1
in
=
. 2
2
2N
22
1(2
xxC42
2
22
+
112
12
+ N 11
12
.1
*/-2
2
2
12
1
and using (1-18)
2\
2
1'
12), =1
cl
K2
<2
1
N
yy + 2Nxy
or
Nxx (c< 2
yields:
N 2 2 + 2N1 2 cosO + N1 1 cos2e) or
or, multiplying by a
N
condition (2-1)
=
4 N
2
In terms of F, we get:
yy
+ 2N
xy
12
C12
2
2
to (1-21)
c2
12
1
1
:
2
19
(F
,yy
_2
.2
- J )
x
(FJ
,xx
12
y
*4
2
-2F
,xy
C4 < (2
12 2
or
4
,xx
-
J
2
(o2
2
,yy
~
2
12)
,xy
12
B2
_4
2
_c44
2
y
2
2
(2-3)
2)
This partial differential equation for
hyperbolic, since its discriminant is
AC
2
2)
12
F is always
2 )2
/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 projected 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 continuous up to the second derivative, or the shape function of the membrane be continuous up to the third.derivative.
This increased difficulty in the definition of
the fictive shape, as well as the need to define a nonregular 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
N
= N
xx
yy'
F
,xx
- F
F ,i.e
or
=
,yy
J
y
- J
(2-7),
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 practical 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 allowable stress in concrete will be limited to a small fraction of the total strength
R
nbr
i.e.
with
5
Rbr
b
'4
n
-
10 , for instance.
Using equation (1-56) with the minus sign yields:
h
n
N
2Rbo
br
+ N2
IV.
1
[1xx
+ 4N 2
<2
-
o 12 021 N xy N xx + 4
o
yy
+ 4
2 + N
o44
2
xy
C2
1
2
2 + 2N
yy
2N
2
xy
N
2
xx yy
12
_
N2
xx
4
2
12)
N
02 N
2 xy yy
Determination of the shape
-
12
(2-8)
Z.
As it was seen in chapter I, the determination of
the shape will require the resolution of the two nonlinear 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
- N2
N
N
xx yy
xy
1
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 .
For equation (1-32) considered as an equation in X
b)
Z,
- z 2
the discriminant is AC - B2 = Z
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 X
along 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 condition 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 related 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 presence of the edge beam of the top, and allows non negligible inextensional bending to occur.
We prefer to use the other alternative:
We prescribe a condition which is compatible with
the presence of the edge beam and involves negligible
inextensional bending by imposing that the tangential
b)
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. 1 5.
We have:
(2-10)
yy = U2,y - Z'yy uz
where u and u 2 are defined by
u = u
vector.
u = u
i + U2
j+
uz rn
,
u being the displacement
If we write also :
rx
+ u 2 r,
+ un rn
,
we have the relation
24.
( ref. III, p.10 ).
U2
U
u
12 U1
+ 0422 '
+ C 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:
(2-12)
(-2
S = (ei12 u~)
1 '9y -- Z,9yy
yy
2= 0 at y = 0 and
Finally, if we notice that
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
have negligible derivative,
dary condition for
-
X
to remain small and
We get finally the boun-
:
E y(2-13)
'yy
It should be noticed that this condition is not contradictory with the discussion developed in paragraph
I-3 .
In fact, the final shape and its normal displa-
cement X will satisfy both
the top of the dam.
N
= ( and
u2 = 0
at
We can even conclude that this
approach eliminates the choice between a completely
free or a fully restrained top edge, by satisfying simulteneously conditions relative to these alternatives.
Now, because of the complexity of the equations
25.
at hand, an analytic approach to the solution is unthinkable. We must devise a numerical solution and the
way in which the problem is formulated suggests obviously the use of finite-difference techniques.
However. we must notice that there are two unknowns 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 linearized 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 differential 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 conditions would be small enough to guarantee convergence.
Assuming that a starting value of Xis known
along the abutments, we can solve the differential equation of equilibrium (1-31) for Z, with boundary conditions (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 equation of compatibility
and (2-13).
(1-32)
with boundary conditions(1-40)
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)
b)
c)
d)
e)
f)
g)
Start from an arbitrary surface Without any knowledge of the bolution, a good starting shape can be
close to actual dam shapes.
From this shape, compute J and J .
y
.x
Integrate the hyperbolic equation (2-7), with boundary conditions (2-4) and (2;-5).
N
, N
, and N .xy
Determine the thickness from formula (2-8).
Compute the strains E
xx ' E yy and E xy by formulas (1-33) to (1-35).
Determine initial values of X on the valley slopes,
From
F(x,y), compute
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
i)
Z.
Determine a new distribution of thickness and of
strains, without changing stress resultants.
j)
Solve equation (1-32) with boundary conditions
1)
(1-49) and (2-13) for X.
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 procedure, skipping steps e), f),
n)
and using the
last values of X as new starting value.
Do the above cycle until the whole procedure converges towards a final shape.
VI.
and g)
Remarks.
27.
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 starting shape), the applicability of the method rests
upon two important assumptions, i.e.:
I. The stress function will be such that INxy(4Nxxi
in order for the problem to be mathematically welldefined.
2. The final shape will not be too different from actual
shapes. More precisely, its curvatures should remain 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 practical solution.
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.
b)
of this problem, it
If we have found a solution Z
is obvious that any other function Z + ax , with a
arbitrary, will also satisfy the equations. We can specify 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
value of
Z = Zb, where
Zb is the prescribed
Z at the bottom.
We have to define a criterion in order to determine when the iteration can be stopped. This will be
c)
done in the following way.
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
S =
E(Z)
and
22
- ZI
where the summation is done
for all points of the mesh where the value of Z is
R
zi+1
Computed. The iteration will be stopped if R/S becomes smaller than a chosen convergence tolerance.
d)
An easier problem could be to determine a shape
which satisfies the differential equation of equilibrium and where the stress resultants are such that
Nyy
x=
Ny
, 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 compatibility 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 convergence.
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 difference methods, we must define their coefficients at
the nodes of the mesh where the finite difference equations will be written. Besides, with the shape continuously varying, an analytical computation of these derivatives 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 program to compute all the quantities appearing in the partial differential equations.
2.
Determination of
Z,
.
We shall have to develop differet formulas according to the position of the point where the derivative
is computed.
In order to be consistent, we shall select 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 parallel to the y-axis.
This requires a somewhat different
approach for the determination of derivatives of boundary points, compared to generally used formulas.
30,
a)
Interior node
(fig.7)
Using the well-known central difference formula,
we write:
(Z,
)
- Zi-1,j) /
= (Zi+1,
(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
with the notation of figure 8.
yi+1,b ii,j
Using a Lagrange interpolation formula through
points (i,j-2) , (i,j-1) and A, we can write:
1'.-n
Z,x(ij)
1+n
Zx(ij-2)
x(A) +
(+n)
+
2(1-n)
~ n
z~
0(k
3
Z,x
_j)
(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
-
(2+m)(1+m-n')
n
Z'
'x(
Z~x(ijj-)
.
+ 0(k 3 )
jj-1 )
(2+m)(1+m) 'X
Z~(A)
+ n(1+n)
(3-3)
Ztx(i,j-2) and Z 'x(i,j-1) have already been computed
formula(3-1); Zx(A) will be computed
from
31.
from
1
2h
Z,x(A)
Z(i+lb) - z
) + O(h2)
(B)
, where
Z(B) is determined by a T-agrange interpolation formula through points (i-1,j-1) , (i-1,j) and (i-1,b).
This yields:
Z(
'x(A)
= - 41[Z (i+1,b)
2h
+
Z .
(n-1 )-(n-1 -e-)
1 +e
Z(i-19j-l)
(+,)
-
j)
n(n-1-e)
e
-
Z(il1
,b)
n(n+1)
e (1+e)
+ O(h2)
(3-4)
Since all Z,
are determined with a truncation error
of O(h 2), formulas (3-2) and (3-3) will have this same
order of error.
M3 Case of _f iuE2_9.
In this case, Z,x(i j) is already computed by
formula (3-1).
Pro ceeding as before, we write:
m(m-n
Z,x(ijb)
+
+
where
A)
Zx (A)
Z,x(A)
-
=
+
z ,x(i,
1+n
1
7il
h
(1+m)(
-n
j-1)
M Z,x(A)
+ 0(k
3
-n)
Zx,)
(3-5)
)
is given by:
+
Z(i+1,b)
,a)
- Z(i-1,j-1)
.(1+n) (n-e)
e
-
z
n(n-e)
1+e
(1+n)n
(i-1,b) (1+e) e J + O(h2 )
(3-6)
32.
Here again, formula (3-5) is of order O(h2 ).
Note: In formulas ( 3-6) and (3-4), it could happen
that e be higher than 1 (see fig.10). In this
case,
we set
e=1 and replace
Z (i-1,b) by Z(i-1,j+1) in form ulas (3-4) and (3-6).
The quantities m and n are always smaller or
equal to 1.
c)
Nodes on the crest (fig.11)
Let us determine the interpolation polynomial
passing through points (1,j), (2,j), (3,j), and (4,j).
We have:
(X2(x-x2
3)
-6h
(X-X
+
-4)
3
Z(1 tj)
(x-x
1)
) (X-x )
2h3
(x-x 1 ) (x-x
-r
2)
(x-x 4)
-2h3
(x-x)
+
(x-x 2 ) (x-x7)
+
6h3
Differentiating and replacing x by xi
,j)
=
-
11Z(
+ 0(h 3 ) .
- 18Z(
2
,j)
0(h)
we get:
+ 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:
33.
Z 'x(1,k.) = 3Zk(lk-1)
-
3 Zix(1,k-2)
+ Z'x(1,k-3)
(3-8)
+ 0(k)
Here, the truncation error is of order O(h3 ) because
the derivatives are derived from extrapolation formulas whose accuracy is smaller.
d)
Special nodes.
The above formulas can be used at a majority of
nodes, for afbitrary boundary curves. However, depending 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-
Zlx(s,1)
6h
F
2Z
-2Z
+ 11Z(s,1
,1
1) - 18Z(Sl1
+ 9Z(s-2,1
- 89s11
+ 0(h3 )
with similar formulas for
(3-9)
Zx(S,2 )
and
Z,
.
Finally, using a horizontal extrapolation, we
get:
Z~x
m(1+m)
ZY,
Z,x(sb)
x(s
+ Z
1
)
2
(2+m)(1+m)
-Zf
-
Z'x(s,2).m.(2+m)
+ 0(k3 )
(3-10)
The order of the error is again 0(h3 ), for the s.ame
reason as for the top nodes.
34.
3.
Determination of
a)
Interior mode (fig.14)
z,y
We have the usual formula:
z y(i,j)
Z(i, j+1) - Z
2k
-
+ O(k 2 )
j1
(3-11)
b)
Nodes on the y-axis
on the y-axis:
Because of symmetry, we have,
z
c)
y(i,1)
(3-12)
=0
Boundary nodes (fig.15)
Writing the interpolation polynomi-al through
we have:
and (ib),
points (i,j-1), (i,j)
z
)
( ~-1 -) (Y~7b9
b)
=Y~
=~.1)
(1+m)k2
(ij-
2Z)(y-y )
+
m(1+m)k2
t
Z(i,j)
- 2
.mk
(ib) + 0(k3)
Deriving and writing successively y=yj
we get:
Z
1
= kg-
i, j)
+
Z9y(ijb)
m
-1+
1
Z(ib)
m+
1
kE
ZM
(i
m
1+m
I
Z
-
1-m
Z(i
-m
+ O(k2 )
)
(3-12)
1+m
35.
+
Remark:
2
(ib) + + 0(k )
1 +2mm
m (1+m)
(3-13)
If the boundary node is one node of the rectangular 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 compatibility, 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
a)
x=0.
Interior node (fig.7)
We have the well-known central difference formula:
Z-
'Z(i,
9j)
- 2Z
+ Z(i-1,j)
+ O(h 2
(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
36.
points (i,j-2) , (i,j-1)
and A , we have:
Z xx(i ,b) = Z'xx(i, j-2 )
(1+m) (1+m-n)
1+n
(2+m) (1+m-n)
-n
+ Z'xx(ij-1)
(2+m) (1+m)
(1+n) n
+ Z xx(A)
Writing now
+ 0(k
2 Z(A)
Z'xx(A) =[Z(i+1,b) -
3
)
1
+ Z(B)
h2
and expressing Z(A) and Z (B) by an horizontal interpolation through points (i, j-1) , (i,9j) , (ijb) ,
and (i-1,j-1) , (i-1,j) , (i-1,b) respectively, we get
the final formula:
Z xx(i,b)
(+M)
( 1+M-11)
1+n
.(2+m) (1+m-n)
-n
+ Z~j
+ Z,j)
{
[
h2
+
[z(+1 i j-1 )
+
n(n-1 -m)
-M
,j-1)
n (n-1
(i-1,b) e (1+e)
-
2Z
(ib)
(n-I-e)
(i-1, j)
1+e
)1
)j
(2+m)
(1+m)
n (1+n)
(n-1)
.j-
_(n-1 -m)
+m
)
)
(n-1) n
(1+m) m
+ Zb
i-1, j -2)
. .
(i,j-1) t Z .
2 (Z
Z(i+1 b) -~(Z
(i,j-1)
(n-1)
+(Z(i-1
2Z(i, j-2) +
i+1, j-2)
n (n-1ie)
-e
+ O(h2)
(3-15)
37.
The truncation error has the order of the error
of each Z,
, since all other interpolating formulas are of higher order.
Using an horizontal interpolation formula, we
can now write:
Zxx(i,
j)=-Z~xx(ij
+ Z,xx(ib)
m
2)
+ 2Z
2+m
m
'xx(i, j-1)
+in
16)
(2+m)(1+m)(
where the order of the truncation error is still the
same as in (3-15).
Case ngeat
s
(fig,
Using exactly the same procedure as above, we get:
(I+m) (m-n)
-n
m(m-n)
+ Z'xx(ij)
1+n
xx(i,b) = Z,
x,
Z
m(1+M)
'xx(A) n(1+n)
0(k3 )
+
or, in termes of Z:
1
Z xx(i,b)
m(m-n)
1+n
fZ( i+1 , j-1)
+ Z(i+1 ,j)
+[ZZ(i+1,b)
-2
(1+n)n
+ Z(i~b) (1+m)m
()
( Z
)
)
-
2Z
-
2Z
(ij-1
)
( i-i, j -1
(1+m) (m-n)
-n
(i, j)
n(n-m)
1+m
+
Z (-1,j-1
)
(ij)
n(n-e)
1+e
(1+n)-(n-m)
-M
+
Z(i
1
I)
38.
(1+n)(n-e)
-e
(1+m)m
(1+n) n
n (1+n)
e (1+e)
+ Z(il b)
+ 0 (h2)
After simplification, we get:
Z xx(ib)
1Zi+1
=
+
EZ(i+1
+
+
S
- 2Z(i
Z(i+1, b)
S(1+n) (n-e)
Z (i
j)
-e
+m
+ 0 (h
2
b(
I-("(i-1, j-1
+
Z (i
)
n(n-e)
1+e
(1+n)n
(1+e)e
,b)
)
(3-17)
When we have the situation depicted in fig.10,
the previous formulascan be used, provided
we set
c)
b)
)I
m(m-n)
1+n
(1+m)(m-n)
-n
+) (i-1,)]
(1+n )n
Remark:
(i-1, j-1
j-1)
e=1
and replace
Z(i-1,b)
by
j+1).
Special points
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)
(x,2) =
, (s-2,2)
,
(s-1,2) and (s,2)
-6h3 ^
:
(s-3,2)
39/
S)
s -1-
(1~s -
+
2h3
(x
+
s-3) (x-x s-2) (x-xs)
-2h
(xs-3
+
3
s-2) (X-x
Z(s-1
1)
6h3
Deriving twice and setting x=x.
Z xx (s,2)
*2
FZ(s-3 ,2)
+ 0(h4 )
, we get
+ 4Z(s-2,2) - 5Z (S-1,2)
+ 0 (h 2 )
+ 2Z(s,2)]
,2)
(3-18)
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)
- Z xx (s, 2)
2
(2+m) (1+m)
2
+ 0(k3)
(3-19)
Here again, the total error is of order h 2
(truncation error on each term.)
5.
Determination of
a)
Interior node
Z,yy .
(fig. 14)
We use the central difference formula:
Z, y ~ i,j)
1
2 I'Z(ij-1)
- 2Z (ij)+
Z(i,j+l)
+ O(k 2 )
(3-20),-
40.
b)
Boundary points
(fig.15)
Let us write the Lagrange interpolation polynomial passing through points (i, j-2) , ( i,.j -1) , ( i, j )
and (ib).
We have:
(y-y2_) (y-y
Z(i,y)
(y-y)
Z(i, j-2)
-2 ( 2+m)k0
(y-yJ-2) (Y~y)
(1+m) k0
(Y~7Y)
+
-2mk3
+
_1)
7-Y
(Y-y i-2)
(Y-y 1 )
+ 0 (k )
Z(ib)
m(2+m) (1+m)k3
Deriving twice and setting successively
we get:
1
y
'Iyy(it
j)
k2
-
-m
rn
SI'
Z
(i,
+
4-2m
1+m
Z(i j-2)
2+m
+
y=y. and y yb
)
m(2+m)(+n
Z
(ij-1)
+6
Z(ib)
+ O(k 2 )
(3-21)
and
Z yy(i, b)
1
k2
U
3+2m
m~
1 +2m
Z(i, j-2)
2+m
Z
+
(it j)
6
+ 4Z(ijl)
Z (ib)]
+ O(k2 )
(3-22)
41.
c)
Nodes on the x-axis
Using symmetry considerations and formula (3-20),
we have:
2
Z(i,2) - Z (i)7j
k
d)
2)
+ 0 (Ic
(3-23)
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 formula (3-8),
we have:
Zlyy(s+1,1)
= Ztyy(s-2,1)
- 3Z 1yy(s-1,1
)
S3Z, yy(s,1)
+ 0 (h3 )
(3-24)
such that the truncation error is still of order K2
6.
a)
Determination of
Interior node
Zxy
(fig.16)
We have the well-kn own formula:
Zxy(
i )
Z(i+l
4hk
+ Z
b)
9j-1
,
)I
j+1
)
- Z (i+1, j-1 )
+ 0 (h2)
y(i1 Lb
For point
(i,j)
y
Z(i-1,j+1)
+ 0 (k2 )
(3-25)
Boundary node
c) Case
-
.
we write:
(fig.9)
42.
Z'y(i+19j)
1
2h
zxy(i, j)
-
j
1
- Z(i+1
,
j-1
)
n
n+1
1
n (n+1)
b)
=2hk
-
Z(i-1
- Z (i+1,j)
1
e(e+1)
,b)
0 (h2 )
we can write:
Using now a formula similar to (3-12),
Z, xy( i )
+
Zy(i-19i)j
1-n
n
(i-1i j)
1-e
e
+ 0 (h2 ) + 0 (k )
(i-1 , -1) -e+1
(3-26)
For points (i,b), using a horizontal
extrapolation, we
write:
Ztxy(i,b)
z(i,j-1)
= Z,
(1+m)m
2
+ Z
(2+m) (1+m)- +
.xy(ij)
- Z,xy(i j-1)
c
(2+m)m
(k3 ),where Z xy(ij-2)
are given in terms of Z by formulas (3-25)
and where Z,x
The total truncation error is still of
and (3-26).
2
order L0 (h ) + 0 (k2
This yields:
Z xy(ib)
i+1 , j-1)
14hk
-
7 i+1,
j-3)
- (1+m)m
.lj..
+ (i-1,9 j-3)1
2
-
Z(i+1, j)
(2+m)
+
Z(i+1 ,b) n (n+1-)
+ Z
+ Z (i-1
j-2) m
- (i+1,t j -2)
i-1 , j-1)
~ Z(i-1, j)
(?i+1, j)
n
43.
n
- Z (i+1 ,j-1)
+ Z
,j-1)
1
e(1 +e)
- Z(i-1,b)
1+n
(2+m)
-1+e
(1+m)}
(i-1, j)
(1-e)
e
+ 0 (h2) + 0 (k2 )
(3-27)
Remark:
If we have the situation of figure 10, we use
the same formulas as before, with e=1 , replacing
z (i-1,b)
Case
with
by
Z(i-1,j+l).
y(i+1 L
2J' L1L
(fig. 8)
Z,xy( i -1) will be determined from formula (3-26)
e=1 and Z(i-1,b) replaced
by Z(i-lj+1)
Now, using a vertical interpolation for Z,
through
points (i-2,j)
(i-1,j) and (i,j) , we have:
-(XXi-
Zy (xvj)
1)
(x-xi)
Z,y(i-2, j)
2h 2
(X-xi-2)(X-Xj)
h2
z 'y(i-1 , j )
(x~x i-2) (x-Ai- 1)
2h 2
z 'y(i,j)
+ 0 (h3 )
Deriving and setting x=xi, we get:
Zxy(I, j)
2h
Z,y(i-2, j) - 4Z, y~
+ 0 (h2 )
Expressing Z,y in terms of Z
+ 3Zy(ij)]
(3-28-a)
,
we get finally:
44.
Z, xy(i j)
+
1
d (1 +d)
4
~ e(1+e)
+
2kh
Z (i-2,b)
d1
~ 1+d-
+
'3
Z
MO +iM) (ib)
e
1 +e
3m
Z(i-1 ,b)
1d
Z
(i-2, j-1
Z(i-1, j-1 )
+
4(1~e
e
3 (1-m)
Z(i, j-1 )
1+m
( - 9j,
d
Z
-m
+ 0 (h2 ) + 0 (k2 )
(3-28-b)
This formula can always be adapted to different situations by setting, if need be, d or e = 1
and replacing Z(i-2,b) and/or Z(i-1,b) by Z(i-2,j+1) and/or Z (i-1, j+1)
Finally, using a horizontal
Z xy(i,b)
'xy(i, j-2)
trapolation, w e write:
(1+m)m
(2+m)
xy(i, j)
-
Z, xy(itj1)
(1+m)
2
(2+m)m
+ 0 (k3)
(3-29)
where Z~xy(ij-2), Z,xy(ij-1) and Z,
j
are given
in terms of Z by formulas (3-25),(3-26) and (3-28)
respectively. The truncation error is still of order
o(h 2)
+ O(k 2 ).
c) Nodes on the x-axis
By symmetry, we have
ZXY(il) = 0
45.
d)
Special points
(fig.13)
Writing a vertical extrapolation, we have :
Ztxy(s,2) = - Ztxy(s-2,2) + 2Z xy(s-1,2) + O(h 2 )
In terms of Z ,we
Z,xy(s, )
2
-
cam write:
4hk
+ 2Z
(s-2,1)
s-1 3)
-
) - 2Z (s-2,3)
- 2Z
2Z (S9)
+ Z_(s-3,)
- Z(s-3,1 )+ Z (s-1,
11)
+ 0 (j2) + 0 (k2)
(3-30)
Using a similar formula for the next column, we get:
Finally,
Z xy(s,b)
-
41
4h k
'xy(s,3)
m
1+m
Z(sb) - 2Z(s-
+
4
m (17+T)
-
Z(s-1,4)
+
0 (h2
)
(s-1 ,2)
+ 0 (k2
4(1-m)
m
Z(0 2) -
2
,4 )
(s-3,4)
+ 0 (k3)
-
)
an horizontal extrapolation
= - m(2+m)
+ 2Z(s-2,2)
Z (s-3,2)j
(3-31)
yields
Z xy(s, 2) + (2+m) 2(1 +i) Z 'xy(s,3)
(3-32)
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 computed only at nodes on the boundary curve.
a)
Case
Y(ij)
y(i+1,b)
(fig.8)
Using horizontal extrapolation, we have:
(1+m-n)
Z 'xxx(ib) 2 Ztxxx(i,j- ) (1+m) 1+n
2
+ Zxx
+ Z
i9
x
xxx(A)
_)_
(2+m) (1+m-nl
+ 0 (k3 )
(2+m)
(1+m)
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)
(1+m) (1+m-n)
1+n
(2+m) (1+m-n)
-n
+
-
xx(i-1,j-2)
+
Z
+Z'xx(i+1,j-1)
Z'xx(i-1,j-1)
Zgxx(i+)
x1,
(n-1)(n-1-e)
1+e
47
n(n-1 -e)
e
+ Zxx (ij)
5-
(2+m) (1+m)
nl(1+n
b)
+
0 (h2)
Case Y(i+1,b)>
n(n-1)
Z,xx(i-1,b)
-
e (1 +e)
+ 0 (k2 )
(iI
(3-33)
(fig.9)
j)
Using the same kind of extrapolation through
points (i,j-1).
1
h
Z xxx(i,b)
+LZ,xx(i+l
+
,j)
and(A)
(i,j)
-
Z,xx(i+1 , j-1
'xx(i-1,
Z,
~ 'xx(i-1l ,b)
-
)
n(n-e)
1 +e
-1+n n
(1+e)e
+ Zxx(il
Special points
c,,,) Node (s,
j)
I
m (1+m)
n (1+n)
+ 0 (h2 ) + 0 (k2)
c)
m (M-n)
1+n
Z,xx(i-l
(1+m)(m-n)
-n
Z xx(i+1
,b) - Z'xx(i'1, j-1)
(1+n)(n-e)
e
we have:
(3-34)
(fig.13)
j
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, ) and Zxxx(s3)
2
48.
Using now a horizontal extrapolation formula,
we have finally, in terms of Z,
:
Z
+
xxx(s,b)
'xx(s-2 ,1)
2h
1) + 3Z,xx(sl)
Z ,xx(s-3,
-3Zxx(s-1, 1)
Z'xx(s-4,1
-
- 3Z,xx(s-2,1)
2
'xx(s-2,2) - Z'xx(s-4,2) - 3Z,xx(s- 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
m(m+2) +IZ xx(s-2,3)
Z qxx(S-193)
+ 3Z,xx(s-3,3)
(1+m)
2
+0
+ NZ,
(k3) + 0 (h)
(3-35)
The final error is still of order
Node
0(h2 )
1.r(s+
u
Using a formula similar to (5-9),' we get:
Z,xxx(s+1,1)
- 18Z
,xx (s,
6h
)+
-2Zxx(s-2, ) + 9Z,xx(s-1,1)
1
11Z,xx(s+1
,1)]
+ 0(h3)
such that the total error is still
of order
(3-36)
0 (h2)
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
xxx(1,p-2)p-2)
--NZx(1p
- Z xx(3,p- 2 )
+ 4Z~x(p.
2
xx(2,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 formula similar to the one used in formula (3-35)
Z,xxx(i~b) == Zxxx(I,p- 2 ) m(+m)
+ Z XXX(lp) (2+m)2 (1+m)
-
Z,XXX(Pl) m(2+m)
+ 0(k5 )
(3-37)
such that the total error is still of order 0 (h2 )
7.
Computation of
Z,
xx ; Z xayy
; Z,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,
a)
Z7,y
and Z~yy respedtively.
Ordinary point
Let
f
denote one of these last functions.
Using fig.15 and writing the Lagrange interpolation
50.
polynomial through points ( i, j-2 ) ) (i, j-1 ) / ( i, j)
and (ib), we get:
(Y-yj-1) (y -YJ) (Y-yb)
f(i,y)
( y-y j-2)
f(i, j-2)
3
-2(2+m)h
~7yj
Y-b)
(1 +m)h3
y-y~j -2)
Y ~7 J -1)
Y-b
- 2mh 3
m(y-y-2) (~-2+1)
m(1+m) (2+m)h3
4
Deriving and setting
'y(i,b)
+ 0 (k)
, we get finally:
m 1+m)
2 (2+m)
h
(1+m) (2+m)
2m
f(i, j) +
f(i,
j-2) +
m(2+m)
1+m
3m2 + 6m + 2
m(1+m) (2+m)
+ 0 (h 3 )
(3-38)
Thsi' formula can be used at any point with f
or Z,yy.
b)
Special point
and
Zyyy
(Z, Y)
Z
,
, we cannot use the above formu-
la.
from symmetry considerations:
z Sxxy - (Zxy),
z
Z,
(fig.13)
At point (s+1,1)
We have,
f(ib)
=0
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,
Zx
we have, using the antisymmetry of
:
Zxyy(s+1,1) =1
Z,xy(s-2,2) - 3Z,xy(s-1,2) + 3Zxy(s,2)
+ 0 (k2 )
8.
(3-39)
Conclusion
All the previous formulas determine, within a
maximum truncation error of 0 (h2) or 0 (k2) (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 equations, 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 differential equation (2-7) at any point of the dam.
We
also know that this fictive surface should be continuous 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
Z
.
'xyy(i,b)
2
equation (4-1)
2
and
2
Z
.
'yyy(i,b)
to
and (4-2) respectively to meat the same
continuity requirements.
53.
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 characteristic lines are straight lines inclined at
the x and y axes.
450 on
Hence, using a square mesh (h=k),
all mesh nodes will be on characteristic lines, providing 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
2
212 +22d<1
x
d
dy
x
which yields:
d
y
~
12+
2
X2
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,
412±
(
_2
C2
_Xk
h
but such that
54.
Sincd we do not know a priori the final shape,
h
this 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:
'6 x Z,
=
J
(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 secondorder Runge-Kutta method, with the general equation:
J
.
=I
x(n+l,j)
+
for all
+ --x(n,j)
=
h
Wxz,
2
(6xZ, ) (n+1, j)
.
(n,j)
(4-5),
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).
55.
b)
Determination of
J
Deriving (1-32) with respect to y , we get:
J
=
yy
(4-6)
' xz,y
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)
= 0
(4-7)
Using the same Runge-Kutta method, we have:
=
y(i,n)
+ -
2
-k
1
'y
(i-n) +
,i, n) +
'y (in+1))
(4-8)
and for values of y within the same
valid for all i
triangular domain as before.
4.
Determination of F by the method of characteristics.
a)
Introduction
Let us look at the equation
a u,
+ b u,
+ c u
= f
This equation, together with expressions of differencan be written in matrix form:
and u,
tials of uf
r
a
b
c
u,
f
dx
dy
0
uxy
d (u, )
0
dx
dy
u,7 v
d(u,
y
)
(4-9)
56.
Investigating under which condition the derivatives
, u,
ux
at a point
and
u,
P
by values of
yields the condition:
a
b
c
dx
dy
dx
0
0
a(dy) 2
are uniquely determined
ux and u,y on a curve
,
'
or
dy
- 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 conditions for (4-9) to be compatible can be written:
a
f
c
dx
d (u, )
d(u,
0
0
)
=
0 , which gives:
dy
ady d(u,x) - f dx dy + c dx d(uY)= 0
a
d(u, )
--
+ c d(u, y)
or
- f dy = 0
(4-11)
This equation is valid along each characteristic line,
b)
Integration of equation (2-7)
With
(d
=1,
a = 1 , b =.0,
and
equations
(H
c = -1
u 3F
(4-11) can be written:
(3
'
=-1
,
f = J
-
yx
J
57.
d(F. )
-
d(F, ) = (Jy
-d(F, )
-
d(F,
Sy) = (Jy - J )dy
on lines
J )dy
-
a<
on lines
With the notation of fig.20, and integrating along the
characteristic lines, we wil 1 have the equations:
,j
42k
-
(F
y
- J X)i+1
-
k
- J )i+1
-
(F
-
,
+ (F, ).
y )i+ 1
+ (y
,
) i,j+1
+ (F
(
-
i,j-1
)i+
X)~i~j-1~
(4-12)
+ (F9, )ij+
,j
+ ((y - J)i,
j+1)
(4-13)
(
Solving for
(FX)
i+1,j
and
1
=
+ (Fy)i,j+1
+(J
2
it j-1
kL 2 (Jy
+
- J)i,
L(F,
(F
)i+1
+
j
yields:
9,)ij+1
- J X)i+1, j
+j1 I
+ (Jy - J)
(4-14)
and
1
y i+1, j
+ (F,
)
2
-4-
_
+F9,
-,(F
)i,j+1
) ij-1
+ (F,)i,j+
y~
x
-
(F,y )
~ y
i, j+1
(4-15)
x
j-1
i,
58.
c)
(2-5)
Initial conditions
It is easily verified that conditions (2-4) and
can be expressed in terms of F, and F, by:
FX
= 0
F1y
= py
on x
=0
(4-16)
It should be noted that this condition is compulsory 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,y
at each point of the triangular domain of integration
Using formulas (1-33) to (1-35) and computing
by
and F,
numerically the first derivatives of F,
formulas similar to (3-1) and (3-11), the stress resultants can easily be determined at each mesh node.
59.
Chapter V.
1.
Determination of Z and
\
Determination of Z
It has been shown in chapter II that
Z was determined 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 interior 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:
Coefficients:
Unknown
Z. .N
(i-1,j-1)
Z(i+-11)
Nxy
xy
/y2hk
2hk
60.
/ h2 - 2N
-2N
Z (i+1,
/ h2
N
j)
-N
Z(i-1, j+1 )
Z(i,j+1 )
/
/ 2hk
yy /
N xy/ 2hk
Z(i+1 , j+1)
Ix oe2
Constant term
All other coefficients will be equal to zero.
Boundary node.
b)
Let us consider the boundary node in the situation of fig.9. Introducing formulas (3-17) , ( 3-22)
and (3-27) into equation (1-48) will involve the f ollowing coefficients
Unkn owm
Coefficient
Z(i
-2 Xsin{ Cos{im (1+m) / 8hk
,
j3)
Z (i1j-23)
il,
-3)
Z (i, j-2)
Z(i+1, j-2)
Z(i-1
0
+2 \sin
cos{ m (1+m)
+2 X sin
cos
- X cos
2
>M
1+ 2m
{
(2+m)
/
8hk
/
4hk
/
- 2\ sin4 cos4 m(2+m) / 4hk
>sin2
j-1 )
m(m-n)
1+n
h2
\sink
+
cosi
2hk
+ m(1+m) (n-e )
+
1+n) (l+e)
m(1+m)
2
e (1+m) (2+m)
1 +e
61.
4 \ cos 2 j/
k2
sin 2 1
h2
j-1 )
m (-n
1+n
sin §cosi
m(1+m)
2hk
n(1+m) (2+m)]
1+n
J
2
(1+m)(m-n) _ m(1+m)(n-e)
_ Xsin 2 %
n
en
1
h2
sinf cosi
[m(2+m)
2hk
Z(i+1, j)
2
3+2m
kcos
2
2
Ssin
(1+m) (M-n)
Xsini Cos
2hk
n
h2
[-m(2+m)
Z (i-1 ,b)
+ (1+m)(2+m)(1-e
e
-
Xsin 2 k
h2
(1-n) (2+m) (1+m
n
m (1+m) +\sin!
e(1+e)
cosf
2hk
(1+m) (2+m)
e
+e)
2
Z(ib)
1b)
sin2 % B (1+m)
h 2-
1 +n)
+XCos29
k2
m(2+m)
6
X sin 2
n
n (1+m)
h2
r 1 +n)
sin i cos
2hk
Constant term
- Exx(i,b)sin21
(1 +m) (2+m)
n(n+1
-
Eyy(ib)cos2
+ 2 Exy(i,b)sin§ cos
All other coefficients are equal to zero.
c) Other cases
Any other situation can be handled in a similar way,
using formulas established in Chapter ITT.
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
Constant term:
N
F1
k
-
Zj1)
+ ZN(o,+1)
-
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 eliminated in a way similar to what was done in the previous
remark, it has been left as above, setting the coefficient of Z(i+1 1) equal to 1 and acll other coefficients
equal to zero, while the constant term was set equal
to Zbe
The 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
Z
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 equation (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.
1.
Programming considerations.
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 necessary, 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 roundoff error in the formulas of chapter III, the pt-ogram
modifies the boundary coordinate y(i,b) when
y(ib) - y.j)
It sets y(ib)
0.05 k (with the notation of fig.8)
Y(ij) , but still considers two
different memories for this double point to which correspond now equal values of the function.
For functions not defined on the top edge such
that
Zx
7...,
another auxiliary matrix
INS
has
is referred to as
been defined, such that Zxx(it)
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;
: computes second derivatives with respect to
COXX
x.
COXY
:
COYY
computes second cross derivatives.
: computes second derivatives with respect to
DERIV
y.
: computes first derivatives with respect to
x
and to
y
COXXXB
: computes
Z, yy
COYB
PARINT
: computes
Z,
: computes
J
x
HART
:determines F,
on the boundary
, Z,,yy,
and
Z,Yyy
on the boundary
J
y
and F,y
by te _method of
characteristics.
FORCES
: computes
NxJ Nyy
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
SECSYS
SYSTEM
SYSTEM
:does the same work for X.
: 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.
1.
Boundary curve.
Results and conclusions
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 meterkilogram 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
A square mesh size of
and
- =
1
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 :
67.
Z(x=O,y=O) e
Z(xo0;=)
These shapes were descri-
bed by equations such that
Z = 8.56 ( \A71
- 0.63x
-1)
-
15
3
- 0.012y 2
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 iterations had a slope Zy at x=O which was continuously
increasing and seemed to tend to Z,
2)
=00
All iterated shapes had very great curvatures Zy
and 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 changing very much.
3)
Contrarily to the assumption done in chapter II, after
the first step of the iteration,
INxy I has been
( in some parts of the
found higher that INXXI = 1T
xx
yy
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 mathematically 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
=
max
yy
xx
dary curve was prescribed.
of equation (1-31')
Z = Zbound
R
, but for which the boun-
br
This needed the resolution
for Z, with the boundary condition
(7-2)
The displacement X could then be found by integration 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 transgressed. 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 determine 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
Z.
=Z.
+ 0.20 ( Zt-
Zg and Z_
i-1
Z.)
(7-3)
The final results are given in Appendix 2.
They confirm all the conclusions drawn in the previous paragraph, i.e. that
69.
1)
2)
the obtained shape has much higher curvatures to
provide a practically usable dam shape;
The slope Z,9x at x=O is almost horizontal;
, the problem
being still higher than Nx
Ny
3)
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 the
xxyy
difference in the structural behavior of vertical elements (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 determination of Z , probably based on some methods of linear
programming and optimization. This is beyond the scope
of this work and was not examined.
Remarks:
INXXI , we have studied the dbove problem by introducing shear stresses
on the top edge, acting in the direction opposite to the
1) In order to decrease Ny
vs..
shear stresses derived from (2-7), expecting these boundary stresses to propagate throughout the dam. This hope
was not fulfilled and the same features as before appeared 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 JY
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 finding 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 exactly 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 =
+ z
xx
+xx
J
(7-5)
x
with the boundary condition:
- 2
>12
+
2
+ FS-
xy
+ (Fy
in cos
2
cos 2 )
(1+v)<(
-
2 c1
2
12 1
sin 2
- (1+v)
2 CV
J )
(F
sin 2
nr
- 2
- J )
-
2 #12
,
o12
12
2
2
+
2 cos2
b
sin2
sincos
2
Cos2
71.
L2
g12
(1+v)o
-
cos 2 j - 2Z,
+ Z,
sin
Eh.<X(Z,
Cos
sin
sin2
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 convergence factor.
are
F =
On the top edge, the boundary conditions
(77)
2
for F
Z~yy
=
-
and
Z(7-8)
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 problems did not yield any result:
assuming
Xinitial = 0 , and
Finitial
[py2 (1+0.05 x2 )
-
10x2(25 +1.25x)
+ xy 2 (50 + 3.75x) + 150.10 6 x
j
/2
(stress fu]ction yielding a distribution of stress resultants 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 diverge 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 Jy). Hence,
it is quite normal that the derived stress resultants
remained too small and required high curvatures for differential 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 determination 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 iterative 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.
zy
x
-fig.
c
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tJ
C4 u
Ixxa6 1
N IoD
4
x
74
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77
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78
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Appendix I
Determination of. Z for the general problem: NewtonRaphson method. --------------------
---------------
The Newton-Raphson method consists here in wri-
- 9Zi-1
ting Z _g1= Z i
Ni-1
and
i-1
yields the iteration
ieplacing into equation (1-31)
equation:
Nxx(i-1)
(Z.
+ 2 Nxy(i..
- SZ
1
x-1
=sZi-
Z
z.
,vxx
+ Z2i-
xy
iyy
z
+ Z2
i-
1
1
xy(i-1) Z.1,xy
+
+ N x(i)SZ(i- ),xx
2
+ 2 N xy(i-1 )
yy
2
yy(i-1)
1,x X
(Zi -EZ.i)
i
1-1
yy(i-1 )
i-1 )
(Zi
)
or, assuming
N xx(i-1
i-1
yy(i-1) Z(i-2),yy
(i-2),xy
Equation (1-32) for
can be treated in a similar way
and yields:
Z(i1)x
-X
,y
xx(i-1 ),7y
( i-i
),
y
x.
yy(i-1),xx + 2
-2
z (i-
) ,xy
xy(i-1),xy
i~j xy
84.
+ Z (i-l),xx
.X
-
2Z(i- 1 ) ,xy
.
(i-2),yy
+S.O.
.(i-1),yy
(i-2.),xx
(-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.
'VEL
C
MOD 0
DATE = 68147
MAIN
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),
X(175Tv,EPY (175 ) ,EPXY ( 175 ),EXB(2)i,EPXY B(-21),I, I
3~IN(271PHI D (2 1EP
4S(157) ,ZXYB (21), ZYYB(21) ,ZXX(157) ,ZXY(157) ,ZYY(157), EPXDX( 175),
5E PXDY ( 175)~,iPYiX (175) ,PYDPYt 1!5),EPXYDX i7 5) -EPXYDY (175), 5Y
( 21f
6ZXXX(20),ZX)XY(20),ZXYY(2C),ZYYY(20),SEC(175b),ZPR(175),EPYXX(157),
7EPXYY(157) ,EPXYXV(157)
COMMON FXFY
EUVALENCE
(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
*6
PINT(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
IGFORMA T(E 14.7,1XE14.7,1Xi12,1X,E14.7.,-X,-E14.7,l-X,-E14.7)
INDC
DE L X=HM
DELY=DELX
MP=M+I
DO 12 I=1,MP
12 Y(I)=DELY*(I-1)
MP=M+1
A=- W**2 / (4 .*H)
B=W** 214.
INTMi) =0
DO 13 I=1,MP
YB(I)=(A*X(I)+B)**0.5
INT(I+1)=INI(I)+YB(I)/DELY+2.05
JR=INT( I+1 1- NTI
IF(A13 StYB (I)-Y(JR-))-0.
05*DELY)
16
16,13
16 YB(I.J=Y(JR-1)
13 CONTINUE
00 14_I=1,MP
JR=INT(I+i)-INT(I)-1
DO 15_K=1,JR
JS=INT(I )+K
Y(K),X(I))
15 Z(JS)=SH(
J S= IN T (1+1)
14 Z(JS)=SH(
Bfl),vXH))
INS(1)=0
DO 31_1=1PM
L=0
JX=INT(I+2)
IF(Z(JX)-Z(JX-1)) 32,33,32
- - - ---
-
- -
- - - - -
- - - - -
86
V L
Q
--
-MAIN
-OD
INDATE
6814722/46/24
33 L=1 _
32 INS(1+1)=IN S(I )+INT( I+2)-INT (I+1)-L
31 CONTINUE
DO
30 1=1,0
PHIII)=1.57C796-ATAN(H/(8.*YB(L)))
30 PHID(I)=8.*H**2/(H**2+64.4YB(I)**2)**1.5
PHI(MP)=C.
PHID(MP) 8. /H
WRITE(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)
,'zxx'/I>3'+
100 FORMA(x
CALL PRI(ZXX,INSM,0)
++
WRITE(6,2C)
20 FOR MATI'',64X ,'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 ,., DEELX ,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
WR ITE (6,18)
DELY,0.)
CALL PRI(ZXINTMP,0)
19 FORMAT(l't63Xr'LY
CALLPRLZ
(M/M)',/62Xt,'+++++++++//)
NAPAO--------------
JP=INT( 2)
ZXYi(1) =PINTZX ( JP-4),ZX (JP-3) ZX ( JP-2,ZX ( J P-1)
ZYYB(l)=PINT(ZY(JP-4) ,LY(JP-3) ,ZY(JP-2),ZY( JP-1)
1=2_.,M P
JP=INS(I)
- - Z X Y B (I)-=Z X Y P) ---36 ZYYB(I)=!ZYY (JP)
DU_ 36
PALLPARINT
IYBI
JXRJYIPOINTX,
DEL X)-----DELX
_
_
_
_
X GA DELX Y ZXYB ZY 0ELYZYYBS-
_
_
_
_
87
EVELC, MOD C
MAIN
CALL HART(
CALL FORCES(
DATE
--
68147
22/46/24
I POIS ,MP ,RJX ,RJYDELXPY)
IP ,1
TRNXRNYRNXYOELXDELYpRJXRJYpMMP)
WRITE (6,24)
24
>161X,'NXX (KG/M)'/60X,....+++++++/
FORMAT(
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
TP
WRITE(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)**2
R=ZSEC (1)**2
Sq=( LSEC (1I)-,Z (1)) 4'*2
MS=INT(MP)+1
DO 50 1=2,M-'U=U+Z(I_)**2
R=R+Z SEC (I) **2
50 S=S+(ZSEC(lI)-Z(I))**2
R=R**0.5
U=U**C.5
SS*
C.5
T=S/R
WRITE(6,61)
SRL , T
6 FORMATI4(3XE14. 7)-//)
IF(S/R-0.05) 54,54,51
IND=IND9(
F
IF(IND-3)
JC,70,90
7C MT=INT(MP)+2
DO 71 I=1,MJ
)=Z SEC (I )O.40*Z
GO TO 91
90 IF(IND-6) 9202,93
11 Z(J
SEC
92 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, 91,56
ITE(6t17)
L
91WR
CALL PRI(ZINTMP,0)
-
-
-
-
-
--
-
-
EVEL---0, MOD
C
MAIN
------------------
DATE
------------
=68147
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)
--.........
--
22/46/24
2_24.
_------
WRITE (6,2C)
PRI (ZX. tINS,M,0)
GU TO 57
5-6 WRIIE(.6,5E)
58F0 RMA
1I/
I
M
CONVE RGENCE
HAS BEEN OBTAIND' )
-
JS=INT(MP+1)
WRITE(7,1C1)
(Z(
),I I i1
1 JS)
FORMAT( 5(E15. 8 1 X) I
GO TO 59
54 WRITE(6,6C)
60 F0RMA T
/
END- OF'1 TERLTfI
JS=INT(MP+1)
WRITE( 7, 101)
(Z(I),I
LST SURFCE
IS
FINAL SHAPE')
1 JS)
DO 35 l=1,MP
JX=INTII fli
JV=INT( 1+1)-i
DO 35 K=JXJV
AP=1.+ZX(K)**2
AS=1.+ZY(K )**2
A T=ZX (K)* ZY (K)
AL=(AP*A-'-A1**2) **0.5
A Z=RNX ( K) *A P+RNY (K) *A S+2. *RNX Y( K) *AT
EPX(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 CONTINUE
.......
JX=INT(2)-i
37 DISHI )=EPY(I)/0.024
DIS(JX+1)=DIS(JX)
CALL BOUND(EPX,EPXBINTYYBZDELX)
CALL BOUD(EP
PYBI NT,,YB ZOELX
CALL 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 ITE (6,44)
44 FORMAT('1',63X,'EPXY',/62X
---------------
++++++//)
------------
EVEL
C,
0
MOD
MAI N
DATE
68147
CALL PRI(EPXYINTMP,0)
CALL DERIV(EPXEPXDXEPXDYDELXYBINTMPY,
C0AL
DRIQ_(PYEPYDX,EPYDY,DErLX,YE3,[NTMPY,
DELYO.)
_
CALL DERIV(EP) YEPXYDXEPXYDYDELXYBINTMPY
CALL COXX{EPY,EPYXXINTINSMDELXYYB)
CALL COYY EPXEPXYY~,
22/46/24
DELY,O .
DELY, 1.)
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'ALLEXIT
END
ei.im=m
4,
m
asa..i~
a ta--iee.e= 4 i..
-- -. .- - - - .m--.
.
,11- . .
.-.
-.
,.9 .. . .
. .~a .,.,
.... , .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~S
JR= IN T ( I+1)- 2
JP= INT (I )+2
QQ 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 LY
ZY(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)/RLA
1+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)+l
JP=JINT (IP)+I.
ZY j p+ 1 -ZY ( jp)
DO030 K=2,MS
IR=INI(K+l )-TNT( K-1
IF(YBM$-Y(JR))
50,51,50
LJ&aJ&s-------------------------------------------------------------------------------0 DO 31 I=1,JR
~J~aN~lK1+L------------------------------------------------------------------------JP= INT (K-1)+I
J1O= INT iK-+1I )+I
IF(YEC(+l)-Y(I)) 32v3203
*GO
----------------------------------------------
TO 31
RMA=(YB(K)-Y(I) )/DELX
-REA= (YB (K-
) -Y (I I ) DFLX
A=(1.-PNA)/( l.+RNA)
C=2./(RNA*(1.+RNA))
RI--(l .- +REA-) +Z( JPR)tRNAA ( RNA-__
11.-REA)/REA-Z(JP+1)*RNA*(RNA-1.)/(REA*(l.+REA)))/(2.*OELX)
---------------------------------------------------------------------------------------
---------------------------------------------------------------- 91 -----------------------------
31
5
36
37
38
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
) +B
CONTINLE
IF(YB(9+1)-Y(JR) ) 30,3CO5
RNA=(YB(K+I)-Y(JR) )/DELX
RMA= (YB
IF(YB(X-1)-Y(JR)-DELX) 36737,37
REA=(YB ( Kl)-Y( JR)
GO TO 38
REA=.l.
A=RMA*(RPA-RNA)/(I.+RNA)
--------- D1132J3-L ------------
----------------------------
&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+JR
AUX=(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-at-K i-=Y-(-j-R±-LLi-ao-*-5?-,-ac -----------------------------------------------------52 ZX(JS+2)=ZX(JS+l)
0 - COA-LLN L F
JP=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
DATE
COXX
1
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)/S
S=DELX** 2
M S=M- 1
DO 1C I=1,MS
JX=INS(I)+1
JV=INS(1+1)-i
00 11 K=JXJV
JA=K-INS[I)
IFI Y(JA)-YB( I+2))
12 JS=INT(I)+JA
JT=INT(I+1)+JA
12,12,11
JZ=INT (I+2)+JA
AXX(K )=(A (JS)-2,*A (JT)+A(JZ))
/S
11 CONTINUE
10 CONTINUE
MS=M-2
DO 13 I=1,MS
JV=INS(I+1)
JA=INS(I+11-INS(I)-i
RMA=( YB ( I+1)-Y(JA) )/DELX
JS=INT(I )+JA-2
JT=INT( I+1)+JA-2
JZ=INT( I+2)+JA-2
IF(Y(JA)-YB(1+2)) 14,15,15
15 REA=( YB'(I)-Y(JA)) /DELX
RNA=(YB( I+2)-Y(JA-1))/DELX
B=( 1.+RMA) *(1.+RMA-RNA )/(S* (1.+R NA )
(S*RNA)
C=(2.+RMA) *(1.+RMA-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(JS
1+2)*D*RNA*(1.+REA-RNA)/REA+A(JS+3)*D*RNA*(RNA-1.)/(REA*(1.+REA ))-A
2(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)*D
GO TO 13
14 RNA=(YB (I+2)-Y(JA) ) /DELX
IF(YB(I)-Y(JA)-DELX)
16,16,17
16 REA=(YB(I)-Y(JA))/DELX
GO 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*
2+A (JZ+1) *B+A(JZ+2) *C+A(
JZ+3) *D
I.+RNA)/( REA*(1.+REA) )-A(JT+3 )*2./S
93
EVEL
1, MOD
1
COXX
DATE
.68157
21/31/59
13 CONTINUE
JV=INS(M)
JA=JV-INS(M-1)-1
RMA=(YB (M)-Y(JA)) /DELX
B=RMA* 11.+RMA ) /2.
C=-( 2.+RMA )*RMA
D=( 2.+RM A).* (1.+RMA )/2.
JS=INT(M-3)
JT = INT (M- 21
JZ=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)+1
AXX(JV)=PI(A(JT+1),A(JZ+1),A(JW+i1),A(JX+1),S)
DO 20 I=1,MS
JA=INS( 1+1)-IN S( I)-l
RMA=(YB(I+1)-Y(JA)) /DELX
JV=INS( I+1)-i
IF(Y(JA)-YB(L+2))
21
20,20,21
AXX(JV)=-RMA*AXX(JV-2)/(2.+RMA)+2.*RMA*AXX(JV-1)/(1.+RMA)+2.*AXX
1(JV+1)/1(2.+RMA)*(1.+RMA))
20 CONTINUE
RETURN
END.
EVEL
COXY
1, MOD 1
DATE=
68157
21/31/59
SUBROUTINE COXY(A ,A XYINT ,INS ,M,DELXYYB)
DIMENSION
P 1 (4B,
A(175) ,AXY(157)
,D)
,INT (22) ,I NS(21), Y(25), YB( 21
=-D*A /(I.+ D) - (1.-D) *B/D+C/
(Dlw,-1.+D)
S=DELX**2
MS=M- 2
Do
10
I=1,MS
JV=IN S(I)+1
AXY(JV)=0.
JV=JV+1
JX=INS(I+1)-1
DO 11 K=JVJX
JA=K- INS( I)
IF(Y(JA+1)-YB (I+2))
12 JS=INTII+2)+JA
JT=INTLI )+JA
12 ,12 ,13
AXY(K)=(A(JS+1)-A(JS-1)-A(JT+1)+A(JT-1))/(4.*SI
GO TO 11
13 [F(Y(JA)-YB(I+2)) 14,11,11
14 RL A=(YB(I+2)-Y(JA ) /DELX
IF(YB(I)-Y(JA)-DELX) 16,17,17
16 REA=(YB(I)-Y(JA))/DELX
GO TO 18
17 REA=1.
18 J S=IN T(I+2) +JA
J T=INT( I) +JA
AXY( 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.+EA
2))/(2.*S)
11 CONTINUE
10 CONTINUE
DO 20 1=1,MS
JV=IN.S 1+ 1)
JA =JV-INS( I )-1
RMA=(YB(I+1)-Y(JA)) /DELX
B=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 ,24
23 RDA=(YB (I-1)-Y(JA)) /DELX
GO TO 25
24 RDA=1.
25 REA=(YB(I)-Y(JA))/DELX
RNA=( YB( 1+2)-YIJA-1) )/DELX
JZ=I.NT( I-1)+JA+ 1
JS=INT (I )+JA+1
JT=INT( I+1)+JA+1
J 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))/DELX
IF(Y3(I)-Y(JA)-DELX) 26,25,27
26 REA=(YB (I )-Y(JA) ) /DELX
GO 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( J
2W+1)/(RNA*(1.+RNA))+2.*REA*A(JS-1)/(1.+REA)+2.*(1.-REA,*A(JS)/REA
3-2.*AIJS+1)/(REA*(1.+REA)))
20 CONTINUE
JV=INS(M)
JA=INS(M)-INS(M-1)
-1
RMA=(Y3 (M)-Y(JA) ) /DELX
B=RMA* (2.+R MA)
C=( 2.*RMA )*(1.+RMA) /2.
JZ=INT(M-'3)+1
JS-=INT( M-2)+1
JT=INT( M-1) +1
JW=INT(M)+1
AXY(JV- 3)=0.
AXYIJV-2)=(2.*(A(JW+2)-A(JW)-A(JS+2+A(JS)-A(JT+2 +A(JT)+A(JZ+21
1-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
2+1))/(4.*S)
AXY(JV)=-B*AXY( JV-2)+C*AXY(JV-1)
JV=IN S(M)+1
AXY(JV)=0.
RETURN
END
JT + 3) +AM
J T+11+ A( JZ+3-A ( J Z
------------ 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)+l
JX=INT(I+I)+l
AYY(JV)=(A(JX+I)-A(JX))*2./S
---------------------------------------------------------------------JA = I N S ( I + I
I NS ( I
--- LE-LKUAJ-- YULLItIII-1-1.,12A.1 -----------------------------------------------------12 JV=INS(1+1)-l
QO TC 13
11 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(=JXIJV
JS=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)))/S
LOL-LLLLIMLE -----------------------------------------------------------------------JT=INS(P)+l
--- dXaJ-bLT-i.L=2-L± L -------------------------------------------------------------------JV=INT(M-1)+l
.17=INIfV)+I
AYY(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-1
DO 1. I=1,MS
JA=INS(I+)-I NS( 1)-l
NJ f I )+JA
JR =IN S (1+1)
IF(YB(I+2)-Y(JA))
11,11,12
2 RMA=( YB( I+1)-Y(JA) ) /DELX
GO TO 13
) /DELX
1 RMA=(YB (I+l)-Y(JA-1)
IF(I-MS)
14,15,14
4 T=A (JP-3)
JP=JP-1
GOTO 13
5 T=A(JP-1)*L
JP=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.)/(RMA
o
2*( 1.+RMA)*(2.+RMA) )J/DELX
CONTINUE
JP=INS(M-3)+1
JR=I NS (M-2)+I
JQ=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
END
=P-3.*Q+3. *R
__
98
L 0, MOD 0
COXXXB
15/52/13
DATE = 68130
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 )/D
PINT(A ,B,C ,D,E) =(-2.*A+9..*B-18. *C+11 .*D)/(6.*E)
P1 (A ,B ,C ) =A-3.*B+3.*C
POL (A ,B ,C ,D,E)=A*D*(D-E)/(1.+ E)-B*(1.+D)*(D-E)/E+C*D*(
1+E))
MS=M-3
00 10 I=2,M-r
1.+D)/( E(1.
_
JP=INS(1-1)+JA
JQ=INSI1 )+JA
JT=INS(
1)
JR=IN Sf I+1)+JA
RMA=(YB(_+1-Y(JA )/DELX
REA=( YB (I)-Y (JA) /DEL X
IFIYB(I +2)-Y(JA)) 14,14,15
4R NAYfB f(+ 2)-Y(JA-1) ) /DELX
D=4 1.+RMA)*(1.+RMA-RNA) / (1.+RNA)
B=-(2.+RMA)*(1.+RMA-RNA)/RNA
C=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*R
GO TO 10
5 RNA=( YB(12-(A)DL
R=(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.+RAP=(A(JR-2)-A(JP-2))/(2. *DELX)
Q=(AIJR-1)-A(JP-1)) /(2.*DELX)
AX ( I ) =D*P-B*Q+C*R
GO TO 10
7D=RMA*(RMA-RNA)/(1.+RNA)
B=( 1.+RMA)*(RMA-RNA) /RNA
C=RMA*(1.+RMA) /( RNA*(.+RNA)
P=(A(JR-1)-A(JP-1))/(2.*DELX)
o
Q=(AJRI-A(JP) )/(2.*DELX)
AX( I) =D*P-B*Q+C*R
CONTINUE
JA=I NSfM-1) -INS( M-2 )-1
JP=INS(M-5) +JA
JQ=INS(M-4J)+JA
_
_
_
_
_
_
_
L -0,1 -MD
C - - - -- - -COXXXB - - - -- - -DAT E- =-68 130
15/5 2/1-3
JR=INS(M-3)+JA
JS =I NS( M- 2) +J A
RMA(YB(M-1.)-Y(J A)) /DEL X
P-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-')+JA
JR=INS(II-3)+JA___________
JS=INS( M-2)+JA
JT=INS(M-1)+JA
B=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
,D)
BAJ R) -A (J)/2*E
C =( A(J S)-A (JQ))/(2. *DL X)
RMA= (YB (M)-Y(JA) ) /DELX
D=I-A (JT-2)*RMA/(2.+RMA)+A(JT-1 )*2.*RMA1
IRMA)*(2.+RMA))-A.(JR))/(2.*'%DE LX)
---
--
-~Q=PI(BtC
-------
(1.+RMA)+A( JT+i)*2,/(i.
KRKA=(YB()(A-)/EX
AM M-1) =POL( PQ,RRMA,1.)
JP=INS IM-3?+1
JQ=INS(M-2)+1
JR=IN 5(1-1)+1
J S =1N-S(M)+ 1
AX(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) ) /DELX
B=-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
SUBROUTINEPARINT(MRJXt,RJY,
DATE
68118
06/08/07
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)=0
DO 22 I=1,M
JP=Is-I+1
IPO(1+1)=IPO(I)+JP
-JL = I NT(U1+1')-i
JS=INT(1+2)-INT(I+1)-i
DO 23 K=1 ,JS
JT= INT( I+1)+K
JW=1P0 ( 1+1)+K(
JV=INT(I)+K
)1+K
JX=IPO(
23 RJX(JW)=RJX(JX)+(ZX(JV)*X(I)+ZX(JT)*X(I+1))*GA*DELX/2.
JS=JS+1
IQ=INT(I+2)+INT(I)-2*INT(1+1)
IF(IQ)
26,25,25
-----.------
26_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( I
1+-1)_ *GA*DELX/2.
25 JQ=JQ+1
DO__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))
i1-LU+1A*XiB~i±+1U)XA ±+1)GA*DEL X/?
*X(I)+(ZX(JV)+(Y(K)
--.-
22 CONTINUE
DO
30 1=1,MP
J=IT 1±I1)-INI(
)-i
JX=IPO(I)+1
RJY(JX)=O.
DO 29 K=2,JS
~JX=IPO(U±+K--
--
*.-.-
-
JV=INT(I )+K
29_RJYJX )=8J-(JX-)+(ZY.(JV-1)+ZY(JV))*GA*X(I)*DELY/2.JS=JS+1
________
Q---JP= pLS-
-- P.R N-
--
L
E
681.18-61-0810-0-AT
- ---
+_
00 31 K=JS,JP
JX= IE ( ii) + K
JV= INT ( I+1)-1
IE 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 31
4Q.LYJ4l=R J
X-_URA
1-2
ZYY l
*1YKiA-YZY(JV±
1---YBA*2--
---
1*Xf( I)*DELY/2.
31_COtUIE
30 CONTINUE
RETURN
__ED.
i se
~
. ===
msir
me.m
@.1.1
sim-,edme
e e m-la
eem
eeehum
mewe
unne
es.
uimm
~a~a
s.am~
i o
*- "mn.
..
,=..e,am
-= ,-
.@=11
*
,mse
-
_
-
-- --------------- ----- ----
102.
-----------------------------
1EL_0vMOD_ 0 ------
7 ------------------------------------------------------
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 ills
FX(K)=O -- ------------------------------------------------------------------------10 FY(K)=P*Y(K)
DO 11 1=2,MP
JV=IPO(I)+.l
JX=IPO(1-1)+l
FY(JV)=O.
FX(JV)= (RJY(JX+1)-RJX(JX+I)+RJY(JV)-RJX(JV) U EULZ,,-trljx+l)
JT=IPO(1+1)
JV=JV+l
DO 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)-R
IJX(K)))+2.*(FX(JX+1)+FX(JX-I)+FY(JX+I)-FY(JX-1)))/4.
,11--CONTINUE -------------------------------------------------------------------------RETURN
------------------------------------------------------------------------.___END
---------------------------------------------------------------------
------------------------------------------------------------------------------------
---------------
---------------------------------------------------------------------------------
FORCES
iEL 0,MOD0
SUBROUTINE
FORCES(
19/57/45
DATE = 68094
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)
,RNY(
175),RJX(320),RJY(320)
COMMON FXtFY
DO
10 I=1,MP
JX INTTI+2
JV=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,JV
JW=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)+I
13 RNY I)P
LAG FXJT
DO 14 I=2,M
FX(J W)FX(JV),FX(I),DELX)-RJY(
I)
JX=INT(I )+I
JV=INT( 1+l)
DO 15 K=JX,JV
JZ=IPO (I )+K-INT( I)
IP 10(I -1 +IK INT I.
-I=
JW=IPO(I+1)+K-INT(I)
15RNY(K)=(FX(JW)-FX(JT))/(2.*DELX)-RJY(JZ)
[4 CONTINUE
JZ= IPO( MP)+1
JX=I NT(TMP) +1
JT I P0(MP-I)+ 1
JW=IPO(MP-2)+1
1V=I(M-P-.3)+1
RNY(JX)=PLAG(FX(JV),FX(JW),FX(JT),FX(JZ),DELX)-RJY (JZ)
RE TURN
EN D
175) PRNXY(1
104
1, MOD 1
FILL
DATE =
03/59/19
68168
SUBROUTINE FILL(AUXVECRNXRNYRNXYNAINTS,MINSZ,YYB,DELXW)
"DIMENS ION AUX 124649T"VECT157T, RNX (175)tRNY( 175) ,RNXY(
1INS(21),Z(175),Y(25),W(175),YB(21)
DO 10 11,24649
0 AUX(I)=0.
DO 11 I=1,NA
1 VEC(I)=0.
2+T~-JX= I NT (2
AUX(1)=-(RNX(JX)+RNY(JX))*2./S
IJ=1I+NS2 *NA
--AUX( IJ)=RNX(JX)/S
AUX(158)=RNYfJXT*2./S
VEC I1)=-RNX( JX)*WUl)/S
-M=M-1
DO 12 I=2,MS
JX=INS(I)+1
IJ=JX+(JX-1)*NA
JV= INTII+1)'+I
AUX(IJ)=-(RNX(JV)+RNY(JV))*2./S
IJ=IJ +NA...
AUXI IJ)=2.*RNY(JV)/S
IJ=JX+INS(I-1)*NA
AUX( IJ)=RNX(JV)/S
1J=JXINS(It*NA--........
2 AUX(IJ)=RNX(JV)/S
MS=M-2
DO 13 I=1,MS
~
JX=1NS (If2
JA=INT(1+2)
IF I ZIJA)YZLJA
20,21720
l JV= INS (I+1)-1
GO TO 22
0 JV=INS(I+1)-2
2 -DO
_i4
JX,~JV
IJ=K+(K-1)*NA
-----
. .--
.-..-.--
.
-
.
.
.
.----.-..
..---..--..--.
JT=INT1I+1)+K-lNS
AUX(IJ)=-(RNX(JT)+RNY(JT) )*2./S
IJ=IJ+A
AUX( IJ)=RNY(JT)/S
IJ=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) *NA
AUXIIJ)=RNXY(JT)/(2.*S)
IJ=I J+NA
AUX( IJ)=RNX(JT) /S
-
'Os
L 1, MOD
FILL
DATE = 68168
03/59/19
IJ=IJ+A
AUX ( IJ)---RNXYJTV/12.*S
0 JA=K-INS(I)+1
IJ=K+(K+INS(I+1)-INS(I)-2)*NA
IF(Y(JA)-Y8(I+2)) 18,18,19
8 AUX IJ) =-RNXY(JT) /2.*S
IJ=IJ+NA
AUX-(J)=RNXTJTT7IJ=IJ+NA
AUX(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))
-iA+K
IJ=I
AUX(IJ)=(RNX(JT)-RNXY(JT)*(1.-RLA)/RLA)/S
AUX(IJ)=RNXY(JT)/(RLA*(I.+RLA)*S)
4 CONTINUE
-~
3 CONTINUE
MS=M-2
--
-
-
-
--
-
-
~
00 25 I=1,MS
JA=.INT(I+2)-
IFlZ(JA)-Z(JA-1)) 26,25,26
6 Jv=INSTI-1)+r
JA=INS(I+1)-iNSlI)-l
JR= INT{ITliT+JA
------
IF(Y(JA)-YB(1+2)) 27,28,28
8 RMA=(YBtI+1T-YTJATT/DEEX
RNA=(YB( I+2)-Y(JA-1) )/DELX
RE A=C(YB(IY-Y(J~fl/DELX IF(YB(I-l)-Y(JA)-DELX) 500,501,501
0 RDA=(YB(I 4)-YTJAy)/oELX
GO TO 502
1 RDA=T.
-
-
2 JT=INS(I-2)+JA-1
IJ=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
+N
AUX(IJ)=RNXY(JR)/(RDA*(1.+RDA)*S)
JT=INS(I-1)+JA 2
--
IJ=JV+(JT-1)*NA
AUX(IJ)TRNX(JR)*T RMA/(2.+RMAT+2.*(1.
RMA-RNA)/(
2. -RMA)*(1.+RNA0)
1)/S
IJ=IJ+NA
AUXIIJ)=RNXY(JR)*4.*REA/((1.+REA)*S)+RNX(JR)*(-2.*( 1 .+RMA-RNA)/
I(I1.+RMA)*RNA)TRNA
.j*(RNA
-l.-REA)*2./((1.+REA*( 1.+RNA)*RNA)
L 1, MOD
DATE
FILL
1
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\A
AUX(IJ)=_RNXY(JR)*4./1REA*(1.+REA)*S)+RNXIJR)*2.*(RNA-I./(REA*
It 1.+REA)*S*(1.+RNA)
JT=INS VI)+JA-2
IJ=JV+(JT-1)*NA
AU X IJ)=RNX IJR)*(2~.*RMA/C(2.+RMA)-4.*(1.+RMA-RNA)/((RNA)*T2.+
1RMA)))/S-RNY(JR)*(I.-RMA)/(2.+RA)*S)
IJ= IJ+NAAUX[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)))/S
IJ=1J+NA
AUX(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+NA
AUX ( 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)+JA2
IJ=JV+(JT-1)*NA
AUX'( 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)))/S
IJ=IJ+NA_
--
AUX(IJ)=RNX(JR)*2./(RNA*(1.+RNA)*S)
GO
TO~25
7 IJ=JV+(JV-3)*NA
JX=INT ( 1+2)-i
AUX.J...Y.....R.A..-.T...-....-.
IJ=IJ+NA
AUX(IJ)=RNYIJX) *14.-2.4*RLA)/((1.T4RLA )*S)IJ=IJ+NA
AUX(IJ)=-(2.*RNXJX)-RNY(JX)*(3.-RLA)/RLA)/S
IJ=IJ+NA
12.+RLA)*S)
AUXLTJ)=6.*RNY(JX)/IRLA* (1.+RLA)-*
RLA=(YB(I+2)-Y(JA))/DELX
IJ=JV+(INS(I1+2)-3)*NA..-
AUX(IJ)=-RNXY(JX)*RLA/U(1.+RLA)*S)
AUX(IJ)=(RNX(JX)-RNXY(JX)*(l.-RLA)/RLA)/S
IJ=IJ+NA
-
-
AUX(IJ)=RNXY(JX)/(RLA*(i.+RLA)*S)
I F (11'141-,41,42
-.-
101
EL 1, MOD 1
FILL
DATE =
68168
41 VE C (JV)= (-RNX(JX )*W (JV)*2.+RNXY ( JX) *( W(J V+1) -W (JV-1 ) )/2.*S)
/)
GO-T0-25
42 IJ=JV+(INS CI-1)+JA-2)*NA
IF(YIJA+I)-YB(I)")W29,29,30
29 AUXIIJ)=RNXY(JX)/(2.*S)
IJ=IJ+NA
AUX(IJ)=RNX(JX)/S
I J= IJ+NA
~~
AUX(IJ)=-RNXY(JX)/(2.*S)
GD To ~2530 RLA=tYB(l)-YIJA))/DELX
AUX ( IJ)-=RNXY TJXT*RLA/ (I1.+RL
A )*S)
IJ=IJ+NA
AUX(IJ])(RNX(JX)+RNXYJXT*T1.-RI~AT/RLA)/
IJ=IJ+NA
AUX(IJ)=-RNXY-JXI/(RLA*u7;+RLA)*S
25 CONTINUE
JV=INS(M)-2
-IJ=JV+INS(M-2)*NA
-
~~
JX=iNTfMJ+2
AUX(IJ)=(RNXY(JX)+RNX(JX))/S
IJ=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)/S
IJ = Ij4-NAAUX(IJ)=(RNXY(JX)+RNY(JX))/S
IJ=J V +NS(M)*N A......- ..
AUX(IJ)=RNX(JX)/S
RMA =(Y B(M)-Y(3)/D[ELX
JR=INT(M)+3
JV=INS(M-1
-
JT=INS(M-3)+2
AJ=JV+CJT-1) *NA
AUX( IJ)=-RNXY(JR)/(2.*S)
IJ= IJ+NA
AUX(IJ)=-RNX(JR)/S
IJ=iJ+fNA
AUX(IJ)=RNXY(JR)/(2.*S)
JT= INS (M-2T+2
IJ=JV+(JT-1)*NA
AUX(IJ)=RNXYT3RT7S
IJ=IJ+NA
AUX(IJ)=RNXLJR)*4./S
IJ=IJ+NA
AUX ( IJ =-RNXY( jR )/S
*,*
03/59/19
J08
L 1, MOD
l
DATE =
FILL
68168
03/59/19
JT=INS(M-1)+2
IJ=JV+(JT-ITYNA
---
AUX(IJ)=RNXYIJR)/(2.*S)
-IJ=IJ+NA
AUX(IJ)=-RNX(JR)*5./S
IJ=1J+NAAUX( IJ)=-RNXY(JR)/(2.*S)
IJ=JV+INS(M*NA
AUX(IJ)=RNY(JR)*(RMA-1.)/(t2.+RMA)*S)
IJ=IJNA
AUX(IJ)=RNY(JR)*(4.-2.*RMA)/f(1.+RMA)*S)-RNXY(JR)*2.*RMA/((1.+RMA)
1*S)
IJ=IJ+1'A
AUX(IJ )=-RNYTJR)*T3.-RMA)/(RMA*S) -RNXY(JR)*2.*(l.-RMA)/(RMA*S)
1+RNX(JR)*2./S
IJ=I J+NA
AUX(IJ)=RNY(JR)*6./(RMA*(1.+RMA)*(2.+RMA)*S)+RNXY(JR)*2./(RMA*(l.
1+RMA)*S)
RETURN
END
-----------
03
--
OD
DATE
-FIRSYS
~~~--~~~
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),
1t-7tr~t-7trt15ttN(2)-----NS
ZXY(157),AUX(24649), VEC(157), X(21),ZX
(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**2
B&8-1+ZYJXt**2
BC=ZX (JX)*ZY(JX)
--BDAtBA+BBi-.**O5A=SINIPHI(1+1) )**2
- 8=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*C
90~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 )**2
61 VEC1~K-TVEtTfGA*RA *X1+1
60 CONTINUE
XX K) *R JX I JT )+2 YY C K) 1*RJY U JT )
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)
) I 1I)+R-JYB-----Y iJV )*S I N (PHI (I +1) )*COS PHI-(
-- I4CO SIPHl-ii-)-**2-2*ZX
2(I+1)*CDEII)+RJXB(
3S=M2--
00 32 I=1, MS
-
I+1)*FJHII)
---
--
--
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 X
AUXIJ)=(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
3-REAVYBt
)-Y(JA)-DELX)
j-Y"IJAX)IIO/Df
35,35,36
X
__
-----.--..-.----...-.--
-.
IFI1-1)-38#391,38
3 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.+
TRJf1)*L.....-2--RN)71 RNA*REA)-)-ST-RI-J-A-)4- 1.-- +RM A )4*RMA/ (-(-10e- +REA-) *RE A)
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+
1-0-
(REARNA)*RMA*
!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
f
AUX(
RNA
)=(1.+RM.A)*IRI4A-
J*RL
A/
(-RNA*S)
-1AIJ)=
(
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)
J)=-4**T
AUXLI
AUXL IJ)=-4,P*V
AUXLIJ)
(1.+RMA)*RLA/
=RMhA*
(2.*S)
IJ=IJ+NA
--AU'XCUJ-P)-5,e*1
-------------
-----
-
-
----
-
-
IJ=IJ+NA
U XtI-JTw5-,*4V
,-A
-
_
_
_
_
_
_
_
_
-
--
_
_
_
_
_
_
_
__
_
_
_
_
_
_
_
IJ=JV+( INS-1 M-1)+JA-3)*NA
IAU=I
J
NA
-
-
-
-
~
-
-
-
-
-
----------
-
-
-
I J=I J+NA
-AUX
IJ=JV+t
(IJ-);-2
INS
MS=M-1
JZ=INTI
1+2)
V
,
(N
-_
)+JA-3)*NA
_
_
_
-
_
_
_
_
_
_
_
_
_
_
_
_
ELIFMOUA------------FISYS
RLA=FJ-(
--- DATE--= -68167
---
1
-
/
1/l
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
--- - - - ----------------AUX(IJ)=AUX(IJ)-13+2.*RMA)*RLA/RMA*S)
-I-J=J+NA--
---
---
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
R LA =H I J I)
50
1
4. *S J
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
- I--.-----__-_._
-- 7UXVIII--=AUX (-IJ-)+ (-IY +RMA-)*RMA*RLA-/2-.------ .. l_. -__.
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+NA
tAUUJ-)=-AtJX(-J)-* 2-+RMA)*(1- RMA)*RLA/(RE A*(1 + R EA)
IJ=JVNl INS 11+1 )-3 ) *NA
-
-.
.
..
.
.----
-----
EL~vMOB4-----------IRSY
-~
--
---- A-UXtIJ)-AUXt--J)-3-*RMA*(2+RMA)*RL-A
--
DATE--=--68 L67-------
1T3-5t------------_
________
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
IJ=JV+ A INS 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)
GO TO 48
*RLA/ ( -RN-A* U-+RNAf-)--
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 L
CO 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,
---
48
t4OO--------FIRSYS
-----
DATE
~
=68167
17/31/51
-
( RNA* ( j.+R NA))-
AUX-1-J.AVX/--+A*2-/
---
--
-_
CONTINUE
JZ=i NTLH+1lJV=INSIM)
JA=JV -NS-IM-1i--RLA=HIJ( M-1)/14.*S)
-fMA1-Y-BM-)-Y4Jk)t/DEL-x-
---
-
-
__
_
T=-RMA*(2.+RMA)*RLA
V(i2+R MAI*1.+ RMA) *RL A/-2.-IJ=JV+IINS ( M-4)+JA-3)*NA
tX( t-1J=AUf~ti)-1IJ=IJ+NA
AUXtP-J+=AUXt-JJ-V
-_________
---
IJ=IJ+NA
IJ=IJ+NA
AUXJI-=AUXtIJ)+
IJ=JV+t INS (M-3)+JA-3)*NA
+2---------AUX-=AUX
I
J=I J+ NA
IJ=IJ+NA
AUXtIJ=AUX (i-J4+2.*V--------------------------I J=I J+NA
AUXVI-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
-
---
-
-
--
-
t21rn-------
IJ=IJ+NA
IJ=JV+
INS(M-1)RJA--3)*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
-
----
--
---
AUXThJAAUX1TTV-
IJ=IJ+NA
AUXLTTj)=AU-
-
-
-
_
--
------
-
~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--
- -
- -
- ----
--
-
-
_
_
_
_
_
- DAT -Ei-OA--------E
-Y 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-iNT1INS,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
A* Q .*BCm-AC--) *Z-XXYI [)+B*t24A Bt
-AUX- (-I-JJ-=AU X-4-1-J1-i-A C*B *ZX-XX-UI
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
IF(FI-MS)
0D~I-----------SECYS
DATE
---
68167
1
/
15~
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+NA
AUX 4 I -)--AUX-IJ )RL-A*t2.fRMAI-*4
GO TO 20
IF(YBiI)-YtJA)J'DELX)
GO TO
1.-+R M Ail1I RNA*- 110+ R NA-)--
16916,17
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
SC-SDATE---68L67---7
VE~jMQA
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(LIJ)=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+4iRMA))
JQ=-INS(M ).I
IJ=JQ+1NS{ M-3)*NA
IJ=JQ*I NSI M-2)*NA
IJ=JQ+ INS (M)*NA
TM*fl+M----
-1---
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
l(JR))
-IFH-ND
110
)-14
.
l
10
J Ri*2.-EPYD X (JR) -P HID (M+1)* (EPX (JR) -2. *EPY
-
REWIND 8
IJ=1+(I-1)*157
120 READ(8) (AUX(K),K=IJJT)
1-CAL-.SYST-EM-157i-AUXVEC-SOL ,NAKKK)
IFfIND) 112,113,112
113-00 421-L41, 57
IJ=1+ (1-1)*157
-
121 WRITEL8)
--
___-__
LAUX(K)PK=IJJT)
100 DIA(I)=DIS(I)
DO-30-I1,9M-JX=INStI)+1
---3V=INStI~elDG 31 K=JXJV
1T=1NT IT4+KT)tNNST-~
31 DISIJT)=SOL{K)
IF(Y(JA)-YB(1+1))
-
------
30,35,30
35-J11rNT-(~+t)INSAI*14-INS (I)---DISiJT+1)=DIS(JT)
-------------3--ONTINUE---RETURN
- END.---
- -
-
__
120
EVEL
1, MOD
1
DATE
SYSTEM
SUBROUTINE
=
68164
01/12/27
SYSTE H( NA ,C ,X ,NALLL)
DTMEN5TON-A24-649TC-(57) ,D (157) T I POS (157)
DOUBLE PRECISION D
GL~ TO T,400 F)~.LUL5 DO 65 I=lN
00 68 J=2,N
J- 1)*NAIF(ABS(Q)-ABS(AIJ))
,
X(157)
ITJT+
67,68,68
~ET T-=~AT1~JT
68 CONTINUE
IJ=I+(J- ])*NA
~~- -- ~ ~
69 -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,Nf
DO 20 J=1,N
20 D(J)=A(IA)
JAUX~=I
DO
T
~~
-
-
~.. -
----.-----.-.
50 K=1,JAUX
IA=K+( I-1)*NA
D(IP ) = (K)
JXP=K+1~DO 50_KL=JAUXP,N
IAL KL + (K ~i NA
50 D(KL )=D(KL)-AI IAL)*A (IA)
28
--
II.=+(I-1)*NA
IPOS(L )=IP
Ix=l+1
IFT-D6NT03T,3i0,
31 DO 60 KM=I X,N
IAM=KM+TI-1)T*NA.-
-
12
1,
-VEL
MOD
SYSTEtM
1
DATE
=
01/12/27
68164
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
CVT0~T~=TiN
110 D ( I )=G I)
400UO
~
~-
H
-
-
,E14.7//)
-_______
DO 200 l=1,NM
-IPETPSTT~
C( I )=D( IP)
T
UTiTPJDT
-___-___-_
IJK=I+1
L JK---N
-DO- 200
JI=J+ (I-I )*NA
2OA i~TJ=DtJ~FAT JI) *C(INN=N+(N-1)*NA
-
XTNT-DTNT7/ATNNT
.----
--------
__
_______
NAUX=N-1
ITU -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 )* NA
AT 2
LLL =2
5XTZ
CT1T1T
ENTUN
E ND
-
-I--
_
_
_
_
_
_
_
-------------- --- -------- - --------- --- ----- ---- -- ---- ---------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=IB41
DU 10 I=IBPIE
-----------------------
11 DUMAX=X(l)
10 CONTINLE
20 UMAX=DUPAX
RETURN
---------------------------------------------------------------------------------------
------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
------------------------------------------------------------------------------------
123
'I
VEL 0,
MOD
0
PRI
SUBROUTINE PRI(Z ,INTMP,LR)
DIMNSN5ION Z (175) , INT ( 22)
DO 10 I=1,MP
JR= INT ( I )+1
JP=INT( I+1) -L R
WRITE(6,11) (Z(K ),K=JRJP)
11 FORMAT(8(2X,E14. 7))
WRITE(6,T12)
12 FORMAT(/)
10 CONTINUE
RETURN
END
DATE =
68094
19/57/45
-------
--
-----
-----
-----
------..
.
.-
~
S~~i~VFU1CTP
?R-Ny-t N)( Y I,7 X ZY', SI ,THNMff
D IMENSI ON RNX( 1*75) ,RNY( 175) ,RNXY(175) ,ZY(175)
On 10toI?1MP
JV=INT( 1+1)
A = I+7 X fK
)4?
C=ZX (K BZY( K.
20.5)
12 THfK)=1.
10 CONTINUE
RE-TURN
FND
~---------fa
,7Yf 175)IT Ht 175), TNT(
-----------
------------------------------------------------------------- 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
II
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
REIURN
END
----------
----------------------------------------------------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
-----------------------------------------------------------------------------------
j
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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, Proceedings 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",
Book company, 1965
IX.
"Theory of Arch Dams",
1965
S.H.
Crandall,
J.R. Rydzewski,
Mc.G-aw Hill
Pergamon Press,
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 Departement, R65-57, November 1965.