STRESSES AND DISPLACEMENTS IN SEMI-INFINITE MEDIA
by
ARISTIDES BRYAN DOMINGUEZ
Ingeniero Mecdnico
Universidad Nacional de Buenos Aires
ARGENTINA
1962
Submitted in partial fulfillment
of the requirements for the degree of
Master of Science
at the
Massachusetts Institute of Technology
1966
(2
Signature of Author
....
.
.
....
Department of Civil Engineerin
Certified by . . .
. . .
.
Thesis Supervisor
Accepted by . . . .
. . .
. . . .
. . .
. . .
. . .
. .
Chairman, Departmental Committee on Graduate Students
- 1 -
ABSTRACT
STRESSES AND DISPLACEMENTS IN SEMI-INFINITE MEDIA
by
ARISTIDES BRYAN DOMINGUEZ
Submitted to the Department of Civil Engineering on
in partial fulfillment of the requirements for the degree of
Master of Science
This thesis presents a general computer oriented
method for the determination of the components of the stress
tensor and the displacement vector, the principal stresses,
and the principal directions of stress at any point of a
semi-infinite elastic medium subjected to static normal and
shearing surface loads.
This method has been programmed in FORTRAN IV for an
IBM / 360 digital computer and the program, with slight
improvements, will also provide the solution for the
homogeneous linear-viscoelastic half-space and static
loadings.
Thesis Supervisor:
Title:
Fred Moavenzadeh
Associate Professor
- 2 -
ACKNOWLEDGMENT
The preliminary investigations and the
preparation of the method herein described have been sponsored by the Inter
American Program of the Civil Engineering
Department of the Massachusetts Institute
of Technology to which the author feels
indebted.
-3-
TABLE OF CONTENTS
Page
CHAPTER
...........
1.
INTRODUCTION
2.
LOAD AND LOADED AREA .
3.
COMPUTATION OF STRESSES AND DISPLACEMENTS
16
4.
IMPROVEMENT OF ACCURACY
..
20
5.
COORDINATE SYSTEMS
..
43
6.
PRINCIPAL STRESSES . .
. .
7.
PRINCIPAL DIRECTIONS .
.
8.
COMPUTER PROGRAM . . .
. . .
9.
RESULTS
10.
.
.
.....
.
.
.
.
..
.
.
.
.
.
.
5
.
.
.
.
.
.
.
.
. .
. . . .
11
49
.
52
.........
.
56
. . . .
70
............
CONCLUSIONS AND SUGGESTIONS FOR
FUTURE WORK. .
.
81
APPENDIX
A.
DEFINITION OF SYMBOLS .. .........
.
82
B.
C.
D.
LIST OF FIGURES . . . ...
.......
LIST OF TABLES . . . . . . . . . . .
STRESS AND DISPLACEMENT COMPONENTS
FOR UNIFORM NORMAL AND SHEARING
85
87
COMPUTER PLOTTING OF VERTICAL STRESSES 125
129
LIST OF PROGRAMS . . . . . . . . ...
CIRCULAR LOAD
E.
F.
86
. .
-4-
. . .
.
.
.
.
.
.
CHAPTER
1
INTRODUCTION
Analysis of stresses and displacements in semi-infinite
elastic media is of special interest in several problems of
soil mechanics, such as the design of foundations for structures, highways and machinery.
In such cases, a load is
distributed over a relatively small area of a body which is
only limited in one direction by a plane surface, it is
generally assumed that this body is weightless, homogeneous,
isotropic and its behavior is linear elastic, Figs. 1, 2,
and 3.
This is a restricted case of the more general problem of; analysis of stresses and displacements in layered,
non-homogeneous, non-isotropic semi-infinite media with
time-dependent properties, subjected to arbitrarily distributed time-varying moving normal and shearing surface
loads and subjected to different interphase conditions
between the layers, Fig. 4.
Most methods of analysis ( see references pages 223
and 124) have attempted to present the solutions of the
homogeneous half-space in the closed form, and due to the
mathematical complications in obtaining this-type of solUtion for arbitrary load distributions, most of them have
been limited to axisymmetric cases whereby the load is
-5-
distributed over a circle.
Closed form solutions for dis-
tributed loads over asymmetrical shapes generally involve
elliptic integral expressions which have limited their
application.
The solution of the basic problem of a normal point
load on an elastic homogeneous half-space was obtained by
Boussinesq in 1885.
Terazawa in 1916 developed the solu-
tion for the stresses and displacements at any point in
a semi-infinite elastic body under distributed normal loads.
This solution is in the form of infinite integrals involving Bessel-Fourier expansions and is applicable to any
distributed axi-symmetric loading.
Love solved the same problem through the use of potential functions in 1929.
His work was summarized and exten-
ded by Fergus and Miner in 1955.
Love's work was used to
solve the problem of a uniform load distributed over an
elliptical area by Deresiewicz in 1959.
A fairly extensive tabulation of stresses, strains,
and deflections has been given, for arbitrary Poisson's
ratio by Ahlvin and Ulery in 1962.
The most practical method of solution up to date is
based on the influence charts developed by Newmark
( 10 and 11 ) Which are general and have sufficient accuracy for most of the engineering applications but are in
disadvantage of requiring excessive time for each point of
.6 .
the semi-infinite medium.
Design is thus somewhat limited
by the lack of a flexible and accurate method that enables
the design engineer to describe the characteristics of the
problem (geometry of the loaded area, load distribution,
properties of the materials), to obtain in real time the
desired distributions of stresses and displacements at any
point of the body, to modify in turn his original conception, and to introduce new data and thus follow the steps
of successive approximations of the design process.
These
can be summarized by stating that the most desired characteristics of any method of solution for an engineering
problem are:
a.
Clear, accurate and simple description of the data.
b.
Great flexibility of the method itself to accept
and produce different types of information corresponding to different situations.
c.
Known and adjustable limits of accuracy.
The purpose of this study is to present a general
computer oriented method for the determination of the
components of the stress tensor and the displacement vector,
the principal stresses and the principal directions of
stress at any point of a semi-infinite elastic medium subjected to static normal and shearing surface loads.
F/d. /
4 to,Y, Z)
PbrYJ
AI77FTimhrx
x
PIG. 2
A (x, y, z)
9 (XYJ
x
f/6. 3
A (xy, L'
- 8 -
V(X, y, a')
x
L area
F/G. 4..
-9
LA vq
2
LA
r 3
lood
/oddD's//bu/oo
'oP/ 11/ /ow1 0I.M.s/,14/o
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-
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p(~w;y, z,
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Orz(I6
Cz
6
w
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-
10
CHAPTER
2
LOAD AND LOADED AREA
To provide the method with a fair amount of generality,
two basic types of load distributions are considered and
superimposed:
I.
II.
Distribution of normal surface stresses p(x,y)
Distribution of shearing surface stresses q(x,y)
In the case of the shearing surface stresses it is
assumed that they are parallel to the X axis.
This assump-
tion does not imply any limitation for the following reasons:
1.
2.
The X axis can be selected arbitrarily.
The expressions of the stress components can be
modified if it is desired to have the shearing
stresses parallel to the Y axis.
3.
In case of stresses parallel to another axis in
the plane X,Y they can always be decomposed in
X and Y components and their total effect can be
calculated using the principle of superposition.
2a.
LOADED AREA
The loaded area shall be enclosed within a grid of
M x M square elements, as shown in Figure 6.
element will be characterized by two numbers:
i = row number
J = column number
_ll
-
Each grid
,9*91Y
z
z
/Z7
I-7Z
A7Z
z
/Ov
z
2b.
LOAD FUNCTION
The load function is defined by two square matrices
of order M, P(I,J) and Q(I,J), associated with the grid.
A.
Normal load function
Each element iJ of the first matrix will contain a
number pij equal to the average intensity of the normal
surface stresses applied on the element of area Aij of
the grid (Fig. 7).
B.
Shearing load function
Similarly, each element ij of the second matrix will
contain a number qij equal to the average intensity of
the shearing surface stresses applied on the element of
area Aij of the grid (Fig. 8).
The elements of normal and shearing loads acting on the
grid shall be treated in two different ways:
a.
As elements of distributed loads of intensities
PiJ and qij respectively.
b.
As equivalent point loads applied at the center of
each element ij, of intensities
Pij=
Qij
ij x Aij
(1)
qij x Aij
(2)
-1I3
-
MiP ampij.
4
a
y
74k
f/c.
7
,oo-l-z
-/:!3(m
- 14 -
This discrimination shall be done according to the
distance from the point P(x,y,z) of the semi-infinite
medium to the center of the element iJ of the grid.
This
distance is a measure of the error incurred in considering
distributed loads as point loads ( see article 4e ).
- 15-
CHAPTER
3
COMPUTATION OF STRESSES AND DISPLACEMENTS
Due to the assumption of linear elastic behavior it is
possible to use the following principle:
3a.
PRINCIPLE OF SUPERPOSITION
The stresses and displacements that occur at a given
point P(x,y,z) of the semi-infinite medium are equal to
the sum of the stresses and displacements produced by each
individual element of the load matrices P(I,J) and Q(I,J).
Let as before:
piJ = average normal surface stress on element ij
qiJ = average shearing surface stress on element 13iJ
G k
= component of stress tensor at point P(x,y,z)
due to total load.
where
k = x,y,z
L = x,y,z
= component of stress tensor at point P(x,y,z)
AGkL
due to load element iJ
AU k
=
component of displacement vector at point P(x,y,z)
due to load element 1J.
- 16 -
Then by the principle of superposition:
jal
jV
U,,ff" L. *Ip
K.. J
(2
M = order of load matrices P(I,J) and Q(I,J).
3b.
MATHEMATICAL PROCEDURE
For the computation of the components
G61
of the
stress tensor and U, of the displacement vector, the
elements of distributed load piJ and qiJ are substituted
by the equivalent point loads given by expressions (1)
and (2) page 13, and the numerical values of JG and
are then obtained by Boussinesq's expressions for
ALUk
normal and shearing point loads (Appendix D, formulas (1)
to (12) ).
This representation of the load function by a finite
number of point
loads has inherent the following sources
of inaccuracies:
a.
Point loads produce infinite discontinuities in
the stress and displacement distributions at the
point of application.
b.
The distributions of the stresses and displacements
for the distributed loads and the equivalent point
- 17 -
loads vary significantly in the vicinity of the
point of application of the load; this variation
gradually reduces as the distance to the point of
application is increased.
Fig. 9 shows a compari-
son of the distributions of 6Z for a point load of
radius R = 1 at a relative depth Z/R = 1.
The rate of decrease of this difference varies
with the intensity and the radius assigned to the
equivalent distributed load.
c.
As a consequence of b, the degree of accuracy will
vary with the size of the grid elements or equivalently, with the number M of subdivisions of the
loaded area and with the distance from the point of
the semi-infinite medium.
This point is discussed
further in article 4e of the next chapter.
- 18 -
15
Ps p..34'
K
"iI
R
f0
•x
z
f1. 9
0
I
1
- 19 -
I
2
Xln
CHAPTER
4
IMPROVEMENT OF ACCURACY
The evaluation and the improvement of the accuracy of
this method requires further analysis of points (a),
(b),
and (c) of article 3b.
4a.
ERROR VOLUME ASSOCIATED TO A LOAD ELEMENT
It was mentioned in section 2b, page 15, that a
measure of the difference in the distributions of stresses
and displacements corresponding to point and distributed
loads is the distance from the point of the semi-infinite
medium that is being considered to the load element ij.
Assuming that the magnitude if the accepted error is
t 4faZ, it is possible to define for each load element
ij a volume of the half-space outside of which the error
is less than
It±4..4.I .
This volume shall be designed as
the "error volume" associated to the load element iJ.
Where the point P(x,y,z) of the half space is within this
error volume, the element ij is treated as a distributed
load.
The actual shape and dimensions of this volume
will depend on items (a),
section.
(b), and (c) of the previous
After the computation of the preliminary results
of article 4e, it was found convenient and simple to
adopt as an error volume a square prism of side
depth ZM1 .
- 20 -
2p
and
z//7
1-41,
7
i
i
-
m
-7
m
7
I
,
7
z7
7 -- 77
7
7
x
"1
00
Z
z
x
1000"
"orZ
7
7.1-1-
/
/
...,
70
10
Zmz
O-e
/
Z
7
7
z
7
1-
000
7
17
IY "e,
/, XI /-/
-,
z
--
-
r--
I
m J
I
fIG. /O
of load
z0
m
'00
z
m
d
/__-- _=8=__
Vo/me
1,0
z
1-
Error
11
e
"OrI
Wz
7_7
z
'4
- -
Zlir
Z
7
0
C7~7L7Z
z
L
pIzIIZ
A
f- -
jI
7
717
I
Aczzrz/zztI~
1
17 l
z71-71
7
A
Y
"I
Z
A
z
7
7
1
'
7
7o
7
-z
7
1-
z
/
yAol
z
Yo
m
z
7
-7
m
7
/Remem ij
m
IY
YLIM
7---
; I I I 1I
L
I
I
I1 -
I I I II I I 1I I I
I
2
P,
I
ATFFFFFF o
-----------------
,-
--
ZIA
w le
1~~~~~~~
iz
PIG. If.
frror vo/lme of fW ka,..
- 2e
7N
I
II
II
II
I
x
4b.
ERROR VOLUME ASSOCIATED TO THE TOTAL LOAD
The error volume as corresponding to the total load is
defined as the integral of the error volumes of all the
load elements ij as shown in Fig. 11, and its dimensions
are 2XLIM, 2YLIM and ZLIM.
4c.
NUMBER OF DISTRIBUTED LOAD ELEMENTS
The procedure to determine these elements is very sim-
ple, and is used only when the point P(x,y,z) of the semiinfinite medium is within the error volume associated with
the total load.
Drawing the error volume of side 2 f and depth ZLIM
with its longitudinal axis passing through P(x,y,z) as
shown in Fig. 12, the intersection with the grid encloses
the elements to be counted (only those whose centers are
within the square of side
-23
2
P).
-
*91J
4d.
PROCEDURE TO HANDLE THE DISTRIBUTED LOAD ELEMENTS
Assuming that the load elements shown in Fig. 12 are
those to be treated as distributed loads, when considering
them one by one with their actual shape, the stress and
displacement components djM and 4J4
have, for arbitrary
points of the semi-infinite medium, complicated expressions
generally involving elliptic integrals.
Even the expres-
sions for uniform circular loads and arbitrary points of
the half space involve elliptic integrals, but the latter
can be reduced to rather simple forms for points on the
vertical axis passing through the center of the circle.
These expressions have been obtained by integration of
expressions (1) to (12) in Appendix D.
This point suggests the following approximation,
which has been programmed and used successfully in this
study.
Considering the results obtained in article 3c, in
order to maintain the error to
eimsx = ± 1%, the length
of the side of the error volume must be:
2
p
= 0.4 x R
R = maximum diameter of the loaded area.
- 25 -
The size of the grid divisions must be:
i
= 0.5 xp
The number of grid elements contained in the side of
a square of side length 2P is:
N=
P
0.4,R
a
0. X
0.
0.8
MR IV2
o.2
0.4aR
These results are illustrated in Fig. 13.
-1 --
,
-.
U
I LJ
I
A
PR
FIG. 13.
As it will be explained in Chapter 5, the center
of coordinates will be located at the center of the grid
of 20 x 20 elements and position of each grid element is
specified by the coordinates of its center, it was found
more convenient to assign to the side of the error volume
a length equal to 5 grid elements as shown in Fig. 14.
- 26 -
Grid Order M, 20
A
PIG. j
1;
I.,"
Square
I LOO •I
".p
ABCD = error volume associated with element ij.
This square is divided into three concentric square rings
as shown in Fig. 15 such that:
F/G. f5.
- 27 -
RI
=i
/2
= radius of ring 1
R2 =3Ap/2 = radius of ring 2
R3 = 5 p/2 = p = radius of ring 3
KlMAX = 1 = number of elements in ring 1
K2MAX = 8 = number of elements in ring 2
K3MAX = 16 = number of elements in ring 3
Kl = number of elements enclosed by ring 1
in a particular case ( 06K16 KlMAX )
K2 = number of elements enclosed by ring 2
in a particular case ( 06 K2_K2MAX )
K3 = number of elements enclosed by ring 3
in a particular case ( OA K3:K3MAX )
The approximation can be described as follows:
1.
Check if point P(x,y,z) is within error volume
corresponding to the whole load function.
If NO use expressions for point loads (expressions (3) to
(20) ).
If YES use algorithm 2.
2.
Draw prism of radius RI =
with longitudinal
axis passing through P(x,y,z).
3.
Count the numbers Kl, K2 and K3 of load elements
that fall within rings RI, R2 and R3 respectively.
- 28 -
Calculate the averages
4.
It/
a~
Z
E P
Al
jul
It
Z
P'O .9
K2
X(2
x3
SPu
avJ =
X3!
5.
4.3
Calculate the fractions Kl/KlMAX, K2/K2MAX,
K3/K3MAX.
6.
Place equivalent uniform normal and shearing
circular ring loads of intensities:
on top of the square rings 1, 2, and 3 respectively.
7.
Compute 46,
and
4k
using formulas for points
on the vertical axis of circular loads (Appendix D).
8.
Multiply the results by the fractions obtained
in (5) respectively.
9.
Add these results to those corresponding to the
other elements of the grid considered as point loads.
- 29 -
This method of handling the elements within the error
volume involves two types of approximations:
a.
It considers a number K of square loads as a
fraction K/KMAX of a circular load.
b.
It reduces the actual distribution of surface
stresses pij and qij within the three square
rings to the six average intensities shown in
page 29.
The determination of the errors introduced by these
two approximations leads to the following considerations:
With respect to approximation (1),
the diagram,
Fig. 16 ( see also reference 12, page 124), shows the
distribution of vertical stresses 6z
for equivalent circu-
lar and square loads ( for the same loaded area and the
same load intensity ) as a function of the depth Z.
For
X=O and Z=O they are coincident and for depths Z/R
between 0 and 1 the percentage error is less than 0.01%.
With respect to approximation (2), the results obtained
by this method have been compared to those obtained with
Newmark's influence charts for the same points of the
error volume of the total load, and an accuracy of the
order of 99% is obtained for well behaved load functions.
- 30 -
jerQriC4L P*Ar$JV,9,f
i QUOqr (S;Oe
r
FIS16
i
ZIR
- 31 -
-.pwivs Q)
4e.
SIZE OF THE GRID ELEMENTS AND RADIUS OF THE ERROR
VOLUME.
In order to determine the influence of the size of the
grid elements on the accuracy of the procedure and in order
to determine the radius of the error volume corresponding
to one grid element, the following procedure was developed:
a.
Consider, for the purpose of this discussion, the
loaded area as a circle of radius R,
form circular load of intensity
b.
with a uni-
D = 1.
Calculate the stress and displacement components
for points on the Z axis (vertical axis passing
through the center) using formulas obtained in
Appendix D.
c.
Draw a circle of radius P
concentric to the first,
and consider the load on top of this circle as
uniformly distributed.
d.
Divide the anular ring of width (R-f) into m concentric rings of width 4p, where Ap will be equiva-
lent in this case to the length of the side of
one square element of the grid in a real case.
e. Divide the anular ring (R-f) into n equal sectors,
n being calculated as follows:
2
p 1p
- 32 -
Place on the centers of each of the m x n elements
f.
of area Aij an equivalent point load of intensity
P1
Ai
=
i - A1
Calculate the components of the stress tensor and
g.
the displacement vector for the same points of
the semi-infinite medium as was done in (b) above.
Calculate the difference between results of h and
h.
b, and calculate the percentage of error based on
values in part (b) above.
i.
Repeat this process for different values of
dp/p
and w/q"
Define a value of the accepted error,jewex %I and
j.
select the number m that produces this error.
Take this value of m one half of the order of
the load matrices or equivalently as one half of
the number of divisions of the grid sides.
This process was carried out only for the stress
component 61 and the following data was used:
0.0$ /o 0.20
f /I
/
= 0.-/ lo 0.
wA/ incremenos of
Wh ;remen/s
w5
- '33 -
0.025
of 0.
Then for any value of
:
d;*r //i
. o.
a
'P IPi -= o. 2
I
I
I
PiI/R
Ip
1
I
I
-
I
0.
I
,I
I5
For each case Z/R was varied from 0 to 4 with intervals
A Z/R = 0.1
This process was programmed for an IBM 360 computer, the
flow chart and the program are shown in pages
36 and 37
repectively.
Table II
(page 39) shows in a compact form the final
selection of M (order of the load matrices).
Adopting a magnitude of the error
It£ma', %XI =I %
the value of M that produces the closest error interval is
(see Table II):
M = 2m = 20
and for M = 20
p
= 0.2 x R
o3 = 0.5 x
=
Radius of error volume
=
Magnitude of a grid element.
- 34-
TABLE
I
SYMBOL
SYMBOL
IN
PROGRAM
Intensity of distributed load
q
Q
Radius of loaded area
R
R
DESCRIPTION
RHO(J)
Radius of central element
Width of the concentric rings
fPjk
DRHO(JK)
Coefficient of width of the rings
a
A(J)
Number of divisions in each ring
n
FN(J)
Number of rings
m
M
Depth
z
Z
Vertical component of stress for
each depth, corresponding to
uniform load distribution
Idem for mixed distribution
zi
6
SIG(I)
S
Error
EPS.
Depth subscript
I
Center element radius subscript
J
Ring width and width coeff.subscript
k
K
Ring subscript
1
L
of circle
d
D
Int.radius of the rings
r
RI
Distance from point loads to center
- 35 -
REA/) J
iJ.I R
.I
a.~ J
E" t}z" /IYAX J.?~"~ .lj4"?-I J /'s"ff,,j}
Q2/
a.:J"
a~
YES
m
R- ./'(J)
=
L1jJ(.Lk)
1 = I - /I1AX. iJZ
C=
yEs
6-G(J)
./00
6(1)
J
Q.5
/JOB
/FTC
C
GO
10
1
2
16
30
3
4
20
5
6
DOMINGUEZ A B
6032
BPR
STRESS DISTRIBUTION STUDY.
DIMENSION SIG(200),DRHO(10,10),RHO(10),A(10),FN(10)
READ (5.1) QRIMAXoZ,DZ
FORMAT(F2.09F2.0,14,F5.1,F5.1)
READ (5#2) (RHO(I)Il=1,7)
FORMAT(7F5.3)
READ(5*16) (A(K)*K=1,5)
FORMAT(5F53)
WRITE(6,30)
FORMAT(11X,1HI10OX,3HZ/Rl3X,6HSIG(I).//)
DO 4 I=1,IMAX
SIG(I)=Q*(1.-(Z**3.)/(SQRT((R**2.+Z**2.)**3.)))
WRITE(693) IZ,SIG(I)
FORMAT(I12,8XF5.1,F15.8)
Z=Z+DZ
WRITE(6.20)
FORMAT(1H )
DO 5 J=1#7
DO 5 K=1,5
DRHO(JK)=A(K)*RHO(J)
DO 6 K=1,5
FN(K)=3.14/(2.0*A(K))
DO 9 J=1#7
WRITE(6,25)
25 FORMAT(1H )
DO 9 K=1,5
WRITE(6,26)
26 FORMAT(1H )
WRITE(6,31)
31 FORMAT(11X,1HJ,3XIHK,1OX,6HRHO(J),1OX,9HDRHO(JK),/)
WRITE(6,15) J,K,RHO(J),DRHO(JK)
15 FORMAT(10X,12,2X,12,10X,F6.3,4XF15.8,//)
WRITE(6,32)
32 FORMAT(11X,1HIt11X,1HZ,15X,1HS,19X,3HDIF,17X,3HEPS,/)
M=(R-RHO(J))/DRHO(J*K)
AIMAX=IMAX
Z=Z-AIMAX*DZ
DO 9 I=1IIMAX
RI=RHO(J)
S=Q*(1.-(Z**3.)/(SORT((RHO(J)**2.+Z**2.)**3.)))
DO 7 L=1,M
P=Q*3.14*((RI+DRHO(JK))**2.-RI**2.)/FN(K)
FL=L
D=RI+DRHO(J,K)/2.
S=S+FN(K)*3**P/(2**3.14*(Z**2*)*(SQRT(1.+(D/Z)**2,)**5*))
7 RI=RI+DRHO(JK)
DIF=S-SIG( I)
EPS=DIF*100/SIG( I)
WRITE(6,8) I#Z,S,DIF#EPS
8 FORMAT(112,8X,F5.1,5X,Fl5.8S5XF15.8,5XF15.8)
9 Z=Z+DZ
GO TO 10
END
- 37 -
/DATA
0.1
40
0.1
1.1.
0.0500.0750.1000.1250.1500.1750.200
0.1000.2000.3000.4000.500
/END OF FILE
- 38 -
TAB/I 7
&
0.050
.
r
0. 100
O./2
o.f
O. 2
0.3
0o.
0.
o0050
0.5
to
o. oZ0
400
0o
/.3
/00
0o
8
0.
0. o0075
267
0.2
0.3
0. o/Jo
0 . 0225
0.4
o. 5
0.0300
o037.50
0. ofo o
o. o/1,o
. 0200
0.175
o. 0/?1
0. 0000
. 0.53
0.0/00
C.
/0.
0.46o3
O. 000o
/34
B9
5
3.3J
/. 8427
o. 456
O.0398
67
S
2.s
,
4. s547
4 . s 23
0..20
. 063/
/0
0. o,8
a 0/47
5
. 33
.
a0.0457
2oL9
. ooo00
. o o00
200
0.2
0. 0 200
/00
5
03
0. /329
o. oaoo
0. o277
67
3.33
0.
0. OS47
0.4
0 odoo
5o
.5
0./
0.2
2.5
0. Osoo
0.0/25
4o
/6o
0.3
4
0o.o0so
0.0375
05oo
0.
o.5
o. o 62
3. 666/
. oo00o
o. 4f 7
. o0374
0 . /04 /
0. o0479
5
0. /0
2. 290
4. .5r/0
. o000oo
0. o0oo
. 9536
o. 16 03
o
~5
4o
J.2
3.33
2.5
/34
/o
5
0.3
o.4
os
67
5
o.
0.2
o. o07
0.0450o
1/4
57
0.3
4
0 o 55
o0-oo
29
/
Ca2
0.3
0.
0.
_
o.o
0. 0200
0. 0400
00oo00
. 0too
a,
27
38
3
/oo00
0
33
25
o. /ooo
00
3. 33
.
2
/0
3
2.5
d.6323
/. d2'/
. 27o1
0.o 37
4. ss3
.oooo
0. ooo
0. oooo
o. ooo
a. oooo
0o. 44is
. 6S85
.o.ooo
0. 000ooo0
6. 83 05
7 . ?5.
. o o0.0000
O. oo000
/. f/,.2
.132
0.0000
/0
o. o27
O. 0 49
5
3.33
o./28c4
3. 64 0
2.5
0.355
o. d so
o. o 7.j
0 0000
0.1773
2
- 39 -
ol/f
2
/0
o0. o/50
o.
02 0300
a oo
o.00oo
O. o'5o
o._
0.
0.200
0.0017
0. 0002
0. 9237
1 4/
0. /
/
0. /10
7.
m
o. 2
95
4f.
DEPTH OF THE ERROR VOLUME
In order to determine the value of ZLIM, the circle
of radius R is divided into concentric rings of width
dP/p
= 0.5 and n sectors, n being as before:
Equivalent point loads Pi
=
Pi
x Aij are placed at the
center of each element and the values of the vertical
stress Gz are calculated for points on the Z axis and
compared with those obtained in item (b) of the previous
section.
The program corresponding to this process is shown
in page 41 .
The substitution of distributed loads by equivalent
point loads can be started, as shown in Table III, at a
depth
Z/R = 0.4 in order to maintain an error or ± 1%.
But for purposes of practical convenience it is an
accepted value:
Z/R = 1
ZLIM
=
- 40 -
R
BPR
6032
DOMINGUEZ A B
GO
/JOB
/FTC
STRESS DISTRIBUTION STUDY
C
10 READ(5,1) QR,IMAX,ZDZtRHO#ADRHO
1 FORMAT(2F3.0159,2F5.1,F12.8,F5.1F5.1)
WRITE(6#2)
2 FORMAT(11Xt1HI,10X3H3Z/R2XHSIG20X1HS19X3HDIF,17X,3HEPS/)
DO 9 I=1,IMAX
SIG=Q*(1.-(Z**3.)/(SQRT((R**2.+Z**2.)**3.)))
FN=3.14/(2,*A)
M=(R-RHO)/DRHO
RI=RHO
S=0.
DO 7 L=1M
P=Q*3.14*((RI+DRHO)**2.-RI**2.)/FN
D=RI+DRHO/2.
S=S+FN*3**P/(2.*3.14*(Z**2.)*(SQRT(1.+(D/Z)**2o)**5-))
7 RI=RI+DRHO
DIF=S-SIG
EPS=DIF*100./SIG
WRITE(698) I Z9SIGsSsDIF9EPS
8 FORMAT(112S8XF5.1,5XF15.8t5XF1l58,5XF15.8,5XF15.8)
9 Z=Z+DZ
GO TO 10
END
/DATA
1el
0.5
0.00000001
0.1
0.1
100
1. 1.
/END OF FILE
- 41 -
TABLE
III
Z/R
ERROR
DISTRIBUTED
POINT
LOAD
LOADS
I
0.1
O.9990
1.1773
17.84
0.2
0.9924
1.0284
3.63
0.3
0.9762
0.9910
1.51
0.4
0.9487
0.9569
0.86
0.5
0.9105
0.9159
0.58
0.6
0.8638
0.8676
0.44
0.7
0.8114
0.8143
0.36
0.8
0.7562
0.7585
0.31
0.9
0.7006
0.7025
0.27
1.0
0.6464
0.6480
0.24
I-
-
42 -
I
CHAPTER
5
COORDINATE SYSTEMS
This method basically utilizes two systems of coordinates:
1.
A main cartesian system of fixed axes X,Y,Z.
ItS
origin 0 is located at the center of the square grid
of 20 x 20 elements.
The X and Y axes are parallel
to the sides of the grid and the Z axis is perpendicular to the grid plane, their positive directions
are shown in Fig. 17.
2.
An auxiliary cartesian system of movable axes
X',Y',Z'.
Its origin O' is, at each step ij of the
process, located at the center of the corresponding
ij element of the grid (Fig. 17).
These axes are parallel to the X,Y,Z axes respectively
and have the same positive directions.
5a.
COORDINATES OF A GRID ELEMENT
The coordinates XA, YA' ZA of a grid element ij with
respect to the main system (Fig. 18) are given by the
following relations:
- 43 -
dv
7
/
77/'
zz
/
z
z
/
z,
/
'777777
1
7
/
7
.7/
I7
/
-7
/7
/
/
7
/
S7
z z
/
/
/
/
7
7
/
/
/
7
T
/
z
/
z
/
z
7
zz
z
/
z
/
zz
7
/z
z
7
z
7/
/z
/
L/
/
l
7
/
/
A
z
/
/
/
7=7
7/
Ir
777777
A
/
/
7e
7
S/
SI
/
/
z
z
7
/
/
z
,'7,',r7,'
/
/
/
A
7
/
7
/
Az/
/
/
-
z
I/
|
/ 0 !,/ I/ V
FIG. f7
7
I/
/
/
z'
77/7
/
/
7
771
7
Z
7
z
7
A
7
Z
,/
7
/
z
z
-
400
A
/
As
4
I
I
- - - - -
4
YaaI
- - - -
- - - - -
- - - -
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Il
i//6. 18 .
- 45 -
X
a.
Quadrant 1
1I
.O
j
XA = - lo-j)i dp -
-
j-
/o.). Ap
p
p.
Y== (rZA
b.
-)j- 1= .d
2
= (/o
2
I
- i4m (/a. - p.
-- O
Quadrant 2
r/ ~_/ - 20
xA , (y - /o). dp -
YA = fo
-
., p +
.(- o- 1.di . (j- o.5). dp
2
2
AP = (/O, _. i)
2
2
ZA = O
-46
-
o =( /4. 5 -U
i
c.
Quadrant
li
1:6 20
xA. -Cto -j).4p. _
2
XA = -==to) Jedp
YA-(i
-o )dp
=(j- 'o.s).4p
2
APeMlp-(,--
= (o.5-j).4p
2
ZA =0
d.
Quadrant 4
f f_ t f 20
/fje
XA =
20
(j-
j/- o). dp -
/0.5).
d4J
2
YA = - t - 0o). d
2
-
ZA = 0
- 47 -
(/o. 5- ij. d
This shows that for the expressions for XA, YA'
ZA
are equal in the four quadrants:
XA =
(j-
YA -
(/0.5 - i). 4p
/0.).
Ap
(4]
(al)
ZA =O
5b.
(c )
COORDINATES OF POINT P(x,y,z) WITH RESPECT TO SYSTEM
X', Y',
Z'.
From Fig. 17 it can be seen that the coordinates of
any point P(x,y,z) of the half-space with respect to the
auxiliary system X',Y',Z' are:
X'
=
X
-
XA
Y'
=
Y
-
YA
2'
=
Z
-
O =
- 48 -
CHAPTER
6
PRINCIPAL STRESSES
The principal stresses are the characteristic values
of the matrix:
L
G1
The three characteristic values of these matrices
can be obtained solving the following determinantal
equation:
6 -6
'Tay
'rz
'trx vy -6
O
,'Iy
Tx
2z
=o
6,-6
Developing this determinant and rearranging terms, the
following characteristic equation is obtained:
6 I- 12 6
where Il'
ae
6 - 13 = o
12, and 13 are the stress invariants.
- 49 -
1,: 6,
* 6y
w' 6
2
I
=3-6 6 yz
1 r,
r
r
- 66z
-
2
A simple direct method of solution for this cubic equation
is to reduce it to the form:
This is the cubic equation for the principal stresses
expressed in their deviatoric form and in terms of the
deviatoric stress components, where
I =
(, 1-+12
6)
I
= 6, -
rx'y
II
rz U>TX
Yzr
z
-~
=Zz-
62
- 50 -
The principal stresses expressed in their deviatoric
form are:
G x; G -J;
Ga
Ggfg
These three stresses are equal to:
O
a
c
co,
's..) (-
where
C X
This solution has been programmed into a subroutine
together with the method for the determination of the
principal direction cosines explained in section 5.
- 51 -
7
CHAPTER
PRINCIPAL DIRECTION COSINES
The direction cosines ilmln,,
the principal stresses
6j, 6jg
,
i
2m2 n2
, and ( m 3n3 of
are the solutions
of the following system of homogeneous linear equations:
+
(6-).1
(6*.
I/
ZdY./
-
T-.
m
(G y- ).m
,
'y. m
Ts.
=0
?tz.
= O
(6z-6).
[1]
0O
When
6j
is substituted the solution will be I~my ni
When
6E
is substituted the solution will be
When
6, is substituted the solution will be / m
mywn2
j
The direction cosines of each principal stress must
satisfy the following relations:
1,+
14
10
,m, +n,
- 52 -
-l
can be written in the following form:
System [1]
SI i *
* /ri
b Yi
Scj Zi
xi +
d
.0
Ja xi f
b, yi
Scj Zi
(1)
I 0O
(2)
i- o
(3)
and relations [2] can be written as:
X
X2
Pt
2
2i
i 1PtZ -A
(4)
where i = 1,2,3
Xi - A
Assuming
from
(5)
(1)
bl
(- c Z - a 4A)
from
(2)
CS*i
fi= t
from (5) and
(6)
-*jz,4
6s/ A Cd, - S,)
Zi =
6, e,substituting in
,i .
i*
(7)
S, c,
(5)
(8)
,4
y/ A
(6)
A- C
6, 6, A (, - 4)
- 53 -
(9)
(10)
The constant A can be determined making use of relation
(4):
A /b
I
.4C,
-c , 6,
I
I
-it)
6,,
- ,cI
,,
Substituting A into (8),
and Zi .
t
b,?(d
- 11
ad
b/
62t4s
s
6/
bA,-
(9), and (10) we obtain Xi, Yi'
In case that the denominator
C,/- Ct a
0
it is necessary to prepare another combination in the
program, for example using equations (2) and (3).
From (2) and (3)
4 -6C)
(a
z2i ,
x,
A bei /
62
c6,-b
' .l. .-"
- 54 -
J
The constant A again can be determined from relation (4):
'
L
C,
/
A I-.
,6,- aJ').
t
i
A."
ei - ba,,
m.Cb,
1
da r
b
ciS
j
- 6CJ, [
(
- 55 -
a r,)
b
b,
- )
2-,
JJ
c1
44(7,
=
CHAPTER
8
COMPUTER PROGRAM
The method described in the previous chapters was
programmed in FORTRAN IV for an IBM/360 digital computer.
The flow chart and the program are shown in pages 57 to
69 .
Table IV
(page 78) shows the output form for each
point of the half-space.
Compilation time for the program is of the order of
0.96 minutes and execution time for each point is approximately 15 seconds.
- 56 -
Fl-
NORQMAI
lOAD ,'YAYRt.
5//Afr4?ING LOA
P,
A.r,
R,
RZy QsIi, Z,
ar
P,, Q,
po,, 0
P0R4e
AlCtW~.w,VA
Of
rgff
TA'oe
rw
vwiAr/.
/JOB
/FTC
C
C
C
C
C
C
C
C
C
C
C
GO
C
3
30
C
988
100
789
C
C
C
DOMINGUEZ A B
6032
BPR
STRESSES AND DISPLACEMENTS IN SEMI-INFINITE MEDIA.
MAIN PROGRAM.
CAPABILITIES OF THE SYSTEM.
FOR ANY POINT OF THE SEMI-INFINITE MEDIUM
FOR ANY SHAPE OF THE LOADED AREA
STRESS TENSOR DUE TO NORMAL LOADS.
STRESS TENSOR DUE TO SHEARING LOADS.
STRESS TENSOR DUE TO SUPERIMPOSED NORMAL AND SHEAR ING LOADS.
VERTICAL COMPONENT OF DISPLACEMENT DUE TO NORMAL L OADS.
VERTICAL COMPONENT OF DISPLACEMENT DUE TO SHEARING LOADS.
TO SUPERIM POSED NORMAL
VERTICAL COMPONENT OF DISPLACEMENT DUE
AND SHEARING LOADS.
PRINCIPAL STRESSES.
PRINCIPAL DIRECTION COSINES.
DIMENSION P(2020),Q(2020)
DIMENSION Ul(25).U2(25) U3(25)#T1(25) 9T2(25) T3(25)
DIMENSION AP(3)tAQ(3),AT(3)
DIMENSION 5(3)tB(393)
COMMON SBgSXX9SYY*SZZoSXY#SXZ#SYZ
READ(5#2) NMAX9L9POISS.ELAST9R
FORMAT(21 6,F7.392F15e5)
READ(5,3) ((P(IsJ)sJ=1,10)I=1920)
READ(593) ((P(IJ),J=1120)#I=1920)
READ(5,3) ((Q(IJ),J=1qlO)1=120)
PEAD(593) ((Q(I9J),J=11920)#I=1#20)
FORMAT(lOF7.2)
READ(5S30) X#YoZtDX*DYtDZ
FORMAT(6F10.5)
PRINT DATA*
WRITE(692) NMAXtLoPOISS.ELASToR
WRITE(69988)
WRITE(6100) ((P(IJ)9J=1,20),I=1920)
WRITE(6,988)
WRITE(69100) ((Q(19J)J=1920)I1120)
WRITE( 6988)
FORMAT(lH )
FORMAT(20F6*0)
WRITE(60789) XtYgZDXDYgDZ
FORMAT(6FIC.5,///////////////////)
ELAST=MODULUS OF ELASTICITY*
POISS=POISSONIS RATIO
FL AND GL=LAME CONSTANTS.
V=1.-2**POISS
FL=POISS*ELAST/((POISS+1.)*V)
E628=6.28*ELAST
GL=ELAST/(2.*(POISS+1.))
RHO=0.2*R
DRHO=O.5*RHO
R1=DRHO/SQRT (3.14)
R2=3.*R1
R3=5.*R1
RHO3=3.*DRHO
DRHO2=DRHO**2
- 58 -
4
C
C
C
C
43
45
46
44
47
C
C
XLIM=12.*DRHO
YLIM=XLIM
ZLIM=R
DO 40 N=1sNMAX
INITIAL VALUES OF STRESS AND DISPLACEMENT COMPONENTS
AT POINT (XtY*Z) OF HALF-SPACE.
P1XX=O.
PlYY=O.
PlZZ=0.
P1XY=O.
P1XZ=0.
P1YZ=0.
P2XX=0.
P2YY=0.
P2ZZ=O.
P2XY=0.
P2XZ=0.
P2YZ=0.
Q1XX=0.
Q1YY=0.
Q1ZZ=0.
Q1XY=0.
Q1XZ=0.
Q1YZ=0.
02XX=0.
Q2YY=O.
Q2ZZ=O.
Q2XY=0.
Q2XZ=0.
Q2YZ=.O
WPI=0.
WQ1=0.
WP2=0.
WQ2=O.
IS POINT (XYtZ) WITHIN SENSITIVE VOLUME (2*XLIM*2YLIM*2ZLIM).
IF YES M=1
IF NO M=O
IF(Z-ZLIM) 43943944
IF(ABS(X)-XLIM) 45,44944
IF(ABS(Y)-YLIM) 4694444
M=1
GO TO 47
M=0O
Kl=1
K2=1
K3=1
DO 9 I--=120
DO 9 J=120
COORDINATES OF LOAD ELEMENT (IJ)
WITH RESPECT TO MAIN SYSTEM.
ZA=0.
HI=I
HJ=J
XA=(HJ-10.5)*DRHO
YA=(10.5-HI)*DRHO
COORDINATES OF POINT (XYZ)
WITH RESPECT TO LOAD ELEMENT (IJ).
XP=X-XA
- 59 -
6
7
5
77
78
YP=Y-YA
ZP=Z
IF(M-1) 5,5.6
IS POINT (XtYZ) WITHIN SENSITIVE ZONE OF LOAD ELEMENT (1,J)
IN EITHER CASE GO TO CORRESPONDING ROUTINE.
IF(ABS(XP)-R3) 7,5#5
IF(ABS(YP)-R3) 895*5
POINT(X9YZ) OUTSIDE OF SENSITIVE ZONE OF LOAD ELEMENT (IJ)
ROUTINE FOR CONCENTRATED LOADS.
D=SQRT(XP*XP+YP*YP+ZP*ZP)
DZP2=(D+ZP)**2
R2=XP*XP+YP*YP
A=D**2
AA=D**3
C=D**5
CCC=6.28*C
E=XP*XP
F=YP*YP
G=ZP**2
H=ZP**3
IF(P(I.J)) 77,78977
PP=P(IJ)*DRHO2
AIXY=O0
TTT=3.*PP/CCC
A1YZ=TTT*YP*G
AlXZ=TTT*XP*G
A1ZZ=TTT*H
UUU=3.*ZP/C
VVV=R2*D* (D+ZP)
SSS=ZP/(R2*AA)
A1YY=PP*(F*UUU-V*((F-E)/(VVV+E*SSS) ))/628
A1XX=PP*(E*UUU-V*((E-F)/(VVV+F*SSS)))/6.28
DP1=PP*(1.+POISS)*(G/AA+V*2*/D)/(6.28*ELAST)
GO TO 79
AIXY=O.
AIXX=O.
A1YY=O,
AlZZ=O.
AIXZ=O.
A1YZ=O.
DP1=O.
79 IF(Q(IJ))
87,88987
87 QQ=0( IJ)*DRH02
B1XZ=(-3*QOQ*E*ZP)/CCC
B1YZ=(-3**QQ*XP*YP*ZP)/CCC
B1XY=QQ*YP*(-3**E/A+V* (-A+E+2.*D*E/(D+ZP))/ (DZP2))
1/(6.28*AA)
BlZZ=(-3**QQ*XP*G)/CCC
B1YY=QQ*XP*(-3.*F/A+V*(3.*A-E-2.*D*E/(D+ZP))/
(DZP2))/
1(6.28*AA)
DQ1=QQ*(XP*ZP/(GL*AA)+XP/((FL+GL)*D*(D+ZP)))/(4.*3*14)
GO TO 89
88 B1XY=O0
B1XX=O,
B1YY=0.
- 60 -
89
C
8
101
103
102
104
106
105
91
92
93
9
C
50
B1ZZO,.
B1XZuO.
81YZ=0.
DQ= 0.
PlXX=PlXX+AlXX
P1YY=P1YY+A1YY
PlZZ=PlZZ+AlZZ
PlXY=P1XY+A1XY
PlXZ=P1XZ+AlXZ
P1YZ=P1YZ+A1YZ
Q1XX=Q1XX+B1XX
Q1YY=Q1YY+B1YY
Q1ZZ=Q1ZZ+BZZ
Q1XY=Q1XY+B1XY
Q1XZ=Q1XZ+81XZ
Q1YZ=Q1YZ+B1YZ
WP1=WP1+DP1
WQ1=WQ1+D1
GO TO 9
POINT (X#Y,Z) INSIDE OF SENSITIVE ZONE OF LOAD ELEMENT
IF(ABS(XP)-R1) 101,1019102
IF(ABS(YP)-R1) 103,103,102
U1(K1)=P(IJ)
TI(KI)=Q(I J)
GO TO 91
IF(ABS(XP)-R2) 104,104,105
IF(ABS(YP)-R2) 106,106,105
U2(K2)=P(ITJ)
T2(K2)=Q(I9J)
GO TO 92
U3(K3)=P(ITJ)
T3(K3)=Q(IJ)
GO TO 93
Kl=Kl+1
GO TO 9
K2=K2+1
GO TO 9
K3=K3+1
CONTINUE
ROUTINE FOR DISTRIBUTED LOADS.
K1MAX=Kl-1
K2MAX=K2-1
K3MAX=K3-1
IF(KlMAX+K2MAX+K3MAX) 50,50951
A2XX=0.
A2YY=O.
A2ZZ=O.
A2XY=0.
A2XZ=0.
A2YZ=O.
B2XX=.O
B2YY=0.
B2ZZ=0.
B2XY=0.
B2XZ=O.
- 61 -
(IJ)
51
600
601
602
900
901
902
903
904
905
906
907
908
82YZ=O.
DP2=0.
DQ2=0.
GO TO 52
SUMP1=O.
SUMP2=0.
SUMP3=O.
SUMQ1=0.
SUMQ2=O.
SUMQ3=0.
DO 600 KI=1.KlMAX
SUMP1=SUMPI+U1(K1)
SUMQ1I=SUMQ1+T1(K1)
DO 601 K2=1,K2MAX
SUMP2=SUMP2+U2(K2)
SUMQ2=SUMQ2+T2(K2)
DO 602 K3=1K3MAX
SUMP3=SUMP3+U3(K3)
SUMQ3=SUMQ3+T3(K3)
R1MAX=KlMAX
R2MAX=K2MAX
R3MAX=K3MAX
IF(R1MAX) 9009900*901
AP(1)=O.
AQ(1)=0.
GO TO 902
AP(1)=SUMP1/RIMAX
AQ(1)=SUMQ1/R1MAX
IF(R2MAX) 90399039904
AP(2)=O0
AQ(2)=0.
GO TO 905
AP(2)=SUMP2/R2MAX
AQ(2)=SUMQ2/R2MAX
IF(R3MAX) 906,906,907
AP(3)=0.
AQ(3)=0.
GO TO 908
AP(3)=SUMP3/R3MAX
AQ(3)=SUMQ3/R3MAX
A=ZP**2
AA=ZP*A
C1=R1**2
C2=R2**2
C3=R3**2
D1=CI*R1
D2=C2*R2
D3=C3*R3
E1=SQRT(C1+A)
E2=SQRT(C2+A)
E3=SQRT(C3+A)
F1=E1**3
F2=E2**3
F3=E3**3
G1=ALOG((R1+E1)/ZP)
- 62 -
G2=ALOG((R2+E2)/
G3=ALOG((R3+E3)/
Hl=ATAN(Rl/ZP)
H2=ATAN(R2/ZP)
H3=ATAN(RI/ZP)
DO 753 Mw192
IF(M-2) 30093019
100 AT(1)=AP(l)
AT(2)=AP(2)
AT(3)=AP(3)
DIXX=(AA/Fl-2**( lo+POISS)*ZP/El+V+6*)/2*
D2XX=(AA/F2-2**( lo+POISS)*ZP/E2+V+69)/2*
D3XX=(AA/F3-2**( 1*+POTSS)*ZP/E3+V+6o)/2*
DlXY=O*
D2XY=Oo
D3XY=Oo
DlXZ=2o*Dl/(3914D2XZ=2o*D2/(3*14D3XZ=2#*D3/(3*14DlYY=O*
D2YY=Oo
D3YY=Oo
DlYZ=O*
D2YZ=Oo
D3YZ=Oo
DlZZ=1*-AA/Fl
D2ZZ=1*-AA/F2
D3ZZ=lo-AA/F3
DlWz(lo+POISS)*(RI/El+2o*V*Gl)/EL AST
D2W=(lo+POISS)*(R2/E2+2**V*G2)/EL AST
D3W=(I*+POISS)*(R3/E3+2**V*G3)/EL AST
GO TO 700
101 AT(1)=AQ(l)
AT(2)=AQ(2)
AT(3)=AQ(3)
DlXX=(-4**(l*-6**POISS)*Rl/El+8*/ (3**Fl)-8**Gl+V*(8**El/RHO3
1-32**ZP/(3**Rl)-16**Hl/3*))/6*28
D2XX=(-4**(lo-6**POISS)*R2/E2+8*/ (3**F2)-8**G2+V*(8**E2/RHO3
1-32**ZP/(39*R2)-16**H2/3*))/6*28
D3XX=(-4**(lo-6**POISS)*R3/E3+8*/ (39*F3)-8s*G3+V*(8**E3/RHO3
1-32**ZP/(3**R3)-16#*H3/3*))/6*28
DlXY=((4*+20**V/3*)*RJ/El+4o*Dl/( 3**Fl)-4**Gl+V*(-80*El/RH03+
132**ZP/RH03+16**Hl/3*))/6*28
D2XY=((49+20e*V/3*)*R2/E2+4**D2/( 3o*F2)-4**G2+V*(-8**El/RH03+
132**ZP/RH03+16**H2/3*))/6*28
D3XY=((4*+20e*V/3o)*R3/E3+4**D3/(3**F3)-4**G3+V*(-8**E3/RH03+
112o*ZP/RH03+16s*H3/3*))/6*28
DlXZ=Oo
D2XZuOo
D3XZzOo
DlYY=(8**POISS*Rl/El+4**Dl/(3**Fl)-4**Gl+V*(-20**El/RH03+
144**ZP/RH03+4**Hl))/6928
D2YY=(8o*POISS*R2/E2+4**D2/(3o*F2)-49*G2+V*(-20**E2/RH03+
144**ZP/RH03+4**H2))/6928
- 63 -
D3YY=( 8.*POISS*R3/E3+4.*D3/ (39*3 )-4.*G3+V*( -20. *E3/RHO3+
144**ZP/RHO3+4**H3) )/6*28
D1YZ=-3**ZP*t 2./(3.*ZP)+A/(3.*Fl).-l./E1)/3.14
D2YZ=-3 .*ZP*( 2./( 3.*ZP )+A/( 3.*F2)-1 ./E2 )/3.14
D3YZ=-3**ZP*( 2./(3.*ZP)+A/(3.*F3)-1./E3) /3.14
DlZZ=-2**Dl/( 3. 14*F 1)
D2ZZ=-2**D2/( 3*14*F2)
D3ZZ=-2**D3/( 3*14*F3)
DlW=2.*(l.+POISS)*(V*Gl/(V+2.*ELAST)-ZP*Rl/El+V*(2.*ZP/Rl
1-2.*El/R1)/(V+2.*ELAST)),(3.14*ELAST)
D2W=2e*(l.+POISS)*(V*G2/(V+2.*ELAST)-ZP*R2/E2+V*(2.*ZP/R2
l-2.*E2/R2)/(V+2.*ELAST) )/(3.14*ELAST)
D3W=2.*(l.,POISS)*(V*G3/(V+2o*ELAST)-ZP*R3/E3+V*(2.*ZP/R3
1-2**E3/R31)/(V+2.*ELAST) )/(3*14*ELAST)
700 IF(AT(1)-AT (2)) 710.7109711
710 IF(AT(1)-AT (3)) 712,712,713
711 IF(AT(2)-AT (3)) 71407149715
712 IF(AT(2)-AT (3)) 716 #716,9717
C
MINwAT(1)
INT=AT(2)
MAX=AT(3)
716 ALFA=AT(1)
FETAzAT(2 )-AT( 1)
GAMMA=AT(3)-AT(2)
C1XX=ALFA*D3XX
C2XX=BETA*(D3XX-DlXX)
C3XX=GAMMA* (D3XX-D2 XX)
C1XY=ALFA*D3XY
C2XV=BETA*(CD3XY-D1XY)
C3XY=GAMMA*(D3XY-D2XY)
C1XZ=ALFA*D3XZ
C2XZ=BETA*(D3XZ-D1XZ)
C3XZ=GAMMA* CD3XZ-D2XZ)
C1VV=ALFA*D3YY
C2YY=BETA*(CD3YY-D1YY)
C3YV=GAMMA*(CD3YY-D2YY)
ClYZ=ALFA*D3YZ
C2YZ=BETA* CD3YZ-D1YZ)
C3YZ=GAMMA*(CD3VZ'-D2YZ)
C1ZZ=ALFA*D3ZZ
C2ZZ=BETA*(D3ZZ-DlZZ)
C3ZZ=GAMMA* (D3ZZ-D2ZZ)
C1W=ALFA*D3W
C2W=BETA*(D3W-D1W)
C3W=GAMMA* (D3Wa-D2W)
GO TO 750
C
MIN=AT(1)
INT=AT(3)
MAX=AT( 2)
717 ALFA=AT(1)
BETA=AT(3)-AT( 1)
GAMMA=AT( 2)-AT(3)
C1XX=ALFA*D3XX
C2XX=BETA*(D3XX-D lXX)
C3XX=GAMMA*(D2XX-D1XX)
ClXY=ALFA*D3XY
C2XY*BETA* CD3XV-D1XY)
C3XY=GAMMA* (D2XY-D1XY)
C1XZ=ALFA*D3XZ
-
64
-
C2XZ=BETA*(D3XZ-D1XZ)
C3XZ=GAMMA*( D2XZ-DlXZ)
C1YY=ALFA*D3YY
C2YY=BETA*( D3YY-D1YY)
C3YY=GAMMA* (D2YY-D iVY)
ClYZ=ALFA*D3YZ
C2YZ=ALFA*(D3YZ-DlYZ)
C3YZ=GAMMA*(D2YZ-D1VZ)
ClZZ=ALFA*D3ZZ
C2ZZ=BETA*(D3ZZ-DlZZ)
C3ZZ=GAMMA* (D2ZZ-DlZZ)
ClWxALFA*DlW
C2WaBETA* (D3W-D1W)
C
C3WuGAMMA* CD2W-DlW)
GO TO 750
M!N=AT(3)
INT=AT(l)
MAXwAT(2)
713 ALFA=AT(3)
BETA=AT(l)
GAMMAuAT(2)-AT( 1)
CIXX=ALFA*D3XX
C2XX=BETA*D2XX
C3XX=GAMMA*(D2XX-DlXX)
C lXY=ALFA*D3XY
C2XY=BETA*D2XY
C3XY=GAMMA* (D2XY"D1XY)
Cl XZ=ALFA*D3XZ
C2XZ=BETA*D2XZ
C3XZzGAMMA* (D2XZ-DlXZ)
Clv V=ALFA*D3YY
C 2YY= BETA*D2VY
C3YY=GAMMA* (D2YY-D1YY)
ClYZ=ALFA*D3YZ
C2YZ*BETA*D2YZ
C3YZ=GAMMA*(CD2YZ-DlYZ)
ClZZ=ALFA*D3ZZ
C2ZZ=BETA*D2ZZ
C 3ZZ =GAMMA* CD2ZZ-D1ZZ)
C1WuALFA*D3W
C2W=BETA*D2W
C3W=GAMMA* CD2W-DlW)
GO TO 750
C
714 IF(ATC1)-AT3) 720,7209721
MINzAT(2)
!NT=AT(l)
MAXwATC3)
720 ALFA=ATC2)
BETA=AT(2)-AT( 1)
GAMMA=AT( 3)-ATC2)
ClXX=ALFA*D3XX
C2XX= BETA*D lxx
C3XX=GAMMA* (D3XX-DlXX)
ClXY=ALFA*D3XY
C2XY=BETA*D1XY
C3XY=GAMMA* CD3XY-DlXY)
ClXZ=ALFA*D3XZ
C2XZ=BETA*DlXZ
C3XZ=GAMMA*(CD3XZaDlXZ)
-
65
-
CIY~zALF'A*D3YY
C 2Y V.BET A*DlYY
C3Y~zGAMMA* CD3YY-D1 VY)
C1YZ=ALFA*D3YZ
C2YZ=BETA*DlYZ
C 3YZz=GAMMA*(CD3VZ-DIYZ)
C lZZ=ALFA*D3ZZ
C2ZZ=BETA*D1ZZ
C3ZZ=GAMMA*( D3ZZ-D1ZZ)
ClWzALFA*D3W
C2W=BETA*D1W
C3W=GAMMA* (D3W.-D1W)
GO TO 750
C
MIN=AT(2)
INT=AT(3)
721 ALFA=AT(2)
BETAzAT(1 )-AT(2)
GAMMA=AT(3)-AT(2)
ClXX=ALFA*D3XX
C2XX=8ETA*D lxx
C3XX=GAMMA* CD3XX-D2XX)
ClXY=ALFA*D3XY
MAX=AT( 1)
C2XY=BETA*DlXY
C3XY=GAMMA*(CD3XY-D2XY)
ClXZ=ALFA*D3XZ
C2XZ=BETA*DlXZ
C3XZ=GAMMA*(CD3XZ-D2XZ)
C 1YY= ALFA*D3YY
C2YY= BETA*D1YY
C3YY=GAMMA*(CD3YY-D2YY)
ClYZ=ALFA*D3YZ
C2VZ=BETA*DlYZ
C3YZ=GAMMA*CD3YZ-D2YZ)
ClZZzALFA*D3ZZ
C2ZZ=BETA*D1ZZ
C3ZZ=GAMMA*(CD3ZZ-D2ZZ)
C1W=ALFA*D3W
C2W=BETA*D1W
C3W=GAMMA* CD3W-D2W)
GO TO 750
C
M!N=AT(3)
INT=AT(2)
715 ALFA=AT(3)
BETA=AT(2)-AT(3)
GAMMA=AT(l1)-AT(2)
ClXX=ALFA*D3XX
C2XX=BETA*D2XX
C3XX=GAMMA*D1XX
ClXY=ALFA*D3XY
C2XY=BETA*D)2XY
C3XY=GAMMA*Dl1XV
ClXZ=ALFA*D3XZ
C2XZ=BETA*D2XZ
C3XZ=GAI4MA*D1XZ
ClYY=ALFA*D3VV
C2YY=BETA*D2Y
C3YV=GAMMA*D1YY
MAX=AT ( 1)
-
66
-
750
751
752
753
C
52
C
C
C
C
CIYZ=ALFA*D3YZ
C2YZ=BETA*D2YZ
C3YZ=GAMMA*DIYZ
C1ZZ=ALFA*D3ZZ
C2ZZ=BETA*D2ZZ
C3ZZ=GAMMA*D1ZZ
C1WzALFA*D3W
C2W=BETA*D2W
C3W=GAMMA*D1W
A2XX=ClXX+C2XX+C3XX
A2XY=C1XY+C2XY+C3XY
A2XZ=C1XZ+C2XZ+C3XZ
A2YY=C1YY+C2YY+C3YY
A2YZ=C1YZ+C2YZ+C3YZ
A2ZZ=ClZZ+C2ZZ+C3Z Z
DW=C1W+C2W+C3W
IF(M-2) 751,752,752
A2YY=A2XX
A2YZ=A2XZ
P2XX=P2XX+A2XX
P2XY=P2XY+A2XY
P2XZxP2XZ+A2XZ
P2YY=P2YY+A2YY
P2YZ=P2YZ+A2YZ
P2ZZ=P2ZZ+A2ZZ
WP2=WP2+DW
GO TO 753
A2XZ=A2YZ
Q2XX=Q2XX+A2XX
Q2XY=Q2XY+A2XY
Q2XZ=Q2XZ+A2XZ
Q2YY=Q2YY+A2YY
Q2YZ=Q2YZ+A2YZ
Q2ZZ=Q2ZZ+A2ZZ
WQ2=WQ2+DW
CONTINUE
STRESS COMPONENTS DUE
PXX=PlXX+P2XX
PYYmP1YY+P2YY
PZZ=PlZZ+P2ZZ
PXYVP1XY+P2XY
PXZ=P1XZ+P2XZ
PYZ=PlYZ+P2YZ
VERTICAL DISPLACEMENT
WP=WP1+WP2
STRESS COMPONENTS DUE
QXX=Q1XX+Q2XX
QYY=Q1YY+Q2YY
QZZnQ1ZZ+Q2ZZ
QXY=Q1XY+Q2XY
QXZ=Q1XZ+Q2XZ
QYZ=Q1YZ+Q2YZ
VERTICAL DISPLACEMENT
WQ=WQ1+WQ2
STRESS COMPONENTS DUE
TO NORMAL LOAD.
DUE TO NORMAL LOAD.
TO SHEARING LOAD.
DUE TO SHEARING LOAD&
TO SUPERIMPOSED NORMAL AND SHEARING LOADS.
- 67
C
C
11
12
13
14
15
SXX=PXX+QXX
SYY=PYY+QYY
SZZ=PZZ+QZZ
SXYmPXY+QXY
SXZaPXZ+QXZ
SYZBPYZ+QYZ
VERTICAL DISPLACEMENT DUE TO SUPERIMPOSED NORMAL AND SHEARING LOADS.
W=WP+WQ
PRINT OUTPUT.
WRITE(6911) X9Y9Z
FORMAT(20X,2HX=F15.8,20X,2HY=F15.820X2HZzF15o8//)
WRITE(6#12)
FORMAT(20X,2HXX,18X,2HYY,8X,2HZZ,18X,2HXY1l8Xt2HXZ,18X2HYZ,//)
WRITE(6913) PXXPYYPZZPXY#PXZPYZ
FORMAT(5X# 1HPo6F20.8//)
WRITE(6.14) QXX QYY QZZ9QXYoQXZ QYZ
FORMAT(5X 1HQ6F20.8 //)
WRITE(6lS5) SXXSYY#SZZoSXY.SXZ.SYZ
FORMAT(5X,1HS96F20.8#//)
WRITE(6#16) WP,WQ*W
C
16 FORMAT(19X,3HWP=F15*8,19X,3HWQ=F15.81o9X,3HW =F15.8,///)
IF(L-2) 17.18918
18 CALL MHOR
PRINCIPAL STRESSES AND PRINCIPAL DIRECTIONS OF STRESS.
WRITE(6955)
55 FORMAT(55X,18HPRINCIPAL STRESSES)
WRITE(619) S(1),S(2),S(3)
19 FORMAT(17X,5HS(1)=F15.8,17X,5HS(2)=F15.8,17X,5HS(3)=F15.8//)
WRITE(6956)
56 FORMAT(50X,27HPRINCIPAL DIRECTION COSINES)
WRITE(6,20) B(1,1)B8(2.1) 8(3,1)
C
WRITE(6.21) 8(1,2).8(2,2)B(32)
WRITE(6,22) B(1,3),B(2,3)B(3s3)
20 FORMAT(15X,7HB(191)=F15.4,15X,7HB(2,1)=F15.4.15X,7HB(3,1)=F15.4,/)
21 FORMAT(15X,7HB(192)=F15.4.15X,7HB(2,2)=F15.4t15X,7HB(3,2)=F15.4,/)
22 FORMAT(15X,7HB(193)=F15.4,15X,7HB(2,3)=F15.4,15X,7HB(3,3)=F15.4,
1////)
INCREASE XYtZ.
17 X=X+DX
Y=Y+DY
40 Z=Z+DZ
CALL EXIT
END
- 68 -
/FTC
C
C
C
C
C
C
C
C
C
SUBROUTINE MHOR
SUBROUTINE FOR PRINCIPAL STRESSES AND PRINCIPAL DIRECTIONS.
DIMENSION S(3)tB(3,3)
COMMON S*BSXXSYYSZZSXSXSXZSYZ
STRESS INVARIANTS.
SNV1=(SXX+SYY+SZZ)/3.
SNV2=(SXX*SYY+SYY*SZZ+SZZ*SXX-SXY**2 -SXZ**2 -SYZ**2 )
SNV3=-(SXX*SYY*SZZ + 2.*SXY*SYZ*SXZ-SXX*SYZ*SYZ - SYY*SXZ*SXZ
1-SZZ*SXY*SXY)
SOLUTION OF THE CUBIC EQUATION.
X=3.*(SNV1**2 )-SNV2
Y-SNV3-SNV2*SNV1+2*(SNV1**3 )
Z=SQRT(2**X/3.)
C3A=Y*SQRT(2.)/(Z**3)
S3A=SQRT(1.-(C3A**2 ))
T3A=S3A/C3A
ALFA=(ATAN(T3A))/3.
C=Z*SQRT(2.)
VOLUMETRIC PRINCIPAL STRESSES.
SS1=C*COS(ALFA)
SS2=C*COS(ALFA+6.28/3.)
SS3=C*COS(ALFA+12.56/3.)
PRINCIPAL STRESSES.
S(1)=SS1+SNV1
S(2)=SS2+SNV1
S(3)=SS3+SNV1
PRINCIPAL DIRECTIONS*
B(191),B(291)9B(3,1) = DIRECTION COSINES OF PRINCIPAL STRESS S(1).
B(192),B(2.2)9B(3#2) = DIRECTION COSINES OF PRINCIPAL STRESS S(2).
B(1,3),B(293).B(393) = DIRECTION COSINES OF PRINCIPAL STRESS S(3).
DO 5 J=1 9 3
Ul=(SYY-S(J))*SXZ-SXY*SYZ
U2=SYZ**2 -(SYY-S(J))*(SZZ-S(J))
IF(U1) 10,11#10
10 V1=SXZ*SXY*(SYY-S(J))*(SXY-(SXX-S(J)))
W1=SXY*(SYY-S(J))*(SXY-(SXX-S(J)))
A=1./SQRT(l.+((-V1/U1-(SXX-S(J)))/SXY)**2 +(W1/U1)**2
GO TO 12
11 V1=-SXY*(SYY-S(J))*SYZ*(S-SXY)
W1=(SYY-S(J))*SYZ*(SXZ-SXY)
A=1./SQRT(1.+((V1/U2-SXY)/(SYY-S(J)))**2 +(W1/U1)**2
GO TO 13
12 B(1,J)=A
B(3,J)=A*SXY*(SYY-S(J))*(SXY-(SXX-S(J)))/U1
8(2J)(-SXZ*B(3,J)-(SXX-S(J))*A)/SXY
GO TO 5
13 B(19J)=A
B(3,J)=A*(SYY-S(J))*SYZ*(SXY-SXZ)/U2
8(2,J)=(-SYZ*B(3J)-SXY*A)/(SYY-S(J))
5 CONTINUE
RETURN
END
- 69 -
CHAPTER
9
RESULTS
The program was tested several times with different
data, an example is shown in pages 72 to 7
in which the
following was utilized:
Number of points to be analyzed
Use subroutine for principal stresses
NMAX = 10
=
2
Poisson's ratio
POISS =
0
Modulus of Elasticity
ELAST =
1
Maximum diameter of loaded area
R =
1
Initial Coordinates
x =
0
y =
0
L
z = 0.2
Increments
- 70 -
Dx =
0
Dy =
0
Dz =
0
Load Matrices:
The two load matrices were identical and
had the following form:
Jao ws
I
I
I
b1
I
-
.
.
-------
.
I
-. .
os ..
I
.
.
I
-
- 0
IAOa.
20
COL UMNJ
- 71 -
XI
Y-
0.0
1s
0.0
YY
ZZ
P
0.51012242
0.58316684
0.97635388
Q
0.0
-0.00000002
0.00000010
-0.00000448
-J.GI72'3
S
0.51012242
0.58316678
0.97635394
-0.0000U448
-j.6
S5()-
1.41631317
8Il1l)-
0.0000
841121
0.u00O
8(2,2)1
881,3)-
060000
8(2,3)*
x-
P
0.30598438
0.28801090
0.87176800
Q
0.0
-0.00000011
0.00000015
S
0.30596436
0.28001078
0.87176812
Sill-
1.14083327
1.0269987
OWQ
J.JJJ33294
1.1544i41>
J.STbb[66:
J.33J
-. 3J.J
u13,31-
Z.
0.0
zz
2
50L~
J(3,2)1
1.0000
YY
7
-0.03Q0001
d13,I-a
-1.0000
Y.
-
S(-):
PRINCIPAL DIRECTION COSINES
-1.0000
8(2,1)-
xx
P=-
.
PRINCIPAL STRESSES
-0.02467448
S2)-
0.0
0.30003294
-J.J3vjjJul
0.0
-0.00000024
wQ'
1.33483410
y
XZ
XY
xx
MPS
4
J.14I999
J.J
i
j.39*9YPI
yz
xi
xy
-3.033Jj.31
0.00000401
-0.00003364
-3.424'465
-0.00000000
-0.00000364
-3.42414677
0.0
-0.00000028
PRINCIPAL STRESSES
S(2)i
-0.04928577
W
513).
0.00000401
1.14883737
J.48S35
95
PRINCIPAL DIRECTION COSINES
94191)
0.0000
8621tll
-1.0000
S13,1-
0.0000
81,2)
0.0000
842,21=
1.0000
813,2)
-0.0000
SI1l3)
0.0000
8(2,3)-
-1.0000
5(3,3)a
0.0000
X-
0.0
Y.
0.0
Z=
J.59999995
xx
Yy
zz
P
0.17173719
0.15050429
0.72884732
Q
0.0
-0.00000010
0.00000023
-0.00000313
-J.259tf351
-0. 00000000
S
0.17173719
0.15050417
0.72864750
-0.00000333
-J.2591736
0.00000386
-
WP-
0.98864913
S(il)
0.79404873
WO
=
XY
0.0
yL
XL
-u.03vJJj1
-0.00000025
* =
PRINCIPAL STRESSES
S42)
-0.03682834
513)-
PRINCIPAL DIRECTIbN CODSIES
8(2t1)-1.0000
0.00000387
J.29315555
8(ts|Il
O.u000
b(ltl2
0.0000
8(2921-
1.000u
(3,Z)=
811,3)=
0.0000
8(2,3)-
-1.0000
d(3,31=
J.3
7=
.. 799i9
XM
0.0
Y=
xx
YY
zI
P
0.09370512
0.08837312
0.59389663
0
0.0
-0.00000004
0.00000023
S
0.09370512
0.08837306
0.59389681
MoQ
t.331)=
0.0
Xy
0.0
J
y
-J.3jJJ)3?2
0.00000275
-0.00000030
-J.1593/418
0.00000000
-0.00000030
-0.159j243.
0.00000275
0.85669965
Sills
0.62271565
681,1)=
0.0000
81,2)
0.0000
(2,2)-
1.0000
B6(32)-
0.0000
8(2931,
-1.0000
313,3)-=
=
,
-. 3:p
XZ
wPm
811,3)
5
,.98o5
-0.00000022
PRINCIPAL STRESSES
SiZ()
-0.02399778
PRINCIPAL DIRECTION COSINES
8(2,1.
-1.0000
.bbX569941
53)-
1(3,11-
J.17573695
.33)
-J.00'j
.309J
X=
0.0
Y=
0.0
XX
YY
ZZ
P
0.05042163
0.05800239
0.4A160243
0
0.0
-0.00000001
0.00000022
S
0.05042163
0.05800238
0.49160261
WP=
0.74955177
Sll)=
0.49494243
7
x
wQ=
8(2,2)=
L.000
811,3)-
0.0000
8(293)-
-1.0003
0.0
Y=
Yy
LZ
P
0.02663846
0.04128749
0.39230436
0
0.0
-0.00000001
0.00000018
S
0.02663846
0.04128748
0.39230454
0.39895099
-
0.
-0.30033J
-.
,1
Qo=
=
7=
XY
3.U000
B(122)=
0.0
8(1l,3)
0.0000
B(2,3)-
1
-J.
JJJJJI
-0.OOOuu32b
-.
,
-0.0000032
-J.J
DIRECTION CDSIES
-1.JOOJ
--
4)1l
1.11)
X
S(U)=
8(1t2)=
.,
13,)=J
PRIN:IPAL STRESSES
S121)
-0.0092756Z
PRINCIPAL
B2,11)-
,
l,z)=
=
0.0000
.,
(3,1)=
-0.00000017
8(1,1)=
0.000032
-,
)=
0.0
XX
S(il)
-0.000033
PRINCIPAL DIRECTION CJSIES
=
8(291)
-1.0000
0.00O00
0.66251713
0.00000o32
PRINCIPAL STRESSES
S12)-0.0151488?
BIl2)=
WP-
-J.JJsJJ
-0.0000023
0.0000
X=
//
O.J
811,1)-
1
.94
(13,1=
0.000000J3
-0.00000000
0.00000033
3Ji
l5 :
.z02
J.JC',,
t
J.
1.000
3(3,2)=
-J.)"
-1.000J
d(3,3)
J.
j,
X=
0.0
Y=
0.0
I
Z=
S91;)
XX
YY
zz
P
0.01346328
0.03100809
0.32233775
Q
0.0
-0.00000002
0.00000012
-0.000032J
-J. '.125!)
S
0.01346328
0.03100806
0.32233787
-0.0000002J
-..
WP=
S(L)
=
0.0
0.0000
B(1,2)-
0.0000
B(2,2)=
1.0000
8(1,3=
O.u000
8(2.3)=
-1.000u
X=
Y=
0.0
ZZ
P
0.00606450
0.02414056
0.26758820
0
0.0
S
0.00606450
-0.oo0oooo0000002
0.02414054
-0.00000000
0.00000032
,
.l
J..j(.i'
3(3,11)=
.))
t(,?)=
-
3(3,)=
Z=
XY
YY
0.00000032
1c~i
.(3)=
0.0
XX
l
=
PRINCIPAL DIRECTION COSINES
B(2,11=
-1.0000o
B(i1lI=
/
-J.JJ
PRINCIPAL STRESSES
Si(2)
-0.00525391
0.32601237
XI
-0.00000012
WQ=
0.59133977
XY
0.0
1.59:04#9o
vL
x;
-J.
J JJ
0.00000033
0.00000000
-0.00000002
-J. 2
0.00000000
0.26758820
-0.00000002
-3.j21/S4i
0.00000031
-0.00000001
=
PRINCIPAL STRESSES
=
S(2)
-0.00235498
5()=
WQ-
WP=
0.53256720
SIll)
0.26988083
8l1,1)=
0.000
81,2)=
0.U000
8(2,2)=
1.000
811,3)=
0.0000
8(2,3)1
-1.0000
PRINCIPAL DIRECTION COSI1ES
B(21)-1.0000
J.5355714
3.333J35
t(3l)-.
13,2)-
a(3I
3-.'
=
-..
X-
0.0
Y=
0.0
1=
XX
YY
ZZ
P
0.00186374
0.01928173
0.22452271
0
0.0
-0.00000002
0.0f000001
-0.0000
S
0.00186374
0.01928771
0.22452271
-0.0003002
WP-
0.48352551
Sill=
0.22612745
8(1,1)-
0.0000
0(1,2)-
0.0000
B(292)-
B(1,31-
0.0000
B12,31
Xe
0.0
0
5
-0.00052694
0.0
-0.00052694
XY
0.0
002
-j.33,J3j1
0.00000029
-J.J
0.0000000
l'5
-J.JSii151
PRINCIPAL
STRESSES
=
S(2)
-0.00008625
S()=
PRINCIPAL DIRECTION C3SIES
B(21)s
-1.0000
(31)-
1.0000
.J
(13,21)
13,3)m
-1.0000
0.0
0.0000
Z=
XY
1.49,j
0.01572476
0.19035697
-0.00000002
0.00000001
-0.00000002
-.
0.01572474
0.19035697
-0.00000002
-J.13
Sill-
0.19158775
WOu
PRINCIPAL STRESSES
S(2)=
0.00206238
4)
Al
0.0
-0.00000001
h
-.
ZZ
0.44216603
0.00000029
J.14i?,
YY
WP-
yL
X1
-0.00000001
Y=
XX
P
WQU
1.79-9)'4
iL
-J.3Jui 331
a
513)=
0.00000030
0 Id
J.
I67J
0.00000030
.44.21
.J173?3
PRINCIPAL DIRECTION COSINES
311,11
0.0000
8(1,2)=
0.0000
8(1,3)-
0.0000
84(21)x
-1.0000
(13l,)
B(2t2)-
-1.0000
6d3,2)"-.
812,3)1
-1.0000
6(3,3)-
3Jx
3.333,
O
02
0.6
0.4
0.8
0.6
08
10
/2
PL O
ON
'f4
16
FIG. 19
18
RELA T/ YI
DIPT
Z/IR
- 77 -
THE
Or
STRESSES
ViR 'T/C4Z AX/S
to0
37
JS
7"A LE / V
COO R O/NA TES
z
y
x
COnPO NENTS
OF
xx
STRSS
TfENSO0
ZZ
YY
xy
rZ
Due /o norna/ /o0d
Doe /o sAesarngy losl
Superpos ;oo
IE£i
Due lo
T/iCA I
COHfPOv/NT Of DI/JPIA CiNrT
oDe /a shertnS
~orm l/ load
losd
WQ
WP
P RINC/IPA
PRI/AC/P41
DIRECTION
S TRESSES
COSiNiS
opCrposi bn
W
yZ
COMPARISON OF RESULTS WITH NEWMARK'S METHOD
The results obtained in this example can be verified
The pro-
by Newmark's influence charts (Reference 13).
cedure, for the case of the vertical stress
Gz
due to
the normal load, is shown in Figs. 20 and 21 which corresponds to points Pl(0,0,1) and P2 (0,0,2) respectively.
POINT
P(0,O,1)
i/h/uence va/ue
of block =oo02
I
r'c.
£24*
of.2/
Depth
3'
?O
z 0.48
The result obtained with the computer is Gt = 0.48160243
- 79 -
POINT
P(0,0,2)
influence v /ue
of 6lock - 0O02
o
0
SDepl
1-2
Zc~P
r1G. 2
Gz
=
-
2o/Aear
0.02 A fi
.
1o
*
sfie
-/-o
/re
iA.Fei.
0.20
The result obtained with the computer is
80 -
= 0.19035697
CHAPTER
10
CONCLUSIONS AND SUGGESTIONS FOR FUTURE WORK
The advantages of this method (generallty, speed and
accuracy) become evident after the example and verifications
shown in Chapter 9.
The results produced are in very
good agreement with those obtained using Newmark's influence charts.
To these advantages also can be added the
possibility of producing the output of this program in
graphical form using any type of graphical device coupled
to the computer.
Appendix E shows an example of computer
plotting of the vertical stress 6Z along a vertical axis.
Considering the actual stage and future evolution of
computer science and technology it seems important to
extend the capabilities of this method towards the solution of the general problem
mentioned in the introduction
(layered half-space with time dependent properties, subjected to time-varying moving loads) and develop the necessary structure of commands to present it definitely in the
form of a problem oriented language.
- 81 -
APPENDIX
A
DEFINITION OF SYMBOLS
VARIABLE
Element of normal load
SYMBOL IN
PROGRAM
SYMBOL
giJ
P(I,J)
Q(I,J)
i
I
Pij
Element of shearing load
Row number of load matrices
J
Column number of load matrices
Coordinates of point of half-space
x,y,z
Coord. of load elem ij with resp.
to main system
xA,YA, zA
XA' YAZA
Coord. of points of half-space witi
resp. to our system.
x',y ', z
XP,YP, ZP
X,Y,Z
Equiv radius of Square ring 1
R1
Rl
Equiv radius of Square ring 2
R2
R2
Equiv radius of Square ring 3
R3
R3
Side of error prism
RHO
jodP
Side of grid element
DRHO
Number of grid elements in Ring 1
K1
K1
Number of grid elements in Ring 2
K2
K2
Number of grid elements in Ring 3
K
K3
3
Max. No. of grid elems in Ring 1
K1MAX
KlMAX
Max. No. of grid elems in Ring 2
K2MAX
K2MAX
Max. No. of grid elems in Ring 3
K3MAX
K3MAX
Average intensity of normal load
in rings 1, 2, 3
AP I,/ AP),AP (3
%Frir~r~
Average intensity of shearing load
in rings 1, 2, 3
Increments of Coordinates
AO,
Ay
Dy,
0 djz
I
- 82 -
Ao),40
, DZ
SYMBOL IN
PROGRAM
PROGRAM
SYMBOL
VARIABLE
Order of grid and load matrices
M
M
xI/IV, ri/, ZI/f
Limits of total error volume
E
Modulus of Elasticity
ELAST
POISS
Poisson's Ratio
/-2Q
1 - 2x Poisson's ratio
V
S, T
Lame constants
Ary, P/XZ, P/fz
Stress components due to normal
point loads . (1)
P/XX,P/Yy,PI/Z z
Stress components due to normal
distribution loads .(2)
PIY, PY7XPJII
px, iyX,
r/Zn
Stress components due to shearing
point loads.(3)
oNx, /1Y,o/zz
Oyz,0/?Z
S/X/,
02X1, 02YI 021
Stress components due to shearing
distribution loads. (4)
Sum of
1
and
2 .(5)
Sum of
3
and
4
PAy, Pz, PYZ
OxA, 9 YY, o1z
(6)
or ¥,61,
Sum of
5
and
6
ofz
sxx,SYY,
S51
sxy,syz, s/i
Principal Stresses
61, 6,i
Direction cosines of Princ. Stress
Xi"i ' Z
Z
J(1)
J:
Index of Principal stresses
(2) $(J)
2,3
Vertical Displacement Components
1
Due to normal point loads
2
Due to normal dist. loads
3
Sum of 1 and 2
wP f
WP2
WP
_____________________________________________________________________________
- 83 -
I ~
VARIABLE
SYMBOL IN
PROGRAM
SYMBOL
4
Due to shearing point load
WQ1
5
Due to shearing distrib. load
WQ2
6
Sum of
4
and
5
WQ
7
Sum of
3
and
6
w
Number of points to be analyzed
that program is to be repeated
nmax
NMAX
I
- 84 -
APPENDIX
B
LIST OF FIGURES
FIGURE
PAGE
1 .
.
.
.
.
.
. .............
2 . . ..
3
5
..............
8
8
.....................
.
. 10
....................
6 .
. ............
.......
12
14
......................
7
.....................
8
10 . .
. . . . . .
14
.....................
21
........
........
11 .....
12
8
22
4
......................
14 . .
19 .
26
.....................
13
. . . .
. .
.
20 . .
. ..
21 . .
........
. .
. . .
..
. . .
. . .
. . . . . ..
.............
.
. 27
. 77
.....................
79
............
80
- 85 -
APPENDIX
C
LIST OF TABLES
TABLE
I .
II
PAGE
. . .
. .
. . . .
.
. . . .
. .
III .
. . . .
. . .
IV . . . . . .
. . .
. . . . .
. . .
. . ....
. .
. . . .
. .
. .
. . . .
.
35
. . . 39
. . .
. . 42
. . . . . . . . . . . ....
78
- 86 -
APPENDIX
D
UNIFORM CIRCULAR LOAD
Expressions of the Cartesian components of the stress
tensor and vertical components of the displacement vector for
points on the vertical axis for normal and horizontal loads.
ARISTIDES BRYAN DOMINGUEZ
1966
NORMAL LOAD.
For a sin/e normn,/
sj/essfretsor
of /e
3zx
2J,
pJo
rcompone-/
rd :
z2 ]
)[
S(12
()
p' ( I Z)
.D
3z'.l
P
/oad, lbe cr/esin
(/-2 )[
.y2 -
fD (,
2
Zx
7)
P.
2
(2)
3)
3Jp z 3
2-= 3P
2/.
yzx
Zx
jp DZ X
(5)f
(5)
Zx v _
Tx v ,
Th.
(6)
o
cor &p on,' y
7
lod (for poe4 oi
*iprfEj/o~ M A'
e z.
isr)
/hbse /or / e whole circ/le
-1-
/I*
,e, o&L.oe .'
uf~li/Of
circ U/sr
.y
/er./on of
dP=- o dA
2
I
~ ~fCOJ)P
y~"
Io Y/
/z0
p
dip
-2
-., f Pdy
-
4.- 6t .
/,;r mav
D&IPcon//liolw~p
6USS//06#7y
~i*
3,a
zJ
2S
ZN
2 -7
1010e,01sY o er
//
jo/
=
=
o
:
0
2;
= p.z I
6Z = P[
Ofn ,p..
69V
2.
JO
Jo
do
. dp
-
2,OL Z 2) 11
eJ1f
-.1
WF/*C c
/4
2
e
e* 1
Sy 4 , 14e
u~~/
P/, = 0
Y.? =
6z = 4 27
/0 P
rspP/
* j. do
-412
2S
67
wi.*
eA,/DrE&/0/r
3
..
(3)
1
a
- Z2/
I
(i??
-(fI
(42jZY'? 32
-
+ f/3
z
2
-3-
z V s1'
T±z
r
-p[,-
1
z
(R2zjZZ)1
Pi~Te'edfl~P/M~
Tx*.
~.-
(4)
,4fDPW*
~/0
IV4
p
P .040tv
J4E'
M6
4F&i f A *Aw/ .oiy
dZ>z
22
z2
=
2.71
(,p
tool
.cos50
zV
2
*dp.d~
2
Zx= ez
2
.z[s~ospa
P2
/rz
2/1
(.OP 2 f
S~
g.i
6eloes#a'/'#/6vo
p3
?-,r
(g2 ~
2-72
z
7
(Am*
-. 4 -
Y/
Z 2-)-
92'-
Ip
46v4:
C.-
Z y .
Uress/vI
4S1;y
dt2
/.O~wW.I (s}
C-D'fftr~a~i~,
on pfoe.
2R7
s, 2
/
2:
P.dlO
'".Zi
yz -
P
f2
Z [C-osyJ
6s/i 1*V/A;
rd,/As/s
2.7
1
R
(.
=)12.g
77
2*
3_
2
(4 A')
-5-
At,
sw//,oyoyd
' 46
CO-I
2P
y
3
Z)
237
So
/0P
2
1
72
rS
~~.'
c/f. dy~
=
(/,
w
3
Z 2 -1
.d
wi*
/
/o
ovspvc/
p eaw
10
Oo pd.
Ul~~
cp/rr/bu
2:
Jz
2 / *
J,
2p2
Co 2
os IP-
p
dG,,= 2S7
'o
(f
.ri
zp)
s2
fff s
"t
7
3 .2
ZpR'7
I'2
dp. d 9
It2, /
i
3z{
d 6, =
I
coZ
cos5
p
p.
iA(pt, g
(pIt, Z
)Y
, (p*
22/ Fe
.2
+
Z.
61/
(p
/s
f5rm
2
P
*0
z9) 12
14
p
dp. do
.
50J (p
IJ
f' 2)
.Pf
s4 2
212
.50
s//li44/
371 Z
(p'i
72)
Z
+Z ) 1
i
(R
-6-
z
zz
1- d(J2,ZV/
4y 4:
4~-4 Ddmub
/,~U
J(
?
s Z'A"
f
2 1
jZ
P2
1
2Rd. /dr
.
)o , z
cos 2 o
-
f
cos 2$0
P
- ZP
J 9z
S(p
z 2)
P2
4)Cos
f
s/pP501.
(Pf
-
~nd
ZV yt
/op
fo-
2
3rd. /efm.
,~
z + (pS Z#
jO
mu~5;9~yr;rg ~
z/
.PI
is/,;
70
|
2
P2
230
ip
z 2 -
A
.dp- ,o
si29 J
9sis
I
fol.
-7-
=
] dp.do
(- z1
zY2
4,2.
d o
d/. do
Z
/1.
I
d.
-/
()o2 t Z2) Y2
I)I"P
d:
)I0
=o
12~
vz
+
1rj;z f
(U
-r
-F.
X/jr
I
z fi)
11 r
z
/I wc
AI,
f
(J'
Ite
0 a'
z
zX
~(tZ #~~)
I-
7-
1.C
.
e.- 6
$046S
6
d~j
A-"""
'urn""ek
D
y1
IjJ YV
/,~U'
P
z
2p 2,p
#
('t
21/
or
AC
*12 p
/ /l2.(P-r
/ /o P
D/t'
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BIBLIOGRAPHY
1.
Love, A. E. H. A Treatise on the Mathematical Theory
of Elasticity. Dover Publications, N. Y.
2.
Timoshenko, S.
Theory of Elasticity.
McGraw Hill, 1951
3. Ford, H. Advanced Mechanics of Materials. Wiley, 1963
4. Bland, D. R. The Theory of Linear Viscoelasticity,
Pergamon Press, 1960.
5. Terazawa, K. On the elastic equilibrium of a semiinfinite solid under given boundary conditions.
Journal of the College of Science, Imperial University, Tokyo. Vol. XXXVII, article 7, 1916.
6.
Deresiewicz, H. The half-space under pressure distributed over an elliptical portion of its boundary.
ASME Transactions, Vol. 27, 1960.
7.
Stoll, U. W. Computer Solution of Pressure Distribution Problem. ASME Transactions. Paper 3219.
8.
Westergaard, H. M. Effects of change of Poisson's
ratio analyzed by twinned gradients. ASME Transactions, September 1940, p. 4-115.
9.
Sanborn, J. L., Stress and Deflections in an Elastic
Mass under Semi-ellipsoidal Loads, Joint Highway
Research Project, May 1963, No. 13.
10.
Newmark, N. M. Influence Charts for Computation of
Stresses in Elastic Foundations. University of
Illinois. Eng. Exp. Station. Bulletin Series No. 335.
11.
Newmark, N. M. Influence Charts for Computation of
Vertical Displacements in Elastic Foundations.
UniverSity of Illinois, Eng. Exp. Station.
Bulletin Series No. 335.
- 123 -
12.
Newmark, N. M.
University of Illinois Bulletin.
Circular No. 24, Sept. 24, 1935.
13.
Barksdale, R. D. An Influence Chart for Vertical
Stress Increases due to Horizontal Shearing Loadings,
144th Annual Meeting, Highway Research Board,
Washington, D. C., 1965.
- 124 -
APPENDIX
E
COMPUTER PLOTTING VERTICAL STRESSES
The program shown in the following pages is a simplified form of the general method described in this paper
and was prepared using an IBM 1620 digital computer and a
GERBER plotter.
The general method was reduced accordingly
to the difference in capacity and speed between this computer and the IBM/360.
- 125 -
3400032007013600032007024900500010520002900100
*
C
C
C
SUPERMONITOR COLD START
ID DOMINGUEZ A B
BPR
6032
SUBSET
VERTICAL STRESSES IN SEMI INFINITE MEDIA.
PLOT OF VERTICAL STRESSES DUE TO NORMAL LOADS.
DIMENSION B(1)#C(2)
CALL GPLOT(2,0,-100)
DIMENSION P(10,10) U1(25) AP(3)
READ 32#NMAX#R
32 FORMAT(I6,F15.5)
READ 3 ( (P(IJ),J= ,10)sI=l,10)
3 FORMAT(10F7.2)
READ 30#CX,CYCZDXDYDZ
30 FORMAT(6F10.5)
PLOT AND NUMBER X AXIS.
CALL GPLOT(11,0,0)
DO 1 I=500,2500#500
1
C
10
C
2
C
11
4
CALL GPLOT(12,1,0)
CALL GPLOT(12,1,20)
CALL GPLOT(11,1-159-30)
F=I
H=F/500.
CALL GNUM(I-15,-30,H,3.1,0.1,1.,0O)
CALL GPLOT(11,I,0)
LABEL X AXIS.
IX=2530
IY=0O
READ 10B8(1)
FORMAT(A4)
CALL GCHAR( IX,IY,B(1),1,20,1s,0)
PLOT AND NUMBER Y AXIS.
CALL GPLOT(11,0,0)
DO 2 I=15091500,150
CALL GPLOT(12,0,1)
CALL GPLOT(12,20,I)
CALL GPLOT(11,-50*I-5)
F=I
H=F/1500.
CALL GNUM(-75,I-5,H,3.1,0.1,1.,0.)
CALL GPLOT(110sI)
LABEL Y AXIS.
IX=0
IY=1530
READ 11,C(1),C(2)
FORMAT(A49A4)
CALL GCHAR(IX,IYC(1),2,20,1s,0)
RHO=0.2*R
DRHO=0*5*RHO
R1=5.*DRHO/SQRT(3*14)
RHO3=3**DRHO
DRHO2=DRHO*DRHO
XLIM=5.4*DRHO
YLIM=XLIM
ZLIM=R
DO 40 N=1NMAX
- 126 -
43
45
46
44
C
C
C
C
C
C
C
C
INITIAL VALUE OF VERTICAL STRESS AT POINT P(X#YZ)
PlZZ=0.
P2ZZ=O.
IS POINT P(XtY,Z) WITHIN SENSITIVE ZONE (XLIM,YLIMZLIM)
IF YES M=1
IF NO M=O
IF(CZ-ZLIM) 43,43,44
IF(ABS(CX)-XLIM) 45,44,44
IF(ABS(CY)-YLIM) 46.44,44
M=1
GO TO 47
M=O
47 K1=1
DO 9 I=1#10
DO 9 J=1O10
COORDINATES OF LOAD ELEMENT (I,J) WITH RESPECT TO MAIN SYSTEM.
ZA=0.
HI=I
HJ=J
XA=(HJ-5.5)*DRHO
YA=(5.5-HI)*DRHO
COORDINATES OF POINT (X,Y,Z) WITH RESPECT TO L OAD ELEMENT (I,J),
XP=CX-XA
YP=CY-YA
ZP=CZ
IF(M-1) 5,6,6
WITHIN SENSITIVE ZONE OF LOAD ELEMENT (IJ)
IS POINT (XYZ)
IN EITHER CASE GO TO CORRESPONDING ROUTINE.
6 IF(ABS(XP)-R1) 7,5,5
7 IF(ABS(YP)-R1) 8,5,5
POINT (XoY,Z) OUTSIDE OF SENSITIVE ZONE OF LOA D ELEMENT (I,J)
ROUTINE FOR CONCENTRATED LOADS.
5 A=XP*XP+YP*YP+ZP*ZP
D=SQRT(A)
DZP2=(D+ZP)**2
AA=A*D
CCC=6.28*A*AA
E=XP*XP
F=YP*YP
G=ZP*ZP
H=G*ZP
IF(P(IJ)) 77,78,77
77 PP=P(ItJ)*DRHO2
TTT=3.*PP/CCC
A1ZZ=TTT*H
GO TO 89
78 AlZZ=O.
89 P1ZZ=P1ZZ+AlZZ
GO TO 9
POINT (X*YVZ) INSIDE SENSITIVE ZONE OF LOAD ELEMENT (I J)
8 UI(K1)=P(I,J)
91 K1=Kl+1
9 CONTINUE
IF(M-1) 52,60,60
ROUTINE FOR DISTRIBUTED LOADS.
60 K1MAX=Kl-1
- 127
IF(K1MAX) 50L50951
A2ZZ=Oo
GO TO 52
51 SUMPl=0.
DO 600 K1=1lKIMAX
600 SUMP1=SUMP1+U1(K1)
R1MAX=KlMAX
IF(R1MAX) 9009900901
900 AP(1)=O.
A2ZZ=0.
GO TO 750
901 AP(1)=SUMP1/R1MAX
IF(CZ) 1509150,908
908 A=ZP*ZP
AA=ZP*A
C1=R1*R1
D1=C1*R1
El=SQRT(C1+A)
F1=E1*E1*E1
201 A2ZZ=AP(1)*(1l-AA/F1)
750 P2ZZ=P2ZZ+A2ZZ
52 PZZ=PlZZ+P2ZZ
C
PLOT OUTPUT*
X=5.*CZ
Y=10.*PZZ
GO TO 151
150 X=5.*CZ
Y=10.*AP(1)
CALL GPLOT(119XY)
GO TO 39
151 CALL GPLOT(129X#Y)
39 CX=CX+DX
CY=CY+DY
40 CZ=CZ+DZ
CALL EXIT
END
DATA
41
1.
0.
0.
0.
0.
0.
0.
0.
0.
1.
1.
1.
1.
1.
1.
1.
1.
1.
1.
I.
1.
1.
1.
1.
1.
1.
1.
1.
1.
50
0.
0.
0.05000
+Z/R
SIGMAZ/P
ZZZZ
0.
0.
0.
0
0.05000
0.
0.
0.00000
END OF JOB
0.
0.
1.
1.
1.
1.
1.
0.
0.
1.
1.
1.
1.
1.
0.
0.
0.
0.
0.00000
1620
- 128 -
0.
0.
1.
1.
1.
1.
1.
0.
0.
0.00000
0.
0.
1.
1.
1.
1.
1.
0.
0.
0.10000
0.
0.
1.
1.
1.
1.
1.
0.
0.
1.
1.
1.
1.
1.
O.
0.
0.
O0
APPENDIX
F
LIST OF PROGRAMS
Page
1.
Flow chart
36
*................
Program
37
0
2.
Program
3.
Flow chart
0
0
0
0
a
Lq
...............
. .
.
.
.
.
0
.00
.
.
.f
.
Program
.
a.
.
. . .
.
.
.
.
.
. . . .
57
58
- 129 -