Finite element analysis of circular cell bulkheads

AN ABSTRACT OF THE THESIS OF

LIKHIT KITTISATRA for the degree of DOCTOR OF PHILOSOPHY

in CIVIL ENGINEERING presented on 16e.A.A4seiv- 13. lg7Y

Title: FINITE ELEMENT ANALYSIS OF CIRCULAR CELL

BULKHEADS

Abstract approved:

Redacted for Privacy

Dr. Laursen

A mathematical model is developed for the analysis of circular

cell bulkheads subjected to gravity loads of cell fill, backfill, lateral

pressure, and surcharge load using the finite element method. The

finite element circular cell model consists of axisymmetric triangular

and/or quadrilateral ring elements for soil and cylindrical shell ele-

ments for the steel cell. To formulate and derive the governing

equations for such elements, the Ritz displacement functions and the

theorem of minimum potential energy are used. The loadings of the

bulkhead system and corresponding displacements are expanded in

Fourier series. The soil fill and foundation are assumed to be

elastic, isotropic and non-homogeneous materials. It is assumed that

there is no slippage in the soil-steel shell interface.

The element matrices and the Fourier harmonic analysis are

verified by comparing results for several structures for which

classical solutions are available.

The circular cell bulkhead is analyzed for stresses in the soil

elements and shell elements and deformations for the isolated filled

cell and backfilled cell cases. Results of the study are presented,

discussed and compared with field measurement data obtained by

other investigators.

Finite Element Analysis of Circular Cell Bulkheads

by

Likhit Kittisatra

A THESIS

submitted to

Oregon State University

in partial fulfillment ofthe requirements for the

degree of

Doctor of Philosophy

June 1976

APPROVED:


Professor of Civil Engineeringin charge of major


Associate Professor of Civil Engineering


Head gf Department of Civil Engineering


Dean of Gradu4te School

Date thesis is presented NioVeyn0-er ICl'75

Typed by Clover Redfern for Likhit Kittisatra

ACKNOWLEDGMENT

The author wishes to express sincere appreciation and gratitude

to his major professor, Dr. H.I. Laursen, for his guidance, under-

standing and encouragement during the preparation of this thesis.

Thanks are due to Dr. W. L. Schroeder for providing the author

with invaluable guidance and philosophical influence throughout the

study.

A special thanks is expressed to his wife, Supaporn, for her

patience, encouragement and understanding during the year of this

study.

Throughout the author's entire academic career, he has

received the most enthusiastic support from his parents and his

brother. This thesis is dedicated to them.

TABLE OF CONTENTS

Chapter Page

I. INTRODUCTION 1

1.1 Statement and Scope of the Problem 2

1.2 Method of Solution 4

II. FINITE ELEMENT FORMULATION 102.1 Triangular Axisymmetric Ring Element Matrix

Equations 122.2 Quadrilateral Element Matrix Equations 272.3 Shell Element Matrix Equations 292.4 System Equations and Solution Process 40

III. TESTING OF COMPUTER PROGRAM 42

IV. ANALYSIS OF CIRCULAR CELL BULKHEADS 564.1 Isolated Circular Cell 584.2 Circular Cell Bulkhead 58

V. DISCUSSION OF RESULTS 1055.1 Discussion of the Isolated Circular Cell 1055.2 Discussion of Circular Cell Bulkhead 108

VI. SUMMARY AND CONCLUSIONS 118

BIBLIOGRAPHY 121

APPENDICES 124Appendix A: Element Matrices 124Appendix B: Fourier Harmonic Coefficients 133Appendix C: Gaussian Quadrature Numerical

Integration Procedure 135Appendix D: Modulus of Elasticity of Soil 138Appendix E: User's Manual for Circular Cell Bulkhead

Program 141Appendix F: Description of Computer Program 158Appendix 0: Program Listing 167

LIST OF FIGURES

Figure

Typical circular cell bulkhead structure.

Typical triangular soil element and shell element.

Finite element simulation of a circular cell bulkheadsystem.

Triangular axisymmetric element.

Quadrilateral element.

Shell element and coordinates.

Finite element model of rigid base soil system.

Vertical stress in soil due to surface load.

Radial stress in soil due to surface load.

Tangential stress in soil due to surface load.

Shear stress in soil due to surface load.

Finite element model of circular tank.

Radial displacement of circular tank.

Hoop force and longitudinal moment of circular tank.

Edge load applied to the cooling tower shell.

Approximate loading diagram for cooling tower.

Cooling tower stress resultants.

Radial and axial displacements for cooling tower at 0 =

Radial and axial displacements for cooling tower at0 = 22.5°.

An isolated circular cell.

Page

1. 1.

1.2.

1.3.

2.1.

2.2.

2.3.

3.1.

3.2.

3.3.

3. 4.

3.5.

3.6.

3. 7.

3. 8.

3.9.

3. 10.

3. 11.

3.12.

3.13.

4.1.

0°.

5

7

8

13

28

30

44

45

46

47

48

49

49

50

51

52

53

54

55

59

Figure Page

4.2. A circular cell bulkhead. 60

4.3. Circular cell finite element model. 61

4.4. Radial displacement, hoop force and sheet stressesdue to gravity load of cell fill (axisymmetric loadingcase). 62

4.5. Vertical displacement of circular cell due to gravityload of cell fill. 63

4.6. Contours of vertical stress in soil due to gravityload of cell fill.

4.7. Contours of radial stress in soil due to gravity loadof cell fill.

64

65

4.8. Contours of circumferential stress in soil due togravity load of cell fill. 66

4.9. Contours of shear stress in soil due to gravity loadof cell fill. 67

4.10. Principal stresses in soil due to gravity load of cell fill. 68

4.11. Equivalent loading diagrams due to backfill. 69

4.12. Assumed arc tension loads. 71

4.13. Fourier harmonic loading expansions for arc tensionload Ta (Case I). 72

4.14. Fourier harmonic loading expansions for uniformdistributed pressures pa and qb (Case II).

4.15. Reference diagram for displacements and stresses.

4.16. Horizontal displacements of cell due to arc tension loadsat levels H5 and top of cell (Case I).

73

76

77

4.17. Horizontal displacements of cell due to arc tensionloadings at levels H6 and H8 (Case I). 78

Figure Page

4.18. Horizontal displacements of backfilled cell at levelsH5 and the top of cell.

4.19. Horizontal displacements of backfilled cell at levelsH6 and H8.

4.20. Radial displacement of backfilled cell at 0 = 0 and180 degrees.

4.21, Radial displacement of backfilled cell at 0 = 30 and150 degrees.

79

80

81

82

4.22. Radial displacements of backfilled cell at 0 = 60 and120 degrees. 83

4.23. Radial displacement of cell at 0 = 90 and 270 degrees. 84

4.24. Deformed shape of circular cell bulkhead. 85

4.25. Vertical and hoop stresses of steel cell at 0 = 0 degree. 86

4.26. Vertical and hoop stresses of steel cell at 0 = 30 degrees. 87






4.32. Hoop force in cell vs. position. 93

4.33. Shearing stress in cell vs. position. 94

4.34. Contours of vertical stress in soil. 95

4.35. Contours of radial stress in soil. 96

4.36. Contours of shear stress (TrZ) in soil. 97

Figure Page

4.37. Contours of circumferential stress in soil. 98

4.38. Radial and vertical stresses in soil vs. depth of cell. 99

4.39. Radial and vertical stresses in soil vs. depth of cell. 100

4.40. Coefficients of lateral earth pressure inside the cell fill. 101

4.41. Comparison of radial deformation of backfilled cellat level H5. 102

4.42. Settlement of the top of the steel sheet piles vs. position. 103

4.43. Comparisons of hoop force in steel cell. 104

Appendix

E. 1. Example of finite element mesh showing node and elementnumber scheme. 143

E. 2. Boundary pressure sign convention. 152

F.1. Flow diagram for circular cell bulkhead analysis. 159

LIST OF TABLES

Table Page

4.1. Mechanical properties of soil. 57

5.1. Comparisons of results. 114

Appendix

B. 1. Fourier force coefficients. 134

C. 1. Gaussian weighting functions and stations. 137

D. 1. Modulus of elasticity of soil. 140

LIST OF TERMS

A Cross sectional area of triangular element

{A} Generalized displacement coordinate in soil

an Fourier coefficient for harmonic number n

ar Radial acceleration

a Axial acceleration

{B} Generalized displacement coordinate in shell

C Shell contour (circle)

[C], [Ciiki] Material constant tensor field

D Diameter of cell

d Distance from the mid -plane to the surface of shell

E Young's modulus

e Void ratio

{F} System load vector

{f} Element load vector

{Fb}, {fb} Body force vector field

{Fs}, {fs} Surface traction force vector field

G Shear modulus of elasticity

Gs Specific gravity

H Height of cell above dredge line

Hd Embedment depth of cell

h Depth of soil above point

Volume integrals

K Coefficient of lateral earth pressure

Ka Rankine coefficient of active earth pressure

[K] System stiffness matrix

[k] Element stiffness matrix

Length of shell element

M,. Shell moment tensor field

M Longitudinal moment in shellzz

00Hoop moment in shell

Mz0 Twisting moment in shell

N,. Shell stress resultant tensor field

N Longitudinal force in shellzz

00Hoop force in shell

Nz0 Shearing force in shell in z-0 plane

n Harmonic number

Lateral earth pressure due to backfillPa

PiPrescribed surface traction on surface S

qb Vertical pressure due to backfill

r Global radial coordinate (in cylindrical coordinate system)

ra Radius of connecting arc

rc Radius of circular cell

S Portion of surface on which stresses are prescribed

[5] Stiffness matrix of soil element in generalized coordinates

[SS] Stiffness matrix of shell element in generalized coordinates

Ta Tension in connecting arc per unit length

t Shell thickness

U Strain energy

{U} Nodal point displacement for entire structure

{u} Displacement vector field

u Radial displacement

V Volume of the soil (solid) element

v Circumferential (tangential) displacement

W Strain energy density

w Axial displacement

Z Global axial coordinate (in cylindrical coordinate system)

z Local longitudinal coordinate of shell

a Rotation of the shell surface in longitudinal axis

13 Transverse shell coordinate

Total unit weight of soil

Buoyant unit weight of soil

Yd Dry unit weight of soil

Unit weight of water

ZShearing strain in r-Z plane

Yr0 Shearing strain in r -8 plane

ZOShearing strain in Z-0 plane

5 Variation symbol

E, {E..} Strain tensor field

E rr' E E00 Radial, vertical and circumferential strains,

respectively

{o-..} Stress tensor field

o-o Average confining pressure

err' ZZ' o-00

Radial, vertical and circumferential stresses,

respectively

Cr Zi' CrZoVertical stresses in shell on inside surface and outside

surface, respectively

Lei' 0-00 Hoop stresses in shell on inside surface and outside surface,

respectively

Shearing stress in r-Z planeTrZ

Shearing stress in r-0 planeTr0

Shearing stress in Z-0 planeTZO

X Shell curvature change tensor field

Shell curvature change in longitudinal directionXzz

X 00Shell curvature change in hoop direction

Xz0 Shell curvature change due to twisting moment

v Pois son's ratio

p Density

Angular velocity

[X 0Shell coordinate transformation matrix

[],{(1)} Displacement expansion tensor field

[v]

(i)

(i)

Strain generalized coordinate tensor field

Inclination angle between local and global coordinates of shell

element

Angle of internal friction of soil

[f0 ] [10] Displacement transformation matrix

Total potential energy of the entire systemTrp

Total potential energy of element eTrpe

[(POI' ['POO] Shell displacement transformation matrix

0 Circumferential coordinate (in cylindrical coordinate system)

FINITE ELEMENT ANALYSIS OF CIRCULARCELL BULKHEADS

I. INTRODUCTION

A cellular bulkhead is a waterfront retaining structure formed

from a series of interconnected straight web steel sheet pile cells and

filled with soil, usually sand or sand and gravel. The combination of

steel and soil fill, which individually are unstable, forms a stable unit

offering resistance to its own gravity loads, lateral loads of water and

earth and surcharge loads. Current cellular bulkhead design methods

have adapted methods used for cellular cofferdams which are still

essentially empirical. Although various theories have been suggested

to derive analytical solutions for the stresses in the cell, so far most

designers in this field still rely heavily on past practice and experi-

ence.

Terzaghi (23) proposed an important method for design. In

using this method, a cellular cofferdam on a rock foundation is first

considered a rigid gravity structure. He proposed essentially a

simple bending theory in which a linear distribution of normal stress

on the base of the cofferdam is assumed and lateral pressures on

vertical shearing planes are obtained from an assumed coefficient of

earth pressure. The shearing stress in the fill and friction in the

interlocks was considered as a critical factor in cell stability.

2

Terzaghi also included other possibilities of failure: interlock pullout,

sliding on the base, and foundation bearing capacity failures.

The Tennessee Valley Authority (22) follows the same general

method of design as proposed by Terzaghi with some modifications.

In the TVA method it is assumed that the maximum lateral pressure

for interlock design occurs at a point one-fourth of the exposed height

of the cell above the dredge line. The full value of pressure at the

dredge line is not used because of the restraint provided by

embedment of the sheet piles.

Cummings (6) has proposed a method of cellular cofferdam

analysis known as the interior sliding theory where the resistance of a

cell to failure by tilting is gained largely through horizontal shear in

the cell fill. Cummings' conclusions are based on model tests which

indicate that the plane of rupture goes from the top of the pressure

side to the bottom inner corner (toe of cofferdam). The cell fill in

the rupture region acts essentially as a surcharge and only the soil

below the failure plane will develop shear resistance. The Cummings'

method for determining interlock tension and sliding on the base is

the same as the Terzaghi method.

1. 1 Statement and Scope of the Problem

Analyzing the soil-steel sheet pile interaction of the cellular

bulkhead problem is difficult because the structure consists of two

3

very different materials. In addition, the bulkhead system is

subjected to poorly defined non-axisymmetrical loading and the

boundary conditions of the system are complicated.

The objective of this research is to develop a rational approach

for stress analysis of circular cell bulkheads and to compare

analytical results with data obtained in full scale field studies by other

investigations. To do so would not only provide an aid for predicting

the most likely mode of failure, but would also provide an insight to

the elastic stress distribution throughout the structure.

To deal with the circular cell bulkhead problem, certain

assumptions have been made. Specifically, they are as follows:

1) The soil is elastic and isotropic.

2) The thickness of steel sheet piles is uniform throughout the

whole structure and no slippage occurs along the interlocks.

3) The soil-structure interface is perfectly rough with no

possibility for slip.

4) The actual continuous circular bulkhead system can be

represented by a single circular cell which is considered as

an axisymmetric structure subjected to non-axisymmetrical

loading.

5) At some finite distance beneath the bottom edge of the sheet

piles, the soil foundation is rigid and rough.

4

The scope of the study includes development of a method for

analyzing circular cell bulkhead problems and the development of a

computer program to make the necessary calculations. Two load

conditions are studied:

1) the gravity load of the cell fill and

2) external loading due to backfill.

These loading conditions are considered as axisymmetrical and

non-axisymmetrical loads, respectively.

1. 2 Method of Solution

A typical circular cell bulkhead is shown in Figure 1.1 wherein

the structure is exposed to water on the river side and soil fills on the

shore side. The system is assumed an axisymmetric structure sub-

jected to non-axisymmetrical loading. The loading, however, is

symmetric about a vertical plane containing the axis of the cell.

Because the bulkhead system is not a continuum, an assumption

concerning the characteristics of the soil-sheet pile interface is

needed. It is assumed that the interface is perfectly rough, with no

possibility for slip between the soil and the sheet pile. Also, the cell

itself is assumed to be a continuous circular cylindrical shell. The

structure is thought of as being composed of two substructures:

solid soil and the cylindrical steel shell enclosing it.

0°. -

Back

I) APlan view

Steel sheet pile

- Inside fill

Dredge line

Section A-A

Backfill

Figure 1.1. Typical circular cell bulkhead structure.

5

6

The basic concept of the finite element method is the idealization

of the actual continuum as an assemblage of discrete structural ele-

ments, interconnected at a finite number of joints or nodal points.

For structures that are physically axisymmetric, i. e. , geometrically

axisymmetric and possessing material properties that are axisym-

metric, the nodes are actually circles and are called nodal circles.

Figure 1.2 illustrates a typical triangular element and a shell ele-

ment. The generalized displacements and loadings of a nodal circle

can be expressed in terms of finite Fourier-series and the problem is

uncoupled in each harmonic (27).

For each harmonic the system stiffness [Kn ] and system load

vector {Fn} are formed by summing appropriately the stiffness

[kn] and force {fn} for the discrete elements of the structure,

where subscript n denotes the nth harmonic. After the system

stiffness and load vectors are obtained, the structure is analyzed by

the standard stiffness method (14). Since the bulkhead system is

assumed as linear and elastic, the principle of superposition allows

a solution for the specified load by simply adding the separate solu-

tions that are obtained from the separate Fourier harmonic terms of

separate load components.

The circular cell bulkhead can be represented by shell elements

and quadrilateral elements as shown in Figure 1.3. Boundary 1 is

assumed to be a firm foundation and can be considered fixed. Boundary

7

Z (Axis of

(a) Triangular ring elenlent

Figure 1.2. Typical triangular soil element and She element.

0

4-Axis of symmetry

Shell element

Soil element

Interface condition

Boundary 2 --,...,.

il

8

r01"Boundary 1

Figure 1.3. Finite element simulation of a circular cell bulkheadsystem.

9

2 is assumed to be far from the axis of symmetry OZ and beyond

the zone of failure. Therefore, boundary 2 can be assumed to be on

rollers.

The finite element analyses yield the distribution of stresses

in the soil elements, the steel sheet stresses and nodal point displace-

ments.

10

II. FINITE ELEMENT FORMULATION

The basic finite elements used in this study are quadrilateral

axisymmetric ring elements for the soil and cylindrical shell-of-

revolution elements for the steel sheet pile with constant cross -

section. The quadrilateral element is composed of four sub-triangles.

To formulate such elements, the Theorem of Minimum Potential

Energy is used.

To apply the minimum potential energy theorem it is necessary

to assume a displacement field in terms of a set of unknown Ritz

parameters (coordinate functions) that satisfy the hypotheses of the

theorem (5). The restrictions on assumed displacement functions are

that they be continuous over the entire body and possess piecewise

continuous first partial derivatives. The method of analysis is to

subdivide the domain into an assemblage of discrete elements and

assume appropriate kinematic functions within each element such that

the compatibility conditions across the element interfaces are

satisfied.

The contribution to the total potential energy of one element

can be written as

Tr pe = U .SSS fiu.dV p.u.dS (2. 1)

11

where

U= irSicWdV

V

U is the element strain energy and W is the strain energy density

which is assumed to be positive definite. The body force vector field

f. and the prescribed surface traction p. on the portion of the

surface S are positive if they act in the direction of positive coordi-

nate axes. u. is the displacement vector field in the volume V.

The strain energy density can be written as

W 1 (2.2)3

-

where o-.. is the stress tensor field and E.. is the strain tensor

field.

The total potential energy of the entire system is the sum of the

potentials of the individual elements. Thus, for a system of L

elements

L

IT = ITp pe

1

(2. 3)

An absolute minimum potential energy is sought by taking the

variation of the potential energy function with respect to the discrete

displacement variables and set it equal to zero

8Tr = 0P

12

(2.4)

Then a set of matrix equations are formed accordingly.

2. 1 Triangular Axisymmetric Ring Element Matrix Equations

A triangular axisymmetric ring element is used in the study as

shown in Figure 2.1. Matrix notation and cylindrical coordinates are

used in the analysis, that is, radial distance r, axial distance Z,

and circumferential angle 0. The right handed system is used in

the coordinate system. Since the soil is assumed a linear elastic

material, the stress-strain relationship may be expressed in the

form of the constitutive equation (7)

IT = Ci. Elj jki k/

or, rewritten in matrix form

(2. 5a)

{a} = [C] {E} (2.5b)

where Cijki is a fourth-order material constant tensor.

If Equations (2. 2) and (2. 5b) are substituted into Equation (2. 1),

the element potential energy becomes

13

Figure 2.1. Triangular axisymmetric element.

14

Trpe = S S S 2 {E(r, Z, 0)}T[C]r{E(r, Z, 0)}dV

V

gir {u(r, Z, 0)}T{f(r, Z, 0)}dV

V

SS {u(r, Z, 0)}T{P(r, Z, 0)}dS (2. 6)

in which

{ } denotes a column or a row matrix

[ ] denotes a rectangular or square matrix

The superscript T denotes the transpose of the matrix.

For an isotropic material with Young's modulus E and Poisson's

ratio v , the constant C will be

1-v v

1-v

0 0

0 0

0 0

v

v

1-v

0

0

0

0

0

0

1-2v

0

0

0

0

122v

0

0

0

0

0

1-2v

2. 7)[C] = [C]T (1 +v)(1-2v)2

0

0 02

The displacements within each triangular element are assumed

to be a linear function of the coordinates r and Z (27).

un = Aln + A2nr + A3nZ

wn A4n + A5nr + A6nZ

vn = A7n + A8nr + A9nZ

15

(2. 8a)

(2. 8b)

(2. 8c)

where u, w, and v are radial, axial and circumferential displace-

ment components, respectively. Subscript n is an integer called

the harmonic number of the Fourier series. As are constant coef-

ficients that represent the generalized displacement coordinates of

the element.

In general cases, an axisymmetric structure subjected to

arbitrary loadings, displacements and the body and surface loads are

represented by Fourier series as follows:

N

u(r, Z, 0) = un r, Z)cos n0 +) un(r, Z)sin nO (2. 9a)

n=0 n=1

N N

w(r, Z, 0) = wn

r, Z)cos ne + wn(r, Z)sin ne (2. 9b)

n=0 n=1

N N

v(r, Z, 0) = vn r, Z)sin nO + r, Z)cos nO (2. 9c)

n=1 n=0

body forces are:

N N.....

fr (r, Z, 0) = f (r, Z)cos nO + f (r, Z)sin nern rnn=0 n =1

fz(r, Z, 0) =

fe(r, Z, 0) =

n=0

N

n(r, Z)cos nO + f Zn(r, Z)sin nO

n=1

r, Z)sin nO +

n=1

N

n=0

16

(2. 10a)

(2.10b)

r, Z)cos nO (2. 10c)

and surface loads are:

pr(r, Z, 0) = prn(r, Z)cos nO + prn(r, Z)sin nO

n=0 n=1

(2. 11a)

N N

pZ

(r, Z, 0) = pZn (r, Z)cos no + / pzn(r, Z)sin nO

n=0 n=1

p0(r, Z, 0) = L pen(r, Z)sin nO + TiOn (r, Z)cos no

n=1 n=0

(2.

(2.

11b)

11c)

where the single barred ( - - ) and double barred ( -) quantities

represent functions of r, Z and n only but not 0.

Since the circular cell bulkhead system as shown in Figure 1. 1

has symmetric loading with respect to the rZ plane at 0 = 0, the

double barred series in Equations (2.9), (2. 10) and (2. 11) will not be

17

used in this study. The axially symmetric case is represented by use

of only the n = 0 term of the single barred series.

Substituting Equations (2.8) into Equations (2. 9), the

displacements become

u(r, Z, 0)

w(r, Z, 0)

v(r, Z, 0)

n=0

where

{4)(r, Z) }Tcos nO 0 0

0 {4(r, Z)}T cos nO 0

0 0 {4)(r, Z) }Tsin nO

{4(r, Z)}T = {1 r Z}

Equation (2. 12) can be written symbolically as

N

(2. 12)

i{u(r,Z, 0)} = [l(r, Z, 0)] {A. = 1, 2, , 9 (2. 13 )in

n=1

Thus the displacement vector field for each Fourier term may be

written as

{u (r, Z)} = ['(r, Z)]{A. } (2. 14)

18

Let u., w., and v. refer to the r, Z and 0 direction

displacement components of any corner node j of the element as

shown in Figure 2.1. Using the displacement boundary conditions for

u at each corner, the following matrix expression is obtained:

1 r Aln

1 r2 Z2 A2n (2. 15a)

Similarly, the displacements w and v can be expressed in terms

of generalized coordinates A. as follows:in

and

l

{f2nrnv

v3n

rl

r2

r3

Z2

Z3

5n

A6n

(2. 15b)

1

r2

r3

Zr1

Z2

Z3 A9n

(2. 15c)

Equations (2. 15) can be combined together and written in symbolic

form as

fu on} [ Y{Ain} (2. 16)

where uOn is nodal displacement and /0

A.

19

is as shown in Appendix

The generalized coordinates A. are obtained by inversionin

of Equation (2. 16) and expressed in terms of nodal point displacements

as

{Ain} = [ 0 1}{uOn}

1 is shown in Appendix A.

(2. 17)

If Equation (2. 17) is substituted into Equation (2. 14), the

displacements at any point in the triangular element can be written

in terms of the nodal displacements as

w

n1

0

0

0

n1

0

0

0

n1

n2

0

0

0

n2

0

0

0

n2

n3

0

0

0

n3

0

02A

where

n1 al + blr + c1Z

n2 a2 + b2r + c2Z

n3 = a3 + b3r + c3 Z

(2. 18)

(2. 19a)

20

al = r2Z3 - r3Z2, b1 = Z2 - Z3, c1 = r3 r2

a2 = r3Z1 - riZ3, b2 = Z3 Z1, c2 = r1

r3

(2. 19b)

a3 = riZ2 r2Zi, b3 = Zi Z2, c3 = r2

- r1

and A is the cross-sectional area of the triangular element.

The element strains are obtained by differentiating Equation

(2,18) to obtain (15)

au

rr araw

'zz az1 , aV

+E u)OA r ae

= 2E =au

+8w

rZ rZ 8Z ar

1 au2E = v) + av

r8 r0 r 80 8r

= 2e ay 1 awze ze az + ae

(2. 20)

The strain tensor can be expressed in terms of generalized

coordinates

{E (r, Z, 0)} = (r, Z, 9)1{A. } (2. 21a)

n=0

where Is' is shown in Appendix A.

If Equation (2. 17) is substituted into Equation (2. 21a) a relation

between element strain tensors and the nodal displacements is

obtained

{E (r, Z, e)} =

n=0

P (r, Z, 0)Dvillu }On

21

21b)

Substituting Equation (2.21b) into Equation (2.6), the total

potential energy of a single triangular element can be written in terms

of its nodal displacements

N M

ape= OiSic(uif

2 On L 0 n(r,Z,O)]T[C][Vm(r,Z,0)][fiNu

Om}

0

n=0 m=0 V

IJS

{ti0n}T[0-1]r[tt.n(r, Z, 0)]T{fm(r, Z, 0)})dV

u n}T

-1 ]T['/n(r, Z, {pm (r, Z, 0)}dSj (2.22)

0

where n and m are harmonic numbers.

Since the 0-dependence in the integrals in Equation (2. 22) is

known explicitly from variables r and Z it can be carried out

directly. At the same time by making use of the orthogonal proper-

ties of the harmonic functions the integrals are integrated from

0 = -Tr to 0 = Tr. These integrals are a well-known type in Fourier

analysis. The general forms are stated as follows:

22

sin m0 sin nOdO =-Tr

for m = n 0

for m n and for m = n = 0

2Tr for m = n = 0

.51 Trcos me cos n0d0 =-Tr

1T for m = n 0 (2. 23)

0 for m mn

iTsin m0 cos n0d0 = 0 for all m and n-Tr

Thus, the sum in Equation (2. 22) exists only for m = n and it

becomes

N

wpe US'S( 2 Onfu }TW0 lir[V (r, Z)]T[CPn'(r Z)P0 1] {u0n}

n=0 V

Let

and

TTS

{uOn

T0

I1]T[n(r, Z) ]T {fn(r, Z)})dV

n}T -1 T Z)]T{pn(r, Z)}dS)

{fbn(r, Z)} = [n(r, Z)]T{fn(r, Z)}

{fsn(r, Z)} = [n(r, Z)]T{Pn(r, Z)}

[sn(r, Z)] = [nt(r, Z)]T[C][P (r, Z)]

Substituting Equation (2. 25) into Equation (2. 24) we obtain

(2. 24)

(2. 25a)

(2. 25b)

(2. 25c)

T -1 T -1IT ssy(1, } [s (r, Z)P, 1{u }

pe 2 On 0 u On

n=0 V

ffS

= {uOn}T

[/-1 ]T{fbn(r, Z)})dV

0

T 1 Tn}

[.10 ] If (r, Z) }dSsn

The body force vector will be (21)

{f(r, Z, 0)}T = {prw2-par cos 0, -pa -par sin 0}

where p is the mass density of the body,

23

(2. 26)

(2. 27a)

w is the angular velocity,

and ar and az are radial and axial acceleration, respectively.

After substituting Equation (2. 27a) into Equation (2. 25a) and

dropping the 0-dependence, the body force components of the triangu-

lar element are

and

{fbn(r, Z)} T{prw

2, pr2 w2,

prZw2, -pa z' -prat, pZa Z' 0, 0, 0}

for n = 0 (2.27b)

{fbn(r, Z)}T = {-par r r r r, -pra, -pZa, 0, -pa, -pra, -pZa

for n = 1 (2. 27c)

{fbn(r' Z)} = 0 for n > 2 (2. 27d)

24

The body force components in the direction of r and 0 are

included only for generality and will not be used in the analysis of

circular cell bulkheads. The surface traction vectors will be inte-

grated explicitly for each loading because they are arbitrary. The

area integration is performed over the triangular area in the r-Z

plane by denoting

[Sn] =Cg [sn(r, Z)]dV

Matrix [Sn

] is shown in Appendix A.

The system body force and surface traction vectors are

SIS {fbn(r, Z)}dV

V

{F sn} = g{fsn(r, Z)}dS

The following notation is used for the various integrals:

11 ,acciv, 16 as dV

V V

12 =ffidV, 17 = siscrdVV V

(2. 28a)

(2. 28b)

(2. 28c)

(2. 29)

25

I3

= S.11c dV, I8

= ZdV

V

I4

= .17 dV, 19

= ygr ZclV (2.29cont. )

V

15 SYS dV, 110 = SSSrZdV

V

After the integration in Equation (2. 28b) is carried out, the body

force vector may be written as

{Fbn}T = {pw2I p2I9'

pw2I10'

Z-pa_I

7, -pa

ZI

8' 0, 0, 0} (2. 30a)

for n= 0

{Fbn}T = {-pa rIF -parI 7'-par 0, 0, 0, -pa rIV -par - parI8} (2. 30b)

for n = 1

For a single finite element the potential energy becomes

N

pen=0

On}T L 0 [ 5 ] 0 ] {u0n}

{uOn}T 0 {Fbn}+{Fsn} }) (2. 31)

Thus for the whole system of L such elements, the total

potential energy in Equation (2.3) is

Tr

p

L N

ff =1 n=0

{Un 0 0}T[OVn[S(1 )][0)]{Un}

{U }T[ 0(i)]rf {F" ")}+{F)} })n bn sn

26

(2. 32)

where {un} is the discretized displacement vector for the entire

system and [1. (/) ] is the generalized coordinate transformation

matrix of element f to the displacements in the entire system.

Taking the variation of the stationary potential energy with

respect to the discrete displacement variables as stated in Equation

(2.4) we obtain

where

L N

=1 n=0

([%-1(1 )]T[S(1)P0-1(1)1{Un}-[%-i"V{F")}) = 0 (2. 33)

{F(i)} = {FM} + {Fsn)}bn

From Equation (2. 33) a set of governing simultaneous equations

can be written for each Fourier harmonic term as

[Kn] {lin} = {Fn} for n = 0,1,2, N (2. 34)

where Kn is the stiffness of the assembled system and Fn is the

system load vector. They may be expressed as follows:

and

L

[Kn] = [0-101)]TisCPn I- 0

f =1

{Fn} =

L

1=1

-1(.e) T ( )] {Fn

)}0

in which [1.01(/)] is shown in Equation (A. 2) in Appendix A.

From Equation (2. 35a), the stiffness of element

Fourier harmonic n can be expressed as

[k(rif )] 01(1)]r[s(f)][

)1(.1 )]

2.2 Quadrilateral Element Matrix Equations

27

(2. 35a)

(2. 35b)

in any

(2. 36)

The use of the quadrilateral as the discrete element to idealize

the system is desirable since it reduces the required input in the

computer program and the resulting set of equilibrium equations has

fewer unknowns for a given number of triangular elements. A typical

quadrilateral element is composed of four triangles as illustrated in

Figure 2.2. The coordinates of the center node are computed as the

average of the four corner point coordinates.

In the case of non-axisymmetric loads, the 12 degrees of

freedom quadrilateral element matrix is formed by first combining

the four 9-degree-of-freedom triangular element matrices into a 15

degree of freedom element matrix. Using a process of static

condensation (5, 28) the three internal displacements are eliminated,

resulting in a 12 degree of freedom quadrilateral element matrix.

4(10, 11, 12)

1(1,2,3)

28

Numbers in bracketscorrespond to unknownat each node

Figure 2.2. Quadrilateral element.

The four triangular element stiffness are combined by the code

number technique (14, 24). If the load vectors {F0} for each of the

triangular elements are similarly superimposed, a partitioned matrix

equation is obtained

kaaS

ab

lc_b a kbb_

(2.37)

where subscript a is associated with nodal points 1 to 4 and

subscript b is associated with point 5. Equation (2. 37) may be

written as two matrix equations

{Fa} [kaa]{ua/ kabilub/ {F0a}

{Fb} = [kba] {ua} + [kbb] {ub} + {F0b}

Equation (2. 38b) can be solved for the displacements ub:

{ub} -[kbb]-l[kba]1

}- {F0b }}

29

(2. 38a)

(2. 38b)

(2. 38c)

If Equation (2. 38c) is substituted into Equation (2. 38a), an

expression is found relating the forces at points 1 to 4 to the

unknown displacements at points 1 to 4 and the known loads as

where

{Fa} [k*aa] lua}{4-a}

J.

[k ] = [k]

[k ilkaa aa ab bb ha]

the quadrilateral element stiffness matrix, and

{F0a} = {F0a} [kab][kbb] -1{FOb

}

the modified load matrix.

2.3 Shell Element Matrix Equations

Finite element analysis of shells of revolution has been

(2. 39a)

(2. 39b)

(2. 39c)

developed and used for axisymmetrical loadings (9, 10, 19). For non-

axisymmetrical loadings, the displacements and loadings are expanded

in Fourier series (12, 18, 21, 27). A typical shell element is

illustrated in Figure 2.3.

Z, w

p,

30

Note: v is positivewhen directedinto the paper

9,v r, u

Figure 2.3. Shell element and coordinates.

For thin shells of revolution, the potential energy of a single

element may be written as (7)

where

1Trpe = {

2(N E

j+1\4..X

1.)ds} p.1

1u.dC

1 13 3( 2.40a)

t /2N.. = o-..df3 (2. 40b)

-t/2

t/2

"

M.. = cr..Pd13 (2. 40c)13 -t /2

31

N,., M.., E.. and x.. are stress resultant, moment,13 13 13 13

extensional strain and curvature tensor fields, respectively, acting

on the portion of shell surface S. p. and u. are nodal loads and

nodal displacements along the nodal circle C, and t is the shell

element thickness assumed as constant throughout the whole cell, p

is the local transverse coordinate (see Figure 2.3).

The displacement field of the conical shell may be assumed as

w'(z) = B1 + B 2z

v'(z) = B3 + B4z

u'(z) = B5 + B 6z + B z 2+ B8z3

3

7

(2. 41a)

(2.41b)

(2. 41c)

The shell displacements of non-axisymmetrical loading case

also can be expanded into Fourier series:

w'(z, 0) =

n=0

yr' (z)cos nOn(2. 42a)

v'(z, 0) =

n=0

v' (z)sin non

(2. 42b)

u'(z, 0) = un(z)cos no (2. 42c)

n=0

where v' and w' are the transverse, circumferential or

tangential, and longitudinal shell displacements, respectivey, with

32

respect to the load element coordinates. z is the longitudinal shell

coordinate and n is the Fourier harmonic number.

The coefficients B1, B2, ..., B8 represent the generalized

displacement coordinates of the shell element. The number of con-

stants B are assumed equal to the number of degrees of freedom

of the element. Each nodal point has four degrees of freedom that are

denoted by u', v', w' and a. Translation u', v', w' have been

defined previously; a is the rotation of the meridian in the shell

surface in a plane which passes through the nodal point and contains

the axis of revolution of the shell. The rotation a is positive if it

corresponds to a positive value of aultaz along the meridian.

By using the boundary conditions, the nodal displacements in

Equation (2.41) can be written as

un' (1)-1

wi(1)n

v.' (1)n

au' (1)

az.

u' (2)n

WI (2)n

vI (2)n

au' (2)n

az

0 0 0 0 1 0 0

1 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0

n

2n

3n

n

n

n

7n

n

(2. 43a)

or in symbolic form

{uon} = [90]{Bin} i = 1 2, , 8

where I is the length of the shell element.

The constants {B.}

are obtained by inversion asin

33

(2. 43b)

{Bin} = [(po- l]{uol

n} (2.43c)

where [cp0

1] is shown in Equation (A. 12) of Appendix A.

Introducing a shell displacement transformation matrix, the

relationship between local and global coordinate displacements are

where [x0]

Appendix A.

{uon} = [xo]{uon)(2. 44a)

is the shell transformation matrix and is shown in

{uOn}

is the global coordinate displacements.

If we substitute Equation (2. 44a) into Equation (2. 43c), the

generalized coordinates become

{Bin} = [(P0 1][X 0i{u

On}

or

{Bin} [`P00]{u0n}

where

[(p00] [C201][X0]

(2. 44b)

(2. 44c)

[9001]is shown in Equation (A. 15) of Appendix A.

According to the Novozhilov theory (17) of thin shells, the

strain-displacement relationships are

aw'E -

ZZ aZ

1 , ayae 'E00 r ( + L.1+w'1 cosCOS +w' sin (0)

.Yz0

ay' 1 aw'az + (r De

a2u,Xzz =

az 2

- sin 4:v')

1 ay' a2 12'X = (cos

ae00 r 2 4)-30 2 )

1 au'- r az sin (I)

1 au' 1 a 2u' lay' 1Xz0 = 42 sin 4) ae

- 1-7azae +(r.az

-r r

34

(2.45)

sin 4)Ocos 4)]

where E zz' E 00' and Ez0 are the longitudinal, normal and shear

strains; xzz , X00 and Xze are the curvature changes of the mid-

dle surface of the shell.

The general stress-strain relationships for a linear isotropic

elastic thin shell are (17)

Nzz

00

Nz0

M zz

MOO

Mz0

Et

1

v

0

0

0

0

v

1

0

0

0

0

0

0

1-v

0

0

0

t 2

0

0

0

vt2

2

0

0

0

21-v 122

vt

122t

12

0

12

0

or written symbolically as

where

{o} = [c]{ E}

0

0

0

0

35

(2.46a)

(2. 46b)

The potential energy in Equation (2. 40a) can be expanded as

13 13

N N

{Sln("T[Nnm(0)] {Slm(z)}

N

3M13 ..X1... =

m=0

N

n=0 m=-0

n(z)}T[Mnm(0)] {S2m(z)}

avI v'{S (z)}

T n n n n nln az az ' r r

a2 av' au' V'

{s(z)}T{n 1 n 1 nnni2n az2 r az r az r2 r2

[Nnm(0)] and [Mnrn(0)] are developed in Appendix A.

(2. 47a)

(2. 47b)

(2.47c)

(2. 47d)

36

Due to the orthogonal properties of the Fourier harmonic

functions in the period of -Tr < 0 < Tr, the sum of the potential

energy exists only for m = n.

If Equation (2. 41) is differentiated and substituted into Equation

(2.47c), {S in(z)}

{S1(z)} =

will become

0 1 0 0

0 0 0 1

10 0r r1

0 0r r

0 0 0 0

0

0

0

0

1

r

0

0

0

0

zr

0

0

0

0

z2r

0

0

0

0

z3

In

B2n

B3n

B 4n

B5n

B6n

B7n

8n

(2.48a)

r_

or written in symbolical form

{S1(z)} = [Xi(z)1{Bid (2. 48b)

Substituting Equation (2. 44b) into Equation (2.48b), {S1(z)}

may be written in terms of nodal displacements as follows

{S1(z)} [X1(z)][4900]{ti0n} (2.48c)

Similarly, if Equation (2.41) is differentiated and substituted into

Equation (2. 47d), {S2n(z)} can be expressed in terms of

generalized coordinates and nodal displacements, respectively, as

{s 2(z ) }=

0

0

0

0

0

0

0

0

0

0

0

0

0

1

0

1

r0

0

0

0

0

1

0

0

1

r

0

z

2

0

2z

6z

0

3z 2

r

0

z2

0

z3

2r0

2r0

2r

2r 2r 2r

or in symbolical form

and

{S2(z)} = [X2 i n

(z)] {B }

)} [x2(z)][900] lu0n}

37

(2. 49a)

(2. 49b)

(2. 49c)

The terms [X 1(z)] and [X2(z)] in Equations (2. 48b) and

(2. 49b) can be expanded further as

where

X1 (z).= G.. x.(z)

X2(z). = Hijkx (z)j

(2. 50a)

(2. 50b)

2, T 1 z z z

31 z z

2z

3{ zx.( )1 =

r,z, 2 2 2 2

, (2.50c)r r r r

38

and

G112 = G214 = G321 = G332 = G423 = 0434 = G525 = G536

= G547 = G558 = 1

H224 = H326 = H473 = H484 = H575 = H586 = H597 = H5108 = 1

H117 = H337 = 2

H348 = 3

H168 = 6

All other G's and H's are zero.

After substituting stress and strain vector fields into Equation

(2. 40a) and using the advantage of 0-dependence being known explicitly,

the 0-integration can be performed directly and the potential energy

becomes

apepen=0

({110n}

T[90-013T [ [X1 z)]

T [Nn][X i(z)]

+ [X2(z)3T[Mn][X2(z)1][9003 {uOn})di

{u0n}T [C900}T{Fzn}(2. 51)

where Q is the integration over the length of the shell element

which is carried out by using numerical integration of the Gaussian

Quadrature Formula (4, 16).

39

Introducing

[Ssn] = S([X1(z)]T[Nn][X1(z)] +[X2(z)]T[Mn][x2(z)])d/ (2.52a)

Equation (2.. 51) can be rewritten as

is

N

/Trpe 2(1 {u

On}T[900-1] T[SS

n ][900-1

] {u n}-{uOn

}T [yo001

] {Fzn})

n=0 (2. 52b)

For the assemblage of L elements, the total potential energy

L N

( fun

}T[q)00

-1(0]T[ss(nikp00 -1(/ )llu2

1=1 n=0

{un}T[P00(fT

{Fzn} ) (2.53)

The governing equations are obtained by performing a variation

on the total potential energy of the system with respect to the discrete

displacements and setting them equal to zero. Accordingly

Ear = 0p

which yields

N L

([901(1 )]r[SSn(1 )1[490-01(1)]{un) = [(P001(1)}{Fzn )})

n=0 1=1

(2. 54)

40

For each Fourier harmonic term

[Kn]{Un} = {Fri}, n = 0,1,2, . . . , N (2. 55a)

where [Kn

] is the system stiffness matrix of the shell element and

is given by the expression

L

[Kn] [(P 1" )]T[SSn" )1[9010" )]00.Q=1

[Fn] is the system load vector of the shell element

L

n1(1) T Fzn}

L 1-1' 001 = 1

The corresponding shell element stiffness matrix is

(1)[k ] [q)

-1( )][SS(1)][0-1()1n 00 n '00

2.4 System Equations and Solution Process

(2. 55b)

(2. 55c)

(2. 55d)

As the element stiffness matrices [kn] for each Fourier

harmonic term are generated, they are appropriately superimposed

into a system matrix [Kn]. The superposition is accomplished by

using the code number technique (14, 24).

A structural system that contains a large number of elements

will involve a large amount of data preparation. This includes a code

41

number for each element. In order to reduce preliminary work of

this nature, a subroutine was written to generate the code numbers

for each element.

The load vector {Fn} for each harmonic is assembled at the

same time as the structural stiffness matrix is formed. They are

generated using the code numbers. The total system of simultaneous

algebraic equations in the unknown {Un} is represented as

[Kn] {un} =n}

(2. 56)

The primary concern in the solution of this system is the

conditioning of the system matrix. It is a banded symmetrical matrix

where the band width is dependent on the direction of numbering the

nodal points. The nodal points should be numbered in such a way as

to minimize the difference between the largest and smallest nodal

point numbers for any element. By taking advantage of symmetry,

the coefficients are stored as an upper triangular matrix.

The solution of the equations is obtained using the linear

equation solver BANSOL. This subroutine uses the Gaussian elimina-

tion method. The band-width is automatically computed prior to

solving the system as it is required input to the solution subroutine

along with the system load vector.

42

III. TESTING OF COMPUTER PROGRAM

The purpose of this chapter is to present verifications of the

finite -21c.rresnt formulation of this investigation and also the computer

program. T3ecause the circular cell bulkhead system consists of

quadrilateral ring elements (4 triangles) and cylindrical shell elements

and Fourier harmonic series are used, three different types of prob-

lems are analyzed and their results are compared with those of

known classical solutions.

First, an elastic soil layer which overlies a rough rigid base is

analyzed. The soil system is subjected to a constant surface loading

q which is applied uniformly over a circular area of radius a as

shown in Figure 3.1. The soil is assumed to be a linearly elastic,

isotropic and homogeneous material. Thirty-five rectangular ring

elements are used to model the system and the results are shown in

Figures 3.2, 3.3, 3.4, and 3.5. It is seen that there is excellent

agreement with the classical solutions proposed by Burmister (3).

Second, a circular cylindrical tank filled with water is

considered in demonstrating the applicability of the shell element

formulation. The shell is idealized as 12 cylindrical elements and 13

nodes as shown in Figure 3. 6. The tank is subjected to hydrostatic

pressure only. The displacements and stress resultants are shown

in Figures 3.7 and 3.8, respectively. It is seen that there is good

agreement with the exact solution (25) except that the displacements

43

and hoop force at the top of the cylinder are somewhat on the high

side. This is due to the assumed point load at the top nodal point of

the tank.

The third structure to be analyzed is a circular cylindrical thin

shell cooling tower. It is included in this investigation in order to

verify and demonstrate the power and versatility of using the Fourier

harmonic analysis of the shell portion of the finite element method.

The cooling tower rests on 8 columns and supports its own weight.

The finite element model is shown in Figure 3.9(a). Five elements

are shown in Figure 3.9(a). A second model, not shown, with ten ele-

ments was also considered in which each element was one-half the size

of the first model. The second model was prepared because the first

model proved to be very coarse. Flugge (8) has developed a classical

solution for this problem with the total edge load as shown in Figure

3. 9(b). The approximate loading diagram which is based on a 5 term

Fourier expansion of the given stress function is shown in Figure 3.10.

The results for force resultants and displacements are shown in Fig-

ures 3.11, 3.12 and 3.13, respectively.

From the results it can be seen that the hoop force of the

5-element model is on the high side but the corresponding axial force

is in good agreement with the classical solution. A significant

improvement is achieved with the ten element model. Therefore it is

concluded that the results are quite dependent on the number of elements.

10'

It.

H

H = 10 ft

a = 2 ft

q = 1 ksf.

E = 30 ksi

v = 0.4

///AW//4Y/WwW7ACTUA

Rigid base with rough surface

35 element model

Roller

r

Fixed20'

Figure 3.1. Finite element model of rigid base soil system.

r

44

4-1

2

4

Vertical stress in percent of applied pressure20 40 60 80 100

8

10

Numbers on curves indicate radialdistances in radii

Figure 3. Z. Vertical stress in soil due to surface load.

Radial stress in percent of applied pressure10 20 30 40 50 60 70

0.75 0.25

Numbers on curves indicate radialdistances in radiiExact solution (3)

O a D e Finite element

Figure 3.3. Radial stress in soil due to surface load.

0Tangential stress in percent of applied pressure

10 20 30 40 50 60 70

0.75

0.25

Numbers on curves indicate radialdistances in radiiExact solution (3)

0 p Finite element

Figure 3.4. Tangential stress in soil due to surface load.

Shear stress in percent of applied pressure4 8 12 16 20 24

Nkinabers on curves indicate radialdistances in radii

Figure 3. 5. Shear stress in soil due to surface load.

D

13

12

Figure 3.6. Finite element model ofcircular tank.

9

8

7

6

5

4

32

Node number

4-3a)(1)

30

25

20

15

10

Exact solution (25)

0 Finite element

1 10-30 1 2 3 4 xRadial displacement, feet

Figure 3. 7. Radial displacement ofcircular tank.

50

Longitudinal moment, MZZ

in kips -ft/ft

-0. 4 -0. 2 0 0. 2 0. 4

Exact solution (25)

0Finite element

Note: Positive momentcauses compressioninside tank

-40 -30 -20 -10 0 10 20 30

Hoop force, N00, kips /ft40

Figure 3. 8. Hoop force and longitudinal moment of circular tank.

3a

V

a

Nodalnumber

5-element model

4t

5

43

21

Section C-D

(a) Cylindrical cooling tower

51

Section A-B

E= 30 x 103 ksiv = 0t = 0.24 incha = 3 ftP= 0.4 k/ft

0° 45°

1412°).1 33°

(b) Total edge load at the bottom

90° 135°

2. 75P

Figure 3.9. Edge load applied to the cooling tower shell.

A loading

- Approximation by 5non-zero Fourier terms

2.75P\rn=8

n=32

-P

/

rn.16 /

..I. -\ /

/ =40 // /// /

/ n.24 // /

\ .e-\ / I, ,-, // "4"\ '\ 4/ ` \ i ; \ ;,/ ,r\N I

is .' Ii \ 'i i I r 0 / / V

% i,0 1..... / \ 4eii---4,, \\/ ,,/ 4. iI\i' -/1 /I \\I A', \iAZI,i'..,4t -.4.1./ 11-,_,,,, \I .......,- 1

' /Ni

i

\ I, \ I\ Ai i' 1 ' /\l/ 1 I ...,/ \ / .... I v

_...-- \ 1 / % / ....1

% ,' I

\,A //III I

WVA\

lir Alik VidaV/

\ /\ /\ /\ /

Figure 3.10. Approximate loading diagram for cooling tower.

9 0 °

0 = 22.5° N/P

Exact solution (8, p. 231)

A 5 Element solution

0 A 10 Element solution

0.5

0.4

0.3

0.2

0.1

0

0 = 0°

Figure 3.11. Cooling tower stress resultants.

N /P

4

0.7

0.6

0.5

0.4

0.3

0.2

0.1

w (axial displacement)

100 200 300 400 500

Eu/P

Figure 3.12. Radial and axial displacements for coolingtower at A = 0°.

600

54

55

0. 7

0. 6

0. 5

0. 4

0.3

0. 2

0. 1

u (radial displacement

-w (axial displacement)

100 200 300 400 500 600

Eu/P

Figure 3. 13. Radial and axial displacements for coolingtower at 0 = 22. 5°.

56

IV. ANALYSIS OF CIRCULAR CELL BULKHEADS

Because the mathematical model and computer program are

applied to the analysis of axisymmetrical cell structures whose con-

figurations are shown in Figures 4.1 and 4.2, no difficulty was

experienced in generating the necessary input data. A user's manual

for the program which describes all necessary operations for gen-

erating the data is included in Appendix E. Appendix F describes the

function of the primary subroutines and Appendix G contains a listing

of the program.

Two analyses of the circular cell are presented in this chapter.

They are:

1. An isolated circular cell subjected to gravity load of the cell

fill which simulates the behavior of the cell after the inside

is filled completely as shown in Figure 4.1.

2. In addition to the weight of cell fill, the circular bulkhead is

subjected to the weight and lateral pressure due to backfill.

The general configuration is shown in Figure 4.2.

The circular cell structure used in this study was of similar

overall dimensions to one constructed at Port of Portland Terminal

No. 4 on the Willamette River near Portland, Oregon. In that bulk-

head, the front sheet piles were 18 feet longer than the back ones.

The web thickness of the front sheet piling was 0.532 inch and of the

57

back was 0.406 inch.

Since the computer program was developed for geometrically

axisymmetric structure, the average web thickness of 0. 469 inch was

assumed to be the uniform shell thickness. The average height of the

cell (from the dredge line) was assumed to be 58.5 feet and the depth

of embedment was uniformly 29.25 feet. The water level was

assumed to be at the top of the cell both inside and outside.

Because of the lack of soil test data, some parameters were

assumed and these properties are summarized in Table 4.1.

Poisson's ratio for the soil was assumed to have a constant value of

0.35 throughout the system. The modulus of elasticity was assumed

constant only within each individual layer of elements. Values were

obtained by using Richart's formula (20) as shown in Appendix D.

The modulus of elasticity of the steel was assumed to be 30 x 10 3 ksi.

and Poisson's ratio was assumed to be 0.3.

Table 4.1. Mechanical properties of soil.

TotalDensity

pcf(1)

Degree

Degree ofSaturation

%

VoidRatio

SpecificGravity

Poisson'sRatio

117.5 34 100 0.94 2.71 0.35

Soil is classified as medium dense to dense fine sand.

A 48 soil element and 9 shell element model was generated for

the structure as shown in Figure 4.3. The vertical boundary line and

58

the base boundary line were selected 60 feet and 29.25 feet from the

cell, respectively. The selection of these boundary lines and the

number of elements were based on the storage limitations of the

computer and estimated zone of influence for the cell.

4.1 Isolated Circular Cell

The isolated circular cell as shown in Figure 4. 1 is subjected

internally to the gravity load of the cell fill and externally by water

pressure. Since the water table inside the cell is at the same level

as the adjacent water, the submerged unit weight of the soil is used to

determine the net pressure loading. The system is obviously axisym-

metrically loaded, so only one Fourier harmonic term of n equal

to zero is necessary for the analysis.

The computer results for the isolated circular cell are shown

in Figures 4. 4 to 4. 10.

4. 2 Circular Cell Bulkhead

The general configuration of the circular cell bulkhead is shown

in Figure 4.2. The phreatic line both inside the cell fill and backfill

was observed at the same' elevation of the outside water (11). In this

study, it was assumed at the top throughout the bulkhead system.

Figure 4.11 illustrates the equivalent loadings due to backfill.

The quantity pa is the active earth pressure and can be calculated

Water

60'

Circular cell

D =65. 8'1.4

Plan

Axis of symmetry

59

60'

H=58.5'

H/2

H12

Boundary line1

/VANYZAN,

Cell

)0,y^NY,,,,ANyA

Steel sheet piles

Dredge line-,

ffy-,/,/,//m/,e,//mrt /7//f/f/ ////////////7 i/77/////7///////////////////////4Boundary line Soil foundation

Figure 4.1. An isolated circular cell.

F

r-Water--)

60'

Circular cell

D=65.8'

'Backfill--4

Axis of symmetry

Plan

60'

60

Water --)

H =58. 5'

Dredge line

H/2

H 2

'>/'/".K

1

1

(-Cell fill)

Soil foundation

Y/A \ssr/Ass

sheet piles

Boundary lines

Figure 4.2. A circular cell bulkhead.

//7/

61

Axis of symmetry

O00

Element layer number

Shell element

Soil elements cRoller

r"

r/7/7 /./ 7.7/7/7////////7/7/7/7/,'-o"

Fixed

Figure 4. 3. Circular cell finite element model.

Hoop force, N kips/in.-2 0 00 2

Radialdispl.

Dredge line

-1 0

0

19.5

39.0a)

a)

58.5

78.0

10 outsideo- ZZ1 inside

( : o. utsdidessufraface

surfacesurface

OR.

-

Li4

to

M1/4St1 \-

A/

I_ _ I _______-%-7.5 -5.0 -2.5 0 2.5 5.0

Displacement, inch Sheet stress, kips per square inch(a) (b)

Figure 4.4. Radial displacement, hoop force and sheet stresses due to gravity load of cell fill(axisymmetric loading case).

ts.)

63

0.1 ftDisplacement scale

---- Deformed shape

- -'I

Figure 4. 5. Vertical displacement of circular cell dueto gravity load of cell fill.

64

-0. 5

-1. 5

-z

- 3

-4

- 5

- 6

-7

Axis of symmetry

N.N. \

..... \.... N...... \ N.....

7-Steel sheet pile

Numbers on curves indicatestress contour in ksf.

5

-- - -- =i N.___ ....... `\- .

, \.....--

.... --

...... \

-1.0

\ ....

\N \ \

s. \ \..... \ \

\ \\

1\ \1 \ \

I %

% 1 \I 1 1 1

-7.0 -4.0 -3.0 -2.0

Figure 4. 6. Contours of vertical stress in soil due togravity load of cell fill.

-0. 9

- 0. 6

-0. 4

- 0. 5

- 1. 0

-2. 0

- 3. 0

Axis of symmetry

i Ir if

I

I/ 1

/ r \,.._

/. i

Steel sheet pile

-0. 3

65


-1.0 -1.5

4.0

.----- - , \i, -1.0 , 1,

1

...-- ----, N.../ ../ ...,..

-4. 0

0

Figure 4.7. Contours of radial stress in soil due togravity load of cell fill.

Axis of symmetry

Steel sheet pile

66

-1. 0

-0.7

-0. 4

-0. 5

-1.0

-2. 0

-3.0

Figure

- .A.

I/

//.../

,/

/

A

II

I/

,/ /

I-

I

II

I/

////

//

//

/

//

I

I

1

I

%

A

AA

///

/rI

I1

1

I

I

I

I

1

1

i

1

1

I

I

I

I

s

-0 3

Numbers onstress contour

-0. 1

-0.

curves indicatein ksf.

5

in soil

..,"---',./ ....

/ I

-0. 7 // //....

... /...,____.... /I ...

...

/

/'

5'

I/

///

4.8.

-3.0 -1.0

Contours of circumferential stressdue to gravity load of cell fill.

5Akis of symmetry

0 -------

0.20.5

1.0

1.4

0

67

Steel sheet pile


I // I

//

//I

1. 0

Figure 4. 9. Contours of shear stress in soil due togravity load of cell fill.

68

Axis of symmetry

A

A

Steel sheet pile

Compression (-)

10 ksf

4-4" .Arrow indicates5 ksf tens ion (+)

k k 4

Figure 4.10. Principal stresses in soil due to gravityload of cell fill.

Front

Steel sheet piles-)Boundary line

Dredge

(Axis of symmetry

Back

69

pa

qb

VV

//////////7777////////////////7/ /////////////////////////////7/Side view

T

Backfill) Ta

Plan view

Figure 4.11. Equivalent loading diagrams due to backfill.

by the following equations (13)

where

pa = Y'hKa

Ka = tan2(45°-)

2

Ka is the Rankine coefficient of active earth pressure

y' is submerged unit weight of soil

h is depth of soil

(I) is angle of internal friction of soil.

70

(4. 1)

(4.2)

The quantity qb in Figure 4.11 is assumed to be the surcharge

load which is equivalent to the dead load of the backfill. pa and qb

are uniformly distributed loads applied from one connecting arc to

the other on the back side of the cell.

The connecting arc tension load was assumed to be Ta and

can be obtained by

Ta = pra a

where ra is the radius of the connecting arc.

(4.3)

In order to represent the arc tension load in a Fourier expansion,

the arc tension load was assumed as a trigonometric loading function

which subtended a small angle at the center of the cell. The area of

the assumed loading must equal to Ta as shown in Figure 4.12.

71

Figure 4. 12. Assumed arc tension loads.

Since the Fourier coefficients for the connecting arc tension

load and backfill distributed loads are different, the computer

analyses were done separately and the results superimposed. The

loadings are divided into two cases in the following manner:

Case I due to connecting arc tension loads. The approximate

loading diagrams are represented by Fourier expan-

sions and are shown in Figure 4. 13.

Case II - includes the gravity load of the system itself and the

distributed load of pa and qb. The Fourier expan-

sion of these loads is shown in Figure 4.14.

The corresponding Fourier coefficients used in the analyses are

shown in Appendix B. Because of the very high cost of a computer

Actual loading

Approximate loading

\ / .........

,... ./ /V, 1C:...'

/."--- s_-,.........4,.<4:41,,,,N-,..., ...--.4 .,z..1-7....:),....:"'

\ / '....... s./ /././ N./ \

',......._ ....,'

30 60

0 in degrees

\

90 120

Line of symmetry

--4 - -\<

150 180

Figure 4. 13. Fourier harmonic loading expansions for arc tension load Ta (Case I).

Approximate loading

n=1

n=4

/A

0 60

Actual loading

8 in degrees

120

" / 1>,

/ / \-.."- ic./_,, ).0\ .... .

180

,

240

,

- -**

--"

300

V

360

Figure 4.14. Fourier harmonic loading expansions for uniform distributedpressures pa and qb (Case II).

74

run, the Fourier analyses were carried out using only 11 harmonics

for Case I loading and 8 harmonics for Case II loading.

An extensive amount of data is obtained (see Appendix E) from

the computer output for each run. In order to present it clearly, it

has been divided into two separate categories. First, the data per-

taining to the shell elements (sheet piles) is presented. This data

includes displacements and stresses in the steel sheet piling cell.

Second, the data pertaining to the soil portion of the structure is pre-

sented. A reference diagram for subsequent graphs is given in Fig-

ure 4.15.

Figures 4.16 and 4.17 show the horizontal displacement of the

cell due to connecting arc tension loads (Case I). The horizontal dis-

placement due to gravity load and uniform distributed pressures

(Case II) and combined displacement are shown in Figures 4.18 to

4.23. The deformed shape of the bulkhead is shown in Figure 4.24.

Figures 4.25 to 4.31 show the vertical stress and hoop stress

on both outside and inside surfaces of the cell at various angles.

Figures 4.32 and 4.33 show the variation of hoop force and

shear stress in the sheet pile along the circumferential direction of

the cell for each level as indicated.

The contours of stresses in soil elements are plotted in Figures

4.34 to 4.37. The radial stress and vertical stress distribution vs.

75

the depth of the cell in sections V1, V2 and V3 are shown in Figures

4.38 and 4.39.

Figure 4.40 is a plot of the coefficients of lateral earth pressure

K in the cell fill against the depth of the cell. The K values were

obtained directly from the ratio between the computed radial stress

and vertical stress in soil elements. The radial deformation, cell

settlement and hoop forces in the cell are plotted in Figures 4.41,

4.42 and 4.43, respectively, to compare with the values obtained in

the field (11).

Top

H1

H2

H3

H4

H5

H6

H7

H8

H9

Center line of cell

V1 V2 V3

I

Arc 0=90°

120° 60°Cell

Plane of

Front

symmetry 1,°Back0=0

150° °

0 =180°

Soil

.4Shell

0=270°

Plan

V1 V2 V3

Vertical section

76

Figure 4.15. Reference diagram for displacements and stresses.

0 9 Harmonics

0 11 Harmonics

Deformed cellOriginal cell

0=0°

At level H-5

Acr

0=180°

Displacement scale

0.2 inch0=0°

At top of cell

Figure 4.16. Horizontal displacements of cell due to arc tension loads at levels H5 and topof cell (Case I).

0=180°

0- 9 Harmonics

0 11 Harmonics

Deformed cellOriginal cell

0=0°

At level H-8

Displacement scale

0.1 inch0=0°

At level H-6 (dredge line)

Figure 4.17. Horizontal displacements of cell due to arc tension loadings at levels H6 andH8 (Case I).

0=1180°

Original cell------ Case II-- Final deformed cell (Case I + Case II)

0=180°

0=0°

At level H-5

Displacement scale

1 inch 0=0*

At top of cell

Figure 4.18. Horizontal displacements of backfilled cell at levels H5 and the top of cell.

Original cell

Case II0=180.

0=0°

At level H8

Final deformed cell(Case I + Case II)

Displacement scale

1 inch 0=0°

At level H6 (dredge line)

Figure 4.19. Horizontal displacements of backfilled cell at level H6 and H8.

Dredge linewANyx6N

1(1?

155 -19. 5Case II

Combined displacement(Case I + Case II)

1I//11'

I

,P - -39.0 48I

, A I

cp. ,a)

I

0-0°

\CI 58.5 0

0 =180°

-78. 0

/NNX/XNY

2 1 0 1 2 2 1 0 1 2

Displacement, inches

Figure 4.20. Radial displacement of backfilled cell at 8 = 0 and 180 degrees.

Dredge line

Case II


0=150° 0=30°

/1),N(//).Nx,

-78.0

2 1 0 1 2 2 1 0 1 2


Figure 4.21. Radial displacement of backfilled cell at 0 = 30 and 150 degrees.

Dredge line

1

------ Case II

0=120°


0=60°

II

II

19. 5

58. 5

78.0

2 1 0 1 2 2 1 0 1 2


Figure 4.22. Radial displacements of backfilled cell at 0 = 60 and 120 degrees.

2

/ANYANNe /"X/X\'V/>NN,t

1 0

_______ Case II

0=270°

1


2 2


0=90°

1

-19.5

-39.0 aa))

-58.50

-78.0

1

Figure 4.23. Radial displacement of cell at 0 = 90 and 270 degrees.

Dredge line

r11\Nxi./>\<

- - ________r--

L_

J

,,,,(Nriv/C

----- Deformed shape

Displacement scale

10.1 ft

-- _ _ _L, -- _ __ i- - -.i-I_ _ _ __------------------_ ___ ___ _ --- .,__

_I_ _ _ _ _ 1 -- - --- I

I

I-

_ ____ I(.... _ _ _ __ t_ -- - + - - -- _ _ _ _ _I_ _ - - - -

0=180° 0=0°

Figure 4.24. Deformed shape of circular cell bulkhead.

0

19.5

a)

039.0

4.,

a)

58.5

78.0

1, 1

outside surfacea- ZZ 0 inside surface

outside surfaceo-X00 Li inside surface

r

1"

cr

4o

Dredgeline level

.

o

Zi

Or.Al

Plillt

N.

-10 -5 5

Stress, kips per square inch-5 0

Figure 4.25. Vertical and hoop stresses of steel cell at 0 = 0 degree.

0

19.5

58.5

78.0

0 outside surfaceo-

aZZ o inside surface

o-00

outside surface

A inside surfaceC

0- 0-Zo

Dredgeline level-mg

0"0

A

Z i

icr

A i

-10 -5 5 -5Stress, kips per square inch

Figure 4.26. Vertical and hoop stresses of steel cell at A = 30 degrees.

5

0

19.5

a)

39.0

58.5

78.0

Ili

r0 outside surface

o- Ai

Z 0 inside surface1

outside surfaceo-

A inside surface

o-o

it

Zo

o-Zi

--Dredgeline level

11.11

-10 -5 0 5


Figure 4.27. Vertical and hoop stresses of steel cell at 0 = 60 degrees.

5

0

19.5

a) 39.0

58.5

78.0

41. 0 outside surfaceo-ZZ o inside surface

outside surface6

00 A inside surface

o-ZoUzi

Dredgeline level

o-eiLeopr

1

111.

iir

-10 -5 5


Figure 4. 28. Vertical and hoop stresses of steel cell at A = 90 degrees.

5

0

19. 5

39. 0

a)

58.5

78. 0

o- ZZ

X0 800

0 outside surface

inside surface

outside surface

A inside surface

I

0-

o-

Dredgeline level-

Cr

ooo

EliZ i

s,.

.

-10 -5 0 5

Stress, kips per square inch


0

0

19. 5

(Do 39.0

58. 5

78. 0

o- ZZ

o-130

0 outside surface

0 inside surface

outside surface

A inside surface

I

2

Cro

Dredgeline level

cro

o-Zi

3

-10 5

Stress, kips per square inch-5

Figure 4.30. Vertical and hoop stresses of steel cell ate = 150 degrees.

0

19. 5

76.0

-

1

o-ZZ

I

0 outside surface

inside surfacea

outside surfacecr

A inside surface2

Dredgeline level

_____a 0-0

. o-o 0-

eiA

All).-10 -5 5



5

2

1

n

'CDCD

za)c.) -10

'4-4

17-14

00

- 2

- 3

Maximum hoop force due to cell ---7fill only at level H5 AN

I

fl II

E.

v 411101111111111111111

)' 1

I

vV

Level

H2H3H4

0 H5a H6V H80H9

1

0

1n An 9n 1 zn 150 11

Position 0, degrees

Figure 4.32. Hoop force in cell vs. position.

H5H4

H3

H2

H6Line ofsymmetry

H8

H90

H3

V H4

Level p H5A H6

0 H9

Line of anti-symmetryof shearing stress

30 60 90 120Position 0, degrees

150

Figure 4.33. Shearing stress in cell vs. position.

180 210

C)

- 1.0

- 1.5

- 2. 0

...-/ ....1 ---,... - .-/ -5. 0 ..... --__.... / /- -./ / JP \ \ .. , .... ...... ...... ......

, /,/

/

...... ,,,,,,..

//11 --- ........... ....-

.... ,,J .// / / ......--

....- --- ....- \ \/ / / .--I \ 5'/ . \I . \ \I

III I / I

/ \ \ \I

-1.0N./ N N./ \

// -1 5 \ or---Steel sheet pile/ ,.-. .

/ / . ./ / 1\,1

// \ II/ / . I. /. / -2.,0 \ \

// \ \/ / ....'\ \/ / \

\N,/...-

/ .. \ N

-3.0. \/ / \\

Stress contour in ksf.

- 3.

- 4.

0

0

/// ../

/< /. % -4. 0-3.0)

-4.0 -5.0 -6.0 -5.0CL of cell

Plane 0 = 180° IN. Plane 9 = 0°

Figure 4.34. Contours of vertical stress-in soil.

Steel sheet pile-0.5-0.5


Plane 0 = /80°

Figure 4.35.

Plane 0 = 0°

Contours of radial stress in soil.

1.0

0.2

0.3

/., ..0 . //

/L 0._/ 5 / /

I

\\

1

1

1

1

0.5

1.3

1.0

Steel sheet pile


\

_ 0.50.3

O. 5

10 5/.

/J,// -/ , ,

/ ,. /. /..II /

Plane 0 = 180° 41

0

Plane 0 = 0°

Figure 4.36. Contours of shear stress (TrZ)

in soil.

0.8

1.0

-0. 5

-1.0

-0 5

-1. 5

I

.-__ ----

. ,.'\\ =2. 2 0

i // [/ /\\

% /...

/ / - N.3. 0 . \\\ \/ ----

1

.__ \ \..- \

/ ---- \.., ,

N\ \. ---

\..- / /-- --- - 4 . 0 N . .\ N

/ / -- r- \ .I / ,.

I\ .

-2.0 q., -2.0

NY/7/(\\Y//AN

Steel sheet pile

-2.0


Plane 0 = 180° 411 Plane 0 = 0°

Figure 4.37. Contours of circumferential stress in soil.

-1. 5

0

19.5

39.0

a)a)

58.5a)

78.0

97.5

117.0

Overburden

Dredge line level

Pile tip level

60 2 4 8Stress, ksf.

(a) Soil stress along section V1 at 0 = 0° (center)

0c

19. 5

39.0

58.5

78.0

97.5

117.002 4 6

Stress, ksf.(b) Soil stress along section V2 at 0 = 0°

Overburden

`Dredge line level

z z \ Pile tip level

Figure 4.38. Radial and vertical stresses in soil vs. depth of cell.

8

19.5

4)41)

39.0

58. 5

78.0

Position

Dredge line level

0.5K

1 0

Front

Position 1 -

Position 2 -

Position 3 -

Front position of the cell(section V3, 0 = 180°) andalso the front position of theisolated cell caseNear center position of thecell (section V1, 0 = 0°)Back position of the cell(section V3, 0 = 0°)

Figure 4. 40. Coefficients of lateral earth pressure inside the cell fill.

+8

cd

P,A ....

/i

..- . /.... .

..... . i." N.N. /

/A\,/Field measurement results

Finite element solution

after backfill (11)

60 120 180 240 300 360Position 0, degrees

Figure 4.41. Comparison of radial deformation of backfilled cell at level H5.

00. 4

0. 5

U

0. 6a)

0. 7

0. 8

Position 0, degrees60 120 180 240 300 360

Total settlement /5. 5

Total settlement/8,P., .

7.---, 7.,.., ./ ..., ki----A\ \. /--a----,

/ \ / // \\/ \\ \ // //\ \ ,,

Elastic settlement

-4\

/

0 just after backfill{Field measurement results (11)

A 2-1/2 months after backfill

O Finite element solutionsI I

Figure 4.42. Settlement of the top of the steel sheet piles vs. position.

10

20

4-3

a)

-30

40

50

60-1

\ \ \\ \\ \ \ =\ \ \\ \ \

\ \ \

120°

\\\\\.\ .\ \

1\\\

Finite element solutionsmeasurement results

of final fill (11)Fieldat stage

\ \\ \\

\

\\\

%

\

-Iiiii

\

,

\

= 90°

\ \I \ \1 \ \.I \ \,, .

\\\\_\0 180°

0 = 60°

\\

MILFr

\\

\\ \\\\ = 150°

0 2 0 1

Hoop tension force, kips per inch

Figure 4.43. Comparisons of hoop force in steel cell.

4

105

V. DISCUSSION OF RESULTS

5.1 Discussion of the Isolated Circular Cell

Figure 4.4 shows the radial displacement, hoop force and

stresses in steel sheets for the isolated circular cell subjected to the

gravity load of cell fill only. The radial displacement and hoop force

of the cell gradually increase as the depth of cell increases up to the

maximum bulging point and then drop off. The maximum bulging of

the cell is 0.57 inch which occurs at a point one-sixth (0. 17) of the

exposed height above the dredge line. The maximum bulging point

that was observed in the field (11) ranged from 0.18 to 0.28 of the

exposed height. The value of 0.18 corresponds to field measurements

on the front side of the cell where sheet pile length was the same as

in the present study.

From Figure 4.4 it can be seen that all vertical stresses in the

steel sheet are compressive with the maximum value of 5.3 ksi at

the bottom edge of the cell. This indicates that the steel sheet trans-

mits the downdrag due to the weight of soil adjacent to the pile, to the

soil foundation. The maximum hoop stress occurs on the outside

surface of the cell at a level one-sixth of the exposed height. There

are also some hoop stresses in the embedment zone.

The elastic settlement of soil fill inside the circular cell is

shown in Figure 4.5. It can be seen that the soil at the center of the

106

cell settles more than that away from the center. This may be

explained by recognizing that near the sheet piles there is friction

(actually no slip) between the soil fill and the wall. The stiffer steel

element deforms less. The maximum settlement of the soil element

at the center of the cell is 1.48 inches, while the top of the steel sheet

settles only 0. 66 inch.

The upward displacement along the vertical boundary line

outside the cell is due to the outward movement of the cell pushing

against the boundary. This is unreasonable for an actual installation.

In order to eliminate the upward displacement in the theoretical model

it is necessary to increase the distance of the vertical boundary from

the cell. By increasing this distance no significant change in the dis-

placement and stresses along the sheet pile node is expected. The

deflected shape and soil element stresses in the current boundary line,

however, may be changed considerably.

The horizontal boundary line at the base of the system as used

in the example is justified by observing the settlement of the nodal

points directly above which indicate the uniform settlement of the soil

element along the horizontal cross section. This indicates that the

horizontal boundary line is deep enough from the bottom edge of the

steel sheets.

Figures 4.6, 4.7 and 4.8 illustrate the contours of vertical,

radial, and circumferential stresses in the soil elements for the cell

107

filled, gravity loading case. In the body of the circular cell, all of

the stresses are compressive with the vertical stresses Cr greaterzz

than the radial stresses T and circumferential stressesrr creer

Outside the cell, just below the dredge line, the radial stress is con-

siderably higher than the vertical stress (Figures 4. 6 and 4. 7). This

distribution of normal stresses thus corresponds to development of a

passive pressure distribution where the sheet piles are being forced

into the soil.

The contours of shearing stress T in the soil are plotted inrz

Figure 4.9. It indicates that the maximum shearing stress occurs on

a plane just behind the sheet piles near mid-height above the dredge

line. Planes away from the sheet piles display a dramatic decrease

in shear stress near mid-height, while the top and bottom portions of

these planes show a slight decrease only.

Figure 4. 10 shows the principal stresses in the soil for the

circular cell under the gravity load of the cell fill. The stresses

inside the cell are all compressive stresses. The directions of the

maximum principal stress above the dredge line all incline toward the

sheet piles. This shows the tendency of the soil fill inside the cell to

bulge the sheet piles. Outside the cell the principal stresses tend to

resist such lateral movement.

108

5.2 Discussion of Circular Cell Bulkhead

As mentioned in Chapter 4 the loadings on circular bulkheads

are separated into two cases, therefore, the computed displacements

are presented separately and then combined.

The deformed shape of the circular cell due to connecting arc

tension loads (Case I) for various levels is shown in Figures 4.16 and

4.17. The radial displacement and tangential displacement data for

9 harmonic terms are plotted on the left side of the circular sections,

while the data for 11 harmonic terms are plotted on the right side of

the sections. It was found that going from 9 to 11 harmonics gave an

approximately 5% increase in the maximum displacement resultant.

The maximum movement occurs on the top level of the cell with 0.2

inch leaning toward the front side.

The radial displacement and tangential displacement due to

Case II loading are shown in Figures 4.18 and 4.19. The displace-

ment data for 6 and 8 harmonics are plotted on the opposite half of

the circular sections in the same manner as in Case I. The dis-

crepancy for displacement between the two harmonics was not greater

than 6% in any section.

The combined displacements are plotted in Figures 4.18 to

4.23. It can be seen that the connecting arc tension loading con-

tributes approximately 10 to 25% of the total horizontal displacements.

109

The deflected shape of the cell indicates that the cell was pushed

toward the water side with the front portion of the embedment sheet

piles forced into the front dredge line and the back portion moved

into the cell.

The maximum bulging point for the circular cell bulkhead occurs

on level H5 (0. 17 of the exposed height above the dredge line) on the

front side (0 = 180°) of the cell. This can be compared to the sug-

gested design value of one-fourth (0. 25) for TVA (22) and the field

measurement value (11) of 0.18 at the front sheet position. The maxi-

mum bulging on the back portion of the cell occurs at the mid-height

level (H3). This is in good agreement with the results from field

measurements (11).

Figure 4.24 shows the deformed shape of the circular cell

bulkhead on the front-back (180° -0° ) cross-section. It indicates that

the settlement of the fill in the center of the cell is consistently

greater than at the edges. The shape of settlement in the cell fill has

a pattern similar to that of field measurement results obtained by

White (26). From Figures 4.5 and 4.24, it can be seen that placing of

the backfill does not significantly change the settlement of the soil fill

inside the body of the cell.

Figures 4.25 to 4.31 show the steel sheet surface stress data

for various angles around the cell. The membrane stress is the

average of the outside surface and inside surface stresses, whereas

110

the bending stress is observed by noting one-half the difference

between these two stresses. It can be seen that the bending stress in

the vertical direction is more significant than that in the hoop direc-

tion. It also appears that both the vertical stress and the hoop stress

are maximum on the front sheet (0 = 180°) of the cell.

Figures 4.32 and 4.33 show how the hoop tension force and

shearing stress in the steel sheet vary around the cell at the levels

indicated. The hoop forces on the level above the dredge line appear

to increase from the back side (0 = 0°) of the cell to a maximum at

the arc connection (0 = 120°). They remain almost constant from this

point to the front sheet of the cell. From Figure 4.32, it can be

observed that in placing the backfill, the hoop forces in the back

portion decreases sharply since the back fill lateral pressures counter

the internal forces from the cell fill. At the front side of the cell,

the hoop forces increase to a very small degree. Moreover, it is

found that the maximum hoop force does coincide with the maximum

bulging point on level H5. Consequently, the TVA design rules (22)

in which maximum hoop tension occurs just inside the arc connection

at a point H/4 above the dredge line are appropriate.

The shearing stress TZ

in the steel sheet as shown in

Figure 4.33 appears to increase from zero at the back side of the cell

to a maximum at the arc connection. It then decreases gradually to

zero at the front sheet of the cell. The shearing stress diagrams are

111

anti-symmetric about the front to back cross-section. The maximum

shearing stress occurs on the dredge line level (H6).

The contours of vertical stress, radial stress, shearing stress

and circumferential or tangential stress in the soil on vertical cross-

sections of planes 0 = 0° and 0 = 180° are plotted in Figures

4.34 to 4.37, respectively. The stress contour data on other planes

of the bulkhead were also computed but they are not included here

because the variations from those shown here were small. The soil

stresses inside the cell appear to change very slightly due to the

effect of back fill (see Figures 4.6 to 4.9 and Figures 4.34 to 4. 37).

Outside the body of the cell, the application of backfill pressure

greatly increases the vertical and radial stresses around the cell.

The noticeable increase in radial stress in front of the cell results

from the passive resisting force due to the outward movement of steel

sheet piles in that area. The vertical stress is quite uniformly dis-

tributed across the base of the cell and approximately equal to the

overburden pressure.

Backfill lateral pressure slightly affects the shear stress in the

cell fill as shown in Figure 4.36. The shear stresses in the middle

are much lower than near the sheet piles. It therefore seems improb-

able that shear failure would start in the midplane and go towards the

steel sheet walls as suggested by Terzaghi (23). Instead it appears

that the failure should start near the dredge line of the sheet pile and

112

progress from there up to the top free surface and down through the

base of the sheets along a path of maximum shear stress. This type

of internal failure mode was also observed in photoelastic analysis (1).

Figures 4.38 and 4.39 show the vertical and radial stresses in

the soil elements inside the cell vs. the depth of the cell. Results for

radial stresses show that they increase more on the back side than the

front due to the lateral pressure of backfill trying to push the back

sheets inward. It also can be seen that levels above the dredge line

experience a small increase in stresses, while the lower portions of

the cell show a considerable increase. Specifically, there is an

abrupt increase in both vertical and radial stresses under the bottom

edge of the sheet piles. This is due to the weight of soil fill transmit-

ted by the sheet piles to the soil foundation. This observation may

help to explain larger than anticipated settlements which have been

experienced by structures of this type in Portland (11) and Long

Beach (26).

The coefficient of lateral earth pressure K shown in Figure

4.41 indicates that at the center plane of cell K decreases slightly

with depth. At the front plane, K is almost constant all the way

down the cell. At the back plane, it increases moderately due to

backfill lateral pressure. The mean values of K in the front posi-

tion (unloaded side), center position and back position (loaded side)

are 0.441, 0.412 and 0. 64, respectively. They indicate that the

113

coefficient of lateral earth pressure of the fill in the cell is higher

than the Rankine earth pressure coefficient. The maximum average

value of K in the field was 0.45 (11).

The results from the study and the field measurements obtained

by Khuayjarernpanishk (11) are compared with the one obtained by

using Cummings', TVA's and Terzaghi's recommended formulae as

shown in Table 5. 1. For the circular cell bulkhead case the maxi-

mum hoop tension force in the steel cell obtained in this study is about

0.75 of the values obtained from Cummings' and Terzaghi's formulae

and 0. 67 of the value obtained from TVA's formulae, for the isolated

cell case. At the same time, it is only 0. 61 of the field measurement

result for the bulkhead case. The computed vertical pressures at the

base of the cell are almost uniform and approximately equal to the

overburden pressure. This indicates that the bending effect due to

backfill results in a very small vertical pressure when compared to

the effect of the gravity load of the cell fill. Consequently, they are

not in the same trend as Terzaghi's concept which states that the

bending stress is linearly distributed across the horizontal section of

the cell by considering the cell as a rigid body. The calculated coef-

ficients of lateral earth pressure in the cell fill are in good agreement

with the values recommended by other investigators and those obtained

from field measurements.

Table 5. 1. Comparisons of results.

MaximurnHoop

Tensiono- ZZ at front base o- ZZ at back base

KSources kips /inch crZZ overburden crZZ overburden Front Center Back

Cummings (6) 2.53* -- -- - - -

TVA (22) 2.85** -- -- - 0.523***

Terzaghi (23) 2.53* 1.25 0.743 0.4 0.4-0.5 -

Kittisatra 1 1.90 1.04 1.05 0.441 0.412 0.64

Lacroix (13) -- -- -- 0.4 0.5 -0. 6 0.7-1.0

Khuayjarernpanishk (11) 3. 10+ -- 0.45+

"At dredge line level for isolated cell.At H/4 above dredge line just inside arc connection for isolated cell.

*** 2 2The Krynine factor, K = (cos 4)/(2-cos O.+Field measurement results for bulkhead1 Finite element results for bulkhead.

115

As mentioned in Chapter IV, the circular cell bulkhead used in

this study is not exactly the same configuration as the one instru-

meni-ed in the field (11). Nevertheless, the radial deformation at

level. H5, the settlement of the top of the cell and the hoop tension

force in the cell have been plotted in Figures 4. 41, 4. 42 and 4.43,

respectively, to compare and show some correlations.

Figure 4.41 shows that the computed deformations are in good

agreement with the general trend of observed values in the field

except for the positions near the left arc (0 = 1200). The exagger-

ated distortions at such points in the field during construction were

presumed to result from wave action, compaction of fill and tightening

of interlock slack (11). They are not associated with elastic

deformation.

Figure 4. 42 illustrates how the edge settlement varied around

the cell for the top level. It appears that the settlement of the back

portion of the cell is greater than the front. This is due to the back

sheet piles being subjected to a higher gravity effect of the backfill

than the front. The observed settlements in the field immediately

after backfilling and 2-1/2 months after backfilling are approximately

5.5 and 8 times the computed elastic settlement. This seems to

indicate that the total settlement in the field resulted from consolida-

tion of the fill in response to stresses due to fill weight in addition to

the immediate elastic distortion due to the same cause. The observed

116

settlements of the cell are in similar pattern to the computed results.

From Figure 4. 43, the computed hoop tension forces in the cell

have a similar pattern to the field measurement values but do

illustrate again significant differences in magnitude. All hoop forces

obtained from the field measurements are greater than in this study.

The main factors associated with the assumptions used in this

study that influence the results are that:

1. The elastic properties of a given soil in a cell depend to a

large extent on the method of filling the cell. There is no

laboratory procedure that would yield reliable advance

information on the state of density of the soil in the cell and

on the corresponding elastic properties of the soil. The

estimated moduli of elasticity E used in this study seemed

to be too high. They are approximately 7 times larger than

that used by Brown (2). If lower values of E had been

used, the nodal point displacements would have been signifi-

cantly greater. Since the steel sheet pile stress is a function

of displacements at the nodes, a corresponding increase in

sheet stress would also be expected.

2. The assumed interface condition (full friction) between the

soil and the steel sheets probably resulted in higher than

realistic axial stress in the steel sheets. The rougher the

steel sheet surface, the better the agreement will be between

117

analytic and field results. It is quite likely that a boundary

element could be developed to represent the actual state of

friction. Such an element is not now available. Additional

research and formulation would have to be undertaken to

develop such an element. It is believed that modification of

the analysis and the computer program can be made to

account for the interface friction force, once the appropriate

element is introduced and the experimental boundary inter-

action data are available.

3. The assumed cylindrical shell-like structure is much stiffer

than the actual interlocked sheet piling cellular bulkhead.

This assumption is a very essential factor, reducing the

horizontal displacements in the cell because no interlock

slack can exist in the shell. But on the other hand, interlock

friction in the actual cell probably reduces slippage to

negligible values for usual cell service conditions. The

interlocks would be expected to slip if the cell were caused

to distort a large amount (fail by bending).

118

VI. SUMMARY AND CONCLUSIONS

The finite element model and the corresponding computer

program developed in this investigation can be used to compute dis-

placements and stresses in a circular cell structure subjected to

gravity loads of cell fill, backfill and surcharge loads. The structure

can be founded on soil foundation or rock foundation. The loading

functions are expanded in Fourier harmonic series. For different

sets of Fourier force coefficients, the analyses must be carried out

separately. The components of displacement and stress for each

harmonic and each set of loadings are superimposed to form the total

solution.

The following conclusions are drawn from this study.

1. An almost uniform vertical pressure distribution on the base

of the circular cell bulkhead was obtained. The magnitude is

approximately equal to the overburden pressure.

2. Horizontal stresses in the soil within the cell are mobilized

as normal active and passive earth pressure.

3. The maximum elastic soil settlement occurred in the center

of the cell, with smaller settlements at the edges. It also

was found that the total elastic settlement at the back side of

the cell is greater than at the front.

4. The vertical shear stress in the cell fill for the bulkhead is

119

maximum near the front and back sheet piles and not on the

center plane. Therefore, the mid-plane shear failure pro-

posed by Terzaghi (23) seems unlikely.

5. Through shear transfer to and from the soil, the sheet piles

appear to transmit an appreciable amount of the load to the

foundation.

6. Placement of the backfill increases the hoop tension forces

in the front portion of the cell. A decrease in hoop tension

in the sides and back portion was found.

7. Both maximum hoop tension force and maximum bulging of

the cell occur on a level 1 /6 of the exposed height above the

dredge line. The maximum hoop tension occurs behind the

connections of the cell and arcs, whereas maximum bulging

occurs at the front sheet of the cell. This indicates that the

TVA (22) suggestion to compute the maximum hoop force in

the cell at the arc connecting sections is justified.

8. Both vertical and circumferential bending stresses do occur

in the steel sheet piles. These bending stresses are lower

than the membrane stresses. However, such stresses should

be taken into account in designing.

9. The coefficients of lateral earth pressure in the cell fill are

0.441 at the front, 0.412 at the center and 0.640 at the back

of the cell. These values agree closely with the values

120

recommended by other investigators.

10. The total eleven Fourier harmonics used in the analysis is

considered adequate for the loading conditions in this study.

121

BIBLIOGRAPHY

1. Burki, N.K. and Rowland Richards, Jr. Photoelastic analysisof a cofferdam. Proceedings of the American Society of CivilEngineers, Journal of the Geotechnical Division, pp. 129-144.February 1975.

2. Brown, P.P. Discussion of: Field study of cellular bulkhead,by A. White, J. A. Cheney and C.M. Duke. Transactions of theAmerican Society of Civil Engineers 128:503-506. 1963.

3. Burmister, D. M. Stress and displacement characteristics of atwo layer rigid base soil system.: Influence diagrams and practi-cal applications. Proceedings of the Highway Research Board35:773-814. 1956.

4. Carnahan, B. , H. A. Luther and J.O. Wilkes. Applied numericalmethods. New York, Wiley, 1969. 604 p.

5. Cook, R.D. Concepts and applications of finite element analysis.New York, Wiley, 1974. 402 p.

6. Cummings, E.M. Cellular cofferdams and docks. Proceedingsof the American Society of Civil Engineers, Journal of theWaterways and Harbors Division 83:13-45. September, 1957.(Also, Trans. ASCE, Vol. 125, 1960)

7. Dym, C. L. and I.H. Shames. Solid mechanics: VariationalApproach. New York, McGraw-Hill, 1973. 556 p.

8. Flugge, W. Stresses in shells. 2nd ed. New York, Springer-Verlag, 1973. 525 p.

9. Grafton, P.E. and D.R. Strome. Analysis of axisymmetricshells by the direct stiffness method. American Institute ofAeronautics and Astronautics Journal 1:2342-2347. October, 1963.

10. Herrmann, L.R. , R. L. Taylor and D.R. Green. Finite elementanalysis for solid rocket motor cases. Berkeley, 1967. (Uni-versity of California, Structural Engineering Laboratory. Reportno. 67-4)

122

11. Khuayjarernpanishk, T. Behavior of a circular cell bulkheadduring construction. Doctoral dissertation. Corvallis, OregonState University, 1975. 156 numb. leaves.

12. Kleins, S. A study of the matrix displacement method as appliedto shells of revolution. In: Proceedings of the First Conferenceon Matrix Methods in Structural Mechanics, Report no. AFFDL-TR-66-80, Wright-Patterson Air Force Base, 1965. p. 275-298.

13. Lacroix, Y,, M.I. Esrig and U. Luscher. Design, constructionand performance of cellular cofferdams. Proceedings of theSpecialty Conference, Lateral Stresses in the Ground and Designof Earth-Retaining Structures, The American Society of CivilEngineers, Soil Mechanics and Foundation Division. June, 1970.p. 271-328.

14. Laursen, H.I. Structural analysis. New York, McGraw-Hill,1969. 486 p.

15. Love, A.E.H. The mathematical theory of elasticity. 4th ed.Cambridge University Press, 1927. 643 p.

16. Milne, W. E. Numerical calculus Princeton, PrincetonUniversity Press, 1949. 393 p.

17. Novozhilov, V.V. Theory of thin shells (Translation).P. Noordhoff, Groningen, The Netherlands, 1959. 376 p.

18. Percy, J. H. , T. H. H. Pian, S. Kleins and D . R. Narvaratana.Application of the matrix displacement method to linear elasticanalysis of shells of revolution. American Institute of Aero-nautics and Astronautics Journal 3:2138-2145. November, 1965.

19. Popov, E.P., J. Penzien and Z. A. Lu. Finite element solutionfor axisymmetric shells. Proceedings of the American Society ofCivil Engineers, Journal of the Engineering Mechanics Division90:119-145. October, 1964.

20. Richart, F. E. , Jr. , J. R. Hall, Jr. and R . D. Woods. Vibrationsof soils and foundations. Englewood Cliffs, Prentice-Hall, 1970.414 p.

123

21. Speare, J. C. Discrete element static analysis of core-stiffenedshells of revolution for asymmetric mechanical loadings.Cambridge, 1968. 229 numb. leaves. (Massachusetts Instituteof Technology, Dept. of Aeronautics and Astronautics, Aero-elastic and Structures Research Laboratory. TR 146-5)

22. Tennessee Valley Authority. Steel sheet piling cellular coffer-dams on rock. TVA Technical Monograph no. 75, Vol. 1.

Knoxville, Tennessee, 1957.

23. Terzaghi, K. Stability and stiffness of cellular cofferdams.Transactions of the American Society of Civil Engineers,110:1083-1202. 1945.

24. Tezcan, S.S. Discussion of: Simplified formulation of stiffnessmatrices, by P.M. Wright. Journal of the Structural Division,American Society of Civil Engineers 89:445-449. December,1963.

25. Timoshenko, S. and K.S. Woinowsky. Theory of plates andshells. 2nd ed. New York, McGraw-Hill, 1959. 580 p.

26. White, A., J. A. Cheney and C. M. Duke. Field study of acellular bulkhead. Transactions of the American Society of CivilEngineers 128:463-508. 1963.

27. Wilson, E. L. Structural analysis of axisymmetric solids.American Institute of Aeronautics and Astronautics Journal3:2269-2274. December, 1965.

28. Zienkiewicz, 0. C. The finite element method in engineeringscience. London, McGraw-Hill, 1971. 521 p.

APPENDICES

APPENDIX A

Element Matrices

A. 1 Triangular Axisymmetric Ring Element

The [%] matrix of Equation (2. 16) is (also see Figure 2. 1)

1 rl Z1 0 0 0 0 0 0

0 0 0 1 rl Z1 0 0 0

0 0 0 0 0 0 1 r1 Z1

124

(A. 1)

125

The matrix

1.11L-0

[c3,1]

a1

b1

c1

0

0

0

0

0

0

of Equation (2. 17) is

0 0 a2

0 0

0 0 b2

0 0

0 0 c2

0 0

a1

0 0 a2 0

b1

0 0 b2

0

C10 0 c2 0

0 al 0 0 a2

0 b1

0 0 b2

0 0 01 c2

a3

b3

c3

0

0

0

0

0

0

0

0

0

a3

b3

c3

0

0

0

0

0

0

0

0

0

a3

b3

c3

(A. 2)2A

where A is the cross sectional area of the triangular element and

a, b, c are nodal coordinate functions as shown in Equation (2. 19b).

The matrix W(r, Z, 0)] of Equation (2.21) is shown below

where n is the Fourier harmonic number.

126

[I' (r, Z, 0)]

0

0

cos nO

cos nO

0

cos nO

0

n sin nO

0

0

cos nO

0

0

0

nZ sin ne

0 0

0 0

Z cos nO

0

0

0

cos nO

0

-n sin ne

0

nZ cos nO

(A. 3)

r

0

n sin nO

0r

cos nO 0

nZ sin nOr0

The matrix

0rn sin ne

0 - r

0 0

0 0

n cos nO n cos nOr0 0

sin ne

r0

Z sin nO0 _

r r0 0 sin nO

(2. 28a) can be found by rewriting

r

[Sn ] in Equation

the material constant C in Equation (2. 7) as

C11 C12 C13 0 0 0

C12 C22 C23 0 0 0

[C] =C13 C23 C33 0 0 0

(A. 4)0 0 0 C44 0 0

0 0 0 0 C55 0

0 0 0 0 0 C66

where

11C11

C1212

44C44

22

13

55

C33

C 23

C66

E(1-v)(l+v)(1-2v)

Ev(l+v)(1-2v)

E2(1+v)

127

Substituting Equations (A. 3) and (A.4) into Equation (2. 25c) and inte-

grating Equation (2. 28a), [Sn

] can be expressed as

[SUUn] [SUWn]

[SUVn]

[sn] = [SWU n] [SWW n] [SWVn] (A. 5)

[SVUn] [SVWn] [SVVn]

where

[SUU n]

(C33+n C55)I3 (C13+C33+n2C55)I1 (C

33+n2C

55)15

(C11+2C13+C33+n2 C55)I1 (C13+C33+n 2

C55)14

Symmetric

[SUWn [SWUn

=

7

(C33 )I +C I33 55 6 44 1

(A. 6)

(A.7)

[SUVn

] = [SVUn

IT

_n(C

33+C

55)13

= n(C 13+C 33 +C 55)12

n(C33+C55)1.5

nC33I2 n(C33

+C 55)15

n(C13

+C33)11

n(C13 +C 33+C55)I4

nC3314

2 2Cn C66I3 n 66I2

(C +n2C

[SWWn] = 44 66)I

1

[SVITn] =

Symmetric

_(n

2C33+C55 )13

Symmetric

[SWV n] = [SVWn]T =

0

0

n2C 33C3312

n2C33I1

0

0

n(C 33+C

55)16

n2C6615

n2C6614

C -,1 +n2C6616

2-. 1

(n 2C

33+C55 )15

n2C33I4

(n2C33 +C55)16 +C6611

-nC6612

- nC66I1

128

(A. 8)

(A. 9)

(A. 10)

(A. 11)

129

A. 2 Shell Element of Revolution

Matrix

o

1[co

0] of Equation (2.43c) is

1 0 0 0 0 0 0

0 - 1 /I 0 0 0 1// 0 0

0 0 1 0 0 0 0 0

0 0 -1// 0 0 0 1/1 0

(A. 12)1 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0

-3 /Q2 0 0 -2// 23// 0 0 -1/1

2// 30 0 1/12 -2 // 3

0 0 1/.e2

Matrix [X0] in Equation (2.44a) is

cos i) sin (I) 0 0 0 0 0 0

-sin 4 cos 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 1 0 0 0 0(A. 13)

0 0 0 0 cos 41) sin (I) 0 0

0 0 0 0 -sin g) cos g) 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 1

The matrix [(6,00

1] of Equation (2. 44b) may be shown by

denoting

and

a = Z2 Z1 (see Figure 2.3)

b = r2

r1

cos (I) = a //

sin (I) = -b

where i is the length of the shell element.

130

(A. 14)

['PO

Substituting Equation (A. 14) into Equation (2.44c),

becomes

[(Poo- l]

b /1 a/1 0 0 0 0 0 0

-b // 2 -a // 20 0 b/i 2 a /i 2

0 0

0 0 1 0 0 0 0 0

0 0 - 1 /.12 0 0 0 1/i 0

a te -b /.2 0 0 0 0 0 0

0 0 0 1 0 0 0 0

-3a /23 313/13 0 -2/i 3a/f 3 -3b/13 0 -1/i

2a // 4- 2b/.24 0 1/12 - 2a/.24 2b if

40 1 /I

2

(A. 15)

Matrices [Nn(0)] and [Mn

(0)] in Equations (2. 47c) and

(2.47d ) are:

[N n(0)] =

and

E11

cn0 0 E12sOcn0 nE 12cn0 E

12c (1)cnO

E 44 sn0 -nE44 snO -E 44si)sn0 0

E 22 s2cknO+n

zE 44 snO nscHE22cnO+E44 snO) E 22 sOccOcnO

Symmetric nE cnO22 +E44sziOsne nE

22ccOcne

E22c2.0cnO

D11

cnO 0 D12 sc)cn0 -nD12

c)c nO -112D12cri°

D 44 c 2One nD 44cOn0 -D 44 c2cOscpsne -nD 44s (1)c On°

2 -n2 sci)(D22cnOD 22s2cl)cnO+n D44snO -nsizOcc1)(DzzcnO+D44snO)

(A. 16)

+D44sn0)

c2on2D

222crie+D44sOnO) nc(1)(nzDzzcnO+s ct)D44snO)Symmetric

n4D22

cn0 +n D 44c cl)snO

(A. 17)

132

where

cl) -= cos 4, c24) = cos 24)

s4) = sin (I), s24) = sin24)

cnO = cos2

nO, snO = sin2nO

Et EvtEll = E22E

121-v 2 1-v2

EtE 44 2(1+v)

Et3 Et 3

D11

= D2212(1-v2) 12(1 -v2))

Et3v Et3

D12 D12(1-v2)

44 12(1+v)

where E is Young's modulus, v is Poisson's ratio.

133

APPENDIX B

Fourier Harmonic Coefficients

There are two sets of Fourier force coefficients in this study;

the first set contains the coefficients due to the connecting arc tension

load as shown in Figure 4.13, the second set contains the coefficients

due to uniformly distributed lateral load and gravity load of backfill

as shown in Figure 4.14.

then

Let

p(0) = ancos nO

n=0

1a0

= p(0)d0-Tr

1Tr

an Tr= p(0) cos ned0, n = 1,2, ,N

-Tr

(B. 1)

(B. 2)

(B.3)

The resulting Fourier coefficients for each case of loadings are

shown in Table B. 1.

134

Table B. 1. Fourier force coefficients.

Coefficient

Case I Case II13Tr 11Tr

2.Tr 2Tr< 0 < -

18 18p(e) = p cos 90

11Tr 13Trp(0)=p, --3 <O<

2Tr 2Tr

< 0 <18 18

p(0) = 0, elsewhere<p(0) = 0, T < 13

3

a00.07073 0.66666

al -0.07053 0.55133

a2

-0.06992 -0.27566

a3

0.13783 0

a4

-0.06752 0.13783

a5

-0.06577 -0.11026

a6 0.12732 0

a7

-0.06124 0.07876

a8-0.05853 -0.06892

a9 0.11111 0

a10

-0.05236 0.05513

all -0 04899

135

APPENDIX C

Gaussian Quadrature Numerical Integration Procedure

Numerical integration by the Gaussian quadrature method is

chosen because high accuracy is obtained with relatively few nodal

stations and singularities along the boundary can easily be handled (21).

Reference (16) gives the following Gaussian quadrature formula:

where

and

1

f(X)dX = g(x)945dxa -1

i=1

66 (C. la)

g(x) = f(h(x)) (C. lb)

X = h(x)

w. is the Gaussian weighting functions given in Table C. 1.

x,1 , y. are the Gaussian integration stations given in Table C. 1.

N is the number of Gaussian stations.

is the differential transformation containing terms to change

the limits of integration to the interval [-1, 1] and terms

from the differential dX.

The extension to integrals involving more than one variable is

straightforward. For example, in the case of two variables, we

obtain

where

S

1

X, Y)dS = g(x, y),95dxdy-1 -1

N Nw.w.g(x.,

g(x, y) = f(hx(x, y), hy(x, y))

X = hX '

(x y)

Y = h (x, y)

136

(C. 2a)

(C.2b)

(C.2c)

(C.2(1)

Values of Gaussian weighting functions and stations are given

in Table C. 1.

137

Table C. 1. Gaussian weighting functions and

stations .

N i Station, xi.VV w.eight, .

I

2 1 -0.57735027 1.00000000

2 0.57735027 1.00000000

3 1 -0.77459667 0.55555555

2 0.00000000 0.88888889

3 0.77459667 0.55555555

4 1 -0.86113631 0.347854852 -0.33998104 0.652145153 0.33998104 0.652145154 0.86113631 0.34785485

5 1 -0.90617985 0.236926892 -0.53846931 0.47862867

3 0.00000000 0.568888894 0.53846931 0.478628675 0.90617985 0.23692689

6 1 -0.93246951 0.171324492 -0.66120939 0.36076157

3 -0.23861919 0.46791393

4 0.23861919 0.46791393

5 0.66120939 0.36076157

6 0.93246951 0.17132449

10 1 -0.97390653 0.066671342 -0.86506337 0.149451353 -0.67940957 0 219086364 -0.43339539 0.269266725 -0.14887434 0.29552422

6 0.14887434 0.295524227 0.43339539 0.269266728 0.67940957 0.21908636

9 0.86506337 0.14945135

10 0.97390653 0.06667134

138

APPENDIX D

Modulus of Elasticity of Soil

The modulus of elasticity of soil used in this study was obtained

by using Richert's formula (20). For an elastic soil as assumed in

Chapter I, the modulus of elasticity E can be expressed in terms of

shear modulus G:

E = 2(1+v)G (D. 1)

where v is Poisson's ratio of soil.

The shear modulus of angular grained materials can be

estimated from the empirical equation

1230(2.97-e) 2 ( 0.51+e

G cr )o

in which both G and T0

are expressed in psi, where

=1

(0- +0- +0- )o 3 xx yy zz

To is the average confining pressure of the soil element

(D.2)

(D. 3)

0- , 0- , T are normal stresses in the X, Y, and Z direction,xx yy zz

respectively.

e is void ratio which can be obtained from

e -GsYw

Yd1 (D.4)

where

where

139

Gs is specific gravity of soil

Yd is dry unit weight of soil.

The dry unit weight can be computed from the following equation:

Gs(Y--Vw)

Nd G -1

-y is total unit weight of soil

y is unit weight of water.

(D. 5)

Once the values of Gs, y, yw of 2.71, 117. 5 pcf. and 62.5 pcf.

respectively, are substituted into Equations (D.5) and (D.4), the

void ratio will be 0.94.

The vertical stress, 0 , in Equation (D. 3) was assumed tozz

be equal to the overburden pressure. Since the water level in this

study was at the top of the cell, o- can be written aszz

0- (y-yw)h

where h is depth of soil above point.

(D.6)

The horizontal stresses 6 and o- may be written asxx yy

V0- = 0-

XX yy -1) zz (D.7)

140

The moduli of elasticity of soil for the circular cells (Figures

4.1 and 4. 2) and the finite element model (Figure 4.3) are shown in

Table D. 1.

Table D. 1. Modulus of elasticity of soil.

Element Layer No. Depth, ft o-o, lb ift 2 E, kip/in 2

1 4.875 185 7.982 14.625 556 13.863 24.375 927 17.904 34.125 1298 21.185 43.875 1670 24.026 53.625 2041 26.557 63.375 2412 28.878 73.125 2783 31.019 82.875 3155 33.02

10 95.063 3619 35.3611 109.687 4176 38.00

141

APPENDIX E

User's Manual for Circular Cell Bulkhead Program

The program consists of a main program and 18 subroutines.

This appendix is included in order to explain the use of the computer

program developed for the analysis of circular cell bulkheads as well

as the capability of the computer program and restrictions concerning

the preparations of the finite element model and input data.

The computer program for the non-axisymmetrically loaded

axisymmetric solid and shell was written in FORTRAN IV for the CDC

3300 computer at Oregon State University.

E. 1 Program Capability

The computer program is used to analyze a circular cell

bulkhead which was filled with cohesionless soil. The cell can be

founded on rock or a sand foundation. The program described herein

does compute principal stresses on a vertical plane with 0 = 0,

stresses in soil elements on any desired vertical plane, shell forces

and deformations for a circular cell due to the gravitational effects of

the fills inside and/or outside the cell. The boundary axisymmetric

or non-axisymmetric surcharge loads can be added simultaneously.

The program can also analyze a shell of revolution or an

axisymmetric structure subjected to external forces. Present

142

limitations on the finite element model's size are as follows: 66

nodal points, 57 elements, 12 different material properties, 50

Fourier terms for the boundary force condition, a maximum band-

width of 28 and maximum degrees of freedom of 182. If the user

wishes to increase these limits for other computers the following

named common areas must accordingly be changed in the program:

COMMON A2, COMMON A3, COMMON A6, COMMON C7, COMMON

D2, COMMON D3 and COMMON F2. In addition, the indices on both

DO loops in subroutine INITL, the DIMENSION and EQUIVALENCE

statements in the main program and subroutines INPUT, SOISTR,

SHLSTR and TSTRES must also be changed.

E. 2 Preparation of the Finite Element Mesh

By making use of the axis symmetry of the circular cell system

and taking advantage of orthogonal properties of the harmonic func-

tions as mentioned in Chapter II, only one plane of revolution of the

structure needs to be discretized. Figure E. 1 is used as a model

structure to show how to prepare the finite element mesh. Two types

of elements are used to construct the model: quadrilateral elements

for the soil, and shell elements for the steel sheet pile. The quadri-

lateral element is sub-divided by the program into four triangles.

Therefore, a quadrilateral element will give the same results as four

triangular elements, but in the mean time a quadrilateral element has

A ZfAxis

63 64 65

59

55

51

47

43

36

29

22

15

8

1

of symmetry

60

56

52

48

44

37

30

23

16

9

61

57

53

49

45

38

31

66

143

@762

58

54

50

Steel sheet

46

39

Soil element number

37 Shell element number Roller40 41 42

Oa 032 33 34 35

24

17

10

O25g

26 27 28

18 19 20 21

11 12 13 14

O O O O O6 7

///////tifr//7///,77/////7//////,/,-//e/7/////7///7/7///7/.77///77/777///)>Fixed

2 3 4 5

Figure E. 1. Example of finite element mesh showing node andelement number scheme.

144

fewer sets of equilibrium equations in the final system than four

triangular elements.

Numbering of the nodal points should begin after the finite

element mesh has been established. Numbering of these nodes can

begin at any corner of the system and proceed either horizontally or

vertically. The primary objective of the numbering scheme is to

minimize the numerical difference between nodal point numbers

associated with the elements. The nodal numbering in this study

(Figure E. 1) was done horizontally.

After the numbering of the nodes has been accomplished, all

elements must be numbered sequentially.

E. 3 Preparation of Input Data

The following information describes the data cards which form

the necessary input data for the program.

E. 3. 1 Number of Problems Card (I5)

Columns

1-5 Number of sets of input (NPROBS)

145

E. 3.2 Control Card (815,3F10.4,215)

Columns

1-5 Total number of nodal points (NUMNPT, 66 max.)

6-10 Total number of elements (NUMELT, 57 max.)

11-15 Total number of different materials (NUMMAT, 12 max. )

16-20 Total number of pressure boundary conditions (NUMPC)

21-25 Total number of Fourier tersm (NUMFOU, 50 max.)

26-30 Total number of output print angles (NANGLE, 7 max.)

31-35 Total number of constrained boundary points (NCONP)

36-40 Problem type (NCUT)

Enter a 0 if no soil elements are present

Enter a 12 if both soil and shell elements are present

Enter a 13 if no shell elements are present

41-50 Radial acceleration (ACELR)

Enter a 0 or leave it blank

51-60 Axial acceleration (ACELZ)

Enter a -1.0 if own weight of soil elements is accounted

61-70 Angular velocity (ANGFQ)

Enter a 0 or leave it blank

71-73 Mesh check tag (ISTOP)

Enter a 0 on control card. The values of all essential

parameters are checked against their allowable values. If

146

ISTOP > 0, a mesh check is obtained then the program is

stopped and no further action is taken.

74-76 Total number of shell elements (NSHELT)

77 -80 Total number of nodal points with axisymmetric surcharge

loads (NAXISF)

E. 3. 3 Material Properties Cards (15, 3F15. 4)

Columns

1-5 Material identification number (MTYPE)

6-20 Material density (DENS, in kips per cu. ft. )

21 -35 Modulus of elasticity (EMOD, in ksi. )

36-50 Poisson's ratio (PR)

Number of material property cards must equal to value of

NUMMAT in Columns 11-15 of CONTROL CARD.

E. 3 . 4 Nodal Po int Card s (13,12, 6F 10.4 213, F9. 3)

Columns

1-3 Nodal point number (N)

4-5 Common node (NPCOM(N))

This entry indicates the number of elements being

connected at this node.

6-15 Radial coordinate (COOR(N, 1) > 0, in feet)

16 -25 Axial coordinate (COOR(N, 2) in feet)

147

26-35 Radial nodal force (FORCE(N, 1) in kips)

36-45 Axial nodal force (FORCE(N, 2) in kips)

46-55 Tangential nodal force (FORCE(N, 3) in kips)

56-65 Nodal moment (FORCE(N, 4) in kips-ft)

Columns are left blank if it is a soil nodal point.

66-68 Nodal point condition code (MCODE(N))

69-71 Type of node (NPTYPE(N))

This entry indicates the degree of freedom the program

will assign to the node. A3 is assigned all nodes except

the nodes on the shell elements which are assigned a 4.

72-80 Shell thickness (T in inch)

Columns are left blank if it is a soil nodal point.

In general every nodal point must be defined but since the

program has an automatic mesh generation feature, a minimum of two

nodal points per row need be input and the intervening points will be

assigned coordinates based on a linear interpolation procedure. For

example, if nodal point 1 is the first point in a row with coordinates

(0, 0) and nodal point 4 is the next point defined with coordinates

(33, 0), then the nodal point 2 will be located (11, 0) and etc.

The nodal point condition code (MCODE) will be set 0 unless

points 1 and 4 have the same MCODE, in which case all intervening

points will be assigned the same MCODE as the two end points.

148

The common node (NPCOM) and type of nodal point (NPTYPE)

of the omitted intervening points will be assigned the same NPCOM

and NPTYPE as the preceding point on that row.

The radial, axial, tangential forces and moments of all omitted

intervening points will be assigned 0 in all cases.

The shell thickness at each nodal point should be non-negative

and it is assumed constant over the shell length in each element.

All loads are total forces acting on a one-radian segment.

The nodal point condition code (MCODE in columns 66-68) is

interpreted in the following manner:

MCODE

Constrained Condition X

Radial (r) Axial (Z) Tangential (0)

0

1

2 X -3 X X X4 - X5 X X -6 X x7 - X X

The nodal point cards must be in order, starting with nodal

point number 1. The last nodal point card must be supplied.

E. 3.5 Connectivity Cards or Element Cards (615)

This data connects the element number with the nodal point

numbers on the element parameters. The nodal points for a

149

quadrilateral and triangular elements are listed counter-clockwise

sequentially around the element. The nodal points for a shell element

are arbitrary but all shell elements' nodal points must be in the

same pattern.

Columns

1-5 Element number (M)

6-10 Nodal point I (ICONN(M, 1))

11-15 Nodal point J (ICONN(M, 2))

16-20 Nodal point K (ICONN(M, 3)). Columns are left blank if it is

a shell element

21-25 Nodal point L (ICONN(M, 4)). Columns are left blank if it is

a shell element or a triangular element

26-30 Material identification (ICONN(M, 5))

In general, every element must be defined but with the

automatic mesh generation feature a minimum of 1 element per row

need be input. The program generates the omitted elements by

incrementing by 1 the preceding I, J, K and L. The material identifi-

cation code for the generated elements is the value specified on the

first generated card. For example if element 1 is read with values

I = 1, J = 2, K = 13, L = 12, material type = 11 and the next element

card read is element 11 with values I = 12, J = 13, K = 24, L = 23,

material type = 10 then element 2 would be assigned value 2, 3, 14,

150

13, 11, element 3 values 3, 4, 15, 14, 11 and etc. The last element

card must be supplied.

The nodal point array on the element cards must of course

correspond to nodal points on the nodal point cards and the material

identification must correspond to the materials in the material cards.

The element cards must be in order, starting with element

number 1.

E. 3. 6 Constrained Boundary Cards (515)

This data is applied primarily to nodal points on the boundaries

of the finite element model in the non-axisymmetric loading case.

A 1 is used to indicate the constraint of a nodal point.

Columns

1-5 Nodal point number of constrained node (KNOVA(N, 1))

6-10 Radial direction constraint (KNOVA(N, 2), 1 or blank)

11-15 Axial or vertical direction constraint (KNOVA(N, 3), 1 or

blank)

16-20 Tangential direction constraint (KNOVA(N, 4), 1 or blank)

21-25 Rotational constraint (KNOVA(N, 5), 1 or blank)

Enter 1 if the nodal point is not on the shell element

The number of cards must equal to NCONP, columns 31-35 of

control card. The program has an automatic constrained nodal point

151

generation i.e. , it will assign a 1 to tangential displacement

(KNOVA(N, 4)) of all nodal points of the system and radial displace-

ment (KNOVA(N, 2)) of all nodal points which are on Z-axis when

Fourier harmonic number equal to zero. The program will also

assign a 1 to rotation displacement (KNOVA(N, 5)) of all soil nodal

points.

E. 3.7 Boundary Pressure and /or Shear Cards (315,2F10.4)

If NUMPC in columns 16-20 of control card is zero or blank,

then omits these cards.

Columns

1-5 Element number (LBC)

6-10 Boundary nodal point I (IBC)

11-15 Boundary nodal point J (JBC)

16-25 Normal traction (PN in kips per sq. ft. )

26-35 Shear traction (PT in kips per sq. ft. )

Nodes I and J must be chosen such that the sequence of nodal

points on the traction cards are in the same manner as on the element

cards. The tractions are assumed constant over the length of the

element boundary. The positive senses of normal pressure and shear

are shown in Figure E. 2.

152

Figure E. 2. Boundary pressure sign convention.

The number of cards must correspond to the value of NUMPC

input in columns 16-20 of CONTROL CARD.

E. 3.8 Axisymmetric Nodal Surcharge Loading Cards (15, 4F15. 5)

If NAXISF in columns 77-80 of CONTROL CARD is zero, then

omits these cards.

Columns

1-5 Nodal point number (NPAX(N))

6-20 Radial axisymmetric nodal force (FORC(N, 1) in kips)

21-35 Axial axisymmetric nodal force (FORC(N, 2) in kips)

36-50 Tangential axisymmetric nodal force (FORC(N, 3) in kips)

51-65 Axisymmetric nodal moment (FORC(N, 4) in kips-ft)

153

If the surcharge loads are distributed over the boundary surface

of the elements, then the surface integral must be carried out

explicitly for the desired loading. The value of the nodal point forces

are total forces acting on a one-radian segment.

The number of these cards must correspond to the value of

NAXISF.

E. 3.9 Fourier Force Coefficient Cards (8F10. 5)

The Fourier force coefficients of the surface tractions and nodal

point forces of the non-zero displacement boundary conditions are as

follow:

Card 1 Columns 1-10 Fourier coefficient a0 (FORCOF(1))

11-20

71-80

Card 2 (if need) 1-10

71-80

Card 3 (if need) 1-10

rr It

r.

a1

(FORCOF(2))

a7

(FORCOF(8))

a8 (FORCOF(9))

a15

(FORCOF(16))

a16 (FORCOF(17))

aN-1 (FORCOF(N))

The value of N must correspond to the value of NUMFOU input

in columns 21-25 of CONTROL CARD.

154

E. 3.10 Angles Station of Displacements and Stresses PrintOut Card (7F10.5)

If value of NANGLE in columns 26-30 of CONTROL CARD less

than or equal zero, no angle station card is needed.

Columns

1-10 Angle station 1 (XANG(1) in radians)







E.3. 11 Output Control Card (215)

Columns

1 -5 IPRIN1. Enter a 1 if the output summation of total nodal

displacements are needed to be punched on cards, otherwise,

these columns are left blank.

6-10 IPRIN2. Enter a 1 if the output summation of total soil

element and shell nodal stresses are needed to be punched on

cards, otherwise, these columns are left blank.

155

E. 4 Printed Information and Data

The input data is printed first. This includes the control

numbers, the material properties, the nodal point coordinates, forces

and properties, the element connectivity, the constraint data, the

Fourier coefficients, and the angles of the results to be printed out.

For each non-zero Fourier coefficient harmonic, some computed

information is printed out next. First, the generated displacement or

rotation numbers for each nodal point in the model are printed. For

each soil nodal point there are two possible displacements for Fourier

numbers equal to zero and three for the succeeding Fourier terms.

For each shell nodal. point there is one more rotational displacement

in addition to the translational displacements. Each number repre-

sents one equation in the system. Secondly, the code numbers for

each element are printed. This information is generated from the

input connectivity and the displacement numbers and is used to

assemble the system of simultaneous equations. Finally, the maxi-

mum half-bandwidth for the system of simultaneous equations for each

Fourier harmonic is printed.

The output data is printed next. First, for every nodal point the

global displacement components and rotations (for shell node) are

printed. The units are feet for displacements and radians for rota-

tions.

156

The soil element stress coefficient on plane 0 equal zero

are printed next. A quadrilateral element will have four sets of com-

ponents, one for the center of each sub-triangle. A triangular ele-

ment will have one set of components. Listed for each element are

the radial stress Cr , the vertical stress Cr , the tangential orrr zz

circumferential stress SAO, , the shearing stresses: , TrO and

the maximum stress, the minimum stress, the maximumTZO'

shearing stress, and the direction of maximum stress from r-axis.

The unit of stresses is kips per square foot, the unit of angle is

degrees.

The next data printed is the shell nodal resultant coefficients

on plane 0 equal zero. Listed for each shell nodal point are the

vertical membrane force NzZ, the hoop membrane force Nzz , 00'

the shearing force Nz0, the vertical moment

moment M 00' and the cross moment Mz0.

M,zz the circular

Finally, the summation of nodal displacements, soil element

stresses, shell nodal stresses which were contributed from each

non-zero Fourier term for different desired angles are printed.

The shell stresses can be computed from the following equations:

N 12M dzz zzo- ,zz 12t

+

t3(E. 1)

where

0012M

00dTOO 12t t 3

Nz0 12MzOd

T= +z0 12t t3

o- ' 00, o- are vertical, hoop and shearing stresses,zz z0

157

(E. 2)

(E. 3)

respectively, in ksi.

Nzz,N00,Nz0 are shell resultant forces, in kips /ft.

Mzz' MOO' Mz0 are shell resultant moments, in kips - ft /ft.

d is the distance along the normal from the mid-plane to the

surface of shell, in inch.

t is the thickness of shell, in inch.

158

APPENDIX F

Description of Computer Program

Eighteen subroutines comprise the body of the computer

program. They are controlled by call statements. The main program

modifies the initial input boundary constrained nodal point data and

input nodal forces for each non-zero Fourier term. The calling

sequence of the subroutines is as follow:

1 INPUT 7 FVECTR

2 SUBCOD 8 BANWID

3 GENCOD 9 BANSOL

4 INITL 10 SOISTR

5 STIFF 11 SHLSTR

6 BIGK

Recycles subroutines 3-11 for each Fourier term

12 TSTRES

Recycles subroutines 1-12 for each problem.

The flow diagram for circular cell bulkhead analysis is

illustrated in Figure F. 1. The functions of the subroutines are as

follows:

Read input dataCALL INPUT

Generate quadrilateral element code numbersCALL SUBCOD

Store boundary conditions and boundary nodalforces on LUN 15

Begin Fourier loopNFOUR = 1

V

Read boundary conditions from LUN 15 and modify themfor each non-zero Fourier term

Yes

Is this the firstsucceeding Fourier term? Yes

159

Generate and print nodaldisplacement code numbersand element code numbersCALL GENCOD

Figure F. 1. Flow diagram for circular cell bulkhead analysis.

160

Initialize the structure to the initial conditionCALL INITL

Read boundary nodal forces from LUN 15 andmodify for each non-zero Fourier term

Is there anysurcharge load?

NAXISF > 0

Start DO-Loop (NN =1, NUMELT) to calculateand assemble element stiffness and load vectorCALL STIFF

Compute shell elementstiffnessCALL SHLSTFCALL SHLING

CALL SHLTRN

Compute triangular elementstiffness and body forcesCALL TRISTF, CALL SOILIN

Compute four sub-triangular elementstiffness and body forces andassemble into one quadrilateralelement CALL QUAD

Figure F. 1. Continued.

161

Assemble element stiffness matrices intosystem stiffness matrixCALL BILK

Assemble system load vectorsCALL FVECTR

NN=NN-I-1

Is DO-Loop satisfied?

Is this thefirst Fourier term or first

succeeding Fourier term?Calculate band widthCALL BANWID

Solve the system governing equationsCALL BANSOL

Compute and output the magnitude of nodaldisplacement and soil element stresscoefficients CALL SOISTR

Compute and output the magnitudeof shell nodal force coefficientsCALL SHLSTR


162

Yes

Yes

Is there anynext problem?

NPROB <NPROBS

Compute and output the totaldisplacements and total stressesfor the required 0 -stationsCALL TSTRES


163

1) Subroutine INPUT. The input data are read and written.

This subroutine generates a mesh of quadrilateral or shell

elements in the r-Z plane, checks the magnitude of all

essential variables against their maximum permissible sizes.

This subroutine also computes geometric properties of shell

elements and store on file LUN19. It calculates the nodal

point force from the boundary tractions.

2) Subroutine SUBCOD. This subroutine generates the 4 x 9

matrix of code numbers used subsequently to combine four

triangular elements into one quadrilateral element.

3) Subroutine GENCOD. This subroutine generates a code

number for each element in the finite element model. These

are subsequently used to assemble the system matrix of

algebraic equations.

4) Subroutine INITL. This subroutine initializes a given

problem to the initial conditions.

5) Subroutine SOILIN. This subroutine calculates the 10

required volume integrals for a triangular element and stores

these data on file LUN16.

6) Subroutine TRISTF. This subroutine generates the 9 x 9

triangular element stiffness matrix and computes the ele-

ment body force vectors for the non-zero Fourier term.

Subroutine SOILIN is called for first non-zero Fourier term,

164

or read the corresponding volume integral data of triangular

element from LUN16 for succeeding Fourier term.

7) Subroutine QUAD. This subroutine computes the center nodal

point of each quadrilateral element using the mean value of

the 4 nodes of that element and then uses code numbers to

assemble four triangular elements into a quadrilateral ele-

ment. The 15 x 15 quadrilateral stiffness matrix and

associated load vector are obtained. Subroutine TRISTF is

called.

8) Subroutine SHLING. This subroutine computes shell volume

integrals using a 10 point Gaussian Quadrature Formula and

stores the results on file LUN17.

9) Subroutine SHLTRN. This subroutine generates 8 x 8

displacement transformation matrix for shell element.

10) Subroutine SHLSTF. This subroutine computes 8 x 8 shell

element stiffness matrix referred to the local coordinates

system, then transforms element stiffness matrix to global

coordinates by calling subroutine SHLTRN.

11) Subroutine STIFF. This subroutine performs the static

condensation process which reduces the 15 x 15 quadrilateral

stiffness matrix to a 12 x 12 element matrix. Subroutine

QUAD is called.

165

12) Subroutine BIGK. This subroutine uses the code number

technique to assemble the element matrices into a system

matrix.

13) Subroutine FVECTR. This subroutine uses the code number

technique to assemble the system load vector.

14) Subroutine BANWID. This subroutine computes the band

width of the system stiffness matrix for each Fourier term.

15) Subroutine BANSOL. This is a standard subroutine used to

solve the system of simultaneous equations by Gaussian

elimination.

16) Subroutine SOISTR. This subroutine outputs the magnitude of

the Fourier coefficients of the nodal point displacements,

stores the nodal point displacement coefficients of each

Fourier term on file LUNG. It computes and outputs the

magnitude of the Fourier coefficients of the element stresses

at the center node of each element, then stores the element

stress coefficients on file LUN10. This subroutine also

computes the principal stresses and their directions at the

plane 0 = 0 (plane of symmetric loading).

17) Subroutine SHLSTR. This subroutine computes and outputs

the magnitude of the Fourier coefficients of the generalized

shell stresses at the shell nodal points. The shell element

stress coefficients for each Fourier term are stored on

166

file LUN12.

18) Subroutine TSTRES. This subroutine computes and outputs

the total displacement for each of the 3 components at each

soil nodal point and the 4 components at each shell nodal

point for the required 0-stations by summing the contribution

of each Fourier term at each nodal point. It also computes

and outputs the total stress components for each element.

167

APPENDIX G

Program Listing

PROGRAM BULKHEADCC

COAmON /41/ WINPT,NUmELT. NCONP.NCOT,NSMEL'TCOMMON /12/ COOR(57,2),IS0NNI57,5),NPTYPE(66),NPCOM(66)

IF BOTH SOIL ELEMENT AND SHELL ELEMENT ARE PRESENTNCuT=12

COMMON /a3/ KNOVAC66,51.N3O0m(56,4) GO TO 13GoMmuN /91/ ACELR.ACELZ.ANGFQ,DENS(12)1 00 2 NP.1.NUNNPTCOMMOU /89/ MAX/SF,NPAX(1311FORC(13.4)

COMmr:N /C1/ FORCE(66.4) 00 3 NC=1.NCONDIF(NP.NE.4.NTEM(NC,1)) GO TO 3COMM's:: /0../ N3DO 5 L.1.5COMMON /F1/ NFOIIR,NFOF2,4UMFOUTNF.NFTIPMF1

5 KNOVA(NP,L)=KNTEM(NC.L):JIMON /F2/ FCRCOF(50),XAMG(7),NANGLEC . GO TO 6

DIMENSION KNTEM(6615) 3 CONTINUE

gg513:1=tPREAL NF, NF21 210C

KNOVA(NP,3)=0READ(60,1:0) NPROBS6 KN04A(NP,4)=1100 FORMATIT5) .

IF(NPTYPE(NP).E0.3) KNOVA(NP,5)4.1NPRO0 =1. IF (COOR(NP,1).NE.0.3) GO TO 210 CALL INPUTKNOVA(NF.2)=1CALL 5j9C0OCOORINR.1,=0.00001C

2 CONTINUEC "4' STSRE BOUNDARY CONDITION DATA ON FILE UNIT NO. 15 NCONR=NUMNPTCGO TO 45REMIND 19

7 00 B NP.1.NOMNPTkRITE115!NCONP,I(KNOVATAII),I21.5),N.1,NCONP)C 00 4 NC=1,NCONPIF(NP.NE.KNTEM(NC,1)) GO TO 9C STORE NODAL BOLINGARY LOAD:: ON FILE LON 15 DO 11 1=1.9C

11 KNOVA(NP.L)=KNTEM(NC,L)ARITE(15) IfFORCE(HP,I),I=1,4),NP=1,NUMNPT) G0 TO 12CC

Ti00AL DISPLACEMENT ANO ELEMENT STRESS-COEFFICIENTS 9 cOlTI.:DENNoVA(NP,i)=NPWILL 8E STORED ON FILES, LJN 6.7.8,10,12,20,21,22,23

C WAZ:i1:8REWI41 6KNOJAINP.4)=0VIN; 7 12 KNOV.INP,5)=1RE-41ND B

PEqINO 1) 8 CONTINUEPE-41;t1 1? talT8=a14PTg.,,.,:tip 20

C;'.Editt.) 21C ** BOTH SOIL ELEMENT AND SHELL ELEMENT ARE PRESENTAit':0 22 CE4r.r.) 23

13 DO 14 N=1,NUINPTDO 14 N=1.5

CC ".C

NF1=;i

BEGIN FOURIER LOOP

NFOUQ=1

14 KNO4A(N,M1=0NCN.GDO 16 NP.I.,NumNPTDO 17 NC=1,NCONPIFtNp.NE.KNTEm(NC,11) GO TO 1720 x--0-0,;.coFINF0uR1 NCN=NGN+1IF(X.FU.S.0) GO TO 310 00 15 L=1,5NF.rr;i?-1

hF2.',T.NF 18 KNOVA(NON,L)=KNTEM(NG,L)GO TO 16'iEMIN1 19

RE.i1":17 16 17 CONTINUEIF(NFTYPEINP).NE.3) GO TO 16RFHI,ris 17

kEwIN7i 11 NCN.NCN1KNOVaiNCN,1)=NP00 25 PL=1.NUMNPTKNOVA(NCN,2)=0GO 25 IG=1,51043vA(Nem,3)=66N)/1(IL.I)=0KN)vA(NON,41 .025 KliEm('IL,ID)=0KNO4A(Nr;N,5)=1ND14.=0

kEAD(15) NCW4PI(KSOVAIM,I/0.1,05),N31,NCONP) 16 CONTINUENDONP=NCNC

C MC017.1. E3',JNOARY CONOITIONS FOR EACH NON-ZERO CC FOLItIER Ti:Rm OF DIFFERENT SYSTEASC * GENERATE ELEMENT CODE NUMBERS AND INITIALIZE THECC00 25 NC=1.NCONP

45 CALL CEN000DO 26 K=1.5 91 CALL INITL26 P.NTEMOiC.K)=KNOYA(NC,K) CCC +" READ NODAL POINT BOUNDARY LOADS FROM FILE LUN 15IF(NrOUR.EC.1) Go TO 1 CIF(NFi.E0.11 GO TO 91 READC151 t( FORCE(NP,I1.I.1,4).NP.1.NUMNPT)C ,

8 ... IF NO SOIL ELEMENTS ARE PRESENT, NCuTwO cC P MODIFY SOLJNOARY NODAL FORCES FOR FOURIER TERN

IFINeuT.EO.0) GO TO 45 DO 109 N.1.NUMNPTDO 109 M=1,4i ". IF NO SHELL ELEMENTS ARE PRESENT, NCUT.13 109 FORCE(N,M)=FORCE(NOOX

IFINCUT.EO.13) GO TO 7 C IFIMAXISF.E0.0.0R.NFOUR.CT.13 GO TO 108

STRUCTURES

C ADD AxiSYmmETRIC SURCHARGE LOADS DATA (mAxNP.66).(mAxEL=57).(MAXmATs12).(NAxFOu$501u0 1'17 N=1,NJmNPTco 1,:6 J=1.NAX/SF REAL NF, NF2IFENFAx(JE.NE.N) GO TO 1EE1 8 READ AND PRINT OF CONTROL DATADO 1,5 K.1.4

C105 FORCE(N.K)=FORCE(N.K)+FOR:(.J.K)WPITE(61.1)GO TO 10'

1 F044AT(11 INPuT DATA t)106 :oNT:',JEREAD(60.100)NumNPT,NUmELT.NUmmAT.hUmPC.NUHFOU,NANGLE.107 :D4TINJE

1 NCO4P,NrJT,ACELROCELZ.AVSFO.ISTOP,NSHELT,NAXISF108 RERIND 19100 FORMAT(815.3E111.4.213,I4)C

IFEISTOP.E0.01 GO TO 5C ,GENERATE SYSTEM STIFFNESS MATRIX AND SYSTEM LOAD VECTOR WRITE(61.238)Cco llo sp4-1.NumELT 238 FO4mAT( /t MESH CHECK ONLY$,111

5 IF(NJMNPT.LE.MAXNP) GO TO 10CALL STIFEENN)wRITE(61.234) MAXNPCALL BIGK(NN)(STOP.1CALL Fs;:TE.ENN)

234 FORMAT( /) huHDER OF NODAL POINTS EXCEEDS THE MAXIMUM OF $.14)110 CCNTIN.,E10 IF(4omELT.LE.NAXEL) GO TO 20RE4I,.,' 15

RRIT-(51.235) MAXELRE4P,D 19235 FORmPT(/$ NUMBER OF FLEME4TS EXCEEDS THE MAXIMUM OF$,I4)IF1'(r1.E1.1) GO TO 111 ISTOP=iC20 IF(vJmrou.LE.mAxFoul GO T) 30C CALCULATE BAND WIDTH

wRITE(61,23.) MAXFOUC236 FOPHAT( /t NUMBER OF FOURIER TERMS EXCEEDS THE*,CALL BANKI0

1 $ MAXIMUM c1F-$,/A)111 RRITE(1.120) N9ISTOP=1120 FORMAT(//,t BAND WIDTH =$./5)

33 wRITE(u1,20c) NumNPT.NumE.T.NSNELT.N,immAT.NUmPC,NumF0o,NANGLE.C1 N00NP.NGuI,ACELR.A,EL!.AN:fuC 4P4". SOLVE THE STRUCTURE FOR EA:H NON-ZERO FOURIER TERM

200 FOR9AT(//$ NUMBER CF NODAL POINTS $$.14/C1$ NUMBER OF ELEMENTSX* NUMBER OF SHELL ELEMENIS2$ NUMBER OF DIFF. MATEIA.s---:ii

CALL EANSOL

i

CALL SoI:TRIF(N,)T.NE.13) CALL SH!STR 3t NUMBER OF PRESSURE CARDSIFENPCuP.ST.1.ArC. XNE.E0. 1.0) NF/s1 4$ NUM0FR OF FCUPIER TEPiSIF(NFOUR..SE.NUMFOU1 GO TO 400 5$ NU'^EER OF PRINT ANGLESC

6$ NumnfR OF CrNSTPAINEo NODES-AC SOLVE THE NEXT FOURIER TERM7$ SHELL CUT OFF NomiFRC8$ RA9IAL ACCELERATION

:i.M.t/9* AXIAL ACCELERATION300 NcoiR.---t.';oJR.1AC=L7.2.3 Xy ANGULAR VELOCITY st,F10.4/$4CtLa=J.0

CANGF0=0.0 CAN=1.0 8 PROPERTIES/F(Irovl.:T.NUmFOU) GO TO 400GO TO 20

00 50 M=1.NUHEIATCREA3(50.1a11 EiTyPE.DENS(RTYPF),EPODOSTYPE),PR(PTyPE)C ". SUM THE FOURIER COMPONENTSIE(HTYPE.LE mTyPE.GT.0) GO TO 40

C430 IF(NANILE.GE.0) CALL TSTRES

ITE(61.237) .lAx.ATRP/TF(61.250)237 PO4 IAT(/* NUMBER OF MATEPIALS EXCEEDS THE MAX. OF$.14)250 FORMAT( / //,t ENO OF THE PROBLEM$) IST0P=1IFiNPPoDS-NPROd) 30.90.80 40 wFITE(51.2a11 mTyPE,OENS(NTYpE).EN0D(NTyPE),PRImTYPE)C201 FORMAT(/* MATERIAL NuHLER - -4,15/C

.. SOLVE THE NEXT PROBLEM1$ MATERIAL DENSITY -$,FIC.3,t KIP.PER CU.FT.*/2$ m0OUtUS OF ELASTICITY--=$,F12.4,, KS1s$80 NPRoB=NPP00.13$ POISSON RATIOCO TO 10 C =*.F10.3./)

CC CALCULATE MATERIAL CONSTAmTS90 51 ±^C

=1=E).144.ENO

CC v

TE10...x(1.-y)/(11.tv)(1.-2.v))C

. Su3ROuTINE INPUTE(1.mTYPE)=TEmPEt2.mTYPEI=TEmPY/(1.-Y1COMMON /AI/ Num.:PT,NUMELT. NCONPOICUTINSHELTEl1.9TYPF)=E(201TYPE)COMMON /12/ Cno,,(67.2),ICONN(57.5).NPTYPE(66).NPCOM(66)E14.1TYPE)=E(1.mTYPE)COMMCN /AZ/ KNovA(66.51.NOOOm(66,4)E(5.mTYPFE=E(2.mTYPE)COMMON /A7/ OP .07 ,x4. ,S3 .13 ,XTt(5,MTYPE)=Ell.MTYP.)C144',4 191/ scELP,AGEL$014;$0.DENS(12)L(7.,ITYPI)=J.5TEmo(1.-2.Y)/(1.-Y)COMMON /9Z/ E(9.12)E(8,..TyPEI.E(7.11TY?E)COmoN /15/ NAXIS= NPAX(13),FORC(1394)E (4. EITYPF) =E I 7. mTyPE1COMMON $C1$ FCCE(5.4)

50 COvTiNuECcm114 /E1/ h$ou,R.NP.I.F2.mumFouoNF,NFT.NF1CDOmm.-..4 /F7/ Fo;cos(50).xAmG(7),NANGLEC READ AND PRINT OF NODAL POINT DATACOMMON $G1/ IPRINI,IPPIN2CC

wRITE(6E.2131DIMENSION MCOOE(661, T(65), roctis), J1C(15).LBC(15),213 FORmAT($1400Es.5X.$ R-oRn.s.5x,s z-oRn.:,:xy: NODE TYPEs.2X.I pNc15), pr(15). 11400112), PP(12) is NODE COOE$.2x,s SHL TNI:Ks0x.t R-LOAls,5x,s 2-LOAD:.EQUIVALENCE (mCODE(1).N000mt1.1)),(IOC(1).NODOm(1.2)1. 2 5X,* T-LOA0*.5X.$ MOMENT COMMON NODES. /)I ($8C(11.NODON(16.21).(L8:11).NODON(31.2)).(T(1)0000M(1.3)) L0

60 REA1(60.103) N.NRCOM( N),COOR(N.1),COOR(Na),(FORCE(N.J)sJwis4)01 M7.0 DE(N),NPTYPE(N),T(N)

103 FORMAT(I3,I2,6E10.4,213,F3.3)NN6=L+12X=N-LIF(L.LE.01 GO TO 70D=.(r-,C.7?(N.1)-COORIL,i))/IX02=(COD(A.2)..COOR(L,2))/ZX

70 L=L*1IF(9-1) 95.90.80

80 MC)]E(L)=5IF(mC0CE(NNL-1).EC.MCOOEI4)) MCODE(L)=MCODE(N)NPTYP=AL)=NPTYRE(L...1)IF(NPTYPE(NNL-1).ED.NPTYPE(N)) NPTYPE(L)=NPTYPE(N)NRCOm(L)=NPG0M(L...1)0)R(1.1)=CODRIL-1.1)+OR

CO-P(L.2)=000P(L-1,2)+02F0RCE(L.1)=0.0FORCE(L92)=0.0FORCE(L.3)=C.0FO,.OE(L.4)=0.0T(L)=T(L-1)CO TO 70

90 ,4,1ITE(61.203) (KI000R(K.1).COOR(K,2),NPTYPE(KI,MCOOE(K),TIK),1 (FOCE(K,J),J=1,4).NPCOMM.K=NNLIN)

203 FOR.AT(I5t2F12.4,2(7X.I5).5F12.4.I10)Ir(NUmPT-.N) 95,110.60

95 ii2ITE161.231) N239 FORMAT(/* NODAL POINT CARD ERROR N=$,I5)

ISTOP=1110 CONTINUE

C ""' REA] AN, PRINT ELEMENT PRDPERTIESC

WRITE(51.214)214 FORM:.T(ti ELEHENT*,4X,* Ity4X,* J*04X,* X$04Xs* LX.

1 ,.X,1 MATERIAL:/)N=0

130 ;4EAD(50.154) M. (ICONNIM.D.I.115)104 FORMAT(6I5)140 4.441

IF(M-N/ 185,170.150150 ICCIN(N,1)=I0ONN14-..1,1)+1

F;ONN(N.2)=ICONN(N-1,2)+1.AND. ICONN(N..1.4).NE. 0) GO TO 151ICONN(N,3).)

IXT1I(T0 ='151 1',0iN(N,T)=ICONN(N.-1,3)+1

ISOV(N,4)=ICCNN(N-1,4)41152 I.7,-.(N(N.5)=ICONN(N-1,5)170 KRiTE(151.234) N, (ICONN(NpI),I=1,5)204 F344AT(7/.515.19)

IFIN-N) 185.110.141150 IF(IUMELT -N) 190.193,130185 mRITE(51,231) N231 FORMAT(/* ELEMENT CARD EPR3R*.a5)

ISTOP=1190 CONTINUE

CC COMPUTE 54ELL ELEMENT RRORERTIESC

IFINCUT.E1.3) GO TO 199

C ". STORE SHELL ELEMENT PROPERTIES ON LUN 19PEmIND 10CO 111 N=1,NUMELTIF(ICON(9.5(.17.NCUT) GO TO 191I*ICr,NN(N.1)J*IGCNN(..,2)AT =;.5.(T(I)+T(J))/12.CR =';00Q(J.1)..LOOR(I11)CZ =CGO'(ls2)..COCP(I.2)XL =S1RTIDR 'OR 02 02 )SS =-02 /XLCS =02 /XLWRITE(19) N,XT,DROZ,XL.SS.CS

191 CONTINUE

8 READ ANO PRINT CONSTRAINED BOUNDARY NODAL POINTS

C199 WRITE(61,222)222 FORMAT(//t CONSTRAINED NODAL POINTS*,/

1 15X,* NODE:.GX,: R*,6X.* Z*,6X,* r*,2x,* ROTATION *)DO 223 M=1,NCONPREA0(60,224) (KNOVA(M,J),J=1,5)

224 FORMAT(5T5)223 WRITE(61.225) M, (KNOYA(M,J),J=1,5)225 FORMATUI5,7X15I8)

C * READ PRESSURE AND/OR SHEAR BOUNDARY STRESSES

IF(NUMPC.LE.0) GO TO 250WRITE(51.215)

215 FORMAT(//: PRESSURE AND -02 SMEAR BOUNDARY CONOTTIONS3//1 3 ELEMENT31143.3 Itg4Xs* .1314X,3 PRESSURE(KSF)3,4X,* SHERD/SF/Si)00 115 L=LINUMPCREA0(60,155) LBC(L), IBC(L),JBC(L),PN(L),PTIL)

105 FORMAT(3I5,2F10.4)195 WRITE(61.205) LBC(L), IOC(0,JBC(0,PN(L),PT(L)205 FORMAT(Iig216.2F12.4)

C * COMPUTE SURFACE INTEGRALSC

C

IF(NUMPC.LE.0) GO TO 250DO 253 N=1,NUMELTDO 260 M4.101DMPCIF(L6C(MM).NE.N) GO TO 26000 230 I=1,4J=I+1IF(ICCNN(4,4).NE.0) J4=4IF(ICONN(9.4).E0.0) J4=3IF(ICON9(4.3).EQ.0) J4=2IF(I.E1.J4) J=1K=ICONN(N,I)L=ICONN(N,J)

ICK=MCO5E(K)IF(K.NE.T3C(Mm).0R.L.NE.JBCOH)) GO TO 230

ICL=MOU0E(L)X=000R(1,11-.COOR(K,1)Y=COO4(1,2)*COOR(K,2)XX=X16.0YY=0.5'CO0R(K,1)+XXXX=1.7+XX/YYFRK=YYIPTIMM)X+PNIMM)Y)FR,=XX+FRKF2K=YY.(0T(MM),Y.-PN(MM)X)F21=XX.F7<

FORCFIK.1)=FORCE(K,t)+FRKFORCE(K,2)=FORCE(K.2)+F2KFORCE(L,1)=FORCF(L.1) +FRLFoRCE(L,2)=FoRCE(1,2)FZLIF(ICK.EJ.1.0R.ICK.EQ.6) ;0 TO 311IF(ICK.EO.2.OR.ICK.EQ.7) ;) TO 312IF(ICK.E"4.3.0R.ICK.E0.51 OD TO 313.GO TO 314

311 FORSE(K.1)=0.0GO TO 314

312 FORCE(K,2)=0.0CO TO 314

313 FORCE(,<,1)=0.0FORCE(K,21=0.0

314 IF(ICL.E1.1.0R.ICL.E0.5) 03 TO 315IF(ICL.E0.2.0R.ICL.E(1.7) GO TO 316IF(ICL.E0.3.0R.ICL.E0.5) GO TO 317GO TO 250

315 FoRCE(1,0=0.0Go TO 250

316 FORGE(1,2)=0.0GO TO 250

317 FORCEIL,11.0.0FORGE(1,2).0.0GO TO 250

'230 CONTINUE260 CONTINUE250 CONTINUE.. READ AX/SYMMETRIC SURCHAR;E LOADS

F.+

O

C COMMON /AI/ NumNPT,NUmELT, NCONP,NCUT,NSHELTIF(44.(ISF.P2.0) GO TO 410 COmmON /42/ CooR(67.2),ICO4N(57.5).NPTYPET661.NPCOm(66)NRITC((.1.411) COMMON /05/ KNOVA166.51.N330M(6C.4)

411 F04.1AT(///* AxiSrmmETRIC SURCHARGE LOAOS*// COMMON /44/ LADOMIt N07,Et.5X.: R-LOADIE.5x.A Z-L04C4,5X,t TLOAC4,5X, COMMON /46/ NC00E(57.12)2$ m3HENT//) COMMON /F1/ NFOUR,NFOF2,NDNFOU,XNE,NFT,NF1DO 412 N.i,NAXISF REAL NF,NF2RE40t2,413) NPAX(N),(FOR:TN,LT.L.1,4) C

413 FORmAT(T5.4F15.5) NFR= NFOUR -1412 HPITET61.4141 RAx(N),(FORC(N,L),L=1,4) C414 FORIAT( /i5.4F12.) C 4"Ps INITIALIZE

C . READ FOURIER COEFFICIENTS CC DO 23 1.1,NUmELT410 wRITE(61.216) 00 23 J.1.12216 FORMAT(t1 FOURIER NO.t,Sx,t FORCE COEFFICIENTS/) 23 NCCOE(I,J).0

READ(6C,106) (FOROOF(N),N.I.,NUmFOU) CO 30 I.2.NUMNPTG3 32C N.1,NumFOu DO 30 J=1,4M =N -1 30 NODOM(I.J)=0

379 wRITE( 51,206) M, FCRCOFTN1 K.0106 FG.!4Alt0r15.5) C

C *** TO GENERATE CODE NUMBERS IN TERM OF NODAL POINT NUMBERFORMAT(1110E2G.5)C

. READ ANGLES CF STRESS OUTPUT DO 12 I=/.NUMNPTC DO 2 N.1,NCONP

IFINANGLF:.GT.G) GO TO 350 IF(KNOVAtN.11.E0.I) GO TO 3311GO 11,, 3 00 4 m.2,5

350 ;FAN!,NIL.E.CT.7! NATILE.7 IF(KNOVA(140).E0.11 GO TO 4pFt.,,5,J.1).1 (XANG(N).N=IgNANGLE) K.K.I.1.,ITE151.5071 (XANG(NI.N.INANGLE) J=4-Im,207 FOR4fT(//A ANGLES OF STRESS PRINT OUT IN RADIANSt,///TF20.6/1)

C4 ENOOI0N(I

J)=K

nC * READ CuTRUT CONTROL CARDS V TOC 2 CONTINUE

IF(NRTYPE(I).GE.4) GO TO 3350 REA3(6C.J11) IPRINI.IPRIN2C111 FO?.44T(2I51

C C 4"14, THIS IS 4 SOIL NODAL POINTC CHECK FIRST FOURIER COEFFICIENT CC J=3

00 410 4=1,NUmFOU GO TO 11IFtFo4SOFIN).E0.0.0) CO TO 409 CN0 ITo

453C ." THIS IS A SHELL NODAL POINT

"400 CCNTINUE

C0 J=4

1,/TE(61,2331 11 00 31 1.1,J233 FOR4AT(//t BAD FOURIER COIFFICIENTSt) K=K+1

ISTOPTI450 IF(ISTOP.EO.0) RETURN 31

NODOM(I,L)=KCONTINUE

STOP 12 CONTINUEE NC, LADOm.K

Cc 8 .. TO GENERATE ELEMENT CODE NUMBERS

Su3RCUTINE SJBCOO CC 00 17 L=1,NUmELT

COMMON /A5/ ICODE(40) N.0C IF(ICONN(L.3).E0.0) GO TO 18

DO 111 0=1.4 IF (ICONN(L.4).E0.0) GO T3 19DO 199 J.1.9 C

CE(I,J1.0C199 NO S ,P,* THIS IS A SOIL ELEMENTI031 2c0 1.1.3CO 241 J.1.6 n10 20

201 1C30E(I.JI.J.N 19 4P.3200 N.'., 20 NOOF=3

GO TO 21CO 212 1=1.4103)1(1,7) .13 CI0 7:3F(I.9)=14 g ... THIS IS A SMELL ELEMENT

7)202 IC1Eli';)=151G3qE4,!1 .10

16ICi0E(4.2)=11 tHW.4. 1C3)C(4.31 .12

21 n414 LOU4=1.NP1C33C(4,4)ziICODE(4,,,)= NTE4P=ICONN(L.LDUM)2IC00.,:A4,61 3 UO 15 I=1, HOOFRETURN NOJI.I.K

15 NCODE(L,N3UM).NODOM(NTEMP,I)K=K+NDOF

En)

S 14 CONTINUE

17 52M711,50SUBROUTINE GENCOD

C / NFR

50 FORMAT(t1 OISPLACEMENT AND COOS NUMBERS FOR FOURIERY,30 TERmt.15,///It 103E1.18X,t DISPLACEMENTSt,/.22)(1* Rt,6X02$ 74,3x,t Portummt./1DO 51 I=IINUmNPT

51 WRITE(51.52) I. (N0)0M(I.J)1J=1.4)52 FOR4A1(15.14X.15.2(3X.I5).5X,15./)

wRITE(51.5J)53 FIRMAT(//,t ELEMENT*.28X,t CODE NUMBEPSW)

CO 5,- I=1.NUMELT54 WRI1E(51.55) I. (NC03E(I.AsJ.1,12)55 FORMAT(I5.5X.12I6,1)

RETURNEN)

8 10 P(CPIN)=7;(WI(NN.M)SU:ROUTINE INITL NP(5)=NUMNPT.1

DO 15 N=1,15CO4mON /Al/ NUMNAT.MUMELT,NCONP .NCUT 0(41=0.0COMMON /A4/ LA2OH DO )5 M4=1.15COMMON /C1/ FcRcE(66,4) 15 SYSK(MOM)=0.0COMMON /32/ 3CKitS2,28) J=NP(1)COMMON /01/ E(182) K=N0(2)00 3,1 I.,-1.tApol L=1P(3)F(L)=100

30 M=1.28 KK=NP(5)M=NP(4).0

30 iisvIL,H)z0.0 CO1RIKK,1).(COOR(J,1).0002(K,1)+COOR(L,1)+COOR(4,1))/44.0ETU CO3R(KK,2).(COOR(J,2)+COOR(K,2).COOR(L,2)+COOR(N,2))/4.0

END 00 25 N,..1A0=1.4C

j5:=Ang«ilIFINOUAD.EU.4) JJ=NP(1)SO N1B'OUT:',E STIFF()KK=NP(5)

C THIS SUBROUTINE PERFORMS THE STATIC CONDENSATION PROCESS CALL TRISTF(NN,II,JJ,KK)'C OF QUADRILATERAL ELEMENT

USE CODE NUMBER TECHNICUE TO ASSEMBLE QUADRILATERAL ELEMENTCOMMON /02/ COOR(57,2),ICONN(57,5),NPTYPE/66/0PCOM(56)COMMON /C2/ 2F(9).0(15).5TF(1) 00 17 M=1,9LGAMON /01/ SI9.51.STSK(15.15).SA(9.9) IF(TCOOC(q0000.M).E0.0) G) TO 17

C K=ICOCE(NQUAD,N)IF(ICONN(' :N.3).E:.°) GO TO 7 U(K)=0(K).0F(M)IF (ICONN(1,4).EQ.)) GO 10 8 01 17 N=1.9

IF(ICO)E(NOUAD.N).E0.0) CD TO 17". TMIS IS 4 ''.:UORILATERAL ELEMENT L.I0o0E(N2uAD,N)

Al(!4N)CAL!. SYSK(K.LI.SYSK(K,L).S(M,N)IT CONTINUE4 11,14

CO=STS<(1II.15)/SYSK(15.15) 25 CONTINUE..i(II)=O(II)-CEQt151 RETURN00 4 JJ=1.14 END

4 SYS<( II*JJ)=SYSK(II,JAC:SYSK(15,JJ)00 5 II=I.13CG.2YSK(II.141/SYSK(14.14) SUBROUTINE TRISTF(NN,II,JJ.KK)Q(II1,-;(r:)-cnc(14)

8CC

DO 5 J.1=1,13 THIS SOBROJTINE FORMS THE TRIANGULAR ELEMENT STIFFNESS MATRIX5 SYSK(II7jJ)=SYSK(II,JJ).-CC.SYSK(14,JJ)

8AND BODY FOPCE VECTOR FOR THE NON -ZERO FCURIER TERM

03 11 11=1.12YSvA II.13)/SYSKI13,13/ COMMON /A2/ COOR(57,2).1C11N157.51,NaTYPE(66)0PCOM(66)

COMMON /II/ ACELR,ACELZ,ANGFQ,DENS(12). COMMON /a2/ E(9,12)L JJ=1.12It STO S<1C1 II.J.H.SYSK(II,JJ1..C:SYSK(131JJ) COMMON /53/ C11.C12,C13.C22,C23,C33,C44,C55466

00 TO 15 COMMON /C2/ 5F(9).0(15).BTF(9)8 CONTINUE COMMON /01/ S(9.0).SYSK(15.15),SA(9.9)

C COMMON /El/ XI(10),A(9.91C ". THIS IS A TRIANGULAR ELEMENT COMMON /F1/ NFOUROF,NF2,NUMFOU,ANF,NFT,NFA

C '0**REAL NE, NF2

KK=ICO110.1.3) C NEOUR.GT.NFT, READ SOIL ELEMENT INTEGRAL INFORMATIONSLOLL TPI;TFINN.II.JJ.KK) C FROM FILE LUN 1633 15 4,1,4L.(1)=0.0

15 GI3 N)TO =3(M)BF(M) 8 INITIALIZE GEOMETRY ANO MATERIAL PROPERTIESG 10

7 CONTINUE MTYRE=ICONN(NN,5)C C11.E(1,MTYPEI

THIS IS A SMELL ELEMENT C12=E(2,MTYPE1CALL SHLSTF(NN) C13=C1203 i M.1,6 C22=C1t

9 Q(MI=0.0

ECC23.C1233.10 kNO ETURN44CmE(7cti04TYPEA

C

SUBROUTINE QUAO(NN)

C '" THIS SUBROUTINE USES THE "DOE NUMBER TECHNIQUE TO ASSEMBLEC FOUR SUB-TRIANGULAR ELEMENTS INTO A QUADRILATERAL ELEMENT

COMMON /Al/ NUMNPT.NUMELT. NCONP,NCUT,NSHELTCOMMON /02/ COOR(67,2),ICONN(57,5),NPTYPE(66),NPCON(65)COMMON /A5/ ICOOE(4.9)COMMON /C2/ OF(9),Q(15),BTF(9)COMMON /01/ S(919),SYSK(15,15)ISA(9,9)DIMENSION NP(5)

C55=C44c6c,.n44IF(N,:oiR.GT.NFT1 GO TO 15

CC FORM STIFFNESS MATRICES IN GENERALIZED CCORDINATES

CALL SOILIN(NN,II.J..1000GO TO 17

C15 REA3(1,) NN,II,JJ,KK, (XI(4),N=1,10),TIA(ItOtI*119)1,9)17 X=::33NF2C55

s(1.1)=,.xx(3)St1.2)=1C13,x)xI(2)S(1.3)=x°xI(5)S(2,2)=IC112.0C13.X1XI(1)(2.3)=C-.13.).X1(4)

5(3,3) =xxI(61.C44XI(1)Y.,1.72.C5Sf4.4)=Yxi(3)5o4.)=Y.x1(217.(,6)=Yx0(5)sc,;.5)=(C44.Y1XI(11:;(5.6)=Yxl(41S(5,1',)=Y"(I(6).C22XI(1)x=7.5s.Nr?o33Y=C33NF2S(7.71=xxI(3)S(7.A)=Y.XIC21517.9)=KxI(5):.,(4,,,)=Y.x:(1)519,=r.sx1(4)Stl ))=1,,x1(614.C66XII11x.NFC=55*NFV.K.C134=S(2,!4)=T*X1(1)V=Y+CC=44CS(1,7)=C*XII3)sf1,81=Xx112)S(1.,)7C.XI(5)5(2,7) =rX1121512.31=vx1(4)5(3,7)=C'xI(5)511,=)=X.xI(4)st3,1)=C.xI(6)511.41=0.05(I.5)=0.0S(1,4,1=C23.XI(2)S(2,4)=0.0512,51=3.0St2.61 =(C12.023)*XI411sts.41 =2.3St!.:,i:44.X1(1)S(3.:)=C22.XI141x,":654,4FY=:23.iFS(4.7)=3.0S(4.91=0.05(4,4).-xx1(2)S(i,7)=0.3515,51=6.L5(5,11=-rxI(115(0,7) =1.X1(2)S(6.8)=1.x1(1)SI6.91=tr-xlXI(4)03 10 1=2.330 13 J=1.I

10 S(/..1)=S(J.1)

FIRM BODY FORCE VECTORi

MTYDE=ICON4(NN,5)AFC=ACELR'OENS(MTTPE)OF2=-ACELZOENS(MTTPE)BFR=3ENSt4TYPE)ANSFQANGCDIF(NFOU-C.GT.11 GO TO 60caFt1)=BFRXI(7)BTF(2)=9FP'XIlliDTF(1)=3,:.XI(11BTF(4)z8FRX1(9)

C

C

CC

CCCCC

C

CCC

60

20

4,,0*

70

40

50

'6"

10

4

20

BTE(51=9FVXI(7)OTF(6)=BFC.XIl7)BTF(7)=8FRX1110)OTF18)=HFZ.x1181BTF(9)=gFC.xI(8)GO TO 70CONTINUEDO 20 1=1,9BTFII) =0.0IF(NFOUR.GT.2) GO TO 70BTF(1) =BTF(.31=BFC.XI(1)BTF141=3T=I6)=OFCXI(71UTF(7)=BTF(91=DFCXI(8)

TRANSFORM THE STIFFNESS MATRIX AND 800Y FORCEVECTOR TO GLOBAL COOROINATES

DO 40 K=1.9CO 40 L=1,9SA(K,L)=0.0DO 40 M =1.9SA(KIL)=SAIK,L)S(K,M)A01,L)DO 50 K=1.9BF(K1=0.0DO 50 L=1,9S(K,L)=3.0HF(K)=0F(<14A(L,BTF(L)DO 50 m=1.9SIK.L)=SOC,L)+A(M,K)SA(M,L)RETURNEND

SUBROUTINE SOILININN,II.JJ,KK)

THIS SURROUTINE CALCULATES THE TRIANGULAR VOLUME INTEGRALSANO FORMS THE DISPLACEMENT TRANSFORMATION MATRIX FOR THESTIFFNESS MATRIX AND BODY FORCE VECTOR

COmmoN /A2/ CooP(67.21,ICONN(57,51,NPTyPE(661,NPCOmf661CORMON /E1/ x1(10),A(9.9)COMMON /F1/ NFOUR,NF,NF2.4UMFOU,XNFOFT,NF1DIMENSION XM(6),R(6),Z(6)REAL NF,NF2

00 10 J=3.10xI(J)=0.0

FORM THE AREA AND VOLUME INTEGRALS USING THEEXACT EXPRESSIONS

R11)=COOR(II.1)2(1)=Cw3R(I1.2)R(?)=CO(J.J.1)Z(2)=CocR(JJ,2)R(31=CcO(!K.1)1(3)=CO3R(KK,2)xI(2)=5.5cR(1)C2(2)-2(3))+R(2)(2(31-2(1))

1 2(3)(2(1)-2(2)11C.xI(2)/3.0XI(I)=C.(R11113(214.4(311X1(4)=C(2(111.2(21.2(3))C=0.25CR(4)=0.544R(114.R(2))R(5)=0.5sOR(2).R(3)1R(5)=0.5(R(3).R(11)Z(4)=0.5.(/(1).2(2)12(5)=0.5(2(2)+2(311Z(6)=0.5(2(3).Z(1)10=3.,::.0DO 20 1=1,3J=143XM(I)C.RtI)XMIJI=0*RIJ)DO 30 1=1,6A=XMII)B=R(I)C=Z(I)D=ABxim=xt(7).0

1 .

...4

to

XI(8)=XI(8).A°C JJ=ICONN(NN.2)/I(1)=X1(9).0.3 HTYPE=ICONN(NN.5)XI(10)=AI(10)0C RFA011.)) N.xToR,OZ,xL,SS,CSIF(9.E7.4.).L) 8=0.00001 IFINFOUR.GT.NFT) GO TO 33U.1/03.31 IFICOOR(II.1).NE.0008(J).11.28.A113)=x14314.n i C00-2(11.2).NE.COOR(JJ.2)) GO TO 1021(5)=xE(5)+0CxI(6)=x/(6)4,0CIT 10

ISToP=1IF(GOW:(II,11.LT.0.0) IST30=1

30 CONTINUE IF(COOR(i..1.1).LT.0.1) ISTOD.1. 1F(XT .LE.O.G) IST3P=1

i '09" FORM THE DISPLACEMENT TRANSFORMATION MATRIX IF(ISTOP.EC.0) Go TO 30WRITE(61.20) NN

C=0.5/x/(2) 20 FORMAT(il BAO SHELL ELEmENT.14/1CO 4C 1.1,9

30 V1TA/215. mTYPE)00 4C J=1,940 A(I. ;)=S.0 E11=XT u(1,HTypE)-E(30TyPE)*E(30TyPE)gx)

A1.R(2).7(3)-P(3).2(2) E12=XT "(E(2.MTYPE)-E(301TYPE).E(5,MTYPE)")e)A2.1(3).2(1)-R(1)Z(3) E22=YT "(E(4.MTYPE)-E(501TYPE)*E(5.MTYPE).X)A3=8(1).?(2)-R(2)"2(1) 144=XT E(701TYPE)b1=Z(2)-Z(3)

:::::::::44X=XT "Xi /12.82=2(3)-7111

83=Z(1)-7(2) Di1=X"Et1C1 rR(3)-8(2)C2=R(1)-2(3) 022=X*E22C3=8121-'<(1)4(1.11=It4,21 =A(7.31 =11"O 39

044=X"E44DO 40 1=1.8

A(2,11=A(5,21=A(5.31=11"C 0)1)=0.3113.1)=1(6,21.1(9,3)=01"C DO 40 J=1.84(1,4)=A (4.5)=A(7.,3)=A2*C 40 0(1,J)=0.0A(2,41./(i.:1=118,01.3,2"C CAry,4)=A(6,,),AC).E0=C2.0 8.. FCP.M STIFFNESS MATRIX IN SENERALIZEO COORDINATESA(1,7)=1(4.C)=4(7,9)=13CA(2.7)=1(5.81 =1(8.91=93C 81(1,1)=E11A(3.7)=A(6.8)=1(9,9)=C3*C gilt:in°"E12IF(NrOUR.ST.NFT) G) TO 50WRITE(16) NN.II,JJ.KK, (XI(N),N=1,10),((A(I,J) 131(1,4)=11,-.E12

1 ,I=1,9),J=1,9) 61(1,5)=CS *E1250 RETURN 131(2.21=E44

ENO 01(2.3)=-NFE44C 8112,4)=-SS "E44C

I 81(2,5)=0.0SU3ROUTINE SHLSTFOIN) al(3,3)=0S 'SS *E22NF24144

01(3,41=NF"SS "(E22.E44)C... THIS SUBROJTINE CALCULATES SHELL STIFFNESS MATRIX 81(3.5)=CS "SS "E22

C. 81(4.4)=N2'E22+SS 'SS E44COMMON /12/ CC07(67.2).1C)4N(57.5),NPTYPE(66)0000M(661 131(4.5)=Nr"CS "E22COMMON /A7/ OR ,07. .XL .SS .CS ,XT 61(5,5)=GS 'CS "122COMMON /82/ E(9.12) CC04404 /341 Eti,E12.E22.E44.011.012,022.044 02(1.1)=011COmMON /C2/ OF(9).Q(15)0TF(8) 82(1,2)=0.0COMmON /S1/ S(90),SYSK(15.151,S1(9.9) 02(1,3)=!;S 112cc44:-,m /E2/ xrs(10.15) 02(1,4)=-NFcS '012COi4on /F3/ A'..:(807) 82(1.5)=-4F2.012COMMON /F1/ NFOLP.OF0F2.4UMFOU.XNF.NFT.NF1 de(2.2),xs ."-,S "044

C 82(2,3)=4F"CS "044DIMENSION IG(10), J5110),KG(101.IH(12),JH(12). 8212.4)=-CS 'CS 'SS "044

1 01(12).1(12).01)5.5).32(5,5) U2(2,5).-NFCS "SS "044MEAL NF. NF2 62(3,3).ss 'OS 022fNr2044

C 02(3,4)=-4Fcs 'OS 0122+744)OATA (1G(1)=1),(IG( 2)=2),(IG(3)=3),(IG(4)=s),(IG(5)=4), b2(3.5).-4F2.SS '(022,044)1 (IG(6).41.(VA7)=5).(IG(i)=5),(IG(91=5),(IG(101 =5), B2(4.4)=CS 'CS "INF20224SS *SS '044)2 IJC(1)=11.1.1G(2).1),IJG(3)=2),(JG(4)=3),(JG15)=2),IJG(6)=3), 02(4,51=NFcs INF2.022.SS 'SS '044)3 (J;(7)=2),(JGtm)=3),(JG(3)=4),(JG(10)=5). (KC(1)=2) 82(5 X 6)=4F2*(NF2022fSS 'SS '044)OATH (KG(2)=4),(KG(3)=1),(KG(4)=2).(KG(5)=3), U0 4., I=2.5

5 (G(6)=4).(<6(7).5),(G1I)=6),(KG(0)=7),(KG(1C)=8). CO 10 J=1.I6 ) I1 (1)=11.(IM(2)=1).(I8(3)=21(TH(4) .3).(!m(5)=3). B1(I.J)=n1(J.I)7 (1,4(6)31.(I4'(7)=4).(1M(4)=410I1(9)=5).(IH(10)=5).(IH(11)=61 90 62(I,J)=112(..),I)DATA (In(12)=5), (JH(1).11,(JH(2)=6).(.1H(3)=2), C

9 (JH(4)=2),(Jm(5)=31.(JH(5)=4),(JH(7)=7),(JH(8)=4), X=XT.XT/12.1 (Ji1(9)=7),(JH(10)=1),(im(11) =9),(JH(12)=10),2 (01(1)=7),(KH(2)=8).(xH(3)=4),(HH(4)=6),44H(5)=7),(KH(6)=8),

E44=1.44044=X.E41,

3 (<4(7)=3),(104(8)=4),(x)4(91=51,(KH(10)=6),(xH(11)=7),(KH(12)*!) IF(NFOUR.ST.NFT) GO TO 91DATA (H(11z2.0),IM(2)=6.0),(M(3)=1.0),(H(4)=1.0), CALL SHLINGINN,IIIJA

5 (M(5)=2.3),(M(6)=3.0Ig(M(7)21.0),(M(8)41.0),(M(9).1.0)41 GO TO 926 (m(10)=1.0).(m(11)=1.0).(H(12)=1.0) 91 READ(17) NM.II,JJ,((XIS(IgAvIa1,10)oJ=1,10)

C 92 DO 100 NGz1,10

CINITIALIZATION GEOMETRY AND MATERIAL PROPERTIES I.KG(NG)

X=JG(NG)/STOP=0 N=IGINGI

DIIgICONMNN,1) O 100 MGz1,10

J=tG(mG)L=.11-.(mS1N=IG(HS) .

70DO 70 .1.1.1xISII.J)=XIS(J,I)IF(NFOUR.GT.NET) GO TO 90

100 S(I.J).31I.J1+81(M,H)XISCK.L1 wRITE(17) NN.II.JJ.C1XISII.A.I=1.101.J=1.1C)CO 110 NS=1.12I .KH(NS)

80 RETURNEND

0,.....,H(N3 )

m=IH(N3) CDO 111 MO=1,12 SUBROUTINE SHLTRN(NN)J.KHIm3) CL=.11(43) FORM DISPLACEMENT TRANSFORMATION MATRIXFCR SHELL.

(Mil . C110 SII,J)=S(I,J)+HING102(MoN)H(MG)XISIK,L) COMMON /A7/ OR gOZ ,XL ,SS .CS ,XT

C COMMON /E3/ AS(8.8)C TRANSFORM STIFFNESS MATRIX TO GLOBAL COORDINATES COMMON /F1/ NFOUROIFOF2.NUmFOU.XNFOFTOF1.C REAL NF.NF2

IF(NrCjR.ST.NFT) GO TO 111CALL =HLT;?N(NN)

CxLI.I.C/XL

,,D TO 112 OCS=xLIo2ill READ(1E) Mo((AS(MoN),M=108),N=1,8) OSS=xLI1R112 OD 11: 1..1.8 X=XL1xLI

G3 130 i=1.4 DO 120 1=1.8SA(I.J)=S.G 00 120 J=118DO 133 K.1.5 120 AS(I.J)=0.2

130 SA(I.J)=SA(I.J)+S(I.K)AS(K.J) AS(1,1)=OSSDO 140 1=1.8 AS(1.2)=1CSDO 140 J=1.8 As(a.t).-DssnrSYS<II,J).0.3 AS(2.2)=-OCSxLI'IX) 141 K=1.8 AS(2.5)=-AS(2,1)

140 SYSK,:.Ji=SYSK(I.J)+AS(K.1)SA(K,J1 AS(2,G)=-AS(2.2)9 =ET.iN AS(3.3)=1.0

END AS(4.3)=-x1I

SAS(4,7)-.xLIAS(5.1)=AS(1,21

SUBROUTINE SHLING(NN,II.JJ) AS(5,2)=-AS(1.1)C AS(6,4)=1.0C THIS SUBROUTINE CALCULATES THE SHELL INTEGRALS AS(7,1)=-3.00CSXC AS17,21= :..3.0SSX

,30m-ION /A2/ COOR(57,2).13344(57.510PTYRE(6610PCOM(86) AS(7.4)=-2.0xLI,com01 /qz/ op .oz .XL .SS .O .XT AS(7,5)=-AS(7.1)

com..^N ,Z2/ xIS(10.10) AS(7.6)=-AS(7.2)ce,,:::4 /F1/ FOUR,NF.NF2.NUmFOU,XNF,NFT,NF1 AS17,91=AS14.3IGImE;4SION xS(13). XN(10). Y(10) AS(3.1)=2.0.)CS.X.XLI

C AS(5.21=-2.5USSxXLIEOJIVALEN3E (r(2).RS).(Y(1.S) AS(3.4)=X;;ATA (YS(1)=-C.,17393E53),(xs(2)=-0.95505337),(XS(3). AS18.5)=-AS(8,1)I -0.N7940,57),(xS(4)=-:,.433395391,(xS15).-0.14987434).(XS16). AS(5,6)=-AS(8,2)2 0.1 494743'.), (XS(7)=0.43334535), tXS(81=3(7940957), (XS(N I= AS(6,81=3 Ogib5V,i171,(AS(/0).0.97390653), ( %M(1)= 0.066571 4),1 %M12)= IF1NFOJR.GT.NFT) GO TO 104 1.1494;135).(XN(3)=0.21918636).(XW(4)=0.26926672). WRITC(18) NNol(AS(ItJ),I=1,8),J=1.0)5 W14(51.1.2955242Z/o(XW(61.0.29552422),(XM(7)=0.26926672),6 114.4011.2.21906636),(XW191=0.14945125)11 M(10)=0.06667134),

10 EUURN7 (Y(1).1.6) CREA.L. NroF2 C

C SUBROUTINE BIGK(NN)00 50 I=1.10 C0.1 AO J.1110 8 THIS SUBROUTINE USES CODE NUMBER TO ASSEMBLE

50 Ais(I.J)=0.0 STIFFNESS MATRIX OF THE SYSTEMA=0.5xL Cx9=3.5OR COMMON /82/ COOR(6712),ICONN(57,5)OPTYRE(66),NPCOM(66)kI=C0OR(I1o1) COMMON /85/ NCODE(57,12)

COMMON /nl/ S(9,9),SYSK(15,15),SA(9,91S 4 `HELL INTEGRALS COMMON /02/ BGK(182.28)

DO 51 K=1.10 Cxicri.Ov,;(K) IF(ICONN(VNg3).E0.0) GO T3 3S.Xxx IFAICONN(NN,41.E12.01 GO T3 4RS.41.x8XX LN.12RS=1.0/RS GO TO 5Y(3)=S47: 3 LN.8Y441=SY(31 GO TO 5Yt5)=SY(4) 4 LN=9T(7)=ASRS 5 00 1 M.I.LNY(5).SY(7) DO 1 N.MILNV(91=5Y(51 X=NCODE(NNIM)Y(11)=ST(9) L.NCODE(NNoN)XXsX/RS IF(K.EQ.0.0R.L.E12.0) GO TO I00 60 I=1,10 IF(K.LE.L1 GO TO 6DO 63 JIc10 TENP=K

60 XIS(I,J).XISII.J)0WL

IXNIK)*Y(I)*Y(J) R=LDO 2U 1=2,10 =TEMP

6 HN34=L-K#1IF(10oRN(NN.4).NE.0) CO TO 7IF(IcoN4(4,4,3).E0.0) Go TO 7

IF(NCOOE(NIT).E0.0) GO TO 20DO 30 L=I.12IF(N000E(N.1).E0.0) GO TO 30

HSK(K03ANP=BGK(K.maAN)+S(401 KK=IA4s(N000E(N,I)-NCOOE(4,01GO To 1 IFIKK.GT.J) J=KK

7 !s)(K,111,11=BGK(X,M3AN)+SYSK(4,41 30 CONTINUE1 G1,)T1uuE 20 COAT INUE

.2E-TORN 10 CONTINUEEND NB=J#1

C RETURNC END

SUBROUTINE FVECTR(NN) CC CC ** THIS SUORouTINE USES CODE NUMBERS TO ASSEMBLE SUBROUTINE BANSOL

THE SYSTEM LOAD VECTCR CCOMMON /A4/ LADOM

CommoN /A2/ CooP(67,2),ICONN(57.5),NPTYPE166),NPCON(66) COMMON /02/ BGK(182,281COm1cN /A3/ NNovA(66.5)04300m(66,4)commN /56/ N000E(57,12)

CGmm0N /03/ F(182)COMMON /D4/ NB

C0mmN /DS/ NAKISF,NPAK(13).FORC(13.4)commoi /01/ F0RCE(66.41 C

DIMENSION T(28)

00,.1111 /02/ NF(9)0(15).BTF(9) NO.LAOom00m1ON /03/ F(182) N=0CO,m3N /F1/ NFouP,NFOF2,NUMFOU,KNF,NFT,NFL 500 N.N+1CO,.10N /F2/ FORCOF(50),XAVG(7),NANGLEREAL NF,NF2 C 'ogy DIVIaE RIGHT SIDE BY DIAGONAL ELEMENT

C F(N)=F(N)/OGK(N.1)

!AllgMl%;1; C " CHECK FOR LAST EQUATIONIF(N-ND) 950.700,700

IF(KK.EO.2) GO TO 5 C '"" DIVIDE N TH EQUATION 9Y DIAGONAL ELEMENTiFtLL.ED.0) GO TO 6 550 00 6J0 K=204L4.1? T(K)=EGK(v.K)CO TO 7 600 BGK(vo()=BGK(N,K)/BGK(N,1)LN=P.GO TO 7 C REDUCE REMAINING EQUATIONS

6 04.9 DO 660 L=2097 U0 1 N././NK=N;N':E(NN.m) IF(NO-I) 660,640.640IF (( 3) GO TO 1 640 J.0CO 2 LP=1,4 DO 650 K=1.001L II , LP) J.J4.1IFtL.E0.01 GO TO 3 650 6GKII.J)=BGK(I,J)-T(L)BCK(N,K)IF( KL) 3,9,3 F(I)=F(I)-T(L)+F(N)

9 .;-,Tcm.IT 660 CONTINUETO 15 GO TO 500

3 L=Nc::-;m(JJ,LP)IF(L.E3.u) GO TO 4 C BACK SUBSTITUTION1F(K-L) 4,8,4 700 N.N1

8 NPTEM=JJGO TO 10 C CHECK FOR FIRST EQUATION

4 IF(K.K.E0.0) GO TO 2 IF(N) 900000.7501.0004(KK,LP) CIF(L.E74.0) GO TO 11 C ** CALCULATE DISPLACEMENTSIF(K -L) 11,12.11 150 Do 800 K=2.N3

12 NPTEM=KK L=N.K-1GO TO 13 IF(NO-L) 800.770.770

C

C

C

C

it IFIL.1.13.0) GO TO 2

IF( -1.) 2.14.214 (.PTEM.LL10 T=PCOm(NDTEm)

FIK)=F(K).0(m),FORCE(NPTE1,LP)/YGO TO

2 CONTINUECONTINUE

RETJRNEN)

SUBROUTINE 0ANNIO

COMMON /AL/ NumNpT,NumELT,COMMON /86/ NCOOE457,12)COMMON /04/ N8

J000 10 N.1,NUMELTBB 20 181,12

.770 F(N)=F(N)-BGK(N.K)*F(L)800 CONTINUE

GO TO 700500 RETURN

ENO

C.CC

Su3ROUTIN2 SOISTR

COMMON /Ai/ NumNPT.NumFLT, NCONP,NCUT,NSHELTCOMMON /A2/ COOR(G7,2),I.70NN(67,5),NPTYRE(66),NRCOM(66)COMMON /A3/ NNovA(60.5).NO]om(66.4)COMMON /AL/ N030E(57,12)COMMON /12/ F(9.12)COMMON /93/ C11,C12.C13.C22,C23,C33,C44455,C66

NCONP,NCUT,NGMELT COMMON /Cl/ FoRGE(660)COMMON /C2/ 3F(9).0(15)0TF(9)COMMON /01/ 5(9,9),SYSK(15,15).S8(9,9)COMMON /03/ F(182)COMMON /FL/ NFOUR,NF,NF2.NUMFOU,XNF,NFTOF1COMMON /F2/ FORCOF(50)0(A4G(7)0NANGLE

DIMENSION TOISP(151.ST(12) ,DISP(66,41 C 4"s THIS IS A QUADRILATERAL ELEMENTEQUIVALENCE IFORCE(1.11.0ISP(1,1)) CREAL NF. Nr2 CALL OUAOINN)

C 00 12 N=1,12C "P PRINT Or NODAL POINT VARIABLES IF(NC00E(NN.N).E0.0) GO TO 105C ,

K=NSOOE(NN.N)DO 41 I=1,NUMNPT TOISF(N)=F(K)CO 41 J=1,4 GO TO 12

41 DISP(I.J)=0.0 105 TOIsR(N)=0.0KFP.FcuR*1MPRINT=3 12 CONTINUEDo 50 I=1.NUMNIRT ST(K)=Q(0.121IF(CooR(I.1).LT.0) GO TO 50 DO 14 L=1.1200 4 J=1,4 14 ST(K)=ST(K)-SYSK(Kf12,L)*TDISP(L)K=N1001(I,J) Al1=SY3K(14,14)*SYSKI15.15)-SYSK(15,14)*SYSK(14.15)IF(K.E7.01 GO TO 4 A21=SY1K(13,14).SYSK(15,15)-SYSK(15,14)*SYSK(12,151DISR(I,J)=DISP(I,J)+FIK) A31=SYSK(13,14) SYSKI14.15)..SYSK(13,15).SYSK(14,14)

4 CONTINUE '422=SYSK(11,13)*SYSK(15.15)..SYSK(12,15)*SYSK(15,13)IFIHRRINT.GT.0) GO TO 40 A32=Sr1K112,13).SYSK(14,15).-SYSK(14,13)*SYSK(13,15)rIppt%T=5c, A33=SYSK(13,13).SYSK(14,14)-..SYSK(14.13)*SYSK(13.14)NR1TE(b1,2C0) NFR DETA=SYSK(13,13).411-SYSK(14,13)*A21+SYSK(15,131431

200 FoRmr.T(s1 CISFLAC7HcNTS FOR THE FOURIER TFRm, NumBER*,/3// DET1=ST(1),,Ali-ST(2),A21+3T(3)*A3iIs 432Eg,3X,* R-OISPLACEmE4Tx.3x.* Z-DISPLACENENTA,3Y. DET2=-ST(1).421,ST(2).A22-ST(3)A32Zs T-)ISPLACEmENTs.9X,* OTATIONs./) OET3=ST(1)4A3L-ST(2),A324.3Tt3)*A33

40 oRITEA.2,,,I) I, (DISPtI.J),J=1.4) TOISP(13)=DET1 /DETA201 FOl1Arli5.4E18.5/1 TDISP(14)=DET2/DETA50 miARI4T=sPRINT-1 TOISP(15)=DET3/DETA

AR4I,cf=0 CC

1,',7123nM:11C . IF LOADINGS ARE AxISY..mETRIC, NANGLE LESS THAN ZEROC LL=ICONN(NM,3)

IF(NAN;LE.LT.C) GO TO 51 HM=Ii:CNUINN,4)C KK=NUPT4.1C .0.* STORE NOOAL DISPLACEHENT COEFFICIENTS ON FILE, LUN 6,7,0 CO3', (IOC, 1.) = (COOR My 1) +COOR WO.) +COOR (LL, 1)+COOP (Mt1.1) )/ 4.C COOR(KK,2)=KOOR(II.2).+COOR(JJ,2)*COORILL.2)+

IFCNFOJR.LT.Bi GO TO I.I SOOR(M4,2))/4.

IF(NFouR.LT.15) GO TO 2iIRITEI81 claisP(K,L).L=1.4)K=1NUmNPT)

DO 16 NuILJAO=1,4II=ICONN(NN,NQUAC)

CO 1,-) ii JJ=ICONN(NN.NQuA041)1 ARITE(b) C(OISP(K,L),L=10),K=1,NumNPT) IF(Nou40.EQ.4) JJ=ICoNN(NN,L)

CO TO 51 AJ=C00R(JJ.1)-CoOR(II.1)2 wRITF(s) l(DISP(K.L),L=1.:).<=1.NUmNPT) AK=CO1Rtp(K.11-COoR(TI,1)

0J=cuoR(JJ,2)-COOR(II,2)C ... IF NO SOIL ELEMENT IS PRESENT ,NCUT=0 BK=CooR(KK.2)-COOR(II.2)

51 IF(NSUT.E.C) GO TO 300 A.AJ-AKNR1TF(61,36) NFR

36 FORHAT(g1 ELEMENT STRESSES FOR FOURIER TERM NUMBER*,U=3J-HKD=AJK-..AK*F5,1

1 13,//i.119Y,, OIR.OF mAx.s/s EL NOnE RR-STRESS*, RI.,COOP(II,1)2t 2/-STRESS TT-STRESS RZ-STRESS RI-STRESS*, 2I=C00R(II.213s 2Y-STRESS MAX-STRESS MIN-STRESS MAX-SHEAR$1 RJ.CooR(JJ,1)4$ PRIN:.sTRESSsis NO NO NO*,106X, 2J=CODRAJJ.2)5$ FROM R-AXISCOEGW) RK=COORCK<,1)

C 2K=CoOklY,<.2)C *4'0 CALCULATION OF CENTER NODE VARIABLES RC=CRI+RJ+RK)/3.C

DO 300 N4=1.NUMELT2C=17I+7J.2i0/3.I=NluAl

mTyRF=ICoNN(NN,5) J=NOUGOtiIF(ITYPE.SE.NCUT) GO TO 300 IF(NQUAO.EQ.4) J=1C11=FlI,HTYPE) K=5C12=E(2.ITYPEI CC13=n12 IZT=-A TOISP(31)-AKT015(3.J1+AJTOTSP(3,01!1C22.C11 EPRRr(2 TIISP(3.1-2)r0KT)ISP(3*J-2)-RJ+TnIS0(3,0K-211/0C?1=C12 ERZZ=I-A TOI;R( 3' I1)-AK.TOISP(3J.1).AJTCISPI3K1))/0C33=C11 ERRZ=l-A"TDISIR(3 1..214.9T)ISP(3.1...1).-AK.Tn/SF(2.J2)C44=E(704TYPE) 1 ,0,0,T0Isp(3 J-1).AJ.TOISP(3,1(-2)-BJ.TOISp(3K-1))/0C55=C44 ER7T=TZT/0C65=C44 EPRT=-((RJ2K-RK.2J)*TOISD(3.I1s(RK.ZI-RI,21414TOISP(3J)

C 1 +IRI 2.1-RJZI)*TOISP(3.K)+20TZT)/(RC0)C ". INITIALIZE STRAINS u=(TOIsR(3.1-2)+TOISR(3J-2).ToISP(31<-2))/3.C W=( TDISP13.1-1).TDISP(JJ-114,TOISR(3.4(-1))/3.

EP?R=0.0 V=( TiIS2(3°I).TOISR13.J)44OISP(3K))/3.EP22=0.0 EPTI=(u+Nrv)/RCEPTT=0.0 EPRT=EPRT-NF.usRCERR2=0.0 EP2T=EPZT-NFm/RCEPRT=0.0 8 . CALCULATE STRESSESEP2T=G.000 15 N=1,9 ST(1)=C11EPRR*Cl2'EPZZ.C13EPTT

1S ST(N)=0.0 ST(2)=C12.EPRR4,C2EP224,CE3EPTT ,-

IFIICONN(NN.4).E0.0) GO T3 104 ST430=C13EPRR.C23EPEZ*C33EPTT -.4

C ST(4)=C44EPRZ -.1

ST(5)=C55EPRT EPRR=EPRP+ERRST(6)=C66*EPZTIF(NFOUR.GT.1) GO TO 26

EPZZ=EP17+E77EPTT=ITOISR(10)+NFTOISR(1211/RK+EPTT

ST(5)=0.0 LPRZ=FPR7+ERZST(6)=6.0 EPRT=EPRT+CERT-NFTOISP(10)1/RK

26 IF(NAIiSLE.LT.0) GO TO 24 EPZT=EPZI+EZT-NFTOISP(111/RKIFINFOj2....T.5) GO TO 5IF(NOUR.LT.9) GO TO 6 E g" CALCULATE STRESSESIF(NOUR.LT.131 GO TO 7 CIr(4FOUR.LT.17) GO TO 8 ST(1)=C11.EPRIR+C12*EPZZ+C13EPTTwRITE(23) NN,II,JJ.(ST(N)04=1.6) ST(2)=C12EPRR+C22EPZ74:23,EPTIGO TO 24 ST(3)=C13+EPRP+C23EPZZ+C33EPTT

5 WRITE(101 NN.II.JJ.(ST(111,14=1.6) ST(4)=C44.EPRZGO TO 24 ST(5)=G55+ERRT

6 NRITE(23) NN.II.JJ.(ST(N).4=1.6 ST(b)=066+EP2TGO TO 24 IF(N0 '!GLE.LT.0) GO TO 20

7 WRITE(211 NN.II,JJ. (ST(N).N=1,61 IF(NFUR.LT.5) GO TO 42GO TO 24 IF(NF)OR.LI.91 GO TO 43

8 wRITE(22) NN.II,JJ. (ST(N).N=1.6) IF(NFOUR.LT.131 GO TO 44C IF(NFOUR.LT.171 GO TO 45C ". CALCULAT7 PRINCIPAL STRESSES AND DIRECTIONS WRITE(23) NN.II,JJ.(ST(N),N=1.6)

24 C414'4.(ST(1)+ST(211/2. CO TO 23C4O1R=(3T(11-ST(2))/2. 42 WRITE(101 NN,II.JJ,(ST(N).N=1,61Rml4QT(0T141'..2+0m0HP'.2) GO TO 20ST(7)=OMOIRPm0HR 43 WRITE(2C) NN.II.JJ.(ST(N).N=1.6)sT(8)=c4nAR-RmomR GO TO 23ST13)=RYCHR WRTE21AN.-,Li:=CAIINFIST(4)/ST(1)157.2957795)/2.

44GO

ITO (20

) NN.II.JJ.(ST(N1.4=116)

IF(GT(11.LT.S1(2)) GO TO 22 45 wRITE(22) NN,II.JJ.(ST(N),4=1.6)G7 TO 23

22 1,.(ST(4).LT.O.C) GO TO 25C8 ... CALCULATE PRINCIPAL STRESSES ANO DIRECTIONS

ANGLE=17.7-A8SF(ANSLE)GO TO 23 20 CMOHR=AST(1)+ST(21)/2.0

25 ANGLE=-90.C+ATISF(ANGLE) OMOHR=(ST(11-ST(2))/2.023 W?ITE(51.37) NN,II JJ, (ST(L),L=1.9).ANGLE RMOHR=SORT(ST(4)**2+0MOMR"2)37 F1=-441t 513 .9E1Z .4.F1,33/1 ST(71=0M0HcaRm0H2lE C'4TINVE ST(A)=CMO1R-RMOHR

GO TO 370 ST(9)=RMOH2NF(ST(41/T(91

C " THIS IS A TRIANGULAR ELEMENT 1ANGLE=( ASI *57.295745S)/.10

C IF(ST(1).LT.ST(2)1 GO TO 32104 CONTINUE GO TO 33

00 10 N=1.9 32 IFIST(4).LT. 0.0) GO TO 35W=NCOr2E(U4,N) ANGLE= 90.47-AOSF(ANGLE)IF(K.E0.01 GO TO 110 GO TO 33T:ISR(N)=F(K) 35 ANGLE= -90.0+ABSF(ANGLE)GO TO 10 C

110 T3IS('(N)=0.0 C '4', PRINT STRESSES FOR EACH F)URIER COEFFICIENT10 COATINJE C

00 11 L=1,3 33 WRITE(61.37) NN,II,JJ, (ST(1.),L=1.91.ANG;-EI=3+LJ=3+L

300RETURN

K=O+L ENDIt TO/SP(I)=(TOISP(L)(1TOISP(J)+TOISF(K)1/3.0 C

II=ICONN(NN.11 CJJ=IO7f41(44.2) SU1ROUTINE SHLSTR10,..ICON(NN,3)RT=100

CA(II.1) COMMON /01/ NUMNPT.NUmCLT, NCONP,NCUT NSHFLT

zr-4,2^;'(I(.2) COMMON /A2/ COOR(67,21.IGO4N(57.510PTYPE(6610PCOM(661RJ....)7:'(JJ,1) COMMON /05/ ICODE(4,9)2J=C774(JJ,2)R<=01)4(K<.11 Eglg Wi; W5P;;L2:55. ,CS ,XT2.(=C00 4(K.2) COMMON /R4/ E11,E12,E2Z,E44,011,012022.044AJ=RJ-RI COMMON /02/ OGK(182,28)AK=RK-RI COMMON /03/ F(18216.1=2J-ZI COMMON /ES/ AS(8.8)BK=ZK-zI COMMON /F1/ NFOUR,NFOF2.4UMF0J.XNE.NFT.NF/A=AJ-AK COMMON /F2/ FORCCF(50),XANG (7) ,NANGLE.8=3J-9K0. A J `,3<- l< 3..1

CDIMENSION TOISP(15), PN(8). ST(12),STR(66.6)

T7T=-A.TOISP(3I-4K,TOTSP(5)+AJTDISP(9) (UI.PN(5)). (WI,PN(1)). (VTIPN(3)) , (T10214(6)3EP.2.(5.T) isP(1)35K.TOIsP(4)-9JTOTSp(71)/13 (EUstliw(2)),(ETT,PNC4Ii,(STRI1,11,0Gx(1,11)EZ2=(-4T:ISP(2)-AKTOISP(51+AJTOISP(811/0LR2=(-AT3I5P11)+8 TOISP(2)-AK.TOISP(4).0KTDISP(5)

1 .11J+TDISP(7)-8JTDIS2(811/0 C '4'4' INITIALIZATIONEZT=TZT/0ERTs-c(RJ lx-RKZJ).TDISP(3).(RK,o2I-RI.ZK)*TOISR(6) 00 Li NR=1,NUmNPT

1 .(RI.FZJ-RJZI)*TOISR(11.2KTZT)/0 00 11 Im=1,6,RK=(RI.R.DRK)/3.0 11 STR(NR,Im) =0,8.2KsIZI42.1.210/3.0 NFR4NFOUR.1

ML=0 1* NODE 8,6X,8 LONG STRESS 8,3X,8 CIRC. STRESS 822X,00 500 NN=1,NUMELT 2$ SHEAR-STRESS 8,3X.8 LONG MOMENT *,3X,* CIRC MOMENT *.mTyPE.ICCNN(NN,5) 3 2X,8 CROSS MOMENT 81IFINTYPE.LI.NCUT1 GO TO 500 00 1.40 Np=1,NumNFT4L=m1.41 IF(NPTYPE(NP1.NE.4) GO TO 140II=ICON4(NN,1) NRITE(61.201) NP,(STR(NP,N),N=1.E)JJ=ICCN4(NN,2) 201 FORMAT(/3X,I5,6E18.5)RSI.1.0/DOORIII,1) 140 CONTINUERSJ=1.0/CCOP(JJ.1) CREAD(111 I,XT,OR,OZ,XL,SS.CS C IF LOADINGS ARE AXISYMMETRIC , NANGLE LESS THAN ZERODO 6 N=1.9 CK=NO3OEINN,N) IFINANGLE.LT.0) GO TO 310IF(K.E0.01 GO TO 103 CTDISP(N).F(K) C STORE SHELL NODAL STRESS COEFFICIENTS ON FILE, LUN 12GO TO 6 C

103CONTINUE 310

3 TDISP(q)=0.0 WRETURRITE(121 ((STRINP,N),N=1,6).NP=1,NUMNPT)

CN

C ." COMPUTE GENERALIZED DISPLACEMENT COORDINATES, PN8

END

CREA0(181 NN, c(AS(I,J),I=1,8),J=1,61 SUBROUTINE TSTRES00 100 K=1,8 CPN(4)=0.: COMMON /01/ NUMNPT.NUMELT, NCONP,NCUT,NSHELTDO 100 L=1.8 COMMON /32/ COOP(67,2),IC1VN(57,5),NPTYPE(66),NPCOM(66)

100 PN(K)=0N(C)+ASIK,L)*TOrSP(L) COMMON /43/ KNrjVA(66,3).N030M(6°,4/C COMMON /C1/ FORCE(b6.4)C COMMON /02/ OCK(182.24)C "4. COMPUTE 7,TPAINS COMMON /F1/ NFOUR,NP,NF2.4UMFOU,XNE,NFT,NF1

UJ=UIXL'lTI.XL0IPN(7),XLPN(8))/ COMMON /F2/ *014.DoF(50),*A4G(7),NANGLEwJ=WI,xL:253I COMMON /01/ IPRIN1,IPRIN2 .VJ=1I.XLETT CTJ=TI,XL(2.0.PN(7).3.0*XLPN(8)) DIMENSION 8ANGLEC7), DIS(66,4),ST(6),0/SP(66,4).EDIS.I.EP;SI 1 STR(66.6).STRES(66,241EoTTI=R31*(NF*V19.CS.UI.SSWI) EQUIVALENCE (FORCE41.DOISP(1,1/),ISTR(1.11,0GK(1,11).EPTTJ=F,SJ*(N,"'VJ.CS.UJI.SS.WJ) 1 (01.5(1,1),OGK(33,3)),(STRES(1,1),OGK(115,41)

DATA (RAD=57.2957795)GAMIETT-6I.ISS.VI.NP.WI)G14J=ETT-7JolSS*VJ.NF.WJ) REAL NF, NF2X<S7,I,-2.3.RN(7) Cki',;CJ,..8KCS1-6.0*PN(9)XL NF3=NUMF0J-3X<TTI.S.I.(74SI*(NF 2fOI.CS.NF.vI)-SS*TI))(KTTJ=R;J.(.--zsd(Nr2..uJ,csN*4,c))-SSTJ)

C 04. SUM THE DISPLACEMENTSC

x*STI.2.0*PSI*(NF0T/+0SI*NF.SSUI+CS(ETT-RSI.SSVI)) DO 200 NA=1,NAUGLEXK3TJ=2.0.RSJ.INF.TJ.RSJ*14F*SSUJ.CS*(ETT-RSJ4SS.VJ)) D3 19 NP=1,NUMNPT

C 00 19 J=1,4C 'A" COMPUTE STESSES 19 DIS(NP,J1=0.0C XANGLE(NA)=XANGINA)RAO

03 110 N=1,12 NRITE(61.15) XANGLE(NA1110 STIN1=0. IF(IPRINI.E0.01 GO TO 52:

ST(11=E11EPCSI.E12'EPTTI PUNCH 17, XANGLEINA)5T17) =E11.FPCSJtE12.ERTTJ 52 DO 50 N=1,NUMNPTST(2)=E124EPSSI.E22'EPTTI CO 53 L=1,4ST(3)=E12°F_PCSJ.E22EPTTJ 50 OIS(N.L) =0.0ST(3)=E4'.'54mI REWINO 6ST111=E44oC,AmJ REWINO 7ST(4).011*KSS1+012**TTT REWIND 8ST(10)=011**SSJf.D12**TTJ DO 100 NFOUR=1,NUMFOUST(5)=312*XKCSI.022XXTTI 1*(*DRDortNFouR).E0.0.0) GO TO 100ST(11)=012YWSSJ.022.XKITJ NFR=NFOUR-1ST(6)=C4'.XKSTI IFIIFRIN1.E0.0) GO TO 55$1*(121=0XR;TJ PUNCH 54,NFR03 12,3 N=1.6 54 FO;MATUA SUM OF NODAL DISPLACEMENTS UPTO FOURIERf,I5,H.4.6 18 HARmONICSx./1IF(mL.NE.11 GO TO 121 55 IF(NFOUR.LE.NF3) GO TO 57STF'CII,N)=STRIII,N1.ST(N) NRITF(61.56) NFRGO TO 121 56 FORMAT(///8 SUM OF NODAL DISPLACEMENTS UPTO FOURIER*,

121 ST(II,t()=5TR(II,N)+ST(N)/2. 1 15,8 HARmoNICS*,/)123 IF(mL.NE.mSHELT) GO TO 122 C

C C . READ DISPLACEMENT COEFFICIENTS FROM LUN 6,701STR1JJ01/=STR(JJ,N).ST(11)GO TO 120 57 IFINFOUR.LT.I/ GO TO 1

122 STR(JJ.N1=STR(JJ.N).STIM)/2. IF(NFOUR.LT.151 GO TO 2120 CoNTLNUF READ(*) ((DISP(N,I),I=1,4),N=1,NumNPT)

IF(mL.LT.NSHELT) GO TO 503 GO TO 3

500GGONTO To

INUE125 1 REA71(6) 11DISP(N,I),I.1.41,N.I.NUMNPT)

2CGO TO 3RE40171 ((01SP(N,I),I.1.41.H141NUOINPT)

C + PRINT SMELL STRESSE 3 NF=NFOUR-1S FOR EACH FOURIER COEFFICIENT1....'

C CO,,COS(NFXANG(NA))125 WRITE161,200/ NFR $I.SIN(NFxANG(NA)) -4200 FORMAT(A1 SHELL STRESS FOR FOURIER TERM NUMBER*,I3//0 00 90 N1914UNNPT .0

00 83 L=1,4 250 STRES(N.M)=STRES(N.4)+ST(4)00K=41';04(N.L) DO 26G M=5.6IF(K.E0.)) GO TO 70 260 STRE7(N,M)=STRES(N.M)*ST(4).S1IF(L.E0.3) GO TO 60 IF(IPRIN2.EQ.0) GO TO 411DISL)=DIS(N.L).0ISP(N.L)*C0 PUNCH 23. N.(STRES(N.m).m=1.6)GO TO 50 411 F(NFOUR.LE.NF3) GO TO 270

60 DIS'N.L)=01S(N.L)+OISP(N.L)SIr.3 T7 eo

WRITE(61.410) N,II.JJ.(STRES(N01)021,6)

70 OIG(N.L)=0.0 CGO TO 270

80 COVITY),- C THIS IS A QUADRILATERAL ELEMENTIF(N5nR.LE.NF31 GO TO 58 6 DO 7 N)UAD=1.4WRITE(51.25) N.(0IS(N.L).L.1.4)

58 ;F(IRRINI.E0.0) GO TO SOPUNCH 15. N. tO/Stu,L).L=1.4)

90 CONTINDE100 CONTINUE READ(23) 4.II.JJ.IST(J).J=1.6)17 FCRMAT(t TOTAL FOURIER OISPLACEMENTS ATY,F7.3.1 DEGREES*, GO TO 361//A NGOEA.6Y.g R-OISPL.X,iX.$ ZDISPL./.6)(si T-OISPL.F, 31 REA0(10) M.II.JJ.(ST(J).J=1.6)2 61,.1 ',70TATIGt) GO TO 36

15 FORMAT(A1 TOTAL FOURIER DISPLACEMENTS AT :.117.3,* OEGREESS. 32 REA0(20) M,II.JJ.(ST(J).J=1.6)1/1/A 'IO0Es.67,8 R-DISPi.8.6X,$ 2-DISPL.$6)(,* T-DISPL.*, 60 TO 352 68,: R77ATIONA/) 33 READ(21) M.II,JJ.(ST(J).J.1.6)

25 F0,4aT(I0,,E15.5/) 00 TO 3610 F7 ?4LT(Iti,4E15.5) 34 REA0(22) M.II.JJ.(ST(J).J=1.6)200 004TINUE 36 NO1= NOUAO -1

C =1C ", IF NO SOIL ELEMENT IS PRESENT , NCUT=0

DO

CJ=NO4 1.I6.I 6IF(I.GE.5) GO TO 12

IF(NCUT.F.O.G) GO TO 131 STRES(N.J)=STRES(N,J)+ST(I).00

C ... SU4 SOIL ELEMENT STRESSESGO TO 9

12 STFES(NO)=STRES(N.J)+ST(I)SIC 9 CONTINUc

IF(NFOUR.LT.5) GO TO 31IF(NFOUR.LT.9) GO TO 32IF(mFouR.LT.13) GO TO 33IF(NFOUR.LT.17) GO TO 34

29

61

6362

641

CCC

65

CC

23

20

21

2224

00 5)C NA=1.NANGLEXAq0LE( N.)=Y1NG(NA)*RAO00 2R N.1,NDmELT3C 2'? H=1,24STREct.,,m),,c.0rf,(IPL,I.Eo.o) GO TO ElPLir:h 27, XANGLE(NAI%1TEIS1.35) XANGLE(NA)

REwl'i 1]REAINO 2.?REdINO 21RE4I.1l 22REM:NO 2300 TYL ric7u,?=1,NumFOUIFIFO,'C?5(NF)UR).EO.G.0) ;0 TO 3004FR-..:J-;-1IF(I0,IN2.EC.G) GO TO 62PJ.U.:H 53. NFRFO,MAT(z SUM IF ELEMENT STRESSES UPTO FOURIERS,I4,sHARMONICStlIF(NFOUR.LE.NFS) GO TO 65m2ITE1;1,:,4) NFRFORMAT( / //t SUM OF SOIL ELEMENT STRESSES UPTO FOURIER:,IS,i HimONICSss)

READ SOIL ELEMENT STRESS :3EFFICIENTS FROM LUN 13,20,21,22,23

hi= m.:F0;J-100=CCS)Nr7ANG)N41)SI=3:uW"x.:04G(NA)l00 27G N=1,NlmELTLF(/GT.N(N,5).;E.NCUT) GO TO 270IFAICONN)N,41.NE.C) GO TO 5

THIS IS A TRIANGULAR ELEMENTIF(N=OJR.1T.51 GO TO 23IF(F0jR.LT.91 GO TO 20IF(NFOjR.LT.131 GO TO 21IF(NGiR.LT.17) GO TO 22REA11231 H.II.JJ,(ST(J).J=1.6)GO Ti Z4T,Ei0(1,;) 4.II,JJOST(A.J.1.6)GO 73 24READCZC) M.II,JJ.(ST(J).J=1,6)GO TO 24REA0(211 M.II.JJ.(ST(J),J.1,61GO TO 24RE40(22) 4,II.J.1,(ST(J),J=1.6)00 250 M=1,4

CCc

CCC

CCC

J1=J-5IFiNrOUR.LE.NF3) GO TO 66hRITE(61,410) N.II.JJ.(STRES(N.K).K=JI.J)

66 IFLIPRIN7.E0.01 GO TO 7PUNCH 28, N.(STRES(N.K1.K=J1,J)

7 CONTINUE270 C3N7INJE300 CONTINUE35 FORMAT(g1 TOTAL FOURIER SOIL ELEMENT STRESSFS AT A,

1 F7.3.8 DEGREESt,///A ELEM NODES:OX,: RP-STRE.'33:,57.: 72-ST RFS8,2 07.8 TT-STRESSS,SX's RZ-STRESSx.5X,t RT-STFFSSi,5X.X 2TSTRESST,3 1)

27 FORMAT): TOTAL FOURIER SOIL ELEMENT STRESSES ATs,F7.3.IS DEGRrrSt//t ELEM RR-STRESS 77-STRESS$,2A TI-STRtS., R7-STRESS RI-STRESS ZT-STRESS:)

410 FORAT(315,61:15.5/)28 FORmaT(15,6E12.41500 CONTINUE

"' IF NO SHELL ELEMENT IS PRESENT , NCUT3

131 IF(NCUT.E0.13) GO TO 10

s" SUM SHELL ELEMENT STRESSES

DO 600 NA.I,NANGLEXANGLF(NA=xANG(NA).RAOIF(IPRIN?.EQ.0) Go TO 601PUNCH 43, XANGLE(NA)

601 WRITE(61,45) XANGLE(NA)00 600 N=1,NUMNPTCO 610 m=1,6

610 sTREs(N.4).0.0REHINO 12DO 70.0 NFOUR=1.NUMFOUIF(FORCOP(NFOUR).EQ. 0.0) ;0 TO 700NER=NFOUP-1IF(IPRIN?.E0.0) GO TO 682PUNCH 603, NFP

603 FORMAT): SUM OF SHELL N000L STRESSES UPT0*.I5,is FOURIER HARmONICSt)

602 IF(NFOUR.LE.NF3IGO TO 67WRITE(51.604) NFR

604 FORMAT( / //t SUM OF SHELL NODAL STRESSES UPTO..I5.1$ FOURIER HARMONICS: /) 1....

CO. READ SHELL STRESS COEFFICIENTS FROM LUN 12 C)

Finite element analysis of circular cell bulkheads

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