Analysis of Industrial Slabs-On-ground

ANALYSIS OF INDUSTRIAL FLOOR SLABS-ON-GROUND FOR DESIGN PURPOSES

by

KANCHEEPURAM N. 6UNALAN, B.E., M.E.

A DISSERTATION

IN

CIVIL ENGINEERING

Submitted to the Graduate Faculty of Texas Tech University in

Partial Fulfillment of the Requirements of

the Degree of

DOCTOR OF PHILOSOPHY

December, 1986

ACKNOWLEDGEMENTS

The author expresses his deep sense of gratitude to Dr. Warren

K. Wray for his encouragement and guidance throughout the course of

this work. He also thanks Dr. C.V.G. Vallabhan, Dr. James R.

McDonald, and Dr. Necip Guven for their valuable guidance. Thanks

are also due to Dr. K.C. Mehta for willing to be an examiner and to

Dr. D. Chou for all his help. The author is indebted to Dr. E.W.

Kiesling, Chairman of the Department of Civil Engineering, for the

financial assistance throughout the course of his study at Texas

Tech University.

The author thanks his parents for their constant encouragement,

guidance and moral support. He also thanks his wonderful wife,

Duru, for her unselfish support, patience and encouragement.

The author wishes to thank everyone at Terra Testing, Inc. for

their encouragement and assistance. Finally, thanks are also due to

Mrs. Sherry Smith for the wonderful job of typing this manuscript.

n

ABSTRACT

Slab-on-ground foundations refer to ground supported floor

slabs used to transfer loads safely to the subgrade without under

going distress themselves. These foundations have been used in

residential, light commercial and industrial buildings for many

years. Since the loading conditions and magnitudes in industrial

buildings are wery different from those in residential and light

commercial buildings, their design must be approached differently.

Various design procedures have evolved for the thickness design of

industrial floor slabs, but most of them have been developed for a

specific use or have certain limitations. Therefore, there is still

a need for a rational design procedure which will eliminate some of

these limitations.

In order to develop a rational design or analysis procedure, a

parametric study involving slab length, slab width, slab thickness,

modulus of elasticity of soil, aisle width between stacks, stack

loading, and forklift loading was conducted to study their influence

on deflections, bending stresses, bending moments, and shear forces

occurring in the slab. The study was conducted by considering the

influence of stack and forklift loadings, both separately and

together. The values used in the study were over a realistic range.

Regression analysis was performed on the results of 618 individual

problems and equations for the thickness design of industrial floor

ii i

slabs were developed. Correlation coefficients for these equations

ranged from 0.78 to 0.99.

As the modulus of elasticity of the soil is an important

parameter in the design of industrial floor slabs, a survey of eight

practicing engineers and commercial testing laboratories was con

ducted to determine the most practical and economical means of

evaluating the modulus of elasticity of soil.

Equations rather than nomographs have been developed for

maximum bending stresses, maximum shear forces and maximum differen

tial deflection for each loading condition. These equations permit

solving for the required slab thickness to resist the imposed loads.

TV

CONTENTS

ACKNOWLEDGEMENTS ii

ABSTRACT i i i

LIST OF TABLES viii

LIST OF FIGURES x

1. INTRODUCTION 1

1.1 Concrete Floor Slabs 1

1.2 Winkler Foundation 2

1.3 Previous Work 4

1.4 Scope of Research 22

2. A BRIEF INTRODUCTION TO APPLICABLE THEORY 24

2.1 Introduction 24

2.2 A Brief Review of

the Theory of Plates 25

2.3 Finite Element Method 26

2.4 Finite Element Computer

Program, SLAB4 27

3. EVALUATION OF MODULUS OF ELASTICITY OF SOIL 32

3.1 Introduction 32

3.2 Characteristics of the Modulus of Elasticity of Soil 33

3.3 Testing Procedures for Determining Modulus of Elasticity of Soil 37

3.4 Evaluation of the Most Practical and Economical Testina Procedure ^4

4. PARAMETRIC STUDY 49

4.1 Introduction 49

4.2 Material Parameters 50

4.3 Structural Parameters 53

4.4 Utility Parameters 55

4.5 Accomplishment of Parametric Study 66

5. DEVELOPMENT OF REGRESSION EQUATIONS ".. 74

5.1 Introduction 74

5.2 Regression Analysis 74

5.3 Development of Regression Equations 75

5.4 Discussion on Regression Equations 81

5.5 Limitations of Using the Regression Equations 83

5.6 Analysis of the Regression Equations 86

6. DESIGN PROCEDURE USING

THE REGRESSION EQUATIONS 92

6.1 Introduction 92

6.2 Soils Investigation 92

6.3 Safety Factor 93

6.4 Design Procedure 94

7. CONCLUSIONS AND RECOMMENDATIONS 98

7.1 Introduction 98

7.2 Conclusions 98

7.3 Recommendations 101

VI

LIST OF REFERENCES 103

APPENDICES

A. USER' S GUIDE FOR COMPUTER PROGRAM SLAB4 108

B. LISTING OF PROGRAM SLAB4 WITH SAMPLE OUTPUT 122

C. QUESTIONNAIRE FOR SURVEY 184

D. VALUES OF DESIGN PARAMETERS DUE TO FORKLIFT LOADING AT 15 LOCATIONS USED IN QUASI-STATIC ANALYSIS 188

E. COMPARISON OF RESULTS FOR DIFFERENT ASPECT RATIOS OF FINITE ELEMENT 205

F. STRESS, MOMENT, SHEAR, AND DIFFERENTIAL DEFLECTION RESULTS FROM PARAMETRIC STUDY 207

G. MOMENT EQUATIONS 237

H. DESIGN EXAMPLE USING EXISTING PROCEDURES 238

I. DESIGN EXAMPLE USING THE REGRESSION EQUATIONS 243

J. A COMPARISON BETWEEN THE PCA METHOD AND THE REGRESSION EQUATIONS 252

v n

LIST OF TABLES

1.1 Vehicle Categories 17

1.2 Traffic Categories for Design Index 18

3.1 Means of Weights of Relative

Importance Assigned to Variables 46

3.2 Computed Weights and Their Sums 47

4.1 Variation in Values of Design Parameters Due to Variation in E 52

4.2 Maximum Deflection Values Correspondi ng to E Val ues 54

4.3a Comparison of Values of Design Parameters with Stacks Oriented Along the Long and Short Directions of a Slab with an Aspect Ratio of 1.33 57

4.3b Comparison of Values of Design Parameters with Stacks Oriented Along the Long and Short Directions of a Slab with an Aspect Ration of 4.0 58

4.4 Details of Forklifts Used in the Analysis 60

4.5 Ratios of Values of Design Parameters for Forklift Loading Alone at Locations Where Maximum Values Occurred with Respect to Values at Position 1 63

4.6 Ratios of Values of Design Parameters for Stack and Forklift Loadings at Locations Where Maximum Values Occurred with Respect to Values at Position 1 64

4.7 Ratios of Values of Design Parameters with Forklift Truck Oriented Perpendicular to the Aisle with Respect to Values with Forklift Truck Oriented Along the Aisle 67

4.8 Values of Parameters Used in the Parametric Study 70

viii

5.1 R-Squared Values from Regression Analysis for Principal Equations 84

5.2 Increase in Design Parameters Due to Increase in Slab Thickness 91

6.1 Comparison Between Regression Equations and Existing Procedures 97

IX

LIST OF FIGURES

1.1 Subgrade and Slab Stiffness Relationship 5

1.2 Uniform Load Design and Slab

Tensile Stress Graphs 6

1.3 Wheel Loading Design 7

1.4 Design Graph for Axles with Single Wheels 9

1.5 Design Graph for Axles

wi th Dual Wheel s 10

1.6 Design Graph for Post Loads 12

1.7 Effective Load Contact Area Depends on Slab Thickness 13

1.8 Design Curves for Concrete Slabs: Warehouse Fl oors and Open Storage Areas 15

1.9 Design Curves for Concrete Slabs: Warehouse Floors and Open Storage Areas (Category VI, Forkl i fts) 16

1.10 Design Curves for Slab-on-Grade - Central Loadi ng 19

1.11 Design Curves for Slab-

on-Grade - Edge Loading 20

2.1 Def 1 ecti ons Due to Loadi ng 29

2.2 Stresses Due to Loading 30

3.1 Gibson Soil Model 35

3.2 Mixed Stratigraphy Model 36

3.3 Load-settlement Curve from Plate Bearing Tests 39

3.4 Variation of f. with Plasticity Index 42

x

3.5 Impact Value versus Elastic Modulus from Theory and Tests 43

4.1 Locations of Forklift for

Quasi-Static Analysis 62

4.2 Forklift Along and Perpendicular to Aisle 65

4.3 Combination of Parameters for Stack Loading Condition 71

4.4 Combination of Parameters for Forklift Loading Condition 72

4.5 Combination of Parameters for Stack Plus Forklift Loading Condition 73

5.1 Comparison Between Computer Analysis and Regression Equations 87

XI

CHAPTER 1

INTRODUCTION

1.1 Concrete Floor Slabs

A need for an economical foundation for residential and light

commercial buildings following World War II led to the use of ground

supported slabs, usually referred to as slab-on-ground, or slab-on-

grade if the subgrade has been prepared. The term "slab-on-ground"

is applied to both unreinforced and reinforced floor slabs. These

slabs have been grouped (7)* into four categories based on the

amount of reinforcement provided. The four categories are:

1. Plain concrete slabs.

2. Plain, nonstructurally reinforced slabs.

3. Structurally reinforced slabs.

4. Post-tensioned slabs.

Although plain concrete slabs have the advantages of economy

and ease of construction, it has become a practice to provide a

minimal percentage of reinforcement in all plain concrete slabs to

compensate for shrinking effects. Thus, in practice the first two

categories complement one another and therefore when plain concrete

floor slabs are referred to both here and elsewhere, it implies

concrete slabs with shrinkage reinforcement. Plain, nonstructurally

*Numbers in parentheses refer to entries in the List of References

1

2

reinforced slabs have also been found to be economical and have been

successfully used for a wide variety of loading and site conditions.

Structurally reinforced and post-tensioned slabs have been used

where unusual loading or very poor site conditions were anticipated.

In general, a concrete floor is expected to give good service

for many years without deteriorating. In particular, large area

concrete floors for industrial buildings must be designed and

constructed with the greatest possible economy to give trouble free

service. However, in spite of these requirements and in spite of

being in use for many years, their design has been more of an art

than a science. Various design procedures have evolved in recent

years for slab-on-ground foundations in residential and light

commercial buildings (7,15,25,46,53). On the contrary, less atten

tion has been paid to the development of a design procedure for

these foundations in industrial buildings. The few procedures

available to assist in the thickness design of industrial floor

slabs have a shortcoming in common, i.e., the inappropriate modeling

of the soil. All of them model the soil as a Winkler foundation.

Why modeling the soil as a Winkler foundation is inappropriate is

explained in the following section.

1.2 Winkler Foundation

Winkler, in 1867 (52), proposed that the deflection of the soil

surface can be modeled by a simple equation:

p = kw (1.1)

where

p = pressure acting on the soil surface, psi

k = proportionality constant called modulus of subgrade

reaction, pci

w = deflection of loaded region, in.

Over the years engineers have made use of Winkler's model because of

its simplicity in representing the soil in soil-structure inter

action analyses. Many empirical formulae were proposed for k, even

though there is no unique value of k for soil. The modulus of

subgrade reaction is not a property of the soil alone; it also

depends on the rigidity of the structure, duration of loading,

loading type and depth of soil medium. Other limitations of

modeling the soil as a Winkler foundation are discussed below.

For instance, in the field when a load is applied on a semi-

infinite elastic half space the surface deflects not only under the

load but it also deflects in the neighborhood of the load with the

magnitude of deflections diminishing with distance away from the

load. With the Winkler model only the surface under the loaded

region will deflect.

The differential equation for slab on elastic foundation is:

Dv w + kw = p(x,y) (1.2)

where D is the flexural rigidity of the plate. Consider a uniform

slab carrying a constant uniform load, p , over the entire slab.

The solution of the governing equation (1.2) for a free edge condi

tion (zero bending moments and shear forces) is a constant displace

ment indicating that there is no bending moment or shear force in

the slab. We know that this is not true because under any given

loading, the slab will experience some bending moment and shear

4

force. Therefore, the design of slabs on ground with the Winkler

assumption will yield unsafe results if the slab carries fairly

large and uniformly distributed loads.

1.3 Previous Work

Design procedures for industrial floor slabs-on-ground were

primarily based on experience or they were accomplished using one of

the following:

1. The Corps of Engineers method (12).

2. The American Concrete Institute method (2).

3. The Concrete Reinforcing Steel Institute method (10).

These methods were found to be more empirical (based on per

formance experience) than rational by researchers in this area

(29,30) and that there was a need for a rational design procedure.

As a result, a number of attempts have been made and there are a few

procedures available today to assist in the thickness design of

industrial floor slabs. These procedures are briefly discussed

here.

1.3.1 Panak's Method (29,30)

The first reported rational design procedure was that of Panak.

He studied deflections, stresses, and moments in large area concrete

slabs using the discrete element slab theory (21) over a range of

practical variables. From the analysis, he developed graphs shown

in Figures 1.1, 1.2 and 1.3. The graph shown in Figure 1.1 provides

a ratio of the slab stiffness to the modulus of subgrade reaction

D/k, required in subsequent graphs. It is based on assumed

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m.

8 thickness of the slab and known values of concrete modulus and

effective subgrade modulus. The graph shown in Figure 1.2 provides

design bending moments in terms of aisle width, D/k ratio, and

magnitude of uniform loading. The graph shown in Figure 1.3 pro

vides design bending moments for a loaded forklift on the slab in

terms of single wheel load, distance between load wheels, effective

tire footprint diameter and D/k ratio. The work of Panak was the

first step towards developing a rational design procedure for

industrial floor slabs-on-ground and has been incorporated into the

Wire Reinforcement Institute (WRI) and the American Concrete Insti

tute (ACI) methods.

Though still considered to be the most rational design pro

cedure available, Panak's method has shortcomings. The primary

shortcoming is that the soil has been modeled as a Winkler founda

tion. The other limitation of this method is that short term values

of m.odulus of subgrade reaction have been used in the analysis and

even if long term value (needed in the case of clayey soils) can be

determined, the design graphs do not permit their use in design.

Also, it has been reported by Wray (53) that the length and width of

the slab influence the values of design parameters such as moments

which have not been considered by this and other procedures dis

cussed in the following sections.

1.3.2 Portland Cement Association (PCA) Method (28)

The design graphs included in Figures 1.4 and 1.5 are based on

computerized solutions by Packard (27) and on Pickett's formulae

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(31). The graphs were developed for a Poisson's ratio of 0.15 and a

concrete modulus of elasticity of 4000 ksi. The graphs for concen

trated wheel loads were developed considering the flexural strength,

factor of safety, wheel spacing, effective wheel contact area and

modulus of subgrade reaction. Post load graphs, such as the one

shown in Figure 1.6, were developed for other modulus of subgrade

reaction values of 100 pci and 200 pci, based on Hetenyi's method

(17). Figure 1.7 provides effective contact area which is based on

load contact area and assumed thickness of the slab. When analyzing

for dual wheels. Figure 1.5 is used to obtain an equivalent load

factor which is based on dual wheel spacing, effective contact area,

and assumed thickness of the slab. The equivalent load factor is

multiplied with the dual-wheel axle load to obtain the equivalent

single-wheel axle load, which is then used with other values in

Figure 1.4 to arrive at the required thickness of the slab.

In this procedure, the soil beneath the slab is modeled as a

Winkler foundation and also, long term values for the modulus of

subgrade reaction have not been included. Panak (30) reports in his

work that the negative moments in the middle of the aisle for a

uniform loading condition with aisleways approximately 10 ft or less

in width were found to control the design, indicating that the aisle

width and uniform loading are important parameters that need to be

considered. But these parameters have not been considered in

developing this procedure.

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13

LOAD CONTACT AREA, sq in.

Figure 1.7: Effective Load Contact Area Depends on Slab Thickness (after [28])

14 1.3.3 Corps of Engineers Design

Curves (4,13,14)

The design curves shown in Figures 1.8 and 1.9 were produced

from computer solutions based on Westergaard's formula (48) for free

edge stress with some joint transfer ability. The curves were based

on a transfer coefficient of 0.75, an impact factor of 25 percent, a

concrete modulus of elasticity of 4000 ksi and a Poisson's ratio of

0.20. The variables considered are: modulus of rupture at 28 days,

wheel spacing, axle loading, wheel contact area and modulus of

subgrade reaction. To account for different types of vehicles and

traffic volumes. Category I, II, III, IV, V and VI have been ex

pressed in terms of equivalent operation of a basic axle loading.

The basic loading was assumed to be a 25,000 lb single-axle load

with two sets of dual wheels spaced 52 inches apart with 11 inches

between dual wheels. These categories are included in Table 1.1.

Also, a parameter called "design index" has been included to express

various axle loads and traffic volume in terms of relative severity.

Table 1.2 contains these design index values.

In developing the curves, the soil beneath the slab has been

modeled, as a Winkler foundation and, therefore, requires a modulus

of subgrade reaction value. The procedure considers only the wheel

loading condition and even for this, the categories used are based

on some arbitrarily selected loading.

1.3.4 Corps of Engineers Curves (34,35)

The curves shown in Figures 1.10 and 1.11 were developed by

Ringo using Westergaard's formula (48). These curves were plotted

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Storage Areas (after [13])

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Figure 1.9: Design Curves for Concrete Slabs: Warehouse Floors and Open Storage Areas ' (Category VI, Forklifts) (after [13])

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S A F E T Y FACTOR AND MODULUS OF RUPTURE NOT INCLUDED WITHIN THE GRAPH

CONTACT AREA 25 sq in. SHRINKAGE COMPESATING CEMENT CONCRETE SOLID TIRES; TWO PER AXLE

SUBGRADE MODULUS

K 50. pel 100.

S=30 in.

S=30ln.

S=40in.

60 40 30 (4)

WHEEL SLAB THICKNESS H, in. SPACING,in.

Figure 1.10: Design Curves for Slab-on-Grade - Central Loading (after [34])

20

THE GRAPH

SUBGRADE MODULUS K 50. pc»

100. 200. 300.

S=30 in.

5=40 in.

S=60 in.

60 40 30 (4)

WHEEL SPACING, in.

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SLAB THICKNESS H, in.

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r.irvps for Slab-on-

21

for specific use and were based on a transfer coefficient of 0.25,

Poisson's ratio of 0.20, wheel area of 25 sq in. and a concrete

modulus of elasticity of 4000 ksi. The variables that were con

sidered are allowable concrete tensile stress per 1000 lbs of axle

load, wheel spacing, and modulus of subgrade reaction.

The curves have been developed by modeling the soil as a

Winkler foundation and do not include the long term values. Also,

the curves consider only one contact area, namely, 25 sq. in.,

implying that they have been developed only for a specific use. The

curves further consider only the wheel loading condition and do not

include uniform (stack) loading condition.

1.3.5 Interactive Computer Solution (36)

The program has been developed by Ringo and Steenken on a soft

disk for a microcomputer. Westergaard's six loading cases (48,49)

based on the classical Hertz solution have been programmed. The

variables that were considered are modulus of rupture, factor of

safety, joint transfer coefficient, Poisson's ratio, axle load,

wheel spacing, wheel area, aspect ratio of wheel area, and modulus

of subgrade reaction. The program provides the thickness required

for the concrete slab based on the above variables.

The solution is based on modeling the soil as a Winkler founda

tion and requires a value for the modulus of subgrade reaction. It

also considers only the wheel loading condition.

22

1.4 Scope of Research

Although there are a number of design procedures already

available to assist in the thickness design of large area or indus

trial type floor slabs, most of these procedures have been developed

for a specific application or by considering only a limited number

of variables over a limited range of values. Therefore, the purpose

of this dissertation is to analyze the slab-on-ground used for

industrial appilcations and to develop a more rational design or

analysis procedure which will overcome the shortcomings of the

existing design procedures.

Based on extensive literature review, it was found that slab

length, slab width, slab thickness, modulus of elasticity of soil,

aisle width between stacks, stack loading and forklift loading are

considered to be important in the design of an industrial floor

slab. So, in order to develop a rational design or analysis

procedure, it was proposed to conduct a parametric study involving

the above parameters to study their influence on deflections,

bending stresses, bending moments, and shear forces occurring in the

slab.

The study would be conducted over a realistic range of values for

the parameters mentioned above. The soil was modeled as an elastic

continuum and a finite element FORTRAN program called SLAB4 was used

for the analyses. The theory and structure of this program is

discussed in Chapter 2.

In order to model the soil as an elastic continuum, the values

of Poisson's ratio and the modulus of elasticity of the soil are

23

used in the analysis. To enable the design engineer to decide on a

method of evaluating the modulus of elasticity of the soil to be

used for the analyses, a survey was conducted on some of the com

monly available testing procedures. Based on the survey, the most

practical and economical means of evaluating this property is

recommended. Details of the testing procedures considered, the

survey, the analysis, and the recommendation are presented in

Chapter 3.

Details of the parametric study, such as parameters considered,

combinations of these parameters, and range of values used for these

parameters are presented in Chapter 4. Discussion on the regression

analysis used to develop the regression equations and the equations

themselves are presented in Chapter 5. A different design procedure

based on regression equations rather than nomographs is presented in

Chapter 6. Finally, conclusions regarding the results of this study

and how well the principle objectives were achieved, along with some

recommendations, are presented in Chapter 7.

CHAPTER 2

A BRIEF INTRODUCTION TO APPLICABLE THEORY

2.1 Introduction

Application of the principles of soil mechanics to the behavior

of structures in practice has explained the behavior of structures

supported on soil reasonably well. The reason for this is that the

behavior of a structure and the underlying soil are interdependent

and, therefore, they need to be analyzed together. This understand

ing has changed the approach to solving soil-structure interaction

problems. For instance, slab-on-ground foundations are being

analyzed by representing the problem as a plate resting on an

elastic foundation (8,19,20,47,53). The problem of bending of a

plate resting on an elastic foundation can be solved in closed form

only for a relatively small number of boundary conditions and,

therefore, approximate numerical methods have to be used.

A finite element program (a numerical method) was developed

(8,19,20,47,53) to analyze concrete slabs-on-ground by representing

the problem as a plate on an elastic continuum. As the objective of

the study reported herein was to develop a rational design procedure

for thickness design of industrial floor slabs and not to develop a

numerical procedure to conduct the parametric study, the existing

finite element program, with suitable modifications and now called

SLAB4, was used. The structure of the program is explained in

24

25

Section 2.4. Also, a brief introduction to the theory of plates and

the finite element method are given in the following sections.

2.2 A Brief Review of the Theory of Plates

To a large extent, bending properties of a plate depend on its

thickness. There are two types of plates: thin plates and thick

plates. A plate is said to be thin if its ratio of thickness to the

smaller span length is less than 1/20; otherwise it is said to be a

thick plate. Associated with these two types of plates are three

types of problems, namely:

1. Thin plates with small deflections.

2. Thin plates with large deflections.

3. Thick plates.

By small deflections it is meant deflections that are smaller than

or equal to the thickness of the plate.

Analysis of thin plates subjected to lateral loads (loads

applied perpendicularly to the plane of the plate) are commonly

accomplished by using the linear theory which assumes that the

lateral displacements due to loads are small in comparison to its

thickness. This theory has been found to apply very well to rein

forced concrete slabs (43,55).

The linear theory, sometimes referred to as the classical

Kirchoff's theory of plates, is based on the following assumptions:

1. The middle plane is free from deformation.

2. Forces normal to the middle plane of the plate before

deformation remain normal after deformation.

26

3. The normal stresses in the direction perpendicular to the

plane of the plate can be disregarded.

Based on the above assumption, Kirchoff developed a theory in

which all stress components can be expressed by a single variable,

w, the deflection of the middle surface of the plate. The develop

ment and final equations for both isotropic and orthotropic plates

can be found in any standard textbook, (e.g., 43).

2.3 Finite Element Method

The advent of high speed digital computers with the aid of

approximate numerical methods have made it possible to solve prob

lems which were once not possible to solve by hand. Of the various

numerical methods known, the finite difference and the finite

element methods have been used extensively for plate bending prob

lems. The philosophy of the finite element formulation is consid

erably different from that of the finite difference formulation. In

the finite difference method, a numerical approximation is made to

the exact mathematical differential equation governing the problem

by concentrating on a number of selected values of the unknown

function at specified mesh points. In the finite element method,

the plate is divided into a series of small elements and these

elements are assumed to be joined only at specified nodal points.

Continuity, together with equilibrium, are established at these

points.

Of the two methods, the finite element method is often found to

be more adaptable because of such things as variations in material

properties, geometry, etc., can be more conveniently handled. The

27

formulation of this method can be found in any standard textbook

(e.g., 55).

2.4 Finite Element Program, SLAB4

After the development of the finite element method, it did not

take yery long to find applications for this method in the field of

civil engineering. The simplicity of formulation with capabilities

of handling odd geometric shapes and varying material properties has

made this method quite popular. Also, the systematic way in which

the procedure reaches a solution makes it well suited for program

ming on a digital computer.

The method of finite elements has been extended to the problems

of bending of slabs and a computer program was developed by

Zienkiewicz and Cheung (54) to analyze elastic, isotropic and ortho-

tropic slabs. Later it was used by Cheung and Zienkiewicz (8) to

analyze plates and tanks on an elastic (semi-infinite half space)

foundation. For their analyses, they assumed that the plates

remained in contact with the subgrade at all times. They demon

strated that there was very little additional difficulty in modeling

the subgrade as an elastic continuum. Boussinesq's equation (44) is

used to obtain the flexibility matrix of the subgrade, which is then

inverted to obtain the stiffness matrix. The stiffness matrix of

the subgrade is added to the stiffness matrix of the plate and the

unknown displacements are solved for, first and then the stresses

obtained. The program was then used by Wang, Sargious, and Cheung

(47) for the analyses of rigid pavements. The program was modified

to analyze two slabs (simulating two lanes of pavement)

28 simultaneously which were assumed to be connected by dowels that

were 100 percent efficient. Effects of temperature differentials

such as warping (cupping) were also included. This modified program

can handle loss of contact between slab and subgrade due to warping

and pumping. However, a major difficulty in terms of large computer

storage (memory) was required because the overall stiffness matrix

was not banded (due to the stiffness matrix of the subgrade). Huang

(19) developed an iterative scheme which makes the stiffness matrix

banded and solves the excessive storage problem. Huang (20) also

incorporated a scheme which makes use of symmetry, thereby further

reducing the storage required and also the time required to solve a

problem. The reliability of the results of an analysis using this

program was compared by Huang (19) with analytical and experimental

results. Based on the comparison, Huang reports that the deflec

tions predicted by the program checked reasonably well with experi

mental measurements and that the edge stresses checked within 6% of

analytical solution (Figs. 2.1 and 2.2). Wray (53) used the program

for analyzing slab-on-ground foundations for residential and light

commercial buildings supported on expansive soils. He modified the

program to be able to handle non-constant rectangular sections

(slab-on-ground foundations with stiffening beams), and also in

cluded a subroutine to calculate bending moments and shear forces.

The program with the above features was named SLAB2 by Wray (53).

Although all of the features described above are not required

for the analysis of an industrial floor slab, additional modifica

tions were made by the writer to make this program more efficient

29

LOAD y-FREE EDGE

t LOAD 'I UONGITUDINAL

JOINT

^H4iW4^ O

0.1 0.2

0.3

0 4

0.5

E E

I -o ui _i b-liJ O

7in.(l80mm)SLAB, 7,000 lb (31 kN) LOAD

' O

o UJ -J

UJ

o

lO 'O

8 in.(200mm) SLAB, 12,000 lb (53 kN) LOAD

z o » -o UJ - I u.

0

5

10

15

20

i

A ô

X 7^ w

<

r Ôin.

/ (51 cm) 1 1

^ ^ ^

1 •^ i> , i

\

0 .

kf M

) ^

> 0

0.1

0.2

0.3 0.4

0 5

E E

O I -

- I b . UJ O

9 in.(230mm)SLAB, 15,000 lb (67 kN) LOAD

FULL CONTACT -—PARTIAL CONTACT

o EXPERIMENTAL

Figure 2.1: Deflections Due to Loading (after [19])

30

UJ s

&> CO

(/>

< «

to K

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z ^

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400

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7 in. (180 mm) SLAB, 9,000 lb (40 kN) LOAD

• 9'

v" / f

r

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20i!

'^ ^

1.

(51 cm) 1 i . . . .

0 Q -SL, *

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2 3 4

8 in. (200 mm) SLAB, 12,000 lb(53 kN) LOAD

± 9_C

L , ^ - Q D

^ / V

«

rr^ ' ' ' *

••

u Tf

J

I

0

I

2

3

4

9 in. (230mm) SLAB, 15,000 lb (67 kN) LOAD

CO

H CO

-ICJ

I* - t CO

CM U J ^

i

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O S gco t CO (!) UJ Z cc

UJ - ^

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^ ^

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_ l < ^

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THEORETICAL - FULL CONTACT - PARTIAL CONTACT

EXPERIMENTAL o TRANSVERSE '^ LONGITUDINAL

Figure 2.2: Stresses Due to Loading (after [19])

31

when used in a parametric study such as the one being reported

herein. In order to study the behavior of a slab under both the

stack loadings representing the stored materials and the forklift

loadings, the program was modified to permit two loading conditions

of different intensities to be imposed simultaneously. Differential

deflections are an important aspect of the analysis and so an

additional routine to calculate the twenty greatest differential

deflection over distance ratios (A/1) has been added. These results

are now printed along with deflections, bending stresses, bending

moments and shear forces. Deflections, bending moments and shear

forces are printed both in the order of the finite element nodes and

also in ascending order of magnitudes. This program has now been

named SLAB4. A user's guide for program SLAB4 is included as

Appendix A and the program source listing with a sample output is

included as Appendix B.

CHAPTER 3

EVALUATION OF MODULUS OF ELASTICITY OF SOIL

3.1 Introduction

Analysis of slab-on-ground foundations have been carried out in

the past by representing the soil beneath the slab by a system of

springs (a Winkler foundation) having a constant modulus of subgrade

reaction, k^. Because of the simplicity of the model, it has been

quite popular and is still used by many engineers. Even though the

concepts of elasticity apply to soils only in a wery approximate

range, it would be more appropriate to model the soil as an elastic

continuum rather than modeling it as a Winkler foundation, for

reasons discussed in Section 1.2.

To represent the soil as an elastic continuum, the two basic

elastic properties, modulus of elasticity of soil, E and Poisson's

ratio of soil, v^ aJ e required. Despite trying to represent the

soil more realistically, the results of an analysis are only as

meaningful as the values of E chosen. Unfortunately, tabulated

values based on simple correlation are too often used. Therefore,

in order for any analysis to be meaningful, the value of the modulus

of elasticity of the foundation soil needs to be evaluated for the

site rather than assumed.

A number of testing procedures have evolved by which the value

of E can be estimated. Thus, an engineer is faced with the task of

32

33

choosing the type of testing procedure to make this empirical

evaluation. An attempt is made below to present the characteristics

of the modulus of elasticity of soil, some of the commonly available

testing procedures and, in a general way, to recommend the most

practical and economical way of evaluating E .

3.2 Characteristics of the Modulus of Elasticity of Soil

Although in many situations soil is assumed to be homogeneous,

isotropic and linearly elastic, in reality it is far from being so.

It is very difficult to represent the soil with all its complexities

and even with the assumptions, the complexity of the material has

its influence on E in the form of (22):

1. Stress history.

2. Stress level.

3. Soil type.

4. Time (thixotropic, aging and strain-rate effects).

5. Type of loading.

6. Soil disturbance.

Thus, it is clear that there can be no general value of E^ and,

hence, the modulus needs to be evaluated on a site-by-site basis.

The evaluation of E is not an easy task because of the above

factors. Associated with the complexity is its variability in both

the horizontal and vertical directions. The variation of E^ with

depth was first considered by Gibson (16). He considered the

influence of variation of E with depth on the stresses and dis

placements in an isotropic elastic half-space subjected to loading

34

normal to its plane boundary. The model proposed by Gibson, now

called Gibson's model, is given by:

E3 = E^ + n,.z (3.1)

where

E = modulus of elasticity of soil

E^ = modulus of elasticity of soil at the surface

m = slope

z = depth

The model is represented in Figure 3.1. Recently a study of fifteen

buildings in the Houston area for foundation settlements during

construction showed that the Gibson model successfully provided

elastic solutions for a soil (overconsolidated Beaumont clay)

exhibiting increasing undrained modulus with depth (50,51). For the

study, an equivalent constant modulus value was used. The equiv

alent constant modulus value will tend to increase with foundation

size, which is supportive of the Gibson model. However, large mat

and combined footings impact deeper soil masses and, typically,

require higher values of normalized modulus as a function of

increasing footing size. This concept of equivalent Gibson model,

which has shown to be useful for cohesive foundation media in the

Houston area, has subsequently been expanded to consider the

presence of sand layers within the supporting layers (Fig. 3.2).

At the present, variations in the horizontal directions cannot

be handled in any way but to use judgment tempered with experience.

35

f f / ^ / //?/y/////////

CLAY

EQUIVALENT CONSTANT Eg MODEL

GIBSON MODEL Ec= EQ+m-z

Figure 3.1: Gibson Soil Model (after [16])

36

• ^ ^ ^ ^ ^

CLAY

wzm^ SAND

CLAY

Eg FOR CLAY MODEL

Eg FOR MIXED MODEL-

GIBSON MODEL

Figure 3.2: Mixed Stratigraphy Model (after [51])

37

3.3 Testing Procedures for Determining Modulus of Elasticity of Soil

Generally, sophisticated testing procedures are used for

research purposes; on the other hand, simple and commonly available

testing procedures are typically used for obtaining soil properties

to be used for design purposes. Therefore, only the commonly

available testing procedures have been considered here. These

procedures include uniaxial compression test (triaxial test),

unconfined compression test, plate bearing test, California bearing

ratio test, pressuremeter test, static cone test, standard penetra

tion test, and vane shear test. The results of these various tests

have been correlated by others in earlier work to the modulus of

elasticity of soil. The various correlations with references are

given below.

3.3.1 Uniaxial Compression Test

The modulus of elasticity of soil is obtained from the stress-

strain relationship of the soil. The slope of the tangent drawn to

the initial point is called the initial tangent modulus and the

slope of the line joining any two separate points is called the

secant modulus. The initial tangent modulus is reported to be used

quite often, but it has been recommended (22) that the secant

modulus value obtained by picking the initial point and a point

corresponding to 1/2 or 1/3 of the peak deviator stress be used.

38

3.3.2 Unconfined Compression Test

The modulus of elasticity value from this test is said to

correspond to one-half the ratio between the failure stress and its

corresponding strain (38).

3.3.3 Plate Bearing Test

From a series of plate load tests using plates of the same

shape but of different size, a curve is drawn between measured

settlement and loading (Fig. 3.3). Then the slope of this line is

related to E by (5):

[(1 - v^s^/^s^ ^w = '/ ^ ( • '

where

s = settlement, in.

q = loading on footing, psi

B = width of footing, in.

E = modulus of elasticity of soil, psi s

V = Poisson's ratio of soil s

I = influence factor which depends on shape of footing and w

its rigidity (refer to Table 5.4 in Ref. 5)

3.3.4 California Bearing Ratio Test

In situ California bearing ratio value is approximately related

to E^ by (11):

E = 500 X CBR (3.3) s

where

E = modulus of elasticity of soil, psi s

CBR = California Bearing Ratio

39

i

qB ^ Es >^ w

0 0

qB

Figure 3.3: Load-Settlement Curve from Plate Bearing Tests (after [5])

40

3.3.5 Pressuremeter Test

The modulus of elasticity of soil is obtained from the pres

suremeter test using the following relationship (24):

h = ^sp (3.4)

where

a = structural coefficient = 2/3 for clays

= 1/2 for silts

E^ = modulus of elasticity of soil, tsf

E^ = spherical modulus of elasticity, tsf

3.3.6 Static Cone Test

The static cone results are related to E by (37):

E3 = 2q^ (3.5)

where

, Kg/crr

2

E = modulus of elasticity of soil, kg/cm'

q = cone resistance value, kg/cm

3.3.7 Standard Penetration Test

The blow count values of the standard penetration test are

specifically related to E by (5):

E = 10 (N + 15) for sands (3.6)

E = 6 (N + 5) for clayey sands (3.7)

where

E = modulus of elasticity of soil,ksf

N = field blow count/foot

and is generally related to E by (41):

E^ = 130 f. N (3.8)

41

where

2 E = modulus of elasticity of soil, kN/m

f. = constant value

2

The value of f. ranges from 4.0 to 6.0 kN/m , based on the plasti

city of the soil. The relationship between f. and plasticity index

is illustrated in Fig. 3.4.

3.3.8 Vane Shear Test

The relationship between the results of the vane shear test and

E is given by (26):

E, = (T/e I d^) (3.9)

where

T = torque required to generate a rotation of

e = angular rotation

d = diameter of blades

IQ = 2/N H/d ([S^]"^{r*})^r*}

The undrained modulus of a saturated soil is said (26) to be easily

obtained from T vs. e plot.

3.3.9 Clegg Impact Test

The relationship between the results of the Clegg impact test

and E is given by (9): s

L = u.u/ ^Liv;

where

E = 0.07 (CIV)"^ (3.10)

E = modulus of elasticity of soil, MPa s

CIV = Clegg impact value

The above relation is illustrated in Fig. 3.5.

8 i

42

6

4H

2-

• _ • .

• LONDON CLAY o BOULDER CLAY o LAMINATED CLAY D BRACKLESHAM BEDS -I- KEUPER MARL • FLINZ

. 4 (kN/tn )

10 20 30 40 50

— • • .

60 70 PI(%)

A OXFORD CLAY AKIMMERIDGE CLAY • WOOLWICH a READING SUPPER LIAS CLAY

lo 20 30 40 50 60 70 PI(%)

Figure 3.4: Variation of f| with Plasticity Index (after [41])

43

o CL

CO

O O

O

CO < -J Ld

500

200

100

50

20

10

10 psi FROM VAN TIL et aL (45)

#

-107/psi / / / FROM U.C. TESTS ON CEMENT STAB, CRUSHED ROCK.

FROM THEORY Es = .07(IV)'

E 2 ' = 1 . 6 6 ( I V ) ^ - 2 . 9 7

•16* psi (AVE. CONFINING STRESS 85kPa)

I I '

20 40 60 80 100

IMPACT VALUE Figure 3.5: Impact Value versus Elastic Modulus from Theory and Tests (after [9])

44

This procedure, which is relatively new in the United States,

came to the attention of the writer much after the survey, explained

in Section 3.4, was conducted. Therefore, this procedure will not

be found in the survey but has been included here to enable the

reader to be up to date on the procedures available to evaluate the

modulus of elasticity of soil.

3.4 Evaluation of the Most Practical and Economical Testing Procedure

The testing procedures described above have their advantages

and disadvantages, but their relative merits are not apparent to

most users when choosing one testing procedure over another.

Therefore, to enable the design professional to choose an appropri

ate testing procedure, a survey was conducted among eight geotech-

nical testing laboratories located across the state of Texas and

four competent individuals from universities.

The survey included an evaluation procedure based on the

following variables (a) availability, (b) reliability, (c) famili

arity, (d) cost of equipment, (e) cost of test, (f) interpretation

of results, and (7) ease of performance. The relative importance of

each of these seven variables was weighted by three knowledgeable

individuals according to the following scale:

1. Very important.

2. Important.

3. Neither important nor unimportant.

4. Unimportant.

5. ye.ry unimportant.

45

The means of these weights were computed and tabulated. The results

are shown in Table 3.1.

The individual variables considered in the evaluation procedure

were weighted on a scale of 1 to 5 in relation to a testing pro

cedure as shown in the sample questionnaire survey included in

Appendix C. For example, this writer would weigh the availability

of unconfined compression test as "1", since it is very likely that

most geotechnical testing laboratories will have the capability to

conduct this test. On the other hand, this writer would weigh the

availability of the pressuremeter test as "5", since it is wery

unlikely that everybody will have the facility to conduct this test.

The results of the survey were compiled and the means of the

weights of all the variables corresponding to each testing procedure

were calculated. These means were then multiplied by the corres

ponding means of weights of relative importance given in Table 3.1.

For example, the mean of weights corresponding to availability of

unconfined compression test was found to be 2.400. This was mul

tiplied by 1.333, which is the mean of weights of relative impor

tance of availability in the evaluation procedure, to give a value

of 3.199 (which can be rounded off to 3.2). The values thus

obtained were tabulated and are included in Table 3.2. The values

obtained for each variable were numerically added for each testing

procedure.

The lowest sum obtained corresponded to the most practical and

economical testing procedure.

Table 3.1: Means of Weights of Relative Importance Assigned to Variables

46

Variable

Availability

Reliability

Familiarity

Cost of Equipment

Cost of Test

Ease of Interpretation

Ease of Performance

Means of Weights of Relative Importance

1.333

1.333

2.667

3.667

2.333

2.333

2.667

47

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48

From this analysis, the unconfined compression test was found

to be the most practical and economical testing procedure. The

standard penetration test and the uniaxial compression test ranked

second and third, respectively. However, it should be understood

that the findings are based on the survey which reflects the common

practices in this geographic region (southwestern U.S.). Therefore,

even though any one of these procedures could be used to evaluate

the modulus of elasticity of soil, it is recommended that their

respective applicabilities and reliabilities in relation to existing

conditions be clearly understood. For instance, the unconfined

compression test can be performed only on samples with some

cohesion. In addition, disturbance resulting from sampling

procedures in slightly cohesive soils can result in large errors in

the value of E . s

V

"c V

c

CHAPTER 4

PARAMETRIC STUDY

4.1 Introduction

The parameters involved in designing an industrial floor slab'

on-ground can be grouped into: (1) material parameters, (2) struc

tural parameters, and (3) utility parameters. Specifically, these

parameters include:

1. Material parameters

a. Modulus of elasticity of concrete, E

b. Poisson's ratio of concrete,

c. Modulus of elasticity of soil, E^

d. Poisson's ratio of soil, v

2. Structural parameters

a. Slab length, L

b. Slab width, W

c. Slab thickness, h

3. Utility parameters

a. Stack loading, p^

b. Forklift loading, p^

c. Aisle width between stacks, A^

In accomplishing this study, three of the material parameters

were assumed to be constant throughout the analysis. These three

parameters were the modulus of elasticity of concrete, Poisson's

49

50 ratio of concrete, and Poisson's ratio of soil. The values of these

three parameters, together with other assumptions made regarding the

range values and the reasoning behind them, are explained below.

4.2 Material Parameters

4.2.1 Modulus of Elasticity of Concrete, E c

A minimum compressive strength of 4,000 psi at 28 days is

usually recommended for any type of industrial or commercial floors

(40). Although lower strengths have been found to be adequate for

supporting the loads on these floors (40), the additional strength

is required to provide satisfactory resistance to wear. But to

account for variations in weather conditions at time of placing,

poor placing or finishing practices, variations in gradation of

aggregates, etc., a low value of 2,770 psi for compressive strength

was used in Eq. (4.1) to determine the modulus of concrete (3):

E = 57,000 VT (4.1)

where

f = compressive strength of concrete, psi

This provided a modulus of elasticity value of 3,000,000 psi for the

concrete used in the analysis model. The empirical formula given by

Eq. (4.1) can be used for general construction grade concrete,

usually exhibiting a 28 day compressive strength of 2,500 psi or

more (3).

A study involving other modulus of elasticity values namely,

3,600,000 psi and 4,000,000 psi, indicate higher values of design

parameters as compared to those obtained with a modulus of

51

elasticity of 3,000,000 psi. These values are normalized with

respect to values obtained with E = 3,000,000 psi and are reported

in Table 4.1. The equations presented in Chapter 5 are based on an

E value of 3,000,000 psi. However, to enable the use of other

values of E in the range between 3,000,000 psi and 4,000,000 psi, a

modulus factor, E or E has been introduced in the equations for

maximum bending stresses, maximum bending moments and maximum shear

forces. The modulus factor, E or E , can be calculated from Eqs. X y' ^

4.2a and 4.2b for the range, 3000 psi ^ f' ; 5000 psi:

E = 0.0167 W + 0.1208 (4.2a) X c

E = 0.0141 ^ + 0.2565 (4.2b)

y c

However, it should be recognized that its application will be appropriate only within the range of E discussed.

4.2.2 Poisson's ratio of concrete. V c

Poisson's ratio of concrete is known to range between 0.15 and

0.20 (32). A conservative value of 0.15 was used in this study.

4.2.3 Modulus of Elasticity of Soil, E^

A wide variation of this property is reported in the literature

(5,22), ranging from 50 psi to 2,000,000 psi. In order to narrow

this range, a study of maximum slab deflection was conducted by

varying only the value of the modulus of elasticity of soil while

holding all other parameters constant. Percent change in maximum

deflection due to changes in E^ were computed. For example, percent

change in maximum deflection due to change in E^ from 2000 psi to

10,000 psi was calculated as (2.84 in. - 0.57 in.)/2.84 in. x 100 =

Table 4.1: Variation in Values of Design Parameters Due to Variation in E

52

Maximum Differential Deflection

Maxmum Stress in x-direction

Maximum Stress in y-directi on

Maximum Moment in x-directi on

Maximum Moment in y-directi on

Maximum Shear Force in x-direction

Maximum Shear Force in y-Directi on

3,000,000

1.00

1.00

1.00

1.00

1.00

1.00

1.00

E^. psi

3,600,000

1.00

1.18

1.15

1.18

1.15

1.18

1.14

4,000,000

1.00

1.29

1.24

1.29

1.24

1.29

1.22

NOTE: Results were obtained using

slab length, L = 150 ft slab width, W = 100 ft slab thickness, h = 6 in. modulus of elasticity, E = 7500 psi aisle width, A = 5 ft ^ stack loading,^p = 8.0 psi

53

80%. These results are given in Table 4.2. The values of other

parameters used in this study are noted at the end of the table.

For values of E lower than 1,500 psi, the total deflections

were found to exceed 5 in. which were considered to be unreasonable.

Although the percentage change in maximum deflection values appeared

to be significant for values of E greater than 15,000 psi, there

really was no practical change in magnitudes of maximum deflection.

Consequently, as a result of this study, values of 1,500 psi and

15,000 psi were selected as the lower and upper bounds, respec

tively. A value of 7,500 psi was also used as an intermediate

value.

4.2.4 Poisson's Ratio of Soil, v

Poisson's ratio of soil is known to range typically (42) from

0.15 to 0.50, where a value of 0.50 corresponds to a compressible

medium. For similar analysis, values of 0.35 or 0.4 have been used

(5,22,53). Because the magnitude of computed deflections are not

highly sensitive to changes in Poisson's ratio, a constant value of

0.40 was adopted in this study.

4.3 Structural Parameters

4.3.1 Slab Length, L, and Slab Width, W

Slab length and width are usually governed by the requirements

of the user. For industrial warehouses, the width may sometimes

exceed 300 ft and the length may exceed 1000 ft. These dimensions

are much larger than those for slabs in residential and light

commercial buildings. However, in this study, clear dimensions

54

Table 4.2: Maximum Deflection Values Corresponding to E Values

Modulus of Elasticity of Soil, E

(psi) '

1,000

2,000

10,000

20,000

30,000

50,000

Maximum Deflection

(in.)

5.00

2.84 1

0.57

0.28

0.19

0.11 •

Maximum Change in Deflection due to Change in E

(%) '

43

80

51

32

42

NOTE: Results were obtained using

slab length, L = 150 ft slab width, W = 50 ft slab thickness, h = 10 in. aisle width, A = 5 ft stack loading,^p = 4 psi

''^-

I,

n n

tm

55

ranging from 50 ft to 250 ft, which commonly occur in practice, are

considered. The specific combinations of lengths and widths are

assumed to cover the range of aspect ratios (1 to 5) normally

encountered.

4.3.2 Slab Thickness, h

Slab thickness contributes to the stiffness of the system and

is therefore an important parameter in the design of floor slabs.

Generally, the preliminary design involves determination of the

optimum uniform thickness of the slab-on-ground for a specific

loading condition. Based on this information, the decision is then

made as to whether the slab needs to be stiffened by grade beams to

reduce the uniform thickness or to structurally reinforce it.

Because a uniformly thick slab is usually the design objective., in

this study only a constant thickness slab is considered. Even

though a minimum thickness of 6 in. has been recommended (3) for

plain concrete floor slabs-on-ground in industrial warehouses, slab

thicknesses of 4 in. have also been found in the literature (33).

Therefore, in order to cover the ranges most likely to occur in

practice, slab thicknesses of 4 in. and 10 in. were selected as the

lower and upper bounds, respectively, for this study. Slab thick

nesses of 6 in. and 8 in. were also used as intermediate values.

4.4 Utility Parameters

4.4.1 Stack Loading

Stack loadings are due to stored materials. Their magnitudes

and locations are usually determined by the user. However, for this

56 study, the following assumptions were made:

1. Stacks are 5 ft wide when access is from one side alone,

and/or 10 ft wide when access is from both sides.

2. Stacks are assumed to be continuous and parallel to the

long dimension of the slab.

3. Concentrated loads due to rack posts cause only a punching

shear problem (34), which can be readily analyzed using conventional

reinforced concrete design procedures.

The lower and upper bounds of the loading intensity due to stacks

were computed based on information obtained from local warehouses.

The values used were 2 psi and 8 psi, respectively. An intermediate

value of 4 psi was also used.

In order to compute the difference in values obtained between

having the stacks oriented parallel to the long dimension of the

slab and parallel to the short dimension of the slab, a separate

analysis was conducted. The values obtained for the various design

parameters with stacks oriented along the long and short directions

of the slab, respectively, are reported in Tables 4.3a and 4.3b.

The analysis indicates that there will be a significant change

(increase and decrease) to the order of about 500% in the values of

design parameters between having the stacks oriented along the long

side of the slab and along the short side of the slab.

Therefore, if any layout other than having the stacks along the

long direction of the slab is contemplated, care should be exercised

in applying the results of this study.

57

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4.4.2 Forklift Loading

The types and rated capacities of forklifts used in this study

are based on manufacturer's information obtained from local ware

houses and forklift truck dealers and are given in Table 4.4.

Forklifts with rated capacities of 2,000 lbs and 15,000 lbs are used

as lower and upper bounds, respectively. To avoid the problem of

colinearity (which means that a nearly exact linear relation among

the predictor variables exists, causing coefficient estimates to be

inflated and to predict unreasonable values), a third type of

forklift with an intermediate rated capacity of 4,000 lbs was used

in sixteen cases, the details of which are also included in Table

4.4. It is recommended (28) that if information regarding wheel

contact area is not available, it can be roughly approximated for

solid or cushion tires by using:

contact area = tire width x 3 or 4 (4.3)

Therefore, the contact area of the rear tires of the forklift was

calculated by multiplying the tire width by 4. The contact pressure

is then obtained by dividing the single wheel load by the contact

area of the tire. In order to obtain uniform contact pressure under

all four wheels, the contact area of the front wheels was adjusted.

Additionally, forklifts are assumed to operate only along the aisles

between the stack locations, even when the slab is not loaded with

stack loadings.

4.4.3 "Worst Case" Forklift Truck Location

A separate, quasi-static analysis was carried out by systemati

cally placing the forklift loading at different points on one

Table 4.4: Details of Forklifts Used in the Analysis

60

Rated Capaci

Front Cushion Tires

Rear Cushion Tires

Loads, lbs: Empty Condition

Loads, lbs: Fully Loaded Condition

Contact Area, sq. in.

1

Axle Width, in.

Distance between

ty, lbs

O.D., in.

I.D., in.

Width, in.

O.D., in.

I.D., in.

Width, in.

Front Wheels

Rear Wheels

Front Wheels

Rear Wheels

Front Wheels

Rear Wheels

Axles, in.

Basic minimum aisle for right angle stacking, in.

2,

16

11

• 5

13

8

4

2,

2,

4,

2,

31

18

34

43

119

000

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700

700

700

700

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4,

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350

150

350

150

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15.

28

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500

100

500

100

.23

.00

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.00

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NOTE: Details obtained from manufacturer's specifications provided by local dealers.

61

quadrant of a slab and assuming it to be symmetrically loaded in

order to determine the location of the loaded forklift truck that

produced the most severe or "worst case" loading condition. The

locations studied are shown in Fig. 4.1. As the maximum values of

deflection, stress, bending moment and shear force are critical for

the design, locations where maximum values of these design param

eters occurred were identified. The values were then converted into

fractions with respect to values obtained at the first or

"benchmark" location. Because of their relevance, only these

maximum values from the "benchmark" location were tabulated and are

included in the text (Tables 4.5 and 4.6). However, the values of

the design parameters obtained at all 15 locations shown in Fig. 4.1

are included in Appendix D for reference by the interested reader.

These results were then carefully analyzed to arrive at the most

critical location which will produce the largest combination of

values for all the design parameters. Based on the analysis for

aisle width of 10 ft, location 12 was used for forklift loading

alone, and location 14 was used for the combined (stack plus fork-

lift) loading. Similarly, for aisle width of 15 ft, location 6 was

used for forklift loading alone, and location 5 for the combined

(stack plus forklift) loading.

A separate study was also accomplished with the forklift

directed or oriented along the aisle, as well as oriented perpen

dicular to the aisle as in loading or unloading from the stack (Fig.

4.2). The results of this study were then normalized with respect

to values obtained with the forklift oriented along the aisle, and

62

AISLE

O - J CO

<

10 8

AISLE -15 -14

posi"noN I

•13-- L i z l II

AISLE u - J CO

<

50 ft

AISLE WIDTH = 10 ft

Figure 4 .1 : Locations of Forklift for Quasi-static Analysis

63

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65

AISLE

UJ - J en

STACK LOADING

AISLE 1

1—1

i2,

|1

STACK LOADING

AISLE

STACK LOADING

AIS

LE

1 ALONG AISLE 2 PERPENDICULAR TO AISLE

Figure 4.2: Forkl i f t Along and Perpendicular to Aisle

66 are included in Table 4.7. As can be seen, the values obtained with

the forklift truck oriented perpendicular to the aisle (as if it

were loading or unloading the stack) are greater than when the

forklift is oriented along the longitudinal axis of the aisle.

Therefore, the study was conducted with the forklift truck per

pendicular to the aisle. It was also found that the values obtained

for forklift trucks facing the stack with both sides of the aisle

loaded was lower than when there was no load behind the forklift due

to stacks; however, analysis for forklift loadings was carried out

with both sides of the aisle being loaded as it is wery difficult to

generalize such a situation.

4.4.4 Aisle Width Between Stacks

An aisle is a passage between stacks, provided to permit

movement of forklift trucks which are used to handle the material

stored. The width of the aisle provided is governed by the turning

radius of the forklift in use and also by the storage requirements.

Aisle widths of 5 ft, 10 ft, and 15 ft were used in the study

because they are known to be commonly used in warehouses. Although

an aisle width of 5 ft does not permit passage of most forklifts, it

was used in this study as a lower bound to understand the influence

of aisle width on results for stack loading conditions alone.

4.5 Accomplishment of Parametric Study

The parametric study was accomplished with the aid of the

finite element program SLAB4, described in Chapter 2. The study was

conducted in three phases: (1) analysis with stack loading alone.

67

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68

(2) analysis with forklift loading alone, and (3) analysis with

stack plus forklift loading.

These assumptions were made for the study:

1. The slab is monolithic and of constant thickness.

2. The slab is in full contact with the subgrade at all times.

3. The slab is subjected to maximum loading and symmetry could

be used.

4. Since the slab will not be exposed to severe weather, there

will be no significant temperature differential across the thickness

of the slab. Also, perimeter (structural) loading was not con

sidered in this study. The reason it was not considered because the

exterior wall loads are typically carried directly by continuous

perimeter footings and generally the slab is isolated from these

footings by isolation joints. Because of symmetry, only the top

right quadrant of the slab was analyzed. The longest (length)

dimension of the slab was always assumed to be along the y-axis.

The slab was discretized into rectangular elements with aspect

ratios less than three, as larger aspect ratios gave unreasonable

values for the design parameters. Comparison of results for similar

problems with different aspect ratios of element is included in

Appendix E. For more details on computer code, input information,

the reader is referred to the user's guide to SLAB4, given in

Appendix A.

With these basic assumptions, the parametric study was con

ducted in a systematic manner. Although the reasons for and

selected ranges of values of parameters have already been discussed,

69

they have been summarized for convenient reference in Table 4.8.

The parametric study was conducted over the range of values for

parameters indicated in Table 4.8. It was accomplished by varying

one parameter at a time in a specific manner for the three loading

conditions, thereby including all possible combinations. This is

indicated schematically in Figures 4.3, 4.4 and 4.5.

The scope of the study used in the overall problem analysis and

subsequent development of the equations was quite extensive. The

study included a total of 618 cases for all three loading

conditions. The values of the several design parameters as cal

culated by the computer code for each of the 618 cases studied are

included in Appendix F. The results of this analysis included

values of deflections, bending stresses, bending moments and shear

forces. Deflections, bending moments and shear forces were listed

both in the order of finite element nodes and in the ascending order

of magnitude in the computer output for each problem. The results

also included the twenty greatest ratios of differential deflection

over distance (distance between corresponding nodes). However, only

the absolute maximum values of these design parameters were used in

the regression analysis described in Chapter 5.

Table 4.8: Values of Parameters Used in the Parametric Study

70

Parameter Symbol Unit Value

Modulus of Elasticity of Concrete

Poisson's Ratio of Concrete

Modlulus of Elasticity of Soil

Poisson's Ratio of Soil

Slab Length

Slab Width

Slab Thickness

Aisle Width

Wheel Spacing

Stack Loading

Forklift Loading

V

V

psi

psi

3,000,000

0.15

1500, 7500 ,15000

0.40 s

L

W

h

A w

S

Ps

Pf

ft

ft

in.

ft

ft

psi

psi

50, 150, 250

50, 100, 200

4, 6, 8, 10

5, 10, 15

2.5, 3.26, 3.5

2, 4, 8

75, 103, 178 _i

71

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CHAPTER 5

DEVELOPMENT OF REGRESSION EQUATIONS

5.1 Introduction

The results of the parametric study described in Chapter 4

included deflections, bending stresses, bending moments and shear

forces. Having acquired all of this information, the next step was

to relate the design parameters as dependent variables to other

parameters involved, such as slab length, slab width, slab thick

ness, etc., as independent variables. To do this, a regression

analysis, which is a statistical technique for modeling and

investigating the relationship between two or more variables was

used. The details of the regression analysis are described in the

following sections.

5.2 Regression Analysis

Regression analysis was accomplished using a select calling

computer program (18,23). The program is designed for variable

selection in least squares regression models. The selection proce

dure is the Hocking-LaMotte-Leslie method as implemented in

LaMotte's select subroutine. Select subroutine is the core of this

program and the method used guarantees selection of the minimum mean

square error (MSE) model. The program may read or generate a

variable pool not to exceed 80 variables. Variables may be

74

75 generated or read directly by a user written subroutine named

"INPUT." Various output options are available to the user. The

result is either a linear equation of the form:

y = a + ^^1 ^ h \ ^ ^ S- i f - '

or a logarithmic equation of the form:

B- Bp B.- B. y = a^x^ X2 X3 x . "• (5.2)

For more information on the regression analysis, the reader is

referred to References 18 and 23.

5.3 Development of Regression Equations

The regression analysis was carried out by representing the

following parameters as independent variables:

1. Slab length, L, ft

2. Slab width, W, ft

3. Slab thickness, h, ft

4. Modulus of elasticity of soil, E , ksf

5. Aisle width between stacks, A , ft

6. Stack loading, p , ksf

7. Forklift loading, p., ksf

8. Wheel spacing, S, ft

and the following design parameters, one by one, as dependent

variables:

1. Maximum differential deflection, A

2. Maximum bending stress in x-direction, a^

3. Maximum bending stress in y-directi on, o

76 4. Maximum shear force in x-direction, V,

X

5. Maximum shear force in y-directi on, V

The variables are read directly with the aid of an INPUT

subroutine. The logarithmic regression, Eq. (5.2), was used in this

3 -3 study. The magnitude of variables ranged from about 10 to 10 .

If the linear regression (Eq. 5.1) was used, the importance of

_3 variables with magnitudes of 10 would be very much suppressed by

variables with larger magnitudes. In order to avoid this problem,

the logarithm of the variable value was used to bring the magnitude

of each variable to within a unit, whereby the importance of one

variable will not override the importance of another just because of

magnitude. Correlation coefficients for a linear regression were in

the order of 0.55 compared to 0.98 for a logarithmic regression,

indicating that the logarithmic regression is more suited for the

data on hand. Therefore, in the INPUT subroutine, the variables

were first converted to a consistent system of units (kips and feet)

and then converted in terms of their logarithms (base 10) before

being used for the regression analysis.

The results of the regression analysis include regression on

the full model, along with optimal regression for a specified number

of subset sizes. For this study, as an input parameter, the number

of subset sizes was specified as four. This means after regression

on the full model, the program will eliminate variables one by one

in a specific manner up to four variables and regress on the remain

ing variables. These results, along with the corresponding corre

lation coefficients are used in deciding on an appropriate model to

77 represent the data. Analysis of all these results for this study

indicated a decrease in the value of correlation coefficient with

fewer variables, reinforcing the fact that all variables considered

are important in the thickness design of industrial floor slabs and

need to be included. Therefore, the results of the regression on

the full model alone have been taken into account.

Regression equations (stack loading condition alone) were

developed by considering (1) ratio of loaded area to area of slab

instead of aisle width, (2) ratio of loaded area to area of slab for

cases with stacks back to back, (3) aisle width for cases with

stacks back to back, (4) lesser data sets (deleting data sets with

radical values of design parameters), (5) thickness h, in inches,

(6) no thickness, etc. These equations were not superior (based on

correlation coefficient) to those reported in this section; on the

contrary, in some cases they were found to be far less superior.

Therefore, none of these equations have been included here. For the

design procedure, stress equations were found to be suitable and

easy to compare directly with the allowable value. Therefore, only

the stress equations have been included in the text. However,

equations for bending moments for all three loading conditions were

developed and are included in Appendix G. Also, if a reinforced

floor slab is contemplated, the moment equations could be used to

determine the design moments.

The study was carried out for the three loading conditions and,

therefore, there are three sets of equations involved. Equations

78 are listed by the loading condition and in turn by the dependent

variables in the following sections.

Stack Loading Condition

(L)0.44 („)0.62 ( ,0.20 ^^ ^0.78 A = (0.47) 5 (5.3)

(E ,0.99 ( )0.29 ^ s' ^ w'

(L)0.14 (y)O.Ol ( ,0.30 ( )0.27 ( ,0.97 a = (EJ(1026) !!! 5 (5.4)

(, ,0.74

(„)0.18 ( ,0.59 ( ,0.91 o„ = (EJ(30,045) 5 (5.5) y y (L}0-*s (E/-^^ (Ajo-18

(,)0.30 (,)1.82 ( ,0.02 ( ,0.98 V^ = (E J(15.14) ^ 5 (5.6)

(„)0.09(,)1.83( )0.74 V = (E )(1467)- ^ (5.7) y y (,)0.83 (,^)0.57 (, ,0.44

Forklift Loading Condition

(L)°-21 (W)O-" (h)°-^l (,,)'•'' A= (0.12) (5-8)

(, ,0.98 (,y.21 (s)1.81

79

(L)°-l^ {h)0-30 (pj2-04 a = (E )(22.25) 1 / c QX

{W)0-OÊjO-72 (A )0-02 (s)2.04 ^^'^^ J w

(W)°-0^ (h)0-34 (pj2-39 a = (E ){347.71) 1 (5 JQ)

^ {L)0-15 (EjO-72 (A )0-51 (s)3.48 w W

(,)0.28 (, 1.80 ( )0.52 ( ,2.80 V = (E )(13.33) "H 'J. (5.11)

(L)0-21 (E )0-57 (s)3.96

(L)C.03 (,)0.11 (,)1.90 ( .0.22 ( ,2.13

\ = (EJ(0.19) i f (5.12) ' ' (E^)^-^^ (S)2-64

Stack Plus Forklift Loading Condition

(, 0.41 (, 0.62 (,)0.19 )0.15 ( 0 . 7 0 ( 0.19

(0.12) ^ 5 ^ (5.13) ( )0.97 (3)0.26

(L)0.25(,)0.02(,)0.14()1.02()0.55 (3)1.81 a^ = (EJ(10.17) "^ ^ (5.14)

(, )0.69 (p^)0.08

(,)0.06 (,)0.18 ( )0.70 ( ,0.50 (^0.68 (3^0.42 a = (E )(131.92) "^ ^ '- (5.15) y y (L)°-28 (E^)0-'2 ^ ^

80

„)0.39,„0.23„)1.77(,0.81(,0.66( ,0.21,5,1.29 V, = (EJ(C.03) ^ 5 i (5.16)

(„,0.01 (,,1.79 (, ,0.84 ( ,0.59 , ,0.84 ,s,0.19 V = (EJ(3.09) !! 5 ! (5.17)

y y ,,,0.48(^^,0.59

where

L = length of slab, ft

W = width of slab, ft

h = thickness of slab, ft

A = aisle width between stacks, ft w E = modulus of elasticity of soil, ksf s

p = stack loading, ksf

p^ = forklift loading, ksf

S = wheel spacing, ft

E = modulus factor in x-direction x E = modulus factor in y-directi on

A = maximum differential deflection, ft 2

a = maximum bending stress in x-direction, kips/ft X

2 a = maximum bending stress in y-direction, kips/ft

V = maximum shear force in x-direction, kips/ft X

V = maximum shear force in. y-direction, kips/ft y

x-direction corresponds to short direction (width) of slab

y-direction corresponds to long direction (length) of slab

81 5.4 Discussion on Regression Equations

Regression equations in terms of independent parameters have

been developed based on studies conducted. A brief discussion

relating to the importance of a parameter in an equation, its

position in the equation, and the reasoning behind it, are included

here.

The relative significance of a parameter in an equation is

measured by the magnitude of its regression coefficient or exponent

in the equation. For example, in Eq. (5.3) the most significant

variable would be the modulus of elasticity of soil with an exponent

of 0.99 and the least significant parameter would be the thickness

of the slab with an exponent of 0.20. Although exponents of magni

tude less than 0.1 can be considered to be insignificant and the

corresponding parameters excluded from the equations, they have been

included so that the user will feel that all the variables of

significance have been considered and included. Therefore, the user

does not have to contemplate on the relative significance of a

parameter, but could directly use it in the equation. The position

of the parameter (whether it is in the numerator or in the denom

inator) relates whether it would increase or decrease the magnitude

of the design parameter, corresponding to an increase or decrease in

its magnitude. In Eq. (5.3) slab length, slab width, slab thickness

and stack load are in the numerator, implying that an increase in

their magnitude will cause the magnitude of maximum differential

deflection to increase, whereas an increase in the magnitude of

modulus of elasticity of soil and aisle width would tend to decrease

82 the maximum differential deflection. The reason for this is that

larger slabs are more flexible, i.e., deflect more; thicker slabs

contribute to greater deflection (due to different model responses

involved), and stack loadings add to the deflection. On the other

hand, as the soil gets stiffer, i.e., greater modulus of elasticity,

the magnitude of deflection decreases. Also, as the stacks are

placed further apart, i.e., greater aisle width, the magnitude of

deflection midway of aisle width decreases.

Equations for design parameters in the x and y directions

exhibit a certain trend due to plate action. For design parameters

in the x-direction (short direction), an increase in slab length

would increase the magnitude of the design parameters; whereas for

parameters in the y-direction (long direction), an increase in the

magnitude of slab length would decrease their magnitude. This

appears to be logical as the aspect ratio of the entire slab in

creases, magnitudes of design parameters in the x-direction would

increase and those in the y-direction would decrease. However,

there appears to be exceptions to this general rule in Eqs. (5.4),

(5.11), (5.12), (5.14) and (5.16). In Eqs. (5.4), (5.12) and

(5.14), the exponents of variables (W, L and W, respectively) are

less than 0,1, indicating that they very well could be in the

denominator. Also, an exponent less than 0.1 indicates that the

variable is of no great significance to the equation and that it

would not change the magnitude of the design parameter significantly

(in Eq. (5.4) when W = 100 ft, (W)°-°^ = 1.047 and when W = 300 ft,

(H)O-Ol = 1.059), so as not to cause any serious error. However, in

83 Eqs. (5.11) and (5.16), this could be attributed to data used from

the analysis of slabs with aspect ratio of 1.

In Eq. (5.14) the forklift loading appears in the denominator

instead of the numerator, implying that an increase in its magnitude

would decrease the magnitude of the design parameter. This is due

to the reduction in deflections and bending moments in aisles as a

result of forklift loading at the same location in certain cases.

An identical finding is reported by Panak (30).

The correlation coefficient, expressed as "R-squared," measures

how well the regression model fits the data. R-squared values near

zero are expected for completely random data, whereas an R-squared

value of 1.0 would imply all data to fall evenly about the curve of

best fit. R-squared values for the regression equations resulting

from the analysis of the various SLAB4 data sets are listed in Table

5.1 and indicate equations provide a good to very good fit.

5.5 Limitations of Using the Regression Equations

The equations presented in Section 5.3 are for thickness design

of industrial floor slabs subjected to normal loadings anticipated

in a warehouse. These equations have been developed based on

certain assumptions and considerations. Therefore, it is essential

that the limitations on the validity of these equations be clearly

understood before attempting to use them. The assumptions made in

developing these equations are:

1. Stacks are 5 ft wide when access is from one side alone,

and/or 10 ft wide when access is from both sides.

1 m 'I

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85

2. Stacks are assumed to be continuous and parallel to the

long dimension of the slab.

3. Concentrated loads due to rack posts cause only a punching

shear problem which is not considered here.

4. Stresses caused at edges due to moving loads and high stack

loads are recognized but have not been considered here (as these

could not be considered on an individual basis).

5. Aisle width is considered to be uniform.

6. Forklift truck is assumed to operate only in the aisles,

even when the slab is unloaded.

Other features that need to be recognized are:

1. A unit analysis on these equations will not yield conven

tional units for bending stress, shear force or differential

deflection. This appears to be an inherent problem when using

regression analysis to develop equations, but the unit "problem" is

accommodated in the coefficient of regression equation.

2. Input values of all parameters must be in consistent units

of kips and feet. Substituting parameters into equations in any

units other than in kips and feet will produce incorrect results.

However, if units of pounds and inches, for example, are desired,

the conversion must be made after obtaining solutions to the

regression equations.

3. It should be recognized that the equations have been

developed using parameters over a range of magnitudes typically

encountered in warehouses. Therefore, the equations will only

predict reliable values for ranges given in Table 4.8. For values

86

outside the ranges indicated in Table 4.8, the equations may not

give reasonable results.

5.6 Analysis of the Regression Equations

A number of examples were worked out using the regression

equations presented in Section 5.3. The results obtained and the

observations made during this exercise are discussed in detail

below.

As a first step, the reliability of the regression equations

was established. As was reported in Section 2.4, the reliability of

the results of an analysis using the finite element program SLAB4

was established by Huang (19) by comparing results predicted by the

program to field measurements. The results obtained by Huang using

the program were shown to compare reasonably well with experimental

measurements. Therefore, it can be concluded that the results from

the analyses used in developing the equations reported in this study

also can be considered to be reasonable reliably and that the

equations themselves will be equally reliable. However, as a check

on the ability of the regression equations to reproduce the data

used in their formulation, results (deflections, moments and shear

forces) from approximately fifteen cases (formed 3 to 18 percent of

the total cases analyzed for the different loading conditions) of

computer analyses were compared to those predicted by the regression

equations. The results were then plotted as shown in Fig. 5.1. As

can be seen, the points plot reasonably well about the 1:1 line

indicating that the equations represent the results of the analyses

reasonably well. Thus, having shown the equations to produce

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87

0.0 0.1 0.2 0.3 0.4

REGRESSION EQUATION

FOR MAXIMUM DIFFERENTIAL DEFLECTION

0 1000 2000 3000 4000

REGRESSION EQUATION

FOR MAXIMUM STRESS IN X-DIRECTION

Figure 5.1: Comparison Between Computer Analysis and Regression Equations

T ^

88

acceptable results, examples were worked out to analyze the

predictions from these equations. An example used to analyze the

equations is discussed below.

Magnitude of the various parameters used are as follows:

Slab length, L = 250 ft

Slab width, W = 125 ft

Aisle width, A = 10 ft

Modulus of elasticity of soil, E = 1805 psi = 260 ksf

s

Stack loading, p = 8 psi = 1.152 ksf

Forklift loading, p^ = 110 psi = 15.84 ksf

Wheel spacing, S = 37 in. = 3.08 ft

Concrete compressive

strength, f' = 5000 psi

Factor of safety, FS = 1.5

Allowable tensile stress, F. = 7.5 = 353.55 psi = 50.91 ksf

FS

Allowable shear stress, v = 4 \ ^ = 188.56 psi = 27.15 ksf

Concrete modulus of elasticity factors:

E = 0.0167 V 5 0 M + 0.1208 = 1.30 X

E = 0 .0141 VFOOO'+ 0 .2565 = 1.25 y

The design parameters for the three loading conditions, assuming a

slab thickness, h, of 4 in. were found to be:


a = 76.42 ksf > 50.91 ksf N.G. X

a = 26.04 ksf < 50.91 ksf O.K. y

'^m

89

V = 0.23 ksf < 27.15 ksf O.K.

Forklift Loadino Condition

a^ = 16.39 ksf < 50.91 ksf O.K.

a, = 13.12 ksf < 50.91 ksf O.K. y

V = 6.65 ksf < 27.15 ksf O.K.


a = 74.37 ksf > 50.91 ksf N.G. a = 39.63 ksf < 50.91 ksf O.K.

V = 0.20 ksf < 27.15 ksf O.K.

As the 4 in. thick slab was found to be insufficient for the stack

loading condition and stack plus forklift loading condition, the

slab thickness was increased to 6 in. The stresses in a 6 in. thick

slab for the three loading conditions were found to be greater than

those in the 4 in. thick slab. This will appear to be contrary to

what one would intuitively expect, but Panak reported an identical

finding (30). Panak reports that as the slab thickness increases,

bending moments and stresses also increase (bending moment being a

product of bending stiffness and curvature). One reason for this is

at the stack-aisle interface there is a reversal in bending moments

and, as the slab gets thicker, the values of the bending moments

increase. To the contrary, a thinner or more flexible slab will be

able to accommodate this reversal better. Panak (30) reports that

while the phenomenon of moment and stress reductions seems to lead

to the conclusion that a very thin slab is ideal for uniform

loading, the necessity for forklifts and trucks to operate on the

slab and the presence of cracks due to shrinkage and other causes

90

limit the advantage to be gained. But he does not discuss the

effect of increasing thickness for forklift loading on the stresses

due to uniform loading. Results of other procedures discussed in

Section 1.3 indicate that for the forklift loading (which do not

discuss uniform loading condition either), as the slab thickness

increases, stresses decrease. This is due to the difference in

modeling of the subgrade. The other procedures model the subgrade

as a Winkler foundation rather than an elastic half-space and use a

single value for the modulus of subgrade reaction. With such a

model, as the slab becomes more rigid, the deflection becomes more

uniform (tending toward rigid body motion) and, therefore, result in

lower bending moments and stresses. It is also not known if the

other procedures take into account the additional load due to the

increased weight of the slab (due to increased thickness) or con

sider the additional thickness for stiffness calculations alone.

This increase in values of design parameters with an increase

in thickness was observed in all the examples attempted. However,

for every problem, there is a certain "threshold" value of soil

modulus below which even slab thicknesses in excess of 10 inches

will not provide acceptable results for the given loading. For

values of soil modulus above this "threshold" value, any thickness

(4 <^h i l O in.) of slab will provide acceptable stresses. The

above findings with the example problem discussed above are

summarized in Table 5.2.

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CHAPTER 6

DESIGN PROCEDURE USING THE REGRESSION EQUATIONS

6.1 Introduction

A step-by-step procedure is presented in this chapter to assist

the user in using the regression equations. Parameters considered

to be important in the thickness design of industrial floor slabs

are presented in Chapter 4 and the regression equations for maximum

bending stresses, maximum shear forces, and maximum differential

deflection for stack loading, forklift loading, and stack plus

forklift loading conditions are presented in Chapter 5.

The design process begins with the determination of field

conditions and the required site preparation. As this step is

critical to a successful design of an industrial floor slab, a brief

introduction to a soils investigation is presented in the following

section.

6.2 Soils Investigation

A soils investigation at the proposed site should be conducted

to determine: (1) type or types of soil at the site, (2) depths and

distribution of each soil type, (3) consistency of clay soils and

density of granular soils, and (4) modulus of elasticity of the

soils.

92

93 The modulus of elasticity of soil which was shown in Chapter 5

to be the most important of the parameters, can be determined by

using the unconfined compression test which has been recommended in

Chapter 3 as the most practical and economical means of evaluating

this property of the soil. If cohesionless soils are encountered,

then the second choice, namely the standard penetration test, may be

used. However, it is left to the judgment of the geotechnical

engineer to decide on a testing procedure based on availability,

types of soil present and site conditions. It is also left to the

judgment of the geotechnical engineer to decide on the locations and

types of soil to be tested so as to provide realistic values for the

design. Further, it is recommended that special preparation of the

site be carried out if unstable soil conditions are present.

6.3 Safety Factor

A safety factor is incorporated into a design procedure to

account for deviations that are bound to occur between conception

and completion. The selection of a value to use as safety factor

depends on other assumptions, but should result in a design that is

reasonably conservative but not excessively so.

With this concept, various design procedures use different

values for safety factors for different loading conditions. But all

of the values fall within the range between 1.25 and 2.0 (32), where

a safety factor of 2 is generally used for unlimited repetitions of

wheel loads. For normal warehouse conditions, safety factors in the

range between 1.4 to 1.7 are used. In the design examples using the

94

regression equations in Appendix I and Appendix J, a safety factor

of 1 has been assumed.

6.4 Design Procedure

The first step towards design would be to gather information

regarding the soil, structural and utility parameters described in

Chapter 4. Also, the compressive strength of concrete, f , and the

allowable tensile stress of the concrete, f., need to be specified.

Slabs of irregular shape (e.g., L-shaped) should be divided into

rectangles and each rectangle designed individually.

The design process begins by first considering the stack

loading condition. The design steps are as follows:

1. Assume a trial slab thickness. Based on construction

practices, it is recommended that it would not be practical to

construct such a floor slab less than 4 in. thick (33). Therefore,

as a starting point, the trial slab thickness could be assumed as 4

in.

2. Calculate the allowable tensile stress from (1),

f. 11 KT = 7.5^71 (6.1) t allowable c

where f ' = compressive strengh of concrete, psi

3. Maximum shear force was found to occur under the load and

toward the middle of the slab. Therefore, it can be assumed to be a

two-way action and the maximum allowable shear stress for two-way

action is determined using (1)

y = ^ \rr' (6.2) c c

where

V = maximum allowable shear stress, psi c

95 f^ = compressive strength of concrete, psi

4. Calculate the stress that the slab would experience in both

directions due to the known imposed loads using Eqs.(5.4) and (5.5.)

If the allowable value is exceeded, then consider improving the

subgrade (with higher modulus of elasticity value) or increasing the

required strength of concrete, namely, f;. Analysis presented in

Chapter 5 showed that if the soil modulus is inadequate, that even

slab thicknesses in excess of 10 in. will not satisfy bending stress

requirements.

5. Calculate expected maximum shear force in each direction in

the section using Eqs. (5.6) and (5.7).

6. Calculate the maximum design shear stress using

V 1000 v = b;d^-l44 (6.3)

where

V = maximum design shear stress, psi

V = maximum shear force, kips/ft

b = minimum perimeter = (1 + 2d) x 4, ft

d = effective depth = 0.8h, ft

h = thickness of slab, ft

If the maximum design shear stress value is greater than the maximum

allowable shear stress, then consider improving the subgrade or

increasing the required strength of concrete, namely, f'. Analysis

presented in Chapter 5 showed that increasing the slab thickness

will not solve the problem.

96 7. Following the same procedure, check the adequacy of the

section for forklift loading condition and stack plus forklift

loading condition.

8. If there is a limit on the allowable differential deflec

tion for the floor slab, calculate the maximum differential

deflections for the three loading conditions using Eqs. (5.3),

(5.8), (5.13) and compare it to the allowable value. If this

allowable value is exceeded, then consider improving the subgrade,

as increasing the thickness will not solve the problem.

A design example illustrating the procedure is included in

Appendix I. Also, a comparison of the required slab thickness

resulting from the equations to the required slab thicknesses from

three other methods has been made and shown in Table 6.1. Details

of design steps of the other procedures are included in Appendix H.

The design example used to illustrate the PCA method (28) has been

worked out using the regression equations in Appendix J to make a

one-to-one comparison between the two. However, such a one-to-one

comparison was not possible with Panak's method (29,30) and the

Corps of Engineers method (4,13,14) because some of the variables in

the examples fell outside the ranges indicated in Table 4.8.

The design procedure presented here is for the required thick

ness of the slab section to satisfy the expected service loading

conditions. However, to control shrinkage cracking it is often

recommended (13) to provide 0.10 percent (of the cross-sectional

area) distribution steel in both directions.

97

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Chapter 7

CONCLUSIONS AND RECOMMENDATIONS

7.1 Introduction

The principal objective which initiated the work presented here

was to develop a design or analysis procedure which would:

1. Be rational.

2. Be capable of producing values of bending stress, shear

force, and differential deflection resulting from expected service

loads.

3. Be applicable to all industrial floor design in general.

4. Be easy to use.

There are certain limitations to the work presented here. A

brief discussion of these limitations are included. Conclusions

based on analysis of the regression equations presented in Chapter 5

are also summarized here.

7.2 Conclusions

7.2.1 Rational Procedure

All the properties needed for the equations presented in

Chapter 5 can be either directly measured or computed and none have

to be assumed.

98

. ».£«

99 7.2.2 Specific Design Values

The design equations developed in Chapter 5 and incorporated

into the design procedure in Chapter 6 will produce values of

bending stress, shear force, and differential deflection. By this it

is meant that these values can be directly compared with the allow

able values, whereas in the other existing procedures it is not

possible to study the variation in any of these design parameters

corresponding to a change in any of the parameters discussed in

Chapter 4. These procedures yield only the thickness required for a

particular set of parameters.

7.2.3 General Application

The procedure presented herein was not developed for a specific

problem or by considering only specific values for parameters and,

therefore, facilitates its use for a wide range (within those

indicated in Table 4.8) of parameters. These values (values of

design parameters) are not dependent on any type of reinforcing

system and therefore can also be used to design a reinforced or a

post-tensioned floor slab.

7.2.4 Ease of Use

The equations are simple, easy to use, and can be easily pro

grammed for a personal computer or a hand calculator.

7.2.5 Limitations

As discussed in section 5.5, the equations developed here have

been based on certain assumptions, which were made in order to

facilitate the analysis. The assumptions made should be clearly

1 in

100 understood before attempting to use the equations. Also, the

equations were developed by considering a wide range of values for

parameters. Therefore, it should be recognized that the use of

these equations for values of parameters outside the limits given in

Table 4.8 will likely yield unreasonable results.

7.2.6 Conclusions Based on Analysis of the Regression Equations

1. At any given value of soil modulus, with all other

parameters held constant, bending stress is shown to increase as

slab thickness increases.

2. Above some "threshold" value of soil modulus, any thickness

of slab will prove to be acceptable with respect to bending stress

(4 £ h £lO in., E^ £ 15,000 psi).

3. Below this "threshold" value of soil modulus, no slab

thickness ( 4 ^ h £ 10 in.) will provide acceptable results.

4. Conclusions 2 and 3 are contingent upon the given set of

parameters, i.e., the "threshold" value of soil modulus varies with

the magnitudes of the parameters and, thus, the "threshold" value is

not a constant value from one problem to the next.

5. Conclusion 1 is consistent with the previously reported

findings of Panak.

6. With all other parameters held constant, as the value of

soil modulus increases, the resulting bending stresses decrease.

Further, as the value of soil modulus increases, the magnitude of

the resulting bending stress becomes less sensitive to variations of

the ratio of E /h.

101

7. From analyzing the same problem where the only varying

parameters are slab thickness and soil modulus, increasing slab

thickness will not result in an acceptable design if the problem

exceeds the allowable bending stress at a given value of soil

modulus. This is due solely to the increase in section stiffness

resulting from an increase in slab thickness. The only way to

reduce bending stress at any given slab thickness (assuming all

other parameters are held constant) is to improve the value of soil

modulus.

7.3 Recommendations

As with any research work, the work presented here has room for

improvement. Based on the experience gained from this study, the

following recommendations are made:

1. Units of design parameters, namely bending stress, shear

force and differential deflection produced by the regression equa

tions are not of the conventional form typically expected, i.e., a 2

dimensional analysis would not strictly result in units of F/L for

stress, e.g. kips/ft^. This is due to the use of regression

analysis to derive the equations (which considers only the numerical

values and not the units for the solution). It should be recognized

that the unit "problem" is accommodated (or accounted for) in the

coefficient of the regression equation. Thus, it is y/ery important

that only the units specified in Chapter 5 be used in solving the

equations. If other units are desired, the conversion must be made

following the solution of the equation and not before. If this

inconsistency in units is not desired, then it is recommended that

102

the design equations be developed in terms of nondimensional param

eters.

2. Constant thickness monolithic slabs alone were considered

and, therefore, the design parameters at an edge need to be fully

analyzed. This analysis would improve the flexibility of this

procedure.

3. The forklift loading analysis was a quasi-static analysis.

This approach was taken to save computer time without sacrificing

principles of statics or dynamics. However, for future use in a

similar analysis, it is recommended a table of values of design

parameters be developed for various loadings, locations, dimensions,

soil conditions and properties.

4. The work presented here needs to be extended to include

larger stack and forklift loading.

LIST OF REFERENCES

1. American Concrete Institute, "Building Code Requirements for Reinforced Concrete," ACI Committee 318, September 1983.

2. American Concrete Institute, "Recommended Practice for Concrete Floor Slab Construction," ACI Committee 302, August 1968.

3. American Concrete Institute, "Manual of Concrete Practice, "Floor and Slab Construction," Part 1, 1978, pp. 302-303.

4. American Concrete Institute, "Design of Slabs on Grade: State of the Art," Preliminary Draft, 1984.

5. Bowles, J.E., Foundation Analysis and Design, McGraw-Hill Book Co., New York, 1968.

6. Brown, P.T. and Gibson, R.E., "Surface Settlement of a Deep Elastic Column Whose Modulus Increases Linearly with Depth," Canadian Geotechnical Journal, Vol. 9, No. 4, 1972, pp. 467-476.

7. Building Research Advisory Board, "National Research Council Criteria for Selection and Design of Residential Slabs-on-Ground," U.S. National Academy of Sciences, Publication No. 1571, 1968.

8. Cheung, Y.K. and Zienkiewicz, O.C, "Plates and Tanks on Elastic Foundations--An Application of Finite Element Method," International Journal of Solids and Structures, Vol. 1, 1965, pp. 451-461.

9. Clegg, B., "Design Compatible Control of Basecourse Construction," Australian Road Research 13(2), June 1983, pp. 112-22.

10. Concrete Reinforcing Steel Institute Design Handbook, 1968, pp. 426.

11. Crossley, Robert W. and Beckwith, George H., "Subgrade Elastic Modulus for Arizona Pavements," Executive Summary submitted to Arizona Dept. of Transportation, Highways Division, March 1978.

103

104

12. Department of the Army, "Rigid Pavement for Roads, Streets Walks and Open Storage Areas," Technical Manual No. 5-822-6 April 1969.

13. Department of the Army, "Concrete Floor Slabs on Grade Subjected to Heavy Loads," Technical Manual 5-809-12, Construction Engineering Research Laboratory, Champaign, Illinois, April

14. Department of the Army, "Engineering and Design of Pavement for Roads, Streets, Walks and Open Storage Areas," Technical Manual TM 5-822-6, April 1977.

15. Frazer, B.E. and Wardle, L.J., "The Analysis of Stiffened Raft Foundations on Expansive Soil," Proceedings, Symposium on Recent Developments of the Analysis of Soil Behaviour and Their Application to Geotechnical Structures, University of New South Wales, Kensington, N.S.W., Australia, July 1975, pp. 89-98.

16. Gibson, R.E., "Some Results Concerning Displacements and Stresses in a Non-Homogeneous Elastic Half Space," Geotechnique 17, 1967, pp. 58-67.

17. Hetenyi, M., "Beams on Elastic Foundations," The University of Michigan Press, Ann Arbor, Michigan, 1946.

18. Hocking, R.R. and Leslie, R.N., "Selection of the Best Subset in Regression Analysis," Technometrics, Vol. 9, 1967, pp. 531-540.

19. Huang, Y.H., "Finite Element Analysis of Slabs on Elastic Solids," Transportation Engineering Journal, May 1974, pp. 403-416.

20. Huang, Y.H., "Analysis of Symmetrically Loaded Slab on Elastic Solid," Technical Notes, Transportation Engineering Journal, May 1974, pp. 537-541.

21. Hudson, William R. and Hudson, M., "Discrete Element Analysis for Discontinuous Plates," Proceedings of the American Society of Civil Engineers, Journal of the Structural Division, ASCE, October 1968, pp. 2257-2279.

22. Lambe, T. W. and Whitman, R. V., Soil Mechanics, SI Version, 2nd Edition, John Wiley and Sons.

23. LaMotte, L.R. and Hocking, R.R., "Computational Efficiency in the Selection of Regression Variables," Technometrics, Vol. 12, 1970, pp. 83-93.

IF»^

105 24. Lukas, R.G. and de Bussey, B. L., "Pressuremeter and Laboratory

Test Correlations for Clays," Proceedings of the American Society of Civil Engineers, Journal of the Geotechnical Engineering Division, No. GT9, Vol. 102, September 1976, pp. 945-962. ^

25. Lytton, R.L., "Design Criteria for Residential Slabs and Grillage Rafts on Reactive Clay," Report for the Australian Commonwealth Scientific and Industrial Research Organization, Division of Applied Geomechanics, Melbourne, Australia, November 1970.

26. Madhav, M.R. and Ramakrishna, K.S., "Undrained Modulus from Vane Shear Test," Technical Note, Proceedings of the American Society of Civil Engineers, Journal of the Geotechnical Engineering Division, Vol. 103, No. GTll, November 1977, pp. 1337-1340.

27. Packard, R.G., "Computer Program for Airport Pavement Design," SR029P, Portland Cement Association, 1967.

28. Packard, R.G., "Slab Thickness Design for Industrial Concrete Floors on Grade," IS 195.OlD, Portland Cement Association, 1976.

29. Panak, J.J., McCullough F.B. and Treybig, H. J., "Design Procedure for Industrial Slabs Reinforced with Welded Wire Fabric," an interim report prepared for the Wire Reinforcement Institute by Austin Research Engineers, Inc., March 1973.

30. Panak, J.J. and Rauhut, J.B., "Behavior and Design of Industrial Slabs on Grade," American Concrete Institute Journal, May 1975, pp. 219-224.

31. Pickett, G., "A Study of Stresses in the Corner Region of Concrete Pavement Slabs Under Large Corner Loads," Concrete Pavement Design, Portland Cement Association, 1951, pp. 77-86.

32. Pierce, D.M., "A Numerical Method of Analyzing Prestressed Concrete Members Containing Unbonded Tendons," Dissertation presented to the University of Texas at Austin, Texas in partial fulfillment of the requirements for the degree of Doctor of Philosophy, 1968.

33. Rice, P.F., "Design of Concrete Floors on Ground for Warehouse Loadings," Journal of the American Concrete Institute, Vol. 29, No. 2, August 1957, pp. 105-113.

34. Ringo, B.C., "Desian, Construction and Performance of Slabs-on-Grade for an Industry," American Concrete Institute Journal, 1978, pp. 594-602.

»»»>< '

106

35. Ringo, B.C., "Planning, Design and Construction of Slabs on Grade," Design of Industrial Floors, American Concrete Institute, SCM-5, 1983.

36. Ringo, B.C. and Steenken, J.M., "Industrial Floor Slabs: A Thickness Solution," American Concrete Institute, COM 1/83, 1983.

37. Schmertmann, J.H., "Static Cone to Compute Static Settlement Over Sand," Proceedings, American Society of Civil Engineers, Journal of Soil Mechanics and Foundations, Vol. 96, SM3, 1011, May 1970.

38. Simons, N., "Settlement Studies on Two Structures in Norway," Proceedings of the 4th International Conference on Soil Mechanics and Foundation Engineering, London, 1957, p. 431.

39. Singer, F.L., Strength of Materials, 2nd Edition, Harper and Row, New York, 1962.

40. Spear, R.E., "Concrete Floors on Ground," Portland Cement Association Journal, 1978.

41. Stroud, M.A., "Standard Penetration Test in Insensitive Clays and Soft Rocks," Proceedings of the European Symposium on Penetration Testing, Stockholm, June 1974, pp. 367.

42. Terzaghi, K. and Peck, R.B., Soil Mechanics in Engineering Practice, John Wiley and Sons, Inc., New York, 1948.

43. Timoshenko, S. and Woinowsky-krieger, S., Theory of Plates and Shells, 2nd Edition, McGraw-Hill Book Co., New York, 1968.

44. Timoshenko, S. and Goodier, J.N., "Theory of Elasticity," McGraw-Hill Book Co., New York, 1951.

45. Vantil, C.J., McCullough, B.F., Vallerga, B.A. and Hicks. R.G., "Evaluation of AASHO Interim Guides for Design of Pavement Structures," NCHRP Rep. 128, Highway Res. Board, 1972.

46. Walsh, P.F., "The Design of Residential Slabs-on-Ground," Division of Building Research Technical Paper No. 5, Commonwealth Scientific and Industrial Research Organization, Highett, Victoria (Australia), 1974.

47. Wang, S.K., Sargious, M.A., and Cheung, Y.K., "Advanced Analysis of Rigid Pavements," Proceedings of the American Society of Civil Engineers, Transportation Engineering Journal, February 1972, pp. 37-44.

107 48. Westergaard, H.M., "Computation of Stresses in Concrete Roads,"

Proceedings, Highway Research Board, Vol. 5, Part I, 1925, pp. 90-112.

49. Westergaard, H.M., "Stress Concentrations in Plates Loaded Over Small Areas," American Society of Civil Engineers, Transactions, Paper No. 2197, 1943.

50. Williams, C.E. and Focht, J.A. Ill, "Initial Response of Foundations on Stiff Clay," ASCE National Convention, New Orleans, Louisiana, October 1982.

51. Williams, C.E. "Initial Response of Foundations on Mixed Stratigraphies," Transportation Research Board Meeting, Washington, DC, January 1986.

52. Winkler, E., "Die Lehre von Elastizitat und Festigkeit," Prague, The Netherlands, 1867, pp. 182.

53. Wray, W.K., "Development of a Design Procedure for Residential and Light Commercial Slabs-on-Ground Constructed Over Expansive Soils," Dissertation presented to Texas A&M University, College Station, Texas in partial fulfillment of the requirements for the degree of Doctor of Philosophy, 1978.

54. Zienkiewicz, O.C. and Cheung, Y.K., "The Finite Element Method for Analysis of Elastic Isotropic and Orthotropic Slabs," Proceedings, Journal of Institute of Civil Engineers, Vol. 28, 1964, pp. 471-488.

55. Zienkiewicz, O . C , The Finite Element Method in Engineering Science, 2nd Edition, McGraw-Hill Book Co., London, 1971.

APPENDIX A

USER'S GUIDE FOR COMPUTER PROGRAM SLAB4

108

109

A.l Comments on the Use of the Guide for Uata Input for Program SLAM

A.1.1 General Program Notes

1. The data cards must be stacked in the proper order for the

program to run.

2. A consistent set of units must be used for all input data,

e.g., pounds and inches, except where specified otherwise.

3. The input data cards, or sets of data cards, are presented

in the order in which they would be required to be arranged if the

analysis option selected (Option 1, 2, 3 or 4) required that par

ticular data. For example. Option 2 would require the nodal numbers

at which subgrade reaction is assumed to be zero as input data; this

data would be Data Set 7. Option 1 does not require nodal numbers of

zero reaction and, thus, the input data deck for this option would

not contain a Data Set 7.

Data Set 1

Defines the number of problems to be solved (data decks to read

in) without recompiling the source deck.

Data Set 2

Defines the rated capacity of the forklift used in the analysis

and the aisle width in which the forklift operates.

Data Set 3

This data set identifies the problem: slab length and width,

beam depth, edge moisture variation distance, maximum amount of soil

110 movement, the swelling mode, the exponent in the exponential equation

that describes the shape of the swelling soil profile, whether the

structural stiffness will be determined as a stiffened slab section

or as a constant depth section, and if there is a second loading on

the slab.

Data Set 4

1. Provides dimensions of the stiffening beams and their

spacing, dimensions of the slab portion of the total cross-section,

and moments of inertia in both directions.

2. This data set is not included if a constant-depth section is

being analyzed.

Data Set 5

1. The first card of this two-card data set identifies the

number of slabs to be analyzed (1 or 2), material properties

(Poisson's ratio and modulus of elasticity of soil and concrete), and

several programming constants, e.g., type of symmetry, if any;

presence of non-contact locations; if pre-calculated deflections are

to be input; and if punched output is desired.

2. The second card defines the number of x and y coordinates

occurring in each slab, the number of iterations to be allowed in

establishing subgrade contact, the number of nodes and the nodal

numbers at which output calculations are to be printed.

3. The coordinates in the x-direction begin at zero and in

crease from left to right. Coordinates at the joint must be counted

twice if two slabs are being analyzed. Y coordinates also begin at

zero and increase from bottom to top. The number and coordinates of

n im

111

the y-direction nodes is the same whether one or two slabs are being

analyzed.

Data Set 6

The coordinates of the x- and y-direction nodes are read in

according to F8.3 format. If two slabs are being analyzed, coor

dinates for the nodes at the joint must be read in twice. More than

one card may be necessary to read in all of the coordinates.

Data Set 7

If NOTCON is not zero, the nodal numbers at which there is no

soil-slab contact are input in accordance with 15 format. More than

one card may be necessary to define all nodal numbers.

Data Set 8

This single card data set identifies the total number of nodes

at which a gap exists between the soil and the slab, the number of

elements experiencing loading, the amount of loading (as a pressure,

e.g., psi or psf), whether temperature is to be considered in the

solution, the temperature difference between the slab top and bottom,

the number of nodes at which convergence will be checked, and itera

tion and tolerance limits.

Data Set 9

Input the nodal numbers at which convergence will be checked in

accordance with 15 format. More than one card may be required to

input all check nodes.

Data Set 10

This Data Set reads in the amount of gap previously calculated

as existing between the slab and subgrade at each node where a gap

112 exists. More than one data card may be required to read in all gap

data. If NREAD = 0, there are no previously calculated gaps and this

Data Set will not be included in the problem data deck.

Data Set 11

If gaps exist at specified nodes, this Data Set identifies those

nodes in 15 format. More than one card may be necessary to identify

all nodal locations. If NREAD = 2, this Data Set is not needed and

will not be included in the problem data deck.

Data Set 12

This Data Set assigns the magnitude of the gap at each location

specified in Data Set 11. The input format is F8.4 and more than one

card may be required to input all data. There must be as many gap

values as there are gap locations in Data Set 11. If NREAD = 2, this

Data Set is not needed and will not be included in the problem deck.

Data Set 13

If a uniformly distributed load acting over the entire slab

surface in addition to any live loading specified in Data Set 8 is to

be considered in the analysis, set NWT = -1 and the uniformly dis

tributed load will be read in on a single card in F8.3 format.

Data Set 14

1. The elemental numbers which are to receive some loading 0 are

identified in 15 format. More than one card may be necessary to

identify all loaded elements.

2. Elemental numbers increase from bottom to top and from left

to right. There are (NX-1) x NY-1) elem.ents.

113 Data Set 15

1. The area being loaded by Q is described by XDA and YDA. The

load is distributed over the elemental area from the center of the

rectangular element on a fractional basis in both directions ranging

from -1.00 to +1.00. For example, if the load were to be applied

over the rightmost 75 percent of the element in the x-direction, the

input values would be: XDA(l) = -0.50 and XDA(2) = +1.00. If the

same load were to cover only the bottom half of the element in the

y-direction, the input values would be: YDA(l) = -1.00 and YDA(2) =

0.00. If the entire element is to be loaded, the load distribution

would be -1.00 and +1.00 in both directions.

2. The distribution cards must be stacked in the proper order

to ensure distribution of load over the corresponding element, NL(I).

Data Set 16

Defines the number of elements on which the second loading Q2,

acts in 15 format and the loading intensity in F10.5 format.

Data Set 17

The elemental numbers which are to receive the second loading Q2

are identified in 15 format. More than one card may be necessary to

identify all loaded elements.

Data Set 18

1. The area being loaded by Q2 is described by XDA2 and YDA2.

The load is distributed over the elemental area from the center of

the rectangular element on a fractional basis in both directions

ranging from -1.00 to +1.00 (Refer to data set No. 15).

114

2. The distribution cards must be stacked in the proper order

to ensure distribution of load over the corresponding element,

NFL(I).

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APPENDIX B

LISTING OF PROGRAM SLAB4 WITH SAMPLE OUTPUT

122

123

/ / JOB l M K y » V 0 , U ^ 9 , 2 O , 3 0 , 0 9 l , • 6 U N A L A N • , C L A S S = ^ , l » e 6 I O N = 5 1 2 0 I C / / EKEC FORTGCL»PARM,FORT=»MOSOURCE» / / S V S I N 00 *

C C' c c c c c c c c c c c c c c c c c c c c c< c c c c c c c c c c c c c c c c c c c c c c

THE ORIGIMAL APPLICAFION OF THIS PROCtAN I S DESCRIBED I N TWO PAPERS BV THE AUTHOR, V .HUANG: 111 - F I < I I T £ ELEHENT ANALrS IS OF SLABS ON E L A S T I C S O L I D S * . TRANSPORTATION ENGINEERING JOURNAL OF ASCEf VOL 1 0 0 , NO. T E 2 8 , HAV, 1 9 7 4 , PAGES 4 0 3 - 4 0 6 ; AND C2I -RECTANGULAR PLATES PARTIALLY SUPPORTED ON AN ELASTIC HALF S P A C E - , PRO-CEEDINGSt 1ST INTERNATIONAL CONFERENCE ON COMPUTATIONAL METHODS IN NONLINEAR ^tECHANICSt 1 9 7 4 , PAGES 4 0 5 > 4 1 4 .

THE ORIG INAL PROGRAM BY HUANG WAS MODIF IED AND GIVEN THE HUANG WAS M O D I F I E D AND GIVEN THE NAME - S L A B 2 " BV W. K. WRAV AND R. L . LYTTON, J U L Y , 1 9 7 7 .

TSLAB2" WAS nOOIF IED TO I T S PRESENT F3RM AND GIVEN THE NAME - S L A B 4 - BY K. N . GUNALAN AND W. K. WRAY* DECEMBER, 1 9 8 6 . A DESCRIPTION OF THE MODIF ICATIONS AND EXAMPLE PROBLEMS ARE GIVEN I N K. N . GUNALAN'S DOCTORAL DISSERTATION E N T I T L E D : - ANALYSIS OF I N D U S T R I A L FLOOR SLABS-ON-GROUND FOR DESIGN PURPOSES-, AT TEXAS TECH U N I V E R S I T Y , LUBBOCK, TEXAS, 1 9 8 6 .

THIS PROGRAM PROVIDES SOLUTIONS FOR THE DEFLECTIONS, STRE S S E S , BENDING MOMENTS, AND SHEAR FORCES DUE TO LOADING AND/OR WARPING I N A SINGLE RECTANGULAR SLAB, OR TWO SLABS CONNECTED BY DOWEL BARS AT THE J O I N T , RESTING ON A FOUNDAI ION OF THE ELASTIC SOLID TYPE.

THE SLAB VARIOUS S I Z E S . THE F I N I T E ELE THE SOLUTION I CONSECUTIVELY TO RIGHT ALONG BARS AT THE JO T W I C E , ONE FOR THE DOWELS ARE THE J O I N T ARE E ITHER OR BOTH SLAB MAY BE CO DEFLECTIONS DU E ITHER COMBINE

S ARE DIVIDED HOWEVER* FOR

MENTt THE SOLU S TERMINATED. FROM BOTTOM TO THE X AXIS. INT, EACH NODE THE LEFT SLAB ASSUMED loot

THE SAME FOR B SLABS, AND TH

MPUTED. THE P E TO DEAD LOAD D OR SEPARATEL

INTO RECTAN ASPECT RAT

TIONS WOULD THE ELEMEN TOP ALONG

IF TWO SLAB AT THE DOW AND THE OT

EFFICIENT, OTH SLABS. E STRESSES ROGRAN CAN V TEMPERATU Y.

GULAR FINITE ELEMENTS OF IO*S SREATER THAN THREE FOR BE ERRONEOUS AND THEREFORE

TS AND NODES ARE NUMBERED THE Y AXIS AND FROM LEFT S ARE CONNECTED BY DOWEL ELLED JOINT MUST BE NUMBERED HER F3R THE RIGHT SLAB. SO THAT THE DEFLECTIONS AT LOADS MAY BE APPLIED TO AT ANY NODE IN EITHER DETERMINE THE STRESSES AND RE WARPING* OR LIVE LOAD*

r*00000010 00000020

r*000O003O *00000040 *OO00O0SO #00000060 *00000070 *00000080 *0 0000090 *00000100 *00000110 *00000120 *00000130 *00000140 *000001S0 *00000160 *00000170 *00000180 *00000190 *00000200 *00000210 *00000220 *00000230 *00000240 *000002SO *00000260 *00000270 *00000280 *00000290 *00000300 *00000310 *000 00320 *00000330 *00000340 *000003S0 *00000360 *00000370 *00000380 *00000390 *00000400 *00000410 *00000420 *00000430 *00000440 *00000450 *00000460

124

c

c

r I.e. w I " ! '•"OSRAM CAN NOW HANDLE TWO L3ADING CONDITIONS S I H U L T A N E O - » 0 0 0 0 0 4 7 0 C " S L Y * E . G . , STACK LOADING CONDITION TOGETHER WITH A FORKLIFT * 0 0 0 0 0 4 8 0 I L O A D I N G . * 0 0 0 0 0 4 9 0

. , „ , *00000500 C THE PROGRAM L I S T S DEFLECTIONS* MOMENTS AND SHEAR FORCES BOTH * 0 0 0 0 0 S 1 0 C I N THE ORDER OF NODES AND I N THE ASCENDING ORDER OF NAGNITUDE. • 0 0 0 0 0 S 2 0 C T H I S I S DONE WITH THE AID OF A IHSL LIBRARY CALLED VSRTR AVAILABLE * 0 0 f t 0 0 5 3 0 C AT TEXAS TECH U N I V E R S I T Y . * 0 0 0 0 0 5 4 0

*00000550 C THE >»ROSRAM NOW PROVIDES THE TWENTY MAXIMUM DIFFERENTIAL * 0 0 0 0 0 5 6 0 C DEFLECTION R A T I O ' S TOGETHER WITH NODE NUMBERS* DISTANCE BETWEEN T H E * 0 0 0 0 0 S 7 0 C NODES, D I F F E R E N T I A L DEFLECTION BETWEEN THE N3DES ETC. * 0 0 0 0 0 5 8 0

* 0 0 0 0 0 S 9 0 C THE PROGRAM PROVIDES THE FOLLOWING FOUR OPTIONS: * 0 0 0 0 0 6 0 0 C OPTION 1 : SLAB AND SUBGRADE ARE I N FULL CONTACT* AS * 0 0 0 0 0 6 1 0 C ORIGINALLY ASSUMED BY P I C K E T T . — > SET NOTCON»00000620 C TO 0* NWT TO 0 * AND CYCLE TO 1 . * 0 0 0 0 0 6 3 0 C OPTION 2 : SLAB AND SUBGRADE ARE I N FULL CONTACT AT * 0 0 0 0 0 6 4 0 C SOME POINTS BUT COMPLETELY OUT OF CONTACT AT * 0 0 0 0 0 6 5 0 C THE REMAINING POINTS BECAUSE OF LARGE GAPS * 0 0 0 0 0 6 6 0 C BETWEEN THE SLAB AND TME SUBGRADE. — > SET * 0 0 0 0 0 6 7 0 C NOTCON TO THE NUMBER OF POINTS NOT I N CON- * 0 0 0 0 0 6 8 0 C TACT* NGAP TO 0 * NWT T3 0 * AND NCYCLE TO 1 . * 0 0 0 0 0 6 9 0 C OPTION 3 : SLAB AND SUBRAOE MAY OR MAY NOT BE I N CON- * 0 0 0 0 0 7 0 0 C TACT BECAUSE OF WARPING OF THE SLAB. WHEN * 0 0 0 0 0 7 1 0 C THE SLAB I S REMOVED* T)« E SUBGRADE WILL FORM * 0 0 0 0 0 7 2 0 C A SMOOTH SURFACE WITH « 0 DEPRESSIONS OR I N I - *0O0O073O C T I A L GAPS. — > SET NOTCON TO 0* NGAP TO 0 * ^ 0 0 0 0 0 7 4 0 C NCYCLE TO MAXIMUM NUMBER OF CYCLES FDR CHECK-*000007SO C ING CONTACT* * 0 0 0 0 0 7 6 0 C OPTION 4 : WHEN THE SLAB I S REMOVED* THE SUBGRADE WILL * 0 0 0 0 0 7 7 0 C NOT FORM A SMOOTH SURFACE BUT SHOWS IRREGULAR*00000780 C DEFORMATION. — > SET NOTCON TO 0 * NGAP TO * 0 0 0 0 0 7 9 0 C NUMBER OF NODES WITH I N I T I A L GAPS* NCYCLE TO * 0 0 0 0 0 8 0 0 C MAXIMUM NUMBER OF CYCLES FOR CHECKING C a N T A C T * 0 0 0 0 0 8 1 0 C * 0 0 0 0 0 8 2 0 C * 0 0 0 0 0 8 3 0 C * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 0 0 0 0 0 8 4 0 C * 0 0 0 0 0 8 5 0 C — — I W P U T DATA: * 0 0 0 0 0 8 6 0 C NORMALLY* INPUT DATA I S I N CONSISTENT UNITS OF POUNDS* * 0 0 0 0 0 3 7 0 C I N C H E S * POUNDS PER SOUARE INCH ( P S I l * E T C . HOWEVER* SLAB WIDTH A N O * 0 0 0 0 0 8 8 0 C LENGTH OF THE SLAB AND THEIR RESPECTIVE NODAL DISTANCES ARE INPUT * 0 0 0 0 0 8 9 0 C I N F E E T . THE INPUT DIMENSIONS I I N F E E T I ARE CONVERTED INTERNALLY * 0 0 0 0 0 9 0 0 C TO INCHES FOR PROGRAM CALCULATIONS. * 0 0 0 0 0 9 I 0 C * 0 0 0 0 0 9 2 0 C * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ^ * * * * * * * * * * * * * * * * * * * * * 0 0 0 009 30 C * 0 0 0 0 a 9 4 0 C NOTES ON DIMENSIONS TO BE S P E C I F I E D BY THE USERS: * 0 0 0 0 0 9 5 0

125

THE DIMENSIONS OF C AND G SHOULD NOT BE LESS THAN:

THE DIMENSION

THE THE

DIMENSION DIMENSION

THE DIMENSION

N 0 B ( 2 I = INX14-NX2 I *NY*NB OF H AN HD SHOULD NOT BE LESS THAN:

NOO s CNONY4- l l *NONY/2 ; I F N S L A B ^ l * NONY = N X l * N Y ; I F NSLAB»2* NONY = I N X 1 ^ N X 2 - 1 I * N Y

OF CO SHALL NOT BE LESS THAN N r * N B * 3 OF F SHALL NOT BE LESS THAN N 0 2 * WHERE:

N02 - I N X 1 « - N X 2 I * N Y * 3 OF D F * GAP* P P F * PF* CURL* FO* DEF* AB* AND NCC

SHOULD NOT BE LESS THAN INX1«-<I X2 l *NY

THE DIMENSIONS INDICATED ABOVE ARE A MINIMUM* THOUGH LARGER DIMENSIONS MAY BE REQUIRED.

P R I N C I P A L NOTATION FOR SLAB2

ABI I AOB ADB2 AISLWD ASPACE

ATB ATB2 B

BDA BDA2 BEAML* BEAMS

BEAMLL

BEAMLW

BEAMSL

BEAMSW

BSPACE

CC I

C0( I CURLtNGAPI)

DISTAN ELEMEN SLAB A Af I ADB SO AISLE CENTER STIFFE Af I ATB SO DISTAN ELEMEN B( I BOA SO LONG 0 BEAM* LONG D SECTIO LONG 0 SECTIO SHORT SECTIO SHORT SECTIO CENTER STIFFE (It SL STIFFN STIFFN AMOUNT

CE BETWEEN MIDPOINTS OF ADJACENT TS* IN X - D I R E : T I O N REA INFLUENCED BY EACH NODE DIVIDED BY B( I UARED WIDTH IN FEET -TO-CENTER SPACING OF LONGITUDINAL NIN6 BEAMS M U L T I P L I E D BV B( I UARED CE BETWEEN MIDPOINTS OF ADJACENT TS* I N Y - D I R E C T I O N D I V I D E D BY A l I UARED IMENSION AND SHORT DIMENSION OF GRAD IN I N C H E S . IMENSION OF LONGITUDINAL BEAM CROSS-N IMENSION OF TRANSVERSE BEAM CROSS-N DIMENSION OF LONGITUDINAL CROSS-N DIMENSION OF TRANSVERSE BEAN CROSS-N -TO-CENTER SPACING OF TRANSVERSE NING B EAMS AB STIFFNESS MATRIX* OR 121 OVERALL ESS MATRIX OF SYSTEM ESS COEFFICIENTS AT THE JOINT

OF GAP BETWEEN SLAB AND SUBGRADE

* 0 0 0 0 0 9 6 0 * 0 0 0 0 0 9 7 0 * 0 0 0 0 0 9 8 0 * 0 0 0 0 0 9 9 0 *00001000 *00001010 *00001020 *00001030 *00001040 *000O10S0 *00001060 *00001070 *00001080 * 0 0 0 0 1 0 9 0 *00001100 *00001110

00001120 00001130 00001140 00001150 0 0 0 0 1 1 6 0 00001170 00001180 0 0 0 0 1 1 9 0 00001200 00001210 00001220 00001230 00001240 00001250 0 0 0 0 1 2 6 0 00001270 00001280

E 0 0 0 0 1 2 9 0 00001300 00001310 00001320 00001330 00001340 00001350 0 0 0 0 1 3 6 0 000J)1370 00001380 0 0 0 0 1 3 9 0 00001400 0 0 0 0 1 4 1 0 00001420 00001430 00001440

126

c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c

OEFI 1

OEFF

DEL

DELF

OFI 1

OKI 1

OXYFAC* DVFAC

DYI 1

EIX EIY Fl 1 FLRCAP FOI 1

FRX

HI 1 ICL

ICLF

lER

ISOTRY

LIFT

MOIX

HOIY

MMM

MXOIFI 1

UNIT DEFLECTIONS FROM SUBGRADE FLEXIBILITY 0 0 0 0 1 4 5 0 MATRIX 0 0 0 0 1 4 6 0 WORKING VARIABLE WHOSE VALUE IS DEPENDENT 0 0 0 0 1 4 7 0 UPON DEF( I ANO LOCATION IN FINITE ELEMENT00001480 6RID 0 0 0 0 1 4 9 0 TOLERANCE TO CONTROL CONVERGENC* GENERAL- 0 0 0 0 1 5 0 0 LY USE 0 . 0 0 1 FOR COARSE CONTROL 0 0 0 0 1 5 1 0 TOLERANCE TO CONTROL CONVERGENCE* GENERAL- 0 0 0 0 1 5 2 0 LY USE 0 . 0 0 0 1 . FOR FINE CONTROL 0 0 0 0 1 5 3 0 DIFFERENCE IN DEFLECTIONS AT THE SAME NODE 0 0 0 0 1 5 4 0 BETWEEN TWO CONSECUTIVE ITERATIONS 0 0 0 0 1 5 5 0 DISTANCE BETWEEN ADJACENT NODES IN THE 0 0 0 0 1 5 6 0 X-DIRECTION 0 0 0 0 1 5 7 0 FORMS OF THE ELASTIC CONSTANTS OF AN ORTHO-00001580 TROPIC SLAB IWHOSE PRINCIPAL DIRECTIONS OF 0 0 0 0 1 5 9 0 OF ORTHOTROPY COINCIDE WITH THE X- AND Y- 0 0 0 0 1 6 0 0 AXESI 0 0 0 0 1 6 1 0 DISTANCE BETWEEN ADJACENT NODES IN THE 0 0 0 0 1 6 2 0 V-OIRECTION 0 0 0 0 1 6 3 0 FLEXURAL RIGIDITY IN LONGITUDINAL 0IRECTION00001640 FLEXURAL RIGIDITY IN TRANSVERSE DIRECTION 0 0 0 0 1 6 5 0 DEFLECTION AT A NODE 0 0 0 0 1 6 6 0 FORKLIFT'S RATED CAPACITY IN POUNDS 0 0 0 0 1 6 7 0 FORCE DUE TO WEIGHT OF SLAB AND/OR UNI- 0 0 0 0 1 6 8 0 FORMLY DISTRIBUTED LIVE LOAD ON ELEMENT 0 0 0 0 1 6 9 0 FLEXURAL RIGIDITY PER UNIT WIDTH* IN L 0 N G I - 0 0 0 0 1 7 0 0 ITUOINAL DIRECTION 0 0 0 0 1 7 1 0 SUBGRADE STIFFNESS MATRIX 0 0 0 0 1 7 2 0 MAXIMUM NUMBER OF ITERATIONS ALLOWED* GEN- 0 0 0 0 1 7 3 0 ERALLV USE 1 0 FOR COARSE CONTROL. 0 0 0 0 1 7 4 0 MAXIMUM NUMBER OF ITERATIONS ALLOWED* GEN- 0 0 0 0 1 7 5 0 ERALLV USE 30 FOR FINE CONTROL 0 0 0 0 1 7 6 0 INSTABILITY ERROR OCCURRING DURING MATRIX 0 0 0 0 1 7 7 0 INVERSION 0 0 0 0 1 7 8 0 SWITCH TO DETERMINE IF STIFFNESS OF CON- 0 0 0 0 1 7 9 0 STANT THICKNESS SLAB OR OF STIFFENED SLAB 0 0 0 0 1 8 0 0 IS TO BE USED IN PROBLEM SOLUTION—> ASSIGN00001810 0 IF CONSTANT THICKNESS; ASSIGN 1 IF 0 0 0 0 1 8 2 0 STIFFENED SLAB 0 0 0 0 1 8 3 0 DETERMINES MODE OF SOIL SWELLING: ASSIGN 0 0 0 0 1 8 4 0 1 IF CENTER LIFT* AND ASSIGN 2 IF EDGE L I F T 0 0 0 0 1 8 5 0 MOMENT OF INERTIA OF STIFFENED SLAB SECTION00001860 IN LONGITUDINAL DIRECTION 0 0 0 0 1 8 7 0 MOMENT OF INERTIA OF STIFFENED SLAB SECTION00001880 IN TRANSVERSE DIRECTION 0 0 0 0 1 8 9 0 EXPONENT OF EXPONENTIAL EOUATIDN DESCRIBING00001900 PROFILE OF SWOLLEN SUBGRADE. EQUATION IS 0 0 0 0 1 9 1 0 OF THE FORM V=C*X**M 0 0 0 0 1 9 2 0 NUMERICAL DIFFERENCE BETWEEN MOMENTS AT 0 0 0 0 1 9 3 0

127


MXVDIFI 1

MYDIFI 1

MYXOIFI 1

NB

NCK

NCYCLE

NELEM NFLINFLOADI

NFL OAD

NGAP .

NGINGAPI

NKENT NLINLOADI NLOAD

NO! 11

N0I2I

NOB! 11 N08t2l NONY

NOO NOOCKINCKI

NOTCON

NOl N013 N02 N023 NPINPRINTI NPRINT

ADJACENT NODES IN THE X -O IRECTION NUMERICAL DIFFERENCE BETWEEN TWISTING MOMENTS AT ADJACENT NODES NUMERICAL DIFFERENCE BETWEEN MOMENTS AT ADJACENT NODES I N TME Y -D IRECT ION NUMERICAL DIFFERENCE BETWEEN TWISTING MOMENTS AT ADJACENT NODES HALF BAND WIDTH* EQUAL TO OR GREATER THAN I NY • 2 1 * 3 NUMBER OF NODAL POINTS FOR CHECKING THE CONVERGENCE MAXIMUM NUMBER OF CYCLES FOR CHECKING SUB-GRADE CONTACT* GENERALLY USE 10 NUMBER OF ELEMENTS I N F I N I T E ELEMENT GRID ELEMENT NUMBERS OVER WHICH SECOND LOAD I S APPLIED NUMBER OF ELEMENTS ON WHICH SECOND LOAD I S APPLIED TOTAL NUMBER OF NODES AT WHICH A GAP EXIST BETWEEN SLAB AND SUBGRADE. ASSIGN 0 IF NO GAP EXISTS OR THE GAP IS VERY LARGE NODAL NUMBER OF THE NODE AT WHICH THE GAP GAP BETWEEN SLAB AND SUBGRADE IS S P E C I F I E D COUNTER TO CALL SUBROUTINE •SHEAR* ELEMENT NUMBER OVER WHICH LOAD I S APPLIED NUMBER OF ELEMENTS 3N WHICH LOAD I S APPLIE USE 0 IF THERE I S NO LOAD TOTAL NUMBER OF COMPONENTS CONTRIBUTING TO INTERNAL WORK I N SLAB t l TOTAL NUMBER OF COMPONENTS CONTRIBUTING TO INTERNAL WORK I N BOTH SLABS N 0 I 1 I * N B N 0 I 2 ) * N B NUMBER OF NODES I N SLAB S i OR NUMBER OF NODES IN BOTH SLABS LESS NY INONY • I I * (NONY) / 2 NODAL NUMBER OF THE POINTS FOR CHECKING CONVERGENCE TOTAL NUMBER OF NODES AT WHICH REACTIVE PRESSURE I S PRESUMED TO BE 0 . I F NCYCLE-1

WILL NEVER BE I N CONTACT; I F THESE N3DES MAY OR MAY NOT BE DEPENDING ON CALCULATED RESULT

THESE NODES NCYCLE > 1 * IN CONTACT* N X 1 * N Y * 3 TOTAL NUMBER NX*NV»3 TOTAL NUMBER OF NODES IN BOTH SLABS NODAL NUMBER OF THE PRINTED POINTS. NUMBER OF NODES AT WHICH STRESSES ARE

OF NODES IN SLAB tl

TO B

00001940 00001950 00001960 00001970 00001980 00001990 00002000 00002010 00002020 00002030 00002040 00002050 00002060 00002070 00002080 00002090 00002100 00002110 S000021.20 00002130 00002140 00002150 00002160 00002170 00002180 0000 02190 00002200 00002210 00002220 00002230 00002240 00002250 00002260 00002270 00002280 00002290 00002300 00002310 00002320 *00002330 00002340 00002350 S00002360 00002370 00002380 00002390 00002400 00002410 E00002420

128


NPROB

NPUNCH

NREAD

NSLAB NSLOAD

NSVM

NTEMP

NWT

NX NXl HK2

NY NYNB NZINOTCONI

PA PB PFf 1 PPF I 1 PR PRS Q

PRINTED NUMBER OF PROBLEMS TO BE SOLVED. I NOTE: OPTIONS 3 AND 4 MAY BE SOLVED IN TWO SEPARATE PROBLEMS. THE F I R S T PROBLEM COMPUTES THE GAPS ANO PRECOMPRESS IONS DUE TO WEIGHT OF SLAB ANO TEMPERATURE BY SETTING NWT TO 1 * ANO THE SECOND PROBLEM COMPUTES STRESSES AND DEFLECTIONS DUE TO L I V E LOAD ONLY BY SETTING NWT TO 0 . 1 ANY GAPS OR PRECOMPRESSION TO BE PUNCHED7 ASSIGN 1 I F YES AND ASSIGN 0 I F NO. ANY GAPS OR PRECOMPRESSIONS TO BE READ IN7 ASSIGN 1 I F YES* ASSIGN 2 IF GAPS ANO PRE-COMPRESSIONS ARE FROM THE PREVIOUS AND ASSIGN 0 I F OTHERWISE.

0 0 0 0 2 4 3 0 0 0 0 0 2 4 4 0 0 0 0 0 2 4 5 0 0 0 0 0 2 4 6 0 0 0 0 0 2 4 70 0 0 0 0 2 4 8 0 000024 90 00002500 00002510 00002520 000 02530 00002540 00002550

PR0BL£M*00002560 0 0 0 0 2 5 70

NUMBER SLABS* EITHER 1 OR 2 0 0 0 0 2 5 8 0 SWITCH TO DETERMINE IF THERE I S A SECOND 0 0 0 0 2 5 9 0 LOADING ON THE SLAB. ASSIGN 0 WHEN THERE I S 0 0 0 0 2 6 0 0 NO SECOND LOADING AND ASSIGN 1 WHEN THERE I S A SECOND LOADING ON THE SLAB CONDITION OF SYMMETRY. ASSIGN 1 WHEN NO SYMMETRY E X I S T S * 2 WHEN SYMMETRIC WITH RESPECT TO Y A X I S * 3 WHEN SYMMETRIC SPECT TO X A X I S * 4 WHEN SYMMETRIC SPECT TO BOTH X ANO Y AXES* ANO 5 SLABS SYMMETRICALLY LOADED. CONDITION OF WARPINS. ASSIGN 0 IF

0 0 0 0 2 6 1 0 0 0 0 0 2 6 2 0 0 0 0 0 2 6 30 0 0 0 0 2 6 4 0 0 0 0 0 2 6 5 0 0 0 0 0 2 6 6 0 0 0 0 0 2 6 7 0 0 0 0 0 2 6 8 0

TEMPERA-00002690

WITH R E -WITH R E -FOR FOUR

TURE GRADIENT I S 0 AND ASSIGN 1 I F I S NOT 0 METHOD EMPLOYED. ASSIGN NOT CONSIDERED* ASSIGN 1

GRADlENTOOO02700 00002710

0 WHEN WEIGHT IS 00002720 WHEN WEIGHT IS 00002730

CONSIDERED FOR SLAB OF NON-CONSTANT CROSS- 0 0 0 0 2 7 4 0 SECTION* ANO ASSIGN - 1 WHEN SLAB I S OF C 0 N - 0 0 0 0 2 7 5 0 STANT* RECTANGULAR CROSS-SECTION TOTAL NUMBER NODES I N THE X DIRECTION NUMBER OF NODES I N X D IRECTION FOR SLAB H NUMBER OF NODES I N X D IRECTION FOR SLAB 12 ASSIGN 0 WHEN THERE IS ONLY ONE SLAB. NUMBER OF NODES I N Y D IRECTION N Y * 3 * N B

0 0 0 0 2 7 6 0 0 0 0 0 2 7 7 0 0 0 0 0 2 7 8 0

. 0 0 0 0 2 7 9 0 0 0 0 0 2 8 0 0 0 0 0 0 2 8 1 0 0 0 0 0 2 8 2 0

NODAL NUMBER OF THE NODES AT WHICH R E A C T I V E 0 0 0 0 2 8 3 0 PRESSURE I S I N I T I A L L Y SET TO ZERO WORKING VARIABLE* EaUAL TO A WORKING VARIABLE* ERUAL TO B A PREVIOUS NODAL DEFLECTION A PREVIOUS NODAL DEFLECTION POISSON'S RATIO OF THE CONCRETE POISSON'S RATIO OF THE SOIL LOADING ON SLAB* EXPRESSED AS A PRESSURE

0 0 0 0 2 8 4 0 0 0 0 0 2 8 5 0 0 0 0 0 2 8 6 0 0 0 0 0 2 8 7 0 0 0 0 0 2 8 8 0 0 0 0 0 2 8 9 0 0 0 0 0 2 9 0 0 0 0 0 0 2 9 1 0

129

c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c

c c

Q2

OK QK2

QO QSLAB

RFJ RM SMI 1

STRI * 1

T TEMP

TEX

TEY

TRBL

TRBW

V L * VM

V X I 1 VXMAXN VXMAXP VVI 1 VYMAXN VVMAKP XT! 1 Y T I 1 WK X I N K 1 ^ N X 2 I

XOAINLOAO* 1 ) XOAINLOAO* 21

X 0 A 2 I N F L 0 A D * 1 > X D A 2 ( N F L 0 A D » 2 I

AND

ANO

SECOND LOADING ON PRESSURE FORCE DUE TO APPLI FORCE DUE TO APPLI ELEMENT WK WEIGHT OF SLAB EXP D ISTRIBUTED LOAD. SLAB I S NOT CONSTA CROSS-SECTION ANO RELAXATION FACTOR R I G I D I T Y MODULUS STIFFNESS MATRIX F OF AN ORTHOTROPIC CALCULATED STRESSE SUPPORT CONDITIONS THICKNESS OF THE C DIFFERENCE I N TEMP BOTTOM OF SLAB. U UPWARD AND NEGATIV EQUIVALENT CONSTAN LONGITUDINAL DIREC EQUIVALENT CONSTAN TRANSVERSE D I R E C T I TORSIONAL R I G I D I T Y DIRECTION

TORSIONAL R I G I D I T Y DIRECTION LOAD MATRICES USED STIFFNESS MATRIX SHEAR FORCE IN THE MAXIMUM NEGATIVE S MAXIMUM P O S I T I V E S SHEAR FORCE IN THE MAXIMUM NEGATIVE S MAXIMUM P O S I T I V E S TEMPORARY STORAGE TEMPORARY STORAGE WEIGHT OF CONCRETE X COORDINATES STAR ING FROM LEFT TO R READ TWICE AT JOIN NATE DISTANCES ARE LOWER AND UPPER L I X D I R E C T I O N . USE ERS THE WHOLE LENG LOWER AND UPPER L I SECOND LOADING IN IF THE LOAD COVERS

SLAB* EXPRESSED AS A

ED LOAD ON AN ELEMENT ED SECOND LOAD ON AN

RESSED AS A UNIFORMLY

0 0 0 0 2 9 2 0 0 0 0 0 2 9 3 0 0 0 0 0 2 9 4 0 0 0 0 0 2 9 5 0 0 0 0 0 2 9 6 0 0 0 0 0 2 9 7 0 0 0 0 0 2 9 8 0

( T H I S INPUT I S USED WHEN00002990 NT DEPTH RECTANGLULAR N t f T s l . l AT THE J O I N T * USUALLY 0 .

OR A RECTANGULAR ELEMENT MATERIAL S DUE TO LOADING AND

ONSTANT DEPTH SLAB EKATURE BETWEEN TOP AND SE P O S I T I V E IF CURLED E IF CURLED DOWNWARD.

0 0 0 0 3 0 0 0 0 0 0 0 3 0 1 0

5 0 0 0 0 3 0 2 0 0 0 0 0 3 0 30 0 0 0 0 3 0 4 0 0 0 0 0 3 0 5 0 0 0 0 0 3 0 6 0 0 0 0 0 3 0 7 0 0 0 0 0 3 0 8 0 0 0 0 0 3 0 9 0 0 0 0 0 3 1 0 0 0 0 0 0 3 1 1 0

T THICKNESS SLAB DEPTH I N 0 0 0 0 3 1 2 0 TION 0 0 0 0 3 1 3 0 T THICKNESS SLAB DEPTH I N 0 0 0 0 3 1 4 0 ON

OF BEAM I N LONGITUDINAL

OF BEAM I N TRANSVERSE

TO DETERMINE ELEMENTAL

X - O I R E C T I O N

0 0 0 0 3 1 5 0 0 0 0 0 3 1 6 0 0 0 0 0 3 1 7 0 0 0 0 0 3 1 8 0 0 0 0 0 3 1 9 0 0 0 0 0 3 2 0 0 0 0 0 0 3 2 1 0 0 0 0 0 3 2 2 0

HEAR FORCE I N X - 0 I R E C T I O N 0 0 0 0 3 2 3 0 HEAR FORCE IN X - 0 I R E C T I O N 0 0 0 0 3 2 4 0

Y - D I R E C T I O N 0 0 0 0 3 2 5 0 HEAR FORCE I N V - 0 I R E C T I O N 0 0 0 0 3 2 6 0 HEAR FORCE IN V - D I R E C T I O N 0 0 0 0 3 2 7 0 F I L E FOR XI 1 F ILE FOR V I 1

I N P S I TING FROM 0 ANO INCREAS-I G H T . I N O T E : X MUST BE T IF TWO SLABS. COOROI-

INPUT I N F E E T . I H ITS OF LOADED AREA I N X

0 0 0 0 3 2 8 0 0 0 0 0 3 2 9 0 0 0 0 0 3 3 0 0 0 0 0 0 3 3 1 0 0 0 0 0 3 3 2 0 0 0 0 0 3 3 3 0 0 0 0 0 3 3 4 0 0 0 0 0 3 3 5 0

- 1 TO • I I F THE LOAD C O V - 0 0 0 0 3 3 6 0 TM OF ELEMENT. 0 0 0 0 3 3 7 0 HITS OF LOADED AREA FOR 0 0 0 0 3 3 8 0 X - O I R E C T I O N . USE - 1 TO « -100003390

THE WHOLE LENGTH OF THE 0 0 0 0 3 4 0 0

130

c c c c c c c c c c c c c c c c c c c c c c c c c c< c c

XEC XMOM XYMOM XXIll XXI2I XXL xxs XYMX

YINYI

VDAINLOAD* I I YOAINLOAD* 2 1

Y D A 2 I N F L 0 A 0 * 1 I Y 0 A 2 ( N F . 0 A D * 2 I

YM YMOM VMS W i l l Y Y I 2 I

ANO

AND

ELEMENT EDGE PENETRA BENDING NOME TWISTING MOM X-COORDINATE X-COORDINATE SLAB LENGTH* SLAB WIDTH* AMOUNT OF DI INCHES

V COORDINATE CREASING TO LOWER ANO UP D I R E C T I O N . WHOLE WIDTH LOWER ANO UP SECOND LOADI IF THE LOAD ELEMENT YOUNG'S MODU BENDING NOME YOUNG'S MODU V-COORDINATE V-COORDINATE

TION OIS NT I N X-ENT S OF CEN S OF CEN

I N FEET IN FEET FFERENTIAL SHRINK OR SWELL* I N

TANCE* I N F E E T . •DIRECTION

ITROIO OF SLAB B l ITROIO OF SLAB C2

S STARTING FROM ZERO AND I N -TOP* I N FEET PER L I M I T S OF LOADED AREA I N V USE - 1 TO -1-1 I F THE LOAD COVER OF ELEMENT. PER L I M I T S OF LOADED AREA FOR NG IN Y - D I R E C T I O N . USE - 1 TO * COVERS THE WHOLE WIDTH OF THE

LUS OF THE CONCRETE NT IN V-OIRECTION LUS OF THE SOIL S OF CENTROID OF SLAB • ! S OF CENTROID OF SLAB 12

DIMENSION S N ( 9 0 1 * S M 1 ( 9 0 ) * S M 2 ( 9 0 I » S N 3 ( 9 0 I * S M * 0 I *XYMOM(6S0 I * MXOIF I 6 5 0 1 * MYDIF I 6 5 0 1 f N X Y O I F I * » V I 6 5 0 t * X T I 6 S 0 l * Y T I 6 5 0 1 * 0 X 1 6 5 0 1 * DYI 6 5 0 1 *HDI * * N O O I 4 I * N O B ( 2 I * C ( 1 5 0 0 0 0 I * G A P I 6 5 0 I * H ( 2 0 8 0 0 0 I * O O O I f C O ( 5 5 0 0 l » P F ( 6 5 0 l * C U R L ( 6 5 0 l * N O D C K I 7 5 l * X

* S O I DIMENSION X D A ( 5 0 0 * 2 I * V D A ( 5 0 0 * 2 1 * K O I 4 1 t V O I 41

* D C F I 6 5 0 I * A B ( 6 5 0 I * N C C ( 6 5 0 I * N Z I 6 5 0 I * S T R I 6 5 0 * 6 * V X I 6 5 0 1 * V V I 6 5 0 1 * I R I 6 5 0 1 * G U N A ( 6 5 0 1 * X D A 2 I 5 0 0 *

4 ( 9 0 I * X M O M 1 6 5 0 I * V M O M I 6 65 01 *MYXDIF I 65 01 * X ( 6 5 0 2 0 8 0 0 0 1 * O F I 6 5 0 1 « N L I 6 5 0 * P P F ( 6 5 0 I * F I 2 1 0 0 I * 6 I 1 5 X I 2 I * Y V I 2 I * Y Y Y I 2 1 I * F 0 I

* N P I 6 5 0 I * N 0 I 2 I * Z Z I 2 1 I * l * N G 1 6 5 0 l * Y X n O M ( 6 5 0 l * 2 1 * Y 0 A 2 ( 5 0 0 * 2 1 * N F L ( 6 5 0

COMMON C » F * G » N 0 * N 8 * X » V * S T R » N P * X M 0 M * V M 0 M * X Y M 0 M * X T * V T * D X * D Y » M X 0 I F * * M V 0 I F » M K Y O I F » M Y X D I F * V X M O M * V X * W * N X Y * N Y X * VXMAXN* VXMAXP* *VVMAXN»VYMAXP

REAL n X D I F * M Y D I F f MXYDIF* MVKOIF* MOIX* MO lY

—FOR REFERENCE ON DATA BLOCKS* SEE Z I E N K I E W I C Z * O . C * -THE F I N I T E —ELEMENT METHOD I N ENGINEERING S C I E N C E " * CHAPTER 1 0 * MCGRAW-HILL —BOOK C O . * 1 9 7 1 .

DATA SMl/60.* 0.* 30.*5*0.* 20.* 30.* 0.* 15.* 3*0.* 15.* 0.*

00003410 00003420 00003430 00003440 00003450 00003460 00003470 00003480 00003490 00003500 00003510 00003520 00003530 S00003540 00003550 00003560 100003570 00003580 00003590 00003600 00003610 00003620 00003630 00003640 00003650 *00003660 00003670 00003680 500003690 100003700 100003710 000003720 600003730 00003740 00003750 00003760 100005770 00003780 00003790 00003800 00005810 00003820 00003830 00003840 00003850 00003860 00003870 00003880 00003890

131

* 1 0 . * - 6 0 . * 0 . * 3 0 . * 3 » 0 . * - 3 0 . * 0 . * 1 0 . * - 3 0 . * 0 . * 1 5 . * 3 * 0 . * 00003900 * - 1 5 . * 0 . * 5 . * 6 0 . * 0 . * 3 0 . * 5 * 0 . * 2 0 . * - 3 0 . * 0 . * I S . * 3 * 0 . * - 1 5 . * 0 0 0 0 3 9 1 0 * 0 . * 5 . * - 6 0 . * 0 . * 3 0 . * 3 * 0 . * - 3 0 . * 0 . * 1 0 . * 6 0 . * 0 . * - 3 0 . * 5 * 0 . * 00003920 * 2 0 . * 3 0 . * 0 . * - 1 5 . * 5 * 0 . * - 1 5 . * 0 . * 1 0 . * 6 0 . * 0 . * - 5 0 . * 5 * 0 . * 2 0 . / 0 0 0 0 3 9 3 0

00003940 DATA S M 2 / 6 0 . * - 5 0 . * 0 . * 0 . * 2 0 . * 4 * 0 . * - 6 0 . * - 5 0 . * 0 . * 5 0 . * 1 0 . * 00005950

* 4 * 0 . * 5 0 . * - l S . * 0 . * - 1 5 . * 1 0 . * 4 * 0 . * - 5 0 . * - 1 5 . * 0 . * 1 5 . * 5 . * 00005960 * 4 * 0 . * 6 0 . * 5 0 . * 0 . * 0 . * 2 0 . * 4 * 0 . * - 5 0 . * I S . * 0 . * - 1 5 . * 5 . * 4 * 0 . *00003970 * 5 0 . * 1 5 . * 0 . * 1 5 . * 1 0 . * 4 * 0 . * 6 0 . * - 5 0 . * 0 .» 0 . * 2 0 . * 4 * 0 . * - 6 0 . * 0 0 0 0 5 9 8 0 * - 5 0 . * 0 . * 5 0 . * 1 0 . * 4 * 0 . * 6 0 . * 5 0 . * 0 . * 0 . * 2 0 . * 4 * 0 . ^ 00003990

00004000 DATA S M 5 / 5 0 . * - 1 5 . * 1 5 . « 0 . * 0 . * - 1 5 . * 5 * 0 . * - 5 3 . * 0 . * - 1 5 . * 5 * 0 . * - 1 5 . 0 0 0 0 4 0 1 0

* * 0 . * 0 . * - 5 0 . * 1 5 . * 0 . * 1 5 . * 5 * 0 . * 5 0 . * 8 * 0 . * 5 0 . * I S . * 1 5 . * 0 . * 0 0 0 0 4 0 2 0 * 0 . * 1 5 . * 5 * 0 . * 5 0 . * 8 * 0 . * - 5 0 . * - 1 5 . * 0 . * - 1 5 . * 5 * 0 . * 5 0 . * - 1 5 . * 00004050 * - 1 5 . * 0 . * 0 . * 1 5 . * 5 * 0 . * - 5 0 . * 0 . * 1 5 . * 5 * 9 . * 1 5 . * 0 . * 0 . * 5 0 . * 00004040 * 1 5 . * - 1 5 . * 0 . * 0 . * - I S . * 5 * 0 . / 00004050

00004060 DATA SN4/ 8 4 . * - 6 . * 6 . * 0 . * S . * 5 * 0 . * 8 . * - 8 4 . * - 6 . * - 6 . * 6 . * - 2 . * 0 0 0 0 4 0 7 0

* 0 . * - 6 . * 0 . * - 8 . * - 8 4 . * 3 * 6 . * - 8 . * 0 . * - 6 . * 0 . * - 2 . * 8 4 . * 6 . * 00004080 * - 6 . * - 6 . * 2 . * 0 . * 6 . * 0 . * 2 . * 8 4 . * 6 . * 6 . * 0 . * 8 . * 5 * 0 . * 8 . * 8 4 . * 0 0 0 0 4 0 9 0 * —6 . * - 6 . * 6 . * 2 . * 0 . * 6 . * 0 . * 2 . * - 8 4 . * - & . * 6 . * - 6 . * - 8 . * 0 . * 00004100 * - 6 . * 0 . * - 2 . * 8 4 . * - 6 . * - 6 . * 0 . * 8 . * 5 * 0 . * 8 . * - 8 4 . * - 6 . * 6 . * 6 . * 0 0 0 0 4 1 1 0 * - 2 . * 0 . * 6 . * 0 . * - 8 . * 8 4 . * 6 . * - 6 . * 0 . * S. * 5 * 0 . * 8 . / 00004120

00004130 101 FORMAT l / * 4 0 X * ' F I N I T E ELEMENT ANALYSIS OF INDUSTRIAL FLOOR SLABS'*00004140

* / 4 0 X * 4 9 ( ' - ' l « / / 5 l X * ' S L A B 4 * BY K.N.GUNALAN* 19 8 5 . ' * / 5 1 X * 2 7 ( • - ' I I 00004150 102 FORMAT l / / 5 X * ' T H I S INDUSTRIAL FLOOR SLAB PROBLEM HAS: ' * 00004160

* / 5 X * 3 9 l ' - ' l l 00004170 103 FORMAT (/ /*SX*'SUMMARY OF VAR I A B L E S ' * / 5 X * 2 1 ( ' - • 1 1 00004180 104 FORMAT I 1 4 I 5 I 00004190 105 FORMAT l 4 F l 0 . 4 * 4 l 5 t 00004200 106 FORMAT I 9 F 8 . 5 I 00004210 107 FORMAT I 6 E 1 5 . 6 I 00004220 108 FORMAT I I 5 9 2 F 1 0 . 4 * 2 E 1 0 . 5 * F I 0 . 4 » S I 5 I 00004250 109 FORMAT l / * 5 X « ' S L A 6 L E N G T H * ' * 1 7 X * F 7 . 2 * 7 X * ' F T ' * / / 5 X * ' S L A B WIOTHxt, 00004240

* 1 8 X * F 7 . 2 * 7 X * ' F T ' * / / 5 X * ' T H I C K N E S S OF S L A B = « * 9 X * F 1 0 . 4 * 6 X * ' I N ' * / / 5 X * 00004250 *'MODULUS OF CONCRETEs ' *10X*E10 .5*2X* 'PSI ' *^ /5X* 'POISSOi lS RATIO OF'00004260 * * • CONCRETE-««F10.4*/ /5X**V.MODULUS OF SUB.RADE' ' *8X*E10 .5* 00004270 * 2 X * ' P S I ' * / / 5 X « ' P O I S S O N S RATIO OF SUBGRADE" *F10 .4 I 00004280

110 FORMAT l / / 5 X * ' D I M E N S I 0 N OF GRADE BEAMS IN I N C H E S ' » / 5 X * 5 4 l ' - • » * 00004290 * / / 4 0 X * ' 0 £ P T H ' * 1 0 X * ' W I D T H ' * 1 0 X * ' S P A C I N G ' * / 4 0 X * 5 l ' - ' l » 1 0 K * 5 l ' - ' l * 00004500 * 1 0 X » 7 ( ' - ' I * / / 1 0 X * » T R A N S V E R S E GRADE BEAM'*6X*F10 .5*5X*F10 .5*7X* 00004310 * F 1 0 . 5 . / / 1 0 X * ' L O N G I T U D I N A L GRADE BEAM'*4X*F10. 5*5X*F10 .5 *7X* 00004320 * F 1 0 . 5 , / / I 00004350

111 FORMAT I 6 I 5 * 4 F 1 0 . 5 * F 5 . 2 * 1 5 1 00004540 112 FORMAT l / / *5X* 'PR0GRAM C0NSTANTS'*/5X* 1 7 1 ' - ' I I 00004550 115 FORMAT l / / 5 X * ' N S V M » ' * 2 X * I 5 * / / 5 X * ' N B « ' * 4 X * I 5 l 00004560 114 FORMAT | / / 5 X * « N X 1 « ' * 5 K * I 5 * / / 5 X , ' N V - ' * 4 X * I 5 * / / 5 X * ' N C Y C L E = ' * I 5 * / / 5 K * 0 0 0 0 4 5 7 0

* ' N P R I N T . ' * I 5 I 00004380

132

115 *s??2iT^^5I ' *7 ' -? /c : * * ' " '^^ ' * ' " ' • ' • • " • "•'^5X*'NCK='.5X*I5*/, w i : l " ^ ^ • " • " • ' ^ * * ' ' * • * * ^ " ° * 5 ' 2 X * ' P S I ' * / / 5 K * ' 0 E L . ' * 2 X * F 1 0 . 5 * • / / 5 K * ' 0 E L F = ' * 1 X * F 1 0 . 5 , / / 5 X * ' R F J = ' * 2 X * F 1 0 . 5 * / / 5 X , ' I C L F « ' * 1 X * I 5 I

/ / 0 0 0 0 4 5 9 0 00004400 00004410

OF X IN INCHES ARE: • * / * 1 1 I 4 X * F 8 . 5 I I 00004420 OF V I N INCHES ARE: ' * / * 1 1 I 4 X * F 8 . 5 1 1 00004430

FOLLOWING NODES ARE USED TO CHECK CONVERGENCE:'*00004440 00004450

ARE APPLIED ON THE FOLLOWING ELEMENTS'*

I INCHESI IN THE ORDER OF N0DES'* /5X»

116 FORMAT l / / 5 X * ' V A L U E S 117 FORMAT l / / 5 X * ' V A L U E S 118 FORMAT l / / 5 X * ' T H E

• / / * 1 2 I 1 0 I 119 FORMAT I 9 F 8 . 4 I 120 FORMAT I F 7 . 5 I 121 FORMAT ( 7 F 1 0 . S I 122 FORMAT ( / / * 5 X * ' L 0 A D S

* 'AND COORDINATES:'! 123 FORMAT I l O X * I 5 * 4 F 15 .51 124 F0RMAr(/ /5X»'DEFLECTIONS

* 4 2 ( ' - ' l l 125 FORMAT I / / * & X* 'NODE'*7X* 'DEFLECTION' *6X* 'NODE'*7X* 'DEFLECTION' *

*6X* 'NODE' *7X* 'DEFLECTION' *6X* 'NODE' *7X* 'DEFLECTION* I 126 FORMAT 1 4 1 4 X * I 5 * 5 X * E 1 3 . 6 I I 127 F0RMATI/ /5X*'DEFLECTIONS I INCHESI I N THE ASCENDING ORDER OF**

* • M A G N I T U D E ' * / 5 X * S 5 l ' - ' I I 128 FORMAT l / / 3 5 X * ' M I N I M U M DEFLEC T IONs '«E15 .6*2 X* * I N ' * / / 5 5 X *

* • MAXIMUM D E F L E C T I 0 N s ' * E 1 5 . 6 * 2 X * ' I N ' I 129 F0RMATI/ /5X* 'STRESSES I L B S / S O . I N I IN THE ORDER OF N00ES ' * / 5X*

* 4 2 l ' - * l l 130 FORMAT I / / * 6 X * ' N 0 0 E * * 8 X * * S T R E S S X**10X*'STRESS V ' *10X*

*'STRESS XY'* 12X* 'MAJ0R ' *15X* *MIN0R**15X* *SHEAR** / I 151 FORMAT I 5 X * I 5 * 6 I 5 X * E 1 5 . 6 I I 152 FORMAT I / / / / / I 155 F0RMATI//5X**M0MENTS ( I N . L B S I 154 F0RMAT( / /6X* 'N00E ' *8X* 'M0MENT 155 F0RMAT(//5X*'MOMENTS I I N . L B S )

* * OF M A G N I T U 0 E ' * / 5 X * 6 0 ( * - * I )

IN THE ORDER OF NODES • * / 5 X * 2 9 l ' - ' I I X**10X*'MOMENT Y'*9X*'MOMENT X Y ' * / ) K AND Y IN TNE ASCENDING ORDER**

X* *6X *• NODE **8X**MOMENT Y**6X**NO0E**8X**MOMENT

X * * / l Y * * / )

OF N O D E S * * / 5 X * 5 3 l * - * l l DIRECTION I L B S / I N I * * / / * 6 I 6

136 F 0 R M A T I / / 6 X * * N0DE**8X**M0MENT 157 FORMAT(//6X*•NODE•*8X**MOMENT 138 F 0 R M A T ( 2 ( 5 X * I 5 * 5 X v E 1 3 . 6 l l 139 FORMAT!'1*1 140 F0RMAT(/ /5X**SHEAR FORCES I N THE ORDER 141 FaRMATI//10K**CALCULATED SHEAR IN LONG

* X * ' I N C R ' * 3 X * ' S H E A R X ' l * / I 142 F 0 R M A T I 6 ( 6 X * I 5 * 5 X * F 8 . 5 I I 145 FORMAT(//10X*'CALCULATED SHEAR IN SHORT DIRECTION ( L B S / I N I • * / / * 6 (

*6X* ' INCR**5X**SHEAR Y * l * / I 144 F0RMATI/ /5X**SHEAR FORCES I N THE ASCENDING ORDER OF MAGNITUDE'*

* / 5 X * 4 7 ( ' - * l ) 145 FORMAT(/ /55X*'MAX NEGATIVE SHEAR FORCE IN K -0 IRECTION* ' *F8 .5» •

* L 8 S / I H * * / 5 5 X * * M A X POSITIVE SHEAR FORCE IN K-D IRECTION** *F 8 . 5 * • * L B S / I N * * / / 5 5 X * * M A X NEGATIVE SHEAR FORCE IN Y -D IRECTION** *F8 .3 * * • L B S / I N * * / 3 5 X * * M A X POSITIVE SHEAR FORCE IN Y-DIRECTION=**F8 .5* * • L B S / I N * I

00004460 00004470 000044 80 00004490 00004500 00004510 00004520 00004550 00004540 00004550 00004560 00004570 00004580 00004590 00004600 00004610 00004620 00004630 00004640 00004650 00004660 00004670 00004680 00004690 00004700 00004710 00004720 00004730 00004740 00004750 00004760 00004770 00004780 00004790 00004800 00004810 00004820 00004830 00004840 000048 50 00004860 00004870

133

c c c c

c< c

c c-c

c c-c-c

c

c c-c

c* c c-

146 FORMAT ( F 7 . 1 * F 1 0 . 4 I

'''*Â;;LTirD?;::;j5x'!^;.;:jj;.p"Tr;'="^=-"'^^-^'''''-^«-'^'"-14? FSSMJT aslFfi ' .s?"**'"" ° ' ""*"* ^°''°''*'^ CONDITION.,/5X*37(.-.) }?? IZltl ;^^V*^'^*-°'*°=*''^»^^5*'*«^=*'5X*Fia.5*2X*.pSI.| 1 5 1 FORMAT ( / / * 5 X * * S E C 0 N 0 LOADING APPLIED ON THE FOLLOWING ELEMENTS*.

** ANO C O O R D I N A T E S : * , / 5 X * 6 5 ( * - * I * / I .-• ."-*•«». C L t n t n i b ,

READ I N THE NUMBER OF PROBLEMS TO BE SOLVED ON THIS RUN

READ ( 5 * 104 1 NPROB

READ 1 5 * 1 4 & I FLRCAP* AISLWD WRITE ( 6 * 1 0 1 1 WRITE ( 6 * 1 3 2 ) WRITE ( 6 * 1 4 7 1 FLRCAP* AISLWD

WRITE ( 6 * 1 0 3 )

— B E G I N PROBLEM SOLUTION

DO 9 5 0 LLL - 1* NPROB —READ I N SLAB DIMENSIONS AND OTHER DESCRIPTIVE VARIALBE OF SLAB-ON-—GROUND TO BE ANALYZED

READ ( 5 * 1 0 5 J XXL* XXS* XEC* XYMX* MMM, I S 3 T R Y * L I F T * NSLOAO

NKENT = 0

I F ( I S O T R Y . E Q . O ) GO TO 8

-READ I N DIMENSIONS OF GRADE BEAMS AND GRADE BEAM SPACINGS

READ ( 5 * 1 0 6 ) BEAMLW* BEAMSW* BEAMLL* BEAMSL* ASPACE* BSPACE

- — R E A D IN MOMENT OF INERTIA I N LONG AND SHORT DIRECTIONS READ 1 5 * 1 0 7 1 MOIX* MOIY

C * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * i C C READ I N SLAB GEOMETRY* ELASTIC CONSTANTS* ANO OTHER CONSTANTS C

3 READ ( 5 * 1 0 8 1 NSLAB* PR* T* YM* VMS* PRS* NSYM* NOTCON* NREAD* *NPUNCH* NB

C* c

00004880 00004890 00004900 100004910 00004920 00004930 00004940 00004950 00004960 00004970 00004980 00004990 00005000 ^00005010 00005020 00005030 00005040 00005050 00005060 00005070 00005080 00005090 00005100 00005110 00005120 -00005130 00005140 00005150 00005160 '00005170 00005180 00005190 00005200 00005210 00005220 00005230 00005240 00005250 00005260 00005270 00005230 00005290 00005300 00005310 00005320 00005330 00005340 00005350 00005360

134

c-c

c c-c

WRITE 1 6 * 1 0 9 ) XXL* XXS* T* YM* PR* VMS* PRS

CALCULATE R I G I D I T Y MODULUS

RM * Y M * T * * 3 / ( 1 2 * ( 1 - P R * * 2 ) I

CALCULATE E L A S T I C I T Y OF S O I L

YMSPRS = Y M S / I 1 . - P R S * * 2 I

I F I I S O T R Y . E Q . O I GO TO 9 WRITE ( 6 * 1 1 0 1 BEAMLW* BEAMSW* BSPACE* BEAMLL* BEAMSL* ASPACE

CALCUALTE FLEXURAL R I G I D I T Y

EIX = YM*MOIX EIY = YM*MaiY FRX = E I X / ( 1 2 . * X X L I

C c-c

CALCULATE CONCRETE WEIGHT CONSTANT FOR SLAB OF CONSTANT THICKNESS

9 WK = 0 . 0 3 7 * T IF IISOTRY.EQ.OI GO TO 7

—CALCULATE TORSIONAL R I G I D I T Y OF TEE SECTION

TRANSVERSE GRADE BEAM TRBW - ( B E f t M L W * B E A M S W * * 5 / 1 6 . 1 * ( ( 1 6 . / 5 . I - ( 5 . 3 6 * B E A M S W / B E A M L W I

* * ( 1 . - ( B E A M S W * * 4 / I 1 2 . * B E A M L W * * 4 I I I I / B S P A C E -LONGITUDINAL GRADE BEAM

C C-c

z* c

TRBL = I B E A M L L * B E A M S L * * 5 / 1 6 . l * l l l 6 . / 5 . 1 - 1 5 . 56*BEAMSL/BEAMLL) * * I 1 . - ( B E A M S L * * 4 / ( 1 2 . * B E A M L L * * 4 ) ) I I / A S P A C E

DXYFAC s I I 6 . * X X L * R M * ( 1 . - P R | | 4 ^ T R B L « - T R B W 1 I ^ E I X OVFAC - ( X X L / X X S I * ( M O I Y / M O I X I

CALCULATE AN EQUIVALENT SLAB DEPTH

TEX = ( ( l . -PR**2l*MOIX/XXLI**0.555555 TEY « TEX*(0YFAC**0.555555I

READ IN NUMBER OF NODES IN EACH DIRECTION ANO NODAL PRINTING INFO

7 READ (5* 104 1 NXl* NX2* NY* NCYCLE* NPRINT, (NP(I)*I-1* NPRINTl

NX - NX14-NX2 N0(2I * NX*NY*5 N02 = NOl 21 N025 - Na2/3

00005370 00005380 00005390 00005400 00005410 00005420 00005430 00005440 00005450 00005460 00005470 00005480 00005490 00005500 00005510 00005520 00005530 00005540 00005550 00005560 00005570 00005580 00005590 00005600 00005610 00005620 00005630 00005640 00005650 00005660 00005670 00005680 00005690 00005700 00005710 00005720 00005730 00005740 00005750 00005760 00005770 00005780 00005790 '00005800 00005810 00005820 00005830 000058 40 00005850

135

c c-C

C C C-

c

N O I l l = N X 1 * N Y * 3 NOl = NOl I I N 0 1 5 s N O l / 3

— R E A D I N COORDINATES OF F I N I T E ELEMENT GRID SYSTEM

READ 1 5 * 1 0 6 1 I X I I I * 1 = 1 * N X l * f V C I l * 1 = 1 * NYI

—CONVERT GRID SYSTEM COORDINATES FROM FEET TO INCHES

DO 1 6 0 I = 1 * NX X ( I ) = X ( I I * 1 2 .

160 CONTINUE 0 0 1 6 1 1 = 1 * NY Y d ) = Y( 1 1 * 1 2 .

1 6 1 CONTINUE

: CALCULATE THE VARIOUS CONSTANTS USED THROUGHOUT THE SOLUTION

LA = NB-1 N O B d l = N0( 1 ) * N B N 0 B ( 2 I = N 0 ( 2 I * N B NOBl « N O B d ) N0B2 3 N0B(2 ) N013P = N013«- l NOIP = N014-1 N015NY = N015*NY

: I N I T I A L I Z E AND SET TO ZERO THE MATRIX OF NODES NOT I N CONTACT I N I

0 0 5 2 0 0 1 = 1 * N025 5200 N C C d l = 0

I F ( N O T C O N . E Q . O I 60 TO 5 2 2 0

: READ IN THE NODAL NUMBERS AT WHICH SUBGRADE REACTION IS PRESUMED : TO BE ZERO

READ (5* 1341 (NZdl* 1 = 1* NOTCONI

DO 5210 1 = 1 , NOTCON 5210 NCC(NZ(I1I = 1 5220 IF (NSLAB-ll 11* 11* 15

11 NONY « Nai3 GO TO 15

13 NONY s N023-NY 15 NOO = ( NQNY«-ll*N0NY/2

00005860 00005870 00005880 00005890 00005900 00005910 00005920 *00005930 00005940 00005950 00005960 00005970 00005980 00005990 00006000 00006010 00006020 00006030 00006040 00006050 00006060 00006070 00006080 00006090 00006100 00006110 00006120 00006130 00006140 00006150 T00006160 00006170 00006180 00006190 00006200 00006210 00006220 00006230 00006240 00006250

*00006260 000062 70 00006280 00006290 000 06300 00006310 00006320 00006330 00006340

136

c C INITIALIZE VARIABLES ANO MATRICES C

ICC = 0 ICCC = 0 NIC - 0 DO 9 5 5 1 = 1 * N023

9 5 5 F ( ( I - 1 1 * 3 > 1 I = 0 C C DETERMINE NUMBER OF ELEMENTS IN FINITE ELEMENT MESH C

NELEM = ( N Y - l ) * ( N X - N S L A B t NYI = NV-1

I F THE SLAB AND SUBGRADE ARE NOT ASSUMED IN FULL CONTACT*

UNTIL THE SAME CONTACT CONDITIONS ARE OBTAINED

965 NIC = NIC4-1

INITIALIZE SUBGRADE STIFFNESS MATRIX

ITERATE

0 0 16 I = 1* NOO 16 H i l l s 0

IC = 0 0 0 55 I = 1* N023 P P F d ) « F ( ( 1 - 1 1 * 3 * 1 1

55 A B d l = 0

C INITIALIZE SLAB STIFFNESS MATRIX

00 19 I = 1 * N0B2 19 C d l s 0

PA s 0 PB = 0

GENERATE STIFFNESS MATRIX OF EACH ELEMENT

DO 2 0 0 K = 1* NELEM 11 = ( K - l l / N Y l 12 = K-I1*NY1 IF (NSLAB-21 2 1 * 22f 22

21 N O D d l = K«-I1 GO TO 2 7

22 IF I K - N Y 1 * ( N X 1 - 1 ) ) 2 1 * 2 1 * 23 23 I I = 11*1

NOO(l ) = K*I14^NY1 27 N0D(2 I = NaD( 11 *1

N0D(3 I = N0D(1I4-NY N0D(4) = N3D(3)«-1

00006350 00006360 00006370 00006380 00006390 00006400 00006410 00006420 00006430 00006440 00006450 00006460 00006470 00006480 00006490 00006500 00006510 00006520 00006530 00006540 00006550 00006560 00006570 00006580 00006590 00006600 00006610 00006620 00006630 00006640 00006650 00006660 00006670 00006680 000066 90 00006700 00006710 00006720 00006730 00006740 00006750 00006760 00006770 00006780 00006790 00006800 00006810 00006820 00006830

137

A = ( X d l * 2 ) - X ( I l 4 ^ 1 ) l / 2 00006840 B = ( V d 2 4 - l l - Y ( I 2 } l / 2 00006850

C 00006860 C TEST FOR DUPLICITY OF CALCULATIONS 00006870 C 00006880

I F ( A B S ( A - P A ) . L T . O . O O l . A N D . A B S ( B - P B I . L T . 0 . 0 0 1 1 GO TO 100 00006890 PA = A 00006900 PB = B 00006910 ATB = A*B 00006920 ATB2 * ATB**2 00006930 AOB = A/B 00006940 BOA = B/A 00006950 A0B2 = AD8**2 00006960 B0A2 = B0A**2 00006970

C 00006980 C COMPUTE THE ELEMENT FLEXIBILITY MATRIX OF SUBGRADE 00006990 C 00007000

XXX = 0 00007010 Y Y Y f l l = A L a 6 ( 2 . 0 l 00007020 00 24 I = 2 * 21 00007030 XXX = XXX4-0. 05 00007040

24 Y Y Y d l = A L 0 G ( l * S O R T ( l * B D A * * 2 * X X X * * 2 l l 00007050 CALL OSF ( 0 . 0 5 * YYY* ZZ* 211 00007060 0EF(K I = ( Z Z ( 2 1 I - A L 0 G ( B D A I * 1 I / ( 4 * 5 . 1 4 1 5 9 5 * A I 00007070 XJ s 0 00007080

C c-c

00007090 —FORMULATE LOAD MATRICES VL AND VM 00007100

00007110 00 95 I = 1* 4 00007120 00 95 J = I* 4 IJ = IJ^l

50 VL = 2*B GO TO 50

IF (M-2) 60* 70* 80 60 VM = 1

GO TO 90 70 VM = 2*B

GO TO 90 80 VM = 2*A 90 CONTINUE

00007130 00007140

DO 95 L = 1* 3 00007150

IF (L-2) 20* 50* 40 JJSJIJJS 20 ui - » 00007170

GO TO 50 00007180

40 VL = 2*A 50 00 95 M = 1* 5

IJLM = M«-(L-11*3*(IJ-11*9 „ « « « , , . n IF d.EO.J.ANO.L.GT.M) GO TO 95 °AÂliti

00007190 00007200 00007210 00007220 00007230

00007250 00007260 00007270 00007280 00007290 00007300 00007310 00007320

138

CALCULATE I S O T R Y = l :

THE ELEMENTAL I F SLAB I S OF

STIFFNESS <4ATRIX ( I F SLAB HAS GRADE BEAMS CONSTANT THICKNESS* I S O T R Y = 0 . )

I F ( I S O T R Y ) 9 5 0 * 9 2 * 93 92 S M d J L M ) = R M * ( B 0 A 2 * S M l d J L M ) * A 0 B 2 * S M 2 d J L M I « - P R * S N 3 ( I JLM)

* * 0 . 5 * ( 1 . - P R ) * S M 4 ( I J L M ) ) * V L * V M / ( 6 0 . * A r B ) GO TO 95

93 S M d J L M ) = F R X * ( B 0 A 2 * S M 1 ( I J L N I « - 0 Y F A C * A O B 2 * S M 2 ( I J L M l * D X Y F A C * S M 4 ( * I J L M ) I * V L * V M / ( 6 0 . * A T B )

95 CONTINUE GO TO 189

100 D E F ( K I = D E F ( K - 1 I 189 I J = 0

DO 200 1 = 1 * 4 A B ( N O O d ) ) = A B ( N 0 0 d ) ) « - A T B

C C-c-c

—SUPERIMPOSE THE ELEMENT STIFFNESS MATRIX TO FORM THE OVERALL —STIFFNESS MATRIX

C c-c-c

DO 200 J = I * 4 IJ = IJ^l 00 200 L = 1* 3 DO 200 M = 1* 3 IJLM » M*(L-1)*3«-(IJ-1)*9 IF (I.EQ.J.ANO.M.LT.L) GO TO 200 IH s (N0D(I)-l)*Na*3*l*(L-l)*NB*(N3D( J)-NO0(I ))*34-M-L C(IH) = C(IH)*SM(IJLM)

200 CONTINUE

SUPERIMPOSE THE ELEMENT FLEXIBILITY MATRIX TO FORM THE OVERALL FLEXIBILITY MATRIX OF THE SUBGRADE

1502 502 504

506 508

510

512

DO II 12 IF IF H(( 60 IF IF DEF 60 IF DEF GO DEF 60 IF

541 1 = 1 * NONY = d-l)/NY*l = I-(I1-1 )*NY (I.GT.NX1*NY( 11=11*1 (NCC(dl-l)*NY*l2).EQ.O) GO TO 1502 I*l)*I/2) = H((I*1I*I/2I*1 TO 541 (I-(I-1I/NV*NY-1) 512* 502* 512 d-1) 506* 504* 506 F = 4*0£F(1) TO 600 d-NONY«-NYl) 510* 508, 5 10 F = 4*DEF((NX-1-NSLAB1*NY1+1) TO 600 F = 2*(DEF(I-I/NY)*0EF(I-I/HY-NY1)) TO 600 (I-(I-1)/NY*NY-NY) 522, 514* 522

*00007330 00007340 00007350 00007360 00007370 00007380 00007390 00007400 00307410 00007420 00007430 00007440 00007450 00007460 00007470 00007480 00007490 00007500 00007510 00007520 00007530 00007540 00007550 00007560 00007570 00007580 00007590 00007600 00007610 00007620 00007630 00007640 00007650 00007660 00007670 00007680 00007690 00007700 00007710 00007720 000077 30 00007740 00007750 00007760 00007770 00007780 00007790 00007800 00007810

139

514 I F ( I - N Y ) 5 1 8 * 516* 518 516 OEFF = 4*0EF(NY1)

GO TO 600 518 I F d-NONY) 510* 520* 510 520 OEFF = 4*0EF( INX-NSLAB)*NY1)

522 I F d - N Y ) 5 2 4 * 600* 526 524 OEFF = 2 * ( 0 E F d ) * D E F ( I - l ) )

00007820 00007830 00007840 00007850 00007860

?? !? ^°? ... 00007870 00007880 00007890

00007910 00007920

< ° TO 600 00007900 526 IF d-NONY* NVl) 550* 600, 52 8 528 OEFF = 2*(0EF(I-I/NY-NY1)*0EF(I-I/NY-NY))

GO TO 600 00007950 530 OEFF = DEFd-I/NYI*OEF(I-I/NY-ll*DEFd-I/NY-NYlI*DEF(I-I/NY-NY) 00007940 600 00 542 L = 1, NSYM 00007950

IF (L.EQ.5) 60 TO 542 00007960 IF (L-2) 270* 272* 274 00007970

270 SIGX = 1 00007980 SIGY = 1 00007990 GO TO 290 00008000

272 IF (NSYM.Ea.3l GO TO 542 00008010 SIGX = -1 00008020 SI6V = 1 00008030 GO TO 290 00008040

274 IF (L-3) 276* 276* 278 00008050 276 SIGX = 1 00008060

SI6V = -I 00008070 GO TO 290 00008080

278 SIGX = -1 00008090 SIGV s -1 00008100

290 00 540 J = I* NONY 00008110 Jl = (J-ll^NY-fl 00008120 J2 = J-(J1-1)*NV 00008130 IF (J.GT.NK1*NY) Jl = Jl*l 00008140 IF (L.EQ.1.AN0.I.EQ.J.0R.L.EQ.2.AND.I.LE.NY.AND.I.EQ.J.0R.L.E0.5. 00008150

*ANO.I.EQ. (I-1)/NY*NY*1.AND.I.EQ.J.0R.L.EQ.4.ANO.I.EQ.1.ANO.J.EQ.1100008160 00008170 00008180 00008190 00008200

224 AAA = l/(5.141595*SQRT((Xdl)-SIGX*X( J1))**2*(Y(I2)-SISY*Y(J2))** 00008210 *2)) 00008220

226 H((J*l)*J/2-J*I) = H((J*l)*J/2-J*I)*AAA 00008230 540 CONTINUE 00008240 542 CONTINUE 00008250 541 CONTINUE 00008260

C 00008270 C STORE THE FLEXIBILITY MATRIX OF THE SUBGRADE 00008280 C 00008290

00 9541 I = 1* NOO 00008300

*G0 TO 60 TO

222 AAA 3 GO TO

222 224 OEFF 226

140

9 5 4 1 C

C

190

1 9 1 192

171 172

C c-c

H O d ) = H I D / Y M S P R S

• INVERT THE F L E X I B I L I T Y MATRIX T3 OBTAIN THE STIFFNESS MATRIX OF -THE SUBGRADE

CALL SINV I H * NONY* l . O E - 0 7 * l E R * NOOl I F I I E R . N E . O ) GO TO 6 0 0 0 0 0 545 1 = 1 * NOO H I D = HI I ) * Y M S P R S

-ST IFFNESS MATRIX OF THE SUBGRADE IS ADDED TO THE STIFFNESS MATRIX -OF THE SLAB TO OBTAIN THE STIFFNESS MATRIX OF THE SYSTEM

DO 1 7 2 1 = 1 * N 0 2 * 3 I F ( N C C ( ( I - l ) / 3 * l ) . N E . O ) GO TO 1 7 2 I F d . 6 T . N O l . A N O . I . L E . N 0 1 * N Y * 3 ) GO TO 172 I F d - N O l l 1 9 0 * 1 9 0 * 191 IK = d - l ) / 3 * l GO TO 192 I K = d - l ) / 5 * l - N Y DO 1 7 1 J = 1 * NB* -3 16 = IK*( J-1 ) / 3 I F ( I . L E . N 3 1 . A N 0 . I G . G T . N 0 1 3 . 0 R . I G . G T . N O N Y I GO TO 1 7 1 I F ( N C C K I - l ) / 3 * l * ( J - l ) / 3 ) . N E . O ) GO TO 1 7 1 I H = ( I - 1 ) * N B * J C ( I H ) = C I I H l * H ( ( I G * l ) * I G / 2 - ( J - 1 ) / 3 ) CONTINUE CONTINUE I F ( N S L A B . E Q . l ) GO TO 2 0 7 NYNB = NY*3*NB

—STORE THE S T I F F N E S S COEFFIC IENTS AT THE JOINT

0 0 2 0 5 1 = 1 * NYNB C O d l s C ( N O B l * I » 203

C C ADJUST THE S T I F F N E S S COEFF IC IENTS AS A RESULT OF SYMMETRY C

0 0 2 0 5 I = 1 * NY 205 C K I - 1 1 * N B * 3 * 1 * N 0 B 1 I = l . O E 2 0 207 I F ( N S Y M . E Q . l . O R . N S Y M . E Q . 5 1 GO TO 203

I F I N S Y M - 3 1 1 7 3 * 1 7 5 * 173 173 DO 174 I = 1 * NY 174 C K d - l >*3«-2 l * N B * l l = l . O E 20

I F ( N S Y M . E Q . 2 I GO TO 2 0 8 175 DO 176 I = 1 * N 0 2 3 * NY 176 C(((I-l)*3*l)*NB*1) = l.OE 20

C C APPLY GAUSS ELIMINATION TO FORM AN UPPER TRIANGULAR COEFFICIENT

00008310 00008320 00008330 00008340 00008350 00008360 00008370 00008380 00008390 00008400 00008410 00008420 00008430 00008440 00008450 00008460 00008470 00008480 00008490 00008500 00008510 00008520 00008530 00008540 00008550 00008560 00008570 00008580 00008590 00008600 00008610 00008620 00008630 00008640 00008650 00008660 00008670 00008680 00008690 00008700 00008710 00008720 000087 30 00003740 00008750 00008760 00008770 00008780 00008790

141

C M A T R I X * WHICH WILL BE USED LATER FOR ITERATION C

208 00 215 N = 1, NSLAB 215 CALL TRIG(N)

IF (NIC.GT.l) GO TO 931

DETERMINE THE COORDINATES OF THE CENTER OF EACH SLAB FOI — W A R P I N G

IF (NSYM-2) 7020* 7030* 7040 7020 X X d ) = X ( N K l ) / 2 .

Y Y l l ) = Y l N Y ) / 2 . 60 TO 7070

7050 XXIl l = 0 . Y V d ) = Y ( N Y ) / 2 . GO TO 7 0 7 0

7040 I F (NSYM-41 7 0 5 0 * 7 0 6 0 * 7 0 2 0 7050 XX( 1) = X ( N X l ) / 2 .

Y V d ) = 0 . 6 0 TO 7 0 7 0

7 0 6 0 X X d l = 0 . Y Y d l = 0 .

7070 I F ( N S L A B . E Q . l ) 6 0 TO 7 0 8 0 X X ( 2 ) = ( X ( N X I * X ( N X l l l / 2 . Y Y ( 2 ) = Y ( M Y ) / 2 . I F ( N S Y M . E 3 . 3 . O R . N S Y M . E Q . 4 ) Y Y ( 2 ) = 0 .

C C READ I N GAP, TEMP, AND LOAD DATA C

7080 READ ( 5 * H I ) NGAP* NTEMP* NLOAD* I C L * NCK* NWT * TEMP, * O E L F , R F J , I C L F

WRITE ( 6 , 1 1 2 ) WRITE 1 6 , 1 1 3 1 NSYM* NB WRITE ( 6 * 1 1 4 1 N X l * NY* NCYCLE* NPRINT WRITE ( 6 * 1 1 5 1 NLOAD* I C L , NCK, NWT, Q* DEL* DELF* R F J , WRITE ( 6 , 1 1 6 1 ( X ( I 1 , I = 1 * NXl WRITE ( 6 * 1 1 7 ) ( Y ( I I * I > 1 * NY)

C C — — R E A D I N NODE LOCATIONS USED TO CHECK CONVERGENCE C

READ (5* 104) (NOOCKdl* 1 = 1* NCKl C< C

C

c-

WRITE (6* 1181 (NOOCKdl* 1=1* NCK) If INREAO-1) 678, 677, 981

—READ IN LOCATIONS OF SPECIFIED OR PRE-CALCULATED CURL

00008800 00008810 00008820 00008830 00008840 00008850

COMPUTINC00008860 00008870 00003880 00008890 00008900 00008910 00008920 00003930 00008940 00008950 00008960 00008970 00008980 00003990 00009000 00009010 00009020 00009030 00009040 00009050 00009060 00009070 00009080 00009090 00009100 *00009110 00009120 00009130 00009140 00009150 00009160 00009170 00009180 00009190 00009200 00009210 00009220 *00009230 00009240 00009250 00009260 00009270 00009280

0, DELi

ICLF

7 C 74

142

6 7 7 READ 1 5 , 1 3 7 ) ( C U R L I I ) , 1 = 1 , N 0 2 3 )

6 0 TO 983 6 7 8 0 0 6 8 0 1 = 1 * N 0 2 3 6 8 0 C U R L I I ) = 0 .

I F ( N G A P . E a . O ) GO TO 1 0 6 9 C C' c

—READ I N GAP LOCATIONS

READ 1 5 * 1 3 4 ) ( N G d ) * 1 = 1,NGAP)

C READ I N LOCATIONS OF S P E C I F I E D OR PRE-CALCULATED CURL C

READ ( 5 , 1 1 9 ) ( C U R L ( N G ( I ) I , 1 = 1 * NGAPI

C COMPUTE THE CURLING OF SLAB DUE TO TEMPERATURE DIFFERENTIALS AND C ADO I T TO THE CAP TO FORM TOTAL CURLING ANO GAP C

1069 I F ( N T E M P . E Q . O ) GO TO 9 8 1 DO 3 I = 1 * NX I F ( I - N X l ) 7 1 0 0 * 7 1 0 0 * 7 2 0 0

7 1 0 0 N = 1 GO TO 7 3 0 0

7200 N = 2 7300 0 0 3 J = 1 * NY

5 C U R L K I - 1 ) * N Y * J I = O . O 0 0 0 0 2 5 * T E M P * ( (X d ) - X X (N ) ) * * 2 * ( Y( J ) - Y Y ( N) ) * * * 2 ) / T * C U R L ( d - l ) * N Y - t - J )

9 8 1 CONTINUE I F (NWT) 2 ? 6 * 9 8 3 * 292

C C I f NUT IS NOT ZERO* COMPUTE THE VERTICAL N30AL FORCES DUE TO THE C WEIGHT OF SLAB C C . . . R E A D I N WEIGHT OF SLAB FOUNDATION OF N0N-C3NSTANT CROSS-SECTION C AND/OR L I V E LOAD AS UNIFORMLY DISTRIBUTED LOADING

C 292 READ ( 5 * 1 2 0 ) QSLAB

c ***************************************************************** *****' C

OQ = QSLAB 00 295 I = 1* N023

295 FOd ) = QSLAB*AB( I) GO TO 931

C DISTRIBUTE WEIGHT OF SLAB OF CONSTANT* RECTANGUALR CROSS SECTION

00009290 00009300 00009310 00009320 00009330 00009340 00009350 00009360 00009370 00009380 00009390 00009400 '00009410 00009420 00009430 00009440 00009450 00009460 00009470 00009430 00009490 00009500 00009510 00009520 00009530 00009540 00009550 00009560 00009570 00009580 00009590 00009600 00009610 00009620 00009630 00009640 00009650 00009660 00009(»70 00009680 00009690 '00009700 00009710 00009720 00009750 00009740 00009750 00009760 00009770

143

c c-c

296 DO 2 9 7 I = 1 * N 0 2 3 2 9 7 F O d I = W K * A B d l

0 0 = WK GO TO 931

I F NWT IS ZERO* I N I T I A L I Z E THE NODAL FORCES TO ZERO

9 8 3 0 0 9 5 2 I = 1 * N023 9 5 2 F O d I = 0 9 3 1 0 0 1 9 5 1 1 = 1 * N023

1 9 3 1 P F d ) = 0 . DO 18 I = 1 * N02

18 F i l l = 0 I F I N I C . G T . I ) GO TO 8 7 1 I F I N L O A O . E Q . O ) GO TO 9 3 3

—COMPUTE NODAL FORCES DUE TO APPLIED LOADINGS

— R E A D I N LOCATION OF APPLIED LOADS

READ 1 5 * 1 3 4 ) ( N L ( I ) * 1 = 1 * NLOAD)

C

c

c

c * * * *

c

c

c 201

C**** c

DO 2 0 1 I . = 1 * NLOAD

-READ I N D I S T R I B U T I O N OF LOAD ON EACH LOADED ELEMENT

READ ( 5 * 1 2 1 ) X D A ( I * 1 I * X D A ( I * 2 l * Y D A ( I * 1 ) * Y 0 A d * 2 )

C c-c

WRITE (6* 122) 00 202 1 = 1 * NLOAO

202 WRITE 16* 123) N L d ) * XDA(I*1)* XDA(I*2)* Y0A(I*1)* YDA(I*2)

FIND FORCE DUE TO APPLIED LIVE LOAD

DO 300 K = 1 * NLOAO IF (NSLAB.EQ.l) GO TO 803 IF (NL(K)-(NX1-1)*NY1I 803* 803* 801

801 II = (NLIK)-11/NY1*1 12 = NL(KI-(I1-11*NY1 NODdl = NL(K)*NY1*I1 GO TO 804

803 II = (NL(K)-1I/NY1 12 = NL(K)-I1*NY1 NOOd) = NL(K)*I1

804 N00(2) = Nao( 1)*1 N00(3) = N0D(1)*NY

00009780 00009790 00009800 00009810 00009820 00009830 00009840 00009850 00009860 00009870 00009880 00009890 00009900 00009910 00009920 00009930 00009940 00009950 00009960 000099 70 00009980 00009990 ^00010000 00010010 00010020 00010030 00010040 00010050 00010060 ^00010070 00010080 00010090 00010100 00010110 00010120 00010130 00010140 00010150 00010160 00010170 00010180 00010190 00010200 00010210 00010220 00010230 00010240 00010250 00010260

144

N0D(4) = N U D ( 3 ) * 1 C 0 0 0 1 0 2 7 0

^ f * D CONTRIBUTING LOADED AREA FOR EACH NODE 00010290 ^ 00010300

00010310 00010320 00010330

C DETERMINE FORCE ACTING ON NODE DUE TO APPLIED LOADING 00010340

A = (X(Il*2)-Xdl*l))/2 B a (Y(I2*l)-Y(I2)l/2

C

OK = Q*A*B 00010350 00010360

00010450 00010460

N0D13 = (N00(l)-ll*3 00010370 NOD33 = (N30(3I-11*3 00010380 X O d ) = XDA(K*2I-XDA(K,1) 00010390 YD(1) = YDA(K,2)-YDA(K,1) 00010400 X0(2) = (XDA(K*2)*X0A(K*l))/2 00010410 YD(2) = (YDA(K,2)*YDA(K,l))/2 00010420 DO 300 I = 1 , 4 00010450 IF (1-2) 210, 220* 230 00010440

210 XI = -1 YI = -1 GO TO 260 00010470

220 XI = -1 00010480 YI = 1 00010490 60 TO 260 00010500

250 IF d-5) 240* 240* 250 00010510 240 XI = 1 00010520

YI = -1 00010530 60 TO 260 00010540

250 XI = 1 00010550 YI « 1 00010560

2 6 0 F O I N O O d ) ) = F O ( N O O ( I ) I * 0 . 2 5 * Q K * ( 1*XI*XD(2I ) * ( 1 * Y I * Y D ( 2 ) 1 * X D ( 1 ) * 0 0 0 1 0 5 7 0 * y O ( l ) 0 0 0 1 0 5 8 0

300 CONTINUE 00010590 C 00010600 C CHECK FOR SECOND L0ADIN6 ON SLAB 00010610 C 00010620

If (NSLOAO.EQ.O) 60 TO 935 00010630 C 00010640 C——COMPUTE NODAL FORCES DUE TO SECOND LOADING 00010650 C -READ IN INTENSITY OF SECOND LOADING AND NUMBER OF ELEMENTS LOADED 00010660 C BV THIS 00010670 C 00010680

READ ( 5 * 149 ) NFLOAD* 02 00010690 C * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 0 0 0 1 0 7 0 0 C 000 10710

WRITE (6* 143) 00010720 WRITE (6* 150) NFLOAD* 02 00010730

C 00010740 C READ IN LOCATION OF SECOND SET OF APPLIED LOADS 00010750

145

READ ( 5 * 104 ) ( N F L ( I ( * 1 = 1 * NFLOAOI

C c-c

——-READ IN DISTRIBUTION OF SECOND LOADING ON EACH LOADED ELEMENT

DO 401 1 = 1 * NFLOAD 401 READ 15* 121 ) X D A 2 ( I * 1 1 * X 0 A 2 ( I * 2 I * Y D A 2 ( I , 1 I , YDA2( 1,21

C * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * c

WRITE ( 6 * 1 5 1 1 00 402 1 = 1 * NFLOAD

402 WRITE ( 6 * 1231 N F L d I * X 0 A 2 ( I * 1 1 * X 0 A 2 ( I * 2 I * V D A 2 ( I * 1 I * YDA2( I *21 C C FIND FORCE DUE TO APPLIED SECOND LOADING C

DO 405 K = 1 * NFLOAD I F ( N S L A B . E Q . l l 60 TO 404 I F I N F L I K 1 - I N X 1 - 1 I * N Y 1 I 404* 404* 405

405 I I = I N F L I K l - 1 1 / N V l * ! 12 = N F L I K I - I I 1 - 1 1 * N V 1 N O O d l = N F L I K 1 * N V 1 * I 1 6 0 TO 406

404 I I = I N F L I K l - l l / N Y l 12 = N f L I K l - I l * N Y l N O D d l = N F L I K I * I 1

406 N0D(2 I = N 0 0 ( 1 I * 1 N00(51 = NOOdl^NY N0D(41 = N0DI514-1

C C-

c -FIND CONTRIBUTING LOADED AREA FOR EACH NODE

A = I K d l * 2 1 - X d l * l l l / 2 B « I Y d 2 * l l - Y ( I 2 1 1 / 2

C C DETERMINE FORCE ACTING ON NODE DUE TO APPLIED LOADING C

0K2 « Q2*A*B N0015 = ( N 0 D ( 1 I - 1 1 * 5 N0055 = ( N 0 D ( 5 1 - 1 1 * 5 X D d l = XDA2(K*2 I -XDA2(K*11 V O I l l = V 0 A 2 ( K * 2 ) - Y D A 2 I K * 1 ) X 0 I 2 ) = I X 0 A 2 I K * 2 ) * X D A 2 I K * l ) ) / 2 Y 0 I 2 ) = I Y D A 2 I K * 2 ) * Y 0 A 2 I K * l l l / 2 00 405 1 = 1 * 4 I F 11 -21 4 0 7 * 4 0 8 * 409

407 X I = - 1 Y I = - 1 6 0 TO 410

00010760 00010770 ^00010780 00010790 00010800 00010810 00010820 00010830 ^00010840 00010850 00010860 00010870 00010880 00010890 00010900 00010910 00010920 00010930 00010940 00010950 00010960 00010970 00010980 00010990 00011000 00011010 00011020 00011030 00011040 00011050 00011060 00011070 00011080 00011090 00011100 00011110 00011120 00011130 00011140 00011150 00011160 00011170 00011180 00011190 00011200 00011210 00011220 00011230 00011240

146

00011250 408 XI = - 1 YI = 1 00011260 GO TO 410 00011270

409 IF 11-31 411* 411* 412 00011280 411 XI = 1 00011290

YI = - 1 00011300 60 TO 410 00011510

412 XI = 1 00011320 YI = I 00011350

410 FOINOOIIII = F O I N O 0 I I l l * 0 . 2 5 * O K 2 * l l * X I * X D l 2 l l * l l * V I * Y D I 2 l l * X D t l l * 00011540 •VOID 00011550

405 CONTINUE 00011560 955 I F INSLAB.EQ. l l 60 TO 871 00011570

DO 875 I = 1 * NY 00011580 875 FOIN015-NY4-I1 = F 0 I N 0 1 3 - N Y * I 1 *F 01 N 0 1 3 * I I 00011390

C 00011400 C FIND D E F L E : T I 0 N DUE TO SLAB WEIGHT AND/OR UNIFORMLY DISTRIBUTED 00011410 C LIVE LOAD 00011420 C 00011430

871 DO 502 1 = 1 * N025 00011440 502 F I ( I - 1 1 * 5 * 1 1 = F O d I 00011450

I F (NLOAD.EO.O.AND.NWT.EQ.O) GO TO 1505 00011460 C 00011470 C COMPUTE DEFLECTIONS AS IF ONLY ONE SLAB EXISTS 00011480 C 00011490

CALL LOAOM ( I I 00011500 1503 CONTINUE 00011510

C 00011520 C UNDER A GIVEN CONTACT CONDITION* APPLY ITERATION PROCESS UNTIL THE00011550 C——-DEFLECTIONS CONVERGE 00011540 C 00011550

510 IC s IC*1 00011560 0 0 2 I = 1* N025 00011570 IF ( IC.EQ.1 .AN0.I .GT.N0131 CO TO 1952 00011580

2 P F d l « F ( ( I - 1 I * 5 * 1 1 00011590 1952 I F INSLAB.EQ. i l 60 TO 526 00011600

IF l i e . E Q . l l 60 TO 511 00011610 0 0 945 I = NOIP* N02 00011620

945 F i l l = 0 00011650 00 947 I = N015P* N025 00011640

947 F I ( I - 1 1 * 5 * 1 1 = F O d l 00011650 C 00011660 C If TWO SLABS EXIST* DETERMINE THE NODAL FORCES IN THE RIGHT SLAB 00011670 C DUE TO THE DEFLECTIONS OF THE SUBGRADE. 00011680 (. 00011690

DO 797 I = N015P* N025 00011700 IF (NCC(I).NE.O) GO TO 797 00011710 IF (I.LE. NOl 3*NY) GO TO 797 ???ff!?? K = I-NY

00011730

http://INSLAB.EQ.il

147

00 795 J = 1 * N023 00011740 I F ( N C C ( J ) . N E . O ) GO TO 795 00011750 I F ( J . 6 T . N 0 1 5 . A N O . J . L E . N 0 1 5 * N Y I GO TO 795 00011760 I F ( J - N 0 1 5 ) 1050* 1050* 1052 00011770

* ° " an'ri i « . . 00011780

, o « - ^°. IV^ 00011790 1052 M « J-MY 00011800 1054 IF I J . 6 T . I -NB/5 .ANO.J .LT.I*NB/3 .AND.J .6T.N013I 60 TO 748 00011810

I f IK-M) 7 i 2 * 742* 744 00011820 742 HH = HIIM*l)*N/2-M*K) 00011850

« 0 TO 754 00011840 744 HH = HIIK*l)*K/2-K*MI 00011850 754 I f INTEMP.EQ.O.ANO.NLOAD.NE.O.ANO.CURLIJI .LE.01 60 TO 746 00011860

f K I - l l * 5 * l l s f ( I I - 1 I * 5 * 1 1 - ( P F I J I - C U R L ( J ) ) * H H 00011870 6 0 TO 795 00011880

746 F ( ( I - 1 ) * 5 * 1 ) = F I I I - 1 I * 5 * 1 1 - P F I J I * H H 00011890 60 TO 795 00011900

748 I F I K - M I 7 8 6 * 786* 788 00011910 786 HH = H ( I M * l l * M / 2 - M * K I 00011920

60 TO 792 00011950 788 HH = H I ( K * l l * K / 2 - K * M l 00011940 792 IF INTEMP.EO.O.ANO.NLOAO.NE.O.AND.CURLIJI .LE.01 60 TO 795 00011950

F I I I - 1 I * 5 * 1 I = F l d - l l * 5 « - l l * C U R L I J}*HH 00011960 795 CONTINUE 00011970 797 CONTINUE 00011980

C 00011990 C——EQUATE THE DEFLECTIONS OF THE JOINT AT THE RI6HT SLAB TO THOSE AT 00012000 C THE LEFT SLAB 00012010 C 00012020

511 00 512 I = N015P* N015NY 00012050 512 F I I I - 1 I * 5 * 1 1 = l .OE 2 0 * I P F I I I * R F J * l f l l I - N Y - l l * 5 * l l - P f ( I l l l 00012040

C 00012050 C — - C O M P U T E DEFLECTIONS OF THE RIGHT SLAB 00012060 C 00012070

CALL LOAOM ( 2 1 00012080 C 00012090 C DETERMINE THE NDOAL FORCES IN THE LEFT SLAB DUE TO THE DEFLECTIONS00012100 C Of THE SUB6RA0E 00012110 C 00012120

526 00 515 I = 1* NOl 00012150 515 f i l l = 0 00012140

00 1100 1 = 1 * N015 00012150 1100 F l d - l l * 3 * l l = F O d l 00012160

001595 I = 1* N015 00012170 IF (NCC(I) .NE.Ol 60 TO 1395 00012180 DO 395 J = 1 * N023 00012190 I F (NCC(J1 .NE.01 GO TO 595 00012200 I F ( J . G T . N 0 1 3 . A N O . J . L E . N 0 1 5 * N Y I 60 TO 395 00012210 IF (J -N015) 2050* 2050* 2032 00012220

148

2050

2052 2054

542

544 554

546

548 586

588 592

595 1595

>AN0.J .LE.N015I 60 TO 548

M = J 6 0 TO 2054 M = J-NY I f ( J . 6 T . I - N B / 5 . A N 0 . J . L T . I * N B / 5 . I f d - M ) 5 4 2 , 542* 544 MM = H((M*l)*M/2-M*II 60 TO 554 HH = H ( ( I * l l * l / 2 - I * M l I f «NTEMP.tO.O.ANO.NLOAD.NE.O.AND.CURL(JI .LE.01 60 TO 546 f 1 ( 1 - 1 1 * 5 * 1 1 = F ( ( I - 1 1 * 5 * 1 I - ( P F ( J I - C U R L ( J I 1 * H H 6 0 TO 595 f ( d - l l * 5 * l l = f ( ( I - l ) * 5 * l ) - P f ( J | * H H 6 0 TO 595 I f (I-MI 586* 586* 588 HH = H((M*ll*M/2-M*Il 6 0 TO 592 HH = H ( ( I * l l * I / 2 - I * M l I f (NTEMP.EQ.O.AND.NLOAD.NE.O.AND.CURL(Jl.LE.O) 60 TO 395 f ( ( I - l ) * 3 * l ) = f ( ( I - 1 I * 3 * 1 I * C U R L I J 1 * H H CONTINUE CONTINUE I f INSLAB.EQ. l ) 60 TO 524

COMPUTE THE VERTICAL NODAL fORCES AL0N6 THE JOINT IN THE LEfT SLA DUE TO THE DEfLECTIONS OF THE RI6HT SLAB

00 516 I = 1 * NY J = I I - 1 1 * N B * 5 * 1 N5 = I N 0 1 5 - N Y - 1 * I 1 * 5 * 1 DO 514 K = 1 * NB JLBK = J - I L A * 1 - K 1 * L A I f I J L B K . L E . O I 60 TO 514 f l N 5 l = F I N 5 1 - C O I J L B K l * F l d - l l * 5 - L A * K * N 0 1 I

514 CONTINUE DO 516 K = 1 * LA I f I I . E 0 . N Y . A N D . K . 6 T . L A - 5 I GO TO 524

516 f l N 5 1 « F I N 5 1 - C 0 ( J * K 1 * F I I I - 1 I * 5 * 1 * K * N 0 1 1 C c-c

c c-c-

COMPUTE DEFLECTIONS OF THE LEFT SLAB

524 CALL LOAOM I I I

CHECK CONVERGENCE UNDER A GIVE CONTACT C0N9ITI0N AND REPEAT THE ——PROCESS IF DESIRED ACCURACY IS NOT OBTAINED. 551 0 0 540 1 = 1 * NCK

I F ( F ( ( N O O C K ( I ) - 1 ) * 3 * 1 ) . E Q . 0 . ) 60 TO 340 If( ABS(OF(NODCKd))/F((NODCK( 11-11* 5*111.6T .OEL.AND. IC. LT.

*ICLI 60 TO 310 540 CONTINUE

00012250 00012240 00012250 00012260 00012270 00012280 00012290 00012500 00012510 00012520 00012550 00012340 00012350 00012360 00012570 00012580 00012590 00012400 00012410 00012420 00012450 00012440 00012450 B00012460 00012470 00012480 00012490 00012500 00012510 00012520 00012550 00012540 00012550 00012560 00012570 00012580 00012590 00012600 00012610 00012620 00012650 00012640 00012650 00012660 00012670 00012680 00012690 00012700 00012710

149

c c-c

I F ( I C C C . E O .

— A S S I 6 N A NEW

DO 54 50 I = C A P d l = F ( ( I F ( 6 A P ( I 1 1

5410 N C C d l = 1 6 0 TO 5 4 3 0

5420 N C C d l = 0 5450 CONTINUE

1.OR.NCYLE.EQ.il 60 TO 9541

CONTACT CONDITION

1* N025 I-11*5*11-CURLI II 5410* 5420* 5420

— C H E C K C0NVER6 REPEAT THE PR

00 5452 I =

IF (F(INOO:K I F l A B S K F d

* . 6 T . O E L . A N D . 54 52 CONTINUE

; UNDER THE F l ; DELF TO OBTA 9 5 4 1 I f d C C C . E Q .

ICCC = I C C C * I C L = ICLF DEL = DELF 6 0 TO 510 I C C 3 I C C * 1

ENCE BETWEEN TWO DIFFERENT CONTACT CONDITIONS AND OCESS I F DESIRED ACCURACY IS NOT OBTAINED

1 * NCK ( I I - l ) * 5 * l ) . E Q . O . I CO TO 5452 NOOCK ( I ) - l ) * 5 * 1 1-PPF ( N O O C K d l I I / F ( I N O O C K d l - 1 1 * 5 * 1 1 1 N I C . L T . N C Y L E l 6 0 TO 965

NAL CONTACT C O N D I T I O N , CHAN6E ICF TO ICLF AND DEL TO I N MORE ACCURATE RESULTS 11 GO TO 5 4 1 1

541

IF I f

( I C C . E Q . l (NUT . E Q . O

— — D E T E R M I N E TH — C U R L ANO THE

958 0 0 9 5 8 1 = 1 CURL I I I = F l

I GO TO 350 ) GO TO 550

E F I N A L CURL AS THE DIFFERENCE BETWEEN THE I N I T I A L DEFLECTION AT EACH NODE

, N025 ( I - 1 ) * 5 * 1 ) - C U R L ( I )

C c-c-c

ON VALUE OF CURL: NEGATIVE SIGN INDICATES GAP BETWEEN —SLAB ANO SUBGRADE —CHANGE SIGN

C c« c

DO 5820 I = 5820 CURLII) = -C 550 CONTINUE

OMPUTE THE

IF (ISOTRY)

1* N025 URLdl

STRESSES AT DESIGNATED NODAL POINTS

950* 8000* 8100

00012720 00012730 00012740 00012750 00012760 000127 70 00012780 00012790 00012800 00012810 00012820 00012830 00012840 00012850 00012860 00012870 00012880 00012890 00012900 00012910 00012920 00012930 00012940 00012950 00012960 00012970 00012980 00012990 00013000 00013010 00013020 00015050 00015040 00015050 00013060 00015070 00013080 00013090 00013100 00013110 00013120 00013130 00013140 00013150 00013160 00013170 00013180 00013190 00013200

http://OR.NCYLE.EQ.il

150

C CALL SUBROUTINE SOLID I F SLAB IS CONSTANT THICKNESS SLAB C

8000 CALL SOLID (NPRINT* NY* N X l * NX* PR* RM* T* NSYM* NKENTl GO TO 8500

C C CALL SUBROUTINE TEE IF SLAB IS STIFFENED SECTION SLAB C

SlOO CALL TEE (NPRINT* NY* NXl* NX* OXYfAC* fRX* TEX* TEY* NSYM* *OVfAC* NKENTl

8500 NKENT = 1 I f d C C . L T . 2 1 GO TO 510

900 CONTINUE 950 CONTINUE

6000 CONTINUE C C PRINT DEfLECTION RESULTS C

WRITE 16* 1241 WRITE ( 6 * 1251 WRITE ( 6 * 1261 ( I * f ( d - l l * 5 * l l * 1 = 1 * N0151 DO 999 1 = 1 * N015

999 C U N A d I = F ( d - l ) * 5 * l l OEFLMX = 0 . 0 DO 996 1 = 1 * N013

996 I F ( O E F L M X . L T . 6 U N A d l l OEFLMX = 6UNA( 11 OEFLMN = GUNAI l l DO 997 1 = 1 * N013 I F IDEFLMN.LT.GUNAdl l 60 TO 997 DEfLMN = 6 U N A d l

997 CONTINUE 0 0 998 1 = 1 * N015

998 I R d l - I CALL VSRTR (6UNA* N015* I R l WRITE 16* 1271 WRITE 16* 1251 WRITE ( 6 * 1261 ( I R d l * 6 U N A ( I 1 * 1 = 1 * N0131 WRITE ( 6 * 1281 DEfLMN* DEfLHX

C c-c

-CALCULATE AND PRINT DIFFERENTIAL DEFLECTIONS

CALL MSLP (N0151

C — — P R I N T STRESS RESULTS C

WRITE 16* 1291 WRITE (6* 1301 DO 887 1 = 1 * NPRINT

887 WRITE 16* 1311 NPdl* (STR(I*J1* J* 1»61

00013210 00013220 00013230 00013240 00013250 00013260 00013270 00013280 00013290 00013500 00015510 00015520 00015550 00015540 00015550 00015560 00015570 00015580 00015390 00013400 00013410 00015420 000154 50 00015440 00015450 00015460 00015470 00015480 000154 90 00015500; 00013510 00015520 00015550 00015540 00015550 00015560 00015570 00015580 00013590 00013600 00013610 00013620 00015630 00013640 00013650 00015660 00015670 00015680 00013690

151

c

c

00 888 I « 1* NXY

C c-c

00013700 C — PRINT MOMENT RESULTS 00015710 C 00015720

WRITE (6* 1321 00015750 WRITE (6* 153) 00013740 WRITE 16* 154) 00015750 DO 886 1 = 1 * NPRINT 00015760

886 WRITE (6* 151) NP(I)* XMOM(I)* YMOMd)* XYMOMd) 00015770 00 776 1 = 1 * NPRINT 00015780

776 I R d ) = I 00015790 CALL VSRTR (XMOM* NPRINT* IRl 00015800 WRITE (6* 1521 00015810 WRITE (6* 1551 00015820 WRITE (6* 1561 00013830 WRITE (6* 1381 (IRdl* XMOMIII* 1= 1* NPRINTl 00015840 00 775 1 = 1 * NPRINT 00015850

775 IRdl = I 00015860 CALL VSRTR IYMOM* NPRINT* IRl 00015870 WRITE (6* 1521 00015880 WRITE (6* 137) 00015890 WRITE (6*1531 (IRdl* YMOMdl* 1= 1* NPRINTl 00015900

00015910 C PRINT SMEAR RESULTS 00015920 C 00015950

WRITE (6* 1591 00015940 WRITE (6* 1401 00013950 WRITE (6* 14110021130 00013960 WRITE (6* 1421 (I* VXdl* I* 1* NXYI 00015970 WRITE (6* 1451 00015980 WRITE (6* 1421 (I* VYdl* 1= 1* NYX1 00015990 WRITE (6* 1591 00014000 WRITE 16* 1441 00014010 WRITE 16* 141) 00014020

00014050 888 I R d l = I 00014040

CALL VSRTR IVK* NXY* I R l 00014050 WRITE 16* 1421 l I R d l * V X I I l * I « 1 * NXYI 00014060 WRITE 16* 1451 ! ? ! } * ! ' ; DO 777 I = 1 * NYX 00014080

777 I R d l = I 00014090 CALL VSRTR IVY* NYX* I R l ° ° ° ! t l ? « WRITE 16* 1421 l I R d l * V Y d l * 1= 1 * HYXl 00014110

00014120

PRINT MAXIMUM SHEAR VALUES ?°°!tlf« 00014140

WRITE 16* 1451 VXMAXN* VXMAXP* VYMAXN* VYMAXP J J J J t J ! °

00014180 END

152

SUBROUTINE SOLID INPRINT* NY* NXl* NX* PR* RM* T* NSYM* NKENTl 00014190 C 00014200 C SUBROUTINE CALCULATES STRESSES FOR SLAB OF CONSTANT THICKNESS 00014210 C 00014220

DIMENSION C11500001*FI21001*6(1500001 *NOl21*XI6501*YI6501«STRI650*00014250 *6>*NPI 6501 *XMOMI6501*YMOMI6501*XYM0M(6501*MXDIF(6501*NYDIF(6501* 00014240 *XT( 6501*VT( 6501*0X16501*DYI6501*MXYDIFI 6501tNVXOIFl6501*VX16501* 000142 50 *VYI6501*YXMOM16501*IRl6501*6UNAI6501 00014260

C 00014270 COMMON C*F*6*N0*NB*X*Y*STR*NP*XM0M*YM0M*XYM0M*XT*YT*DX*0Y*MX0IF* 00014280

*M YD IF*MXYDIF*MYXDIF*YXMOM*VX*VY*NXY*NYX*VXMAXN*VXMAXP*VYMAXN* 00014290 *VYMAXP 00014500

C 00014510 REAL MXOIF* MYOIF* MXYDIF* MYXDIF* MOIX* MOIY 00014520

C 00014330 C 00014540

NOl = NX1*NY*5 00014550 N013 « NOl/3 00014560

C

557 N s NOl

C c-c

00014570 C — — B E 6 I N S0LVIN6 STRESS MATRIX FOR INTERNAL MOMENTS 00014580 C 00014590

DO 450 I = 1* NPRINT 00014400 11 = I N P d l - l l / N Y * l 00014410 12 = N P I I I - I l l - l l * N Y 00014420 IF I N P d l - N 0 l 5 l 555* 555* 557 00014450

555 N = 0 00014440 Ml = 0 00014450 NXX = NXl 00014460 NPl = I N P d l - l l * 5 00014470 GO TO 558 00014480

00014490 Nl = Nal3 00014500 NXX~= NX 00014510 NPl = ( N P ( I I - N 0 1 5 - 1 » * 5 ° ° ° J t H S

558 NP2 = NP1*5*NY S J J l t " ? NP5 = NP1-5*NY Tr.^ltr^

00014550 ZERO OUT OLD STRESS MATRIX 00014570

00014580 00014590 0 0 1359 J = 1* 5

1559 STR(I*J1 = 0 . ftnft,i.*.An C DETERMINE IF NODE IS ON PERIMETER OF SLAB " J J l t ^ ? ; C CORNER N0DE7 OoSiJtJS

IF (NP(II.Ea.NY*NXXI 60 TO 450 J « « , ! t * n C TOP ED6E NODE? 2 S 2 } r l ! S

IF ( N P ( I l . E Q . ( N P d l - l l / N Y * N Y * N Y l 60 TO 575 J S J I t t t o C SIDE ED6E NODE? S S J } ! t ! n

I f (NP(I ) .6E.NV*(NXX-1)*1) 60 TO 420 S J S u t J o C

153

I T " ? " " ; ; ? ^ , " ? ! ? " " ^ " ^ * " " ' • DIMENSIONS AND SOLVE FOR M( I ) 0 0 0 1 4 6 8 0 « - 1 XI 11*1 l - X d l ) ) / 2 0 0 0 1 4 6 9 0 : • ^y[\l*l*'y*l2n/2 0 0 0 1 4 7 0 0 * ° " " • < • 0 0 0 1 4 7 1 0 "O** = B/« 0 0 0 1 4 7 2 0 S T R d * l ) = ( 6 * ( B D A * P R * A D B ) * F ( N * N P 1 * 1 ) - 8 * A * P R * F ( N * N P 1 * 2 ) * 8 * B * F ( N * 0 0 0 1 4 7 3 0

* N P l * 5 l - 6 * A 0 B * P R * f ( N * N P l * 4 l - 4 * A * P R * f ( N * N P 1 * 5 » - 6 * B 0 A * F I N * N P 2 * 1 I * 4 * 0 0 0 1 4 7 4 0 * B * F I N * N P 2 * 5 1 1 / I 4 * A * B 1 0 0 0 1 4 7 5 0

S T R d « 2 1 = ( 6 * ( A 0 B * P R * B 0 A I * F ( N * N P 1 * 1 I - 8 * A * F ( N * N P I * 2 1 * 8 * B * P R * F I N * 0 0 0 1 4 7 6 0 * N P l * 5 l - 6 * A D a * F l N * N P l * 4 1 - 4 * A * F l N * N P l * 5 l - 6 * B D A * P R * F I N * N P 2 * l l * 4 * B * 0 0 0 1 4 7 7 0 * P R * F ( N * N P 2 * 5 I I / ( 4 * A * B 1 0 0 0 1 4 7 8 0

S T R ( I * 5 1 = 0 . 5 * ( 1 - P R I * ( - 2 * F ( N * N P 1 * 1 1 * 4 * B * F ( N * N P 1 * 2 I - 4 * A * F ( N * N P 1 * 0 0 0 1 4 7 9 0 * 5 I * 2 * F ( N * N P 1 * 4 1 * 4 * A * F ( N * N P 1 * 6 1 * 2 * F I N * N P 2 * 1 I - 4 * B * F I N * N P 2 * 2 1 - 2 * F I 0 0 0 1 4 8 0 0 * N * N P 2 * 4 1 1 / I 4 * A * B 1 0 0 0 1 4 8 1 0

——DETERMINE LOCATION OF NODE IN NODAL 6R ID WORK 0 0 0 1 4 8 2 0 I f I N P I I I - 1 - N l l 5 7 0 * 5 6 0 * 5 7 0 0 0 0 1 4 8 5 0

560 CON = 6 0 0 0 1 4 8 4 0 « 0 TO 4 5 5 0 0 0 1 4 8 5 0

I N P I I I . E 0 . I N P I I 1 - 1 I / N Y * N Y * 1 1 6 0 TO 420 0 0 0 1 4 8 6 0 0 0 0 1 4 8 7 0

DETERMINE C0NTRIBUTIN6 AREAL DIMENSIONS ANO SOLVE FOR MIJl 0 0 0 1 4 8 8 0 0 0 0 1 4 8 9 0

A « I X ( I l * l l - X ( I l l } / 2 0 0 0 1 4 9 0 0 B » ( Y ( I 2 l - Y I I 2 - l l l / 2 0 0 0 1 4 9 1 0 AOB = A/B 0 0 0 1 4 9 2 0 BOA = B/A 0 0 0 1 4 9 5 0 S T R d * l l = S T R I I * 1 1 * I - 6 * A D B * P R * F I N * N P 1 - 2 I * 4 * A * P R * F I N * N P 1 - 1 1 * 6 * I 0 0 0 1 4 9 4 0 *B0A*PR*ADB}*FIN*NP1*1I*8*A*PR*FI N*NPl*21*8* B*FI N*NP1*51-6*B0A*FI 000149 50 *N*NP2*1I*4*B*FIN*NP2*511/I4*A*BI 00014960

S T R d * 2 l = S T R I I » 2 1 * I - 6 * A D B * F I N * N P 1 - 2 I * 4 * A * F I N * N P 1 - 1 1 * 6 * I A 0 B * P R * 0 0 0 1 4 9 7 0 *BDAI*f l N * N P l * l l * 8 * A * f I N*NP1*2 l*8*B*PR*f IN*NP1 *5 1-6* BDA*PR*f I N*NP2*000 1 4 9 8 0 * 1 I * 4 * B * P R * F I N * N P 2 * 5 I I / I 4 * A * B 1 0 0 0 1 4 9 9 0

STRI 1*5) = S T R l I * 5 1 * 0 . 5 * d - P R l * ( - 2 * F ( N * N P l - 2 1 - 4 * A * F ( N * N P l l * 2 * F ( 0 0 0 1 5 0 0 0 * N * N P l * l l * 4 * B * F ( N * N P l * 2 l * 4 * A * F ( N * N P l * 5 1 * 2 f t F ( N * N P 2 - 2 1 - 2 * F ( N * N P 2 * l l - 0 0 0 1 5 0 1 0 * 4 * B * F ( N * N P 2 * 2 1 I / ( 4 * A * B I 0 0 0 1 5 0 2 0

C — — — — D E T E R M I N E LOCATION OF NODE IN NODAL 6RI0W0RK 0 0 0 1 5 0 5 0 I f ( N P d l . E O . N Y * N l l GO TO 5 6 0 0 0 0 1 5 0 4 0 I F ( N P d l . E a . ( N P ( I I - l l / N V * N Y * N Y I GO TO 4 5 0 0 0 0 1 5 0 5 0 I f ( N P d I - N Y - N l l 4 1 0 * 4 1 0 * 4 2 0 0 0 0 1 5 0 6 0

4 1 0 CON = 5 0 0 0 1 5 0 7 0 GO TO 455 0 0 0 1 5 0 8 0

C 0 0 0 1 5 0 9 0 C — — — D E T E R M I N E CONTRIBUTING AREAL DIMENSIONS AND SOLVE FOR M(Kl 0 0 0 1 5 1 0 0 C 0 0 0 1 5 1 1 0

4 2 0 A « ( X ( I l l - X ( I l - l l l / 2 0 0 0 1 5 1 2 0 B » ( Y ( I 2 * l l - Y d 2 l ) / 2 0 0 0 1 5 1 3 0 ADB « A/B 0 0 0 1 5 1 4 0 BOA « B/A 0 0 0 1 5 1 5 0 S T R I I * 1 1 * S T R I I * 1 1 * I - 6 * B 0 A * F I N * N P 3 * 1 1 - 4 * B * F I N * N P 5 * 5 1 * 6 * I B 0 A * P R * 0 0 0 1 5 1 6 0

154

c c-c

* 4 * A * f l N * N P l * 6 1 l / l 4 * A * B l I f I N P I I ) . E Q . N V * I N X X - 1 I * 1 1 GO TO 560 I f l N P d l - l N P I I l - l ) / N Y * N Y - l ) 4 3 0 * 4 1 0 4 3 0

—DETERMINE CONTRIBUTING AREAL DIMENSIONS AND SOLVE FOR M I L )

430

00015240 00015250 00015260 00015270 00015280 00015290 00015300 00015310 00015320 00015330 00015340 00015350 00015360

B = IY(I2)-V(I2-l))/2 A = IX(ri)-X(Il-l))/2 AOB = A/B BOA = B/A

STRd*l) = STR(I*1)*(-6*BDA*F(N*NP3*1)-4*B*F(N*NP3*3)-6*A0B*PR*F( *N*NP1-2)*4*A*PR*F(N*NP1-1I*6*(BDA*PR*AD3)*F(N*NP1*1)*S*A*PR*F(N* *NP1*2)-8*B*F(N*NP1*3))/(4*A*B) STRd*2) = STR(I*2)*(-6*8OA*PR*F(N*NP5*ll-4*B*PR*F(N*NP3*31-6*AOB*00015370 • F (N*NPI-2)*4 *A*F(N*NP1-11*6*(A0B*PR*B0AI*F( N*NP1*11*8*A*F(N*NP1* 000153 80 *21-8*B*PR*F( N*NP1*311/(4*A*B1 STR(I*31 = STR(I*31*0.5*(1-PRI*(-2*F(N*NP3-21*2*F(N*NP3*1I*4*B*F(

*N*NP5*21*2*F(N*NP1-21-4*A*F(N*NP1I-2*FIN*N*1*1I-4*8*FIN*NP1*21* *4*A*FIN*NP1*5)1/(4*A*B)

DETERMINE LOCATION OF NODE IN NODAL GRID WORK IF (NP( I).Ea.NY*NXX) 60 TO 360 IF (NP(I).EO.(NP(I)-1)/NY*NY*NY.OR.NP(I).GE.NY*(NXX-11*11

*6 0 TO 410 CON = 1.5

SOLUTION OF STRESS MATRIX COMPLETED

CONVERT INTERNAL MOMENT TO STRESS

455 DO 4 56 J = 1 * 5 4 5 6 S T R ( I * J 1 ' R M * C 0 N * S T R ( I * J 1 / T * * 2

I F ( N P ( I I . E a . ( N P ( I I - 1 1 / N Y * N Y * 1 . A N 0 . N S Y M . N E . 5 . A N D . N S Y M . N E . 4 . a R . * N P d l . E Q . ( i l P ( I l - l l / N Y * N Y * NYI S T R d * 2 1 = 0 .

I F ( N P d I . L E . N Y . A N D . N S Y M . N E . 2 . A N D . N S Y M . N E . 4 . 0 R . N P ( I 1 . G T . ( N X 1 - 1 ) * N V 0 0 0 1 5 5 70

00015390 00015400 00015410 00015420 00015430 00015440 00015450 00015460 00015470 00015480 00015490 00015500 00015510 00015520 00015530 00015540 00015550 00015560

9 8 7 6 4 50

* . A N D . N P d I . L E . ( N X 1 * 1 I * N Y . O R . N P ( I I . S T . ( N X - 1 I *NY) S T R ( I * 1 I = I F ( S T R ( I * 1 ) . E Q . O . O R . S T R d * 2 ) . E Q . O ) S T R ( I * 3 ) = 0 .

—DETERMINE MAJOR ANO MINOR STRESSES AND HORIZONTAL SHEAR X-Y S T R ( I * 6 ) = S Q R r ( 0 . 2 5 * ( S T R ( I * 1 ) - S T R ( I * 2 ) I * * 2 * S T R ( I * 3 ) * * 2 ) S T R ( I * 4 ) = I S T R ( I * l ) * S T R ( I * 2 ) ) / 2 . - S T R ( 1 * 6 ) S T R ( I * 5 ) = ( S T R ( I * l ) * S T R d * 2 ) ) / 2 . * S T R ( I * 6 ) CONTINUE CONTINUE

0.

PLANE

00015580 00015590 00015600 00015610 00015620 00015630 00015640 00015650

155

^ 0 0 0 1 5 6 6 0 C CONVERT STRESS TO MOMENT 0 0 0 1 5 6 7 0 ^ 0 0 0 1 5 6 8 0

PZOP = T * T / 6 . 0 0 0 1 5 6 9 0 DO 475 LP = 1 *NPRINT 0 0 0 1 5 7 0 0 XMOM(LP) = S T R ( L P * 1 ) * P Z 0 P 0 0 0 1 5 7 1 0 YHOM(LP) = S T R ( L P » 2 ) * P Z Q P 0 0 0 1 5 7 2 0 XYMOM(LP) = S T R ( L P * 3 ) * P Z Q P 0 0 0 1 5 7 3 0

475 CONTINUE 0 0 0 1 5 7 4 0 I F ( N K E N T . E Q . O ) GO TO 6 0 0 0 0 0 1 5 7 5 0

C 0 0 0 1 5 7 6 0 C CALCULATE SHEAR FORCES 0 0 0 1 5 7 7 0 C 0 0 0 1 5 7 8 0

CALL SHEAR (NX* NY) 00015790 C 00015800

600 RETURN 0 0 0 1 5 8 1 0 END 0 0 0 1 5 8 2 0 SUBROUTINE TEE ( N P R I N T * NY* N X l * NX* DXYFAC* FRX, TEX, TEY, NSYM, 0 0 0 1 5 8 3 0

* O Y F A C , NKENT) 0 0 0 1 5 8 4 0 C 0 0 0 1 5 8 5 0 C SUBROUTINE CALCULATES STRESSES FOR SLAB WITH GRADE BEAMS 0 0 0 1 5 8 6 0 C 00015870

DIMENSION C( 1 5 0 0 0 0 ) , F ( 2 1 0 0 I * G ( 1 5 0 0 0 0 1 * NO(21 *X ( 6 5 0 1 * Y ( 6 5 0 ) » S T R ( 6 5 0 , 0 0 0 1 5 8 8 0 * 6 1 * N P ( 6 5 0 I * X M O M ( 6 5 0 ) * Y M O M ( 6 5 0 ) * X Y M O M ( 6 5 0 1 * M X O I F ( 6 5 0 ) * M Y O I F ( 6 5 0 ) * 0 0 0 1 5 8 9 0 * X T ( 6 5 0 ) * Y T I 6 5 0 1 * 0 X ( 6 5 0 ) * 0 Y ( 6 5 0 ) * M X Y D I F ( 6 5 0 ) * M V X D I F ( 6 5 0 ) * V X ( 6 5 0 ) * 0 0 0 1 5 9 0 0 *V Y ( 6 5 0 ) , y X M 0 M ( 6 5 0 ) * I R ( 6 5 0 ) *CUNA(65.0) 0 0 0 1 5 9 1 0

C 0 0 0 1 5 9 2 0 COMMON C * F * G * N O , N B * X * Y * S T R * N P * X M O M * Y M O N * X Y M O M * X T * Y T * O X , 0 V , M X 0 I F , 0 0 0 1 5 9 30

*MYD I F * M X Y D I F *MYXOIFtVXMOM*VX*VY*NXY* NY X*VXMAXN*VXMAXP,VYMAXN, 0 0 0 1 5 ^ 4 0 *VYMAXP 0 0 0 1 5 9 5 0

C 0 0 0 1 5 9 6 0 REAL MXOIF* HYDIF* MXYDIF* MYXDIF* MOIX, M3 lY 00015970

C 00015980 C 00015990

NOl = NX1*NY*3 00016000 N015 = NOl/5 00016010

C 00016020 C BEGIN SOLVING STRESS MATRIX FOR INTERNAL MOMENTS 0 0 0 1 6 0 3 0 C 0 0 0 1 6 0 4 0

0 0 5 0 0 1 = 1 , NPRINT 0 0 0 1 6 0 5 0 11 = INPI I I - 1 1 / N Y * 1 00016060 1 2 = N P d l - l I 1 - 1 1 * N Y 0 0 0 1 6 0 7 0 I F ( N P ( I ) - N 0 1 3 1 5 * 5* 10 0 0 0 1 6 0 8 0

5 N = 0 N l = 0 NXX = NXl N P l = ( N P d ) - l ) * 3 GO TO 15

10 N = NOl

00016090 00016100 00016110 00016120 00016150 00016140

156

Nl = N013 NXX = NX NPl = (NP(I)-N013-1)*3

15 NP2 = NP1*3*NY NP3 = NP1-3*NY

C C ZERO OUT OLD STRESS MATRIX

00016150 00016160 00016170 00016180 00016190 00016200 00016210 00016220 00016230 00016240

.,, 00016250 C DETERMINE IF NODE IS ON PERIMETER OF SLAB 00016260

C 00 50 J = 1*

50 STR(I*J) = 0 C

c C — — — C O R N E R NODE?

00016270 00016280

IF (NPd) .Ea.NV*NXX) 60 TO 300 00016290 C TOP EDGE NODE? 00016300

IF (NP(I).EO.(NP(I)-1)/NY*NY*NY) 60 TO 120 00016310 C SIDE E06E N00E7 00016320

IF INP(I).6E.NV*(NXX-1(*1) 60 TO 210 00016330 C 00016340 C DETERMINE CONTRIBUTING AREAL DIMENSIONS AND SOLVE FOR M d ) 00016350 C 00016360

A « (Xdl*l)-X(Il))/2. 00016370 B = (Y(I2*l)-Y(I2)l/2 00016380 DENOM = 4.*A*B 00016390 ADB = A/3 00016400 BDA = B/A 00016410 STR(I*11 = I6*BDA*F(N*NP1*1 )*8*B*F(N*NP1*31-6*BDA*F(N*NP2*11*4*B* 00016420 *F(N*NP2*3ll/OENOM 00016430 STR(I*2I = (6*ADB*F(N*NP1*11-8*A*F(N*NP1*21-6*A0B*F(N*NP1*41-4*A* 00016440

* F ( N * N P l * 5 1 l * (0YFAC/0ENaNI 000164 50 S T R ( I * 3 1 = D X Y F A C * ( - 2 * F ( N * N P 1 * 1 I * 4 * B * F ( N * N P 1 * 2 I - 4 * A * F ( N * N P 1 * 3 1 * 2 * 00016460

* F ( N * N P 1 * 4 1 * 4 * A * F ( N * N P 1 * 6 ) * 2 * F ( N * N P 2 * 1 ) - 4 * B * F ( N * N P 2 * 2 ) - 2 * F I N*NP2* 000164 70 * 4 ) 1 / ( 4 * A * B ) 00016480

C — D E T E R M I N E LOCATION OF NODE I N NODAL GRIOWORK 00016490 I F ( N P ( I ) - l - N l } 110* 100* 110 00016500

100 CON = 6 00016510 6 0 TO 400 00016520

110 I F (NPl I ) . E Q . ( N P ( I ) - 1 ) / N Y * N Y * 1 ) 60 TO 210 00016530 C 00016540 C DETERMINE CONTRIBUTING AREAL DIMENSIONS ANO SOLVE FOR MIJ I 00016550 C 00016560

120 A = ( X ( I l * l ) - X ( I l ) ) / 2 00016570 B = ( Y d 2 ) - Y ( l 2 - l ) ) / 2 00016580 DENOM = 4 . * A * B 00016590 AOB = A/B 00016600 BOA = B/A 00016610 S T R ( I * 1 ) = S T R d * l ) * ( 6 * B D A * F ( N*NP1*1) *8*B*F ( N * N P 1 * 3 ) - 6 * BDA*F ( N*NP2000 16620

* * 1 ) * 4 * B * F ( * 4 * N P 2 * 3 ) ) / D E N O M 000166 50

157

S r R d * 2 ) = S T R ( I * 2 I * ( - 6 * A 0 B * F ( N * N P 1 - 2 1 * 4 * A * F ( N * N P 1 - 1 ) * 6 * A D B * F ( N * 0 0 0 1 6 6 4 0 * N P 1 * 1 ) * 8 * A * F ( N * N P 1 * 2 ) ) * ( 0 Y F A C / 0 E N 0 M 1 0 0 0 1 6 6 5 0

S T R d * 3 } = S T R ( I » 3 1 * a X Y F A C * ( - 2 * F ( N * N P l - 2 1 - 4 * A * F ( N * N P l l * 2 * F ( N * N P l * 0 0 0 1 6 6 6 0 *l)*4*B*F(N*NPl*2t*4*A*F(N*NPl*5l*2*F(N*NP2-21-2*F(N*NP2*ll-4*B* 00016670 *F(N*NP2*?I1/(4*A*B1 00016680

C — ~ ~ DETERMINE LOCATION OF NODE I N NODAL GRIOWORK 0 0 0 1 6 6 9 0 I F ( N P ( I l . E a . N Y * N l l GO TO 100 0 0 0 1 6 7 0 0 I F ( N P ( I ) . E Q . ( N P ( I ) - 1 ) / N Y * N Y * N Y ) GO TO 5 0 0 0 0 0 1 6 7 1 0 I F ( N P ( I ) - N Y - N 1 ) 2 0 0 * 2 0 0 * 2 1 0 0 0 0 1 6 7 2 0

200 CON = 3

C

0 0 0 1 6 7 3 0 50 TO 400 00016740

0 0 0 1 6 7 5 0 C DETERMINE CONTRIBUTING AREAL O IMENSI f f iS ANO SOLVE FOR M(K) 0 0 0 1 6 7 6 0 ^ 0 0 0 1 6 7 7 0

210 A = ( X ( I l ) - X ( I l - l ) ) / 2 00016780 B = ( Y d 2 * l ) - Y ( I 2 ) ) / 2 0 0 0 1 6 7 9 0 DENOM = 4 . * A * B 00016« i00 AOB = A/B 0 0 0 1 6 8 1 0 BOA = B/A 0 0 0 1 6 8 2 0 STRI 1 * 1 ) = S T R d * 1 ) * ( - 6 * B 0 A * F ( N * N P 3 * 1 ) - 4 * B » F ( N * N P 3 * 3 ) * & * B O A * F ( N * 0 0 0 1 6 8 3 0

* N P 1 * 1 ) - 8 * 3 * F ( N * N P 1 * 3 ) ) / D E N 0 M 0 0 0 1 6 8 4 0 S T R d * 2 ) = S T R ( I * 2 ) * ( 6 * A D a * F ( N * N P l * l ) - 8 * A * F ( N * N P l * 2 ) - 6 * A0B*F ( N * N P 1 0 0 0 168 50

* * 4 ) - 4 * A * F ( N * N P l * 5 ) ) * ( 0 Y F A C / D E N 0 M l 0 0 0 1 6 8 6 0 S I R ( 1 * 5 ) = S T R d *31 *DXYFAC* ( - 2 * F ( N * N P 3 * 1 1 * ( * B * F ( N * N P 3 * 2 ) * 2 * F ( N * N P 3 0 0 0 168 70

* * 4 ) * 2 * F ( N * N P 1 * 1 ) - 4 * B * F ( N * N P 1 * 2 ) - 4 * A * F ( N * N P 1 * 3 ) - 2 * F ( N * N P 1 * 4 ) * 4 * A * 0 0 0 1 6 8 3 0 * F ( N * N P 1 * 6 ) ) / ( 4 * A * B ) 0 0 0 1 6 3 9 0

I F ( N P d ) . E a . N Y * ( N X X - l ) * l ) 60 T3 100 0 0 0 1 6 9 0 0 I F ( N P d ) - ( N P d ) - l ) / N Y * N Y - l ) 3 0 0 * 2 0 0 * 3 0 0 0 0 0 1 6 9 1 0

C 0 0 0 1 6 9 2 0 C DETERMINE CONTRIBUTING AREAL DIMENSIONS AND SOLVE FOR MIL) 00016930 C 00016940

300 B = ( Y ( I 2 ) - Y ( I 2 - l ) l / 2 0 0 0 1 6 9 5 0 A = ( X d l l - X ( I l - l l l / 2 0 0 0 1 6 9 6 0 DENOM s 4 . * A * B 0 0 0 1 6 9 7 0 ADB = A/B 0 0 0 1 6 9 8 0 BOA = B/A 0 0 0 1 6 9 9 0 S T R d * l l = S T R d * 1 I * ( - 6 * B 0 A * F ( N * N P 3 * 1 1 - 4 * B * F ( N * N P 3 * 5 } * 6 * B D A * F ( N * 0 0 0 1 7 0 0 0

* N P 1 * 1 1 - 8 * S * F ( N * N P 1 * 5 I I / 0 E N 0 M 0 0 0 1 7 0 1 0 S T R ( I * 2 1 = S T R ( I * 2 1 * ( - 6 * A 0 B * F ( N * N P 1 - 2 1 * 4 * A » F ( N * N P 1 - 1 I * 6 * A D B * F ( N * 0 0 0 1 7 0 2 0

* N P 1 * 1 1 * 8 * A * F ( N * N P 1 * 2 I I * ( O Y F A C / D E N O M I 0 0 0 1 7 0 50 S T R ( I * 5 I = STRI I * 5 1 * D X Y F A C * ( - 2 * F ( N * N P 3 - 2 I * 2 * F ( N * N P 3 * 1 1 * 4 * B * F ( N * N P 3 0 0 0 1 7 0 4 0

* * 2 ) * 2 * F (N*»|P 1 - 2 ) - 4 * A * F I N*NP l l - 2 * F ( N * M P l * l 1 - 4 * B * F ( N * N P l * 2 ) * 4 * A*F I N * 0 0 0 1 70 50 * N P 1 * 3 1 I / ( 4 * A * B ) 0 0 0 1 7 0 6 0

C 00017070 C DETERMINE LOCATION OF NODE I N NODAL GRIOWORK 0 0 0 1 7 0 8 0 C 0 0 0 1 7 0 9 0

IF INP(II.Ea.NY*NXXl GO TO 100 00017100 IF (NP(I).EQ.(NP(I)-1)/NY*NY*NY.0R. NPd).6£.NY*(NXX-l)*l) 00017110

*C0 TO 200 00017120

158

CON = 1.5 00017150 C 00017140 C SOLUTION OF STRESS MATRIX COMPLETED 00017150 C 00017160 C — — C O N V E R T INTERNAL MOMENT TO STRESS 00017170 C 00017180

400 STR(I*1) = ( FRX*C0N*STR(I*1 ))/dEX**2) 00017190 STR(I*2) = (FRX*CON*STR(I*211/(rEY**21 00017200

450 STR(I*3) = I FRX*CON*SrR(I»3))/(TEX**2} 00017210 IF (NP( I).Ea.(NP( I)-1)/NY*NY*1.AN0.NSYM.NE. 3. AND. NS YM.NE. 4.0K. NP( 1000172 20

*I.EQ.(NP( I)-1)/NY*NY*NV) STR(I*2) = 0. 00017230 IF (NPl II.LE.NY.AN0.NSVN.NE.2.AND.NSYM.NE.4.3R.NP(I).CT.I NX 1-1)*NY000172 4O

*.AND.NP(I).LE.(NXl*l)*NY.OR.NPd).CT.(NX-l) *NYI STR(I*1) = 0. 00017250 IF (STR(I*1) .EQ.0.0R.STR(I*2I.EQ.01 STR(I*31 = 0. 00017260

C 00017270 C —DETERMINE MAJOR ANO MINOR STRESSES AND HORIZONTAL SHEAR (X-Y PLANE00017230 C 00017290

STRd*6) = SQRT(0.25*(STR(I*l)-STRd*2))**2*STR(I*3)**2) 00017300 STRd*4) = ( STRd*l)*STR(I*2) )/2.-STR(I*6) 00017310 STRd*5) = ( STR(I*1)*STR(I,2) )/2.*STRd*6) 00017320

9876 CONTINUE 00017330 500 CONTINUE 00017340

C 00017350 C CONVERT STRESS TO MOMENT 00017360 C 00017370

00 490 1 = 1 * NPRINT 00017380 XMOMd) = ( S T R d * l ) * T E X * * 2 ) / 6 . 00017390 YMOMd) = ( S T R d » 2 ) * T E Y * * 2 ) / 6 . 00017400 XYMOMd) = ( STRI I , 5 l * T E X * * 2 ) / 6 . 00017410

490 CONTINUE 00017420 I F (NKENT.EQ.O) GO TO 600 00017430

C 00017440 C CALCULATE SHEAR FORCES 00017450 C 00017460

CALL SHEAR (NX* NY) 00017470 (. 00017480

600 RETURN 00017490 gÔ 00017500 SUBROUTINE MfSO (A* N* EPS* lER* NT I 00017510

g 00017520 C SUBROUTINE IS ALGORITHM TO FACTOR A SYMMETRICAL POSITIVE DEFINITE 00017530 C MATRIX ••.'"> ^ DIMENSION A(NT) °'!!i\lltl - 00017570

DOUBLE PRECISION OPIV* OSUM* DSQRT, 08LE So017590 C TEST ON WRONG INPUT PARAMETER N 2°21It?«

00017540 00017550

C 00017610

159

1 lER = 0 0 0 0 1 7 6 2 0 C 0 0 0 1 7 6 3 0 C I N I T I A L I Z E DIAGONAL LOOP 0 0 0 1 7 6 4 0 C 0 0 0 1 7 6 5 0

K P I V = 0 0 0 0 1 7 6 6 0 DO 11 K = 1 , N 0 0 0 1 7 6 7 0 K P I V = K P I V * K 0 0 0 1 7 6 8 0 INO = KPIV 0 0 0 1 7 6 9 0 LEND = K - 1 0 0 0 1 7 7 0 0

C 0 0 0 1 7 7 1 0 C CALCULATE TOLERANCE 0 0 0 1 7 7 2 0 C 0 0 0 1 7 7 3 0

TOL = A B S ( E P S * A ( K P I V ) ) 0 0 0 1 7 7 4 0 C 00017750 C START F A C T 3 R I Z A T I 0 N LOOP OVER THE KTH ROW 0 0 0 1 7 7 6 0 C 0 0 0 1 7 7 7 0

DO 11 I = K, N 0 0 0 1 7 7 8 0 OSUM = 0 . 0 3 0 0 0 1 7 7 9 0 I F (LEND) 2 , 4 * 2 0 0 0 1 7 8 0 0

C 0 0 0 1 7 8 1 0 C START INNER LOOP 0 0 0 1 7 8 2 0 C 00017830

2 DO 3 L = 1 , LEND 0 0 0 1 7 8 4 0 LANF = K P I V - L L I N O = I N D - L

3 OSUM = D S U M * D B L E ( A ( L A N F ) * A ( L I N O I ) 0 0 0 1 7 8 7 0 000 17830

END OF INNER LOOP 0 0 0 1 7 8 9 0 0 0 0 1 7 9 0 0

C 9 OP IV = DSQRTIOSUMl

A ( K P I V 1 = D P I V O P I V 2 l . D O / O P I V GO TO 1 1

C C CALCULATE TERMS I N ROW

0 0 0 1 7 8 5 0 0 0 0 1 7 8 6 0

C c-c C TRANSFORM ELEMENT A I I N D I ^ ^ ' ^ I ' ^ l ^ g 0 0 0 1 7 9 2 0

4 OSUM = 08LE I A d N O U - D S U H ? ? ? ^ I ? ? ? IF d - K l 10* 5* 10

C C TEST FOR NEGATIVE PIVOT ELEMENT ANO FOR LOSS OF S IGNIF ICANCE C

5 I F I S N 6 L I O S U M l - T O L l 6 * 6* 9 6 I F (OSUM) 1 2 * 1 2 * 7 7 I F ( l E R l 8* 8 * 9 8 lER = K-1

C C COMPUTE PIVOT ELEMENT

00017940 00017950 00017960 00017970 00017980 00017990 00018000 00018010 00018020 00018030 00018040 00018050 00018060 00013070 00018080 00018090 00013100

160

C 0 0 0 1 8 1 1 0 10 AIINDI * DSUM*DPIV 0 0 0 1 8 1 2 0 11 INO = IND*I 0 0 0 1 8 1 3 0

C 0 0 0 1 8 1 4 0 C — — E N D OF DIAGONAL LOOP 0 0 0 1 8 1 5 0 C 0 0 0 1 8 1 6 0

RETURN 0 0 0 1 8 1 7 0 C 0 0 0 1 8 1 8 0 C NEGATIVE PIVOT ELEMENT FOUND 0 0 0 1 8 1 9 0 C 0 0 0 1 8 2 0 0

12 lER = -1 00018210 RETURN 00018220 END 00018230 SUBROUTINE TRIG (Nl 00018240

C 00018250 C THIS SUBROUTINE APPLIES THE GAUSS ELIMINATION METHOD TO FORM A 00018260 C UPPER TRIANGULAR BAND MATRIX. FOR A GIVEN CONTACT CONDITION* THIS00018270 C TRIAN6ULATI0N IS PERFORMED ONLY ONCE AND THE RESULTS CAN BE USED 00018280 C REPEATEDLY. 000182 90 C 0 0 0 1 8 3 0 0

DIMENSION C( 1 5 0 0 0 0 I * F ( 2 1 0 0 I * 6 ( 1 5 0 0 0 0 I * N O ( 2 1 * X ( 6 5 0 1 *YI6501 * S T R I 6 5 0 * 0 0 0 1 8 3 1 0 *6 l*NP(650)*XMOM(650l*YMOM(6501*XYMOM(650l*MXDIFI650)*MY0IFI650l* 0 0 0 1 8 3 2 0 *XTI 6501*YTI 6501*OKI 6501*DYI6501*MXVOIF 16501*MYXDIFI650)*VX1650)* 0 0 0 1 8 5 50 *VVI650)*YXM0MI65 0 ) * I R I 6 5 0 ) * 6 U N A I 6 5 0 I 0 0 0 1 8 5 4 0

C 0 0 0 1 8 3 5 0 COMMON C*F*G*NO*NB*X*Y*STR*NP*XMOM*YMOM*XYMOM*XT*YT*OX*OY*MXDIF* 00018360 *M VO IF *MXYDIF* MYXDIF* YKMOM*VX*VY* NXY *NYX* VXMAXN* VXMAXP* VYMAXN* 00018570 *VYMAXP 00018580

C 00018590 MOIX* MOIY 00018400

00018410 00018420 000184 50 00018440 00018450 00018460 00018470 00018480 00018490 00018500

fi'j* ' ciJi ;;;!!cis 6 i j * K i = c ( j * K i ; « ! ; ! c ? ;

10 c i j * K i = c i j * K i / 6 i j i ? ? „ ? , : : LC = LA DO 50 L = 1* LA MI = J*L*NB-1 DO 2 0 K = 1* LC

20 CIMI*KI = CIMI*K1-CIJ*K*LA-LC1*6IJ*L1

REAL MXOIF* MYOIf* MXYOIf*

NOE - NOINI LA = NB-1 LB = NOE-LA I f I N - 1 1 2* 2 * 4

2 I B = 1 6 0 TO 6

4 I B = N 0 ( 1 1 * 1 6 00 50 I = I B * LB

J = NB*t-LA

MYXOIf *

00018550 00018560 00018570 00018580 00018590

161

1 S ~ ô"I 00018600 1 J * , \ 00018610 LO = LA-1 00018620 00 90 I = LB* NOE 00018650

C

J » NB*I-LA 00018640 6IJ)= CIJI 00018650 If (1-LDI 52* 52, 56 00018660

52 DO 54 K = 1* LD 00018670 54 6(J*K1 = C(J*KI 00018680 56 DO 40 K = 1* LA 00018690 40 C(J*KI = C(J*KI/6(J1 00018700

LC = LA-1 00018710 If (I-NOEI 50* 90* 50 00018720

50 DO 70 L = 1* LD 00018750 MI = J*L*NB-1 00018740 00 60 K = 1 * LC 00018750

60 C ( M I * K I = C ( M I * K > - C ( J * K * L A - L C - 1 I * 6 ( J - H . I 00018760 LC = LC-1 00018770 I f d - L C l 7 0 * 7 0 * 80 00018780

70 CONTINUE 000 18790 80 LO = LO-1 00018800 90 CONTINUE 00018810

RETURN 00018820 END 00018850 SUBROUTINE LOAOM (N l 00018840

00018850 SUBROUTINE USED THE TR lANGULARIZ ED MATRIX FROM SUBROUTINE 00018860

-TRIG AND COMPUTES THE DEFLECTIONS OF THE SLAB. 00018870 00018880

DIMENSION C ( 1 5 0 0 0 0 1 * F ( 2 1 0 0 1 * G ( 1 5 0 0 0 0 1 * N 0 ( 21 * X ( 6 5 0 1 * Y ( 6 5 0 ) * S T R ( 6 5 0 * 0 0 0 1 8 8 9 0 *6 l *NP(6S01*XMOMI6S0l*YMOMI650l*XYMOMI650l*MXDIF(6501*MYDIf ( 6 5 0 1 * 00018900 * X T I 6 5 0 I * Y T ( 6 5 0 1 * 0 X ( 6 5 0 1 * D Y ( 65 01 * MXYOIf (6501 *MYXOIf( 6501 *VX (6501 * 00018910 * V V I 6 5 0 1 *YXMaMI6S0 l * IR(6501 •6UNA(6501 00018920

00018950 COMMON C,f,G*NO*NB*X*Y*STR*NP*XMOM*VMOM*XYMOM*XT*YT*DX*DY*MX0IF* 000189 40

*HYD IF *MXVDIF*MYXDIF*VXMOM*VX*W* NXY *NYX* VXMAXN* VXMAXP* VYMAXN* 0001895 0 *VYMAXP 00018960

0 0 0 1 8 9 7 0 REAL MXOIF* MYOIF* MXYDIF* MYXOIF* MOIX* MOIY 0 0 0 1 8 9 8 0

00018990 NOE = NOINI 00019000 LA = NB-1 00019010 LB = NOE-LA 00019020 I F ( M - l l 2* 2 * 4 00019030

00019040 C ONLY ONE SLAB TO BE CONSIDERED 00019050 C 00019060

2 IB = 1 00019070 60 TO 6 00019080

162

c c-c

c C ' c

c c-C" c

c c-

— T W O SLABS TO BE CONSIDERED

4 IB = N0(11*1 6 DO 20 I = I B * LB

J * NB*I-LA f d l = f ( I t / 6 ( J l DO 20 L = 1 * LA

20 f d * L l = F d * L l - F I I I * 6 l J*L1 LB = LB*1 LD = LA-1 0 0 60 I = LB * NOE J = NB*I -LA F d l s F ( I 1 / G ( J 1 LC = LA-1 I F ( I - N O E l 5 0 * 7 0 * 50

50 DO 40 L = 1 * LD F d * L I = F I I * L l - f d l * 6 l J*L1 LC = LC-1 I f l l - L C I 4 0 * 4 0 * 50

40 CONTINUE 50 LO = LO-1 60 CONTINUE 70 0 0 80 IK = I B * NOE

1 = N 0 E - I K * 1 * I N - l l * N O d l J s NB* I -NB*2 00 80 K = 1 * LA I f I I * K . 6 T . N 0 E 1 60 TO 80 f d l = f ( I l - f ( I * K I * C ( J * K - l l

80 CONTINUE RETURN END SUBROUTINE SINV (A* N* EPS* lER* NT I

THIS SUBROUTINE INVERTS A SYMMETRICAL POSITIVE DEf lNITE MATRIX

DIMENSION AINTI

DOUBLE PRECISION DIN* WORK* DOLE

fACTORIZE GIVEN MATRIX BY MEANS OF SUBROUTINE MFSD — A » TRANSPDSE(T1*T

CALL MFSD (A* N* EPS* lER* NT 1

CHECK FOR INSTABILITY IF (lERl 9* 1* 1

INVERT UPPER TRIANGULAR MATRIX "T*

00019090 00019100 00019110 00019120 00019150 00019140 00019150 00019160 00019170 00019180 00019190 00019200 00019210 00019220 00019250 00019240 00019250 000192 60 00019270 000192 80 00019290 00019500 00019510 00019520 00019550 00019540 00019550 00019560 00019570 00019580 00019590 00019400 00019410 00019420 00019430 00019440 00019450 00019460 000194 70 00019480 00019490 00019500 00019510 00019520 00019530 00019540 00019550 00019560 00019570

163

c c-c

c c c-c

c c-c

c c-c

c C' c

c c-c

c c-c c-c-c c-

— P R E P A R E I N V E R S I O N LOOP

1 IPIV « N * ( N * l l / 2 . INO « IPIV

— I N I T I A L I Z E I N V E R S I O N LOOP

DO 6 I = 1 * N D I N = 1 . 0 0 / O B L E ( A d P I V l l A d P I V l = D I N M I N = N KEND = I - l LANF = N-KEND I F (KENOl 5 * 5 * 2

2 J = INO

I N I T I A L I Z E ROW LOOP

0 0 4 K = 1 * KEND WORK = 0 . 0 0 MIN = M I N - 1 LHOR = I P I V LVER = J

— S T A R T INNER LOOP

0 0 5 L = LANF* MIN LVER » L V E R * 1 LHOR s LHOR*L

5 WORK s W 0 R K * 0 B L E ( A ( L V E R 1 * A ( L H 0 R I 1

— E N D OF INNER LOOP

A I J I = -WORK*DIN

4 J « J-MIN

END OF ROW LOOP

5 IPIV * IPIV-MIN 6 INO = IND-1

EMO OF INVERSION LOOP

CALCULATE INVERSEIAI BY MEANS OF INVERSEdl* WHERE INVERSEIAI INVERSEdl * TRANSPOSE( INVERSE( Tl 1.

—INITIALIZE MULTIPLICATION LOOP

00019580 00019590 00019600 00019610 00019620 00019650 00019640 00019650 00019660 000196 70 00019680 00019690 00019700 00019710 00019720 00019750 00019740 00019750 00019760 00019770 00019780 00019790 00019800 00019810 00019820 00019850 00019840 00019850 00019860 00019870 00019880 00019890 00019900 00019910 00019920 00019950 00019940 00019950 00019960 000199 70 00019980 00019990 00020000 00020010 00020020 00020030 00020040 00020050 00020060

164

c c« c

0 0 8 I = 1* N IPIV * I P I V * I J = IPIV

— INITIALIZE ROW LOOP

0 0 8 K = I* N WORK = 0 . 0 0 LHOR = J

C C START INNER LOOP C

0 0 7 L = K* N LVER = LHOR*K-I WORK = W 0 R K * 0 B L E ( A ( L H 0 R I * A ( L V E R 1 I

7 LHOR = LHOR*L C C-c

c c-c

c c-c-c

END OF INNER LOOP

A I J I - WORK 8 J X J4.IC

END OF ROW ANO M U L T I P L I C A T I O N LOOP

9 RETURN END SUBROUTINE OSF ( H , Y* Z * N D I M I

- — - T H I S SUBROUTINE COMPUTES THE VECTOR OF INTEGRAL VALUES FOR A 6 I V E N E Q U I D I S T A N T TABLE OF FUNCTION VALUES.

C C-C C-C

DIMENSION YINDIMI* ZINDIMI

HT « .5555555*M

If INOIM-51 7* 8* 1

—NOIM IS GREATER THAN 5.

PREPARATION Of INTEGRATION LOOP 1 SUMl = V(2)*Y(21

SUHl = SUM1*SUM1 SUMI = HT*(Y(11*SUM1*Y(51I AUXl s Y(41*Y(41 AUXl = AUX1*AUX1 AUXl s SUM1*HT*(Y(51*AUX1*Y(5 11 AUX2 = HT*(Y(11*5.875*(Y(21*Y(5I1*2.625*(Y(5I*Y(411*Y(61I SUM2 > Y(51*Y(51

00020070 00020080 00020090 00020100 00020110 00020120 00020150 00020140 00020150 00020160 00020170 00020180 00020190 00020200 00020210 00020220 00020250 00020240 00020250 000202 60 00020270 00020280 00020290 00020500 00020310 00020320 00020350 00020540 00020550 00020560 00020570 00020580 00020590 00020400 00020410 00020420 00020450 00020440 00020450 00020460 00020470 00020480 00020490 00020500 00020510 00020520 00020550 00020540 00020550

165

c c-c

c c-

SUM2 s SUM2*SUM2 SUM2 = A U X 2 - H T * ( V ( 4 I * S U M 2 * Y ( 6 I I Z d l = 0 . AUX = V ( 5 1 * Y ( 5 1 AUX = AUX*AJX Z ( 2 1 = SUM2-HT*( Y ( 2 I * A U X * V ( 4 I 1 Z ( 5 I = SUHl Z ( 4 1 = SUM2 I f (NDIM-61 5* 5* 2

-INTEGRATION LOOP

00 4 I SUMl = SUM2 « AUXl = AUXl ' AUXl = Z ( I - 2 1

NOIM* 2 = 7* AUXl AUX2 Y ( I - 1 1 * Y ( I - 1 I AUX1*AUX1 S U M 1 * H T * ( Y ( I - 2 ) * A U X 1 * Y ( I I 1 = SUMl

I f d - N D I M l 5* 6* 6 AUX2 = Y( I I * Y d l AUX2 = AUX2*AUX2 AUK2 = S U M 2 * H T * ( V ( I - 1 I * A U X 2 * Y ( I * 1 1 I Z d - l l = SUM2

C

c

U

c-c

END OF INTEGRATION LOOP 5 Z ( N D I M - 1 I = AUXl

Z INOIMl = AUX2 RETURN

6 Z ( N 0 I M - 1 I s SUM2 Z INOIMl = AUXl RETURN

NOIM IS LESS THAN 5

7 IF (NOIM-51 12* 1 1 * 8

—NOIM I S EQUAL TO 4 OR 5

SUM2 « 1 . 1 2 5 * H T * ( V d l * Y ( 2 1 * Y ( 2 1 * Y ( 2 l * Y ( 5 l * Y ( 5 l * Y ( 5 1 * Y ( 4 1 l SUMl = y ( 2 l * Y ( 2 1 SUMl s SUM1*SUM1 SUMl = HT* (Y( 1 I *SUM1*Y I51 I Z d l = 0 . AUXl = Y ( 5 I * Y ( 5 ) AUXl = AUX1*AUX1 ZI2I = SUM2-HT*(Y(21*AUX1*VI41I If INOIM-51 10* 9* 9

00020560 00020570 00020580 00020590 00020600 00020610 00020620 00020650 00020640 00020650 00020660 00020670 00020680 00020690 00020700 00020710 00020720 00020750 00020740 00020750 00020760 00020770 00020780 00020790 00020800 00020810 00020820 00020850 00020840 00020850 00020860 00020870 00020880 00020890 00020900 00020910 00020920 000209 50 00020940 00020950 00020960 00020970 00020980 00020990 00021000 00021010 00021020 00021030 00021040

Ifi6

10

A U X l = A U X l « Z I 5 I » Z d l « Z ( 4 1 = RETURN

Y I 4 I * Y I 4 1 AUX1*AUX1 S U M 1 * H T * I V I 5 1 * A U X 1 * V I 5 I 1 SUMl SUM2

C C NOIM I S EQUAL TO 5 C

11

12

SUMl SUM2 SUM2 Z ( 5 I Z l l l Z I 2 I RETURN END SUBROUTINE

H T * l l . 2 5 * Y d l * Y I 2 1 * Y ( 2 l - . 2 5 * Y ( 5 1 1 Y I 2 1 * Y ( 2 1 SUM2*SUM2 H T * 1 Y I 1 1 * S U H 2 * V I 5 I I 0 . SUMl

SHEAR I N X * NYI C c-c-c C' c-c-c

—SUBROUTINE USES A NUMERICAL INTEGRATION TEC —SHEAR f O R C E . UNITS ARE fORCE PER LENGTH* E

—SUBROUTINE WAS WRITTEN S P E C I f l C A L L V FOR USE —SLAB SYMMETRICAL TO BOTH THE X AND Y AXES. —FOR SLABS WITH OTHER SYMMETRY.

HNIQUE TO CA . 6 . * LBS PER

WITH A RECT I T HAS NOT

DIMENSION : i 1 5 0 0 0 0 I * F I 2 1 0 0 I * 6 I 1 5 0 0 0 0 ) * N O I 21 * 6 1 * N P I 6 5 0 I * X M O M I 6 5 0 1 * Y M O M I 6 5 0 I * X V M O M I 6 5 0 1 * M * X T I 6 5 0 l « V T I 6 5 0 l * 0 X l 6 5 0 1 * 0 V I 6 5 0 1 * M X Y D I f l 6 5 01 * V Y I 6 5 0 1 * V X M 0 N I 6 5 0 I * I R I 6 5 0 I * 6 U N A I 6 5 0 I

COMMON C * f * 6 * N 0 * N B * X * Y * S T R * N P * X M 0 M * V M 0 M * X Y M * M Y D I f * M X Y D I F * M Y X O I F * V X M O M * V X * V Y * N X Y * N Y X * V X M *VVMAXP

* X I 6 5 0 1 * Y I 6 5 X D I F ( 6 5 0 1 * M Y * M Y X D I F I 6 5 0 1

OM*XT*YT*DX* AXN*VXMAXP*V

LCULATE INCH.

AN6ULAR BEEN TE

0)*STRI DIFI650 *VXI650

DY*MXDI YMAXN*

REAL MXOIF* MYOIF* MXYOIF* MYXOIF* MOIX* M3 lY C C STORE NODAL COORDINATES C

DO 6 0 0 I = 1 * NX X T I I ) = X d l

6 0 0 CONTINUE 0 0 6 0 5 I = 1 * NY Y T d l * Y d l

605 CONTINUE C c-c

—ESTABLISH W0RKIN6 VARIABLES

NZl = NX*NY

00021050 00021060 00021070 00021080 00021090 00021100 00021110 00021120 00021150 00021140 00021150 00021160 00021170 00021180 00021190 00021200 00021210 00021220 00021250 00021240 00021250 00021260 00021270

STED00021280 00021290 00021500

650*00021510 I* 00021520 1* 00021550

00021540 00021550

F* 00021560 00021570 00021580 00021590 00021400 00021410 00021420 00021450 00021440 00021450 00021460 00021470 00021480 00021490 00021500 00021510 00021520 00021550

167

6 0 8 C c c-c

NXX = N X - 1 NYY = N Y - 1 NYX = ( N Y - 1 I * N X NXY = ( N X - 1 1 * N Y HZl = ( N X - 1 I * ( N Y - 1 1 KX = N X * 1 KV = N Y * 1 KZ = KX*KY KXV = N Y * ( N X * 1 I KYX = N X * ( N Y * 1 I 0 0 6 0 8 1 = 1 * N Z l Y X M O M d l X - l . * X Y M O M ( I l

C c-c-c-c-c C'

NUMBER THE COLUMNS AND ROWS OF THE F I N I T E ELEMENT 6R ID

0 0 6 1 0 J = 1 * NX 0 0 6 1 0 I = 1 * NY

6 1 0 Y d * ( J * N Y l - N V l = Y T d l L = 0 0 0 6 1 5 J = 1 * NX DO 6 1 5 I = 1 * NY L * L * l

615 X ( L 1 = X T ( J I J J J = 0

B E 6 I N SOLUTION FROM UPPER RI6HT CORNER (NODE N Z I I OF UPPE — — Q U A D R A N T . NUMBERIN6 OF DIFFERENCE ELEMENTS INCREASES FRO — T O BOTTOM OF 6 R I 0 PATTERN* THEN FROM R I6HT TO L E F T . NODE — — I N 6 REMAINS THE SAME.

C-C-

——CALCULATE DELTA-Y*S 00 700 I « 1* NYY

700 OVdl » YCNZ1-I*1I-V(NZ1-II DISTRIBUTE OELTA-Y*S THR0U6H0UT

— OELTA-Y*S BELOW HORIZONTAL LINE OYINYI = OY(NV-ll DO 710 J = 1* NY 00 710 I = 1* NX OYdI*NYYI*J*Il = OVIJI

710 CONTINUE CALCULATE DELTA-X* S

00 720 J = 1* NXX OXIJl = X(NZl-(J*NYl*NYl-X(NZl-(( J*1)*NY)*NY)

720 CONTINUE DISTRIBUTE 0ELTA-X*S

00 750 J = 1* NXX DO 750 I = 1* NY OK(I*(J*NV)-NV) = DX(J)

6RI0. ALSO GENERATE OF SYMMETRY.

THROUGHOUT GRID.

00021540 00021550 00021560 00021570 00021580 00021590 00021600 00021610 00021620 00021650 00021640 00021650 00021660 00021670 00021680 00021690 00021700 00021710 00021720 00021750 00021740 00021750 00021760 00021770 00021780 00021790

R RIGHT 00021800 M TOP TO 00021810 NUMBER- 00021820

00021830 00021840 00021850 00021860 00021870

IMAGINAR00021880 00021890 00021900 00021910 00021920 00021930 00021940 00021950 00021960 00021970 00021980 00021990 00022000 00022010 00022020

168

730 CONTINUE GENERATE IMA6INARY DELTA-X*S TO THE LEFT OF VERTICAL L NT = (NY*NXX)*1 wcKiit*!!. L

NPP s NT*NYY 00 740 I = NT* NPP • X d ) = O X d - N V )

740 CONTINUE CALCULATE OELTA-XMOM'X ANO DELTA-XYMOM'X

L = 0 00 750 J = 1 * NXX 00 750 I = 1 * NY L = L * l MXDIF IL) = X M O M ( N Z l - ( J * N Y I * N V - I * l l - X M O M ( N Z l - ( J * N Y I - I * 1 I MXYOIf (L l = X Y M 0 M ( N Z 1 - ( J * N Y I * N Y - I * 1 I - X Y M 0 M ( N Z 1 - ( J * N Y 1 - I * 1 I

750 CONTINUE CALCULATE IMA6INARY DELTA-XYMOM'S

DO 755 I = NT* NPP M X Y D I f d l = M X Y O I f d - N Y l

755 CONTINUE ——CALCULATE DELTA-VMOM'S

L = 0 DO 765 J = 1 * NX DO 760 1 = 1 * NYY L = L * l MYDIF IL I = Y M O M ( N Z l - ( J * N Y ) * N Y - I * l l - Y M O M ( N Z l - ( J * N V I * N V - I I

760 CONTINUE MVOI f (J *NYl = MYOIf ( ( J * N Y 1 - 1 1 L = L * l

765 CONTINUE —CALCULATE DELTA-YXMOM'S

770

775

L * 0 DO 775 J = 1 * NX 00 770 I = 1 * NYY L « L * l MYXOIf ( L l » YXM0M(NZ1- (J*NYI*NY- I *1 I -YXM0M( N Z 1 - ( J * N Y I * N V - I 1 CONTINUE MYXOI f (J*NYl = M Y X O I f l l J * N V l - l l L » L * l CONTINUE

——CALCULATE SHEARS

-INCREMENTS OVER WHICH SHEARS ARE CALCULATED ARE NUMBERED BE6 -AT THE UPPER RIGHT CORNER Of THE SLAB PORTION IN THE UPPER R -OUAORANT. INCREMENT NUMBERING INCREASES fROM TOP TO BOTTOM -fROM RIGHT TO LEfT. NODAL NUMBERING REMAINS UNCHANGED.

00022030 INE OF 00022040

00022050 00022060 00022070 00022080 00022090 00022100 00022110 00022120 00022150 00022140 00022150 00022160 00022170 00022180 00022190 00022200 00022210 00022220 00022250 00022240 00022250 00022260 000222 70 00022280 00022290 00022300 00022310 00022320 00022350 00022540 00022550 00022560 00022570 00022380 00022590 00022400 00022410 00022420 000224 30 00022440

INNIN600022450

I6HT ANO

867 — C A L C U L A T E SHEAR DO 870 1 = 1 * NXY

IN X-DIRECTION AND WRITE RESULTS

00022460 00022470 00022480 00022490 00022500 00022510

169

c c-c

8 7 0 V X I I l = I M X 0 I F I I l / O X I I l l * l M Y X 0 I F d l / D Y I I l l CALCULATE SHEAR IN Y-DIRECTION

K = 0 L = 0 0 0 875 J = 1 * NX 00 875 I = 1 * NY L « L * l I f d . E Q . N Y l 60 TO 875 K = K * l V Y I K l = ( M Y D I F ( L I / D Y ( L 1 1 - ( M X Y D I F ( L I / D X ( L 1 1

875 CONTINUE J J J = 1

SORT TO FIND MAXIMUM SHEAR FORCE IN EACH DIRECTION

VXMAXN = 0 . 0 VXMAXP = 0 . 0 VYMAXN = 0 . 0 VYMAXP = 0 . 0 DO 880 1 = 1 * NXY I F ( V X M A X N . 6 T . V X ( I I 1 VXMAXN = V X d l

880 I F ( V X M A X P . L T . V X d l l VXMAXP = V X I I l 00 885 1 = 1 * NYX I F (VYMAXN.6T.VY( I11 VYMAXN = V V ( I 1

885 I F ( V Y M A X P . L T . V Y d l l VYMAXP = V Y d l RETURN END SUBROUTINE MSLP (N015I

C C——SUBROUTINE CALCULATES DIFFERENTIAL DEFLECTION C

DIMENSION C ( 1 5 0 0 0 0 1 * F ( 2 1 0 0 1 * 6 ( 1 5 0 0 0 0 1 * N D ( 2 1 * X ( 6 5 0 1 * Y ( 6 5 0 ) * S T R ( 65 *6 I *NP(650 I *XMOMI650 I *YMOMI6501*XYMOMI650 I *MXDIF (650 I *MYOIF I650 I * * X T 1 6 5 0 1 * Y T I 6 5 0 1 * D X I 6 5 0 1 * D Y I 6 5 0 I * M X Y D I F I 6 5 0 ) * N V X O I F I 6 5 0 ) * V X I 6 5 0 ) * * V Y ( 6 5 0 ) * Y X M O M ( 6 5 0 I » I R ( 6 5 0 I * 6 U N A ( 6 5 0 I * A ( 2 0 I * B ( 2 0 I * C X ( 2 0 1 * L X ( 2 0 I * * 0 I 2 0 I * I A I 2 * 2 0 1

C COMMON C*F*G*N0*NB*X*Y*STR*NP*XMOM*YMOM*XYMOM*XT*YT*DX*0Y*MXOIF*

*MYD IF*MXYOIF*MYXDIF*VXMOM*VX*VY*NXY*NYX*VXMAXN*VXMAXP*VYMAXN* *VYMAXP

C 7 FORMAT I / /SX* 'TWENTY MAXIMUM DIFFERENTIAL DEFLECTION RATIOS'*

* / 5 X * 4 5 l ' - * l ) 8 FORMAT I / / 6X* *N0DE1* *8X* *N0DE2* *12X* *DELTA* * 1 2 X * * D I S T * , 1 2 X , * D I ST

* 1 2 X , * 0 E L T A / L * , 1 2 X , * L / D E L T A * , / 5 6 X * * I N C H E S * * 1 O X * * I N C H E S * * 1 2 X * * F T * 1 9 FORMAT I / 6 X * I 5 * 1 0 X * I 5 * 1 0 X * E 1 5 . 6 * 5 X * E 1 5 . 6 * 5 K * F 6 . 2 * 8 X * E 1 3 . 6 * 1 1 X * I 3

DO 11 IM = 1 * 20 11 A f l M ) = 0 .3

00 1 I = 1* N015

00022520 00022550 00022540 00022550 00022560 00022570 00022580 00022590 00022600 00022610 00022620 00022650 00022640 00022650 00022660 00022670 00022680 00022690 00022700 00022710 00022720 00022750 00022740 00022750 00022760 00022770 00022780 00022790 00022800 00022810 00022820

0*00022850 00022840 00022850 00022860 00022870 00022880 00022890 00022900 00022910 00022920 00022950 00022940

* *00022950 00022960

) 00022970 00022980 00022990 00023000

170

IF IK IF 00 I I I J

GO TO 4

60 TO 5

TO

DO 2 J = 1* N015 I F ( J . E Q . l ) 6 0 TO 2

DIST = S Q R T ( ( X ( J ) - X ( I ) ) * * 2 * ( Y ( J I - Y ( 1 1 1 * * 2 1 ODIFF = F ( ( J - 1 I * 5 * 1 I - F ( ( I - 1 1 * 5 * 1 1 SLP « OOIFF/OIST FOIST = D I S T / 1 2 0 0 5 K = 1* 2 0 I F ( S L P . 6 T . A ( K 1 I CONTINUE 6 0 TO 2

( A ( K ) . E Q . O . O l = 20-K ( I K . E Q . O l 6 0 6 IL = 1* IK = 20- IL = 11*1

A d J I = A d i l B d J l = B d l l o d j i = o d i i cxdji = cxdii IA(1*IJ1 = IA(1*III IA(2*IJ1 = IA(2*III A(KI = SLP BIKl = ODIFF OIK) = DIST CX(K) = FOIST IA(1*K) = I IA(2*K) = J CONTINUE CONTINUE WRITE (6*71 DO 10 M = 1* 20 NSLP = 1/A(M1*0.5 LX(M) « NSLP CONTINUE WRITE (6*8) WRITE (6*91 RETURN END

//LKEO.SYSLIB DO // DO // DO DSN=EDR.IHSL.V90.SINGLE.LIB*DISP=SHR // DO OSN=EOR.IMSL.V90.DOUBLE.LIB*DIS»=SHR //LKED.SYSLMOO OO DSN=WYL.VO.WKW.LOADLIB(GUNA)* 01SP=OLD*UNIT = //

6 5

2 1

10

((IA(J*I)*J = 1*2)*8(I)*0(I)*CX(I )*Ad)*LX ID* 1 = 1*20)

00025010 00023020 00023050 00025040 00025050 00025060 00025070 00025080 00025090 00023100 00023110 00025120 00025150 00025140 00023150 00025160 00023170 00023180 00023190 00023200 00025210 00025220 00025230 00023240 00023250 00025260 00025270 00023280 00023290 00025500 00025310 00023520 00025330 00023540 00025350 00023560 00025570 00025380 00023390 00023400 00023410 00023420 00023430 00023440 00023450

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APPENDIX C

QUESTIONNAIRE FOR SURVEY

184

185

Texas Tech University Department ol Civil Engineering

May 23, 1985

Dear Sir:

I am a graduate student in the Department of Civil Engineering at Texas Tech University, working on my Ph.D. under the guidance of Dr. Warren K. Wray. Part of my dissertation requires that I evaluate the most efficient/economical way of determining the modulus of elasticity of soil. This will be based on statistical analyses that I will carry out on some of the commonly available testing procedures. To carry out these analyses, I am conducting a survey in order to gain the necessary reliable data. Each of the testing procedures listed on the enclosed form will be evaluated based on the following factors:

1. Availability 2. Reliability 3. Familiarity 4. Cost of Equipment 5. Cost of Test 6. Interpretation of Results 7. Ease of Performance

Each of these factors is to be individually weighted on a scale of 1-5 as indicated below.

Availability 1. Very likely that everybody will have 2. Likely that everybody will have 3. Not sure 4. Unlikely that everybody will have 5. Very unlikely that everybody will have

Reliability 1. Reliable 2. Fairly reliable 3. Not sure 4. Not very reliable 5. Unreliable

Familiarity ^ .,. . ^ , , 1. Very likely that everybody is familiar with the procedure Z. Likely that most persons are familiar with the procedure 3. Note sure ^ ^ 4. Likely that most persons are unfamiliar with the procedure 5! Very unlikely that everybody is familiar with the procedure

Bo« 4069/Lubbock, Texas 79409 (8061 742-3523

186

Pace 2

Cost of Equipment (from testing laboratory's viewpoint) 1. Very inexpensive 2. Inexpensive 3. Neither expensive nor inexpensive 4. Expensive 5. Very expensive

Cost of Test (as perceived by the client/user) 1. Very reasonable 2. Reasonable 3. Neither reasonable nor unreasonable 4. Unreasonable 5. Very unreasonable

Interpretation of Results 1. Very simple 2. Simple 3. Neither simple nor difficult 4. Difficult 5. Very difficult

Ease of Performance of Test 1. Very easy 2. Easy 3. Neither easy nor difficult 4. Difficult 5. Very d i f f i c u l t

If you are familiar with at least five of the testing procedures and are also familiar with at least five of the factors listed above, please fill out the enclosed table, using the indicated scale, by placing the appropriate number in the applicable box. The survey should take less than five minutes of your time. If you are unable to complete the survey, please return this letter and the uncompleted survey in the envelope provided. I would appreciate it if you could return your survey by June 15, 1985.

Thank you for your assistance,

Sincerely yours.

K.N. Gunalan

KNG/sh

Enclosure

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VALUES OF DESIGN PARAMETERS DUE TO FORKLIFT

LOADING AT 15 LOCATIONS USED

IN QUASI-STATIC ANALYSIS

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01 U X S- 3»

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c i- o

Maximum

Shea

in x-Directi

(lbs

/in.

)

p>» ^ «a-l - H

p.. I - H

. i n 1

CTI •

0 0 l - H

^• i n

• i n 1

<T> 0 0

• CO l - H

i n VO

" i n 1

CTI CO

i n 1

OJ o > ,

o LU •

c ( . o

Maximum

Shea

in

y-Directi

(lbs

/in.

)

192

CO t — l

CSJ I - H

l - H

1—1

o I - H

at

00

r«>.

Position

o o

• o in

o o

• o m

o o

• o m

o o

• o tn

o o o in

o o o tn

o o

50.

Length,

L, ft

o o

• o in

o o

• o in

o o

• o tn

o o

• o in

o o

• o in

o o in

o O

50.

Width,

W,

ft

o o

• ^

o o

• "*

o o

• t r

o o

• ^

o o

• t r

o o

• ^•

o o t f

c

Thickness,

h o o

• o l - H

o o

• o l - H

o o

• o l - H

o o

• o 1—1

o o

• o 1—1

o o

• o I - H

o o

t - l

«4-

• k

Aisle

Width,

o o

• o o m I-H

o o

* o o m l -H

O o

• o

150

o o

• o o tn l -H

O o

• o o . in <-<

o o

• o o m l -H

o o

1500.

> i

icit

m m IO o.

Modulus

of E

of Soil,

E ,

CO l - H

. CO p«. t - H

CO l - H

« 00 PN. t -H

CO I—1

. 00 p^ l -H

CO t - H

. 00 1—1

CO l - H

. CO p-» I-H

CO l - H

. CO I - H

CO l - H

00 p^ I-H

M

ding

CTI CVJ

• o

CTI CVI

• o

CO Csl

• o

at CSI

• o

O l CVI

• o

00 CVi

• o

00 CVi

• o

c • -

on.

ecti

Forklift

Loa

Pf.

psi

Maximum

Defl

»-l CVi

• o

l - H

CSJ •

o

o Csi

• o

I - H

CSJ •

o

l - H

CVJ •

o

l - H

CVI •

o

t - H

CVI

o

• 1— c IO • ! -

•r-•u> «

CTI tn

• o l - H

CO O l

• t r l - H

cn o

• o t - H

t - H

^ • CSJ

l - H

CO i n

• CO l - H

CTI Csl

• CSi t - H

p>. CO

t r t—l

C X — » OJ IO « r s- ^ ' OJ <J o

Maximum

Diff

Deflection,

l -H

X

X

<o E

X

c

in 1/1

o p«.

• o CO t -H

t - H

CVI •

CO VO CVi

CO l - H

. in CO t-l

at to

• VO CVi I—1

CO i n

• CTI i n CVi

CTI p~

• p«« CSI l -H

r»» «r

260.

X

Maximum

Stre

Direction,

o (I

bs/s

q in.)

p». VO

« CSi CTI CSJ

CVi 00 CO m CO

CTI VO

• CVJ o CO

o CSJ

• p». VO CSi

i n VO

• o CVi CO

CVJ l - H

. p^ VO CVJ

CO in

320.

>> c

•t—

l/l 1/1 > ,

Maximum

Stre

Direction,

o (I

bs/s

q in.)

^ in CO t r CO

CO 00 I - H

o 1 ^

to CO

• o VO CO

<r 00

. p» CO CO

o CVI

• CVI CTI VO

p^ r»»

• o t r CO

CO m

694.

c X

+f 2 :

Maximum

Mome

x-Direction,

(in.-lbs/in.

m ^

0 00

CVJ tn

« CO

0 1

VO t - l

. p^ 0 CO

CO tn

• Csl I-H P^

VO 0

• i n i n 0 0

Csl CO

• CVJ l - H

00 CO

854

c

* . 2 : ^ c - ^

Maximum

Mome

y-Direction,

(in.-lbs/in.

CVJ CO

i n

t - H

in a

at 1

CVJ CVI

• CO l - H

VO l - H

. i n

CO CO

• 0 I - l

CJI CO

" i n

t r 0

CTI

OJ 0 X

0 LU -

c s- 0

Maximum

Shea

in x

-Directi

(lbs/in.)

CO p-»

. in 1—1

1

CTI 0

a

CJt l -H

1

tn <j>

a

t o I-H

1

r». 0

. CO l -H

1

t r in

. tn l -H

1

00 0

* CO

1

t r

in I-H

1

0) 0 > 1

LU •> c

i~ 0

Maximum

Shea

in y-Directi

(lbs/in.)

193

i n

t r

CO

CVi

I - H

i n t — t

^ I -H

tion

Po

si

o o

• o m

o o

• o i n

o o

• o t n

o o o t n

o o

a

O t n

o o

50.

o o

50.

L, ft

th.

Leng

o o

• o m

o o

• o t n

o o a

O i n

o o

• o i n

o O

a

O i n

o

50.

o o

50.

* 3

m

Widt

o o

• ^

o o

• t r

o o a

<a-

o o ^

o o

a

^

o o «*

o o ^

s, h,

in

kn

es

Thic

o o

• i n t - H

O o

. i n t - H

o o

• i n l - H

O o

« t n l - H

o o

a

m 1—1

o o

10.

o o

• o t - H

4-> t»-

dth,

A

3

OJ

Aisl

o o

• o o m I-H

O o

« o

150

o o a

O

150

o o a

O O m l -H

o o

a

O O m l -H

o o

1500.

o o

1500.

>,

4 J •«-l/l m lO o .

1 1 1 . .

I/I I f - UJ

o

lus

oil,

Modu

of S

o o

• i n r

O O

• m r

o O

. i n i*»

o o

• t n p>.

o o

a

i n p^

CO l - H

178.

CO t-l

0 0 p -I-H

«

Load

ing

lift

ps

i

i n l - H

. o

t n l - H

. o

t n l - H

. o

^ t - l

. o

CO I - H

a

o

o CO

o

o CO

o

^ c

. t —

« c o

Defl

ecti

mu

m

J- - X O *4- IO

Lu Q . Z

CTI o a

O

CO o

a

O

CO o a

O

p". o a

O

ps. O

• O

CVI CVJ

o

CVI CVi

o

• t— c lO '^

•t—

t r o a

i n

CTI CTI

a

^

t - H

CTI

• t r

«r p^

. t n

t - H

CO

• «r

0 0 VO

oi

CO CO

CO t -H

C X - - ^ OJ IO ^

^ s '^ 0) < o

14- ^ 14- •> X • 1 - c ^

o o

mum

ecti

Ma

xi

Defl

lO

<"

VO VO

• t n p«.

CTI O

a

CO p>«.

CO VO

. i n p».

p>» CVJ

• CO

r

t r CVI

• ^ p~

CVI i n

129,

o t—l

292

1 X

c

Maximum

Stress

i Di

rect

ion,

°

(Ibs

/sq

in.)

^

CO CVJ

a

CO CO

r p *

• t r t o

t o t n

• t r CO

CJI VO

• CJI to

r-» ^

• CJt p -

CO i n

292

.87

353

1

>> c

<A 1/1 ^ > » 01 o -—• c • ! - > » £ : 1/1 C t -o E •<- CT 3 •»-) m E o-- .

•I— OJ l/l X s- . a l O • • - t —

S. Q — '

p * r

. t - H

O CVJ

l - H

CT> a

t r CTI l -H

CO t o

. I - H

o CVi

CO CO

. i n as l -H

CO CTI

• P^ CTI l -H

o ^

345

.92

698

c

Moment

i ti

on,

M s/in.)

^

imum

ir

ec

.-lb

X o c IO 1 • < -

Z X w

i n CJI

. l - H

CVi CVi

CVJ p *

. CVi 1 ^ l -H

CTI " T

. t n CVI CVi

CO 00

. t n 0 0

. -H

CTI

• t - H

I-H CVJ

0 0

o

780

.65

943

c

OJ ' ' E c = 0 0 •<-

2 : — ^

imum

ir

ec

.-lb

X 0 c 10 1 -t-

z > 1 - ^

CJI to

• CVI

p». t o

. CVI

CTI VO

• CVI

i n VO

• CVI

VO VO

' CVJ

.07

i n

.20

0 l -H

OJ U X

Shear

Fon

rect

ion,

V

.)

imum

x-

Di

s/in

X J3 10 C 1—

2 : - r - - ^

1— 0

• H T

CSJ CSJ

. CO 1

CO CJt

• CO 1

CO I - H

. CO

t — l

r s

^ 1

p~

..

-15

.90

CO t -H

1

OJ

1 Shear

F(

rect

ion,

.)

imum

y-

Di

s/in

X . a 10 c >—

2 : - I - - ^

194

C V i

l - H

o l - H

at

00

p ^

t o

c o

Posi

ti

o o o i n

o o

• o i n

o o

• o t n

o o

• o i n

o o a

O i n

o o a

O m

o o o m

4 ^

•+-^

_ l

Length

o o o i n

o o

• o m

o o

• o m

o o

• o i n

o o

• o i n

o a

O i n

o o o m

4 - J t ^

3

^

Width

o o

• ^

o o

• ^

o o

• ^

o o

• ^

o o

• « •

o o a

t r

o o t r

, c

m

.c m

1 Thickness

o o m l - H

o o

• i n l - H

o o

• i n l - H

O o

• m l - H

O o

• m I -H

o o a

m l - H

O o

« t n l - H

H.J I4_

m

3 <c

^ ^J

• o

3

Aisle

O o

• o o in l - H

O o

• o 091

o o

• o

150

o o

• o

150

o o

• o o in l - H

o o

• o

150

o o

• o o tn t - H

>» 4-> • r -

o • r -4 J - t -

l / l l / l

(O ^

11 i * 1/1

I f - U J

o t .

m t— •^ •!—

Modu

li

of So

CO f*

CO p ^ f H

CO t - H

. 00

t - H

O o

• t n p ^

o o

. i n p>«

o o

• i n p ^

o o

• t n p ^

o o a

i n

p *

A

O l

c "O I O

o _ i

4-> 1 * - •<-• 1 - 1/1

Fork

l Pf.

P

p>» CSJ

o

V O CSJ

• o

t o I -H

. o

VO l - H

. o

VO l - H

. o

i n t - l

. o

t o l - H

. o

c •^

«k

c o

• 1 —

4 J

u 0)

14-OJ

o E 3

Maxi

mi

o i t - H

o

CO l - H

• o

o t - H

. o

CTI

o • o

CTI

o • o

CO o

« o

at o

• o

• 1 — c IO T -

+ J «

C X

CVJ CO

t r l - H

CO CO

a

o I - H

VO t r

• t r

C O CO

• t r

CO t r

• ^

CO CTI

• i n

r» CJI

• i n

^—^ O J 1 0 < r

S- E OJ <

1 4 -

t 4 - " • I - C

o o E -^ 3 U

Maxim

Defle

1

o r - H

X - -

X IO E

>

1 X

c '^ m m OJ

H - l

E 3

CO VO

C3 CO CVJ

i n I - H

. i n

o CVI

i n

CO

• C O

0 0

t r CO

. CVI

(^

00 o

• o CTI

l - H

f ^

. P ^ P«.

VO to

• l - H

CTI

X D ^ ~

« c o

• r - CT 4 - > m

Maxim

Direc

(lbs

/

I - H

tn

V O C V i

CVJ

t n o

• t n o l - H

i n

CTI a

0 0

C O

i n

o • 00

00

p « .

00 •

0 0

G O

i n

o •

0 0

C O

t n

p »

• C V I C V i l - H

1

>> c

• t —

l / l

• «>» OJ o ^—*

H-> " C t / 1 c • . -

o E -r- CT 3 H J l / l

Maxim

Direc

(lbs

/

CO l - H

m l - H

to

m o p ^

t r i n

o to

a

t n CO CVI

o CJI

. CVJ CT^ I - H

o CVI

• o CVJ

CVJ CVJ

. p ~ .

o CVJ

f l

t r

• t r

t r CVI

c •^ X

+-»£ C ' — OJ t l .

E c c o o ••-

2 : ••- - ^ H ^ m

E CJ - O

3 OJ t —

Maxim

x-Dir

(in.

-

^ o t r o VO

CO l - H

. o CO CVJ

o CVJ

a

p>«

C O CVJ

CTI P ~

. ^ CO CVJ

00 CJI

• to CO CVI

o 00

• t r CO CVJ

t r CO

• p * C V I

C O

c • t —

>» • u > "S.

c --^ OJ •» .

E c c o o •>-

2: ••- — 4 -> m

E u .ca 3 0 ) I —

Maxim

y-Dir

(in.

-

p ^

VO

0 0

CTI P *

p>«

CTI t n

. CVI

0

. CVI

<J1 i n

• CVJ

< j i CSJ

. t r

CVJ

r^ • t o

O J 0 X S - 5 >

0 C u •

c L. 0 1 0 f OJ H->

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1;=; '= 3 Q - 1 -

Maxim

in X-

(Ibs

/

0 0

VO t - l

i n

t n a

i n

1

CO CVJ

a

t r 1

to CO

. t r

1

CO CVI

• t r

1

VO i n

. ^

00 p ~

• VO 1

OJ 0 > , 4 . ^ 0

L U -C

L- 0 ro •< -01 H->

. S U 1/1 QJ - — •

i . • E — C

3 Q - ^

Maxi

IT

in y-

(Ibs

/

195

at

00

p>.

t o

m

^

CO

c o

Posi

ti

o o

• o m

o o

• o m

o o

• o i n

o o

• o t n

o o a

O m

o o o t n

o o o m

4-> <4-

Length,

L,

o o

• o i n

o o

• o m

o o

• o i n

o o

• o i n

o o a

o m

o o i n

o o o m

t 4 -

3

Width:

O o

• ^

o o

• t r

o o

• t r

o o

• ^

o o a

t r

o o ^

o o t r

c • t ^

I .

- C

-

Thickness

o o

• m l - H

O o

* t n l - H

o o

• i n l - H

O o

• t n t - H

O o a

i n f*

o o a

m l - H

O o

• i n t - H

^J «4-

• 1

3 « t

J ^

Widi

Aisle

o o

• o 150

o o

• o o m l -H

O o

• o

150

o o

a

O

150

o o

• o

150

o o

a

O O i n l-H

O o

a

O

150

>» 4.>

CJ . r -4-J • ! -I / I t /1 fO CL

11 1 « .

t/1 14— L U O

Modu

li

of So

CO t - H

. 00 p ^ t -H

CO I-H

. CO

I—1

CO t - H

. CO p>» l-H

CO l - H

. CO p« . l -H

CO l - H

. CO 1 ^ I-H

CO t - H

. CO p -l -H

CO l - H

. 00 p ^ t - l

• k

O l c

•^ "O IO

o 1

14- " -'t— l / l f— ca.

I . » O <4-

Lu ca.

CTI CVI

• o

00 CSJ

• o

CO CVi

a

o

VO CO

• o

CO CVJ

• o

.00 CSJ

a

o

p * CSJ

• o

^ c

• r *

«k

c o

• r -

••-> (J 0)

14-OJ o E 3

Maxi

mi

l - H

CSi a

O

00 CSJ

• o

l - H

CVI

« o

CO CVi

• o

o Csi

• o

o CVI

a

o

o CVJ

• o

• 1— c IO -r-

• f j • C X

CTI t — l

• CO t — l

o t r

a

O t - H

O 00

a

t r l - H

O l t r

a

p«» t - H

CSJ 00

a

o l - H

C O

p *

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p^ CO

a

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< — . OJ IO ^ t- ^ E OJ <

• 4 -Vf- « • t - c

um D

ctio

i Ma

ximi

De

fies

1 o l-H X

max

'(l/v)

1 X

c •^ m m OJ

H-> t / 1

E 3

l - H

O a

t o t r CVI

CVJ CO

• 00 CVJ t -H

CVJ CTI

a

VO t r CVJ

CTI 00

a

CO CTI CVI

i n i n

• VO

o CVJ

o f ^

. o CO CVI

VO t n

• t o

o CVI

^x o.-.. " c C - t -

o •r- CT H-> m

Maxim

Direc

(lbs

/

o CTI

. t—l

at Csi

CTI

^ • CO

t r CVI

CVI CO

. I - H

CTI CVi

CTI P^

. o o t r

o p«.

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CO CTI

• t n CVI CVi

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CVi t - 4

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m <A > , 0) D ' - ^

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um S

tion

sq i

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CO

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CTI 1 ^

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lbs/

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x-Di

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at CO

CO

p *

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CVJ t o VO

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in X-

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1

t r r>«.

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1

VO CTI

a

t r l - H

CO CTI

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t n <J>

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t n 1

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Maxim

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196

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c o • . •4->

l / l

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COMPARISON OF RESULTS FOR DIFFERENT

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205

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APPENDIX G

MOMENT EQUATIONS

236

237


(L)0.18 ( 2.27 (, 0.27 ( 0.97

{W)°-^^ {h)2-57 (p )0.91 M = (E^)(7752) 'l (G.2)

(L)0-52 (E )0.82 (, )0.07


(L)°-^2 (,)2.30 ( 2 . 0 4 M, = (EJ(3.66) f (G.3)

(,)C.04 (^0.72 ( )0.02 (3)2.03

(,)0.16(,)2.57 0.41

M = (E )(56.84) 5 (G.4) y y (,)0.15 ( )0.72 (, )0.50 (3)3.47

5 W


(L)0.26(^^)0.06(^)2.18(^ )1.09(p )0.63( ,0.17(3)1.30

M^ = (EJ(1.25) "^ ^ i (G.5) (, )0.72

(,)0.09(j^)2.22()0.75(i0.54( 0.78(3)0.24

ri = (Ej(2i.2i) ^ ^ (G.6) y y (L)0.30 (, )0.73

APPENDIX H

DESIGN EXAMPLE USING EXISTING PROCEDURES

238

239

H.l Introduction

A design example is used here to compare the results obtained

from three procedures which are in common use among practicing

professionals. The three procedures are:

1. Portland Cement Association method (PCA).

2. Panak/Wire Reinforcement Institute method (Panak/WRI).

3. Corps of Engineers method.

Brief descriptions of each of the procedures are given here.

However, for more details, the reader is asked to refer to the

specific references indicated.

H.2 Design Data

A warehouse floor slab, 100 ft x 50 ft, needs to be designed for

the following conditions:

Aisle width between stacks, A^ = 7.5 ft

Modulus of elasticity of soil, E^ = 5000 psi = 720 ksf

Stack loading, p^ = 8 psi = 1.15 ksf

Design Axle Load = 10*000 ^^s

Contact area = 27 sq in.

Forklift truck loading, p^ = 185 psi = 26.64 ksf

Wheel Spacing, S 31 in. = 2.58 ft

Compressive strength of concrete, f^ = 5000 psi = 720 ksf

240 The three procedures stated here use a modulus of subgrade

reaction, k^, value as opposed to the modulus of elasticity of the

soil. Therefore, in order to be able to really make a comparison,

all parameters involved should be identical. Therefore, the modulus

of elasticity value is converted to an equivalent modulus of sub-

grade reaction value using the following equation (47)

k = 60 (E /1000)°-^^^ (H.l)

where

E = modulus of elasticity of soil, psi

k = modulus of subgrade reaction, pci

The value of k corresponding to an E value of 5000 psi is

calculated using Eq. (H.l) as follows:

,865

= 241.41 pci

H.3 Portland Cement Association (PCA) Method

k^ = 60 (5000/1000)'

The design chart used is shown in Figure 1.4. A safety factor

value of 1.5 is used. The design steps involved in thickness design

for single-wheel axle loads are:

1. Flexural strength of concrete, MR:

MR = 7.5 VT" = 7.5 VÔOO c

= 530.33 psi

2. Concrete working stress, WS:

WS = MR/SF = 533.33/1.5 = 353.55 psi

241

3. Slab stress per 1000 lb of axle load:

WS/axle load, kips = 353.55/10 = 35.35 psi/1000 lb

4. Enter Fig. 1.4 at left with stress of 35.4 psi; move right

to effective contact area of 27 sq in. down to wheel spacing of 31

in.; right to read an approximate slab thickness of 4.5 in. on the

imaginary line of subgrade modulus, k , of 241 pci.

H.4 Panak/Wire Reinforcement Institute Method (Panak/WRI)

The design charts used here are shown in Figs. 1.1, 1.2 and 1.3.

The design steps involved are:

1. Assume a slab thickness, say 6 in.

2. Enter Fig. 1.1 at left at 6 in. on the line of concrete

modulus, E , of 4 X 10 psi; move right to effective subgrade Vf

5 modulus, k of 241 pci; then down to read D/k ratio of 3.75 x 10

. 4 in. .

3. Enter Fig. 1.2a at bottom with aisle width of 90 in.; move

5 4 Straight up to D/k of 3.75 x 10 in. ; move right, cross over to

Fig. 1.2b with uniform load of 1.15 ksf to intersect maximum slab

bending moment at 350 Ib-ft/ft; move right to intersect tensile

stress curve of 354 psi (7.5 VTr/1.5); down to read an approximate

slab thickness of 4.0 in.

4. Calculate the equivalent loaded diameter from

d = \/(4/7T) (contact area) (H.2)

d = \/(4/7T) (27) = 5.86 in.

5. Enter Fig. 1.3 at bottom with equivalent loaded diameter

value of 5.86 in.; move straight up to intersect D/k curve at

5 . 4 ^ ^ 3.75 X 10 in. ; move left to read unit moment corresponding to 1000

lb wheel, which is 270 Ib-in./in.

6. Enter Fig. 1.3b at bottom with wheel spacing of 31 in.; move

up to intersect D/k curve of 3.75 x 10^ in.^; move left to read

additional unit moment value of 25 Ib-in./in.

7. Add values from steps 5 and 6 and multiply this sum by total

load on wheel in kips.

8. Enter Fig. 1.2b with a maximum slab bending moment value of

1475 Ib-ft/ft; move right to tensile stress curve of 354 psi; down to

read a slab thickness of approximately 5.5 in.

9. Compare values obtained in Step 3 and in Step 8. Use the

larger of the two, which would be 5.5 in this case.

H.5 Corps of Engineers Method

The design chart used is shown in Fig. 1.8. The type of fork-

lift truck used corresponds to Category I and Design Index 1 as given

in Tables 1.1 and 1.2. As the values used here fall outside the

bounds of the curve, the curves will have to be extrapolated in order

to come up with a value for the thickness of the slab. The design

steps involved are:

1. Enter chart at left with a flexural strength value of 354

psi; move right to subgrade modulus, k , value of 241 pci; move to

curve corresponding to DI 1; move right to read a thickness value of

7.75 in.

The same example is worked out using the regression equations in

Appendix I, and the result is compared with those obtained here and

are tabulated in Chapter 6.

APPENDIX I

DESIGN EXAMPLE USING THE

REGRESSION EQUATIONS

243

244

1.1 Introduction

A design example is used here to demonstrate the use of the

regression equations and to compare the result with those obtained

using three existing procedures included in Appendix H.

1.2 Design Data

A warehouse floor slab, 100 ft x 50 ft, needs to be designed for

the following conditions:

Aisle width between stacks, A = 7.5 ft

Modulus of elasticity of soil, E^ = 5000 psi = 720 ksf


Design Axle Load = 10000 lbs

Contact Area = 27 sq in.

Forklift truck loading, p^ = 185 psi = 26.64 ksf

Wheel Spacing, S = 3 1 in. = 2.58 ft

Compressive strength of concrete, f ' = 5000 psi = 720 ksf

1.3 Design Calculations

1. Assume trial slab thickness of 4 inches.

2. Calculate allowable stresses:

f = 7.5 Vf^ = 530.33 psi = 76.37 ksf

V = 4 ^ = 282.84 psi

245 Stack Loading Condition:

3. Calculate design stresses:

(L)°-lV)0-°l{h)°-30(A )0-27(p )0-97

°x = (Ex)(l°26.00) TT^^^'" s

W' "S

nnn^0.14/(.«x0.01/./Tî0.30,7 t-N0.27/H icÔ.97 = (1.29)(1026.00) !^^ ^-^ Wl^) (7.5) [WIS)

(720)^*^^

(1.29)(1026.00) (1.906)(1.040)(0.719)a^

28.60 k ips / f t2 < f^ ^^^^^^^^^

(^^^0.18(,)0.59( .0.91

ay = (Ey)(30,045.00)^^^Q^^3^^ ^032^^ ^0.18

, . n v 0 . 1 8 , . / H ^ v 0 . 5 9 , H 1 0 0 . 9 1 = (1.24)(30,045.00)^^Q^ n ,^n^^^^\ .,^^'^^1 o

(100)°-^^(720)°-^^(7.5)°--^^

/ I o>.^/on nAC rtn^ ( 2 . 0 2 2 ) ( 0 . 5 2 3 ) ( 1 . 1 3 6 ) (1.24)(30,045.00)(^^^^Q)J330 g99)(l 437)

= 15.50 k i p s / f t ' < f^ allowable

4. Calculate expected shear forces: ( L ^ 0 . 3 0 ( , ) 1 . 8 2 ( , y . 0 2 , p ^ ) 0 . 9 8

V, = (E,)(15.14) ( , )0.n(E^)0.6U

246 = M oo>M. ..^(100)Q-^Q(4/12)^-B^f7.5^0-0^M 1.^0.98

(50)^-^^(720)°-^°

= (1 29)(15 i/|)(3.981)(0.135)(1.041)(1.147) ^ ^ ^ -^^^ (1.538)(51.808)

= 0.157 k i p s / f t

(W)0.09(^)1.83( jO.74

V^^V(^^^^-^\,)0.83(,^)0.57j^)0.44

= (1.24)(1467.0) (50)Q;0 ; (4 /12)^>^3, ,^^ , ,0 .74 (100)0.83(720)0.57(7 5)0.44

= f l ?4UUfi7 n^ (1.4222)(0.134)(1.109) ^^'^^^^^^^^'^) (4b.709)(42.529)(2.427)

= 0.082 kips/ft

5. Calculate maximum design shear stress:

„ , „ 0.157 '- 1000 (1+2x0.8x0.3333)x4x0.27 ^ 144

0.66 psi < V ,, ^ c all owable

6. Calculate maximum differential deflection

A = (0.47)

(L)0.44(^)0.62(^)0.20(p ,0.78 s

(. )0.99(^ ,0.29

._ (0 47) (100)Q-^^(50)Q'^^(4/12)Q-^Q(1.15)Q-^^

(720)^-^^(7.5)°'^^

(n Ai\ (7.586)(11.307)(0.803)(1.115) ^ ^^'^^^ (674.154)(1.794)

= 0.030 ft



(L)0.12(^)0.30( ,2 .04

a , = ( E J ( 2 2 . 2 5 ) ^

247

X > X" • ' („)0 .04(^ )0.72(^ )0 .02(3)2.04

= (1 .29)(22.25) ^ ^ Q Q ] ' ' ' ' ( y ^ ^ ) ' ' ' ' l ^ ^ ; ^ ^ ) ' - " ' (50)0.04(720)0.72(7^5)0.02(2^33)2.04

(1.738)(0.719)(809.263) [l.d^)[dd.db} ( i . i 6 9 ) ( l l 4 . 0 9 8 ) ( l . 0 4 l ) ( 6 . 9 1 4 )

= 30.24 k ips / f t ^ < f . , , . , ^ t allowable

(^)O.04(^)0.34(p^)2.39

oy = (Ey)(347.71)^^ô.l5(^ )0.72(^ )0.51(3)3.48

î;n^0.04,. , ^ ^ . 0 . 3 4 , . . ^>,N2.39 = (1 24)(347 71) (50) (4/12) (26.64)

(100)°-^^(720)°-^^(7.5)°-^^(2.58)3-^^

n ÂU- ziT 7^) (1.69)(0.688)(2552.859) [i..d^)K:i^/./i) (1.995) (194.098) (2.794) (27.067)

43.70 kips/ft^ < f, ^^^^^,^^

8. Calculate expected shear forces:

(„,0.28(,,1.80(,^)0.52(p^,2.80

^x = (Ex)(13.33) ^^j0.21(^ ,0.57(5,(3.96)

n „ w n 331 (50)°-^^(4/12)^-^°(7.5)°-^^(26.64)^-^° = (1.29)(13.33) (ioO)0-21(720)°-"(2.58)3-96

248 n PQUi? - ^ (2 .990)(0.138)(2.851)(9806.089) U . ^ y n i J . ^ s j ; (2 .630)(42.528)(42.659)

= 41.57 k i p s / f t

( L ) 0 . 0 3 ( , ) 0 . 1 1 ( , ) 1 . 9 0 ( , )0.22( )2.13

\ = (E ) (0 .19) n .Q ;> ^4 ^ y y (^^)0.59(3)2.64

= (1 .24) (0 .19) (100)^'^3(50)Q-^^(4/12)^-^Q(7.5)Q-^^(26.64)^-^3

(720)°*^^(2.58)^-^^

- n oA\fn 1Q^ (1.148)(1.538)(0.124)(1.558)(1087.396) " U . 2 4 ) ( 0 . 1 9 ) (48.509)(12.209)

= 0.148 kips/ft


41.57 ^ 1000 ^ " (1+2x0.8x0.3333)x4x0.27 ^ 144

= 174.33psi < v^ ,n,^,t)le

10. Calculate maximum differential deflection: (^,0.21(„)0.53(^^,0.71(p^)1.12

A = (0-12) (E^)0.98(,^)0.21(3)i.81

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251 13. Calculate maximum design shear stress

^ ^ ^ 0.14 100 (1+2x0.8x0.3333)x4x0.27 ^ 144

= 0.59 psi < V TT ^ c all owable

14. Calculate maximum differential deflection:

(L)0.41(^)0.62(^)0.19(^ )0.15( )0.70( )0.19

^ = (0-12) 0 .7 0 26 " ' (^^)0.97(3)0.26

__ (o^^2)(100)"'^'(50)Q-^^(4/12)Q-^^7.5)Q'^^(1.15)Q-^Q(26.64)^-^^

(591.034)(1.279)

.^ ,^^(6.607)(11.307)(0.812)(1.353)(1.103)(1.866) " ^^'^'^^ (591.034)(1.279)

= 0.03 ft

Therefore the assumed slab thickness of 4 inches would be suf

ficient to sustain the imposed loading.

A safety factor of 1 was assumed in using the regression

equations. However, even with a safety factor of 1.4 to 1.5, a 4 in.

thick slab will be found to be sufficient.

APPENDIX J

A COMPARISON BETWEEN THE PCA METHOD

AND THE REGRESSION EQUATIONS

252

253

A design example is used to illustrate the PCA method and to

compare the PCA method with the regression equations on a one-to-one

basis. However, in order to adopt the equations, certain assumptions

had to be made. For example, the PCA method does not require the

length or the width of the slab, but these dimensions are required in

the equations and therefore a reasonable value for these had to be

assumed. Design data used are given in Section 0.1. The design

example as worked out to illustrate the PCA method is presented in

Section J.2 and the example is worked out using the equations in

Section J.3.

J.l Design Data

Axle load = 25 kips

Wheel spacing, S = 3 7 in. = 3.08 ft

Number of wheels on axle = 2

Tire inflation pressure, p^ = 110 psi = 15.84 ksf

Tire contact area = 114 sq in.

Subgrade modulus, k = 100 pci

Concrete flexural strength, MR = 640 psi C« 28 days

Compressive strength of concrete, f^ = 4000 psi = 576 ksf

Assumed values are as follows:

Slab length, L = 150 ft

Slab width, W = 100 ft

Aisle wid th , A w

10 f t


Calulated values are as fo l lows:

254

k^ = 60 1000

0.865

E = 10

hog 6.55917 k 0.865

= 10

nog 6.55917 x lOo" L 0.865

= 1805 psi

= 260 ksf

J.2 Design Example Using PCA Method

1. Safety factor, SF:

For frequent operations of this forklift truck in

channelized aisle traffic, select a safety factor of 2.0 (permits

unlimited stress repetitions).

2. Concrete working stress, WS:

,,c MR 640 WS = fTF = 3P = 2:0 = 320 psi

3. Slab stress per 1,000 lb of axle load

WS axle load, kips

320 25 = 12.8 psi

4. Enter Fig. 1.4 with stress of 12.8 psi; move right to

contact area of 114 sq in.; then down to wheel spacing of 37 in.;

255 then right to read a slab thickness of 7.9 in. on the line for

subgrade k^ of 100 pci (use 8 in. thick slab).

J.3 Design Example Using the Regression Equations

1. Assume trial slab thickness of 4 inches.

2. Calculate allowable stresses:

^t " '^'^ ^ " 474.34 psi = 68.31 ksf

V = 4 VT~ = 252.98 psi

Stack Loading Condition:


(L)0-l^W)0-01(h)°-30(A ) 0 - " { p )0-97 \ = (E,)(1026.00) 0 .74" ^

S

= (1.18)(1026.00)(^^°'°- ' ' ( l°°)°-°'(^/ l^)°i^°>°-' ' (°-^^^)'- ' ' (260)°-'*

= (1.18)(1026.00) (2-017)(1.047)(0719)(1.862)(0.586) (61.246)

= 32.75 k ips/ f t^ < f t allocable

,„)0.18(,,0.59( ,0.91

"y = (Ey)(30,045.00) ^^ô.48(g ,0.82(ft ,0.18

(150)°-*^(260)°-^2(10)°-l^

M iOr-5n riA^ nnK(^-^91j(0.523)(0.605) - (1.15) (30,045.00)^^_pgo)(g5_5go)(i_514j

= 15.63 kips/ft2< f^ allocable

256 4. Calculate expected shear forces:

(L)°-30(h)l-82(A )0-02(p )0-98 ^x = (Ey)(15.14) ,, . > \ ^;'' X y- j y jO . i i ^g ,0.60

/ ,f .pix0.30/./ .^vl.82,,nx0.02,^ C7c^0.98 = (1.18) (15.14)^^-5^^ WlLl (10) (0.576)

(100)°-^-(260)°-^°

= (1 18H15 l/l^(4.496)(0.135)(1.047)(0.582) • ^^•^«Hib. i4j (l.660)(28.118)

= 0.142 k ips/ f t

(,)0.09(^)1.83(p^)0.74

\ - (^)(1^67-0) (^)0.83(,^)0.57(;^)0.44

= (1 15)(1467 0) (100)^'^^(4/12)^-^3(0.576)Q-^^ (1.15)11467.0) (^50)0.83(260)0.57(10)0-^^

- n iRUiAfi7 n^ (1.514)(0.134)(0.665) (63.996)123.798)(2.754)

= 0.054 kips/ft


0.142 1000 ^ " (1+2x0.8x0.3333)x4x0.27 ^ 144

= 0.593 psi < v^ allowable

257 6. Calculate maximum differential deflection:

(L)0.44(^)0.62(^)0.20(p^)0.78

^ (0 47) (150)Q-^^100)0'^2(4/12)Q-2Q(0.576)Q'^^

(260)°-^^(10)°-2^

(0 A7\ (9.067)(17.378)(0.803)(0.650) ^ • ' (245.937)11.950)

= 0.081 ft



(L)0.12(^,0.30( ,2.04

o„ = (EJ(22.25) 'x - ^^x'^"-"' („)0.04(E^,0.72(^y.02(s,2.04

- (1 18U22 25) '150l°-l^(4/12)°-^°(15.84)2-°^ - (1.18)(22.25) (ôo)0-0^260)°-"(10)0-02(3.08)2-04

/, ,o\^,o ,c^ (1.825)(0.719)(280.221) = (1.18)(22.25) •(i.202)(54.800)(l.047)(9.923)

= 14.107 kips/f t2< f t allowable

(w)°-°^(h)°-3^P,)2-^^

258 = (1.15)(347.71) (100)^-Q^(4/12)Q-34(i5^84)2.39

(150)0-15(260)0.72(10)0.51(308)3.48

= (1.15)(347 71) (1.202)(0.688)(736.911) ^ ^^^ - ^ (2.120)(54.80)(3.236)(50.137)

= 12.928 kips/ft^ <f. ,, ., t allowable

8. Calculate expected shear forces:

(W)0-28(h)l-80(A )0.52( )2.80 V = (E )(13 33) ^ w ^^f X xH13-33) (L)U.21(,^)0.57(3)(3.96)

= (1.18)(13.33) (100)^-^^(4/12)l-^Q(lO)0-52(15.84)2-^0

(150)0-21(260)0-5^3.08)3-^^

= (^ ifiUi- o'^\ (3.631) (0.138) (3.311) (2287.255) U.iOMi^.oJJ (2.864)(23.798)(86.032)

= 10.179 k ips/ f t

(L)0.03(^)0.11(^)1.90(^ ) 0 . 2 2 ( ) 2 . 1 3

\ - ' ^y^ ( ° -^^ ' (E^)0.59(3)2.64

= (1 i5)(o 19) (150)Q-03(100)0-11(4/12)1-^0(10)°-22(15.84)2-13

(260)0-5^(3.08)2-^^

, , -^w^ ,^t (1.162) (1.660) (0.124) (1.660) (359.318) = (1-15)(0.19) (26.597)(19.488)

= 0.060 k ips/ f t

9. Calculate maximum design shear stres:

^ _ ^ 10.179 1000 (1+2x0.8x0.3333)x4x0.27 ^ 144

259

= 42.501 psi < V ,, ^ c all owable

10. Calculate maximum differential deflection: (L)0-21(w)0-53(h)0.71(p^)1.12

(E )0-^^(A )0-21(s)1.81

,n ION (150)0-21(100)0-53(4/12)0-^1(15.84)1-12

" ^ ^ ^ (260)0-98(io))0.21(3.o8)l-81

f^ m^t (2.864)(11.482)(0.458)(22.066) " ^ -- ^ (232.634)(1.622)17.661)

= 0.014 ft

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Documents

Analysis of Industrial Slabs-On-ground