deflections of reinforced concrete beams and columns

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ACI-318-89 ve TS-500 şartnameleri, ve literatürdeki test verileriyle ...... 437. 30.3. 186.4. 63.5 a b. 3.1. 7.9. 0.80. 1.10. 20C3a. 506. 89. 40. 22.4. 233.5. 31.8 a b.
DEFLECTIONS OF REINFORCED CONCRETE BEAMS AND COLUMNS

A THESIS SUBMITTED TO THE GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCES OF THE MIDDLE EAST TECHNICAL UNIVERSITY

BY

MEHMET HALİS GÜNEL

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY IN THE DEPARTMENT OF CIVIL ENGINEERING

SEPTEMBER 1995

Approval of the

Graduate

School of

Natural and Applied

Sciences ______________________ Prof. Dr. İsmail Tosun Director

I certify that this thesis satisfies all the requirements as a thesis for the degree of Doctor of Philosophy.

__________________________ Prof. Dr. Doğan Altınbilek Head of Department

This is to certify that we have read this thesis and that in our opinion it is fully adequate, in scope and quality as a thesis for the degree of Doctor of Philosophy.

____________________ Prof. Dr. Uğur Ersoy Supervisor Examining Committee Members

Prof. Dr. S. Tanvir Wasti

_____________________

Prof. Dr. Uğur Ersoy

_____________________

Prof. Dr. Fuat Erbatur

_____________________

Assoc. Prof. Dr. Uğur Polat

_____________________

Assoc. Prof. Dr. Hüsnü Can

_____________________

ii

ABSTRACT

DEFLECTIONS OF REINFORCED CONCRETE BEAMS AND COLUMNS

Günel M. Halis Ph. D., Department of Civil Engineering Supervisor: Prof. Dr. Uğur Ersoy

September 1995, 167 pages

This study presents the development of a computer program

based

on

a

method

derived

by

the

author

for

predicting the short-term and long-term deflections of reinforced

concrete

members.

Deflections

of

beams

and

slender columns computed using this program are compared with the results obtained from ACI-318-89, TS-500, and available test data in the literature. In the design codes some empirical equations are given for beams but

iii

these cannot be used for slender columns or beam-columns. The calculated deflections have shown good agreement with 63

test

data

from

11

references

reported

in

the

literature. The program consists of structural analysis subroutines

with

an

iteration

technique

to

take

into

account the nonlinear behavior of reinforced concrete, and

moment-curvature

subroutine

with

defined

stress-

strain relations for concrete in tension and compression for flanged sections with up to five layers of steel. The moment

diagram

of

the

member

is

determined

by

an

iterative procedure in which the moments are computed from the flexural rigidities assumed initially. Moments and

flexural

acceptable

rigidities

error

limit.

are The

recomputed defined

up

to

an

stress-strain

relations for short and long-term periods can be modified easily , since they are entered as statement functions in the program.

Keywords: Deflection, Reinforced Concrete Beams, Reinforced Concrete Columns, Stress-Strain Relation for Reinforced Concrete, Creep, Shrinkage

iv

ÖZ

BETONARME KİRİŞ VE KOLONLARIN SEHİMİ

Günel M. Halis Doktora, Inşaat Mühendisliği Bölümü Tez Yöneticisi: Prof. Dr. Uğur Ersoy

Eylül 1995, 171 sayfa

Bu

çalışmada

araştırmacının

geliştirdiği

bir

metot esas alınarak betonarme elemanların kısa ve uzun süreli

yükler

altındaki

bigisayar

programı

kolonların

program

sehimini

hesaplamak

hazırlanmıştır. kullanılarak

Kiriş

hesaplanan

için ve

bir

narin

sehimleri,

ACI-318-89 ve TS-500 şartnameleri, ve literatürdeki test verileriyle karşılaştırılmıştır. Tasarım şartnamelerinde, kirişler

için

bazı

ampirik

denklemler

verilmektedir,

fakat bu denklemler narin kolonlar ve kiriş-kolonlar için

v

kullanılamamaktadır. 11

referanstan

sağlamaktadır. davranışını

Hesaplanan

alınan

63

Program,

hesaba

sehimler,

test

sonucuna

betonarmenin

katarak,

literatürdeki iyi

doğrusal

deneme-yanılma

uyum

olmayan

tekniğiyle;

yapı analizi altprogramları, ve tablalı ve beş seviyeye kadar donatılı beton kesitlerin çekme ve basınç altında tanımlanan

zamana

ilişkilerini

bağlı

içeren

oluşmaktadır.

gerilme

ve

moment

yanılmaların

kabul

birim

moment-eğrilik

Elemanın

rijitlikleri

-

deformasyon

altprogramından

moment

diyagramı,

eğilme

diyagramı

arasındaki

deneme-

sınırları

arasında

edilebilir

hata

kalmasıyla kesinlik kazanır. Beton için tanımlanan zamana bağlı gerilme - birim deformasyon ilişkileri programda kolayca değiştirilebilinir.

Anahtar Kelimeler: Sehim, Betonarme Kirişler, Betonarme Kolonlar, Betonarmede Gerilme - Birim Deformasyon Ilişkisi, Sünme, Rötre

vi

ACKNOWLEDGMENTS

This study was suggested by Prof. Dr. Uğur Ersoy and the work has been carried out under his supervision.

The appreciation

author to

wishes

Prof.

Dr.

to Uğur

express Ersoy

his for

grateful his

close

supervision, guidance and encouragement throughout the study. It has been a pleasure to have studied under the direction of Dr. Ersoy.

The author would also like to thank Dr. Uğur Polat,

Dr.

Ergin

Atımtay,

Dr.

Fuat

Erbatur,

and

Dr.

Tanvir Wasti for their invaluable help and comments.

This thesis is dedicated to Prof. Dr. Uğur Ersoy.

vii

TABLE OF CONTENTS

ABSTRACT .............................................iii ÖZ .....................................................v ACKNOWLEDGMENTS ......................................vii TABLE OF CONTENTS ...................................viii LIST OF TABLES ........................................xi LIST OF FIGURES ......................................xii LIST OF SYMBOLS....................................... xv CHAPTER I.

INTRODUCTION ......................................1 1.1 General .......................................1 1.2 Object and Scope of This Study ................3

II. REVIEW OF RELATED INVESTIGATIONS ..................6 2.1 Analytical Studies and Code Requirements ......6 2.1.1 ACI-318-89 (1989) .......................6 2.1.2 TS-500 (1985) ..........................11 2.1.3 CEB-FIB Model Code (1990) ..............11 2.1.4 Method of Yu and Winter (1960) .........12 2.1.5 Method of Corley and Sozen (1966) ......14 2.1.6 Method of Branson (1963,1977) ..........16

viii

2.1.7 Analytical Study of Roger Green for Unrestrained (Pin Ended) Columns .......18 2.2 Experimental Studies on Beams ................19 2.2.1 Washa and Fluck (1952) .................19 2.2.2 Washa and Fluck (1956) .................20 2.2.3 Yu and Winter (1960) ...................20 2.2.4 Corley and Sozen (1966) ................21 2.2.5 Bakoss et al. (1982) ...................22 2.2.6 Clarke et al. (1988) ...................23 2.2.7 Christiansen (1988) ....................23 2.2.8 Al-Zaid et al. (1991) ..................24 2.3 Experimental Studies on Columns ..............25 2.3.1 Viest et al. (1956) ....................25 2.3.2 Green (1966) ...........................26 2.3.3 Hellesland and Green (1971) ............27 III. METHOD OF ANALYSIS ...............................29 3.1 Introduction .................................29 3.2 Basic Assumptions and Methodology ............31 3.2.1 General ................................31 3.2.2 Stress-Strain Relationship of Concrete and Steel .....................35 3.2.2.1 Stress-Strain Relationship of Concrete Under Compression ...........35 3.2.2.2 Time-Dependent Behavior of Concrete ..37 3.2.2.3 Stress-Strain Relationship of Concrete Under Tension ...............44 3.2.2.4 Stress-Strain Relationship of Steel ..46

ix

3.3 Moment-Curvature Relationship ................47 3.4 Method of Analysis ...........................54 IV. NUMERICAL APPLICATIONS ...........................60 4.1 Introduction and Additional Information About the Available Data .....................60 4.1.1 Experimental Beam Deflection Studies and Comparisons ........................60 4.1.2 Experimental Column Deflection Studies and Comparisons ........................61 4.2 Discussion of the Results ....................93 4.3 Reconsideration of TS-500 and ACI-318-89 Codes ........................................98 V. CONCLUSIONS .......................................99 5.1 General Remarks ...............................99 5.2 Conclusions ..................................100 5.3 Recommendations ..............................103 REFERENCES .........................................105 APPENDIX A.1 Fortran Listing of Program BECODE ............113 A.2 Data Input to Program BECODE .................169 VITA ...............................................171

x

LIST OF TABLES

TABLE 4.1 Properties of Test Beams .......................72 4.2 Comparison of Calculated and Measured Deflections (Beams) ............................78 4.3 Properties of Test Columns and Deflections (Measured and Calculated) ......................86

xi

LIST OF FIGURES

FIGURE 2.1

Typical Test Beam of Washa and Fluck (1952) ....19

2.2

Typical Test Beam of Washa and Fluck (1956) ....20

2.3

Typical Test Beam of Yu and Winter (1960) ......21

2.4

Typical Test Beam of Corley and Sozen (1966) ...22

2.5

Typical Test Beam of Bakoss et al. (1982) ......22

2.6

Typical Test Beam of Clarke et al. (1988) ......23

2.7

Typical Test Beam of Christiansen (1991) .......24

2.8

Typical Test Beam of Al-Zaid et al. (1991) .....25

2.9

Typical Column Specimen Tested by Viest et al. (1956) ............................26

2.10 Typical Test Specimen of Green and Breen (1969) 27 2.11 Typical Test Specimen of Hellesland and Green (1971) ...................................28 3.1

Stresses in a Cracked Beam Under Service Loading ........................................30

3.2

Deflection of a Column .........................34

3.3 Hognestad's (1951) Stress-Strain Diagram ........36 3.4 Time Versus Strain Corresponding to Maximum Stress (6 Months Duration) ..............42

xii

3.6

Proposed Time-Dependent Stress-Strain Diagram for 2½ Years ...........................43

3.7

Assumed Stress-Strain Diagram for Concrete in Tension ............................45

3.8

Elasto-Plastic Stress-Strain Diagram for Steel in Tension or Compression ................47

3.9

Strain Distribution, Stress Block for Concrete in Compression and Tension, and Stress-Strain Diagram for Steel in Tension and Compression ...50

3.10 Typical Moment-Curvature Curves ................52 3.11 Influence of Concrete in Tension on Moment-Curvature Curves for a Column ...........53 3.12 Tracing the Moment-Curvature Diagram ...........57 3.13 Flowchart for the Computer Program (Main Steps) ...................................58 3.14 Flowchart for the Computer Program (Structural Analysis Part) .....................59 4.1

Comparison of Computed and Observed Deflections - Washa and Fluck (1952) ...........62

4.2

Comparison of Computed and Observed Deflections - Washa and Fluck (1956) ...........64

4.3

Comparison of Computed and Observed Deflections - Yu and Winter (1960) .............66

4.4

Comparison of Computed and Observed Deflections - Corley and Sozen (1966) ..........69

4.5

Specimen 1B2 - Comparison of Computed and Observed Deflections - Bakoss et al. (1982) ....71

xiii

4.6

Bar Charts for Short-Term Deflections of Beams-ACI ......................................95

4.7

Bar Charts for Short-Term Deflections of Beams-Calculated ...............................95

4.8

Bar Charts for Long-Term Deflections of Beams-ACI ......................................96

4.9

Bar Charts for Long-Term Deflections of Beams-Calculated ...............................96

4.10 Bar Charts for Short-Term Deflections of Columns-Calculated .............................97 4.11 Bar Charts for Long-Term Deflections of Columns-Calculated .............................97

xiv

LIST OF SYMBOLS

As

: area of tension steel

As' : area of compression steel b

: width of beam at the compression side

bw

: width of beam at the tension side

Ct

: creep coefficient (ratio of creep strain to initial (elastic) strain) at any time t

Cu

: ratio of ultimate creep strain to initial (elastic) strain

d

: effective depth of reinforced concrete beam

d"

: distance between tension and compression steel layers

Ec

: modulus of elasticity of concrete

Es

: modulus of elasticity of steel

EI

: flexural rigidity

e

: load eccentricity

fc

: compressive strength of concrete

fc' : compressive strength of concrete (28 day cylinder strength) fct' : tensile strength of concrete h

: total depth of beam

xv

Icr : moment of inertia of cracked transformed section Ie

: effective moment of inertia

Ig

: moment of inertia of gross concrete section

l

: span length

P

: concentrated load

p

: uniformly distributed load

N

: eccentric load

t

: loading duration in days

Δ

: maximum deflection

εc

: concrete compressive strain

εco : concrete compressive strain corresponding to maximum stress εcu : ultimate concrete compressive strain εcot: time dependent concrete compressive strain corresponding to maximum stress εct : concrete tensile strain εcto: concrete tensile strain corresponding to maximum stress εctu: ultimate concrete tensile strain εsh : shrinkage strain σc

: concrete stress in compression

σct : concrete stress in tension ACI : American Concrete Institute TS

: Turkish Standards

xvi

CHAPTER I

INTRODUCTION

1.1 General

Reinforced material.

The

concrete

is

proportions

a

widely

of

used

flexural

building

members

are

generally chosen so as to restrict the deflections to acceptable limits. However, in recent years, the tendency of using higher strength concrete and reinforcing steel has

resulted

result,

the

in

the

use

problem

of shallower sections. As a

of

predicting

deflections

of

reinforced concrete members has gained more importance.

Both

short-term

and

the

long-term

deflections

must be considered in design. Since more slender members are

possible

deflection design. occurring

today,

becomes

Short-term

due

a

to

high

controlling

deflections

instantaneously

upon

strength aspect

are the

materials,

of

the

total

defined

as

those

application

of the

load. The time element is assumed to be unimportant,

provided the load is applied within a matter of hours. Long-term deflections refer to those existing at some time

interval

after

the

initial

load

application.

A

combination of shrinkage and creep, or plastic-flow of concrete under sustained load, gives rise to the timedependent

phenomenon.

Long-term

or

time-dependent

deflections occur only under sustained loads.

The computation of deflections within reasonable accuracy is a difficult task. Reinforced concrete is a nonhomogeneous, material.

anisotropic,

Therefore

classical

nonlinear,

and

methods

for

inelastic deflection

calculation will not yield accurate results.

The modulus of elasticity of concrete changes as a function of time under sustained loads. This change is a function of many different variables, such as, time, age of the concrete when loaded, load level, humidity, temperature,

type

of

aggregate,

type

of

cement

etc.

Therefore it is very difficult to estimate the modulus of elasticity, Ec, for concrete with reasonable accuracy. The change in the modulus of elasticity with time can be as much as 300%.

Since the tensile strength of concrete is very low,

reinforced

concrete

members

crack

under

service

loads. The moment of inertia of a cracked section is very

2

different from that of an uncracked section and depends on the length of the crack. However, the concrete section between

the

cracks

should

still

be

considered

as

uncracked. In conclusion, it may be stated that, due to cracking, the moment of inertia varies along the span of the member.

In clear

the

that

it

light is

of

above

very

discussion,

difficult

to

it

becomes

calculate

the

deflection of reinforced concrete members.

1.2 Object and Scope of This Study

In

the

previous

section,

the

difficulties

encountered in calculating the deflection of reinforced concrete members have been discussed. Up until now, a few analytical

studies

deflections.

While

have the

been

made

deflections

for

predicting

predicted

by

the such

methods agreed quite well with the results obtained from a certain test series, the agreement was not as good with other test series.

In the design codes, some simplified version of these methods are given. Such methods are only applicable to

beams

addition

and to

cannot bending,

be

used

carry

3

for

columns,

axial

loads.

which Also

in the

prediction of beam deflections by such methods can result in errors by as much as ±40%.

The object of this study was to develop a general method based on the real behavior of reinforced concrete to predict the instantaneous and long-term deflection of simple

beams,

continuous

beams,

and

columns,

with

rectangular or flanged cross section. For this purpose, a computer

program

was

developed

based

on

the

finite

element method. In the method developed, first, a momentcurvature program is written, based on realistic models for steel and concrete, taking into consideration the tensile strength of concrete. The deflections are found by

an

iterative

relations

technique,

developed.

For

using

the moment-curvature

long-term

deflection,

a

different concrete model is used which also considers the creep and shrinkage deformations.

It was also intended to make an extensive review of

previous

research

related

to

the

deformation

of

reinforced concrete beam and column members, both under short

and

long

term

loading.

In

Chapter

II,

Related

sections of ACI-318-89 (1989) and TS-500 (1985) Building Codes, and some empirical methods about deflections are summarized, together with the detailed review of previous experimental studies.

4

In Chapter III, the method of analysis and the computer

program

developed

by

the

author

for

the

prediction of the deflections are presented.

The accuracy of the method developed was checked by

comparing

the

deflections

obtained

by

the

method

developed and available test results presented in Chapter IV.

In

this

chapter,

this

comparison

is

given

in

a

tabulated form together with the results of ACI-318-89 (1989) and TS-500 (1985) Building Codes.

Conclusions

based

on

Chapter V.

5

this

study

are

given

in

CHAPTER II

REVIEW OF RELATED INVESTIGATIONS

2.1 Analytical Studies and Some Code Requirements

2.1.1 ACI-318-89 (1989)

ACI-318-89 (1989) Building Code Requirements for Reinforced prediction

Concrete, of

gives

short-term

and

empirical

equations

for

long-term

deflections of

reinforced concrete beams.

For

calculating

short-term

deflections

of

partially cracked reinforced concrete beams under service loads, an effective moment of inertia, Ie, suggested by Branson (1963) is used. This approach provides a smooth transition value between the moment of inertia Icr of the transformed,

fully

cracked

section

and

the

moment

inertia Ig of the gross uncracked concrete section.

6

of

Ie = (Mcr/Ma)3Ig+1.0-(Mcr/Ma)3Icr  Ig

(2.1)

where Mcr : cracking moment = frIg/yt Ma

: maximum moment at the stage where deflection is computed

fr

: modulus of rupture of concrete = 0.7fc' (in MPa.)

yt

: distance from centroidal axis of gross section, to extreme fiber in tension

For continuous members, the effective moment of inertia can be taken as the average of values obtained from Eq. (2.1) for the critical positive and negative moment

sections.

For

prismatic

members,

the effective

moment of inertia can be taken as the value obtained from Eq. (2.1) at midspan for simple and continuous spans, and at the support for cantilever beams or slabs (The use of midspan properties for continuous prismatic members is considered

satisfactory

in

approximate

calculations

because the midspan rigidity has the dominant effect on deflections).

7

Short-term

deflection

is

then

computed

by

substituting Eq. (2.1) into Eq. (2.2).

i = kMal2/(EcIe)

(2.2)

where k

: a deflection coefficient that depends on the load distribution and support conditions.

Ec : modulus of elasticity for concrete = 4700fc' (in MPa.) l

: span length

According to the ACI-318-89 (1989), additional long-term

deflection

of

reinforced

concrete

beams

resulting from creep and shrinkage is to be found by multiplying

the

immediate

deflection

caused

by

the

sustained load by the factor

 = /(1.0+50')

(2.3)

where ' is the compression steel ratio at midspan for simple and continuous spans and at the support for cantilevers, and  is a time dependent factor to be taken as follows : time 5 years or more  = 2 ; 12 months  = 1.4 ; 6 months  = 1.2 ; and 3 months  = 1.0

8

Subsequently,

the

total

deflection

due

to

sustained load is

t = (1.0+)i

(2.4)

Some variables such as age at loading and ambient relative humidity which influence deflections are ignored for simplicity.

The method was introduced with the 1983 ACI Code. An earlier method, presented first in the 1963 ACI Code, used the ratio of compression steel to tension steel areas, As'/As to account for the influence of compression steel. Both approaches are empirical, based largely on tests of beams made in the 1950s and 1960s.

In

the

case

of

unrestrained

and

eccentrically

loaded reinforced concrete columns, secondary deflections from the additional second order moments must be taken into consideration. The deflection increases with time, since

creep

of

concrete

tends

to

reduce

the

bending

stiffness, and hence increases the deflection. In the ACI-318-89 (1989) Code, for hinged columns with equal end moments, the effect of additional moments is taken into account by the moment magnifier method (End moment M2b is magnified by a factor b). In this method, the creep effect is considered by dividing the bending stiffness by

9

1.0+d , where d is the ratio of dead load moment to total load moment. The critical load Nc (a column with an effective length lk, and axial load Nu at a certain eccentricity) required in the calculation of the moment magnifiers is then based on this reduced value for the bending

stiffness.

deflection

is

Namely,

taken

additional

into

moment

consideration

by

due

to

moment

magnification, and the long-term effect, by reducing the bending

stiffness.

However

there

is

no

recommended

practical deflection equation for columns as in the case of beams. The equations related to moment magnification are given below.

Mc = bM2b

(2.5)

b = 1.0/(1.0-Nu/Nc)  1.0

(2.6)

=

2EI/lk2

EI = EcIg/2.5/(1.0+d)

(2.8)

Nc (2.7)

There are two stiffness EI definitions, but Eq. (2.8) is more conservative.

10

2.1.2 TS-500 (1985)

In

TS-500

(1985)

Turkish

Building

Code

Requirements for Reinforced Concrete, effective moment of inertia, and short-term deflection calculations are very similar

to

ACI-318-89

(1989)

Code.

However,

the

calculation of additional long-term deflection is found by multiplying the immediate deflection of the sustained load by,

 = 2.5/(1.0+0.7As'/As)

(2.9)

Also, modulus of elasticity (Ec = 3250fc'+14000 MPa.), cracking moment (Mcr = 1.3fct'Ig/yt), and direct tensile strength of concrete (fct'=0.35fc' in MPa) are defined differently from the ACI (1989) Code.

2.1.3 CEB-FIB Model Code (1990)

According deformations

due

to

CEB

Code

to

bending

(1990),

are

the

calculated

short-term from

the

curvatures by applying appropriate procedures such as the principle of virtual work or double integration. For the calculation of long-term deflections, empirical equations are given.

11

Both for the short-term and long-term deflection calculations,

impractical

and

complicated

empirical

equations are proposed. However, this approach does not account for variations in flexural rigidity along the beam, and no special reference is made to continuous beams.

2.1.4 Method of Yu and Winter (1960)

In

this

empirical

method

for

calculating

beam

deflection, moment of inertia of the cracked transformed section,

Icr

at

midspan

is

used as a constant value

throughout the length of the span for simple spans, but the average value of the positive and the negative moment regions

is

used

in

the

deflection

calculation

for

continuous beams.

For short-term deflections;

Method

A

:

Moment

of

inertia

of

the

cracked

transformed section can be used in the well known elastic deflection equations.

Method B : To account for the participation of concrete

in

tension,

deflections

in

Method

A

decreased by multiplying by a correction factor,

12

can

be

 = 0.1(fc')2/3bh(h-kd)/Ma  1.0

(2.10)

where fc' : compressive strength of concrete b

: width of beam at tension side

h

: total depth of member

kd

: depth of neutral axis

Ma

: maximum moment in member

For long-term deflections;

Method

C

:

An

empirical

'modified

sustained

modulus of elasticity' value, and dependent moment of inertia of the cracked transformed section is used in the elastic deflection formulas (Branson,1977:202).

Ect' = Ec/(1.0+EcYc't1/3)

(2.11)

where Y

: multiplier according to the number of years under load

t

: duration of loading in days, but not to exceed 365

c' : modified coefficient of creep and shrinkage in millionths

13

Method D : Short-term deflection as calculated from

Method

A

is

multiplied

by

an

empirical

factor,

considering the duration of loading and the reinforcing condition (Branson,1977:202).

2.1.5 Method of Corley and Sozen (1966)

Empirical deflections

of

curvatures, beams

are

for

short

proposed,

and

and

long-term

a

numerical

procedure is then used to compute deflections.

The instantaneous curvature is,

i = M/(EcIcr) = i/(kd)

(2.12)

where M

: design moment

Icr : moment of inertia of cracked transformed section i

: concrete strain in extreme fiber in compression

kd

: depth of neutral axis

If the product of the tension steel ratio and the modular ratio n is greater than about 0.18, the member is considered to be fully cracked; if n is less than

14

about 0.08, the member is considered to be partially cracked; if n is between 0.18 and 0.08, the member is considered to be fully cracked if the applied moment is more than twice the cracking moment. For T-beams, the value n should be calculated using the effective width as that at the extreme fiber in tension.

The shrinkage curvature is,

sh = 0.035(-')/d

(2.13)

where  and ' : tension and compression steel ratios respectively d : effective depth of section

The creep curvature is,

c = ikd/d*m

(2.14)

where m = 3-'/

(for rectangular sections)

m = 2(3-'/) (for flanged sections with kd  2t/3 where t is the thickness of the top flange)

15

2.1.6 Method of Branson (1963,1977)

For short-term deflections of beams, Eq. (2.1) and Eq. (2.2) are used as in ACI-318-89 (1989). Long-term deflection, cs is computed using Eq. (2.15) which is based on the recent improvement made by the ACI Committee 209 (1982), in the prediction of creep and shrinkage. The procedure

is

also

recommended

by

ACI

Committee

435

(1966,1978).

cs = kr(CF)a(CF)hCti+shl2

(2.15)

where kr : compression steel factor (CF)a , (CF)h

: correction factor for the age at

loading, and ambient relative humidity respectively Ct : creep coefficient (ratio of creep strain to initial strain at any time) 

: a constant related to the boundary conditions 0.5 for cantilever beams, 0.125 for simply supported beams, 0.086 for beams continuous at one end only, 0.063 for beams continuous at both ends.

sh : shrinkage curvature l

: span length

16

kr = 0.85/(1.0+50')

(2.16)

Ct = t0.60 / (10+t0.60)*Cu (2.17)

where t

: duration of loading in days

Cu : ultimate creep coefficient (recommended value is 2.35)

(CF)a = 1.25ta-0.118

(2.18)

(CF)h = 1.27-0.0067H

(2.19)

where ta : age at loading in days H

: ambient relative humidity, percent

for (-')  3% , sh = 0.7sh/h(-')1/3(-')/1/2

(2.20)

for (-')  3% , sh = sh/h

(2.21)

where sh : shrinkage strain

17

sh = t/(35+t)*(sh)u

(2.22)

where (sh)u : ultimate shrinkage strain (recommended value is 780*10-6 in/in )

2.1.7 Analytical Study of Roger Green for Unrestrained (Pin Ended) Columns

An prediction

analytical of

unrestrained difference

method

deformation

columns. method

was

A

was of

in

for

eccentrically

numerical used

developed

solution

the

by

computer

the

loaded finite program

developed for rectangular sections with two layers of steel. Concrete was assumed to carry no tension. Shortterm stress-strain relationship was similar to Hognestad (1951); but maximum fiber resistance was taken as 95% of the standard compressive cylinder strength for concrete (Hognestad proposed 85%). Based on the work of Rüsch (1960),

a

time-dependent

stress-strain

diagram

for

concrete was used by increasing the strains in the shortterm diagram.

18

2.2 Experimental Studies on Beams

A considerable number of studies related to timedependent deflection of beams have been conducted.

2.2.1 Washa and Fluck (1952)

In 1952, Washa and Fluck (1952) reported their experimental experimental

results. research

The was

main to

objective

study

the

of

this

influence

of

compressive reinforcement on the long term behavior of reinforced

concrete

flexural

members.

Single

span

rectangular beams were uniformly loaded at the age of 14 days

and this load was kept on the beams for for 21/2

years

(Fig.

2.1).

Environmental

conditions

were

not

controlled.

p=1.56or5.52N/mm l=6.1m

h=203or305mm bw=152.5or203mm

Fig. 2.1 Typical Test Beam of Washa and Fluck (1952)

19

2.2.2 Washa and Fluck (1956)

In 1956, again Washa and Fluck (1956), reported the results of an experimental study, similar to the one published

in

1952.

However

in

this

study,

two-span

continuous reinforced concrete beams were tested (Fig. 2.2). At the supports, sections were heavily reinforced in tension and in compression. Tests were made under uncontrolled environmental conditions, and age at loading was 14 days. Load was kept on the beams for 21/2 years. This research is considered to be the most compressive test

program

related

to

the

long

term

deflection

of

statically indeterminate reinforced concrete beams.

p=2.77N/mm l=2*6.1m

h=203mm bw=152.5 mm

Fig. 2.2 Typical Test Beam of Washa and Fluck (1956)

2.2.3 Yu and Winter (1960)

Yu and Winter (1960) in 1960, have reported the results of tests on flanged beams. The main parameters investigated were the shape of the cross-section, the

20

amounts

of

span/depth distributed

tensile ratio

and

and

loading

compressive

the was

concrete applied

to

reinforcement, the strength. Uniformly single

span

beams

(Fig. 2.3), at about 28 days, and was kept for 270 days. Relative humidity varied between 15 and 53%.

p=3.8,6.4,11.7,12.3 N/mm l=4.3or6.1m

b=305or610mm t=51or63.5mm h=203or305mm bw=152.5mm

Fig. 2.3 Typical Test Beam of Yu and Winter (1960)

2.2.4 Corley and Sozen (1966)

Corley and Sozen (1966), made tests to study the long term deflections of reinforced concrete beams. Tests were made in laboratory under controlled environmental conditions, with a relative humidity of 50%, on simply supported and single reinforced rectangular beams (Fig. 2.4), loaded at the age of 28 days. They were loaded at the quarter points and loading was kept on the beams for 2 years.

21

P=5KN

5KN

p=0.20or0.27N/mm

h=109.5or152.5mm

l=1.8m

bw =76mm

Fig. 2.4 Typical Test Beam of Corley and Sozen (1966)

2.2.5 Bakoss et al. (1982)

In 1982, Bakoss et al. (1982) reported results of long-term tests on single reinforced, rectangular, and simply supported beams (Fig. 2.5). Third points loading was applied at 28 days and sustained for 500 days. Tests were executed with relative humidity varying between 35 and 75%. P=2.6KN

2.6KN

p=0.35N/mm

h=150mm

l=3.75m

bw=100mm

Fig. 2.5 Typical Test Beam of Bakoss et al. (1982)

22

2.2.6 Clarke et al. (1988)

Clarke et al. (1988) reported tests on simply supported loading months.

rectangular

was

applied

Tests

beams at

were

(Fig.

28

days

carried

2.6). and

out

Third

points

sustained

under

for

6

uncontrolled

environmental conditions, where relative humidity varied between

35

horizontal

and

60%.

plane

(so

The

loads

that

the

were

applied

self-weight

in

effect

the is

eliminated). P=5KN

5KN h=153mm

l=2.1m

bw=100mm

Fig. 2.6 Typical Test Beam of Clarke et al. (1988)

2.2.7 Christiansen (1988)

In 1988 Christiansen (1988) reported results of tests Simply

with

the

longest

supported,

load

rectangular

duration beams

ever

were

recorded.

subjected to

sustained loading for about 8 years (Fig. 2.7). Midspan and quarter point loading was applied to the test beams. The parameters investigated were the age at loading (20

23

or 174 days) and the compressive reinforcement ratio. Experiments

were

conducted

under

fluctuating

environmental conditions. The average relative humidity given for the first two years was 55%. P  1.5or6.0KN p=1.12N/mm

h=280mm

l=7.5m

bw=170mm

Fig. 2.7 Typical Test Beam of Christiansen (1988)

2.2.8 Al-Zaid et al. (1991)

Al-Zaid et al. (1991), published the results of an

experimental

program

which

included

the

tests

on

simply supported beams with rectangular cross sections having under

compressive point

loading

reinforcement. applied

at

Beams

midspan

were and

tested

uniformly

distributed loading. All beams were tested under shortterm loading. Loading was applied at the age of 28 days.

24

P=8.1-33.7KN p=1.5-6.1N/mm

h=200mm

l=2.5m

bw=200mm

Fig. 2.8 Typical Test Beam of Al-Zaid et al. (1991)

2.3 Experimental Studies on Columns

Data concerning the time-dependent deflections of columns under sustained loading are limited.

2.3.1 Viest et al. (1956)

Viest et al. (1956) in 1956, reported results of tests on sustained load strength of eccentrically loaded unrestrained

(pin

(Fig.

The

2.9).

ended)

reinforced

parameters

studied

concrete were

the

columns age

at

loading, the concrete strength, and the load magnitude and eccentricity. Duration of loading varied between 457 and 933 days. Short-term deflection data are available for those which failed under sustained loading.

25

N=91.6-246.9KN e=31.8-96.8mm

A

.

.

1m .

127mm A

127mm Sect. A-A

Fig. 2.9 Typical Column Specimen Tested by Viest et al. (1956)

2.3.2 Green (1966)

In 1969, an experimental investigation related to the

sustained

load

response

of

eccentrically

loaded

unrestrained columns (Fig. 2.10) was reported by Green and Breen (1969). This study was based on the doctoral dissertation were

the

of

load

Green

(1966).

magnitude

and

Parameters

investigated

eccentricity.

Average

relative humidity was 75%, during the first 8 months. Columns were loaded at about 50 days, and loading was kept on the specimens for 11/2 years.

26

N=8.2-24KN e=3.8-43.2mm

152.5mm

1.9m A

.

.

101.5mm

A

Sect. A-A

Fig. 2.10 Typical Test Specimen of Green and Breen (1969)

2.3.3 Hellesland and Green (1971)

In Hellesland and Green's (1971) study, published in 1971, results of tests on unrestrained columns (Fig. 2.11) subjected to eccentric loading were reported. Tests were carried out in an uncontrolled environment. Average relative humidity during the test period was about 50%. Main parameters investigated were the age at loading, and load magnitude and eccentricity.

27

N=101.4-222.4KN e=12.7-50.8mm

178mm

1.9m A

.

.

127mm

A

Sect. A-A

Fig. 2.11 Typical Test Specimen of Hellesland and Green (1971)

28

CHAPTER III

METHOD OF ANALYSIS

3.1 Introduction

In general, the moment along the length of a beam or column varies. At sections with small bending moments, concrete works both in tension and compression. At places of greater moments, the conrete in tension fails at the outer fibers, and cracks are formed at random intervals. At places of still higher moments, the tensile failure in the concrete is more extensive, cracking extends closer to the neutral axis and the cracks widen (Fig. 3.1). Because of the extent of cracking varies along the span length

of

the

member,

the

flexural

rigidity

is

not

constant. Variation of reinforcement along the span also causes variation of the moment of inertia. A cracked member behaves, in general, as a member of variable cross section. Because of this behavior, in determining the deflections, the prediction of variation of the moment of inertia,

or

the

flexural

rigidity

29

of

the

reinforced

concrete

member

through

the

span

length

is

of

prime

importance.

Mmax Service load moment diagram

M cr

loading ___ ___ ___ ___

___

___

Fig. 3.1 Stresses in a Cracked Beam under Service Loading

A

theoretically

correct

method

for

computing

deflections of continuous members must use the actual moment pattern on the member. The moment pattern can be determined by an iterative procedure in which the moments in the frame are computed from an assumed set of flexural rigidities.

The

actual

flexural

rigidities

at

each

section, taking into account cracking, are then computed using

the

moments

determined

in

the

previous

step.

Moments and flexural rigidities are recomputed over a sufficient number of cycles to allow both quantities to converge

to

within

acceptable

margins

of

error.

This

procedure applies equally well for obtaining long-term

30

deflections if a suitable time-dependent stress-strain relationship for concrete is used.

Obviously,

computation

of

deflections

by

this

method necessitates the use of computers. It should be recognized that the above so called 'theoretically exact' method

will

not

necessarily

predict

exact

actual

deflections. This is because the mathematical model of the

theoretical

approach

may

overlook

other

considerations such as statistical variations in concrete properties,

namely,

the

uncertainty

of

the

material

parameters such as stress-strain relationship, modulus of elasticity,

creep

coefficient,

shrinkage,

and

tensile

strength of concrete.

3.2 Basic Assumptions and Methodology

3.2.1 General

The total deflection of a beam or a column under the service load is computed in two components, shortterm

and

long-term

deflections.

These

two

deflection

components are added to get the total deflection.

In order to predict the deformations accurately, shrinkage and creep, and other material properties such

31

as, strength and deformation properties under short and long-term

loading

of

concrete

and

steel,

should

be

realistically modelled.

The principal factors which affect the initial or short-term

deflection

of

reinforced

concrete

flexural

members under service loads are : compressive and tensile stress-strain relationship and modulus of elasticity of concrete, load distribution, support conditions, length of

the

span,

axial

(in

the

case

of

columns)

and

transverse loads and moment of inertia considering degree of cracking along the member.

Time-dependent or long-term deflection is caused by

creep

and

shrinkage

of

concrete.

When

the

reinforcement is unsymmetrical, the resulting nonuniform strain distribution due to shrinkage strains increase the curvature

and

thus

the

deflection.

Under

sustained

loading, the deflection due to creep occurs due to the increase in compressive strains which lead to an increase in curvature. Creep is a stress-dependent deformation, while shrinkage is not. Although creep and shrinkage are not independent phenomena, it is convenient to assume that they are additive in analysis. Compression steel significantly reduces both creep and shrinkage, namely long-term

deflections.

In

general,

both

shrinkage and

creep increase with time, although at a very low rate at

32

later

ages.

Experimental

results

show

that

about

two

thirds of the total shrinkage and one half of the total creep takes place within the first 3 months (Troxell et al., 1958).

For sectional design and stress analysis, it is common to neglect the small contribution of the concrete in

tension.

However,

for

deflection

computations, the

contribution of the uncracked part of concrete in tension below the neutral axis should be taken into consideration since

it

increases

the

stiffness

of

the

member

considerably.

In

the

case

of

reinforced

concrete

slender

columns, the behavior is affected significantly by the lateral deflection (Fig. 3.2). For an unrestrained (pin ended) column, axially loaded at a certain eccentricity, external moment is the sum of the first order and second order moments, namely, M = N(e+y).

33

e

N

ymax

N

Fig. 3.2 Deflection of a Column

At shown in

low

levels

of

load,

the

mid-height

moment

Fig. 3.2 is mainly due to the primary moment

Ne. As the load is increased, the deflections increase, and the secondary moment Nymax starts to gain importance. Therefore

in

computing

the

column

deflection,

the

influence of second order moments should be taken into consideration. Obviously the importance of second order moment is a function of the slenderness of the column.

34

3.2.2 Stress-Strain Relationship of Concrete and Steel

3.2.2.1 Stress-Strain Relationship of Concrete Under Compression

In

general,

the

behavior

of

concrete

in

the

compression zone of a flexural member is assumed to be similar to that obtained from uniaxial compression tests. Stress-strain models proposed by different researchers simplify

the

actual

stress-strain

relations

obtained

experimentally. These models do not differ much for the ascending portion. Since descending portion is not used in deflection calculations and since ascending portion is almost same in all models, only the Hognestad (1951) model will be discussed. Hognestad (1951) has developed a stress-strain compression,

diagram from

the

for tests

concrete of

in

vertically

flexural cast

short

columns subjected to combined bending and axial load. The proposed stress-strain curve and limiting strain value have been widely accepted by other researchers. The model proposed by Hognestad (1951) is shown in Fig. 3.3.

35



co 0.15f" c

Stress , 

c

f" = 0.85f' c c

 0

0

1

2 Strain , 

c

= 0.0038 cu

3 4 ( Thousandths )

Fig. 3.3 Hognestad's (1951) Stress-Strain Diagram

The initial part of the curve is a second degree parabola, expressed by Eq. (3.1),

c = fc"2c/co - (c/co)2

(3.1)

where fc" = 0.85fc' c , co : concrete strains at c and fc" respec. co = 2fc"/Ec Ec = 0.006895(1800000+500fc")

Between stress, strain

co

and

the the

relationship

strains

corresponding

the

maximum

cu,

the

stress-

ultimate is

strain

assumed

straight line.

36

in MPa

to

be

a

descending

c = fc"1.0-0.15((c-co)/(cu-co))

(3.2)

where cu = 0.0038

It should be pointed out that simplified stress blocks such as a rectangle or a trapezoid developed to predict

the

ultimate

strength,

cannot

be

used

for

deflection prediction.

3.2.2.2 Time-Dependent Behavior of Concrete

Time-dependent behavior of concrete, namely time effect in this study, is introduced into the formulation of

concrete

behavior,

by

defining

a

time-dependent

stress-strain relationship.

Since time-dependent deformations, namely creep and

shrinkage

result

in

an

increase

in

strain

under

constant stress, a time dependent stress-strain diagram can

be

obtained

by

multiplying

the

strains

in

the

reference diagram by a factor (1.0+), where  takes care of the creep and shrinkage effects (equals to the creep coefficient Ct for the case when shrinkage is neglected).

37

According stress-strain

to

the

analysis

relationships

of

obtained

time-dependent

experimentally

by

Rüsch et al. in 1962, Sargýn (1971) argued that a timedependent strain corresponding to maximum stress could be expressed as,

cot = co*0.05(log102t + 0.5log10t + 18) (3.3)

where cot : time dependent strain corresponding to maximum stress t

: loading duration in minutes

Based on the test results of Rüsch (1960), Weil (1959), and Pauw and Meyers (1964) on plain concrete under

sustained

loading,

Green

(1966)

proposed

the

following relationship for the strain corresponding to maximum stress.

cot = co + 0.0006log10(t+1.0) + 0.0015log10(t+1.0)-2

where t

: loading duration in hours

cut = 2cot

38

(3.4)

Green relationship sustained

(1966) between

loading

and

also

proposes

maximum

fiber

time,

based

the

following

resistance on

works

of

under Rüsch

(1960), Sell (1959), and Freudenthal (1958).

fc't = fc'A+B/(1.8+log10(t+1.0))

(3.5)

where A,B : constants so that fc't = 0.95fc' for t=0. Indicated values of A and B are, 0.75 and 0.36 respectively. A limiting strength of 0.75fc' results for concrete as t tends to go to infinity. fc' : standard cylinder strength of concrete at the time when sustained loading is applied.

ACI

Committee

209

(1982)

has

recommended

the

hyperbolic-type of equation proposed by Branson et al. (1971,1977) for the creep coefficient Ct. Ct = ta/(b+ta)*Cu

(3.6)

where Ct : ratio of creep strain to initial strain at any time t

: loading duration in days

39

a, b : normal ranges and recommended values of the constants a, b, and ultimate creep coefficient Cu are : 0.40a0.80, a=0.60; 6b30, b=10; 1.30Cu4.15, Cu=2.35

ACI Committee 209 (1982), also recommended the equation of Branson et al. (1971,1977), for unrestrained shrinkage of concrete as a function of time.

sh = t/(d+t)*(sh)u

(3.7)

where sh

: shrinkage strain

t

: loading duration in days

d

= 35 (for moist cured concrete)

(sh)u : 415*10-6(sh)u1070*10-6 in/in (recommended value is 780*10-6 in/in )

In this study, based on the time dependent creep coefficient and shrinkage strain equations recommended by ACI Committee 209 (1982), and taking into consideration experimental results in literature (Al-Zaid et al., 1991; Bakoss

et

al.,

Sozen,

1966;

1982;

Green,

Christiansen, 1966;

Green

1988; and

Corley

Breen,

and

1969;

Hellesland and Green, 1971; Viest et al., 1956; Washa and Fluck, 1952, 1956; Yu and Winter, 1960), the following

40

time-dependent relationship is proposed for the strain which corresponds to the maximum stress.

cot =

co*(1.0+Ct+St)

(3.8)

where Ct

= t0.6/(18+t0.6)*4.15

St

= t/(35+t)*0.001070*(-')//co

t

: loading duration in days

(3.9) (3.10)

, ': tension and compression steel ratios respectively

Time stress

curves

versus based

strain on

corresponding

recommendations

to made

maximum by

two

researchers and the one proposed by the author are shown in Fig. 3.4 and Fig. 3.5.

41



cot

(Thousandths)

10 9 8 7 6 5 4 6 months

3 2 1

0

50 Green

100 Time (days) Sargýn

150

200

Proposed

Fig. 3.4 Time Versus Strain Corresponding to Maximum Stress 

cot

(6 Months duration)

(Thousandths)

10 9 8 7 6 5 20 years

4 3 2 1

1

10 100 Time (days), (t+1) Sargýn Green

1000 (Log scale) Proposed

Fig. 3.5 Time Versus Strain Corresponding to Maximum Stress (20 Years duration)

42

10000

Concrete strength increases with time, but on the other hand, the effect of sustained loading reduces the strength. For a young concrete, the strength reduction caused

by

sustained

loading

is

counteracted

by

the

strength increase with time (Rüsch, 1960). Time-dependent stress-strain diagram for a 2½ years period proposed by the author (based on the Eq. 3.8, 3.9 and 3.10) is shown in Fig. 3.6. 2½ years is taken as a limit since test data beyond 2½ years is rare. Also it is believed that after 2½

years,

the

effects

of

shrinkage

and

creep

become

negligible.

f" c ( 0.85f' ) c

t=0 Stress  c

0

t=2.5 years

0.005

0.01 Strain , 

0.015

0.02

c

Fig. 3.6 Proposed Time-Dependent Stress-Strain Diagram for 2½ Years

43

3.2.2.3 Stress-Strain Relationship of Concrete Under Tension

Since the tensile strength of concrete is very low

and

since

concrete

cracks,

tensile

strength

is

generally neglected in strength calculations. However in serviceability checks, such as deflection calculations, tensile strength of concrete should be considered.

Three kinds of tests have been used to determine the

tensile

strength

of

concrete:

the

direct

tension

test, the beam test (modulus of rupture test), and the splitting test (split cylinder test). Among these methods of

testing,

direct

tension

test

represents

the

basic

tensile strength of concrete better than others.

The shape of the concrete stress-strain diagram in tension depends heavily on the testing procedure used. Only the direct tension test can provide the complete stress-strain diagram in tension, namely with ascending and descending branches. When concrete is tested under direct

tension,

where

the

strain

rate

can

not

be

controlled, a linear diagram is usually obtained. But experiments have shown that using a testing machine in which the strain rate is controlled, the stress-strain diagram in tension is nonlinear and has well defined

44

ascending

and

descending

branches

(Carreira

and

Chu,

1986). In the light of Rüsch's test results, the author has decided to use the stress-strain curve shown in Fig. 3.7 for concrete in tension, which is similar in nature to the relationship for compression proposed by Hognestad (1951).

0.15f" ct Stress , ct f" = 0.85f' ct ct  cto 0.1

0

 ctu 0.2

Strain ,  ( Thousandths ) ct

Fig. 3.7 Assumed Stress-Strain Diagram for Concrete in Tension

The direct tensile strength of plain concrete is conservatively given in ACI Committee 209 (1982), for normal weight concrete as, fct'  0.33fc' (in MPa.).

45

Values for the strain corresponding to maximum stress cto, and the ultimate strain ctu may be assumed as

0.0001

0.00015

and

and

0.0002

0.00030

respectively respectively

(Ersoy,

1985)

or,

(Carreira

and

Chu,

1986). The author has decided to use Ersoy's proposal since it checks better with the comprehensive tests made by Rüsch (Rüsch and Hilsdorf, 1963).

3.2.2.4 Stress-Strain Relationship of Steel

The

steel

reinforcement

used

in

reinforced

concrete can be grouped in two classes, hot rolled or cold worked, depending on the manufacturing process. Hot rolled steel has a definite yield plateau, while for the cold worked steel, the yield strength is defined as the stress corresponding to a permanent set of 0.002. This stress can be found by a line drawn from a strain of 0.002 parallel to the initial slope of the stress-strain curve.

In

general,

stress-strain

relation

for

the

reinforcing steel is assumed to be elasto-plastic, and identical

under

compression

and

tension

(Fig.

3.8).

Although elasto-plastic model agrees very well with the behavior of hot rolled steels, it does not represent the true

behavior

of

cold

worked

46

steels.

However,

in

deflection calculations, since the service load level is the case, this assumption will not create any problem even for cold worked steels.

Unlike

concrete,

the

modulus

of

elasticity

of

steel varies little with its strength. ACI-318-89 (1989) Code

recommends

the

use

of

Es

=

200000

MPa

for

the

modulus of elasticity for nonprestressed reinforcement.

Stress,f s

fy



sy

Strain, 

s

Fig. 3.8 Elasto-Plastic Stress-Strain Diagram for Steel in Tension or Compression

47

3.3. Moment-Curvature Relationship

In beam and column analyses, the moment-curvature relationship (or more generally load-moment-curvature) is very

important,

calculations.

especially

It

in

should

be

the

case

recalled

of

deflection

that,

if

the

curvature distribution along a member is known, then the deflection can be obtained by integrating the curvature twice.

Assuming cross

section

a

linear

with

strain

perfect

distribution

bond

between

in the

the two

materials, and suitable stress-strain diagrams for the materials

(namely

concrete

and

steel),

the

following

procedure can be used for the computation of moments and corresponding

curvatures

in

obtaining

the

moment-

curvature curves.

1- Referring to Fig. 3.9, for a given extreme compressive fiber strain of ci, make an assumption about the

depth

of

the

neutral

axis,

'c'.

This

assumption

completely defines the strain distribution.

2strains,

Using

similar

stresses

triangles,

(si=siEs

fy),

determine and

forces

(Fsi=Asisi) in the reinforcement at different levels.

48

the

3- Enter the - diagram of concrete with ci, and determine the stress block for the compression zone of the member (shaded area).

4- Enter the - diagram of concrete in tension with extreme uncracked fiber strain in tension cti and determine the stress block for the tension zone. When the strain exceeds the tensile strain capacity of concrete, that part of concrete is neglected (cracked).

5- Using the stress blocks determined in steps (3) and (4), compute the concrete forces in tension and compression (volume of the stress block)

6- Check force equilibrium, Fint.=N. If force equilirium is not satisfied, assume another value of 'c' (taking into account the force equilibrium check) and repeat steps (2)-(6).

7- When force equilibrium is satisfied, compute 'M',

by

taking

moments

of

internal

forces

about

the

centroidal axis. Also calculate the curvature, ci/c = .

By repeating steps (1) through (7) and carrying out the calculations for a range of strain values, the moment-curvature curve can be plotted.

49

N

N 

ci

A s5 A

c

F

c

s4

s4

c.g.

Fs5

s5

Fs4

A s3



F

A s2 A

 s2 s1

F Fct s2 Fs1

s3

s1



s3

cti

Stress ,c (compression)

Stress,ct(tension)

 ci

Strain,  c

Strain,  ct

 cti

Stress,fs fy

sy

Strain,s

Fig. 3.9 Strain Distribution, Stress Block for Concrete in Compression and Tension, and Stress-strain Diagram for Steel in Tension and Compression

50

The moment-curvature curve for zero axial load level

(namely

pure

flexure

case),

increases

almost

linearly with two different slopes, uncracked and cracked (O-A and A'-B respectively in Fig. 3.10). After yielding of

tension

reinforcement,

the

curve

runs

almost

horizontal up to very large values of curvature. For axial

load

levels

below

the

balanced

value

(tension

failure), the curve initially rises steeply (O-C in Fig. 3.10). Slope decreases when concrete cracks beyond this point. After the yielding of tension steel, the curve flatens. The shape of the curve can be approximated by a trilinear graph. When the axial load level exceeds the balanced value (compression failure), an almost linear portion (O-E in Fig. 3.10) is followed by a curve of lower stiffness, as tension strains develop and concrete cracks. In case of compression failure, ductility is very limited.

51

N (Axial load)

comp. fail.

tens. fail. pure flex. M (Moment)

tension failure D

Moment

pure flexure B E C

A O

compression failure

A' Curvature

Fig. 3.10 Typical Moment-Curvature Curves

Extra stiffness caused by the tension of concrete is quite significant for beams and columns at low levels of load. This can be seen easily by comparing the two increasing slopes O-A and A'-B in Fig. 3.10 for the pure flexure case. Neglecting the tension in concrete would

52

cause the curve to begin with the second slope, namely A'-B up to the yielding of steel.

In

case

of

members

carrying

significant

axial

load in addition to bending, the effect of tension is shown in Fig. 3.11. The figure shown is the portion of OC-D the upper curve shown in Fig. 3.10.

Moment

Neglecting tensile strength of concrete

C

Considering tensile strength of concrete

C'

O

Curvature

Fig. 3.11 Influence of Concrete in Tension on Moment-Curvature Curves for a Column

53

3.4 Method of Analysis

The computer program, based on the finite element method, computes deflections by an iterative technique. Since

deflections

affecting

the

are

accuracy

time-dependent, of

the

the

analysis

key element

is

the

time-

dependent stress-strain relationship of the concrete.

The solution technique in the analysis is based on the following assumptions:

1- For the concrete, the stress-strain relations proposed in section 3.2.2 and for the steel, elastoplastic relationship shown in Fig. 3.8 are assumed to be correct.

The

time-dependent

stress-strain

curve

for

concrete is obtained by multiplying the strains in the Hognestad (1951) model by a factor which takes care of the creep and shrinkage effects (Sec. 3.2.2). Proposed relation between concrete strain and time is,

cot =

co*(1.0+Ct+St)

(3.8)

where cot

: time-dependent strain corresponding to maximum stress

54

Ct

= t0.6/(18+t0.6)*4.15

St

= t/(35+t)*0.001070*(-')//co

t

: loading duration in days

(3.9) (3.10)

, ': tension and compression steel ratios respectively

2-

Strain

distribution

is

linear

across

the

steel

and

section.

3-

There

is

perfect

bond

between

concrete, namely no slip occurs.

Based on the proposed stress-strain relationships for concrete and steel, load-moment-curvature diagrams for rectangular or flanged cross sections, up to five layers of steel are computed at different sections with different reinforcement and geometry along the length of the member. Tensile strength of concrete is taken into consideration by the proposed stress-strain relationship, and

the

moment-curvature

diagrams

are

constructed

accordingly.

Using

a

numerical

solution

by

finite

element

method, the member is divided into n number of segments along which the flexural rigidity is assumed to remain constant. The moment pattern is determined by assuming a set of flexural rigidities for each segment, through the

55

member

length.

A

new

set

of

flexural

rigidities

are

computed by entering the corresponding moment-curvature diagrams

using

the

flexural

rigidities

moments are

calculated.

recomputed

up

Moments to

a

and

certain

acceptable error limit. (This process of iteration is continued until the set of flexural rigidities in two successive

iterations

agree

within

0.1

percent).

The

nonlinear behavior of reinforced concrete necessitates the use of such an iterative solution.

Since the moment-curvature diagrams are sometimes curvilinear or bilinear, the secant (O-C in Fig. 3.12) or tangent

(tangential

line

at

C

in

Fig.

3.12)

modulus

values of flexural rigidity are not used. In the computer program for a better trace (O-A-B-C in Fig. 3.12), the load was applied in increments and tangential flexural rigidities portions.

were This

nonlinearity,

so

computed

for

procedure

takes

the

member

realistically.

56

the

corresponding

care

of

the

behavior

is

studied

load

material more

C Moment

B

A

O

Curvature

Fig. 3.12 Tracing the Moment-Curvature Diagram

The given

in

flow Fig.

diagram 3.13

for

and

the

3.14.

computer

program

Structural

is

analysis

subroutines are based on the Program SAPIV (A General Stress Analysis Program) which is originally developed by Wilson et al. (1974) in Fortran Language for CDC Large System Computer. The computer program of this study is developed in FORTRAN Language, and modified to IBM PC Compatible microcomputers.

57

INPUT, MEMBER AND SECTIONAL PROPERTIES, ASSUMED SET OF FLEXURAL RIGIDITIES, AND LOADING DURATION

COMPUTE, MOMENT-CURVATURE DIAGRAMS FOR DIFFERENT CROSS SECTION TYPES

COMPUTE, MOMENT VALUES, AND CORRESPONDING FLEXURAL RIGIDITIES BY ENTERING MOMENT-CURVATURE DIAGRAMS

NO

COMPARE THE FLEXURAL RIGIDITIES WITH THE PREVIOUS SET WHETHER WITHIN ACCEPTABLE ERROR LIMIT OR NOT YES SOLVE FOR THE DISPLACEMENTS

Fig. 3.13 Flowchart for the Computer Program (Main Steps)

58

CALCULATE , STIFFNESS MATRIX OF EACH ELEMENT USING, GROSS SECTIONAL PROPERTIES OR UPDATED FLEXURAL RIGIDITY VALUES

ASSEMBLE STRUCTURAL STIFFNESS MATRIX AND SOLVE THE STRUCTURE

CALCULATE FLEXURAL RIGIDITY OF EACH ELEMENT , BY ENTERING THE MOMENT-CURVATURE DIAGRAMS OF THE ELEMENTS

3.14 Flowchart for the Computer Program (Structural Analysis Part)

59

CHAPTER IV

NUMERICAL APPLICATIONS

4.1 Introduction and Additional Information About the Available Data

Deflections of 37 beams (rectangular and flanged sections

single

span

and

continuous

spans)

and

26

columns, obtained experimentally from eleven references have been compared with the deflections determined using ACI-318-89

(1989)

and

TS-500

(1985)

Codes

and

the

computer program developed in this study. The computer program

is

DEflections).

designated S.I.

Units

as have

BECODE been

(BEam

used

in

COlumn all

case

studies.

4.1.1 Experimental Beam Deflection Studies and Comparisons

Beam 2.2)

are

deflections

compared

with

obtained the

60

experimentally

results

of

the

(Sec.

computer

program developed, ACI-318-89 (1989), and TS-500 (1985) Codes. These (maximum beam deflection) values, together with the properties of test specimens, are summarized in Tables

4.1

variation

and of

experimentally

4.2.

In

maximum are

Fig.

4.1-4.5,

deflection

compared

with

time-dependent

values

the

measured

results obtained

using the computer program proposed. Values calculated using

the

proposed

computer

program

are

marked

as

'calculated' in these figures. In each case, the last figure given is a summary of all test results of the researcher considered.

4.1.2 Experimental Column Deflection Studies and Comparisons

Column deflections obtained experimentally (Sec. 2.3)

are

compared

with

the

results

computer

program.

Maximum

together

with

properties

the

column

summarized in Table 4.3.

61

of

of

the

deflection test

proposed values,

specimens,

are

Max. deflection (mm.) 80 B1,B4

60

40

20

0

0

100 200

300 400 500 600 700 Time (days)

experimental

800 900 1000

calculated

(a) Specimen B1,B4 (average) Max. deflection (mm.) 80 B2,B5 60

40

20

0

0

100 200

300 400 500 600 700 Time (days)

experimental

800 900 1000

calculated

(b) Specimen B2,B5 (average) Fig. 4.1 Comparison of Computed and Observed Deflections - Washa and Fluck (1952)

62

Max. deflection (mm.) 80

60

B3,B6

40

20

0

0

100 200

300 400 500 600 700 Time (days)

experimental

800 900 1000

calculated

(c) Specimen B3,B6 (average) Max. deflection (mm.) 80 60 40 20 0

0

100 200 300 400 500 600 700 800 900 1000 Time (days) calculated experimental

B1B4 B2B5 B3B6

B1B4 B2B5 B3B6

(d) Summary of all Tests Fig. 4.1 (cont'd)

63

Max. deflection (mm.)

X1,X4

30

20

10

0

0

100

200

300

400 500 600 Time (days)

experimental

700

800

900 1000

calculated

(a) Specimen X1,X4 (average) Max. deflection (mm.)

X2,X5

30

20

10

0

0

100

200

300

400 500 600 Time (days)

experimental

700

800

900 1000

calculated

(b) Specimen X2,X5 (average) Fig. 4.2 Comparison of Computed and Observed Deflections - Washa and Fluck (1956)

64

40 Max. deflection (mm.) 30

X3,X6

20

10

0

0

100

200

300

400 500 600 Time (days)

experimental

700

800

900 1000

calculated

(c) Specimen X3,X6 (average) 40 Max. deflection (mm.) 30

20

10

0

0

100

200

X1X4 X2X5 X3X6

300

400 500 600 Time (days)

experimental

(d) Summary of all Tests Fig. 4.2 (cont'd)

65

700

800

900 1000

calculated X1X4 X2X5 X3X6

Max. deflection (mm.) 100 80

A

60 40 20 0

0

50

100 150 Time (days) experimental

200

250

300

calculated

(a) Specimen A Max. deflection (mm.) 100 80 B 60 40 20 0

0

50

100 150 Time (days) experimental

200

250

calculated

(b) Specimen B Fig. 4.3 Comparison of Computed and Observed Deflections - Yu and Winter (1960)

66

300

Max. deflection (mm.) 100 80 C

60 40 20 0

0

50

100 150 Time (days) experimental

200

250

300

calculated

(c) Specimen C Max. deflection (mm.) 100 80

D

60 40 20 0

0

50

100 150 Time (days) experimental

(d) Specimen D Fig. 4.3 (cont'd)

67

200

250

calculated

300

Max. deflection (mm.) 100 80 60 E

40 20 0

0

50

100 150 Time (days) experimental

200

250

300

calculated

(e) Specimen E Max. deflection (mm.) 100 80

F

60 40 20 0

0

50

100 150 Time (days) experimental

(f) Specimen F Fig. 4.3 (cont'd)

68

200

250

calculated

300

Max. deflection (mm.) 15

10

C1

5

0

0

100

200

300 400 500 Time (days)

experimental

600

700

800

calculated

(a) Specimen C1 20 Max. deflection (mm.) 15

C3

10

5

0

0

100

200

300 400 500 Time (days)

experimental

600

700

calculated

(b) Specimen C3 Fig. 4.4 Comparison of Computed and Observed Deflections - Corley and Sozen (1966)

69

800

Max. deflection (mm.) 15 C4 10

5

0

0

100

200

300 400 500 Time (days)

experimental

600

700

800

700

800

calculated

(c) Specimen C4 20 Max. deflection (mm.) 15

10

5

0

0

C1

100

C3

C4

200

300 400 500 Time (days)

experimental

(d) Summary of all Tests Fig. 4.4 (cont'd)

70

600

calculated C1

C3

C4

Max. deflection (mm.) 30

1B2

20

10

0

0

100

200 300 Time (days) experimental

400

500

600

calculated

Fig. 4.5 Specimen 1B2 -Comparison of Computed and Observed Deflections - Bakoss et al. (1982)

71

Table 4.1 Properties of Test Beams Ref.  Span Spec. no Length

Section Dimen.

l

bw*h

(mm)

(mm)

Flange Dimen.

Tension Steel

b*t (mm)

Compres. Effec. Steel Depth

As

As'

d

(mm2)

(mm2)

(mm)

Distance Age Concrete Betw. St -@ Load. Strength Layers -@ Final @ Load. d" fc (mm)

(days)

(MPa)

Washa and Fluck (1952), Single span rectangular beams A1,A4

6096

203*305

852

852

257

210

14 915

25.9

A2,A5

6096

203*305

852

400

257

211

14 915

25.9

A3,A6

6096

203*305

852

14 915

25.9

B1,B4

6096

152*203

400

400

157

111

14 915

20.8

B2,B5

6096

152*203

400

200

157

111

14 915

20.8

B3,B6

6096

152*203

400

14 915

20.8

257

157

Table 4.1 (cont'd) * Ref.  Span Spec. no Length

Section Dimen.

l

bw*h

(mm)

(mm)

Flange Dimen.

*

Tension Steel

b*t (mm)

*

*

Compres. Effec. Steel Depth

As

As'

d

(mm2)

(mm2)

(mm)

Distance Age Concrete Betw. St -@ Load. Strength Layers -@ Final @ Load. d" fc (mm)

(days)

(MPa)

Washa and Fluck (1956), Double span rectangular beams X1,X4

2*6096

152*204

400 684

400 600

157 157

111 111

14 915

22.3

X2,X5

2*6096

152*204

400 684

200 600

157 157

111 111

14 915

22.3

X3,X6

2*6096

152*204

400

157

111

14

22.3

157

111

915

684

600

* In the marked columns, the top value refers to the span (positive moment section) and the bottom value to the mid support (negative moment section).

Table 4.1 (cont'd) Ref.  Span Spec. no Length

Section Dimen.

l

bw*h

(mm)

(mm)

Flange Dimen.

Tension Steel

b*t (mm)

Compres. Effec. Steel Depth

As

As'

d

(mm2)

(mm2)

(mm)

Distance Age Concrete Betw. St -@ Load. Strength Layers -@ Final @ Load. d" fc (mm)

(days)

(MPa)

Yu and Winter (1960), Single span T-beams A

6096

152*305

305*64

400

259

B

6096

152*305

305*64

400

200

259

C

6096

152*305

305*64

400

400

259

28 270

25.4

219

28 270

26.8

219

28

24.4

270 D

6096

152*305

610*64

774

246

28 270

25.4

E

4267

152*305

305*64

400

249

28 270

29.4

F

6096

152*203

305*51

400

157

28

29.4

270

Table 4.1 (cont'd) Ref.  Span Spec. no Length

Section Dimen.

l

bw*h

(mm)

(mm)

Flange Dimen.

Tension Steel

b*t (mm)

Compres. Effec. Steel Depth

As

As'

d

(mm2)

(mm2)

(mm)

Distance Age Concrete Betw. St -@ Load. Strength Layers -@ Final @ Load. d" fc (mm)

(days)

(MPa)

Corley and Sözen (1966), Single span rectangular beams C1

1829

76*152

142

137

28 715

24.1

C3

1829

76*110

142

92

28 715

24.1

C4

1829

76*110

213

92

28

24.1

715 Bakoss et al. (1982), Single span rectangular beam 1B2

3750

100*150

226

130

28 525

39.0

Table 4.1 (cont'd) Ref.  Span Spec. no Length

Section Dimen.

l

bw*h

(mm)

(mm)

Flange Dimen.

Tension Steel

b*t (mm)

Compres. Effec. Steel Depth

As

As'

d

(mm2)

(mm2)

(mm)

Distance Age Concrete Betw. St -@ Load. Strength Layers -@ Final @ Load. d" fc (mm)

(days)

(MPa)

Christiansen (1988), Single span rectangular beams L1,L2

7500

170*280

452

39

249

221

20 2933

28.2

L3,L4

7500

170*280

452

39

249

221

20 3052

29.9

L5

7500

170*280

452

39

249

221

23

27.1

3025 L7

7500

170*280

452

226

249

218

20 3028

25.5

L8

7500

170*280

452

226

249

218

174 2856

30.7

L9,L10

7500

170*280

452

452

249

218

22

33.1

3123

Table 4.1 (cont'd) Ref.  Span Spec. no Length

Section Dimen.

l

bw*h

(mm)

(mm)

Flange Dimen.

Tension Steel

b*t (mm)

Compres. Effec. Steel Depth

As

As'

d

(mm2)

(mm2)

(mm)

Distance Age Concrete Betw. St -@ Load. Strength Layers -@ Final @ Load. d" fc (mm)

(days)

(MPa)

Clarke et al. (1988), Single span rectangular beams A1,A2

2100

100*153

157

B1,B2

2100

100*153

157

131 157

28 200

25.9

131

111

28 200

25.9

175

150

28

31.4

Al-Zaid et al. (1991), Single span rectangular beams BCU1-10

2500

200*200

400

79

Table 4.2 Comparison of Calculated and Measured Deflections (Beams) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Washa and Fluck (1952), Single span rectangular beams A1,A4

5.52

a b

13.5 23.6

0.90 1.03

1.25

0.87 1.15

A2,A5

5.52

a b

15.7 32.3

1.00 0.90

1.40

0.97 1.30

A3,A6

5.52

a b

17.0 44.7

1.06 1.01

1.45

0.99 1.25

B1,B4

1.56

a b

23.4 51.1

0.93 1.35

1.23

0.82 1.25

B2,B5

1.56

a b

24.9 65.0

0.98 1.32

1.28

0.85 1.35

B3,B6

1.56

a b

26.4 86.4

1.02 1.22

1.35

0.86 1.29

Table 4.2 (cont'd) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Washa and Fluck (1956), Double span rectangular beams X1,X4

2.77

a b

14.2 29.0

0.80 1.09

1.09

0.80 1.12

X2,X5

2.77

a b

14.5 32.3

0.81 0.93

1.11

0.80 1.14

X3,X6

2.77

a b

15.7 37.8

0.87 0.76

1.18

0.85 1.10

Table 4.2 (cont'd) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Yu and Winter (1960), Single span T-beams A

6.42

a b

34.0 67.3

1.13 0.97

1.61

1.06 1.40

B

6.44

a b

31.5 56.6

1.06 0.88

1.53

1.00 1.34

C

6.41

a b

30.2 51.8

1.02 0.85

1.44

0.96 1.30

D

11.73

a b

32.3 67.6

0.99 0.90

1.44

0.95 1.30

E

12.29

a b

13.0 29.5

0.94 0.92

1.30

0.85 1.29

F

3.79

a b

55.9 100.3

1.06 0.82

1.53

1.00 1.14

Table 4.2 (cont'd) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Corley and Sözen (1966), Single span rectangular beams C1

0.27

quarter 2*4982

a b

3.0 7.4

1.00 0.91

1.36

0.91 1.14

C3

0.20

quarter 2*4982

a b

7.9 17.3

1.00 0.85

1.44

0.90 0.92

C4

0.20

quarter

a

6.1

1.00

1.42

0.88

2*4982

b

15.5

0.96

0.93

Bakoss et al. (1982), Single span rectangular beam 1B2

0.35

third 2*2600

a b

8.9 25.0

0.72 0.80

1.11

0.67 1.04

Table 4.2 (cont'd) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Christiansen (1988), Single span rectangular beams L1,L2

1.12

quar+mid a 3*3750 b

5.6 39.0

0.90 2.12

0.76 1.57

0.82 1.93

L3,L4

1.12

quar+mid a 3*5750 b

44.9 101.0

0.94 0.73

1.35 0.90

0.90 1.15

L5

1.12

quar+mid a

14.2

0.75

0.93

0.67

3*1510

b

56.0

1.01

1.09

1.37

L7

1.12

quar+mid a 3*5970 b

47.5 88.0

1.01 0.72

1.37 0.89

0.91 1.17

L8

1.12

quar+mid a 3*5975 b

45.8 76.0

0.98 0.63

1.39 0.81

0.92 1.05

L9,L10

1.12

quar+mid a

41.6

0.92

1.31

0.86

3*6120

69.0

0.67

0.88

1.10

b

Table 4.2 (cont'd) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Clarke et al. (1988), Single span rectangular beams A1,A2

0.36

third 2*5000

a b

5.1 11.9

1.00 1.05

1.38

0.94 1.30

B1,B2

0.36

third 2*5000

a b

4.8 8.8

0.98 1.02

1.92

0.96 1.42

Table 4.2 (cont'd) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Al-Zaid et al. (1991), Single span rectangular beams BCU1

1.47

midspan 8090

a b

1.5

1.13

1.13

1.23

BCU2

1.96

midspan 10790

a b

2.6

1.12

1.35

1.12

BCU3

2.45

midspan

a

3.7

1.15

1.48

1.09

13480

b

BCU4

2.94

midspan 16180

a b

4.6

1.12

1.53

1.07

BCU5

3.43

midspan 18880

a b

5.6

1.15

1.56

1.08

Table 4.2 (cont'd) Ref.  Uniform Spec. no Load

(N/mm)

Point Load(s)

a:Inst. b:Final

(N)

Deflec. (Test)

Test/ACI Test/TS

Test/Cal

(mm)

Al-Zaid et al. (1991), Single span rectangular beams BCU6

4.17

midspan 22920

a b

7.2

1.18

1.67

1.09

BCU7

4.66

midspan 25610

a b

8.3

1.20

1.69

1.12

BCU8

5.15

midspan

a

9.3

1.20

1.72

1.12

28310

b

BCU9

5.64

midspan 31010

a b

10.4

1.24

1.76

1.14

BCU10

6.13

midspan 33700

a b

11.3

1.23

1.77

1.14

Table 4.3 Properties of Test Columns and Deflections (Measured and Calculated) Ref.  Total Spec. no Reinf.

(*)

Ast

d"

(mm2)

(mm)

Age Concrete Axial -@ Load. Strength Load -@ Final @ Load. fc (days)

(MPa)

(KN)

Eccena:Inst. tricity b:Final of Axial Load (mm)

Deflec. (Test)

Test/Cal

(mm)

Viest et al. (1956), Span Length l=1.0 m, Section Dimen. b*h=127*127 mm, 20B3a

506

89

35 505

14.4

107.6

92.2

a b

3.6 8.8

0.80 1.09

20B4a

506

89

34 513

16.3

97.9

92.2

a b

3.6 8.7

0.97 1.71

20B4b

506

89

83 457

12.2

91.6

96.8

a b

4.0 7.3

0.98 1.22

35B3a

506

89

47

30.7

194.4

63.5

a b

4.1

0.98

35B4a

506

89

48

29.8

193.0

63.5

a b

4.1

0.98

(*) d" is the distance between the centroids of reinforcement on opposite faces

Table 4.3 (cont'd) Ref.  Total Spec. no Reinf.

(*)

Ast

d"

(mm2)

(mm)

Age Concrete Axial -@ Load. Strength Load -@ Final @ Load. fc (days)

(MPa)

(KN)

Eccena:Inst. tricity b:Final of Axial Load (mm)

Deflec. (Test)

Test/Cal

(mm)

Viest et al. (1956), Span Length l=1.0 m, Section Dimen. b*h=127*127 mm, 35B4b

506

89

271 664

32.4

187.3

57.2

a b

3.4 5.5

1.10 0.96

50B4a

506

89

41 548

35.4

231.3

57.2

a b

4.2 10.2

1.02 1.19

50B3b

506

89

52

30.3

186.4

63.5

a

3.1

0.80

b

7.9

1.10

437 20C3a

506

89

40

22.4

233.5

31.8

a b

2.1

0.95

20C4a

506

89

34 567

17.7

213.9

31.8

a b

2.3 9.5

1.00 1.64

(*) d" is the distance between the centroids of reinforcement on opposite faces

Table 4.3 (cont'd) Ref.  Total Spec. no Reinf.

(*)

Ast

d"

(mm2)

(mm)

Age Concrete Axial -@ Load. Strength Load -@ Final @ Load. fc (days)

(MPa)

(KN)

Eccena:Inst. tricity b:Final of Axial Load (mm)

Deflec. (Test)

Test/Cal

(mm)

Viest et al. (1956), Span Length l=1.0 m, Section Dimen. b*h=127*127 mm, 20C3c

506

89

50 933

15.1

196.2

31.8

a b

1.8 5.7

0.82 1.06

35C4b

506

89

87 457

27.2

213.5

44.5

a b

2.6 7.8

0.93 1.34

(*) d" is the distance between the centroids of reinforcement on opposite faces

Table 4.3 (cont'd) Ref.  Total Spec. no Reinf.

(*)

Ast

d"

(mm2)

(mm)

Age Concrete Axial -@ Load. Strength Load -@ Final @ Load. fc (days)

(MPa)

(KN)

Eccena:Inst. tricity b:Final of Axial Load (mm)

Deflec. (Test)

Test/Cal

(mm)

Green and Breen (1969), Span Length l=1.9 m, Section Dimen. b*h=101.5*152.5 mm, S1

284

63.5

50 143

27.6

235.7

3.8

a b

1.4 10.6

1.00 3.21

S2

284

63.5

50 143

37.9

122.3

25.4

a b

4.7 16.8

1.52 2.07

S3

284

63.5

50

37.9

189.0

3.8

a

1.0

1.43

b

3.3

1.74

550 S5

284

63.5

50 550

27.3

184.6

10.8

a b

3.5 21.7

1.25 3.10

S7

284

63.5

50 550

27.3

129.0

15.2

a b

2.3 8.3

0.92 1.34

(*) d" is the distance between the centroids of reinforcement on opposite faces

Table 4.3 (cont'd) Ref.  Total Spec. no Reinf.

(*)

Ast

d"

(mm2)

(mm)

Age Concrete Axial -@ Load. Strength Load -@ Final @ Load. fc (days)

(MPa)

(KN)

Eccena:Inst. tricity b:Final of Axial Load (mm)

Deflec. (Test)

Test/Cal

(mm)

Green and Breen (1969), Span Length l=1.9 m, Section Dimen. b*h=101.5*152.5 mm, S8

284

63.5

50 550

28.3

133.4

27.9

a b

4.7 26.3

0.71 1.61

S10

284

63.5

50 250

23.8

80.1

43.2

a b

12.6 25.7

1.47 1.81

(*) d" is the distance between the centroids of reinforcement on opposite faces

Table 4.3 (cont'd) Ref.  Total Spec. no Reinf.

(*)

Ast

d"

(mm2)

(mm)

Age Concrete Axial -@ Load. Strength Load -@ Final @ Load. fc (days)

(MPa)

(KN)

Eccena:Inst. tricity b:Final of Axial Load (mm)

Deflec. (Test)

Test/Cal

(mm)

Hellesland and Green (1971), Span Length l=1.9 m, Section Dimen. b*h=127*178 mm, C1

284

89

169 117

50.3

181.0

12.7

a b

0.8 1.6

1.14 0.85

C2

284

89

111 145

37.2

222.4

12.7

a b

1.7 3.4

1.30 1.13

C3

284

89

112

35.9

215.7

12.7

a

2.0

1.54

b

3.4

1.13

123 C4

284

89

14 128

29.0

101.4

50.8

a b

4.2 10.2

0.95 1.20

C5

284

89

28 122

31.4

104.5

50.8

a b

4.3 9.1

0.96 1.08

(*) d" is the distance between the centroids of reinforcement on opposite faces

Table 4.3 (cont'd) Ref.  Total Spec. no Reinf.

(*)

Ast

d"

(mm2)

(mm)

Age Concrete Axial -@ Load. Strength Load -@ Final @ Load. fc (days)

(MPa)

(KN)

Eccena:Inst. tricity b:Final of Axial Load (mm)

Deflec. (Test)

Test/Cal

(mm)

Hellesland and Green (1971), Span Length l=1.9 m, Section Dimen. b*h=127*178 mm, C6

800

89

14 131

31.7

141.0

50.8

a b

5.0 8.2

1.25 1.24

C7

800

89

28 123

36.5

153.5

50.8

a b

4.9 8.4

1.20 1.22

(*) d" is the distance between the centroids of reinforcement on opposite faces

4.2 Discussion of the Results

In the case of short-term deflections of beams, comparing the predictions made by ACI-318-89 (1989) and TS-500 (1985) Codes, and the computer program developed in this study, it was found that, for ACI-318-89 there is approximately deflections

an

will

89 be

percent within

possibility

the

range

of

that 20

the

percent

scattering around the measured (test) value, and for the computer program, almost the same claim can be made since there is a 91 percent possibility that the will be within

the range of

deflections

20 percent (Table 4.2).

However, TS-500 underestimates the deflections, and there is

approximately

a

90

percent

probability

that

the

computed deflections will be 70 percent or less below the measured

(test)

value

(Table

4.2).

Test/ACI,

and

Test/Calculated values for beams are also presented as bar charts, in Fig.4.6-4.9. In the bar charts, TS-500 is not included since the deviation is very large.

In case of long-term deflections of beams, it was found that there is more than 90 percent possibility that the deflections

will be

within the range of 40 percent

for ACI-318-89, and -40 percent for the computer program proposed by the author. TS-500 Code predicts final longterm deflections, not the intermediate values. Therefore comparison with test data is not possible since very

93

little data is available for long-term deflections beyond 2½ years.

In ACI-318-89 and TS-500 Codes, as opposed to the case of beams no provisions are given for calculating the deflection of columns. For columns, the computer program proposed by the author predicts short-term deflections within value,

the with

range 85

of

±30

percent

percent

around

reliability.

the For

measured long-term

deflections, ±35 percent scattering is in the 75 percent confidence interval (Table 4.3). The confidence intervals were reduced to 85% and 75% due to limited test data. Test/Calculated values for columns are also presented as bar charts in Fig. 4.10 and 4.11.

94

% of tests

+ 20%

30

20

10

0

0.7

0.8

0.9

1.0

1.1

1.2

1.3 Test/ACI

Fig. 4.6 Bar Charts for Short-Term Deflections of Beams-ACI

% of tests

+ 20%

30

20

10

0

0.7

0.8

0.9

1.0 1.1

1.2

1.3 Test/Cal.

Fig. 4.7 Bar Charts for Short-Term Deflections of Beams-Calculated

95

% of tests + 40%

30

20

10

0

0.6

0.8

1.0

1.2

1.4 Test/ACI

Fig. 4.8 Bar Charts for Long-Term Deflections of Beams-ACI

% of tests

+40%

30

20

10

0

0.6

0.8

1.0

1.2

1.4 Test/Cal.

Fig. 4.9 Bar Charts for Long-Term Deflections of Beams-Calculated

96

% of tests 30

20

+ 30%

10

0

0.6

0.8

1.0

1.2

1.4 Test/Cal.

Fig. 4.10 Bar Charts for Short-Term Deflections of Columns-Calculated

% of tests 30 + 35% 20

10

0

0.4 0.6 0.8 1.0 1.2 1.4 1.6 Test/Cal.

Fig. 4.11 Bar Charts for Long-Term Deflections of Columns-Calculated

97

4.3 Reconsideration of TS-500 and ACI-318-89 Codes

The

procedures

calculations

of

for

ACI-318-89

short-term

and

TS-500

deflection

Codes

are

very

similar, but the modulus of elasticity of concrete and the modular ratio values are defined differently. The modulus

of

elasticity

of

concrete

in

TS-500

is

approximately 20 percent greater than that of ACI-318-89. On

the

other

hand,

the

moment

of

inertia

of

the

transformed fully cracked section calculated using TS-500 leads to values approximately 12 percent greater than the ACI-318-89 value, since the modular ratio of TS-500 is usually greater than that of ACI-318-89 Code. These two factors are the main causes of differences between the two

codes

and

disagreement

results.

98

of

TS-500

with

the

test

CHAPTER V

CONCLUSIONS

5.1 General Remarks

The purpose of this study was to develop a method for

predicting

the

short-term

and

time-dependent

deflections of reinforced concrete members. It was also intended to check the accuracy of the methods proposed in ACI-318-89 and TS-500 Codes taking into consideration the available test results. A computer program for predicting the

short

and

long-term

deflections

of

continuous

reinforced concrete flexural members and slender columns was developed. The program uses the finite element method with iteration in moments and flexural rigidities. The actual moment pattern - which is necessary for computing deflections - is determined by an iterative procedure, beginning with an assumed set of flexural rigidities. The iterative nonlinearity

process in

is

necessary

reinforced

because

concrete.

The

of

the

solution

accuracy mainly depends on the assumed short and long-

99

term stress-strain diagrams for concrete. Based on the assumed

stress-strain

relationships,

load-moment-

curvature diagrams are obtained along the span length of the member for rectangular or flanged sections, up to five layers of steel. Predicted values of deflections using this program were found to be in good agreement with available test data.

Unfortunately,

not

all

tests

published

in

the

literature could be taken into account, since some of the references lacked important data needed for deflection calculations. The database has been reduced to 63 results from 11 different experimental programs.

5.2 Conclusions

The

following

conclusions

are

based

on

the

evaluation of the computer program using the available test data.

1-

The

computer

predicts

and

concrete,

simply

program

evaluates

BECODE

deflections

supported

slender columns

100

and

satisfactorily of

continuous

reinforced beams

and

2- The accuracy of the method proposed for both short-term and long-term deflections depends on the short and lond-term stress-strain relation for concrete.

3- The stress-strain relationship input must be time-dependent to take into account the sustained load behavior of reinforced concrete member.

4-

In

the

computer

program

developed,

the

contribution of concrete in the tension zone was taken into

consideration.

When

the

beam

has

a

high

reinforcement ratio, neglecting tension does not change the results much. However when the reinforcement ratio is low, neglecting tension introduces error in the results.

5- Whether the tensile resistance of concrete is taken into consideration or not, the use of compression reinforcement

is

effective

in

decreasing

compressive

strains due to plastic-flow.

6-

The

deflection

of

beams

tested

by

various

researchers (short and long-term) were computed using the computer program developed and compared with the test results. It was found out that for short-term deflections there was a 9 percent probability of having an error in deflection prediction of more than ±20 percent.

101

For

long-term

probability

of

deflections

more

deflections,

less than

than 40

10

percent

there

percent less

was of

than

a

having

the

ones

predicted by the program. Considering numerous variables which affect the long-term deflections, this error may be considered within acceptable limits.

7- Comparing the deflections calculated using the ACI-318-89 with the test values, it was concluded that for short-term deflections, ACI method was almost as good as

the

computer

deflections

there

program would

developed. be

less

For

than

long-term 10

percent

probability of having an error greater than ±40 percent.

The

evaluation

given

above

clearly

shows

that

deflection of beams can be predicted with a reasonable accuracy using the ACI procedure. This is not surprising, since the ACI method was developed empirically using the available test results. However the values of the latest experimental study of Al-Zaid et al. in 1991 (namely after the derivation of the empirical ACI Code equation for short-term deflections) agree better with the results of the computer program than with ACI Code results.

102

8- Although the TS-500 method is based on ACI, significantly greater errors were obtained when compared with the test data. Sources of this error are the modulus of elasticity and modular ratio which are very different in TS-500 as compared to ACI.

In

the

revision

of

TS-500,

the

modulus

of

elasticity and the modular ratio should be changed to agree with the ACI values.

9- The main advantage of the computer program developed

is

the

inclusion

of

the

axial

load

which

enables it to predict the column deflections. The ACI Code and TS-500 have no provisions for the calculation of column deflections.

5.3 Recommendations

In the computer program (App. A) developed in this study, the chosen function of time-dependent stressstrain

diagram

can

be

modified

by

carrying

out

experimental research. To be considered adequate, stressstrain diagrams must consider time as a variable. Since the accuracy mainly depends on the assumed stress-strain diagram, different statement functions can be assumed for columns and beams based on the experimental results. The

103

author recommends testing of two span continuous T-beams under

sustained

loading.

Concrete

cylinders

under

sustained uniaxial compression should also be tested for predicting time-dependent stress-strain diagrams. It is also recommended to test columns under sustained loading.

104

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106

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Green,

R.,

1966.

"Behavior

of

Unrestrained

Reinforced Concrete Columns Under Sustained Load", Ph.D. Dissertation, University of Texas, Austin.

Hellesland, J., and Green, R., 1971. "Sustained and

Cyclic

Loading

of

Concrete

ASCE, V.97, pp.1113-1128.

109

Columns",

Proceedings

Hognestad, E., 1951. "A Study of Combined Bending and Axial Load in Reinforced Concrete Members", Univ. of Illinois

Engrg.

Experiment

Station

Bulletin,

No.399,

Urbana.

Pauw, A., and Meyers, B. L., 1964. "Effect of Creep

and

Shrinkage

on

the

Behavior

of

Reinforced

Concrete Members", Symposium on Creep of Concrete, ACI Publication SP-9.

Rüsch, H., Grasser, E., and Rao, P. S., 1962. "Principes de Calcul du Béton Armé Sous des Etats de Contraintes

Monoaxiaux",

Bulletin

d'Information

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Comité Européen du Béton, Luxemburg.

Rüsch, H., 1960. "Researches Towards a General Flexural

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Structural

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Rüsch,H., Characteristics Stress",

Hilsdorf, of

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M.,

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"Sustained

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112

APPENDIX A FORTRAN LISTING OF PROGRAM BECODE AND DATA INPUT

A.1 Fortran Listing of Program BECODE C C C C C C C C C C C C C C C C C C C

AN ANALYSIS PROGRAM FOR STATIC RESPONSE OF REINFORCED CONRETE CONTINUOUS BEAMS AND SLENDER COLUMNS BY M.H. GUNEL MIDDLE EAST TECHNICAL UNIVERSITY , ANKARA BASED ON SAP4 A STRUCTURAL ANALYSIS PROGRAM FOR STATIC AND DYNAMIC RESPONSE OF LINEAR SYSTEMS K.J. BATHE , E.L. WILSON , F.E. PETERSON UNIVERSITY OF CALIFORNIA , BERKELEY CHARACTER*4 HED COMMON /JUNK / HED(12),JUK(408) COMMON /ELPAR/ NPAR(14),NUMNP,MBAND,NELTYP,N1,N2,N3,N4,N5,MTOT,NEQ 113

COMMON /EM/ QQQ(2846),IFILL2(33) COMMON /EXTRA/ MODEX,NT8,N10SV,NT10,KEQB,NUMEL,T(10) COMMON /SOL/ NBLOCK,NEQB,LL,NF,IDUM,NEIG,NAD,NVV,ANORM,NFO COMMON /CHAL1/ HMOM(100),EIK(100),CUR(10,2,200),RMOM(10,2,200) *,EI(100),HMOMO(100),SIGK(100,12),SIGKO(100,12),IB(20),INBE COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) COMMON /CHAL3/ EHAL3,ELIN(100,2) COMMON /CHAL30/ RRL(200,2),RRLO(200,2) C C C

PROGRAM CAPACITY CONTROLLED BY THE FOLLOWING TWO STATEMENTS ... COMMON A(15000) DIMENSION IA(15000) EQUIVALENCE(IA(1),A(1)) OPEN(1,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(2,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(3,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(4,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(7,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(8,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(9,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(10,ACCESS='SEQUENTIAL',FORM='UNFORMATTED') OPEN(11,FILE=' ',STATUS='NEW') OPEN(5,FILE=' ',STATUS='OLD') OPEN(6,FILE=' ',STATUS='NEW') MTOT=20000 ICN=0

C C USE THE IBM FORTRAN EXTENDED ERROR HANDLING FACILITY TO C ELIMINATE PRINTOUT OF UNDERFLOW ERROR MESSAGE (ERROR NUMBER 208) C C CH : MODIFICATIONS C C NT8 = 8 REWIND NT8 NT10= 10 REWIND NT10 N1=1 C C PROGRAM CONTROL DATA C CH IMHG1=1

114

IHC1=0 5 REWIND 11 IHC1=IHC1+1 IF(IHC1.GT.1) GO TO 73 CH READ (5,100,END=990) HED,NUMNP,NELTYP,LL,NF,NDYN,MODEX,NAD, 1 KEQB,N10SV,IMHG WRITE (11,100)HED,NUMNP,NELTYP,LL,NF,NDYN,MODEX,NAD,KEQB,N10SV GO TO 74 73 READ (11,100)HED,NUMNP,NELTYP,LL,NF,NDYN,MODEX,NAD,KEQB,N10SV 74 IF(MODEX.GT.0) MODEX = 1 IF (NUMNP.EQ.0) STOP IF(IHC1.GT.1) GO TO 731 WRITE (6,200) HED,NUMNP,NELTYP,LL,MODEX 731 IF(KEQB.LT.2) KEQB = 99999 IF(LL.GE.1) GO TO 10 WRITE (6,300) STOP C*** DATA PORTHOLE SAVE 10 IF(MODEX.EQ.1) *WRITE (NT8) HED,NUMNP,NELTYP,LL,NF,NDYN C KDYN = IABS(NDYN) +1 IF(KDYN.LE.5) GO TO 14 STOP C C C INPUT JOINT DATA C 14 N2=N1+6*NUMNP N3=N2+NUMNP N4=N3+NUMNP N5=N4+NUMNP N6=N5+NUMNP IF(N6.GT.MTOT) CALL ERROR(N6-MTOT) C CALL INPUTJ(IA(N1),A(N2),A(N3),A(N4),A(N5),NUMNP,NEQ) C CALL INPUTJ(IA(N1*2-1), .... FOR DOUBLE PRECISION C C FORM ELEMENT STIFFNESSES C C MBAND=0 NUMEL=0 REWIND 1

115

REWIND 2 C DO 900 M=1,NELTYP IF(IHC1.GT.1) GO TO 75 READ (5,1001) NPAR WRITE (11,1001) NPAR NPAR(5)=NPAR(2) GO TO 76 75 READ(11,1001) NPAR CH NPAR(5)=NPAR(2) CH C*** DATA PORTHOLE SAVE 76 IF(MODEX.EQ.1) WRITE (NT8) NPAR WRITE (1) NPAR NUMEL=NUMEL+NPAR(2) MTYPE=NPAR(1) C CALL ELTYPE(MTYPE) C 900 CONTINUE C C DETERMINE BLOCKSIZE C C ADDSTF C NEQB=(MTOT - 4*LL)/(MBAND + LL + 1)/2 C C OVER-RIDE THE SYSTEM MATRIX BLOCKSIZE WITH THE INPUT (NON-ZERO) C VALUE, KEQB. C THIS OVER-RIDE ENTRY IS TO ALLOW PROGRAM CHECKING OF MULTIC BLOCK ALGORITHMS WITH WHAT WOULD NORMALLY BE ONE BLOCK DATA. C IF(KEQB.LT.NEQB) NEQB = KEQB C C STATIC SOLUTION C 690 CONTINUE NEQB1=(MTOT - MBAND)/(2*(MBAND+LL) + 1) NEQB2=(MTOT - MBAND - LL*(MBAND-2))/(3*LL + MBAND + 1) IF (NEQB1.LT.NEQB) NEQB=NEQB1 IF (NEQB2.LT.NEQB) NEQB=NEQB2 NBLOCK = (NEQ-1)/NEQB +1 IF(NEQB.GT.NEQ) NEQB=NEQ GO TO 790

116

C 790 CONTINUE C C C

INPUT

NODAL

LOADS

N3=N2+NEQB*LL N4=N3+6*LL IF(IHC1.GT.1) GO TO 732 WRITE (6,201) NEQ,MBAND,NEQB,NBLOCK C C 732 CALL INL(IA(N1),A(N2),A(N3),A(N4),NUMNP,NEQB,LL) C C C FORM TOTAL STIFFNESS C NE2B=2*NEQB N2=N1+NEQB*MBAND N3=N2+NEQB*LL N4=N3+4*LL NN2=N1+NE2B*MBAND NN3=NN2+NE2B*LL NN4=NN3+4*LL C CALL ADDSTF (A(N1),A(NN2),A(NN3),A(NN4),NUMEL,NBLOCK,NE2B,LL,MBAND 1,ANORM,NVV) C C C SOLUTION PHASE C 20 GO TO (30), KDYN C C STATIC SOLUTION C 30 IF(MODEX.EQ.0) GO TO 32 DO 31 I=6,10 31 T(I) = T(5) GO TO 90 C 32 CALL SOLEQ DO 33 I=7,10 33 T(I) = T(6) C CH 90 IF(IMHG.EQ.0) GO TO 990

117

CALL MOMCUR DO 4914 I=1,INBE IF(IHC1.EQ.1) GO TO 4920 IF(EIK(I)/EI(I).LE.1.001.AND.EIK(I)/EI(I).GT.0.999) * GO TO 4914 GO TO 4920 4914 CONTINUE IMHG1=IMHG1+1 DO 347 IH53=1,INBE DO 347 IH54=1,12 347 SIGKO(IH53,IH54)=SIGKO(IH53,IH54)+SIGK(IH53,IH54) DO 351 IH53=1,NUMNP DO 351 IH54=1,6 351 DKO(IH53,IH54)=DKO(IH53,IH54)+DK(IH53,IH54) IF(IMHG1.GT.IMHG) GO TO 990 WRITE(6,552) 552 FORMAT(2X,'*** SOLUTION FOR THE NEXT LOAD INCREMENT ***') DO 346 IH52=1,INBE EIK(IH52)=EI(IH52) HMOMO(IH52)=HMOMO(IH52)+HMOM(IH52) RRLO(IH52,1)=RRL(IH52,1) 346 RRLO(IH52,2)=RRL(IH52,2) GO TO 5 4920 DO 4917 I=1,INBE 4917 EIK(I)=EI(I) CH GO TO 5 990 IF(IMHG.EQ.0) STOP CH WRITE(6,2003) DO 352 IH52=1,NUMNP IH55=NUMNP+1-IH52 352 WRITE(6,3003)IH55,L,(DKO(IH55,I),I=1,6) WRITE(6,2002) DO 349 IH52=1,INBE 349 WRITE(6,3002) IH52,L,(SIGKO(IH52,I),I=1,12) CH STOP 2002 FORMAT(/29H1.....BEAM FORCES AND MOMENTS// . 10H0BEAM LOAD 5X 5HAXIAL 2(6X,5HSHEAR),4X 7HTORSION . 4X,7HBENDING 5X,7HBENDING/ 10H NO. NO. 8X 2HR1 9X 2HR2 9X . 2HR3 9X 2HM1 9X 2HM2 10X 2HM3) 3002 FORMAT(I5,I4,5E11.4,E12.5/9X,5E11.4,E12.5/) 2003 FORMAT (1H1,38HN O D E D I S P L A C E M E N T S / , 1 17HR O T A T I O N S,// 1X,4HNODE,2X,4HLOAD,7X,2HX-,

118

2 9X,2HY-,9X,2HZ-,9X,2HX-,9X,2HY-,9X,2HZ-, /, 3 1X,4HNUM.,2X,4HCASE,3(2X,9HTRANSLAT.), 4 3(3X,8HROTATION), / 1X) 3003 FORMAT (1H0,I4,I5,6E11.4 / (10X,I5,6E11.4) ) C 100 FORMAT (12A4/10I5) 200 FORMAT(1H1,12A4/// 1 38H C O N T R O L I N F O R M A T I O N, // 4X, 2 27H NUMBER OF NODAL POINTS =, I5 / 4X, 3 27H NUMBER OF ELEMENT TYPES =, I5 / 4X, 4 27H NUMBER OF LOAD CASES =, I5 / 4X, B 27H SOLUTION MODE (MODEX) =, I5 / 4X, C 19H EQ.0, EXECUTION, / 4X, D 20H EQ.1, DATA CHECK, / 4X) 201 FORMAT (38H1E Q U A T I O N P A R A M E T E R S, // * 34H TOTAL NUMBER OF EQUATIONS =,I5, 1 /34H BANDWIDTH =,I5, 2 /34H NUMBER OF EQUATIONS IN A BLOCK =,I5, 3 /34H NUMBER OF BLOCKS =,I5) C 300 FORMAT (// 48H ** ERROR. (AT LEAST ONE LOAD CASE IS REQUIRED) ) 320 FORMAT (// 47H ** WARNING. ESTIMATE OF STORAGE FOR A DYNAMIC, 1 32H ANALYSIS EXCEEDS AVAILABLE CORE, // 1X) C 1001 FORMAT (14I5) END SUBROUTINE INL(ID,B,TR,TMASS,NUMNP,NEQB,LL) C C C CALLED BY: MAIN C C INPUT NODAL LOADS AND MASSES C DIMENSION ID(NUMNP,6),B(NEQB,LL),TR(6,LL),TMASS(NEQB) COMMON / JUNK / R(6),TXM(6),IFILL1(408) COMMON /EXTRA/ MODEX,NT8,IFILL4(14) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) C NT=3 REWIND NT KSHF=0 IF(IHC1.GT.1) GO TO 733 WRITE (6,2002) 733 IF(MODEX.EQ.1) GO TO 50 DO 750 I=1,NEQB

119

TMASS(I)=0. DO 750 K=1,LL 750 B(I,K)=0.0 C 50 DO 900 NN=1,NUMNP C DO 100 I=1,6 TXM(I)=0. DO 100 J=1,LL 100 TR(I,J)=0.0 C IF(NN.EQ.1) GO TO 300 150 IF(N.NE.NN) GO TO 400 DO 200 I=1,6 IF (L) 180,180,190 180 TXM(I)=R(I) GO TO 200 190 TR(I,L)=R(I) 200 CONTINUE 300 IF(IHC1.GT.1) GO TO 73 READ (5,1001) N,L,R WRITE (11,1001)N,L,R WRITE(6,2001)N,L,R IF(IMHG.EQ.0) GO TO 74 DO 554 IEYB=1,6 554 R(IEYB)=R(IEYB)/IMHG GO TO 74 73 READ(11,1001)N,L,R IF(IMHG.EQ.0) GO TO 74 DO 556 IEYB=1,6 556 R(IEYB)=R(IEYB)/IMHG 74 IF (N.EQ.0) GO TO 150 GO TO 150 C 400 IF(MODEX.EQ.1) GO TO 900 DO 800 J=1,6 II=ID(NN,J)-KSHF IF (II) 800,800,500 500 DO 600 K=1,LL 600 B(II,K)=TR(J,K) TMASS(II)=TXM(J) 610 IF(II.NE.NEQB) GO TO 800 WRITE (NT)((B(I3,I4),I4=1,LL),I3=1,NEQB),(TMASS(I3),I3=1,NEQB) KSHF=KSHF+NEQB DO 700 I=1,NEQB

120

TMASS(I)=0. DO 700 K=1,LL 700 B(I,K)=0.0 800 CONTINUE 900 CONTINUE C IF(MODEX.EQ.1) RETURN C WRITE (NT)((B(I3,I4),I4=1,LL),I3=1,NEQB),(TMASS(I3),I3=1,NEQB) C RETURN 1001 FORMAT (2I5,7E10.4) 2001 FORMAT (2(1X,I4),1X,6E11.4) 2002 FORMAT (47H1N O D A L L O A D S (S T A T I C) ,/// B 1X,4HNODE,2X,4HLOAD, 1 2(5X,6HX-AXIS,5X,6HY-AXIS,5X,6HZ-AXIS), /1X,4HNUM.,2X,4HCASE, 2 3(6X,5HFORCE), 3(5X,6HMOMENT), / 1X) END SUBROUTINE BEAM C C C CALLS: TEAM,STRSC C CALLED BY: ELTYPE C COMMON /ELPAR/ NPAR(14),NUMNP,MBAND,NELTYP,N1,N2,N3,N4,N5,MTOT,NEQ COMMON /JUNK/ LT,LH,L,IPAD,SIG(20),N6,N7,N8,N9,N10,IFILL1(391) COMMON /EXTRA/ MODEX,NT8,N10SV,NT10,IFILL4(12) COMMON /CHAL1/ HMOM(100),EIK(100),CUR(10,2,200),RMOM(10,2,200) *,EI(100),HMOMO(100),SIGK(100,12),SIGKO(100,12),IB(20),INBE COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) COMMON /CHAL3/ EHAL,ELIN(100,2) COMMON /CHAL30/ RRL(200,2),RRLO(200,2) COMMON A(1) DIMENSION IA(1) EQUIVALENCE (A(1),IA(1)) C CH INBE=NPAR(2) CH IF(NPAR(1).EQ.0) GO TO 500 N5A=N5+NUMNP N6=N5+NPAR(5) + NUMNP N7=N6+NPAR(5) N8=N7+NPAR(5) N9=N8+12*NPAR(4)

121

N10=N9+6*NPAR(3) N11=N10+NPAR(5) IF(N11.GT.MTOT) CALL ERROR(N11-MTOT) c DO 6246 IHH1=1,NUMNP RRL(IHH1,1)=A(N2+IHH1-1) 6246 RRL(IHH1,2)=A(N3+IHH1-1) C CALL TEAM(NPAR(2),NPAR(3),NPAR(4),NPAR(5),IA(N1),A(N2),A(N3), 1 A(N4),A(N5A),A(N6),A(N7),A(N8),A(N9),A(N10), 2 NUMNP,MBAND) C RETURN C CH 500 WRITE (6,2002) NUME=NPAR(2) 4914 DO 800 MM=1,NUME CALL STRSC (A(N1),A(N3),NEQ,0) CH WRITE (6,2001) DO 800 L=LT,LH CALL STRSC (A(N1),A(N3),NEQ,1) CH WRITE(6,3002) MM,L,(SIG(I),I=1,12) CH DO 346 I=1,12 346 SIGK(MM,I)=SIG(I) HMOM(MM)=(-SIG(6)+SIG(12))/2 IF(SIG(1).LE.0.) GO TO 5253 NHI=ELIN(MM,1) NHJ=ELIN(MM,2) HC2=(DK(NHI,1)+DK(NHJ,1))/2 HC3=(DK(NHI,2)+DK(NHJ,2))/2 HC4=SQRT(HC2*HC2+HC3*HC3) HC5=RRL(NHJ,2)-RRL(NHI,2) HC6=RRL(NHJ,1)-RRL(NHI,1) HC7=SQRT(HC5*HC5+HC6*HC6) HC8=(HC3/HC4*HC6/HC7+HC2/HC4*HC5/HC7)*HC4 HC1=HC8*SIG(1)*IMHG WRITE(6,348) HMOM(MM),SIG(1),HC8,HC1 348 FORMAT(1X,4F12.4) IF(HMOM(MM).LT.0.) HMOM(MM)=HMOM(MM)-ABS(HC1) IF(HMOM(MM).GE.0.) HMOM(MM)=HMOM(MM)+ABS(HC1) CH

122

C*** STRESS PORTHOLE 5253 IF(N10SV.EQ.1) *WRITE (NT10) MM,L,(SIG(I),I=1,12) 800 CONTINUE CH WRITE(6,6245) (RRL(IH1,1),IH1=1,NUMNP), CH * (RRL(IH1,2),IH1=1,NUMNP) 6245 FORMAT(2x,5F10.4) RETURN 2001 FORMAT (/) 2002 FORMAT(/29H1.....BEAM FORCES AND MOMENTS// . 10H0BEAM LOAD 5X 5HAXIAL 2(6X,5HSHEAR),4X 7HTORSION . 4X,7HBENDING 5X,7HBENDING/ 10H NO. NO. 8X 2HR1 9X 2HR2 9X . 2HR3 9X 2HM1 9X 2HM2 10X 2HM3) 3002 FORMAT (I5,I4,5E11.4,E12.5/9X,5E11.4,E12.5/) END SUBROUTINE TEAM (NBEAM,NUMETP,NUMFIX,NUMMAT,ID,X,Y,Z,E,G,RO, 1 SFT,COPROP,WGHT,NUMNP,MBAND) C C CALLS: NEWBM,SLAVE,CALBAN C CALLED BY: BEAM C C FORMS 3-D BEAM STIFFNESS AND STRESS ARRAYS C COMMON/EM/LM(24),ND,NS,ASA(24,24),RF(24,4),XM(24),SA(12,24), 1 SF(12,4),XWT(24),IFILL2(1797) COMMON /NEWB/ LC(4),T(3,3),JK(6),MELTYP,MATTYP,DL COMMON /EXTRA/ MODEX,NT8,IFILL4(14) COMMON /CHAL1/ HMOM(100),EIK(100),CUR(10,2,200),RMOM(10,2,200) *,EI(100),HMOMO(100),SIGK(100,12),SIGKO(100,12),IB(20),INBE COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) COMMON /CHAL3/ EHAL,ELIN(100,2) DIMENSION X(1),Y(1),Z(1),ID(NUMNP,1),E(1),G(1),SFT(NUMFIX,1) 1 ,COPROP(NUMETP,1),RO(1),EMUL(3,4),WGHT(1) DIMENSION ILC(4),TI(3,3),TJ(3,3),STIF(722),TS(2,2),LS(4) EQUIVALENCE (STIF(1),LM(1)) C C C INITIALIZATION C IF(IHC1.GT.1) GO TO 731 WRITE (6,2005) NBEAM,NUMETP,NUMFIX,NUMMAT 731 N=0 DO 5 I=1,1058 5 STIF(I)=0. C

123

C C CH

4916 773 4915

READ AND PRINT MATERIAL PROPERTY DATA WRITE (6,2001) IF(IHC1.EQ.1) GO TO 4915 WRITE(6,2014) DO 4916 I=1,INBE READ(11,1101)IAR,E(IAR),G(IAR),RO(IAR),WGHT(IAR) E(IAR)=EI(IAR) NUMMAT=INBE DO 773 I=1,INBE WRITE(6,2020) I,E(I) GO TO 4917 WRITE(6,2001) READ (5,1001) I,E(I),G(I),RO(I),WGHT(I) WRITE(6,2002) I,E(I),G(I),WGHT(I) EHAL=E(I) NUMMAT=INBE G(I)=0.5*E(1)/(1.+G(1)) DO 10 I=1,INBE E(I)=E(1) RO(I)=RO(1) WGHT(I)=WGHT(1) G(I)=G(1) WRITE(11,1101)I,E(I),G(I),RO(I),WGHT(I)

10 CH C*** DATA PORTHOLE SAVE 4917 IF(MODEX.EQ.1) *WRITE (NT8) (E(N),G(N),RO(N),N=1,NUMMAT) C C READ AND PRINT GEOMETRIC PROPERTIES OF COMMON ELEMENTS. C IF(IHC1.GT.1) GO TO 733 WRITE (6,2003) 733 DO 30 I=1,NUMETP IF(IHC1.GT.1)GO TO 113 READ (5,1002) N,(COPROP(N,J),J=1,6) WRITE(11,1102) N,(COPROP(N,J),J=1,6) GO TO 114 113 READ(11,1102) N,(COPROP(N,J),J=1,6) CH COPROP(N,5)=1. COPROP(N,6)=1. CH 114 CONTINUE

124

IF((COPROP(N,1).NE.0.0).AND.(COPROP(N,4).NE.0.0).AND. 1 (COPROP(N,5).NE.0.0).AND.(COPROP(N,6).NE.0.0)) GO TO 20 WRITE (6,2013) STOP 20 IF(IHC1.GT.1) GO TO 30 WRITE (6,2004) N,(COPROP(N,J),J=1,6) 30 CONTINUE C*** DATA PORTHOLE SAVE IF(MODEX.EQ.1) *WRITE (NT8) ((COPROP(N,J),J=1,6),N=1,NUMETP) C C ELEMENT LOAD MULTIPLIERS C IF(IHC1.GT.1) GO TO 83 READ (5,1006) ((EMUL(I,J),J=1,4),I=1,3) WRITE (11,1106)((EMUL(I,J),J=1,4),I=1,3) GO TO 84 83 READ(11,1106) ((EMUL(I,J),J=1,4),I=1,3) C*** DATA PORTHOLE SAVE 84 CONTINUE IF(IHC1.GT.1) GO TO 742 WRITE(6,2006) ((EMUL(I,J),J=1,4),I=1,3) 742 IF(MODEX.EQ.1) *WRITE (NT8) ((EMUL(I,J),J=1,4),I=1,3) C C READ AND PRINT FIXED END FORCES IN LOCAL COORDINATES C IF(NUMFIX .EQ. 0) GO TO 56 IF(IHC1.GT.1) GO TO 744 WRITE (6,2010) 744 DO 55 I=1,NUMFIX IF(IHC1.GT.1) GO TO 93 READ (5,1005) N,(SFT(N,J),J=1,12) WRITE(11,1105) N,(SFT(N,J),J=1,12) WRITE(6,2011) N,(SFT(N,J),J=1,12) IF(IMHG.EQ.0) GO TO 55 DO 558 IEYB=1,12 558 SFT(N,IEYB)=SFT(N,IEYB)/IMHG GO TO 55 93 READ(11,1105) N,(SFT(N,J),J=1,12) IF(IMHG.EQ.0) GO TO 55 DO 557 IEYB=1,12 557 SFT(N,IEYB)=SFT(N,IEYB)/IMHG 55 CONTINUE C*** DATA PORTHOLE SAVE

125

IF(MODEX.EQ.1) *WRITE (NT8) ((SFT(N,J),J=1,12),N=1,NUMFIX) 56 CONTINUE C C C

READ AND PRINT ELEMENT DATA. GENERATE MISSING INPUT.

747 60

103 104

15 65

66 67

90

91 68 69

IF(IHC1.GT.1) GO TO 747 WRITE (6,4000) L=0 KKK=0 IF(IHC1.GT.1) GO TO 103 READ (5,3000) INEL,INI,INJ,INK,IMAT,IMEL,ILC,INELKI,INELKJ,INC WRITE(11,3000)INEL,INI,INJ,INK,IMAT,IMEL,ILC,INELKI,INELKJ,INC GO TO 104 READ(11,3000) INEL,INI,INJ,INK,IMAT,IMEL,ILC,INELKI,INELKJ,INC IF (INEL.NE.1) GO TO 15 NI=INI NJ=INJ NK=INK IF (INC.EQ.0) INC=1 L=L+1 KKK=KKK+1 ML=INEL-L IF (ML) 66,67,68 WRITE (6,4003) INEL STOP NEL=INEL NI =INI NJ =INJ NK=INK MATTYP=IMAT MELTYP=IMEL DO 90 I=1,4 LC(I)=ILC(I) NLOAD=LC(1)+LC(2)+LC(3)+LC(4) NEKODI=INELKI NEKODJ=INELKJ DO 91 I=1,3 T(2,I)=TI(2,I) GO TO 69 NEL=INEL-ML NI =IN+KKK*INCR NJ =JN+KKK*INCR CONTINUE IF(IHC1.GT.1) GO TO 748

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CH MATTYP=NEL CH WRITE (6,4001) NEL,NI,NJ,NK,MATTYP,MELTYP,LC,NEKODI,NEKODJ ELIN(NEL,1)=NI ELIN(NEL,2)=NJ 748 CONTINUE C*** DATA PORTHOLE SAVE CH MATTYP=NEL CH IF(MODEX.EQ.1) *WRITE (NT8) NEL,NI,NJ,NK,MATTYP,MELTYP,LC,NEKODI,NEKODJ C 74 DX=X(NJ)-X(NI) DY=Y(NJ)-Y(NI) DZ=Z(NJ)-Z(NI) DL=SQRT(DX*DX+DY*DY+DZ*DZ) IF(DL) 75,75,76 75 WRITE (6,4005) NEL STOP C C FORM GLOBAL TO LOCAL COORDINATE TRANSFORMATION. C 76 T(1,1)=DX/DL T(1,2)=DY/DL T(1,3)=DZ/DL C C COMPUTE DIRECTION COSINES OF LOCAL Y-AXIS C A1=X(NJ)-X(NI) A2=Y(NJ)-Y(NI) A3=Z(NJ)-Z(NI) B1=X(NK)-X(NI) B2=Y(NK)-Y(NI) B3=Z(NK)-Z(NI) AA=A1*A1+A2*A2+A3*A3 AB=A1*B1+A2*B2+A3*B3 U1=AA*B1-AB*A1 U2=AA*B2-AB*A2 U3=AA*B3-AB*A3 UU=U1*U1+U2*U2+U3*U3 UU=SQRT(UU) IF (UU.GT.0.) GO TO 40 WRITE (6,4002) INEL

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STOP 40 CONTINUE IF(MODEX.EQ.1) GO TO 185 T(2,1)=U1/UU T(2,2)=U2/UU T(2,3)=U3/UU T(3,1)=T(1,2)*T(2,3)-T(1,3)*T(2,2) T(3,2)=T(1,3)*T(2,1)-T(1,1)*T(2,3) T(3,3)=T(1,1)*T(2,2)-T(1,2)*T(2,1) C C C

CHECK IF NEW STIFFNESS NEEDED IF (NEL.GE.1) GO TO 80 IF (ABS(DS-DL) .GT. DL/100.) GO TO 80 IF ((MT.NE.MATTYP).OR.(ME.NE.MELTYP)) GO TO 80 IF ((JK(1).NE.NEKODI).OR.(JK(2).NE.NEKODJ)) GO TO 80 DO 81 I=1,4 IF (LS(I).NE.LC(I)) GO TO 80 81 CONTINUE DO 82 I=1,2 DO 82 J=1,2 IF (ABS(TS(I,J)-T(I,J)) .GT. ABS(T(I,J)/100.)) GO TO 80 82 CONTINUE GO TO 185

C 80 DS=DL MT=MATTYP ME=MELTYP DO 77 I=1,2 DO 77 J=1,2 77 TS(I,J)=T(I,J) DO 78 I=1,4 78 LS(I)=LC(I) JK(1)=NEKODI JK(2)=NEKODJ C C C

FORM NEW STIFFNESS CALL NEWBM(E,G,RO,WGHT,COPROP,SFT,NUMFIX,NUMETP)

C C C

ADD GRAVITY LOADING ... POINT LOADS ONLY COMPUTED DO 180 I=1,3 DO 180 J=1,4 RF(I,J)=RF(I,J)+EMUL(I,J)*XWT(I)

128

180 RF(I+6,J)=RF(I+6,J)+EMUL(I,J)*XWT(I+6) C C C

FORM ELEMENT LOCATION MATRIX 185 CONTINUE DO 170 M=1,6 LM(M)=ID(NI,M) LM(M+12)=0 LM(M+18)=0 170 LM(M+6)=ID(NJ,M)

C NS=12 ND=12 C C C C

TRANSFORM TO MASTER DEGREES OF FREEDOM CALL SLAVE(X,Y,Z,ID,NUMNP,NI,NJ)

C C C

WRITE ELEMENT INFORMATION ON TAPE NDM=24 CALL CALBAN (MBAND,NDIF,LM,XM,ASA,RF,ND,NDM,NS) IF(MODEX.EQ.1) GO TO 300 WRITE (1) ND,NS,(LM(I),I=1,ND),((SA(I,J),I=1,NS),J=1,ND), 1 ((SF(I,J),I=1,NS),J=1,4)

C C C

CHECK FOR LAST ELEMENT 300 IF(NBEAM-NEL) 66,500,260 260 CONTINUE IF (ML.GT.0) GO TO 65 IN =INI JN =INJ INCR=INC GO TO 60 500 RETURN

C 1001 1101 1002 1102 1005 1105 1006

FORMAT(I5,4F10.0) FORMAT(I5,4E10.4) FORMAT(I5,6F10.0) FORMAT(I5,6E10.4) FORMAT(I5,6F10.0/F15.0,5F10.0) FORMAT(I5,6E10.4/F15.4,5E10.4) FORMAT (4F10.0)

129

1106 FORMAT (4F10.4) C 2001 FORMAT (/// 20H MATERIAL PROPERTIES, // 5X,8HMATERIAL,8X, 1 7HYOUNG*S,6X,9HPOISSON*S,9X,6HWEIGHT, / 7X, 2 6HNUMBER,8X,7HMODULUS,10X,5HRATIO,8X,7HDENSITY, / 1X) C 2014 FORMAT (/// 20H MATERIAL PROPERTIES, // 5X,8HMATERIAL,7X, 1 8HFLEXURAL,/, 7X, 2 6HNUMBER,7X,8HRIGIDITY, /,1X) 2002 FORMAT (8X,I5,E15.4,F15.4,2E15.4) 2020 FORMAT (8X,I5,E15.6) 2003 FORMAT (/// 26H BEAM GEOMETRIC PROPERTIES, // 1X,7HSECTION,2X, 1 10HAXIAL AREA,2(2X,10HSHEAR AREA),4X,7HTORSION,2(5X, 2 7HINERTIA),/ 2X,6HNUMBER,8X,4HA(1),8X,4HA(2),8X,4HA(3), 3 7X,4HJ(1),8X,4HI(2),8X,4HI(3), / 1X) 2004 FORMAT (2X,I5,6E12.4) 2005 FORMAT (34H13 / D B E A M E L E M E N T S, /// . 36H NUMBER OF BEAMS =,I5/ . 36H NUMBER OF GEOMETRIC PROPERTY SETS=,I5/ . 36H NUMBER OF FIXED END FORCE SETS =,I5/ . 36H NUMBER OF MATERIALS =,I5) 2006 FORMAT(///25H ELEMENT LOAD MULTIPLIERS / 20X,1HA,14X,1HB,14X,1HC, 1 14X,1HD,/6H X-DIR4E15.6/ 6H Y-DIR4E15.6/ 6H Z-DIR4E15.6/ ) 2010 FORMAT(1H1,1H , 1 40H FIXED END FORCES IN LOCAL COORDINATES 2//46H TYPE NODE FORCE X FORCE Y FORCE Z 3 31H MOMENT X MOMENT Y MOMENT Z) 2011 FORMAT(1H ,I3,5X,1HI,1X,6F11.3/1H ,8X,1HJ,1X,6F11.3/) 2013 FORMAT(1H0/ 1 60H SECTION PROPERTIES OTHER THAN SHEAR AREAS MAY NOT BE SPECIF 2 34HIED AS ZERO. EXECUTION TERMINATED.) 3000 FORMAT (10I5,2I6,I8) 4000 FORMAT (22H13/D BEAM ELEMENT DATA, /// 3X,4HBEAM,3(2X,4HNODE),3X, 1 8HMATERIAL,3X,7HSECTION,3X,17HELEMENT END LOADS,3X, 2 9HEND CODES, / 7H NUMBER,4X,2H-I,4X,2H-J,4X,2H-K,1X, 3 2(4X,6HNUMBER),4X,1HA,4X,1HB,4X,1HC,4X,1HD,4X,2H-I,4X, 4 2H-J, / 1X) 4001 FORMAT (1X,4(1X,I5),6X,I5,5X,I5,4I5,2I6) 4002 FORMAT (9H0BEAM NO ,I5, 26H K NODE ON BEAM X-AXIS , . 26H......EXECUTION TERMINATED ) 4003 FORMAT(36H0ELEMENT CARD ERROR, ELEMENT NUMBER= I6) 4004 FORMAT(1H ,31HNODAL POINT NUMBERS FOR ELEMENT,I5,36HARE IDENTICAL. 1 EXECUTION TERMINATED.) 4005 FORMAT(8H0ELEMENT,I5,39H HAS ZERO LENGTH. EXECUTION TERMINATED.) END

130

SUBROUTINE NEWBM(E,G,RO,WGHT,COPROP,SFT,NUMFIX,NUMETP) C C C C C

CALLED BY: TEAM FORM NEW BEAM STIFFNESS DIMENSION E(1),G(1),RO(1),COPROP(NUMETP,1),SFT(NUMFIX,1),WGHT(1) COMMON/EM/LM(24),ND,NS,ASA(24,24),RF(24,4),XM(24),SA(12,24), 1 SF(12,4),XWT(24),IFILL2(1797) COMMON /NEWB/ LC(4),T(3,3),JK(6),MELTYP,MATTYP,DL COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) COMMON /CHAL3/ EHAL,ELIN(100,2) DIMENSION R(12),S(12,12),C(12)

C DO 5 I=1,12 DO 5 J=1,12 5 S(I,J)=0.0E0 AX=COPROP(MELTYP,1) AY=COPROP(MELTYP,2) AZ=COPROP(MELTYP,3) AAX=COPROP(MELTYP,4) AAY=COPROP(MELTYP,5) AAZ=COPROP(MELTYP,6) SHFY=0.0 SHFZ=0.0 ZY=E(MATTYP)/(DL*DL) EIY=ZY*AAY EIZ=ZY*AAZ IF(AY.NE.0.0) SHFY=6.*EIZ/(G(MATTYP)*AY) IF(AZ.NE.0.0) SHFZ=6.*EIY/(G(MATTYP)*AZ) COMMY=EIY/(1.+2.*SHFZ) COMMZ=EIZ/(1.+2.*SHFY) C C C

FIXED END FORCES IN LOCAL COORDS

70 71 72 73

DO 73 N=1,4 M=LC(N) IF (M.GT.0) GO TO 71 DO 70 I=1,12 SF(I,N)=0. GO TO 73 DO 72 I=1,12 SF(I,N)=SFT(M,I) CONTINUE

C

131

C C

FORM ELEMENT STIFFNESS IN LOCAL COORDINATES S(1,1)= E(MATTYP)* AX/DL

CH IF(IHC1.GT.1) S(1,1)=EHAL*AX/DL CH S(4,4)= G(MATTYP)*AAX/DL S(2,2)= COMMZ*12./DL S(3,3)= COMMY*12./DL S(5,5)= COMMY* 4.*DL*(1.+0.5*SHFZ) S(6,6)= COMMZ* 4.*DL*(1.+0.5*SHFY) S(2,6)= COMMZ* 6. S(3,5)=-COMMY* 6. DO 102 I=1,6 J=I+6 102 S(J,J)=S(I,I) DO 104 I=1,4 J=I+6 104 S(I,J)=-S(I,I) S(6,12)= S(6,6)*(1.-SHFY)/(2.+SHFY) S(5,11)= S(5,5)*(1.-SHFZ)/(2.+SHFZ) S(2,12)= S(2,6) S(6, 8)=-S(2,6) S(8,12)=-S(2,6) S(3,11)= S(3,5) S(5, 9)=-S(3,5) S(9,11)=-S(3,5) DO 106 I=2,12 K=I-1 DO 106 J=1,K 106 S(I,J)=S(J,I) C C C C

MODIFY ELEMENT STIFFNESS AND ELEMENT FIXED END FORCES FOR KNOWN ZERO MEMBER END FORCES. IF ((JK(1)+JK(2)).EQ.0) GO TO 145 DO 140 K=1,2 KK=JK(K) KD=100000 I1=6*(K-1)+1 I2=I1+5 DO 140 I=I1,I2 IF (KK.LT.KD) GO TO 140 SII=S(I,I) DO 125 N=1,12

132

125 R(N)=S(I,N) DO 130 M=1,12 C(M)=S(M,I)/SII DO 130 N=1,12 130 S(M,N)=S(M,N)-C(M)*R(N) DO 135 N=1,4 SFI=SF(I,N) DO 135 M=1,12 135 SF(M,N)=SF(M,N)-C(M)*SFI 136 KK=KK-KD 140 KD=KD/10 145 CONTINUE C C C C

OBTAIN SA(12,12) RELATING ELEMENT END FORCES (LOCAL) AND JOINT DISPLACEMENTS (GLOBAL). DO 31 I=1,12 DO 31 J=1,24 31 SA(I,J)=0.0E0 DO 150 LA=1,10,3 LB=LA+2 DO 150 MA=1,10,3 MB=MA-1 DO 150 I=LA,LB DO 150 JM=1,3 J=JM+MB XX=0. DO 151 K=1,3 151 XX=XX+S(I,K+MB)*T(K,JM) 150 SA(I,J)=XX

C C C

ELEM STIFF ASA(12,12) AND FIXED END FORCES RF(12) IN GLOBAL COORDS DO 32 I=1,24 DO 32 J=1,24 32 ASA(I,J)=0.0E0 DO 160 LA=1,10,3 LB=LA-1 DO 160 MA=1,10,3 MB=MA+2 DO 160 IL=1,3 I=IL+LB DO 160 J=MA,MB XX=0. DO 161 K=1,3

133

161 XX=XX+T(K,IL)*SA(K+LB,J) 160 ASA(I,J)=XX C DO 165 LA=1,10,3 LB=LA-1 DO 165 IL=1,3 I=IL+LB DO 165 N=1,4 XX=0. DO 162 K=1,3 162 XX=XX-T(K,IL)*SF(K+LB,N) 165 RF(I,N)=XX C C C

FORM MASS AND GRAVITY LOAD MATRIX XXM=RO(MATTYP)*AX*DL/2. WTM=WGHT(MATTYP)*AX*DL/2. DO 180 M=1,3 XWT(M)=WTM XWT(M+3)=0. XWT(M+9)=0. XWT(M+6)=WTM XM(M)=XXM XM(M+3)=0. XM(M+9)=0. 180 XM(M+6)=XXM RETURN END

C SUBROUTINE SOLEQ C C C C C C

CALLS: SESOL,PRINTD,STRESS CALLED BY: MAIN STATIC SOLUTION PHASE COMMON A(1) DIMENSION IA(1),TT(4) EQUIVALENCE (A(1),IA(1)) COMMON /ELPAR/ NP(14),NUMNP,MBAND,NELTYP,N1,N2,N3,N4,N5,MTOT,NEQ COMMON /SOL / NBLOCK,NEQB,LL,NF,IFILL3(6) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6)

C C C

SOLVE FOR THE DISPLACEMENT VECTORS

134

C NSB=(MBAND+LL)*NEQB NSBB=NEQB*LL*(2+(MBAND-2)/NEQB) IF(NSBB.LT.NSB) NSBB=NSB N4=N3+NSBB MI = MBAND + NEQB -1 CALL SESOL (A(N1),A(N3),IA(N4),NEQ,MBAND,LL,NBLOCK,NEQB,NSB,MI *, 1 4,3,2,7) C C C

PRINT DISPLACEMENTS N2=N1+NUMNP*6 N3=N2+6*LL CALL PRINTD (IA(N1),A(N2),A(N3),NEQB,NUMNP,LL,NBLOCK,NEQ,2,1)

C C C

COMPUTE AND PRINT ELEMENT STRESSES N2=N1+4*LL N3=N2+NEQB*LL LB=(MTOT-N3)/(NEQ +12) CALL STRESS(A(N1),A(N2),A(N3),NEQB,LB,LL,NEQ,NBLOCK)

C RETURN END SUBROUTINE STRSC(STR,D,NEQ,NTAG) C C C

CALLED BY: TRUSS,BEAM,PLANE,THREED,SHELL,BOUND,PIPE DIMENSION STR(4,1),D(NEQ,1) COMMON /JUNK/ LT,LH,L,IPAD,SG(20),SIG(7),EXTRA(186),IFILL1(203) COMMON /EM/ NS,ND,B(42,63),TI(42,4),LM(63) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6)

C IF (NTAG.EQ.0) GO TO 800 LL=L-LT+1 DO 300 I=1,NS SG(I)=0.0 DO 300 J=1,4 300 SG(I)=SG(I)+TI(I,J)*STR(J,L) DO 500 J=1,ND JJ=LM(J) IF(JJ.EQ.0) GO TO 500 DO 400 I=1,NS 400 SG(I)=SG(I)+B(I,J)*D(JJ,LL)

135

C 500 CONTINUE GO TO 900 800 READ (1) ND,NS,(LM(I),I=1,ND),(( B(I,J),I=1,NS),J=1,ND), 1 ((TI(I,J),I=1,NS),J=1,4) 900 RETURN END SUBROUTINE CALBAN (MBAND,NDIF,LM,XM,S,P,ND,NDM,NS) C C CALLED BY: RUSS,TEAM,PLNAX,BRICK8,TPLATE,CLAMP,ELST3D,PIPEK C C-----CALCULATES BAND WIDTH AND WRITES STIFFNESS MATRIX ON TAPE 2 DIMENSION LM(1),XM(1),S(NDM,NDM),P(NDM,4) COMMON /EXTRA/ MODEX,NT8,IFILL4(14) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) MIN=100000 MAX=0 DO 800 L=1,ND IF (LM(L).EQ.0) GO TO 800 IF (LM(L).GT.MAX) MAX=LM(L) IF (LM(L).LT.MIN) MIN=LM(L) 800 CONTINUE NDIF=MAX-MIN+1 IF (NDIF.GT.MBAND) MBAND=NDIF IF(MODEX.EQ.1) GO TO 810 C LRD=ND*(ND+1)/2+5*ND WRITE(2) LRD,ND,(LM(I),I=1,ND),((S(I,J),J=I,ND),I=1,ND), 1 ((P(I,J),I=1,ND),J=1,4),(XM(I),I=1,ND) RETURN C 810 WRITE (1) ND,NS,(LM(I),I=1,ND) RETURN C END SUBROUTINE ELTYPE(MTYPE) C C C CALLED BY: MAIN,STRESS C GO TO (2,7),MTYPE C C THREE DIMENSIONAL BEAM ELEMENTS C 2 CALL BEAM

136

GO TO 900 C C 7 CALL BOUND GO TO 900 C 900 RETURN C END SUBROUTINE PRINTD (ID,D,B,NEQB,NUMNP,LL,NBLOCK,NEQ,NT,MQ) C C CALLED BY: SOLEQ,SOLEIG,RESPEC C DIMENSION ID(NUMNP,6),B(NEQB,LL),D(6,LL) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) DATA Q11,Q21,Q12,Q22,Q13,Q23/'LOAD','CASE','EIG.','VEC.', 1 'MODE','NUM.'/ C REWIND 8 READ (8) ((ID(I3,I4),I4=1,6),I3=1,NUMNP) M=NEQ NN=NEQB*NBLOCK C REWIND NT Q1=Q11 Q2=Q21 CH 60 IF(IHC1.EQ.1) WRITE (6,2003) Q1,Q2 C N=NUMNP C DO 500 KK=1,NUMNP C I=6 DO 250 II=1,6 DO 100 L=1,LL 100 D(I,L)=0. IF(M.GT.NN) GO TO 150 IF (M.EQ.0) GO TO 150 READ (NT) ((B(IHG,J),J=1,LL),IHG=1,NEQB) NN=NN-NEQB 150 IF(ID(N,I).LT.1) GO TO 250 K=M-NN M=M-1 C

137

DO 200 L=1,LL 200 D(I,L)=B(K,L) 250 I=I-1 C CH IF(IHC1.EQ.1) WRITE (6,2004) N,(L,(D(I,L),I=1,6),L=1,LL) C CH DO 346 IH52=1,6 346 DK(N,IH52)=D(IH52,1) CH 500 N=N-1 C RETURN C 2003 FORMAT (1H1,38HN O D E D I S P L A C E M E N T S / , 1 17HR O T A T I O N S,// 1X,4HNODE,2X,A4,7X,2HX-, 2 9X,2HY-,9X,2HZ-,9X,2HX-,9X,2HY-,9X,2HZ-, /, 3 1X,4HNUM.,2X,A4,3(2X,9HTRANSLAT.), 4 3(3X,8HROTATION), / 1X) 2004 FORMAT (1H0,I4,I5,6E11.4 / (10X,I5,6E11.4) ) C END C SUBROUTINE ERROR(N) WRITE(6,2000)N 2000 FORMAT(// 20H STORAGE EXCEEDED BY I6) STOP END SUBROUTINE STRESS(STR,B,D,NEQB,LB,LL,NEQ,NBLOCK) C C CALLS: ELTYPE C CALLED BY: SOLEQ C DIMENSION D(NEQ,LB),B(NEQB,LL),STR(4,LL) COMMON /ELPAR/ NPAR(14),NUMNP,MBAND,NELTYP,N1,N2,N3,N4,N5,MTOT,MEQ COMMON /JUNK/ LT,LH,IFILL1(418) COMMON /EXTRA/ MODEX,NT8,N10SV,NT10,IFILL4(12) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) C READ (8)((STR(I3,I4),I4=1,LL),I3=1,4) NT=(LL-1)/LB +1 LH=0 C*** STRESS PORTHOLE IF(N10SV.EQ.1)

138

*WRITE (NT10) NELTYP,NT C DO 1000 II=1,NT C LT =LH+1 LLT=1-LT LH=LT+LB-1 IF(LH.GT.LL) LH=LL C C C

MOVE DISPLACEMENTS INTO CORE FOR LB LOAD CONDITIONS

REWIND 2 C*** STRESS PORTHOLE IF(N10SV.EQ.1) *WRITE (NT10) LT,LH NQ=NEQB*NBLOCK DO 200 NN=1,NBLOCK READ (2)((B(I3,I4),I4=1,LL),I3=1,NEQB) N=NEQB IF (NN.EQ.1) N=NEQ-NQ+NEQB NQ=NQ-NEQB DO 200 J=1,N I=NQ+J DO 200 L=LT,LH K=L+LLT 200 D(I,K)=B(J,L) LK=LH-LT+1 C C CALCULATE STRESSES FOR ALL ELEMENTS FOR LB LOAD CONDITIONS C REWIND 1 DO 1000 M=1,NELTYP READ (1) NPAR C*** STRESS PORTHOLE IF(N10SV.EQ.1) *WRITE (NT10) NPAR MTYPE=NPAR(1) NPAR(1)=0 CALL ELTYPE(MTYPE) 1000 CONTINUE C RETURN END C SUBROUTINE INPUTJ(ID,X,Y,Z,T,NUMNP,NEQ)

139

C C C C

CALLED BY: MAIN DIMENSION X(1),Y(1),Z(1),ID(NUMNP,6),T(1)

C COMMON /EXTRA/ MODEX,NT8,IFILL4(14) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) C C---- SPECIAL NODE CARD FLAGS C C IT = COORDINATE SYSTEM TYPE (CC 1, ANY NODE CARD) C EQ.C, CYLINDRICAL C IPR = PRINT SUPPRESSION FLAG (CC 6, CARD FOR NODE 1 ONLY) C EQ. , NORMAL PRINTING C EQ.A, SUPPRESS SECOND PRINTING OF NODAL ARRAY DATA C EQ.B, SUPPRESS PRINTING OF ID-ARRAY C EQ.C, BOTH *A* AND *B* C DIMENSION IPRC(4) C DATA IPRC/' ','A','B','C'/ C IPR = IPRC(1) RAD = ATAN(1.0E0)/45.0E0 C C C---- READ OR GENERATE NODAL POINT DATA-------------------------------IF(IHC1.GT.1) GO TO 731 WRITE (6,2000) WRITE (6,2001) 731 NOLD=0 10 IF(IHC1.GT.1) GO TO 73 READ (5,1000)IT, N,JPR,(ID(N,I),I=1,6),X(N),Y(N),Z(N),KN,T(N) WRITE(11,1100)IT, N,JPR,(ID(N,I),I=1,6),X(N),Y(N),Z(N),KN,T(N) GO TO 74 73 READ(11,1100) IT, N,JPR,(ID(N,I),I=1,6),X(N),Y(N),Z(N),KN,T(N) 74 IF(IHC1.GT.1) GO TO 732 WRITE (6,2002) IT,N,JPR,(ID(N,I),I=1,6),X(N),Y(N),Z(N),KN,T(N) 732 IF(N.EQ.1) IPR = JPR IF(IT.NE.IPRC(4)) GO TO 15 DUM = Z(N)* RAD Z(N) = X(N)*COS(DUM) X(N) = X(N)*SIN(DUM) 15 CONTINUE

140

IF(NOLD.EQ.0) GO TO 50 C-----CHECK IF GENERATION IS REQUIRED----------------------------------DO 20 I=1,6 IF(ID(N,I).EQ.0.AND.ID(NOLD,I).LT.0) ID(N,I)=ID(NOLD,I) 20 CONTINUE IF(KN.EQ.0) GO TO 50 NUM=(N-NOLD)/KN NUMN=NUM-1 IF(NUMN.LT.1) GO TO 50 XNUM=NUM DX=(X(N)-X(NOLD))/XNUM DY=(Y(N)-Y(NOLD))/XNUM DZ=(Z(N)-Z(NOLD))/XNUM DT=(T(N)-T(NOLD))/XNUM K=NOLD DO 30 J=1,NUMN KK=K K=K+KN X(K)=X(KK)+DX Y(K)=Y(KK)+DY Z(K)=Z(KK)+DZ T(K)=T(KK)+DT DO 30 I=1,6 ID(K,I)=ID(KK,I) IF (ID(K,I).GT.1) ID(K,I)=ID(KK,I)+KN 30 CONTINUE C 50 NOLD=N IF(N.NE.NUMNP) GO TO 10 C C---- PRINT ALL NODAL POINT DATA---------------------------------------C IF(IPR.EQ.IPRC(2) .OR. IPR.EQ.IPRC(4)) GO TO 52 IF(IHC1.GT.1) GO TO 52 WRITE (6,2003) WRITE (6,2001) WRITE (6,2005) (N,(ID(N,I),I=1,6),X(N),Y(N),Z(N),T(N),N=1,NUMNP) 52 CONTINUE C C-----NUMBER UNKNOWNS AND SET MASTER NODES NEGATIVE--------------------C NEQ=0 DO 60 N=1,NUMNP DO 60 I=1,6 ID(N,I)=IABS(ID(N,I))

141

IF(ID(N,I)-1) 57,58,59 57 NEQ=NEQ+1 ID(N,I)=NEQ GO TO 60 58 ID(N,I)=0 GO TO 60 59 ID(N,I)=-ID(N,I) 60 CONTINUE C C---- PRINT MASTER INDEX ARRAY C IF(IPR.EQ.IPRC(3) .OR. IPR.EQ.IPRC(4)) GO TO 62 IF(IHC1.GT.1) GO TO 62 WRITE (6,2004) (N,(ID(N,I),I=1,6),N=1,NUMNP) 62 CONTINUE IF(MODEX.EQ.0) GO TO 70 C*** DATA PORTHOLE SAVE WRITE (NT8) ((ID(N,I),I=1,6),N=1,NUMNP) WRITE (NT8) (X(N),N=1,NUMNP) WRITE (NT8) (Y(N),N=1,NUMNP) WRITE (NT8) (Z(N),N=1,NUMNP) WRITE (NT8) (T(N),N=1,NUMNP) ENDFILE NT8 C REWIND 2 WRITE (2)((ID(I3,I4),I4=1,6),I3=1,NUMNP) C RETURN C 70 CONTINUE REWIND 8 WRITE (8) ((ID(I3,I4),I4=1,6),I3=1,NUMNP) C RETURN C 1000 FORMAT (2(A1,I4),5I5,3F10.0,I5,F9.0) 1100 FORMAT (2(A1,I4),5I5,3F10.4,I5,F9.4) 2000 FORMAT (//23H NODAL POINT INPUT DATA ) 2001 FORMAT (5H0NODE 3X 24HBOUNDARY CONDITION CODES 11X . 23HNODAL POINT COORDINATES / 7H NUMBER 2X 1HX 3X 1HY 3X 1HZ 2X . 2HXX 2X 2HYY 2X 2HZZ 8X 1HX 11X 1HY 11X 1HZ 13X 1HT ) 2002 FORMAT(1X,A1,I4,A1,I3,5I4,3F12.4,I5,F8.3) 2003 FORMAT (//21H1GENERATED NODAL DATA) 2004 FORMAT (//17H1EQUATION NUMBERS/ 1 35H N X Y Z XX YY ZZ /(7I5))

142

2005 FORMAT(1X,I5,6I4,F13.4,2F14.4,F8.3) END SUBROUTINE SESOL (A,B,MAXA,NEQ,MA,NV,NBLOCK,NEQB,NAV,MI,NSTIF, 1 NRED,NL,NR) C C CALLED BY: SOLEQ C DIMENSION A(NAV),B(NAV),MAXA(MI) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) C MM=1 MA2=MA - 2 IF(MA2.EQ.0) MA2=1 INC=NEQB - 1 NWA=NEQB*MA NTB=(MA-2)/NEQB + 1 NEB=NTB*NEQB NEBT=NEB + NEQB NWV=NEQB*NV NWVV=NEBT*NV C N1=NL N2=NR REWIND NSTIF REWIND NRED REWIND N1 REWIND N2 C C MAIN LOOP OVER ALL BLOCKS DO 600 NJ=1,NBLOCK IF (NJ.NE.1) GO TO 10 READ (NSTIF) A IF(NEQ.GT.1) GO TO 100 MAXA(1)=1 WRITE(NRED) A,MAXA IF(A(1)) 1,174,3 1 KK=1 WRITE(6,1010) KK,A(1) 3 DO 5 L=1,NV 5 A(1+L)=A(1+L)/A(1) KR=1+NV WRITE(NL) (A(KK),KK=2,KR) RETURN 10 IF (NTB.EQ.1) GO TO 100 REWIND N1

143

REWIND N2 READ (N1) A C C 100

120

130 136 140 160 110 C 174

172 C C 176 210

220

FIND COLUMN HEIGHTS KU=1 KM=MIN0(MA,NEQB) MAXA(1)=1 DO 110 N=2,MI IF (N-MA) 120,120,130 KU=KU + NEQB KK=KU MM=MIN0(N,KM) GO TO 140 KU=KU + 1 KK=KU IF (N-NEQB) 140,140,136 MM=MM - 1 DO 160 K=1,MM IF (A(KK)) 110,160,110 KK=KK - INC MAXA(N)=KK IF (A(1)) 172,174,176 KK=(NJ-1)*NEQB + 1 IF (KK.GT.NEQ) GO TO 590 WRITE (6,1000) KK STOP KK=(NJ-1)*NEQB + 1 WRITE (6,1010) KK,A(1) FACTORIZE LEADING BLOCK DO 200 N=2,NEQB NH=MAXA(N) IF (NH-N) 200,200,210 KL=N + INC K=N D=0. DO 220 KK=KL,NH,INC K=K - 1 C=A(KK)/A(K) D=D + C*A(KK) A(KK)=C A(N)=A(N) - D

C IF (A(N)) 222,224,230

144

224

222 C 230

280

300 240 C

440 430 C 200 C C

KK=(NJ-1)*NEQB + N IF (KK.GT.NEQ) GO TO 590 WRITE (6,1000) KK STOP KK=(NJ-1)*NEQB + N WRITE (6,1010) KK,A(N) IC=NEQB DO 240 J=1,MA2 MJ=MAXA(N+J) - IC IF (MJ-N) 240,240,280 KU=MIN0(MJ,NH) KN=N + IC C=0. DO 300 KK=KL,KU,INC C=C + A(KK)*A(KK+IC) A(KN)=A(KN) - C IC=IC + NEQB K=N + NWA DO 430 L=1,NV KJ=K C=0. DO 440 KK=KL,NH,INC KJ=KJ - 1 C=C + A(KK)*A(KJ) A(K)=A(K) - C K=K + NEQB CONTINUE CARRY OVER INTO TRAILING BLOCKS DO 400 NK=1,NTB IF ((NK+NJ).GT.NBLOCK) GO TO 400 NI=N1 IF ((NJ.EQ.1).OR.(NK.EQ.NTB)) NI=NSTIF READ (NI) B ML=NK*NEQB + 1 MR=MIN0((NK+1)*NEQB,MI) IF(MA.EQ.1) ML=MR MD=MI - ML KL=NEQB + (NK-1)*NEQB*NEQB N=1

C DO 500 M=ML,MR

145

510

520 530

550

575 540 C 580

620 610 C 505 500 C

570 560

NH=MAXA(M) KL=KL + NEQB IF (NH-KL) 505,510,510 K=NEQB D=0. DO 520 KK=KL,NH,INC C=A(KK)/A(K) D=D + C*A(KK) A(KK)=C K=K - 1 B(N)=B(N) - D IF (MD) 580,580,530 IC=NEQB DO 540 J=1,MD MJ=MAXA(M+J) - IC IF (MJ-KL) 540,550,550 KU=MIN0(MJ,NH) KN=N + IC C=0. DO 575 KK=KL,KU,INC C=C + A(KK)*A(KK+IC) B(KN)=B(KN) - C IC=IC + NEQB KN=N + NWA K=NEQB + NWA DO 610 L=1,NV KJ=K C=0. DO 620 KK=KL,NH,INC C=C + A(KK)*A(KJ) KJ=KJ - 1 B(KN)=B(KN) - C KN=KN + NEQB K=K + NEQB MD=MD - 1 N=N + 1 IF (NTB.NE.1) GO TO 560 WRITE (NRED) A,MAXA DO 570 I=1,NAV A(I)=B(I) GO TO 600 WRITE (N2) B

146

C 400 C

590 C 600 C C 700

CONTINUE M=N1 N1=N2 N2=M WRITE (NRED) A,MAXA CONTINUE VECTOR BACKSUBSTITUTION DO 700 K=1,NWVV B(K)=0. REWIND NL

C

820 810

850 855

870

DO 800 NJ=1,NBLOCK BACKSPACE NRED READ (NRED) A,MAXA BACKSPACE NRED K=NEBT DO 810 L=1,NV DO 820 I=1,NEB B(K)=B(K-NEQB) K=K - 1 K=K + NEBT + NEB KN=0 KK=NWA NDIF=NEQB IF (NJ.EQ.1) NDIF=NEQB - (NBLOCK*NEQB - NEQ) DO 855 L=1,NV DO 850 K=1,NDIF B(KN+K)=A(KK+K)/A(K) KK=KK + NEQB KN=KN + NEBT IF(MA.EQ.1) GO TO 915 ML=NEQB + 1 KL=NEQB DO 860 M=ML,MI KL=KL + NEQB KU=MAXA(M) IF (KU-KL) 860,870,870 K=NEQB KM=M DO 880 L=1,NV KJ=K

147

890 880 860

920

DO 890 KK=KL,KU,INC B(KJ)=B(KJ) - A(KK)*B(KM) KJ=KJ - 1 KM=KM + NEBT K=K + NEBT CONTINUE N=NEQB DO 910 I=2,NEQB KL=N + INC KU=MAXA(N) IF (KU-KL) 910,920,920 K=N DO 930 L=1,NV KJ=K DO 940 KK=KL,KU,INC KJ=KJ - 1 B(KJ)=B(KJ) - A(KK)*B(K) K=K + NEBT N=N - 1

940 930 910 C 915 KK=0 KN=0 DO 950 L=1,NV DO 960 K=1,NEQB KK=KK + 1 960 A(KK)=B(KN+K) 950 KN=KN + NEBT C WRITE (NL) (A(K),K=1,NWV) 800 CONTINUE C 1000 FORMAT (// 46H STOP *** ZERO DIAGONAL ENCOUNTERED DURING, 1 18H EQUATION SOLUTION, / 2 13X,18H EQUATION NUMBER =, I6 ) 1010 FORMAT (/ 50H WARNING *** NEGATIVE DIAGONAL ENCOUNTERED DURING, 1 18H EQUATION SOLUTION, / 2 13X,18H EQUATION NUMBER =, I6, 5X, 7HVALUE =, E20.8 ) C RETURN END SUBROUTINE ADDSTF (A,B,STR,TMASS,NUMEL,NBLOCK,NE2B,LL,MBAND,ANORM, 1NVV) C C C CALLED BY: MAIN

148

C C C

FORMS GLOBAL EQUILIBRIUM EQUATIONS IN BLOCKS DIMENSION A(NE2B,MBAND),B(NE2B,LL),STR(4,LL),TMASS(NE2B)

C COMMON /EM/ LRD,ND,LM(63),IPAD,SS(2331),IFILL2(482) COMMON /EXTRA/ MODEX,NT8,IFILL4(14) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) C NEQB=NE2B/2 K=NEQB+1 X=NBLOCK MB=SQRT(X) MB=MB/2+1 NEBB=MB*NE2B MM=1 NDEG=0 NVV=0 ANORM=0. NSHIFT=0 REWIND 3 REWIND 4 REWIND 9 C C C

READ ELEMENT LOAD MULTIPLIERS

IF(IHC1.GT.1) GO TO 751 WRITE (6,2000) 751 CONTINUE DO 50 L=1,LL IF(IHC1.GT.1) GO TO 73 READ (5,1002) (STR(I,L),I=1,4) WRITE(11,1102) (STR(I,L),I=1,4) WRITE(6,2002) L,(STR(I,L),I=1,4) GO TO 50 73 READ(11,1102) (STR(I,L),I=1,4) IF(IHC1.GT.1) GO TO 50 WRITE (6,2002) L,(STR(I,L),I=1,4) 50 CONTINUE IF(MODEX.EQ.0) WRITE (8)((STR(I3,I4),I4=1,LL),I3=1,4) C C 65 IF(MODEX.EQ.1) RETURN C C FORM EQUATIONS IN BLOCKS ( 2 BLOCKS AT A TIME)

149

C DO 1000 M=1,NBLOCK ,2 DO 100 I=1,NE2B DO 100 J=1,MBAND 100 A(I,J)=0. READ (3) ((B(I,L),I=1,NEQB),L=1,LL),(TMASS(I),I=1,NEQB) IF (M.EQ.NBLOCK) GO TO 200 READ (3) ((B(I,L),I=K,NE2B),L=1,LL),(TMASS(I),I=K,NE2B) 200 CONTINUE C REWIND 7 REWIND 2 NA=7 NUME=NUM7 IF (MM.NE.1) GO TO 75 NA=2 NUME=NUMEL NUM7 =0 C 75 DO 700 N=1,NUME READ (NA) LRD,ND,(LM(I),I=1,ND),(SS(I),I=1,LRD) MSHFT = ND * (ND+1)/2 +4 *ND DO 600 I=1,ND LMN=1-LM(I) II=LM(I)-NSHIFT IF (II.LE.0.OR.II.GT.NE2B) GO TO 600 IMS=I+MSHFT TMASS(II)=TMASS(II)+ SS(IMS) DO 300 L=1,LL DO 300 J=1,4 KK = ND *(ND+1)/2 + ND*(J-1) 300 B(II,L)=B(II,L)+SS(I+KK)*STR(J,L) DO 500 J=1,ND JJ=LM(J)+LMN IF(JJ) 500,500,390 390 IF(J-I) 396,394,394 394 KK = ND*I-(I-1)*I/2 +J-ND GO TO 400 396 KK =ND*J -(J-1)*J/2+I-ND 400 A(II,JJ)=A(II,JJ)+SS( KK) 500 CONTINUE 600 CONTINUE C C C

DETERMINE IF STIFFNESS IS TO BE PLACED ON TAPE 7

150

IF (MM.GT.1) GO TO 700 DO 650 I=1,ND II=LM(I) -NSHIFT IF(II.GT.NE2B.AND.II.LE.NEBB) GO TO 660 650 CONTINUE GO TO 700 660 WRITE (7) LRD,ND,(LM(I),I=1,ND),(SS(I),I=1,LRD) NUM7=NUM7+1 C 700 CONTINUE DO 710 L=1,NEQB ANORM=ANORM + A(L,1) IF (A(L,1).NE.0.) NDEG=NDEG + 1 IF (A(L,1).EQ.0.) A(L,1)=1.E+20 IF (TMASS(L).NE.0.) NVV=NVV + 1 710 CONTINUE C 716 WRITE (4) ((A(I,J),I=1,NEQB),J=1,MBAND),((B(I,L),I=1,NEQB),L=1,LL) 718 WRITE (9) (TMASS(I),I=1,NEQB) C IF(M.EQ.NBLOCK) GO TO 1000 DO 720 L=K,NE2B ANORM=ANORM + A(L,1) IF (A(L,1).NE.0.) NDEG=NDEG + 1 IF (A(L,1).EQ.0.) A(L,1)=1.E+20 IF (TMASS(L).NE.0.) NVV=NVV + 1 720 CONTINUE C 726 WRITE (4) ((A(I,J),I=K,NE2B),J=1,MBAND),((B(I,L),I=K,NE2B),L=1,LL) 728 WRITE (9) (TMASS(I),I=K,NE2B) C IF (MM.EQ.MB) MM=0 MM=MM+1 1000 NSHIFT=NSHIFT+NE2B IF (NDEG.GT.0) GO TO 730 WRITE (6,1010) STOP 730 ANORM=(ANORM/NDEG)*1.E-8 C RETURN 1002 FORMAT (4F10.0) 1102 FORMAT (4F10.4) 1004 FORMAT (5I5,3F10.0) 1010 FORMAT (51H0STRUCTURE WITH NO DEGREES OF FREEDOM CHECK DATA 2000 FORMAT (/// 10H STRUCTURE,13X,7HELEMENT,4X,4HLOAD,4X,

151

)

1 11HMULTIPLIERS,/ 10H LOAD CASE,12X,1HA,9X,1HB,9X,1HC,9X,1HD,/ 1X) 2002 FORMAT (I6,7X,4F10.3) END SUBROUTINE BOUND C C CALLS: CLAMP,STRSC C CALLED BY: ELTYPE C COMMON A(1) DIMENSION IA(1) EQUIVALENCE (A(1),IA(1)) COMMON /ELPAR/ NPAR(14),NUMNP,MBAND,NELTYP,N1,N2,N3,N4,N5,MTOT,NEQ COMMON /JUNK/ LT,LH,L,IPAD,SIG(20),IFILL1(396) COMMON /EXTRA/ MODEX,NT8,N10SV,NT10,IFILL4(12) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) C IF (NPAR(1).EQ.0) GO TO 500 C CALL CLAMP (NPAR(2),IA(N1),A(N2),A(N3),A(N4),NUMNP,MBAND) C RETURN C 500 IF(IHC1.GT.1) GO TO 742 WRITE (6,2002) 742 NUME=NPAR(2) DO 800 MM=1,NUME CALL STRSC (A(N1),A(N3),NEQ,0) IF(IHC1.GT.1) GO TO 743 WRITE (6,2001) 743 DO 800 L=LT,LH CALL STRSC (A(N1),A(N3),NEQ,1) IF(IHC1.GT.1) GO TO 761 WRITE (6,3002) MM,L,(SIG(I),I=1,2) 761 CONTINUE C*** STRESS PORTHOLE IF(N10SV.EQ.1) *WRITE (NT10) MM,L,SIG(1),SIG(2) 800 CONTINUE RETURN C 2001 FORMAT (/) 2002 FORMAT (48H1B O U N D A R Y E L E M E N T F O R C E S /, 1 14H M O M E N T S, // 8H ELEMENT,3X,4HLOAD,14X,5HFORCE, 2 9X,6HMOMENT, / 8H NUMBER,3X,4HCASE, // 1X) 3002 FORMAT (I8,I7,4X,2E15.5)

152

END SUBROUTINE CLAMP (NUMEL,ID,X,Y,Z,NUMNP,MBAND) C C C C

CALLS: CALBAN CALLED BY: BOUND COMMON/EM/LM(24),ND,NS,S(24,24),P(24,4),XM(24),SA(12,24),TT(12,4), 1 IFILL2(1821) DIMENSION X(1),Y(1),Z(1),ID(NUMNP,1) COMMON / JUNK / R(6),RM(4),IFILL1(410) COMMON /EXTRA/ MODEX,NT8,IFILL4(14) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6)

C IF(IHC1.GT.1) GO TO 741 WRITE (6,2000) NUMEL C 741 NS=2 ND=6 C IF(IHC1.GT.1) GO TO 73 READ (5,1005) RM WRITE (11,1105)RM GO TO 74 73 READ(11,1105) RM 74 IF(IHC1.GT.1) GO TO 763 WRITE (6,2005) RM 763 CONTINUE C*** DATA PORTHOLE SAVE IF(MODEX.EQ.1) *WRITE (NT8) RM C C INITIALIZATION C DO 30 NI=1,ND XM(NI) = 0.0 DO 20 NJ=1,ND 20 S(NI,NJ)= 0.0 30 CONTINUE DO 50 NK=1,NS DO 40 NL=1,ND 40 SA(NK,NL) = 0.0 DO 50 NI=1,4 TT(NK,NI) = 0.0 50 CONTINUE C

153

NE=0 IF(IHC1.GT.1) GO TO 210 WRITE (6,2007) 210 KG=0 MARK=0 C 200 IF(IHC1.GT.1) GO TO 83 READ (5,1000) NP,NI,NJ,NK,NL,KD,KR,KN,SD,SR,TRACE WRITE(11,1100)NP,NI,NJ,NK,NL,KD,KR,KN,SD,SR,TRACE GO TO 84 83 READ(11,1100) NP,NI,NJ,NK,NL,KD,KR,KN,SD,SR,TRACE 84 IF (TRACE.EQ.0.) TRACE=1.0E+10 IF (KG.GT.0) GO TO 550 C C C

COMPUTE THE DIRECTION COSINES OF THE ELEMENT*S AXIS

250 260

3000 270

C C C

KG=KN IF(MODEX.EQ.1) GO TO 530 IF(NJ.EQ.0) GO TO 250 X1=X(NJ)-X(NI) Y1=Y(NJ)-Y(NI) Z1=Z(NJ)-Z(NI) X2=X(NL)-X(NK) Y2=Y(NL)-Y(NK) Z2=Z(NL)-Z(NK) T1=Y1*Z2-Y2*Z1 T2=Z1*X2-Z2*X1 T3=X1*Y2-X2*Y1 GO TO 260 T1=X(NI)-X(NP) T2=Y(NI)-Y(NP) T3=Z(NI)-Z(NP) XL=T1*T1+T2*T2+T3*T3 XL=SQRT(XL) IF(XL.GT.1.0E-6) GO TO 270 WRITE (6,3000) FORMAT (32H0*** ERROR ZERO ELEMENT LENGTH, / 1X) STOP CONTINUE T1=T1/XL T2=T2/XL T3=T3/XL DISPLACEMENT PRESCRIPTION

154

IF (KD.EQ.0) GO TO 300 SA(1,1)=T1*TRACE SA(1,2)=T2*TRACE SA(1,3)=T3*TRACE S(1,1)=T1*T1*TRACE S(1,2)=T1*T2*TRACE S(1,3)=T1*T3*TRACE S(2,2)=T2*T2*TRACE S(2,3)=T2*T3*TRACE S(3,3)=T3*T3*TRACE PP=TRACE*SD R(1)=T1*PP R(2)=T2*PP R(3)=T3*PP GO TO 350 300 DO 310 I=1,3 R(I) = 0.0 SA(1,I) = 0.0 DO 310 J=I,3 310 S(I,J) = 0.0 C C C

ROTATION PRESCRIPTION 350 IF (KR.EQ.0) GO TO 400 SA(2,5)=T2*TRACE SA(2,4)=T1*TRACE SA(2,6)=T3*TRACE S(4,4)=T1*T1*TRACE S(4,5)=T1*T2*TRACE S(4,6)=T1*T3*TRACE S(5,5)=T2*T2*TRACE S(5,6)=T2*T3*TRACE S(6,6)=T3*T3*TRACE PP=TRACE*SR R(4)=T1*PP R(5)=T2*PP R(6)=T3*PP GO TO 450 400 DO 410 I=4,6 R(I) = 0.0 SA(2,I) = 0.0 DO 410 J=I,6 410 S(I,J) = 0.0 450 DO 500 I=1,ND DO 500 J=I,ND

155

500 S(J,I) = S(I,J) DO 520 I=1,ND DO 520 J=1,4 520 P(I,J)=R(I)*RM(J) 530 NN = NP NNI=NI NNJ=NJ NNK=NK NNL=NL NKD=KD NKR=KR SSD=SD SSR=SR TTR=TRACE GO TO 560 550 MARK=1 555 NN=NN+KG NNI=NNI+KG 560 KEL = NE+1 IF(IHC1.GT.1) GO TO 766 WRITE (6,2010) KEL,NN,NNI,NNJ,NNK,NNL,NKD,NKR,KN,SSD,SSR,TTR 766 NE=NE+1 C*** DATA PORTHOLE SAVE IF(MODEX.EQ.1) *WRITE (NT8) NE,NN,NNI,NNJ,NNK,NNL,NKD,NKR,SSD,SSR,TTR C DO 600 I=1,ND 600 LM(I)=ID(NN,I) C NDM=24 CALL CALBAN (MBAND,NDIF,LM,XM,S,P,ND,NDM,NS) IF(MODEX.EQ.1) GO TO 650 WRITE (1) ND,NS,(LM(L),L=1,ND),((SA(L,K),L=1,NS),K=1,ND), 1 ((TT(L,K),L=1,NS),K=1,4) C 650 CONTINUE IF (NE.EQ.NUMEL) RETURN IF (NN.LT.NP) GO TO 555 IF (MARK.EQ.1) GO TO 210 GO TO 200 C 1000 FORMAT (8I5,3F10.0) 1100 FORMAT (8I5,3F10.4) 1005 FORMAT (4F10.0) 1105 FORMAT (4F10.4)

156

C 2000 FORMAT (34H1B O U N D A R Y E L E M E N T S, /// 1 27H ELEMENT TYPE = 7, / 2 21H NUMBER OF ELEMENTS =,I6 /// 1X) 2005 FORMAT (30H ELEMENT LOAD CASE MULTIPLIERS, // 8X,7HCASE(A),8X, 1 7HCASE(B),8X,7HCASE(C),8X,7HCASE(D),/ 4F15.4 /// 1X) 2007 FORMAT (53H ELEMENT NODE NODES DEFINING CONSTRAINT DIRECTION, 1 3X,38HCODE CODE GENERATION SPECIFIED,6X, 2 22HSPECIFIED SPRING, / 3 53H NUMBER (N) (NI) (NJ) (NK) (NL), 4 3X,38H KD KR CODE (KN) DISPLACEMENT,6X, 5 22H ROTATION RATE, / 1X) 2010 FORMAT (1X,2(2X,I5),2X,4(4X,I5),2(2X,I5),7X,I5,2E15.4,E13.4) END SUBROUTINE SLAVE (X,Y,Z,ID,NUMNP,NI,NJ) C C CALLED BY: TEAM C C PERFORMS SLAVE...MASTER DISPLACEMENT TRANSFORMATION C ( FOR NODES CONNECTED TO BEAM ELEMENTS ONLY) C DIMENSION X(1),Y(1),Z(1),ID(NUMNP,1) COMMON /EM/ LM(24),ND,NS,S(24,24),R(96),XM(24),SA(12,24) ,TT(12,4) 1 ,IFILL2(1821) COMMON /EXTRA/ MODEX,NT8,IFILL4(14) COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) C DETERMINE REQUIRED TRANSLATION DEGREES OF FREEDOM C DO 54 NF=1,12,6 NOD=NI IF (NF.EQ.7) NOD=NJ DO 30 K=1,3 I=K+NF-1 IF (LM(I).GE.0) GO TO 30 M=-LM(I) LM(I)=ID(M,K) IF(K-2) 35,45,55 35 D1=-(Y(NOD)-Y(M)) D2= Z(NOD)-Z(M) LM(ND+1)=ID(M,6) LM(ND+2)=ID(M,5) GO TO 50 45 D1=-(Z(NOD)-Z(M)) D2= X(NOD)-X(M) LM(ND+1)=ID(M,4)

157

LM(ND+2)=ID(M,6) GO TO 50 55 D1=-(X(NOD)-X(M)) D2= Y(NOD)-Y(M) LM(ND+1)=ID(M,5) LM(ND+2)=ID(M,4) 50 CONTINUE IF (MODEX.EQ.1) GO TO 80 C TRANSFORMATION...ARRAYS INCREASE IN SIZE C DO 60 II=1,ND S(ND+1,II)=S(I,II)*D1 S(ND+2,II)=S(I,II)*D2 S(II,ND+1) = S(II,I) *D1 S(II,ND+2) = S(II,I) *D2 60 CONTINUE XM(ND + 1) = XM(I)*D1*D1 XM(ND + 2) = XM(I)*D2*D2 C DO 70 II=1,NS SA(II,ND+1)=SA(II,I)*D1 70 SA(II,ND+2)=SA(II,I)*D2 C S(ND+1,ND+1)=S(I,I)*D1**2 S(ND+2,ND+2)=S(I,I)*D2**2 S(ND+1,ND+2)=S(I,I)*D1*D2 S(ND+2,ND+1)=S(ND+1,ND+2) 80 ND = ND + 2 30 CONTINUE C C SET ROTATIONS C DO 54 J=1,3 K=NF+J+2 IF(LM(K).GE.0) GO TO 54 M=-LM(K) LM(K)=ID(M,J+3) 54 CONTINUE C RETURN END C SUBROUTINE MOMCUR C C MOMENT CURVATURE PROGRAM

158

C C FOR RECTANGULAR OR T SECTIONS, C WITH FIVE LAYERS OF STEEL C C STRESS STRAIN RELATION IS GIVEN AS STATEMENT FUNCTIONS C COMMON /CHAL1/ HMOM(100),EIK(100),CUR(10,2,200),RMOM(10,2,200) *,EI(100),HMOMO(100),SIGK(100,12),SIGKO(100,12),IB(20),INBE COMMON /CHAL2/ IHC1,IMHG,IMHG1,DK(200,6),DKO(200,6) COMMON /CHAL30/ RRL(200,2),RRLO(200,2) REAL N C FE0TS(TS)=0.002*(1+ * (TS**HA/(HB+TS**HA)*HC) * *(1.25*TLA**(-0.118)) * *(1.27-0.0067*ARH) * *(1.00/(1.+HD*ROP)) * +(TS/(35+TS)*HE*(RO-ROP)/RO)/0.002) C C FUNCTION OF FIRST PART OF THE STRESS STRAIN CURVE OF CONCRETE C IN COMPRESSION F1(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*(2.*ECH/E0H-ECH*ECH/E0H/E0H) C C F1 * ECH F2(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*(2.*ECH**2/E0H-ECH**3/E0H**2) C C INTEGRAL OF F1 F3(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*(ECH**2/E0H-ECH**3/3/E0H**2) C C INTEGRAL OF F2 F4(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*(2*ECH**3/3/E0H-ECH**4/4/E0H**2) C C F11(ETH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * * FTP*(ETH**2/ET0H-ETH**3/3/ET0H**2) F12(ETH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FTP*(2*ETH**3/3/ET0H-ETH**4/4/ET0H**2) C C C FUNCTION OF SECOND PART OF THE STRESS STRAIN CURVE OF CONCRETE C IN COMPRESSION

159

F5(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*((0.15*EUH-0.15*ECH)/(EUH-E0H)+0.85) C F6(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*((0.15*EUH*ECH-0.15*ECH**2)/(EUH-E0H)+0.85*ECH) C F7(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*(0.15*EUH*ECH/(EUH-E0H)-0.15*ECH**2/2/(EUH-E0H) * +0.85*ECH) C F8(ECH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FCP*(0.15*EUH*ECH**2/2/(EUH-E0H)-0.15*ECH**3/3/(EUH-E0H) * +0.85*ECH*ECH/2) C C F15(ETH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FTP*(0.15*ETUH*ETH/(ETUH-ET0H)-0.15*ETH**2/2/(ETUH-ET0H) * +0.85*ETH) F16(ETH)= (0.75+0.18/(1.8+log10(ts2*24+1))) * *FTP*(0.15*ETUH*ETH**2/2/(ETUH-ET0H)-0.15*ETH**3/3/(ETUH-ET0H) * +0.85*ETH*ETH/2) C C IF(IHC1.GT.1) GO TO 4914 C READ(5,73)IST 73 FORMAT(I5) READ(5,102)(IB(I),I=1,2*IST) 102 FORMAT(20I4) DO 74 IHG=1,IST C READ(5,101)AS1,AS2,AS3,AS4,AS5 C STEEL AREAS FROM TOP TO BOTTOM IF(AS1.LT.-0.5) GO TO 652 GO TO 653 652 IAS1=ABS(AS1) DO 654 I=1,IAS1 READ(5,1101)RMOM(IHG,1,I),CUR(IHG,1,I),RMOM(IHG,2,I),CUR(IHG,2,I) CH WRITE(6,1102)CUR(IHG,1,I),RMOM(IHG,1,I),CUR(IHG,2,I),RMOM(IHG,2,I) 654 CONTINUE 1102 FORMAT(2X,4E15.8) 1101 FORMAT(4E15.0) GO TO 74 653 READ(5,101)D1,D2,D3,D4,D5 C DISTANCE OF STEEL LAYERS FROM TOP TO BOTTOM

160

READ(5,101)B1,B2,H,T,N C B1 : FLANGE WIDTH , B2 : WEB WIDTH , H : BEAM HEIGHT C T : FLANGE THICKNESS , N : AXIAL FORCE READ(5,101)FCP,FY,EM,E0,EU,TS,TLA,ARH C PARAMETERS IN STRESS STRAIN CURVES FOR CONCRETE AND STEEL READ(5,101)FTP,ET0,ETU,TS2,HE C PARAMETERS IN STRESS STRAIN CURVE FOR CONCRETE IN TENSION READ(5,101)ECI,RIEC,HA,HB,HC,HD C ECI : INITIAL STRAIN , RIEC : STRAIN INCREMENT 101 FORMAT(8F10.0) IF(TLA.LT.6.9) TLA=7. C TLA : LOADING AGE (DAYS) C TS : LOADING DURATION (DAYS) C ARH : AMBIENT RELATIVE HUMIDITY IF(ARH.LE.40.) ARH=40. CGH=((B1-B2)*T*T/2+B2*H*H/2)/((B1-B2)*T+B2*H) T1=0. C EY=FY/EM IHC2=0 4915 IHC2=IHC2+1 IF(TS.LT.1.)GO TO 365 ROP=AS1/(B1*D5) RO=AS5/(B1*D5) IF(IHC2.EQ.2) ROP=AS1/(B2*D5) IF(IHC2.EQ.2) RO=AS5/(B2*D5) IF(RO.LE.ROP.OR.RO.EQ.0) RO=ROP E0=FE0TS(TS) EU=2*E0 365 EC=ECI II=1 C C COMPUTATION OF MOMENT AND CURVATURE FOR A STRAIN VALUE 7 C=H RI=1. C 6 ECH=C E0H=E0*C/EC EUH=EU*C/EC 20 ES1=EC*(C-D1)/C ES2=EC*(C-D2)/C ES3=EC*(C-D3)/C ES4=EC*(C-D4)/C ES5=EC*(C-D5)/C C

161

IF(ES1.GT.0..AND.ES1.GT.EY) ES1=EY IF(ES1.LT.0..AND.ES1.LT.-EY) ES1=-EY IF(ES2.GT.0..AND.ES2.GT.EY) ES2=EY IF(ES2.LT.0..AND.ES2.LT.-EY) ES2=-EY IF(ES3.GT.0..AND.ES3.GT.EY) ES3=EY IF(ES3.LT.0..AND.ES3.LT.-EY) ES3=-EY IF(ES4.GT.0..AND.ES4.GT.EY) ES4=EY IF(ES4.LT.0..AND.ES4.LT.-EY) ES4=-EY IF(ES5.GT.0..AND.ES5.GT.EY) ES5=EY IF(ES5.LT.0..AND.ES5.LT.-EY) ES5=-EY C C CONCRETE COMPRESSIVE STRENGTH CALCULATION C FC1=0. FC2=0. FC3=0. FC4=0. XFC1=0. XFC2=0. XFC3=0. XFC4=0. C IF(EC.GT.E0) GO TO 1 C FC1=F3(ECH) XFC1=F4(ECH)/FC1 FC1=FC1*B2 IF(T.EQ.0.) GO TO 10 C IF(C.LE.T) GO TO 12 C FC2=F3(ECH)-F3(ECH-T) XFC2=(F4(ECH)-F4(ECH-T))/FC2 GO TO 13 C 12 FC2=F3(ECH) XFC2=F4(ECH)/FC2 13 FC2=FC2*(B1-B2) GO TO 10 C 1 FC1=F3(E0H) XFC1=F4(E0H)/FC1 FC1=FC1*B2 FC2=F7(ECH)-F7(E0H) XFC2=(F8(ECH)-F8(E0H))/FC2

162

FC2=FC2*B2 IF(T.EQ.0.) GO TO 10 C IF(C.LE.T)GO TO 88 C IF(E0H.LE.ECH-T) GO TO 11 C FC3=F3(E0H)-F3(ECH-T) XFC3=(F4(E0H)-F4(ECH-T))/FC3 FC3=FC3*(B1-B2) FC4=F7(ECH)-F7(E0H) XFC4=(F8(ECH)-F8(E0H))/FC4 FC4=FC4*(B1-B2) GO TO 10 C 11

C 88

FC3=F7(ECH)-F7(ECH-T) XFC3=(F8(ECH)-F8(ECH-T))/FC3 FC3=FC3*(B1-B2) GO TO 10 FC3=F3(E0H) XFC3=F4(E0H)/FC3 FC3=FC3*(B1-B2) FC4=F7(ECH)-F7(E0H) XFC4=(F8(ECH)-F8(E0H))/FC4 FC4=FC4*(B1-B2) GO TO 10

C C CONCRETE TENSILE STRENGTH CALCULATION C 10 FT1=0. FT2=0. FT3=0. FT4=0. XFT1=0. XFT2=0. XFT3=0. XFT4=0. IF(FTP.EQ.0.) GO TO 1010 ET=(H-C)*EC/C IF(ET.LE.0.) GO TO 1010 ETH=H-C ET0H=(H-C)*ET0/ET ETUH=(H-C)*ETU/ET UC=(H-C)*ETU/ET

163

IF(ET.GT.ETU) GO TO 2000 IF(ET.GT.ET0) GO TO 1001 C

1012 1013 1001

1011

1088

2000

FT1=F11(ETH) XFT1=F12(ETH)/FT1 FT1=FT1*B2 IF(T1.EQ.0.) GO TO 1010 IF(H-C.LE.T1) GO TO 1012 FT2=F11(H-C)-F11(H-C-T1) XFT2=(F12(H-C)-F12(H-C-T1))/FT2 GO TO 1013 FT2=F11(H-C) XFT2=F12(H-C)/FT2 FT2=FT2*(B1-B2) GO TO 1010 FT1=F11(ET0H) XFT1=F12(ET0H)/FT1 FT1=FT1*B2 FT2=F15(H-C)-F15(ET0H) XFT2=(F16(H-C)-F16(ET0H))/FT2 FT2=FT2*B2 IF(T1.EQ.0.) GO TO 1010 IF(H-C.LE.T1) GO TO 1088 IF(ET0H.LE.H-C-T1) GO TO 1011 FT3=F11(ET0H)-F11(H-C-T1) XFT3=(F12(ET0H)-F12(H-C-T1))/FT3 FT3=FT3*(B1-B2) FT4=F15(H-C)-F15(ET0H) XFT4=(F16(H-C)-F16(ET0H))/FT4 FT4=FT4*(B1-B2) GO TO 1010 FT3=F15(H-C)-F15(H-C-T1) XFT3=(F16(H-C)-F16(H-C-T1))/FT3 FT3=FT3*(B1-B2) GO TO 1010 FT3=F11(ET0H) XFT3=F12(ET0H)/FT3 FT4=F15(H-C)-F15(ET0H) XFT4=(F16(H-C)-F16(ET0H))/FT4 FT3=FT3*(B1-B2) FT4=FT4*(B1-B2) GO TO 1010 ETUH=UC ET0H=UC*ET0/ETU IF(T1.EQ.0..OR.UC.LE.H-C-T1) GO TO 2002

164

IF(H-C.GT.T1) GO TO 90 FT1=F11(ET0H) XFT1=F12(ET0H)/FT1 FT1=FT1*B1 FT2=F15(UC)-F15(ET0H) XFT2=(F16(UC)-F16(ET0H))/FT2 FT2=FT2*B1 GO TO 1010 90 FT1=F11(ET0H) XFT1=F12(ET0H)/FT1 FT1=FT1*B2 IF(H-C-T1.LT.ET0H) GO TO 93 IF(H-C-T1.EQ.ET0H) GO TO 236 FT2=F15(H-C-T1)-F15(ET0H) XFT2=(F16(H-C-T1)-F16(ET0H))/FT2 FT2=FT2*B2 236 FT3=F15(UC)-F15(H-C-T1) XFT3=(F16(UC)-F16(H-C-T1))/FT3 FT3=FT3*B1 GO TO 1010 93 FT2=F11(ET0H)-F11(H-C-T1) XFT2=(F12(ET0H)-F12(H-C-T1))/FT2 FT2=FT2*(B1-B2) FT3=F15(UC)-F15(ET0H) XFT3=(F16(UC)-F16(ET0H))/FT3 FT3=FT3*B1 GO TO 1010 2002 FT1=F11(ET0H) XFT1=F12(ET0H)/FT1 FT1=FT1*B2 FT2=F15(UC)-F15(ET0H) XFT2=(F16(UC)-F16(ET0H))/FT2 FT2=FT2*B2 GO TO 1010 C 1010 FC=FC1+FC2+FC3+FC4-FT1-FT2-FT3-FT4 FT=FT1+FT2+FT3+FT4 FCC=FC1+FC2+FC3+FC4 C FS1=ES1*EM*AS1 FS2=ES2*EM*AS2 FS3=ES3*EM*AS3 FS4=ES4*EM*AS4 FS5=ES5*EM*AS5 FS=FS1+FS2+FS3+FS4+FS5 91

165

C IF(N.EQ.0.) GO TO 9 C R=FC+FS IF(R/N.GT.0.999.AND.R/N.LT.1.001) GO TO 5 C IF(R.GT.N) GO TO 3 C GO TO 58 C 9

IF(FS.EQ.0..AND.FTP.EQ.0.) GO TO 49 IF(FS.EQ.0..AND.FTP.NE.0.) GO TO 577 GO TO 578 577 IF(FCC/FT.GT.0.999.AND.FCC/FT.LT.1.001) GO TO 5 IF(FC.GT.0.) GO TO 3 RI=RI*2 CKEEP=C C=C+H/RI IF(C/CKEEP.GT.0.999.AND.C/CKEEP.LT.1.001) GO TO 5 GO TO 4 578 IF(FC/FS.GT.0.) GO TO 49

C IF(ABS(FS/FC).GT.0.999.AND.ABS(FS/FC).LT.1.001) GO TO 5 C 49 C 58

C 3

4 C 5 C C

R=FS+FC IF(R.GT.0.) GO TO 3 IF(N.NE.0..AND.RI.EQ.1.) GO TO 7001 RI=RI*2 CKEEP=C C=C+H/RI IF(C/CKEEP.GT.0.999.AND.C/CKEEP.LT.1.001) GO TO 5 GO TO 4 RI=RI*2 CKEEP=C C=C-H/RI IF(C/CKEEP.GT.0.999.AND.C/CKEEP.LT.1.001) GO TO 5 GO TO 6 II=II+1 CUR(IHG,IHC2,II)=EC/C RMOM(IHG,IHC2,II)=(FS1*(CGH-D1)+FS2*(CGH-D2)+FS3*(CGH-D3)+FS4*

166

+(CGH-D4)+FS5*(CGH-D5)+FC1*(XFC1+CGH-ECH)+FC2*(XFC2+CGH-ECH)+ +FC3*(XFC3+CGH-ECH)+FC4*(XFC4+CGH-ECH)+ +FT1*(XFT1-(CGH-C))+FT2*(XFT2-(CGH-C))+ +FT3*(XFT3-(CGH-C))+FT4*(XFT4-(CGH-C))) IF(RMOM(IHG,IHC2,II).LT.RMOM(IHG,IHC2,II-1)) GO TO 7565 IF(MARK.EQ.1) GO TO 7567 GO TO 7566 7567 II=II-1 MARK=0 GO TO 7001 7565 II=II-1 MARK=1 GO TO 7001 C CHP 7566 WRITE(6,552)CUR(IHG,IHC2,II),RMOM(IHG,IHC2,II),FS5,EC,C 552 FORMAT(1X,5E15.8) CHP 7001 EC=EC+(EU-ECI)/RIEC IF(EC.GT.EU) GO TO 333 C GO TO 7 C 333 T1=T T=0. TAS1=AS1 TAS2=AS2 AS1=AS5 AS2=AS4 AS5=TAS1 AS4=TAS2 TTD1=D1 TTD2=D2 D1=H-D5 D2=H-D4 D3=H-D3 D4=H-TTD2 D5=H-TTD1 IF(IHC2.EQ.2) GO TO 74 GO TO 4915 C 74 CONTINUE 4914 DO 4918 I=1,INBE DO 5000 I1=1,2*IST,2 IF(I.GE.IB(I1).AND.I.LE.IB(I1+1)) GO TO 5001

167

5000 CONTINUE 5001 I1=I1/2+1 DUMMY=HMOM(I)+HMOMO(I) IF(DUMMY.GE.0.) GO TO 4919 DO 4922 K=1,200 IF(-DUMMY.GE.RMOM(I1,2,K).AND.-DUMMY.LE.RMOM(I1,2,K+1)) * GO TO 4923 4922 CONTINUE 4923 RRL(I,1)=-DUMMY RRL(I,2)=CUR(I1,2,K+1)* (CUR(I1,2,K+1)-CUR(I1,2,K))*(RMOM(I1,2,K+1)+DUMMY)/ * (RMOM(I1,2,K+1)-RMOM(I1,2,K)) GO TO 4918 4919 DUMMY=HMOM(I)+HMOMO(I) DO 4920 J=1,200 IF(DUMMY.GE.RMOM(I1,1,J).AND.DUMMY.LE.RMOM(I1,1,J+1)) * GO TO 4921 4920 CONTINUE 4921 RRL(I,1)=DUMMY RRL(I,2)=CUR(I1,1,J+1)* (CUR(I1,1,J+1)-CUR(I1,1,J))*(RMOM(I1,1,J+1)-DUMMY)/ * (RMOM(I1,1,J+1)-RMOM(I1,1,J)) 4918 CONTINUE IF(IHC1.GT.1) GO TO 597 DO 596 I=1,INBE 596 EI(I)=RRL(I,1)/RRL(I,2) RETURN 597 IF(IMHG1.GT.1) GO TO 593 DO 594 I=1,INBE EI(I)=((RRL(I,1)/RRL(I,2))+EIK(I))/2 594 CONTINUE RETURN 593 DO 595 I=1,INBE IF(RRL(I,1).EQ.RRLO(I,1)) GO TO 741 EI(I)=(ABS((RRL(I,1)-RRLO(I,1))/(RRL(I,2)-RRLO(I,2)))) IF(IHC1/2*2.EQ.IHC1) EI(I)=(EI(I)+EIK(I))/2 IF(IHC1/2*2.NE.IHC1) EI(I)=(EI(I)+3*EIK(I))/4 GO TO 742 741 EI(I)=((RRL(I,1)/RRL(I,2))+EIK(I))/2 742 CONTINUE 595 CONTINUE RETURN END

168

A.2 Data Input to Program BECODE Data input to Program BECODE will be the same as Program SAPIV (1974), but the following additions have to be done. 1. In the Master Control information line of SAPIV input data, number of increments of load (explained in Sec. 3.4) is entered in columns 45-50 as an integer variable. 2. Following group of lines have to be added at the end of SAPIV input data. Line 1

Column 1-5

2

1-4 9-12 17-20 25-28 1-10 21-30 41-50 1-10 21-30 41-50 1-10 11-20 21-30 31-40 41-50 1-10 11-20 21-30 31-40

3 4 5

6

7

8

41-50 51-60 1-10 11-20 21-30 41-50 1-10

, , , , , ,

Entry Number of different section types (Integer) Initial and final element numbers of different sect. types (Integer)

5-8 13-16 21-24 29-32 ... 11-20 Areas of steel layers 31-40

, 11-20 , 31-40

Distance of steel layers Flange width Web width Beam height Flange thickness Axial force Concrete strength at loading Steel strength Modulus of Elasticity of steel Concrete compressive strain corresponding to maximum stress Ultimate concrete comp. strain Duration of loading (days) Concrete tens. strength at loading Concrete tensile strain corresponding to maximum stress Ultimate concrete tensile strain 0.00107 (Eq. 3.10) 0.0001 : Initial strain (Sec. 3.3) 169

11-20 21-30 31-40 41-50

100 : (Ultimate conc. comp. strain -initial strain)/100 (Sec. 3.3) 0.60 (Eq. 3.9) 18 (Eq. 3.9) 4.15 (Eq. 3.9)

170

VITA

Mehmet

Halis

Günel

was

born

in

İzmir,

Turkey.

He

received his BSc degree in Civil Engineering, and MSc degree in Structures from Middle East Technical University. He worked in the same university as a research assistant from 1982-1984 and 1986-1989. From 1989-1990, he worked in Prokon Consultants A.Ş., Ankara as Project Manager for ERDEMIR TAŞ Privatization Project. Since then he has been an Instructor in Faculty of Architecture in the Middle East Technical University.