ACI-318-89 ve TS-500 Åartnameleri, 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)3Icr 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 = 0.7fc' (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 = 4700fc' (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 = 3250fc'+14000 MPa.), cracking moment (Mcr = 1.3fct'Ig/yt), and direct tensile strength of concrete (fct'=0.35fc' 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)hCti+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 = 0.7sh/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"2c/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) + 0.0015log10(t+1.0)-2
where t
: loading duration in hours
cut = 2cot
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 : 0.40a0.80, a=0.60; 6b30, b=10; 1.30Cu4.15, 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)u1070*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' 0.33fc' (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=Asisi) 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
REFERENCES
ACI Shrinkage
Committee and
209,
1982.
Temperature
Structures","Designing
for
"Prediction
Effects Creep
in
and
of
Creep
Concrete
Shrinkage
in
Concrete Structures", SP 76, American Concrete Institute, Detroit.
ACI Requirements
Committee for
318-89,
Reinforced
1989.
Concrete
"Building and
Code
Commentary",
American Concrete Institute, Detroit.
ACI
Committee
435,
1973.
"Deflections
of
Continuous Concrete Beams", American Concrete Institute, Detroit. (Also, ACI Manual of Concrete Practice 1987.)
ACI Reinforced
Committee Concrete
435, Flexural
1966.
"Deflections
Members",
ACI
of
Journal
Proceedings, V.63, No.6. (Also, ACI Manual of Concrete Practice 1987.)
105
ACI Committee 435, 1978. "Proposed Revisions by Committee
435
Provisions
on
to
ACI
Building
Deflections",
ACI
Code
and
Journal
Commentary Proceedings,
V.75, pp.229-238.
Al-Zaid,
Rajeh
Z.,
Al-Shaikh,
Abdulrahman
H.,
Abu-Hussein, Mustafa M., 1991. "Effect of Loading Type on the Effective Moment of Inertia of Reinforced Concrete Beams", ACI Structural Journal, V.88, No.2, pp.184-190.
Atımtay,
E.,
and
Cengizkan,
K.,
1975.
"A
Numerical Approach to Creep of Reinforced Concrete", METU Journal
of
Pure
and
Applied
Sciences,
Vol.8,
No.3,
pp.365-343.
Bakoss, S. L., Gilbert, R. I., Faulkes, K. A., Pulmano, Reinforced
V.
A., Concrete
1982.
"Long-Term
Beams",
Deflections
Magazine
of
of
Concrete
Research, London, V.34, pp.203-212.
Branson, Dan E., 1963. "Instantaneous and TimeDependent Deflections of Simple and Continuous Reinforced Concrete Beams", Part 1, Report No.7, Alabama Highway Research Report, Bureau of Public Roads, pp.1-78.
106
Branson,
Dan
E.,
1968.
"Design
Procedures
for
Computing Deflections", ACI Journal Proceedings, V.65, pp.730-742.
Branson, Dan E., Christiason, M. L., 1971. "TimeDependent Strength
Concrete and
Properties
Elastic
Related
Properties"
in
to
Design
"Designing
for
effects of Creep, Shrinkage and Temperature in Concrete Structures", ACI Publication SP 27-13, pp.257-277.
Branson, Dan E., 1977. "Deformation of Concrete Structures", Mc Graw-Hill Book Co., New York.
Carreira, D. J., and Chu, K. H., 1986. "StressStrain Relationship for Reinforced Concrete in Tension", ACI Journal Proceedings, V.83, pp.21-18.
CEB "CEB-FIB
Comite
MOdel
Euro-International
Code",
Bulletin
Du
Beton,
D'Information,
1990. No.
213/214"
Christiansen, K., 1988. "Eight-Year Deformation Tests
on
Reinforced
Concrete
Beams",
Structures (RILEM), V.21, pp.172-178.
107
Materials
and
Clarke, G., Sholz, H., Alexander M., 1988. "New Method
to
Reinforced
Predict
the
Concrete
Creep
Flexural
Deflection Members",
of
Cracked
ACI
Journal
Proceedings, V.85, pp.95-101.
Corley, W. G., and Sozen, M. A., 1966. "TimeDependent Deflection of Reinforced Conrete Beams", ACI Journal Proceedings, V.63, pp.371-386.
Ersoy, Uğur, 1985. "Betonarme - Temel İlkeler ve Taşıma Gücü Hesabı", Ankara.
Ersoy, Elemanlarda
Uğur,
Sehim
1973.
Hesabı",
"Eğilmeye İnşaat
Maruz
Betonarme
Mühendisleri
Odası
Teknik Bülten, pp.53-
Espion, B., 1988. "Long-Term Sustained Loading Tests on Reinforced Concrete Beams : A Selected Data Base", Bulletin du Service Genie Civil, No.88-1.
Espion, Deflections Reconsideration
B., of of
Halleux, Reinforced Their
P.,
1990.
Concrete
Variability",
Journal, V.87, pp.232-236.
108
ACI
"Long-Term Beams
:
Structural
Freudenthal, A. M., and Roll, F., 1958. "Creep and Creep Recovery Under High Compressive Stress", ACI Journal Proceedings, V.54.
Ghali,
A.,
and
Favre,
R.,
1986.
"Concrete
Structures: Stresses and Deformations", Chapman and Hall, New York.
Goyal, B., B., and Jackson, N., 1971. "Slender Concrete Columns under Sustained Load", Proceedings ASCE, V.97, pp.2729-2750.
Green,
R.,
and
Breen,
J.,
E.,
1969.
"Eccentrically Loaded Concrete Columns Under Sustained Load", ACI Journal Proceedings, V.66, pp.866-883.
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
No.36,
Comité Européen du Béton, Luxemburg.
Rüsch, H., 1960. "Researches Towards a General Flexural
Theory
for
Structural
Concrete",
ACI Journal
Proceedings, V.57.
Rüsch,H., Characteristics Stress",
Hilsdorf, of
Concrete
Materialpprüfuagsant
H., Under für
1963.
"Deformation
Concentric des
Tensile
Bauwesen
der
Technischen Hochschule München, Report No. 44 (Translated into English by Roger Green).
110
Sargın, for
Concrete
Sections",
M.,
and
1971.
the
Solid
"Stress-Strain
Analysis
Mechanics
of
Relationships
Structural
Division
Concrete
University
of
Waterloo, Canada.
Sell,
R.,
1959.
"Investigations
of
Sustained
Loads on the Strength of Concrete", RILEM Bull., No.5.
Troxell, G. E., Raphael, J. M., and Davis, R. E., 1958. "Long-Time Creep and Shrinkage Tests of Plain and Reinforced Concrete", Proceedings ASTM, V.58, pp.1-20.
TS
500
(Türk
Standardları),
1985.
"Betonarme
Yapıların Hesap ve Yapım Kuralları", Türk Standardları Enstitüsü, Ankara.
Viest, I., M., Elstner, R., C., and Hognestad, E.,
1956.
"Sustained
Load
Strength
of
Eccentrically
Loaded Short Reinforced Concrete Columns", ACI Journal Proceedings, V.52, pp.727-756.
Washa, G. W., and Fluck, P. G., 1952. "Effect of Compressive Reinforced
Reinforcement Concrete
on
Beams",
V.49, No.8, pp.89-108.
111
the ACI
Plastic
Journal
Flow
of
Proceedings,
Washa, G. W., and Fluck, P. G., 1956. "Plastic Flow (Creep) of Reinforced Concrete Continuous Beams", ACI Journal Proceedings, V.52, pp.549-561.
Weil, G., 1959. "Influence des Dimensions et des Contraintes sur le Retrait et le Fluage du Beton", RILEM Bull., No.3
Wilson, E. L., Bathe, K. J., and Peterson, F. E., 1974. "SAPIV - A Structural Analysis Program For Static and Dynamic Response of Linear Systems", Report No. EERC 73-11 University of California Berkeley, California.
Yu, W. W., Winter, G., 1960. "Instantaneous and Long-Time Deflections of Reinforced Concrete Beams Under Working Loads", ACI Journal Proceedings, V.57, pp.29-50.
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
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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
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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
126
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)
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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)
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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
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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
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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
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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
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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
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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
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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)
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*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)
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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))
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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.