L Shaped Column

 

004s7949/925.00 0.00 Q 1992 ergamon ress t d

Com purer s Swucr ures Vol. 44. No. 5,Pp. 112141 11214138. 38. 1992 Printed n GreatBritain. GreatBritain.

TECHNICAL

NOTE

COMPUTE R AIDED ANALYSIS OF REINFORCED COMPUTER CONCRETE COLUMNS SUBJECTED TO AXIA AXIAL L COMPRESSION COMPRESS ION AND BENDING-I L-SHAPED

MALLlKARJUNAt

Karnataka

SECTIONS

and P.

MAHADEVAPPA

Regional Engineering College, Suratkal-5 Suratkal-574 74 157, India (R ecei cei ved

July 199 1991) 1)

Abstract-Numerical investigations on the strength of L-shaped short reinforced concre concrete te columns columns subjected to combined axial load and bending were undertaken for th thee purpose of providing design aaids ids for structural engineers. The use of a computer lends itsel itselff naturally to the solution of the pro problem blem which generally requires an iterative process. Therefore, an attempt has been made in this paper to computerize the analysis procedure for L-shape L-shaped d sections and in the accompanying paper (part 11)s for T-shaped column sections. The ACI-318, CP CP-110 -110 and IS-456 codes present presented ed design aids only for square/rectangular and circular columns. Apparently this study constitutes the first to present the’ interaction curves for L-shaped and T-shaped column sections with the limit state analysis.

INTRODUCTION

ASSUMPTIONS AND

The analysis and design of of L-shaped corner columns are complicated and cumbersome cumbersome.. Next to rectangular and circular shapes, L-section L-sectionss may be the most most frequently encountered reinforced concrete columns, since they can be be used at outside and re-entrant building corners. Nevertheless, information for their analysis and design is not generally available to structural engineers. either in working stress or ultimate strength theories. There are so some me design approaches in which the desig design n effort is reduced by app approxiroximated shape of strength envelopes (e.g. Bresler [2], Parme et al. [3], CP-110 [4 [4], ], ACI-318 [5] and IS-456 [6,7]), and the use of simplifying approximations (e.g. the m method ethod of superposition [8] and the method method of equivalent uniaxial eccentricity IS]). Ramamurthy 191 1 91 developed simple equations to closely represent the load contours in square and rectangular columns. He also illustrated how the they y can be used to determine the appropriate interaction diagram for given eccentricities of the load. Although several noteworthy noteworthy articles [2,3,8-171 on biaxial bending of square/rectangular column sections, which contributed greatly to the understanding of this subject, have appeared in recent years, significant gaps in the area of de design sign aids for biaxial bending still exist. exist. To lessen these gaps, a number of com comprehensiv prehensivee design aids are presented in the present investigation. This paper deals with the limit state analysis of L-s L-shaped haped reinforced concrete (R.C.) columns. The aim of limit state design is to aachieve chieve acceptable probabilities that the structure will not become unfit for the use fo forr which it is intended, that is, that it will not reach a limit state. To ensure the above objectives, the des design ign should be based on characteristic values for material strengths and applied loads, which takes into account the variations in the material strengths and in the lo loads ads to be supported. 7 Present address: Engineering Mechanics and Design Group, Department of Mechanical Engineering, University of Toronto, Toronto MSS lA4. Canada. See ref. [l].

MATERIAL

PROPERTIES PROPERTIES

In the analysis, the following assumptions [6], which are almost the same as those those codified in CP-110 CP-1 10 [4], are made: compression (4 the strain distribution in the concrete in compression

and the strain in the re reinforcement, inforcement, whether in tension or compressio compression, n, are derived from the assumption that plane sections normal to the axis remain plane after bending, and that there is no bond-slip bond-slip between the reinforcement reinforceme nt and the concrete, the tensile strength of concrete is ignored, the relationship between stress-strain distribution in concrete is assumed to be parabolic as shown in Fig. 1. The maximum compressive stress is equal to 0.67&/l ..5 (see Fig. 2), stresses in reinforcement reinforcement are derived from the (4 the stresses representative stress-strain curve for the type of steel

I” ’

used. Typical curves are shown in Figs 3 and 4,

maximum compress compressive ive strain in concrete in axial (4 the maximum

compression is taken as 0.002, compression com(f) the maximum compression strain at the highly compressed extreme fibre in concrete subjected to axial compression compressi on and bending, but wh when en there is no tension tension on the section, is taken as 0.0035 minus 0.75 times the strain at the least compressed compressed extreme fibre (see Fig. 2a), highly (9) the maximum compressive strain at the highly compressed compres sed extreme fibre in concrete subjected to axial compression compressi on and bending, when part of the section is 111 tension, is taken as 0.0035 (s (see ee Fig. 2b). In the limiting case, when the neutral axis lies along one edge of the section, the strain varies from 0.0035 0 .0035 at the highly compressed edge to zero at the opposite edge. METHOD

OF ANALYSIS

The criteria determining the ultimate strengthgenerally of R.C.proposed memb members ers forsubjected to axial compression compressio n combined with be bending nding are based on limiting the maximum strain ((or or stress) in the concrete concrete to some some prescribed value. The load-carrying capacities discussed 1121

 

Technical Note

22

Parabolic

0.67

curve

fck E =200000N/mm2

0

Strain

Fig. 3. Idealized stress-strain

0

0.0035

o.DD2

Strain

Fig. 1. Idealized stress-strain

curve for for concrete.

here apply to relatively short columns for whic which h the ef effect fect of lateral deflections on the magnitude of bending moments is negligible. Also effects of sustained load and reversal of bending moments are not considered. In the the pres present ent investigation, a square section of size B x B is considered. considered. If a small square of size B, x B, (B, < B) is removed from the co corner rner of a original square section then it will become a symmetric L-section as show shown n in Fig. 5. For different B, the the L-sectio L-sections of various sizessection. can be obtained. values If B, =of0.0, L-section n become becomes s square Since the ratio of B, /B for all practical purposes varies from 0.3 to 0.6, the present present work is limited to the ratios of B,/B equal to 0.3,0. 0.3,0.4,0.5 4,0.5 and 0.6. The parameters considered are symmetric L-shaped column sections and reinforcement is assumed to be uniformly distributed as a thin strip alon along g all the sides sides with effectiv effectivee cover to depth ratio (E/B) as 0.1. Design charts for co combined mbined axial compressi compression on and bending are obtained in the for form m of interaction diagrams in which curves for P, B* versus M B 3 are plotted for different values of p/f,. When bending bending moments are acting

curve for mild steel bars. bars.

in addition to axial load, the points points for plotting plotting the interaction diagrams are obtained by assuming different positions of neutral axis. For each po position sition of the neutral axis, the the strain dist~bution across the se section ction and the stress block parameters are determined. The sstresses tresses in the reinforcementt are also calculated from the known strains. reinforcemen Thereafter the resultant axial force and the moment about the centroid centroid of the section are calculated as follows. To find the forces forces and moments due to concrete concrete in the the L-section subjected to axial compression compression and bending (both uniaxial and biaxial bending with equal eccentricities e, = eY= eY= e), the following following procedure is used in the analysis. The stress block (see Fig. 2) is divided into number of strips. First the width of each strip is calculated. This strip width is multiplied by corresponding width of the section and depth of the strip, which gives the force force in that strip of concrete. The algebraic sum of all such elemental forces gives the the total force in concrete. This force in concrete concrete multiplied by the distance between centroid of the stress stress block and centroid of the section section gives the moment due to concrete. The forces and mome moments nts due to rei reinforcement nforcement (both for uniaxial and biaxial bending) bending) are determined determined as follows: force in the reinforcement reinforcement = i cf,, -f,)p,A,/lOO ,=I

(la)

moment of resistance with respect to steel = i (~i-fCi)piA,~,/lOO i=I

I[ ‘$ stain

’

(lb)

E =200000N/mm2

2srass

b) neutral axi \ insi de the the se m x,,,~= depth

of neutral

axis

Fig. 2. Stress and strain diagrams.

Fig. 4. Idealized stress-strain curve for high yield strength deformed bars.

 

1123

Technical Note

“\

Major principal axis b

v

‘1. Y-8-“ Y-8-“ ’

reinforcement reinforceme nt in the ith row, yi is the distance of the ith row of reinforcement reinforcement measured from the centroid of the section. section. It is positive towards the highly compres compressed sed edge and negative towards the least compressed compressed edge. INTERACTION

v

/

Minor principal axis

Fig. 5. L-shaped column section. in which n is the number of rows of reinforceme reinforcement, nt, but in the present present work, since the reinforcem reinforcement ent is assumed to be. distributed uniformly as thin strip along all ssides ides of the L-section, the notation n here refers to the number of unit length of the reinforcement strip at which the stresses in steel and concrete (at that level) are to be determ determined, ined, JSi is the stress in the ith row of steel (com (compression pression being positive and tension negative), & is the stres stresss in concrete at the level o off the ith row of reinforcement, A, is the area of concrete, may be taken equal to the gross area, pi = AJAJlOO is the percentage of steel in the ith row, A, is the area of

DIAGRAMS

Because of symmetry, in a square section, the eccentricity on either side of the centre of gravity makes no difference in the approach to interaction curves except when steel is not symmetric, whereas, in the case case of L-shaped column sections the same is not true. For the L-shaped section considered here, the minor principal axis UU-U U and major principal axis V-V are shown shown in Fig. 5, and the interaction curves have been prepared by considering the axis of bending as explained below: Case 1. Uniaxial bending parallel to edge l-2 in compressio compression. n. Case 2. Uniaxial bending parallel to edge l-2 in tension. Case 3. Biaxial bending with equal corner 2 in compressio compression. n. Case 4. Biaxial bending with equal corner 2 in tension.

edge edge 1-2, by treating edge l-2, by treating eccentricities treating eccentricities treating

The four compute computerr programs, namely UNIAXI, UNIAXI, UNIAXZ, BIAXl, and BIAXZ for the above-ment above-mentioned ioned cases 1-4, respectively, are developed by using FORTRAN and are presented in the Appendix. These prog programs rams were used to obtain the ultimate bload (PJ and mome moment nt (M,,) as

Fig. 6. Axial com compression pression with uniaxial bending.

 

1124

Technical Note

 

112s

Technical Note

I 2i i

i i i iY

 

1126

Technical Note

r

d

I I I

I

I

I

 

Technical Note

27

8

5

t CM U/SPA

 

 

28

Technical Note

 

1129

Technical Note

Fig. 17. 17 . Axial co compression mpression with biaxial bending. output for different positions of neutral axes. The input data consists of the square size (B,) of the rremoved emoved portion from the original square section, depth (B) of the section, cover depth (B’), characteristic strength of the concrete (&) and steel (&), and modulus of elasticity of steel (E,). All the loads and moments so obtained are graphically represented in terms of non-dimensional non-dimensional parameters (P&B* versus

compression and biaxial bendin compression bending g with equal eccen eccentricities, tricities, the curves for two cases are shown shown in Figs 1414-15 15 and Figs 16-17. 16-1 7. It is hoped that the charts which are included in this paper, will be useful aids for designers designers and also will bring some attention to the the particular form of resistance exhibited by these cross-sections. cross-sections. It of offers fers the possibility possibility of economizing economizin g and can compleme complement nt the existing design

M&B’)

procedures.

and are shown in Figs 6-17. LIMIT

TIONS

The number of variables considered in this paper are restricted as th there ere was limited space. In tthe he present investigation, the following parameters were considered:

(i) symmetric L-shaped column sections with rein reinforceforcement as a thin strip along all sides, (ii) effective cover to depth ratio (B’/B) is taken as 0.1 in all the cases, (iii) B, /B ratios are 0.0, 0.3, 0.4, 0.5 and 0.6, (iv) the modulus of elasticity of mild steel is taken equal to 200 kN/mm2. CONCLUSIONS The

analysis of reinforced concrete L-shaped column

sections subjected to axial ccompressio ompression n and bending (uniaxial and biaxial) has been computerized. Interaction curves for L-shaped L-shaped column sections under axial axia l compression and uniaxial bending for two two ccases ases are pr presented esented in Figs 6-9 and Figs 10-13. 10-13 . For columns under axia1

Acknowledgemenrs-Partial Acknowledgeme nrs-Partial support of this research by the Structural Engineering Research Centre, Madras, is gratefully acknowledged. The authors would like to express their gratitude to Mr G. V. Surya Kumar, Mr K. Rama Raju and P. Shivakumar, Scientists, SERC, Madras, for their help help in developing the program on a microcomputer. REFERENCES

Mallikarj una and P. Mahadevappa, Computer aided Mallikarjuna analysis of reinforced concrete columns su jec ted to axial compression compression and bending-II. T-shaped se sections. ctions. Comput. Strut., in press. B. Bresler, Design criteria for reinforced concrete columns under axial load and biaxial bending. ACI Slrucf. Jnl SI, 481-490 (1960). A. L. Parme, J. M. Nieves Nieves and A. Gouwens. Capacity of reinforced rectangular columns subject to biaxial bending. AC1 S@uct. Jnf 63 91 l-923 (1966). CP-110, Code of Practice for the Structural use of Concrete: Part 1: Design, Materials and Workmanship;

 

1130

5.

6. 7. 8. 9.

10.

Technical Note

Part 3: 3 : Design Charts for Circular Columns and Prestressed Beams. British Standards Institution, London (1972). ACI-3 18, Building Code Requirements for Reinforced Concrete. American Concrete Institute, Detroit (1983). IS-456, Code of Practice for Plain and Reinforced Concrete. Indian Standards Institution, New Delhi (1978). IS-456, Design Design Aids for R Reinforced einforced Concrete to IS-4561978. Indian Standards Institution, New Delhi (1978) (1978).. R. Park and T. Paulay, Reinforced Concrete Structures, Chap. 5, pp. 158-16 158-160. 0. John Wiley. L. N. Ramamurthy, Investigation of the ultimate strength of square and rectangular columns under biaxially eccentric loads. Symposium on Rei Reinforced nforced Concrete Columns, Detroit, ACI-SP-13, Paper No. 12, 263-298 ( 1966). K. H. Kwan and T. C. Liauw, Computer aided design of reinforced concrete members subjected to axial

1. 12. 13. 14. 15. 16. 17.

compression and biaxial bending. Slrucf. Engineer, 63B, compression 34-40 (1985). E. Czerniak, Analytical approach to biaxial eccentricity eccentricity.. Proc. ASCE, J. Struct. Div. 88. 105-158 (19621. P. W. R Reed, eed, Simplified analysis.for thrust gnd moment moment of concrete sections. ACI Sbuct. Jnl77,195-200 1980). L. L Lachance, achance, Stress distribution in reinforced concrete sections subjected to biaxial bending. AU Struct. Jnl 77. 116-123 (1980). A..K. Jain, J ain, R‘einfoked Concrete. Nem Chand, Roorkee (1983). Mallikarjuma, Mallikarjum a, Computer aided analysis and design design (in structural analysis). Research Report, Structural Struct ural Engineering Research Centre, Madras (1985). Mallikarjuna, Mallikarju na, A computer program for de design sign of reinforced concrete members. Indian Concrete Journal 63, 101-105 (1989). P. Purushothaman, Reinforced Concrete Slructural (1984) . Elements. McGraw-Hill, New Delhi (1984).

APPENDIX c*****************************************~~~~**~~~~~~~~~~~~~~~~

ANAIYSIS OF L-SHAPED COLUMN SECTIONS UNDER AXIAL C COMPRESSION AND UNIAXIAL BENDING (See Figs. 6-9) C**********************************************~****~~*-*~~~~~~~~ C

OPEN (S,FILE=‘UNIAXl .DAT’ ) OPEN (6, FILE=’ UNIAXl . UT’ , STATUS=’ DO 299 I=l,lO

C 10

20 30

40

45 50

60

65

NEW’ )

RFAD(5,5,END=3OO)Bl,B,DC,FCK,FY,ES FORMAT(SF6.2,F10.2) P=O.O P=P+O.5 XU=DC xu=xu+5.0 FK=O. 446*FCK Y1=(O.5*B*(B-Bl)+Bl*(B-O.5*Bl))/(B+B1) A=B* (B-Bl) +Bl* (B-Bl) AS=P*A/lOO.O RL=4.O*B-8.0*DC TS=AS/RL DETERMINATION DETERMINATIO N OF FORCES DUE TO CONCRETE IF (XU-B) 10,10 10,10,20 ,20 x1=3.0*xu/7.0 X2=4.O*XU/7.0 GO TO 30 X1=3.0*B/7.0 X2=4.0*B/7.0 PC=O.O AMc=o.o B5= (B-Bl) IF (Xl-B5 (Xl-B5)) 40,70,70 Cl=B*Xl*FK BMCl=Cl* (Yl-0.5*X1) x=0.0 x=x+1.0 BG=B-Bl-Xl IF (X-B6) 50,50,60 F=FK-FK*X*X/ ( (XU-Xl) * (XU-Xl) PCl=B*F AMcl=Pcl* (Yl- (x1+x) ) GO TO 65 F=FK-FK*X*X/ ( (XU-Xl) * (XU-Xl) PCl= (B-Bl) *F AMc1=pc1* (Yl- (x1+x) ) PC=PCtPCl AMC=AMC+AMCI IF(X.LT.X2) IF(X.LT .X2) GO TO 45 TPC=PCtCl

)

)

 

Technical ote

70

75

C C

80 90

95

96 97

100 110

115

C

116

117 118

TAMC=AMC+BMCl GO TO 80 Cl=(B-Bl)*B*FK BMCl=Cl*(Yl-O.S*(B-Bl)) C2=(B-Bl)*(Xl-(B-Bl))*FK AMC2=C2*fY1-(0.5*Xl-0.5*(B-B1))) x=0.0 x=x+1.0 F=FK-FK*X*X,'((XU-Xl *(XU-Xl)) PCl=(B-Bl)*F AMc1=Pc1*(Y~-(x1+x)) PC=PC+PCl AMC=AMC+AMCl IF(X.LT.X2) GO TO 75 TPC=Cl+C2+PC TAMC=BMCl+AMC: tAMC DETERMINATION OF FORCES IN COMPRESSION REINFORCEMENT B7=B-DC IF(XWB7) 90,160,160 PSC=O.O AMsc=o.o PSCl-0.0 AMsc1=0.0 D=O.O D=D+l.O EC=O.O035*D/XU IF(FY.EQ.250.0)GO TO 96 IF(EC.GE.O.0038) GO TO 96 IF(EC.LE.0.00145) GO TO 96 FC=FY/1.15-12831145.0*~0.0038-EC)*(O.O038-EC) GO TO 97 FC=EC*ES Fl=FY/1.15 IF(FC.GE.Fl) FC=Fl B8=XU-(B-Bl-DC) IF(D-B8) 110,100,110 PSCl=Bl*TS*(FC-FK) AMSCl=PSCl*(Yl-(XU-D)) GO TO 115 PSC2=2.0*TS*(FC-FK) AMSC2=PSC2*(Yl-(XU-D)) PSC=PSCtPSC2 AMsc=AmctAMsc2 B9=XU-DC IF(D.LT.B9) GO TO 95 PSC3=(B-2.0*DC)*TS*(FC-FK) ~SC3=PSC3*(Yl-(XU-D)) TPSC-PSCtPSCltPSC3 TAMSC=AMSCtAMSCltAMSC3 DETERMINATION OF FORCES DUE TO TENSILE REINFORCEMENT Zl=(B-DC-XU) PST=O.O AMST=O.O PSTl=O.O AMSTl=O.O z=o.o Z=Ztl.O ET=O.O035*Z/XU IF(FY.EQ.250.0) GO TO 117 IF(ET.GE.0.0038) GO TO 117 IF(ET.LE.0.00145) GO TO 117 FT=FY/1.15-12831145~0*(0.~038-ET)*(0.0038-~T) GO TO 118 FT=ET*ES Fl=FY/l.lS IF(FT.GE.Fl) FT=Fl Z2=fB-Bl-DC-XU) IF(Z-Z2) 120,130,120

1131

 

1132 130 120

125

160

165

168

Technical ote PSTl=-Bl*TS*FT AMSTl=PSTl*(Yl-(XU+Z)) GO TO 125 PST2=-2.0*TS*FT ~ST2=PST2*(Y~-(XU+Z)) PST=PST+PSTZ AMST=AMST+AMST2 XF(Z.LT.21) GO TO 116 PST3=-(B-2.O*DC)*TS*FT AMST3=PST3*(Yl-(XU+Z)) TPST=PST+PSTl+PST3 T~ST=~STi~STl~~ST3 GO TO 200 FT=O.O PSC=O.O AMsc=o.o PSC2=0.0 AMsc2=0.0 Dl=(XU-(B-DC)) D=Dl-1.0 D=D+l.O EC-O. 002*D/(XU-Xl~ 002*D/(XU-Xl~ IF(FY.EQ.250.0) IF(FY.EQ.250. 0) GO TO 168 XF(EC.GE.0.0038) XF(EC.GE.0.00 38) GO TO 168 IF(EC.LE.0.00145) IF(EC.LE.0.00 145) GO TO 168 FC=FY/l.l5-~2831145.0*(0.00 FC=FY/l.l5-~2 831145.0*(0.0038-EC)*(O.O03 38-EC)*(O.O038-EC) 8-EC) GO TO 169 FC=EC*ES Fl=FY,'1.15

299

IF(FC.GE.Fl)FC=Fl IF(D-Dl) x66,167,166 PSC3=(3-Bl-Z.O*DC)*TS*(FC-FK) AMsc3=Psc3*(Yl-(xu-D)) GO TO 165 Bll=XU-(B-Bl-DC) XF(D-Bll) 170,180,170 PSCZ=Bl*TS*(FC-FK) AMSCZ=PSC2*(Yl-(XU-D)) GO TO 175 PSC4=2.0*TS*(FC-FK) ~SC4=PSC4*(Yl-(XU-D)~ PSC=PSC+PSC4 AMSC=AMSC+AMSC4 B12=XU-DC IF(D.LT.BlZ) GO TO 165 PSCl=(B-2.0*DC)*TS*(FC-FK) AMSCl=PSCl*(Yl-(Xu-D)) TPSC~PSC+PSC1+PSc2+Psc3 TAM~C=AM~~+F,M~~~+?~~~~+AMSC~ TPST=O.O TAMST=O.O TF=TPSC+TPST+TPC TM=TAMSC+TAMST+TAMC E=TM/TF XC~TM/(FCK*B*B*B~ YC=TF/ (FCK"l3"I.S) PI-P/FCK WRITE(6,6) xU,Pl,FT,XC,TC xU,Pl,FT,XC,TC Fo~(~X,'X~=;',F~.~,~X,'P~= Fo~(~X,'X~=;' ,F~.~,~X,'P~=',F~.~,~X,'FT ',F~.~,~X,'FT=',F~.~, =',F~.~, SX,'XC=', 5.4,5X,'YC=',F5.4/) XuMAX=2*0*B ~F(XU.LE.XU~) GO TO 8 IF(P.LT.4.0) GO TO 7 CONTLNUE

300

STOP END

169 167 166 180 170

175

200

6

 

Technical ote C************************************* C************************ ***********************~*** **********~*************** ************ ANAIXSIS OF L-SHAPED COLUMN SECTIONS UNDER AXIAL COMPRESSION AND UNIAXIAL BENDING (See Figs. 10-13) ; c************************ c*********** ************************** *************************** ************************** ************ OPEN(5,FILE='UNIAX2.DAT') OPEN(6,FILE='UNIAX2.OUT',ST OPEN(6,FILE='U NIAX2.OUT',STATUS='NEW') ATUS='NEW') DO 299 1=1,10 READ(S,S,END=300)Bl,B,DC,FCK,FY,ES FORMA.T(SF6.2,F10.2) P=O.O P=P+O.S XU=DC xu=xu+5.0 FK=O. 446*FCK A=B*(B-Bl)+Bl*(B-Bl) Y1=B-(O.5*B*(B-Bl)+Bl*(B-O Y1=B-(O.5*B*( B-Bl)+Bl*(B-O.5*B1))/(B+Bl .5*B1))/(B+Bl) ) AS=P*A/100.0 RL=4.0*B-8.0*DC TS=AS/RL DETERMINATION OF FORCES DUE TO CONCRETE C IF(XU-B) 10,10,20 10 x1=3.0*xu/7.0 xz=xu-Xl GO TO 30 20 X1=3.0*B/7.0 X2-B-X1 30 TPC=O.O TAMC=O.O x-o.0 35 40 50 7": 100 130

C

140 150 160

168 169

x=x+1.0 IF(X-X1)40,40,50 F=FK GO TO 60 F=FK-FK*(X-X1)*(X-Xl)/(XU F=FK-FK*(X-X 1)*(X-Xl)/(XU-Xl)/(XU-Xl) -Xl)/(XU-Xl) IF(X-Bl) 70,100,100 W=B-Bl GO TO 130 W=B PC=W*F AMc=Pc*(Yl-X) TPC=TPC+PC TAMC=TAMC+AMC BS=Xl+X2 IF(X.LT.B5) GO TO 35 DETERMINATION OF OF FORCES IN COMPRESSION COMPRESSION REINFORCEMENT TPSC=0.0 TAMSC=O.O B6=B-DC IF(XU-B6)140,150,150 D=O.O GO TO 160 D=XU-B6-1.0 D=D+l.O EC=O.O02*D/(XU-Xl) IF(FY.EQ.250.0) IF(FY.EQ.250. 0) GO TO 168 IF(EC.GE.0.0038) IF(EC.GE.0.00 38) GO TO 168 IF(EC.LE.0.00145) IF(EC.LE.0.00 145) GO TO 168 FC=FY/l.lS-12831145.0*(0. FC=FY/l.lS-1 2831145.0*(0.0038-EC)*(0.0 0038-EC)*(0.0038-EC) 038-EC) GO TO 169 FC=EC*ES Fl=FY/l.lS IF(FC.GE.Fl) FC-Fl B7=XU-B+DC B8=XU-Bl-DC Bll=XU-DC IF(D.EQ.B7) GO TO 180 IF(D.EQ.Bll) GO TO 200 IF(D.EQ.BE) GO TO 190 WS=2.0*TS

1133

 

1134

Technical ote

180 190 200 210

c

215

216 217

230 240 260

265

6

299 300

GO TO 210 WS=(B-2.0*DC)*TS GO TO 210 WS=Bl*TS GO TO 210 WS=(B-Bl-2.0*DC)*TS PSC=WS*(FC-FK) AMSC=PSC*(Yl-(XU-D)) TPSC=TPSC+PSC TAMSC=TAMSC+AMSC IF(D.LT.Bll) GO TO 160 DETERMINATION OF FORCES IN TENSION STEEL FT=0.0 TPST=O.O TANST=O.O IF(XU.GE.BG)GO TO 265 z-0.0 Z=Z+l.O ET=O.O02*Z/(XW-Xl) IF(FY.EQ.250.0) GO TO 216 IF(ET.GE.0.0038) GO TO 216 IF(ET.LE.0.00145) GO TO 216 FT=FY/1.15-12831145.0*(0.0038-ET)*(O.O038-ET) GO TO 217 FT=ET*ES Fl=FY/1.15 IF(FT.GE.Fl) FT-Fl B12=Bl+DC-XU B14=B-DC-XU IF(Z.EQ.Bl2) GO TO 230 IF(Z.EQ.Bl4) GO TO 240 WS=2.0*TS GO TO 260 WS=Bl*TS GO TO 260 WS=(B-2.0*DC)*TS PST=-WS*FT AMST=PST*(Yl-(XUtZ)) TPST=TPSTtPST TAMST=TAMSTtAMST IF(Z.LT.Bl4) GO TO 215 TF=TPCtTPSCtTPST TM=TAMCtTAMSC+TAMST E=TM/TF Pl=P/FCK XC=TM/(FCK*B*B*B) YC=TF/(FCK*B*B) WRITE(6,6) XU,Pl,FT,XC,TC FO~(5X,'XU=',F6.l,SX,'Pl=',F5.4,5X,'FT=',F5.1, SX,'XC=',F6.4,5X,'YC=',F6.4/) XUMAX=2.0*B IF(XU.LT.XUMAX) GO TO 8 IF(P.LT.4.0) GO TO 7 CONTINUE STOP END

c********************************* c**************** ********************************** ******************************** ***************

C

ANALYSIS OF L-SHAPED COLUMN SECTIONS SUBJECTED TO AXIAL COME'RESSION AND BIAXIAL BENDING WITH EQUAL ECCENTRICITIES (See Figs.14 and 15) ***************************x****xx************************~***~~*

5 7

OPEN(5,FILE='BI~l.DAT'~ OPEN(6,FILE='BI~l.OUT',STATUS='NEW') DO 299 I=l,lO READ(5,5,END=300)Bl,B,DC,FCK,FY,ES FORMAT(5F6.2,F10.2) P=O.O P=P+O.5

 

Technical ote

8

C

10

20 30 35 40 50 60 70 80 90 100 110 120 130

140 150 160

168 169

180

XU=DC xu=xu+5.0 FK=0.446*FCK A=B*(B-Bl)+Bl*(B-Bl) Y1=(B*B*B*0.707-B1*Bl*Bl* Y1=(B*B*B*0. 707-B1*Bl*Bl*O.7O7-B1*B1*( O.7O7-B1*B1*(B-B~)*~.4~4)/A B-B~)*~.4~4)/A AS=P*A/lOO.O RL=4.0*B-E.O*DC TS=AS/RL DETERMINATION DETERMINATIO N OF FORCES DUE TO CONCRETE B2=(2.0*B-B1)*0.707 IF(XU-B2) 10,10,20 x1=3.0*xu/7.0 x2=xu-Xl GO TO 30 X1=3.0*B2/7.0 X2=B2-Xl TPC=O.O TAMC=O.O x=0.0 x=x+1.0 IF(X-X1)40,40,50 F=FK GO TO 60 F-FK-FK*(X-X1)*(X-Xl)/(XUF-FK-FK*(X-X1 )*(X-Xl)/(XU-Xl)/(XU-Xl) Xl)/(XU-Xl) B3=1.4142*(B-Bl) B4=0.707*B IF(X-B3) 70,70,100 IF(X-B4) 80,80,90 w=2.o*x GO TO 130 W=2.0*(1.4142*B-X) GO TO 130 IF(X-B4) 110,110,120 W=2.0*1.4142*(B-Bl) GO TO 130 W=(2.0*B-Bl-X*1.4142)*2.0*1.4142 PC=W*F AMC=PC"(Yl-X) TPC=TPC+PC TAMC=TAMC+AMC B5=Xl+X2 IF(X.LT.B5) GO TO 35 DETERMINATION OF FORCES IN COMPRESSION COMPRESSION REINFORCEMENT TPSC=O.O TAMSC=O.O B6=(2.0*B-Bl-2.0*DC)*O.7071 IF(XU-B6)140,150,150 D=O.O GO TO 160 D=XU-B6-1.0 D=D+l.O EC=O.O02*D/(XU-Xl) IF(FY.EQ.250.0) GO TO 168 IF(EC.GE.0.0038) GO TO 168 IF(EC.LE.0.00145) GO TO 168 FC=FY/l.l5-12831145.0*(0. FC=FY/l.l5-1 2831145.0*(0.00~8-EC)*(O.O 00~8-EC)*(O.O038-EC) 038-EC) GO TO 169 FC=EC*ES Fl=FY/l.l5 IF(FC.GE.Fl) FC=Fl B7=B*0.7071 BE=(B-Bl-DC)*1.4142 B9=XU-B7 BlO=XU-BE Bll=XU-1.4142*DC IF(D-BlO) 180,180,190 WS=4.0*TS*1.4142 GO TO 210

35

 

1136

Technical ote 190 210

C

215

216 217

230 240 260

265

6

299 300

WS=2.O*TS*1.414 PSC=WS*(FC-FK) AMSC=PSC*(Yl-(XU-D)) TPSC=TPSCtPSC TAMSC-TAMSCtAMSC IF(D.LT.Bll) GO TO 160 DETERMINATION OF FORCES IN TENSION STEEL FT=O.O TPST=O.O TAMST=O.O IF(XU.GE.BG)GO TO 265 z=o.o Z=Z+l.O ET=O.O02*Z/(XU-Xl) IF(FY.EQ.250.0) GO TO 216 IF(ET.GE.0.0038) GO TO 216 IF(ET.LE.0.00145) GO TO 216 FT=FY/1.15-12831145.0*(0.0038-ET)*(O.O038-ET) GO TO 217 FT=ET*ES Fl=FY/1.15 IF(FT.GE.Fl) FT=Fl B12=-B9 B13=-BlO B14=B6-XU IF(Z-B13) 230,230,240 WS=2.O*TS*1.4142 GO TO 260 WS=4.0*TS*1.4142 PST=-WS*FT AMST=PST*(Yl-(XUtZ)) TPST=TPST+PST TAMST=TAMSTtAMST IF(Z.LT.Bl4) GO TO 215 TF=TPCtTPSCtTPST TM=TAMCtTAMSC+TAMST E=TM/TF Pl=P/FCK XC=TM/(FCK*B*B*B) YC=TF/(FCK*B*B) WRITE(6,6) XU,Pl,XC,TC FORMAT(5X,'XU=',F7.1,5X,'Pl=',F8.5, 5X,'XC=',F6.4,5X,'YC=',F6.4/) XUMAX=1.5*B IF(XU.LT.XUMAX) GO TO 8 IF(P.LT.4.0) GO TO 7 CONTINUE STOP END

C

c********************************************************* ANAIXSIS OF L-SHAPED COLUMN SECTIONS SUBJECTED TO AXIAL COMPRESSION AND BIAXIAL BENDING WITH EQUAL C ECCENTRICITES (See Figs.16 and 17) C c***********************************************************

5 7 8

OPEN(5,FILE='BIAX2.DAT') OPEN(6,FILE='BIAX2.0UT',STATUS='NEW') DO 299 I=l,lO READ(5,5,END=300)Bl,B,DC,FCK,FY,ES FORMAT(5F6.2,F10.2) P=O.O P=P+O.5 XU=DC xu=xu+5.0 FK=0.446*FCK A=B*(B-Bl)tBl*(B-Bl) Y1=(B*B*B*0.707-B1*Bl*Bl*O.707-Bl*Bl*(B-Bl)*l.4l4)/A AS=P*A/lOO.O

 

Technical Note

10 20 30

35 40 50 60

70 80

90

100

110 120 130

140 150 160

168 169

180 190 210

RL=4.0*B-8.0*DC TS=AS/RL DETERMINATION OF FORCES DUE TO CONCRETE B2=(2.O*B-Bl)*O.707 Y2=B2-Yl IF(XU-B2) 10,10,20 x1=3.0*xu/7.0 x2=xu-Xl GO TO 30 x1=3.0*B2/7.0 X2=B2-Xl TPC=O.O TAMC=O.O x=0.0 x=x+1.0 IF(X-X1)40,40,50 F=FK GO TO 60 F=FK-FK*(X-X1) * (X-Xl) / (XU-Xl) /(XU-Xl) B3=1.4142*(B-Bl) B4=0,707*B B31=B2-B3 B41=B2-B4 IF(X-B31) 70,70,100 IF(X-B41) 80,80,90 w=4.o*x GO TO 130 W=(B-B1)*1.4142*2.0 GO TO 130 IF(X-B41) 110,110,120 W=(1.4142*B-B2tX)*2.0 GO TO 130 W=(B2-X)*2.0 PC=W*F AMc=PC*(Y2-X) TPC=TPCtPC TAMC-TAMCtAMC BS=XltX2 IF(X.LT.BS) GO TO 35 DETERMINATION OF FORCES IN COMPRESSION REINFORCEMENT TPSC=O.O TAMSC=O.O B6=(2.0*B-Bl-2.0*DC)*O.7071 IF(XU-B6)140,150,150 D=O.O GO TO 160 D=XU-B6-1.0 D=Dtl.O EC=O.O02*D/(XU-Xl) IF(FY.EQ.250.0) GO TO 168 IF(EC.GE.0.0038) GO TO 168 IF(EC.LE.0.00145) GO TO 168 FC=FY/l.lS-12831145.0*(0.0038-EC)*(O.O038-EC) GO TO 169 FC=EC*ES Fl=FY/l.lS IF(FC.GE.Fl) FC=Fl B7=B*0.7071 B8=(B-Bl-DC)*1.4142 B9=XU-B7 BlO=XU-B2tB8 Bll==XU-1.4142"DC IF(D-BlO) 180,180,190 WS=2.0*TS*1.4142 GO TO 210 WS=4.0*TS*1.414 PSC=WS*(FC-FK) AMSC=PSC*(Y2-(XU-D)) TPSC=TPSC+PSC

37

 

1138

Technical Note

215

216 217

230 240 260

265

6

299 300

TAMSC=TAMSC+AMSC IF(D.LT.Bll) GO TO 160 DETERMINATION OF FORCES IN TENSION STEEL STE EL FT=O.O TPST=O.O TAMST=O.O IF(XU.GE.B6)GO TO 265 Z=O.O Z=Z+l.O ET=O.O02*Z/(XU-Xl) IF(FY.EQ.250.0) GO TO 216 IFfET.GE.0.0038) GO TO 216 IF(ET.LE.O.00145) GO TO 216 FT=FY/l.l5-12831145.0*(0.0038-ET)*(O.O038-ET) GO TO 217 FT=ET*ES Fl=FY/l.lS IF(FT.GE.Fl) FT=Fl B12=-B9 B13=B2-B8-XU Bl4=B6-XU IF(Z-B13) 230,230,240 WS=4.O*TS*1.4142 GO TO 260 WS=2.O*TS*1.4142 PST=-WS*FT ~ST=PST*(Y2-(XU+Z)) TPST=TPST+PST TAMST=TAMST+AMST IF(Z.LT.Bl4) GO TO 215 TF=TPC+TPSCtTPST TM=TAMCtTAMSC+TAMST E=TM/TF Pl=P/FCK XC=TM/(FCK*B*B*B) YC=TF/(FCK*B*B) WRITE(6,6) XU,Pl,XC,YC FORH&T(5X,'XU=',F7.1,5X,'Pl=',F8.5, 5X,'XC=',F6.4,5X,'YC=',F6.4/) XUMAX=1.5*B IF(XU.LT.XUM?iX) GO TO 8 IF(P.LT.2.0) GO TO 7 CONTINUE STOP END