curvature ductility of rc sections based on eurocode: analytical procedure

KSCE Journal of Civil Engineering (2011) 15(1):131-144DOI 10.1007/s12205-011-0729-4

− 131 −

www.springer.com/12205

Structural Engineering

Curvature Ductility of RC Sections Based on Eurocode: Analytical Procedure

Srinivasan Chandrasekaran*, Luciano Nunziante**, Giorgio Serino***, and Federico Carannante****

Received October 12, 2008/Accepted March 16, 2010

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Abstract

Correct estimate of curvature ductility of reinforced concrete members has always been an attractive subject of study as itengenders a reliable estimate of capacity of buildings under seismic loads. The majority of the building stock needs structuralassessment to certify their safety under revised seismic loads by new codes. Structural assessment of existing buildings, byemploying nonlinear analyses tools like pushover, needs an accurate input of moment-curvature relationship for reliable results. Inthe present study, nonlinear characteristics of constitutive materials are mathematically modelled according to Eurocode, currently inprevalence and analytical predictions of curvature ductility of reinforced concrete sections are presented. Relationships, in explicitform, to estimate the moment-curvature response are proposed, leading to closed form solutions after their verification with thoseobtained from numerical procedures. The purpose is to estimate curvature ductility under service loads in a simpler closed formmanner. The influence of longitudinal tensile and compression steel reinforcement ratios on curvature ductility is also examined anddiscussed. The spread sheet program used to estimate the moment-curvature relationship, after simplifying the complexities involvedin such estimate, predicts in good agreement with the proposed analytical expressions. Avoiding somewhat tedious hand calculationsand approximations required in conventional iterative design procedures, the proposed estimate of curvature ductility avoids errorsand potentially unsafe design.Keywords: analytical solutions, concrete, curvature ductility, elasto plastic, reinforced concrete, seismic, structures, yield

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1. Introduction

The focus of earthquake resistant design of Reinforced Concrete(RC) framed structures is on the displacement ductility of thebuildings rather than on the materials like reinforcing steel.Critical points of interest are the strain levels in concrete andsteel, indicating whether the failure is tensile or compressive atthe instant of reaching plastic hinge formation (Pisanty andRegan, 1998). Studies show that the estimate of ductility demandis of particular interest to structural designers to ensure effectiveredistribution of moments in ultra-elastic response, allowing forthe development of energy dissipative zones until collapse (see,for example, Pisanty and Regan, 1993). In areas subjected toearthquakes, a very important design consideration is the ductilityof the structure because modern seismic design philosophy isbased on energy absorption and dissipation by post-elastic defor-mation for survival in major earthquakes (Paulay and Priestley,1992). Many old buildings show their structure unfit to supportseismic loads demanded by the structural assessment requests ofthe revised international codes (see, for example, Chandrasekaranand Roy, 2006; Chao Hsun Huang et al., 2006). Further, Sinan

and Metin (2007) showed that the deformation demand pre-dictions by improved Demand Capacity Method are sensitive toductility as higher ductility results in conservative predictions.Estimate of moment-curvature relationship of RC sections hasbeen a point of research interest since many years (Pfrang et al.,1964; Carrreira and Chu, 1986; Mo, 1992); historically, moment-curvature relationships with softening branch were first intro-duced by Wood (1968). Load-deformation characteristics of RCstructural members, bending in particular, are mainly dependenton moment-curvature characteristics of the sections as most ofthese deformations arise from strains associated with flexure(Park and Paulay, 1975). As seen from the literature, in well-designed and detailed RC structures, the gap between the actualand design lateral forces narrows down by ensuring ductility inthe structure (see, for example, Luciano and Raffaele, 1988;Pankaj and Manish, 2006). With regard to RC building frameswith side-sway, their response assessment is complicated notbecause of the influence of second order deformations, but alsodue to the fact that considerable re-distribution of moments mayoccur due to plastic behaviour of sections. Plastic curvature istherefore a complex issue mainly because of interaction of various

*Associate Professor, Dept. of Ocean Engineering, Indian Institute of Technology Madras, Chennai 600036, India (Corresponding Aughor, E-mail:[email protected])

**Professor, Dept. of Structural Engineering, University of Naples Federico II, 21 via Claudio, 80125, Naples, Italy (E-mail: [email protected])***Professor, Dept. of Structural Engineering, University of Naples Federico II, 21 via Claudio, 80125, Naples, Italy (E-mail: [email protected])

****Visiting Researcher, Dept. of Structural Engineering, University of Naples Federico II, 21 via Claudio, 80125, Naples, Italy (E-mail: [email protected])

Srinivasan Chandrasekaran, Luciano Nunziante, Giorgio Serino, and Federico Carannante

− 132 − KSCE Journal of Civil Engineering

parameters namely: i) constitutive material’s response; ii) mem-ber geometry; as well as iii) loading conditions. Observationsmade by Challamel and Hjiaj (2005) on plastic softening beamsshow that the correct estimate of yield moment, a non-localmaterial parameter, is important to ensure proper continuitybetween elastic and plastic regions during the loading process.Experimental evidences on moment-curvature relationship ofRC sections already faced limited loading cases and supportconditions (see, for example, Ko et al., 2001). While Mo (1992)suggested classical approach to reproduce moment-curvaturerelationship with the softening branch carried out elastic-plasticbuckling analysis using finite element method, an alternativeapproach proposed by Jirasek and Bazant (2002) uses a simpli-fied model where this complex nonlinear geometric effect isembedded in the nonlinear material behaviour of the crosssection. Experimental investigations also impose limitations inestimating the plastic rotation capacity. For instance, studiesshow that experimental results obtained from rotation-deflectionbehaviour show good agreement with the analysis in elasticregime; but for phase of yielding of reinforcing steel, theoreticalresults do not agree with the experimental inferences (see, forexample, Lopes and Bernardo, 2003).

Studies reviewed above show that there exists no simplifiedprocedure to estimate curvature ductility of RC sections. While re-sponse of RC building frames under ground shaking generallyresults in nonlinear behaviour, increased implementation of displa-cement-based design approach lead to the use of nonlinear staticprocedures for estimating their seismic demands (ATC, 2005;BSSC, 2003). An estimate of moment-curvature relationship be-comes essential for performing non-linear analyses. Therefore, inthis study, an estimate of curvature ductility of RC sections, usingdetailed analytical procedure is attempted. Calculations of moment-curvature relationship are based on their nonlinear characteristicsin full depth of the cross section, for different ratios of longitudinaltensile and compression reinforcements. They account for the vari-ation on depth of neutral axis passing through different domains,classified on the basis of strain levels reached in the constitutivematerials, namely concrete and steel. Obtained results, by employ-ing the numerical procedure on example RC sections, are verifiedwith expressions derived from detailed analytical modelling.

2. Mathematical Development

Significant nonlinearity exhibited by concrete, under multi-axial stress state, can be successively represented by nonlinearcharacteristics of constitutive models capable of interpretinginelastic deformations (see, for example, Chen 1994a, 1994b).Studies conducted by researchers (Sankarasubramanian andRajasekaran, 1996; Fan and Wang, 2002; Nunziante et al., 2007)describe different failure criteria in stress space by a number ofindependent control parameters while the non-linear elasticresponse of concrete is characterized by parabolic stress-strainrelationship in the current study, as shown in Fig. 1. Elastic limitstrain and strain at cracking are limited to 0.2% and 0.35%respectively, as prescribed by the code, currently in prevalence(DM 9, 1996; UNI ENV, 1991a, 1991b; Ordinanza, 2003, 2005;Norme tecniche, 2005). Tensile stresses in concrete are ignoredin the study. Design ultimate stress in concrete in compression isgiven by:

(1)

where, γc and Rck are the partial safety factor and compressive cubestrength of concrete, respectively. The stress-strain relationship forconcrete under compressive stresses is given by:

(2)

where, parameters a, b and c in Eq. (2), are determined byimposing the following conditions:

(3)

By solving, we get:

(4a)

σc00.83( ) 0.85( )RcK

γc--------------------------------------=

σc εc( ) aεc2 bεc c+ +=

σc εc( ) σc0=σc εc( ) 0=

0 εc εc0≤ ≤

εc0 εc εcu≤ ≤

εc 0≤

σc εc 0=( ) 0=σc εc εc0=( ) σc0=

dσc

dεc--------

εc εc0=0=

c 0=

aεc02 bεc0 σc0=+

2aεc0 b 0=+

⇒

aσc0

εc02

------- b2σc0

εc0---------- c 0=,=,=

Fig. 1. Stress-strain Relationships: (a) Concrete, (b) Steel


Vol. 15, No. 1 / January 2011 − 133 −

Stress-strain relationship for concrete is given by:

(4b)

Stress-strain relationship for steel, an isotropic and homogene-ous material, is shown in Fig. 1. While the ultimate limit strain intension and that of compression are taken as 1% and 0.35%respectively, elastic strain in steel in tension and compression areconsidered the same in absolute values (see, for example, DM9,1996). The design ultimate stress in steel is given by:

(5)

where γs and σy are partial safety factor and yield strength ofreinforcing steel, respectively. Stress-strain relationship for steelis given by:

(6)

The fundamental Bernoulli’s hypothesis of linear strain overthe cross section, both for elastic and for elastic-plastic responsesof the beam under bending moment combined with axial force,will be assumed. The interaction behaviour becomes critical whenone the following conditions apply namely: i) strain in reinforc-ing steel in tension reaches ultimate limit; ii) strain in concrete inextreme compression fibre reaches ultimate limit; as well as iii)maximum strain in concrete in compression reaches elastic limitunder only axial compression. In the following section, only re-ctangular RC sections under axial force, P and bending moment,M will be considered.

2.1 Moment-curvature in Elastic RangeIt is well known that the bending curvature is the derivative of

bending rotation, varying along the member length and at anycross section, it is given by the slope of the strain profile. It dependson the fluctuations of the neutral axis depth and continuouslyvarying strains. The moment-curvature relationship, in elasticrange, depends on both the magnitude and nature of the axial forceas well. Fig. 2 shows the variation of curvature with respect tostrain variation in constitutive materials. Magnitude of axial forceis assumed to vary in the range:

− (Asc + Ast)σs0 < P < {bDσc0 + (Asc + Ast)σs0} (7a)

Nature of axial force shall vary as: i) tensile axial force (con-sidered as negative in this study); ii) zero axial force; as well asiii) compressive axial force (considered positive). Stress andstrain in concrete and steel, in elastic range are given by:

(7b)

2.1.1 Tensile Axial ForceTensile axial force results in reduced curvature for which axial

force and bending moment, in explicit form, are given by:

(8)

(9)

Percentage of steel, in tension and compression zones, is givenby:

(10)

By solving the Eq. (8) respect to xc, we obtain the followingrelationship:

(11)

By substituting the Eq. (11) in Eq. (9), moment-curvature re-lationship is given by:

(12)

where, φ0 is the limit curvature for xc = 0; by imposing this con-dition in Eq. (11), we get:

(13)

As curvature is influenced by percentage of tension reinforce-ment, by imposing the conditions: inEq. (8) and solving with respect to pt, for a specified range of:

(14)

Eq. (12) is defined in the total range [0, φE], where φE is the

σcσc0

εc02

-------εc2 2σc0

εc0----------εc+–=

σs0σy

γs-----=

σs εs( ) Esεs=σs εs( ) σs0=σs εs( ) σ– s0=

0 εs εs0≤ ≤

εs0 εs εsu tensile,≤ ≤

εsu compressive, εs εs0–≤ ≤– εsu tensile, εsu=( )

εc φe xc y–( ) ;= εsc φe xc d–( );= εst φe D xc d––( ) ;=

σcxc y–( ) 2εc0 xc y–( )σc0φe–[ ]

εc02

---------------------------------------------------------------- ;= σsc Esφe xc d–( ) ;=

σst Esφe D xe– d–( ) =

Pe σ– stAst σscAsc b d D–( )=+=d pc pt–( ) Dpt pc pt+( )xc–+[ ]Esφ

Me σstAst σscAsc+( ) D2---- d–⎝ ⎠⎛ ⎞ 1

2---b D 2d–( ) D d–( )==

pc pt–( )xc Dpt d pc pt+( )–+[ ]Esφ

Ast ptb D d–( ) ;= Asc pcb D d–( )=

xcPe b d D–( ) d pc pt–( ) Dpt+[ ]Esφ+

b d D–( ) pc pt+( )Esφ-------------------------------------------------------------------------------=

MeD 2d–

2 pc pt+( )--------------------- [Pe pc pt–( ) 2b(D2 d2 3dD)Espcptφ]–++=

φ 0 φ0,[ ]∈∀

φ0Pe

b d D–( )Es Dpt d pc pt–( )+[ ]-----------------------------------------------------------------=

xc 0 & φ εs0 D d–( )⁄==

pt Pe bdEspcεs0+( ) b d D–( )σs0( )⁄<

Fig. 2. Curvature Profile for Strain Variation in Concrete and Steel



limit elastic curvature and is derived in following section.For further increase in curvature more than ϕ0, concrete also

contributes to the compression resultant and the expressions foraxial force and bending moment take the form, as given below:

(15)

,

(16)

where, the coefficients Ai (for i = 0 to 3) and Bi (for i = 0 to 4), asa function of curvature are given by:

(17)

By solving Eq. (15) with respect to variable xc, three roots ofthe variable are obtained as:

(18)

where,

(19)

Out of the above, only one root, namely xc3, closely matcheswith the numerical solution obtained and hence by substitutingthe root xc3 in Eq. (18), moment-curvature relationship in elasticrange is obtained as:

(20)

2.1.2 Axial Force Equal to ZeroThe moment-curvature relationship is given by Eq. (20) for the

complete of [0, φE].

2.1.3 Compressive Axial ForceExpressions for axial force and bending moment are given by:

(21)

(22)

where, the coefficients Ei= 0,1,2 and Fi=0,1 are given by:

(23)

By solving the Eq. (21), position of neutral axis is determinedas:

(24)

By substituting the Eq. (24) in Eq. (22), we get:

(25)

where,

(26)

By imposing the condition (xc = D) in Eq. (24), limit curvatureφ0 is determined as given above. Further increase in the curvaturechanges the equilibrium conditions due to the contributions toresultant compressive force by concrete. For curvature more thanφ0, moment-curvature relationship is given by Eq. (20).

Pe bσc εc y( )[ ]dy σst Ast σscAsc+( )–0

xc

∫=

A0 φe( ) A1 φe( )xc A2 φe( )xc2 A3 φe( )xc

3+ + +=

Me bσc εc y( )[ ] D2---- y–⎝ ⎠⎛ ⎞dy σstAst σscAsc+( ) D

2---- d–⎝ ⎠⎛ ⎞+

0

xc

∫=

Me B0 φe( ) B1 φe( )xc B2 φe( )xc2 B3 φe( )xc

3 B4 φe( )xc4+ + + +=

A0 φe( ) b d D–( ) Dpt d pc pt–( )+[ ]Esφe ;=A1 φe( ) b D d–( ) pc pt+( )Esφe ;=

A2 φe( )bσc0φe

εc0--------------- A3 φe( )

bσc0φe2

3εc02

--------------- ;–=;=

B0 φe( ) 12---b 2d2 3dD D2+–( ) Dpt d pc pt+( )–[ ]Esφe ;=

B1 φe( ) 12---b 2d2 3dD D2+–( ) pc pt–( )Esφe B2 φe( )

bDσc0φe

2εc0-------------------- ;=;=

B3 φe( )bσc0φe 2εc0 Dφe+( )

6εc02

-------------------------------------------- B4 φe( )bσc0φe

2

12εc02

--------------- =;–=

xc1 Pe φe,( ) 16A3 φe( )------------------ 2A2 φe( )–( ) [=

2.5198 A22 φe( ) 3A1 φe( )A3 φe( )–( )C1 φe Pe,( )

--------------------------------------------------------------------------- 1.5874C1 φe Pe,( )+ +

xc2 Pe φe,( ) 112A3 φe( )---------------------=

4A2 φe( )– 2.5198 4.3645i+( ) A22 φe( ) 3A1 φe( )A3 φe( )–( )

C1 φe Pe,( )-------------------------------------------------------------------------------------------------------–

1.5874 2.7495i–( )C1 φe Pe,( )–

xc3 Pe φe,( ) 112A3 φe( )---------------------=

4A2 φe( )– 2.5198 4.3645– i( ) A22 φe( ) 3A1 φe( )A3 φe( )–( )

C1 φe Pe,( )-----------------------------------------------------------------------------------------------------–

1.5874 2.7495i+( )C1 φe Pe,( )–

C1 φe Pe,( ) 4 A22 3A1A3–( )3– 2A2

3 9A1A2A3– 27A32 A0 Pe–( )+( )2++

2A23 9A1A2A3 27A3

2 A0 Pe–( )–+–

1 3⁄

=

Me B0 φe( ) B1 φe( )xc3 φe Pe,( ) B2 φe( )xc32 φe Pe,( )+ +=

B3 φe( )xc33 φe Pe,( ) B4 φe( )xc3

r φe Pe,( )+ + φ∀ φ0 φE,[ ]∈

Pe bσc εc y( )[ ]dy σstAst σscAsc E0 E1xc E2xc2+ +=+–

0

D

∫=

Me bσc εc y( )[ ] D2---- y–⎝ ⎠⎛ ⎞dy σstAst σscAsc+( ) D

2---- d–⎝ ⎠⎛ ⎞+

0

D

∫=

F0 F1xe+=

E013---bφ 3d d D–( )Espc 3 d D–( )2Espt– D2σc0 3εc0 Dφ+( )

εc02

----------------------------------------– ,=

E1bφ dEs pc pt+( )εc0

2– D Es pc pt+( )εc02 σc0 2εc0 Dφ+( )+( )+[ ]

εc02

------------------------------------------------------------------------------------------------------------------------------------- ,=

E2bDσc0φ

2

εc02

-------------------- ,–=

F0bφ12------ [6d D 2d–( ) D d–( )Espc 6 d D–( )2 2d D–( )Espt+–=

D3σc0 2εc0 Dφ+( )εc0

2---------------------------------------- ,–

F1bφ 3 D2 2d2 3dD–+( ) pc pt–( )Esεc0

2 D3σc0φ–[ ]6εc0

2----------------------------------------------------------------------------------------------------------=

xcE1– E1

2 4E2 E0 Pe–( )–+2E0

----------------------------------------------------------=

Me F0 φ Pe,( ) F1 φ Pe,( )xc+= φ∀ 0 φ0,[ ]∈

φ03bεc0 D d–( )Es Dpc d pt pc–( )+( ) D2σc0+

2bD3σc0

------------------------------------------------------------------------------------------------=

εc0 3b 3b D d–( )Esεc0 Dpc d pt pc–( )+( )( D2σc0)2+[ ] 4PeD

3σc0–2bD3σc0

----------------------------------------------------------------------------------------------------------------------------------------------------------–


Vol. 15, No. 1 / January 2011 − 135 −

2.2 Elastic Limit Bending Moment and Curvature The limit elastic curvature, depending on the magnitude of

axial force and percentage of reinforcing steel in tension andcompression, results in four possible cases namely: i) strain intension steel reaches yield limit and stress in concrete vanishes;ii) strain in tension steel reaches yield limit but stress in concreteis present; iii) strain in compression steel reaches elastic limit; aswell as iiv) strain in extreme compression fibre in concretereaches elastic limit value.

2.2.1 Case (i): Strain in Tension Steel Reaches Yield Limitand Stress in Concrete Vanishes

This case is verified when . By imposing σst = σs0 and recalling the Eq. (8), depth of

neutral axis can be obtained as given below:

(27)

By substituting the Eq. (27) in Eq. (8), elastic limit curvaturecan be determined as:

(28)

By substituting Eq. (28) in Eq. (9), elastic limit moment isobtained as:

(29)

2.2.2 Case (ii): Strain in Tension Steel Reaches Yield Limitand Stress in Concrete Not Equal Zero

Depth of neutral axis is given by:

(30)

By substituting Eq. (30) in Eq. (15), expression for limit elasticcurvature can be obtained as:

(31)

where, the coefficients Li=0,1,2,3 are given by:

(32)

By solving Eq. (31), which is of a third degree polynomial,only one real root (third root) gives the limit elastic curvature:

(33)

where,

(34)

By substituting Eq. (33) in Eq. (16), limit elastic bending mo-ment is obtained as:

(35)

where, super script (ii) represents the second case; constants ofEq. (35) are given by:

,

(36)

2.2.3 Case (iii): Strain in Compression Steel Reaches Elas-tic Limit Value

Depth of neutral axis is given by:

(37)

By substituting Eq. (37) in Eq. (15), expression for limit elasticcurvature is obtained as:

(38)

where the constants Hi (for i = 0 to 3) are given by:

,

,

,

(39)

By solving Eq. (38), only one real root (the second one) givesthe limit elastic curvature as:

(40)

where,

(41)

pt Pe bdEspcεs0+( ) (b d D–( )⁄<σs0)

xci( ) D d

εs0

φE------––= xc 0<∀

φEPE b D d–( ) pc pt+( )σs0+bEspc D2 2d2 3dD–+( )

---------------------------------------------------------=

MEi( ) D 2d–

2--------------- PE 2b D d–( )ptσs0+[ ]=

xcii( ) D d εs0

φE------––= xc∀ 0 D d–,[ ]∈

L0 L1φE L2φE2 L3φE

3 0=+ + +

L0bεs0

2 3εc0 εs0+( )σc0

3εc02

------------------------------------------ ,=

L1PEεc0

2 b D d–( )εs0 2εc0 εs0+( )σc0 Es pc pt+( )εc02+[ ]+

εc02

---------------------------------------------------------------------------------------------------------------------–=

L2b D d–( ) 2d D–( )Espcεc0

2 d D–( ) εc0 εs0+( )σc0+[ ]εc0

2-----------------------------------------------------------------------------------------------------------------=

L3b d D–( )3σc0

3εc02

-----------------------------=

φEii( ) 1

12L3----------- [4L2

2.5198 4.3645i–( ) L22 3L1L3–( )

λ------------------------------------------------------------------------––=

1.5874 2.7495i+( )λ]–

λ [ 2L23– 9L1L2L3 27L3

2L0–+=

4 L22 3L1L3–( )

3– 2L23 9L1L2L3– 27L3

2L0+( )2++ ]

1 3⁄

MEii( ) b

2εc02

--------- M1ii( )

φEii( )2

---------- M2ii( )

φEii( )

--------- M3ii( ) M4

ii( )φEii( ) M5

ii( )φEii( )2+ + + +=

M1ii( ) εs0

3 4εc0 εs0+( )σc0

6--------------------------------------- M2

ii( ) D 2d–( )εs02 3εc0 εs0+( )σc0

3----------------------------------------------------------- ,–=,=

M3ii( ) D d–( )εs0 D 2– d( )Es pt pc–( )εc0

2 d 2εc0 εs0+( )σc0–[ ] ,=

M4ii( ) D d–( ) 3 D 2d–( )2Espcεc0

2 D d–( ) 2d D+( ) 2εc0 εs0+( )σc0+[ ]3

------------------------------------------------------------------------------------------------------------------------------------------=

M5ii( ) d D–( )3 D d+( )σc0

6-------------------------------------------=

xciii( ) d εs0

φE------+=

H0 H1φE H2φE2 H3φE

3 0=+ + +

H0bεs0

2 3εc0 εs0+( )σc0

3εc02

------------------------------------------=

H1PEεc0

2– bεs0 D d–( )Es pc pt+( )εc02 dσc0 2εc0 εs0–( )+[ ]+

εc02

---------------------------------------------------------------------------------------------------------------------------=

H2 b 3dD D2 2d2––( )Esptd2σc0 εc0 εs0–( )

εc02

----------------------------------+=

H3bd3σc0

3εc02

---------------–=

φEiii( ) 1

12H3------------ [4H2– 2.5198 4.3645i+( ) H2

2 3H1H3–( )ω

----------------------------------------------------------------------------–=

1.5874 2.7495i–( )ω]–

ω 2H23 9H1H2H3 27H3

2H0–+–[=

4 H22 3H1H3–( )

3– 2H23 9H1H2H3– 27H3

2H0+( )2+ ]

1 3⁄+



By substituting Eq. (40) in Eq. (16), limit elastic bending mo-ment can be obtained as follows:

(42)

where,

(43)

2.2.4 Case (iv): Strain in Extreme Compression Fibre inConcrete Reaches Elastic Limit Value

Now, the depth of neutral axis is given by:

(44)

By substituting Eq. (44) in Eq. (15), expression for limit elasticcurvature is obtained as:

(45)

where the constants Ri (for i = 0 to 2) are given by:

(46)

By solving Eq. (45), the only real root (in this case, first root)gives the limit elastic curvature as:

(47)

By substituting Eq. (47) in Eq. (16), limit elastic bendingmoment, ME, can be obtained as follows:

(48)

where,

(49)

It may be easily seen that for percentage of tension steelexceeding the maximum limit of 4%, as specified in many codes(see for example Indian code (IS 456, 2000), case (iv) shall never

result in a practical situation. For the case (xc>D), the limits ofthe integral in Eq. (15) will be from (0, D), which shall also resultin compression failure and hence not discussed. Expressions forlimit elastic moments are summarised as below:

(50)

where pt,el, for tow cases namely: i) axial force neglected; and ii)axial force considered are given by the following equations:

(51)

(52)

2.3 Percentage of Steel for Balanced SectionPercentage of reinforcement in tension and compression for

balanced failure are obtained by considering both the conditionsnamely: i) maximum compressive strain in concrete reachesultimate limit strain; and ii) strain in tensile reinforcementreaches ultimate limit. Balanced reinforcement for two cases isconsidered namely: i) for beams where axial force vanishes; andii) for beam/columns where P-M interaction is predominantlypresent. For sections with vanishing axial force, depth of neutralaxis is given by:

(53)

For vanishing axial force, governing equation to determine thepercentage of reinforcement is given by:

(54)

In explicit form, Eq. (53) becomes:

(55)

By solving, percentage of steel for balanced section is obtainedas:

(56)

For a known cross section with fixed percentage of compres-sion reinforcement, Eq. (56) gives the percentage of steel for abalanced section. It may be easily seen that for the assumedcondition of strain in compression steel greater than elastic limit,Eq. (56) shall yield percentage of tension reinforcement forbalanced sections, whose overall depth exceeds 240 mm, whichis a practical case of cross section dimension of RC beams usedin multi-storey building frames. For sections where axial force ispredominantly present, percentage of balanced reinforcementdepends on the magnitude of axial force. By assuming the samehypothesis presented above, depth of neutral axis is given by Eq.(53); but Eq. (55) becomes as given below:

MEiii( ) b

2εc02

--------- M1iii( )

φEiii( )2

----------- M2iii( )

φEiii( )

----------- M3iii( ) M4

iii( )φEiii( ) M5

iii( )φEiii( )2+ + + +=

M1iii( ) εs0

3 εs0 4εc0–( )σc0

6-------------------------------------- M2

iii( ) D 2d–( ) 3εc0 εs0–( )εs02 σc0

3---------------------------------------------------------- ,=,=

M3iii d D–( )εs0 D 2d–( )Es pt pc–( )εc0

2 d 2εc0 εs0–( )σc0–[ ] ,=

M4iii( ) D d–( ) D 2d–( )2Esptεc0

2 d2 3D 2d–( ) εc0 εs0–( )σc0

3--------------------------------------------------------- ,+=

M5iii( ) d3 d 2D–( )σc0

6--------------------------------=

xciv( ) εc0

φE------=

R0 R1φE R2φE2 0=+ +

R02bεc0σc0

3-------------------- R1 Pe b D d–( )Esεc0 pc pt+( ) ,+–=,=

R2 b D d–( )Es Dpt d pt pe–( )–[ ]–=

φEiv( ) R1 R1

2 4R0R2–+2R2

---------------------------------------–=

MEiv( ) M1

iv( )

φEiv( )2

---------- M2iv( )

φEiv( )

---------- M3iv( ) M4

iv( )φEiv( )+ + +=

M1iv( ) bDεc0σc0

3---------------------=

M2iv( ) 1

4---bεc0

2 σc0–=

M3iv( ) 1

2---b D2 2d2 3dD–+( )Es pc pt–( )εc0=

M4iv( ) 1

2---b D2 2d2 3dD–+( )Es Dpt d pc pt+( )–[ ]=

MEME

ii( ) if pt pt el ,<

MEiii( ) if pt pt el ,<⎩

⎨⎧

=

pt el, peD2 D 3εc0 εs0–( ) 6dεc0–[ ]σc0

6 D d–( ) D 2d–( )2Esεc02

-----------------------------------------------------------------+=

pt el, =6 D 2d–( )2εc02 PE+b d D–( )Espcεs0[ ] bD2εs0 6dεc0+D εs0 3εc0–( )[ ]σc0+

6b D d–( ) D 2d–( )2Esεc02 εs0

---------------------------------------------------------------------------------------------------------------------------------------------------------------

xcεcu

εcu εsu+-----------------⎝ ⎠⎛ ⎞ D d–( )=

P bσc εc y( )[ ]dy Asc Ast–( )σs0 qbσc0 0=+ +q

xc

∫=

b d D–( ) σc0εc0 3εcuσc0– 3 pc pt–( ) εcu εsu+( )σs0–[ ] 0=

pt bal, pc3εcu εc0–( )σc0

3 εcu εsu+( )σs0--------------------------------+=


Vol. 15, No. 1 / January 2011 − 137 −

(57)

By solving, percentage of steel for balanced section is obtainedas:

(58)

where, P0 is the axial force (P0 > 0 if it is compression). For theknown cross section with fixed percentage of compressionreinforcement, Eq. (58) gives the percentage of steel for balancedsection. In the similar manner, percentage of compression rein-forcement for a balanced section, by fixing pt, can be obtained byinverting the relationship given in Eqs. (56) and (58) for respec-tive axial force conditions.

2.4 Ultimate Bending Moment-curvature RelationshipStudy in this section is limited to RC sections imposed with

tension failure as the compression and balanced failures do nothave any practical significance in the displacement-based designapproach, in particular. Let us consider two possible cases: i)neutral axis position assumes negative values; and ii) neutral axisposition assumes positive values.

2.4.1 Neutral Axis Position Assuming Negative ValuesBy imposing the conditions: and solv-

ing Eq. (8) respect to pt, for a specified range of tension steel per-centage, , depth of neutral axis isgiven by:

(59)

At collapse, the equilibrium equations become:

(60)

(61)

By solving Eq. (60) with respect to φu, we obtain the ultimatecurvature, as reported below:

(62)

By substituting Eq. (62) in Eq. (61), ultimate bending momentcan be determined as:

(63)

It may be noted that the ultimate bending moment in this caseis similar to one given by Eq. (29) for elastic range.

2.4.2 Neutral Axis Position Assuming Positive ValuesUnder this condition at collapse, four different cases of tension

failure of RC sections are possible, namely:

(a)

(b)

(c)

(d) (64)

As the strain in tensile steel reaches its ultimate value (tensilefailure), in all the four cases mentioned above, equation forcomputing the depth of neutral axis, as function of ultimatecurvature, will remain unchanged and is given by:

(65)

Axial force and bending moment in the cross section atcollapse, for case (a) are given by:

(66)

(67)

By substituting the Eq. (65) in Eq. (66) we get:

(68)

where the constants Ji (for i=0 to 3) are given by:

(69)

By solving Eq. (68), the real root (in this case, the third root)gives the ultimate curvature as:

(70)

where,

(71)

By substituting Eq. (70) in Eq. (67), ultimate moment is givenby:

(72)

where the super-script (a) stands for the case (a); the constants ofEq. (72) are given by:

,

b d D–( ) σc0εc0 3εcuσc0– 3 pe pt–( ) εcu εsu+( )σs0–[ ] P0=

pt bal, pc3εcu εc0–( )σc0

3 εcu εsu–( )σs0-------------------------------- P0

b D d–( )σs0---------------------------–+=

xc 0 & φ εsu D d–( )⁄==

pt Pu bdEsεsu+( ) b d D–( )σs0( )⁄<

xc D d εsu

φu------––= xc 0<∀

Pu σs0Ast– σscAsc b d D–( ) ptσs0 Espc d xc–( )φu+[ ]=+=

Mu σs0Ast σscAsc+( ) D2---- d–⎝ ⎠⎛ ⎞=

b D 2d–( )2

----------------------- D d–( ) ptσs0 Espc xc d–( )φu+[ ]=

φuPu b D d–( ) σs0pt Espcεsu+[ ]+

bEspc D2 2d2 3dD–+( )--------------------------------------------------------------------=

MuD 2d–

2--------------- Pu 2b D d–( )ptσs0+[ ]=

εst εsu εsc εs0 ,<,= εc max, εc0 ,<

εst εsu εsc εs0 ,<,= εc0 εc max, εcu ,< <

εst εsu εs0 εsc εsu εc max, εc0<,< <,=

εst εsu εs0 εsc εsu εc max, εcu<,< <,=

xca d–( ) D d εsu

φu------––=

Pu bσc εc y( )[ ]dy σs0Ast– σscAsc+0

xc

∫=

Mu bσc εc y( )[ ] D2---- y–⎝ ⎠⎛ ⎞dy σs0Ast σscAsc+( ) D

2---- d–⎝ ⎠⎛ ⎞+

0

xc

∫=

J0 J1φu J2φu2 J3φu

3 0=+++

J0bεsu

2 3εc0 εsu+( )σc0

3εc02

------------------------------------------=

J1Puεc0

2– b d D–( ) Espcεc02 σc0 2εc0 εsu+( )+( )εsu ptσs0εc0

2+[ ]+εc0

2--------------------------------------------------------------------------------------------------------------------------------------=

J2 b D2 2d2 3dD–+( )Espcd D–( )2 εc0 εsu+( )σc0

εc02

------------------------------------------------+=

J3b d D–( )3σc0

3εc02

-----------------------------=

φua( ) 1

12J3---------- 4J2

2.5198 4.3645i–( ) J22 3J1J3–( )

α----------------------------------------------------------------------– 1.5874 2.7495i+( )α––=

α 2– J23 9J1J2J3 27J3

2J0–+[=

4 J22 3J1J3–( )

3– 2J23 9J1J2J3– 27J3

2J0+( )2+ ]

1 3⁄ +

Mua( ) b

2εc02

--------- M1a( )

φua( )2

--------- M2a( )

φua( )

--------- M3a( ) M4

a( )φua( ) M5

a( )φua( )2+ + + +=

M1a( ) εsu

3 4εc0 εsu+( )σc0

6--------------------------------------- M2

a( ) 2d D–( )εsu2 3εc0 εsu+( )σc0

3-----------------------------------------------------------=,=



,

,

(73)

Axial force and bending moment in the cross section at collapse,for case (b) are given by:

(74)

(75)

By substituting the Eq. (65) in the (74), we get:

(76)

where, the constants Qi=0,1,2 are given by:

,

(77)

By solving Eq. (76), the first root of the quadratic, representingthe ultimate curvature is given as:

(78)

By substituting Eq. (78) in Eq. (75), ultimate moment is ob-tained as:

(79)

where,

,

(80)

Axial force and bending moment in the cross section atcollapse, for case (c) are given by:

(81)

(82)

By substituting the Eq. (65) in Eq. (81), we get:

(83)

where, the constants Wi=0,1,2,3 are given by:

(84)

By solving Eq. (83), the real root (in this case, it is the thirdroot) gives the ultimate curvature as:

(85)

where,

(86)

By substituting Eq. (85) in Eq. (81), ultimate moment is ob-tained as:

(87)

where,

,

(88)

Axial force and bending moment in the cross section atcollapse, for case (d), are given by:

(89)

(90)

By substituting the Eq. (65) in Eq. (89) and solving, the ulti-mate curvature is obtained as:

(91)

By substituting Eq. (91) in Eq. (90), ultimate bending momentis obtained as:

(92)

M3a( ) D d–( ) 2d D–( )Espcεsuεc0

2 dεsu–[=

2εc0 εsu+( )σc0 D 2d–( )ptεc02 σs0]+

M4a( ) D d–( ) 3 D 2d–( )2Espcεc0

2– d D–( ) 2d D+( ) εc0 εsu+( )σc0+[ ]3

-------------------------------------------------------------------------------------------------------------------------------------------=

M5a( ) d D–( )3 D d+( )σc0

6-------------------------------------------=

Pu bσc εc y( )[ ]dy Astσs0– Ascσsc qbσc0+ +q

xc

∫=

Mu bσc εc y( )[ ] D2---- y–⎝ ⎠⎛ ⎞dy Astσs0 Ascσsc+( ) D

2---- d–⎝ ⎠⎛ ⎞ qbσc0

2------------- D q–( )+ +

q

xc

∫=

Q0 Q1φu Q2φu2 0=+ +

Q0bσc0 εc0 3εsu+( )

3------------------------------------–=

Q1 b D d–( ) σc0 Espcεsu– σs0pt+( ) pu–=

Q2 bEsPc D2 2d2 3dD–+( )=

φub( ) Q1 Q1

2 4Q0Q2–+–2Q2

---------------------------------------------=

Mub( ) b

2--- M1

b( )

φub( )2

--------- M2b( )

φub( )

--------- M3b( ) M4

b( )φub( )+ + +=

M1b( ) εc0

2 4εc0εsu 6εsu2+ +( )σc0

6-----------------------------------------------------=

M2b( ) D 2d–( ) εc0 3εsu+( )σc0

3----------------------------------------------------=

M3b( ) D d–( ) d 2Espcεsu σc0 2ptσs0–+( ) D ptσs0 Espcεsu–( )+[ ]=

M4b( ) D d–( ) D 2d–( )2Espc=

Pu bσc εc y( )[ ]dy σs0 Asc Ast–( )+0

xc

∫=

Mu bσc εc y( )[ ] D2---- y–⎝ ⎠⎛ ⎞dy Ast Asc+( )σs0

D2---- d–⎝ ⎠⎛ ⎞+

0

xc

∫=

W0 W1φu W2φu2 W3φu

3 0=+ + +

W1Puεc0

2– b d D–( ) 2εc0 εsu+( )σc0εsu pt pc–( )σs0εc02+[ ]+

εc02

--------------------------------------------------------------------------------------------------------------------------=

W2b d D–( )2σc0 εc0 εsu+( )

εc02

--------------------------------------------------- W0 J0 W3 J3=,=,=

φuc( ) 1

12W3-------------=

4W2– 2.5198 4.3645i–( ) W22 3W1W3–( )

β----------------------------------------------------------------------------- 1.5874 2.7495i+( )β––

β 2W23 9W1W2W3 27W3

2W0– + +–[=

4 W22 3W1W3–( )

3– 2W23 9W1W2W3– 27W3

2W0+( )2+ ]

1 3⁄

Muc( ) b

2εc02

--------- M1c( )

φuc( )2

--------- M2c( )

φuc( )

--------- M3c( ) M4

c( )φuc( ) M5

c( )φuc( )2+ + + +=

M3c( ) d d D–( )εsuσc0 2εc0 εsu+( ) D2 2d2 3dD–+( ) pc pt+( )σs0εc0

2+=

M4c( ) d D–( )2 2d D+( ) εc0 εsu+( )σc0

3--------------------------------------------------------------------=

M1c( ) M1

a( ) M2c( ) M2

a( ) M5c( ) M5

a( )=,=,=

Pu bσc εc y( )[ ]dy Asc Ast–( )σs0 qbσc0+ +q

xc

∫=

Mu bσc εc y( )[ ] D2---- y–⎝ ⎠⎛ ⎞dy Ast Asc+( )σs0

D2---- d–⎝ ⎠⎛ ⎞ qbσc0

2------------- D q–( )+ +

q

xc

∫=

φud( ) bσc0 εc0 3εsu+( )

3 b D d–( ) σc0 σs0 pc pt–( ) Pu–+( )[ ]--------------------------------------------------------------------------------=

Mud( ) bσc0

12---------- 6 D d–( ) d D 2d–( ) pc pt+( )

σs0

σc0-------+⎝ ⎠

⎛ ⎞=

2 D 2d–( ) εc0 3εsu+( )φu

IV( )------------------------------------------------ εc0

2 4εc0εsu 6ssu2+ +

φuIV( )2

-----------------------------------------–+


Vol. 15, No. 1 / January 2011 − 139 −

For the condition of , ultimatemoment, derived above takes the following form:

(93)

where,

,

(94)

Percentage of tension reinforcements are determined by im-posing the conditions: i) pt

(1) is determined by imposing εst = εsu,εc,max = εc0 and solving Eq. (67) with respect to pt ; as well as ii)pt

(2) is determined by imposing the εst = εsu, εsc = εs0 and solvingEq. (75) with respect to pt.

For the other condition, namely ,ultimate moment now takes a different form as give below:

(95)

where,

(96)

Percentage of tension reinforcements are determined by im-posing the conditions: i) pt

(3) is determined by imposing the εst

= εsu, εc,max = εc0 and solving Eq. (82) with respect to pt; and ii)pt

(4) is determined by imposing the εst = εsu, εsc = εs0 and solving(90) respect to pt.

For the condition:

(97)

ultimate moment is given by:

(98)

3. Numerical Studies and Discussions

An example RC section of 300 × 500 is considered for thestudy. The section is reinforced on both tension and compression

zones whose percentage is varied to study their influence on thecurvature ductility. Concrete with compressive cube strength of30 N/mm2 and steel with yield strength of 415 N/mm2 are con-sidered. Fig. 3 shows the variation of elastic moment with tensionreinforcement for a constant compression reinforcement con-sisting 4Φ22. It is seen from the figure that the limit elasticmoment increases linearly for the case of strain in tensile steelreaches yield limit while strain in concrete is within elastic limit(see the curve governed by Eqs. (29 & 35)). For other casesnamely: i) strain in compression steel reaches elastic limit (seethe curve governed by Eq. (42); as well as ii) crushing failurewhere strain in extreme fibre in concrete reaches elastic limit(see the curve governed by Eq. 48), the influence of percentageof tension reinforcement on the limit elastic moment is marginal;though there is a sharp rise for lower percentage of reinforce-ments, this increase becomes marginal for higher percentagevalues. The point of intersection of moment profiles governed byEqs. (29) and (35) with that of Eq. (42) give the limit value of per-centage of tensile reinforcement (pt,elastic); percentage of tensilesteel reinforcement, lesser than this value results in yielding oftensile steel while greater values result in yielding of compres-sion steel. The point of intersection of moment profiles governedby Eqs. (29) and (35) with that of Eq. (48) is not of significant im-portance as the latter results in crushing failure of concrete. It isevident that percentage of tensile reinforcement influences limitelastic moment considerably in case of ductile failure only. Itmay be noted that Fig. 3 plots the moment variation based on thesame governing equations used subsequently for estimating mo-ment-curvature relationship. It can be seen from the figure thatlimit elastic moment is given by the minimum of the four valuesgiven by the Eqs. (29), (35), (42) and (48), respectively. The traceof the point along the hatched line gives the minimum limit elasticmoment, thus obtained. Fig. 4 shows the moment-curvature plotsfor the RC section reinforced with 4Φ22 on tension face, butvarying the compression steel. It can be seen from the figure thatfor a fixed percentage of tensile reinforcement, influence of varia-

D d 2εc0 εs0 εsu+–( )<( ) εc0 εs0–( )⁄

Mu

Mua( ) if pt pt

1( )<

Mub( ) if pt

1( ) pt p< t2( )<

Mud( ) if pt

2( ) pt<⎩⎪⎨⎪⎧

=

pt1( ) 3 εc0 εsu+( ) Pu bEspc d 2εc0 εsu+( ) Dεc0–( )+[ ] 2b d D–( )εc0σc0+

3b d D–( ) εc0 εsu+( )σs0---------------------------------------------------------------------------------------------------------------------------------------------=

pt2( ) =

3 εc0+εsu( ) Pu+b d D–( )Espcεs0[ ] bσc0 D εc0 3εs0–( )+d 3εs0 2εc0– 3εsu–( )[ ]+3b d D–( ) εs0 εsu+( )σs0

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------

D d 2εc0 εs0 εcu+–( )( ) εc0 εs0–⁄>

Mu

Mua( ) if pt pt

3( )<

Muc( ) if pt

3( ) pt p< t4( )<

Mud( ) if pt

4( ) pt<⎩⎪⎨⎪⎧

=

pt3( ) pc

3Pu εc0 εsu+( ) 2b d D–( )εc0σc0+3b d D–( ) εc0 εsu+( )σs0

-------------------------------------------------------------------------+=

Pt4( ) Pc

Pu

b d D–( )σs0---------------------------+=

Dεs0 d εsu εs0–( )+[ ]2 D εc0 3εs0–( ) d 6εc0 εs0 εsu+–( )+[ ]σc0

3εc02 εs0 εsu+( ) D 2d–( )2 d D–( )σs0

-------------------------------------------------------------------------------------------------------------------------------------+

D d 2εc0 εs0 ε+ su–( )( ) εc0 εs0–⁄=

MuMu

a( ) if pt pt*<

Mua( ) if Pt

* Pt<⎩⎨⎧

= pt* pt

1( ) pt2( ) pt

3( ) pt4( )= = = =

Fig. 3. Variation of Elastic Moment with Percentage of TensileSteel Reinforcement Relationship



tion of compression reinforcement on moment-curvature is onlymarginal. Also, there exist at least one critical value of percent-age of both tensile and compression reinforcement, which reducesthe curvature ductility to the minimum. The proposed analyticalexpressions are capable of tracing this critical value, so that it canbe avoided for a successful design of the section.

The effect of axial force on moment-curvature is also studiedby subjecting the RC section reinforced with 4Φ22, both oncompression and tension sides. The section is subjected to com-pressive axial force only as the tensile force limits the curvatureand cannot be helpful in predicting the desired behaviour. Fig. 5shows the moment-curvature for different axial forces consi-dered. For all the four cases shown in the figure, there is a mar-ginal increase in ultimate moment with respect to their corres-ponding limit elastic moment. It is seen that the variation in themagnitude of axial force does not influence the ductility ratio incomparison to their influence on limit elastic and ultimate mo-ments, as well, for the numerical cases examined. However, higheraxial forces tend to reduce the curvature ductility. The criticalvalue of axial force, beyond which, a reduction is caused in cur-vature ductility, can also be obtained from the proposed analyti-cal hypothesis. The moment-curvatures seen in the figure, showslinear response in elastic range and hardening-like response inelasto-plastic range.

Influence of percentage of reinforcing steel on ductility ratio,

for varying the axial forces, is also studied. Two cases are consi-dered namely: i) by varying steel percentage in tension, with4Φ22 on compression side; as well as ii) by varying the percen-tage of compression reinforcement, with 4Φ22 on tension side.Figs. 6 and 7 show the influence of tensile and compression rein-forcements on curvature ductility, respectively. It is seen from Fig.6 that plastic softening behaviour is observed in the section underlarge curvature amplitudes. This may be attributed to the expect-ed failure pattern (local collapse mechanism) of the structuralmembers of building frames located in seismic areas. Largerductility ratios for reduced tensile reinforcement prompt the designof members initiating ductile failure, as better ones. However,tensile reinforcement closer to pt,bal will result in more curvatureductility as there is a marginal reduction seen due to the kink inthe curve for (lesser) values closer to pt,bal. It can be seen fromFig. 7 that maximum curvature ductility is obtained for com-pression reinforcement equals pc,bal, when the section is subjectedto axial compressive force. However, for tensile axial forces,percentage of compression steel as same as that of tension steel(pc=pt), gives the maximum curvature ductility. It can be sum-

Fig. 4. Variation of Moment-curvature with Percentage of Com-pression Reinforcement

Fig. 5. Moment-curvature Relationship for Different Axial Forces

Fig. 6. Variation of Curvature Ductility with Percentage of TensileSteel Reinforcement

Fig. 7. Variation of Curvature Ductility with Percentage of Com-pression Steel Reinforcement


Vol. 15, No. 1 / January 2011 − 141 −

marized that percentage of tension reinforcement influences cur-vature ductility to a larger extent and therefore demands goodductile detailing in the members of building frames located inseismic areas. Recent development in codes (see, for example,IS:13920, 2003) also insist the same for a safe distribution ofearthquake forces without complete collapse of the building.

Spread sheet program is used to estimate the moment-cur-vature by iteration, after simplifying the complexities involved insuch estimate. The values are estimated in two ranges, namely i)elastic; and ii) elasto-plastic, separately. Tables 1 and 2 show thevalues of the points traced along the M-Φ curve, obtainednumerically, for two cases namely: i) no axial force; and ii) axialforce of 200 kN, respectively. The shaded rows show the valuesat limit elastic and ultimate states, in order, respectively. Stepsinvolved in the numerical procedure are now discussed. Firstly,to predict the moment-curvature relationship in elastic range,steps followed are namely: i) an arbitrary value is assumed forthe limit elastic curvature; ii) fixing axial force to the desiredvalue, depth of neutral axis is determined. The strains in con-crete, compressive and tensile steel are examined for their elasticlimit values. Value of limit elastic curvature is now changed untilstrain in one of the above, reach their elastic limit. For example,as seen in Table 1, for the limit elastic curvature of 0.005780 rad/

m, strain in tensile steel reaches its elastic limit (0.00172), forzero axial force, causing a tensile failure in this case. Fixing thisvalue as the limit elastic curvature and by sub-dividing it equally,moment-curvature values for the first five rows are now obtainedby repeating the above steps. Secondly, for estimating the valuesin elasto-plastic range, following steps are adopted: i) an arbitr-ary value is now assumed for the limit ultimate curvature; ii)fixing axial force to the desired value, depth of neutral axis isnow determined. The strains in concrete, compressive and tensilesteel are further examined for their ultimate limits. Curvaturevalue is changed until strain in one of the above, reach theirultimate limit. For example, as seen in Table 1, for the ultimatecurvature of 0.025276 rad/m, strain in tensile steel reaches itsultimate limit (0.01), for zero axial force, causing a tensilefailure. Fixing this value as the ultimate curvature and by sub-dividing thisvalue equally, moment-curvature values in the elasto-plastic range are now obtained by repeating the above steps.Based on the results obtained, moment-curvature relationship ofthe RC section, reinforced with 4Φ22, both in tension and com-pression sides, is now plotted for different axial loads (onlycompressive). The curves are compared with those obtained byusing the proposed analytical expressions. Fig. 8 shows thecomparison of the curves obtained by employing both numerical

Table 1. Moment-curvature Relationship of RC Section 300×500 for No Axial Force (pt = 1.08%, pc = 1.08%, Rck = 30 N/mm2, fy = 415 N/mm2)

P(kN)

φ(rad/m)

xc(m) εc,max εsc εst

σc,max(kN/sq.m)

σsc(kN/sq.m)

σst(kN/sq.m)

q(m)

M(kN-m)

0.00 0.000010 0.165 0.00000 0.000001 0.00000 22 284 640 0.00 0.41

0.00 0.001166 0.167 0.00019 0.000159 0.00035 2444 33437 74301 0.00 48.07

0.00 0.002322 0.168 0.00039 0.000320 0.00070 4657 67283 147269 0.00 95.20

0.00 0.003478 0.169 0.00059 0.000485 0.00105 6648 101874 219493 0.00 141.76

0.00 0.004634 0.171 0.00079 0.000654 0.00139 8408 137269 290913 0.00 187.72

0.00 0.005780 0.173 0.00100 0.000825 0.00172 9909 173216 360856 0.00 232.62

0.00 0.007080 0.153 0.00108 0.000872 0.00224 10457 183156 360870 0.00 234.86

0.00 0.008379 0.139 0.00116 0.000911 0.00278 10906 191230 360870 0.00 236.41

0.00 0.009679 0.127 0.00123 0.000943 0.00332 11283 197947 360870 0.00 237.55

0.00 0.010979 0.118 0.00130 0.000970 0.00386 11603 203635 360870 0.00 238.40

0.00 0.012279 0.111 0.00136 0.000993 0.00441 11879 208523 360870 0.00 239.07

0.00 0.013578 0.105 0.00142 0.001013 0.00496 12118 212773 360870 0.00 239.61

0.00 0.014878 0.099 0.00148 0.001031 0.00552 12325 216505 360870 0.00 240.04

0.00 0.016178 0.095 0.00153 0.001047 0.00607 12504 219811 360870 0.00 240.40

0.00 0.017478 0.091 0.00159 0.001061 0.00663 12659 222762 360870 0.00 240.69

0.00 0.018777 0.087 0.00164 0.001073 0.00719 12792 225415 360870 0.00 240.95

0.00 0.020077 0.084 0.00169 0.001085 0.00775 12904 227814 360870 0.00 241.16

0.00 0.021377 0.081 0.00174 0.001095 0.00831 12999 229997 360870 0.00 241.34

0.00 0.022677 0.079 0.00179 0.001105 0.00887 13075 231994 360870 0.00 241.50

0.00 0.023976 0.076 0.00183 0.001113 0.00944 13136 233829 360870 0.00 241.64

0.00 0.025276 0.074 0.00188 0.001122 0.01000 13180 235525 360870 0.00 241.77



and analytical procedures. By comparing, it can be seen thatthere is practically no difference between the curves in the elasticrange, whereas there exist a marginal difference in the plasticrange. However, both the procedures estimate the same ultimatecurvature and the ultimate moments as well. Also the curvatureductility ratio obtained by both the procedures, remains same.With regards to their close agreement, the proposed closed formexpressions for moment-curvature relationship, accounting fornonlinear characteristics of constitutive materials according toEuro code, are thus qualified for using them in seismic designand structural assessments as well.

It can be inferred from the above discussions that detailed traceof moment-curvature relationship is inevitable for successfulseismic design of structures. The relationship is however verycomplex due to many factors namely: i) constitutive material’snonlinear response; ii) magnitude of axial load and their nature;as well as iii) cross sectional properties and percentage of rein-forcement (tensile steel, in particular). The numerical studiesconducted lead to useful design guidelines of multi-storey RCbuildings. The upper floor elements (beams, in particular) shallbe designed to have ductile failure, which in turn shall permitlarge curvature ductility. This, in fact, helps the formation ofplastic hinges at upper floors (on beams, in particular with a

strong column-weak beam design concept) first and enablingeffective redistribution of moments, resulting in formation ofplastic hinges at lower floors, subsequently. On the contrary, acolumn member, usually subjected to larger axial force, shall bedesigned without much increase in compression reinforcement,

Fig. 8. Comparison of Moment-curvature by Analytical and Numeri-cal Procedures

Table 2. Moment-curvature Relationship of RC Section 300×500 for 200 kN Axial Force (pt = 1.08%, pc = 1.08%, Rck = 30 N/mm2, fy = 415N/mm2)

P(kN)

φ(rad/m)

xc(m) εc,max εsc εst

σc,max(kN/sq.m)

σsc(kN/sq.m)

σst(kN/sq.m)

q(m)

M(kN-m)

200.00 0.000010 7.990 0.00008 0.000080 -0.00008 1036 16715 -15791 0.00 0.71

200.00 0.001286 0.274 0.00035 0.000314 0.00025 4250 65892 52935 0.00 68.08

200.00 0.002562 0.227 0.00058 0.000505 0.00062 6582 106147 130582 0.00 119.98

200.00 0.003838 0.211 0.00081 0.000696 0.00099 8557 146245 208386 0.00 170.62

200.00 0.005114 0.204 0.00104 0.000891 0.00136 10208 187085 285449 0.00 220.32

200.00 0.006380 0.201 0.00128 0.001089 0.00172 11515 228645 360867 0.00 268.70

200.00 0.007708 0.180 0.00139 0.001154 0.00224 11979 242397 360870 0.00 272.23

200.00 0.009036 0.164 0.00148 0.001209 0.00277 12335 253949 360870 0.00 274.77

200.00 0.010364 0.151 0.00157 0.001256 0.00330 12609 263846 360870 0.00 276.67

200.00 0.011692 0.141 0.00165 0.001297 0.00385 12819 272456 360870 0.00 278.14

200.00 0.013020 0.132 0.00172 0.001334 0.00440 12976 280040 360870 0.00 279.31

200.00 0.014348 0.125 0.00180 0.001366 0.00495 13091 286791 360870 0.00 280.25

200.00 0.015676 0.119 0.00186 0.001395 0.00550 13168 292853 360870 0.00 281.03

200.00 0.017004 0.114 0.00193 0.001421 0.00606 13212 298340 360870 0.00 281.67

200.00 0.018332 0.109 0.00199 0.001444 0.00662 13228 303340 360870 0.00 282.21

200.00 0.019660 0.105 0.00206 0.001466 0.00718 13228 307925 360870 0.00 282.67

200.00 0.020988 0.101 0.00212 0.001486 0.00775 13228 312145 360870 0.01 283.07

200.00 0.022316 0.097 0.00217 0.001505 0.00831 13228 316042 360870 0.01 283.41

200.00 0.023644 0.094 0.00223 0.001522 0.00888 13228 319653 360870 0.01 283.71

200.00 0.024972 0.092 0.00229 0.001538 0.00945 13228 323007 360870 0.01 283.97

200.00 0.026300 0.089 0.00234 0.001553 0.01002 13228 326131 360870 0.01 284.21


Vol. 15, No. 1 / January 2011 − 143 −

as this does not help to improve its curvature ductility.

4. Conclusions

In this paper, a new analytical procedure for estimating curva-ture ductility of RC sections is proposed. The purpose is to esti-mate moment-curvature relationship under service loads, in asimpler closed form manner. Analytical expressions for moment-curvature relationship of RC sections, accounting for nonlinearcharacteristics of constitutive materials according to Eurocode,are proposed in elastic and elasto-plastic ranges as well. Percent-age of tension reinforcement influences curvature ductility to alarger extent. There exist at least one critical value of percentageof both tensile and compression reinforcements, which reducesthe curvature ductility to the minimum. The proposed analyticalexpressions are capable of tracing this critical value, so that it canbe avoided for a successful design of the section. Tensile rein-forcement, closer to pt,bal, will result in more curvature ductilityas there is a marginal reduction seen due to the kink in the curvefor (lesser) values closer to pt,bal. Maximum curvature ductility isobtained for compression reinforcement equals pc,bal, when thesection is subjected to axial compressive forces; for tensile axialforces, percentage of compression steel as same as that of tensionsteel (pc=pt), gives the maximum curvature ductility.

The spread sheet program used to estimate moment-curvaturerelationship simplifies the complexities involved in such estimate,thus encouraging the designers and researchers to use it instantlyand with confidence. With regards to their close agreement withthe analytical procedure, the proposed expressions for moment-curvature estimate are thus qualified for using them in design andstructural assessments as well. Avoiding somewhat tedious handcalculations and approximations required in conventional iterativedesign procedures, the proposed method avoids errors and po-tentially unsafe design. It is felt that enough experimental evidenceis not available to be more conclusive on the topic, but the pro-posed closed form solutions of the unknown curvature ductilityratios is confident of giving reliable and safe estimate of the saidparameter. With due consideration to the increasing necessity ofstructural assessment of existing buildings under seismic loads,the proposed expressions of moment-curvature relationship shallbecome an integral input while employing nonlinear static pro-cedures.

Notations

Asc : Area of compression reinforcement (mm2)Ast : Area of tension reinforcement (mm2)b : Width of the beam (mm)D : Overall depth of the beam (mm)d : Effective cover (mm)

Es : Modulus of elasticity in steel (N/mm2) M : Bending moment (N-m)

Me : Elastic bending moment (N-m)ME : Limit elastic bending moment (N-m)

Mu : Ultimate bending moment (N-m)P : Axial load (N) pc : Percentage of compression reinforcement Pe : Elastic axial load (N) PE : Limit elastic axial load (N)pt : Percentage of tensile reinforcementPu : Ultimate axial load (N)q : Depth of plastic kernel of concrete (mm)

Rck : Compressive cube strength of concrete (30 N/mm2)xc : Depth of neutral axis measure from extreme compression

fibre (mm)εc : Strain in generic fibre of concrete

εc,max : Maximum strain in concreteεc0 : Elastic limit strain in concreteεcu : Ultimate limit strain in concreteεs0 : Elastic limit strain in reinforcement εsc : Strain in compression reinforcement εst : Strain in tensile reinforcementεsu : Ultimate limit strain in reinforcement φ : Curvature (rad/m)

φ0 : Curvature for xc = 0 (rad/m)φe : Elastic curvature (rad/m)φE : Limit elastic curvature (rad/m)φu : Ultimate curvature (rad/m) γc : Partial safety factor for concreteγs : Partial safety factor for steelη : Curvature ductility ratio = ϕu/ϕE

σc : Stress in generic fibre of concrete (N/mm2)σc,max : Maximum stress in concrete (N/mm2)

σc0 : Design ultimate stress in concrete in compression (N/mm2)

σs0 : Design ultimate stress in steel (N/mm2) σsc : Stress in compression reinforcement (N/mm2) σst : Stress in tensile reinforcement (N/mm2) σy : Yield strength of steel (415 N/mm2)

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