Design of RC Columns Using Glass FRP ReinforcementHany Jawaheri Zadeh, Ph.D.1; and Antonio Nanni, F.ASCE2
Abstract: Current guidelines do not cover the subject of the fiber-reinforced polymer (FRP) reinforced concrete (RC) members subjected tosimultaneous flexural and axial loads. The obvious implication is that, presently, RC columns cannot be designed with FRP longitudinalbars and ties. This paper suggests the consideration of this type of application of glass FRP (GFRP) reinforcement for the next revision ofguidelines by (1) showing the theoretical approach at the basis of the behavior of GFRP-RC members subject to simultaneous flexural andaxial loads; (2) providing revised design and analysis provisions for GFRP-RC columns similar to those in practice for steel-RC columns;(3) showing how interaction diagrams can be developed for both rectangular and circular cross sections; (4) explaining the rationale for a newformulation of the strength reduction factor for simultaneous flexural and axial resistance that is consistent with current guidelines, and;(5) reformulating the shear strength computation. Recognizing the limited availability of experimental data on GFRP-RC column behavior,the design provisions proposed in this paper are not applicable to structures in seismic zones and may be applicable only to buildings oflimited size and height. An example of the design of a GFRP-RC square column for a hospital facility concludes the paper. DOI: 10.1061/(ASCE)CC.1943-5614.0000354. © 2013 American Society of Civil Engineers.
CE Database subject headings: Fiber reinforced polymer; Glass; Concrete columns; Design; Axial loads.
Author keywords: Column; Design; Glass FRP; Interaction diagram.
Introduction
Presently, fiber-reinforced polymer (FRP) bars are a competitiveoption in reinforced concrete (RC) members subjected to flexureand shear because of their compelling physical and mechanicalproperties, corrosion resistance, and electromagnetic transparency.The use of FRP reinforcement is attractive for structures thatoperate in aggressive environments, such as in coastal regions, orfor buildings that host magnetic resonance imaging (MRI) unitsor other equipment sensitive to electromagnetic fields. However,the behavior of FRP bars as longitudinal reinforcement in com-pression members is still a relevant issue to be addressed andnot yet covered by ACI 440.1R-06 (ACI 2006). A brief summaryof publications covering this topic as available in the technicalliterature follows:• Glass FRP (GFRP). This is the most commonly used FRP
material system for internal reinforcement of FRC mem-bers. Experimental research studies (Paramanantham 1993;Alsayed et al. 1999; De Luca et al. 2009, 2010; Issa et al.2011; Deiveegan and Kumaran 2009, 2011) have investigatedthe effect of the compressive behavior of longitudinal GFRP
bars by testing RC columns. The overall conclusion of thesestudies is that GFRP-RC columns are a doable application.
• Aramid FRP (AFRP). Kawaguchi (1993) tested 12 concretemembers reinforced with AFRP bars and subjected to eccentrictension or compression. He reported that AFRP-RC columnscan be analyzed using the same procedure as for steel-RC col-umns. Similarly, Fukuyama et al. (1995) tested a half-scalethree-story AFRP-RC frame under quasistatic loading. AramidFRP bars were used for columns, beams, and slabs. The frameremained elastic up to a drift angle of 1=50 rad, and no substan-tial decrease in strength took place after rupture of some main-beam AFRP bars because of the high degree of indeterminacyof the frame.
• Carbon FRP (CFRP). Kobayashi and Fujisaki (1995) testedcolumns reinforced with CFRP grids and determined thatstrain compatibility was maintained up to the crushing of con-crete. Carbon FRP grids were later used by Grira and Saatcioglu(1999) as transverse reinforcement for columns tested undercyclic loading, and it was concluded that their performancewas comparable to that of columns reinforced with steelstirrups.The outcomes of these experimental studies may ultimately pro-
vide a convincing case to allow for the limited use of FRP bars incolumns, particularly when corrosion resistance or electromagnetictransparency is sought. Similarly, given the novelty of the FRP-RCtechnology for column design and construction, limitations suchas the ones adopted in ACI 314-11 (ACI 2011b) may be considered.Specifically, recommendations described in this paper and pro-posed for design apply to buildings with five or fewer stories aboveground and no more than one basement level. Although someexperimental and analytical work has been performed in FRP-RCbeam-column joint performance (Said and Nehdi 2004; Saravananand Kumaran 2011), because of the limited availability of datain this field, the design provisions proposed in this paper are notapplicable to structures in seismic zones. This is a limitation thatmay be removed in the future on the basis of the outcomes of moreexperimental evidence.
1Dept. of Civil, Architectural, and Environmental Engineering, Univ. ofMiami, 103 McArthur Engineering Bldg., Coral Gables, FL 33146-0630(corresponding author). E-mail: [email protected]
2Lester and Gwen Fisher Endowed Scholar, Professor and Chair ofthe Dept. of Civil, Architectural, and Environmental Engineering, Univ. ofMiami, 323 McArthur Engineering Bldg., Coral Gables, FL 33146-0630;Professor, Dept. of Structural Engineering, Univ. of Naples, Federico II,Via Claudio 21, Naples, P.O. Box I-80125, Italy. E-mail: [email protected]
Note. This manuscript was submitted on August 20, 2012; approved onDecember 14, 2012; published online on December 17, 2012. Discussionperiod open until November 1, 2013; separate discussions must besubmitted for individual papers. This paper is part of the Journal of Com-posites for Construction, Vol. 17, No. 3, June 1, 2013. © ASCE, ISSN1090-0268/2013/3-294-304/$25.00.
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General Design Considerations
For the purpose of this paper only GFRP bars are covered in detail.The same considerations can be extended to the other types ofFRP systems.
Minimum Longitudinal Reinforcement
Since 1936, the Building Code [ACI 318 (ACI 2011a)] has requiredthat the minimum reinforcement ratio be 0.01 of the gross area ofconcrete section. This minimum reinforcement area was intendedto prevent “passive yielding” of the steel, which occurs when loadis transferred gradually from concrete to the reinforcement as theconcrete creeps under sustained axial load (Lin and Furlong 1995).Although it appears to have become an outdated restriction formodern concrete and steel (Ziehl et al. 1998) and notwithstandingthe consideration that GFRP does not yield, this requirement hasbeen retained also for the case of GFRP reinforcement for analogy.
Equivalency between GFRP under Compression andConcrete
With reference to the behavior of FRP bars in compression, it isknown that their testing is complicated by the anisotropic andnonhomogeneous nature of the FRP material, which can lead toinaccurate measurements (Mallick 1993). Different modes of fail-ure (transverse tensile failure, fiber microbuckling, or shear failure)may affect the response of FRP bars in compression, depending onthe type of fiber, fiber volume fraction, and type of resin (Mallick1993). For the case of GFRP bars in particular, reductions in thecompressive strength and elastic modulus by up to 45% and 20%with respect to the values in tension, respectively, have beenreported (Wu 1990). Similar results were reported for GFRP barsby Deitz et al. (2003), who indicated that the compressive to tensilestrength and modular ratios were approximately 50% and 100%,respectively. Accordingly, GFRP mechanical characteristics exceedthose of concrete in compression and, therefore, the equivalencybetween the two materials when performing analysis and designis justifiable (Choo 2005; Choo et al. 2006a, b).
Limit on Maximum Tensile Strain in GFRP
The tensile rupture strain of GFRP bars exceeds 2%. Such a strainleads to unacceptably large deformations if the full tensile capacityof the bars is to be achieved. To avoid this, it is recommended that,for design purposes, the ultimate design strain not exceed a fixedlimit. One suggestion may be defining the ultimate design strain,εfd and strength, ffd, as
εfd ¼ Minðεfu; 0.010Þ and ffd ¼ Minðffu; 0.010EfÞ ð1Þ
In other words, the tensile design strain of the GFRP bars islimited to 0.01.
Limit on Maximum Spacing of TransverseReinforcement
For steel-RC columns, this spacing is limited by ACI 318-11 (ACI2011a) to the least dimension of the column, 16 longitudinal-baror 48 tie-bar diameters. The spacing between ties can be related tothe diameter of the longitudinal bars by a simplified model thatassumes that the bar is a compressive member simply supportedbetween the two supports provided only by the ties. The lateralsupport provided by the concrete cover is neglected, as at the point
of failure, the loss of cover is very probable. For such a memberto reach a strain level of ε without buckling this condition mustbe upheld:
π2EIs2
≤ AEε ð2Þ
where s = spacing of the ties; I and A = moment of inertia and thearea of the longitudinal bars; and E = their modulus of elasticity.For a round bar of diameter db this can be rewritten as
smax ¼πdb4
ffiffiffiε
p ð3Þ
For steel bars to yield before buckling, ε ¼ εy ≈ 0.002,which leads to smax ≈ 17.5db, which is in good agreement withACI 318-11 (smax ¼ 16db). For GFRP bars to avoid buckling be-fore concrete crushing, ε ¼ εcu ¼ 0.003 and, therefore, smax ≈ 14db.
The spacing is also related to the diameter of tie bars to achieve adesired level of concrete confinement at the core of the column.From this viewpoint, De Luca et al. (2009) suggest that forGFRP-RC columns the spacing of ACI 318-11 must be halved(i.e., the other limitation is 24 tie-bar diameters). Considering allthese points, the authors suggest this set of limitations for thespacing of transverse reinforcement:
smax ¼ Min(least dimension of the column, 12 longitudinal-bardiameters, 24 tie-bar diameters), noting that the theoretical smax ≈14db is rounded conservatively to smax ≈ 12db.
Modified Column Stiffness
In the analysis stage of ordinary steel-RC structures, the flexuralstiffness of members is modified to account for the cracking of thesections. When a GFRP-RC structure is analyzed, because of thedifferent mechanical properties of GFRP and steel, these modifica-tions need to be adjusted to incorporate GFRP reinforcement. Thissection develops modification factors for the moments of inertiaof GFRP-RC members that are styled after those of ACI 318-11(ACI 2011a).
ACI 318-11 recommends that the internal forces and the lateraldeflections of RC building systems resulting from factored loads becomputed by linear analysis with modified moments of inertia, I, ofthe members, as listed:• Steel-RC flexural members:
Ibeam ¼ 0.35Ig ð4Þ
Islab ¼ 0.25Ig ð5Þ
Alternatively, for any flexural member a more accurateformulation is
Iflexure ¼ ð0.10þ 25ρÞð1.20 − 0.20bw=dÞIg ≤ 0.50Ig ð6Þ
• Steel-RC columns:
Icolumn ¼ 0.70Ig ð7Þ
Again, a more detailed approach is
Icolumn ¼ ð0.80þ 25ρstÞ½1 −Mu=ðPuhÞ − 0.5Pu=P0�Ig≤ 0.875Ig ð8Þ
The aim is to obtain simple expressions like Eqs. (4), (5),and (7) for GFRP-RC members.
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• GFRP-RC flexural members:It can be reasoned that for GFRP-RC flexural members
Eq. (6) may be generalized as
Iflexure ¼ ð0.10þ 25ρfEf=EsÞð1.20 − 0.20bw=dÞIg ≤ 0.50Igð9Þ
Therefore, from the comparison of a GFRP-RC flexuralmember to a geometrically similar steel-RC member [Eqs. (6)and (9)], it can be concluded that
ðIFRP−RC=Isteel−RCÞflexure ¼ ð0.10þ 25ρfEf=EsÞ=ð0.10þ 25ρÞð10Þ
The ratio of the moments of inertia decreases with thereinforcement ratio if ρf ¼ ρ. Thus, to obtain a conservativeestimate independent of the reinforcement ratio, a heavily rein-forced beam or slab may be considered. Substituting: Isteel−RC ¼0.35Ig [Eq. (4)], and an average of ρf ¼ ρ ¼ 1.5% for beamsand Isteel−RC ¼ 0.25Ig [Eq. (5)], and an average of ρf ¼ ρ ¼0.5% for slabs into Eq. (10) and rounding the results, the modi-fied moments of inertia for GFRP-RC flexural members can beexpressed as
Ibeam ¼ ½0.075þ 0.275ðEf=EsÞ�Ig ≤ 0.35Ig ð11Þ
Islab ¼ ½0.10þ 0.15ðEf=EsÞ�Ig ≤ 0.25Ig ð12Þ
• GFRP-RC columns:Again, the generalized form of Eq. (8) for GFRP-RC flexural
members is
Icolumn ¼ ð0.80þ 25ρfEf=EsÞ½1 −Mu=ðPuhÞ− 0.5Pu=P0�Ig ≤ 0.875Ig ð13Þ
and from Eqs. (8) and (13), if two columns are compared thatonly differ in the type of reinforcement:
ðIFRP−RC=Isteel−RCÞcolumn ¼ð0.80þ25ρfEf=EsÞ=ð0.80þ25ρstÞð14Þ
Substituting: Isteel−RC ¼ 0.70Ig [Eq. (7)] and an average ofρf ¼ ρst ¼ 2.5%, after rounding, Eq. (14) can be simplified as
Icolumn ¼ ½0.40þ 0.30ðEf=EsÞ�Ig ≤ 0.70Ig ð15Þ
Similar to ACI 318-11, lateral deflections resulting fromservice lateral loads may be computed by a linear analysis using1.4 times the flexural stiffness previously defined.
Slenderness Effects
According to ACI 318 (ACI 2011a), the effects of slendernessfor compression members in a sway frame may be neglectedwhen the slenderness ratio (klu=r) is less than 22. Mirmiran et al.(2001) showed that, for GFRP-RC columns not braced againstside-sway, this limit should be reduced to 17. The authors suggestapplying a similar limit change for compression GFRP-RC col-umns in a nonsway frame, so that the effects of slenderness maybe neglected when
klu=r ≤ ðklu=rÞmax ¼ 29 − 12ðM1=M2Þ ≤ 35 ð16Þ
Eq. (16) is the modified form of the limit set by ACI 318-11:
klu=r ≤ ðklu=rÞmax ¼ 34 − 12ðM1=M2Þ ≤ 40 ð17Þ
where M2 = larger end moment; and M1 = smaller end moment;M1 and M2 are factored end moments obtained by an elastic frameanalysis, and the ratio M1=M2 is positive if the column is bent insingle curvature, negative if bent in double curvature.
Combined Effect of Flexure and Axial Load
The strength of a GFRP reinforced section under combined flexureand axial load is calculated by satisfying the strain compatibility,assuming that GFRP bars in compression are equivalent to concreteat any level of stress, and assuming that GFRP bars fail in tensionat their design strain, εfd. For the simplicity of calculations, barsunder compression are replaced by an equivalent area of concrete.
The combined nominal axial force and moment strength(Mn, Pn) is then multiplied by the appropriate strength reductionfactor, ∅, to obtain the design strength (∅Mn, ∅Pn) of the section.The design strength must be equal to or greater than the factoredmoment and axial load:
ð∅Mn;∅PnÞ ≥ ðMu;PuÞ ð18ÞThe factored pair of moment and axial load (Mu, Pu) denotes the
structural effects of the various combinations of loads to which astructure is subjected.
Similar to steel-RC columns a “strength interaction diagram”can be generated by plotting the design axial force strength,∅Pn, against the corresponding design moment strength, ∅Mn. Thisdiagram defines the strength of a section at different eccentricitiesof the load that must encompass all the points associated with(Mu, Pu). With such assumptions, typical design load-momentstrength interaction diagrams, (Mn, Pn), may be formulated asshown in the following section.
Interaction Diagram for Rectangular Cross Section
Compressive force is assumed to be positive and the moment to beapplied around the center line (C.L.) of the cross section, as shownin Fig. 1. The interaction diagram can be constructed by locating afew critical points as subsequently listed. The succession of pointsis according to the level of axial load: from maximum tensileload to maximum compressive load or equally according to x, thelocation of the neutral axis (N.A.), from −∞ to þ∞. Failure iseither initiated by GFRP tensile rupture (εf ¼ εfd) or by concretecrushing (εc ¼ εcu), which are referred to, respectively, as tension-controlled and compression-controlled modes of failure.
b
d
x
A1
A2
N.A.
C.L.
L1
LC
L2
1x
Fig. 1. Cross section of a rectangular column
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1. x ¼ −∞The maximum tensile force constitutes the lowermost
point of any interaction diagram and is one of the two pointscorresponding to zero eccentricity. If εf ¼ εfd:
Pn ¼ Tmax ¼ −2ðA1 þ A2Þffd; Mn ¼ 0 ð19Þ2. x ¼ 0
Because the entire cross section is in tension, GFRP bars arethe only component engaged in resisting the load.Therefore, fail-ure is triggered by rupture in the extreme layer of reinforcement(εf ¼ εfd). Furthermore, because of the linear behavior of thereinforcement, the portion of the interaction curve from the point(a) to (b) is linear. The load and moment can be calculated fromthe traditional equations regarding the combined effect of axialload and bending moment on a section with linear behavior:
Pn ¼ −A1 þ A2
γffd ð20Þ
Mn ¼ð2γ − 1Þ2
2γ
�A1 þ
A2
3
�ffdh ð21Þ
where γ ¼ d=h.3. xb ≤ x ≤ d
The two modes of failure are separated by the “balancedfailure,” when GFRP and concrete reach their failure simulta-neously; xb marks the location of the neutral axis for such acondition and can be calculated as
xb ¼d
1þ e; e ¼ εfd
εcu; εcu ¼ 0.003 ð22Þ
If the neutral axis is situated beyond the balanced location(x ≥ xb), the failure mode shifts from tension to compressionand starts with concrete crushing. Assuming that xb lies withinthe reinforced area (i.e., xb ≥ h − d = concrete cover), which isnormally the case, a range can be defined in terms of dimen-sionless parameters that covers both conditions:
Max
�1
γ− 1;
1
1þ e
�≤ α ¼ x
d≤ 1 ð23Þ
For any arbitrary α within this range, the forces in concreteand bars and their corresponding levers may be calculated as
C ¼ 0.85αβ1γf 0cbh; LC ¼ ð1 − αβ1γÞ
h2
ð24Þ
T1 ¼1 − αα
εcuEfA1; L1 ¼ ð2γ − 1Þ h2
ð25Þ
T2 ¼γ
2γ − 1
ð1 − αÞ2α
εcuEfA2; L2 ¼�2
3ð2þ αÞγ − 1
�h2
ð26Þ
Pn ¼ C − T1 − T2 ð27Þ
Mn ¼ CLC þ T1L1 þ T2L2 ð28Þ4. d < x ≤ h
If the neutral axis lies within the concrete cover on the ten-sion side, then the contribution of the reinforcement vanishesfrom the equations presented in condition 3. For example, ifx ¼ h or α ¼ 1=γ:
C ¼ 0.85β1f 0cbh; LC ¼ ð1 − β1Þ
h2
ð29Þ
T1 ¼ T2 ¼ 0 ð30ÞPn ¼ C; Mn ¼ CLC ð31Þ
5. x ¼ þ∞The maximum compressive force is the uppermost point
of any interaction diagram and is the other point correspondingto zero eccentricity:
Pn ¼ P0 ¼ 0.85f 0cbh; Mn ¼ 0 ð32Þ
These points provide a set of nominal moment-axial loadcouples, (Mn, Pn), that allows for the generation of a suffi-ciently smooth and accurate interaction diagram.
Interaction Diagram for Circular Cross Section
Fig. 2 displays the assumed parameters and the geometry of thesection. Similar to a rectangular cross section, the interaction curvemay be calculated at certain points based on the position of theN.A., x.1. x ¼ −∞
Maximum tensile force:
Pn ¼ Tmax ¼ −Afffd; Mn ¼ 0 ð33Þ2. x ¼ 0
If failure is initiated by rupture in the extreme layer ofreinforcement (εf ¼ εfd):
Pn ¼ −Afffd2γ
ð34Þ
Mn ¼ð2γ − 1Þ2
8γAfffdD ð35Þ
where γ ¼ d=D.3. xb ≤ x ≤ d
Again, xb marks the location of the neutral axis forbalanced failure:
xb ¼d
1þ e; e ¼ εfd
εcu; εcu ¼ 0.003 ð36Þ
Also for the neutral axis to be located in the reinforcedregion:
D d C.L.
N.A.
CT
x 1x
Afh
Fig. 2. Cross section of a circular column
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Max
�1
γ− 1;
1
1þ e
�≤ α ¼ x
d≤ 1 ð37Þ
For any arbitrary α within this range, the forces andmoments in concrete and bars may be calculated as
C ¼ 0.85f 0cðθC − sin θC cos θCÞD2
4ð38Þ
MC ¼ 0.85f 0csin3θCD3
12ð39Þ
where cos θC ¼ 1 − 2αβ1γ and 0 ≤ θC ≤ π.
T ¼ π cos θT − θT cos θT þ sin θTπð1þ cos θTÞ
1 − αα
εcuEfAf ð40Þ
MT ¼π− θT þ sinθT cosθT
2πð1þ cosθTÞ1−αα
�γ− 1
2
�εcuEfAfD ð41Þ
where cos θT ¼ ð1 − 2αγÞ=ð2γ − 1Þ and 0 ≤ θT ≤ π.
Pn ¼ C − T ð42Þ
Mn ¼ MC þMT ð43Þ4. d < x ≤ h
If the neutral axis lies within the concrete section, butbeyond the reinforced area, then the contribution of thereinforcement vanishes from the equations presented in con-dition 3. For example, if x ¼ D or α ¼ 1=γ:
Pn ¼ C ¼ 0.85f 0cðθC − sin θC cos θCÞD2
4ð44Þ
Mn ¼ MC ¼ 0.85f 0csin3θCD3
12ð45Þ
where cos θC ¼ 1 − 2β1 and 0 ≤ θC ≤ π.5. x ¼ þ∞
Maximum compressive force:
Pn ¼ P0 ¼ 0.85f 0cπD2
4; Mn ¼ 0 ð46Þ
P�M Diagram Example
To illustrate these formulas, the interaction diagrams of a squarecolumn and a circular one are plotted and discussed:Square column: bxh ¼ 600 × 600 mm ð24 × 24 in:ÞCircular column: D ¼ 600 mmð24 in:ÞConcrete: f 0
c ¼ 28 MPa psi ð4,000 ksiÞ; β1 ¼ 0.85Reinforcement: ffu ¼ 400 MPa ð60 ksiÞ;
Ef ¼ 40 GPa ð6,000 ksiÞ;εfu ¼ ffu=Ef ¼ 0.01 therefore:ffd ¼ 400 MPa ð60 ksiÞ; εfd ¼ 0.01
Square column: A1 ¼ 4Φ25 ¼ 1,963 mm2 ð48 ¼ 3.14 in:2Þ;A2 ¼ 2Φ25 ¼ 982 mm2 ð28 ¼ 1.57 in:2ÞAf ¼ 2ðA1 þ A2Þ ¼ 12Φ25
¼ 5,890 mm2 ð128 ¼ 9.42 in:2ÞCircular column: Af ¼ 12Φ25 ¼ 5,890 mm2 ð128 ¼ 9.42 in:2ÞCover to the center of bars: 63 mm (2.5 in.)
The design strength of GFRP, ffd, is deliberately selected sim-ilar to the yield stress of ordinary Grade 60 steel bars, fy ¼ 60 ksi(413 MPa), so that the two materials can be compared. Fig. 3demonstrates the points 1 to 5 on the interaction diagram of therectangular column and compares it with that of a similar sectionwith Grade 60 steel bars. Understandably, the compressive strengthof steel and its yielding ability allow its interaction diagram to sur-round the diagram associated with GFRP. Similarly, Eqs. (33–46)are plotted as an interaction diagram for the circular cross sectionin Fig. 4.
Strength Reduction Factor, ∅As ACI 440.1R-06 (ACI 2006) is silent about columns, a method tocalculate the strength reduction factor for columns is presented onthe basis of the generalization of the current flexural reductionfactors of ACI 440.1R-06. This method follows the approachtaken by ACI 318-11 (ACI 2011a) in unifying the strength reduc-tion factors of columns and beams as a function of the maximum
-4000
-2000
0
2000
4000
6000
8000
10000
12000
0 200 400 600 800 1000 1200
Pn
(kN
)
Mn(kN-m)
a: pure tension
e: pure compression
b: x=0
c: balanced failure
c: x=dd: x=h
(Mn, Pn) FRP reinforced
(Mn, Pn) steel reinforced
Compression controlled failure
Tension controlled failure
Fig. 3. Interaction diagram of GFRP- and steel-RC columns (squarecross section)
-4000
-2000
0
2000
4000
6000
8000
10000
0 100 200 300 400 500 600 700 800
a: pure tension
b: x=0
c: x=dd: x=D
e: pure compression
c: balanced failure
(Mn, Pn) steel reinforced
(Mn, Pn) FRP reinforced
Compression controlled failure
Tension controlled failure
Mn (kN-m)
Pn(k
N)
Fig. 4. Interaction diagram of GFRP- and steel-RC columns (circularcross section)
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tensile strain the reinforcement. For beams, ACI 440.1R-06recommends:• A strength reduction factor of ∅ ¼ 0.55 for tension-controlled
sections. A section is categorized as tension-controlled if itsfailure is initiated by GFRP rupture. For a beam this can betranslated to ρf ≤ ρfb.
• A strength reduction factor of ∅ ¼ 0.65 for compression-controlled sections. A section is categorized as compression-controlled if its failure is initiated by concrete crushing. For abeam this can be translated to ρf ≥ 1.4ρfb.To avoid discontinuity, ACI 440.1R-06 inserts a transitional
range between the two modes (ρfb ≤ ρf ≤ 1.4ρfb) in which thereduction factor is linearly interpolated between 0.55 and 0.65 fromthe reinforcement ratio, ρf. This can be formulated as
∅ ¼
8>>><>>>:
0.55 ∶if ρf ≤ ρfb ¼ 0.85β1f 0c
ffuεcu
εcuþεfu
0.30þ 0.25 ρfρfb
∶if ρfb ≤ ρf ≤ 1.4ρfb
0.65 ∶if ρf ≥ 1.4ρfb
ð47Þ
ρf, however, is not the only decisive factor in determining thefailure mode when axial forces are present. Therefore, similar toACI 318-11, to introduce a unified strength reduction factor appli-cable to both beams and columns, the failure mode is linked to thestrain in the extreme tensile layer of reinforcement (εf in this studyas opposed to εt in ACI 318-11).
For beams, the relationship between reinforcing ratio and itstensile strain is
ρf ¼ 0.85β1
f 0c
ff
εcuεcu þ εf
ð48Þ
Therefore:
ρfρfb
¼ ffuff
εcu þ εfdεcu þ εf
¼ εfuεf
εcu þ εfdεcu þ εf
ð49Þ
And the reduction factor can be reformulated as
0.55 ≤ ∅ ¼ 0.30þ 0.25εfdεf
εcu þ εfdεcu þ εf
≤ 0.65 ð50Þ
This generalized format of the reduction factor is suitable forcolumns. It can also be rewritten in terms of the dimensionlessparameters introduced previously:
0.55 ≤ ∅ ¼ 0:30þ 0:25eð1þ eÞ α2
1 − α≤ 0.65 ð51Þ
Figs. 5 and 6 display the interaction diagrams of the designstrength (∅Mn, ∅Pn) for the square and circular columns pre-viously discussed. The “flat-top” segment of the design strengthcurve defines the limiting axial load strength corresponding to0.8∅Po as per 318-11 (ACI 2011a). As expected, the reductionfactor of 0.65 is applicable to the upper part of the diagram(compression-controlled failure), which covers most of the practi-cal cases.
Shear Strength, Vn
The nominal shear strength of a column, Vn, is the sum of thecontributions of concrete, Vc, and GFRP stirrups in the form ofties, Vf:
Vn ¼ Vc þ Vf ð52Þ
• Concrete contribution, VcAccording to ACI 440.1R-06, the concrete shear capacity,
Vc, of flexural members using GFRP as main reinforcementcan be evaluated as
Vc ¼2
5
ffiffiffiffiffif 0c
pbwc∶ ðSI unitsÞ;
Vc ¼ 5ffiffiffiffiffif 0c
pbwc∶ ðImperial unitsÞ ð53Þ
where c = compressive depth of the neutral axis in the fullycracked section. For members with only one layer of tensilereinforcement and subject to no axial load (Pu ¼ 0), c is calcu-lated according to ACI 440.1R-06 as
cd¼ kc ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðρfηfÞ þ ðρfηfÞ2
q− ðρfηfÞ ð54Þ
where ρf = reinforcement ratio of the longitudinal GFRP bars;and ηf = modular ratio (ηf ¼ Ef=Ec).
-4000
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
0 100 200 300 400 500 600 700 800
(Mn, P
n)
M (kN-m)
P (
kN) (ØM
n, ØP
n)
0.80ØP0
Ø=0.65
Ø=0.55
Fig. 5. Interaction diagram of a GFRP-RC column (square column inFigs. 1 and 3)
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
6000
7000
8000
0 100 200 300 400 500
P (k
N)
M (kN-m)
(Mn, P
n)
(ØMn, ØP
n)
0.80ØP0
Ø=0.65Ø=0.55
Fig. 6. Interaction diagram of GFRP-RC column (circular column inFigs. 2 and 4)
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However, multiple layers of reinforcement, as in columns,render this equation inaccurate. To generalize the formulationfor the rectangular columns, the location of the neutral axisin a fully cracked section can be calculated. Equalizing themoments of area of the compressive part and the transformedtensile part of the section around a separating neutral axislocated at x ¼ kd, one obtains:
k2c ¼ 2ðρf1ηfÞð1 − kcÞ þ 2ðρf2ηfÞγ
2γ − 1ð1 − kcÞ2 ð55Þ
where ρf1 ¼ Af1=bd and ρf2 ¼ Af2=bd.The last term of Eq. (55) accounts for the presence of the
two side layers of reinforcement. Although this equation can besolved for the exact value of kc, the final closed-form solution israther unwieldy. Because the exact value of k is of little practicalimportance, a simpler iterative solution is presented here as analternative. Defining the “effective” tensile reinforcement ratio,ρf , as
ρf ¼ ρf1 þγð1 − kcÞ2γ − 1
ρf2 ð56Þ
and starting with an initial guess for kc (kc ¼ 0.2 is recom-mended), the equation presented by ACI 440.1R-06 can be usedrepetitively to calculate kc.
In the example of the square column previously investigated:d ¼ 600 − 63 ¼ 537 mm ð21.5 in:Þ,γ ¼ d=h ¼ 537 mm=600 mm ¼ 0.895Concrete: Ec ¼ 4.75
pf 0c ¼ 25 GPa ð3,600 ksiÞ
Reinforcement: ηf ¼ Ef=Ec ¼ 40 GPaÞ=ð25 GPaÞ ¼ 1.60ρf1 ¼ ð1,963 mm2Þ=½ð600 mmÞð537 mmÞ� ¼ 0.00609ρf2 ¼ ð982 mm2Þ=½ð600 mmÞð537 mmÞ� ¼ 0.00305Assuming kc ¼ 0.2ρf ¼ 0.00885and kc can be calculated according to ACI 440.1R-06 askc ¼ 0.155
The new value of kc is of sufficient accuracy, but to demon-strate the convergence trend, another step is performed:Assuming kc ¼ 0.157ρf ¼ 0.00901, and according to ACI 440.1R-06:kc ¼ 0.156c ¼ 84 mmð3.4 in:ÞVc ¼ 106.7 kN ð25.8 kipsÞ
This value provides a conservative estimate of the concretecontribution if the column is subjected to compressive axialforces. In the uncommon case of tensile axial loads, the use ofthis equation is not recommended. Instead, Eq. (57) must beemployed, which, with negligible approximation, predicts thelocation of the neutral axis under the combined effect of axialload and flexure:
0 ≤ cd¼ kc þ
k2cð1 − kc3Þ
2
�PudMu
�≤ 1 ð57Þ
where Pu andMu = ultimate axial load and moment correspond-ing to Vu. Compressive axial force is positive, tensile axial forceis negative, and the moment is always positive. With c fromthis equation, Eq. (53) may be used to achieve a more accurateassessment of the shear strength provided by concrete.
• Shear reinforcement contribution, VfOnce Vc is calculated, the shear strength provided by ties of
the column, Vf , can be calculated as
Vf ¼ Afvffvd
sð58Þ
Afv is the area of GFRP ties within a spacing of s whose limitswere discussed previously.
Other parameters are briefly discussed, as they are similarto beams. The tensile strength of GFRP for shear design,ffv, is calculated as
ffv ¼ 0.004Efv ≤ ffb ð59Þwhere Efv = modulus of elasticity of the GFRP ties; ffb,strength of bent portion of GFRP stirrups, depends on thatrb=db, the ratio of internal radius of bend in stirrups to theirdiameter, which has to be at least equal to 3, the minimumrecommended by the ACI 440 guideline.
ffb ¼�0.05
rbdb
þ 0.3
�ffu ≤ ffu ð60Þ
ffu = ultimate longitudinal tensile strength of GFRP stirrups.Jawaheri Zadeh and Nanni (2012) showed that the current
shear strength reduction factor of 0.75 may be maintained solong as the maximum effective level of shear resistance of amember with GFRP shear reinforcement does not exceedfour times the strength provided by concrete (Vn ≤ 4Vc or,equally, Vf ≤ 3Vc).
For circular sections, the same equations can be used if anequivalent rectangular section is defined as
bw ¼ D; d ¼ 0.8D; A1 ¼ A2 ¼ Af=4 ð61Þ
The equivalent dimensions replicate the provisions of ACI318 for shear strength of circular sections.Appendix provides a practical example that employs the design
formulations presented in this paper.
Conclusions
The paper shows that, with the existing knowledge, it is possible todevelop a methodology for the design of concrete columns withrectangular or circular cross section using GFRP bars and ties.The design methodology presented in the paper highlights a fewimportant aspects affecting computations and safety:• Interaction diagrams are developed assuming that GFRP long-
itudinal bars are only effective in tension. When subject tocompression, they can be replaced with the equivalent area ofconcrete as if they were not present in the cross section.
• The unified strength reduction factor formulation for the inter-action diagram can be based on the present ACI 440.1R-06values established for flexure (ACI 2006).
• To avoid exaggerated deflections, a limit of 1% is imposed onthe maximum design strain of GFRP longitudinal bars.
• On the basis of ACI 440.1R, the contribution of concrete toshear strength is reformulated to accommodate columns.
• Modification factors for the flexural stiffness of GFRP-RCmembers are proposed. These factors are functions of the mod-ulus of elasticity of GFRP bars and their application incorpo-rates this parameter in the analysis results. These factors are ofspecial importance when lateral deflections and second-ordermoments are of concern.
Appendix. Column Design Example
Design column B3 in the first story of the two-story hospital build-ing with a floor plan shown in Fig. 7. The clear height of the firststory is 4.10 m (13 ft,6 in.) (Fig. 8). Assume that the lateral load
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effects on the building are caused by a wind force of 450 kN(100 kips) applied conservatively at the roof level, and that the deadloads are the only sustained loads. Other pertinent design data forthe building are as follows:Concrete: f 0
c ¼ 35 MPa ð5,000 psiÞReinforcement: ffu ¼ 434 MPa ð63 ksiÞ;
Ef ¼ 46 GPa ð6,700 ksiÞεfu ¼ 0.0094 < 0.010 therefore:ffd ¼ ffu ¼ 434 MPa ð63 ksiÞ
Beams: 750 × 350 mm ð30 × 14 in:ÞSlab: 200 mm ð8 in:ÞColumns: 500 × 500 mm ð20 × 20 in:ÞRoof uplift: 1.75 kN=m2 ð35 psfÞ
Wind loads are computed according to ASCE 7-10 (ASCE2010). Loading of the first floor is according to Fig. 7, with MRIequipment weight included as dead load. The self-weight of thestructural is applied separately.1. Axial force and bending moments for column B3 in the
first storyTo analyze the structure and calculate the lateral deflec-
tions, the stiffness of each member type is modified asIbeam ¼ ½0.075þ 0.275ðEf=EsÞ�Ig ¼ 0.139Ig [Eq. (11)]Islab ¼ ½0.10þ 0.15ðEf=EsÞ�Ig ¼ 0.135Ig [Eq. (12)]Icolumn ¼ ½0.40þ 0.30ðEf=EsÞ�Ig ¼ 0.469Ig [Eq. (15)]
Table 1 presents the results of the structural analysis ofcolumn B3 of the frame displayed in Fig. 8. It is assumed that
the column is subject to monoaxial bending around the B-axisline. Negative values of axial load signify tension. The differ-ence in the moment sign at the top and bottom of the columnindicates double curvature.
2. Load combinationsThe forces and moments of Table 1 can be combined ac-
cording to ACI 318-11 to obtain the ultimate factored loadsthat are used for design. Table 2 presents the ultimate loads.
3. Determining the story type (sway or nonsway)According to ACI 318-11, a column in a structure can be
considered nonsway if the column end-moments attributableto second-order effects do not exceed 5% of the first-orderend-moments. It is also permitted to assume a story within astructure is nonsway if
Q ¼ ðΣPuΔ0Þ=ðVulcÞ ≤ 0.05 ð62Þwhere Q = stability index; ΣPu = total vertical load in thestory; Vu = total story shear;Δ0 = first-order relative displace-ment between top and bottom of the story; lc = length of thecolumn center to center of joints; and lc ¼ 4,850 − ð750=2Þ ¼4,475 mmð176 in:Þ
Table 3 shows that under no load combination is the framesway type (Q ≤ 0.05).
4. Slenderness effectsFor compression members in a nonsway frame, effects of
slenderness may be neglected whenklu=r ≤ ðklu=rÞmax ¼ 29 − 12ðM1=M2Þ ≤ 35 [Eq. (16)]k ¼ 1.0: effective length factor of for nonsway frameslu ¼ 4,100 mmð162 in:Þ: clear height of the columnr ¼ ð0.3Þð500 mmÞ ¼ 150 mmð6.0 in:Þ: radius of the gyra-tion of the columnklu=r ¼ 27.0
From Table 2, it can be concluded that the minimum valueof the ðklu=rÞmax is associated with the case of single curva-ture, thereforeðklu=rÞmax ¼ 29 − 12ð15.5=154.6Þ ¼ 27.8klu=r ¼ 27 < ðklu=rÞmax
Fig. 7. First-story slab plan and loading
Fig. 8. Frame geometry
Table 2. Analysis Results for Load Combinations (Column B3)
Load combinationAxial load:Pu (kN)
Bendingmoment: Mu(kN · m) Shear:
Vu (kN)Top Bottom
1.4D 1,164.0 99.3 −68.8 40.91.2Dþ 1.6L 1,865.0 85.0 −58.9 34.71.2Dþ 1.6Wþ þ 1.0L 1,293.5 15.5 154.6 −33.81.2Dþ 1.6W− þ 1.0L 1,293.5 154.6 −272.5 103.60.9Dþ 1.6Wþ 501.7 −5.8 169.5 −42.30.9Dþ 1.6W− 501.7 133.4 −257.9 94.7
Note: 1 kN ¼ 0.225 kip; 1 kN · m ¼ 0.74 ft · kip.
Table 1. Analysis Results for Individual Load Cases (Column B3)
Load caseAxial load:P (kN)
Bending moment:M (kN · m) Shear:
V (kN)Top Bottom
Dead (D) 831.3 70.9 −49.1 29.4Live (L) 542.2 0.0 0.0 0.0Wind (Wþ) −154.3 −43.5 133.6 −42.7Wind (W−) −154.3 43.5 133.6 43.1
Note: 1 kN ¼ 0.225 kip; 1 kN · m ¼ 0.74 ft-kip.
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Hence, slenderness effects may be disregarded.5. Bars selection and section details
Fig. 9 presents the designated longitudinal and transversereinforcement of the column.Af ¼ 8Φ25 ¼ 3,927 mm2ð88 ¼ 6.28 in:2Þρf ¼ 1.57%
6. Interaction diagramUsing the equations previously presented, the interaction
diagrams of the cross section detailed in Fig. 9 is plotted inFig. 10. The points denote the demand, (Mu, Pu) according toTable 2. As it can be seen, demand is totally enveloped by thedesign capacity and, therefore, it can be stated that the sectionis of sufficient capacity.
7. Shear strength controlAccording to Table 2, Vu;max ¼ 103.6 kNð23.3 kipsÞ.
• Concrete contribution, Vc:Square column:b × h ¼ 500 × 500 mmð20 × 20 in:ÞCover to the center of bars: 2.5 in.d ¼ 500 − 63 ¼ 437 mmð17.5 in:Þγ ¼ d=h ¼ 437=500 ¼ 0.874Concrete:f 0c ¼ 35 MPa ð5,000 psiÞ
Ec ¼ 4.75pf 0c ¼ 28 GPa ð4030 ksiÞ
Reinforcement:Ef ¼ 46 GPa ð6,700 ksiÞηf ¼ Ef=Ec ¼ 46 GPaÞ=ð28 GPaÞ ¼ 1.64A1 ¼ 1,473 mm2 ð2.36 in:Þ,ρf1 ¼ ð1,473 mm2Þ=½ð500 mmÞð437 mmÞ� ¼ 0.00674A2 ¼ 491 mm2ð0.78 in:Þ,ρf2 ¼ ð491 mm2Þ=½ð500 mmÞð437 mmÞ� ¼ 0.00225Assuming kc ¼ 0.2ρf ¼ 0.00882 [Eq. (56)]and k can be calculated according to ACI 440.1 askc ¼ 0.157 [Eq. (54)]c ¼ 69 mm ð2.75 in:Þ [Eq. (54)]Vc ¼ 81.6 kN ð19.4 kipsÞ [Eq. (53)]Since the column, under all combinations, is subject to com-
pressive axial force, this is a conservative estimate of Vc. It canalso be obtained with more details from Eqs. (53) and (57):
Pu ¼ 1; 293.5 kN ð290.8 kipsÞ (Table 2)Mu ¼ 272.5 kN · m ð200.4 ft · kipÞ (Table 2)c=d ¼ 0.182 [Eq. (57)]c ¼ 81 mmð3.18 in:Þ [Eq. (54)]Vc ¼ 95.8 kN ð22.5 kipsÞ [Eq. (53)]
• GFRP ties contribution, Vf:Vf is calculated from Eq. (43), substituting:Afv ¼ Φ10@200 mm ð3@8 in:Þ ¼ 157 mm2 ð0.22 in:2Þs ¼ 200 mmð8 in:Þd ¼ 437 mm ð17.5 in:Þffv ¼ 0.004Efv ≤ ffb [Eq. (59)]ffb ¼ ð0.05rb=db þ 0.3Þffu ≤ ffu [Eq. (60)]Efv ¼ 46 GPa ð6,700 ksiÞffv ¼ 184 MPa ð26.8 ksiÞrb=db ¼ ðrb=dbÞmin ¼ 3.0ffb ¼ 0.45ffu ¼ 195 MPa ð28.4 ksiÞ > 84 MPa ð26.8 ksiÞffv ¼ 184 MPa ð26.8 ksiÞVf ¼ 63.1 kN ð12.9 kipsÞ [Eq. (58)]Therefore, the total shear capacity can be computed accord-
ing to Eq. (52) and the factored capacity is derived consider-ing ∅ ¼ 0.75:
Vn ¼ 158.9 kN ð35.4 kipsÞ∅Vn¼119.2kNð26.6kipsÞ>Vu¼103.6kNð23.3kipsÞ OK
Acknowledgments
The authors gratefully acknowledge the financial support for thisresearch provided by the National Science Foundation (NSF) underGrant No. IIP-0933537 as well as the contribution of the industrymembers to the NSF Industry/University Cooperative ResearchCenter based at the University of Miami.
Notation
The following symbols are used in this paper:Af = area of GFRP reinforcement, mm2 (in:2);
Φ10@200 (#3@8)
8Φ25 (8#8)
500 mm (20 in.)
500 mm (20 in.)
63 mm (2.5 in.)
Fig. 9. Section details (Column B3)
-2000
-1000
0
1000
2000
3000
4000
5000
6000
7000
8000
0 100 200 300 400 500 600
(Mn, Pn)
(ØMn, ØPn)
(Mu, Pu)
M(kN-m)
P(k
N)
0.80ØP0
Fig. 10. Nominal and design interaction diagrams of the section inFig. 9 versus demand
Table 3. Calculation of Stability Index, Q
Load combinationΣPu(kN)
Vu(kN)
Δ0
(mm) QFrametype
1.4D 1,164.0 40.9 0.0 0.00 Nonsway1.2Dþ 1.6L 1,865.0 34.7 0.0 0.00 Nonsway1.2Dþ 1.6Wþ þ 1.0L 1,293.5 −33.8 17 0.039 Nonsway1.2Dþ 1.6W− þ 1.0L 1,293.5 103.6 −17 0.039 Nonsway0.9Dþ 1.6Wþ 501.7 −42.3 17 0.015 Nonsway0.9Dþ 1.6W− 501.7 94.7 −17 0.015 Nonsway
Note: 1 kN ¼ 0.225 kip; 1 kN · m ¼ 0.74 ft · kip; 1 mm ¼ 0.04 in.
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Afv = area of GFRP reinforcement in shear, mm2 (in:2);A1, A2 = area of GFRP reinforcement along sides 1 and 2 of
a rectangular section, mm2 (in:2);b = width of rectangular cross section, mm (in.);
bw = width of the web, mm (in.);C = compressive force, kN (kip);c = distance from extreme compression fiber to the
neutral axis of a fully cracked section, mm (in.);D = diameter of circular cross section, mm (in.);d = distance from extreme compressive fiber to
centroid of tensile reinforcement, mm (in.);Ec, Es = modulus of elasticity of concrete and steel, GPa
(ksi);Ef , Efv = modulus of elasticity of GFRP longitudinal bars
and ties, GPa (ksi);e = ratio of εcu over εfu;
ffu = tensile strength of the GFRP reinforcement, MPa(ksi);
ffv = tensile strength of the GFRP reinforcement inshear, MPa (ksi);
ffb = strength of bent portion of GFRP ties, MPa (ksi);fy = steel yield strength, MPa (ksi);f 0c = compressive strength of concrete, MPa (ksi);h = height of rectangular cross section, mm (in.);I = moment of inertia of section about centroidal axis,
mm4 (in:4);Ig = moment of inertia of gross concrete section,
neglecting reinforcement mm4 (in:4);k = effective length factor;kc = ratio of depth of neutral axis to reinforcement
depth;Lc, lu = length of the column center to center of joints, clear
height of the column, mm (in.);L1, L2, LC = arm in moment computation, mm (in.);
MC, MT = contributions to nominal moment capacity forcircular section, kN · m (ft · kip);
Mn, Mu = nominal moment capacity and demand, kN · m(ft · kip);
Po = nominal axial capacity for zero eccentricity, kN(kip);
Pn, Pu = nominal axial capacity and demand, kN (kip);Q = stability index;r = radius of gyration of the column, mm (in.);
rb, db = radius of bend and diameter of tie, mm (in.);s = tie spacing, mm (in.);
T, Tmax = tensile force and maximum tensile force, kN (kip);T1, T2 = tensile force corresponding to A1 and A2, kN (kip);Vc, Vf = contributions of concrete and GFRP to nominal
shear capacity, kN (kip);Vn = nominal shear capacity, kN (kip);
x, xb = distance of N.A. from compression edge and atbalanced condition, mm (in.);
α = ratio of x over d;β1 = factor relating depth of equivalent stress block to
neutral axis depth;γ = ratio of d over h or of d over D;
Δ0 = first-order relative displacement between top andbottom of the story, mm (in.);
εc, εcu = compressive strain and ultimate compressive strainfor concrete;
εf, εfd, εfu = tensile, design, and ultimate strain for GFRPreinforcement;
εy = yield strain of steel;θC, θT = angles measured according to Fig. 2 (rad);
ηf = modular ratio Ef over Ec;ρ = reinforcement ratio;
ρf , ρfb = reinforcement ratio and balanced reinforcementratio for GFRP-RC members;
ρf1, ρf2 = reinforcement ratios corresponding to A1 and A2;ρst = reinforcement ratio for steel-RC columns; and∅ = strength reduction factor.
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