# Mathematical Model for Analysis of Uniaxial and Biaxial Reinforced Concrete Columns

Authors: Mohammed Salem Al-Ansari; Muhammad Shekaib Afzal
Journal: Advances in Civil Engineering
 (2020)

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## Abstract

This paper presents a mathematical model for the analysis of reinforced concrete (RC) uniaxial and biaxial columns. This proposed model is a quick and faster approach for the analysis and design of reinforced concrete rectangular columns without going through the interaction charts procedure as well as other iterative methods for the computation of required axial load capacity (Pc) and moment capacity (Mc). A simplified flow chart has also been developed to find the required column capacity using this mathematical model. Eight uniaxial columns (C-1 to C-8) and seven biaxial columns (CB-1 to CB-7) are analysed in this study. Each column is analysed having different steel reinforcement ratiosρ with different loading conditions. In addition, the studied columns are subjected to both tension and compression failures. The detailed examples for both uniaxial and biaxial columns (one for each case) are also presented in this study. The studied columns are also analysed using computer software spColumn. The average variation of the mathematically computed values to the finite element software is not more than 10%, showing promising computational results.


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## Body

## 1. Introduction

Columns are the vertical compression members, which transmit loads from the upper floors to the lower levels and to the soil through the foundations [1]. Based on the position of the load on the cross section, columns are classified as concentrically loaded (Figure 1) or eccentrically loaded columns (Figure 2). Eccentrically loaded columns are subjected to moments, in addition to axial force. The moments can be converted to a load P and eccentricities eXandeY. The moments can be uniaxial, as in the case when two adjacent panels are not similarly loaded, such as columns A and B in Figure 3 [2]. A column is considered as biaxially loaded when the bending occurs about the x- and y-axis, such as in the case of corner column C in Figure 3. In a recent study [3], Al-Ansari and Afzal also presented an analytical model for generating interaction diagram charts for biaxial columns.Figure 1
Concentrically loaded columns.Figure 2
Eccentrically loaded column.
(a)(b)Figure 3
Uniaxially and biaxially loaded column.The strength of reinforced concrete columns is normally expressed using interaction diagrams to relate the design axial load2∅Pn to the design bending moment ∅Mn [4, 5]. Each control point on the column interaction curve ∅Pn−∅Mn represents one combination of design axial load, ∅Pnand design bending moment, ∅Mn, corresponding to a neutral-axis location (Figure 4) [6].Figure 4
Control points for column interaction curve∅Pn−∅Mn.Extensive studies have been carried out on the interaction diagrams (uniaxial and biaxial columns) of reinforced concrete (RC) rectangular columns [6–12]. Several studies have also been performed on providing numerical approaches for the analysis and design of reinforced concrete columns. Furlong et al. [13] provided an overview of the analysis and design of reinforced concrete columns subjected to biaxial bending. They reviewed several methods of analysis that use traditional design methods and compared their results with the obtained data from physical tests of normal strength concrete columns subjected to short-term axial loads and biaxial bending’s. They concluded that the elliptic load contour equation [14] and the reciprocal equation [15] are the simplest to use, as they do not require complicated calculations.Chen et al. [16] proposed an iterative numerical method for rapid section analysis and design of short concrete composite columns subjected to biaxial bending. Wang and Hsu [17] proposed the numerical method approach for the determination of load-moment curvature relationship for short and slender columns. This numerical method approach is also applicable for columns, made of different materials, and shows good agreement with the different experimental results obtained in their study.Whitney [18] and Hsu et al. [19] provided major research studies on numerical method approaches. Whitney suggested an approximate equation to estimate the nominal compressive strength of columns subjected to compression failure. Hsu in different research projects [10, 17, 20, 21] also presented the results of experimental and analytical studies on the strength and deformation of biaxially loaded short and tied columns with L−, channel, and T-shaped cross sections. In another study, Hsu [22] suggested a general equation for the analysis and design of reinforced concrete short and tied rectangular columns.This study proposes a mathematical model to analyse and design the uniaxial and biaxial columns based on ACI building code of design [23]. This model is a quick and easy approach for analysing and designing the reinforced rectangular columns without going through the interaction charts for the computation of the required axial load capacity (Pc) and moment capacity Mcx,Mcy. A simplified flow chart has also been developed to find the required column capacity using the proposed model approach. The previous research studies of the mathematical model approach are limited to columns having compression failure only. This study includes the numerical examples of columns using the proposed mathematical model approach for both compression and tension failure cases. This relatively new approach will also be useful to the undergraduate and graduate students as well as researchers to calculate the required column capacities using this approach in their research-related activities.Numerical examples for the selected reinforced concrete columns (uniaxial and biaxial columns) are also illustrated to check the adequacy of this proposed model. Eight uniaxial columns (C-1 to C-8) and seven biaxial columns (CB-1 to CB-7) are analysed in this study. These columns are analysed having different steel reinforcement ratiosρ, different values of steel yield strength fy, concrete compressive strength fc′ , and different load capacity conditions. Moreover, the results obtained from this proposed model are compared with computer software spColumn 2016 [24].


## 2. Mathematical Model Formulation: ACI Code Design

The stress and strain distribution of a rectangular column section (uniaxial column) for the calculation of Pn and Mn is given in Figure5.Figure 5
Calculation of Pn and Mn for a given strain distribution. (a) Section. (b) Strain. (c) Stress. (d) Internal forces. (e) Resultant forces.The resultant forcePN is equal to the summation of all internal forces:(1)PN=CCon−Ts+CS.Similarly, the resultant momentMN is equal to the summation of all internal moments:(2)MN=Mconc+MT+MCS.The following steps revealed the calculation of the required internal forces and internal moments for a rectangular uniaxial RC column.


### 2.1. Plain Concrete Section

The Internal concrete compressive forceCconc is computed as(3)Cconc=0.85fc′ba,Cconc=0.85fc′bβc,where Cconc = internal concrete compression force, fc′ = compressive concrete strength, b = column width, a = depth of the compression stress block, β=0.85−0.008fc′−30≥0.65, and c = distance from extreme compression fiber to the neutral axis.Referring to Figure5, the moment about the midpoint of the section Mconc can be computed as(4)Mconc=Ccd−a2−d″,Mconc=0.85fc′bad−a2−d″,where h = column total depth, d″=h/2−d′, d = column effective depth h−d′, and d′ = distance from extreme compression fiber to centroid of top reinforcing steel.


### 2.2. Tension Steel Section

The internal tensile forceTs is computed as(5)Ts=Asfy,where As = area of tensile steel reinforcement and fy = yield stress of reinforcing steel.The internal moment MT is(6)MT=Asfyd″.


### 2.3. Compression Steel Section

The internal compressive forceCs is computed as [25](7)Cs=As′fs′,Cs=As′fs′−0.85fc′,where As′ = area of compression steel reinforcement and fs′=fy (if the compression steel yields).The internal momentMT is(8)MT=As′fs′−0.85fc′d−d′−d″.


## 2.1. Plain Concrete Section

The Internal concrete compressive forceCconc is computed as(3)Cconc=0.85fc′ba,Cconc=0.85fc′bβc,where Cconc = internal concrete compression force, fc′ = compressive concrete strength, b = column width, a = depth of the compression stress block, β=0.85−0.008fc′−30≥0.65, and c = distance from extreme compression fiber to the neutral axis.Referring to Figure5, the moment about the midpoint of the section Mconc can be computed as(4)Mconc=Ccd−a2−d″,Mconc=0.85fc′bad−a2−d″,where h = column total depth, d″=h/2−d′, d = column effective depth h−d′, and d′ = distance from extreme compression fiber to centroid of top reinforcing steel.


## 2.2. Tension Steel Section

The internal tensile forceTs is computed as(5)Ts=Asfy,where As = area of tensile steel reinforcement and fy = yield stress of reinforcing steel.The internal moment MT is(6)MT=Asfyd″.


## 2.3. Compression Steel Section

The internal compressive forceCs is computed as [25](7)Cs=As′fs′,Cs=As′fs′−0.85fc′,where As′ = area of compression steel reinforcement and fs′=fy (if the compression steel yields).The internal momentMT is(8)MT=As′fs′−0.85fc′d−d′−d″.


## 3. Mathematical Model Analysis

The following steps should be revealed to calculate the design axial load and moment capacity of the required rectangular RC column section. Columns may be subjected to tension failure or compression failure depends on the balanced eccentricity valueeb:(9)eb=MbPb,(10)Mb=Ccbd−ab2−d″+Csd−d′−d″+Ts×d″×10−3,(11)Pb=Cs+Ccb−Ts,where ab=β×cb, cb=600×d/600+fy, Ccb=0.85×fc′×ab×b×10−3, and Cs=As′fs′−0.85fc′×10−3 (if the compression steel yields, then fs′=fy).


### 3.1. Tension Failure Analysis

Tension failure will occur when the balanced eccentricity valueeb is less than load eccentricity e. Substituting the values of Cc,Cs, and Ts in equation (1) and solving for a will be a second-degree equation [14]:(12)PN=CCon−Ts+CSEquation1,Aa2+Ba+C=0,where A=0.85×fc′×b/2, B=0.85×fc′×b×e′−d, C=As′fs′−0.85fc′e′−d+d′−fyAse′, and e′=e+d″, e′=e+d−h/2whenAs=As′.Solve fora to get(13)a=−B±B2−4AC2A.Substitute the value ofa in equation (3) to calculate Cc and from equations (5) and (7) to compute Ts and Cs values. These obtained values are substituted in equation (1) for PN=CCon−Ts+CS,(14)PN=1e′Ccd−a2+Csd−d′,(15)MN=PN×e.The column axial load capacity and moment capacity can therefore be computed as(16)Pc=∅PN,(17)Mc=∅MN,(where ∅ is the column reduction factor having the value of 0.65).


### 3.2. Compression Failure Analysis

Compression failure will occur when the balanced eccentricity valueeb is bigger than the load eccentricity (e). Substituting the values of Cc,Cs, and Ts in equation (1) and solving for (a) will be a cubic equation [14]:(18)Aa3+Ba2+Ca+D=0,where A=0.85×fc′×b/2, B=0.85×fc′×b×e′−d, C=As′fs′−0.85fc′e′−d+d′+600Ase′, and D=−600Ase′β′d,where e′=e+d″e′=e+d−h/2whenAs=As′.Once the values ofA,B,C,andD are calculated, the value of a can be determined by the trial method or directly by using MATLAB or any scientific calculator. Moreover, the cubic equation can also be solved using different numerical methods, for example, Newton Raphson Method. After getting the required value of a, similar equations from (14) to (18) (as mentioned in the Tension Failure Analysis) should be used to get the required value of column axial load capacity Pc and moment capacity Mc.The following flow chart (Figure6) can be followed to find the required capacity of the rectangular uniaxial column section.Figure 6
Flow chart of the mathematical model for the rectangular uniaxial rectangular column.


## 3.1. Tension Failure Analysis

Tension failure will occur when the balanced eccentricity valueeb is less than load eccentricity e. Substituting the values of Cc,Cs, and Ts in equation (1) and solving for a will be a second-degree equation [14]:(12)PN=CCon−Ts+CSEquation1,Aa2+Ba+C=0,where A=0.85×fc′×b/2, B=0.85×fc′×b×e′−d, C=As′fs′−0.85fc′e′−d+d′−fyAse′, and e′=e+d″, e′=e+d−h/2whenAs=As′.Solve fora to get(13)a=−B±B2−4AC2A.Substitute the value ofa in equation (3) to calculate Cc and from equations (5) and (7) to compute Ts and Cs values. These obtained values are substituted in equation (1) for PN=CCon−Ts+CS,(14)PN=1e′Ccd−a2+Csd−d′,(15)MN=PN×e.The column axial load capacity and moment capacity can therefore be computed as(16)Pc=∅PN,(17)Mc=∅MN,(where ∅ is the column reduction factor having the value of 0.65).


## 3.2. Compression Failure Analysis

Compression failure will occur when the balanced eccentricity valueeb is bigger than the load eccentricity (e). Substituting the values of Cc,Cs, and Ts in equation (1) and solving for (a) will be a cubic equation [14]:(18)Aa3+Ba2+Ca+D=0,where A=0.85×fc′×b/2, B=0.85×fc′×b×e′−d, C=As′fs′−0.85fc′e′−d+d′+600Ase′, and D=−600Ase′β′d,where e′=e+d″e′=e+d−h/2whenAs=As′.Once the values ofA,B,C,andD are calculated, the value of a can be determined by the trial method or directly by using MATLAB or any scientific calculator. Moreover, the cubic equation can also be solved using different numerical methods, for example, Newton Raphson Method. After getting the required value of a, similar equations from (14) to (18) (as mentioned in the Tension Failure Analysis) should be used to get the required value of column axial load capacity Pc and moment capacity Mc.The following flow chart (Figure6) can be followed to find the required capacity of the rectangular uniaxial column section.Figure 6
Flow chart of the mathematical model for the rectangular uniaxial rectangular column.


## 4. Numerical Examples for Uniaxial Columns

Eight reinforced rectangular columns (C-1 to C-8) having different column sizes are analysed using the numerical method approach. These columns are having different reinforcement ratiosρ in addition to different failure types, both tension and compression failures. The design input load data for these columns are illustrated in Table 1.Table 1
Uniaxial column input data.
Column identifierPu (kN)Mu (kN·m)As (mm2)As′ (mm2)fc′ (MPa)fy (MPa)ϕd′ (mm)e (mm)C1 (200 × 400)30060400400303000.780200C2 (200 × 400)20050400400203000.760250C3 (300 × 500)80020010001000304150.775250C4 (400 × 400)40010010001000304150.6580250C5 (300 × 450)1300157.415301530253000.6575121C6 (200 × 400)40020402402303000.76050C7 (300 × 500)60010015001500304150.6590167C8 (400 × 800)100020020002000304150.6580200The above eight columns C1 to C8 are analysed using the mathematical model approach to find the required values of axial load capacity, Pc, and moment capacity, Mc. Moreover, these values are also compared with the computer software spColumn. The results obtained are depicted in Table2.Table 2
Uniaxial column design results.
Column identifiereb (mm)Failure conditionMathematical modelspColumnPc (kN)Mc (kN·m)Pc (kN)Mc (kN·m)C1140.7Tension377.675.5378.375.6C2141.5Tension234.858.7247.461.84C3233Tension10332581036259C4181.8Tension650.3162.6650163C5222Compression13101581425172.5C6138Compression119159.5181090.52C7276Compression14952501677279.5C8364Compression36067223975795These above columns are also analysed with different available methods, Whitney’s 1st approximation method [18], Whitney’s second approximation method [18], and the method provided by HSU [19]. These available methods are only available for the columns having the compression failure. There are no examples available for the columns with the tension failure cases. The results comparison is mentioned in Table 3.Figure 7
Column C-4.Table 3
Column design results comparison with different methods.
Column identifierFailure conditionMathematical modelWhitney 1st approximationWhitney 2nd approximationHSUPc (kN)Mc (kN·m)Pc (kN)Mc (kN·m)Pc (kN)Mc (kN·m)Pc (kN)Mc (kN·m)C1Tension377.675.5——————C2Tension234.858.7——————C3Tension1033258——————C4Tension650162——————C5Comp.1310158158019112851551580191C6Comp.119159.5187794112556187894C7Comp.1495250167728014822471677280C8Comp.3606722430086035827164300860


### 4.1. Detailed Numerical Example for Column C-4 (400 × 400)

Input Data: Figure7Pu = 400 kNMu = 100 kN·mfc′ = 30 MPafy = 415 MPaAs = 1000 mm2As′ = 1000 mm2d′ = 80 mmϕ = 0.65Solution:(1)
Finding the value ofe=Mu/Pu=100/400 = 250 mm(2)
cb=600×d/600+fy=600×320/600+415 = 189.16 mm(3)
ab=β×cb=0.85×189.16 = 160.788 mm(4)
Ccb=0.85fc′abb=0.85×30×160.788×400 = 1.64 × 103 kN(5)
Cs=As′fy−0.85fc′=1000415−0.8530 = 3.895 × 102 kN(6)
Ts=Asfy=1000×415 = 4.15 × 102 kN(7)
Pb=Cs+Ccb−Ts = 1.615 × 103 kN(8)
Mb=Ccd−ab/2−d″+Csd−d′−d″+Ts×d″d″=h/2−d′=120mmMb=2.927×102 kN·m(9)
eb=Mb/Pb=181.3mm<e250mm (TENSION Failure)(10)
Finding the value of (a) using the Quadratic EquationAa2+Ba+C=0A=0.425×fc′×b=5.1×103B=0.85×fc′×b×e′−d, where; (e′=e+d″ = 370 mm)B=5.1×105C=As′fy−0.85fc′e′−d+d′−fyAse′=−1.029×108a=−B+B2−4AC/2Aa=100.59mm(11)
Check if the tension steel has yielded;c=a/β=118.34mmεs=d−c/c×0.003=0.005,εy=fy/Es=0.002εs>εy, steel yields, ∴fs=fy(12)
PN=1/e′Ccd−a/2+Csd−d′×10−3Cc=0.85fc′ab=1.026×106kNPN=1000.55kNMN=PN×e=250.1 kN·m(13)
Pc=∅PN = 650.35 kN(14)
Mc=∅MN = 162.56 kN·m


## 4.1. Detailed Numerical Example for Column C-4 (400 × 400)

Input Data: Figure7Pu = 400 kNMu = 100 kN·mfc′ = 30 MPafy = 415 MPaAs = 1000 mm2As′ = 1000 mm2d′ = 80 mmϕ = 0.65Solution:(1)
Finding the value ofe=Mu/Pu=100/400 = 250 mm(2)
cb=600×d/600+fy=600×320/600+415 = 189.16 mm(3)
ab=β×cb=0.85×189.16 = 160.788 mm(4)
Ccb=0.85fc′abb=0.85×30×160.788×400 = 1.64 × 103 kN(5)
Cs=As′fy−0.85fc′=1000415−0.8530 = 3.895 × 102 kN(6)
Ts=Asfy=1000×415 = 4.15 × 102 kN(7)
Pb=Cs+Ccb−Ts = 1.615 × 103 kN(8)
Mb=Ccd−ab/2−d″+Csd−d′−d″+Ts×d″d″=h/2−d′=120mmMb=2.927×102 kN·m(9)
eb=Mb/Pb=181.3mm<e250mm (TENSION Failure)(10)
Finding the value of (a) using the Quadratic EquationAa2+Ba+C=0A=0.425×fc′×b=5.1×103B=0.85×fc′×b×e′−d, where; (e′=e+d″ = 370 mm)B=5.1×105C=As′fy−0.85fc′e′−d+d′−fyAse′=−1.029×108a=−B+B2−4AC/2Aa=100.59mm(11)
Check if the tension steel has yielded;c=a/β=118.34mmεs=d−c/c×0.003=0.005,εy=fy/Es=0.002εs>εy, steel yields, ∴fs=fy(12)
PN=1/e′Ccd−a/2+Csd−d′×10−3Cc=0.85fc′ab=1.026×106kNPN=1000.55kNMN=PN×e=250.1 kN·m(13)
Pc=∅PN = 650.35 kN(14)
Mc=∅MN = 162.56 kN·m


## 5. Numerical Examples for Biaxial Columns

Seven reinforced biaxial rectangular columns (CB-1 to CB-7) having different column sizes are also analysed using the proposed model. These columns are also having different reinforcement ratiosρ in addition to different failure types, that is, tension-tension, compression-compression, and tension-compression failures. The design input load data for these columns are illustrated in Table 4. The column cross section subjected to biaxial bending is shown in Figure 8. A similar flow chart has to be adopted (as discussed in the uniaxial column sections), once for the case of eccentricity in the x-direction ex and later for the eccentricity in the y-direction ey to obtain the required values of load capacities in x- and y-direction ∅Px,∅Py. These values are later used in Bresler’s formula [15] (equation (19)) to find the value of Pc. Moreover, the Mcx and Mcy values can be found by using equations (21) and (22) accordingly:(19)Pc=11/∅Px+1/∅Py−1/∅PNmax,(20)∅PNmax=0.8ϕ0.85f′cAg−Ast+fyAst,where ∅PNmax = maximum permissible column load, Ast = total area of steel, and Ag=Gross area of cross section−sectional area of concrete member.Table 4
Biaxial column input data.
Column identifierPu (kN)Mux (kN·m)Muy (kN·m)As (mm2)As′ (mm2)fc′ (MPa)fy (MPa)ϕd′ (mm)ex (mm)ey (mm)CB-1 (300 × 600)3001008012321232304000.6580267333CB-2 (200 × 400)2004020628.4628.4203000.740100200CB-3 (300 × 300)250025012012251225304000.657048100CB-4 (375 × 500)170020010021002100304150.656059118CB-5 (400 × 500)800200501413.81413.8304150.656062.5250CB-6 (350 × 700)400604016381638203000.6545100150CB-7 (400 × 1200)150030030030803080203000.6560200200Figure 8
Biaxial column cross section.The moments in thex- and y-direction can be found as(21)Mcx=PC×euy,(22)Mcy=PC×eux.The above seven columns (CB-1 to CB-7) are analysed with mathematical model approach to find the required values of axial load capacity Pc, using reciprocal formula. Moreover, the values of Pc are also compared with the computer software spColumn. The results obtained are depicted in Table5.Figure 9
Column CB-7 (X-X axis).Figure 10
Column CB-7(Y-Y axis).Table 5
Biaxial column design results.
Column identifierFailure conditionMathematical modelspColumnPc (kN)Pc (kN)CB-1Tension and tension280343CB-2Tension and tension262228CB-3Compression and compression480415CB-4Compression and compression21322035CB-5Tension and compression12911142CB-6Compression and compression28522805CB-7Compression and tension19932081


### 5.1. Detailed Numerical Example for Column CB-7 (400 × 1200)

Input Data: Figure9Pu = 1500 kNMux = 300 kN·mMuy = 300 kN·mfc′ = 20 MPafy = 300 MPaAs = 3080 mm2As′ = 3080 mm2d′ = 60 mmϕ = 0.65Solution: (Solving for theX-direction)(1)
Finding the value ofey=Mux/Pu = 200 mm(2)
cb=600×d/600+fy=600×1140/600+300 = 760 mm(3)
ab=β×cb=0.93×760 = 706.8 mm(4)
Ccb=0.85fc′abb=0.85×20×706.8×400 = 4.81 × 103 kN(5)
Cs=As′fy−0.85fc′=3080300−0.8520 = 8.71 × 102 kN(6)
Ts=Asfy=3080×300 = 9.24 × 102 kN(7)
Pbx=Cs+Ccb−Ts = 4,754 kN(8)
Mbx=Ccd−ab/2−d″+Csd−d′−d″+Ts×d″d″=h/2−d′=540mmMbx=2.155×103 kN·m(9)
eby=Mbx/Pbx=453mm>ey200mm (COMPRESSION Failure)(10)
Finding the value of (a) using the Cubic EquationAa3+Ba2+Ca+D=0A=0.425×fc′×b×10−3=3.4B=0.85×fc′×b×e′−d×105, where; (e′=e+d″ = 740 mm)B=−2.72×103C=As′fs′−0.85fc′e′−d+d′+600Ase′=1.071×106D=−600Ase′β′d=−1.45×109a=944.47mm(11)
Check if the tension steel has yielded;c=a/β=1016 mmεs=d−c/c×0.003=0.0028,εy=fy/Es=0.0015εs>εy, steel yields, ∴fs=fy(12)
PNX=1/e′Ccd−a/2+Csd−d′×10−3Cc=0.85fc′ab=6.422×103kNPNX=7068kNMNX=PNX×ex=1414 kN·m(13)
PCX=∅PNX = 4594 kN(14)
MCX=∅MNX = 918.79 kN·m(Solving for theY-direction) (Figure 10)(1)
Finding the value ofex=Muy/Pu = 200 mm(2)
cb=600×d/600+fy=600×1140/600+300 = 226.67 mm(3)
ab=β×cb=0.93×226.67 = 210.8 mm(4)
Ccb=0.85fc′abb=0.85×20×210.8×1200 = 4.3 × 103 kN(5)
Cs=As′fy−0.85fc′=3080300−0.8520 = 8.71 × 102 kN(6)
Ts=Asfy=3080×300 = 9.24 × 102 kN(7)
Pby=Cs+Ccb−Ts = 4.248 × 103 kN(8)
Mby=Ccd−ab/2−d″+Csd−d′−d″+Ts×d″d″=h/2−d′=140mmMby=6.582×102 kN·m(9)
ebx=Mby/Pby=154.94mm>ex200mm (TENSION Failure)(10)
Finding the value of (a) using the Quadratic EquationAa2+Ba+C=0A=0.425×fc′×b×10−3=10200B=0.85×fc′×b×e′−d×105, where; (e′=e+d″ = 340 mm)B=0C=As′fs′−0.85fc′e′−d+d′−fyAse′=−2.62×108a=160.227mm(11)
Check if the tension steel has yielded;c=a/β=172.287 mmεs=d−c/c×0.003=0.00292,εy=fy/Es=0.0015εs>εy, steel yields, ∴fs=fy(12)
PNY=1/e′Ccd−a/2+Csd−d′×10−3Cc=0.85fc′ab=3.27×106kNPNY=3216kNMNX=PNY×ey=643.25 kN·m(13)
PCY=∅PNY = 2091 kN(14)
MCY=∅MNY = 418.12 kN·mFinding the value of Pc using Bresler’s equation (19):(23)Pc=11/∅Px+1/∅Py−1/∅PNmax,where ∅PNmax=0.8ϕ0.85f′cAg−Ast+fyAst=5150kN, Pc=1/1/4594+1/2091−1/5150=1993kN.


## 5.1. Detailed Numerical Example for Column CB-7 (400 × 1200)

Input Data: Figure9Pu = 1500 kNMux = 300 kN·mMuy = 300 kN·mfc′ = 20 MPafy = 300 MPaAs = 3080 mm2As′ = 3080 mm2d′ = 60 mmϕ = 0.65Solution: (Solving for theX-direction)(1)
Finding the value ofey=Mux/Pu = 200 mm(2)
cb=600×d/600+fy=600×1140/600+300 = 760 mm(3)
ab=β×cb=0.93×760 = 706.8 mm(4)
Ccb=0.85fc′abb=0.85×20×706.8×400 = 4.81 × 103 kN(5)
Cs=As′fy−0.85fc′=3080300−0.8520 = 8.71 × 102 kN(6)
Ts=Asfy=3080×300 = 9.24 × 102 kN(7)
Pbx=Cs+Ccb−Ts = 4,754 kN(8)
Mbx=Ccd−ab/2−d″+Csd−d′−d″+Ts×d″d″=h/2−d′=540mmMbx=2.155×103 kN·m(9)
eby=Mbx/Pbx=453mm>ey200mm (COMPRESSION Failure)(10)
Finding the value of (a) using the Cubic EquationAa3+Ba2+Ca+D=0A=0.425×fc′×b×10−3=3.4B=0.85×fc′×b×e′−d×105, where; (e′=e+d″ = 740 mm)B=−2.72×103C=As′fs′−0.85fc′e′−d+d′+600Ase′=1.071×106D=−600Ase′β′d=−1.45×109a=944.47mm(11)
Check if the tension steel has yielded;c=a/β=1016 mmεs=d−c/c×0.003=0.0028,εy=fy/Es=0.0015εs>εy, steel yields, ∴fs=fy(12)
PNX=1/e′Ccd−a/2+Csd−d′×10−3Cc=0.85fc′ab=6.422×103kNPNX=7068kNMNX=PNX×ex=1414 kN·m(13)
PCX=∅PNX = 4594 kN(14)
MCX=∅MNX = 918.79 kN·m(Solving for theY-direction) (Figure 10)(1)
Finding the value ofex=Muy/Pu = 200 mm(2)
cb=600×d/600+fy=600×1140/600+300 = 226.67 mm(3)
ab=β×cb=0.93×226.67 = 210.8 mm(4)
Ccb=0.85fc′abb=0.85×20×210.8×1200 = 4.3 × 103 kN(5)
Cs=As′fy−0.85fc′=3080300−0.8520 = 8.71 × 102 kN(6)
Ts=Asfy=3080×300 = 9.24 × 102 kN(7)
Pby=Cs+Ccb−Ts = 4.248 × 103 kN(8)
Mby=Ccd−ab/2−d″+Csd−d′−d″+Ts×d″d″=h/2−d′=140mmMby=6.582×102 kN·m(9)
ebx=Mby/Pby=154.94mm>ex200mm (TENSION Failure)(10)
Finding the value of (a) using the Quadratic EquationAa2+Ba+C=0A=0.425×fc′×b×10−3=10200B=0.85×fc′×b×e′−d×105, where; (e′=e+d″ = 340 mm)B=0C=As′fs′−0.85fc′e′−d+d′−fyAse′=−2.62×108a=160.227mm(11)
Check if the tension steel has yielded;c=a/β=172.287 mmεs=d−c/c×0.003=0.00292,εy=fy/Es=0.0015εs>εy, steel yields, ∴fs=fy(12)
PNY=1/e′Ccd−a/2+Csd−d′×10−3Cc=0.85fc′ab=3.27×106kNPNY=3216kNMNX=PNY×ey=643.25 kN·m(13)
PCY=∅PNY = 2091 kN(14)
MCY=∅MNY = 418.12 kN·mFinding the value of Pc using Bresler’s equation (19):(23)Pc=11/∅Px+1/∅Py−1/∅PNmax,where ∅PNmax=0.8ϕ0.85f′cAg−Ast+fyAst=5150kN, Pc=1/1/4594+1/2091−1/5150=1993kN.


## 6. Validation of the Mathematical Model

In order to validate the proposed mathematical model approach, the model is validated with the existing experimental results of columns subjected to uniaxial and biaxial loadings. The experimental results data has been extracted from the test results provided by HSU [22]. Two uniaxial columns as provided by Bresler (B-1 and B-2) and one biaxial column as provided by Anderson and Lee (SC-4) are selected from the research article [22] to compare the results with the mathematical model.Table6 illustrates the experimental testing data provided by HSU. The data and the results are provided in imperial units (Kips-ft) units. Therefore, they are converted to metric units accordingly to compare the values with our results.Table 6
Experimental testing data [22].
Experimental investigatorColumn identifier with size (inx in) b (mm x mm)As (In2) (mm2)fc′ (Ksi) (MPa)fy (Ksi) (MPa)d′(In) (mm)ex (in) (mm)ey (in) (mm)Bresler (uniaxial column)B-1 (6 × 8) (152 × 203)1.24 (800)3.7 (25.6)53.5 (369)1.75 (44.5)6 (152.4)0Bresler (uniaxial column)B-2 (6 × 8) (152 × 203)1.24 (800)3.9 (27)53.5 (369)1.75 (44.5)3 (76.2)0Anderson and Lee (biaxial column)SC-4 (4 × 4) (102 × 102)0.8 (516)5.435 (37.5)45.6 (314.6)0.75 (19)2.82 (71.63)2.82 (71.63)Table7 provides the experimental test results as well as the validation of the test data with the proposed mathematical model. The column capacity (Pc) results obtained from the experimental data are quite close to the mathematical model results, showing satisfactory computational results.Table 7
Validation of experimental data [22].
Experimental investigatorColumn identifier with size (inx in) (mm x mm)ϕceb (in) (mm)Failure conditionExperimental resultsMathematical modelPexp/PMathPc (kips) (kN)Pc (kips) (kN)Bresler (uniaxial column)B-1 (6 × 8) (152 × 203)0.654.67 (118.5)eb<ex (tension)24 (107)29 (132)0.83Bresler (uniaxial column)B-2 (6 × 8) (152 × 203)0.654.6 (117)eb>ex (compression)60 (267)59 (263)1.01Anderson and Lee (biaxial column)SC-4 (4 × 4) (102 × 102)0.652.7 (69)eb<ex (tension)13.5 (60)11 (49)1.22


## 7. Results and Discussions

The results obtained from the mathematical model approach for both uniaxial and biaxial columns showed a safe and conservative column design method. The results of eight uniaxial column sections (C-1 to C-8) using the proposed model are also compared with different available mathematical models, provided by Whitney's 1st approximation method, Whitney's second approximation method, and the method provided by HSU. Columns C1 to C-4 were subjected to tension failure, whereas columns C-5 to C-8 were the compression failure cases. The other three mathematical studies (Whitney's 1st approximation, Whitney's second approximation, and the method provided by HSU) are only limited to the case where the columns are subjected to the compression failure only.These studied columns (C-1 to C-8) are also analysed using the computer software spColumn and the comparison results for axial load capacities (Pc) and moment capacities (Mc) are displayed in bar charts (Figures11 and 12).Figure 11
Axial load capacity comparison for uniaxial columns (C1–C8).Figure 12
Moment capacity comparison for uniaxial columns (C1–C8).For the biaxial columns (CB-1 to CB-7), the axial load capacity results for mathematical model approach using Bresler’s formula and the computer software spColumn are displayed in the bar chart (Figure13).Figure 13
Axial load capacity comparison for biaxial columns (CB-1–CB-7).The values of Pc obtained using the mathematical model are quite close to the computer software results, showing relatively satisfactory computational results.


## 8. Conclusion

In this study, the mathematical model is presented to analyse and design the uniaxial and biaxial columns without going through the column interaction charts to find the required axial load capacities and moment capacities. A simplified flow chart has also been developed to solve the required column section following the mathematical model steps.Eight (RC) uniaxial columns (C-1 to C-8) and seven (RC) biaxial columns (CB-1 to CB-7) are analysed in this study. These columns are analysed having different steel reinforcement ratiosρ, different values of steel yield strength fy, concrete compressive strength fc′, and different load capacity conditions. Moreover, the studied columns are subjected to both tension and compression failures.For the uniaxial columns, the proposed mathematical model results are also compared with the different available numerical approaches done by Whitney's 1st approximation, Whitney's 2nd approximation, and the method provided by HSU. All of these three methods were formulated based on the case of compression failure only. These uniaxial columns are also analysed using the computer software spColumn. The results obtained showed that this proposed mathematical approach showed good agreement with the computer software spColumn showing relatively satisfactory results.The studied biaxial columns are subjected to different failure conditions, that is, tension-tension failure, compression-compression failure, and tension-compression failure. Bresler’s formula was used to find the required capacity (Pc) after finding thePxandPy from the mathematical model approach. The biaxial columns were also analysed with the computer software. The average variation of the mathematically computed values for biaxial columns to the finite element software was not more than 10%. Moreover, the results obtained for the columns subjected to tension failure are quite close with the computer software spColumn. Moreover, this mathematical model has also been validated with the existing experimental results conducted by HSU.In short, this newly proposed mathematical model is a good and quick approach to analyse the reinforced concrete uniaxial and biaxial columns. This model can also help the students and the academic researchers to find the column capacities without going through the column interaction charts and other long iterative approaches.



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*Source: 8868481-2020-11-25.xml*