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ORIGINAL ARTICLE Optimal reinforcement of RC columns for biaxial bending Luisa Marı ´a Gil-Martı ´n Enrique Herna ´ndez-Montes Mark Aschheim Received: 4 May 2009 / Accepted: 7 December 2009 / Published online: 15 December 2009 Ó RILEM 2009 Abstract The Reinforcement Sizing Diagram (RSD) approach to determining optimal reinforce- ment for reinforced concrete beam and column sections subjected to uniaxial bending is extended to the case of biaxial bending. Conventional constraints on the distribution of longitudinal reinforcement are relaxed, leading to an infinite number of reinforce- ment solutions, from which the optimal solution and a corresponding quasi-optimal pragmatic is determined. First, all possibilities of reinforcement arrangements are considered for a biaxial loading, including sym- metric and non-symmetric configurations, subject to the constraint that the reinforcement is located in a single layer near the circumference of the section. This theoretical approach establishes the context for obtaining pragmatic distributions of reinforcement that are more suitable for construction, in which distributions having double symmetry are considered. This contrasts with conventional approaches for the design of column reinforcement, in which a predeter- mined distribution of longitudinal reinforcement is assumed, even though such a distribution generally is non-optimal in any given design. Column and wall sections that are subjected to uniaxial or biaxial loading may be designed using this method. The solutions are displayed using a biaxial RSD and can be obtained with relatively simple algorithms imple- mented in widely accessible software programs such as Mathematica Ò and Excel Ò . Several examples illustrate the method and the savings in reinforcement that can be obtained relative to conventional solutions. Keywords Ultimate strength design Optimal reinforcement Biaxial bending List of symbols A c Cross sectional area of concrete section A s Area of bottom reinforcement A 0 s Area of top reinforcement A p Area of prestressing tendon N Nominal axial strength N d Design value of the applied axial force M Bending moment applied at the center of gravity of the gross section M d Design value of the applied bending moment M x Flexural moment strength about x-axis M xd Design value of the bending moment applied about the x-axis M y Flexural moment strength about y-axis L. M. Gil-Martı ´n (&) E. Herna ´ndez-Montes Department of Structural Mechanics, University of Granada, E.T.S. Ingenieros de Caminos, Campus Universitario de Fuentenueva, 18072 Granada, Spain e-mail: [email protected] E. Herna ´ndez-Montes e-mail: [email protected] M. Aschheim Department of Civil Engineering, Santa Clara University, 500 El Camino Real, Santa Clara, CA 95053, USA e-mail: [email protected] Materials and Structures (2010) 43:1245–1256 DOI 10.1617/s11527-009-9576-x

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  • ORIGINAL ARTICLE

    Optimal reinforcement of RC columns for biaxial bending

    Luisa Mara Gil-Martn Enrique Hernandez-Montes

    Mark Aschheim

    Received: 4 May 2009 / Accepted: 7 December 2009 / Published online: 15 December 2009

    RILEM 2009

    Abstract The Reinforcement Sizing Diagram

    (RSD) approach to determining optimal reinforce-

    ment for reinforced concrete beam and column

    sections subjected to uniaxial bending is extended to

    the case of biaxial bending. Conventional constraints

    on the distribution of longitudinal reinforcement are

    relaxed, leading to an infinite number of reinforce-

    ment solutions, from which the optimal solution and a

    corresponding quasi-optimal pragmatic is determined.

    First, all possibilities of reinforcement arrangements

    are considered for a biaxial loading, including sym-

    metric and non-symmetric configurations, subject to

    the constraint that the reinforcement is located in a

    single layer near the circumference of the section.

    This theoretical approach establishes the context for

    obtaining pragmatic distributions of reinforcement

    that are more suitable for construction, in which

    distributions having double symmetry are considered.

    This contrasts with conventional approaches for the

    design of column reinforcement, in which a predeter-

    mined distribution of longitudinal reinforcement is

    assumed, even though such a distribution generally is

    non-optimal in any given design. Column and wall

    sections that are subjected to uniaxial or biaxial

    loading may be designed using this method. The

    solutions are displayed using a biaxial RSD and can be

    obtained with relatively simple algorithms imple-

    mented in widely accessible software programs such

    as Mathematica and Excel. Several examples

    illustrate the method and the savings in reinforcement

    that can be obtained relative to conventional solutions.

    Keywords Ultimate strength design

    Optimal reinforcement Biaxial bending

    List of symbols

    Ac Cross sectional area of concrete section

    As Area of bottom reinforcement

    A0s Area of top reinforcement

    Ap Area of prestressing tendon

    N Nominal axial strength

    Nd Design value of the applied axial force

    M Bending moment applied at the center of

    gravity of the gross section

    Md Design value of the applied bending moment

    Mx Flexural moment strength about x-axis

    Mxd Design value of the bending moment applied

    about the x-axis

    My Flexural moment strength about y-axis

    L. M. Gil-Martn (&) E. Hernandez-MontesDepartment of Structural Mechanics, University

    of Granada, E.T.S. Ingenieros de Caminos,

    Campus Universitario de Fuentenueva,

    18072 Granada, Spain

    e-mail: [email protected]

    E. Hernandez-Montes

    e-mail: [email protected]

    M. Aschheim

    Department of Civil Engineering, Santa Clara University,

    500 El Camino Real, Santa Clara, CA 95053, USA

    e-mail: [email protected]

    Materials and Structures (2010) 43:12451256

    DOI 10.1617/s11527-009-9576-x

  • Myd Design value of the bending moment applied

    about the x-axis

    fck Characteristic compressive strength of

    concrete

    fyk Characteristic yield strength of reinforcement

    sh Distance between centroids of consecutive

    bars of the top and bottom reinforcement

    sv Distance between centroids of consecutive

    bars of side reinforcement

    x Depth to neutral axis from top fiber of cross

    section

    y Vertical coordinate measures from the center

    of gravity of the gross section

    rc Stress in concrete

    rp Stress in prestressing tendon

    rs Stress in bottom reinforcement

    r0s Stress in top reinforcement

    n Intersection of the neutral axis with the y-axis

    u Angle of the neutral fiber

    U Bar diameter

    1 Introduction

    The approaches, assumptions, and hypotheses com-

    monly used for the design of reinforced concrete

    sections subjected to combinations of axial load and

    moment applied about one or both principal axes of the

    cross section were established many years ago. Even

    though conventional approaches for the analysis and

    design of cross sections make use of pre-determined

    patterns of longitudinal reinforcement distributed

    symmetrically about the cross section, there are

    significant differences in the approaches suggested

    by different authors. For example, Bresler [1] and

    Gouwens [2] have suggested different approximations

    for design of rectangular sections subjected to biaxial

    bending. Marn [3] addressed the biaxial bending of L-

    shaped sections. Approximate methods for biaxial

    bending were evaluated by Lepage-Rodrguez [4] and

    Furlong et al. [5]. A rigorous application of the plane-

    sections remain plane assumption for the sectional

    analysis of rectangular sections is described by Leet

    and Bernal [6]. Even more recently, Leps and Sejnoha

    [7] used genetic algorithms to designRCcross sections.

    The present work applies conventional assump-

    tions and hypotheses at a fundamental level for the

    design of longitudinal reinforcement, but relaxes the

    conventional assumption that the reinforcement

    should be distributed symmetrically. Unlike previous

    approaches, the present work determines longitudinal

    reinforcement solutions for the design problem as a

    function of the neutral axis depth and inclination, and

    thus identifies an infinite number of reinforcement

    solutions from which the optimal reinforcement

    solution may be selected for design. In the present

    case, the fundamental assumptions and hypotheses

    are given as represented in Eurocode 2 [8].

    The present work represents a simplification of the

    more general approach for ultimate strength design

    presented by Aschheim et al. [9], wherein a single

    model applicable to a rectangular section and solution

    approach was used for the design of reinforced

    concrete beams, columns and walls for uniaxial or

    biaxial loading. The model uses a conjugate gradient

    search method to solve the nonlinear optimization

    problem of minimizing the total reinforcement area.

    The solution approach allows additional constraints

    to be imposed such as single or double symmetry of

    the reinforcement distribution.

    The theory for strength design of biaxial columns

    can be found in several references or text books [6,

    10]. Beal and Pannell [11] shows that multibar

    rectangular concrete columns can be designed using

    the published curves for four-bar columns if the

    effective depth of the multibar column is transformed

    into an equivalent four-bar effective depth. Rafiq and

    Southcombe [12] presented an interesting approach to

    optimal design of RC biaxial columns using genetic

    algorithms. Extensions to RC moment frames have

    been made. For example, Liu [13] considered a

    quadratic expression for the relationship between the

    area of longitudinal reinforcement and the fully

    plastic moments of cross-sections in order to mini-

    mize the total volume of reinforcing steel in the frame.

    Once the reinforcement has been determined,

    existing software such as PCAColumn [14] or

    BIAX-2 [15] may be used to analyse the strength of

    the section, producing interaction diagrams that are

    useful for validating trial designs. Ehsani [16] shows

    a valid algorithm that can be used to obtain the

    interaction diagrams. These programs, which analyse

    sections with known reinforcement, do not provide a

    direct solution for the reinforcement required to

    provide a section with adequate strength according to

    a governing building code.

    Recent work by the authors has emphasized a

    unique solution strategy in which reinforcement

    1246 Materials and Structures (2010) 43:12451256

  • solutions are obtained as a function of the neutral axis

    depth, allowing optimal reinforcement solutions to be

    characterized and used for design. The resulting

    reinforcement distributions generally are not sym-

    metric, but conform to building code requirements

    and may result in significant savings of reinforce-

    ment, and thus advance the aims of sustainability in

    construction. In one instance, Reinforcement Sizing

    Diagrams are applied to the design of sections

    subjected to uniaxial bending in conjunction with

    axial load [17]. Characteristics of the optimal solu-

    tions obtained with Reinforcement Sizing Diagrams

    are the basis of the Optimal Domains identified in

    Aschheim et al. [18]. The treatment of multiple load

    combinations is addressed by Lee et al. [19]. A

    general solution to the biaxial bending problem was

    described by Aschheim et al. [9]. This reference

    includes a comparison of an optimal solution with a

    conventional textbook solution for the biaxial bend-

    ing of a rectangular section column.

    The present paper extends the uniaxial Reinforce-

    ment Sizing Diagram concept to sections subjected to

    biaxial bending. Rather than considering the general

    solution that is described in [9], this solution reflects a

    view that rectangular cross sections should have

    reinforcement patterns in which double symmetry is

    present (i.e. parallel sides have equal reinforcement).

    Reinforcement configured in this way, with double

    symmetry, is often used for rectangular sections

    subjected to biaxial bending [20]. The consideration

    of double symmetry forces to a reduction in the

    number of variables, with this restriction, admissible

    solutions for strength design depend only on one

    variable, so the minimum reinforcement area can be

    identified on a 2D (two-axis) diagram, this point will

    be develop later on the paper. In this paper,

    commercial software is used to show that reinforce-

    ment solutions, plotted on biaxial RSD diagrams

    provide the section with sufficient resistance to the

    specified combination of axial load and moment. The

    minimum total reinforcement area solution, under the

    constraint of double symmetry, is readily identified,

    as evident in several examples.

    2 Ultimate strength evaluation of cross sections

    The design problem for combined biaxial flexure and

    axial load involves the simultaneous consideration of

    equilibrium, compatibility, and the constitutive rela-

    tions of the steel and concrete materials at the section

    level. In the present biaxial formulation, the orthog-

    onal axes x and y are aligned with the principal axes

    of the cross section, and the applied moment, M, is

    represented in terms of its components about the x

    and y axes, Mx and My, respectively.

    The compatibility conditions make use of Ber-

    noullis hypothesis that plane sections remain plane

    after deformation (see Fig. 1). The plane sections

    hypothesis allows the distribution of strain over the

    cross section to be defined by two variables in the

    case of uniaxial bending (usually strain at the center

    of gravity of the gross section and curvature, or

    alternatively, depth to the neutral axis and curvature).

    In the case of biaxial bending, three variables are

    sufficient to define the distribution of strain over the

    cross section (usually strain at the center of gravity of

    the gross section, angle of inclination of the neutral

    fiber relative to a principal axis of the section, and

    curvature). For example, the maximum compressive

    strain of the gross section, the height of the neutral

    axis (along the y-axis) from the center of gravity of

    Neutral axis

    Tension

    Compression

    b

    h

    Neutral axis

    Tension zone

    M

    N

    y

    xMx

    My

    C ompression zone

    Fig. 1 Strains at cross section level in biaxial bending

    Materials and Structures (2010) 43:12451256 1247

  • the gross section (n), and the angle between the x-axis

    and the neutral fiber (u) may be used to define the

    strain diagram, as illustrated in Fig. 1. Under the

    constraint of ultimate strength design, one of these

    variables is given, because the maximum compres-

    sive strain is established by code.

    Lets consider the following example to clarify the

    previous paragraph: h = 700 mm and b = 400 mm

    are given for the cross section in Fig. 1. Steel grade is

    500 (fyk = 500 MPa) and concrete strength, fck, is

    30 MPa (C-30). Reinforcement consists of five bars

    along each side of the cross-section and mechanical

    cover is 70 mm. In order to resist Nd = 200 kN,

    Mxd = 400 kN m and Myd = 100 kN m, the bars

    must have a diameter of 16.1 mm (so 20 mm is

    used) and the sectional strains are defined by three

    variables: maximum compression strain of 0.035,

    u = 438 and n = 0.07 m.

    The strain diagram defines the strains at every fiber

    of the cross section. Based on the constitutive

    relationships for the component materials, or useful

    design approximations thereof, the nominal axial

    strength, N, and flexural strengths Mx and My may be

    determined as

    N X

    internal

    N

    Z

    Ac

    rcdAc

    Z

    A0s

    r0sdA0s

    Z

    As

    rsdAs

    Z

    Ap

    rpdAp

    Mx X

    internal

    Mx

    Z

    Ac

    rcydAc

    Z

    A0s

    r0sydA0s

    Z

    As

    rsydAs

    Z

    Ap

    rpydAp

    My X

    internal

    My

    Z

    Ac

    rcxdAc

    Z

    A0s

    r0sxdA0s

    Z

    As

    rsxdAs

    Z

    Ap

    rpxdAp

    1

    where N acts at the point where x and y are measured

    from andP

    internal

    Mx is the summation of internal

    bending moments about the x-axis (idem forP

    internal

    My

    about the y-axis). The convention adopted in this paper

    is to measure x and y from the center of gravity of the

    gross section (see Fig. 2). In Eq. 1, extended from

    Collins and Mitchell [21] to biaxial bending, rc = the

    normal stress on a differential element of concrete,

    Ac = the area of concrete in the cross section,

    r0s = the stress and A0s = the area of the compression

    reinforcement, rs = the stress andAs = the area of the

    tension reinforcement, and rp = the stress and

    Ap = the area of the prestressing steel, if present, with

    x and y being the respective distances of each

    differential area (Ac, As, A0s or Ap) from the center of

    gravity of the gross section, as illustrated in Fig. 2.

    Axial forces and stresses are positive in compression

    and negative in tension. The moments,Mx andMy, are

    considered positive if they produce tensile strain on the

    upper and left portions of the section, respectively.

    3 Reinforcement Sizing Diagrams for uniaxial

    bending

    In the event thatMx orMy is zero, the biaxial problem

    reduces to the simpler case of uniaxial bending in

    conjunction with axial load. In this case bending

    moments are applied about a principal axis of the

    cross section and only two variables are required to

    define the strain distribution, because the neutral axis

    is aligned with one of the principal axes of the cross

    section. Thus, design for uniaxial bending requires

    that only two equilibrium equations be satisfied

    (N and either Mx or My).

    The reinforcement is used most efficiently when it

    is located as far from the neutral axis as possible. Thus,

    for uniaxial bending problems it is desirable to have

    reinforcement at the bottom of the section (As) and/or

    at the top of the section (A0s). In this case, the number of

    unknowns in the two governing equilibrium equations

    is three: As, A0s and the neutral axis depth, x. Since the

    number of unknowns exceeds the number of equilib-

    rium equations, an infinite number of solutions are

    possible, and constraints of some kind must be

    imposed in order to identify a particular solution. For

    beam design, A0s = 0 is often imposed, while for

    column design A0s = As is often imposed. More

    generally, however, one may observe that solutions

    for As and A0s may be determined for any neutral axis

    depth, x. Alternatively, since top reinforcement may be

    used to satisfy strength requirements or for other

    reasons, such as to increase the curvature ductility of

    1248 Materials and Structures (2010) 43:12451256

  • the member, to control long term deformations due to

    creep and shrinkage, or to aid in constructability, one

    may solve for the required value of As given a

    predetermined value of A0s.

    The general solution of As and A0s as a function of y

    for a cross section subjected to a single combination

    of axial load, N, and moment, Mx, can be portrayed

    on a Reinforcement Sizing Diagram. This diagram,

    proposed by Hernandez-Montes et al. [17], provides a

    convenient way to represent the infinite solutions

    given by the equilibrium equations (N P

    internal

    N and

    Mx P

    internal

    Mx) for the case of uniaxial bending. In

    addition to plotting As and A0s as a function of the

    neutral axis depth, y, the sum (As ? A0s) is also

    plotted, in order to allow the minimum total longi-

    tudinal reinforcement solution to be identified. The

    RSD also portrays solutions corresponding to sym-

    metric reinforcement and the solution for one steel

    area given a known value for the other steel area.

    Fig. 3 illustrates the RSD for an example according

    to design provisions of ACI-318 [17]. Cross section

    analyses indicate that ductility capacities are not

    significantly reduced, and may be enhanced in some

    cases, when optimal (minimum) reinforcement is

    selected relative to conventional (symmetric) rein-

    forcement [22].

    The RSD is intended for use in design, that is, for

    determining necessary and sufficient reinforcement

    quantities for a given cross section for the governing

    center of gravity of the gross

    N

    y

    x

    As1 s1

    fcu

    As2 s2

    As3 s3

    1

    2

    s1

    c,max

    s2

    s3

    3

    Neutral axis

    MMy

    Mx

    c

    Fig. 2 Stresses and

    resultants in biaxial bending

    for an arbitrary reinforced

    concrete section

    Neutral axis depth, x (mm)

    Steel area (mm2)

    Total area (As + As)

    As

    As

    225 250 275 300 325 350 375 400

    2000

    4000

    6000

    8000

    10000

    h=406 mm (20 in)

    b= 508 mm (16 in)

    0.75h

    Pn=3559 kN (800 kips) e=178 mm (7in)

    AsAs

    Fig. 3 Example of RSD for uniaxial bending

    Materials and Structures (2010) 43:12451256 1249

  • load case. NM interaction diagrams [23] may then

    be developed for the section having a particular

    combination of As and A0s in order to establish that the

    given design has adequate resistance for the various

    combinations of N and M associated with the relevant

    load combinations. In this way, an interaction

    diagram for an asymmetrically reinforced section is

    developed, in contrast to the conventional use of NM

    interaction diagrams, which are prepared for a

    predetermined reinforcement pattern. Conventional

    interaction diagrams enforce ad hoc assumptions

    about the distribution of reinforcement in the cross

    section and provide no information about the range of

    acceptable solutions, and thus do not aid the designer

    in identifying a truly optimal reinforcement solution.

    4 Optimization problem for rectangular RC

    concrete under biaxial bending

    In the case of rectangular RC cross sections, the

    general problem of reinforcement distribution can be

    considered in terms of independent quantities of

    reinforcement along each side: As,t, As,b, As,l and As,r.

    For strength design of rectangular RC cross section,

    equilibrium provides three equations while there are

    six unknowns (As,t, As,b, As,l, As,r, n, and u), therefore

    an infinite number of satisfactory combinations of

    reinforcement steel areas may be identified.

    Aschheim et al. [9] presented a procedure to easily

    obtain the optimal solutions using nonlinear conju-

    gate gradient search methods; the optimum could be a

    global minimum area or minimum area subjected to

    additional constraints, such as the use of equal

    reinforcement on opposite faces or equal reinforce-

    ment on all faces. Smeared reinforcement was

    considered by using plates along each side. To

    illustrate the application, an example previously

    evaluated by McGregor and Wight [24] was used.

    Aschheim et al. [9] presented minimum reinforce-

    ment required to provide the section with the same

    nominal strength (Nn = 1700 kN, Mnx = 225 kN m

    and Mny = 111 kN m), determined for different val-

    ues of n and u (see Fig. 4). Minimum reinforcement

    solutions are shown for each value of n and u

    (uniformly spaced) where such solutions could be

    obtained, along with contours. The global minimum

    is given by qg = 2.22%, and occurs for n/h = 0.750

    and u = 38.8. Small sketches superimposed on

    15 20 25 30 35

    0.6

    50.7

    00.7

    50.8

    00.8

    50.9

    0

    2.25

    2.25

    2.37

    2.22

    2.64

    2.47

    2.33

    2.25

    2.23

    2.27

    2.39

    2.62

    2.99

    4.75

    5.95

    2.59

    2.43

    2.33

    2.26

    2.25

    2.3

    2.43

    2.66

    3.04

    3.63

    5.14

    2.59

    2.44

    2.36

    2.29

    2.28

    2.34

    2.48

    2.72

    3.11

    3.71

    4.67

    2.66

    2.51

    2.42

    2.36

    2.36

    2.43

    2.58

    2.85

    3.26

    3.9

    4.91

    2.88

    2.72

    2.59

    2.54

    2.55

    2.65

    2.84

    3.16

    3.66

    4.43

    5.67

    3.2

    2.94

    2.81

    2.77

    2.82

    2.97

    3.23

    3.66

    4.33

    5.39

    7.1710.51

    2.41%

    0.20% 0.33%

    0

    0.89%

    0.12%4.38%

    0

    0.04%

    2.11%0

    0.28%

    Symmetric2.72%

    Optimal

    0.09%

    0.78%2.76%

    0

    /h

    Fig. 4 Contours of total

    reinforcement (expressed as

    a percent of gross section

    area) required to provide

    Nn = 1700 kN,

    Mnx = 225 kN m, and

    Mny = 111 kN m as a

    function of n/h and u

    1250 Materials and Structures (2010) 43:12451256

  • Fig. 4 indicate the distribution of reinforcement over

    the section for four combinations of n/h and u.

    Clearly, very different patterns of reinforcement may

    be used to provide a section with a desired nominal

    strength. For comparison, the symmetric reinforce-

    ment solution (equal reinforcement on all faces)

    requires qg = 2.72%, and is obtained for n/h = 0.784

    and u = 30.0.

    Constraints may be imposed on the reinforcement

    areas to obtain solutions mimicking conventional

    column reinforcement distributions. Where equal

    reinforcement areas are desired on opposite faces

    (As,t = As,b and As,l = As,r), the variables As,b and As,rmay be set equal to the values used on the opposite

    faces, and values of As,t and As,l are sought that

    minimize the objective function (As,t ? As,b ?

    As,l ? As,r, for example). Similarly, if equal rein-

    forcement areas on all faces is desired, three

    reinforcement areas may be set equal to a fourth,

    and values of the fourth are sought that minimize the

    objective function.

    5 Reinforcement Sizing Diagrams for biaxial

    bending

    Unsymmetrical bar arrangements are not suitable for

    common practice in columns. Reinforcement arrange-

    ments having double symmetry may be acceptable in

    some circumstances.

    Authors of this paper [25] have investigated

    unsymmetrical reinforcement arrangements for sev-

    eral years. Some configurations have been found to

    be very attractive, such as circular piers for retaining

    walls (protected by international patents property of

    the University of Granada), see Fig. 5.

    These piers can be installed prior to excavation

    and in some circumstances are very cost effective.

    Solutions like this may save more than 50% of the

    reinforcement and 1015% of total (materials and

    labor) cost for the reinforced concrete portions of the

    work.

    Of interest in this paper is the case that a rectangular

    RC section is used with equal reinforcement along

    opposite sides, an arrangement that aids in construc-

    tability of reinforced concrete columns. For this case,

    common computing software can be used for the

    optimization process described above. IECA 3.0 [26]

    is used in this paper for the design verification of

    reinforcement, which assures the fulfilment of Euro-

    code 2 [8] requirements.

    Relative to the case of uniaxial bending, strength

    design for biaxial bending requires an additional

    variable to define the strain distribution. Chosen here

    are the intersection of the neutral fiber with the y-axis

    (n) and the angle between the x-axis and the neutral

    axis (u). The steel areas can be parameterized in

    several ways including some that require a large

    number of variables. Constructability considerations

    require that the reinforcement pattern not be exces-

    sively complex. Thus, a single longitudinal bar size is

    used, and the spacing of the bars along the top and

    bottom faces (sh) is allowed to differ from the spacing

    along the side faces (sv). Consequently, bottom and

    top reinforcement areas are equal, and may differ

    from the reinforcement on each side, as represented

    in Fig. 6a. Consequently, for a given set of assump-

    tions regarding ultimate strength analysis, the strain

    distribution is defined by two unknowns (n and u,

    which define the position of the neutral axis) in

    5 32 @ 64mm + 9 10 @ 270mm

    5 32 @ 64mm

    Fig. 5 Cross section used for tangent wall piles

    Materials and Structures (2010) 43:12451256 1251

  • conjunction with the strain limit specified in the

    applicable design code for ultimate strength analysis,

    while the reinforcement distribution is defined by two

    unknowns (the diameter of the reinforcement, U, and

    the bar spacing ratio, sh/sv). Thus, the equilibrium

    equations can be stated as:

    0

    500

    1000

    1500

    2000

    2500

    3000

    3500

    4000

    4500

    5000

    0 1000 2000 3000 4000 5000 6000

    Top and Bottom Reinforcement (mm2)

    To

    tal

    Sid

    e R

    ein

    forc

    em

    en

    t (m

    m2)

    (b)

    (c) Biaxial Reinforcement Sizing Diagram

    Biaxial Reinforcement Sizing Diagram

    4200

    4300

    4400

    4500

    4600

    4700

    4800

    4900

    5000

    0 1000 2000 3000 4000 5000 6000

    Top and Bottom Reinforcement Area (mm2)

    To

    tal

    Re

    info

    rce

    me

    nt

    Are

    a (

    mm

    2)

    (a) 400 mm

    700 mm

    Neutral Axis y

    x

    sv

    sh

    fck = 30 MPa; fyk = 500 MPa Mechanical cover = 7 mm

    Case 1 Case 2Nd = 200 kN 200 kN Mxd = 300 kNm 400 kNm Myd = 250 kNm 100 kNm

    Fig. 6 Reinforcement

    Sizing Diagrams for biaxial

    bending. a Cross section

    and load cases. b Total side

    reinforcement area plotted

    against top and bottom

    reinforcement area. c Total

    reinforcement area plotted

    against top and bottom

    reinforcement area

    1252 Materials and Structures (2010) 43:12451256

  • N X

    Nn;u;/; sh=sv

    Mx X

    Mxn;u;/; sh=sv

    My X

    Myn;u;/; sh=sv

    2

    where summations are used in place of integrals to

    reflect the discrete locations of the resultant forces,

    determined in conformance with the applicable

    design provisions. The right side of the first equation

    in (2) represents the internal contribution to the axial

    force, and includes concrete and steel. The same

    types of terms contribute to the flexural moments Mxand My, in the second and third equations of (2),

    respectively. The number of variables considered is

    four (n, u, / and sh/sv), which cover all the possible

    cases for double symmetry reinforcement arrange-

    ments. Solutions may be obtained by calculating the

    required bar diameter (together with n and u) for

    different values of sh/sv. In practice, the required bar

    diameter may be determined assuming an integer

    number of bars on the top and bottom faces and a

    different integer number of bars along the side faces.

    This represents a form of discrete optimization, and

    differs from the smeared reinforcement distribution

    (as plates) considered by Aschheim et al. [9]. One

    useful graphical representation, named Biaxial Rein-

    forcement Sizing Diagram (BRSD), plots the total top

    and bottom reinforcement area on the abscissa and

    the total reinforcement area on the ordinate, as

    illustrated subsequently.

    For a set of axial load and biaxial load bending

    moments a check is required to ensure that the cross-

    section resists this given set. This can be done with

    PM interaction diagrams or with the Bresler load

    contour method [1], although the later is a conserva-

    tive approximation.

    6 Biaxial bending example

    The 700 9 400 mm rectangular cross section of

    Fig. 6a was designed with doubly symmetric rein-

    forcement according to Eurocode 2 [8], for a concrete

    strength (fck) equal to 30 MPa and steel strength (fyk)

    of 500 MPa. To illustrate the potential under different

    loading conditions, two different design loadings are

    considered: (1) Nd = 200 kN, Mxd = 300 kN m and

    Myd = 250 kN m and (2) Nd = 200 kN, Mxd =

    400 kN m and Myd = 100 kN m. Column cross

    section width and height are given so these values

    are not included as optimization parameters.

    Eurocode 2 [8] assumptions for ultimate strength

    analysis are used, wherein the maximum usable

    compressive strain is 0.0035, the maximum usable

    tensile strain in the reinforcement is 0.01, and

    admissible strain diagrams are limited to strain

    domains as described by Eurocode 2 [8]. The

    reinforcement is assumed to be elastoplastic, with

    design yield strength fyd equal to the nominal yield

    strength fyk divided by a material safety factor, cs,

    which is equal to 1.15. The parabola-rectangle

    diagram for concrete compressive stress defined in

    Eurocode 2 [8] was used, with a maximum stress of

    0.85 fcd. The design strength fcd is given by the

    nominal strength fck divided by a material safety

    factor cc, which is equal to 1.5. The depth of the

    concrete stress block is given by k times the depth of

    the neutral axis, even where the neutral axis is

    inclined at an angle u. In this example k is taken

    equal to 0.8.

    Table 1 shows that the minimum total reinforce-

    ment consists of 16 /11.6 bars distributed equally to

    the top and bottom layers with 13 bars added to each

    side, resulting in a total reinforcement area of

    4439 mm2 (Case 9). The reinforcement areas shown

    in Table 1 are plotted in Fig. 6b and c. Figure 6b

    shows the variation in side reinforcement area as a

    function of top and bottom steel reinforcement area;

    total reinforcement area is plotted in Fig. 6c. One can

    observe that the optimum (or minimum) reinforce-

    ment area is obtained for a specific amount of bottom

    and top reinforcement area. Because significant

    amounts of reinforcement are required on each face,

    the minimum reinforcement solution is similar to the

    solution obtained for equal reinforcement on each

    face, given by Case 5 of Table 1. The optimum

    reinforcement solution requires 97.3% of the rein-

    forcement area obtained with equal reinforcement on

    each face (4560 mm2). In this procedure the bar

    diameter has been obtained from Eq. 2 and does not

    match the available range of bars in the market, so a

    slightly larger, commercially available, bar would be

    used in construction.

    The second loading case better illustrates the

    potential of biaxial RSDs to save reinforcement. For

    this case, Nd = 200 kN,Mxd = 400 kN m andMyd =

    100 kN m. The optimum reinforcement is given by

    Case 1 of Table 2, for which 16 longitudinal bars are

    Materials and Structures (2010) 43:12451256 1253

  • Table 1 Analyses used to computer the biaxial Reinforcement Sizing Diagram for the example cross section, for Nd = 200 kN,

    Mxd = 300 kN m and Myd = 250 kN m

    Analysis

    case

    Bar diameter (mm) Number

    of top and bottom

    barsa

    Number

    of side barsbsh/sv Total area

    of top and

    bottom bars (mm2)

    Total area

    of side bars

    (mm2)

    Total area

    of bars

    (mm2)

    1 19.9 8 0 0.08 4976 0 4976

    2 18.5 8 1 0.16 4301 538 4839

    3 17.3 8 2 0.24 3761 940 4701

    4 15.6 8 4 0.40 3058 1529 4587

    5 14.4 8 6 0.56 2606 1954 4560

    6 13.4 8 8 0.72 2256 2256 4512

    7 12.6 8 10 0.88 1995 2494 4489

    8 11.9 8 12 1.04 1780 2669 4449

    9 11.6 8 13 1.13 1691 2748 4439

    10 11.9 6 14 1.69 1335 3114 4449

    11 12.6 4 14 2.81 998 3491 4489

    12 13.4 2 14 8.44 564 3949 4513

    13 10.8 2 23 13.50 366 4214 4580

    14 9.6 2 30 17.44 290 4343 4633

    The minimum total reinforcement solution is shown in bold text; that obtained using equal amounts of steel on each face is shown by underlining

    a Number of top and bottom bars is the number of bars at the top layer, which is the same as the number of bars in the bottom layer in this

    implementation of biaxial RSDs

    b The number of side bars, excluding the corner bars (which are counted as top or bottom bars)

    Table 2 Analyses used to computer the biaxial Reinforcement Sizing Diagram for the example cross section, for Nd = 200 kN,

    Mxd = 400 kN m and Myd = 100 kN m

    Analysis

    case

    Bar

    diameter

    (mm)

    Number

    of top and bottom

    barsa

    Number

    of side barsbsh/sv Total area

    of top and

    bottom bars (mm2)

    Total area

    of side bars

    (mm2)

    Total area

    of bars

    (mm2)

    1 15.3 8 0 0.08 2942 0 2942

    2 14.6 8 1 0.16 2679 335 3013

    3 14.0 8 2 0.24 2463 616 3079

    4 12.9 8 4 0.40 2091 1046 3137

    5 12.1 8 6 0.56 1840 1380 3220

    6 11.4 8 8 0.72 1633 1633 3266

    7 10.8 8 10 0.88 1466 1832 3298

    8 10.3 8 12 1.04 1333 2000 3333

    9 10.1 8 13 1.13 1282 2083 3365

    10 10.4 6 14 1.69 1019 2379 3398

    11 11.2 4 14 2.81 788 2759 3547

    12 12.1 2 14 8.44 460 3220 3680

    13 9.8 2 23 13.50 302 3470 3771

    14 8.7 2 30 17.44 238 3567 3805

    The minimum total reinforcement solution is shown in bold text; that obtained using equal amounts of steel on each face is shown by underlining

    a Number of top and bottom bars is the number of bars at the top layer, which is the same as the number of bars in the bottom layer in this

    implementation of biaxial RSDs

    b The number of side bars, excluding the corner bars (which are counted as top or bottom bars)

    1254 Materials and Structures (2010) 43:12451256

  • distributed equally to the top and bottom faces and

    without side bars. The total required reinforcement is

    2942 mm2. This is 91.4% of the total reinforcement

    required if equal reinforcement is provided to each

    face (Case 5). Figure 7 illustrates the reduction in

    total reinforcement area as the top and bottom

    reinforcement area increases.

    7 Conclusion

    The infinite number of reinforcement area solutions

    for the classic flexural design problem has led in

    current and past practice to the imposition of

    constraints on the reinforcement pattern in order to

    make column design practicable. With the advent of

    modern computing tools, however, the constraints on

    the reinforcement pattern can be relaxed, allowing a

    larger number of solutions to be identified. By

    portraying these solutions on a biaxial Reinforcement

    Sizing Diagram, the engineer is given the freedom to

    select the solution that is optimal in any particular

    context. An example illustrated the use of biaxial

    RSDs under the constraint that the same bars are

    used, at one spacing along the top and bottom faces,

    and at a different spacing along the side faces. The

    examples illustrated the potential to reduce the

    reinforcement required by 2.7 or 8.6%, depending

    on the relative values of the moments Mx and My.

    Greater savings may potentially be achieved depend-

    ing on the relative values of the moments, the axial

    load, and the willingness of the engineer or contractor

    to construct reinforcement cages having an uncon-

    ventional layout of longitudinal reinforcing bars.

    Savings in longitudinal reinforcement not only reduce

    the cost of construction, but also reduce the carbon

    footprint, helping to improve the sustainability of

    reinforced concrete construction.

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    To

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    Are

    a (

    mm

    2)

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    Nd = 200 kN, Mxd = 400 kN m and Myd = 100 kN m

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