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trabajo realizado para un diseño optimo de columnas de concreto bajo carga axial y momentos biaxiales
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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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2700
2900
3100
3300
3500
3700
3900
0 500 1000 1500 2000 2500 3000 3500
Top and Bottom Reinforcement Area (mm2)
To
tal
Re
info
rce
me
nt
Are
a (
mm
2)
Fig. 7 Reinforcement Sizing Diagram for biaxial bending, for
Nd = 200 kN, Mxd = 400 kN m and Myd = 100 kN m
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