Struct Multidisc Optim DOI 10.1007/s00158-007-0186-3
INDUSTRIAL APPLICATION
Theorem of optimal reinforcement for reinforced concrete cross sections E. Hernández-Montes L. M. Gil-Martín M. Pasadas-Fernández M. Aschhe Aschheim im &
&
&
Received: 18 September 2006 /Revised: 16 July 2007 /Accepted: 30 August 2007 # Springer-V Springer-Verlag erlag 2007
Abstract A theorem of optimal (minimum) sectional reinforceme for cement nt for ulti ultimate mate stre strengt ngth h des design ign is pre present sented ed for design assumptions common to many reinforced concrete building codes. The theorem states that the minimum total reinforcement area required for adequate resistance to axial load lo ad an and d mo mome ment nt ca can n be id iden enti tifie fied d as th thee min minim imum um admiss adm issible ible sol solutio ution n amo among ng fiv fivee dis discret cretee ana analysi lysiss cas cases. es. Therefore, only five cases need be considered among the infinite set of potential solutions. A proof of the theorem is
E. Hernández-Montes : L. M. Gil-Martín : M. Pasadas-Fernández University of Granada, Campus de Fuentenueva, 18072 Granada, Spain E. Hernández-Montes e-mail:
[email protected] L. M. Gil-Martín e-mail:
[email protected] M. Pasadas-Fernández e-mail:
[email protected] M. Aschh Aschheim eim (*) Civil Engineering Department, Santa Clara University, 500 El Camino Real, Santa Clara, CA 95053, USA e-mail:
[email protected] E. Hernández-Montes : L. M. Gil-Martín Department of Structural Mechanics, University of Granada, Campus de Fuentenueva, 18072 Granada, Spain M. Pasadas-Fernández Department of Applied Mathematics, University of Granada, Campus de Fuentenueva, 18072 Granada, Spain
made by means of a comprehensive numerical demonstration. The numerical demonstration considers a large range of par parame ameter ter val values, ues, which enc encomp ompass ass tho those se mos mostt ofte often n used use d in str structu uctural ral eng engine ineeri ering ng pra practic ctice. e. The des design ign of a reinforced concrete cross section is presented to illustrate the practical application of the theorem. Keywords Reinforced Reinforced concre concrete te . Beams . Columns . . Optimal reinfor reinforcement cement Concr Concrete ete structur structures es Notation Ac cross-sectiona cross-s ectionall area of concr concrete ete section As area of bottom reinfor reinforcement cement 0 As area of top reinfor reinforcement cement N axial force applie applied d at at the the center center of gravity of the gross sectio section n M bending bendin g moment moment applie applied d at the center of gravity gravity of the gross section b width of cross sectio section n d depth to centroid of bottom reinforc reinforcement ement from top fiber of cross section d ′ depth to top top reinforc reinforcement ement from top fiber of cross section f c specified specifie d compres compressive sive streng strength th of concre concrete te f y specified specifie d yield streng strength th of reinfor reinforcement cement h overall depth of cross-s cross-section ection x depth to neutral axis from top fiber of cross section x* x * dept de pth h to to neu neutra trall axis axis co corr rres espo pond nding ing to a com compre presssive strain of 0.003 at top fiber and a tensile strain of 0.01 at bottom reinforcement xb depth to neutral neutral axis corres correspondin ponding g to to a tensile strain of ɛ y at bottom reinforcement 0 xb depth to neutral neutral axis corres correspondin ponding g to to a tensile strain of ɛ y at top reinforcement xbb depth to neutra neutrall axis given by (9)
Hernández-Montes et al.
xc 0
xc1
0
xc2
x0 y Δ x
ɛ ɛ c ɛ c,
depth to neutral axis corresponding to a compressive strain of ɛ y at bottom reinforcement depth of neutral axis corresponding to a compressive strain of ɛ y at top reinforcement and a tensile strain of 0.01 at the bottom reinforcement depth of neutral axis corresponding to a compressive strain of ɛ y at top reinforcement and a compressive strain of 0.003 at top fiber. depth of neutral axis at the minimum of total reinforcement vertical coordinate measures from the center of gravity of the gross section discrete increments in the depth of the neutral fiber strain strain of the concrete maximum compressive strain of the concrete
max ɛ s 0
"s ɛ u ɛ y
σs 0
s
s
strain at bottom reinforcement strain at top reinforcement maximum allowable tension strain of the steel yield strain of the reinforcement stress of bottom reinforcement stress of top reinforcement
1 Introduction
The approaches commonly used for the design of reinforced concrete sections subjected to combinations of axial load and moment applied about a principal axis of the cross section were established many years ago. However, a new design approach was presented recently by HernándezMontes et al. (2004, 2005), which portrays the infinite number of solutions for top and bottom reinforcement that provide the required ultimate strength for sections subjected to combined axial load and moment. Solutions obtained with this new approach allow the characteristics of optimal (or minimum) reinforcement solutions to be identified. The characteristics of these optimal solutions have led to the development of the theorem of optimal section reinforcement (TOSR) presented herein. The longstanding conventional approaches treat the design of sectional reinforcement in one of two ways. One approach utilizes the distinction between large and small eccentricities based on an approach taken by Whitney and Cohen (1956), as described in Nawy’s (2003) textbook. The second approach uses N – M interaction diagrams, which have been widely used since their initial presentation by Whitney and Cohen (1956). These diagrams provide solutions for the reinforcement required to resist a specified combination of axial load, N , and moment, M , under the constraint that the reinforcement is arranged in a predetermined pattern. Typically, a symmetric pattern of reinforce-
ment is used. However, experience with beam design suggests that an asymmetric pattern of reinforcement would be more economical for small axial loads for cases in which the applied moment (or eccentricity) acts in one direction only. The family of solutions for combinations of top and bottom reinforcement required to confer adequate strength to a cross section constitutes an infinite set of solutions that includes the symmetric reinforcement solution obtained using conventional N – M interaction diagrams. The family of solutions is displayed graphically on a “Reinforcement Sizing Diagram” (RSD) as described by Hernández-Montes et al. (2005). Using an RSD, an engineer can rapidly select the reinforcement to be used in reinforced and prestressed concrete sections subjected to a combination of bending moment and axial load. Reinforcement may be selected to achieve whatever may be dictated by the design objectives, such as minimizing cost, facilitating construction, or providing a structure that has a very simple and uniform pattern of reinforcement. RSDs were used in a recent investigation by Aschheim et al. (2007) to characterize the optimal (minimum) reinforcement solutions for a cross section over the twodimensional space of design axial load and moment. This study identified domains in N – M space for which the optimal solution for nominal strength is characterized by 0 either constraints on reinforcement area ( As =0, As ¼ 0, or 0 As ¼ As ¼ 0) or constraints on the strains at the reinforcement locations (ɛ s = ɛ y, ɛ s equal to or slightly greater than 0 −ɛ y, or " s ¼ " y ) for stresses and strains taken as positive in compression), where As =the cross-sectional area of 0 bottom reinforcement, As =the cross-sectional area of top reinforcement, ɛ s =the tensile strain in the bottom rein0 forcement, "s =the compressive strain in the top reinforcement, and ɛ y =the yield strain of the reinforcement. The optimal domains approach uses the characteristics of the optimal solution to solve directly for the minimum reinforcement required for a given combination of axial load and moment. The present paper puts forth a TOSR and demonstrates its validity by argument and computationally using numerical results obtained for a large range of parameter values representative of those commonly encountered in practice (Table 1).
2 Assumptions used in flexural analysis and strength design
The design problem for combined flexure and axial load involves the simultaneous consideration of equilibrium, compatibility, and the constitutive relations of the steel and concrete materials at the section level.
TOSR for reinforced concrete cross sections Table 1 Range of variables studied for rectangular cross sections
Variable
Range
Concrete strength resistance ( f c) Height (h) Width (b) Yield strength ( f y) Yield strain ( ɛ y) Distance from extreme compression fiber to centroid of compression reinforcement (d ′) Mechanical cover condition Modulus of elasticity of reinforcement ( E s) Axial load Flexural moment
Lower limit
Upper limit
25 (MPa) 200 (mm) 200 (mm) 275 (MPa) 275/200,000 h − d
55 (MPa) 2,000 (mm) 2,000 (mm) 500 (MPa) 500/200,000
d ′ ≤ h/4 and d ≥ 3/4 h 200,000 (MPa)
−0.2bhf c
bhf c 0.25bh2 f c
0
Equilibrium equations must be satisfied at the section level (e.g., Fig. 1) for the governing combination of bending moment, M , and axial force, N : 0
0
N ¼ Σ N ¼ ∫ Ac σc dAc þ ∫ A0 σs dAs þ ∫ As σs dAs s
0
ð1Þ
0
M ¼ Σ M ¼ ∫ Ac σc ydAc þ ∫ A0 σs ydAs þ ∫ As σs ydAs s
where M is computed relative to the location that N is applied, with y being the distance of each differential area 0 (dAc, dA s, dA s , or dA p) from the location of the point about which the stress resultants act, as illustrated in Fig. 1. Stresses and axial forces in (1) are positive in compression and negative in tension. Without loss of generality, the axial load, N , and moment, M , that equilibrate the internal stress resultants are presumed to act about the center of gravity of the gross section (see Fig. 1). The moment, M , is considered positive if it produces tensile strain on the bottom fiber. For consistency, in the case that the applied loads cause compression over the depth of the section, the moment is considered positive if the compressive strain at the bottom fiber is smaller than the compressive strain at the top fiber.
Fig. 1 Strains and stresses diagrams at cross section level
The compatibility conditions make use of Bernoulli’s hypothesis that plane sections remain plane after deformation and assume no slip of reinforcement at the critical section. The Bernoulli hypothesis allows the distribution of strain over the cross section to be defined by just two variables. The strain at the center of gravity (ɛ cg) of the gross section and the curvature () of the cross section may be used to define the strain diagram, as illustrated in Fig. 1. For strength design, the reinforcement usually is modeled to have elasto-plastic behavior, and the concrete compression block usually is represented using a rectangular, trapezoidal, or parabolic stress distribution. ACI-318 (2005) allows the use of a rectangular stress block having depth equal to the product of a coefficient, β 1, and the depth of the neutral axis, where β 1 varies between 0.85 and 0.65 as a function of the specified compressive strength of the concrete. Eurocode 2 (2002) specifies that the stress block has a constant compressive stress of η f cd having a depth equal to the λ x, where x =the depth of the neutral axis and f cd =the design strength of the concrete, for concrete whose resistance is between 25 and 55 MPa. The factor λ defines the effective height of the compression zone, and the factor η defines the effective strength. The design strength of the concrete is given as a function of the specified characteristic strength, f ck , where f cd = αcc f ck / γ c. The term α cc accounts for long term effects on strength and the rate at which the load is applied. The term γ c is the partial safety factor for concrete, taken as 1.5. For the range contemplated in Table 1, we have chosen λ =0.8 and η =1.0 and αcc =0.85, as these represent fairly typical values. Perhaps the greatest difference in code provisions for ultimate strength determination is the treatment of cross section strains. The maximum usable strain at the extreme compression fiber is 0.003 in ACI 318 (2005), and there is no limit on the strain in the tensile reinforcement. Consequently, the neutral axis depth is a positive number. In Eurocode 2 (2002), the maximum usable strain in the extreme compression fiber ranges between 0.002 and 0.0035, and the tensile reinforcement strain cannot exceed 0.01. The Eurocode 2 (2002) approach invokes the concept of strain domains, wherein the strain diagram pivots about certain points located on the boundaries between adjacent
A’s
N
ε cg Center of gravity of the gross section
Neutral fiber
M
σc
y
σ’s
x
φ
Ap
σp σs
As Strains due to external loads
Stresses due to external loads
Hernández-Montes et al.
domains; consequently, the neutral axis depth may assume positive or negative values. The Swiss Concrete Code SIA162 (1989) establishes the maximum compressive strain ɛ c,max =0.003 and a tensile steel strain limit of ɛ s,max =0.005. For the demonstration of the theorem, assumptions intermediate between ACI 318 (2005) and Eurocode 2 (2002) were adopted. It is assumed that plane sections remain plane, the maximum usable strain for concrete in compression is given by ɛ c,max =0.003, and the maximum usable strain for steel in tension is given by ɛ s,max = 0.01. These assumptions are similar to those used in the Swiss Concrete Code; the only difference is that the tensile strain limit used for the demonstration is 0.01 and the Swiss Concrete Code uses 0.005. Walther and Miehlbradt (1990) indicate that the choice of a tensile strain limit of 0.005 or 0.01 has little effect on strength design. These strain limits are illustrated in Fig. 2, where three domains are identified. In domains I and II, the extreme
Fig. 2 Possible strain distributions in the ultimate limit state. Stress distributions according to rectangular block assumption and equilibrium of applied N and M with internal stress resultants for x >0
compression fiber is at a strain of 0.003; for domain I, the bottom reinforcement is either in compression or is at a tensile strain less than the yield strain, while in domain II, the bottom reinforcement is yielding in tension. In domain III, the bottom reinforcement is at a strain of 0.01, and the top fiber is either in tension or at a compressive strain less than 0.003. Thus, for domains I and II, the neutral axis depth, x, assumes a positive value, and can approach + ∞ as the strains approach 0.003 over the entire section. The neutral axis depth may be positive or negative in domain III, and approaches − ∞ as the strains approach 0.01 (in tension) over the entire section.
3 Design solutions
The algebraic form of the integrals of (1) allows the internal stress resultants to be determined as the product of the stresses and the corresponding areas. Using the above
TOSR for reinforced concrete cross sections
simplifying assumptions, the compressive force carried by the concrete, N c, can be expressed as
8< ð Þ¼ :
N c x
0 0:85 f c 0:8 x b 0:85 f c hb
0:8 x 0 0 0:8 x < h 0:8 x > h
if if if
As
ð2Þ
for a rectangular section of reinforced concrete with width b and height h. Where the compressive stress block includes the top reinforcement (0.8 x>d ′), the stress assigned to the rectangular concrete block should not be counted in determining the force carried by the top reinforcement, 0 N s . Similarly, if the compressive stress block extends below the bottom reinforcement, the stress carried by the rectangular stress block should not be counted in determining the force carried by the bottom reinforcement, N s. 0 Therefore, N s and N s can be determined as: 0
0
0
N s ð xÞ ¼ σs ð xÞ As ð xÞ N s ð xÞ ¼ σs ð xÞ As ð xÞ
ð3Þ
0
where As =the cross-sectional area of steel located at a distance d ′ from the top of the section and As =the crosssectional area of steel located at a distance d from the top of the section, and 0
ð Þ¼
σs x
σs ð xÞ ¼
0
σs ð xÞ 0:85 f c if 0 σs ð x Þ σs ð xÞ 0:85 f c if σs ð x Þ
0:8 x > d 0 otherwise 0:8 x > d otherwise
sum of moments about the location of the bottom reinforcement results in (7).
0
As
M N ð h2 d 0 Þ N c ðxÞð d 0 0:4 xÞ σs ð xÞð d d 0 Þ
if
*
M ð N 0:85 f c bhÞð h2 d 0 Þ σs ð xÞð d d 0 Þ
*
M þ N ðd h2 Þ N c ðxÞð d 0:4 xÞ 0
σs ð xÞð d d 0 Þ
0:8 x < h
ð6Þ otherwise
if
0:8 x < h
*
M þð N 0:85 f c bhÞðd h2 Þ 0
σs ð xÞð d d 0 Þ
ð7Þ
otherwise
*
The solutions for reinforcement areas given by (6) and (7) are functions of the neutral axis depth, x. Some values 0 of x result in negative values of As and As , and therefore must be considered inadmissible. The admissible solutions 0 for A s and A s , obtained with (6) and (7), are plotted on a RSD. Such a plot clearly indicates that the minimum total reinforcement solution generally differs from the symmetric reinforcement solution that typically is represented using conventional N – M interaction diagrams. More information on RSDs can be found in Hernández-Montes et al. (2005) where an example of the design of a column from ACI Publication SP-17 (1997) is used (see Fig. 3). ACI Publication SP-17 presents only the symmetric reinforceh=406 mm (20 in) 0.75h
ð4Þ
b= 508 mm (16 in)
0
The stress s s and σ s are positive in compression. 0 The internal stress resultants N c, N s , and N s equilibrate the applied load, N , and moment, M . For a given neutral axis depth, material properties, and reinforcement areas As 0 and A s , the internal stress resultants can be determined and equations of equilibrium can be applied to the free body diagram of Fig. 1 to determine the axial load and moment resisted by the section. The equations of equilibrium for N and M applied at the center of gravity of the gross section are:
e=178 mm (7in)
Pn=3559 kN (800 kips)
Steel area (mm 2) 10000 Total area (As + A’s)
As
8000
0
N ¼ N c ðxÞ þ N s ðxÞ þ N s ðxÞ 0 N c ð xÞ 2h 0:4 x þ N s ðxÞ 2h d 0 N s ðxÞ d h2 if 0:8 x h M ¼ 0 N s ðxÞ 2h d 0 N s ðxÞ d h2 otherwise
ð5Þ
6000 4000 A’s
2000
Alternatively, the equations of equilibrium can be solved 0 to determine the steel areas As and As required to provide the section with sufficient capacity to resist the applied loads N and M . In particular, the sum of moments about the location of the top reinforcement results in (6), while the
225
250
275
300
325
350
375
400
Neutral axis depth, x (mm)
Fig. 3 Example RSD
Hernández-Montes et al.
ment solution, but the RSD indicates that a significant savings in cost can be obtained when optimal (minimum total) reinforcement solutions are used in place of traditional symmetric reinforcement solutions. Although it is possible to obtain zero reinforcement solutions from (6) and (7), code-required minimum reinforcement requirements also must be considered in design. The theorem addresses only equilibrium considerations.
4 Theorem of optimal section reinforcement 0
Theorem The top ( A s) and bottom As reinforcement required to provide a rectangular concrete section with 1 adequate ultimate strength for a combination of axial load and moment applied about a principal axis of the cross section has the following characteristics: 0
(1) An infinite number of solutions for A s and A s exists. 0 (2) The minimum total reinforcement area As þ As 0 occurs for one of the following cases: As =0, As ¼ 0, 0 As ¼ As ¼ 0, ɛ s equal to or slightly greater than −ɛ y, 0 0 ɛ = ɛ s = " = ɛ c,max =0.003, and ɛ = ɛ s = " ¼ 0:01. s s
Corollary The minimum reinforcement area for a specific combination of axial load and moment may be determined by: 0
0
(1) Evaluating the cases: As =0, A s ¼ 0, ɛ s = −ɛ y, ɛ = ɛ s = "s = 0 ɛ c,max =0.003, and ɛ = ɛ s = " ¼ 0:01. s (2) Selecting the minimum of the admissible solutions, where an admissible solution is one in which the value 0 of x0 is real and the areas As and As are each nonnegative, subject to the following: 0
0
a. If A s =0 produces a negative value for A s and A s ¼ 0 produces a negative value for As, then the minimum reinforcement solution is given by 0 As ¼ As ¼ 0. b. If ɛ s = −ɛ y produces an admissible solution and As( xb)< As ( xb), the minimum ( x0) will be located at xb < x0 < xbb.
5 Proof
A formal analytical proof requires the treatment of (6) and (7) in a piecewise fashion over more than seven domains of x. As an alternative to this lengthy approach, a numerical proof is provided in the following. The validity of the 1
Within in the range of Table 1.
theorem is established numerically for common material strengths and section dimensions, as specified in Table 1. The range of parameters considered encompasses most practical cases that arise in civil engineering practice. Formal constraints expressing the limits of applicability of the theorem have not been encountered yet, and parameter values beyond those considered in Table 1 may be contemplated. The proof is formulated in terms of the section variables defined in Fig. 2. Here, solutions for the rein0 forcement areas As and As are determined as a function of the depth of the neutral axis, x. Solutions given by (6) 0 and (7) may, for some x, require negative values of A s or A s ; such solutions are physically inadmissible and thus are excluded. To summarize in mathematical terms: x 2 ð1; þ1Þ 0 0 As ¼ As ð xÞ; As ¼ As ð xÞ reinforcement areas for x The optimal reinforcement solution is given for x= x0,
0
Min x As ð xÞ þ As ð xÞ ¼ As ðx0 Þ þ As ðx0 Þ 0 and A s ðx0 Þ 0 and A s ðx0 Þ 0 Conjecture For the simplifying assumptions given previously (and illustrated in Fig. 2), the solution x0 is charac0 terized by A s =0 and/or A s ¼ 0, ɛ s equal to or slightly greater 0 0 than − ɛ y, ɛ = ɛ s = "s = ɛ c,max =0.003, and ɛ = ɛ s = "s ¼ 0:01. Conditions associated with the simplifying assumptions, the geometry of the section, and the elasto-plastic nature of the constitutive relationship for the reinforcement 0 cause the areas As and As to be defined as piecewise functions of x. To define these functions, the parameters x b, 0 0 0 x*, x c, x b , x c1 and x c2 are defined, as given in the Appendix and in Figs. 8 and 9. For all possible combinations of values of the parameters 0 of Table 1, solutions for As and As were determined for x 2 ð1; þ1Þ, and the optimal (i.e., minimum total area) 0 solution parameters x0, As( x0), and As ( x0) were retained. The optimal reinforcement solutions were determined numerically for all combinations of N and M defined by −0.2bhf c ≤ N ≤ bhf c and 0 ≤ M ≤ 0.25bh2 f c, with each range being subdivided into 80 steps. The numerical solutions were determined by considering discrete values of x in increments of 1 mm. To focus attention on significant aspects of the numerical results, only the subset of cases in which h =700 mm, 200 ≤ b ≤ 2,000 mm, 25 ≤ f c ≤ to 55, f y =400 MPa, and h − d = d ′ = 70 mm are considered in detail. Numerical solutions are plotted for this subset of cases in Fig. 4a,b. As( x0) is 0 plotted as a function of x 0/ d in Fig. 4a, and A s ( x0) is plotted in Fig. 4 b. To clarify, (6) and (7) are evaluated in the range
TOSR for reinforced concrete cross sections Fig. 4 a Values of A s( xo/ d) determined for optimal solutions. b Values of determined for optimal solutions
0
of 2 x b ; 2 xc for fixed values of N and M . Discrete values of x are considered in increments of 1 mm, and the 0 minimum value of the total area As þ As is retained and 0 plotted as a function of x0 [ As( x0), and As ( x0)]. Thus, each
point in Fig. 4a,b is the retained (optimal) solution for a fixed value of N and M . Inspection of these plots reveals 0 that there are regions where either As or As are equal to 0 zero, regions where both As and As appear to be nonzero,
Hernández-Montes et al. Table 2 Limits for the piecewise functions used to define A s( x) and 0 As ð xÞ
Variable
Value of the variable (mm)
Value of the variable normalized by d
x b
−70
−0.11
x*0 xc1 0 xc2
145 210 163 378 1,890
0.23 0.33 0.26 0.60 3.00
0
x b xc
and values of x0/d for which either many or no optimal 0 0 0 solutions occur. The values of x b, x*, x c, x b , x c1 and x c2 are identified on Fig. 4a,b and in Table 2. Inspection of Fig. 4a,b reveals several singularities and zones as follows. These are discussed sequentially in order of increasing x 0/ d , with respect to Fig. 4a,b: 0
0
a) x0/ d = −0.11, which corresponds to x0 ¼ x b , (i.e., "s ¼ "y for Grade 400 reinforcement). For this case, both 0 As and As assume positive values. It is interesting to observe that for given values of N and M , any x smaller 0 0 0 0 0 than x b results in As( x) = As x b and As ð xÞ ¼ As x b . Thus, if an optimal solution is found for x0/ d < −0.11, 0 the result obtained is equal to that obtained for x0 = x b 0 (given equivalently by " s ¼ "y ). The same result also 0 can be obtained for ɛ = ɛ s = "s ¼ 0:01. b) x0/ d =1/9, which corresponds exactly to x0 =d ', Here, there is zero strain in the top reinforcement. As the top reinforcement carries no force at x0/ d =1/9, the mini0 mum top reinforcement is given by As ¼ 0. 0 0 c) As ( x0)=0 in the zone x b < x0 < x b, This zone, which includes the singularity at x0 =d ', is shown in white in the bar at the bottom of Fig. 4a,b. Note that the discrete steps used in the computation of x 0 result in very small 0 0 values of A s , but in the limit, as Δ x approaches zero, A s approaches zero. d) x0/ d = 0.6, which corresponds exactly to x0 = x b for Grade 400 reinforcement [see (12) in Appendix]. For 0 x0 = x b =0.6d , both As and A s are positive. e) Figure 4a illustrates a region where values of As are limited to relatively small, positive numbers, where x0/ d is just slightly greater than 0.6. This corresponds to 0 x b < x0 < x bb, for Grade 400 reinforcement. Both As and A s are positive in this region, which is shown hatched on the bar at the bottom of Fig. 4a,b. An explanation for this phenomenon is provided later in the paper. f) x0/ d =1, which corresponds to zero strain in the bottom reinforcement. Because the bottom reinforcement carries no force, one may use A s =0. However, the discrete steps used in the computation of x 0 result in very small, nonzero, strains (and values of As) in the computed
results. In the limit, as Δ x approaches zero, As ap proaches zero. g) As( x0)=0 in the zone x bb < x0 < xc, This zone is shown in gray in the bar at the bottom of each figure. However, the discrete steps used in the computation of x 0 result in very small, nonzero strains (and values of As) in the computed results. h) x0/ d =3, which corresponds exactly to x0 = xc for Grade 400 reinforcement (see (13) in Appendix for xc, for ɛ s = ɛ y). Figures 4a,b demonstrate that for this value of x0/ d , the optimal solution gives positive values of both 0 As and A s . In fact, for any given N and M , any x greater 0 0 than x c results in A s( x) = As( xc) and As ( x) = As ( xc). Thus, if an optimal solution is found for x0/ d >3, the result obtained is equal to that obtained for x0 = xc, which corresponds to ɛ s = ɛ y. For the grades of reinforcement considered, the same result is also obtained for ɛ = ɛ s = 0 "s = ɛ c,max =0.003. A trivial case exists where the axial load N provides the section with sufficient capacity to resist the applied moment, M , without requiring reinforcement to carry tension or compression. For this case, the solution for A s = 0 0 0 results in A s < 0, while the solution for A s ¼ 0 results in As <0, indicating that no reinforcement is necessary. The singularities or zones described in b, e, and f are addressed in greater detail in the following sections. 5.1 Singularities at the reinforcement position To better characterize the singularities that appear at x 0/ d = 1 and x0/ d = 1/9, optimal solutions were obtained using a smaller numerical increment in x. Figure 5a shows As( x0) determined using increments, Δ x, equal to 1 mm (as before), and Fig. 5 b shows the more precise values of A s( x0), determined for Δ x=0.1 mm, for x0/ d ≈ 1. One may observe that the computed values of A s are smaller in Fig. 5 b. This illustrates that greater precision in the calculation results in reduced total steel areas and, moreover, that the singularity at x0/ d =1 is not an artifact of the computational approach. As stated earlier, because the strain at this location is zero, the force carried by the reinforcement at this location is zero, and thus, the minimum reinforcement area is given by As =0 in the case that x 0/ d =1. Similar results were found for x0/ d =1/9. This case, which is equivalent to x0 =d ′ for Grade 400 reinforcement, is parallel to the previous one and does not require further consideration. 5.2 Characterization of optimal solutions in the range x b < x0 < x bb One may observe in Fig. 4a that values of As somewhat greater than zero appear to the right of the singularity at
TOSR for reinforced concrete cross sections
x0/ d = x b/ d =0.6. Figure 6a shows this region for the subset of data represented in Fig. 4a for which b =200 and b =1,000. One may further observe that the minimum values of A s are aligned based on the value of the parameter b. This result is not an artifact of the step size Δ x. Where x b < x0 < x bb, the associated values of A s depend on the value of b as indicated in Fig. 6a. The correct solution for this case can be found by considering the minimum, 0 which is given where d ð As þ As Þ dx ¼ 0 for two different positions of the neutral fiber (at x and x+dx) under the
As ¼
8< ð Þ ¼ ð þ Þ : ð ðÞ Þ¼þ ð ðþÞ ¼Þ
M x M x dx N x N x dx 0 d A x As x 0 s dx
x bb ¼
ð8Þ
Equation (8) may be solved by considering the differential forms of the first two equations and using the third 0 equation to replace dAs ð xÞ in the first two equations with −dAs( x). The first two equations can be used to eliminate dAs, to obtain a single equation that may be solved for As. Solving for A s as a function of x results in:
bf c x 1; 020 E s d 2 þ 1; 836 E s xd þ x 289; 000 f c d 0 340; 000 f y d 0 816 E s x 231; 200 f c x þ 272; 000 f y x
1; 275 f c d 1; 275 f c d 0 1; 500 f y d þ 1; 500 f y d 0 E s d
where the required area of bottom reinforcement in this case is designated by A s , and the minimum total reinforcement may occur anywhere in the range x b < x0 < x bb. Equation (9) is plotted in Fig. 6 b for the same parameter values represented in the computed results of Fig. 6a. Inspection of Fig. 6a reveals that optimal solutions at x b are characterized by A s > As ( x b). Thus, to determine whether the optimal solution occurs at x 0 = x b or in the range x b < x0 < x bb, it is sufficient to compare A s( x b) with A s ( x b), where A s( x b) is
ð9Þ
determined using (6). If As( x b) < As ( x b), then the optimal solution must be located to the right of x b (i.e., x b < x0 < x bb). Also of interest is the fact that (9) indicates that A s ( x b) is a pro per ty of the cro ss sec tion and mat eri als and is independent of N and M. The x intercept of (9) was determined by setting As =0 and solving for x. This intercept is designated x bb. The value of x bb is given by
918 E s d þ 144; 500 f c d 0 170; 000 f y d 0 0:5sqrt 816 E s þ 231; 200 f c 272; 000 f y
where sqrt ¼
q ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi 4; 090d 2 E s 816 E s þ 231; 200 f c 272; 000 f y þ 1; 836 E s d þ 289; 000 f c d 0 340; 000 f y d 0
For any particular case where A s( x b) < As ( xb), the optimal solution occurs for x b < x0 < x bb, and the area of bottom reinforcement may be obtained using (6) or (9). The corresponding value of x0 may be determined by minimizing the total area, obtained with (6) and (7). In this case, the optimal solution corresponds to strain in the bottom reinforcement given by ɛ s = "*, given as y where " * is y "* y ¼
constraint that the sections at these locations resist the same values of N and M . This can be stated as:
0:003ðd x0 Þ x0
ð11Þ
5.3 Design using the minimum of the admissible cases Because the optimal (minimum) reinforcement solution 0 occurs for one of the following cases: A s =0 and/or A s ¼ 0,
2
ð10Þ
0
equal to or slightly greater than −ɛ y, ɛ = ɛ s = "s = ɛ c,max = 0 0.003, and ɛ = ɛ s = "s ¼ 0:01, the other cases must produce 0 non-optimal solutions for As and As . The non-optimal solutions either (1) have total area greater than optimal or (2) are inadmissible because one or both computed areas are negative or imaginary solutions are obtained. Consequently, the optimal solution is producing the smallest total area among the admissible solutions obtained for the possible cases. This may be seen by consideration of Fig. 4 and (6) and (7) and will be illustrated with an example. Consider an example used in Aschheim et al. (2007), using Grade 400 reinforcement, 25 MPa concrete, and a concrete cover of 0.05 m, shown in Fig. 7. This section was designed using RSDs for many combinations of N and M. Values of N and M will be selected to illustrate the design ɛ s
Hernández-Montes et al.
a
a As (mm2)
2
As (mm ) 2000
20000 1750
17500 15000
1500
12500 1250
10000
b = 1000 mm 1000
b = 200 mm
7500 750
5000 2500
500
x/d 0.9
0.95
1.0
1.05
250
1.1
0.60
b
0.65
0.7
0.75
0.8
x/d
As (mm2)
b
20000
2
As (mm ) 2000
17500 1750
15000 12500
1500
10000 1250
7500 1000
5000 2500
750
x/d 0.9
0.95
1.00
1.05
1.1
Fig. 5 a Detail of the singularity at x =d , for Δ x =1 mm. b Detail of the singularity at x =d , for Δ x=0.1 mm
500
250
0.60
0.65
0.7
0.75
0.8
x/d
of the section using the TOSR. The selected values of N and M are those for which the optimal solutions occur for each of the possible cases. These values are presented in normalized form in Table 3. One may observe that in general, the minimum of the admissible solutions corresponds to the optimal solution. Two special cases are described as follows: 1) For the first case, no admissible solutions are found, and in particular, the case As =0 produces a negative 0 0 value for As, and second, the case As ¼ 0 produces a negative value for As. In this case, no reinforcement is required for the section to resist the combined axial load and moment, and the optimal solution is 0 As ¼ As ¼ 0. 2) For the last case of Table 3, three admissible solutions were identified. Among these, x= x b appears to be the optimal solution, but there is a possibility that the true optimal solution is in the range x b < x0 < x bb. To address this possibility, (9) is solved to obtain A s ( x b)=465 mm2. Based on the discussion of Fig. 6a, because As( x b)= 51 mm2 < As ( x b), we expect the optimal solution to be to the right of x b (i.e., x b < x0 < x bb). In fact, the optimal solution occurs for ɛ s = "y*. However, the solution
Fig. 6 a Solutions of A s in the vicinity of x b, for f c =25 MPa, d ′ = h − d =70 mm, h =700 mm and b =200 and 1,000 mm. b Solution of A s given by ( 9)
determined for ɛ s = −ɛ y (or equivalently, x0 = x b) has a total steel area (645 mm2) that is only 4.4% greater than that determined for the optimal solution (618 mm2) while being considerably easier to determine. 500 mm
A’ s
500 mm
A s 50-mm to centroid of reinforcement Fig. 7 Example cross section
TOSR for reinforced concrete cross sections Table 3 Reinforcement area solutions obtained by considering individual cases (mm2)
N Ag f c
Minimum Solution (mm2)
Solutions obtained for individual cases (mm2) =0.003
ɛ
0.4 1.0 Admissible solutions are shown in italics; the minimum of the admissible solutions is shown in bold typeface. a For this case none of the solutions are admissible and, in par0 ticular, As =0 requires As < 0 0 an d A s ¼ 0 requires A s < 0. Thus, no reinforcement is required to support the axial force and moment, corresponding to 0 the solution A s ¼ As ¼ 0. b Imaginary solution
M Ag hf c
0.8 0.2 0.0
−0.2 0.4
0.05 0.02 0.15 0.20 0.10 0.02 0.12
As 0 As As 0 As As 0 As As 0 As As 0 As As 0 As As 0 As
0a 0 825 1,650 0 2,817 3,176 595 1,885 0 1,953 1,172 196 422
To illustrate the comparison of As ( x b) and A s( x b) for a case in which x0 = x b, we consider the fourth case ( N =0.2 Ag f c, M =0.20 Aghf c) of Table 3. To determine if the minimum might be in the range x b < x 0 < x bb, (9) is used to determine As ( x b)=465 mm2. Because As ( x b)=3,176 mm2 > As ( x b), the optimal solution is at x 0 = x b.
−4744 −2682 825 1,650 −3,507 2,682 −9,488 −1,238 −9,076 −4,951 −9,076 −8,250 −6,188 −1,237
0
As =0
ɛ s
= −ɛ y
As ¼ 0
0
−1,317 −849 −6,590
−1761
−2,074 b b
0 2,817 b b b b b b
0 672
3,483 −2,488 4,514 3,176 595 2,785 −3,118 2,785 −6,418 51 594
= −0.01
ɛ
−2,148 −4,102 −7,422 −8,203 −3,320 −9,180
0 b b b b
5,980 0 1,885 0 b b
880 0
2,344 −5,469 1,953 −1,953 1,953 1,172 −781 −5,469
ultimate strength evaluation. More generally, RSD may be used to identify optimal solutions for any cross-sectional shape, for materials with different constitutive behaviors, and for different assumptions regarding ultimate strength evaluation. Acknowledgments The authors wish to recognize the helpful comments of two anonymous reviewers.
6 Conclusion
A TOSR was articulated based on clear patterns evident in the minimum reinforcement solutions obtained in the optimal design of reinforced concrete sections for combinations of axial load and bending moment. According to the theorem, the minimum total reinforcement solution can be determined by examining the following five cases: As = 0 0 0, As ¼ 0, ɛ s = −ɛ y, ɛ = ɛ s = "s = ɛ c,max =0.003, and ɛ = ɛ s = 0 "s ¼ 0:01. Admissible solutions, which may exist for a subset of these cases, are examined to determine the minimum total reinforcement solution, as described in the paper. As a result, minimum reinforcement solutions may be obtained speedily, requiring far fewer analyses than are required to determine optimal solutions using RSD (described recently by Hernández-Montes et al. 2005). The individual cases associated with the minimum reinforcement patterns correspond one-to-one with the optimal domains described by Aschheim et al. (2007). Whereas the optimal domains are described in N – M space and require deductive logic to establish the governing domain, the solution procedure contained within the TOSR may be considered to be more direct and may be easier to implement in numerical solution procedures. The TOSR was established for a large range of material properties and rectangular section geometries commonly used in practice and a particular set of assumptions used for
Appendix 0
For the computation of As and As using (6) and (7), it is 0 necessary to express the stresses σs and s s as explicit functions of the neutral axis depth, x. These expressions are developed in the following based on whether the reinforcement is yielding in tension or compression. Bottom reinforcement Based on the geometry shown in Fig. 8, the depth of the neutral axis for certain states can be defined. The balanced condition is the state at which the bottom reinforcement is at the yield point in tension, given by x b x b ¼
0:003 d 0:003 "y
ð12Þ
where ɛ y assumes a negative value for tension strain. The depth of the neutral axis when the bottom reinforcement is at the yield point in compression is given by the same expression due to the fact that ɛ s changes sign so that x c is given by xc ¼
where
0:003 d 0:003 "y
ɛ y
ð13Þ
assumes a positive value for compressive strain.
Hernández-Montes et al.
B
Therefore, the stress in the bottom reinforcement σ s( x) is *
A’s
x
8< ð Þ¼ :
d
xb
A As
-0.01
xc
-εy
0 ε y 0.003
f y
h d
Fig. 9 Depth of the neutral axis for cases in which the top reinforcement yields in tension or compression
x < x b x b x xc x > x c
ð14Þ
For large values of x and typical configurations of reinforcement, that is, x ≥ 1.25 h and x ≥ xc, the stresses acting on the section are independent of x; therefore, the external N and M required to equilibrate the internal stress resultants are independent of x. For ordinary cover dimensions, the limit x ≥ xc determines the fixed values of N and M , and this limit governs for cover thicknesses such that
Fig. 8 Depth of the neutral axis for cases in which the bottom reinforcement yields in tension or compression
if if if
0:003 E s d x x
σs x
h
f y
"y
0:0006
0:0024
ð15Þ
h
which corresponds to h − d ≤ 0.583 h for Grade 400 reinforcement and 0.792 h for Grade 500 reinforcement. Thus, the fixed values of N and M can be found for x ≥ xc using x= xc for virtually all practical cases of interest. As shown in Fig. 2, the strain diagram used to establish ultimate strength can be located in strain domains I, II, or
Compression yield
x'c2
x'c1
-0.01
-εy 0 εy 0.003 , (a) x=x c1 x*
-εy 0 εy 0.003
-0.01
, x=x c2
(b)
Tension yield
x'b
-0.01
-εy 0 ε y 0.003 , (c) x=x b
-0.01
-εy 0 ε y 0.003 (d)
x>x*
TOSR for reinforced concrete cross sections
III. Within domains I and II, the strain diagram simply rotates about point B, thereby ensuring that the extreme compression fiber is at a strain of 0.003. Within domain III, the strain diagram pivots about point A, thereby ensuring the bottom reinforcement is at a tensile strain of 0.01. The neutral axis depth varies from + ∞ (represented by a vertical line in domain I) to − ∞ (a vertical line in domain III). The neutral axis depth corresponding to the border between domains II and III is given by x = x*, as shown in Fig. 8. For the values of strain used to define points A and B, x* ¼
3 d 13
ð16Þ
a strain of −0.01. The neutral axis depth for top reinforce0 ment yielding in tension, x b , is given by 0
x b ¼ d þ
0
3 d 13 0
0:003 "y 0:003
d ¼
0 :003 "y 0:013
xc1 ¼
0:003d 0 0:003 "y
x*
d
0 :003 "y 0:013
d
ð18Þ
ð19Þ
Thus, for case (b), 0
xc2 ¼ d 0 þ
"y ðd d 0 Þ
0:01 þ "y
8> >< ð Þ¼ >>:
σs x
f y 0 0:01 E s d d x x 0 0:003 0:003 d x E s f y
0
1 < x < x b 0 if x b x x* 0 if x* x xc1 if
0
if
x > x c1
ð22Þ
For case (b), the stress carried by the top reinforcement is 0
8< ð Þ¼ :
σs x
f yd 0 0:01 E s d d x x fy
if if if
0
1 < x < x b 0 0 x b x < x c2
ð23Þ
0
x > x c2
ð17Þ
For Grade 60 (414 MPa) reinforcement, (17) corresponds to d ′ >0.0715d . Case (b) is defined by yielding of the top reinforcement in compression in concert with the bottom reinforcement having a strain of −0.01. More generally, the top reinforcement maybe responding elastically or maybe yielding in compression or in tension while the bottom reinforcement at strain of −0.01 (in tension). Any of these conditions may occur for d 0 <
ð21Þ
0:01 þ "y
References
For case (a), the depth of the neutral axis corresponding 0 to yield of the top reinforcement in compression, xc1 , is given by 0
"y ðd d 0 Þ
Case (d), which represents yielding of the top reinforcement in tension while the extreme compression fiber strain is equal to 0.003, is only possible for large values of d ′, beyond the range contemplated in Table 1 for the cover dimensions d ′ and h − d . Typically, only case (c) will be applicable for top reinforcement yielding in tension. Thus, to describe the stress behavior of the top reinforcement, it is necessary to distinguish between cases (a) and (b), as follows. For case (a), the stress carried by the top reinforcement is:
Top reinforcement Neutral axis depths corresponding to yielding of the top reinforcement in concert with obtaining strains of 0.003 in the extreme compression fiber or −0.01 in the bottom reinforcement can be defined. Although four cases are illustrated in Fig. 9, only cases (a) through (c) are of interest for ordinary section geometries. For top reinforcement yielding in compression, only the two cases shown in Fig. 9 are possible. For case (a), the top reinforcement is yielding in compression, and the extreme compression fiber strain is equal to 0.003. For this case, 0 xc1 ≥ x*. Based on similar triangles, this case applies where
0
ð20Þ
Case (c) of Fig. 9 represents yielding of the top reinforcement in tension in concert with bottom reinforcement having
ACI Committee 340 (1997) ACI design handbook: design of structural reinforced concrete elements in accordance with the strength design method of ACI 318-95, ACI Special Publication SP-17(97). American Concrete Institute, Detroit, MI, USA ACI 318-05 (2005) Building code requirements for structural concrete. American Concrete Institute, Farmington Hills, MI, USA Aschheim M, Hernández-Montes E, Gil-Martín LM (2007) Optimal domains for strength design of rectangular sections for axial load and moment according to Eurocode 2. Eng Struct (in press) Eurocode 2 (2002) Design of concrete structures- Part 1: general rules and rules for buildings prEN 1992-1-1 (July 2002). European Committee for Standardization, Brussels Hernández-Montes E, Aschheim M, Gil-Martin LM (2004) The impact of optimal longitudinal reinforcement on the curvature ductility capacity of reinforced concrete column sections. Mag Concr Res 56(9):499 – 512 Hernández-Montes E, Gil-Martín LM, Aschheim M (2005) The design of concrete members subjected to uniaxial bending and compression using reinforcement sizing diagrams. ACI Struct J 102(1):150 – 158 Nawy (2003) Reinforced Concrete. A fundamental approach, 5th edn. Prentice-Hall, New Jersey SIA 162 (1989) Norme SIA 162: Ouvrages en bétón. Société suisse des ingénieurs et des architectes, Zurich Walther R, Miehlbradt M (1990) Dimensionnement des Structures en Béton. Presses Polytechniques et Universitaires Romandes, Lausanne, Swiss Whitney CS, Cohen E (1956) Guide for ultimate strength design of reinforced concrete. ACI J 28(5):445 – 490 (Nov., Proceedings V.53)
