Most concrete codes have empirical equations to estimate theminimum steel reinforcement requirements for flexural members.High-strength thick concrete plates are used for offshore andcontainment structures for nuclear power generation. An accurateestimate of the minimum steel flexure reinforcement ratio can resultin saving millions of dollars for a single project. The recommendedconcept utilizes the fracture mechanics principles to modify thesandwich panel model and account for thick slabs. A comparisonbetween the proposed equations with experimental tests carried byBattista is conducted; also, a comparison of the proposed equationwith different code formulas is illustrated. The results obtained bythe proposed equations are confirmed by the Norwegian code(NS 3474E) and suggestions by Bosco and Carpinteri. Insummary, the two new main contributions in this research are asfollows: the first contribution considers the size effect throughfracture mechanics, and the second contribution considers thetorsional moment for thick plates in calculating the minimumreinforcement of thick plates.
INTRODUCTIONConcrete structure codes have a minimum steel reinforcement
requirement for flexural members. Steel reinforcementrequirements are intended to prevent excessive cracking atservice loads for aesthetic and durability reasons, tie thestructure together and ensure adequate deflection, andprovide a ductile response and ensure adequate warning ofan impending failure at extreme overloads. While theminimum reinforcement requirements are empirical, there isa definite need to change them for thick concrete structures.It is commonly agreed that prior to failure, the reinforcedconcrete (RC) structure must have ductility to provideenough warning through cracking and visible deformations. Thisconstitutes the requirements for the maximum reinforcementratio. Also, a minimum amount of reinforcement is needed toavoid brittle failure.
Research results suggest that the minimum reinforcementratio is member size dependent. The exact tendency perfor-mance criterion, however, is not very clear. A criterion forevaluating the minimum flexural reinforcement require-ments is to have an ultimate moment; Mu should be greaterthan the cracking moment, Mcr , that is, Mu /Mcr > 1.0.Although all researchers agree that the ultimate momentshould be greater than the cracking moment, there is noagreement on the ratio. Most design codes deal with theminimum steel reinforcement ratio independent of membersize. Only a few codes, such as the NS 3473 E1, suggest thatthe minimum reinforcement ratio is a size dependent factor.The size-dependence of the fracturing stability in lightly-reinforced concrete beams has been analytically explained
by means of fracture mechanics concepts since the 1980s byCarpinteri2 and Bosco and Carpinteri.3
Minimum reinforcement requirements needed to preventexcessive cracking and limit the crack width depends on barbond characteristics and environmental factors, as well asfracture considerations. Tests that determine the amount ofreinforcement required to prevent brittle failure in high-strength concrete slabs was conducted by Battista.4 Theresearcher found that the code requirement of CSA A23.3-945
Clause 7.8.1, which specifies Asmin = 0.002Ag, can be sufficientin producing a ductile failure for slabs with strengths up to85 MPa (12,328 psi). Battista4 found that thicker slabscracked at a lower calculated stress than thinner slabs,implying that the minimum reinforcement ratio for thickermembers can be less than that required for thinner members.The fact that larger members crack at lower values of flexuraltension bars is recognized in the CEB-FIP6 Code, where theflexural cracking stress is assumed to be inversely propor-tional to the fourth root of the depth, up to depths of 1 m(39.4 in.).
Marzouk and Hussein7 reported tests of 17 slabs withvarying concrete strength 30 to 80 MPa (4350 to 11,600 psi),slab thickness 90 to 150 mm (3.5 to 6 in.), tension reinforcementratios (0.64 to 2.37%), and column sizes of 150 to 300 mm (6 to12 in.). Major conclusions derived from this investigationincluded: 1) punching failure of high-strength concreteslabs can be classified into two modes, “flexure-punching”and “punching shear” failure. Flexural-punching occurred inthe slabs with relatively low reinforcement ratio; 2) asreinforcement ratio increased, slab stiffness increased anddeformation capacity decreased; 3) ACI equations based onMoe’s analysis overestimated the shear capacity of a high-strength concrete slab; and 4) relating connection shearstrength to the square root of concrete strength resulted in anoverestimation of the effect of concrete strength. A post-cracking tensile model of reinforced high-strength concretebased on the fracture energy of high-strength concreterecorded from stress-displacement curves (stress-crackwidth curves) was incorporated into the shell element usedby Marzouk and Chen8 to analytically study the gravityloading capacities of high-strength concrete slabs tested byMarzouk and Hussein.7 These square slabs, with a 150 mm(6.0 in.) thickness and 1500 mm (59 in.) span, were simplysupported along their four edges and loaded axially throughthe column stub during the testing. The strength predicted
Title no. 106-S61
New Formula to Calculate Minimum Flexure Reinforcement for Thick High-Strength Concrete Platesby E. Rizk and H. Marzouk
657ACI Structural Journal/September-October 2009
using ABAQUS analysis for lightly reinforced slabs was ingood agreement with the experimental results.
RESEARCH SIGNIFICANCEThe amount of reinforcement that is enough for crack
control is not easily determined. Because the formation ofcracks is a complex behavior, the present minimum reinforce-ment guidelines are empirical and do not normally consider theeffect of section thickness (size effect) or concrete cover. Anaccurate estimate of the minimum steel flexure reinforcementratio can save millions of dollars for a single project. In thispaper, a new equation was developed to calculate minimumflexure reinforcement for thick plates and two-way slabs. Themain contribution of this research investigation is to account forthe torsional moment and the size effect factor in estimatingthe minimum reinforcement of concrete slabs.
PREVIOUS RESEARCHDesign codes specify the minimum flexural reinforcement
for reinforced concrete beams and slabs. With the extensiveuse of thick concrete covers, the empirical expressions usedin the past for minimum flexural reinforcement, whichusually ignored the effect of concrete cover thickness, haveto be revised. Research results suggest that the minimumreinforcement ratio is member size dependent; however,with the lack of experimental data, the effect of member sizeis not clear.
Minimum flexural reinforcementThe minimum flexural reinforcement requirements given
in the CSA A23.3-849 are the same as those in the ACI 318-89,10
which are briefly summarized in the following paragraph. The SIunits are used for all the expressions. The 1984 Canadian standardand the 1989 ACI Code (CSA A23.3-849 and ACI 318-8910) are
(1)
The expression in the Canadian standard5 for minimumflexural reinforcement was revised in 1994 to include theconcrete strength. The expression is very similar to the onegiven in ACI 318-95.11
(2a)
(2b)
ρminAs
bwd---------⎝ ⎠⎛ ⎞
min
1.4fy
-------= =
ρminAs
bth-------⎝ ⎠⎛ ⎞
min
0.2fc′ (MPa)
fy
---------------------------≅=
ρminAs
bth-------⎝ ⎠⎛ ⎞
min
2.4fc′ (psi)
fy
----------------------≅=
The limit of validity for concrete strength is given as 20 MPa(2900 psi) < fc′ < 80 MPa (11,600 psi). The 2004 CanadianOffshore Code (CSA-S474-0412) states that the area ofreinforcement near each face and in each of the two orthogonalreinforcement directions shall not be less than 0.003 times thearea of the concrete section for all exterior elements. Thearea of primary reinforcement near the concrete face of anelement exposed to fluid pressure shall not be less than thearea calculated using the following equation
(mm2) (CSA-S474-0412) (3)
where w is the fluid pressure applied on the face, MPa. Thecenter-to-center spacing of the reinforcing bars near eachconcrete face and in each direction should not exceed 300 mm(12 in.). This equation gives high values for minimumreinforcement, especially for structures subjected to highfluid pressures.
The 2008 ACI Code13 does not account for the membersize effect. The code states that at every section of a flexuralmember where tensile reinforcement is required by analysis,it should not be less than that given by
(mm2) (ACI 318-0813) (4a)
(in.2) (ACI 318-0813) (4b)
where bw is the minimum effective web width, d is thedistance from extreme compression fiber to centroid oflongitudinal tension reinforcement, fc′ is the specifiedcompressive strength of concrete, and fy is the specified yieldstrength. As,min shall not be taken less than 1.4bwd/fy. Forstatically determinate members with a flange in tension,As,min shall not be less than the value given by Eq. (4), exceptwhere bw is replaced by either 2bw or the width of the flange,whichever is smaller. All of the North American codesignore the size effect of the slab and the torsional momentson the minimum required reinforcement.
The Norwegian code1 accounts for the effect of the membersize. The code determined member size transverse to the mainreinforcement and directly on this, a continuous minimum rein-forcement shall be placed. The reinforcement shall have atotal cross-sectional area equal to
As ≥ 0.25kwAc ftk /fsk (NS 3473 E1) (5)
where kw is 1.5 – h/h1 ≥ 1.0 (size effect factor), h is the totaldepth of the cross section and h1 = 1.0 m, ftk is the expectedlower characteristic tensile strength of the concrete in MPa,and fsk is the characteristic reinforcement strength defined asthe yield stress or the 0.2% proof stress in MPa. In structureswhere special requirements to limiting crack widths apply,the minimum reinforcement should be at least twice thevalue given previously.
Eurocode 214 states that if crack control is required, aminimum amount of bonded reinforcement is required tocontrol cracking in areas where tension is expected. The
Asfcr w+
fy
---------------- bhef=
As min,
0.25 fc′ (MPa)
fy
-------------------------------------bwd=
As min,
3 fc′ (psi)
fy
--------------------------bwd=
E. Rizk is a Research Assistant at Memorial University of Newfoundland, St. John’s,NL, Canada, and is an Assistant Lecturer at Menoufiya University, Shebin Elkom,Egypt. He received his BSc and MSc from Menoufiya University in 1999 and 2005,respectively. His research interests include cracking and minimum reinforcement ofoffshore structures and shear strength of two-way slabs.
ACI member H. Marzouk is the Chair of the Civil Engineering Department at RyersonUniversity, Toronto, ON, Canada. He received his BSc from Cairo University, Giza,Egypt, and his MSc and PhD from the University of Saskatchewan, SK, Canada. He isa member of ACI Committees 209, Creep and Shrinkage in Concrete, and 213, LightweightAggregate and Concrete. His research interests include structural and material propertiesof high-strength and lightweight high-strength concrete, creep, fracture mechanics,and finite element analysis.
658 ACI Structural Journal/September-October 2009
amount may be estimated from equilibrium between thetensile force in concrete just before cracking and the tensileforce in reinforcement at yielding or at a lower stress ifnecessary to limit the crack width. Unless a more rigorouscalculation shows lesser areas to be adequate, the requiredminimum areas of reinforcement may be calculated asfollows: in profiled cross sections like T-beams and boxgirders, minimum reinforcement should be determined forthe individual parts of the section (webs, flanges)
As, min = KcK fct,effAct /σs (Eurocode 214) (6)
where As,min is the minimum area of reinforcing steel withinthe tensile zone, and Act is the area of concrete within tensilezone. The tensile zone is that part of the section to be intension just before formation of the first crack, and σs is theabsolute value of the maximum stress permitted in thereinforcement immediately after formation of the crack.This may be taken as the yield strength of the reinforcement,fyk. A lower value, however, may be needed to satisfy thecrack width limits according to the maximum bar size or themaximum bar spacing; fct,eff is the mean value of the tensilestrength of the concrete effective at the time when the cracksmay first be expected to occur, fct,eff = fctm or lower, ( fctm(t)),if cracking is expected earlier than 28 days, k is the coefficientwhich allows for the effect of nonuniform self-equilibratingstresses, which lead to a reduction of restraint forces, (k = 1.0for webs with h ≤ 300 mm (12 in.) or flanges with widths lessthan 300 mm (12 in.), and k = 0.65 for webs with h ≥ 800 mm(32 in.) or flanges with widths greater than 800 mm (32 in.)intermediate values may be interpolated), and kc is a coefficientwhich takes account of the nature of the stress distributionwithin the section immediately prior to cracking and of thechange of the lever arm for pure tension, kc = 1.
Bosco et al.15 evaluated the minimum flexural reinforcementcorresponding to a condition at which the formation of firstflexural cracking and yielding of the steel reinforcementoccur simultaneously. The brittleness of reinforced concretebeams increases as the beam size increases or the steelcontent decreases. The linear elastic fracture mechanics(LEFM) model was able to capture the most relevant aspectsand trends in the mechanical and failure behavior of lightlyreinforced HSC beams in flexure. The brittleness number isderived from LEFM concepts as
(7)
where As /A is the steel ratio based on the gross section ofthe beam, KIC is the critical stress intensity factor calculated as(GfEc)
0.5 where Gf is the fracture energy and Ec is the modulusdetermined by standard methods, fy is the yield strength ofthe steel, and h is the overall depth of the beam.
Carpinteri2 and Bosco et al.16 found that the brittleness ofstructural concrete increases as the size increases and/or thereinforcement ratio decreases. Physically similar behaviorhas been revealed in cases where the brittleness number NPis the same. Different sizes of beams with HSC (compressivestrength 91.2 N/mm2 [13,227 psi]) have been tested. At avalue of NP equal to 0.26, the yielding moment is more orless equal to the first cracking moment of the beam. Thereinforcement corresponding to this condition was considered
As
A-----
NPKIC
fyh0.5---------------=
for predicting the minimum reinforcement in flexuralmembers. The percentage of reinforcement established bymany codes is conservative for large size beams, whereas ittends to be insufficient for small size beams. The minimumpercentage of reinforcement tends to be inversely proportionalto the beam depth, while the values specified by the codes areindependent of the beam depth. Hillerborg17 consideredstrain localization in concrete while analyzing the reinforcedconcrete beams. Strain localization is a fact in tension inconcrete and the stresses pass through the peak. Thedescending portion occurs due to crack formation within thefracture process zone. From the analysis of reinforcedconcrete beams, the balanced reinforcement ratio decreaseswith increasing beam depth.
Bosco et al.18 considered five values of nondimensionalductility numbers, that is, 0.0, 0.10, 0.30, 0.75, and 1.20, byvarying both the size and the percentage of steel reinforcement.For low NP values in lightly reinforced beams or for smallcross sections, the fracture moment decreases while thecrack extends. The peak or the first cracking load is lowerthan the steel-yielding load only at a high brittleness number.
Based on experimental results,3 the following relationshipbetween the critical values of the transitional brittleness numberNPC, corresponding to the minimum reinforcement conditionand the concrete compressive strength, has been determined
NPC = 0.1 + 0.0023fcm (8)
where NPC is the critical value of the brittleness number,which distinguishes two failure modes (brittle and ductile),and fcm is the cylindrical compressive strength of concrete.By substituting this expression in Eq. (7), the followingformula for the evaluation of the minimum reinforcementcan be derived as follows
(9)
Ghali et al.19 recommended that if the reinforcement in across section of a member is below a minimum ratio, ρmin,y ,yielding of the reinforcement occurs at the formation of thefirst crack. Such a crack will be excessively wide and formationof several cracks with limited width does not take place. Thisis true when cracking is induced by applied forces orimposed displacements. The minimum reinforcement cross-sectional area As,min,y , and the corresponding steel ratioρmin,y ensure that wide isolated cracks do not occur due toyielding is given by the following expression
(10)
where ρmin,y = As,min,y /Ac, b is the width of the tension side,d is the height of the cross section, fct is the tensile strengthof concrete determined from split cylinder tests, fy is theyield strength of steel, and Ac is the cross-sectional area. Thisequation is suitable for members subjected to significantamounts of normal tensile forces. Hence, it is suitable foroffshore applications.
ρminKIC
fyh0.5------------ 0.1 0.0023fcm+( )=
ρmin y,As min y,,
bd------------------ 0.24
fct
fy
-----≅=
ACI Structural Journal/September-October 2009 659
Minimum reinforcement for HSC slabsA few tests were conducted to determine the amount of
reinforcement required to prevent brittle failure in high-strength concrete slabs by Battista.4 Twenty-four one-wayslabs subjected to pure flexure were tested. The test variableswere concrete compressive strength, slab thickness, and ratioof ultimate stress to yield stress of reinforcement. While the150 mm (6 in.) thick high-strength concrete slab with areinforcement ratio of 0.002 showed an undesirableresponse, a companion specimen that was 300 mm (12 in.)thick made from the same concrete, also having a reinforcementratio of 0.002, displayed a very ductile response with the post-cracking capacity exceeding the cracking moment byapproximately 60%. After cracking, it is important to guaranteethat the forces resisted by concrete in tension are transmittedto tensile longitudinal steel. The 300 mm (12 in.) thick slabstested by Battista4 cracked at a lower calculated stress thanthe 150 mm (6 in.) thick slabs, which had the same minimumreinforcement ratio; this implies that the minimum reinforcementratio required to carry the applied forces for thicker memberscould be less than that required for thinner members, whichis known as the size effect factor.
The fact that larger beams crack at lower values of flexuraltension bars is recognized in the CEB-FIP,6 where the flexuralcracking stress is assumed to be inversely proportional to thefourth root of the depth, up to depths of one meter. It hasbeen recommended that the code requirement of CSAA23.3-94 Clause 7.8.1,5 which requires Asmin = 0.002Ag , canbe sufficient in producing a ductile failure for slabs withstrengths up to 85 MPa (12,328 psi).
Size effectFor design engineers, the size effect is probably the most
compelling reason for using fracture mechanics. The size ofthe fracture process zone is represented by a material propertycalled the characteristic length lch. It expresses the fractureproperties of the concrete, such as the modulus of elasticityEc , the fracture energy Gf , and the tensile strength ft.
(11)
where Gf is defined as the amount of energy required tocause one unit area of a crack, so that it can be obtained asthe area under the load-crack width curve. It should be notedthat the characteristic length can be calculated using Eq. (11).The characteristic length, however, has no physicalcorrespondence, that is, it is not a real length that can bemeasured. A higher value of lch results in the material beingless brittle. A smaller value of lch results in the material beingmore brittle. In an earlier investigation by Marzouk andChen,20 the fracture energy, Gf, was determined experimentallyfor high-strength concrete to be 160 N/m (11 lb/ft) compared to110 N/m (7.5 lb/ft) for normal-strength concrete. The fractureenergy is calculated as the area under the descending portionof the stress-crack width curve. The characteristic length lch wasestimated to have an average value of 500 and 250 mm (20 and10 in.) for normal- and high-strength concrete, respectively.
ANALYTICAL INVESTIGATIONSlabs may be subdivided into thick slabs with a thickness
greater than approximately 1/10 of the span, thin slabs witha thickness less than approximately 1/40 of the span, and
lchEcGf
ft2
-----------=
medium-thick slabs. Thick slabs transmit a portion of theloads as a flat arch and have significant in-plane compressiveforces, with the result that the internal resisting compressiveforce is larger than the internal tensile force. Thin slabstransmit a portion of the loads acting as a tension membrane.A medium-thick slab does not exhibit either arch action ormembrane action.
An elastic solution is not suitable for an RC structure aftercracking; before the onset of cracking, the behavior of reinforcedconcrete can be based on elastic solutions or based on the studyof plain concrete because the contribution of reinforcementat this stage is negligible. In evaluating minimum reinforcementratio, it is important to guarantee that if the cracking momentMcr is reached due to eventual overloading, the forcesresisted by concrete in tension are transmitted to tensilelongitudinal steel capable of resisting Mcr. Calculation ofcracking moments is usually based on the behavior ofuncracked concrete sections (Stage I), so using elastic solutionsto compute cracking moment Mcr is acceptable.
Effect of torsional moment on minimum reinforcement of thick slabs
An approximate solution for a rectangular plate problemsubjected to uniform loading has been presented by manyauthors. This problem requires the solution of the deferentialequation
(12)
subject to the boundary conditions
w = 0 = 0 (x = –a, x = a) (13a)
w = 0 = 0 (y = –b, y = b) (13b)
The moment components are related to the deflection w asfollows
(14)
(15)
(16)
where D is the bending rigidity of the plate defined as
(17)
with E as the modulus of elasticity, ν as Poisson’s ratio, andh as the plate thickness. The problem of a rectangular plateclamped at four sides with a uniform load is considered. The
∂4w
∂x4--------- 2 ∂4w
∂x2∂y2---------------- ∂4w
∂y4---------+ + q x y,( )
D----------------=
∂w∂x-------
∂w∂y-------
Mx D ∂2w
∂x2--------- ν∂
2w
∂y2---------+
⎝ ⎠⎜ ⎟⎛ ⎞
–=
My D ∂2w
∂y2--------- ν∂
2w
∂x2---------+
⎝ ⎠⎜ ⎟⎛ ⎞
–=
Mxy D 1 ν–( ) ∂2w
∂x∂y------------–=
D Eh3
12 1 ν2–( )-------------------------=
660 ACI Structural Journal/September-October 2009
sides of the rectangular section are at x = ±a and y = ±b, andit is supposed throughout the paper that a ≥ b. Severalapproaches21 exist to approximately solve the biharmonicproblem governing the plate’s deflection w in a convenientmanner for direct use in engineering applications. In all ofthem, the boundary conditions are satisfied identically.
The most common ways of solving Eq. (14), (15) and (16)is to use a finite element analysis. Such an analysis givesvalues of Mx, My, and Mxy in each element where Mx, My,and Mxy are moments per unit width. A portion of an elementbounded by a diagonal crack is shown in Fig. 1(a). Themoments on the x and y faces from the finite element analysisare shown in Fig. 1(b). The moment about an axis parallel tothe crack is Mc given by
This slab is to be reinforced with bars in the x- and y-directionwith positive moment capacities Mrx and Mry per unit width.The corresponding moment capacity at the assumed crack is
(19)
where Mrc must equal or exceed Mc to provide adequatestrength. Equating Eq. (18b) and (19), and solving forminimum, we get
(20)
because Mry must equal or exceed My to account for theeffects of Mxy, (1/k)Mxy ≥ 0, which gives
(21a)
(21b)
where k is a positive number. This must be true for all crackorientations (that is, for all values of k). As k is increased,
Mry goes down and Mrx goes up. The smallest sum of the two(that is, the smallest total reinforcement) depends on the slabin question, but k = 1 is the best choice for a wide range ofmoment values.22, 23
The reinforcement at the bottom of the slab in each directionis designed to provide positive moment resistances of
(22a)
(22b)
If either of these is negative, it is set equal to zero. Similarly,the steel at the top of the slab is designed to provide negativemoment resistances of
(23a)
(23b)
If either of these is negative, it is set equal to zero.In this paper, a simply supported uniformly loaded square
plate and a rectangular clamped plate with different aspectratios, a/b = 1, a/b = 1.5, a/b = 2, and a/b =2.5, were solvedusing the structural analysis program (SAP) to evaluate thevalue of the bending moments Mx and My and the torsionalmoment Mxy within the plate. The value of the torsionalmoment Mxy is added to the value of the bending moment Mxat each point and compared with the maximum positivebending moment Mx,max+ve. It is found that for simplysupported square plates or clamped square plates, the ratio is
(24)
For a clamped rectangular plate with an aspect ratio a/b =1.5, however, the same ratio is found equal to 1.04; for a/b =2.0, the same ratio is found equal to 1.27; and finally, for a/b =2.5, the same ratio is found equal to 1.13. The differencebetween the normalized calculated torsional moment versusthe aspect ratio for clamped rectangular plates is showngraphically in Fig. 2 as (Mxy + Mx)/Mx,max+ve versus a/b. Inthe case of square clamped plates, the location of themaximum bending moment is at the center of the plate,whereas for rectangular clamped plates with an aspect ratioa/b ≥ 1.0, the location of maximum positive bendingmoment Mx,max+ve is shifted from the center of the plate.
Mcdyds------⎝ ⎠⎛ ⎞ 2
Mx k2My 2kMxy+ +( )=
Mrcdyds------⎝ ⎠⎛ ⎞ 2
Mrx k2Mry+( )=
Mry My1k---Mxy+=
Mry My1k--- Mxy+=
Mrx Mx k Mxy+=
Mry My Mxy+=
Mrx Mx Mxy+=
Mry My Mxy–=
Mrx Mx Mxy–=
Mx Mxy+( )Mx max+ve,
--------------------------- 1.0≤
Fig. 1—Resolution of moments.
ACI Structural Journal/September-October 2009 661
This means that there are two points of maximum positivemoment instead of one point at the center of the plate. For anaspect ratio a/b = 2.0, the location of the points of maximumpositive moment is almost at the quarter points of the plate inthe long direction. The drop in the moment ratio for theaspect ratio a/b of 2.5 is due to the fact that the loadtends to be transferred in the short direction (one wayaction) instead of two way action, this minimizes theeffect of torsional moment.
Based on the previous findings, it can be concluded thatneglecting the effect of the torsional moment Mxy in calculatingthe minimum reinforcement ratio for clamped rectangularplates with an aspect ratio a/b = 2.0 results in approximatelya 25% error. Therefore, using an empirical code formula incalculating the minimum reinforcement ratio for clampedrectangular plates with an aspect ratio a/b > 1.0 is unsafe.
Marti shear sandwich model24
Generally, slab elements are subjected to eight stressresultants, that is , the three membrane force components Nx ,Ny , and Nxy = Nyx; two transverse shear force components Vxand Vy; two flexural moments Mx and My; and torsionalmoment Mxy = Myx (refer to Fig. 3(a)). Marti24 introduced asandwich model where the covers are assumed to carrymoments and membrane forces, while the transverse shearforces are assigned to the core (refer to Fig. 3(b)). As asimple approximation, Marti24 assumed that the middleplanes of the cover elements coincide with the middle planesof the reinforcing meshes close to the slab surfaces.Assuming equal cover element thicknesses at top and bottomc, the lever arm of the in-plane forces in the cover elementsdv, equal to the effective shear depth of the core, equal to thedistance from the center of the top concrete cover element tothe center of the bottom concrete cover element, is given bydv = h – c, where h = slab thickness. For the dimensioning ofthe in-plane reinforcement, the well-known limit-designmethod for RC membrane elements can be employed.Accordingly, the necessary resistances of the reinforcementsin the two orthogonal directions x and y are equal to Nx + k[Nxy]and Ny + [Nxy]/k, respectively, where k denotes an arbitrarypositive factor and Nx, Ny, and Nxy , are the applied
membrane force components. Hence, from Fig. 3, therequirements are obtained
(25)
and
(26)
where ax and ay denote the cross-sectional areas of theorthogonal bottom reinforcements per unit width of the slab,and Vo is the principal diagonal shearing force carried bythe core.
Modified sandwich modelBased on the shear sandwich model proposed by Marti24,
a new modified sandwich model is proposed. The new model
ax fyMx
dv
-------Nx
2------
Vx2
2Vo θtan--------------------- k
Mxy
dv
---------Nxy
2--------
VxVy
2Vo θtan---------------------+ ++ + +≥
ay fyMy
dv
-------Ny
2------
Vy2
2Vo θtan--------------------- 1
k---
Mxy
dv
---------Nxy
2--------
VxVy
2Vo θtan---------------------+ ++ + +≥
Fig. 2—Normalized torsional moment versus aspect ratio((Mxy + Mx)/Mx,max+ve versus a /b) for clamped rectangularplates.
accounts for the slab size effect through the term (lch/h)0.33.Based on the results of a finite element parametric study of81 slabs by Marzouk et al.25, it was found that the shearstrength decreases as the brittleness of the concrete slab, h/lchincreases, where h is the slab thickness. The brittleness ratiois not only a reflection of the size factor, but also a functionof the tension and crack width relationship, the concretestrength, and the type of aggregates. The use of the brittlenessratio h/lch is much more meaningful than the use of an empiricalsize factor. The exponent α = 0.33 represents a sensitivitynumber; it was found by Marzouk et al.25 that a value of 0.25may be taken for concrete strength less than 35 MPa, and avalue of 0.33 may be taken for concrete strength of 75 MPa(10,877 psi). For concrete strength between 35 and 75 MPa(5076 and 10,877 psi), a linear interpolation may be takenbetween the values of 0.25 and 0.33. For concrete strengthhigher than 75 MPa (10,877 psi), α may be considered to behigher than 0.33 but without exceeding the limit of 0.5. In thepresent research, it was found that a value of 0.33 will bemore consistent. The researchers25 recommended that theexpression for shear strength be rewritten for high-strengthconcrete application as follows
(27)
A similar semi-empirical expression was developed inGermany on punching shear capacity of point supported, RCnormal- and high-strength slabs by Staller.26 Theresearcher26 confirmed that the exponent α should be takenequal to 0.33 instead of 0.5 as proposed by North Americancodes.13,27 Based on statistical evaluation of punching
vu 0.88ftcd---⎝ ⎠⎛ ⎞ 0.4–
ρlch
h------⎝ ⎠⎛ ⎞
3=
and shear tests, the design value of punching load withoutshear reinforcement, according to DIN 1045-1,28 is givenby the equation
(28)
where ucrit is the length along a control perimeter at thedistance of 1.5d from the column face. This leads to atheoretical punching cone with an inclination of approximately34 degrees compared to the horizontal.
It is important to guarantee that if the cracking momentMcr is reached due to eventual overloading, the forcesresisted by concrete in tension are transmitted to tensilelongitudinal steel capable of resisting Mcr. With reference toFig. 4 shown below, we can calculate the illustrated forces.Taking moments about the point of action of C1
(29)
where As,min is the minimum flexure reinforcement area, ft isthe tensile strength of concrete, fy is the yield strength ofsteel, d is the distance from extreme compression fiber tocentroid of longitudinal tension reinforcement, kd is thedistance from extreme compression fiber to the neutral axiscalculated using the first moment equation of area, s is theheight of uncracked concrete in tension as illustrated in Fig. 4,and b is the width of the tension side.
The second part of the right hand side of Eq. (29) is verysmall, so it can be neglected. From a linear elastic stressdistribution
(30)
where fr is the rupture strength of concrete
fr = 0.6λ (MPa) (CSA-A23.3-0427) (31a)
fr = 7.5λ (psi) (ACI 318-0812) (31b)
where λ is a modification factor reflecting the reducedmechanical properties of lightweight concrete. Neglect the effectof transverse shear components from the equilibrium equationsfor the modified sandwich model as shown in Fig. 5.
Hence, Fig. 5 illustrates the modified sandwich assumptionsand uses Eq. (25) and (26) to calculate the required reinforcementas follows
(32)
and
(33)
VRd ct,
0.12 1 200d
---------+⎝ ⎠⎛ ⎞ 100ρl fck( )1/3ducrit=
Mcr As min ,
fy d kd3
------–⎝ ⎠⎛ ⎞ ft
2---sb 2kd
3--------- 2s
3-----+⎝ ⎠
⎛ ⎞+=
Mcrfrbh2
6------------=
fc′
fc′
ax min,
fy1
lch/h( )0.33-----------------------
Mx
d------- k
Mxy
d---------+≥
ay min,
fy1
lch/h( )0.33-----------------------
My
d------- 1
k---
Mxy
d---------+≥
Fig. 4—Schematic of reinforced concrete section in bending.
Fig. 5—Modified sandwich model.
ACI Structural Journal/September-October 2009 663
where ax,min and ay,min denote the cross-sectional minimumareas of the orthogonal bottom reinforcements per unit widthof the slab, k denotes an arbitrary positive factor taken equalto unity, Mxy is the torsional moment, in the case of aclamped plate with an aspect ratio a/b = 2.0, and Mxy = 0.27Mx.Hence, the value of ax,min can be calculated by substitution inEq. (32) as follows
(34)
where Mx is equal to the cracking moment. After crackingmoment Mcr is reached, it is assumed that all forces resistedby concrete in tension are transmitted to tensile longitudinalsteel capable of resisting Mcr
(35)
substituting for the value of Mcr
(36)
For plates and concrete slabs with 100 to 200 mm (4 to 8 in.)having concrete covers up to 50 mm (2 in.), it can beassumed that h = 1.4d. It can also be assumed that h ≈ 2hef ,where hef is the effective embedment thickness as the greaterof (Cc + dbe′ ) + 7.5dbe′ and a2 + 7.5dbe′ but not greater thanthe tension zone or half slab thickness (Fig. 6). Substitutingin Eq. (36), the expression proposed for minimum steelreinforcement is given as follows
(37)
For plates and concrete slabs with 200 to 400 mm (8 to 16 in.)having concrete covers up to 75 mm (3 in.), we can assumethat h = 1.3d. Substituting in Eq. (36), the expression proposedfor minimum steel reinforcement is given as follows
(38)
COMPARISON OF PROPOSED EQUATION WITH DIFFERENT CODE PREDICTIONS
To verify the validity of the new proposed equation, acomparison between the proposed equation with Battista’s4
ax min,
fy1
lch/h( )0.33-----------------------
1.27Mx
d-----------------≥
ax min,
fy1
lch/h( )0.33-----------------------
1.27Mcr
d-------------------≥
ax min,
fy1
lch/h( )0.33-----------------------
1.27frbh2
6d-----------------------≥
ρmin0.415fr
fy
----------------- lch/2hef( )0.33=
ρmin0.358fr
fy
----------------- lch/2hef( )0.33=
experimental results and with different code formulas forcalculating minimum reinforcement for flexural members ispresented in the following sections.
Comparison between proposed equationswith experimental results
Twenty four one-way slabs subjected to pure flexure weretested by Battista.4 The test variables were concretecompressive strength, slab thickness, and ratio of ultimatestress-to-yield stress of reinforcement. Details of the testedslabs are presented in Table 1. Minimum flexural reinforcementratios, which are required by different codes and proposedequations, are given in Table 2. All slabs had the samereinforcement ratio. Battista4 found that the crackingloads of the high-strength and the very high-strength specimenswere greater than those of the normal-strength specimens, thebrittleness of the very high-strength concrete slabs increasedas the slab thickness increased, and as the concrete compressivestrength increased for the 300 mm (12 in.) thick specimens, thetotal deformation decreased. This means that structuralbehavior for 300 mm (12 in.) thick high-strength concreteslabs could be enhanced by using a smaller reinforcement ratio;this is due to the size effect. This is verified by the minimumreinforcement ratio required by the proposed equation; forexample, investigating the first group of Specimens S1,S13, and S19, which had thicknesses equal to 150, 200, and300 mm (6, 8, and 12 in.), the minimum reinforcement ratiorequired by equation decreased as the slab thickness increased,and as the concrete strength increased the required minimumreinforcement ratio increased. This is confirmed by thevalues required by the NS 3474 E code1 as well as formula
proposed by Bosco and Carpinteri.3 Examining Table 2, itcan be concluded that Eurocode 214 underestimates theminimum reinforcement ratio required by almost 75%; also, itcan be concluded that Eurocode 2,14 as well as the ACI 318-08,13
does not account for size effect factor. ACI 318-0813
overestimates minimum reinforcement ratio required forthick concrete slabs. In general, it can be noticed that theCanadian building code (CSA-A23.3-0427) does not accountfor size effect factor and the torsional moment effect.
Comparison between proposed equationswith different code formulas
To verify the validity of the new proposed equations, acomparison between the proposed equations with differentcode formulas for calculating minimum reinforcement for
flexural members is implemented as shown in Fig. 7 to 9. Allcalculations were made using a specified steel yield strengthvalue of 60 ksi (415 MPa). The new proposed equationaccounts for the size effect; this agrees with the Norwegiancode1 as well as suggestions by Bosco and Carpinteri.3 It canbe concluded from Fig. 7 to 9 that, the American (ACI 318-0813)and Canadian (CSA-A23.3-0427) codes overestimate theminimum reinforcement ratio required for thick concreteslabs greater than 200 mm (8 in.), due to the fact that none ofthese codes contain a size effect factor. This can result in ahuge saving in steel reinforcement. Eurocode 214 underestimatesthe minimum reinforcement ratio required by as much as 75%,using the Eurocode 214 formula and resulting in an unsafedesign for very thick plates.
CONCLUSIONSThe torsional moment (Mxy) factor is very important in
determining the minimum reinforcement calculation forthick two-way concrete plates (slabs) and should be takeninto account.
Using fracture mechanics concepts, the minimumpercentage of flexural steel reinforcement is size dependent,whereas most codes of practice specify the minimumreinforcement as independent of the depth of themember. Therefore, there is a definite need to change codeformulas for thick-plated offshore applications.
ACI 318-0813 and CSA-A23.3-0427 overestimate theminimum reinforcement ratio required for thick concrete slabs.
For continuous or fixed plates, a minimum top reinforce-ment at least twice the value given by the proposed equation
Table 2—Minimum reinforcement ratios required by proposed equation and different code formulas for tested slabs by Battista4
Slab no.Compressive
strength fc′ , MPa fy , MPa
CSA-04ACI 318-08
ρ, %CSA-S474-04
ρ, %NS 3474 E-89
ρ, %EC2-2004
ρ, %Bosco and Carpinteri
(1992)
Proposed equation by authors
ρ, %
S1 35 584 0.25 0.20 0.31 0.08 0.15 0.38
S7 35 584 0.25 0.19 0.29 0.08 0.13 0.34
S13 35 619 0.24 0.18 0.27 0.07 0.12 0.32
S19 35 619 0.24 0.17 0.25 0.07 0.10 0.28
S4 35 545 0.27 0.21 0.33 0.08 0.16 0.40
S10 35 545 0.27 0.20 0.31 0.08 0.14 0.37
S16 35 571 0.26 0.20 0.30 0.08 0.13 0.35
S22 35 571 0.26 0.19 0.27 0.08 0.11 0.31
S2 55 584 0.32 0.25 0.40 0.10 0.23 0.42
S8 55 584 0.32 0.24 0.37 0.10 0.20 0.38
S14 55 619 0.30 0.23 0.36 0.09 0.19 0.36
S20 55 619 0.30 0.22 0.32 0.09 0.16 0.32
S5 55 545 0.34 0.27 0.43 0.11 0.25 0.45
S11 55 545 0.34 0.26 0.40 0.11 0.22 0.41
S17 55 571 0.32 0.25 0.39 0.10 0.21 0.39
S23 55 571 0.32 0.24 0.35 0.10 0.17 0.34
S3 85 584 0.39 0.31 0.52 0.12 0.38 0.40
S9 85 584 0.39 0.30 0.48 0.12 0.33 0.37
S15 85 619 0.37 0.28 0.46 0.11 0.31 0.35
S21 85 619 0.37 0.27 0.42 0.11 0.25 0.30
S6 85 545 0.42 0.33 0.56 0.12 0.41 0.43
S12 85 545 0.42 0.32 0.52 0.12 0.35 0.39
S18 85 571 0.40 0.31 0.50 0.12 0.34 0.38
S21 85 571 0.40 0.29 0.46 0.12 0.27 0.33
Note: 1 MPa = 145 psi.
Fig. 7—Comparison of minimum reinforcement ratioscalculated by different codes (fc′ = 35 MPa [5075 psi]).
ACI Structural Journal/September-October 2009 665
should be provided, extending into the top of the slabs oneach side of the girders.
ACKNOWLEDGMENTSThe authors are grateful to the Natural Sciences and Engineering
Research Council of Canada (NSERC) for providing the funds for the project.
NOTATIONAc,ef = area of concrete symmetric with reinforcing steel divided by
number of barsAg = gross area of cross sectionAs = area of reinforcement within effective embedment thicknessAs,min = minimum flexure reinforcement areaax , ay = cross-sectional areas of orthogonal bottom reinforcements per
unit width of slabb = width of sectionbw = minimum effective web widthCc = clear cover from nearest surface in tension to flexural tension
reinforcementc = concrete coverD = bending rigidity of plated = effective depth to centroid of tensile reinforcementdv = the effective shear depth of core, given by dv = h – cEc = modulus of elasticity of concreteEs = steel modulus of elasticityfc′ = uniaxial compressive strength of concrete (cylinder strength)fcm = cylindrical compressive strength of concretefct = tensile strength of concrete determined from split cylinder testsfr = rupture strength of concretefs = stress in reinforcement due to applied loadfsk = characteristic reinforcement strength defined as yield stress or
as 0.2% proof stress, in MPaftk = expected lower characteristic tensile strength of concrete, in MPafy = yield stress of steelGf = fracture energyh = section heighthef = effective embedment thicknessKIC = critical stress intensity factor calculated as (Gf Ec)
0.5
kw = size effect factor
kd = distance from extreme compression fiber to neutral axis calculatedusing first moment equation of area
Ich = characteristic lengthMc = magnified factored moment to be used for design of compression
memberMcr = cracking bending momentMu = ultimate bending momentMx = bending moment per unit lengthMxy = torsional moment per unit lengthMy = bending moment per unit lengthNP = brittleness numberNPC = critical value of brittleness number, which distinguishes two
failure modes (brittle and ductile)Nx = in-plane axial applied force per unit lengthNxy = in-plane shearing force per unit lengthNy = in-plane axial applied force per unit lengths = center-to-center spacing of flexural tension reinforcement
nearest to surface of extreme tension faceVx = shearing force per unit lengthVy = shearing force per unit lengthvu = concrete shear strengthw = the fluid pressure on face, MPaw = plate’s deflectionα = sensitivity numberν = Poisson’s ratioτbk = lower fractile value of average bond stress
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Fig. 8—Comparison of minimum reinforcement ratioscalculated by different codes (fc′ = 70 MPa [10,150 psi]).
Fig. 9—Comparison of minimum reinforcement ratioscalculated by different codes (fc′ = 90 MPa [13,054 psi]).
666 ACI Structural Journal/September-October 2009
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