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where Ie is computed for the service load moment Ma at thePeter H. Bischoff, FACI, is a Professor in the Civil Engineering Department at theUniversity of New Brunswick, Fredericton, NB, Canada. He is a member of ACI critical section, and the cracking moment Mcr (relative to theCommittees 224, Cracking; 360, Design of Slabs on Ground; 435, Deflection of applied moment Ma) controls the amount of tension stiff-Concrete Building Structures; 544, Fiber Reinforced Concrete; and Joint ACI-ASCE ening in the member response. Equation (3) was adopted byCommittee 408, Development and Splicing of Deformed Bars. His research interestsinclude serviceability behavior of concrete structures. ACI 318 in 1971 and continues to be used for calculating deflection with the exception of slender (tilt-up) wall panelsAndrew Scanlon, FACI, is a Professor of civil engineering at the Pennsylvania State (ACI Committee 318 2008).University, University Park, PA. He is Chair of ACI Committee 435, Deflection ofConcrete Building Structures, and is a member of ACI Committees 224, Cracking; 342, Branson’s equation works well for moderately to heavilyEvaluation of Concrete Bridges and Bridge Elements; and 348, Structural Safety; and reinforced members with a steel reinforcing ratio greaterHis research interests include safety and serviceability of concrete structures. than approximately 1%, but consistently overestimates member stiffness for lightly reinforced members and fiber-member, and effective depth d of the tension steel. Values of reinforced polymer (FRP) reinforced concrete membersthe time-dependent factor ξ prescribed by ACI 318 are 1.0, (Bischoff 2007; Bischoff and Scanlon 2007). Hence, deflection1.2, 1.4, and 2.0 for 3 months, 6 months, 1 year, and 5 years is underestimated as a consequence. An alternative expressionor more, respectively. The deflection multiplier λ has a proposed by Bischoff (2005) for use in ACI 318 is given bymaximum value of 2.0 for the worst case (assuming no Eq. (4) and used in this paper to evaluate the span-depthcompression steel) when loads are sustained over a period of relationship limits, as this expression works equally well for5 years or more. steel- and FRP-reinforced concrete members over a wide This paper uses ACI-prescribed deflection limits to evaluate range of reinforcing ratio.maximum span-depth ratios for one-way flexural membersand compares these to the ACI 318 minimum thickness I crvalues. Requirements for incremental deflection Δincr , I e = --------------------------------------- ≤ I g with η = 1 – I cr ⁄ I g - (4) 2which occurs after attachment of the nonstructural 1 – η ( M cr ⁄ M a )elements, typically govern such that SUPPORT CONDITIONS Δincr = Δlt + Δi,L(add) ≤ l/240 or l/480 (2) End restraint conditions at the supports are taken into account with the restraint factor K used in Eq. (1) to compute elastic deflection. Values given in Table 1 are based on thewhere Δlt is the long-term deflection under sustained loads deflection equation corresponding to uniform loading for thethat occurs after installation of the nonstructural elements support conditions indicated. For a continuous member withand Δi,L(add) equals the immediate deflection from the either one or both ends continuous, the restraint factorremaining live load that is not part of the sustained load. In depends on the assumed value of the end support momentsother words, Δi,L(add) = Δi,D+L – Δi,D+L(sus). When used in and does not account for redistribution of moments as theconjunction with Eq. (1), Eq. (2) can be rearranged to give members crack. Table 1 also gives the moment Ma at the criticalspan-depth ratios as a function of the reinforcing ratio ρ for a section expressed as a ratio of the total static moment Mo = w l2/8,given deflection limit (l/240 or l/480). Comparison is made for and is directly related to the restraint factor K = 1.2 – 0.2/Crectangular and T-shaped sections as outlined later in the paper. (except for cantilevers) given that C = Ma /Mo. For a member with both ends continuous, the value of K is RESEARCH SIGNIFICANCE assumed to vary from 0.8 using the ACI moment coefficient for Design of reinforced concrete structures requires evaluation an interior span (based on a positive midspan moment Mm ofof serviceability conditions related to deflection which is wl2/16 or 0.5Mo) down to 0.7 using Rangan’s (1982)controlled by either choosing a member that meets minimum assumption that the average moment at the end supports equalsthickness requirements or limiting computed values of 0.6Mo. For a member with one end continuous, the value of Kdeflection to some fraction of the member span. Maximum varies from 0.85 using the ACI moment coefficient for an endspan-depth ratios (l/h) are developed using the incremental span with the discontinuous end integral with the support (baseddeflection limits given in ACI 318 and compared to the ACI on a positive midspan moment Mm of wl2/14 or 0.57Mo) downminimum thickness values for one-way members. Results to 0.8 using Rangan’s assumption that Mm equals 0.5Mo. Ashow that, for the most part, members satisfying the ACI member with the discontinuous end unrestrained (Mm =minimum thickness requirements do not necessarily satisfy wl2/11 = 0.73Mo) gives a value of 0.925 for K.the deflection limits prescribed by the ACI 318 Building Code. Equation (1) implies that deflection of continuous members depends on the stiffness of member at the critical EFFECTIVE MOMENT OF INERTIA midspan section with no consideration given to the stiffness Cracking is taken into consideration using an effective at the end supports (unless a weighted average is taken ofmoment of inertia Ie that accounts for nonlinear behavior stiffness at the midspan and end supports). Using the stiffnessarising from a gradual reduction in flexural stiffness as the at midspan alone gives reasonable results and, in most practicalmember cracks and loses tension stiffening under increasing cases, gives a better approximation than deflection computed with a weighted average of member stiffness (Bischoff 2007).load. Branson (1965) introduced the concept of an effectivemoment of inertia using the following expression to modelthe transition from a gross (uncracked) moment of inertia Ig SHRINKAGE RESTRAINT CRACKINGto the cracked transformed moment of inertia Icr. AND CONSTRUCTION LOADS Restraint to shrinkage induces tensile stresses in the concrete that decrease the cracking moment and reduce the M cr 3 M cr 3 flexural stiffness of the member. Recent work by Scanlon I e = ⎛ --------⎞ I g + ⎛ 1 – ⎛ --------⎞ ⎞ I cr ≤ I g - - (3) ⎝ Ma ⎠ ⎝ ⎝ Ma ⎠ ⎠ and Bischoff (2008) suggests using Eq. (4) for Ie along with618 ACI Structural Journal/September-October 2009
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a reduced cracking moment equal to two-thirds the value ofMcr computed with the rupture modulus fr = 7.5 f c′ psi(0.62 f c′ MPa) prescribed by ACI 318-08 and this corre-sponds to a reduced rupture modulus fr = 5 f c′ psi(0.42 f c′ MPa). Decreasing the cracking moment has asignificant effect on computed values of deflection forlightly reinforced members such as slabs, while deflection ofmoderately to more heavily reinforced members is notgreatly affected at full service load because the effectivemoment of inertia Ie is closely approximated by Icr at theseload levels (Scanlon and Bischoff 2008). Long-term deflections computed with the long-term multiplierλ usually depend on calculation of the immediate deflectionunder sustained load using an effective moment of inertiaIe,sus corresponding to the sustained load moment only. Fig. 1—Typical span-depth ratios (l/h) for slabs and beams.Preloading from construction loads prior to installation ofthe nonstructural elements often causes additional crackingthat reduces the member stiffness and increases deflection cracked-to-gross stiffness ratio Icr /Ig = 12(d/h)3[k3cr /3 + nρ(1 –arising from the sustained portion of loading. Scanlon and kcr )2] for a rectangular section with kcr = ( ηρ ) 2 + 2ηρ –Bischoff (2008) recommend computing the immediate nρ, ρ = As /bd, and n = Es /Ec.deflection under sustained loads with an effective moment of The incremental deflection limit Δall /l is equal to either 1/480inertia (Ie,D+L) corresponding to the full (dead plus live) or 1/240 depending on whether or not the nonstructuralservice load moment to account for the reduced stiffness; elements are likely to be damaged by large deflections.this simplifies the calculation procedure considerably. Equations (5) and (6) essentially express the span-depth limit needed for deflection control as a function of the SPAN-DEPTH RATIO RELATIONSHIP reinforcing ratio ρ needed to satisfy strength requirements Relationships for the span-depth ratio developed in this at the critical section and take implicit account of the loadingpaper depend on deflection limits for incremental deflection magnitude w.and allow for a portion of the live load to be sustained. Ageneral expression for the limiting span-depth ratio (l/h) is STRENGTH-CONTROLLED LIMITobtained by equating the incremental deflection from Eq. (2) to Given that the moment at the critical section Ma = (φ/αD+L)Mn,a deflection limit Δall. Mn = Rnbd 2, Mo = w l 2/8, and Ma = CMo (refer to Table 1) leads to the following expressions relating the l/h needed for E c ( d ⁄ h ) ( I e, D + L ⁄ I g ) ( I g ⁄ bd ) 3 Δ all strength control l -- ≤ ----------------------------------------------------------------------- - - -------- (5) h K ( 5 ⁄ 48 )Ω ( φ ⁄ α D + L )R n l 2 l 8 ( φ ⁄ α D + L )R n ( d ⁄ h )Full details of the derivation and explanation of variables are -- = - ---------------------------------------------------- - (7) h C(w ⁄ b)found in the Appendix. This expression is subsequentlysimplified for a rectangular section to give for a slab strip with width b subjected to a uniformly distributed load w, and l 0.8E c ( I e, D + L ⁄ I g ) Δ all -- ≤ ---------------------------------------------------------- - - -------- (6) h 2 l KΩ ( φ ⁄ α D + L ) ( d ⁄ h ) R n 3 l 8 ( φ ⁄ α D + L )R n ( d ⁄ h ) -- = - 3 ---------------------------------------------------- - (8)where the factor Ω = [1 + γ(λ – 1)(Ie,D+L/Ie,sus)] accounts for h C(d ⁄ b)(w ⁄ l)long-term effects with the deflection multiplier λ , the ratio γof sustained load to full service load, and the difference for beams with a d/b aspect ratio and distributed load-to-spanbetween the member stiffness at a sustained and full service length ratio w/l.load when preloading is not taken into account. SettingIe,D+L/Ie,sus = 1 takes account of preloading from construction ALLOWABLE SPAN-DEPTH RATIOSloads such that λMsus + ML,add = ΩMD+L. Using a reduced Figure 1 shows a typical plot of the maximum span-depthcracking moment 0.67Mcr to account for shrinkage restraint ratio (l/h) for deflection control (Eq. (6)) together with plotshas the effect of decreasing the effective moment of inertia Ie. of the limiting l/h controlled by strength for slabs (Eq. (7)) The strength reduction factor φ for flexure is taken equal and beams (Eq. (8)). Intersection of the deflection andto 0.9 for tension-controlled sections, and αD+L is an aver- strength-controlled curves defines the l/h requirement for aaged load factor depending on the ratio of dead to live load. flexural member, with the location of the intersection pointThe flexural resistance factor Rn = Mn/bd 2 = ρfs[1 – ρfs / depending on the loading conditions and aspect ratio of the(2α1 fc′)] for a rectangular section with bar stress fs = fy when member cross section. Loading for the slabs ranges between 100 lb/ft2 (4.8 kPa) for Slab S1 (corresponding to a 4 in.the section is under-reinforced, α1 is a rectangular stress [100 mm] thick slab with a live load of 50 lb/ft2 [2.4 kPa]),block factor equal to 0.85, and I e /I g = (Mcr /M a) 3 + [1 – 200 lb/ft2 (9.6 kPa) for Slab S2, and 500 lb/ft2 (24 kPa) for(Mcr /M a) 3](I cr /Ig) using Branson’s (1965) Eq. (3) or (Icr /Ig)/ Slab S3 (representing a much heavier 2 ft [600 mm] thick slab[1 – η(Mcr /Ma)2] using Bischoff’s (2005) Eq. (4). The with a 200 lb/ft2 [9.6 kPa] live load). Different combinationsACI Structural Journal/September-October 2009 619
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with C = Ma /Mo (Table 1) and Ωw = λwsus + wL,add when Ie,D+L/Ie,sus = 1. This version of the expression also depends on the reinforcing ratio ρ unless simplifying assumptions are made to express Ie as a fixed ratio of Ig (see, for example, Scanlon and Choi 1999). Equation (9) works well for slabs as dependence on the width b of the member is accommodated by using a slab strip of unit width, and the equation is easily rearranged to give an expression more appropriate for beams with a specific d/b aspect ratio for the member cross section. l 6.4E c ( I e, D + L ⁄ I g ) ( d ⁄ h ) Δ all -- ≤ - 4 --------------------------------------------------------- -------- - (10) h KCΩ ( w ⁄ l ) ( d ⁄ b ) lFig. 2—Comparison of l/h limits at full Mcr and no preload. COMPARATIVE STUDY A comparative study is performed to demonstrate the sensitivity of calculated l/h values to various parameters as discussed in this section. The general trend is shown in Fig. 1 as discussed previously. In the results presented further in the paper, the l/h limit is plotted against the reinforcing ratio ρ for a rectangular section using 4000 psi (27.6 MPa) concrete and Grade 60 (414 MPa) reinforcing steel. All members are assumed to have an effective depth-to-height ratio (d/h) of 0.85. Unless otherwise noted, comparisons are made for a simply supported member with dead-to-live load ratio ranging from 0.5 to 2.0, and are based on the l/240 incre- mental deflection requirement using a long-term deflection multiplier λ of 2.0. Calculations, for the most part, use Bischoff’s expression for Ie (Eq. (4)), take account of preload from construction loads, and use two-thirds of Mcr.Fig. 3—Effect of construction loads and reduced cracking The effect of concrete strength, steel grade, support condi-moment. tions, and use of T-shaped sections is also considered.of slab thickness and live load are possible for each slab Effective moment of inertialoading case that depends, of course, on the member span. Figure 2 compares the l/h limit for deflection computed withBeam B1 has an assumed d/b aspect ratio of 1.0 for the both Branson’s (Eq. (3)) and Bischoff’s (Eq. (4)) expressionsmember cross section and w/l equal to 75 lb/ft2 (3.6 kPa), while for Ie. Results are presented for a simply supported memberBeam B2 has a higher aspect ratio of 2.0 and a larger w/l with a deflection limit of l/240. Calculation of incrementalof 300 lb/ft2 (14.4 kPa). deflection, in this instance, is based on the full cracking The maximum l/h limit drops as the reinforcing ratio moment Mcr with no account taken of preloading. Memberincreases (particularly for lightly reinforced members typical thickness is heavily dependent on the proportion of sustainedof slabs) and is heavily dependent on the proportion of loading characterized by the dead-to-live load ratio (D/L),sustained loading. The l/h limit for slabs is greater than that and differences between the two approaches are evident forfor beams as expected. The l/h limit for deflection given by lightly reinforced members with a reinforcing ratio less thanEq. (6) may also not govern under certain conditions for approximately 0.85%, as expected (Bischoff and Scanlonbeams with high d/b and w/l ratios and, in this instance, the 2007). At higher reinforcing ratios, the effective moment ofmember thickness is controlled by strength requirements inertia Ie quickly approaches Icr using either approach.only as observed for Beam B2 in Fig. 1. Deflection requirements are satisfied for most slabs using The solution to the l/h expression given by Eq. (6) is iter- the ACI minimum thickness value (with the exception ofative (depending on ρ, which is unknown until the member heavily loaded slabs having a dead-to-live load ratio [D/L] ofthickness h is chosen). While not suitable for design because of 2.0) but not for beams. Span-depth ratio requirements forits complexity, the developed expression is useful for beams need to be as low as 9 depending on the loadingunderstanding the effect of parameters such as ρ, I e, and conditions and are even more severe as the incrementalMcr ; preloading from construction loads; dead-to-live load deflection limit decreases from l/240 to l/480.ratio; fc′; and fy , on the required member thickness needed tosatisfy deflection requirements. Shrinkage restraint and construction loads Similar types of formulation have expressed the span- Figure 3 shows that preloading from construction loadsdepth ratio as a function of the distributed load w (Rangan and use of a lower cracking moment to account for shrinkage1982; Gilbert 1985; Scanlon and Choi 1999; Scanlon and restraint have a significant effect on computed values ofLee 2006). This approach can be expressed as deflection and corresponding l/h limit for reinforcing ratios less than approximately 1%. Deflection of slabs using the ACI l/h limit of 20 is only satisfied in this instance for slabs l 6.4E c b ( I e, D + L ⁄ I g ) Δ all -- ≤ - 3 --------------------------------------------- -------- - (9) with a service load less than approximately 200 lb/ft2 (9.6 kPa) h KCΩw l when the D/L equals 2. The l/h limit for more heavily rein-620 ACI Structural Journal/September-October 2009
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Fig. 4—Effect of sustained live load. Fig. 6—Effect of concrete strength. Fig. 7—Effect of steel reinforcing grade. shoring. Adding 1% compression steel decreases the long- term deflection multiplier to give λ = ξ/(1 + 50 ρ′) = 2.0/(1 + 50 × 0.01) = 1.33 when partitions are installed immediately upon removal of the shoring, while a 3-month delay in installation of the partitions gives λ = (2.0 – 1.0)/(1 + 50 × 0.01) = 0.67 to allow for long-term deflection that occurs during the first 3 months before the partitions are installed. None of the live load is assumed to be sustained in this example. In both cases, the addition of compression steel and delay in installation of the partitions increases the l/h limit needed to satisfy ACI incremental deflection requirements. Even though changes to the l/h limit can be significant, beams using the existingFig. 5 Effect of compression steel (λ =1.33) and delay in ACI minimum thickness values still do not satisfy deflectionpartition installation (λ = 0.67) for: (a) D/L = 2.0; and (b) requirements in many cases except when the deflectionD /L = 0.5. multiplier λ = 0.67.forced beams is controlled by Icr and remains at 9 (D/L = 2)and 13 (D/L = 0.5) compared to the present requirement of 16 Concrete strength and fyfor beams. Figures 6 and 7 indicate that flexural members with lower- strength concrete and higher-strength steel require greaterSustained live load minimum thickness values as expected (with a larger influence Figure 4 shows the effect of having part of the live load for lower D/L ratios). While the l/h limit for the lower, 4 ksisustained. Increasing the sustained load increases incremental (27.6 MPa) strength concrete, is less than the 10 ksi (69 MPa)deflection and leads to a decrease in the l/h limit with a concrete at reinforcing ratios up to approximately 2%, thesubsequent increase in the minimum member thickness limit for the 10 ksi (69 MPa) concrete eventually drops off towhen deflection governs. a value lower than the 4 ksi (27.6 MPa) concrete at higher reinforcing ratios. Recent work by Tang and Lubell (2008)Long-term deflection multiplier provides more detailed information on the effect of reinforcing Figure 5 shows the effect of first adding compression steel grade on deflection and minimum thickness requirementsand then installing partitions 3 months after removal of the for slabs.ACI Structural Journal/September-October 2009 621
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Fig. 8—l/h limits for T-beam sections.Fig. 9—Effect of incremental deflection limit (l/240 versusl/480).T-shaped sections Figure 8 plots l/h requirements for T-shaped sections withthe reinforcing ratio expressed relative to the web width(ρw = As /bwd). Although the minimum thickness requiredto satisfy the incremental deflection limit decreases as theflange to web width (bf /bw) increases, incremental deflectionrequirements are still not satisfied using ACI 318 minimum Fig. 10—l/h limit for: (a) one end continuous; (b) both endsthickness values for beams (with λ = 2.0). continuous; and (c) cantilever.l/240 versus l/480 Figure 9 shows the effect of decreasing the allowable significant effect on the magnitude of service load for slabsvalue of incremental deflection from l/240 down to l/480 where the self-weight comprises a significant portion ofwhen damage to attached nonstructural elements needs to be the loading.considered. Minimum thickness requirements can doublewith a corresponding 50% decrease in the l/h limit. For the Support conditionssimply supported case shown, heavily reinforced beams Minimum thickness requirements in Eq. (6) are directlywould require an l/h of 4.5 (D/L = 2) and 6 (D/L = 0.5), proportional to the restraint factor K taken from Table 1 towhereas the l/h for slabs can be as low as 12.5. reflect support conditions of a uniformly loaded member. Deflection requirements, however, are not as severe as Once again, any changes to the minimum thickness requiredthey first appear in Fig. 9 because any required increase in for a member are tempered by changes in the reinforcingmember thickness is accompanied by a decrease in reinforcing ratio required to satisfy strength for a given span andratio needed to maintain the same member capacity, and this, in loading condition.turn, has the effect of increasing the limit for the l/h. Hence, Requirements for l/h are plotted in Fig. 10 for a memberthe increase in member thickness will most often be less than with one end continuous (K = 0.85), both ends continuous (K =twice the value required to satisfy the l/240 deflection limit 0.8), and for a cantilever (K = 2.4). Figures 3 and 4 give the case(particularly for lightly reinforced members such as slabs for a simply supported member (K = 1.0). In all cases, thewhere the l/h limit changes rapidly with reinforcing ratio) ACI minimum thickness requirement for beams does notand this is demonstrated in deflection examples presented satisfy incremental deflection requirements when using alater in the paper. Changes in member thickness also have a long-term deflection multiplier λ of 2.0, whereas incremental622 ACI Structural Journal/September-October 2009
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Table 2(a)—Simply supported beam example: 4000 psi (27.6 MPa) concrete and 60 ksi (414 MPa) steel* l/h ρ, % Ma/Mcr Beam height h, in. (mm) Beam width b, in. (mm) Deflection, in. (mm)† ACI 318 minimum thickness 16 — — 15 (381) — — Strength at 0.5 ρmax ‡ 11.8 1.0 3.4 20.3 (515.0) 11.5 (292.0) 1.13 (28.8) Bischoff Ie with no preload, Mcr and λ = 2 11.2 0.84 2.9 21.4 (543.3) 12.1 (307.9) 1.0 (25.4) Bischoff Ie with preload, 0.67 Mcr and λ = 2 10.6 0.71 3.7 22.5 (572.6) 12.8 (324.5) 1.0 (25.4) λ=1 14.0§ 1.8 8.4 17.1 (435.4) 9.7 (246.7) 0.945 (24.0) fy = 75 ksi and λ = 2 9.7 0.425 2.8 24.7 (626.2) 14.0 (354.9) 1.0 (25.4) fc′ = 10,000 psi and λ = 2 11.4 0.84 2.85 21.1 (535.1) 11.9 (303.2) 1.0 (25.4) Δ = l/480 and λ = 2 8.2 0.31 1.7 29.4 (746.9) 16.7 (423.3) 0.5 (12.7)*l = 20 ft (6096 mm), d/h = 0.85, K = 1.0, D/L = 2.0, and w = 2.1 k/ft (30.6 kN/m).† Δ ≤ l/240 = 1 in. (25.4 mm) for control of deflection.‡Maximum reinforcing ratio ρmax ≈ 0.72ρb.§ Strength-controlled.Note: 1 ksi = 1000 psi; 1 psi = 0.00690 MPa.(l/240) deflection requirements for slabs using the ACIminimum thickness values are mostly satisfied with theexception of those that are heavily loaded with service loadsgreater than approximately 200 to 250 lb/ft2 (9.6 to 12 kPa). Satisfying l/h requirements is important for slabs, as designoften begins by choosing a slab thickness based on minimumthickness requirements. Hence, adequate flexural stiffness ofslabs is ensured by specifying the correct l/h limits. Designof beams, on the other hand, begins by satisfying strengthrequirements where the member cross section is proportionedfor strength. Choosing a relatively low reinforcing ratio in therange of 0.25ρb to 0.40ρb will often result in a member thicknessgreater than the minimum thickness needed to satisfy deflectionrequirements (ACI Committee 435 1995), depending on thew/l ratio, aspect ratio of the member cross section, long-term Fig. 11—Minimum thickness requirements for beamdeflection multiplier, and dead-to-live load ratio. deflection example. DEFLECTION EXAMPLES Using Bischoff’s expression at full Mcr and with no Deflection examples are worked out for: 1) a simply preload (Fig. 11) requires a beam with an l/h limit of 11.2 atsupported rectangular beam; and 2) an interior span of a a reinforcing ratio ρ of 0.84%. The width and height of thiscontinuous one-way slab. Calculations for both examples beam equals 12.1 in. (307.9 mm) and 21.4 in. (543.3 mm),use 4000 psi (27.6 MPa) concrete and Grade 60 (414 MPa) respectively. Accounting for preload and using 0.67Mcrreinforcement to give fr = 474 psi (3.26 MPa), Ec = 3605 ksi decreases the l/h limit from 11.2 to 10.6 at a lower reinforcing(24.39 GPa), and n = 8.04. The d/h is assumed to remain ratio of 0.71% to account for the increased member thickness.constant at 0.85 and λ = 2.0. Note that dimensions are not The width and height of the beam, in this case, is 12.8 in.rounded off in these examples for ease of checking. (324.5 mm) and 22.5 in. (572.6 mm), respectively. In both cases, the member deflection equals the limiting value of l/240Example 1—Rectangular beam equal to 1 in. (25.4 mm). A simply supported beam with a 20 ft (6096 mm) span is Delaying partition installation by 3 months decreases thedesigned to support a uniformly distributed dead load of 1.4 k/ft long-term deflection multiplier to λ = 1. The l/h deflection(20.4 kN/m) and a live load of 0.7 k/ft (10.2 kN/m). This gives a curve no longer intersects the strength curve in this instancew/l of 105 lb/ft2 (5 kPa), D/L of 2.0, service load moment at (Fig. 11) and the l/h limit increases to 14.0 based on strengthmidspan of Ma = 105 k-ft (142.4 kN-m), and factored requirements at a reinforcing ratio of 1.8%. The correspondingmoment Mu = 140 k-ft (189.8 kN-m). Results for this width and height of the beam equals 9.7 in. (246.7 mm) andexample are summarized in Table 2(a) and plotted in Fig. 11. 17.1 in. (435.4 mm), respectively, and the midspan deflection Designing the beam for strength based on an assumed of 0.94 in. (24.0 mm) is less than the maximum value of 1 in.reinforcing ratio ρ = 1% (0.5ρmax) and d/b of 1.5 gives a (25.4 mm) as deflection no longer governs.beam cross section with width b = 11.5 in. (292 mm) and Using reinforcing steel with 75 ksi (517 MPa) yieldheight h = 20.3 in. (515 mm). The corresponding l/h of 11.8 strength decreases the l/h to 9.7 at a reinforcing ratio of 0.425%is less than the ACI 318 limit of 16 for beams, implying that for a beam with width of 14.0 in. (354.9 mm) and height ofdeflection limits should be easily satisfied. The computed 24.7 in. (626.2 mm). The increase in member thickness,value of incremental deflection, however, (using either Eq. (3) compared to a beam with Grade 60 steel, is approximatelyor (4)) equals 1.13 in. (28.8 mm), which is greater than the 10% in this case. A beam with 10,000 psi (69 MPa) concretel/240 limit of 1 in. (25.4 mm). Hence, the member thickness and 60 ksi (414 MPa) steel has a limiting l/h of 11.4 at aneeds to be increased to satisfy ACI 318 requirements for reinforcing ratio of 0.84%, giving a beam with a width ofincremental deflection. 11.9 in. (303.2 mm) and height of 21.1 in. (535.1 mm) that isACI Structural Journal/September-October 2009 623
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Table 2(b)—Continuous slab example: 4000 psi (27.6 MPa) concrete and 60 ksi (414 MPa) steel* l/h ρ, % Ma /Mcr D/L h, in. (mm) Deflection, in. (mm) ACI 318 minimum thickness 28 0.2 0.72 (uncracked) 1.8 8.6 (218) 0.174 (4.4) Bischoff Ie with no preload, Mcr and λ = 2 41.9 0.367 1.27 1.19 5.73 (145.5) 1.0 (25.4) Bischoff Ie with preload, 0.67 Mcr and λ = 2 32.2 0.243 1.31 1.55 7.45 (189) 1.0 (25.4) λ=1 36.8 0.301 1.58 1.35 6.5 (165.5) 1.0 (25.4) fy = 75 ksi and λ = 2 30.8 0.185 1.23 1.62 7.8 (198) 1.0 (25.4) fc′ = 10,000 psi and λ = 2 41.8 0.361 1.20 1.20 5.74 (145.7) 1.0 (25.4) Δ = l/480 and λ = 2 28.8 0.205 1.12 1.74 8.32 (211.5) 0.5 (12.7)* l = 20 ft (6096 mm), d/h = 0.85, K = 0.8, C = 0.5; and live load = 60 lb/ft2 (2.9 kPa).Note: 1 ksi = 1000 psi; 1 psi = 0.00690 MPa. stiffness at this low reinforcement ratio and gives a much smaller deflection of 0.65 in. (16.4 mm). Results are summarized in Table 2(b) and plotted in Fig. 12. Accounting for preload and using 0.67Mcr with Bischoff’s expression for Ie decreases the l/h to 32.2 with a corresponding reinforcing ratio of 0.24%. A slab thickness of 7.45 in. (189 mm) is needed to maintain the computed value of deflection at 1 in. (25.4 mm), and is used as the basis for other comparisons that include a delay in partition installation, increase in concrete strength, higher grade steel, and allowable deflection of l/480 when damage to partitions is of concern. Delaying partition installation by 3 months decreases the long-term multiplier to λ = 1 and this gives a thinner 6.5 in. (165.5 mm) slab based on l/h = 36.8 at a reinforcing ratio of 0.3%.Fig. 12—Minimum thickness requirements for slab Increasing the compressive strength of concrete to 10,000 psideflection example. (69 MPa) decreases the slab thickness considerably to 5.75 in. (145.7 mm) for l/h = 41.8 at a reinforcing ratio of 0.36%,approximately 6% less than the thickness of a beam using while using Grade 75 (517 MPa) steel gives a slightly thicker,4000 psi (27.6 MPa) concrete. 7.8 in. (198 mm) thick slab with l/h = 30.8 at a reinforcing ratio Decreasing the allowable deflection value to 0.5 in. (12.7 mm) of 0.185%. The higher grade steel results in a 5% increase incorresponding to the l/480 incremental deflection limit slab thickness, which is less than the 15% increase computeddecreases the l/h limit to 8.2 at a reinforcing ratio of 0.31%. using the ACI 318 assumption that minimum thickness valuesThe beam has a height of 29.4 in. (746.9 mm) that is 30% for fy other than 60,000 psi (414 MPa) shall be multiplied bygreater than the required thickness for the l/240 deflection limit. 0.4 + fy /100,000. Similar observations were made by Tang and Lubell (2008). Note also the change in dead load and corre- sponding D/L as the slab thickness changes.Example 2—One-way slab Deflection requirements are considered for an interior Finally, decreasing the allowable deflection value to l/480 =span of a continuous one-way slab with a 20 ft (6096 mm) 0.5 in. (12.7 mm) gives a thicker, 8.3 in. (211.5 mm) slab withspan. The slab supports its own self-weight plus a live load l/h = 28.8 at a reinforcing ratio of 0.205%. In this instance,of 60 lb/ft2 (2.9 kPa). None of the live load is sustained and the slab is barely cracked with Ma /Mcr = 1.12. While Eq. (6)proportioning is carried out for an allowable deflection of l/240. suggests that the slab thickness doubles when the allowableMoments are based on the ACI design moment coefficient at deflection drops by one half, increasing the slab thicknessmidspan giving C = 0.5 and a restraint factor K = 0.8 (Table 1). also results in a lower value of ρ and, hence, the correspondingDesign of the support sections is not considered in this example. increase in slab thickness is only approximately 12% in Design begins by assuming l/h = 28 based on existing ACI 318 this case.minimum thickness requirements to give h = 8.6 in. (218 mm) Assumptions made regarding tension stiffening have aand corresponding dead load of 108 lb/ft2 (5.2 kPa). A considerable effect on computed values of deflection whenreinforcing ratio of 0.2% satisfies strength requirements service loads are close to the cracking value, as is the case forwith the 8.6 in. (218 mm) thick slab and this gives an the slab in this example, with Ma/Mcr ranging between 1 anduncracked slab under service loads (at full Mcr) with a 1.5. For beams with a reinforcing ratio greater than 1%, thecomputed deflection of 0.174 in. (4.4 mm) that is considerably Ma/Mcr ratio is typically greater than 3 and the effectiveless than the allowable value of l/240 = 1 in. (25.4 mm). Using moment of inertia is then closely approximated with Icr.Bischoff’s (2005) expression for Ie at full Mcr and with no In closing, it should be noted that deflection requirementspreload gives an l/h limit of 41.9 and a corresponding slab are satisfied in all cases for the slab in this example whenthickness of 5.7 in. (145.5 mm) with ρ = 0.37%. Recall that the using the ACI 318 minimum thickness value, whereas this isreinforcing ratio needs to be increased as the member thickness not the case for the beam in Example 1. The slab wasdecreases to maintain the same strength. The slab is now subjected to a service load of approximately 150 lb/ft2cracked (Ma/Mcr = 1.27) and has a computed value of deflection (7.2 kPa), and a slab supporting a heavier live load of 125 lb/ft2equal to the deflection limit of l/240 = 1 in. (25.4 mm). (6 kPa) or more corresponding to a full service load of at leastBranson’s (1965) expression (Eq. (3)) overestimates member 230 lb/ft2 (11 kPa) would not have satisfied deflection624 ACI Structural Journal/September-October 2009
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2requirements with a slab thickness based on the ACI (l/28) kcr = ( nρ ) + 2nρ – nρminimum value. l = span length Ma = service load moment (at critical section) Mcr = cracking moment CONCLUSIONS MD = dead load moment Span-depth ratio expressions for one-way members have MD+L = full (dead + live) service load momentbeen developed based on deflection calculation equations in ML = live load moment ML,add = part of live load moment that is not sustainedterms of the reinforcement ratio required to meet flexural ML,sus = sustained part of live load momentstrength requirements and deflection control for specified Mm = midspan momentdeflection limits. Mn = nominal moment capacity Results of a comparative study show that the span-depth Mo = static moment capacity (wl 2/8)ratio is highly sensitive to reinforcement ratio (and hence, Msus = sustained (dead + sustained live) moment Mu = factored momentloading) for lightly reinforced members typically associated n = modular ratio (Es/Ec)with one-way slabs (less than approximately 0.5% reinforce- Rn = nominal flexural resistance factor (Mn/bd2)ment). Slabs designed using the ACI minimum thickness w = uniformly distributed loadrequirement do not always satisfy the l/240 incremental wL,add = portion of live load not sustaineddeflection requirement for heavier loaded slabs. wsus = sustained load α1 = rectangular stress block factor (0.85) Values obtained for moderate to high reinforcement ratios αD+L = average load factortypical for beams indicate that the ACI 318 minimum thickness Δall = permissible (allowable) deflection (l/240 or l/480)values for these members often do not satisfy the span/240 Δi = immediate deflectiondeflection limit for additional deflection that occurs after Δi,D+L = immediate deflection from full (dead + live) service load Δi,D+L(sus) = immediate deflection from sustained (dead + sustainedinstallation of nonstructural elements. Despite the fact that live) loadbeams satisfying the ACI minimum thickness requirement Δi,L(add) = immediate deflection from remaining part of live load notmay not satisfy the l/240 requirement for incremental deflec- sustainedtion, strength will often govern in many cases (particularly Δincr = incremental deflection (λΔi,sus + Δi,L(add))for beams with high w/l and d/b) to give a beam with a Δi,sus = immediate deflection from sustained load Δlt = long term deflection (λΔi,sus)deflection less than l/240. φ = strength reduction factor Although the developed expressions are not intended for γ = ratio of sustained to full service load (Msus/MD+L)hand calculation as a design tool, they can be used to check η = 1 – Icr /Igdesign equations for span-depth limits intended for deflection λ = long term (deflection) multiplier ξ/(1 + 50ρ′)control because no simplifying assumptions have been made ρ = reinforcing ratio (As /bd) ρ′ = compression reinforcing ratio (As′ /bd)other than those implicit in ACI 318 calculation procedures. ρb = balanced reinforcing ratio (bar strain of fy /Es at nominal Results of the study indicate limitations inherent in the strength)current ACI 318 span-depth ratios for one-way members, ρmax = maximum reinforcing ratio (bar strain of 0.004 at nominalparticularly for beams. The results suggest that changes are strength) ρw = web reinforcing ratio (As /bwd)required to account for these limitations so that the minimum ξ = time-dependent factorthicknesses specified are consistent with the span/240 limitation Ω = 1 + γ(λ – 1)(Ie,D+L/Ie,sus)for members not supporting or attached to nonstructuralelements likely to be damaged by large deflections. REFERENCES ACI Committee 318, 2008, “Building Code Requirements for Structural ACKNOWLEDGMENTS Concrete (ACI 318-08) and Commentary,” American Concrete Institute, Financial support was provided by the Natural Sciences and Engineering Farmington Hills, MI, 465 pp.Research Council of Canada and is gratefully appreciated. ACI Committee 435, 1995, “Control of Deflection in Concrete Structures (ACI 435R-95),” American Concrete Institute, Farmington Hills, MI, 88 pp. ACI Committee 435 (Subcommittee 1), 1968, “Allowable Deflections,” NOTATION ACI JOURNAL, Proceedings V. 65, No. 8, Aug., pp. 433-444.As = tension steel area Bischoff, P. H., 2005, “Re-Evaluation of Deflection Prediction forAs′ = compression steel area Concrete Beams Reinforced with Steel and Fiber-Reinforced Polymerb = width of member Bars,” Journal of Structural Engineering, ASCE, V. 131, No. 5, pp. 752-767.bf = flange width Bischoff, P. H., 2007, “Rational Model for Calculating Deflection ofbw = web width Reinforced Concrete Beams and Slabs,” Canadian Journal of CivilC = Ma /Mo Engineering, V. 34, No. 8, pp. 992-1002.d = effective depth of tension steel Bischoff, P. H., and Scanlon, A., 2007, “Effective Moment of Inertia forEc = elastic modulus of concrete Calculating Deflections of Concrete Members Containing Steel ReinforcementEs = elastic modulus of reinforcing steel and Fiber-Reinforced Polymer Reinforcement,” ACI Structural Journal,FD/L = dead-to-live load (D/L) ratio (MD /ML) V. 104, No. 1, Jan.-Feb., pp. 68-75.FS = sustained live load ratio (ML,sus/ML) Branson, D. E., 1965, “Instantaneous and Time-Dependent Deflectionsfc′ = specified compressive strength of concrete of Simple and Continuous Reinforced Concrete Beams,” HPR Report No. 7,fr = rupture modulus of concrete Part 1, Alabama Highway Department, Bureau of Public Roads, Alabama,fs = bar stress 78 pp.fy = yield strength of reinforcing steel CSA A23.3, 2004, “Design of Concrete Structures,” CSA Standardh = height or depth of member A23.3-04, Canadian Standards Association (CSA), Rexdale (Toronto), ON,Icr = cracked transformed moment of inertia Canada, 214 pp.Ie = effective moment of inertia Gilbert, R. I., 1985, “Deflection Control of Slabs Using Allowable Span toIe,D+L = effective moment of inertia corresponding to full (dead + Depth Ratios,” ACI JOURNAL, Proceedings V. 82, No. 1, Jan.-Feb., pp. 67-72. live) service load Rangan, B.V., 1982, “Control of Beam Deflections by AllowableIe,sus = effective moment of inertia corresponding to sustained load Span-Depth Ratios,” ACI JOURNAL, Proceedings V. 79, No. 5, Sept.- moment Oct., pp. 372-377.Ig = gross (uncracked) moment of inertia Scanlon, A., and Bischoff, P. H., 2008, “Shrinkage Restraint andK = end restraint factor Loading History Effects on Deflection of Flexural Members,” ACI StructuralACI Structural Journal/September-October 2009 625
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Journal, V. 105, No. 4, July-Aug., pp. 498-506. Msus = MD + ML,sus = γMD+L (A-4) Scanlon, A., and Choi, B.-S., 1999, “Evaluation of ACI 318 MinimumThickness Requirements for One-Way Slabs,” ACI Structural Journal,V. 96, No. 4, July-Aug., pp. 616-621. with γ = Msus/MD+L = (FD/L + FS)/(1 + FD/L). Scanlon, A., and Lee, Y. H., 2006, “Unified Span-to-Depth Ratio Equation Substitution of Eq. (A-2) to (A-4) into (A-1) then givesfor Nonprestressed Concrete Beams and Slabs,” ACI Structural Journal, V. 103,No. 1, Jan.-Feb., pp. 142-148. 5 ΩM D + L l Tang, J., and Lubell, A. S., 2008, “Influence of Longitudinal Reinforcement 2Strength on One-Way Slab Deflection,” Canadian Journal of Civil Δ incr = K ----- ------------------- ---- ≤ Δ all - - - (A-5)Engineering, V. 35, No. 10, pp. 1076-1087. 48 I e, D + L E c APPENDIX: DERIVATION OF SPAN-DEPTH LIMIT with Ω = [1 + γ(λ – 1)(Ie,D+L/Ie,sus)]. ΩMD+L = λMsus + Combining Eq. (1) and (2) gives a general expression for ML,add when Ie,sus = Ie,D+L. Equation (A-5) also gives theincremental deflection. immediate deflection from live load for the case where λ = 0 and the sustained load equals the dead load alone (F s = 0 Δ incr = λΔ i, sus + Δ i, L ( add ) ≤ Δ all (A-1) giving γ = F D/L/(1 + FD/L)). Rearranging Eq. (A-5) in terms of I e,D+L/I g, with 2 M D+L = (φ/αD+L)M n and Mn = R nbd2 leads to 5- M sus l - Δ i, sus = K ----- ----------------- (A-2) 48 E c I e, sus 2 5 Ω ( φ ⁄ α D + L )R n l Δ incr = K ----- ------------------------------------------------------------ ≤ Δ all - (A-6) 48 E d ( I 3 2 2 c e, D + L ⁄ I g ) ( I g ⁄ bd ) 5 MD + L l 5 M sus l Δ i, L ( add ) = K ----- --------------------- – K ----- ----------------- - - - - (A-3) 48 E c I e, D + L 48 E c I e, sus in which φ is the strength reduction factor for flexure and αD+L is an averaged load factor equal to (1.2FD/L + 1.6)/with Δi,L(add) = Δi,D+L – Δi,D+L(sus) and Δall = l/240 or l/480. (FD/L + 1) for a dead load factor of 1.2 and live load factorEquation (A-3) accounts for the case where there is no of 1.6. Rearranging Eq. (A-6) in terms of the span-depthpreloading from construction loads (Ie,sus > Ie,D+L). ratio (l/h) gives A dead-to-live load (D/L) ratio FD/L = MD/ML andsustained live load ratio FS = ML,sus/ML are used to definethe sustained moment Msus (arising from the dead plus 3 l E c ( d ⁄ h ) ( I e, D + L ⁄ I g ) ( I g ⁄ bd ) Δ allsustained part of the live load) in terms of the full service -- ≤ ----------------------------------------------------------------------- - - -------- (A-7)load moment MD+L. h K ( 5 ⁄ 48 )Ω ( φ ⁄ α D + L )R n l626 ACI Structural Journal/September-October 2009
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