TensinACI-318-11Longitud de desarrollo de varillascorrugadas a tensinProyecto:fecha:Sep-13Fy =59684psi4200kg/cm^2f 'c =3013psi212kg/cm^2yt =1ye =1dim. =1inl =1conc. Peso normalFy yt ye / l(f 'c)^.5 =1087.3945453157caso =20200020ld =54.37dimetrosusar44dimetrosACI 12.2.2Factors for Use in the Expressions for Determining Required Development Lengths forDeformed Bars and Deformed Wires in Tension (ACI 12.2.4)(1) t =reinforcement location factorHorizontal reinforcement so placed that more than 12 in. of fresh concrete is cast in the memberbelow the development length or splice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3Other reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0(2) e =coating factorEpoxy-coated bars or wires with cover less than 3db, or clear spacing less than 6db . . . . . . . . . . . 1.5All other epoxy-coated bars or wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2Uncoated and zinc-coated reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0However, the product of te need not be taken as greater than 1.7.Para aplicar la frmula 12-1, seguir el siguiente procedimiento:(3) s =reinforcement size factorNo. 6 and smaller bars and deformed wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8No. 7 and larger bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0In SI unitsNo. 19 and smaller bars and deformed wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8No. 22 and larger bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0(4) (lambda) = lightweight aggregate concrete factorWhen lightweight aggregate concrete is used, shall not exceed . . . . . . . . . . . . . . . . . . . . . . . . . 0.75However, when fct is specified, shall be permitted to be taken as 6.7 *fct = resistencia promedio a la tensin (traccin)Itsin SIbut not greater than . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0When normal weight concrete is used . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0(5) cb =spacing or cover dimension, in.Use the smaller of either the distance from the center of the bar or wire to the nearest concrete surface,or one-half the center-to-center spacing of the bars or wires being developed.In the following paragraphs, all of the terms in ACI Equation 12-1 that have not previouslybeen introduced are described. their values for different situations were given inthe previous page.1. Location of reinforcementHorizontal bars that have a least 12 in.[3] of fresh concreteplaced beneath them do not bond as well to concrete as do bars placed nearer the bottomof the concrete. These bars are referred to as top bars. During the placing and vibrationof the concrete, some air and excess water tend to rise toward the top of the concrete,and some portion may be caught under the higher bars. In addition, there may be somesettlement of the concrete below. As a result, the reinforcement does not bond as well tothe concrete underneath, and increased development lengths will be needed. To accountfor this effect, the reinforcement location factor, t, is used.2. Coating of barsEpoxy-coated reinforcing bars are frequently used today to protect thesteel from severe corrosive situations, such as where deicing chemicals are used. Bridgedecks and parking garage slabs in the colder states fit into this class. When bar coatingsare used, bonding is reduced and development lengths must be increased. To accountfor this fact, the term ethe coating factoris used in the equation.3. Sizes of reinforcingIf small bars are used in a member to obtain a certain total crosssectionalarea, the total surface area of the bars will be appreciably larger than if fewerbut larger bars are used to obtain the same total bar area. As a result, the requireddevelopment lengths for smaller bars with their larger surface bonding areas (in proportionto their cross-sectional areas) are less than those required for larger-diameter bars.This factor is accounted for with the reinforcement size factor, s.4. Lightweight aggregatesThe dead weight of concrete can be substantially reduced bysubstituting lightweight aggregate for the regular stone aggregate. The use of such aggregates(expanded clay or shale, slag, etc.) generally results in lower-strength concretes.Such concretes have lower splitting strengths, and so development lengths will have tobe larger. In the equation, is the lightweight concrete modification factor discussed inSection 1.12.5. Spacing of bars or cover dimensionsShould the concrete cover or the clear spacingbetween the bars be too small, the concrete may very well split, as was previouslyshown in Figure 7.6. This situation is accounted for with the (cb +Ktr)/db term in thedevelopment length expression. It is called the confinement term. In the equation, cbrepresents the smaller of the distance from the center of the tension bar or wire to thenearest concrete surface, or one-half the center-to-center spacing of the reinforcement.In this expression, Ktr is a factor called the transverse reinforcement index. It is used toaccount for the contribution of confining reinforcing (stirrups or ties) across possible splittingplanes.Ktr =40Atr / snwhere:Atr = the total cross-sectional area of all transverse reinforcement having thecenter-to-center spacing s and a yield strength fytn = the number of bars or wires being developed along the plane of splitting. If steelis in two layers, n is the largest number of bars in a single layer.s = center-to-center spacing of transverse reinforcingThe code in Section 12.2.3 conservatively permits the use of Ktr = 0 to simplify the calculations,even if transverse reinforcing is present. ACI 12.2.3 limits the value of (cb + Ktr)/dbused in the equation to a maximum value of 2.5. (It has been found that if values larger than2.5 are used, the shorter development lengths resulting will increase the danger of pullout-typefailures.)The calculations involved in applying ACI Equation 12-1 are quite simple, as is illustratedin Example 7.2.(del libro de Mcormac y Brown, 9a edicin)In SI units,Ktr =Atr fyt / 10sndimetro de estribos =3/8inramas verticales =2Atr =0.221in^2db =1inacero longitudinalcb =1.5inrecub. o dist a c-c de vrs long.ys =1espaciam. Entre estribos "s" =8incant de vrs long "n" =3Ktyr = 40 Atr/sn =0.368Fy yt ye ys / l(f 'c)^.5 =1087.3945453157(cb+Ktr/db) =1.87ino.k.debe ser =< 2.5ld =44dimetros

CASO 1CASO 2CASO 3CASO 4

CompresinACI-318-11, 12.3Longitud de desarrollo de varillascorrugadas a compresinProyecto:fecha:Sep-13l =1Concreto de peso normalFy =411.85Mpaf 'c =35Mpadiam de varillas =1.5in12.3.2SIeq. 1(0.24fy /l(fc)^.5 )dbeq. 2(0.043fy)dbdb =38mmeq. 1eq. 2635673mmEsta longitud puede ser multiplicada por:rel= As req./As realrelL dc0.7471mm0.75505mm0.8538mm0.85572mm0.9606mm12.3.1L dc no debe ser menor de 200 mm12.2.3Si el ref est confinado por una espiral de dimetro. => 6 mm y una separacin =