ACI-318-11

Longitud de desarrollo de varillascorrugadas a tensión

Proyecto:

Fy = 59684 psi 4200f 'c = 3013 psi 212

1

1diám. = 1 in

l = 1 conc. Peso normal

1087.3945453157caso = 2

0 20 0 0

54.37usar 44

ACI 12.2.2

Factors for Use in the Expressions for Determining Required Development Lengths for

(1) ψt = reinforcement location factor

Horizontal reinforcement so placed that more than 12 in. of fresh concrete is cast in the member

below the development length or splice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3

Other reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

(2) ψe = coating factor

Epoxy-coated bars or wires with cover less than 3db, or clear spacing less than 6db . . . . . . . . . . . 1.5

All other epoxy-coated bars or wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

Uncoated and zinc-coated reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

However, the product of ψtψe need not be taken as greater than 1.7.

yt =

ye =

Fy yt ye / l(f 'c)^.5 =

ld =

Deformed Bars and Deformed Wires in Tension (ACI 12.2.4)

CASO 1

CASO 3

Para aplicar la fórmula 12-1, seguir el siguiente procedimiento:

(3) ψs = reinforcement size factor

No. 6 and smaller bars and deformed wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8

No. 7 and larger bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

In SI units

No. 19 and smaller bars and deformed wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8

No. 22 and larger bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

(4) λ (lambda) = lightweight aggregate concrete factor

When lightweight aggregate concrete is used, λ shall not exceed . . . . . . . . . . . . . . . . . . . . . . . . . 0.75

However, when fct is specified, λ shall be permitted to be taken as 6.7 *

fct = resistencia promedio a la tensión (tracción)It’s in SI

but not greater than . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

When 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 previously

been introduced are described. their values for different situations were given in

the previous page.

1. Location of reinforcement—Horizontal bars that have a least 12 in.[3] of fresh concrete

placed beneath them do not bond as well to concrete as do bars placed nearer the bottom

of the concrete. These bars are referred to as top bars. During the placing and vibration

of 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 some

settlement of the concrete below. As a result, the reinforcement does not bond as well to

the concrete underneath, and increased development lengths will be needed. To account

for this effect, the reinforcement location factor, ψt, is used.

2. Coating of bars—Epoxy-coated reinforcing bars are frequently used today to protect the

steel from severe corrosive situations, such as where deicing chemicals are used. Bridge

decks and parking garage slabs in the colder states fit into this class. When bar coatings

are used, bonding is reduced and development lengths must be increased. To account

for this fact, the term ψe—the coating factor—is used in the equation.

3. Sizes of reinforcing—If small bars are used in a member to obtain a certain total crosssectional

area, the total surface area of the bars will be appreciably larger than if fewer

but larger bars are used to obtain the same total bar area. As a result, the required

development lengths for smaller bars with their larger surface bonding areas (in proportion

to their cross-sectional areas) are less than those required for larger-diameter bars.

This factor is accounted for with the reinforcement size factor, ψs.

4. Lightweight aggregates—The dead weight of concrete can be substantially reduced by

substituting 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 to

be larger. In the equation, λ is the lightweight concrete modification factor discussed in

√ f ' c /1 .8 fct

Section 1.12.

5. Spacing of bars or cover dimensions—Should the concrete cover or the clear spacing

between the bars be too small, the concrete may very well split, as was previously

shown in Figure 7.6. This situation is accounted for with the (cb +

development length expression. It is called the confinement term. In the equation, cb

represents the smaller of the distance from the center of the tension bar or wire to the

nearest 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 to

account for the contribution of confining reinforcing (stirrups or ties) across possible splitting

planes.

Ktr = 40Atr / sn

where:

Atr = the total cross-sectional area of all transverse reinforcement having the

center-to-center spacing s and a yield strength fyt

n = the number of bars or wires being developed along the plane of splitting. If steel

is in two layers, n is the largest number of bars in a single layer.

s = center-to-center spacing of transverse reinforcing

The 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)/db

used in the equation to a maximum value of 2.5. (It has been found that if values larger than

2.5 are used, the shorter development lengths resulting will increase the danger of pullout-type

failures.)

The calculations involved in applying ACI Equation 12-1 are quite simple, as is illustrated

in Example 7.2.(del libro de Mcormac y Brown, 9a edición)

In SI units, Ktr = Atr fyt / 10sn

diámetro de estribos = 3/8 ramas verticales = 2

Atr = 0.221db = 1

cb = 1.5

1

8cant de vrs long "n" = 3

Ktyr = 40 Atr/sn = 0.368

(cb+Ktr/db) =

ys =espaciam. Entre estribos "s" =

Fy yt ye ys / l(f 'c)^.5 =

ld =

fecha: Sep-13

kg/cm^2kg/cm^2

20

diámetrosdiámetros

Horizontal reinforcement so placed that more than 12 in. of fresh concrete is cast in the member

below the development length or splice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3

Other reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

Epoxy-coated bars or wires with cover less than 3db, or clear spacing less than 6db . . . . . . . . . . . 1.5

All other epoxy-coated bars or wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

Uncoated and zinc-coated reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

CASO 2

CASO 3 CASO 4

No. 6 and smaller bars and deformed wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8

No. 7 and larger bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

No. 19 and smaller bars and deformed wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8

No. 22 and larger bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

When lightweight aggregate concrete is used, λ shall not exceed . . . . . . . . . . . . . . . . . . . . . . . . . 0.75

but not greater than . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

When normal weight concrete is used . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0

Use the smaller of either the distance from the center of the bar or wire to the nearest concrete surface,

3. Sizes of reinforcing—If small bars are used in a member to obtain a certain total crosssectional

substituting lightweight aggregate for the regular stone aggregate. The use of such aggregates

√ f ' c / fct

Ktr)/db term in the

account for the contribution of confining reinforcing (stirrups or ties) across possible splitting

The code in Section 12.2.3 conservatively permits the use of Ktr = 0 to simplify the calculations,

used in the equation to a maximum value of 2.5. (It has been found that if values larger than

2.5 are used, the shorter development lengths resulting will increase the danger of pullout-type

in

in^2in acero longitudinalin recub. o dist a c-c de vrs long.

in

1087.394545315671.87 in o.k.

debe ser =< 2.5

44 diámetros

ACI-318-11, 12.3

Longitud de desarrollo de varillascorrugadas a compresión

Proyecto:

1 Concreto de peso normalFy = 411.85 Mpaf 'c = 35 Mpa

diam de varillas = 1.5 in

12.3.2 SIeq. 1

eq. 2 (0.043fy)db

db = 38 mm

eq. 1 eq. 2 635 673 mm

Esta longitud puede ser multiplicada por:rel= As req./As real

rel L dc0.7 471 mm

0.75 505 mm0.8 538 mm

0.85 572 mm0.9 606 mm

12.3.1 L dc no debe ser menor de 200 mm

12.2.3 Si el ref está confinado por una espiral de diámetro. => 6 mm y una separación =<100 mmo estribos de 13 mm espaciados a no mas de 100 mmmultiplique por: 0.75

L dc = 505 mm

l =

(0.24fy /l(f´c)^.5 )db

fecha: Sep-13

Concreto de peso normal

L dc no debe ser menor de 200 mm

Si el ref está confinado por una espiral de diámetro. => 6 mm y una separación =<100 mm