Reinforcement Development Length or Bond

Outline

1. Introduction
2. Development Length of (Straight) Deformed Bars
3. Example Calculation
4. Applications
5. Transverse Reinforcement Index
6. Conclusion
7. References

1. Introduction

Loads generally arrive at reinforced concrete members via the concrete, whether across the faces of the concrete, fasteners, anchors, etc., or the weight or mass (mass times acceleration) of the concrete itself. For reinforced concrete to `work' these loads (at least the tension ones) must be transferred to the steel. For cast-in-place concrete the transfer mechanism is the shear resistance of the concrete in contact along the reinforcement via the `ridges and valleys' of the rebar deformations. (Yeah, in this regard `smooth' reinforcement doesn't work.) Hence, transfer is across `surface area of contact'. To ensure there is enough contact area for the load transfer to take place we will specify minimum `lengths' of contact related to bar size (and some other things).

In this regard, if we consider a certain amount of load to be carried by some arrangement of reinforcement, smaller bars will have more surface area per bar, and the `lengths' of contact required will be relatively smaller. Said differently, smaller barfs are more efficiently `developed'. Or, said yet again differently, a big bar may be way stronger than a smaller one, but will also have a greater development length, maybe cumbersomely so.

So, to resist any particular load there may be a number of rebar and development length options. Numerous smaller bars will be nice in that the development lengths are modest and better `fit' into the spaces we are working in, but too many bars may lead to congestions (and eventual code violations with regard to spacing). A few bigger bars may be more efficiently handled, but too big and the development lengths required begin to be issues. Also, at some point the bigger bars themselves become burdens, as they are themselves heavy, and hard to cut, if being cut by hand in the field. Having cut bars myself in the field, I like to stay on the smaller side. And, also, in my mind I begin to lose the sense that the rebar is `holding the thing together' if they are spaced too far apart (regardless of what the `numbers' say.)

2. Development Length of (Straight) Deformed Bars

From Chapter 12 in the ACI Code, our basic equation for development length, ʆ _d , is, ...

ʆ _d = { (3/40) (f_y /√f '_c) (α β γ λ ) / [ (c + k_{t r}) / d_b ] } d_b

where,

f_y = specified yield strength of the reinforcement (psi)

f 'c = specified 28-day compressive strength of the concrete (psi) ... (and where √f '_c carries the units of psi),

α = bar location factor, which is 1.0 except for horizontal bars with more than 12 in. of fresh concrete cast below them,

β = bar coating factor, which is 1.0 for uncoated bars, and 1.2 or so for epoxy coating ... the epoxy somewhat smoothing out the valleys and ridges (?) ...

γ = bar size factor, 0.8 for # 6 and smaller bars; 1.0 otherwise;

λ = lightweight concrete factor, 1.0 for normalweight concrete ...

c = spacing / cover dimension, being the smaller of the distance from the center of the bar to the nearest face of concrete, or half the center-to-center spacing of the reinforcement,

k_{t r} is a transverse reinforcement index ... (see below, but) which may be taken as 0, and

d_b = is the diameter of the bar being developed.

The Code places an upper limit on the quantity (c + k_{t r}) / d_b of 2.5.

Note: the minimum development length is 12 in., regardless of the equation.

In many cases by satisfying other spacing and cover limitations in the Code (Ch. 7) a value of 1.5 can be obtained, and some authors generate tables of development length values using the 1.5.

3. Example

Let's calculate the development length of # 6 bars, Gr. 60 ksi, in 4000 psi normalweight concrete. The bars will not be coated, and let's calculate the development length for both vertical bars and bars that are placed horizontal with the potential for more than 12 in. of fresh concrete to be placed below them. The minimum center-to-center spacing of the bars is 5-1/4 in., and the clear cover of the rebar is at least 3/4 in. (The 5-3/8 in. comes about from 6 in. center-to-center spacing allowing for a lap splice of one of two bars at the location of interest.)

So,

α = 1.0 or 1.3 (depending on vertical or horizontal),

β = 1.0

γ = 0.8 (# 6 or smaller)

λ = 1.0

c = smaller of ¾ in. + ½ of ¾ in. (bar diameter of # 6), which is 1.125 in. ... or, ... ½ of 5-1/4 which is 2.625 in., which gives us ... c = 1.125 in.

And let k_{t r} = 0,

thus ... ( c + k_{t r}) / d_b= 1.125 / 0.75 = 1.5.

So,

ʆ _d = { (3/40) (f_y /√f '_c) (α β γ λ ) / [ (c + k_{t r}) / d_b ] } d_b =

... [(3/40) (60,000 / √4000 ) (1.0 or 1.3) (1.0)(0.8)(1.0) / 1.5] d_b =

= (37.9 or 49.3) d_b ...

The reason I express the above values in terms of `diameters' is because that is the way they (or, I should say, splice lengths, which derive from development lengths) appear in some construction documents. However, expressing as such would only be valid for size # 6 and smaller (due to the bar size factor used).

So, for size # 6 bar ...

ʆ _d = 28.4 in. and 37.0 in. (vertical and horizontal bars, respectively).

Answer: the development lengths for the # 6 bar are: 29 in. vertical and 37 in. horizontal bars with 12 in. or more of fresh concrete cast below.

4. Applications

In the basic sense the development length is the amount (length) of embedment (contact) we need for the reinforcement to be effective. Let's say that the above reinforcement is for the vertical reinforcement in a wall acting as a one-way slab and a certain spacing has been determined based on the maximum moment at a location near the mid-height of the wall. For the reinforcement to be effective at this location it must extend at least the development length in both directions (up and down). Hence, from the point of maximum moment the bars must extend 29 in. both up and down.

As we move away from the point of maximum moment, obviously, less reinforcement will be required. Or, if the amount of reinforcement is unchanged, then less will be required of the reinforcement that is there. For example, theoretically there are locations in the wall where the moment drops to half of the maximum, and thus (roughly) half of the reinforcement would be required. Thus, we could speculate that we could `terminate' half of the reinforcement. And indeed we can. (But there are some rules we need to follow in doing so). Likewise, let's say we don't terminate the reinforcement; we leave it in there. That means that at locations where the moment is half of the value that we used to determine the reinforcement, if we leave it all in, we have excess reinforcement. (Yes, it does mean that.) And, if we're clever, we could argue that we thus don't need to fully develop it, or all of it. (And, indeed, yes, we'll find that the Code does allow us to not fully develop the bond, at locations where it is not fully needed.) We'll handle these issues in a later lesson.

And we'll also run into a couple other things .... first, we may have sections where tension reinforcement isn't needed for tension as such (at the section), but it is needed because it is part of the development length needed somewhere else. And, finally, in case we haven't thought of it already, it is possible that the demand for the development of the rebar changes at a rate different than it becomes available (for example at the end of a simple beam). In such a case we may end up `pulling the rebar out', resulting in no rebar at a section in tension, which would be disastrous!

And there are integrity and safety issues, where we will continue reinforcement beyond where we might think it is no longer needed. So stay tuned.

5. The Transverse Reinforcement Index

Here it the transverse reinforcement index that we talked about above, in brief,

... k _tr = A_tr f_yt / (1500 s n)

where,

A_tr = total cross section area of all transverse reinforcement within the spacing s and that crosses the poetical plane of splitting through the reinforcement being developed,

f_yt = specified yield strength of transverse reinforcement,

s = maximum center-to-center spacing of transverse reinforcement ... and

n = number of bars being developed (or spliced) along the plane of splitting.

Usually I just let k _tr = 0, which is allowed by Code. But I say this admitting most of my designs are for relatively small projects where the benefit to detailed `hair splitting' in the calculations is pretty small.

6. Conclusion (for now)

We have now covered how to calculate the development length, which is the length of contact required between rebar and concrete for the rebar to be (fully) effective. The rebar must extend `the development length' in both directions from where it is needed. Where less is required of the rebar, less length is required, and we will address this in the next lesson. And there are other rebar detailing issues, touched on above, that we need to deal with, and these issues are often `dealt with' in terms utilizing the `development length' ʆ <sub>d</sub> (`term') that we have now learned how to calculate.

(Continuation is ... here.)

7. References

Building Code Requirements for Structural Concrete, ACI 318, American Concrete Institute, P.O. Box 9094, Farmington hills, Michigan, 48333.