Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

ASTM A496 rebar instead of A615 rebar? 1

Status
Not open for further replies.

nicaeng

Civil/Environmental
Oct 4, 2011
21
CA
Can A496 bars be used instead of A615 rebars?, ACI 318 limits the deformed bars specifications to A615, A706, A995, A996 (Sec. 3.5.3.1) and A1035 (Sec. 3.5.3.3). Here in Nicaragua the masonry construction method used is "confined masonry", concrete elements are usually 6"x6" and these A496 bars are quite cheaper than the A615 because of their higher strenght and smaller diameters. These concrete elements are designed following ACI 318 recommendations, but i'm not sure if you can write on the drawings that ACI code 318-xx has been used for the design when A496 is not specified as rebar, although it is specified as deformed wire (Sec. 3.5.3.5). I might be misinterpreting the code thinking deformed wire can be used only for slabs and walls.

Thanks in advance for any help.
 
Replies continue below

Recommended for you

No comment on A496, but 6"x6" concrete elements? What type structure can you build with such small sizes?
 
hokie66, here are a couple of documents that explains what "confined masonry" (CM) is, personally, i don't like this construction system, when building with masonry, reinforced masonry is the way to go, but, CM is the most used in Nicaragua, i'll say almost exclusively, and not even done right. I've read documents from Mexico, Chile, Peru and Colombia on CM and according to them CM "works" as a construction system that resists seismic loads when done the right way. Now that i've mentioned reinforced masonry, do you think A615 rebars can be replaced by A496 bars for one story buildings?
 
A496 is no longer an active specification. It has been replaced by A1064. That aside, wire is not the same as rebar, and the primary differences are ductility, deformation geometry, and the diminished ratio of yield-to-ultimate.
Ductility is not defined in the same ways in A615 and A1064, so a quick comparison cannot be made here.
The development provisions of ACI 318 apply to wires, the same as rebar.
Deformed wire should yield at 75 ksi, with an ultimate of 85 ksi; grade 60 rebar yields at no less than 60 ksi, with an ultimate of 90 ksi, minimum (grade 75 yields at 75 ksi with an ultimate of 100 ksi.)

ACI 318 refers to "deformed reinforcement", with the implication that wire and bar are similar in use, with certain limitations where ductility is critical. There is also a limit on wire diameter, namely that wire larger than D31 is permitted only in welded wire applications.
 
Thank you TXStructural for your response, i believe in my case ductility is very critical since the masonry walls do not contain any steel and the confining elements provide the ductility for the wall, the main supplier for the A496 (now A1064) wires in Nicaragua is advertising it's product claiming that you can substitute A615 rebars with theirs since it has a higher yield strenght, therefore diameters can be smaller hence reducing material and labor costs up to 30%, i do not know if they've actually made tests to back these assumptions up.
 
Regardless of the strength, all steel has essentially the same modulus of elasticity. So if deformation of the bars is the important thing, as I think you are saying is the case with this type system, then decreasing the bar/wire size increases deformation accordingly, provided the bars are in the elastic range.
 
Strength (presumably yield, for our purposes) is critical, but since ductility is critical, keep in mind that the range of yielding is smaller for cold-drawn wire, which is a reason that wire and certain high strength reinforcement is not acceptable in some uses. As Hokie points out, all steels elongate under load, resulting in earlier crack development and greater crack widening as you reduce the area of steel crossing the cracks, regardless of yield strength (d=PL/AE across cracks).
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor

Back
Top