min. development length

[struct_eeyore]

I had a contractor cut off hooks which were vertically restraining a flood proof slab - and of course everything got poured before I had time to digest. Now, it just so happens that I've got just about 12" of embedment of said vertical hooks into the slab. Going thru ACI it looks like we can still reduce development length by As/Arq. If I got 11k of ultimate demand on a #6 bar, (26k design strength tension), I assuming I can use 11/26 as my reduction factor - which brings me down to the 12" min development length from a previous 17" (assuming max Ktr and lowest phi values).
Seems chintzy... any thoughts?

 

[KootK]

It does seem chintzy as it messes with the potential hierarchy of failure modes. Now you might pull the dowels out of the cylindrical voids that they are embedded in before:

1) You would yield the dowel and potentially be able to avail yourself of some local demand redistribution.

2) You would mobilize a concrete breakout frustum.

That said, I would probably still go ahead with the prorated development length scheme in many cases. I'm guessing that alternate "fixes" would be pretty costly. And it would be hard to justify that when there appears to be a code "out" for the situation ready to hand. And, realistically, it will surely be fine.

 

[struct_eeyore]

Koot - thanks as always.

I looked into cylindrical withdrawal, but ACI doesn't appear to provide bond strengths. Based of an internet search, I'm coming up with 350-500 psi - which is roughly what my ultimate load develops. Do you by chance know if there's proper literature that can be used to justify this?

 

[KootK]

Welcome.

I feel that your cylindrical withdrawal thing really IS your development length check. Nothing extra required.

 

[TLHS]

I mean, another sanity check is an Appendix D style breakout check and just see where you're at with it compared to other failures. It's not what code says you need to do necessarily, but it'd give you an idea of potential concrete breakout failure mechanism.

 

[EngDM]

I don't think you can use 11/26 as your reduction factor. Since you're reducing by the ratio of provided As to required As, then isn't Asreq found by back calculating As from setting Tr = Tf? I'm not familiar with ACI, but I'm assuming that there is a phi factor in Tr = phi*AsSteel*Fy?

So your reduction would be more like 11/(26*phi).