Minimum Stirrup Spacing

[labeattie]

Hi all,

In ACI 3018 11.5.5.1, there's a minimum spacing stipulation of d/2 for perpendicular stirrups. This usually has not been a problem for me, but on my current project, I have a little cantilevered sidewalk on h=12" t-beams, making my d value only about 9". Stirrups spaced at 4.5" seems a bit excessive to me. Any advice on this would be appreciated. Thanks!

 

[manstrom]

Read the exceptions for stirrups requirements in ACI.

If the beam is very stout like a slab you can get away from it. Also, if Vu<Vc/2, they aren't require (I think).

If you had to do stirrups, maybe consider using draped mesh instead. 4.5" stirrup spacing is silly.

When I am working on a problem, I never think about beauty but when I have finished, if the solution is not beautiful, I know it is wrong.

-R. Buckminster Fuller

 

[labeattie]

I do need stirrups, and, strictly structurally, #3's @ 12" work. The mesh drape is a good idea and a real possibility. I would prefer to use bars though. I just didn't know if anyone had run across something similar. Is mesh the only way to deal with this, excepting unreasonable 4.5" stirrups spacing?

 

[PMR06]

Can you refine your load calcs and increase beam width and/or f'c such that Vu<Vc/2?

Otherwise 4.5" oc doesn't seem that unreasonable to me. It is what it is.

 

[labeattie]

No I don't think there's a very good way for me to get it so that Vu<Vc/2. And fair enough on it is what it is. I guess those shear cracks have to catch up to a stirrup.

I have one more question regarding this though. On 11.5.5.3 it says "where Vs exceeds 4*sqrt(f'c)*bw*d, maximum spacings given in 11.5.5.1 and 11.5.5.2 shall be reduced by one half". This is another stipulation that has never given me trouble till now, but as Vs is proportional to Av, using more stirrups (which i would generally consider safer) can reduce your minimum spacing by half. As you might imagine, using #3's at 4.5" in my beam as mentioned makes Vs rather large, causing 11.5.5.3 to call for 2.25". What is up with this? Can you reduce the Vs to the Vsreq? That would make sense to me, but I'm just making it up.

 

[manstrom]

Can you widen the T beam so that it is more of a slab than a beam. I forget the aspect ratio, but squat beams don't need stirrups if Vc>Vu

Or bump up the f'c to get more shear capacity.

When I am working on a problem, I never think about beauty but when I have finished, if the solution is not beautiful, I know it is wrong.

-R. Buckminster Fuller

 

[labeattie]

Even if I double the web width and increase f'c from 3 to 4 ksi, Vu is still greater than Vc. This configuration is indeed exempt from Vu<Vc/2, but I need stirrups nonetheless. Thanks for your help though. It's very much appreciated. Any ideas on the 11.5.5.3 I mentioned above? I definitely get the need for a transverse bar to cross every potential 45 degree shear crack (so the d/2 requirement) even thought it results in super tight spacing for me, but I do not understand why have more Av than required should reduce that minimum spacing (d/2) by half.

 

[Tomfh]

We've done a few shallow beams with stirrups at 100mm (4 inches). It didn't seem that unreasonable and no-one complained.

Re. your 2.25' thing. Is that clause dependent on the Av you actually need, as opposed to what you've actually put in simply to satisfy spacing?

 

[labeattie]

That is a suspicion/hope I have, but I don't currently see any support for it. I very well might be missing something though.

 

[PMR06]

I think the intent of 11.5.5.3 is for cases where the Vs required is greater than 4*sqrt(f'c)*bw*d, NOT the Vs recalculated after you meet d/2 spacing. However, I agree it is not explicitly stated so.

In your case if you reduce the spacing by half and recalculate Vs, it probably exceeds 11.5.7.9 Vs max of 8*sqrt(f'c)*bw*d. There's a logical bust there...

 

[labeattie]

I didn't see that as a logic problem, just that you can only count on up to 8*sqrt(f'c)*bw*d from you stirrups. As in I thought you could put enough steel in there that eqn 11-15 gives a result greater than 8*sqrt(f'c)*bw*d, but you can't assume that the stirrups will absorb more shear than 8*sqrt(f'c)*bw*d. Anyways just thought that was interesting. Either way, it doesn't help with the 11.5.5.3 thing. I'm not too worried about it, as i think stirrups at 4" are definitely enough, but I do think the code reads that it would be the Vs after you meet d/2. It doesn't make sense to me to use strength as a cutoff for more stringent spacing requirements, but that's definitely how the code reads. It does seem we all think it should be based off demand rather than capacity.

 

[JAE]

The Vs used to compare with 4sqrt(f'c) is the required Vs in my view.

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[labeattie]

Fair enough. Perhaps it's just my inexperience that was preventing me from making a similar statement. However I wasn't too worried about the 4" spaced stirrups in this particular case, and seeing your all's opinions reinforces mine. Thanks!