Minimum Shear and Torsion Reinf per ACI 318-11

[dcarr82775]

I have a condition where I need to provide minimum shear steel in a beam, and I also need to provide shear steel for torsion. The amount of steel I need to provide for torsion is greater than the 50*bw*s/fy minimum. I believe that I am therefore covered for the minimum shear steel as well per 11.5.5.2. Would you agree?

 

[jayrod12]

does your torsion reinforcement provide enough capacity to resist the applied torsion and shear stresses? Then you are ok. If not you still need to meet that requirement in my mind. It likely wouldn't mean adding extra steel in the full amount of minimum shear steel, but it may be more than you're providing for the torsion alone.

 

[dcarr82775]

The minimum requirement is the same be it shear only or torsion plus shear. My shear is between 0.5Vc and Vc so I need to provide minimum shear steel. The steel I am providing for torsion is greater than that minimum and per 11.5.5.2 I think that means I am good to go. I may just provide more steel because it feels like I should, but I wanted to see how others interpret this condition

 

[structSU10]

shear stresses due to torsion and shear are cumulative, at least on one side of the beam (torsional shear stresses counterbalance general shear stresses on one side). If max shear and torsion do not occur at the same location, you may be able to get away with lesser reinforcement, otherwise the requirement for each should be added together as jayrod stated.

 

[jayrod12]

Ok so you are providing Torsion reinforcement that is in excess of minimum steel. My point was, can the specified steel take the applied torsion and the applied shear? or is it only designed to take the applied torsion?

 

[dcarr82775]

Steel is only for the applied torsion because there is no applied shear, Vu < Vc. But since Vu > 0.5Vc the minimum requirement on shear kicks in.

 

[jayrod12]

Torsion and no shear? Or not enough shear to reach cracking when analyzed alone?

Since the torsion and shear share the same reinforcing, you need to make sure you have enough to resist both simultaneously. Regardless of individual levels. At least that is my opinion.

 

[dcarr82775]

I worded that badly. I needs steel for torsion. As for Shear, Vu is less than Vc, but greater than 0.5Vc so there is no calculated amount of shear reinforcing required, only the minimum.

 

[jayrod12]

Ok I get what you're thinking.

The issue I see is once the torsion sets in (you obviously have an appreciable amount of it) the section will crack, the torsion failure cracks are extremely similar to shear cracks and therefore the available Vc goes out the window in my mind.

The force in your stirrups would be a sum of the torsion and shear stress on one leg and a difference of the stresses on the other leg. Therefore your reinforcing has to be capable of taking T and V/2 in a single leg. Can yours do that?

Maybe I'm being over-conservative... wouldn't be the first time.

 

[JAE]

I've always calculated the "required" shear and "required" torsion reinforcement separately (whether it is calculated or minimum steel)
then added them together....This might be conservative but I'm not sure if ACI ever comments on this.



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[jayrod12]

I feel JAE's method is the quickest way to ensure you're covered and requires the least overall analysis.

If you wanted to trim down on the total reinforcing you could determine the reinforcing required to resist both your torsional and shear stresses and as long as that required amount is more than the minimums than I think it's ok.

 

[dcarr82775]

Jayrod, but you see that is the sneaky bit because there is nothing more than the prescriptive minimum required for shear. So there is no 'amount' to be calculated based on a shear demand. Maybe it is all semantics anyway.

I decided to just provide the minimum for shear plus that required for torsion. I think 318 would let me just provide that steel required for torsion, but it feels wrong. When in doubt throw in some more steel.

 

[jayrod12]

I would be pretending the Vc isn't there and determining the shear and torsion stress in the stirrups assuming a typical spacing.

I would probably start by determining the torsion spacing and size, then tighten the spacing up to some multiple of 4" or 6". Then recheck for the combined stresses.

 

[JAE]

dcarr82775, you might try contacting ACI - send them an email with your question and see how they respond. Then post their answer here.


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[structSU10]

If you have a copy of PCA notes on ACI, they have a design example in the torsion section (for PCA notes on ACI 318-05 its example 13.1), where they check equilibrium torsion, then once they see torsion controls, they calculate the amount of shear and torsion steel is required, and add them together,

 

[KootK]

I'm inclined to tepidly agree with your original hypothesis Dcarr.

1) As you've identified, torsion is really just additional shear, differently distributed.

2) The aims of the minimum steel requirements, as stated in the commentary, have nothing to do with resisting a particular load. So there's little cause to suppose that additional load equates to additional minimum steel.

Given the spirit of the minimum steel provisions, I could imagine a case where neither the shear nor torsion demand, considered separately, warranted minimum steel but the combination of the two would.

This seems like a good thread for TXStructual. Hopefully he surfaces.



I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.