Pryout of anchor in shear - ACI 349 App. B

[bones206]

I am currently reviewing a spreadsheet written by a former employee, which calculates anchor capacities based on ACI 349-01 appendix B. This is very similar to 318 appendix D.

I'm trying to follow his methodology for calculating pryout in shear. The nominal strength equation is:

Vcp = kcp*Ncb, where Ncb is the nominal breakout strength in tension.

There are really 2 issues that I am having trouble reconciling.

1) Instead of calculating Ncb for a single anchor, he calculates the breakout strength of the entire group, Ncbg. This introduces the modification factor ?1 for eccentricity (of the tensile force). He then checks the "group pryout" strength against the total shear force acting on the anchor group. Does this seem like a valid way to analyze pryout?

2) When calculating ?1, he chooses to replace the tensile eccentricty term, e'N, with the eccentricty of the shear force, e'V. I am having trouble conceptualizing what this means in practical terms... A shear eccentricty term is placed in a tension strength modification factor, which ultimately is part of a shear pryout capacity equation. Is this just hybridizing the code for calculation purposes or does it fall outside the code?

I appreciate any thoughts.

 

[Lion06]

1) This does seem like the right way to approach this. ACI does say to calc Ncbg per D-4 and D-5 (I think those are the two equations - I don't have my code in front of me).

2) I don't remember if the eccentricity equations are the same for tension and shear or if they differ. If they are the same, then using the shear eccentricity factor seems fine. If they are different, then I think it's appropriate to use the shear eccentricity in the tension eccentricity factor. It is the shear eccentricity that you are worried about and you could possibly have no tension eccentricity but have a shear eccentricity and you want to account for that.

 

[bones206]

I agree with you that analyzing the pryout with Ncbg is valid. I got my hands on a copy of 349-06, which includes the provision for group pryout (not included in 349-01).

The eccentricity factors are different for tension and shear. Fundamentally, the tensile eccentricity is calculated only for those anchors loaded in tension, whereas the shear eccentricity is calculated only for those anchors loaded in shear in the direction of the free edge. So you are probably going to have 2 different sets of anchors -- depending on the load case -- for which you would calculate the respective tensile and shear eccentricities. Furthermore, if you have tensile eccentricity about 2 axes, you calculate ?1x & ?1y, then ?1 = ?1x * ?1y (using e'Nx & e'Ny). This approach does not correspond to shear eccentricity.

I am trying to track down the theory behind the ACI pryout equation, but not having much luck. They reference a 1995 ACI Structural Journal article on the CCD method. However, the article explicitly does not cover the pryout failure mode. It instead refers a 1993 German PhD thesis... So I don't know if there is a clear cut answer to this, but I'd like to find the original source of the equation -- I may need to learn German first.

 

[Lion06]

Agreed, but if you have a shear force only and it is eccentric it doesn't make sense to me to not take that eccentricity into account.

 

[PackerFan]

The way this is calculated is not correct. The idea behind pryout is that very short stiff anchors simply will pop out a concrete cone in the opposite direction of the shear force, that is why Ncbg for tension is used for a shear calculation. As an example, if you have a 6 anchor base plate with a moment that results in only 4 anchors in tension, Ncbg for your tension calculations needs to be calculated only using those 4 anchors. When you calculate your pryout you would need to recalc Ncbg using 6 anchors since although Ncbg is used in this calc it has nothing to do with how many anchors are loaded in tension. You may take this Vcp and divide it by the number of anchors and compare it to your shear force per anchor, especially if you also have eccentric shear forces which results in different shear forces in each anchor.

So:

1. You do not want to use Ncb for a single anchor for your pryout equation. You must calculate Ncbg for the group. You may need to recalculate Ncbg for your pryout equation if you have an eccentric tension load. You may however take your total Vcp and divide it by the number of anchors loaded in shear and compare it to shear force per anchor.

2. You may not do this. Again, you must recalcuate your Ncbg for your pryout calc and ignore any eccentricty you assumed in your tension calculations.

I hope this makes sense. I have attached a sample problem that has this exact case. Enjoy.

 

[Lion06]

I can't view your example. Can you re-post it?

 

[Lion06]

Can anyone view this example?

 

[Lion06]

I'm not sure I believe that you can simply divide Vcp by the total number of anchors and compare that to the highest anchor load (based on eccentric loading).
ACI clearly says that Vcp is a function of Ncb (which is a group breakout) and takes eccentricity into account through the reduction factor (si,ec) and doesn't require any additional consideration for eccentricity. You wouldn't find Ncb and divide that by the total number of anchors and compare that to your highest anchor load (based on eccentric loading) so why would you do that for Vcp if the code doesn't state it?
It makes sense to do that for the steel strength in both tension and shear as well as the pull-out capacity, but I don't believe that is appropriate for pryout (although, it would be conservative).
My main question is the same as question 2) in the OP.

 

[bones206]

I can't view the attachment either. I have asked many engineers for their opinion's on this and I have yet to get the same answer twice. Since most of the calculations I am checking have already been signed off, they obviously felt that the design was safe. But since these calcs will inevitably come up again in the future (and likely have both eccentric tension and eccentric shear load cases), it's something that I think needs to be cleared up. Many engineer's may approach this part of the code in a safe and prudent way, but the problem is nobody I've talked to has done it the same way.

 

[civilperson]

Multiple methods for this problem does not imply that any of the techniques are incorrect. There is NO ONE CORRECT METHOD to represent the real world with a simplified analysis.

 

[bones206]

civilperson,

I agree with you. I think that the main reason that this part of the code bothers me is that the nuances and gray areas tend to lead one away from a practical analysis and into speculation about how to implement the code correctly.

I'm not so concerned with the ability to design conservatively despite the code, but I am worried about down the road if one of these anchors did pryout and someone looked back at the calcs, demanding justification for my interpretation of the code.

 

[PackerFan]

Please feel free to ask any questions.

 

[PackerFan]

Hmmm...I can't get it to post. Try this.

tp://

http://www.yousendit.com/download/Y2orcmxaYUlwcFUwTVE9PQ

 

[Lion06]

This example doesn't have an eccentric shear load. We were discussing how to account for eccentric shear in pryout.
Is this your example or a Hilti example?