Understanding Shear Friction

[mastruc]

Good afternoon, all:

I had a question regarding the way ACI-318 handles shear friction. My understanding of the concept is that at, say, a retaining wall's base which was poured separately from its footer/slab, shear resistance is provided by the compressive component of the moment couple caused by flexure in the wall. What I don't seem to fully grasp is that in ACI-318, the equation for shear friction resistance is Vn=Avf*fy*mu ..... that is, it doesn't use the actual or factored moment-couple force, it essentially uses the flexural As*fy. This might make sense to me if the wall has just enough steel in it to resist its Mu, but what if you've got additional flexural steel in the wall? Am I to understand, based on this equation, that shear resistance due to friction can be had by adding flexural reinforcement that will never approach its fy under factored loads? And that this shear resistance is available no matter what the flexural loads in the wall are? (i.e., if you apply a horizontal point load very near the wall's base, say.) I suspect that there's a nuance to this that I'm not appreciating, so any light that you fine people can shed on this would be much appreciated.

 

[hokie66]

There is no understanding shear friction...it is just black magic...just my opinion, of course. But the concept is just that reinforcement crossing a joint provides a clamping force which mobilizes the friction. I only think about shear friction in desperation.

 

[mastruc]

I'm glad that my nervousness about the code's treatment of SF isn't isolated to me. No matter what ACI318 says, I don't think I'd ever have the courage to design something based on a normal force that I couldn't justify.

 

[RWW0002]

I think you confusion lies in your understanding of the shear friction mechanism. Shear friction applies when reinforcing crosses a shear plane. As the two parts try to move (parallel to the shear plane) the reinforcement is engaged (in tension) and the two parts are "clamped" together. This results in a net compressive force between the two parts which can then be used to resist the shear forces through the traditional friction resistance equations (Fn*mu). For a retaining wall, the compressive component of the moment couple is canceled out due to the tension in the reinforcing. As explained in the commentary to ACI 318, when there is a moment acting across the shear plane the net compressive force due to the tension and compressive components of moments is zero.

So to answer your question, adding steel above what is required for moment resistance will in fact increase the shear resistance. As the additional reinforcement yields there will be an increased clamping force across the shear plane (Although the commentary does suggest that additional shear friction reinforcement beyond that required for flexure be placed primarily in the flexural tension zone)

 

[mastruc]

RWW002: So, according to the theory behind shear friction -- in my hypothetical, where a load is applied at or near a wall's base (and the wall is experiencing no additional loads), the shear force will be relatively large and the moment relatively small. And despite the lack of an overturning moment, the shearing force will engage the reinforcement not in shear, but in tension? I'd be concerned about breakout of the reinforcement in that case, but at least according to the letter-of-the-theory, this is not a concern?

 

[RWW0002]

I should preface this with the fact that I am no expert in shear friction, and based on past posts, Hokie seems to have much more experience than me in RC design.

But the situation you are describing is much like a corbel bracket (high shear and small moment) in which case the required area of shear friction reinforcement may be much larger than required flexural steel.

Back to your example. If all of the flexural steel is yielded then the clamping force (compressive component of moment couple) equals the tension force in the reinf = As* Fy. If additional steel is added, the clamping force due to flexure would = As* Fs. As the wall and footing try to move relative to one another due to the application of the low horizontal force, additional tension would develop in the reinforcement creating a greater clamping force.

Shear breakout may still be a valid concern with small edge distances. And you should also be careful that you do not end up with an over-reinforced section for flexure.

 

[msquared48]

My Pre-Stress prof in the late 60's was the originator of shear friction and he pushed the concept in the class. J. R. Birkeland if I remember correctly.

As I understood it, and I never use it, mind you, either, it was the theory that in order for the vertically irregular concrete surfaces to slide one against the other horizontally, a very small amount of vertical displacement, or lift, would have to occur between the two concrete elements to reduce the friction to allow the sliding to occur (assuming the concrete surface would not shear off these elements), and this lift was to be restrained by the vertical shear-friction steel, thus maintaining the frictional resistance.

Approved or not, I never rely on it solely on it's own merit. Maybe ass an undocumented safety factor, but that's all.

Mike McCann
MMC Engineering

 

[RWW0002]

Just for reference here is a past thread on this.

http://www.eng-tips.com/viewthread.cfm?qid=122089

Msquared, if you do not rely on shear friction, how do you account for the shear across the shear plane between two non-monolithic surfaces (such as a retaining wall and a footing as described above). Shear keys? Breakout prisms @ dowels per appendix D??

Although many of us no not make use of the shear friction equation everyday, shear friction is still used on a regular basis as the primary mechanism to resist in plane shear across surfaces.

 

[msquared48]

I rely on the shear through the reinforcing steel present, not the interactional surfaces between the concrete slabs.

The increase in friction allowed is only due to the presence of the interstitial reinforcing steel.

Mike McCann
MMC Engineering

 

[Lion06]

I've often wondered about shear friction. I use it when I have to, but it does seem like black magic.

My understanding is in line with Mike's. What seems to not make sense is how it works for steel against concrete. For concrete on concrete, aggregate can cause movement normal to the shear plane to engage the bars in tension and apply the normal force, but for something smooth, like steel, against concrete, I don't see that same mechanism as being realistic.

Mike, when you design the bars for shear through the bars at the base of a wall, how do you account for breakout if you have only 1.5" of edge distance? I've never seen in ACI that using the bars in shear is an acceptable method of transferring shear across a plane - did I miss that?

 

[msquared48]

Lion:

I generally do not rely on the steel, but either bearing against a slab, or a shear key in the wall. I just feel better about it.

Mike McCann
MMC Engineering

 

[msquared48]

I also think that shear friction is used more in the prestress industry where the concrete is under compression, and any movement in any shear plane would be more restrained than normal by the greater frictional component across the shear plane due to the compressive force.

I'll stop right there though... Enough said.

Mike McCann
MMC Engineering