Shear-Friction Design Method for Reinforced Concrete 4

[clemsonC]

I have ACI 318-02 (11.7.4) and a reinforced concrete textbook. Here is my understanding of the shear-friction design method:

This method of determining Vn considers shear capacity of "dowels" as well as the friction between adjacent faces of a plane of concrete, whether the plane is the result of cracking or separate placements of concrete.

1) Is this correct?

2) What qualifies as shear friction reinforcement?

3) What are the limitations of this method, beyond the prerequisite that the bar be tensioned by the shear force?

4) What are some appropriate scenarios for use of this method?

Thanks.

 

[FrustEng]

According to the Canadian standard CSA A23.3-1994:
1) Yes
2) both the longitudinal rebars and stirrups are shear Friction reinforcement,
3) stirrups crossing the failure plane are assumed to be at yield, shear forces are limited so that will not cause compression failure in the concrete, and the beam is as strong as its weakest shear plane, and
4) it is allowed to use the shear friction method to consider interfaces between elements such as webs and flanges, between dissimilar materials and between concrete cast at different times or at existing or potential majot cracks along which slip can occur.

 

[DaveAtkins]

When you pour concrete against existing concrete, and shear must be transferred, dowels are commonly used. The dowels "clamp" the new concrete to the existing concrete, so the dowels must be developed on both sides of the mating surface. So I would say developing the dowels is the biggest drawback/difficulty with shear friction reinforcement.

DaveAtkins

 

[cap4000]

Shear friction is a tried and true method. Corbels are designed with this method. A similar condition is when a retaining wall footing shear key or better yet per ACI a roughen concrete surface is joined together by rebar dowels. ACI probably has tested this unique condition out the whazoo. Good Luck

 

[JAE]

This method of determining Vn considers shear capacity of "dowels" as well as the friction between adjacent faces of a plane of concrete

Is this correct? - I would say no - the shear friction method is all about the friction between two surfaces of concrete - dependent on the roughness of the concrete. The shear friction works due to the ability of the reinforcing to hold the surfaces together and this engages the rough edges such that slippage is minimized.

So the concrete works in shear, the reinforcing works in tension. Shear capacity of dowels are not involved.

 

[cap4000]

Just to add to JAEs excellent response. If you have 2 separate pours, the first pour should have a "rake-roughen joint surface" at 1/4 inch amplitudes per ACI 11.7.9. The ACI premise I believe is that for the concrete to actually slide in shear and move over the 1/4 inch high points the rebar will immediately go into TENSION not shear. The typical 2x4 or 2x6 footing key everybody uses may not be technically as good as the ACI roughen condition. ACI 11.7.5 limits the concrete shear portion of the Vn, above that rebar is required. Dowel shear capacity a moot point.

 

[clemsonC]

Thanks for the quick & helpful replies.

JAE & cap4000: ACI 318-02 defines Vn = Avf*fy*mu. So, you're clarifying that any reinf crossing the shear plane is acting in tension, not shear. Right?

It seems that you've all discussed existing shear planes, due to noncurrent/nonmonolithic placements of concrete. So here's another question:

If a monolithically placed reinforced structure with no shear reinf (say, a 2-way slab with reinf EW) fails/cracks in shear, THEN I can account for the reinf that crosses the shear plane, because now that reinf functions as dowels?

 

[JAE]

humble pie - for me a bit, I think....but in general (here comes my covering statements) the dowel shear is not included. However, having said that - there is this statement in the ACI commentary for section 11.7.4.3:

For concrete cast against hardened concrete not roughened in accordance with 11.7.9, shear resistance is primarily dur to dowel action of teh reinforcement and tests indicated that reduced value of mu = .6 x lambda specified for this case is appropriate.

So for very smooth conditions, they do seem to depend upon shear resistance of the dowels...but not for the other cases where there is a roughness present.

 

[theonlynamenottaken]

Be careful what reinforcement you consider to be providing "clamping" force for shear friction. For example: If you have a pier poured atop a footing where the only loading is a lateral load on the pier then you will have a bending moment and a sliding shear at the pier/footing interface. You must calculate what portion of the bars crossing that interface are being used for flexure... because a bar can't be "double yielded" resisting both flexural tension and the sliding shear friction clamping.

 

[DaveAtkins]

Sorry, theonlynamenottaken, but you are not correct. Because a bending moment generates a tension on one side of the pier, and an equal compression on the other side of the pier, there is no net tension across the interface, and you don't have to add the flexural tension reinforcing to the shear friction reinforcing (see the Commentary on Section 11.7.7 in the 1995 ACI 318).

DaveAtkins

 

[TTK]

Let me clarify what I think DaveAtkins is trying to say...

When you have a bending moment acting on the section in question, it creates a force couple on the section...compression on one side, tension on the other.

These forces are equal and they are due to the bending moment. In this case you are better off just using the compression force for your shear friction and ignore the tensile force of the reinforcing which will be needed for flexure.

If you had the same section but it was only used to resist shear (any moment was insignificant) then you would utilize the AsFy of the steel crossing the section as your shear friction resistance since you would have no sustained compressive force to use.

If you have a significant bending moment AND you want to use the AsFy of the reinforcing (in lieu of the compresive force) for shear friction you DO need to use a combined stress approach....usually the compressive force alone is enough to get you there in my experience.

 

[JAE]

Here's the ACI Commentary on part of section 11.7.7:

When moment acts on a shear plane, the flexural tension stresses and flexural compression stresses are in equilibrium. There is no change in the resultant compression Avf(fy) acting across the shear plane and the shear-transfer strength is not changed. It is therefore not necessary to provide additional reinforcement to resist the flexural tension stresses, unless the required flexural tension reinforcement exceeds the amount of shear-transfer reinforcement provided in the flexural tension zone. This has been demonstrated experimentally.

 

[TTK]

I don't have access to ACI at the moment...does it also allow you to use the compression force IN ADDITION to the AsFy?

I suspect it does not, and what the commentary is really saying is that you can use the compression component of the force couple without having to add any additional AsFy.

AASHTO does a better job of explaining this I believe.

 

[JAE]

TTK - ACI does indeed allow an increase in shear friction for sections with a resultant compressive force applied.

So your SF equation would be:

[φ]Vn = (C + Avf(fy)) x [μ]

 

[TTK]

If this is true then you will need to do a combined stress interaction formula of the reinforcing if you are also using it for flexural resistance...there must be more information in the ACI commentary on this...I will check at the office tomorrow....

...it's a good thread though, shear friction should be better understood by all (including me!)

 

[JAE]

TTK - there's nothing about a combined stress interaction formula in the code or commentary in the Shear Friction section. Check out the commentaries for sections 11.7.7 and 11.7.8

 

[TTK]

JAE

after reviewing ACI and the PCA Notes I agree with your formula but only if the "C" compression force in the formula below is due to a sustained compressive force NOT already utilizing the AsFy.

?Vn = (C + Avf(fy)) x ?

If you have gravity loads that are causing a significant compressive force you can add this to the shear friction capacity....

..but if a bending moment is causing the "compressive force" AND the reinforcing in question is already being used to resist this flexural stress then you cannot use this compressive force in addition to the AsFy.

In summary, unless you have a significant gravity load component then ACI will only allow a shear friction capacity of ?Vn = Avf(fy)) x ?

 

[DaveAtkins]

The Code is clear -- you don't have to add shear friction reinforcing to the reinforcing required for bending. Because the bending causes a compression force that will not allow the shear planes to slide, the flexural reinforcing in effect can resist bending and shear friction at the same time.

DaveAtkins

 

[TTK]

you are absolutely right BUT my point was that you can't double count this compressive force which I beleive may have been stated or alluded to above.

The bending causes a compressive force which can be as high as AsFy but that's all you get without an additional gravity compressive force...period.

You can't use the compressive force due to bending as the "additional compressive force" and add this to AsFy.

The AsFy is already using this bending compressive force based on your own definition above.

 

[JAE]

I think you're right TTK.