While I am not a proponent of the shear-friction theory used in North America, I thought it was primarily intended for perpendicular construction joints. The diagonal tension provisions of various codes have served us well, so there is no need for an additional check for that type shear.

I have to admit that I only used it at cold joints or at column brackets (infrequently used). I only used vertical stirrups, but always wondered if diagonal stirrups would be preferable so as to have more than one stirrup cross any vertical plane. You have obviously thought about this in far more detail and your question regarding the monolithic value is very pertinent. I hope other comments can help clear this up for all of us.

gjc

I have only looked at it at construction joints.

Thanks for your contributions gentlemen. So far, everyone's been answering question #2: when should we check shear friction? And it sounds as though we're quickly coming to a consensus. We've been checking at cold joints.

If you'll indulge me, I desperately want your answers to question #1. I want to know what you think of my conclusion that shear friction needs to be satisfied, although not necessarily checked, everywhere.

To steer the conversation to the heart of the matter, please review the attached sketch. In that sketch, I suggest that shear friction must be at work on a vertical cut through the beam in order for equilibrium to be satisfied. Do you agree or disagree? I love explanations but will also happily accept a yes/no vote.

@Hokie: I realize that you're a not a fan of shear friction, perhaps not even a believer. With that in mind, what shear resisting mechanism do you see at work across the vertical section cut in my sketch? I'm betting that it's shear friction by another name: aggregate interlock, compression block friction, dowel action. I've waited a long time for my chance to try and lure you to the dark side...

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

• http://files.engineering.com/getfile.aspx?folder=b6a7bb8f-5317-4849-a207-81

I pretty much only use it at #1 when trying to connect something new to something existing. I have never used it as described in #2. I may have used it at #3 (I can't recall senator), and never as #4.

I don't think your sketch applies. The shear cracks happen at the 45, not vertically. You do not get vertical shear cracks (principal stresses and all that). I tend to look at shear friction at the construction joints since there is no(?) other codified (ACI anyway) manner to transfer the load across that joint. I actually think shear friction makes perfect sense for the unroughened and roughened cases per ACI.

I agree with dcarr82775. Shear cracks are not vertical--they are diagonal. You have sketched a failure mechanism which is not consistent with experience.

DaveAtkins

Yes. The applicability of my sketch is precisely the issue Dcarr. I'm going to sell it a little harder to see if I can't make a convert out of you.

For any given location along the length of the beam in my sketch, I argue that shear capacity must be satisfied for all possible orientations of the shear crack (15 deg, 30 deg, 45 deg...90 deg). The reason that we check a 45-ish degree crack in practice is simply because that's the one that generally governs as a result of concrete's inherent weakness in tension. Just because a 90 degree shear crack doesn't govern, doesn't mean that a shear mechanism isn't required along a 90 degree plane. And I argue that the only shear mechanism available for the 90 degree plane is shear friction.

Looked at another way, it's just a matter of statics/equilibrium. If you take the FBD that I've drawn, with the section cut at 90 degrees, diagonal tension isn't available as a shear resisting mechanism. Something else must be getting the job done. Again, I think that something is shear friction.

As an aside, note that for a prestressed member, it is entirely appropriate to think of your shear crack at an angle other than 45 degrees.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

I should reiterate that I am not suggesting that shear friction needs to be checked gratuitously in monolithically cast members. Rather, I'm asking:

1) Is shear friction, as a necessary mechanism, not present everywhere that shear is present? It seems to me that it is.

2) We're all confident that shear friction doesn't need to be checked in monolithic concrete. How do we know that other than testing and experience? I pitched my theory at the beginning of this thread.

These two points are just rephrased versions of the two questions that I posed originally.

I've attached a good article by Loov that gets into this a bit. I'm sure that no one will have the time to read it but I thought that I'd throw it out there anyhow for sport.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

• http://files.engineering.com/getfile.aspx?folder=37684da7-0daf-48a9-a10c-91

KootK,
Sorry, but I'm not going to be drawn into debating the METHOD of shear design. Research is good, but it sometimes produces METHODS which are not required. Based on the various threads here indicating confusion with even the intent of the shear friction theory, I will just choose not to use it.

I understand Hokie. Despite the title of this thread, shear friction buy in isn't really necessary for participation. I'll rephrase the question in a way that doesn't point to shear friction:

For the vertical shear plane that I've been harping on in my sketches, what shear resisting mechanism do you think is at work?

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

As you listed above. Aggregate interlock, dowel action, etc., but I just think of a truss analogy.

My little contribution to kootk and Hokie discussion.

Canada Concrete Code required minimum shear reinforcement only where it is required : Vf > Vc (Clause 11.2.8.1)

So in my opinion, you cannot used the 'Interface Shear transfert (11.5)' equation with the reinforcement dowel action effective. Just use Vc (11.3) plain simple !

@Hokie: thanks for clarifying your view. It may interest you to know that it was the truss analogy that got me thinking about this. I've started a separate thread about that: Link. It's related but deserving of it's own discussion I think.

@PicoStruc: thanks for joining in the conversation. Vc is, of course, diagonal tension failure. As such, it will not contribute to shear resistance over the vertical section that I've directed attention to in this thread. I'm not sure how 11.2.8.1 impacts a designer's ability to utilize shear friction. Can you elaborate on that?

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

I think it doesn't control so the limit state doesn't apply. Your sketch still gives you the diagonal tension afterall. I suspect the vertical case you are curious about is so far from controlling that it doesn't matter, you can't practically get there. This is all speculation on my part of course.

@Dcarr: I mostly agree. In the majority practical applications, shear friction on a vertical plane won't govern. I wouldn't say that it's miles away from mattering however. The monolithic value is only 40% stronger than the roughened value after all. And, in the general case for monolithic construction, the designer is not paying any explicit attention to shear friction reinforcement in the form of +ve/-ve steel.

I'm not suggesting for a second that I think that vertical plane shear friction will govern in monolithic construction. In fact, in my initial post, I proposed a theory attempting to "prove" why it won't ever govern. What I'm trying to accomplish here is to build myself a world view of sorts that will allow me to better understand the mechanics of shear friction and concrete in general. Your speculation is exactly what I'm seeking here.

Quote (dcarr82775)

Your sketch still gives you the diagonal tension afterall

Can you expand on that Dcarr? I am of the opinion that, at any given location, shear needs to be addressed on both a diagonal plane and a vertical plane (every conceivable plane really). Having satisfied diagonal tension doesn't automatically mean that shear friction on the vertical plane is satisified. That is, unless you're buying into my "proof" in my original post. And I'm not even sure that I believe my proof.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Kootk,

Yes the Vc+Vs is based a strut/Tie model, but Vc can be used alone in a design point of view (see one way shear) as described in clause 11.2.8.1 to resist shear AND torsion !!!

11.2.8.1 impacts a designer's ability to utilize shear friction because your 'Interface Shear Transfert' resistance that include dewel action cannot be greather than Vc. If it is the case, then minimal shear reinforcement is required.

To me ACI 318 provides very clear guidance on when to apply shear-friction. Per 11.6.1: Provisions of 11.6 are to be applied where it is appropriate to consider shear transfer across a given plane, such as: an existing or potential crack, an interface between dissimilar materials, or an interface between two concretes cast at different times.

The commentary for 11.6.1 states: "With the exception of 11.6, virtually all provisions regarding shear are intended to prevent diagonal tension failures rather than direct shear transfer failures. The purpose of 11.6 is to provide design methods for conditions where shear transfer should be considered: an interface between concretes cast at different times, and interface between concrete and steel, reinforcement details for precast concrete structures, and other situations where it is considered appropriate to investigate shear transfer across a plane in structural concrete."

So, to me shear friction is to be applied only when the normal diagonal shear crack design assumption is not valid.

Maine EIT, Civil/Structural.

@TME: Thanks for joining the discussion.

"...or potential crack...". That leaves rather a lot of room for interpretation wouldn't you say? Will any old thermal restraint or flexural crack do?

"...other situations where it is considered appropriate to investigate shear transfer across a plane in structural concrete...". Equally vague.

Quote (TheMightyEngineer)

So, to me shear friction is to be applied only when the normal diagonal shear crack design assumption is not valid.

I disagree with this statement. For any shear friction plane that you investigate, there's usually going to be a diagonal tension check that needs to be performed either at that same location (cold joint) or right next door (dowelling into existing). Diagonal tension and shear friction are generally checks that need to be performed concurrently, not independently.

Notwithstanding the above nitpicking, I agree with you. I only check shear friction at cold joints. However, this begs the question: if shear friction doesn't need to be checked on monolithic concrete-to-concrete interfaces, then why doe codes provide mu values for that?

Where to check shear friction is really of much less interest to me that the question of where does shear friction need to be satisfied. Care to tackle the question that I posed in my 11:26am post? In your opinion, what's the shear transfer mechanism at work on the vertical plane that I sketched?

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Quote (PicoStruc)

Yes the Vc+Vs is based a strut/Tie model, but Vc can be used alone in a design point of view (see one way shear) as described in clause 11.2.8.1 to resist shear AND torsion !!!

Vc is diagonal tension and would not apply to a vertical shear plane as I have proposed Pico. The diagonal tension shear plane is at roughly 45 degrees to the vertical shear plane of interest. Satisfying Vc isn't always sufficient to guarantee that you've satisfied shear friction, as evidenced by design issues at cold joints.

Quote (PicoStruct)

11.2.8.1 impacts a designer's ability to utilize shear friction because your 'Interface Shear Transfert' resistance that include dewel action cannot be greather than Vc. If it is the case, then minimal shear reinforcement is required.

I'm afraid that I don't follow your logic here Pico. I've got 11.2.8.1 in front of me as I type this post. It doesn't mention shear friction anywhere within that clause so I don't see how it would place limits on its use. Perhaps that limitation is implied in some way that isn't apparent to me.

11.2.8 Minimum shear reinforcement
11.2.8.1
A minimum area of shear reinforcement shall be provided in the following regions:
a) in regions of flexural members where the factored shear force, Vf, exceeds Vc + Vp;
b) in regions of beams with an overall thickness greater than 750 mm; and
c) in regions of flexural members where the factored torsion, Tf, exceeds 0.25Tcr.
Note: Footings and pile caps designed using strut-and-tie models in accordance with Clause 11.4 need not satisfy
the minimum shear reinforcement requirements of Clause 11.2.8.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

I think that the situation illustrated in this slide would be a case to check shear friction with monolithic concrete

• http://files.engineering.com/getfile.aspx?folder=041fe9b6-dfa5-4194-a075-e6

Yes, yes it would. Thanks Pikku.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Quote (OP)

Vc is diagonal tension and would not apply to a vertical shear plane

Vc is diagonal cracking but is used all the time with the vertical shear plane dv x bw, check One way shear in slab, in footing, etc... all the structural element usually don't have any shear reinforcement !

Quote (OP)

I've got 11.2.8.1 in front of me as I type this post. It doesn't mention shear friction anywhere

What i meant is literally if Vf > Vc, by code, you need to add at stirrup. You cannot justify not adding reinforcement by 'proving' using interface shear resistance (with dowel) is enough !!!.


And don't forgot that 'interface shear transfer' suppose a interface... a joint, a crack, etc... so if your interface is not a cold joint but a shear crack.... your member already failed, in a performance point of view.


why risk cracking for some stirrup (that confine concrete and improve global performance of the member !)... but that is my opinion !

It seems like you've thought about this quite a bit so you're unlikely to get a satisfactory answer.

You keep going back to your question #1 about where does shear friction need to be satisfied. I think you're generally correct that along a vertical plane (in your diagram) some type of shear mechanism does need to be satisfied, by basic statics there is no way around that. Whether or not the shear friction provisions of the code are the correct method here is a bit more grey. For one thing the shear friction equations are partially b.s. created to agree with testing "it is therefore necessary to use artificially high values of the coefficient of friction in the shear friction equations so that the calculated strength will be in reasonable agreement with test results". Because the equation doesn't come from mechanics it will be hard to derive a satisfying answer.

A quick attempt to put numbers to it, writing this out as I go so there might be something catastrophically wrong with my logic here but I'll give it a go.
Assume a simply supported beam, leave out phi factors and whether ultimate or not.
M_demand = WL^2/8 (assume k-ft)
V_demand = WL/2 (assume k)
Ast = M/(4d) (using rule of thumb for Ast, takes care of conversion, i.e. sq. in.)

From shear friction:
V_capacity = Ast x 60 (ksi) x 1.4 (kips)
= M/(4d) x 60 x 1.4 (kips)
= WL^2/8 x (1/4d) x 60 x 1.4
= WL/2 x L/4 x (1/4d) x 60 x 1.4
= Vdemand x L/d x (1/16) x 60 x 1.4
= Vdemand x L/d x 5.25

Most likely L/d is always >> 1.0, so V_capacity is >> Vdemand for a simply supported beam.
Or maybe I made a mistake above and it's all nonsense.

@Pico: I'm grateful for your participation in this discussion but we are clearly not on the same page. And that's a result of my failure to communicate my ideas effectively. While I don't agree with your latest comments, I don't think that there's anything further to be gained by my attempting to badger you into seeing things my way. I may just be flat out wrong for all I know. We'll just have to agree to disagree on this one. That's a valid outcome from a healthy debate in my mind.

@Bookowski: I have, without hyperbole, been thinking about this for years. I feel that the answers to the questions that I've posed are fundamental to a unified theory of shear, including shear friction. I want that unified theory to a) satisfy my own intellectual curiosity and b) inform my design decisions going forward. The design decisions bit is much less important to me really. We all seem to have an intuitive feel for what needs to be done even if our understanding of "why" is a bit murky.

And you're right, I have been struggling to tease out an answer that I find satisfactory. Your comment has gone a long way to helping in that regard, however, and I thank you for it. It's great to hear at least one other human agree that a shear mechanism akin to shear friction is likely required across the vertical section that I proposed. I was hoping to ultimately achieve a consensus agreement that wherever there is shear, there is a shear friction mechanism at work that is generally a) taken for granted in design and b) non-critical based on some rational explanation (I attempted one). For me, that "discovery" was a rather significant shift in my thinking and I've been hungry for feedback on the concept.

I believe that a major reason that it has been difficult for me to get a KootK-satisfactory answer to my questions is that the questions themselves are pointless from a practical design point of view. My next step will be to try to generate a numerical scenario where Vc + Vs > Vertical section SF. This will probably take the form of a less elegant version of what you've already done Bookowski. Again, thanks for the effort.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

See attached. A simple FBD ives the answer. Also, remember why reinforcing is there. The reinforcing and concrete have the same stress until the concrete cracks, then the reinforcing acts alone. (compatibility analysis anyone?) Once the concrete cracks, there would be no shear friction since the ability to provide a compressive force (required for shear friction) is no longer present.

• http://files.engineering.com/getfile.aspx?folder=31a6d833-61c5-4540-81e0-a0

Welcome VTPE.

The concrete is just holding the steel in place? The reinforcing acts alone? You've got to be kidding. Have you somehow missed out on the glacially unfolding revolution that is strut and tie design? From a shear perspective, steel ties are just one half of the truss mechanism that depends, unequivocally, on the presence of concrete struts between those ties. The concrete between stirrups matters a great deal.

Quote (VTPE)

The reinforcing and concrete have the same stress until the concrete cracks, then the reinforcing acts alone.

It is strain that the reinforcement and concrete have in common prior to cracking, not the stress.

And you certainly can have shear friction in the presence of shear cracking. Firstly, not all of the beam is in pure shear like the infinitesimal element that you've drawn. There will be a compression block that will do an excellent job of transmitting vertical shear. Secondly, neither dowel action nor aggregate interlock depend on un-cracked concrete. Both mechanisms have been tested and shown to be active post-cracking.

A free body diagram cut between stirrups is every bit as valid as any other FBD. If the concrete is ornamental in your estimation, what is it that you do see transmitting shear between stirrups?

Lastly, conventional shear design is Vc + Vs, not just Vs. Right? Concrete matters?

I think that you've grossly underestimated the importance of concrete "holding the steel in place".

Thanks for posting the sketch VTPE. Taking time out of your day to walk to the scanner = commitment.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

I don't believe that shear friction has to be checked where a "traditional" diagonal failure plane applies because of the following:

1) Shear friction to me seems to be inherently lower in stiffness than concrete and steel shear resistance given the usual assumptions such as diagonal cracking and transverse shear reinforcement. Thus, shear will be resisted by the traditional method first and then shear friction second. Essentially, shear friction does not strike me as another limit state for shear, only another way to reinforce for the same limit state.

As an example. Lets say we fillet welded two overlapping steel plates together. We then also took some clamps and clamped the plates together giving a friction force. If we apply a tensile load on each plate then the load is resisted by both the welds and the clamping friction force. However, by inspection we don't check the clamps because it's also welded as they're resisting the same limit state. If the plates couldn't be welded then of course we check the clamping force. However, with the weld the presence of the clamps are redundant and generally not as optimal a solution. Thus, we would either remove the clamps or ignore them.

2) With the exception of strange circumstances I would imagine that that traditional shear resistance design methods would result in a higher shear strength than those provided under shear friction. Therefore, if traditional concrete shear design is satisfactory then you have resisted the shear load. Any added strength from shear friction will not be needed. Further, because shear friction is "less stiff" shear failure would already have had to occur or be occurring before shear friction could be considered engaged. If that were the case and traditional shear design applied but was purposefully not met in favor of shear friction then I would say it was an incorrect design.

3) Shear friction relies on aggregate interlock, compressive friction, and/or dowel action. With the exception of dowel action; after a diagonal shear failure, cracks will have opened up wide enough to prevent the full application of the shear friction method for shear resistance. Thus, only one check or the other should be used, but not both at the same time.

Thus, I feel personally that shear friction only applies when normal shear design assumptions do not apply. To apply it when those assumptions do apply seems to be unnecessary as shear friction is not an additional limit state but rather a "weaker" method of shear reinforcement but less constrained by design assumptions and more applicable to discrete shear failure planes.

Maine EIT, Civil/Structural.

KootK,
Your response to VTPE was excellent. In the diagonal shear provisions, Vc still exists after flexural/shear cracking occurs. But the rest of this thread, and the obvious continuing confusion about shear friction, has reinforced my view that this artificial method is, as bookowski put it "partially b.s.".

Nice TME -- thank you. If you read my comments in this thread carefully, you will notice that I've been very careful with my language. I have proposed that:

1) Shear friction must be satisfied everywhere and;
2) Shear friction must be checked only at cold joints.

That difference between where SF should be checked and where it must be satisfied is very important. Perhaps I've been too subtle with that semantic distinction. Like you, I also do not believe that SF needs to be checked wherever diagonal tension is evaluated. I do, however, believe that shear friction needs to be satisfied there.

I believe that your latest comments are almost 100% incorrect TME. Or, more precisely, your comments are almost wholly in opposition to my own views. Since I really do not know that I am in the right here, by definition, I also do not know that you are in the wrong. I have my fingers crossed that you'll take my comments here as they are intended: spirited debate between respected colleagues, not combative assholery.

I thought that your weld plate example was a brilliant device for conveying your ideas. I think that it would be a slightly better analogy if we called it a combination bolted/welded connection. It's more real worldy and the principles are the same, namely deformation compatibility. I'm going to use that analogy below.

Quote:

Essentially, shear friction does not strike me as another limit state for shear, only another way to reinforce for the same limit state.

Quote:

If that were the case and traditional shear design applied but was purposefully not met in favor of shear friction then I would say it was an incorrect design.

These two statements seem contradictory to me. I agree with the second one.

Quote:

1) Shear friction to me seems to be inherently lower in stiffness than concrete and steel shear resistance given the usual assumptions such as diagonal cracking and transverse shear reinforcement. Thus, shear will be resisted by the traditional method first and then shear friction second.

I suspect that you are thinking of the two resisting mechanisms -- diagonal tension (DT) and shear friction (SF) -- as being two alternate mechanisms for addressing the same failure mode (shear). This is analogous to bolts and welds used in the same tension connection. DT and SF are independent failure modes, at the same location, and both need to be satisfied independently. The better analogy would be between bolts and net section rupture in a tension connection. To speak machine, the condition isn't <DT OR SF>, it's <DT AND SF>. As such, the differential stiffnesses of the two mechanisms isn't relevant.

Quote:

2) With the exception of strange circumstances I would imagine that that traditional shear resistance design methods would result in a higher shear strength than those provided under shear friction.

Hopefully the inverse is true. We check DT because we expect it will govern. We ignore SF because we expect that it will not govern. For this to be true, SF capacity must generally be greater than DT capacity.

Quote:

With the exception of dowel action; after a diagonal shear failure, cracks will have opened up wide enough to prevent the full application of the shear friction method for shear resistance.

Not so. To quote myself from 2 PM:

Quote:

you certainly can have shear friction in the presence of shear cracking. Firstly, not all of the beam is in pure shear like the infinitesimal element that you've drawn. There will be a compression block that will do an excellent job of transmitting vertical shear. Secondly, neither dowel action nor aggregate interlock depend on un-cracked concrete. Both mechanisms have been tested and shown to be active post-cracking.

ACI specifically directs designers to use shear friction at locations of real and imagined cracks. That wouldn't make much sense if a crack rendered the method ineffective. When shear cracks from, we don't say that resistance is limited to Vs. Rather, we use Vs + Vc because we acknowledge that shear cracks don't nullify concrete shear resistance. The one glaring exception would be shear within plastic hinges in high seismic zones.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Thanks Hokie. I'll admit that the more I look at shear friction, as a method, the less I love it. Still, it's the law of the land in my particular jungle. I was thumbing through a NZ document that CEL pointed me to last night. I think that I'm starting to understand the non-shear friction view of things. Dowels are treated in true "dowel" fashion. Kind of like mortis and tenon construction in wood. It certainly appeals to my intuition.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

Do a google search for "direct shear failure in concrete beams under impulsive loading" by T. Ross, it's a 200 page pdf. It is focused on dynamic (blast) loading but provides a summary of past research for static loading as well.

In general, I would apply shear friction only where reinforcement is used to "clamp" surfaces together. This is the case with old + new surfaces, many cold joints, and typical wall-footing joints. Once the reinforcement acts in direct tension (or compression) with a parallel component of force, that is the preferred design scheme. In a beam, strut and tie is the scheme to achieve a safe design (and is the basis for the code-level shear reinforcement design), and shear friction would only apply when a cold joint is allowed to form.

@TX: I know that you're a code committee guy. So why the heck is there a mu value for monolithic concrete if shear friction only applies over cold joints? Pikku14 has provided one explanation above.

You've weighed in on the second question of my original post. Care to take a swing at the first? It was most succinctly restated in my supplemental post of [29 Sep 14 11:26].

Your comments reminded me of another issue that I'm interested in. We generally supply shear friction reinforcement in the flexural zone of members subjected to bending. If there's rebar in the compression block as well, does that increase or decrease shear friction capacity? I would expect that it would decrease shear friction capacity since it would shelter the compression block concrete from some of the clamping required for shear friction to develop.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

That last question is one of the most ambiguous things about shear friction. Shear friction reinforcement is supposed to provide a clamping force, and there is no way that steel in the compression zone serves to clamp. Thus, those who want to distribute shear friction reinforcement throughout the section are kidding themselves. Distributed reinforcement works for direct tension and can work as dowels, as long as it is not too close to an edge.

Kootk: I did miss that distinction and you're entirely right to point it out.

By all means feel free to disagree with me. I definitely tried to make it clear that the above was my opinions based on my engineering judgement alone and with no qualifying evidence to support it. I enjoy a good spirited debate.

Quote (Kootk)

These two statements seem contradictory to me. I agree with the second one.

I can see how they could be taken that way. Perhaps limit state isn't the right word. My intent is to say that shear friction is just another way to resist shear, like using bolts instead of welds.

My second sentence was meant to say that traditional shear design per chapter 11.2 and 11.4 must always be checked if applicable but shear friction will not control in those cases. Thus, designing for shear friction without checking the traditional equations of 11.4 would be incorrect.

Below I have conceded this may be wrong.

Quote (Kootk)

I suspect that you are thinking of the two resisting mechanisms -- diagonal tension (DT) and shear friction (SF) -- as being two alternate mechanisms for addressing the same failure mode (shear). This is analogous to bolts and welds used in the same tension connection. DT and SF are independent failure modes, at the same location, and both need to be satisfied independently. The better analogy would be between bolts and net section rupture in a tension connection. To speak machine, the condition isn't <DT OR SF>, it's <DT AND SF>. As such, the differential stiffnesses of the two mechanisms isn't relevant.

Hmmmm, I've written about 5 different responses to this and then keep finding a flaw in my argument. I'll concede this only on the basis that I've having trouble refuting it even though it seems wrong to me.

Quote (Kootk)

Hopefully the inverse is true. We check DT because we expect it will govern. We ignore SF because we expect that it will not govern. For this to be true, SF capacity must generally be greater than DT capacity.

Whoops, that was indeed backwards. As you've probably noted SF should never control over DT.

In case anyone doubt this:

Per 11.6.4.2, for SF with a diagonal bar (not perpendicular to the failure plane) you get an increase in strength. Per 11.6.7 you're required to resist net shear across the failure plane. Both occur in a DT situation with traditional vertical stirrups. Let's assume these negate each other for a DT situation (as I expect they do).

Lets assume a typical 45 degree crack for this situation and consider only a single crack with a single vertical #4 bar crossing it. Per 11.4.7.5 we can calculate the DT shear resistance for a single #4 bar using equation 11-17.

If we graph the two equations (or just look at them), SF will never control over DT. This is obviously neglecting theta and plain concrete shear strength.

So, perhaps we do always need to consider SF, but because it will inherently never control we just don't bother.

Maine EIT, Civil/Structural.

Whoops, three paragraphs were out of order. It should go:

"I can see how they could be taken that way..."

"Below I have conceded this may be wrong."

"My second sentence was meant to..."

Maine EIT, Civil/Structural.

As for steel in the compression zone, 11.6.7 and it's commentary appear to address this. It states roughly that flexural tension and compression cancel each other out while net tension requires additional resistance and net compression helps to clamp the surfaces in addition to the shear friction reinforcement. This does seem odd as I would expect that these could not be added together like the code implies it can.

Maine EIT, Civil/Structural.

Please review the attached PDF. It is a numerical example of a beam that is adequately reinforced for flexure, passes a diagonal tension shear check, but fails a shear friction investigation. The second page is some MathCAD scribbling for anyone who wants to check my numbers.

While I've been able to come up with numerical example that demonstrates my point, that example is highly contrived. In fact, it took me the better part of an hour to get something to work out the way that I wanted. And, even at that, the proportions of the beam are ridiculous. It's bordering on deep beam territory and, if properly detailed in that context, would likely be self solving for the shear friction check anyhow.

This supports our expectation -- and my hope -- that a vertical shear friction plane is highly unlikely to ever govern the shear design of a properly detailed concrete member.

@Bookowski: Hat's off to you for your on the fly, 9th Grade Algebra-esque, symbolic proof of the same phenomena. Your work has passed my QC review and is confirmed by my clumsy numerical fiddling.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

• http://files.engineering.com/getfile.aspx?folder=6c013f26-3151-4f05-90b2-f9

@TME: I believe that we're closing in on full agreement here. Thanks for making the effort to respond again.

Regarding 11.6.7., while I agree with your interpretation, it doesn't really address the issue that Hokie and I are concerned about. In a flexural member, some of the rebar often counted on for shear friction winds up being in compression under flexural load. It's hard to imagine how a bar that is in compression can contribute to shear friction clamping. In fact, the force in that bar would resist clamping in my mind. As Hokie has rightly pointed out, a bar experiencing compression could still participate in dowel action if edge distances were appropriate.

Quote (TheMightyEngineer)

Per 11.6.4.2, for SF with a diagonal bar (not perpendicular to the failure plane) you get an increase in strength. Per 11.6.7 you're required to resist net shear across the failure plane. Both occur in a DT situation with traditional vertical stirrups. Let's assume these negate each other for a DT situation (as I expect they do).

It's important to recognize that the situation that you've described here is not shear friction as there is no sliding parallel to the adjacent surfaces. This scenario is just straight up Vc + Vs. Although I agree with your conclusion that vertical plane shear friction is a moot point from a practical perspective, this comparison is not proof of that.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

I don't have the time right now but what happens if you run the more advanced shear strength calculation using equation 11-5 in ACI 318 taking into account flexure?

Maine EIT, Civil/Structural.

Quote (Kootk)

It's important to recognize that the situation that you've described here is not shear friction as there is no sliding parallel to the adjacent surfaces.

Hmmm, yes I see that now and agree.

Maine EIT, Civil/Structural.

I'm not sure what would happen TME. I'm going to leave that check on the table as I'm already satisfied that shear friction on a vertical shear plane is a matter of purely theoretical interest. As I mentioned in my first post of this thread, I believe that there's also a shear-compression field generating clamping force as well. Adding that into the mix would put the capacity way over the top I reckon.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

We're approaching fifty posts with this thread. In my experience here at Eng-Tips, that's about the maximum life span. The end is near. With that in mind, I'd like to wrap up with this:

1) Thank you all for your participation in this largely pedantic discussion.
2) Now for some cake.

Cake you say? Yes... cake. For practical minded folks like yourselves, this thread has probably been a bit like downing a meal of gruel and green beans: frustrating and unsatisfying. Such a meal should be followed by desert. And I have some... sort of.

I believe that there is a very practical application for my "imaginary shear friction plane" concept for those who choose to ascribe to it. Thinking in those terms allows me to be more flexible when considering unplanned cold joints. To understand how that is, please read these two snippets from another thread that is currently active (Link):

Quote (Hokie66)

...Columns don't have inclined joints, and neither should STM struts.

Quote (KootK)

You raise an interesting point regarding columns Hokie. As you know, you've been helping me out with a related shear friction issue lately (Link). While you're no doubt sick of indulging me on the SF front, I believe that the concepts in the other thread have application here. Try his wacky supposition on for size:

I believe that columns -- all columns -- do in fact have inclined joints. Several actually: 15deg, 30 deg, 60 deg... and at every location along the columns too. Those joints are just monolithic, held together by shear friction (or the atheist equivalent), and perhaps... imaginary.

Why would I bother to say something so fru-fru and bizzare? Imagine me in seated lotus position as I type this. For me, the imaginary shear friction plane idea from the other thread has a very practical application. It allows me to be less conservative when dealing with surprise cold joints. Here's how it goes:

1) Contractor calls me up and proposes / tells me about a unplanned cold joint.
2) I freak out. You put a cold joint WHERE? It's the worst possible location.
3) I remind myself that there was always a slip plane at the proposed cold joint location. The only difference is that it was originally a monolithic shear friction joint and now it will be a roughened shear friction joint.
4) I check the numbers (mu = 1.4 vs 1.0) and, if things work out, I carry on.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

It seems that the biggest confusion is coming from 'satisfied' vs 'checked' question. There is no arguing that equilibrium must be satisfied across any section that you draw - vertical, 2 degrees, 10 degrees etc. So some shear mechanism has to be at work across your vertical section. Whether or not that is the ACI shear friction is where it goes off track and I don't think you can derive a true answer as that method is something that is acknowledged to not be entirely correct.

Also note that in my quick derivation previously the M/4d approximation for steel area does a unit conversion so there are mixed units in the final result in that L is in ft. and d is in inches. To make it more useful you can convert and say that for a simply suppt'd beam as long as L/d > 2.3 the flexural reinforcing will force the shear friction equation to work. Because of the different phi factors you need to scale that by 0.9/.75 so it's actually L/d > 2.7 - and that is only an approximation because of the estimate for steel area. This checks out if you look at two beams:
beam 1: Wu = 10k/ft, L = 6ft, b = 12", d = 14", (2)-#6 bars flexural, 5ksi. L/d = 5.1 and shear friction is satisfied (mu = 1.4) by the flexural reinforcing
beam 2: Wu = 35k/ft, L = 6ft, b = 12", d = 28", (3)-#6 bars, 5ksi. L/d = 2.6 and shear friction no longer satisfies Vu.
You could derive similar symbolic results for various beams but I am guessing you'd always get a similar result, that it take a very highly shear controlled beam to be worth a look (i.e. tiny span, large depth, high shear/moment ratio).

Conclusion: There clearly has to be some limit at which you could force a direct shear mechanism, the capacity is not limitless. In 99% of typical cases this is never close to controlling so it is 'satisfied' and passes by inspection (or more likely never considered). In some cases, i.e. very very shear controlled it may require checking. (See link to paper I posted earlier, there are several papers on direct shear actually occurring - mostly it seems to come from dynamic, i.e. seismic or blast).

The "correctness" of the shear friction provisions is another interesting issue.

My understanding is that part of the reason that the mu values were calibrated upwards was to account for cohesion. In Canada, cohesion is a separate term and, as a result, mu values are lower.

To some extent, I wonder if the shear friction provisions weren't calibrated upwards to ensure that monolithic shear planes would never govern. It all just seems to work out way to nicely. Or maybe it's truly connected to some unifying mechanical principle that just jives.

If you look at detail "B" in the sketch that I provided with my initial post, you'll see that my theory predicts that a compression field will develop to automatically take care of the vertical plane shear friction business so long as the angle of the assumed struts doesn't exceed 55 degrees. What's the most commonly quoted upper limit recommended for struts? Fifty five. Freakin'. Degrees! And I didn't fudge the numbers on that; they just fell out of the analysis that way.

I know, I'm starting to sound like an Area 51 conspiracy theorist.

Back to the more pedestrian issue of where SF needs to be checked, the A23.3-94 concrete code stated:

Quote (CSA A23.3-94)

interfaces between elements such as webs and flanges, between dissimilar materials, and between concretes cast at different times or at existing or potential major crack on which slip can occur

"

That definition is a bit more expansive than others that I've seen.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

from Structural Concrete article: "As previously mentioned, direct shear failures can occur in uncracked elements at locations near a support when a static load is applied in its proximity. However, tests have shown that concrete elements can also fail in a direct shear mode under the action of an intense dynamic load distributed along the length of the element [2]. In these tests the roofs of reinforced concrete box structures were subjected to loads from detonating explosive charges. The box structures were cast monolithically with two open ends and, for one series of tests, with dimensions and reinforcement as shown in Fig. 6. The roof slab did not have any prior crack planes through its thickness. The test results show that the slabs failed in direct shear in several cases and that the slabs were completely severed from the walls along vertical failure planes. Most of the slab reinforcing bars were pulled out of the wall, with a few broken bars exhibiting minor necking. It was further observed in several tests that the central portion of the roof slab remained relatively flat, as though no flexural deformations had taken place (shown schematically in Fig. 7).

Little is known about the actual fracturing and direct shearing process in dynamic events and therefore it was assumed in [14] that the direct shear failure in the dynamic case behaves in accordance with the shear transfer mechanism under quasi-static loading conditions. It was stated in [14] that the direct shear failure of the roof slabs is characterized by the rapid propagation of a vertical crack through the element depth. Since direct shear is associated with crack planes perpendicular to the longitudinal axis of the element, such failures are also possible in elements designed for flexural shear. Failure curves for reinforced concrete elements were developed in [14] and used in a parametric study of direct shear failures, see Fig. 8. The failure curve is unique for the specific element in question such that a family of curves could be generated for elements with different properties. A failure curve is constructed such that the combined values of pressure and rise time below the curve relate to no failure, and values above the curve relate to direct shear failure. The increase in pressure for an increasing rise time implies that the element is able to resist direct shear at higher pressure levels if the load is applied more slowly. The shape of the failure curve also indicates that for very small rise times, in this case < 0.1 ms, the maximum pressure appears to be approximately constant.

The analyses in [14] indicate that the resistance to direct shear increases as the element span to effective depth ratio L/d for uniformly distributed loads increases. However, the influence of L/d diminishes for small rise times and disappears at rise times close to zero. Another case is the comparison between elements with fixed support conditions and reduced end restraints. Two failure curves were generated and the difference between these curves diminishes for small rise times. Further results in [14] suggest that strength enhancement due to strain rate effects increases the shear resistance such that the entire failure curve is shifted upwards. Thus, even for the case with zero rise time, the failure curves do not coincide. It was also mentioned in [14] that the load duration does not affect the direct shear failure curve significantly. Other investigations have involved theoretical analyses of the direct shear mode of concrete elements [24–26]. However, this work is not the focus of the present paper."