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Headed Stud / Shear Friction

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slickdeals

Structural
Apr 8, 2006
2,268
Folks,
If there is an embedded plate with headed studs on the soffit of a concrete beam, can the shear-friction design method be used for tension loads in lieu of Appendix D?

The logic I am thinking is that the headed studs will transfer tensile loads through bearing of the head in concrete. Once this load is transferred into concrete, any concrete cone breakout is prevented by adding longitudinal steel in the beam in addition to what is needed for flexure. Calculate As = Tu/ phi fy. In addition, you can always add closely spaced shear ties (say at 3" o.c.) to prevent the cone breakout on the sides.

Does this make sense?
 
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slickdeals-

I would say no, and here is why - The shear friction provisions assume a true shear plane, not the principal tensile stress plane. The reason this is important (IMO) is that the principal tensile stress plane will not have the necessary aggregate interlock to get the coefficient of friction values listed for the shear friction provisions.

Just as a comparison, even if you have the bottom longitudinal reinforcement of a concrete beam fully developed into a support, you can't count on that reinforcement in shear calculations. That's a very similar condition to what you are suggesting.
 
I could be tempted, if there are stirrups local to the studs, and if the studs to be high enough into the concrete so the reinforcing passes through the cone.

Michael.
Timing has a lot to do with the outcome of a rain dance.
 
paddington-
I agree with that, but that's not shear friction, that's just developing the reinforcing per App. D in ACI 318-08.
 
I don't think that the concrete cone pull out if there is reinforcing crossing the breakout cone. I am curious what other people think.
 
If it's vertical reinforcement, I agree. If it's horizontal reinforcement, I disagree. It helps, yes, but it won't prevent the cone from breaking out.
 
See Figure R11.7.4 ACI 318-02. If reinforcing not parallel to an assumed crack plane is allowed, why do you think that longitudinal reinforcing won't prevent the concrete cone from breaking out?

Not looking to pick up an argument/fight, just trying to understand. :)
 
If you check ACI 318-08 Appendix D 5.2.9 they discuss what you are talking about.

There are diagrams as well, and a design method for the reinforcing.
 
slick-
I'll look at the 02 version tomorrow. I just want to ask this question to you. Using your thought pattern, what would preclude one from using bottom longitudinal tension steel as "shear friction" steel at supports?
 
If there was a shear plane (crack) along the face of the support, just like you would have in case of a precast soffit beam, then you would use longitudinal bars in shear friction. In fact, that is what they do in precast soffit beams.

Using longitudinal steel has some sort of benefit in the shear capacity. The use of the rho(w) factor in the shear capacity equations of concrete does allow for this. A flexural tension crack that propagates to a flexural shear crack is definitely going to be held together better if you have reinforcing passing through it.

The mode of shear transfer close to the supports is based on a compression strut being developed. Definitely the bottom steel there acts as a tension tie.

You do bring up a good question about longitudinal steel in shear friction. I believe there was a discussion in this forum on that very topic. I will dig it up and look.
 
Right, IF you had a shear plane along the face of the support, but you don't typically have that condition. In the application you're talking about with the studs in tension and longitudinal steel passing through the cone, you definitely don't have a vertical shear plane, because it's a cone. If it broke out in a cube, maybe shear friction would apply.

You're correct that the longitudinal steel has some shear capacity - it's lumped into Vc (as in a section that is unreinforced for shear, i.e. no stirrups) for the basic case. Vc is a combination of the shear strength of the concrete, aggregate interlock, and dowel action of the longitudinal steel. If you decide to take the long route, you don't get much benefit from calc'ing out the "exact" amount. Additionally, I would say that any additional capacity gained is from dowel action, not from shear friction. If the capacity gained were from a shear friction mechanism the capacity would be substantially higher.

I agree that the bottom steel is seeing some tension as a tie mechanism as you point out, but that's still not a shear capacity.

The bottom line is if you have a section reinforced for tension without stirrups (specifically designed to take the shear force), then your shear capacity is only phiVc. Because phiVc already accounts for the dowel action of the longitudinal steel you don't get any benefit by having it in. If you don't have longitudinal steel (I admit this wouldn't happen in practice, but just for discussion...) then I would take an additional reduction since you don't have all of the conditions that apply to phiVc.

I think I started a thread a while back asking about the shear friction mechanism for longitudinal steel in RC Beams. That thread helped me understand what I'm trying to explain here.
 
I think we are going off on a tangent now, although it is a good discussion.

We are talking about whether longitudinal steel in a beam passing through the concrete cone will prevent such a failure from happening say for a stud that is 8" deep.
 
slickdeals,
I'd first like to comment on your reference to Figure R11.7.4 in ACI 318-02. If you look at the figure closely, you can see that it is a direct shear plane. It's not a tensile plane, not a principal stress plane, not any other kind of plane................ just a shear plane. The condition you describe is not a shera plane. I'm attaching a sketch to make sure we're picturing the same thing (I left out the top reinforcement and the stirrups for clarity).

Why do you think that the longitudinal reinforcement WILL prevent a breakout of the cone in tension? What is the reasoning for this other than shear friction?
 
 http://files.engineering.com/getfile.aspx?folder=97d6dfda-a188-46e2-ab05-efd5febd4d6c&file=3-18-10_non-Shear_Friction_Sketch.pdf
StructuralEIT:
What is the mechanism by which supplementary reinforcing that is provided parallel to the direction of force ACI 318-08 Appendix D 5.2.9 resists the applied tension? That tension reinforcing is developed on either side of the breakout surface (shear plane?). Does it make a difference is this reinforcing is parallel to the force or perpendicular to the force, if it is developed on both sides of the breakout surface?

Maybe I am missing something really important here.
 
Yes, there is a big difference. The reinforcement that is in the direction of the load is in tension. The reinforcement that is perpendicular to the load is in shear, but the shear friction provisions don't apply because the plane that the steel crosses is not a shear plane.

Does that make sense? You can only use shear friction for steel crossing a shear plane. A edges of a breakout cone from anchors loaded in tension is not a shear plane (i.e. there is not shear along that plane, therefore you don't get the clamping force).
 
I should also clarify that the supplemental reinforcement in the direction of the load is developed by the bars developing in tension on both sides of the plane. There is no need for shear friction, it's a simple tension development issue.
 
I think it is beginning to make more sense, but I will need to think and understand the mechanics before I am certain to say that I get my head around it.

Thanks for the explanation.
 
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