Deeper understanding needed for the max allowed shear reinforcement in concrete beam from ACI Code 3

[e104909]

Per ACI 318-14 22.5.1.2, Vu ≤ ф[Vc + 8*SQRT(f'c)bwd]

My understanding is that the 8*SQRT(f'c)bwd is equivalent to maximum shear reinforcement that you are allowed to put in your concrete beam section. But I cannot find any wording in the code that this is the limitation in the shear reinforcement. The commentary R22.5.1.2 didnt mention anything about this either. All I understood is, as long as the combination of the concrete strength and shear reinforcement exceeds the required ultimate shear, then that should be adequate.

My understanding is that the longitudinal rebars has lower and upper limits and were defined as rho min and rho max which is very clear in the code. But what if your Vs = Av x fy x d / s exceeds Vsmax = 8*SQRT(f'c)bwd? what failure could it be? If you provide more stirrups than what is allowed, will it be considered as over reinforced? will it be a sudden and brittle failure?

Do you have any testing or research about this topic? I am very interested to find out about this limitation if it is the same as the flexural bars.

If the statement about this limitation is true, how can we repair such beam that has exceeded this limit?



Regards,
E104909
Civ-Str P.E.

 

[Settingsun]

The concrete fails in diagonal compression, so additional shear reinforcement provides no increase in strength.

This reference might be a bit hard to track down. I assume though that the limit includes an appropriate reduction factor for brittle failure so no riskier than other brittle failure modes.

 

[e104909]

When is this possible to happen? is it when it exceeds the concrete strength фVc and starts to engage the stirrups? The literature you posted says tension in the vertical stirrups.

How do we measure the extent of its capacity until you will notice the cracks in the web?

I'm confused with the sentence "before the stirrups have yielded".

 

[ProgrammingPE]

steveh49 is correct that diagonal compression failure occurs (also mentioned in the commentary R22.5.1.2). I wouldn't interpret this as a maximum reinforcing limit, but rather as a maximum amount that the shear reinforcing can increase your shear capacity.

It doesn't make sense to exceed this amount, but doing so would not be violating any provisions in the code. If ACI wanted to keep the reinforcing below this amount, they could have added Av,max instead which would be similar to how they limit the maximum reinforcing for flexure in beams. I would think that since shear already has a Φ of 0.75 since it is a brittle failure that a diagonal compression failure would be covered since it is another type of brittle failure.

The shear reinforcing doesn't do anything until cracks form. Since the concrete shear capacity is roughly 2*√(f'c)*bw*d and the diagonal compression capacity is then roughly 10*√(f'c)*bw*d, there's a big gap where I would expect to see shear cracks to form.

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[e104909]

ProgrammingPE,

R22.5.1.2 states that the combination of concrete capacity and 8*SQRT(f'c)bwd shall be greater than Vu to minimize the likelihood of diagonal compression failure but did not mention any limitation on the shear reinforcement.

I have the same exact thoughts as you (thats the maximum contribution that the stirrups can give). But how do we measure the possible diagonal compression failure relative to the applied Vu? If your Vu is less than the equation in 22.5.1.2 and you provided a Vs exceeding 8*SQRT(f'c)bwd, will this crack in diagonal compression? How about when Vu is just above фVc but you have provided a Vs exceeding 8*SQRT(f'c)bwd, will this explode or the web will crush like what steveh49 posted?

When I'm reading this book Design of reinforced concrete 10th ed by Jack C McCormac, its more like just a maximum contribution but not a limitation. Although, I still wanted to know if this will cause any sudden failure in concrete like in flexure and why this another author mentions

I dont seem to find what he was referring in ACI

he said:

 

[e104909]

ProgrammingPE,

Where did you get diagonal compression capacity? Can you share where I can find that?

 

[Settingsun]

The diagonal compression force is related to the shear force, not related to the amount of shear reinforcement. When you have a small amount of shear reinforcement, the stirrups are what will fail. Let's say that occurs at shear force = V1. If you add more stirrups but still have shear force = V1, the stirrups don't fail but neither does the concrete.

The issue the code is avoiding is providing a massive amount of shear reinforcement to carry a large shear force when the beam isn't large enough for that shear force (not enough concrete). *If* you apply a shear force beyond the limit, you risk brittle failure (explosive actually). If your design shear force is below the limit, neither the concrete nor the steel will fail. It's the same risk as with over-reinforced beams in bending.

 

[Tomfh]

Maximum shear is the genuine concrete shear capacity, which is a failure point after the stirrups have all yielded and when the concrete struts ruptures in shear/compression. It is a hard limit, and a dangerous brittle failure.


It can be a bit confusing because we call stirrups “shear reinforcing”, but all they are really doing is carrying the diagonal tension component of the shear.

 

[HTURKAK]

I did not look to the previous posts in detail but ST model ( composite model ) at Mr steveh49 (Structural) explains the concept for max. value of the shear reinf.

These are copy and paste from ACI 318 14

22.5.1.2 Cross-sectional dimensions shall be selected to satisfy Eq. (22.5.1.2).
Vu ≤ (Vc + 8 fc √bwd) (22.5.1.2)

and

R22.5.1.2 The limit on cross-sectional dimensions in 22.5.1.2 is intended to minimize the likelihood of diagonal
compression failure in the concrete and limit the extent of cracking.

IMO, This is an implicit limitation for the max value of shear reinforcement.If V > (Vc + 8 fc √bwd) , the cross section shall be increased.

The Eurocode 2 approach is somehow different.The model is similar to Mr steveh49 (Structural) picture and it is assumed that at ultimate loads, the stirrups may yield but concrete struts do not crush.

The excerpt below shows truss model with a cut section perpendicular to the struts;

The angle ( cotθ ) has limiting values ; 1 ≤ cot θ ≤ 2.5 and the maximum value of shear steel is needed when cot θ = 1.0.

 

[e104909]

@Tomfh,

Do you mean to say "after all the stirrups have yielded", when you provide stirrups more than 8*SQRT(f'c)bwd then it will create those diagonal compression cracks?

Lets say Vu = 100 kips. фVc = 90 kips. Vs provided is 500 kips. Vsmax = 400 kips. So Vu = 100 kips < 90 + 0.75*400 = 390 kips. Will those cracks appear?

What if I increase Vu to 380 kips. So Still Vu = 380 < 390 kips. Will those cracks appear? There might be cracks at this point since I provided more stirrups than Vsmax. But I want to know if theres an empirical formula in the code that will tell me that cracks will start to appear like what ProgrammingPE mentioned [10sqrt(f'c)bwd]

My interpretation that explosion type of failure will only appear when you reach 390 kips. But when will cracks and significant cracks start to form?

 

[e104909]

@HTURKAK,

I agree that limitation only applies when your Vu exceeds ф(Vc + 8 fc √bwd) even you provided steel reinforcement that exceeds 8 √fc bwd.

Id like to clarify "the stirrups may yield but concrete struts do not crush". In the case where you provided more stirrups that exceeded Vsmax, the stirrups will not yield and concrete will crush "instantly" once Vu exceeds ф(Vc + 8 fc √bwd).

 

[Tomfh]

Do you mean to say "after all the stirrups have yielded", when you provide stirrups more than 8*SQRT(f'c)bwd then it will create those diagonal compression cracks?

The diagonal cracks are diagonal tension cracks, and they appear before the shear reinforcement yields. The shear reinforcement is there to keep those diagonal cracks closed once the cracks appear, in much the same way that flexural steel is there to keep the bending cracks closed.

What I meant by “after all the stirrups yield” was that shear compression failure is a failure that occurs at higher load levels, at a loading level beyond stirrups yielding (assuming you had limited the amount of shear steel as per the code). The code is safeguarding against that type of shear failure, by providing a maximum limit on shear capacity.