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Pretension Shear Equation 6

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Johns20188

Structural
Jan 26, 2015
14
Hi everyone,

I am looking at using equation 22.5.8.3.1a from ACI 318-14 to calculate shear capacity of topped prestressed hollow core plank. What confuses me is why does the shear capacity depend on Vd, the shear from unfactored dead loads? Why does our capacity increase as Vd increases? It doesn't make sense to me that adding more load increases your shear capacity.


Here's the equation:
Vci=0.6*sqrt(fc')bw*d+Vd+Vi*Mcre/Mmax

Thank you.
 
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See the linked blog, that discredit the ACI procedures/equations for shear capacity in prestressed concrete. Link
 
I might be wrong since I'm not familiar with ACI, but could it be that Vd means "shear strength due to dowel action" and not "design shear force"?
 
For what it's worth, AASHTO LRFD between its 2014 and 2017 publications completely removed the design of prestressed concrete using the simplified shear procedure. This procedure in AASHTO was very similar to your ACI 22.5.8.3.1a. I think there has been a lot of research that has shown that the Modified Compression Field Theory approach is a more accurate predictor of how the member is going to behave in shear. I didn't read retired13's link entirely but it looks like a good one.
 
If you read the commentary for 22.5.8.3.1 you will see they reference Vd as the dead load shear "due to unfactored dead load acting on that part of the section resisting the dead loads acting on the composite member". Later in the commentary they show how, for non-composite members, the Vd goes away.

retired13 - the link you show only questions the shear equation for post-tensioned members - Johns20188 didn't say this was post-tensioned so your link really doesn't apply unless it is post-tensioned.

 
No interest to argue. Does ACI needs two set of equations for prestressed concrete? Maybe, I don't know.

In practice and in teaching prestressed concrete for many years at the university level, we have identified a number of anomalies in this ACI Code shear design procedure, most particularly in beams with both positive and negative moments. That would include most post-tensioned beams, the vast majority of which are multi-span and/or built monolithically with their supports. For that reason, the balance of this article will focus only on the shear design of post-tensioned concrete beams
 
Yes that's the sentence that qualified the whole article to only post-tensioned conditions.

 
Thank you everyone for the responses so far. Sorry, yes this is pre-tensioned topped hollow core planks. I heard the shear equations may not be accurate for deep hollow core planks, however, the one I am checking is 10" deep with 2" topping. I still need to read your article retired13.
The commentary states:
"Vd is then the total shear force due to unfactored dead load acting on that part of the section resisting the dead loads acting prior to composite action plus the unfactored superimposed dead load acting on the composite member."
To me this sounds like Vd is all unfactored dead load, including self weight of the plank and topping.
The confusing part is why shear capacity goes up if my dead loads increase.
 
The confusing part is why shear capacity goes up if my dead loads increase.

Okay, I find something that indirectly pointing to solve your confusion on this issue. See p2-10 on the linked document, and understand the term v[sub]ci[/sub], and its use in design. After all PCI uses the same equation as ACI, but better represent, I believe. [link file:///C:/Users/Kai%20PC/Documents/Engineering/PCI-Hollowcore-Design-Manual-MNL-116.pdf]Link[/url]

Also, I read an article named "PCI Practice", it indicates that shear reinforcement is not required for hollow core slab. I am not positive this is the exact language, or any condition might invalidate it. If I come across the article again, I'll upload for information. (see correction below)
 
See paragraphs below for shear reinforcement for precast double tees, and minimum shear reinforcement requirement for waiving of hollow core and flat slabs.

11.5.6 If Vu is less than φVc at the end region of prestressed double-tees not subjected to point loads, shear reinforcement may be omitted with a nominal minimum provided for 5 ft to 10 ft from the ends. This is based on research by Alex Aswad and George Burnley, “Omission of Web Reinforcement in Prestressed Double Tees.”20 The approach is permitted by Section 11.5.6.2. A reduction in the φ-factor in flexure may be warranted when Po /A levels are low. See recommendations by Aswad, et al, “Load Testing of Prestressed Concrete Double Tees without Web Reinforcement.”21 Since the effective shear performance of the tee stems at the ends is dependent on the prestressing, this exception should only be used when the strand bond has been qualified as meeting ACI transfer and development-length provisions. (Reference PCI Design Handbook, Sections 5.3 and 5.3.4)

Prestressed hollow-core and flat slab units fall under (a) slabs and footings, and require no shear reinforcement, provided Vu ≤ φVc. R11.5.6.1 indicates that deeper hollow-core sections may have reduced web shear strength. ACI 318-05 does not require minimum area of shear reinforcement Av,min in hollow-core units where the untopped depth is not greater than 12.5 in. or where Vu is not greater than 0.5φVcw.

Article link (p.14). Link
 
Thank you very much retired13! I'm interested in reading the pages in your second to last post, however, the link is not working (seems like it's a folder address). When you have some time can you re-link it?

Thank you.
 
Sorry for the bad link. Try this one. Let me know if not work. Link
 
OP said:
The confusing part is why shear capacity goes up if my dead loads increase.

It doesn't, and that's the rub. This is one of many unfortunate instances of code writers over condensing equations at the cost of obfuscating what's actually going on.

If you drill down into mathematics of the sub equations on this, all the way to fd, you will discover that when the second term (Vd) grows, the third term shrinks (not quite one for one). Taken together, the last two terms are really the shear present when the sum of these two components exceeds the concrete cracking stress:

1) the tensile flexural stress induced at the extreme fiber by Md acting on the non-composite section and;

2) the tensile flexural stress at the extreme fiber induced by (Mu - Md) acting on the composite section.

Noodle on that and report back in the morning if it's not clicking for you. I think I can do the derivation but then youl'll owe me $60 for the 30 min.
 
op said:
To me this sounds like Vd is all unfactored dead load, including self weight of the plank and topping

That is exactly right. Unfortunately, you'll find that even some of the pci docs suggest that Vd is only the plank self weight. That's bunk because the wet weight of the topping contributes to the flexural stresses acting on the non-composite section as well.
 
 Fvtge
Johns20188

If you read the ACI318 commentary beside the clause it is explained fully.

For Non-composite members Vd is 0 and Mmax = Mu and Vi = Vu.

For composite,

Vd is the dead load before composite action and
Mmax = Mu - Mud and Vi = Vu - Vd

and then Vd is added separately for its effect at decompression.

This is all because PT shear calculation is treated as basically the same as RC shear calculation above the decompression moment/shear.

The extra term for PT is adding the effect to get the section to decompression.
 
Below is my try to decode the puzzle, see if it makes sense.

V[sub]ci[/sub] = concrete shear strength + shear resistance produced by prestress force for dead load shear + shear resistance produced by prestress force for a reduced shear force, V[sub]i[/sub]*M[sub]cre[/sub]/M[sub]max[/sub] = (V[sub]u[/sub]-V[sub]d[/sub])*M[sub]cre[/sub]/M[sub]max[/sub]

V[sub]i[/sub] is "the factored shear force at section due to externally applied loads occurring simultaneously with Mmax".

Note the latter two terms are resistance forces effected from the prestress, that resulted in the effect dead load shear + fraction of effect maximum total shear force less effect dead load force.

Correction: The V[sub]d[/sub] in the third term is de-compression effect due to dead load, as explained by rapt.
 
Thank you for the replies, everyone. That decompression explanation really cleared it up for me.
Basically, the ratio of your moment to your cracking moment determines how much of that shear is being resisted by the prestress. The remaining shear is checked against the RC-like shear equation/terms (0.6*sqrt(fc')bw*d). As mentioned before in this thread, the second term is a simplification of the actual equation. The terms multiplied by Vd ultimately come out to 1.

Since Mcre=I/yt(6*lambda*sqrt(fc')+fpe-fd) per ACI eq 22.5.8.3.1c,
that means the remaining available cracking moment strength is Mremain=I/yt(fd).

Therefore, the expanded equation 22.5.8.3.1a is:
Vci=0.6*sqrt(fc')bw*d+(Vd*Mremain/Md)+Vi*Mcre/Mmax

Mremain will always equal Md since it's the same thing so Mremain/Md comes out to 1 and you can use the full Vd value.

Sorry if this was obvious to you all. It wasn't immediately clear to me from the original equation but now it makes sense. Thanks again for your help!

Edit: I'm talking like I know for sure that this is correct. But I don't. Please let me know what you all think of this explanation.
 
You made me understand better. Thanks. The class was taken long time ago. :)
 

Johns20188

Remember this is not an exact science.

The logic for flexure shear is based on the assumption that the section has to crack in flexure before extending to shear cracking. Therefore with PT, decompression is the point at which RC becomes similar.

But this is just an empirical logic to try to get a reasonable result.

The method falls down in some areas. eg the PT decompression term has been known to give a very large negative shear capacity, which is illogical. Plus, ACI does not actually use decompression it uses cracking moment which is different to other codes that use this logic.

MCFT tries to overcome the problems like this, but it has its own inherent problems in many of the simplifications added to make it usable.
 
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