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Re: Re: Punching Shear Reinforcing Cutoff Distance

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Yeah, it's bad news for the water.  I think the Cressbrook Dam which I saw at Xmas was even lower.  It's strange to see such a huge dam almost empty, with the boat ramps about 150 m (slope length) above the water line.

I am very interested to see what the code guys say.  I agree about the first perimeter near the column.


On 3/18/07, Gil Brock < gil(--nospam--at)> wrote:

We are down to about 20% capacity and falling and entering our "dry" period of the year. Unfortunately they will not be able to produce the rum if they cannot grow the sugar but we are saved because they are getting lots of rain where Bundaberg rum is made.

I would not think the UDL loading is an average but also would not think it is an upper bound. It would be a statistical weighted average of some sort but there could be areas of no load and some of higher load than the design UDL. I will have a talk to our loading code convener to see what he thinks.

Obviously when a larger area is inside the shear perimeter, the average load is more likely to be applied over the area. But for the shear perimeters near the column I would not think you could guarantee it. For example, in a parking structure it is hard to get a wheel within the first punching perimeter.

At 02:10 PM 17/03/2007, you wrote:
Hi Gil, nice to hear from you again.  How's the water supply in Beenleigh ?  I hear that things are very drastic in that area of the driest continent, and that you will have no water left in a few months.  I hope you get some rain soon.  (Not wishing to make light of this situation, but Beenleigh rum may be an alternative for drinking, at least.)

Your comment implies something I had never considered: that the UDL may really be an equivalent average load, not a maximum intensity, and that the real load may all be outside the shear perimeter. 

I am not convinced, my friend.  At the limit, when the shear perimeters from adjoining columns overlap, there is surely no load left outside them, wouldn't you say ?  ;-)  

I always thought that the variations to be expected were still within the specified load, and if not, then the maximum expected load was to be applied in the expected places.


On 3/15/07, Gil Brock <gil(--nospam--at)> wrote:

This reduction is technically correct if the floor loading is a pure UDL as is assumed for design.

In actual fact the floor loading is never uniform. There are often concentrations of load in one area and no loads in other areas. One of there concentrations of load could be immediately outside the perimeter with no load within the perimeter. For this reason, unless I had a load that I know was fixed within the perimeter (say a column load from above) then I would not do the reduction as you suggest.

At 01:48 PM 15/03/2007, you wrote:
Will, I agree with you.  The shear calcs must take account of the moment.  Actually, I wonder why the examples you will find in most concrete manuals don't also take account of the floor loads (dead and live) that fall within the shear perimeter where you want to discontinue the shear reinforcement.  Probably because of the added complexity.  But once you develop the spreadsheet to take account of these loads, (they reduce the calculated shear at the perimeter), it is no longer complex.

The shear around the perimeter is basically caused by the floor loads over the tributary area around the column (plus the moment in the column).  For the calcs at d/2 from the face of the column, virtually all of that floor load causes shear at the perimeter.  But if you extend that perimeter to some distance from the column, then some of that floor load is now within the shear perimeter, and is not transmitted to the column by shear at the perimeter.


On 3/12/07, Will Haynes <gtg740p(--nospam--at)> wrote:
I have a 2-way slab that I am designing for punching shear that will require shear reinforcing. I am looking at using stud rails or stirrups and need to determine the distance which the shear reinforcing may be discontinued.
In ACI 318-05 section, it states that "The shear stress due to factored shear force AND MOMENT shall not exceed phi*2*sqrt(f'c) at the critical section located d/2 outside the outermost line of stirrup legs that surround the column". The termination distance calculated in the example from the document "Shear Reinforcing for Slabs"-ACI 421.1 R-99 appears to follow the above statement by combining moment and shear stress beyond the column to find the shear reinforcing cutoff location.

However, I have a copy of PCA notes on ACI 318-99 and on page 18-35, there is an example where the stirrup termination distance is determined by the direct shear only, and ignores the shear produced by the moment! It appears to me that PCA has it wrong here, but I was wondering if there is something I am missing?
Will Haynes
Regards  Gil Brock
Prestressed Concrete Design Consultants Pty. Ltd. (ABN 84 003 163 586)
5 Cameron Street Beenleigh Qld 4207 Australia
Ph +61 7 3807 8022               Fax +61 7 3807 8422
email:            gil(--nospam--at)
email:            sales(--nospam--at)
email:            support(--nospam--at)

Regards  Gil Brock
Prestressed Concrete Design Consultants Pty. Ltd. (ABN 84 003 163 586)
5 Cameron Street Beenleigh Qld 4207 Australia
Ph +61 7 3807 8022               Fax +61 7 3807 8422
email:            gil(--nospam--at)
email:            sales(--nospam--at)
email:            support(--nospam--at)