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Limits for Reinforcement of compression members 1

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htex404040

Structural
May 10, 2007
7
Hello all,

I'm looking for a clarification on the minimum reinforcement ratios for tied columns. I know ACI-05 10.9.1 states that the Ast for concomposite compression members shall be not less than 0.01Ag, then in the commentary I see it talks about the report of ACI committee 105 and a minimum reinforcement ratio of 0.005 was recomended for tied columns, then the next sentance says in all editions of the code since 1936 the minimum ratio has been 0.01.

Many of my co-workers say minimum area of steel for these members is 1/2 of 1% (0.005), but I'm just trying to find where in the code it states this and the above section is the only place I've seen this referanced, and from the wording of it it looks to me like the minimum is 0.01.

Any input would be appreciated!

Thanks.
 
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You are correct, the minimum reinforcement for columns is .01Ag.
 
ACI 10.8.4 allows you to reduce the percentage by up to 50%.
 
You probably cannot reduce the area of steel unless the column is truely oversized.
 
Well, you can't go to just 0.005Ag and design your column. 10.8.4 allows you to consider a reduced Ag up to 50%.

For example: if you have a 20 x 20 column and you think it's too large, you can design a column with half the area, Ag, still limited to 1% reinforcing.

20 x 20 = 400 sq. inches.
(min. steel typically = .01 x 400 = 4 sq. inches)

1/2 Ag = 200 sq. inches

sqrt(200) = 14.14 x 14.14 column.
At 1% steel this is 2.0 sq. inches

2.0 sq. inches is of course 0.005Ag of the original 20x20 column.

But you don't check a 20x20 column with 2.0 sq. inches. You must check a 14.14 x 14.14 column with 2.0 sq. inches to see if you can go that low with your reinforcing.

There is a difference.


 
Yup...if you can show that the P/M interaction curve of you concrete column still works using 1/2 the concrete that is actually there, you can take 1% of the concrete you do need, or 0.5% of your actual concrete cross section.

I've heard it called an "architectural column" when you do this.

This may be helpful in a situation where you have bumped up your column size to satisfy punching in the slab above...
 
JAE-
Excellent example. I have a question regarding that now. Now that you have showed that the 14.14 x 14.14 column works with some cover, c (which was the only way to get the reduced area of steel), do you have to detail the column such that the actual cover is 5.86-c (and you are essentially providing 1%Ag fo the "smaller" column), or can you detail identical to a typical column with cover c (and essentially use 0.5%Ag for the actual column)?
 
I want to see JAE's opinion, but I have always interpreted that to mean you can detail the large column like a regular column. I think using a large amount of cover would do more harm than good, from a crack control standpoint.

DaveAtkins
 
It is important for all of us to realize that these lower limits on reinforcing come from an effort to prevent yielding in longitundinal reinforcing (due to creep) under sustained loads. This came from tests ran in the early 30's.
 
jike's comments make me think that these should be detailed with the much larger cover.
 
I believe that you would still detail it with normal cover. Excessive cover could cause cracking or spalling problems on the face.
 
Still with normal cover. Just as long as your "reduced" column has a phiP/phiM curve containing Pu, Mux, and Muy.
 
...sorry for the delay - busy day.

I agree with jike - you design the column with the reduced section - get the steel - verify that it works (i.e. at 14" square) and then detail it as a normal 20" square column with normal cover.

 
minimum column reinforcement in other standards:

British code (BS8110) Asmin=0.004Ac

Euro code (EC2) Asmin=0.1*N/fy>0.002Ac ( N design axial force)

Australian code (AS3600) Asmin =0.01Ac reduced if As*fy>0.15 N

New Zealand code (NZS 3101) Asmin=0.008Ac

 
JAE,

As per Canadian Codes, you can still use 20"x20" column in your calculation for axial capacity with the ratio of steel as 0.5%(instead of 14"x14" column with 1% steel).
But the important thing is that you have to reduce the axial capacity by multiplying with 0.5[1+{actual steel % used)/0.01}]
 
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