2 way slab 2

[DNBEA]

I am reading ACI318 chapter 13 and have a question for everyone here. 13.5.3 states that when forces cause transfer of moment between slab and column, a fraction of the moment shall be transfered through moment and a fraction of it through shear. 13.5.3.2 also tells us where to put the extra bars for the fraction of the moment transfer. But how can we apply this artical?

when we do the actual design with a computer program, the unbalanced moment will be distributed in accordance with the stiffnesses of adjoining elements. and the slab will be reinforced according to the larger of the 2 interior negative moments on each side of the columne. To me it seems like we never need to do 13.5.3.2. Do you agree?

 

[JAE]

If you use the Equivalalent Frame Method in Chapter 13, you will model both the slab, and the columns supporting the slab.

When you complete the analysis of this system, some of the moments extend down into the columns. This column moment is being transferred from the slab to the column and is the subject of section 13.5.3.

The "unbalanced" moment is due to the fact that slab moments on one side of the column are different than the other side of the column and the difference is the transferred column moment.

This moment is brought down into the column in two ways - via bending in a small strip of slab over the column (which is narrower than your overall column strip) and via shear forces around the column which are higher on one side of the column than the other, creating a vertical load eccentric to the column (see the diagram of this in Figure R11.12.6.2).

Your program analyzes the overall column and middle strip moments and provides the required bending reinforcing per those moments. Section 13.5.3 is an additional effort to focus additional reinforcing over and near the column to account for the column moment transfer. Your program may or may not do this for you - you would need to check on its actual features.

 

[DNBEA]

Thanks. I understand what you are talking about. let me give you an example, say we have 100kft in slab on one side of the column and 60kft on the other side. the unbalance would be 40kft. 13.5.3 tells us to add bars in 1.5h area to account for a fraction of 40 which is say 25kft. so one side shall be designed for 60+25 the other side would be 60. but if the program designs both sides for 100kft. why do we need to do 13.5.3. do you agree with me?

 

[JAE]

Because the 60+25 must be resisted by a narrower strip of slab than your full column/middle strip that resists the 100.

 

[JAE]

1.5 x h is the width on either side of the column to use

where h = the depth of the slab+drop panel.

So you would first analyze your equivalent frame, design your column and middle strips for the slab moments (your 100 ft-kips).

Then you would take the column moment (the unbalanced moment), multiply it by [γ]f and design reinforcing for it based on a narrower width of concrete(based upon the 1.5h).

Then, for the (1 - [γ]f) x Mu you would check the shear at the column per 11.12.6 which would be the full slab shear plus the shear caused by the (1 - [γ]f) x Mu
OK?

 

[DNBEA]

I think the 1.5h(2) shall have additional bars for 25kft instead of 60+25. 'cause only this 25 will be transfered to the column by flexure. if we do as you said. I beleive we need to design both sides for 60 then add additional bars in the narrow area on one side for that 25. not as you said design 100 then add bars for 85 which would end up 185kft on one side. maybe I misunderstood you.

also if we use direct design method. according to 13.6.3.4 we shall design for the larger of the two which in my case is 100. 100>60+25. why do we need extra bars for that 25.

 

[JAE]

I checked out an ACI design guide example. What they do is they determine the slab moments and column moments - for our example, you have a slab moment of 100 ft-kips on one side of a column and 60 ft-kips on the other, thus 40 ft-kips into the column.

They determine the required column strip and middle strip moments. The column strip moments result in a calculated amount of flexural reinforcing...this would be using your 100 ft-kips and a concrete width = your column strip width.
Let's call this As(CS) = 6 sq. inches.

Now, you calculate the [γ]f and multiply that times the 100 ft-kips to get an unbalanced moment factor. Lets say that [γ]f = 0.35. So your moment that is transfered by flexure into the column itself would be 0.35 x 40 ft-kips = 14 ft-kips. And therefore, the moment that is transferred by shear into the column is 40-14 = 26 ft-kips.

You would then take the 14 ft-kips and calculate a required amount of flexural reinforcing based on the 1.5h width on either side of the column. Let's suppose this narrow width is 36 inches and the total column strip width is 120 inches.

This gives you a required amount of rebar that must be concentrated in that narrow width (column width + (2 x 1.5h))= 36". This amount of rebar, let's call As(unb) = 0.80 sq. inches.

So in your column strip you would have a total amount of reinforcing = As(CS) = 6 sq. inches. Of this amount, you would concentrate As(unb)= 0.80 sq. inches within a width of 36" centered on the column. And the rest (6.0 - 0.8 = 5.2) would be distributed equally on either side of this 36" width but still within the 120 inch column strip width.

 

[DNBEA]

I agree with most of you said. but I believe we should evenly distribute the 5.2 in the whole 120in and add 0.8 to that 36" which make this 36" strip reinforced heavier than the rest of the strip. dont you agree?

 

[JAE]

DDWDS - that is not what the ACI example showed. They placed the .8 in the narrow strip at the column (36") and evenly distributed the 5.2 everywhere else in the remaining column strip (120-36 = 84").

But by adding additional steel as you suggest, I would say that its conservative, except you'd have to check the max. As allowed.

 

[JAE]

Sorry...you didn't suggest adding additional steel, just spreading it out over the full 120". I think that's OK, but the ultimate intent is to just make sure that there is a certain minimum steel near/at the column.

 

[DNBEA]

thank very much. how do you get the book of examples? what is the name of the book? where can I buy it? thanks

 

[rapt]

Jae

Normal logic is that the section over the column take its share of the averaged column strip moment plus the transferred moment. This is normally achieved by
-- calculating the moment required for transfer,
-- subtract it from the total
-- spread the remainder evenly over the full columns trip width
-- add the transfer moment reinforcement over the column head width.

In your example, your numbers give a very bad example (I know you were only guessing but), you have
.8in2 spread over 36" width = .26666 in2/ft width over the column head and
5.2 in2 spread over 84" width = .743 in2/ft width over the remainder of the column strip

The reinforcement over the column head should be more concentrated than the remainder of the column strip width.

 

[DNBEA]

I agree. the reason I asked the question to begin with is because I always design the slab with the larger moment of the two sides of the column. but 13.5.3 just does a fraction of the unbalanced moment which is obviously less than the total moment.

Also my guts tells me the porpuse of 13.5.3 is to make the width over the column more concentrated. if by doing 13.5.3 we end up with less bars over the column than simply evenly distributing 6 in2 in the whole strip just does not make sense.

Anyway thank you all for clearification

 

[JAE]

rapt - agree that my numbers weren't correctly proportional to the idea that MORE rebar is concentrated at the column than in the overall column strip.

 

[rapt]

DDWDS

Reading clause 13.5.3,
"A fraction of the unbalanced moment given by gfMu shall be considered to be transferred by flexure within an effective slab width between lines that are one and one-half slab or drop panel thicknesses (1.5h) outside opposite faces of the column or capital, where Mu is the moment to be transferred"

and commentary

"This section is concerned primarily with slab systems without beams. Tests and experience have shown that, unless special measures are taken to resist the torsional and hear stresses, all reinforcement resisting that part of the moment to be transferred to the column by flexure should be placed between lines that are one and one-half the slab or drop panel thickness, 1.5h, on each side of the column."

Nowhere does it say that ONLY the moment transferred to the columns is carried by the reinforcement over the width of the shear head. Instead that an amount to resist the transferred moment should be included in this width.

I still interpret this this clause to agree with what I said in my earlier post. The interpretation in the design guide as quoted by JAE is wrong in my opinion.

Also, 13.5.3.4 talks about "concentration of the reinforcement" in this width.

 

[JAE]

rapt - the design example is published by ACI - I would guess that they would be consistent with their own code.

 

[rapt]

JAE,

Not necessarily. The design examples are not done by the code committee as part of the code writing.

They are another engineers interpretation of the code.

 

[DNBEA]

JAE

I think maybe you want to double check the example and make sure you quote or understand it correctly. i agree with rapt. without seeing the example, as I said before, if by doing what you said the strip over column has less bars than not doing it, that will bring us back to my original question. why do we need to do it? To me it is just against common sence. Even the example book does do what you said, I still think it is wrong unless someone can give me a reasonable explaination. But thank you for your time anyway.

 

[WillisV]

The way I see it is as follows:

1. Determine the moment in the column strip and calculate how much reinforcing you need distributed evenly throughout the full column strip width.

2. Seperately determine the unbalanced moment and the portion of this moment which must be transferred by flexure and the required steel reinforcing for this moment which must be placed over the smaller "transfer width."

3. If the amount of reinforcing located within the smaller transfer width ALREADY provided by the reinforcing distributed evenly throughout the column strip in part 1 provides enough reinforcing to meet that required in part 2 then you are finished.

4. If more steel is required within the smaller zone than is provided as an even distribution throughout the column strip in part 1 you can either:

a) shift the spacing of the bars from part 1 around so that more are concentrated within the smaller transfer zone to meet part 2 and the remainder are now spaced a little farther apart within the rest of the column strip or

b) just add enough extra bars within the smaller transfer zone to make up the difference between what is required within that zone and what was provided through the even column strip distribution in step 1.

This methodology is consistant with that provided in:

MacGregor - Reinforce Concrete - 3rd Edition - see Example 13-7 specifically p627

PCA Notes on ACI 318-02 - see Example 20.1 specifically p20-29 - 20-30 as well as Example 19.1 specifically p19-21 - 19-22

 

[JAE]

WillisV - thanks for laying this out. This is what I think I was trying to communicate but you did a great job. I agree with your procedure and it also appears to agree with the ACI example.