Integrity Steel Interpretation

[EngDM]

Hey all,

I'm hoping I can get some thoughts on how to interpret clause 13.10.6.1 of CSA A23.3. It has sum(Asb)=2*Vse/fy.

The symbols glossary out front of the code has Asb being "the minimum area of bottom reinforcement crossing one face of the periphery of a column and connecting the slab to the column or supported to provide structural integrity". So I interpret the sum(Asb) to be the summation of all four faces of my column.

Is this to say, that if I count each shear plane of each integrity bar to make up Asb? So in the image below, I have a total of 9-30M bars (3 vertically, and 6 horizontally). Would my Asb then be 9*2*700 = 12600mm^2, since each of the 9 bars has 2 shear planes?



Reason I am asking this is because in a textbook I have "Reinforced Concrete Design - A practical approach by Brzev and Pao" their design example calculates a required Asb of 1235mm^2, and then they go on to say they require 8-15M bars, which if taken as a single shear plane is 1600mm^2, but if taken as double shear plane as I've done in my example above, it would be 3200, in which case they would only need to have provided 4-15M bars (2 horizontal 2 vertical).

 

[rapt]

The sentence in the code above the equation explains it. The sum on all faces of the column. You should ask for your money back on the text book. ACI code works differently and might result in something like the textbook.

 

[Retrograde]

Would my Asb then be 9*2*700 = 12600mm^2, since each of the 9 bars has 2 shear planes?

The Australian Code has similar provisions and I have always taken it this way - counting the 2 shear planes for each bar.

 

[W460x68]

Reason I am asking this is because in a textbook I have "Reinforced Concrete Design - A practical approach by Brzev and Pao" their design example calculates a required Asb of 1235mm^2, and then they go on to say they require 8-15M bars, which if taken as a single shear plane is 1600mm^2, but if taken as double shear plane as I've done in my example above, it would be 3200, in which case they would only need to have provided 4-15M bars (2 horizontal 2 vertical).

I have the same textbook 3rd edition... check out Figure 12.38. It illustrates 4-15M bars (that is, 2-15M bars each way continuous over the column) providing 8*Ab or 1600 mm2.

 

[EngDM]

I have the same textbook 3rd edition... check out Figure 12.38. It illustrates 4-15M bars (that is, 2-15M bars each way continuous over the column) providing 8*Ab or 1600 mm2.

I guess the example on page 636 is just incorrect then.

 

[W460x68]

I guess the example on page 636 is just incorrect then.

Either that or the example intended to provide 8 discrete integrity bars over the column, similar to the first diagram in Figure 12.38, instead of continuous bars like the second diagram. Regardless, I agree that it isn't clear. A diagram in the example showing the integrity bar layout would've made it clear.