Sections with FRP's

[huffte]

It seems that your strategy should work reasonably well zsaeed.

Use two different elements to define the column, one with a confinement factor based on spirals alone and a second with FRP effect included.

Best of luck.

[zsaeed]

Thanks sir but i want to know that for the lower half portion where confinement due to spiral and CFRP wrapping exist i should add their confinement in simple algebraic fashion or have to follow some rule for it ??

thanks

[fakharifar.mostafa]

Huffte's reply is completely correct. Your question however is not SStruct related, but more of earthquake engineering topic. BTW, just to add some hints for you:

1- Confining pressure of 2.45 is really high! Make sure you calculate it again to get the right value. Modern reinforced concrete columns with seismic detailing fall in the range of 1.2 to 1.4, or maybe 1.8.

2- The implemented confining concrete models in SStruct are not additive, meaning that the FRP confined concrete model in SStruct includes the confining pressure only from FRP jacket. This methodology works well for substandard columns with low transverse reinforcement ratio, where the confining pressure from existing spirals is not substantial. This should work for you, if you are not using SStruct for Reinforced Concrete Filled FRP Tubes.

However, if you are trying to model a column which is really high in transverse reinforcement ratio, you may need to read the below paper to include the confining effect of spiral+CFRP. Confining pressure is not simply algebraically addition of two numbers.

CONSTITUTIVE MODEL FOR FRP-AND-STEEL-CONFINED CIRCULAR CONCRETE COLUMNS IN COMPRESSION.

Two papers:

Hu, H., & Seracino, R. (2013). Analytical Model for FRP-and-Steel-Confined Circular Concrete Columns in Compression. Journal of Composites for Construction, 18(3).

http://www.iifc-hq.org/proceedings/CICE ... NFINED.pdf

Hope this helps.

[huffte]

The points from fakharifar.mostafa are excellent. The stated confinement factor is extremely high. And the second reference should help with modeling the lower element for the combination of FRP and spirals.

[zsaeed]

Thanks Sir for the reference papers. i tried to check the value of confinement factor calculated and attached a pic of it below ..it was coming out to be 2.4 which seems that i am making some big mistake in it
Please find in the following link for snap shots of calculation and respective section

https://drive.google.com/file/d/0B7rekW ... sp=sharing

https://drive.google.com/file/d/0B7rekW ... sp=sharing

I will go through the paper and will discuss with you if i come with some more question .

[huffte]

From your screenshots, I believe that you have mistakenly take a 3mm bar to be equivalent to a #3 bar. This is not the case. A 3mm reinforcement is more like a wire (0.118 inches diameter), while a #3 bar has a diameter of 0.375 inches.

Also the specified longitudinal steel area is extremely small. It appears that you have input the area of a single 7.37mm bar instead of the area of 26 7.37mm bars.

[huffte]

Note that since there is no 3mm bar in the SeismoStruct pull-down box, you will likely need to calculate by hand the confinement factor. Either calculate by hand or refer to the original Mander model theory to come up with an equivalent bar size and spacing using a bar size available in SeismoStruct.

[zsaeed]

quote:Originally posted by huffte

From your screenshots, I believe that you have mistakenly take a 3mm bar to be equivalent to a #3 bar. This is not the case. A 3mm reinforcement is more like a wire (0.118 inches diameter), while a #3 bar has a diameter of 0.375 inches.

Also the specified longitudinal steel area is extremely small. It appears that you have input the area of a single 7.37mm bar instead of the area of 26 7.37mm bars.

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Although the spiral was of 3mm dia but i couldn't find any option for such small area in that dialogue box..the smallest bar available there is #3 bar
As far as longitudinal steel area , you are right i thought we have to enter area of 1 bar instead of 26 bars..
Is there some way to specify 3mm bar for confinement bar ???

[huffte]

As I say, you will need to go back to the equations in the Mander model and either calculate the confinement factor by hand - it isn't very difficult - or, again from the model equations, find a bar size and spacing which gives the same confinement factor as does your 3mm bar at 1.5 inches.

A web search on 'Mander model for confined concrete' should get you enough information on the equations used to calculate the confinement factor.

Best of luck.

[zsaeed]

quote:Originally posted by huffte

From your screenshots, I believe that you have mistakenly take a 3mm bar to be equivalent to a #3 bar. This is not the case. A 3mm reinforcement is more like a wire (0.118 inches diameter), while a #3 bar has a diameter of 0.375 inches.

Also the specified longitudinal steel area is extremely small. It appears that you have input the area of a single 7.37mm bar instead of the area of 26 7.37mm bars.


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Yes Sir i was wrog in putting value of Longitudinal Steel area ...i put area of one bar instead of 26..Thanks