Flat plate analysis

[HoshangMustafa]

Hi

I tried modeling Example 13.3 from Nilson's Design of Concrete Structures SI unit 14th edition textbook. Please find the attached link:

Panel cut A-A1 displays MYY Automatic direction Integral value of -87.93(kN*m/m)*m for ULS case component 5

Table 13.8 of the textbook displays 113kN*m.

Panel cut A-A4 displays MYY Automatic direction Integral value of -84.44(kN*m/m)*m for ULS case component 5

Table 13.8 of the textbook displays 75kN*m.

Panel cut A-A2 displays MYY Automatic direction Integral value of 274.94(kN*m/m)*m for ULS case component 4

Table 13.8 of the textbook displays 271kN*m.

Panel cut A-A3 displays MYY Automatic direction Integral value of 88.52(kN*m/m)*m for ULS case component 4

Table 13.8 of the textbook displays 90kN*m.

Panel cut A-A2 and A-A3 results seem comparable whereas results for panel cuts A-A1 and A-A4 don't seem comparable.

Any thoughts would be highly appreciated.

[Romanich]

Hi @HoshangMustafa ,

I'm not sure, but the mentioned example refers to the equivalent frame method and this method is not implemented in Robot. If you are willing to play with the verification then I would recommend to look at the NAFEMS test results:

https://www.autodesk.com/support/technical/article/caas/tsarticles/ts/551zPIHIzSZBhVZIfm00ru.html

[HoshangMustafa]

Hi @Romanich 

the mentioned example refers to the equivalent frame method.

Yes

this method is not implemented in Robot.

I know. But this is not the intent. The intent is comparisons and reliability. If one design with these low results of ARSA, as compared with the equivalent frame method, the design may not be safe.

If you are willing to play with the verification then I would recommend to look at the NAFEMS test results:

https://www.autodesk.com/support/technical/article/caas/tsarticles/ts/551zPIHIzSZBhVZIfm00ru.html

The links are not working.

Another inquiry:

Panel cut A-A2 displays QXX local direction Integral value of -324.31(kN/m)*m for ULS case component 4

Diagrams for bar 6 displays Fx=214.91kN at node 12. How these results are compared?

[Romanich]

I have attached some documents.

[HoshangMustafa]

Hi @Romanich 

You wrote:

I have attached some documents.

I followed the examples. No example for verification of equivalent frame method with ARSA.

Still waiting for answers on my queries in messages 1 &3.

Any help would be highly appreciated.

[HoshangMustafa]

Hi all,

126 views, no replies.

[HoshangMustafa]

Hi all,

Should one design based on lower ARSA results as compared with equivalent frame method?

[DonBAE]

are you comparing the section cut moments inclusive of Wood/Armer moment transforms or just looking at the Myy integral. The equivalent frame method is calibrated against the elastic solution including twisting moment so you need to make sure that you are comparing with something like Wood/Armer results so that the Mxy twisting moments are included.

[HoshangMustafa]

Hi @DonBAE 

As you can see, no change with Wood and Armer.

What about my Another inquiry in message 3?

[DonBAE]

I have the US customary unit version of the example and the loading pattern they describe is very convoluted I am only able to get results similar to the example by putting no loading in the A panels at all which is counter to the description and common sense as dead load must exist in that panel. 

I would compare the full strip positive and negative moments for the fully loaded case, which I get nearly exact agreement against.

[DonBAE]

Ok the example needs to be viewed in the context of an equivalent frame in that regard I believe the loading to correspond to their load case (b) is actually the 0.75*live load along the central span panels and full dead load in the adjacent spans. This produces a central span positive moment that is much closer to the values obtained in the example:

Using moment distribution without consideration of rigid offsets or the torsional member similar moments are obtained on the assumption that only the central span is fully loaded:

[HoshangMustafa]

Hi @DonBAE 

Can you post your rtd file in ARSA 2019 version?

[DonBAE]

sorry I'm not able to we only have ARSA 2024. You just need to change your loading so that the central strip of panels is loaded with the live load and the flanking panels are loaded with dead load only.

[HoshangMustafa]

Hi @DonBAE 

Thanks. Now it's clear. What about my inquiry in message 3?

[HoshangMustafa]

Hi @DonBAE 

I mean:

Diagrams for bar 6 displays Fx=214.91kN at node 12. How one can get this result? Isn't it the reduced shear in a panel cut?

[DonBAE]

If your trying to compare the axial force in the column to the integrated shear in a panel cut you really won't every get those to match because the column reaction is influenced by a tributary area in two-dimensions and the panel cut is at an infinitesimal slice in a single axis within the strain field of the panel.

[HoshangMustafa]

Hi @DonBAE 

I appreciate your help.

You wrote:

If your trying to compare the axial force in the column to the integrated shear in a panel cut you really won't every get those to match

So, how one can get column axial force from panel results in ARSA?

[DonBAE]

@HoshangMustafa 
"So, how one can get column axial force from panel results in ARSA?"

You can't get results for the column from the panel, that is why you model the bar element representing the column.

[HoshangMustafa]

Hi @DonBAE 

Let me clarify it further:

If I had a 2D frame with beams on columns, one can get axial load of the column from shear force of the beams, and bending moment of the columns from bending moment of the beams.

What if I had a 3D frame with slabs, beams, and columns. How one can get column axial forces?

[DonBAE]

@HoshangMustafa 

important edits to your statement:

If I had a 2D frame with beams on columns, one can get axial load of the column from difference in shear force of the adjacent beams, and bending moment of the columns from unbalanced bending moment of the adjacent beams.

What if I had a 3D frame with slabs, beams, and columns. How on can get column axial forces? 

With just beams and columns the same method above applied in each direction. When a slab is introduced you need to rely on the column bar element to get the axial force, again that is why you need to model the column element.

A fundamental thing you need to understand is that the plate elements representing the slab and the bar elements representing beams/columns are not the same. Each plate element typically only satisfies deformation compatibility between connected nodes the approximating functions for shear and moment diverge from the true mathematical solution and that is where things like nodal averaging come into play.