IDA, Analysis Terminated

[anantparghi]

Hi there;

I am running IDA for the FRP-confined RC bridge columns. The model was running fine for cyclic analysis and results were matched with experimental results, but when I run IDA with same model, its run few minutes and say unable to apply the next load step. Analysis terminated.I guess, there is something wrong in mass or modelling, or may be the convergence problem, but I am not able to figure it out.

Anybody could please help me. Please have a look in the model.

https://drive.google.com/open?id=0B3tbc ... zI0bFBKZ3M
https://drive.google.com/open?id=0B3tbc ... zNHZnNVeEU

kind regards.
Anant

[z.gronti]

Dear anantparghi,

Unfortunately i cannot view your model through the provided links.

[anantparghi]

Hi Zoi;

Thank you very much for your reply. Please see the following link
https://app.box.com/s/8gzy4kaupw4pto2zmmmnke1wrselp8wm
https://app.box.com/s/qof92laoi9lpeo58r95oyjm62bmcdi33

I hope this link would works. I am trying hard to run IDA, but the analysis getting terminated after few mins. Could you please help what is wrong in the model. Your help would greatly appreciated.
with kind regards
Anant

[z.gronti]

Dear anantparghi,

I checked one of your models (C1_NF1-3-Layer-CFRP-Col-IDA_27th Jan-2016.spf) and realized that your analysis stopped due to the defined projects settings. If you apply the Program defaults in the Project Settings module the IDA analysis will be completed successfully.

[anantparghi]

Dear Zoi;

Thank you very much for your help and advise. Now the models running perfectly alright. But, I have a question, how do we provide the rotational mass? I used the same value of lumped mass Mx, My, and Mz in element class properties. But, I don't know what is the meaning of rotational mass here, such as, Mxx, Myy, and Mzz. However, I do calculated these rotational mass using Gere and Timoshenko Mechanics of Materials Book equation Mr^2/2 (where r is the radius of circular column, and m is the mass). Your help and advise would greatly appreciated.
Thanks
Anant

[z.gronti]

Dear Anant,

I think that the available figure in the "Lumped mass" topic of SeismoStruct's Help System will help you to understand the meaning of the rotational masses.

[huffte]

Zoi is correct, Anant. As the figure shows, these are local rotary inertias about the three primary axes. It is a common assumption that these are ignored (set = 0) for the analysis of some structure types. For example, the rotation about any horizontal axis of a floor slab in a building is very, very small compared to the translation and, hence, rotary inertia effects are often ignored in such cases. However, if torsional effects are significant to the extent that rotation of the floor about a vertical axis is possible, that effect should probably be included. Note that this only applies to a 3D analysis. If we assume that the plan dimensions are large compared to the thickness, then the rotary inertia of a rectangular floor is (m/12)(a^2+b^2), where m is the floor mass, a and b are the plan dimensions. But note that this is the rotary inertia of the entire floor slab and modelers frequently break up the floor mass and distribute it to the various nodes. It's a good question and the correct answer depends on the modeling details, Anant. Best of luck.

[anantparghi]

Dear Zoi and Huffte;

Thank you very much for your explanatory answer of my query. Actually, I referred lot of literature, SeismoStruct forum questions, and refer the well known text book Mechanics of Materials. Even, I asked many friends in the University, but did not satisfied the answer. I put the mass Mx, My, and Mz and ignored the rotational mass Mxx, Myy and Mzz, but I like to confirm and get more opinion that I am ignoring the rotational mas, and my assumption is correct, since I am analyzing single column.
kind regards,
Anant

[huffte]

Well, let's take an example, anantparghi. Suppose I have a rectangular floor slab, a long and b wide. As stated above, the exact rotary inertia about a vertical axis is (m/12)(a^2 + b^2). If, on the other hand, I beak that deck mass up into a 5x5 grid, and distribute mass to the grid nodes, then, even with local rotary inertias equal to zero, you will find that the global rotary of the inertia of the 25 discrete masses is equal to (m/12.8)(a^2 + b^2). The rotary inertia effect would still be captured and quite close to the exact value, even with nodal rotary inertia set to zero. This would be an example of a case where setting nodal rotary inertia values equal to zero. I tried to download your file from the above link, but it has been removed. Regardless, the point is that there are many cases where setting the nodal rotary inertia to zero is appropriate. You have to look at each cases individually. Hope this helps, anantparghi.