Thank you for warm feedback :)

Thank you for the feedback man

Thank you for the reply. In the future I plan to make a video on aluminium column modelling. Regards!

Pradipta Biswas , thank you for the nice feedback

Thank you for the kind feedback!

I'm really glad it was helpfull

Expecting more videos from u :)

I appreciate it

It is the magnitude of the imperfection. Since I modeled in millimeters, it means the initial imperfection is 28 millimeters. File=job-1 and step=1 means we are taking imperfection shape from the Job-1 file after loading step 1. Regards!

Thank you my man

Load -> Predefined fields -> Stress Good Luck

thanks a lot very help full

@@SalahM07 You're welcome bro

To simulate the fork boundary conditions as close as possible.

@@ivancudina93 thanks for the quick reply, and yes it makes sense. I thought there was some trick)

It is difficult to say with fem analyses. However, since I conducted parametric analysis later on it would be too much time consuming to make model with true stress strain curve so I did not try. Also, my results were fairly good with bilinear stress strain curve.

I already answered that one Load -> Predefined fields -> Stress Good Luck

After nonlinear analysis is complete you can easily do it. Abaqus has a tool that allows user to obtain stresses from arbitrary point. I recommend using "von Mises" rule for stress calculation. Regards!

@@ivancudina93 buckling stresses and von mises stresses are different I guess..

@@mariajames6737 If you are referring to elastic buckling stresses (which is a theoretical value) than Abaqus does not have a function to determine such value. But you can calculate it using common rules.

I'm sorry, but I wouldn't like to answer that because I am not entirely sure about the answer. I have never studied buckling resistance of steel plates exposed to shear loading.

Hello! Unfortunately, the scope of my research does not cover masonry structures, so I can not assist you in that matter. Regards!

I used the procedure given in EC1993-1-1 (general method). I can not give you that procedure because it's copywritten. I do have to say I chose buckling curve 'd' if I'm not mistaken. Regards!

Ivan Cudina thank you very much for your response you’re the best 👍🏻.

First eigenvalue is the minimum load (or if you wanna be accurate, the smallest number that multiplies load applied to the beam), that would be required to cause buckling of a beam if the behaviour of the beam was ideally elastic. Which means that strength-strain curve of the steel would have to be completely linear and there would not be initial bow of the beam nor any residual stresses. That is only possible in theory that is why eigenvalues have purely theoretical meaning. Similarly, second eigenvalue is the minimum load required to cause the buckling of a beam in the shape of the second eigenmode. Since the first eigenvalue is always lower than the second, beam always buckles in the shape of the first eigenmode. Second eigenmode also have purely theoretical meaning.

Thanks a lot for the explanation!

It is possible. Finite element modeling is a very complex procedure and engineers often encounter illogical results. That is why numerical modeling without experimental background has very little meaning. I personly never investigated such phenomena but I would be interested in your research if you would publish it.

I can't say exactly without additional info. Try playing with the boundary conditions and loading to see if it will help. If not, send me the model, and if I find some time, I'll try to find the error.

In which analysis? Eigenvalue or Riks?

Can you please leave your email? So we can continue there.

Carolina, I am sorry but I don't have enough expertise in that field to give you a reliable answer. Best of luck on your research. Regards!

Yes. For this particular specimen modelled it is the ultimate load.

@@ivancudina93 I tried to do this but my LPF gets bigger than 1.0 - how is this possible?

I have used pure elastic material for your information

@@lw7474 It is completely fine to have LPF>1,0. It means that the resistance of your member is greater than the load you applied in the model. Regards!

@@ivancudina93 but I applied the load in the size of the first eigenvalue. So it shouldn't become bigger than 1.0 I guess...

Load -> Predefined fields -> Stress

I got the mail in previous Comments.

ivancudina93@gmail.com

Thanks Sir

Ivan Cudina i request you to upload a video for calculate shear center using abaqus,i really need that help. i also mail about the same on the e-mail given by you. Thanku in advance....

I will send some info on your email

I took mine from Eurocode 3, table 5.1. But bare in mind it is not real geometrical but the equvivalent geometrical imperfection which takes into account all the imperfections and is a bit conservative. Real initial bow is much smaller and it can be aproximately taken as L/1000

Realistically, the size of geometrical imperfection is equal to L/1000. So, we can include that kind of geometrical imperfection into our model, but then we need to include residual stresses as well. However it is possible to model equivalent geometrical imperfection (larger then the realistic one) which will also include residual stresses. That is a simplified version of modeling nonlinear behaviour of the beam which I used in the tutorial. According to European codes (Eurocode), the size of the equivalent geometrical imperfection should be L/100 for channel sections. In our case, that is 28 mm because the length of the beam is 2800 mm. So we needed to multiply geometrical imperfection by 28 to get 28 mm.

Ivan Cudina Thank you Ivan for replying....I got the point about imperfections... What if we don't add imperfections in our model ?? Another Thing I have want to ask is , how did you decide inputs for Rik's method, that is estimated arc length, maximum increment size etc...

If you do not add imperfections in your model, then you would get much higher value of buckling resistance. But eventually, the beam would buckle even if there are no imperfections included. You can simply try that but assigning the value of imperfection equal to zero. And now the second part of your question. I just took some recommended values for arc length and increment etc. They worked fine so I did not go very deeply into that matter. If you want to understand that topic better, you should do some research on arc length method. Cheers

Ivan Cudina Thank you Ivan...

Because the size of imperfection in any eigenmode is always set equal to unity by Abaqus. I wanted to apply the size of imperfection equal to 28 mm (span/100) because it is the value suggested by the Eurocodes. However, I believe now it is too much on the conservative side. Regards!

@@ivancudina93 Great! thanks!

Yes, LPF multiplied by the applied force respresents the ultimate buckling resistance of the beam. If you are asking why the displacement does not occur more suddenly, that is because channel sections as uniaxially symmetrical cross sections (unlike biaxially symmetrical I or H sections) behave differently. Not all elements buckle suddenly under load. For example, plates tend to show postbuckling resistance. Which means that after they buckle, you can continue to apply force of even greater value before they completely fail. Cheers

Thanks Ivan! So how do you know about completely fail from observing the curve or graph?

The failure usually happens after peak value in LPF curve. There are few exceptions, but not in our case

There is a way to do everything in Abaqus. Please send me more info to ivancudina93@gmail.com

Thank you very much. I just emailed you!

Just partition the model in such a way that you can place shell element on wanted position. In case od midspan, partition the model in midspan and add the shell element there. Alternativly you can try with reference points.

@@ivancudina93 Sir, am not that much good in Abaqus. Can you please make a 5 min short video of a simply supported beam subjected to point load. Please, don't mind.

Check the Abaqus keywords manual. I'll put the link. dsk.ippt.pan.pl/docs/abaqus/v6.13/books/key/default.htm

@@ivancudina93 Thank you very much.

Realistically, the size of geometrical imperfection is equal to L/1000. So, we can include that kind of geometrical imperfection into our model, but then we need to include residual stresses as well. However it is possible to model equivalent geometrical imperfection (larger then the realistic one) which will also include residual stresses. That is a simplified version of modeling nonlinear behaviour of the beam which I used in the tutorial. According to European codes (Eurocode), the size of the equivalent geometrical imperfection should be L/100 for channel sections. In our case, that is 28 mm because the length of the beam is 2800 mm. So we needed to multiply geometrical imperfection by 28 to get 28 mm.

Because I took the size of geometrical imperfection to be Length/100. Since the length of the beam is 2800 mm, that equals to 28 mm. That's why I have to multiply imperfection size with 28, otherwise it would remain 1 mm. The actual geometrical imperfection is much smaller, but this is simplified way of taking into account residual stresses.

Ivan Cudina thanks for your intereset...

I think you messed up keywords. Click on "Discard all edits" and type them again. If you still get an error, just do the whole model from scratch.

OK I WILL TRY

but the same error is still

Because you have not write .fil file (in that example you must write in model-1 keywords: *Node file U

Hello there! At around 3:40 you can see a table taken from Eurocode 3 which states that the size of the imperfection for this kind of cross section and plastic nonlinear analysis should be Length/100 which equals to 28 mm. Regarding the keywords in the tutorial, I suggest you do exactly as I showed because abaqus is very sensitive regarding errors when editing keywords. You can find much more info on keywords in Abaqus manual if you are interested. Finally, I don't plan to make any tutorials in the near future but it is very simple to plot stress-strain curves. However, it is not difficult and I am sure you figure it out on your own. Unfortunately I do not have Abaqus on my computer currently so I can not write exact steps. Regards! Edit: Nonlinearities used in the tutorial do not apply for the rebars!!

@@ivancudina93 Thank you for your reply Ivan! I followed your suggestion in the keywords and my model is finally working! And just to let you know I have figured out the stress strain curve. Thanks so much again your video helped my dissertation a lot!

@@antheaellis17 Thank you for your kind feedback. I am very glad my tutorial helped you!

I explained it in the tutorial 6:35

Give me your mail. I will send you the answer there

mrt.gund@outlook.com

I will put below a link for the paper where you can see the formula and its factors explained. That formula is valid only for doubly symmetric cross sections but I believe it can be used for channel sections if they are loaded centrically, that is through the shear center. eurocodes.jrc.ec.europa.eu/doc/WS2008/SN003a-EN-EU.pdf

Load -> Predefined fields -> Stress Good Luck

Thank you are you have a tutorial about using UMAT ?

No, I don't

I also had problems with reference points. Try to introduce shell element on midspan od the beam and apply load there like I showed in the tutorial.

@@ivancudina93 Sir, I resolved the problem of reference points. Finally it worked, Bam. Thanks.

I don't remember making Job-5 in my tutorial. I think easiest solution for you is to go through tutorial again and follow each step more carefully.

Did you solve the problem? If yes can you share how?

Leave everything the same but instead of "1, 28.0" write "1, 4.0".

Hi, thanks very much, an other question that, is lateral constraint of upper flange necessary in buckling and riks analysis of a pfrp I beam with 4 points loads or no?

@@omidalirostamzade3966 You can play with boundary conditions to achieve best results. There is no universal rule

Thanks a lot.

The size of the force does not have to be 1N. But usually, in eigenvalue analysis, the force of 1N is used. Let's presume you get eigenvalue of 150 for example. That means that critical force is 150N. If you use the force of 30N for example, then you will get eigenvalue of 5 (because 5x30=150) which is a bit confusing. My email is ivancudina93@gmail.com I'll answer you if I find the time.

I am getteing 1.38 eigen value and Mcr as 58kN why is that..tho i followed ur step one by one!!1

That means that your model is showing linear behaviour. It can be number of reasons. It could be that you are not applying enough load for the nonliear behaviour to manifest itself. It is also possible that the size of the imperfection in your model is not large enough. Or you failed to include any imperfections in your model.

@@ivancudina93 I have included 10% imperfection in my model and I am applying load as obtained in the first eigenmode. In the whole analysis, the load is increasing it is not decreasing and in the end, the analysis is aborted due to the convergence of increment.

What does 10% imperfection mean? What kind of model are you working on?

I don't know. That is strange. It should converge in both cases. As long as the load case is the same, the results should also be the same. The only thing that is changing in regards to load value is the LPF. If you assign for example 10 kN and get LPF = 0.8, then for the same load, but with the value of 1 kN, you should get LPF=8.0. So probably you have completely different load cases. But I can't say is it correct or not because you did not give me enough data. If you could send more data to my e-mail about your problem, maybe I could help.

I don't know. That is strange. It should converge in both cases. As long as the load case is the same, the results should also be the same. The only thing that is changing in regards to load value is the LPF. If you assign for example 10 kN and get LPF = 0.8, then for the same load, but with the value of 1 kN, you should get LPF=8.0. So probably you have completely different load cases. But I can't say is it correct or not because you did not give me enough data. If you could send more data to my e-mail about your problem, maybe I could help.

Yes of course, I would appreciate your help! Could you tell me please your email?

Hello Raja I'm really happy that you find my tutorial helpful It's really impressive that you use abaqus on undergraduate level I need more info to solve your problem, so send it on my mail This is my email: ivancudina93@gmail.com In the meantime, check this tutorial for workbench. It will show you very well how to include initial imperfection kzfaq.info/get/bejne/l6idobeW0ZrRo5c.html But honestly, I prefer Abaqus ;)

Unfortunatelly I don't have time now for making any more videos

Ivan Cudina please do I'm also stuck in modeling a plate girder for failure in bending only (elasto-plastic analysis). Should i go through this same process and I'm using a uniformly distributed load instead of point load. Should my udl be 1lb/in as you used a point load of 1N to get an eigenvalue

You can use 1lb/in if you want, but you don't have to. You can use any value in eigenvalue analysis. With plate girders I would recommend using shell elements. I don't know exactly what is the scope of your research but try reading this paper on modeling of web crippling (www.steel-stainless.org/media/1117/25_bock_real_arragay.pdf )

Ivan Cudina I'm using shell. Please is there a way to reach you that way i can explain better. Email perhaps. I'm in dire need of help

Akaolisa Okafor ivancudina93@gmail.com

Unfortunately, I don't have it anymore.

@@ivancudina93 Ok, thanks so much, sir

Son Tung Vy Eigenvalue is easy You have analitical equation that you can use, or some free simple software like LTBeam. For the design moment you use GMNIA analisis (like I showed in tutorial). There are no design rules to get the design moment for channel section yet. Actually, there are some proposed design rules that I can send you

Thank you very much, However, Can you tell me how to calculate the critical moment (Mcr) and momen resistance (Mn) of a bended beam in Abaqus. Because I want to determine for other type of section (in LTbeam does not have this type of section).

Son Tung Vy The eigenvalue gives you the information about critical moment Also, in LTBeam you have option Section by properties It works for any tipe of cross section. To get the design moment you need to include the nonlinearities in your model like I showed in the tutorial, 2nd model (nonlinear stress strain curve of the steel and imperfection) If this doesn't help, send more info on my mail (in description)

Dear Ivan Cudina, I am currently building up a model to determine the LTB critical moment for glass beam and i faced the same problem of determine the critical moment. I had set the model as 3D solid the same as yours therefore cannot obtain Mcr direct from the model. I though the eignenvalue value will allow you the determine the critical load only. How can we determine the Mcr from Critical load? or if there anyway can obtain it from the model?

oh using moment equation such as L x Pcr = Mcr directly will give me the answer?