Ahmed Elkady really looks like you are an engineer with solid knowledge in both the mechanics of materials and the Abaqus software. I really appreciate the knowledge you share. I am convinced that in some way: life, karma (destiny), God, etc; It will return in blessings the good will you have to teach and share what you know. I send you a big hug from Peru (this is a way to say an affectionate farewell when you say goodbye to someone). Sorry if the translation is not good, I used a translator.
Thank you professor for you beautiful and informative lectures on Abaqus and FEM. I have learned a lot with solid theoretical background from your lectures.
Please note that the proper way to compute the plastic strain is to subtract sigma@current strain/E from the total strain. This is actually the method used in the EXCEL file provided in the link in the video description. However, sometimes for simplicity, you can just subtract sigma_yeild/E; the difference is generally minor for steel materials with relatively large E.
Good explanation. I've an observation about what you said when you defined the engineering strain: L0 is the length of the specimen's uniform section named "gauge length". If you consider this uniform section, between the fillets, you have an uniaxial stress state in each point. So the engineering-stress strain curve is correctly defined if and only if we consider these hypotesis: - uniaxial stress state; - small displacement; - volume conservation; Other curves: true stress-strain from elongation and true stress-strain from section, are defined by the violation of the cited hypotesis.
I've got a question. It seems in the Abaqus documentation that the plastic strain must be evaluated not only on the basis of the initial yield stress. You always remove SigmaYield/E whereas this should be updated as SigmaTrue/E at each SigmaTrue level according to the doc. Do you have any point of view on that? This is a difference of understanding of the method between several of my colleagues.
Correct, the proper way to compute the plastic strain is to subtract sigma@current strain/E from the total strain. This is actually the method used in the EXCEL file provided in the link in the video description. However, sometimes for simplicity, you can just subtract sigma_yeild/E; the difference is generally minor for steel materails that have relatively large E.
Dear Mr. Elkady, I was wondering how to get the data of plasticity if we use the temperature-dependent data? I am confused on how to pick the data. Because the strain curve for each temperature differs. Thank you in advance and looking forward for the answer
Thermally-dependant plasticity can be defined through the Johnson-cook mode. There are many publications online describing the model parameters calibration and definition.
Sir I'm doing CFST column steel beam joint under cyclic loading. My model is running with displacement at the beam end But it is showing convergence error when it is subjected to cyclic loading at beam end Will it be because of any error in material property given? Can u please share ur view on the topic Please help
Hello, I don't quite understand why your plastic strain is minus a constant? Shouldn't the elastic strain be changing all the time?Different stresses correspond to different elastic deformations
Correct, the elastic strain value will change with stress. This is how the plastic strain is dedcued in the referenced Excel sheet on GitHub. Sometimes however, it can be acceptable to just subtract the yield strain for simplicity.
Hello sir I want to ask you simple question.. I finished my simulation in abaqus of tensile test now i put a script in python to allow me to change value of young's modulus 10 times si i've now 10 inp and odb files so all what I want is how can I call abaqus to read this odb files automatically to exctract the graph of stress strain or if there is a simple way to do this
Thank you sir, great video and explanation. Just one question: I think the plastic deformation is not difference between total deformation and elastic deformation for yield stress but it is difference between total deformation and elastic component of deformation for the current stress. In video time 16:33 equation "Epsylon_pl = Epsylon_true - Epsylon_y" should be "Epsylon_pl = Epsylon_true - Epsylon_el" => "Epsylon_pl = Epsylon_true - Sigma_true/E". Am I right? If not please explain.
Yes, the proper way to compute the plastic strain is to subtract sigma@current strain/E from the total strain. This is actually the method used in the EXCEL file provided in the link in the video description. However, sometimes for simplicity, you can just subtract sigma_yeild/E; the difference is generally minor for steel materials with relatively large E.
What if we only have engineering stress and not true stress at fracture? In such case we can not calculate the elastic component at the fracture as we lack true stress and ultimately can not calculate plastic strain at material failure. What do we do then? Any suggestions will be appreciated. Thank you.
Thanks for the great explanation. I have a question if you don't mind. If I'm validating an experiment using ABAQUS, the author didn't provide a stress-strain curve he just mentioned the yield and ultimate stresses and their corresponding strain as a result a tensile coupon test. Should I assume that the mentioned points are the true stress-strain points ? And put them directly to ABAQUS?
Commonly, researchers report the engineering values not the true ones; so I would assume its the former. One more way to check is by comparing the reported ultimate stress with the expected value that is reported in the design standards for the same material grade (which is also an engineering one).
Oh no. I just realized that I've made simulations with engineering stress-strain graphs so far. Oops... Thanks for enlightening me on this mistake of mine.
Great video thanks. I have a question if you do not mind. I see that true fracture stress is above the ultimate stress. However, in abaqus documentation, the stress is lower than the ultimate value due to damage accumulation. 130.149.89.49:2080/v6.14/books/usb/default.htm?startat=pt05ch23s02abm18.html (Figure 24.2.3-1) Assuming that I am comparing with physical test, which one is correct? I am bit confused here.
The true stress of metalic materials such as steel increases with increasing strain. The figure you are refering to is concerened with the phenomenological ductile damage model which is a material-property-add-on that degrades the material's undamaged stress (discussed in this tutorial) to simulate strain softening behaviour leading to fracture.
In case, if we use a strain gauge attached to a coupon sample for measuring the strains directly ,,does this strain is considered as Engineering or true strain ???
Thank's, Dr. Ahmed please can you give me answer for my question ? When analyzing metal fatigue, should we prioritize the first plastic strain or the equivalent plastic strain? How might this choice impact the accuracy and efficiency of our results?
Thank you sir. But I have promblem. I am working on V bending 2D plain strain. When I define true stress- true strain graphs all points for plasticty in the material definition, the results of the springback are less than the experiment. However, if I define only the line where the plastic strain is 0 (the first line (yield point)) as you describe in the video 19:46 , the spring-back estimation gets much closer to the experimental results. Is it a correct method to define only the line where the plastic strain is 0? Do you have any other suggestions to increase springback in terms of material? I should point out that; the material is copper and I am using stress strain curves from other articles. Because the experiment I'm analyzing is not mine.
You should model the material's elasto-plastic behaviour to match the material uniaxial tensile test (true stress-strain curve). Perhaps a bilinear elastic-perfectly plastic model matches your material the best.
Hi, your video was awesome. If i plot in "plastic table" only 1 data like your example of MIN 19:40 I will design in elastic domain right? Please can you help me with this doubt.
No, in that case, you have defined an elastic perfectly-plastic material. No need to define a plastic behaviour if you are conducting an elastic analysis.
Yes, it should be true stress-strain. But note that it probably does not have significant impact for rebar; actually, many just model rebar using elastic-perfectly-plastic material.
Ahmed Elkady really looks like you are an engineer with solid knowledge in both the mechanics of materials and the Abaqus software.
I really appreciate the knowledge you share. I am convinced that in some way: life, karma (destiny), God, etc; It will return in blessings the good will you have to teach and share what you know.
I send you a big hug from Peru (this is a way to say an affectionate farewell when you say goodbye to someone).
Sorry if the translation is not good, I used a translator.
Thank you professor for you beautiful and informative lectures on Abaqus and FEM. I have learned a lot with solid theoretical background from your lectures.
Please note that the proper way to compute the plastic strain is to subtract sigma@current strain/E from the total strain. This is actually the method used in the EXCEL file provided in the link in the video description. However, sometimes for simplicity, you can just subtract sigma_yeild/E; the difference is generally minor for steel materials with relatively large E.
Good explanation. I've an observation about what you said when you defined the engineering strain: L0 is the length of the specimen's uniform section named "gauge length". If you consider this uniform section, between the fillets, you have an uniaxial stress state in each point. So the engineering-stress strain curve is correctly defined if and only if we consider these hypotesis:
- uniaxial stress state;
- small displacement;
- volume conservation;
Other curves: true stress-strain from elongation and true stress-strain from section, are defined by the violation of the cited hypotesis.
This tutorial is handy. Keep up the excellent work sir.
thank you for this wonderful lesson,nice explanation sir ...waiting for future vedios..thank you
I've got a question. It seems in the Abaqus documentation that the plastic strain must be evaluated not only on the basis of the initial yield stress. You always remove SigmaYield/E whereas this should be updated as SigmaTrue/E at each SigmaTrue level according to the doc. Do you have any point of view on that? This is a difference of understanding of the method between several of my colleagues.
Correct, the proper way to compute the plastic strain is to subtract sigma@current strain/E from the total strain. This is actually the method used in the EXCEL file provided in the link in the video description. However, sometimes for simplicity, you can just subtract sigma_yeild/E; the difference is generally minor for steel materails that have relatively large E.
@@AElkady thank you for your answer Ahmed
I was having the same confusion. It is clear now. Thank you all.
hi can you please reply, i want to input stress strain curve in abaqus cdp how to do it i am doing project on cfst columns
Thanks a lot for the contents and everything.
Can I use the excel file for extrusion(CEC) plastic property?
Dear Mr. Elkady, I was wondering how to get the data of plasticity if we use the temperature-dependent data? I am confused on how to pick the data. Because the strain curve for each temperature differs. Thank you in advance and looking forward for the answer
Thermally-dependant plasticity can be defined through the Johnson-cook mode. There are many publications online describing the model parameters calibration and definition.
Really helpful thanks!
I am using bilinear elasto plastic curve..I am getting stresses are above ultimate in some area using Abaqus...why kindly explain
This could happen as a result of stress concentration (depends on element type, mesh size, geometry and applied loads)
Sir how to break the tensile sample with dissimilar welding components
Sir
I'm doing CFST column steel beam joint under cyclic loading.
My model is running with displacement at the beam end
But it is showing convergence error when it is subjected to cyclic loading at beam end
Will it be because of any error in material property given?
Can u please share ur view on the topic
Please help
Hello, I don't quite understand why your plastic strain is minus a constant? Shouldn't the elastic strain be changing all the time?Different stresses correspond to different elastic deformations
Correct, the elastic strain value will change with stress. This is how the plastic strain is dedcued in the referenced Excel sheet on GitHub. Sometimes however, it can be acceptable to just subtract the yield strain for simplicity.
@@AElkady Thank you for your response
can you please doctor
provide a vedio explaining modeling posttensioned beam on abaqus
Can we generate the stress-strain data with ultimate stress/crushing stress of concrete only?
I'm bachelor's senior civil engineering student and find how to do a FE modelling for project. Thank you for making videos.^^!
you are awsome. great video. Thank you.
Thanks for sharing sir!!!
What is the density of the material used ?
What's the material name and can we use him in stamping ?
Hello, great video, thanks for sharing. I can't download the excel file. Can you share again please?
he file is available to download on github github.com/amaelkady/ABAQUS-Tools
@@AElkady Ok, thanks.
Hello sir I want to ask you simple question.. I finished my simulation in abaqus of tensile test now i put a script in python to allow me to change value of young's modulus 10 times si i've now 10 inp and odb files so all what I want is how can I call abaqus to read this odb files automatically to exctract the graph of stress strain or if there is a simple way to do this
you can record a marco and process the odb file as usual and then check/modify the python commands in the generated journal file.
Thank you sir, great video and explanation. Just one question:
I think the plastic deformation is not difference between total deformation and elastic deformation for yield stress but it is difference between total deformation and elastic component of deformation for the current stress. In video time 16:33 equation "Epsylon_pl = Epsylon_true - Epsylon_y" should be "Epsylon_pl = Epsylon_true - Epsylon_el" => "Epsylon_pl = Epsylon_true - Sigma_true/E". Am I right? If not please explain.
Yes, the proper way to compute the plastic strain is to subtract sigma@current strain/E from the total strain. This is actually the method used in the EXCEL file provided in the link in the video description. However, sometimes for simplicity, you can just subtract sigma_yeild/E; the difference is generally minor for steel materials with relatively large E.
What if we only have engineering stress and not true stress at fracture? In such case we can not calculate the elastic component at the fracture as we lack true stress and ultimately can not calculate plastic strain at material failure. What do we do then? Any suggestions will be appreciated. Thank you.
Thank you very much!
Thank u Dr.
Thanks for the great explanation.
I have a question if you don't mind. If I'm validating an experiment using ABAQUS, the author didn't provide a stress-strain curve he just mentioned the yield and ultimate stresses and their corresponding strain as a result a tensile coupon test. Should I assume that the mentioned points are the true stress-strain points ? And put them directly to ABAQUS?
Commonly, researchers report the engineering values not the true ones; so I would assume its the former. One more way to check is by comparing the reported ultimate stress with the expected value that is reported in the design standards for the same material grade (which is also an engineering one).
@@AElkady Thanks Dr.Ahmed. Very helpful.
@@AElkady Sir for Abaqus simulation of dampers which data we have to use ,"true or engineering?
Thanks a lot
Oh no. I just realized that I've made simulations with engineering stress-strain graphs so far. Oops... Thanks for enlightening me on this mistake of mine.
Great video thanks. I have a question if you do not mind. I see that true fracture stress is above the ultimate stress. However, in abaqus documentation, the stress is lower than the ultimate value due to damage accumulation.
130.149.89.49:2080/v6.14/books/usb/default.htm?startat=pt05ch23s02abm18.html (Figure 24.2.3-1)
Assuming that I am comparing with physical test, which one is correct? I am bit confused here.
The true stress of metalic materials such as steel increases with increasing strain. The figure you are refering to is concerened with the phenomenological ductile damage model which is a material-property-add-on that degrades the material's undamaged stress (discussed in this tutorial) to simulate strain softening behaviour leading to fracture.
Hello Dr. Ahmed.
I have few queries on Abaqus modelling. Please let me know how can I be able contact you.
thanks a lot for eliminating my confuse when set material property in abaqus.👍
Can you please show in a video how to apply fire load (convection and radiation) on concrete
In case, if we use a strain gauge attached to a coupon sample for measuring the strains directly ,,does this strain is considered as Engineering or true strain ???
strain gauges/extensometers measure strain based on a fixed gauge length (i.e., engineering)
Thanks for sharing!!!!!!! Love from HK!
Thank's, Dr. Ahmed
please can you give me answer for my question ?
When analyzing metal fatigue, should we prioritize the first plastic strain or the equivalent plastic strain? How might this choice impact the accuracy and efficiency of our results?
Your Video help me so much in my Abaqus project, Great job Dr. Ahmed
Where do the numbers esilon engineering and sigma engineering come from?
Thank you sir. But I have promblem. I am working on V bending 2D plain strain.
When I define true stress- true strain graphs all points for plasticty in the material definition, the results of the springback are less than the experiment. However, if I define only the line where the plastic strain is 0 (the first line (yield point)) as you describe in the video 19:46 , the spring-back estimation gets much closer to the experimental results. Is it a correct method to define only the line where the plastic strain is 0? Do you have any other suggestions to increase springback in terms of material? I should point out that; the material is copper and I am using stress strain curves from other articles. Because the experiment I'm analyzing is not mine.
You should model the material's elasto-plastic behaviour to match the material uniaxial tensile test (true stress-strain curve). Perhaps a bilinear elastic-perfectly plastic model matches your material the best.
Hi, your video was awesome.
If i plot in "plastic table" only 1 data like your example of MIN 19:40 I will design in elastic domain right? Please can you help me with this doubt.
No, in that case, you have defined an elastic perfectly-plastic material. No need to define a plastic behaviour if you are conducting an elastic analysis.
Hello! Mr.Ahmed
How can i calculate energy absorption from load displacement curve
By integrating the area under the F-D curve.
For a rc frane analysis i used engineering stress strain curve of steel. Should i change it then to true stress? Oh my god..
Yes, it should be true stress-strain. But note that it probably does not have significant impact for rebar; actually, many just model rebar using elastic-perfectly-plastic material.
Thank u very much!
Exceellent tutorial
very helpful sir.
Awesome! Keep going
Dear Mr. Elkady, do you have the values of engineering stress-strain curve for material S235 maybe, or do you know where can I find it?
You can find several curves reported in litrature. Here is one: www.sciencedirect.com/science/article/pii/S0143974X16305181
@@AElkady Yes, but I can't find the results from some tensile test with the point coordinates in detail as in your example in excel.
The reference above include an easy empirical equation that can be used to generate the stress-strain data.
@@AElkady Thank you very much for help.