After countless hours of research and trial & error I have now finanlly created and validated a working ABAQUS model for a tensile test of a steel specimen with damage evolution. The official ABAQUS documentation is unfortunatly so bad... Is that by design so that I buy the official abaqus course ? There are countless threads on the internet on how to determine the basic input properties for the damage for ductile metals material model - ductile damage. To the best of my knowledge all of them are either wrong, partially wrong or incomplete. Nobody seems to know all the correct details... UNTIL NOW ! One of the main points everyone gets wrong is the Parameter L which is according to the abaqus manual, the characteristic element length. However, I have read the official ABAQUS Course: Modeling Fracture and Failure with Abaqus and here L is the gauge length of the tensile test specimen which belongs to the stress-strain input data ! Also I was not sure about the definition of the fracture strain, because there are many confusing threads on the internet and nobody seems to know "how ABAQUS defines the fracture strain". The fracture strain is the corresponding strain at damage initiation and not the rapture strain of failure strain of the test. I will upload a longer version of this video in the coming days which shows every step in an excel sheet.
hi Dr. Ing. Ronald Wgner, thank you for sharing which deepen my knowledge. Would you mind sharing 《Modeling Fracture and Failure with Abaqus》, You demonstrate 3 methods to calculate characteristic element length in your video, one is element length, one is IVOL^(1/3), another one is the gauge length of the tensile test specimen. they are quite different and confusing, which one is correct?
Thank you very much for this enlightening video. Great work! Regarding the parameter L, the ABAQUS course refers again that "The characteristic length L is computed automatically by Abaqus based on element geometry." and "The damage evolution law can be specified either in terms of fracture energy (per unit area) or in terms of the equivalent plastic displacement and that Both approaches take into account the characteristic length of the element." While somewhere in the beginning the characteristic length is correlated to the ASTM standards provided lengths. That still causes some confusion. It would be very helpful if you could include some additional information regarding the Characteristic length definition as the gauge length of a tensile specimen
Thanks for the video Dr. Wagner. I want to ask a question, why did not you leave the Max Degradation value at default (which is 1) and changed it to 0.95 instead ?
This was a recommendation from an abaqus expert, the last 5% may lead to unnecessary long computation time without extra benefit in accuracy, in my example it makes actually barely a difference. I just want to share some tips and tricks from other professionals :)
Thank you for the information Dr Wagner, it really helps doing my experiments. I have one question. Is stress-strain curve you used in the video obtained from quasi-static condition? Thanks a lot Dr.
@@hnrwagnerDoctor, I apologize for the late question, but I have one more question. Is it necessary to use a round specimen for the ductile damage experiment? I am curious whether it is affected by rolling direction when using a flat specimen. Thank you always for your kind answers.
@@hnrwagner thanks very much Dr. Last question: if that's the case. Is the fracture strain (strain at ultimate stress) attained from smooth tensile test also used for NRB specimens. ? Or would it be different ?
Sir thank you for this informative video about ductile material. It's very helpful for me as I am working on ductile zones . I have a question sir if I will use a rectangular bar (maybe 100*20 scale) then what is the gauge length for my modelling? Could you please explain how to get plastic displacement?
Thank you very much for this amazing video, I could really use some help in determination of Damage variable and plastic displacement, In the sake of finding the damage variable do you use the stress values that co-respond to the same plastic displacement value? how do you determine the offset value for the elastic functions(slope*strain+offset=stress)? Thank you for your help Have a nice day
@@hnrwagner hello again sir thank you for your response, Even tho I was able to solve my issues myself the abaqus is giving errors and acting in a relatively weird way. I assign a max degradation of 0.95 just like you but abaqus does not delete the mesh at 0.95 criteria instead criteria goes up to thousands and analysis gets aborted due to extensive disorientation How could I resolve this issue Best regards Good day
Hi,could I ask some questions?I remember you has a video calculating fracture energy by characteristic length.Could I know how could decide to use the specific way to calculate JC damage evolution fracture energy? And actually for my personal project, the UTS is 128MPa and fracture point is about 126 MPa (strain difference is 0.005).The calculated value seems only 8% of the energy value which I type in Abaqus and think the simulation result graph is close to experimental true stress strain graph.I am very confused about calculating the fracture energy so is it possible to get more suggestions?Here thank you very much for you videos.It makes me understanding corressponding Abaqus manual meaning.Thank you very much!
Thank you for your garatful help to understand Abaqus analysis. I would like to have a question. Is the fracture engergy in this video equivalent with the energy release ratio which can be define from fracture toughness based on material property ?
Thank you for your time and effort. However, it made a big question to me. In some other FE simulations, such as CZM, damage parameters are introduced to be depend to the mesh length. It means that whenever you refine the mesh size your damage evolution parameters should change. Have you ever seen them? or what do you think?
They depend on the mesh BUT internally, so abaqus calculates the values itself, in this model you do not need to adjust the input parameters to the mesh
I learned it that way in the official abacus course, it should be correct. Results also match well with the test. What should the value be in your opinion
that a recommendation from the abaqus course, because the slope of the function you need to determine the plastic displacement is zero, it may cause problems in the analysis (very long computation time)
@@hnrwagner Dr. Wagner, May I ask one more question? can i apply your method to define fracture energy / displacement at failure values in traction separation laws - maxps - damage evolution ? is it just for ductile damage?
Thank you so much for the great tutorial videos. Just a question. Why do you put 0 for stress triaxility? Your model is uniaxial tension loading. I think stress triaxiality should be 0.33 for this case.
I'm wondering if it's still valid to compute true stress/true strain like you shown in Slide 4 in the necking section of the stress-strain curve. Usually, people estimate the true stress-strain behavior in necking region as an extended line to the true rupture strain. It's kind of important as it's related to your plastic displacement calculation later on. thank you.
cant say for sure, the ABAQUS user manual gives no information regarding area after necking. However, as the simulation results is quite close to the test, this approach seems to be accurate enough
It seems that the function used to calculate the true stress does not apply after necking starts (volume is not constant after that, etc.). So. I agree that the true stress-strain curve should continue as a straight line. Actually, after necking starts, the cross section change should be measured/ evaluated with a proper method, like DIC, and then the true stress should be calculated using the measured cross section area.
Thank you for your great lecture, But I have one question, when you get stress-displacement curve. why do you use true strain rather than engineering strain? because I think that using engineering strain is actual displacement by multiplying gauge length.
I understand but I found the related contents in Abaqus documentation, Abaqus analysis user's guide 24.1.1, 24.2.1 and 24.2.3 , they define L as element characteristic length
Thanks for the explanation Sir. I have any question, a. In the simulation you use load control/displacement control/velocity control? And what value did you enter Sir? b. Also do you use mass scaling Sir?
@@hnrwagner Okay Sir. I used a step time of up to 2400 in my project with a displacement control of 10. However, my specimen has not broken Sir. Do you know how to determine the correct step time and displacement control values Sir?
@@hnrwagner sorry but something is wrong. During the linear elastic part of the deformation you don't have Plastic Displacement. Your graphic is saying the opposit.
@@mariogalindoq you are right, I just checked the documents again, it should be "total displacement u" only the last part after damage initiation is defined as plastic displacement
@@hnrwagner I understand that the Plastic Displacement starts just after the yield point at the end of the linear elastic deformation. However, for the damage evolution it is necessary to subtract the Plastic Displacement defined at the damage initiation point. Therefore, for the Damage computation, you should use Up-Up0 being Up the current Plastic Displacement and Up0 the Plastic Displacement at the damage initiation point (normally the Plastic Displacement corresponding to the maximum stress) only when Up>Up0. Do you agree? Thank you for your answer. I found good videos in your channel.
Thank you very much. I would like to ask, is there a formula for rough estimation of fracture energy, in case where the stress-strain curve is not available and only yield stress and ultimate stress data is available?
After countless hours of research and trial & error I have now finanlly created and validated a working ABAQUS model for a tensile test of a steel specimen with damage evolution.
The official ABAQUS documentation is unfortunatly so bad... Is that by design so that I buy the official abaqus course ? There are countless threads on the internet on how to determine the basic input properties for the damage for ductile metals material model - ductile damage. To the best of my knowledge all of them are either wrong, partially wrong or incomplete.
Nobody seems to know all the correct details... UNTIL NOW !
One of the main points everyone gets wrong is the Parameter L which is according to the abaqus manual, the characteristic element length. However, I have read the official ABAQUS Course: Modeling Fracture and Failure with Abaqus
and here L is the gauge length of the tensile test specimen which belongs to the stress-strain input data ! Also I was not sure about the definition of the fracture strain, because there are many confusing threads on the internet and nobody seems to know "how ABAQUS defines the fracture strain". The fracture strain is the corresponding strain at damage initiation and not the rapture strain of failure strain of the test.
I will upload a longer version of this video in the coming days which shows every step in an excel sheet.
waiting
@@hellosushant9471 at least one :)
hi Dr. Ing. Ronald Wgner, thank you for sharing which deepen my knowledge. Would you mind sharing 《Modeling Fracture and Failure with Abaqus》, You demonstrate 3 methods to calculate characteristic element length in your video, one is element length, one is IVOL^(1/3), another one is the gauge length of the tensile test specimen. they are quite different and confusing, which one is correct?
@@hnrwagner i have same question with Jade (above comment)
Do you have any link to the ABAQUS Course: Modeling Fracture and Failure with Abaqus powerpoints/handouts?
Great explanation Dr. Wagner. Could you show your "steps" configurations? Did you set Non-lineartity on?
#abaqus #metal #hnrwagner #ductiledamage
Reference for Thumbnail Figures:
www.researchgate.net/publication/304186947_Effects_of_Workshop_Fabrication_Processes_on_the_Deformation_Capacity_of_S960_Ultra-high_Strength_Steel
Slides of the presentation:
www.researchgate.net/publication/360927265_The_ultimate_quick_guide_to_damage_for_ductile_metals_-_ductile_damage_for_ABAQUS_CAE
Github Link:
github.com/hnrwagner/UMAT_Lecture_1
Google Scholar:
scholar.google.de/citations?user=a4sKEKsAAAAJ&hl=en
Researchgate:
www.researchgate.net/profile/Ronald-Wagner
Timecodes:
0:00 - Intro
0:14 - Engineering Stress-Strain Curve
0:23 - True Stress-True Strain Curve
0:29 - E-Modulus, Yield and Ultimate Stress
0:38 - True Stress-Plastic Strain Curve
0:48 - Fracture Strain
1:05 - True Stress-Displacement Curve
1:56 - Determine Damage Variable and Plastic Displacement
2:21 - Damage Evolution - Tabular Plastic Displacement
2:24 - Damage Evolution - Linear Plastic Displacement
2:29 - NO Damage Evolution - Plastic Displacement = 0
2:34 - Fracture Energy
2:44 - ABAQUS model geometry
2:48 - Material data input for ABAQUS
3:26 - Reference Point - Rigid Body Constraint
3:36 - Boundary Conditions
3:48 - Field Output, STATUS Variable
3:56 - Element Deletion Settings
4:06 - How to plot stress-strain curve ?
4:31 - True Stress-True Strain - no damage evolution -Test/ABAQUS
4:51 - ABAQUS - Tensile test - no damage evolution
5:11 - True Stress-True Strain - linear damage evolution -Test/ABAQUS
5:18 - ABAQUS - Tensile test - linear damage evolution
5:45 - True Stress-True Strain - tabular damage evolution -Test/ABAQUS
5:52 - ABAQUS - Tensile test - tabular damage evolution
6:06 - Comparison of all models with test data
6:30 - How to improve the damage evolution ?
7:36 - Final results
7:51 - References and Ending
Thank you very much Dr Wagner for your time. You have explained damage in details with different test cases. Would like to see some video on Fesafe .
Never used fesafe unfortunately
Great explanation with example. Thanks Dr. Wagner
Thank you very much for this enlightening video. Great work!
Regarding the parameter L, the ABAQUS course refers again that "The characteristic length L is computed automatically by Abaqus based on element geometry." and "The damage evolution law can be specified either in terms of fracture energy (per unit area) or in terms of the
equivalent plastic displacement and that Both approaches take into account the characteristic length of the element."
While somewhere in the beginning the characteristic length is correlated to the ASTM standards provided lengths.
That still causes some confusion. It would be very helpful if you could include some additional information regarding the Characteristic length definition as the gauge length of a tensile specimen
I will add a video with more information regarding the Characteristic length definition as the gauge length of a tensile specimen in the future
really awesome, I am shocked by so many useful videos
thanks
Awesome content. Thanks for sharing your knowledge. Your videos help me a lot.
Good to hear, thanks
Great work! Thank you a lot for sharing your knowledge.
Thanks you for the knowledge
Nice presentation, have you done any high cycle fatigue analysis in ABAQUS. Thank you in advance
No haven't done it yet, do you have a reference, than i may take a look at it
thnak you Dr. Wagner, you're a good help to us
thanks
Thank you very much!
a pleasure for me :)
Thank you very much for this helpful video. I hope you can make tutorial about adhesive element .
Thanks for the video Dr. Wagner. I want to ask a question, why did not you leave the Max Degradation value at default (which is 1) and changed it to 0.95 instead ?
This was a recommendation from an abaqus expert, the last 5% may lead to unnecessary long computation time without extra benefit in accuracy, in my example it makes actually barely a difference. I just want to share some tips and tricks from other professionals :)
Thank you for the information Dr Wagner, it really helps doing my experiments. I have one question. Is stress-strain curve you used in the video obtained from quasi-static condition? Thanks a lot Dr.
Yes it's from quasi static test
@@hnrwagner Thank you sir! Always thanks for your videos!
@@hnrwagnerDoctor, I apologize for the late question, but I have one more question. Is it necessary to use a round specimen for the ductile damage experiment? I am curious whether it is affected by rolling direction when using a flat specimen. Thank you always for your kind answers.
@@user-vl5hi2bs7v the shape of the stress Strain curve is slightly different for a round specimen compared to a flat specimen
@@hnrwagner Thanks for your response sir! It really helps!
Can the same fracture energy be used for smooth tensile specimen and other different notch round bar (NBR) specimen of the same material???
No the energy has to be calculated for each specimen geometry
@@hnrwagner thanks very much Dr.
Last question: if that's the case. Is the fracture strain (strain at ultimate stress) attained from smooth tensile test also used for NRB specimens. ? Or would it be different ?
Sir thank you for this informative video about ductile material. It's very helpful for me as I am working on ductile zones . I have a question sir if I will use a rectangular bar (maybe 100*20 scale) then what is the gauge length for my modelling? Could you please explain how to get plastic displacement?
Thank you very much for this amazing video,
I could really use some help in determination of Damage variable and plastic displacement,
In the sake of finding the damage variable do you use the stress values that co-respond to the same plastic displacement value?
how do you determine the offset value for the elastic functions(slope*strain+offset=stress)?
Thank you for your help
Have a nice day
I will upload a long video on this topic, this will clarify everything
@@hnrwagner hello again sir thank you for your response,
Even tho I was able to solve my issues myself the abaqus is giving errors and acting in a relatively weird way.
I assign a max degradation of 0.95 just like you but abaqus does not delete the mesh at 0.95 criteria instead criteria goes up to thousands and analysis gets aborted due to extensive disorientation
How could I resolve this issue
Best regards
Good day
@@umitjhukov6662 have you enabled the "STATUS" variable in the field output request? Abaqus won't delete elements if the STATUS variable is inactive.
Hi,could I ask some questions?I remember you has a video calculating fracture energy by characteristic length.Could I know how could decide to use the specific way to calculate JC damage evolution fracture energy? And actually for my personal project, the UTS is 128MPa and fracture point is about 126 MPa (strain difference is 0.005).The calculated value seems only 8% of the energy value which I type in Abaqus and think the simulation result graph is close to experimental true stress strain graph.I am very confused about calculating the fracture energy so is it possible to get more suggestions?Here thank you very much for you videos.It makes me understanding corressponding Abaqus manual meaning.Thank you very much!
Thank you so much 😊
Could you explain how the model recognizes the fracture should occur along the maximum shear stress direction (45 Deg)?
Thank you for your garatful help to understand Abaqus analysis. I would like to have a question. Is the fracture engergy in this video equivalent with the energy release ratio which can be define from fracture toughness based on material property ?
slide @ time 2.11 shows stress = 273MPa on graph but in damage computation you use stress = 237MPa ? Is it a typing mistake or is there another reason
its a typo,thanks
Thank you for your time and effort. However, it made a big question to me. In some other FE simulations, such as CZM, damage parameters are introduced to be depend to the mesh length. It means that whenever you refine the mesh size your damage evolution parameters should change. Have you ever seen them? or what do you think?
They depend on the mesh BUT internally, so abaqus calculates the values itself, in this model you do not need to adjust the input parameters to the mesh
In the damage parameters, the failure strain should be the effective plastic strain. I think you input 0.073 incorrectly!
I learned it that way in the official abacus course, it should be correct. Results also match well with the test. What should the value be in your opinion
I think we should subtract the elastic strain, don't you think? After all, there is no damage or plastic strain during the elastic stage@@hnrwagner
thank you for this great video.
i have question why did you calculate u_f at damage variable not d=1 but d=0.99 ?
that a recommendation from the abaqus course, because the slope of the function you need to determine the plastic displacement is zero, it may cause problems in the analysis (very long computation time)
@@hnrwagner thanks
@@hnrwagner Dr. Wagner, May I ask one more question? can i apply your method to define fracture energy / displacement at failure values in traction separation laws - maxps - damage evolution ?
is it just for ductile damage?
@@user-nh9fb8wb4s i do not know, i will check
Thank you so much for the great tutorial videos.
Just a question. Why do you put 0 for stress triaxility?
Your model is uniaxial tension loading. I think stress triaxiality should be 0.33 for this case.
this value is only needed if you have more than 1 fracture strain, alone it has no value
@@hnrwagner Thank you so much
Amazing tutorial, I have a question on why in the fracture energy equation we have a 2 factor ?
Its the area of a triangle and you calculate this using length times witdh divided by 2
Thank you for the useful video. I was wondering how to calculate the displacement at failure, do you have a tutorial about it?
at 2:20 it is shown in the video...
@@hnrwagner Yes, I mean an excel tutorial to calculate the displacement at failure🙏
I'm wondering if it's still valid to compute true stress/true strain like you shown in Slide 4 in the necking section of the stress-strain curve. Usually, people estimate the true stress-strain behavior in necking region as an extended line to the true rupture strain. It's kind of important as it's related to your plastic displacement calculation later on. thank you.
cant say for sure, the ABAQUS user manual gives no information regarding area after necking. However, as the simulation results is quite close to the test, this approach seems to be accurate enough
It seems that the function used to calculate the true stress does not apply after necking starts (volume is not constant after that, etc.). So. I agree that the true stress-strain curve should continue as a straight line. Actually, after necking starts, the cross section change should be measured/ evaluated with a proper method, like DIC, and then the true stress should be calculated using the measured cross section area.
Thank you for your great lecture, But I have one question, when you get stress-displacement curve. why do you use true strain rather than engineering strain? because I think that using engineering strain is actual displacement by multiplying gauge length.
Also I confused that as I know, length L to get damage variables, L is characteristic length of the element.
the video was done all according to official ABAQUS lecture, there true strain was used and so on@@TheSh2589
I understand but I found the related contents in Abaqus documentation, Abaqus analysis user's guide 24.1.1, 24.2.1 and 24.2.3 , they define L as element characteristic length
Thanks for the explanation Sir. I have any question,
a. In the simulation you use load control/displacement control/velocity control? And what value did you enter Sir?
b. Also do you use mass scaling Sir?
Displacement control of 10 mm and no mass scaling
@@hnrwagner Okay Sir.
I used a step time of up to 2400 in my project with a displacement control of 10. However, my specimen has not broken Sir. Do you know how to determine the correct step time and displacement control values Sir?
@@hnrwagner Even my model is not failing I used displacement control of 10mm using amplitude (0,0) and (1,1)
@@akshaydengwani3916 have you the same model as in the video or a different? (material, geometry)
when in doubt use 100 mm
@@hnrwagner Yes Sir I have replicated everything same. I will try with higher value once.
Do you have a book where I can find this theory?
no, its based on an official abaqus course
Hi. i want to know can use this method for all of Ductile material or there is exception ?
Cant say,you should Read the abaqus manual, it works for metals okish,dont know What you want to model
From 1:08 to 2:30 you are drawing Displacement vs Stress but you are labeling the Displacement as Plastic Displacement. Why?
its defined like this in the official abaqus regarding damage models when i am not mistaken
@@hnrwagner sorry but something is wrong. During the linear elastic part of the deformation you don't have Plastic Displacement. Your graphic is saying the opposit.
@@mariogalindoq you are right, I just checked the documents again, it should be "total displacement u" only the last part after damage initiation is defined as plastic displacement
@@hnrwagner I understand that the Plastic Displacement starts just after the yield point at the end of the linear elastic deformation. However, for the damage evolution it is necessary to subtract the Plastic Displacement defined at the damage initiation point. Therefore, for the Damage computation, you should use Up-Up0 being Up the current Plastic Displacement and Up0 the Plastic Displacement at the damage initiation point (normally the Plastic Displacement corresponding to the maximum stress) only when Up>Up0. Do you agree? Thank you for your answer. I found good videos in your channel.
What is the Load Value used in ABAQUS CAE is it Force or Displacement ?
Displacement = 10 mm
@@hnrwagner Danke Professor
How can I get the number 14621, sir?
You get it from curve fitting with excel
Rapture or Rupture?
Rupture :P
Thank you very much. I would like to ask, is there a formula for rough estimation of fracture energy, in case where the stress-strain curve is not available and only yield stress and ultimate stress data is available?
In this case i would assume fracture energy = 0
@@hnrwagner If so, where do we input the fracture toughness (KIC) data of the material being used?
@@jasonf8080 there is no fracture toughness in ABAQUS for damage for ductile metals only fracture energy, see at 3:19
@@hnrwagner Alright, thank you very much
I need to know the properties of aluminum 2024
which properties exactly?
@@hnrwagner strain rate triaxiality stress and displacement
@@kikinano5830 ok and have you already checked Google scholar or only Google? That what i would do