Great explanation with example. Thanks Dr. Wagner

Great work! Thank you a lot for sharing your knowledge.

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.

@hellosushant9471

2 жыл бұрын

waiting

@hnrwagner

2 жыл бұрын

@@hellosushant9471 at least one :)

@Jade-jv6eq

Жыл бұрын

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?

@user-nh9fb8wb4s

Жыл бұрын

@@hnrwagner i have same question with Jade (above comment)

@nabilishes

Жыл бұрын

Do you have any link to the ABAQUS Course: Modeling Fracture and Failure with Abaqus powerpoints/handouts?

Awesome content. Thanks for sharing your knowledge. Your videos help me a lot.

@hnrwagner

2 жыл бұрын

Good to hear, thanks

Thanks you for the knowledge

really awesome, I am shocked by so many useful videos

@hnrwagner

Жыл бұрын

thanks

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 .

@hnrwagner

Жыл бұрын

Never used fesafe unfortunately

thnak you Dr. Wagner, you're a good help to us

@hnrwagner

Жыл бұрын

thanks

Thank you very much!

@hnrwagner

2 жыл бұрын

a pleasure for me :)

Thank you very much for this helpful video. I hope you can make tutorial about adhesive element .

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 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 so much 😊 Could you explain how the model recognizes the fracture should occur along the maximum shear stress direction (45 Deg)?

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

@hnrwagner

2 жыл бұрын

I will add a video with more information regarding the Characteristic length definition as the gauge length of a tensile specimen in the future

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?

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!

Nice presentation, have you done any high cycle fatigue analysis in ABAQUS. Thank you in advance

@hnrwagner

2 жыл бұрын

No haven't done it yet, do you have a reference, than i may take a look at it

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 ?

@hnrwagner

2 жыл бұрын

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 :)

Amazing tutorial, I have a question on why in the fracture energy equation we have a 2 factor ?

@hnrwagner

Жыл бұрын

Its the area of a triangle and you calculate this using length times witdh divided by 2

Great video! But I have a question to 3:00. You say that you can set the Stress Triaxiality and the Strain Rate to zero unless you have more than one Fracture strain. Why is that, or where in the Abaqus Manual is that explained, because normally people set the Stress Triaxiality in uniaxial tests to 1/3?

@dzej113

9 ай бұрын

I think it's just that if you are giving only one row of information for the failure (initiation) strain, there is no point in allocating stress triaxiality or strain rate values associated with it because there is no extrapolation/interpolation to be done no matter what values of stress triaxiality or strain rate is calculated internally in Abaqus... If you have experimental data available for failure strain obtained from different tests (test conditions resulting in different stress triaxiality or strain rate values), then it makes sense to put in more rows of information, and then, if the current simulation parameters (internally calculated strain rate and stress triaxiality values by Abaqus solver) are somewhere in between the values of your input parameters, Abaqus performs interpolation/extrapolation of this data..

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?

@hnrwagner

11 ай бұрын

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

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

2 жыл бұрын

I will upload a long video on this topic, this will clarify everything

@umitjhukov6662

2 жыл бұрын

@@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

@dzej113

9 ай бұрын

@@umitjhukov6662 have you enabled the "STATUS" variable in the field output request? Abaqus won't delete elements if the STATUS variable is inactive.

Thank you for the useful video. I was wondering how to calculate the displacement at failure, do you have a tutorial about it?

@hnrwagner

Жыл бұрын

at 2:20 it is shown in the video...

@arumys9303

Жыл бұрын

@@hnrwagner Yes, I mean an excel tutorial to calculate the displacement at failure🙏

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 ?

@hnrwagner

Жыл бұрын

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)

@user-nh9fb8wb4s

Жыл бұрын

@@hnrwagner thanks

@user-nh9fb8wb4s

Жыл бұрын

@@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?

@hnrwagner

Жыл бұрын

@@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.

@hnrwagner

2 жыл бұрын

this value is only needed if you have more than 1 fracture strain, alone it has no value

@nimash4098

2 жыл бұрын

@@hnrwagner Thank you so much

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.

@hnrwagner

10 ай бұрын

Yes it's from quasi static test

@user-vl5hi2bs7v

10 ай бұрын

@@hnrwagner Thank you sir! Always thanks for your videos!

@user-vl5hi2bs7v

10 ай бұрын

​​@@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

10 ай бұрын

@@user-vl5hi2bs7v the shape of the stress Strain curve is slightly different for a round specimen compared to a flat specimen

@user-vl5hi2bs7v

10 ай бұрын

@@hnrwagner Thanks for your response sir! It really helps!

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.

@hnrwagner

Жыл бұрын

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

@jarmohavula6587

6 ай бұрын

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.

Hi. i want to know can use this method for all of Ductile material or there is exception ?

@hnrwagner

11 ай бұрын

Cant say,you should Read the abaqus manual, it works for metals okish,dont know What you want to model

Can the same fracture energy be used for smooth tensile specimen and other different notch round bar (NBR) specimen of the same material???

@hnrwagner

10 ай бұрын

No the energy has to be calculated for each specimen geometry

@steveokocha9860

10 ай бұрын

@@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 ?

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.

@TheSh2589

9 ай бұрын

Also I confused that as I know, length L to get damage variables, L is characteristic length of the element.

@hnrwagner

9 ай бұрын

the video was done all according to official ABAQUS lecture, there true strain was used and so on@@TheSh2589

@TheSh2589

9 ай бұрын

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

How can I get the number 14621, sir?

@hnrwagner

Жыл бұрын

You get it from curve fitting with excel

What is the Load Value used in ABAQUS CAE is it Force or Displacement ?

@hnrwagner

Жыл бұрын

Displacement = 10 mm

@akshaydengwani3916

Жыл бұрын

@@hnrwagner Danke Professor

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

@hnrwagner

Ай бұрын

its a typo,thanks

Do you have a book where I can find this theory?

@hnrwagner

4 ай бұрын

no, its based on an official abaqus course

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

Жыл бұрын

Displacement control of 10 mm and no mass scaling

@suryanto5125

Жыл бұрын

@@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?

@akshaydengwani3916

Жыл бұрын

@@hnrwagner Even my model is not failing I used displacement control of 10mm using amplitude (0,0) and (1,1)

@hnrwagner

Жыл бұрын

@@akshaydengwani3916 have you the same model as in the video or a different? (material, geometry) when in doubt use 100 mm

@akshaydengwani3916

Жыл бұрын

@@hnrwagner Yes Sir I have replicated everything same. I will try with higher value once.

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?

@hnrwagner

Жыл бұрын

In this case i would assume fracture energy = 0

@jasonf8080

Жыл бұрын

@@hnrwagner If so, where do we input the fracture toughness (KIC) data of the material being used?

@hnrwagner

Жыл бұрын

@@jasonf8080 there is no fracture toughness in ABAQUS for damage for ductile metals only fracture energy, see at 3:19

@jasonf8080

Жыл бұрын

@@hnrwagner Alright, thank you very much

In the damage parameters, the failure strain should be the effective plastic strain. I think you input 0.073 incorrectly!

@hnrwagner

10 ай бұрын

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

@user-nr1qm8ie3h

10 ай бұрын

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

Rapture or Rupture?

@hnrwagner

3 ай бұрын

Rupture :P

I need to know the properties of aluminum 2024

@hnrwagner

2 жыл бұрын

which properties exactly?

@kikinano5830

2 жыл бұрын

@@hnrwagner strain rate triaxiality stress and displacement

@hnrwagner

2 жыл бұрын

@@kikinano5830 ok and have you already checked Google scholar or only Google? That what i would do

From 1:08 to 2:30 you are drawing Displacement vs Stress but you are labeling the Displacement as Plastic Displacement. Why?

@hnrwagner

3 ай бұрын

its defined like this in the official abaqus regarding damage models when i am not mistaken

@mariogalindoq

3 ай бұрын

@@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.

@hnrwagner

3 ай бұрын

@@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

@mariogalindoq

3 ай бұрын

@@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.