Good job. It has been demonstrated in several articles that triaxiality factor, which is correlated to the first deviatoric invariant, and the normalized Lode angle, which is correlated to the second and third deviatoric invariant, are the main parameters which characterize the initiation of ductile damage. The study of fracture propagation is more complex than initiation. There are different models based on Continuum Damage Mechanics or experimental mechanics. A very accurate prediction model for damage initiation in tensile test is the Bao-Wierzbicki Damage Model. It’s very simple to write a subroutine with this model to use with FEM. I suggest to you to try it.
Thanks for suggestions dear. I'll appreciate if you share paper/pdf on the same. (gkn0504@gmail.com) Currently I'm working on different topic but I'll surely explore it.
Nice video sir. As per this theory, if we opt for a coarser mesh, the material will undergo sudden failure. But if the mesh size if finer, then the then the material will undergo a gradual failure. Because the ultimate deformation depends only on the strain and characteristic length. In that, strain is a constant and characteristic length is varying as per the mesh size.. So, same material can have dissimilar stress strain response with mesh refinement which is little confusing..please correct me if I am wrong..
Thank you, sir! Brilliant explanation! So basically, the element deletion highly depends on the characteristic length of the element, yes? It is far preferable to have consistent SIZE of elements in the region of damage-prone (e.g., hex elements) because they provide a better shape and size if we mesh them strategically. More than that, we need to find the smallest element in the meshing part and calculate its characteristic length (must consider what type of element order too). And if we choose the biggest element with characteristic length as input, then it would be troublesome for small elements because they would be instantly deleted and not accurate for the simulation. According to what you've presented, Upl=characteristic length* plastic strain is essentially a failure displacement at the local level or the failure of a single smallest element. When those conditions are met (e.g., 0.15875), any element in the part will be automatically deleted by Abaqus. And also, it would be nice if you could share with us how to calculate the characteristic length for 3D elements (hex and tetra shape). Thank you again for this knowledge.
Yes. Consistency in mesh size is recommended. Yes actually we don't know where crack will propagate...so we consider calculation with smallest size. For 3D element characteristic length of element calculated by considering its volume...there is simple formula...!
@@simtech0598 Hello, I watch this video again and I have a question. At around 32:18, as an example the Upl is calculated by 4 multiplied by 0.05 where 4 is the characteristic length and 0.05 like you said, is the plastic strain. However, is 0.05 the plastic strain for maximum stress (before damage), and should Upl be calculated by 4 multiplied by another strain (greater than 0.05)? Thank you so much. Best regards
Great video, but I think at 27:10 there is a mistake. According to the Abaqus manual e_pl_f is not called fracture strain but equivalent plastic strain at failure. I think the fracture strain would correspond to e_pl_0, but correct me if I am wrong.
Thank you so much, sir. Could you explain how the model recognizes that it should fail in 45 Deg direction which well agrees with the actual ductile material fracture behavior?
In the damage parameters, the failure strain should be the effective plastic strain. How are the parameter determined? Can it be obtained from the actual stress-strain curve?
thank you very much! Excellent video！ I learned lot from it. and I am little confused about one thing about it: at the end of this video, you calculated failure displacement with 0.05*4(characteristic length), but in the plastic table, 0.05 corresponded to 110MPa，which is damage stress, so it means 0.05 is damage strain, not failure strain. the failure displacement is supposed to be calculated by damage strain, am I wrong somewhere?
Dear, 0.05 strain corresponding to 110 stress is damage initiation point...damage start here but not complete damage. So as per Abaqus documentation we have to use damage initiation point (this point is ultimate strain stress point).
Is it possible to simulate polymers using the ductile criteria of metals? If it's yes, does it have a scientifc paper talking about this possibility? Thank you for the video!
Dear, Ideally it should same. However, in FE analysis even size of element change the results. So make sure the mesh dependent study and other influencing parameters.
Thank you for your share! I learned a lot from your video. Please allow me to ask you one question. How do you apply ductile and shear criteria to get different fracture angles? I am working on this problem. Hope to get your kind suggestions. Thanks
Sir you have calculated the displacement at failure as 4*0.05 = 0.2. I have a doubt that u are working in mm coordinate system thats why your element charcteristic length is 4mm. if I m working in SI coodinate system them for this case my displacement at failure will be 0.004*0.05 = 0.2e-3. Am I correct??
I am trying a metal matrix composite system. Can you suggest the material model for matrix and reinforcement(ceramic) under both static and dynamic conditions for a simple compression test modeling ?
Dear, Best of my knowledge - For ceramic you can use the Johnson-Holmquist-Beissel (JHB) and the Johnson-Holmquist (JH-2) ceramic material models, which are available in Abaqus/Explicit as built-in user materials. For metal matrix you can use elastic plastic material model. However, selection of material model also depends on application.
Thank you very much, Sir. @@simtech0598 Can you please suggest the same for quasi-static test conditions for which I am using the General static model.
Material models or criteria present in this video is use to determine ductile fracture dependency on state of stress in terms of stress triaxiality and Lode angle. However, these material model not consider softening effect caused by damage. For many ductile material in which damage is very much pronounced during large plastic deformation Lemaitre Damage model is more accurate. Refrence - www.sciencedirect.com/science/article/pii/S2211812814002983
Excellent video! Keep it up. I'm learning from your videos. Thanks for your contribution to FE modelling.
Absolutely excellent. Thank you for the clear and thorough explanation!
Most welcome dear.
Thank you SimTech05, this is an excellent presentation on the complicated issue of "damage" prediction.
You are welcome sir.
Good job. It has been demonstrated in several articles that triaxiality factor, which is correlated to the first deviatoric invariant, and the normalized Lode angle, which is correlated to the second and third deviatoric invariant, are
the main parameters which characterize the initiation of ductile damage. The study of fracture propagation is more complex than initiation. There are different models based on Continuum Damage Mechanics or experimental mechanics. A very accurate prediction model for damage initiation in tensile test is the Bao-Wierzbicki Damage Model. It’s very simple to write a subroutine with this model to use with FEM. I suggest to you to try it.
Thanks for suggestions dear. I'll appreciate if you share paper/pdf on the same. (gkn0504@gmail.com)
Currently I'm working on different topic but I'll surely explore it.
Most underrated RUclips channel
Share with needy guys.
Sureeee
Great job, valuable video
such a informative video sir please keep it up
Keep watching..!
Nice video sir. As per this theory, if we opt for a coarser mesh, the material will undergo sudden failure. But if the mesh size if finer, then the then the material will undergo a gradual failure. Because the ultimate deformation depends only on the strain and characteristic length. In that, strain is a constant and characteristic length is varying as per the mesh size.. So, same material can have dissimilar stress strain response with mesh refinement which is little confusing..please correct me if I am wrong..
Thanks a lot for a nice explanation.
Thank you, sir! Brilliant explanation! So basically, the element deletion highly depends on the characteristic length of the element, yes? It is far preferable to have consistent SIZE of elements in the region of damage-prone (e.g., hex elements) because they provide a better shape and size if we mesh them strategically. More than that, we need to find the smallest element in the meshing part and calculate its characteristic length (must consider what type of element order too). And if we choose the biggest element with characteristic length as input, then it would be troublesome for small elements because they would be instantly deleted and not accurate for the simulation. According to what you've presented, Upl=characteristic length* plastic strain is essentially a failure displacement at the local level or the failure of a single smallest element. When those conditions are met (e.g., 0.15875), any element in the part will be automatically deleted by Abaqus. And also, it would be nice if you could share with us how to calculate the characteristic length for 3D elements (hex and tetra shape). Thank you again for this knowledge.
Yes. Consistency in mesh size is recommended.
Yes actually we don't know where crack will propagate...so we consider calculation with smallest size.
For 3D element characteristic length of element calculated by considering its volume...there is simple formula...!
Thank you so much !!! it helped !!
thanks for the great video
Glad you enjoyed it
please upload the second part of this videos its very useful
Ok dear.
Nice video. Thank you so much!
Welcome dear.
@@simtech0598 Hello, I watch this video again and I have a question.
At around 32:18, as an example the Upl is calculated by 4 multiplied by 0.05 where 4 is the characteristic length and 0.05 like you said, is the plastic strain. However, is 0.05 the plastic strain for maximum stress (before damage), and should Upl be calculated by 4 multiplied by another strain (greater than 0.05)?
Thank you so much.
Best regards
Great video, but I think at 27:10 there is a mistake. According to the Abaqus manual e_pl_f is not called fracture strain but equivalent plastic strain at failure. I think the fracture strain would correspond to e_pl_0, but correct me if I am wrong.
Dear,
Equivalent plastic strain at failure point (breaking point) is itself fracture strain.
Good video lecture
Thanks and welcome
thank you
Thank you so much, sir. Could you explain how the model recognizes that it should fail in 45 Deg direction which well agrees with the actual ductile material fracture behavior?
Dear,
Ductile material generally fail in cone shape if tension force applied. However in FEA it depends on material model also.
In the damage parameters, the failure strain should be the effective plastic strain. How are the parameter determined? Can it be obtained from the actual stress-strain curve?
Not only by SS curve.
thank you very much! Excellent video！ I learned lot from it. and I am little confused about one thing about it: at the end of this video, you calculated failure displacement with 0.05*4(characteristic length), but in the plastic table, 0.05 corresponded to 110MPa，which is damage stress, so it means 0.05 is damage strain, not failure strain. the failure displacement is supposed to be calculated by damage strain, am I wrong somewhere?
Dear,
0.05 strain corresponding to 110 stress is damage initiation point...damage start here but not complete damage. So as per Abaqus documentation we have to use damage initiation point (this point is ultimate strain stress point).
@@simtech0598 thanks lot! if it is possible for you to sent me some documents about it?
Very good job and very interesting video!
I woul ask you if you know a book where I can learn more about these topics.
Thanks in advance
Dear,
I suggest you to read Abaqus documentations.
Sir, Could you please explain damage criterion for non metals like rock, soil?
Dear,
Till today i not worked on nonmetals.
Awesome video sir, thank you very much for sharing this knowledge. Sir, is it possible to share ppt you explained?
Dear,
I cannot share the PPT. Please listen content and prepare wirh own understanding. Watch carefully..
@@simtech0598 sure sir, thank you 😊
Hi sir, displacement at failure at which direction you mean because we have 3 direction x, y and z ?. if you can explain that I will be very gratefull
Maximum displacement in any direction will be consider.
Sir where is the option to define in plane fracture toughness values in abaquas material property
the displacement at failure is calculated with the ultimate plastic strain or with the plastic strain at maximum stress?
At ultimate plastic strain dear.
Sir, can you please make a video on Disc Brake Noise Complex Frequency Analysis?
I am doing a project on Abaqus and stuck on it.
Dear,
Better to ask your queries instead of video.
Is it possible to simulate polymers using the ductile criteria of metals? If it's yes, does it have a scientifc paper talking about this possibility?
Thank you for the video!
No dear,
This material model only for metals as per my knowledge.
Very nicely explained sir.... please make video on fatigue crack growth using xfem.
Dear,
That topic I'll consider in future videos.
The curve strain stress may change with dimensions of spicemen??
Dear,
Ideally it should same.
However, in FE analysis even size of element change the results. So make sure the mesh dependent study and other influencing parameters.
Sir is there any way by which we can model that material fails only in tension and not in compression
It seems very odd dear,
May i not understood your question properly.
Thank you for your share! I learned a lot from your video. Please allow me to ask you one question. How do you apply ductile and shear criteria to get different fracture angles? I am working on this problem. Hope to get your kind suggestions. Thanks
Dear,
I not understood your question properly.
Sir you have calculated the displacement at failure as 4*0.05 = 0.2. I have a doubt that u are working in mm coordinate system thats why your element charcteristic length is 4mm. if I m working in SI coodinate system them for this case my displacement at failure will be 0.004*0.05 = 0.2e-3. Am I correct??
Yes dear correct.
@@simtech0598 thankyou sir for reply. Please also help me in how to calculate the characteristic length of 3D element
It is curve to Engineering curve??
Dear,
Abaqus work with true SS curve only.
👍
I am trying a metal matrix composite system. Can you suggest the material model for matrix and reinforcement(ceramic) under both static and dynamic conditions for a simple compression test modeling ?
Dear,
Best of my knowledge -
For ceramic you can use the Johnson-Holmquist-Beissel (JHB) and the Johnson-Holmquist (JH-2) ceramic material models, which are available in Abaqus/Explicit as built-in user materials.
For metal matrix you can use elastic plastic material model.
However, selection of material model also depends on application.
Thank you very much, Sir.
@@simtech0598 Can you please suggest the same for quasi-static test conditions for which I am using the General static model.
Dear,
I'm not sure aware to ceramic material model for static step. I'll try to find it out.
Thanks sir!, about theory, what is the difference between the damage model of LEMAITRE and the one presented in the video
Material models or criteria present in this video is use to determine ductile fracture dependency on state of stress in terms of stress triaxiality and Lode angle. However, these material model not consider softening effect caused by damage. For many ductile material in which damage is very much pronounced during large plastic deformation Lemaitre Damage model is more accurate.
Refrence - www.sciencedirect.com/science/article/pii/S2211812814002983
thank you for your good explanation.
When will be released the second part of video?
Max - January month 2022
Min - before end of dec 2021
Can you do a simulation on aluminium round tensile test getting ultimate strengths
Dear,
In this simulation ultimate tensile strength we used as input...we can not find out it by this method.