Excellent video! Keep it up. I'm learning from your videos. Thanks for your contribution to FE modelling.
@naderzamani575210 ай бұрын
Thank you SimTech05, this is an excellent presentation on the complicated issue of "damage" prediction.
@simtech059810 ай бұрын
You are welcome sir.
@jteague2382 жыл бұрын
Absolutely excellent. Thank you for the clear and thorough explanation!
@simtech05982 жыл бұрын
Most welcome dear.
@giuseppebua8955 Жыл бұрын
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.
@simtech0598 Жыл бұрын
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.
@raghu.r4142 жыл бұрын
Most underrated KZfaq channel
@simtech05982 жыл бұрын
Share with needy guys.
@raghu.r4142 жыл бұрын
Sureeee
@filipparczynski4283 Жыл бұрын
Great job, valuable video
@sajidsarabi57489 ай бұрын
such a informative video sir please keep it up
@simtech05988 ай бұрын
Keep watching..!
@anikdasanto2 жыл бұрын
Thanks a lot for a nice explanation.
@arvindrajm686910 ай бұрын
thanks for the great video
@simtech059810 ай бұрын
Glad you enjoyed it
@vishnuos407511 ай бұрын
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..
@is-ig4zh2 жыл бұрын
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.
@simtech05982 жыл бұрын
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...!
@chuandongxie91482 жыл бұрын
Nice video. Thank you so much!
@simtech05982 жыл бұрын
Welcome dear.
@chuandongxie91482 жыл бұрын
@@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
@NarendraBhandava3 ай бұрын
Thank you so much !!! it helped !!
@landrytim85952 жыл бұрын
please upload the second part of this videos its very useful
@simtech05982 жыл бұрын
Ok dear.
@Peter109ful Жыл бұрын
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.
@simtech0598 Жыл бұрын
Dear, Equivalent plastic strain at failure point (breaking point) is itself fracture strain.
@npnjo2 жыл бұрын
Good video lecture
@simtech05982 жыл бұрын
Thanks and welcome
@user-ik5ge4io6k8 ай бұрын
thank you
@xchengzhang54092 жыл бұрын
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?
@simtech05982 жыл бұрын
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).
@xchengzhang54092 жыл бұрын
@@simtech0598 thanks lot! if it is possible for you to sent me some documents about it?
@sejong4672 Жыл бұрын
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?
@simtech0598 Жыл бұрын
Dear, Ductile material generally fail in cone shape if tension force applied. However in FEA it depends on material model also.
@user-nr1qm8ie3h Жыл бұрын
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?
@simtech0598 Жыл бұрын
Not only by SS curve.
@cesari93 Жыл бұрын
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
@simtech0598 Жыл бұрын
Dear, I suggest you to read Abaqus documentations.
@landrytim85952 жыл бұрын
the displacement at failure is calculated with the ultimate plastic strain or with the plastic strain at maximum stress?
@simtech05982 жыл бұрын
At ultimate plastic strain dear.
@alialijamelll50162 жыл бұрын
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
@simtech05982 жыл бұрын
Maximum displacement in any direction will be consider.
@nareshnaik48722 жыл бұрын
Awesome video sir, thank you very much for sharing this knowledge. Sir, is it possible to share ppt you explained?
@simtech05982 жыл бұрын
Dear, I cannot share the PPT. Please listen content and prepare wirh own understanding. Watch carefully..
@nareshnaik48722 жыл бұрын
@@simtech0598 sure sir, thank you 😊
@AbinasBehera-pd9ot Жыл бұрын
Sir, Could you please explain damage criterion for non metals like rock, soil?
@simtech0598 Жыл бұрын
Dear, Till today i not worked on nonmetals.
@vikramroy7471 Жыл бұрын
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??
@simtech0598 Жыл бұрын
Yes dear correct.
@vikramroy7471 Жыл бұрын
@@simtech0598 thankyou sir for reply. Please also help me in how to calculate the characteristic length of 3D element
@joaovictorsantos5660 Жыл бұрын
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!
@simtech0598 Жыл бұрын
No dear, This material model only for metals as per my knowledge.
@kikinano5830 Жыл бұрын
The curve strain stress may change with dimensions of spicemen??
@simtech0598 Жыл бұрын
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.
@vikramroy7471 Жыл бұрын
Sir is there any way by which we can model that material fails only in tension and not in compression
@simtech0598 Жыл бұрын
It seems very odd dear, May i not understood your question properly.
@hitman5601 Жыл бұрын
Sir where is the option to define in plane fracture toughness values in abaquas material property
@zhenzhang66242 жыл бұрын
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
@simtech05982 жыл бұрын
Dear, I not understood your question properly.
@saurabhchauhan35062 жыл бұрын
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.
@simtech05982 жыл бұрын
Dear, Better to ask your queries instead of video.
@muralikartheekss29322 жыл бұрын
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 ?
@simtech05982 жыл бұрын
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.
@muralikartheekss29322 жыл бұрын
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.
@simtech05982 жыл бұрын
Dear, I'm not sure aware to ceramic material model for static step. I'll try to find it out.
@impacthealth4u2 жыл бұрын
Very nicely explained sir.... please make video on fatigue crack growth using xfem.
@simtech05982 жыл бұрын
Dear, That topic I'll consider in future videos.
@user-xu2tu6dm8u Жыл бұрын
It is curve to Engineering curve??
@simtech0598 Жыл бұрын
Dear, Abaqus work with true SS curve only.
@sinaseyedjafari47902 жыл бұрын
thank you for your good explanation. When will be released the second part of video?
@simtech05982 жыл бұрын
Max - January month 2022 Min - before end of dec 2021
@sitarambainsla31792 жыл бұрын
👍
@houssameddinemoussaoui61872 жыл бұрын
Thanks sir!, about theory, what is the difference between the damage model of LEMAITRE and the one presented in the video
@simtech05982 жыл бұрын
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
@mashokgoud56772 жыл бұрын
Can you do a simulation on aluminium round tensile test getting ultimate strengths
@simtech05982 жыл бұрын
Dear, In this simulation ultimate tensile strength we used as input...we can not find out it by this method.