I haven't used Abq for some time. But I imagine this is the criterion for your individual elements. So if the displacement value for a given element exceeds that value it will be automatically deleted. Similarly for strain and stress criterion.

Hi Shalin,

First, at certain equivalent plastic strain, damage is initiated, any additional equivalent plastic strain contributes to plastic displacement at failure.

For simplicity, lets say you have stress-plastic stain result from a uniaxial tensile test and at 0.1 plastic stain (PS) damage initiates. The PS at complete failure is 0.3 also calculated from test result. So this 0.3-0.1=0.2 of PS will contribute to displacement at failure. If you use an element of one unit length (characteristic) then your displacement at failure is 0.2 for linear element and 0.1(half) for quadratic element. If you have an element length of 3 unit then for a linear element, your displacement at failure is 0.2*3=0.6 and so on.

However, it would assume that there is a linear progression of damage from zero damage at initiation to full damage at failure (if not otherwise stated). One can define tabular data of damage variable 'D' and displacement at failure to get arbitrary damage evolution. In a way here you divide your total displacement at failure in few steps and impose damage to the material by the value of 'D' in the pair (D takes values between 0-1, 0-No damage and 1-Complete damage). The failure energy evolution from onset of damage can be used to more accurately calculate these tabular data.

While you select element type, get the element delegation checked. At the displacement at failure (or final displacement if tabular data used) the element will be deleted.

This question fallows the answer given by Md. Shafiqul Islaml

I would like to better understand some of the parts of the answer:

1. Should I understand that the displacement at failure is related with the element size (expressed in mm)?

2. In the answer there is a reference to "tabular data". Is this for Ductile Damage: Fracture strain, Stress Triaxiality and Strain rate? If so, what is the significance of more than one row of data?

Hi Ionut,

Thank you for the interest. Regarding your questions,

1. Yes, displacement at failure is strongly related with element size.

2. Ductile damage evolution can be defined as Type: displacement, softening: Tabular. Progressive damage can be then defined in the table as I tried to explain in my previous reply.

Further, Fracture strain, Stress Triaxiality and Strain rate values are conditions those need to met to start the damage (Damage initiation). Your material may start to fail at different strain (Fracture strain) depending on the Stress Triaxiality and Strain rate values. If you have data on that, you can use more than one row. Otherwise, the material will behave as if it's damage initiation is independent of Stress Triaxiality and Strain rate.

Cheers!

what if the mesh size is not constant over the part , for instance , steel angle with holes for bolts , the meshing around holes around 2mm and angle mesh 6mm , that will produce a random sizes for mesh ?

how can we calculate displacement at failure if the mesh size is not constant over the part , for instance , steel angle with holes for bolts , the meshing around holes around 2mm and angle mesh 6mm

Md Shafiqul Islam Thank you for your detailed answer. However in Abaqus documentation, while describing the "Failure at displacement" it says

Characteristic element length * equivalent plastic strain rate.

Kindly clarify about this.

Here is the link

http://130.149.89.49:2080/v6.14/books/usb/default.htm

Mohamed Kamal Abdelaziz

How do you calculate the characteristic length ? given that the mesh size is 5mm and the total length of the section is 1000 mm.

thanks

i have created here an tutorial on how to determine the strain at fracture & triaxiality stress for ductile damage of metals:

https://youtu.be/qccyQFzIMFM

here are the slides:

Presentation ABAQUS Tutorial: Damage for Ductile Metals - Material Model ...

@Ronald Heinz Norbert Wagner Thank you sir for sharing valuable presentation. Sir could you please share one more video regarding Shear damage Criteria for metals. Thank you sir in anticipation.

Mohamed Kamal Abdelaziz

Did you find out how to calculate the characteristic length for variable mesh size?

Regards,

Kindly check the attached file.

Amr Abdullah Heggab

I am working with a complex geometry with C3D4 elements with fracture of elements at contact. The mesh is variable with sizes ranging from 7.25 units to 0.3 units. What would be the characteristic length in this case as the element size is varying constantly because of the use of single bias edge seeding in the part.

If you could help me in this case, I would be grateful.

Regards.

Dear Shalin P Marathe

The softening response after damage initiation in ductile damage model is characterized by a stress-displacement response rather than a stress-strain response. Thus, the damage evolution law is specified in terms of equivalent plastic displacement or in terms of fracture energy dissipation.

It is suggested to read following papers for more information:

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Dear Mohit Goyal

Did you find out how to calculate the characteristic length for variable mesh size?

Regards

https://youtu.be/VK1qtjMYRo0 Xuemeng Bie