August 24, 2023 at 1:37 pmHywel DaviesSubscriber
Using Explicit Dynamics, I am attempting a simulation of an impact test. This test involves dropping a round-nosed steel projectile onto an enclosure made of glass-reinforced PBT-ASA blend (Crastin). Based on stress/strain curves from the datasheet, I have created a material model for this simulation.
My questions are:
1) To assess whether the part has failed or not, is comparing total strain to the strain-at-break value the best criteria?
2) Can total strain be plotted in Explicit Dynamics? It is not available in the standard result options and I cannot find a suitable expression to input as a User Defined Result.
3) Based on plastic strain values alone, the housing would appear to fail. However, physical testing that has been performed on this part has proven that it will remain intact even when the projectile is dropped from almost double the height specified in the simulation. I therefore find the simulation result dubious. Is the failure criteria I am using incorrect, or does there appear to be something else fundamentally wrong with the setup of the material or simulation?
August 25, 2023 at 3:02 pmArmin_ASubscriber
I have a couple of questions regarding your model:
- How did you obtain the strain-at-break value for this material? Was it determined experimentally in a standard uniaxial tensile test under quasi-static conditions?
- I can see that you employed the Multilinear Isotropic Hardening model to describe the plastic response. By doing so, do you simplify the behavior of your “composite” material to behave like a homogeneous material?
August 25, 2023 at 4:22 pmChris QuanAnsys Employee
First of all, you need to make sure that the hardening curve in Engineering Data has sufficient data to cover the entire strain range of your simulation. Your plastic strain plot shows the plastic strain has reached 13%. However, the maximum plastic stain in the hardening curve seems to be only 1.9%, much less than the actual strain range. If this is true, you need to extend the strain range in your hardening curve so the material yield strength can be accurately predicted.
In your simulations, plastic strain is much larger than elastic strain. Usually, plastic strain failure is used as failure model. You can apply Plastic Strain Failure model to your material in Engineering Data. See attached picture below. If you must use Total Starin, you can apply Principal Strain Failure to your material.
Once the simulation has completed, you can click on Solution in Tree Outline and then click on Worksheet on the top tool bar to show a list of the user-defined results. Among them, you will find the directional total strains like STRAIN_XX, elastic strains like EPELX, and plastic strains like EPPLX, or the Princial Strains like P_Strain_1.
In the following example, the elastic strain (-1.8874E-4) + the plastic strain (-3.0859E-2) = total strain (-3.1047E-2).
August 29, 2023 at 9:03 amHywel DaviesSubscriber
Many thanks to you both for your responses.
1) I am taking the strain-at-break value from the 23°C graph on page 6 of the following datsheet (screenshot also pasted below) http://www.foomx.com/static/pdf/CrastinLW9020NC010_en.pdf I believe this value would have been obtained under the conditions you describe, Armin.
2) In the absence of more detailed information, I am assuming homogenous behaviour of the material.
3) The Multilinear Isotropic Hardening model is approximated from the following graph. The graph looks to become nonlinear from roughly about the 65 MPa stress/1% strain point, so I have assumed that plastic deformation begins from there. I have just noticed that my model is actually slightly incomplete (X-axis for plastic strain should perhaps extend to nearer 2.2%?), but I certainly wouldn't expect strain to reach so high as 13% as indicated by the simulation results.
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