## Simulation Mechanical and Multiphysics

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# Breaking load of an oblong link - Non-linear

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Please find attached a model of a link for chain systems. Also attached is an image of some broken test pieces. I would like to model the failure of the link and determine the magnitude of the force. I need to report this back to our customer.

Would you please review the image and file and advise where I am going wrong. The non-linear crashes out after three load steps.

Thanks,

John

# Re: Breaking load of an oblong link - Non-linear

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Hi John,

I ran design scenario 2 because design scenario 3 did not have a material property when I opened the model. Just to clarify, the analysis did not crash after time step 3 -- it reached what is known as "AutoTM limit" = "automatic time management limit". Because the processor reduced the time step to such a small time increment (2E-7) and was not able to converge, it assumes that there is a problem and stops.

True, autotm versus a crash is just a technicality. The issue is that the analysis is not proceeding to the end. You may want to apply the load more slowly. It looks like the part starts to yield in time step 2. Keep in mind that the load you applied is 1/8th of the total load because you have modeled 1/8th of the full link. So perhaps you have more load on the part that you realized.

Eventully, you may need a finer mesh or at least use midside nodes ("Element Definition > Midside Nodes > Included").

Mechanical Engineer

Pittsburgh, PA

16.9 years experience with Simulation Mechanical

# Re: Breaking load of an oblong link - Non-linear

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Thank you for your response John. I have tried your suggestions and have more questions.

Please find attached “one eighth.jpg”. This graphically indicates that although my model is one eighth, my loads are one quarter. Would you agree with this?

Also find attached, “master link.jpg”. It is from the Gunnebo lifting product catalogue. If you look at the model marked M-19-10, this is the one that is the subject of the attached file - 1874-CA-1211 A.zip. I am trying to move forward by predicting the behaviour of a known structure. These units are tested at twice proof load. I have been speaking with an Australian load testing company and they confirm that they have confirmed that these product to stand up to the two times overload.

For this model, I used 3 mm element size (3848 nodes), time to apply load = 1 second, material = AISI 4130 (Y.S. 1064 and 1172)

run time = 02:20; Max VM 1488 MPa; Max xx tensor +1535 min -1763.

Then I added mid-side nodes

run time = 10:44; Max VM 1610 MPa; Max xx tensor +1726 min -2345

Then removed mid-side, element size = 1.5 mm

run time = 22:07; Max VM 1575 MPa; Max xx tensor +1637 min -1765

Then element size = 3 mm (no mid-side) and time to apply load = 40 second

run time = 4:49; Max VM 1484 MPa; Max xx tensor +1538 min -1778

Would you please explain the very different result with mid-side nodes.

The results still imply that permanent deflection (yielding) of this link is likely under proof test conditions. This is not the case. Would please see if you are able to produce a result where the master link does not yield.

Finally, would you also see if you can produce a result to verify the load at which destruction of the link occurs. This should be at five times the working load limit. Then please guide us on how to set up that model.

Thanks in advance,

John

# Re: Breaking load of an oblong link - Non-linear

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Sorry. You are correct that the total load simulated is 4 times the applied load, not 8 times the applied load that I stated. The diagram makes that very clear.

Midside nodes are a way to obtain more accurate results without needing to create a finer mesh. (A finer mesh is the "normal" way to get more accurate results.) If we think in terms of 2D elements (4-nodes) for a moment, an element without midside nodes can only predict a linear change in displacement between the corner nodes, and therefore a quadratic change in stress (if my memory is correct: stress is related to the change in displacement). By adding midside nodes, there are three points between the corners, resulting in a quadratic displacement between the corners, and a corresponding accuracy in the stress. The larger stresses are presumably closer to the theoretical solution. The von Mises stress isn't too different between the runs (1490 to 1610), so I think the results are not that different.

You mentioned permanent deformation. Is that the result of the physical test that is measured? How accurately are the measurements? Keep in mind that the volume of material above the yield stress when under full load does not necessarily give a qualitative measurement of permanent deformation. You probably should remove the load and compare the simulated permanent deformation with the measured deformation. If the measurements are made close to where the load is applied, it may be important to simulate the loading method more accurately.

Mechanical Engineer

Pittsburgh, PA

16.9 years experience with Simulation Mechanical

# Re: Breaking load of an oblong link - Non-linear

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Hi John

Thank you for explanations.

Critical guidelines for this physical testing are

- Links are loaded by pins not exceeding two thirds of the internal width of the link being tested.
- Test forces are applied gradually to the lifting component in tension along the axis of loading.
- Proof loading needs to take place without any permanent set of more than 1% of a specific dimension. This dimension is across the link as indicated on the attached sketch. The original size is indicated with a centre punch and a scribed arc. After testing, the link is re-scribed at the same radius and compared. (see attached).

This two thirds diameter is more than I had envisaged. As you say, introducing the effect of the load application pin will change things. I did try this with point contact at the intersection of the arcs. I hoped that the two metals would crush to meet each other. This did not occur. Therefore to introduce greater contact, I ended up artificially creating a small arc of conformance of the link to the loading pin. I tried this with linear; but have not tried this with non-linear as yet.

Please comment on the point contact geometry deformation issue since it is undesirable to alter the geometry, however perhaps Hertzian contact pressure is still too much for FEA.

Cheers,

John

# Re: Breaking load of an oblong link - Non-linear

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I'm sure the discrepancy between your simulation and the physical test is the material properties. When trying to capture the effects post plasticity, the yield strength and strain hardening modulus are important.

- The yield strength depends on the heat treatment and material. This information is relatively easy to find. (I looked through one chapter of the catalog but did not see the material.)
- The strain hardening modulus -- essentially the slope of the stress-strain curve after yielding -- is more difficult to locate. The properties in the library are based on just two data points: the yield strength and the ultimate strength. So this is an overall average. If you can find a stress-strain curve, then you can use the material model "von Mises curve with X hardening".

I just eye-balled the maximum size pin (2/3 of the opening) to see if that might impede the deformation of the link. It looks like it would not have a significant effect, especially considering that the simulation gives a much larger deformation (5%) than the physical test (1%). See the attached image.

So it looks as if there will be Hertz contact. Your simulation approximates that by putting the load on the symmetry plane. My gut feeling is that modeling the pin and the contact will not change the overall results that much. Because the Hertz contact occurs at such a small area, it is challenging to get that stress result. But the force distribution and surface contact should be possible to simulate. I suggest meshing the model, then adding a refinement point at the point of contact and remeshing with a "large" divide factor and a "small" radius of influence.

You may have had problems with the surface-to-surface contct if you were running "Static Stress with Nonlinear Materials" and did not do something to make the pin statically stable. I assume you added only the force to the pin, and therefore it would be statically unstable. Swithing the analysis type to "MES with Nonlinear Materials" will introduce the effects of mass and acceleration, so the surface-to-surface contact is easier to be detected.

Mechanical Engineer

Pittsburgh, PA

16.9 years experience with Simulation Mechanical

# Re: Breaking load of an oblong link - Non-linear

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Please find attached model 1874-CA-1211 A. (I have also attached the inventor components)

This one shows how I permit some initial contact between the pin and the link and then as the link deforms, it contacts the pin around its circumference. The model achieves contact penetration and eventually autotm. In the past, this might be where gap elements would have come in, though it would be difficult with the radial contact.

Would you please review and see if there is way for this progressive contact to be permitted in the model.

Cheers

JB

# Re: Breaking load of an oblong link - Non-linear

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Hi John,

Did you make a typo when you wrote model ...1211 A, or did you attach the wrong model ...1212 A? Model 1212 A was set for nonlinear static stress and did not converge at all. When this occurs, I always suggest switching the analysis type to MES because the effects of mass add "stability". (If the two parts are not in contact, then part 2 -- the pin -- is statically unstable. Depending on how the software is written, I do not know if the parts are considered to be in contact at time 0 or only after the first iteration when the processor detects that part 2 has penetrated into part 1.)

When using MES, the first time step showed a displacement of 2E6 or a similar large value. From looking more closely, the Z symmetry constraint on the bottom of the pin (most -Z coordinate) was accidentally set as X symmetry. Hence, neither part was statically stable. :-)

After fixing the constraints and running the simulation, I noticed some strange results near the contact between the two parts. A closer inspection revealed that the two parts were bonded together at some of the nodes because the only surface to surface contact specified was between 1 element in the link (part 1 surface 5) and the pin (part 2 surface 2). So I setup more contact between the link (part 1 surface 4) with the pin (part 2 surface 2) and remeshed the model. (Meshing uses the contact setup to adjust how the mesh is done, so specifying contact first is a good idea.) This ran fairly well, but I still got some slight penetration of the pin into the link. This indicates that the contact stiffness needs to be higher. Right-click on the surface to surface contact entry in the browser and choose "Settings". Then click the "Advanced" button and change the contact stiffness from Automatic to "User-defined" and start with a value of 20000. (The summary file from the analysis shows that the automatic calculated value was 9325, so I increased this by 2.)

Even though the surface contact was not entirely correct in my simulation, it does reveal that the X displacement of the link is about 6 times smaller than the previous analysis where the force was applied directly to the link. (Do I remember correctly that someone predicted the difference would be small? ;-)

Speaking of forces, doesn't the model with the pin and surface to surface contact have a different magnitude (39240 N) compared to the model 1874-CA-1210 A where the force is applied directly (17792 N)? Was that intentional, or is it related to the 1/2 versus 1/4 versus 1/8 symmetry issue? The way I see it, the link (part 1) is the same in both models, so the same load should be applied for comparison purposes. The only difference is how the load gets into the link, whether applied directly or through surface to surface contact.

Mechanical Engineer

Pittsburgh, PA

16.9 years experience with Simulation Mechanical