Calculation of displacement by taking eigenvalues into consideration

[dido32511]

Hi,

So I have conducted an experiment on a real steel beam and I found that maximum displacement is more than 22 depends on the position but when I run the analysis on NASTRAN I get small number only 4 mm and I know that's because the start case of non-linear analysis is not based on eigenvalues.

My question is since I can't import imperfections into analysis , How can I get a realistic calculation for displacement and other values too like stress and strain ?

I think without being able to do an accurate analysis then the software vaules are not realistic.

I attached the material curve of the material that I used and most of the settings I applied

Thank you 

[Roelof.Feijen]

Hi @dido32511 ,

You can import imperfections into your analysis. It's just a little tricky. Especially if you have extra nodes in your model as a result of, for example, a workpoint or sketch point that you used with a Rigid Connector. And very important, you are not allowed to change your mesh between different analyses. I wrote an article about it, maybe it will help you further. If you're a bit handy with programming, you could even automate these steps.

[dido32511]

Hi, thanks a lot for your reply.

I checked the article because I already did some search on how to modify or use the BDF file with nastran but your article seems to have detailed explanation.However , isn't there any method to get the actual or the real displacement from Nastran directly?

It is seems a bit frustrating that to validate a software results you have to go through the whole cycle again and again.

There must be another way to analysis a beam or complex part to get the actual displacement.

But thanks for your detailed article I will try to apply it as I have no other option than keep working on Nastran at this point of my academic research 

[dido32511]

I read the article and I have another question regarding the topic 

"By default a scale factor of 1 is applied to deformed grid point translation. You can change the scale factor to a different value by changing the DISPGEOMSFACT parameter. Review the standard you are going to apply to see if other scale factors should be used.

Geometric Imperfections Sensibility Study

In practice you create 3 or more Bulk Data Output Files with different scale factors. Than you use them in 3 or more Nonlinear Buckling Analyses or Nonlinear Static Analyses, to justify the right choice for a scale factor."

Do you mean that we repeat the steps from (8) till the end with different scale to reach the optimum result ? And if that's so , what's your recommendation to the scale number as I don't understand what is the scale referring to ? Should I try which numbers with scale factor 

Also

I understand that you create two type of analysis but the first one seems to be with no use other than numbering the mesh sizes and nodes since you only do the grid modification to the BDF file coming of step (8)  

Thank you for your support and help

[Roelof.Feijen]

Are you sure you did not over constrain your model, so you made it to stiff?

Can you share the model so one of use can review it?

[dido32511]

Yes of course 

[Roelof.Feijen]

The scale factor is coming from a standard, like ISO, DNV guidelines, etc. I do not know how they have determine the scale factors. 

In step 1 normally I would run several studies with different mesh size, a mesh sensitivity studie. In this step I determine the mesh size to be used to create the BDF file.

[Roelof.Feijen]

I took a quick look at the setup of your analysis, but didn't have time to change it. What I notice is that both ends can't move, can't rotate. I would at least expect that one end can move in the z-axis and both ends can rotate about the x-axis. I think your model is overconstrained. A linear analysis gives exactly the same results as a non-linear analysis. Can you tell what the correct degrees of freedom should be at the ends of the beam? Wouldn't it be better to start from a surface model with shell elements?

[dido32511]

Hi , I think you are right regarding constraints, if you check the attached picture you will see the actual constraint during the experiment which clearly mean the beam can rotate in x axis and z direction as you mentioned.

I will re check it again 

I am sorry but what do you mean by "Wouldn't it be better to start from a surface model with shell elements?"

Do you mean I should model Beam element not solid or what?

Thank you a lot for your time and help

[Roelof.Feijen]

Hello @dido32511 ,

What I meant is building a surface model like shown in the picture below and use shell elements in Nastran.

I have attached a sample. I suggest to review the Rigid Connector and the degrees of freedom applied by the Constraints on the workpoints. Good luck with your academic research.

[dido32511]

Hi, thank you very much for your detailed suggestions.

I think surface solution is suitable but in my case I will prepare another model with the same properties except that the flange of the beam will be coated with epoxy then carbon fiber so I think solid geometry will be more accurate for that case to match the experiment case and to use shell element for composite material

I choose the laminate option to model the composite parts which in my case will be (epoxy covering the steel then FRP sheet on the epoxy. then I applied the shell properties to the solid part face.

If you have any hints or advice regarding the procedures please let me know

Does the arrangement of the material in the laminate part that I followed in the attached picture mean that epoxy is the first layer attached to the steel of the solid element ?

[Roelof.Feijen]

I haven’t got any experience with laminate, other than the standard tutorial. May be someone else can help you with that.

[dido32511]

Hi, I was hoping to find an explanation to the following points :

1-the lateral displacement in X direction despite the load is vertical and the constraints doesn't allow movement in X direction nor rotation. plus the unsymmetric distribution of stresses along middle of the beam despite effecting with a load right in the middle and having the constraints degree of freedom set just as you recommended with permission of rotation in z direction.

 
2-The maximum displacement should happen at the bottom of the beam since the load is vertical but unlike that the bottom has a really small value plus most of the displacement happens in the lateral direction not the vertical direction (X)

3 -Does the type of surface contact (bonded, separation) effect the result in case the beam web , flange and stiffener are welded together so they should be in a "Separation" mode.

I apologize I know I have been asking so much but this is supposed to be a simple loading case with simple setup so I am confused of the results compared to actual experiment which was direct 3 bending moment test and in case of complex design it will be hard to evaluate such results.

[Roelof.Feijen]

I think the lateral displacement in the x direction is caused by the uneven distribution of the nodes. Although we select an edge or face when placing a load, the load is distributed over the nodes that lie on the edge or face. If a side has one or more nodes then that might explain the lateral displacement.

Looking at your constraints, I think it's still over constrained. I think I would apply a Rigid Connector to the faces where you have now applied a constraint. Let the Rigid Connectors automatically create a workpoint themselves and add constraints to these workpoints.

With the constraints you also have to realize that even though we select an edge or a face, we constrain the nodes that lie on that edge or face.

Did you know that the nodes of solid elements can rotate, but we can't control that? So you can't limit the rotation of solid nodes against a twist, and they can't absorb a moment either.

Checking or unchecking Rx, Ry and/or Rz has no effect at all on solid nodes, but it does on the nodes created by a workpoint or a sketch point.

A Separation Contact will ensure that parts can touch each other, but can also separate from each other. In your case, a Bonded Contact seems the right solution.

It is also possible that the contacts make the model too stiff. Which setting do you use for the max. activation distance? These tips and tricks from John Holtz describe how you can determine your max. activation distance.

[John_Holtz]

Hi @dido32511 

Have you made any changes to this model since you posted the original model "Lusas scaled 14-10-2022.ipt"? The original model is created improperly. Instead of creating the CAD model as 18 separate solid bodies, it should be merged together into one solid body. There is no reason to split the flanges (such as 1,2, 5 and 6 in the image below), and certainly no reason to split the stiffener (3,4) into half thicknesses. The problems with creating separate solid bodies (when it is not necessary) is as follows:

• It requires contact to keep everything together. Contact is an approximation of the bonding between the parts and is especially an approximation for bonding half the stiffener to the other half stiffener. 
• Having separate bodies in one part prevents you from hiding the individual bodies when viewing the results. If you were to create an assembly of separate parts, then you can at least hide different parts when viewing the results. (Again, I do not see any advantage to split the model and use contact.)

The X displacement begins because all the contact causes the model to be unsymmetric. The stiffeners on one side of the model are full thickness whereas the stiffeners on the opposite side are half thickness joined by contact. This makes the model unsymmetric.

The magnitude of the X displacement at the end occurs because the model buckles after 70% load.

John

[dido32511]

Hi, thank you for checking the model.

I did try to update the model after your note and made the beam as one modeled element but the best criteria to balance the load to not have displacement in X direction was to use pressure with equal value of load and it gave symmetric  stress distribution. However after assigning contact with type "Bond" the displacement I still can't get the bottom of the beam to gave displacement of value of 20,00 mm (which is the experiment value) but I get almost 5 mm which is very small compared to the stress value which is more than the yield point and fracture point 

I keep changing values and run as many analysis as I can but it seems tricky to get the model to deflect 
1- stress values
2-Displacement
3-Displacement curve
4- stress -strain curve

I also took into consideration applying the rigid connector as advised above 

[dido32511]

I changed the model to one segment  and I assigned the rigid connector to faces with constraints but what really cancelled the X displacement was changing the "Force" type to an equal values of "Pressure".

I am not sure of over constraining edges because when I release any additional movement or rotation the model fails to solve.

Regarding the separation contact , also it fails to solve when I change the stiffeners contact with beam to separation since the rest of the geometry is one solid beam.

I am sorry that this matter took so much but I still can't understand how the stiffness is not reduced after reaching yielding point  

[John_Holtz]

Hi @dido32511 

1. Why did you make the model 7 bodies instead of 1?
2. Why is the model located 13 m from the origin? (Round off errors when the model is far from the origin can be a problem. Perhaps not in this situation, but something to keep in mind for the future.)

I am not sure if we have asked before, and not sure if you answered, but how is the beam supported in the test?

• Is it fixed so that the ends cannot rotate? The model has a constraint applied to a face of the beam. If the nodes over an area cannot translate in the Z direction, the face cannot rotate. You have created an approximation of a fixed beam. 
• Is the test beam simply supported so that the ends can rotate? If the beam is supposed to be simply supported, the constraint should be applied to the center point of the rigid connector. Constraining one point and freeing the R rotation allows the node (and therefore the beam) to rotate.

The next thing to check is the yield stress. 402 MPa (58.3 ksi) seems a little high for "mild steel". Is this value and the tangent modulus entered into Nastran based on a physical test of the material? Only a small volume of the cross-section is yielding in the analysis.

What was the permanent deformation of the test beam after the load was removed? This would give some idea of how much of the volume was yielding. If the permanent deformation was close to 0, then only a small portion of the test model was yielding. If the permanent deformation was significant, then you know that a significant height of the beam was yielding, and therefore the yield stress was lower than 402 MPa.

P.S. Add a Tx constraint at the middle of the span to prevent it from displacing in the X direction. For example, put a Tx constraint on the -x and +x end of the edge where the load is applied. Sometimes it's not worth fighting the math to prevent the model from moving in the X.

John

[dido32511]

Hi , Thank you for your help.

Actually the beam is simply supported as per the attached picture. you can see that the faces are supported not just point so I tried to match the actual case , you may notice that the support can't move in any direction but can rotate around one axis only (its axis) .

Regarding the permanent deformation it was measured at the bottom flange at center point to be near 20 mm 

Regarding material yield stress it is 402 as mentioned because this is a hot rolled member and it is mentioned on the material sheet.

"What was the permanent deformation of the test beam after the load was removed? This would give some idea of how much of the volume was yielding. If the permanent deformation was close to 0, then only a small portion of the test model was yielding. If the permanent deformation was significant, then you know that a significant height of the beam was yielding, and therefore the yield stress was lower than 402 MPa."

I am not sure what you mean but the final maximum load applied was 597 kn so I would say that the stress was more than 402 yield stress 
I 

[John_Holtz]

Thanks @dido32511 . Here is the displacement in the model that you provided, exaggerated by a large magnification factor so that you can see what is happening at the constraints. The constraints applied to the face of the beam are clearly not duplicating the test. The constraint needs to be applied to the center of the rigid connector as I described.

Furthermore, the center of the rotating support is a significant distance below the face of the flange. You need to include that in the analysis, too.

• Add a work point or sketch point in the model at the location of the center of rotation.
• When creating the rigid body connector (not an interpolation type connector), select the work point/sketch point for the center of the connector.
• Put the constraint at the center of the connector.

I am confused about your statement that "the permanent deformation it was measured at the bottom flange at center point to be near 20 mm". The permanent deformation is the displacement after the load is removed. None of your setups include that in the analysis; you are calculating the displacement with the load applied. What displacement are you measuring: with the load applied or after the load is removed? If the 20 mm displacement is with the load applied, what is the displacement with the load removed? (When I changed your model to use simply supported constraints to allow the beam to rotate, the displacement with load was under 5 mm at the top where the displacement is the largest, and the displacement with the load removed was 0.7 mm. Both of these indicate your yield stress is inaccurate in the analysis.)

Also, what is the calculated displacement when you do a hand calculation (ignoring yielding of course)? What is the calculated stress? You need something to compare to the analysis and the physical test.

John