

September 19, 2018 at 7:35 pmAchilleasMilSubscriber
Hello,
I want to do a linear simulation of a three point bending test of a sandwich panel.
According to C393 ASTM norm, the test specimens must have a length of L=200mm and a width of b=75mm.
The specimens are supported on two roller bars under them, which shall allow free rotation and would have a midspan distance from each other of S=150 mm. The load force should be applied by another roller bar positioned in the middle of the specimens on the top of them.
A pressure pad of length l=25mm and throughout the width of the panel: b=75mm , should be positioned between the specimens and the loading bar in order to prevent local damage of the facing.
I try simulating the experiment by applying a force in a middle area of the upper facing with a geometry of lxb=25x75 mm^2 , in accordance to the contact area of the upper pressure pad used for the loading.
My question is how can I best simulate the simply supported area of the contact between the supporting roller bars and the sandwich panel, since the simply supported feature of ansys is only applicable on points or edges..?
By creating two lines accordingly at the center of the supposed roller bars and choosing a midspan distance 150 mm between them in spaceclaim afterwards my results of max deformation have a differnce of more than 20% lower value than the the analytical calculations.
I am using a shell solid shell model for the sandwich panel and use two bonded contacts between them where both have as a contact geometry the solid.
From my understanding the support is not completely simulated here and the line boundary condition is causing this difference.
Can you help me in how to better simulate the simply supported area between the roller pads and the panel, while staying in the linear analysis and not using two other bodies for the under roller pads?
Thank you in advance!
Gratefully,
Achilleas Milios

September 20, 2018 at 12:29 ampeteroznewmanSubscriber
Hello Achilleas,
(1) Take advantage of symmetry and model only one quarter of the sample. You can cut it in half lengthwise so the model is now 150/2 mm long, and you can cut it in half on the breadth and use 75/2 mm. You don't care about the sample that extends beyond the roller.
(2) Use a Remote Force to model a loading bar, without having an actual bar and contact. If you have a 25/2 mm wide pressure pad on the top face panel, you can pick that face and place the coordinate system for the Remote Force on the plane that cut the sample to a half length. The Remote Force is scoped to the pressure pad. Since you have a quarter model, you only apply 1/4 of the force.
(3) Since the two symmetry planes stopped the model from moving anywhere but vertically along the Y axis, you can easily create a "roller" on the 150/2 mm end of the sample by picking the edge of the face place and just applying a Y = 0 displacement support. That is a roller since you are leaving X free.This is a completely linear model so it will solve very quickly.
If you want a pressure pad on the bottom roller bar, then a Remote Displacement with Y = 0 is used and the pressure pad face is the scope of the remote displacement and the coordinate system for the remote displacement is put at X = 150/2 mm.
Regards,
Peter

September 20, 2018 at 6:14 pmAchilleasMilSubscriber
Hi sir,
Thanks for the quick reply!
1) Why do I not need the rest of the sample that extends beyond the supporting roller bars? Isn't that an assumption which affects the realistic solution?
I get different results when I do a model with and without the edges.
3) Supporting a sample on two roller bars, does that mean only Z deformation is limited and all other deformations and rotations are free or does that mean that X,Y and Z deformations are limited and only the rotations are free?
4) Do you maybe have a suggestion video for trying a nonlinear analysis with two supporting bars and a contact between them and the panel, where the extended length of the panel is not overlooked?
Gratefully,
Achilleas Milios

September 20, 2018 at 7:43 pmpeteroznewmanSubscriber
Hi,
1) Maybe I was wrong and the overhang has some effect.Try it with and without the extension and note the difference.
3) Roller bars means Z = 0 only, all other displacements and rotations are free.
4) You will want a plane to split the bodies to place the roller constraint under the sample with the overhang. Put the split bodies in a multibody part to have a connected mesh. Then you can just suppress the extension and solve without to see if it made a difference.
Regards,
Peter

September 21, 2018 at 4:14 pmAchilleasMilSubscriber
Hi again,
I try it and it is different.. it is also different if I use remote displacements and a remote force in contrast to when I use simple force and simple displacement boundary conditions.
I tried doing a non linear simulation with three roller bars (for loading and support) but don't get the contacts settings correct as they are not recognized, the loading bar just goes through the specimen, I tried a settup by a video on youtube by cae worlwide..I did the splitting you suggested but can you give me more specific details? What do I need to specify in the contact region?
Thank you very much,
Achilleas

September 21, 2018 at 5:45 pmpeteroznewmanSubscriber
Leave the extension in the model. I can explain how to split bodies to get a result with and without the extension, but I need to know if you are using DesignModeler or SpaceClaim to edit your geometry, since the steps are different.
What output are you tracking when you try different boundary conditions?
How much difference is there between the different boundary conditions?
Please reply with images of the model and the actual values of the result.
Recall the discussion 4 weeks ago where I showed there was a 35% difference in deformation between a shell model and solid model? I didn't have to worry about that because I chose one model and I adjusted the Shear Modulus until that model matched the experimental data. So just choose a boundary condition and tune the model until it matches the experimental data.
It's great to learn how to use contact, but you don't need it for this model.
Regards,
Peter

September 24, 2018 at 6:21 pmAchilleasMilSubscriber
Hi Sir,
Sorry for my late reply! And sorry for the following text it is a long answer, but can you help me with all these things?
I am using spaceclaim, and since the model is basically simple, running time less than a minute, I didn't use the symmetry to take 1/4th of the panel.
I am tracking the deformation in the center of the specimen as it is supposed to be the maximum at S/2.
General Values:
E_facesheets = 52,6 GPa
G_core = 0,438 GPa
Geometry: 200x75x20 (mm)
Midspan: 150 mm
Here follow some screenshots from the models:
1)
Force Applied in an Area : 25x75 , force direction along Z axis:
Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0:
Results in total deformation, z = 0,1288 upper face at the center / z = 0,123 under face at the center:
We can see that the deformation is not symmetric and there are rotations happening.
2)
Same Force
Changing the BC's:
Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0, X=0:
Results in total deformation, z = 0,102 upper face at the center / z = 0,096 under face at the center:
3)
Changin BC's to remote ones each time from a distance of 12,5mm away from the specimen because of the 25 mm roller diameter suggested in the C393 ASTM.
Remote Force Applied in an Area : 25x75 , force direction along Z axis, Scope from point in the center and 12,5mm above the upper face, Behavior: Deformable:
Remote Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0, Scope from point center of each line and 12,5mm under the lower face of the specimen, Behavior Coupled:
Results in total deformation, z = 0,126 upper face at the center / z = 0,120 under face at the center:
We see again that the deformation is not symmetric..
4)
Changing BC's to limit also X deformation:
Remote Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0X=0, Scope from point center of each line and 12,5mm under the lower face of the specimen, Behavior Coupled..
Results in total deformation, z = 0,102 upper face at the center / z = 0,096 under face at the center:
We see that the deformation now shows symmetry and is close to the normal BC's ( not remote ones).
5)
Not extended model.
Force Applied in an Area : 25x75 , force direction along Z axis:
Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0:
Results in total deformation, z = 0,148 upper face at the center / z = 0,136 under face at the center:
The deformation again is not symmetrical.
6)
Not extended model
Changing BC's toi limit X deformation.
Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0, X=0:
Results in total deformation, z = 0,122 upper face at the center / z = 0,116 under face at the center:
We see that the deformation now is symmetric but we can notice it is different from the extended model.
7)
Not extended model.
Changing the boundary conditions to remote ones.
Remote Force Applied in an Area : 25x75 , force direction along Z axis, Scope from point in the center and 12,5mm above the upper face, Behavior: Deformable:
Remote Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0, Scope from point center of each line and 12,5mm under the lower face of the specimen, Behavior Coupled:
Results in total deformation, z = 0,140 upper face at the center / z = 0,134 under face at the center:
Non symmetric deformation again.
7)
Not extended model.
Boundary conditions to remote ones and limiting also x deformation.
Remote Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0X=0, Scope from point center of each line and 12,5mm under the lower face of the specimen, Behavior Coupled..
Results in total deformation, z = 0,124 upper face at the center / z = 0,115 under face at the center:
In Conclusion:
Prefering the symmetric deformation plotsso limitating both Z and X deformation, we get close results between the models with normal bc's and remote bc's.
W_max = 0,105 mm up, W_max = 0,096 mm down, extended model
W_max = 0,124 mm up, W_max = 0,115 mm down, not extended model
that is a difference close to 18% for the upper face and 16,5% for the lower face, between the extended and non extended model.
Literature:
The Literatur, according to ANALYSIS AND DESIGN OF STRUCTURAL SANDWICH PANELS, Howard G. Allen
for a simple supported beam with a central point load gives the total central deflection as :
Δ = WL^3 / 48D + WL / AG,
where:
 W the loading Force value,
 L the midspan between the supports, so L=S
 D = E_f * b * t_f* h^2 / 2,
 AG = G_c * b * d^2 / c,
flexural rigidity of the facesheets, Ef Youngs Modulus facesheets, b = 75 the width, t_f the thickness of the facesheets here it is = 1mm, h = height/thickness of the panel, here it is 21 mm.
AG: shear rigidity of the core, G_c core sghear strength, b width of the specimen, d thickness of the specimen 21 mm, c thickness of the core 20mm.
In my case, by using the values of :
 W = 994,33 N
 L = 150 mm
 E_f = 52,6 Gpa
 b =75mm
 t_f = 1mm
 h=20mm
 G_c = 0,438 Gpa
 d=21mm
 c=20mm
We get that w_max = 0,132mm according to the literature.
This value is close to the non extended model, under 10%, and I don't know why? Normally it should be the other way around..but it is sure that it plays a role if simulate the extended model or not.
The thing is that I don't have experimental results to relate this with and the shear modulus of the core has come after a homogenization assumption.
I just wanna try doing the most realistic one in the end and that's why I also asked about doing an non linear contact simulation afterwards.
Because in reality the supporting bars will result to an area of contact with the panel and not a line I also did a simulation where the displacements were set on two faces of an area 12,5x75 (diameter of 12,5 rollers bars) and not on a line, and of course I had different results: upper face w_tot = 0,089 mm and lower face w_tot=0,083mm..
9)
After the formulation of that link : https://www.uniulm.de/fileadmin/website_uni_ulm/uzwr/lehreseminare/AdvMatFEM/WiSe201516/FELab1.pdf
I also tried cutting the model in half and fixed the center face which remains vertical as stated in the theory and put a force on a line/ face in a distance of S/2 from the fixed center face with the half value F=497,165 N along the positive Z axis:
I had also different results:
Now I don't know why the theory doesn't much with the extended model, and why I get that bif difference with last number configuration, where it is supposed to be the same.
I have to choose one model but I want it to be as aforementioned the closer to the bending test formulation of the 3 Point Bending Test of C393 ASTM.
So the only thing missing is to run also a simulation with a non linear contact simulation.
Can you help me?
Gratefully,
Achilleas

September 24, 2018 at 9:11 pmpeteroznewmanSubscriber
Dear Achilleas,
Wow! You have done a lot of work here. I have read it top to bottom and have the following comments to each of the quoted statements for the numbered sections.
1)
"since the model is basically simple, running time less than a minute, I didn't use the symmetry to take 1/4th of the panel."
Peter> Symmetry is valuable for applying necessary constraints, not just for faster solve times.
"Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0:"
"We can see that the deformation is not symmetric and there are rotations happening."
Peter> This is not enough constraints to prevent rigid body motion. That is why you are seeing nonsymmetric results and some rotation. There is not enough constraints. See my first point about using Symmetry.
2)
"Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0, X=0:"
Peter> Now you have too many constraints. What you want is a roller on each end so you don't create extra tension between the lines on the bottom as the load is applied, which would not occur with real rollers. That is what you can have when you use two planes of symmetry and build a 1/4 model.
3)
"Remote Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0, Scope from point center of each line and 12,5mm under the lower face of the specimen, Behavior Coupled:"
Peter> This is not enough constraints to prevent rigid body motion. That is why you are seeing nonsymmetric results and some rotation. There is not enough constraints. See my first point about using Symmetry. The 12.5 mm offset is not necessary.
4)
"Remote Displacements applied along two lines at a distance between them of S =150, Displacement condition Z=0X=0, Scope from point center of each line and 12,5mm under the lower face of the specimen, Behavior Coupled.."
Peter> Now you have too many constraints. What you want is a roller on each end so you don't create extra tension between the lines on the bottom as the load is applied, which would not occur with real rollers. That is what you can have when you use two planes of symmetry and build a 1/4 model.
Skipping 5) 6) 7) without the extension for now, and skipping .
9) "cutting the model in half and fixed the center face"
Peter> A fixed support is not the same as a symmetry boundary condition. A fixed support on the cut face overconstrains the nodes on that face. In the real model, while the plane remains vertical, the nodes are allowed to move around on that plane. For example, the nodes on the top face under compression want to expand laterally while the nodes on the bottom face under tension want to contract laterally due to Poisson's ratio. Note that the lateral motion will be symmetric about the center plane, leading to... you guessed it, a 1/4 symmetry model!
Kind regards,
Peter

September 25, 2018 at 8:37 pmAchilleasMilSubscriber
Thank you for your answer sir,
1)
I tried doing a 1/4th of the model where I set the following symmetry on two faces and two edges:
Afterwards applying only a Z=0 displacement on the corresponding line and a normal force of F/4 on the corresponding face I get:
upper face w_tot = 0,147 mm nad lkower face w_tot = 0,14 mm
After changing to remote force and applying a coupled behavior on the face I get:
upper face w_tot = 0,122 mm nad lkower face w_tot = 0,120 mm
Both have a difference from 10% or less from Literature results, but the second one is stiffer and the first one softer in contrast to analytic calculation of w_tot= 0,132 mm.
Is the symmetry settup correct? I wanted to ask, is it enough for the program if I aplly the symmetry just in the edges or do I also have to set it on the faces as I did?
2)
I also tried do one simulation with a honeycomb model made of only surface bodies, but after cutting my model to 1/4 th and hitting run I get constantly the error that:
'At least one body has no elements in it. this may be due to mesh based defeaturing of the geometry. the offending bodies can be identified in the worksheet view on the geometry folder and suppressed if desired'
And I figure out that the following faces positioned exactly on the plane which cut my model in the width direction have not been meshed, and after trying I cannot fix that. They remain unmeshed.
Do you know why that might be happening and how to solve it?
Gratefully,
Achilleas

September 26, 2018 at 12:15 ampeteroznewmanSubscriber
Hello Achilleas,
I'm glad to see that the two planes of symmetry are helping. Yes, you need symmetry on the edges and the faces.
Please show the location of the coordinate system that the Remote Displacement is using. It may not be on the center plane. It may be at the center of the face that was selected when the Remote Displacement was created. You have to move the Remote Displacement origin to the center plane after it is created. If you don't, it is as if you are doing a fourpoint bending test instead of a threepoint bending test, and those will have different deformations for the same force.
For the honeycomb model, go into Mesh Details and turn off Mesh Defeaturing.
Regards,
Peter

September 26, 2018 at 11:02 amAchilleasMilSubscriber
Yes you are right, I had it set on the global coordinate system and the point is shown in the middle of the face:
I tried setting a new coordinate system by choosing the face where the specimen was cut lengthwise:
And then set the remote force by that coordinate system but the result remains the same, and the point seems to be the same also:
So the values remain the same w_tot_up=0,122 mm and w_tot_low = 0,120 mm..
Also for the honeycomb I changed the automatic mesh based defeaturing off but still the faces remain unmeshed.
Is there something else I have to change?
Gratefully,
Achilleas

September 27, 2018 at 3:50 ampeteroznewmanSubscriber
Achilleas,
See where it says 6.25 mm for the X coordinate? Type over that with a 0.0 and the force will be applied on the center plane.
On the honeycomb, Clear the mesh, then with the Mesh branch picked in the Outline, right click on the face that does not normally get meshed and select Mesh Selected Body and see what happens.
Regards,
Peter

September 28, 2018 at 11:05 amAchilleasMilSubscriber
Hi sir,
Ok I changed the scope point for the remote force:
and the results are the same:
As for the honeycomb, I tried doing what you suggested and by showing the mesh in the display it seems that the faces are meshed, but I still got the same message after trying to run the simulation, so I went to the worksheet and again the specific faces had no mesh:
I went to the selected items in the tree:
and then hit show mesh again:
there is a mesh displayed but there is no edge or face connectivity:
The same connectivity is showed in the other side but there the faces have elements in the worksheet:
Do you have any other suggestion? Should I make another post maybe?
Thanks in advance for your help,
Gratefully,
Achilleas

September 28, 2018 at 7:05 pmAchilleasMilSubscriber
Hi again sir,
Found out what was going on with the honeycomb, there were doublicate faces remaining from after the cutting of the model to the 1/4 th.
I deleted the duplicates and it run. I noted the symmetry of the model in these faces and edges which were on the planes who cut the panel lengthwise and widthwise into the final 1/4th of the initial geometry:
I set the Axis normal to which there is the symmetry ( Y or X) and then got the following deformation plot displayed for the whole panel:
It looks correct and is near to the analytical value, but I am not sure if I have set the symmetry totally correct, can you please tell me if I have set anything wrong in the symmetry setup?
As for the remote point from which the force is scoped, I do not get a difference in the results by changing its location, should that be a problem?
Thanks a lot for your help!
Gratefully,
Achilleas

October 28, 2018 at 8:09 amvkumarsingh388Subscriber
Hey,
I also want to do 3 point bending test on honeycomb structure according to astm c393 but i don't know how to do it ,
as i read on upper post i have seen lots of suggestion but i don't understand some of them and also i want to know what is the correct approach.
I have made honeycomb structure using surfaces.
If you can please suggest me steps or suggest me some tutorials on how to do it then it would be best.

October 28, 2018 at 10:58 pmpeteroznewmanSubscriber
Please copy the above post and paste it into a New Discussion, then delete the post from this discussion, which has been marked as Solved. Some members won't look at posts that are marked as Solved.

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