RBE3 problem or something else !? 5

[Davoue]

Hello every one,

I'm new at FEA Advance Simulation in NX9.

I have two parts one ' A ' is fixed and the other ' B ' is on top of it.

I made some RBE2 connections between the two parts.

At this time I can solve my solution.

But the part ' B ' is bending in the middle of it. I want it to lay down on the part ' A ' to see deviation of the part ' A ' when it support the weight of the part ' B '.

I've been told to use RBE3 connections between the two parts because I don't want them to be rigid...

I connected the two parts with one nod on each parts.

The solver in .f06 give me fatal error :

*** USER FATAL MESSAGE 2038 (RBE3D)
RBE3 ELEMENT 51 IS SINGULAR.
USER ACTION: ADD MORE DOFS TO THE CONNECTED POINTS TO INSURE THAT THEY CAN CONSTRAIN ALL 6 RIGID BODY MODES
OF THE ELEMENT.

I tried to manually play with different DOFS configurations in mesh associated Data. I cannot find what is wrong with my RBE3.

Do I use the wrong connection ? Maybe it is a problem my nods are not really co linear?

I tried to find out what I'm doing wrong in this forum and I found nothing.

I hope you can help me folks to find what is my problem and BTW the your eng-tips forum are very helpfuuuulll!!!!

 

[stressebookllc]

From what I know, RBE3 is an interpolation element. It works by resolving applied load into forces and moments (better not choose rotations for the independent grid points, unless you really know what you are doing because it is very tricky), not displacements like RBE2 does.

RBE3 will need enough independent grid points to map the applied loading from the dependent grid point. Try to model Part A with at least three nodes not inline and use a concentrated mass for part B and connect it to the three nodes on Part A and see if that works. It should.

http://www.stressebook.com

Stressing Stresslessly!

 

[rb1957]

"I've been told to use RBE3 connections between the two parts because I don't want them to be rigid" ... the "R" in RBE is for "rigid". if you want controlled stiffness, use a CBUSH.

i think the FEA doesn't like a part of the model completely supported RBEs ... i'd add one nodal constraint (all 6dofs) and check it (it should be zero).

being new, i'd attack the problem more slowly ...
1) "weld" the two pieces together (use the same node on both parts, say three or four should be enough),
2) then use RBEs (to model fasteners ?) ... i don't think you can have an RBE linked directly to another RBE so i'd put a CBUSH between them (to model the stiffness of the fastener).

another day in paradise, or is paradise one day closer ?

 

[nlgyro]

From your description it looks like part B needs to be constrained in the in-plane rotational dof. Add the corresponding rotational dof on the RBE3 element and the element singularity should be solved.

 

[Naffoussi]

"I connected the two parts with one nod on each parts." this is the problem.
as stressebookllc mentioned above, pick more nodes on each parts to form a "plane" this will avoid the rotational dof of the part B...
R means rigid but not for type 3...

Seif Eddine Naffoussi, Stress Engineer

http://www.Innovamech.com

33650 Martillac û France

 

[Davoue]

Thank you guys and Seif who already answer me one month ago on another thread!

Thats it you help me found my error I was using like 1 nod for my first pick of the RBE3 and for my second I picked only 1 nod.

What worked for me is : I used 1 nod for my first pick and 3 nods for my second pick and it worked.


I am using NX Nastran Simulation Fem. My other problem was that I was trying for my first pick on part A to take a lot of nods and a lot of nods for my second pick for part B it didn't work.

It seem in NX Nastran Simulation Fem that I have to pick them each one at the time,( I pick at every 6 inches ) it take lot of time because my parts was big, but it looked to work fine.

Thank you guys it was very helpful and I hope this thread will help others, I'm sure it will !!!





Mechanical designer
STYL&TECH inc.

http://www.stylntech.com

 

[iceblink]

I've tried to reproduce loads in an RBE3 with hand calculations, and I can see that the results are wrong, but I fail to see why. Can someone have a look and tell me what i'm doing wrong?

http://communities.ptc.com/servlet/JiveServlet/download/271683-78849/Mathcad - rbe3.pdf

p1-p3 are the independent nodes.
PL is the dependent node.
F and M is force and moment on the dependent node.
pCG is the calculated "centre of gravity" between the independent nodes.
r1-r3 are the vectors between the CG and the independent nodes.
FF1-FF3 are resultant forces in the indep. nodes due to the force in the CG
FM1-FM3 are resultant forces in the indep. nodes due to the moment in the CG

Unfortunately when i calculate the resulting force and moment of all the forces, the force is okay but the resulting moment is wrong.

By the way the author of the stressbook.com article probably forgot to mention the source of his pictures ;-)

 

[rb1957]

"I used 1 nod for my first pick and 3 nods for my second pick and it worked." ... first pick, secon pick ... very sloppy terminology, maybe "ESL". the nodes being "picked" have specific names, dependent and independent, and it's important to appreciate the difference. i suspect if you'd read the manual that the limitations would have been apparent.

another day in paradise, or is paradise one day closer ?

 

[rb1957]

@iceblink,
your Mcg is wrong ... you haven't calculated the distance between the CG and L, it looks like you've just subtracted their co-ordinates. i figure the Mcg is 6*(2*sqrt(2)) = 16.97.

if this is a student post, pls go to engineering.com

another day in paradise, or is paradise one day closer ?

 

[iceblink]

Hmmm I don't see how Mcg is wrong... My example is almost 2D so it can be drawn on a sheet of paper. Load is applied in (3,3) and is 6 into the paper. CG is in (1,1). That means the moment around vertical is the force multiplied with the horizontal distance, so 6 * (3-1) = 12. Moment around horizontal is force multiplied with vertical distance, so also 12, except it points the wrong way, so -12. These two moments combined give SQRT(12^2+(-12)^2) = 16.97 pointing in 45 deg in the XZ-plane.

 

[rb1957]

you're right, i tried to recant, i looked again and you're using vector components and the resultant is correct (as you show).

i drew a napnkin sketch of your problem. do you realise your getting the applied load distributed over your 3 points (as opposed to the rections). The momentCG will give a -ve (load) on 1 and a +ve on 2 and 3, FM1 will be double FM2; so 1/2 the moment, 12sqrt(2), will be reacted at 1, sqrt(2) from the CG, so the load on 1 due to moment should be -6 and +3 at 2 and 3 (in the 2 direction). ok, now i reckon i've got the answer, lets look at your math.

You're out by a factor of 3. i suspect that FM1 is reacting only 1/3rd of MCG; i suspect that you need to modify "sumsq" by dividing by "n".

another day in paradise, or is paradise one day closer ?

 

[iceblink]

It's not quite that simple...

Using your numbers:
Moment around x-axis in CG: -6 * 1 + 3 * -2 + 3 * 1 = -9
Moment around z-axis in CG: -6 * -1 + 3 * -1 + 3 * 2 = 9
But we should have found the Mcg which is -12 around x-axis and 12 around z-axis.

The correct forces due to the moment are -8 for point 1 and +4 for point 2 and 3.
And it is not just a matter of a simple factor difference, because if the load is (6,6,0) my calculation finds an FM1 = (-1,-2,1) whereas this should be (-1,-8,1).
(Your remarks did make me think about different sumsq factors for each projection plane, so i'm looking into that now.)

You are absolutely right about components and reactions, but I'll worry about the minus sign later

Thank you for looking at it. Much appreciated!

 

[rb1957]

i can't open your attachment now (site maintenance).

is the load at (6,6) ? earlier you said (3,3) which i think is right, applied load is 6, yes? that gives a moment of 12 in both direction (or 12sqrt(2) combined, as we've poted above).

the moment load on 1 is -8, and +4 on 2 and 3 ... i had quickly and incorrectly said (to myself) the moment absorbed at 1 is 1/2 the applied, but it is 2/3 (i can see why, now!?).

another day in paradise, or is paradise one day closer ?

 

[rb1957]

attachment still unavailable "organisation's certificate has been revoked" ? maybe upload through engineering.com ?


another day in paradise, or is paradise one day closer ?

 

[iceblink]

The site moved to a new domain past weekend. You can find the file here now.

Why is the moment absorbed at point 1 two thirds of the total moment? I'm looking for the math behind that.

 

[rb1957]

one way to see why (2/3rd moment reacted at 1) is to note that force 1 and force at 2 and 3 balance (are the same, force 2 = force 3 = force 1/2) but the moment arm for 1 is double the moment arm for 2 and 3, so moment reacted at 1 is double the moment reacted by 2 and 3, ie 2/3rd of the applied.

the other way to see this is to work through the equations of equlibrium.

i think your problem is sumsq. i think you should be summing the distance component in the direction being considered. i'd write the equations out long hand, then re-express as matrix math.



another day in paradise, or is paradise one day closer ?

 

[iceblink]

I wrote down the equations but at the end i'm still off by a factor of 2...

 

[rb1957]

looking at things, your problem is with the YZ plane, MCGx is a value (-12 i think), so there should be Fy loads. they should be the same as your XY plane, and so your factor of two.

another day in paradise, or is paradise one day closer ?