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-   -   how to constrain rigid body motion (https://www.cfd-online.com/Forums/cfx/118379-how-constrain-rigid-body-motion.html)

MUMMED May 26, 2013 14:04
how to constrain rigid body motion
 
Hi
Does anybody know how I can constrain a rigid body motion I mean for like specific rotational angle ?
I can not just stop the solution and re-run for fixed boundaries because my inlet velocity profile is pulsatile.
thanks

Mal May 26, 2013 21:34
when you define a rigidbody,you can define the degree of freedom of it，including Movement and Rotation。
good luck。

MUMMED May 27, 2013 05:32
Quote:
Originally Posted by Mal (Post 430151)
when you define a rigidbody,you can define the degree of freedom of it，including Movement and Rotation。
good luck。
dear MAL
thanks for reply
I knew what you mentioned but i am trying to constrain rotation angle i mean when RB rotated for certain angles then it should be stopped in that position and solution go on with fixed boundaries and i don't know how much time does it take to rotate and i am going to find out by RB simulation?
Tnx

mvoss May 27, 2013 08:17
hey

not sure if i got you right, but if you can express your RBmovement via CCL then you can simply add another expression with respect to simulation "runtime counters" like AITERN,CITERN,T,TSTEP or ATSTEP. e.g.
Code:
      RampLength = 1.0[s]
      Ramp = (-2*(min((t)/RampLength,1))^3+3*(min((t)/RampLength,1))^2)
and then multiply your "movement" by the ramp-value.

MUMMED May 27, 2013 14:25
1 Attachment(s)
dear neewbie
thank for reply
I am afraid i did not really get what you mean but do you have any idea how i can model this rotation with rigidbody

mvoss May 28, 2013 09:56
some thoughts:
1# you can apply a symmetry on the middle ?!
2# never done smth. with RBM but maybe you can use this as a starting point

Code:
New X = ((x-X0)-Total Mesh Displacement X)*cos(alpha)+((y-Y0)-Total Mesh Displacement Y)*sin(alpha)+X0
New Y = -((x-X0)-Total Mesh Displacement X)*sin(alpha)+((y-Y0)-Total Mesh Displacement Y)*cos(alpha)+Y0
New Z = ((z-Z0)-Total Mesh Displacement Z) + Z0
X0 = 0 [mm]
Y0 = 0 [mm]
Z0 = 0[mm]
by replacing the MeshDisplacemt with smth. similar for RBM and give the body the new x,y,z as coordinates. X0Y0Z0 should be the center of rotation.
3# your are running transient cases, right?

MUMMED May 28, 2013 10:58
Dear Matthias
I am really appreciate for spending time to answer.
As you said I am running a transient simulation with symmetric geometry
and my biggest problem is finding rotation angle .
I know the range of rotation but i do not know exact value for rotation angle and i am trying to find out by RBM (i mean applying momentum forces to RB)
and i think that code must work for known rotation angle

mvoss May 28, 2013 11:09
u r welcome.

You apply the forces on the RB (input) and you are interested in the transient behavior and end position(output) of the RB in equilibrium state when also applying the given inlet conditions?!

So to get the real position is totally a topic for post-processing. Is that correct?
First guess--- define a Monitor point at a certain node of the RB and grab the coordinates.

MUMMED May 28, 2013 13:18
1 Attachment(s)
Quote:
Originally Posted by neewbie (Post 430554)
u r welcome.

You apply the forces on the RB (input) and you are interested in the transient behavior and end position(output) of the RB in equilibrium state when also applying the given inlet conditions?!

So to get the real position is totally a topic for post-processing. Is that correct?
First guess--- define a Monitor point at a certain node of the RB and grab the coordinates.
I think I went wrong
this is my problem
the leaflet must stop at 85 degree and simulation go on until the end of the pulse
thank you a lot

singer1812 May 28, 2013 15:12
If you are using the 6DOF solver, as implemented in CFX, you cant constrain a rigid body to rotate about a hingepoint (not coincedent with the body center).

That said, this is easy enough to do with the CEL equations that newbie has layed out (I didnt check for correctness). You can use CEL to limit alpha to what you want, and you can use equations of motion to put d(alpha)/dt in terms of forces (moments), which you can read off the body.

Note (X0,Y0,Z0) will be your hinge point.

MUMMED May 28, 2013 15:37
Quote:
Originally Posted by singer1812 (Post 430603)
If you are using the 6DOF solver, as implemented in CFX, you cant constrain a rigid body to rotate about a hingepoint (not coincedent with the body center).

That said, this is easy enough to do with the CEL equations that newbie has layed out (I didnt check for correctness). You can use CEL to limit alpha to what you want, and you can use equations of motion to put d(alpha)/dt in terms of forces (moments), which you can read off the body.

Note (X0,Y0,Z0) will be your hinge point.
Dear singer thanks for reply
up to now I have been working with 6DOF solver and I do not have much experience to apply momentum with CEL.
Do you have any suggestion?
thanks

singer1812 May 28, 2013 16:39
Are you familiar with CEL?

Can you discretize angular motion equations?

If the answer to both of those is no, I would start with reading up on the CEL section of CFX help docs.

Then read up on angular motion in a dynamics book.

A complete discussion is more than I want to put in this forum.

Specific questions on either of those topics (other than how do i do the entire thing) I can answer.

MUMMED May 29, 2013 02:49
you are right
I read CEL section and angular momentum before and I should take a look again
I would like to thank you again for spending time.
Thanks

ns778 April 13, 2020 13:47
Quote:
Originally Posted by singer1812 (Post 430603)
If you are using the 6DOF solver, as implemented in CFX, you cant constrain a rigid body to rotate about a hingepoint (not coincedent with the body center).

That said, this is easy enough to do with the CEL equations that newbie has layed out (I didnt check for correctness). You can use CEL to limit alpha to what you want, and you can use equations of motion to put d(alpha)/dt in terms of forces (moments), which you can read off the body.

Note (X0,Y0,Z0) will be your hinge point.
Hi, even if you define a new coordinate frame with its origin at the middle of the hingepoint does the 6 DOF solver not give you what you want?


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