# how to constrain rigid body motion

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 May 26, 2013, 14:04 how to constrain rigid body motion #1 Member   Mohamad Alagheband Join Date: Oct 2012 Posts: 41 Rep Power: 13 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

 May 26, 2013, 21:34 #2 New Member   Ma.L Join Date: May 2013 Location: ShangHai,China Posts: 17 Rep Power: 13 when you define a rigidbody,you can define the degree of freedom of it，including Movement and Rotation。 good luck。

May 27, 2013, 05:32
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Quote:
 Originally Posted by Mal when you define a rigidbody,you can define the degree of freedom of it，including Movement and Rotation。 good luck。
dear MAL
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

 May 27, 2013, 08:17 #4 Senior Member   Matthias Voß Join Date: Mar 2009 Location: Berlin, Germany Posts: 449 Rep Power: 20 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.

May 27, 2013, 14:25
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dear neewbie
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
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 May 28, 2013, 09:56 #6 Senior Member   Matthias Voß Join Date: Mar 2009 Location: Berlin, Germany Posts: 449 Rep Power: 20 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 likes this.

 May 28, 2013, 10:58 #7 Member   Mohamad Alagheband Join Date: Oct 2012 Posts: 41 Rep Power: 13 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

 May 28, 2013, 11:09 #8 Senior Member   Matthias Voß Join Date: Mar 2009 Location: Berlin, Germany Posts: 449 Rep Power: 20 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.

May 28, 2013, 13:18
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Quote:
 Originally Posted by neewbie 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
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Last edited by MUMMED; May 28, 2013 at 15:26.

 May 28, 2013, 15:12 #10 Senior Member   Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 511 Rep Power: 20 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 likes this.

May 28, 2013, 15:37
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Quote:
 Originally Posted by singer1812 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.
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

 May 28, 2013, 16:39 #12 Senior Member   Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 511 Rep Power: 20 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.

 May 29, 2013, 02:49 #13 Member   Mohamad Alagheband Join Date: Oct 2012 Posts: 41 Rep Power: 13 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

April 13, 2020, 13:47
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John Applesmith
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Quote:
 Originally Posted by singer1812 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?