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Valued Contributor
3DAli
Posts: 63
Registered: ‎10-14-2008
Message 1 of 6 (562 Views)
Accepted Solution

Applying moment Load on multiple components

562 Views, 5 Replies
05-08-2012 12:46 PM

Hi there,

 

I'd like to know how inventor defines the center of rotation when you apply moment load on multiple faces at once. I'm not basically able to define where exactly I need that axis to be. and I dont understand where inventor assumes that axis is. if you take a look at attached model, obviously the resultant forces caused by the applied moment are bigger on the single pillar located on the extention, thats why its been displaced more than the others, but where exactly is center of rotation. please let me know if I'm missing something or this might not the right approach to study this kind of scenarios. I'm trying to avoid modeling of the section above this which technically transfers the moment thorough shear pins on the structure I have modeled.

 

your help is very much appreciated in advance

Sincerely

Ali

Warm Regards
Employee
henderh
Posts: 649
Registered: ‎06-07-2007
Message 2 of 6 (549 Views)

Re: Applying moment Load on multiple components

05-08-2012 04:23 PM in reply to: 3DAli

Hi Ali,

 

  When a load is applied to multiple faces, the load magnitude is divided according to face area ratio (distributed as the percentage of total area).  You can also verify this by inspecting the reaction forces.

 

  With a moment load, the 'center of rotation' is about the centroid of the face.  In the case of multiple faces selected for a single moment load, the 'center of rotation' is about the centroid of the combined areas...which of course could be off both faces.  I've attached an image to illustrate.

 

  Another method you can try [Edit: to convince yourself of the "distributed load" behavior] is to use individual fixed constraints under each pillar.  You will have to chop the 'base' up so that each pillar is 'stand-alone' (to keep the 1:1 load to constraint relationship).  Again, the reaction moments are about the centroid of the face(s) with the fixed constraint.

 

  What I'm not sure I understand is why the 4th pillar (at 9 o'clock in your image) appears compressed and not 'torqued'.  If you use the chopping method above, you can verify from the reactions if the load is really compressive or a moment.  I don't think it would be compressed if using a single moment load, I think it is the camera view that creates this illusion in your image.

 

Hope this helps.  Please let us know if you have additional questions or comments.



Hugh Henderson
Simulation QA Engineer
DLS
Valued Contributor
3DAli
Posts: 63
Registered: ‎10-14-2008
Message 3 of 6 (528 Views)

Re: Applying moment Load on multiple components

05-08-2012 07:27 PM in reply to: henderh

Hi Hugh

 

Thanks for the reply, I really appreciate it. I guess its the camera view that cause that illusion on that pillar, how ever, In the company I'm with, they used to use Solidworks and since I had a wonderful experience with Inventor, they decided to switch. a good friend of mine there showed me how they could apply torque and choose center of rotation (basically define a work axis and select it as center of rotation) in SW, I wanted to create the same analysis with IV and now its not apple to apple. in reality though, it does matter where the center of rotation is and there should be a way to do this, even if I choose to translate Moment into force and apply it on the faces to create the same effect as moment does. would you mind assist me as this is really important to me and my employer.

 

Thanks a lot

P.S. this is not to compare two softwares, after all , modeling and file management is fantastic in IV and thats the reason why they took my advice to switch.

 

Sincerely

Ali

Warm Regards
Employee
henderh
Posts: 649
Registered: ‎06-07-2007
Message 4 of 6 (501 Views)

Re: Applying moment Load on multiple components

05-09-2012 12:26 PM in reply to: 3DAli

Hi Ali,

 

  There are some options you can use for your simulation situation.  The easiest, most straightforward way I can think of right now is to make a "torque plate helper part" to use where you can vary the moment center of rotation.  This is as close as simulating the real-life situation without having to model then entire assembly.

 

  You mention translating the applied moment into Force.  This is possible using Force, Remote Force [Edit: or Bearing Load] but would require either hand calculations or setting up a different sort of 'helper part' that would have a layout of the center of rotation, load sense (direction), and moment arm distances.  It could be as easy as drawing out a sketch and extruding a surface to use to define the load directions.  It would still require some bit of hand calcs (or entering the force loads as moment/distance) if the center of rotation is not equidistant to where the loads are applied.

 

  Please let us know if the first option works for you.  I've attached a sample assembly in R2012 version as well.

 

Thanks and Best regards, -Hugh

 

Torque_plate_helper_part.png



Hugh Henderson
Simulation QA Engineer
DLS
Valued Contributor
3DAli
Posts: 63
Registered: ‎10-14-2008
Message 5 of 6 (487 Views)

Re: Applying moment Load on multiple components

05-09-2012 02:13 PM in reply to: henderh

Thanks a lot Hugh,

 

to be honest with you, I thought about the same approach and modeled something very similar to what you did. I appreciate your help, just one more question, is there a way to mark that helper part as a rigid body so it doesnt deform itself and transfers all the load onto my actual modeled assembly? if so, this solution would be perfect.

 

Regards

Ali

Warm Regards
Employee
henderh
Posts: 649
Registered: ‎06-07-2007
Message 6 of 6 (485 Views)

Re: Applying moment Load on multiple components

05-09-2012 02:34 PM in reply to: 3DAli

Hi Ali,

 

  You could specify a very, very high stiffness (Young's Modulus) in the material property specification.  This will minimize the deflection of the helper part.

 

  However, I would think all the load has to transfer through from the helper part due to the linear FEA solver  No matter the stiffness, the reactions will be equal and opposite to the applied loads (continuum mechanics)

 

  We appreciate the feedback and follow-up!  As always, let us know if you have additional questions comments or suggestions.

 

Warm regards, -Hugh



Hugh Henderson
Simulation QA Engineer
DLS

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