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deflections/stress due to self weight

5 REPLIES 5
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Message 1 of 6
madg1
2236 Views, 5 Replies

deflections/stress due to self weight

hello

 

suppose there is a large beam say 80 ft cantilivered beam rectangular c/s.

so asim doesnt output results beacuse of the loading caused only by self weight right?

madg1
ASIMM 2012
5 REPLIES 5
Message 2 of 6
John_Holtz
in reply to: madg1

Wrong. The beam has weight, and causes stress and displacement, if the material properties include the mass density and gravity ti turned on.

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉
Message 3 of 6
madg1
in reply to: madg1

wow...I somehow thought that gravitaional force will be added.....we need to turn it on.....thats why i see less stress values......thanks john...

 

madg1
ASIMM 2012
Message 4 of 6
madg1
in reply to: John_Holtz

John one more question

 

Say I got a same cantilevered rectangular c/s beam solid meshed say with tet elements.

say side A is fixed and side B is free.

 

suppose on the top of side B of the cantileverd ,solid meshed beam, i would create a beam element ( yes i will create X to avoid balljoint) and give rectangular section properties.Its a dummy element and I want to give a mass property of my own, not the one calculated by software.Is there an accepted way to this?

 

What I did was, make the density of the Beam element Zero   and add a distributed load value of my required weight.

 

I did that and I see some weird number in the weight of the overall structure.So I am wondering if there is an accepted way to do that?

madg1
ASIMM 2012
Message 5 of 6
John_Holtz
in reply to: madg1

Hi,

 

I do not know how you are getting the weight of the overall structure ("I see some weird number in the weight"), but a distributed load is obviously not a weight. So if you are using the "Analysis > Analysis > Weight and Center of Gravity", the calculated weight will not include the "weight" that you simulated using the distributed load.

 

If you sum the reaction forces in the vertical direction, they should equal the weight of the solid elements plus the distributed load applied to the beam elements.

 

The other way to add weight to a model is with the "Setup > Loads > Weight" command. Smiley Happy

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉
Message 6 of 6
madg1
in reply to: John_Holtz

yes you are right...

 

I theroized that distributed load in global Z direction as weight of the part if you can change the value of density of the part to zero ...obvisouly you can get that as reaction force....Smiley Very HappySmiley Happy

madg1
ASIMM 2012

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