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#51220 - 10/09/12 02:58 AM Definition of a Distributed load
Lorenzo Offline
Member

Registered: 09/28/12
Posts: 19
Loc: Italy
Hy guy,

I have the problem to introduce in my model the distributed weight of some elements not consider in the modeling. Considering that I cannot use uniform loads because i need to apply the distributed weight only in a part of system, I was thinking to introduce this effect by evaluating an equivalent density for the pipe and to force this value in the Pipe density input only for the part of model in which I have to consider the distributed weight. I would like to ask you if you are agree with this way to go on or if in this way i introduce something wrong for the solution code running.

thank you for your help

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#51228 - 10/09/12 07:32 AM Re: Definition of a Distributed load [Re: Lorenzo]
Lorenzo Offline
Member

Registered: 09/28/12
Posts: 19
Loc: Italy
Looking at:

Sl = Slp + Fax/A + Sb

the equivalent density should affect only the resultant bending stress involving a more real value than avoiding the distributed weight introduced by the elements not included in the f.e.m. modeling.
In few word, equivalent density should involve a more real consideration of weight in the sustained condition.
Are you agreed?

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#51231 - 10/09/12 09:57 AM Re: Definition of a Distributed load [Re: Lorenzo]
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
I'm a little confused as to what you want to do. You can use "distributed loads", just set the magnitude to zero where applicable.
_________________________
Regards,
Richard Ay - Consultant

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#51232 - 10/09/12 10:42 AM Re: Definition of a Distributed load [Re: Lorenzo]
Lorenzo Offline
Member

Registered: 09/28/12
Posts: 19
Loc: Italy
Example:

Pipe Outside Diam: 850mm
Wall thickness: 12mm
Pipe density: 0.0083 Kg/cu.cm

If I also have 200kg distributed in 2m of pipe I could do something like this:

density = Mass/Volume

A = volume of 2m of pipe is: 2000mm * 3,14 * (sq(850/2) - sq((850-12-12)/2))
density = 200Kg/A

The value that i obtain is to be added to the 0,0083 of the pipe in order to obtain a density which also consider the weight which is distributed on 2m of pipe.

Could be correct to act in this way? otherwise I should act to uniformo load but the input is N/cm...

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#51240 - 10/09/12 09:03 PM Re: Definition of a Distributed load [Re: Lorenzo]
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
Just convert your value:

200kg * 9.8 m/sec^2 / 200 cm = 9.8 N/cm

Apply this value as a -Y uniform load for the 200 cm length. Then reset the Y uniform load to zero for the remainder of the system.

(Make sure the above conversion is correct - don't use my numbers.)
_________________________
Regards,
Richard Ay - Consultant

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#51242 - 10/10/12 01:14 AM Re: Definition of a Distributed load [Re: Lorenzo]
Lorenzo Offline
Member

Registered: 09/28/12
Posts: 19
Loc: Italy
Thank you Richard, for sure an easier and faster way to go on!
See you soon

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