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#67804 - 12/01/16 04:48 AM Displacement Input in Caesar II (Not as an anchor)
Borzki Offline
Member

Registered: 09/16/04
Posts: 759
Loc: Traz
Hello Stressers!!!

Is there a way in Caesar II to input displacements not acting as an anchor with movements. Basically, as a displacement internally within the pipe.

Is there a work around to handle this?

Many Thanks!!! Cheers!!!

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#67805 - 12/01/16 06:19 AM Re: Displacement Input in Caesar II (Not as an anchor) [Re: Borzki]
Borzki Offline
Member

Registered: 09/16/04
Posts: 759
Loc: Traz
Basically my objective is to super impose the displacement at a node with the thermal displacement but with the pipe free at that node (no restraint at that point).

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#67806 - 12/01/16 08:02 AM Re: Displacement Input in Caesar II (Not as an anchor) [Re: Borzki]
Ali_Raza_Bashir Offline
Member

Registered: 09/05/16
Posts: 19
Loc: Pakistan

Based upon what I have understood from you post, I suggest the following:

Directly input displacement vector value by checking displacement tab in CAESAR II without any Cnode and subsequently add the D in load cases like W+P1+T1+D1.
This will impose the displacement on the system without any restraints.
But keep in mind that this is an imposed displacement and it will not algebraically resolve with piping thermal displacement in W+P+T+D.

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#67807 - 12/01/16 09:05 AM Re: Displacement Input in Caesar II (Not as an anchor) [Re: Ali_Raza_Bashir]
Borzki Offline
Member

Registered: 09/16/04
Posts: 759
Loc: Traz
That's right. I have done it. The one I'm trying to find is the one that resolves into thermal displacement. Anyway, thanks for the info.

Anybody experience the same situation.

Many Thanks,

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#67821 - 12/02/16 09:21 AM Re: Displacement Input in Caesar II (Not as an anchor) [Re: Borzki]
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
A displacement is a boundary condition (the same as a restraint). Stated another way, a restraint is the same as a zero-valued displacement.

You can accomplish what you want to do with a force. Apply a unit force in the direction you're interested in. Run the model with only that force in the load case to obtain the displacements. Finally, scale up the force to obtain the displacement you're after. Use this final force value in your real load cases.
_________________________
Regards,
Richard Ay - Consultant

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#67831 - 12/03/16 05:53 PM Re: Displacement Input in Caesar II (Not as an anchor) [Re: Richard Ay]
Borzki Offline
Member

Registered: 09/16/04
Posts: 759
Loc: Traz
Nice Richard... You've nailed it... Thanks for that info... I'll try to put it in my model...But I know this would yield a more realistic result rather than inputting displacement (since it's a boundary condition)..

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#68407 - 02/21/17 05:14 AM Re: Displacement Input in Caesar II (Not as an anchor) [Re: Borzki]
Ali_Raza_Bashir Offline
Member

Registered: 09/05/16
Posts: 19
Loc: Pakistan
Dear Richard,

I am facing the same problem as stated above. But I am unable to understand what exactly I have to do to resolve these imposed forces and moments. Could you please share any example.

Regards,
Ali

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