Topic Options
#5764 - 06/08/06 05:33 PM Bilinear Support - Y2 .
Bilumon Offline
Member

Registered: 05/18/06
Posts: 8
Loc: Singapore
Hi all

Could anybody throw some light on the BiLinier support - Y2 ..

In short I wwant to know where is it practically we can use this.X2,Y2,Z2.
_________________________
Rama

Top
#5765 - 06/08/06 09:46 PM Re: Bilinear Support - Y2 .
anindya stress Offline
Member

Registered: 04/12/04
Posts: 493
Loc: London, UK
One application of bilinear support is to model sway braces.The other application is to model soil stiffnesses for U/G pipes.

Regards
_________________________
anindya

Top
#5766 - 06/19/06 03:35 AM Re: Bilinear Support - Y2 .
uhloop Offline
Member

Registered: 05/08/06
Posts: 13
Bilinear supports can be used for generating of additional friction forces independing from pipe weight. Such kind of supports may be used as vibration damperners.
One additional recommendation: Don't use bilinear supports and normal supports with friction simultaneously at the same NODE. In cases of inputing bilinear supports and normal supports with friction with the same NODE CAESARII calculates sometimes wrong restraint loads.
_________________________
uhloop

Top
#5767 - 06/19/06 09:44 AM Re: Bilinear Support - Y2 .
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
Quote:
In cases of inputing bilinear supports and normal supports with friction with the same NODE CAESARII calculates sometimes wrong restraint loads.
Hmmmm - I disagree with "wrong" above. If you read the section in the Technical Reference Manual on friction, you'll see that there are multiple equilibrium solutions to a system with friction - some of which may not make practical sense in the world of piping. Knowing this, I would not attempt to put two restraints (in the same direction) at the same node, both with friction. How would you suggest the friction load be shared among them?
_________________________
Regards,
Richard Ay - Consultant

Top
#5768 - 06/20/06 02:27 AM Re: Bilinear Support - Y2 .
uhloop Offline
Member

Registered: 05/08/06
Posts: 13
What I am understanding here when I say "wrong restraint loads" is the following thinking:
The sum of the forces of all external restraint of one model (without CNODE) in one horizontal direction (x or z) shall be zero for one static load case like "W+P+T+D". If such a sum reaches values of approximatly 10kN for small models including about 15 support NODES there is something wrong with the calculated results.
During the past years of my pipe stress analyses I found a few of such "wrong" solution. Everytime they could found only for models where I inputted at one NODE normal rigid supports with friction and bilinear supports.
The most interesting things of this models were - for a part of the static load case of type "W+P+T+D" of one job the result were OK for other not. After the change of the "Friction Normal Force Variation" parameter the "wrong load cases" could show good results, but some "good load cases" became wrong ones.
The sum of the restaint loads were every time OK when I was setting the fiction factor to "0" for the part of the normal rigid support.
A simple example for such a support in question consist of a horizontal pipe section actuating on a steel beam giving a friction force from the weihgt loads in addition with an "U-bolt"-clamp fastened in a force-fitted manner giving an axial resistance due to friction.
_________________________
uhloop

Top
#5769 - 06/20/06 07:57 AM Re: Bilinear Support - Y2 .
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
But, friction is a non-conservative phenomenon. The assumptions above don't necessarily hold true. Get a copy of Sobieszczanski’s ASME Paper paper and read it.
_________________________
Regards,
Richard Ay - Consultant

Top
#5770 - 06/20/06 08:36 AM Re: Bilinear Support - Y2 .
Dave Diehl Offline
Member

Registered: 12/14/99
Posts: 2382
Loc: Houston, TX, USA
You mention that you change the "Friction Normal Force Variation". I think this might be the source of the imbalance. By changing those parameters you are changing the tolerance on the trigger to continue iterating for a solution. For example, that particular parameter (normal force variation) is set at 15% by default. That means that if, in the current iteration, the normal load at that friction support changes more than 15% from the previous iteration, the friction force (mu*N) used in the current iteration is off by more than 15% and the program will recalculate that friction force and run through another iteration to get things (more) right. If you increase that variation limit, you might satisfy this check but your numbers drift even farther from the "right" answer.
_________________________
Dave Diehl

Top



Moderator:  Denny_Thomas, uribejl 
Who's Online
0 registered (), 42 Guests and 2 Spiders online.
Key: Admin, Global Mod, Mod
May
Su M Tu W Th F Sa
1 2 3 4
5 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30 31
Forum Stats
12065 Members
14 Forums
16973 Topics
75151 Posts

Max Online: 303 @ 01/28/20 11:58 PM
Top Posters (30 Days)