Dear all:

I am working with a polyester [img]C:\Documents and Settings\msf\Escritorio\foro 1.tif[/img] 3000mm ID pipe (BS7159 Code), and I have two questions:

a) How can I know forces and moments in the brach point (I must give this data to the piping manufacturer for calculate branch reinforces if necessary)

If I study Global element forces I have the results attached. I do not know how to understand them:

For knowing forces and moments in the branch conection must I sum moments and forces for 30 and 33 nodes and the result of this operation will be forces and moments in the branch conection?

Thanks in advance

I hope I understand you're questions correctly. If not, these answers will maybe not be what you're looking for.

a) You can check for example the local elements output. It however depends on how you've modelled the tee in question. CII works with line models, so the forces and moments in the intersection are those calculated as if you've very small diameter pipe (centerline only). With a 3000mm ID pipe you want to know the forces and moments at 1500mm of the centerline of the run pipe.

b) you've nothing attached, but let's say nodes 10 - 20 - 30 is the run pipe, nodes 20 - 110 is the branch. What you need to know is the forces from piece 20-110 in node 20.

Thanks you very much... but must I check global element forces or local?

Dear Elena,

We can use report "Global Element Forces" or "Local Element Forces" report the result should be the same. But for local element forces the summation should be handle carefully. For me I will use global element force report since more easy to sum.

Let say:
230 - 240 - 245 ; Header
240 - 400 ; branch

There are 3 row report for node 240. Sum all forces and moment in the same X, Y, and Z direction (global).

Global element forces:
ELEMENT ---------Forces( N.)--------- --------Moments( N.m. )--------
NODES FX FY FZ MX MY MZ

230 -3659. 6892. -8156. -6600.8 2147.3 7950.9
240 3659. -6711. 8156. 4988.4 -2147.3 -7176.1

240 2158. 7773. -8598. 75.7 3378.5 5401.0
245 -2158. -7417. 8598. -75.7 20.1 -2315.1
.
.
.
240 -5817. -1062. 441. -5064.1 -1231.3 1775.1
400 5817. 1418. -441. 5064.1 1374.9 -1225.5

BE CAREFUL if your header and branch does not run in MAJOR coordinate since you may need the reaction value relative to your element run. Read also the pdf file relate to local coordinate and global coordinate located in CAESAR directory.

Elena,
I have simple solution for that kind of problem. If I want to know forces and moments at point which is important to you (usually I'm using it at flange connections, but maybe it will not be wrong if we apply it to branch connection). So you can simply insert an ANC restraint at that point with CNODE to point connected, so:

At point 240 put an ANC with CNODE to point 400, also at 240 insert SIF & TEES (there will be warning because Caesar will not find branch connection to this tee (SIF(i) can be different than it should be - but we can calculate SIF(i) without ANC and then manually insert calculated value at node 240 (in and out) in case when we model the ANC at node 240).
Then at point 400 you model your branch etc...
After running calculations you will find loads at point 240 at Restraint Summary.

Like I said - I'm using that kind solution for calculating flange bending moment, but maybe for branch it can be also correct (?)

Regards,
JK

Dear Giaa,

Yes I have ever heard that method but still not tried yet. BTW, why did you use node 400 as Cnode? it is already used as node for branch element. Is it a must for this method?

Using the more slower method explained by me above will make us understand that there are three elements give contribution forces and moments to node 240 that need to be summed to have each resultant in global X, Y, and Z direction.

Please look at the pdf document,

Regards,
JK

Dear Giaa,

Thank you very much for the pdf file. I think for my case above if using your method the elements run should be:
------------------------------------------------------------------------------------
Let say:
230 - 240 - 245 ; Header (in element 230-240 Restrain: ANC 240 Cnode: 400)
400- 410 ; branch instead of 240-400
------------------------------------------------------------------------------------

I will try to compare w/ the conventional method ...thx again.

Uhh! Thank you all very much.

I am starting trying both methods!!

Thx again :-)

Dear all, I have just tried to sum F and M for each component, but alwais obtain the same result: cero (0) So I suppose I am doing something wrong...

Dear Elena,

Ahhh...I'm starting doubt w/ my way. I start remembering Newton Law III and Free Body Diagram (FBD) concept that if we sum all F&M act inside one system = 0. FBD for Inside one system, all forces and moments neglect each other. Dear Corne is right???...:( Apologize me.

I mean in order to know F&M at node 240 the correct free body diagram should be (240 at the edge of the system):

++++++400++++++
+++++++o+++++++
+++++++|+++++++
+++++++|+++++++
+++++++|+++++++
+++++++o+++++++
++++++240++++++

not like this (node 240 inside the system):

++++++400++++++
+++++++o+++++++
+++++++|+++++++
+++++++|+++++++
+++++++|+++++++
o--------o--------o
230+++240++++245

Note:
o-----o : element
o : node
+ : paper