| Applies To | |||
| Product(s): | AutoPIPE, | ||
| Version(s): | All | ||
| Environment: | N/A | ||
| Area: | Report | ||
| Original Author: | Bentley Technical Support Group | ||
Problem:
How to sum forces about a Tee = 0?
Example #1: Balance Forces at Tee node Point
If I wanted to run a separate analysis on that tee (nodes E01, E02, E03, and R02), I would want to apply the observed forces and moments to the faces to recreate the loading. I have tried following the help file convention but seem to be confused on which of the values read from the chart should be taken as their correct value, or if I need to negate what is shown. Specifically, I am applying the loads from E01, E03, and R02.
Example #2: Balance Forces from 3 node points around a Tee node point
Consider an AutoPIPE example system, Class1_tutorial.DAT, found in the AutoPIPE's Example Model folder (i.e. File> Open Example Models).
There is only one tee in this model, A150
Solution:
Example #1: Balance Forces at Tee node Point
AutoPIPE Forces and Moments report displays the pipe internal forces and moments caused by the applied loads. To be able to recreate the loading scenario on the Tee, you would need to apply the forces and moments in the direction opposite to the internal forces and moments displayed in the output report.
Considering gravity load case only for the tee point E02:
Forces and Moments to be applied (Global)
Method #1:
See docuemt here that explains calculation in detail
Method #2::
At the TEE point with no branch connection then the loads would be the same on each side. Adding the branch connection has the effect of introducing a discontinuity (like the supports do at nodes). Therefore, in this instance then the loads do not match and the difference is the load in the branch line!
Remember that the reported loads are in LOCAL coordinates not GLOBAL. The diagram below simplifies it and demonstrates the point.
Local forces and moments from a Forces and Moments report:
|
Point |
Combination |
Fx |
Fy |
Fz |
FR |
Mx |
My |
Mz |
Mr |
|
W12 Branch |
GP1T1{1} |
-706 |
-4835 |
1842 |
5222 |
140 |
-709 |
-2626 |
2724 |
|
W12 - |
GP1T1{1} |
24599 |
-9525 |
-23520 |
35342 |
-6439 |
-67701 |
-125943 |
143131 |
|
W12 + |
GP1T1{1} |
22757 |
-8819 |
-18685 |
30737 |
-3812 |
-67840 |
-125235 |
142480 |
Resolving local Fx (header) and local Fz (branch) at node point W12:
Fx (- header) – Fx (+ header) = Fz (branch)
24599 – 22757 = 1842
In a similar way the other two forces compute:
Fy (-header) – Fy (+header) = Fx (branch)
-9525-(-8819) = -706
Fz (-header) – Fz (+header) = Fy (branch)
-23520-(-18685) = -4835
There are known differences between the General result and the code compliant result. This is down to a number of things but the main reason is illustrated below.
General Pipe Code Compliance uses effective section modulus for branch end of pipe (including out of plan SIF consideration), whereas General Pipe Stress does not use effective section modulus.
Uses Zeffective for reduced tees at branch (pi.r^2.teff where teff is the lesser of SIfout.tb and tm & r = Rm of branch using tb)
Some sample calculation for effective vs natural section modulus for node W12 branch:
|
Header Outside diameter |
Dho |
508 |
mm | |
|
Header wall thickness |
th |
20.62 |
mm | |
|
Header Inside diameter |
Dhi |
466.76 |
mm | |
|
Branch outside diameter |
Dbo |
60.3 |
mm | |
|
Branch wall thickness |
tb |
8.74 |
mm | |
|
Branch inside diameter |
Dbi |
42.82 |
mm | |
|
Mean branch radius |
r2 |
25.78 |
mm | |
|
In plane SIF |
SIFI |
2.106692 | ||
|
Out plan SIF |
SIFO |
2.106692 | ||
|
Natural section modulus branch |
Z |
16051.88 |
mm^3 | |
|
Effective branch wall thickness |
Ts |
18.41 |
mm |
Lesser of th and SIFI*tb |
|
Effective section modulus branch |
Ze |
38443.97 |
mm^3 | |
|
Effective / Natural |
Ze / Z |
2.39 |
Looking at the ratio of effective vs natural section modulus is about 2.39 which is substantial and causing the differences in stresses.
Example #2: Balance Forces from 3 node points around a Tee
Open example model, Class1_tutorial.DAT, in AutoPIPE. A few dialogs may appear but that is OK This example problem will assume any edition of AutoPIPE (i.e. Standard, Advance, Nuclear) will be able to reproduce the same results by following the steps below:
Note:
1. License is set to Standard, and using AutoPIPE CONNECT 10.01.00.08
2. Highly suggest saving this example model to a new folder after opening to maintain original file for future usage.
Step #1: Change Piping code to ASME B31.3 - 2014 on the General Model Options dialog
Step #2: Set Material = A53-A for all pipe properties.
Step #3: Add the following Xtra Data Reference points
Step #4: Analyze > Static Only
Step #5: Highlight node points
a. A140 - A160
b. A150 to B180
Global Axis:
Local Axis:
Step #6: To keep things simple, limit the output report to just Load Set #1. Turn off print option for all load sets except for combinations added to Load Set 1, (ex. Gravity(1)):
Step #7:Create An output report, limit results to highlighted points, to include the following sub reports only: Forces & Moments and Reference Loads.
Step #8: Review Output report,
Step #9: Use Excel to compile the results about node point A150 in the global / local Dx, Dy, & Dz directions.
Note: Consider only the faces of the nodes pointing towards A150
Global Forces: from Reference Point Loads report
Local Forces
Note: be mindful of adding forces in the same direction. Fx ( header) – Fx (header) - Fz (branch) = 0
The forces balance, except for forces in the Y direction. Why is that? think what are we forgetting to consider in this calculation?
Correct, weight of the component and contents.
Step #10: Use the Center of Gravity report to help calculate the weight of pipe and contents.
Note: COG report only includes the currently visible beams and segments. All pipe segments are included. Therefor before creating COG report use one of two options:
Option #1: Delete everything in the model except for the highlighted Tee shown above in step #5.
Option #2: Insert Segment breaks at node points A140, A160, and B180. Hide all other segments except those that make up the tee.
Step #11: Confirm weight of Pipe and Contents from COG report equals that calculated in Step #9. And, it does.
Step #12: Done, all forces at 3 node points around a Tee node point (ex. A150) balance or equals 0.00.
See Also
"Forces & Moments" sub-report using Results> Output Report
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