| Applies To | |
| Product(s): | AutoPIPE |
| Version(s): | ALL; |
| Area: | Results |
| Original Author: | Bentley Technical Support Group |
| Date Logged & Current Version |
April. 2017 11.00.00.22 |
Problem:
How are supports axial force due to xxx load case calculated in AutoPIPE's reports?
Example:
With the give piping arrangement :
Why are axial support forces for all operating load cases not all Fnormal x Friction Coeff (ex. 1000 lbs x 1 = 1000)?
Note:
a. Pipe was assumed to have 0 density
b. Specific Gravity = 0.00
c. V-stop support Friction coefficient = 1.0
d. Rigid anchor at Node point A00
e. 3 Thermal loads have been defined with different thermal expansion values (T1- low, T2 - med, T3 - high)
f. Only weight in system is added Weight = 1000 lbs at node point A02.
Solution:
Basically, pipes with a load case applied horizontal to the support will experience a resistant ‘static friction’ up to the maximum allowed friction, which is the product of the friction factor, f, and the normal force, N. Until the maximum value has been reached, the pipe does not move. After the critical friction force is met (), slippage occurs and ‘kinetic friction’ dominates. The pipe begins to move in the direction of the load. See the following plot created to help visualize, where the blue line is the support force and the red line is the support displacement.
Again, note from the plot above, the pipe displacement remain 0.00 (red line) until the maximum resistance force has been reached (blue line) causing the pipe to move. The pipe may continue moving but the force required cannot be larger than the maximum resistance force calculated.
Detailed Examination of Example:
First look at displacement in the axial direction for the support node point(s):
Notice that the displacement due to operating conditions, ex. GT1 and GT2, are essentially 0.00 inches; while the displacement for GT3 was >0.00. Now look at the forces acting on the support by the respective operating conditions. Understand in order for the pipe to move, a pulling / pushing force = Fnormal x Friction Coeff must be applied. Otherwise the node point will not move.
In the example above,
Fnormal = 1000 lbs
Friction Coef = 1.00
Fpush = Fnormal x Friction Coeff
Fpush = 1000 lbs x 1.00
Fpush = 1000 lbs.
Again, so long as a Fpush force was >1000 lbs, node point will be displaced, conversely any value of Fpush <1000, node point will not move. AutoPIPE's output report indicates a maximum force or portion of the maximum force required to move the piping node point.
Please see the following AutoPIPE help section for more details on this topic:
Help > Contents> Contents Tab> Reference Information> Analysis Considerations > Static Analysis > Nonlinear Analysis > Support Friction Properties
Results are affected by:
Please see the following AutoPIPE help section for more details on this topic:
Help > Contents> Contents Tab> Reference Information> Analysis Considerations > Static Analysis > Nonlinear Analysis > Force Tolerance and Friction Tolerance
Adjust the non-linear settings as required to get closer to the nominal friction force calculated by hand.
Scenario #2:
How can the axial support force be larger that the maximum resistance force and still not cause any pipe movement?
example: Given
- max downward load on the support= -27221 N
- coeff of friction = 0.3
- Maximum resistance force = 27221 N x 0.3 = 8166.3 N
- Analysis type = Non-linear Analysis
How can the maximum load on the support be 15409 N for T1 load-case and with no pipe movement (i.e. displacement = 0.00) ?
Answer:
The most important aspect to remember, this is a Non-Linear Analysis type. As such the only results that should be considered are Operating Condition combinations, ex. GrT1P1, Sus+E1, ect. Do Not consider the individual load-case (i.e. T1, P1, E1, Gr, etc..) unless trying to understand how the results for an Operating Condition combination are being calculated.
The following explanation may help clarify the results of AutoPIPE. Again, for non-linear load sequences the result of an individual load case may not make sense if load sequence was ignored.
1st:
Gravity is the first load case in the load sequence. Downward force at that time is 27,222 N downwards. This when multiplied by the friction coefficient gives 8167 N which is the maximum resistance the friction can give. The direction of this force on the support is +X global.
2nd:
The thermal force acts in the opposite direction (-X global). It first cancels out the existing gravity load, and then continue to act in the -X direction developing friction force on the support in the -X direction. The results for GT1{1} combination would make more sense in this case showing that now the lateral load on the support is 7,301 (still under the downward force * friction coefficient limit), but is now acting in the opposite direction.
3rd:
Pressure load case was applied acting in the +X direction canceling out a portion of the thermal load force (P1 = 91 N)
4th,
The operating condition combination, GT1P1{1}, gives the final state of support forces which was calculated to be 7,210 N in the -X global direction. Again, since the Maximum Resistance Force had not been reached there was no pipe movement.
Scenario #3:
Why the axial Anchor Axial forces are different in Anchor Forces & Moments and Code Stresses?
How software calculates the following values?
-
- Forces and Moments: FY (Axial Force) = 568790 N
- Code Stresses: Axial Force = 667821 N
Note: Zero density considered for pipe material and service fluid.
example: See below details.
Answer:
There are two longitudinal forces acting at the anchor support at the same time
-
-
- Axial force due to pressure end cap, See below formula,
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-
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- Axial force required to prevent pressure extension
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Net Longitudinal (Axial) force is equal to axial force due to pressure end cap minus axial force required to prevent pressure extension.
The axial force due to the pressure end cap (1236611N) and Net axial force (667821N) are used for post-processing (Code Calculations), whereas forces on the anchors (568790N) are the pressure extension forces,
Axial Stresses due to pressure can be calculated by following equations,
Find attached spreadsheet of manual calculation for your reference.