Problem Description

Question #1:

Output report, subsection - Frequency found the Participation Factor-X =-0,  Captured Modal Mass-X = 0 and Cumulative Modal Mass-X = 0, for all calculated natural frequencies, but in anchors there are existing forces greater than 0, FX(Point 10) = 350.87 N and FX(Point 100) = 360.91 N. How can this situation be explained?

Answer #1.

The model was found to be a single segment, 100m long on the Global X axis. The pipe has no bends, expansion loops, or is modeled on any other axis except Global X.

As a result the participation factors were reported in the output report as 0.00 for the same pipe axis (Global X), however the actual value maynot be 0.00. AutoPIPE is only able to report a number with 2 significant digits. use the following procedure to validate the actual value  calcualted:

1. Open the model

2. Analyze the model

3. Select Results> Grids

4. Select Frequency Tab in the grid

5. Select any cell in the Particip. Factor X column, note that the value is not 0.00 but much smaller, ex: -5.2860372e-011. AutoPIPE output report is not able to show such small values.

Suggestion, confrm actual value calculated using the results grid.

Question #2

Given a model with a single pipe, modeled between 2 anchors on the Global X axis, the calculated inertial forces and moments in the TEST1 model are FX, FY, FZ (where, FZ=2258.35 N), MX, MY (where, MY=5588.1 N-m) & MZ, for both anchor points 10 and 100, are significantly lower than the inertial forces and moments FX, FY (where, FY=3030.55) FZ, MX, MY (where, MY=8585.51N-m) & MZ calculated in TEST2 model,  accordingly.

The only difference between these 2 models is that modal analysis cut of freq = 66 hz for TEST1 & 120 hz to TEST2.  Thus, even if more frequencies and their associated modal masses are obtained in the analysis TEST2 model (Fr  eq = 120 Hz and 12 modes obtained), the lesser inertial forces and moments values are obtained in anchor points 10 and 100, com  paring with TEST1  (Freq = 60.2 Hz and 6 modes obtained), respectively. So, how can this situation be explained?

Answer #2:

One issue was that more than a different value for modal analysis was found between the 2 original models. To be absolutely sure that the models are identical,  open the 1st model, saved as 2nd model name, and make the changes as required, ex: change modal cut-off frequency from 33hz to 120 hz. 

To answer the question, might find that the anchor reaction is more for the model with higher cut-off frequency as it has higher level of discretization and thus able to capture higher modes better.

Question #3

Analyzing the 2 models TEST1 and TEST2, we observe that for the first 6 vibration modes (Mode No 1 to Mode No 6) each calculation model gives different values for the first 6 calculated frequencies, as well as for the first 6 Participation Factors (in t he X, Y and Z directions) and for the first 6 Captured Modal Mass (in the X, Y and Z directions) respectively. Obviously, the first 6 Cumulated Modal Masses (in the X, Y and Z directions) are also different in the 2 models (TEST1 and  TEST2). Furthermore,  the first 6 Shape Modes from each model are different. How is this possible, given the identical conditions in the calculation models - except for the different Cutoff Frequencies (66Hz and 120 Hz, respectively)?

Answer #3:

One issue was that more than a different value for modal analysis was found between the 2 original models. To be absolutely sure that the models are identical,  open the 1st model, saved as 2nd model name, and make the changes as required, ex: change modal cut-off frequency from 33hz to 120 hz. Knowing the files are now identical with exception to changes made, performed a model analysis and Static analysis on both models.

Changing the cut-off frequency when you have Tools> Model Options> Edit> Mass points per span = A option enabled  will lead to different level of discretization in the model. That should explain the differences in the frequencies.

Another words, not expect the first 7 frequencies to be same between the two models as they are no longer "equivalent" - because the level of discretization is not same. The higher the cutoff frequency, the higher the level of discretization.