To All,

Company standard practice is to to break the pipe model on the midpoint to get the deflection of the pipe. I tried manual computation using beam formulas to compare but im getting a different result.

Im thinking now that the approach is wrong? I hope someone can give their expert advice. Thank you

What do you mean by "breaking the pipe"? If you mean modeling a 20 ft span between supports by having two 10 ft elements, that approach seems fine.

Manual calculations will result in different results unless you're accounting for all the same loads present in the system modeled.

I recommend sharing your work so that we can critique.

The manual/hand/analytical calculation of pipe span length, pipe sagging deflection and/or dead weight stress is typically performed using simply supported beam or fully anchored beam formulas.

There are some adjusted formulas considering the intermediate situation.

However, the real piping layout corresponds rather to continuously supported beam situation. Moreover, the actual pipe span length may not be uniformly distributed. In addition, you might have direction changes (elbows), branched pipe connections, inserted loops, valves etc.
This is the reason why Caesar II stress model quantifies realistic sagging deflections that may not match the simply hand-calculated values.

As long as you develop the piping system model correctly, including all the relevant piping components weights and following the actual layout and support arrangement, you should not be worried about results accuracy.

The analytical/hand-calculation span formulas are simplified tools, which were extensively used in the past, when piping flexibility programs were not available or hardly accessible.

Regards,

Since my pipe is located on the piperack, I used beam formula with both sides fixed. Is this correct. My computation and CAESAR Result is shown below:

Note that in your CAESAR output that both nodes 400 and 500 have rotation, which is inconsistent with a fixed/fixed beam condition.

Try running a model with 2 anchor points and a center point. Why stop there, though? Put an odd number of points between the 2 anchor points, and you can plot the results, and compare the plotted results with your formula.

Park,

The answer to your question may be found in my previous post. I'm afraid you are not aware of the actual structural model you need to consider.
Pipe-rack piping should typically be assimilated with CONTINUOUSLY SUPPORTED BEAM model. Simply supported beam or double anchored beam models are not relevant for such situation. Such simple models are used for fast and conservative assessment of supporting span length.

Go to a Strength of Materials or Mechanics of Materials book and review what means such CONTINUOUSLY SUPPORTED BEAM model. Then, consider a continuously supported beam with 4...6 supporting spans. One terminal support may be considered fixed point. Then consider the uniformly distributed weight load and calculate the deflections of the center-points of the middle span lengths.
You might need Mathcad for such calculations, since the formulas are complex enough.

You'll find the sagging deflections are closed enough to those given by Caesar II, if your computer model is identical (e.g. same dimensions, same span lengths, no valves etc.) with your analytical model.

Regarding the results you've got with your previous attempt, you have 18.895 mm (analytical) vs 19.829 mm (Caesar II). This means less than 5% difference!!...from engineering point of view, this means full agreement, what do you want more?!

We are engineers, not pharmacists!!...

JR Park,

I understand you expect that someone is able to explain you the difference in deflections 12.5 mm in your hand calculation and 19.8 mm in your model. This is not possible without having details about your model.

But I expect you can have results closed to deflection of 12.5 mm if you model a large number -let's say 11- elements of 18,895 mm (representing the span), add a middle point on each span and consider proper supports.
Looking to the deflection of the middle span of your model,you can expect to have the deflection closed to 12.5 mm, but not this value for marginal and intermediate elements. In fact it is what Michael suggested you previously.

Theoretically, for an infinite continuous beam with equal spans, you have no rotation on supports due to the symmetry; each half of span is balanced by the half of span beyond adjacent support and eventually, in this situation, in which part would be possible a rotation on support, clockwise or counterclockwise? or no rotation?
For this case, because you have no rotation on supports the deflection formula should be identical with "fixed end" one span (even you have just rests as supports).

Because you haven't an "an infinite continuous beam", nobody can guess (but one can hand calculate-there are several methods well known in the old school, however useless in practice for our days) the actual deflection on spans of a continuous beam.

It is true that the "old practice" was to approximate something between simple supported case (when the rotation on supports is maximum) and "fixed end" case (when the rotation on supports is zero). The numerical coefficients for deflection are 5/384 for the first case and 1/384 for the last one. With no math reasons, the engineers assumed a coefficient as arithmetic mean of those coefficients; you can find in books a "generic formula" for deflection with a coefficient of 3/384= 1/128. But this is just a reasonable approach useful to dimension spans, you cannot use it to check a software. For your case, it seems there is a coefficient as 1.58/384 and this is due to some rotation on supports.