Good morning Sirs,
In basic soil modeler, is “Temperature change” the difference between operating and installation temperature of the pipe or of the soil?

I think “Thermal expansion coefficient” is inherent to the soil; because it has a default value, does it take in count heating by pipe?

Thanks for Your help.
Best regards.

Alessio

This temperature change is ONLY used to estimate the virtual anchor length - that length of pipe over which the axial friction matches the axial load due to strain. This axial strain includes pipe extention caused by thermal strain and axial pressure stress and also contraction due to the Poisson effect due to hoop pressure stress.
This temperature change is the difference between operating and installation temperature of the pipe.

Thanks Dave.

If temperature change is the difference between operating and installation temperature of the pipe then "Thermal expansion coefficient” is inherent to the pipe at operating temperature. Did I understand well?

Can you explain me what is the theoretical "virtual anchor lenght" and what is its use in buried analysis? In User guide I found it but there isn't an axplanation

Thanks

Regards

Francesco

Hi Francesco,
as regards virtual anchor length, you can read it in "Longitudinal pipe movement" of this file:

http://www.pipestress.com/papers/UnderGrd-2.pdf

Best Regards.

Alessio

Thank you Alessio, but is it correct what I have written in my last post?

Regards

Francesco

I think that "Thermal expansion coefficient" regards soil, infact I changed pipe material in my calculation but that number didn't change.
Best regards.

Alessio

If you click F1 in "Thermal expansion coefficient" it refers to pipe not to soil.

What is the practical use of the "virtual anchor lenght"?

Thanks

Regards

Francesco

What use is the virtual anchor length (VAL)? The pipe can only push so much (thermal and pressure strain minus Poisson effect of hoop pressure). If you have a VAL run of buried pipe, no information (loads and deflections) from a node upstream of the VAL can be passed to a node downstream of that VAL. Because no information cannot pass thought this VAL run, this whole VAL then serves as an anchor. It's not a point anchor. There is no reason to model beyond this anchor if you are only looking at the upstream piping or the downsream piping.
Knowing the VAL will allow you to terminate your model or at least segregate a pipeline into subsystems.

So your recommendation is just to stop the model where you consider the pipeline is fully restrained? Maybe I was not able to understand well.

My recommendation is to place in that zone an anchor point, otherwise the model shall experience there a "cap effect" due to pressure load.

Best regards.

I guess I could've said that better. Yes, the proper length of pipe (the virtual anchor length, VAL) will disconnect the system into an upstream leg and a downstream leg. But you want to know how much of that upstream or downstream leg must be modeled and whether or not an anchor be placed in yoru CAESAR II model.
Maybe an example will help...
Let's say you have a buried run of pipe between two 90 degree elbows and a given VAL of 1000ft. If that run between elbows is less than 1000ft, then there is no separation and the model should continue beyond that second elbow. If that run between elbows is 5000ft, then a VAL exists and the two elbows can be evaluated in separate models. How do you end these two models? I suggest you run at least VAL beyond the upstream elbow to establish enough axial resrtaint through friction to get a sensible response for the elbow. Focus on the elbow, not on the pipe on the other side of the VAL. Since that elbow may get pushed back (according to the bearing capacity of the soil on the incoming leg of this elbow), the thermal strain and therefore axial load is reduced. With a reduced axial load, a shorter VAL would exist.
So one difficulty is what's going on on the other sides of these bends - that bearing stiffness of the soil and the pipe cantilever stiffness. Another confusing point is how CAESAR II treats this soil friction. We model this friction with springs. These springs have to be moved to generate load. This necessary motion makes the VAL in the CAESAR II model greater than the VAL in theory. (Note that the newer ALA soil model gives you more control over this axial soil stiffness.)
Back to the example...
If I have 5000ft between these elbows, I would probably run through both those elbows. It's just two more pieces of pipe. CAESAR II will do the work of "burying" it. You would then look for motion of each elbow away from the 5000ft run and a section of the 5000ft run that has no axial deflection. The length of pipe between the elbow and the "inert" pipe is the effective VAL for this CAESAR II analysis (and, again, it is also influenced by what's around the bend). If I have a 10,000ft straight run between elbows, I would probably run at least 2000ft beyond the first elbow and stop there. I would let the axial friction balance the thermal strain and I would not add an anchor. If you have the time, you can check the maximum axial load in the run headed to the VAL. Then run the job again with a longer run. If the maximum axial load stays the same, the run is long enough.
That's my opinion, I invite others.

Dear Mr. Diehl,

As usual, your answer is very instructive and, no doubt, the described procedure is correct. Probably I would prefer to avoid "If you have the time"… I mean, I consider is OK to model 2 times VAL beyond the first elbow and stop there, but I would consider as mandatory to check that 2 times VAL length run is enough.

To comment your examples, modeling just 2 times VAL and after that stopping the model, it means to have a run with one end- elbow, and other end- a "virtual" cap. This "virtual" cap is not visible in model, but the "cap effect" disturbs the axial strain distribution on a considerable part of run considered in model. So you need to experiment more by changing the length of run between the real elbow and the "virtual cap" end and see the results are sensitive to these trials. Of course, also in this case one must have some concerns (soil parameters- springs), exactly as you’ve described for a model with a run and 2 elbows\legs (first part of your example)

Some companies prefer – as an alternative- to force in a way the model to consider a zero longitudinal strain in the zone where this situation should be. The "imperfect tool" to do this is a fictitious anchor placed there, and I would admit is unaesthetic and questionable by the Contractor/ Consultant when looking in the model…but in the same time, one can keep in mind that is also true no cap is there.

IMO placing a fictitious anchor at 1.5…2 times VAL will help me to be focused on the elbow, with a good chance to be conservative in this approach. Of course, to be sure, I would prefer to consider different length and soil parameters and see the sensitivity of the results, exactly as you suggested.

I cannot say one method is good and one is bad. For computational effort, shortening in a intelligent way the run would be an advantage, but in our days we have a supercomputer on our desk….

Anyway, my opinion is to consider as worst case modeling just a run with the presumed VAL length, stop the model there and doing nothing more; I consider this approach shall lead to nonconservative results for elbow/leg in the end of the run.

[img]http://www.coade.com/ubbthreads/ubbthrea...0229561194ccdd4[/img]
Maybe this will help...
I have a buried L with a 400 ft run on one side and a variable run on the other side of a 90 degree bend. How long must that variable leg be in order to develop a "virtual anchor length" (VAL)?
The plot shows the axial deflection of the variable leg. The leg on the elbow side moves in the positive direction and the free end moves in the negative direction. When the leg collects enough friction on either side, given a fixed amount of thermal strain (a fixed internal load), it will not be able to move. Clearly, the 2000 ft and 4000 ft legs "lock up"; they are fully restrained as demonstrated by the large portion of zero growth. The 1000 ft run may or may not show a VAL and the 500 foot length is clearly too short to build up the required friction restraint.
If I look at the axial compressive load in each of these four legs I see that the 2000 & 4000 ft runs have THE SAME load of 45566; the 100 ft run has a little less - 45392 and the 500 ft has 41601. So we see that the 1000 ft run does not create a complete VAL.
I could also examine the deflection of the free end. 4000, 2000 & 1000 all achieve -0.1765 inch while the 500 ft run moves only -0.1759. These numbers would be more dramatic if I chose "better" soil properties. But we can see that the longer legs have enough length to collect the friction. In a way, you could look at this defection as the growth of the leg length that is needed to collect all that friction. Once again, the 1000 ft leg is just about long enough.
I am uncertain of what you mean by "virtual" cap. It is the pipe-soil friction in the axial direction that builds up the restraint load. If that free end was another elbow then the soil bearing around that corner would alter these results but as long as the run is long enough, that "free end" will not affect the position of the elbow included in this model. Inert pipe indicates a good location to end this model.

Thanks for the answer.

I guess it's not more than a matter of terminology.

I can imagine your pipeline as a buried L with a 400 ft run on one side and 100,000 ft run on the other side of a 90 degree bend.

And you construct the model as a buried L with a 400 ft run on one side and a variable run on the other side of a 90 degree bend.
You investigate 4 models with the run length of 500, 1000, 2000 and 4000 ft, the goal being to investigate which model is appropriate in terms of actual displacement received by elbow.

You call the terminal point of the modeled run as "free end". You investigate the deflection of the free end and found it's about -0.17 inch.

In reality this point is axially restrained- at least this is true for section of pipeline 4000ft beyond the elbow. So it is not so free in real word as in the calculation.

Probably the point I’ve tried to highlight is: would I consider as wrong practice to place an anchor at 4000 ft, instead to consider a "free end"?
I keep in mind no free end is there and no anchor is there, and the ultimate goal is to obtain actual displacement and stress in elbow and 400 ft leg.

What is your opinion?

Thank you again.

My best regards.

My focus is on the elbow end. The elbow responds the same, I get the same axial deflection, with the 1000, 2000 and 4000 foot run. Growth = 0.1693 inch. The elbow moves only 0.1686 inch for the 500 foot run. (Again, if I chose better soil paramaters, the difference would be more dramatic.)
I don't care about that free end.
Now you ask about adding an anchor. I reran the 2000 foot run with an anchor in that "inert" section and the axial defection is essentially the same. The anchor had a very small load (14 lbf) because the pipe didn't want to move. The elbow had the same axial displacement. I also placed the anchor at the "free" end of the 2000 foot run. Of course that eliminated any axial deflection of that end BUT the elbow still responded the same.
So go ahead and add that end anchor AS LONG AS it does not lie within the moving section near the elbow. Unfortunately, we don't know where the inert pipe begins unless we run the model without the anchor first.
Once again, if you have sufficient straight pipe to build up this virtual anchor, the upstream and downstream segments can be analyzed independently.

Now that Dave has shown it doesn't matter whether or not you put the anchor in the model, I'm going to suggest that you don't add the anchor (contrary to what Dave suggests). The reason being that too many people think there is a real anchor there. Here is a perfect example....

A few years ago someone called up complaining that the pipe in the field didn't behave the way CAESAR II said it would. I asked for an explanation. What they had done was run CAESAR II to obtain the distance to the virtual anchor. They then went a few hundred feet past that location, and dug up the pipe to insert a valve. They were most surprised (and upset) when the pipe jumped out of the ground - after all, they were past the virtual anchor, there shouldn't have been any movement ??

Most people don't understand that you can never reach the virtual anchor. If you're standing at "point A", and you compute the virtual anchor to be at a distance of 1500 feet away, and you walk 500 ft in that direction, the virtual anchor also moves 500 ft.

Perhaps we should remove all references to the term "virtual anchor" and replace it with the phrase "point where it is ok to stop modeling", assuming nothing else changes.

Richard, I don't understand what you mean by saying the mentioned above; I mean, if the VAL is established as the length where enough friction is collected to stop movement, how is this VAL going to change by walking in one direction or another.

Regarding to people who called you up, I understand that pipeline behaved in that way because there was another change of direction close, whose VAL interacted with the VAL formerly obtained. Am I ok?

Because "val" is measured from the point at which you're standing ...

Here is another example - assume you're holding an ordinary shovel (1.5 meters long) horizontally infront of you, such that the sun casts a shadow below the blade of the shovel. Now walk forward 1.5 meters. Are you standing on the shadow ?

Dear Mr. Richard,

In Caesar-II how the thermal displacement is calculated with respect to the ambient temperature (i.e.)

Case-1 - If the ambient temperature is default ambient temperature (21 Deg C)

Design Temp is 84 Deg C

The change in temperature may be 84-21 = 63 deg C and the corresponding thermal coefficient shall be used to calculate the thermal displacement.

Case-2 - If the ambient temperature is reduced to 12 Deg C

Design Temperature is 84 deg C

The change in temperature may be 84 -12 = 72 deg C and the corresponding thermal coefficient shall be used to calculate the thermal displacement.

From the above two cases, the results of expansion stress may vary due to different change in temperatures such as 63 and 72 deg C.

But the displacement must be same, since at ambinet temperature (21 deg C), the thermal coefficient is zero (assumed universally). So when the ambient temperature is 12 deg C, displacement process shall be as follow (i.e.) at 12 deg C, the pipe will be in contraction till it reaches the ambient temperature (21 deg C) and after raising above the ambient temperature the pipe will expand till it reaches the temperature of 84 deg C.

The query is as follows,

How Caesar-II is calculating the displacement (i.e.) Is caesar-II calculates the total displacement from (12 to 84) or from (21 to 84)?

If it calculates from 12 to 84, why it is not showing the vector values such as negative displacement for contraction and positive displacement for expansion.

Please clarify......

The term "Ambient temperature" in caesar is in fact "Installation temperature" or zero expansion.

For convenience we consider 12 deg C as installation temperature and we will have the entire range from 12 to 84 in one single step.

If you want to obtain the real situation you must do the following:

- maintain your "Ambient temperature" in Caesar at 21 deg C.

- insert two temperatures in temperatures field. one at 84 deg C, the other at 12 deg C.

Construct your load cases as follow:

L1=W+P1+T1 (OPE)
L2=W+P1+T2 (OPE)
L3=W+P1 (SUS)
L4=L1-L3 (EXP)
L5=L2-L3 (EXP)
L6=L2-L1 (EXP)

In your L4 you will see displacements from expansion 21 to 84
and in L5 you will see displacements from contraction 21 to 12.

In L6 you will see the total displacement

regards,

Just a side note...
Ambient temperature is set for new jobs from the Configuration file. Once a model is built, changes to the Configuration file will not change that job's ambient temperature. If you wish to change the ambient temperature of an existing job, use the Special Execution Options.

Dear Mr.Danb and Dave,

My query is, if the installation temperature is 21 deg C and the design temp T1 is 84 deg C, for ex. if i get a value of 80mm as the displacement.

Caesar-II must show the same displcement as 80mm, if the installation temperature is 12 deg C and the design temp T1 is 84 deg C, since as i mentioned earlier, the pipe will contract when the temperature is below 21 deg C and expands above 21 deg C.

So irrespective of the installation temperature the displacement must be same, but the Caesar-II calculates the displacement considering the change in temperature what ever we provide in the installation temperature (i.e.), if i provide installation temp as 12 deg C and T1 as 84 deg C, the Caesar-II calculates displacement based on (84-12) = 72 Deg C, this is actually not a correct way of calculating the displacement. Since displacement is a vector quantity. The above case may be applicable for a displacement stress range but not for the displacement only.

My actual requirement must be, if i input installation temp as 12 deg C and T1 as 84 deg C, i must get a displacement both in contraction and expansion considering 21 deg C a zero expansion point.

I think atleast this may be understandable, if so, please clarify....

It is you that control the software.

Entering 12 deg as installation temperature or "Ambient temperature" you instruct Caesar to "forget" about 21 deg and to consider 12 deg as zero expansion point. Caesar do what you ask it to do. If you want to see the results as you want i.e.

"My actual requirement must be, if i input installation temp as 12 deg C and T1 as 84 deg C, i must get a displacement both in contraction and expansion considering 21 deg C a zero expansion point."

you have to do as I already shown:

"If you want to obtain the real situation you must do the following:

- maintain your "Ambient temperature" in Caesar at 21 deg C.

- insert two temperatures in temperatures field. one at 84 deg C, the other at 12 deg C.

Construct your load cases as follow:

L1=W+P1+T1 (OPE)
L2=W+P1+T2 (OPE)
L3=W+P1 (SUS)
L4=L1-L3 (EXP)
L5=L2-L3 (EXP)
L6=L2-L1 (EXP)

In your L4 you will see displacements from expansion 21 to 84
and in L5 you will see displacements from contraction 21 to 12.

In L6 you will see the total displacement"

There is no other way.

Regards,

Dear Richard,

Now, for example a 1500ft (X axis) buried run of pipe between two 90 degree elbows and a given VAL of 1000ft.

is it ok from VAL perspective to separate this buried pipe into 2 parts. part one, 1 elbow up to 1000ft (lets say to +X direction) and another one, 1 elbow up to 1000ft (to -X direction)??

please advise.

Thanks.

In theory - yes exactly. (You should have started a new post. This buried pipe topic has nothing to do with the topic of this thread.)