Longitudinal Pipe Stress Due to Friction at Pipe Supports 1

[MECHPIPENEER]

Hello Folks,

I am trying to better understand how friction at pipe supports contributes to pipe stress due to thermal expansion/contraction on long runs of pipe supported in a rack. I have conceptualized a hypothetical scenario as follows to aid in my understanding:

- A straight run of pipe of OD 'D', thickness 't', and length 'U' anchored at one end (infinite capacity anchor) and fully free to displace axially at the other end
- The pipe is supported on a rack with spans of length 'L', the pipe is free to displace axially along the rack, however there is an axial friction resistance at each rack support of 'Ff'

Does friction due to thermal exp/contraction accumulate from the total qty of pipe supports to produce an axial force = U/L*Ff between the anchor and first support? Or does the disproportionately larger thermal pipe displacements at the supports furthest from the anchor result in friction accumulating and dissipating in a staggered manner between the supports back to the anchor resulting in a lower axial force?

In the end can a pipe as described above be infinitely long without exceeding allowable stress due to thermal exp/contraction or is there a limit to how long the pipe can be due to the accumulation of friction at the supports? I am working to B31.3 and B31.4.

Thanks in advance!

 

[saplanti]

Think about infinitely stiff piping and infinitely stiff rack pipe supports.

The only force can be created is the friction force by the motion of the pipe with the gravity forces on the supports against the motion.

Your end fix support is the only one will take entire friction forces under thermal displacement.

Assume you have n number of supports including the first fixd end support. Each supports willl carry w pipe weight and friction coefficient is m.

Fixed end will have the follwing accumulated force;

F = m w (n-1)

Assume the pipe cross section is A.

Therefore the axial stress will be;

Axial stress= F / A

This will be compression in thermal expansion and tension in contraction. So you can make the decision from now on.

Please note that we assume that the pipe does not have guide and lateral displacements under the friction loads. The span lengths are the same to give the same gravity force w.

 

[1503-44]

I think it is more like the opposite. The initial anchor, in the direction of heat flow, develops an anchor force, The pipe expands towards the next support. The expansion is resisted by the first supports friction. When, or if, the expansion force exceeds the maximum static friction load of the first support, the pipe can begin to expand between support 1 and 2. If the expansion load does not exceed static friction of the first support, support 1 becomes in effect another anchor and as, or if temperature in segment 2 increases, growth of the pipe begins from zero again. If temperature in segment two does not rise high enough to exceed the static friction load of support 3, temperatures tend to decrease with distance from source, support 3, effectively becomes an anchor, and so on. If thermal pipe load at 3 exceeds static friction, the pipe slides into the next segment, leaving the force created by dynamic friction factor at support 3. The anchor force at the beginning equals the load of all supports that slide, including the one at the last point where the pipe does not slide.

If no expansion force in any segment exceeds its corresponding static friction load at any support, then all supports are effectively full anchors.

 

[saplanti]

1503-44,

Expansion/contraction under the thermal motion is unstoppable, by friction forces of the supports since the pipe infinitely stiff, during temperature rise/drop on the metal. This expansion/contraction at the start-up and shut down will be ultimate, and the motion will be on all the supports. I forgot to mention this part perhaps not to cause more complication in the previous post. Therefore all the supports will be under thermal expansion/contraction friction and fixed support (the first support in the direction of flow) will be taking the entire friction load. Let us not forget the discussion was based on hypothetical scenario.

If the fixed support was not there and the pipe is flexible enough the scenario was completely different, I agree with that.

 

[LittleInch]

Ok,

So a few further assumptions need to be made.

1) The pipe is subject to the same thermal expansion / temperature increase along its entire length at more or less the same time
2) The pipe is restricted to axial movement only and has vertical and horizontal guides preventing it from lifting or snaking.

Does friction or axial stress accumulate - Yes, but maybe not quite as uniform as your equation. A decent stress analysis program will show you this very easily.
Or does.... No. Each section of pipe as it expands needs to push the entire pipe that distance and hence it all builds from the free end back to the anchor.
The limit of the length is finite. At some point the axial force available to the pipe from thermal effects to push the increasing length of pipe equals each other. This point is termed the virtual anchor.

For buried pipelines it is often in the region of 150 to 250m from the free end due to the much larger soil friction forces. For above ground piping and subsea lines which rest on the seabed it is more like 3000-4000m.

The reason no one does this is that the pipe is then fully restrained and subject to significant axial stress. If there is any ability for the pipe to go in a different direction ( sideways, upwards) it will do so and if the supports are sufficiently far apart it may buckle. Also the end movement of the pipe at the "free" end becomes unreasonable and is measured in metres.



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

Saplanti, I know you know the theory better than what you said above. Hypothetical still requires a correctly explained theory. "Expansion/contraction under the thermal motion is unstoppable, by friction forces of the supports since the pipe infinitely stiff, during temperature rise/drop on the metal." What you don't explain is extremely important, especially in the case of virtual anchor development by soil surrounding pipelines. Expansion is stopped by friction, anchors and virtual soil anchors. Pipe is not infinitely stiff. Motion need not occur on all supports. Thermal loads in a pipe segment actually decrease by the friction load taken by each prior support. Fully restrained pipe has no motion. Partially restrained pipe will have some, but not all motion. That is important to realise when different span dimensions create different weight and consequential friction loads at their respective supports and thermal forces are not uniformly distributed at all supports. Thermal stress is inversely proportional to movement of the pipe, whereas anchor and support loads are directly proportional to the movement that is restrained at each..

If you simplify, you can only understand the simple.

 

[1503-44]

Little Inch, you were OK up to "the reason nobody does this". That's exactly what a buried pipe is supposed to do. After the virtual anchor, it is fully restrained and fully stressed.

 

[saplanti]

1503-44, LittleInch

I think that you missed the rack piping side, this is aboveground piping.
I trust this is sufficient. Regards.

 

[1503-44]

The axial stress in a pipe at any point, above or below ground, is proportional to ( 1- Actual_Movement/Movement_if_totally_free).

Theoretically even a 100 pound friction load will restrain a sliding steel pipe's expansion by a very tiny amount, or if a rubber pipe, probably restrain it totally. Movement not restrained is carried to the next segment.

 

[LittleInch]

Saplanti - I know it's piperack piping he's talking about - I was just comparing it to buried pipelines. I don't think I've seen a pipe rack long enough to get a virtual anchor situation, hence why everyone usually uses expansion loops if you're getting big temperature issues.



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[saplanti]

It seems that we are talking about the same thing, there will be restraining friction forces but movement will be there. These friction forces may reduce or increase the compression/tension stress on the pipe at each support, at the fixed support ultimately as mentioned. When pipe pressurised and temperature introduced it will reduce the stress in the pipe, however in the shut down the friction will be against the contraction, this may introduce additional tension stress to the pipe if the pressure not released earlier. I was on the hypothetical side, I apologise not to explain everything clearly with all details. 1503-44, thanks for correcting me on the subject.

Unfortunately, these are fictitious assumptions, real life for aboveground piping is totally different for variety type of piping systems. Non of structural member can take this accumulated large lateral forces, and therefore the designer will use sufficient number of expansion joints or change of direction to accomodate the support forces and pipe stresses. In early days engineers used to make too much assumption and solve the analysis by the Kellogs Tables and procedures or similar applications. Today this is a lot more easier, we leave the analysis of the piping system to the software to solve by iterations.

Similar is done for the buried piping, but it is more complex and requires more computer power due to the huge number of nodes, if you model the entire piping, and soil complexity. There are ways to split to model into small parts to solve it easily. But this require checking each model for consistent behaviour of the split locations.

This is short summary, I trust OP is not confused by all these. Regards.

 

[1503-44]

I don't think fictions is the proper word. You should do nuclear power pipe stress anchor and support design. I did that on the South Texas NPS at Brown and Root back in the 70's. Supports are not only designed for extremely high thermal loads, but also for pipe burst and resulting whipping restraint impacts. I've been using computers for this since punch cards and PDP days.

A linear pipe as questioned in the OP, in a rack, or underground is very easily solved using Excel from scratch in a few minutes, if you completely understand how to model it. But neglecting any small thing can have vast implications. I know you know the theory. All I ask is that you drop the generalities and explain it correctly, so the OP doesn't get the wrong idea from over-simplifications that work only when EA >> Wf.

 

[saplanti]

OK, I’l do it. I respect your knowledge on the subject.

Let us assume the pipe pin connected on two adjacent supports.

The thermal expansion on the pipe between two supports shall be;

dL = alpha . L . DeltaT

The thermal axial stress on the pipe due to this thermal expansion will be;

ft = E . e = E dL / L = E . alpha . L . DeltaT / L = E . alpha . DeltaT

The Thermal axial force

Ft = ft . A = E . alpha . DeltaT . A = E . A . Alpha . DeltaT

Where
A : cross section area of pipe
E : elasticity modulus of pipe material
alpha : thermal expansion coefficient of pipe material
DeltaT : temperature change on pipe material

In case the friction force (w.m) is smaller than the thermal axial force, Ft the pipe will move on the support. w is the gravity force on the support, and m is the friction coefficient between pipe and support. Please note that the support is free on one end, and will not introduce the same force for the pipe pinned both ends.

( Ft - w.m ) force will be the force to give the pipe displacement on the free support. Ft is far larger than w.m . I guess 1503-44 wanted to imply this.

However the rest is as given in my previous post first paragraph and my first post. Please note that I have just wanted to be consistent with my notations in my post. So excuse me in case they are not representing yours.

I trust this presentation is clearer.

 

[1503-44]

Looks good now. Thank you.

 

[saplanti]

We are at this stage so far I would like to continue, I posted the last message at 2 am and had to go to bed.

Critical buckling load for the pipe between pin ends at first bay;

Pc = pi^2 . E . I / L^2

Where I is the moment of inertia of the pipe.

Let’s assume Pc = F

pi^2 . E . I / L^2 = m . w . ( n - 1 )

n = 1 + pi^2 . E . I / ( L^2 . m . w ) this is the number of supports will cause buckling by friction at the first bay if the pipe adequately guided along the pipe at support locations.
Small diameter piping may see this kind of snake buckling between supports.


Another scenario is f . A = F = m . w ( n - 1 )
Where f is the allowable stress.

n = 1 + f . A / ( m . w ) this is the number of the support will create allowable stress on the pipe by friction if buckling does not occur and, pressure stresses ignored, under thermal expansion.

Under the thermal contraction the tension stresses occur in the pipe, in this scenario the pressure stresses may exist. So they are additive.

f . A = p . d / ( 2 t ) + m . w ( n - 1 ) can be written.

n = 1 + ( f . A - p . d / ( 2 t ) ) / ( m . w ) will be the number of supports will cause tensile stresses equal to allowable stress by friction under thermal contraction including pressure longitudinal stress.

These scenarios are valid for aboveground piping only. Buried piping does not go into Euler buckling due to the continuous soil support. Friction side is variable with the soil types along the pipe, piping construction system, pipe types etc. I do not go in this subject here for now not to cause confusion.

I trust this finalises the discussion.

 

[1503-44]

Thanks for humouring an old guy. Its complete AFAIC.

As a matter of potential interest, underground pipe can buckle. Usually it occurs at overheads, where eccentricity can be pronounced, contributing to the buckling problem. It may also occur on flat profiles when the weight of soil above the pipe is not sufficient to restrict lateral upward movement initiated by buckling.

MECHPIPENGEER hopefully I didn't mess your thread up too much.