Pipeline Stress Analysis 6

[ONENGINEER]

I am a soil engineer but recently involved in stress calculations for gas pipelines. Could someone introduces to me a classical method of calculation? Going through some calculations by others has not been conclusive to me. In particular:

What factor of sefaty is used in oil&gas projects to calculate maximum allowable stress. Of course as in other disciplines, there may not be a unique f.o.s but still looking for a so-called industry standard.

Is the code b31.8 the only code used in o&g design work? If not what is/are the others?

I have read in another Eng-Tip posting that "max bending stress= maximum allowable stress-0.3*hoop stress-thermal stresses" What about the longtudinal stresses, which would also exist if hoop stress existed. Where does the coefficient 0.3 come from. What is the range of this coefficient used by varios designers in the US practice? I guess shear and tortional stresses could also come to this equation but if so what numerical coefficient shoiuld be associated with them.

I know the above questions are primitive ones for the knowledgeable experts on this site but the responses would be a starting point for me.

 

[1503-44]

When a pipe is bent into an elastic curve matching a specific radius, the stress can be calculated by the formula :
S = E x D /R /2
Where
R= Bend radius inches
E= Young modulus of the material psi
D= the outside diameter of the pipe inches
S = the bending stress psi
To result in an elastic bend, S must remain below yield stress.

------------------
The stress in a member subject to a bending moment is
S = M c / I

M moment inch-lbs
c distance from outer fiber to the neutral axis. Inches. For pipe = outside diameter /2
I = moment of inertia of the cross section. Inches^4

‐------------------

When a pipe is subject to a bending moment it bends. Bending a pipe to a specific constant radius requires that the bending moment be uniformly applied and constant at every point X along the pipe.
Set M c/ I = E D / 2 /Bend_radius. Solve for M = I/c * E * D/2/R
If you apply that M to every point x of the pipe length by bending while holding both ends, it results in a constant radius bend.

A bending moment created by a point load, or uniform distributed load on the pipe, does not result in a constant bend radius, because the moment is different at each point x along the pipe length.

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

 

[pmover]

1ENGINEER,
double check the cell formulas in N16 & N17 in Rev-1.

 

[ONENGINEER]

pmover: the N16 is 0.3 x Sh = 0.3 x 44400 = 13320 and similarly N17 should be 22200 psi. The algorithms were not used properly. Thanks for your note.

 

[Snickster]

Here is how the two bending equations are related.

For a typical beam in bending per textbooks, the maximum moment M = EI/R where R is radius of curvature.

For a beam in bending S = M/Z or M = S x Z

Therefore by equating moments in both cases = S x Z = EI/R or S = EI/ZR

Since I/Z = D/2 Then

S = ED/2R

This is the stress caused by the elastic bending of the pipeline.

 

[Snickster]

The maximum longitudinal stress is stress due to pressure plus bending plus thermal plus any other longitudinal general stresses if any.

Sl = Sp + Sb + St + Sx

Sp pressure stress for restrained pipelines (above or below ground - fully anchored):

Since pipeline is fully restrained then there is no direct longitudinal pressure stress since longitudinal pressure stress is absorbed by anchors, this is assuming that even at bends there are thrust blocks that fully restrain the piping from expanding or contracting. However due to the hoop stress there is hoop strain and per poisons ration there is a resulting longitudinal contraction stain of 0.3 times hoop strain, but since the pipe is restricted from contracting by the anchors then there is a tensile longitudinal strain produced and hence stress of o.3 times hoop stress. This is where the 0.3 Sh comes from for restrained pipelines.

For unrestrained pipe lines the longitudinal stress is PD/4t or 0.5 Sh and produces tension in the pipe due to pressure thrust forces at elbows.

Sb bending stress for pipelines:

Above Ground: due to weight of pipe between supports, wind, earthquake etc. These forces all cause beam bending of pipe.

Buried Pipe: Due only to elastic bending of pipe across varying terrains = ED/2R since pipe is fully supported by ground and no wind or earthquake.

St Axial stress due to temperature for restrained pipelines (above or below ground)

Since fully anchored pipe is restricted from expanding then a compressive axial strain occurs during temperature increases equal to E(alpha(delta T). Therefore for a restrained pipeline all thermal stresses are axial compressive for increases in temperature and axial tensile for decreases in temperature - there is no thermal bending stress for restrained lines.

For unrestrained pipe the thermal axial expansion or contraction of a longitudinal run of pipe expanding into a connected directional pipe change causes bending stresses in that pipe that is perpendicular to the longitudinal run such at changes in direction using 90 deg. elbows. Manual and computer methods are used to determine the bending and axial stresses developed in the piping systems.

Sum of Longitudinal stresses for restrained pipe line:

Sl = Sp + Sb + St

Sp is always tensile for restrained pipes since it is due to stretching of pipe back to original length after hoop stress causes it to contract.
Sb is tensile at top of bend curvature compressive at bottom of bend curvature
St is compressive for temperature increases and tensile for decreases.

Combined Stresses

For pure biaxial tensile stress without shear the maximum principal stresses are the maximum biaxial stresses Sh and Sl and the maximum shear stress is 1/2 the difference in accordance with a Mohrs circle diagram. This is the basis of the allowable combined stress Sh-Sl since failure occurs under the Maximum Shear Stress Theory when principal stresses reach Sh-Sl producing the actual failure shear stress of 1/2 this value.

 

[1503-44]

That's amazing.

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

 

[Snickster]

A good reference book for pipelines that I use is Pipelines Rules of Thumb, E.W. McAllister.

 

[1503-44]

I've heard of that one.

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

 

[ONENGINEER]

Thank you Snickster for the great comments covering all. Trying to visualize the Sp in unrestrained condition (PD/4t) and wonder if a pipeline was assumed to be of a straight infinitive length, would this category of Sp still exist. Also thinking about the concept of unrestrained as each section of pipeline is restrained in longitudinal direction by the adjacent pieces except may be at the inlet and outlets. Nevertheless, my current interest is on HDD pipelines which are globally restrained to different degrees.

 

[Snickster]

For the infinite straight pipeline case it depends on the configuration. In order to hold any pressure the ends of the pipe would need to be capped. If an anchor is placed directly on the cap at each end say by using a big thrust block with the pipe end with cap imbedded into the block then the internal pressure would produce a thrust force on the pipe cap internally that is balanced by the concrete block considering that the block is strong enough to absorb the force without moving. In this case the pressure force will be taken by the block and not the pipe. Without anchor at capped end the pressure force on each cap end would balance each other by transfer of tensile force and stress longitudinally through the pipe wall since by statics all forces and moments must balance if the pipe is not moving. With the concrete block the pressure thrust is balance by each block and without the block the pressure thrust is balance by the pressure thrust at each end transferred through the pipe wall which is what longitudinal stress PD/4t is.

Say you had a long run of pipe with a 90 degree elbow and then continue with another long run of pipe in an L-shape and you put an anchor not directly on the elbow but one say 3 feet distance upstream and one 3 feet distance downstream. Well the pressure thrust will be felt internally on the elbow in both the X and Z directions meaning that the pressure force will point toward the center of the bend from both the upstream and downstream side. In this case the forces will be balance by tension in the pipe (longitudinal stress (PD/4t) which will travel back through the pipe to the anchors 3 feet distance upstream and downstream. Therefore those first anchors would take the load and balance the pressure thrust and that little short L-section between anchors would act similar to unrestrained pipe.

For unrestrained pipe as you mention the a long section of unrestrained pipe as it grows thermally axially is restrained from growing by adjacent connected piping. Since the connected piping is usually long and flexible it will bend to allow axial thermal expansion of the straight run. This is because to hold back just a little axial movement takes hundreds of thousands of pounds verses a few thousand pounds or less required to bend the connecting pipe in beam bending. So you are not going to stop the axial movement of the pipe but what is going to happen is that you are going to bend the hell out of the connecting pipe and cause serious bending stress. This is what stress analysis of above ground piping is all about - making sure that all connecting pipe to a pipe that is axially thermally expanding is flexible enough to bend but not exceed allowable bending stress when added to other stresses (such as internal pressure, wind earhquake, concentrated loads, etc.).

 

[ONENGINEER]

Great! From your description I assume the PD/4t stress cause a force analogous to the kickback a gardener feels in water hoses near the nozzle jet during watering.

 

[1503-44]

Have to take a bit of an issue with that explanation.
"In order to hold any pressure the ends of the pipe would need to be capped."
That is obviously not true.

Just for fun, consider a 20" straight pipeline on a steep slope in the same direction as the pipeline, capped at the bottom, open to an atmospheric pressure tank at the top of the hill. The pipeline is laid on the surface. Only friction of 1 lb/inch length between the pipe and soil underneath keeps the pipe in place. Pressure at the Cap is 500 psig. The pressure at the tank is 0 psig. The connection to the tank is very flexible. What happens to the one-end capped pressure force? Where did it go? Even if you closed a valve, or capped the line at the tank, that end Cap force would be zero. Unbalanced forces? Is the pipe moving? Where did the tension near the bottom Cap go?

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

 

[ONENGINEER]

I'm still in the learning process, however, the use of term "Thrust" in the earlier comment resembled a nozzle thrust in which the force/stress follow fluid mechanics laws, assuming fluid outflow is maintained. In this case the longitudinal force is expected to be resisted by outside restrains.

If the system is capped and no outflows, isn't that the longitudinal stress would be resisted by the pipe wall properties and the design would be similar to design of a compressed air cylinder. The longitudinal stresses would thus be independent of the outside restraining conditions, unless the restraining material's elastic modulus is near to or larger than the pipe wall modulus with a high shear strength interface between the pipe outside perimeter surface and restraining object.

 

[Snickster]

To reply to your previous post, the kickback that a garden hose produces is due to jet reaction force where force F = (M)(A)where pressure head is converted to kinetic energy of velocity increase of the fluid through the nozzle. It is a little complicate how this actually comes about considering the balance of forces since the force required to produce the acceleration is actually twice the force of the pressure acting on the area of the nozzle opening. I would need to draw a diagram to fully show how it works.

To reply to your most recent post follows:

When I referred to the thrust force of internal pressure on the end cap (which is just pressure x internal pipe area)it is because that is how it is usually referred to in piping design when design of an anchor to contain this thrust force. For instance sometimes in long straight runs of pipe above ground to handle thermal axial expansion an expansion joint is used somewhere in the straight length (alternately an expansion loop consisting of a U-shaped pipe segment inserted can be used) When using an expansion joint like this a full anchor is put on either side of the expansion joint down the line where it makes a turn at either end. In this case when the line thermally expands it forces the thermal expansion into the expansion joint which compresses the expansion joint. When the expansion joint compresses it cannot transfer any tension forces through it. The internal pressure times area is then like a thrust force on the anchors at each end which the anchors must be designed for otherwise the expansion joint would stretch apart and fail. So it is referred to commonly as the internal pressure thrust force at the elbows.

So yes it would be just like design of a capped cylinder where there are no external forces transmitted even if there were axial supports along the length. But theoretically I guess if you had a straight length of pipe capped at both ends say 20 feet long with absolute infinitely stiff anchors at each end just before the caps then the internal pressure force on the end caps would stretch the pipe up to the anchors and since the anchors are infinitely stiff, they would absorb the force and transmit to the ground with no stress in the pipe in-between.

 

[Snickster]

1503-44 in response to your post:

I think I get what you are saying: So if there is a long pipe on a steep slope of constant angle filled with water from a atmospheric tank with cap on the far-side end and with end connected to the tank being flexible so it cannot support a load (such as an expansion joint without tie-rods). Well the weight of the water in the pipe produces a downward force which can be broken into components normal and tangential to surface. The force normal to surface produces the friction force you mentioned. The force tangential to the surface produces the pressure force on the end cap where P=F/A. If the friction force due to the sum of the shear stress over the contact surface of the length of pipe back to the tank is not equal to or greater than the force on the end cap the pipe will slide down the incline. In other words the pressure force at the end cap is balance by the sum of the differential friction force acting on the entire length of pipe in the opposite direction. If you take a sectional cut along the pipe anywhere then the sum of the friction forces on the tank side pipe always equals the pressure force on the pipe cap acting in the opposite direction minus the sum of the friction forces on that side of the cut. This produces a tensile stress in the pipe at the cut. Note that this tensile force/stress is maximum at a cut taken at the pipe cap and decreases to zero at tank flexible connection.

 

[1503-44]

IMO, that's correct, except it decreases to zero at the point wherever the summation of friction against the soil equals the end cap force, (Hopefully it would be before the flex connection.) End cap forces are soon transferred in "real" pipelines to soil friction.

You could also have an effective interior cap. A closed block valve in the middle of a pipeline would effectively add an "end cap" load there. If it was buried, I assume that the load is resisted by virtual anchors on either side of the valve.

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

 

[ONENGINEER]

Trying to understand the meaning of the terms used in previous posts. By atmospheric tank situation, does it refer to a pipe with a length of L full of water, open to the atmosphere at the top (under ~ 1 bar atmospheric pressure) while the pipe is closed with a cap near the bottom of slope (angle α with horizon)? Therefore no water flow through the pipe is expected? And also does the p in the p=F/A in previous post refer to internal water pressure on the internal surface of the cap which would be (L).(Sinα).(γw)? Is the cap welded to the bottom of the pipe or is the cap in place by applying a force on the external face of the cap?
Or is the water flowing through the pipe and the tank feeds the pipe while the pressure at top is maintained at ~ 1 bar? In this case what is the function of the cap?

(γw = unit weight of water)

 

[1503-44]

Yes.

A cap and a closed valve are essentially the same thing. A closed, or partially closed valve would create a similar end cap load, proportionally less if the valve was only partially open.

Theoretically any pressure drop from flowing fluid should also be resisted by the pipe, such as at a partially closed valve, or even the summation of all pressure drop fluid friction shear forces on the inside wall. If you draw a free body diagram, there is nowhere else you could put them.

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

 

[ONENGINEER]

I should like to thank you all for your comments on stress analysis subject, which helped me a great deal.

Another question I have is about restraining parameters for HDD alignments. Also what is the recommended model for pipeline-soil interaction? Bearing in mind that the pipelines are pulled through reamed bores of larger diameter, which are filled with drilling fluid (usually bentonite mud) and left "as is" on the completion of pullback operation.

 

[Snickster]

I don't have any experience in HDD. My experience is in process systems design including piping stress analysis of Oil and Gas facilities, refineries, chemical plants, offshore platforms, etc.