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Anchor forces over several 1

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Orefunto

Civil/Environmental
Jun 24, 2011
26
Hello Pipeliners,

Hope I can ask what must be a very basic question regarding anchor forces where I have a number of anchors restraining a line.

Suppose I have say four Anchors as shown below and I calculate anchor forces due to thermal and pressure expansion and find this to be say50lbs for each segment of the line anchor.

Would the total force at each anchor be an addition of the two forces coming from each segment on either side of the anchor or would the forces balance out and cancel each other, making the resultant zero?

[A1]50LB---------50LB[A2]50LB-------50LB[A3]50LB-------50LB[A4]

I hope its not too much trouble.

I used to resolve this kind of question easily with moment distribution when I designed structures several years ago, but not this is not quite the same as moments and I don't want to add the forces if.

Any help would be sincerely and thankfully appreciated
 
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Are you assuming the 50 lbf force is an axial force coming from expansion/contraction between each of the four stations? It does not appear to include the dead weight coming radially downwards at each of the four stations.
 
Hi raccokpe,


I can assume the deadweight is negligible , i.e supported by the ground between anchors and the force is from expansion/contraction

Its an hypothetical question to gain insight into how to allocate the forces over the anchors.

Sincerely

 
Remember a pipeline is not at the same temperature along it's entire length during transients. As the fluid enters the system then the pipe will heat up from that end ( assuming the fluid is hot). So you have to consider these cases as well as the normal operation. Also how can you be sure the friction on each side of the anchor point is the same (exactly) ?
 

The actual amount of the forces I have indicated are hypothetical. I'm wondering if we can consider the forces on either side of the anchor cancelling out and then obtaining a resultant force of zero on an anchor?

What I've seen is people subtracting the forces on either side of the anchor, that is what I'm really trying to understand if that approach is correct?
 
Depends whether you think those anchors are going to stay still or are going to accelerate. A free body diagram ought to help.

- Steve Perry
This post is designed to provide accurate and authoritative information in regard to the subject matter covered. It is offered with the understanding that the author is not engaged in rendering engineering or other professional service. If you need help, get help, and PAY FOR IT.
 
Thanks guys for all your answers so far and I've attached a sketch of the conceptual problem.

In Scenario( A) with Bends/corners between the anchors, its not going to be much of a problem since the bends will relieve the stresses. I believe I will only calculate the anchor force (using the straight lengths just to see what the magnitudes are like. I'll check the deflection and see that it can be absorbed by the legs between the anchor

Scenario (B)is more of a concern.
Assuming the forces on either side appear to balance out, so I believe I'm supposed to subtract but I'm not too comfortable with that mode of thinking.

So if the forces on either side are equal, then hypothetically, the resultant force is .....zero? But I really wonder......



 
 http://files.engineering.com/getfile.aspx?folder=20d88472-ef5c-4c86-8c5c-f00d2bc8f4de&file=anchor_sketch.pdf
We will neglect vertical loads. Theoretically temperature increases in your pipe segments will make a 50 lbs compression load in each pipe element. When you transfer those to your anchors, they will cancel out on interior anchors (those with pipe segments on both sides of that anchor). Your end anchors will have a shear load of 50 lbs at the top, probably creating some kind of shear and overturning moment, which must be balanced by soil reaction loads pushing back on the anchor.

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
Thanks BigInch
The approach of canceling out of expansion forces on interior anchors is what I am a bit wary about. If this is the case, I don't even understand why even use anchors at all in the interior as guides alone would appear to be sufficient.????

Thanks
 
there are other forces besides expansion, such as thrust caused by transient surges or gravity loads on steep slopes. those are not balanced out
 
Ha! you're right CVG.
I'm only doing the static side of things for now so I didn't think of that
 
Still no rule of thumb for this case, what % from one side thrust load (most conservative load) with considering the flow speed, pipe thickness, fluid properties :(
I tend to use substraction-result, assume fluid flow very fast which pipe temperature raise up almost simultaneously.

 
During startup, for example, one pipe segment might be a lot hotter than another one until things get moving. Don't believe the theoretical case describes all situations you might encounter in the field. Its your job now to think of all the situations where the forces might not cancel out.

Anchors are designed to anchor a segment of pipe between them, not for anchoring two adjoining segments. If you design for that force, when the forces happen to balance and cancel out you won't have anything to worry about.

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
Hello again guys
I'd like to beg your indulgence once again on what I thought I had down pat regarding forces and moments acting at end anchors of an L configuration

I have to use the thermal expansion force at the elbow of the L to obtain moments at the end anchors and I'm getting confused about what this force exerted at the bend/elbow of an L configuration would be

Calculating the thermal Force F in say the long leg is quite easy with the well known formula F= A*E*CoefExpansion*(T2-T1) and I think the moment at the short anchor should simply be F*Lenght of the perpendicular short leg.

But when I consider that the deflection imposed by F on the other leg of the L as a guided Cantilever deflection dx = PL^3/12EI from which I can obtain also obtain a P = 12 E*I/L^3

So I'm wondering if the Force I should use to obtain momenst should be
i)F( A*E*CoefExpansion*(T2-T1)) or

ii)should the force be P = 12EI *(dx)/L^3

The reason I'm finding this confusing is that I've calculated F, and P and they are not equal. I expected that they would be equal.

Although I must mention that both legs of the L were given and I merely checked the flexibility and found them adequate.

I'm suspecting that the guided Cantilver is supposed to be used to obtain an opposing leg lenght given an F calculated from the thermal expansion formulae in which case the dx equation of the cantilever would be solved using dx = FL^3/12EI in order to obtain an adequate Lenght for L.

I would appreciate your kind answers.

Yours sincerely but confusedly


 
You are trying to treat this as a statically determinate problem, which it is not.

The deflection of each leg influences the other by the amount that each pushes the other off their axial alignments, thus creating rotation and the resulting moments due to those rotations. You must first calculate the fixed end moments due to those displacements and rotations, then simultaneously solve for the amount of the combined rotation which will be equal on both members.

I think I mentioned that even the easy ones are not so easy even with a spreadsheet.

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
>>You must first calculate the fixed end moments due to those displacements and rotations, then simultaneously solve for the amount of the combined rotation which will be equal on both members.<<

Thanks BigInch, I hear you.

I want to use the moment distribution method you mentioned to distribute the end moments (when I calculate it) using the stifnesses of the legs.

But wondering which formulae to use to obtain the end moments, that is if the end moment at the anchor will be obtained from the canteliver displacement dx= Ml^2/12EI or from F*L where my F is the expansion force.

Sincerely

 
Fixed end moments are those that resist all rotation at both ends when they are assumed totally fixed and there is a relative lateral displacement off the axis. Its FEM = 6 EI[&Delta;]/L, see the diagram 4th one down on the right side here,


Use the axial growth of one pipe to get the displacement that you will use in the above formula to calculate the fixed end moments of the other pipe.

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
Thanks again BigInch,

So the F.E.M is merely 6 EI?/L alone?

I think I get the statical indeterminacy and obtaining the displacement ? as a result of the thermal expansion as well as the thermal expansion force is very easy

I was confused into thinking maybe i needed to use F*L and 6 EI?/L

So in doing my moment distribution e.t.c I will use the F.E.M as simply 6 EI?/L?



Thanks

 
For this problem, yes.

I forgot I had already suggested moment distribution. That is a good way to solve this problem, so it's a good thing you remembered and chose to use it.

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
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