Forces Diff Btw KELLOGG, GRINNELL, and Cantilever ?? 1

[Adam6]

Good day,

I have Just confused when i am doing some calculations of forces and moments affecting on the expansion loop,

Pipe Data:
10" X 0.365", API 5L X52
Moment of Inertia: 160.8 in^4
Diff in temp (btw installation and operating) = 165 C = 329 F
Thermal Coefficient = 6.72 x 10^-6 in/in F
Modulus of elasticity = 28.23 x 10^6 PSI
Pipe All length = 920 m, straight line above ground with two anchors at the ends.
So we will have 3 loops with specific dimensions, with supports, and anchors (please find attached)

========
calculations done by three ways:
(KELLOGG - GRINNELL - CANTILEVER METHOD)

1- Kellogg (Chart C-12)
Fx = 2395 lb
M = 38693.11 Lb.ft

2- ITT Grinnell
Fx = 120.45 lb (its so small, and i consider this force acting on the guide not on the anchor)
as we have (6m span distance) so 24 guides (supports)will be on each side and the acting force on the anchor will be 120.45 X 24 = 2891 lb,, is't TRUE ??!

Sb= 967 Psi ??

3- Cantilever Method:
by dividing pipe into two segments one horizontal (X direction) other is vertical (Y Direction):
Fx = 12 x E x I x Delta / L^3 = 7153.27 lb (this force is the whole force on the vertical leg of loop)
Mx = 6 x E x I x Delta / L^2 = 3576.63 lb.ft

NOTED that
Every method has a value of force and moment,, and the value of cantilever method is very very high comparing with Kellog and Grinell ???

So can any one understand this puzzle and help me, in any way....

sorry for my long POST !!

REGARDS

 

[BigInch]

I really can't comment on the differences without the Kellog chart, ITTG info or a diagram of the pipe you're trying to analyze. Your one-loop diagram doesn't look much like the 3 loop system you describe either, but the canteliver simlification you're using might be far too conservative. I also don't think the 120 lbs is on each and every guide, but I'm not sure what Fx you are really talking about.

You should explain more about the system you have and especially more about the forces and directions you have drawn in there too. These are highly directional and sometimes axial forces in vertical pipe tend to make vertical reactions, so difficult to seprate them without a better analysis diagram showing where those forces are and what global coordinate system directions they all have.

 

[Adam6]

BigInch,
Thanks for reply,

Please find attached:
- Line Digram with 3 loops, before loops its only straight line with two ended anchors.
- C-12 Kellogg Chart
- ITTG, Table
- cantilever

After looking to C-12, ITTG table, you'll recognize the directions of forces and moments.

As for the cantilever method, first please see the attachment )cantilever) I've just divide my loop into half to become (L shape),
then divide it to two legs vertical leg (B-C) and horizontal leg (A-B)
The force that i am talking about Fx =7153.27 lb
is across the vertical leg (B-C) which i thing its the greatest force that loop will bear on X direction, and its the same force which the left anchor (A) must bear, Am I right ?


Please Note:
That all the calculated forces in the three ways above are for to determine the stiffness of the anchor (i.e. What the forces that must anchors bear), besides to check these resultant thermal forces with the code ASME B31.1 to see if the line with this system will fail or not.

and if you kindly have another way to calculate the forces and moments by hand to this system, I'll be very thankful to view them here.

Hope that I clear the whole issue.

Thanks again

 

[Canadieng]

The simplified hand methods (ie, Kellogg, Grinnel, ...) all use differeing assumptions and estimations and are all just approximations.

See the attached graph outlining differing results taken from "Introduction to Pipe Stress Analysis" - Sam Kannappan.

 

[BigInch]

Hard to explain that difference. Looks like they obviously take into consideration the effect of having corners made of elbows, but we don't know what radaii they used, then extended the results into the same overall frame dimensions. For example, you would expect a lot of difference in the relative moment and bending stresses in a 20 ft loop made of 8" diameter pipe and 20 ft loop made of 16" diam.

It's a relationship that does depend on I, but also I/Length and that relationship doesn't seem to be evident.

 

[Adam6]

Canadieng, thanks for the graph,
yes its clear that there are differences between the three ways,
And as I've know and noticed in the graph that KELLOGG is the nearest one to computerize results !!

BigInch, other guys,
If we back to the force calculated by Grinnell
Fx = 120.45 lb, Am I true in my suppose that this force is sooo small so we can say that ?? :
6m span distance so 24 guides (supports)will be on each side and the acting force on the anchor will be 120.45 X 24 = 2891 lb ??
i.e. : could we add the forces like this to get the maximum one which affect on the anchor !!

=================
Last Line
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Can any pipe stress professional guides me to the most correct way to HAND calculate the forces and moments due to thermal effect on the supports and anchors
in this design that i have ???

REGARDS

 

[BigInch]

The easy way to do this would be just to put an anchor between each loop. Is there some reason you don't want to or can't.

Use some structral engineering analysis techniques,

Put a roller support between each bent and at one end to make it a determinate structure. Calculate axial displacements at those rollers and at the end of the structure. Now put a virtual stop at the rollers one at a time, making a separate analysis case out of each separate stop. Then calculate the force required to push the rollers back to the stop position. Then you'll be able to write 3 equations with three unknowns which you can then solve for the displacements at the roller supports.

Or, Use Moment Distribution to calculate all the moments at the rigid joints and anchors directly and then calculate internal shears and axial loads, then the support reactions.

 

[Adam6]

In my case I've put an anchor between each loop, so of course i don't have problem with that.

As for your way in calculations, sorry but I don't get it !!!

You said "Put a roller support between each bent and at one end to make it a determinate structure"

What did you mean by the bent ?? and the end ??
any Illustrated can you make to help me more >>>

You said "Calculate axial displacements at those rollers and at the end of the structure"
You mean by the thermal expansion law : Delta=Alfa x L x (T2-T1) ??

As for the rest I don't get it !!

could you made an example or more explanation, or any reference,,,please...

T really want to understand this way to calculate....

Thanks

 

[BigInch]

I will, but it's Christmas in Spain. If you can be a little patient for the next couple days I will dig/make up an example or two and post them back here.

A "bent" is an "up, over and down-type" structure. I should have stuck with the pipe terminology convention and called it an expansion loop. I don't like "loop" because that implies full circle and I have yet to see one of those.

 

[BigInch]

... made intentionally.

 

[Adam6]

Ahh, so you hate to say LOOP,,made me just google about meaning of bent for a time !!!

Ok, waiting you after couple of days,, with a great examples

Happy Chrismas, hope that you spend it with all your beloved ones.....BUT Don't eat the whole Turkey... ^_^

REGARDS

 

[BigInch]

Here's a spreadsheet showing how to analyze a determinate and indeterminate beam and a pipe between two anchors with a temperature change.

To a more complicated structure is possible, but takes a lot of time as separate cases must be added for each support reaction, deflection calculations made, replacement loads set and more deflections calculated. After deflections are added together to get total deflections at any reaction point, you can "ratio" what then need to be to arrive at your boundary conditions. Once you have all support loads, then you can calculate the internal forces.

 

[BigInch]

Here's a good link to what you might be able to do with moment distribution,

http://www.colincaprani.com/files/notes/SAIII/Moment Distribution.pdf

These structural methods assume rigid joints, which isn't entirely true for pipe ells, so the moments found will be somewhat conservative and axial stresses somewhat less than conservative.

Don't forget the stress intensification factors.

 

[Adam6]

BigInch, really thanks for your support...

your attachments are very useful, but please bear with me with some extra questions:

1- In the spread sheet when you talk about growth temp between two anchors,,, you put a standard straight beam ...
But, in our case we have a loop between the two anchors..
so lets suppose that we have a virtual anchor at the middle of (width of loop = w),
""I put an Virtual Anchor as at this point the forces and moments will become ZERO i think !!"",

How you will manage the area from (a to AV)???, see attached pic,

2- Please view the attachment (Depending on Cantilever) which is a PDF file i found on the net, calculate the thermal forces by another easier way,
Find my comments and issues on it... and give me your reply and opinion.

REGARDS

 

[BigInch]

A virtual anchor in the middle of the upper member won't be very accurate, because there will be some upward bowing of that piece and a centerline deflection which would cause your anchor to move. Any movement of the anchor would reduce the regidity and the resultant moment. You would have to add a case considering the moments and forces resulting in the frame from that support movement and add the two cases together. It is easier to look for where points of zero moment might occur and put artificial pinned joints there making the moment at those points always equal to zero. Virtual pinned joints are more likely to form in the middle of the vertical legs. That would indicate that instead of having the expansion force create a moment of F * H at the top joint, the moments would tend to be equal and opposite at each end of the vertical leg and have a magnitude of 1/2 F * H.

A beam with both ends fixed (no end rotations) and a differential support movement will have a fixed end moment of,
M[sub]FE[/sub]=6EI[Δ]/L[sup]2[/sup]
As less rotation is allowed at the joint, the fixed end moments reduce and the joint moments increase to the point where joint moments and fixed end moments are equal when axial deflections are allowed, but no rotation is permitted.
(I'm pretty sure that's right, but I haven't looked at it for a long time, so check it).

Have a look at this spreadsheet.

I happened to stumble onto this page which I think will be good for you too. It is with straight beams, but well applicable anyway.

http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT Kharagpur/Structural Analysis/pdf/m2l9.pdf