When we model tied universal bellows in simple model we give trans. stiffness and keep bending stiffness blank in simple models ( Page 5-11 of application guide). When we model individual bellows of universal exp. joint we put transverse stiffness as 1 E12 and put value of bending stiffness. ( Page 5-13 of application guide). What is the logic behind this? Why the individual bellow is made stiff in transverse when it would not be actually so?
Same way if I have to model Pressure balanced bellow with individual bellows modeling do I have to put bending stiffness and make trans. stiffness as infinite ? The overall assembly of tied universal and pressure balanced is really not stiff in transverse. Then why to model it so ? At least for pressure balanced joint the trans. stiffness should not be mad infinite when we give bending stiffness.
ANYONE CAN HELP ?

One more thing I wanted to add was- When we model universal bellow using CAESARII modeler it automatically models individaul bellows with trans and bending stiffness values specified. None of this is modeled as 1 E12. This is not matching with page 5-13 of application guide.
Anyone from COADE can throw light ? I feel the expansion joint modeler is correct in this regard and not 5-13.
PLEASE CLARIFY.

Dear Honest Stressman,

Please refer EJMA standard for expansion bellows design.

WOULD THE MODERATOR KINDLY ATTEND TO THIS AND THROW SOME LIGHT ? SOME TRAINEES IN THE FIELD OF STRESS ANALYSIS GET CONFUSED WHEN THEY READ THE LITERATURE AS EXPLAINED ABOVE.
OR IS IT THAT SUCH THINGS ARE CLARIFIED WHEN ONE ATTENDS CAESARII SEMINAR ? TILL THEN HE/SHE HAS TO DRAW HIS OWN CONCLUSIONS ?

Compensators are confusing. I know as I design them. In universal expansion joint the primary movement is lateral movement between the connection ends. Very rarely they compress or rotate.

As I understand the simple unit you model from connection flange to connection flange and the lateral stifness is for the full unit and not for individual elements (page 5-11). That lateral stifness or spring rate is effecting the rest of the piping.

When you analyse the universal item by item you will find that the individual bellows elements never move lateraly. The bending spring rate of one bellows unit is very low compared to lateral. Two elements are connected with nipple (centre pipe), which also rotates. By giving the rotation spring rate for individual element you are giving exactly the value that is needed.

I happen to be busy with same item. Provisionally on page 5-14 restraints on nodes 1036, 2036 and 3036 should be +X and not -X. I am busy with this issue with COADE.

No reason to shout...

For a complete definition of stiffness, zero length expansion joints require definition of both transverse and bending values. When length is specified, simple beam relationships can be used to establish the other value if one is given. For many years, piping programs only provided zero length joints. Our examples show both varieties.

Page 5-11 illustrates a 6 foot joint with only a transverse stiffnes specified. The bending term is calculated.

Page 5-13 shows two zero-length joints where transverse and bending stiffnesses are entered.

I just ran the XJ modeler for Senior/Flexonics/Pathway and that model shows length and transverse stiffness.

Does this explanation satisfy you?

Here are two added points. 1) you can override the transverse/bending relationship with length by entering both of these terms, CAESAR II will use your data 2) many expansion joint catalogs list a bending stiffness that should be used with zero-length joints. I would avoid (or verify) those data in building a finite length joint by hand.

My advice - use the modeler and then replace values with your data, it will yield a good model that others can reproduce. Save your fancy models for detailed work.

The illustration on 5-14 is not correct. The correct illustration shows the tie rods 1033-1034, 1034-1035, and 1035-1036 all running in the -X direction. Node 1036 is attached to node 1015 in the far negative X end of the joint. In that case the technical discussion is correct. We will put the original illustration back into the manual to avoid this misinterpretation.

Thanks to Dave Diehl and all others.
Regards,
Honest

Be carefull with bellows etc.

I usually spend a full day just re-thinking what is going on before i am happy with it.
You have to be 100% happy that the model is reflecting the real world. This is no more so than when assesing pressure thrust on adjacent supports which Caesar also does'nt do correctly.

Tim

David,

My understanding was that bending stiffness from catalogues should be multiplied by 4 (four) when used for finite length ex-joints and by 1 (one) when used for zero length ex-joints. Is this still correct?

Regards,
Veit

Veit,

I think so...

I just checked the Senior Flexonics Pathway catalog using the 8 inch 50 pound 8 convolution joint with an effective area of 70 sq.in.

The effective diameter (Deff) is 9.44 in and the flexible length (L) is 5 in (each 4 addditional convolutions add 2.5 inches to the OAL so 8 cons is 5 in).

Also published K axial=982 lbf/in, K lateral = 5248 lbf/in and "K bending" = 191 in-lbf/deg.

The lateral stiffness can be calculated using the axial stiffness:

K lateral = (3/2)(Deff^2)(K axial)/L^2 = 5250 lbf/in

The BENDING FLEXIBILITY can also be calculated using the axial stiffness:

Bending Flex = (1/8)(K axial)(Deff^2) = 10939 in-lbf/radian
= (10939)(pi)/180 = 191 in-lbf/degree

So these values match the catalog. BUT

This "Bending Flexibility" is the EJMA stiffness calculated for, and to be used at, the center of the bellows. This is too flexible for our expansion joint with length. (If I put a 191 in-lbf moment on the end of the XJ, the end will rotate 1 degree but it will also deflect. This is the EJMA term. For our beam model, we use a bending stiffness that is based on that rotation where no deflection is allowed. I will require a moment four times greater to get that same 1 degree rotation with no deflection. - It's akin to the guided cantilever method where the lateral stiffness of a guided cantilever (with no end rotatation) is 4 times stiffer that a free cantilever.

So,

BENDING STIFFNESS = four times BENDING FLEXIBILITY
= (1/2)(K axial)(Deff^2) in in-lbf/radian

= = = = = = =

This is why I would leave bending empty in a CAESAR II XJ model with finite length. The program will calculate the bending stiffness based on the other data. If you enter a bending term, CAESAR II will use it.

I am unsure if all manufacturer's publish this "Bending Flexibility". You can always run a simple model to check what's going on.

In this connection I would like to take views of some of you on following issue:-

I have a 40 " vent line ( which I have described under topic named VENT LINES in this forum).Would someone tell guide me on following:-
In the horizontal run do I need a hold down stop with gap to make this joinrt effective?
Regards,
H.S.

posting from "VENT LINE"
I am presently working on one 40" vent line running from top of column which is around 30m high. This line is connected to an exchanger which is on a structre around 4.0 m away from column in horizontal direction and structre TOS is 26.0 m. The line operating temperature is 200 Deg. C, Line is SS material. The column temperature varies along its height. At column nozzle lvl I have a vertical expansion of around 62.0 mm The line comes from top of column , runs vertically around 3-4 m. and then gets connected runs horizontally around 4.0 m , then goes down 3.5 m and reaches exchanger top nozzle. I have put a pressure balanced expansion joint in vertical on column nozzle to absorb column growth ( which is vertical).

Would someone tell guide me on following:-
In the horizontal run do I need a hold down stop with gap to make this joinrt effective?
Regards,
H.S.

This is not exactly a CAESAR issue but if you use any catalogues for your calculations there are couple issues you need to consider:
- some "big boys" have special manufacturing methods and they can make compensators with high number of layers or on big sizes 2/3 layers. More layers lower spring rate. Many smaller local suppliers cannot do these and then who ever is ordering cannot get what you specified. (Germany has at least one of these low spring rate manufacturers)


- some catalogues have working spring rates. These cannot be calculated, only tested. EJMA has formula for initial spring rates, which can be calculated. These are substantially higher than working rates. Now again local suppliers cannot always do these tests. Result is that if you used the working rates the buyer is in trouble.

I have done a lot of bellows designs. Requirements come often from stress engineers. Often big problems. Absolutely too low spring rates specified and too short space given.

Remember to consider compensator mass. Sometimes they are several metric tn.

When ever you do critical design get your local supplier on board from day one.