What is effective area Aef for pressure balanced bellows?
Is Aef=0 or Aef=pi*Di^2/4, where Di - pipe inner diameter?

That depends on how you model them. If you model them using only 1 expansion joint element simply set the Aef = 0.

But if I set Aef=0.0001, would It be the same as Aef=0. I'm not sure...

Effective Area consista of 2 part: Aef=A1+A2

Aef=pi*Dm^2/4 - full effective area
A1=pi*Di^2/4 - effective area of pipe
A2=Aef-A1

Dm=(Dib+Do)/2 - is medium diameter of bellows
Dib - inner diameter of bellows
Do - external diameter of bellows
Di - inner diameter of pipe

What effective area I must set in Caesar Aef or A2?!
If Aef then for pressure balanced bellows Aef=pi*Di^2/4
If A2 the for pressure balanced bellows A2=0

Pressure thrust in case of pressure balance bellow to be taken care of by bellow itself.
When previously I was trying to model bellow in Caesar with three convolution piece & two tie rod set as it build.
But Caesar do not give good result. Bellow in model do not take care of pressure thrust as it do for tied expansion bellow.

Then I was told to put effective ID of bellow zero to get zero pressure thrust & feed all stiffness values as supplied by Vendor.

Straight pipe have pressure thrust too. Effective area for pipe is A1=pi*Di^2/4. If we set effective area of pressure balanced expansion joint Aef=0 it will be compressed under pressure thrust from pipes

<-Pipe-> <-bellows-> <-Pipe->
P*A1 P*A1 P*Aef P*Aef P*A1 P*A1

So, if Aef=0 we have P*Aef-P*A1 = P*0-P*A1 = -P*A1

expansion joint is compressed by force = -P*pi*Di^2/4

Is it really right or not? Sorry, I can't understand how effective area could be Aef=0
I think effective area must be pi*Di^2/4 for pressure ballanced expansion joints...

the whole idea of pressure balanced bellows is that all the forces from pressure trust are absorped in the bellow itself. Otherwise making it useless for axial compensation cause of the high pressure trust forces.
Only force that acts on the pipe is the thermal expension (or other displacements) of the pipe * the bellow stiffness while non-compensated bellows always give the pressure trust as force on the pipe.

btw: I've modelled a 3 piece pressure balanced bellow in ceasarII once and that worked correctly. Don't know why it didn't work for shr.

Hi mav
I am not telling pressure force will not be there in bellow portion. I am telling for pressure balance bellow modelling purpose consider thrust force zero at bellow portion.
Pressure thrust force is always there in pipe & that creates pressure component of longitudinal stress.
In a continuous pipe you will get component of longitudinal stress ( internal force due to pressure) but no external force.

Whenever we are using untied bellow then only we get unbalance pressure thrust force in external reference like support equipment otherwise not.
Since pressure balance bellow act like a continuous pipe as far as pressure thrust force is concern. So we can ignore external pressure force on both side of bellow( since it is balance one).

Expert please correct me if I miss some point.

Thanks for corne and shr. I understand how pressure balanced expansion joint works.
Now the second question. What area we must set in Caesar II?

I think we must set not full effective area of expansion joint Aef, we must set only additional effective area of expansion joint in comparsion to pipe - A2.

for ordinary bellows it is:

A2 = Aef-A1 = pi*Dm^2/4-pi*Di^2/4

And for pressure balanced expansion joints it is:

A2 = 0

Is it right?

And third question. What effective area usually written in catalogues for expansion joints Aef or A2?
If it is Aef, we must substract value pi*Di^2/4 from it to set in Caesar II. Right?

1)If you can set it properly in caesar ( I could not make it good for work same as tied bellow )full effective area of pressure balance expansion expansion joint Aef to be feed.

2) bellow vendor also give data for full effective area of expansion joint Aef

3) No need to subtract pi*Di^2/4


If you change effective area only tie rod force will change . It should not have any significant effect on support or nozzle so will not make much different in output result.

Effective area have very significant effect in case of untied bellow , untied universal bellow only. For pressure balance elbow no much effect.

I'm confused...

For untied universal bellow I must set Aef - full effective area
For pressure balanced expansion joint (it consist of 3 bellows and ties, but I use simple model without tie bars) I must set Aef=0

How It may be?

I have 3 variants:

1)
For untied universal bellow I must set area pi*Dm^2/4
therefore for pressure balanced expansion joint I must set pi*Di^2/4

2)
For untied universal bellow I must set area pi*Dm^2/4-pi*Di^2/4
therefore for pressure balanced expansion joint I must set 0

3) illogical variant:
For untied universal bellow I set area pi*Dm^2/4
for pressure balanced expansion joint I set 0
but It's physically not right!

How caesar works? Maybe if we set effective area = 0 It automatically use value of pi*Di^2/4?

I don't know if I can explain this properly, but I'll give it a try:

- a pipe under pressure itself doesn't give forces in axial direction. All forces in axial direction are absorped by the pipe itself (s = P*d / 4*t).

- an untied bellow can't absorp axial forces cause it has no axial stiffness. All axial forces due to pressure will be directed to the neighbouring piping/restraints. (F = pi/4 * Dm^2 * P) The complete cross-sectional area of the bellow gives the acting force. So use the biggest cross-sectional area of the bellow in CeasarII.

- A pressure balanced expension joint (the complete part, existing of 3 bellows and 2 sets of rods) is designed so it doesn't distribute axial forces due to pressure to the neighbouring piping cause all forces are absorped in the rods of the bellow itself.
* You can model the bellow in two ways in ceasarII:

1. Model the three seperate expansion joints in combination with the rods. Each bellow gives axial forces, so Dm can be used for the cross-sectional pressure area. If you make this model without surrounding piping but with an anchor on each side of the complete bellow you'll see that no axial forces are distributed to the restraints.
If you put an anchor on 1 side, and a 1 mm axial displacement on the other side you'll see an axial force in the restraints equal to 1 x the bellow stiffness.
This model is exactly as the bellow operates (physically correct).

2. Model the bellow as a 1 part expansion joint. Now you simplify the model described above. What you want are the same results as above, but with a simpler model. Problem with simplified model is often that you have to trick the conditions to get the correct results.
As we don't want axial forces from the bellow resulting from pressure, and we only want forces due to bellow stiffness resulting from displacements, we have to make sure that pi/4 * Dm * P equals 0. As pi/4 * P is a given expression Dm is the only variable. So we have to set this to 0 in the simplified model.

You can compare this modelling to the modelling of a lateral compensator. You can model it with 4 rods each having only an axial restraint, which is physically correct. But you can also simplify the model by using only 1 rod which has both an axial and a rotational restraint. Which is physically not correct but gives the same results.

I believe most expansion joint (xj) catalogs provide either effective diameter or effective area for their joints. Use this value to specify effective diameter in CAESAR II xj input if you wish the program to include a structural load in any load case with pressure. This structural load is the specified pressure (e.g. P1 or P2) times the effective area of the pipe. This load is placed on either end of the xj, pointing away from the xj.

Dave, are you sure about this?
A balanced bellow doesn't give axial forces to the surrounding piping and restraints. When you model it as 1 xj only, you can't add the area into ceasar imo.

Hi corne
Theoretically pressure force of pressure balance bellow should be absorbed by tie rod. But while modelling it in Caesar or in practical also very little effect will be there in support & nozzle .
One thing I miss in my previous post that I was told to make effect ID zero for In-line pressure balance bellow only.
For Tee type or elbow type pressure balance bellow, it act very fine in caesar with tie rod & effective ID of bellow.
But for in line pressure balance expansion bellow tie rod & effective Id of bellow do work properly in conventional simple modelling, that’s why when I work with in line pressure balance expansion bellow I put effective id of bellow as zero.

I've made drawing to explain my question

http://i024.radikal.ru/0804/67/f4d2a7dc62ca.gif

What variant is right "h" or "j"?

see the comments

shr thanx for your comments!
But I'm afraid you are wrong. I'm not agree that supprot force from pipe with caps will be zero (picture a). I've tested this example in autopipe and I get support force P*Fcap.

I'll try to explain: without supports pipe without expansion joint with end caps will elongate due to pressure on end caps! The elongation will be D=Pcap*L/EF.
Pcap=P*Fcap
L-pipe length
F - pipe crossection area
E - modulus of elasticity
So, if we'll add end supports pipe will can not elongate and support force will be N=D*EF/L
So, N=Pcap*L/EF*EF/L=Pcap!

For pipe with expansion joint will be the same thing, but elongation will be much more. D=Pcap*L/(EFc). EFc - bellows axial stifness.
But support force will be the same:
N=Pcap*L/(EFc)*(EFc)/L=Pcap!

I think It's most misunderstood thing that pipe without expansion joint with end caps hold pressure. It's not right

But if there's no end caps (equipment at the end of the pipe) then you're right. The support force (to nozzle) will be zero!
look:
http://i006.radikal.ru/0804/eb/ae4b2becf823.gif

@shr
Indeed, for inline pressure belanced bellows Dm = 0 when you use a simplified model, but you can also model these with 3 xjs and 8 rods. It will give the same results. A tee variant can be best modelled with 2 bellows and tie-rods and using the actual effective Dm imo.

@mav
Please get some facts straight for yourself. The actual question you're dealing with is:
"Does a pressure in a pipe give axial forces in restraints?"
Check shr's comments for the answer.
Once that's settled I think you'll have the answer for your question about pressure balanced bellows too.

Corne,

We both can be right on this. My comment was that if you specify an effective diameter (Deff) in your CAESAR II xj input, the program will add opposing pressure forces to both ends of the joint equal to P*Aeff.

If you have a pressure balanced joint or if you have some means of containing those forces (hinge plates or tie rods) then you can choose how to handle the pressure aspects of this assembly - 1) enter Deff and all the hardware or 2) ingore Deff and (most of?) the hardware.

I am not addressing the structural response of the xj itself, only the pressure.

I'll add this... I see a few xj's where overall stiffness is affected by the pressure - as if there is a break free load that must be achieved before the tie rods or hinges are able to move. This pressure effect is not automatically considered in CAESAR II.

The key to figures h & j is the effective areas of the different components in this assembly. To quote John Brock (Chapter 4 of Crocker & King's Piping Handbook 5th Ed.) in the caption for figure 46 - "The pressure area of the central portion is twice that of the end portions." The net imbalance area equals zero.

It is unfortunate that the author of your illustration re-used that P*Fcap to identify this effective area of the assembly. Figure h says total effective area is 0; Figure j says total effective area is P*Fcap.

Hi Corne
I was trying to make a complex model for in line pressure balance expansion bellow with 3 xjs and 8 rods as advised by you. But still not able to balance pressure thrust within the bellow.
I shall be grateful if you can send me any sample file in my mail id, how to model exactly in line pressure balance pressure with full effective area of bellow.

Regards

Habib

habibur.sk@simoncarves.com

Habib,

I've send you a small example. I'm sorry I can't provide any more examples, but I'm in the stress analysis business too and I don't think my boss wants me to do someone elses job for free.

Please check the example carefully and remember what mav and Dave Diehl have said above about axial forces resulting from pressure in normal pipe.

Dear Corne
Understand your concern.
Yes I wanted only small inline pressure balance bellow model. Thanks for spending your valuable time.
I find my mistake in previous modelling now . I forgot to change effective ID of middle Xj section (as sqrt root of 2 times end xj Id ) .
Now my model is also working fine with complex bellow model.

Keep in mind that most vendors only supply the stiffness of the complete part, not of the 3 bellows itself. If you model it as a complex model you have to adjust the stiffnesses to get the correct results.

Are you sure that middle effective ID is sqr(2)*end ID?
I haven't seen this information in vendors catalogues. They provide only total assembly stiffness and no information about effective area or effective diameter... That is motive of my first question in this topic!

Actually it isn't cause the stiffness of a larger diameter bellow is different from a smaller one. So in practice it will be a combination of area * stiffness that is decisive for the measurements of the 3 parts.
It was quick modelling by me why I made it that way, with equal stiffness for all three parts.
As said before vendors normally give a stiffness for the complete bellow and thus it can be easiest modelled as a single XJ with vendor stiffness and Dm = 0.

Did vendors gave you information that total assembly Dm=0? They could do xj with Dm<>0.

PS:
I don't see any necessity of modeling such xj as complex model.
1) There's no data available to do it exactly.
2) What is the goal using complex model? To estimate tie strength? This does not make sense.
I think simple model is the best way (total stiffness and Dm). Tie strength is vendor's problem. They must provide allowable axial displacement to prevent any problems with xj. But it's just my opinion

You are right Mav
"I think simple model is the best way"
To do this simple way we need overall stiffness ( vendor to provide)
For pressure balance bellow Dm has no significant effect so ignore it.
Tie rod will take all pressure effect & it's vendor responsibility to ensure that.
In simple model Caesar will understand with overall stiffness & Dm zero value ( I consider it as caesar's way to take care).

If total assemly's Dm <> 0 there will be an axial force remaining. Not the complete assembly but only two parts of it are restrained. So the total Dm must be zero.
You can check it using a complicated model.

Pressure balanced units especially the axial types are one of the most difficult bellows to design and understand correctly. Many manufacturers even do not get them right each time. I would not try to use CAESAR II to do the internal part design so simple modelling should be best. My current standard design calcs for pressure balanced units are anything between 40 to 60 pages of calcs including FEA, special software, MathCAD... Some comments:

Axial pressure balanced:
- axial spring rate is 2* small bellows plus large bellows spring rates
- if there is a lateral movement you have to get the details from the bellows designer. Make sure you have sufficient space
- axial pressure balanced units are heavy. Get mass from the supplier. Remember to consider increased fluid volume in the large bellows area.
- internal friction forces can be so high that the unit doesn't work. Consult bellows supplier
- because of manufacturing limitations units are not necessarily in full pressure balance
- in theory the large centre bellows should have exactly twice the cross section area of the two smaller element
- there are designs where you see only one set of rods

Elbow arrangement:
- axial spring rate is 2* one element spring rate. Elements are normally same
- if there is a lateral movement you have to get the details from the bellows designer.
- internal friction forces can be so high that the unit doesn't work. Consult bellows supplier

In general if no other information is available bellows element's effective diameter can be estimated:
- smaller sizes take pipe OD plus 60 mm
- large duct size bellows take duct OD plus 120 mm

Any spring rate given or calculated has easily a tolerance of 30%

I am a bellows designer and I still say that it is best to design lines without bellows when ever possible.

Thanks Jouko
For giving a clear cut explanation about bellow consideration.
“ I am a bellows designer and I still say that it is best to design lines without bellows when ever possible.”
- - Next time I must be more cautious before considering a bellow as an option.
I just again want to clarify one point
“- internal friction forces can be so high that the unit doesn't work. Consult bellows supplier.”
Can you please elaborate the above point. Are you talking about friction between fluid & corrugated bellow body or friction at tie rod connection. Is the point valid only for pressure balance bellow or any kind of tied or untied bellow.
Your suggestion is really worth to us.

Thanks Jouko! you've answered my question

One more question for Jouko,
Does effective area depends on pressure?

Friction relates to metal parts touching, e.g. normal metal to metal friction. One case is where the unit is horizontal and middle flanges and fluid have to be supported. Each case has to be evaluated by the bellows designer. Just remember these pressure balanced units are expensive and some suppliers only see $ so they will sell you even when it may not work. Valid for pressure balanced units. I have had one case where I did consider using teflon bushes but at the end left them out. If you look at http://www.jat.co.za/worksamples.htm you can see couple pressure balanced units. If I remember correctly that 54" unit was over 4 metric tn.

Effective area doesn't depend on pressure for a given bellows element. Effective area is defined as pi * Dm^2/4 where Dm = Db+w+nt where Db is bellows ID, w convolution height, n is number of layers in the bellows element and t is the thickness of a layer before forming. As you can see there is no pressure. Pressure comes into picture when you design the bellows. Different design pressures may require different convolution geometry to keep the stresses in the bellows element within the allowable limits. When you calculate the pressure thrust then you use the pressure.

Thank you Jouko!