Dear all,
One line from a column reboiler preheater to a tower,the line is very short,so it is not necessary to do the support for this line.
Because of the two vessels temperature is different and the support of the vessels are in different elevation,so i want to put four spring supports to support the column reboiler preheater.
But how to exactly model this vessel.I think inside the vessel is very complex,i want to know how to predigest and also ensure the result close to correct.
Best regards!

Focus on what CAESAR II does - F=KX using beam (bending) elements.
For F it's the deadweight and thermal growth. Your K is the stiffness of the piping, vessels & your boundary conditions (supports). And X is the calculated result - the equilibrium position of the loaded system.
Do not worry about what's inside the vessel if you got the weight right; you aren't analyzing that with CAESAR II. CAESAR II provides system level response, although it has executables to address local stress/load evaluation.

About a year back I did look into putting some HEX on springs. Movements were small so I used cup springs. Not a good idea at the end. When the HEX is drained the load changes a lot. Would have lifted up. Very problematic with short pipe I had.

Look into your case. If the mass of your equipment changes either during operation or maintenance you need to do something. If the mass change is only out of operation locking of springs may be an option. Unfortunately I have to say that there is very small possibility that the plant will block and unblock the supports.

In my case I put compensators and fixed the vessels.

for the ultimate in automated solutions.... Lisega makes an active (Hydraulic) "spring" support....

Thank you for your replied

When i am modelling the vessel If i know the weight,temprature and the dimension.
which is better?
(1) put in the weight as force on one element and others element weight give 0 .
(2) according to the vessel datasheet,put in the actual diameter,wall thickness...

Actual diameters, thicknesses, and lengths would be best - especially if you later apply wind.

One of our clients asking about the allowable stress for operating case.How should i reply them.In B31.3 code,is there any thing about this topic or coade has not considered the in the programme.I have gone through the all topics in the discussion forum.I understood well,but how to reply the client,like giving any reference in the code or in manual.I mean any documentary proof.Please help me.

Read the appendix in B31.3 - 2004 or 2006

And if you want CAESAR II to give you "Code Stresses" and "Code Allowables" in the B31.3 Operating Case, turn this option on using the "SIFs & Stress" tab of the Configuration Module.

Thanks a lot Richard ay.

Thank you richard.
One more thing is, is there any allowable value for hoop stress.
i am getting more value than the code allowable, of course, code stress check has been passed.
what is exactly the 3D max intensity.
krishna

Hoop stress is done in accordance with the code formula for wall thickness, you should do this calculation prior to creating a CAESAR II model YOU are responsible for adequate pressure wall. In other words the pipe wall thickness input should be previously checked for adequate wall thickness.

3D max intensity is Von Mises stresses named after the famous Osterreichen scientist / economist. They are not used by the B31 codes do a a web search on Von Mises stresses for more information

I think the better and simple way to solve the problem is to put an expansion joint.

Regards

"I think the better and simple way to solve the problem is to put an expansion joint."

A bad choice... springs versus bellows sheesh you have to be kidding me.... which owners are you doing work for?? My experience across multiple owners and in power, chemical and hydrocarbons is that most owners won't let you use a joint even if its the best way to solve a problem. Certainly for this case I have used and springs and never ever seen an EJ used to solve this problem... Pressure thrust usually makes the use of joints a complex issue that is seldom well thought out by most people despite loads of information on the matter available on the web.

Hi everyone!

With all respect for Mr. Luf, it should be noted that 3D Maximum Shear Stress Intensity (the actual concept I believe Krishna has referred to in his latest question) represents an equivalent normal stress based upon Maximum Shear Theory (also named as Coulomb-Tresca theory).

The Octahedral Shear Stress represents an equivalent normal stress based upon Von Mises theory (that is also known as a distortion/strain energy theory).

ASME BPV Codes (Sect VIII, Div. 1&2) and ASME B31.3 Piping Code use by default the Maximum Shear Theory (i.e. Tresca theory) to evaluate the max 3D Shear Stress Intensity in order to check the strength criteria for combined loadings.

Regards,

Dorin,

I may be wrong but I hope not....

First if someone from Coade could speak about what the 3d Max intensity is would be helpful.

Second please review the attached document...

The 3D Maximum Shear Stress Intensity and the Octahedral/von Mises stresses are similar in what they represent and how they are computed. (For details on how these values are computed, see the help text for “Yield Criterion” on the “SIF and Stress” tab of the Configuration Module.) Neither stress represents a Code value for B31.1/B31.3 type Codes.

As per such references as "Advanced Strength and Applied Stress Analysis" (Richard Budynas, 1977) and "Design of Piping Systems" (M. W. Kellogg, 1956) octahedral shear stress is defined as 1/3 sqrt[(S1-S2)(S1-S2) + (S2-S3)(S2-S3) + (S3-S1)(S3-S1)], so failure would occur in a uniaxial test specimen when the octahedral shear stress exceeds sqrt(2) * SY / 3. This isn't a very convenient failure point, so von Mises adjusted the constant on both sides of the equation (calculated stress and allowable stress) to look more like the maximum shear stress theory:

Von Mises stress = [1/(2 sqrt(2))] sqrt[(S1-S2)(S1-S2) + (S2-S3)(S2-S3) + (S3-S1)(S3-S1)], with failure occurring when von Mises stress exceeds SY / 2.

We have also seen the constant adjusted further to 1/sqrt(2), in which case the requirement conveniently looks more like that for stress intensity:

Von Mises stress = [1/sqrt(2)] sqrt[(S1-S2)(S1-S2) + (S2-S3)(S2-S3) + (S3-S1)(S3-S1)], with failure occurring when von Mises stress exceeds SY.

When it comes down to it, it really doesn't matter what the constant is, as long as the same constant is used on each side of the equation.

What fascinates me about this thread is how far from the starting question this discussion has gotten. The original post indicated a lack of knowledge as far as modeling techniques is concerned... and we have gone into the modes of failure etc.

As far as I am concerned what is of greatest importance so far as this thread is concerned are the appropriate modeling technique, and practical design solution(s).