Hi,
When calculating anchor under a "classical standpoint" in a change of direction, someone should use the formulation R=2xPxAxsin(alpha/2), being P the pressure, A the internal pipe area and alpha the angle between the pipes. This formulation causes large anchors to be built. However, a simmilar analysis using Caesar in a change of direction causes that sometimes the resulting force R might be neglected.
Of course, a first explanation is obtained by considering the flexibility of the system, allowing displacement of it. But, the differences between one approach and another are so huge (for instance 550 thousands pounds and 22 thousands pounds) that makes me doubt whether Caesar considers all of the variables that it should for a problem like this or I am modeling correctly. Even more, only considering temperature expansion (Load Case Ope=W+P+T or Exp=L1-L2) causes huge forces, but never considering only pressure (Load Case Sus=W+P).

May somebody help me to clear up this issue???. (Maybe recommending/sending me a simple example that obtains forces more or less the same that of those obtained under the "classical approach").
I would appreciate your help.

Patricio Alvarez
Santiago of Chile.

Mr. Alvarej,

The explanation is as follows:

The pressure thrust load=P*A where P is the line pressure and A is the internal area or rather the flow area of the pipe is always there , where there is a change in direction. Until and unless the load is of dynamic nature ( to put it simply it is not simultaneously acting at say two elbows in the same direction) , there is a static balance with no unbalanced load going to the support components.But that is creating a longitudinal stress = PD/4t and this is acknowledged by the code where this stress is added to the longitudinal bending stress and compared against the hot metal allowable.

Now in cases like W+P , this load due to P*A is not considered as the Code does not consider , only the stresses due to the same is computed in the category W+P.

Typically when analyzing bellow systems, this P*A is considered by the program at two ends of the bellows. However it is a misconception to believe that only when bellows are there, there will be thrust load , in fact a tied bellow behaves axially like a rigid pipe and in both cases you will have P*A acting on the system whenever there is a change in direction. In fact if you have an untied bellow in front of a nozzle and the downstream side of the bellow is very stiff in comparison to the upstream side , you may have little or no pressure thrust going to the nozzle.
as by the theorem of moment distribution, stiffer members of a structure take more loads.

Now to compare the effect of the thrust force at the elbows between a conventional and Caesar calulation, try to put the P*A effect on the elbow using the Force/ Moment icon on the input spreadsheet and see the behaviour. The load will be distributed in the system depending on relative stiffnesses of the different components of the system.A conventional hand calulation will only compute the force at the elbow , but for a 3-D system considering the fact that each node is having 6 degress of freedom, the two analysis results cannot be the same . This you can infer considering the enormity of the stiffness matrix.

In order to get a close feel as to how the distrubution of P*A takes palce, use the force P*A at the elbows and check the results.

I hope this clarifies your doubt.

Regards

A. Bhattacharya

Stress Analyst

Bechtel Corporation

Additionally in case of high-pressure long pipeline with considerable pipe dia, there is significant expansion in the line under static pressure. If this expansion is restricted by placing an anchor then there will be significant force on the anchor depending upon the magnitude of pressure, pipe dia and wall thickness. In ordinary piping systems this force is neglected because of insignificant magnitude of the expansion. If desired this force can be address by activating Bourdon Effects in Special Execution Parameters.