Reducer Input overedeterminend

Why that ?
Its enough data input exist that the program calculate angle "alpha" self.
Two diameter and the element length its enough.
Make simple,if you can do it for us

You can't do it that way. There is a straight section at each end of the reducer, such that the diameter transition does not encompass the entire length of the reducer. Furhtermore, the length of the transition and the alpha vary from manufacturer to manufacturer. This is why "alpha" is an input item.

.......and of course, when we are doing the preliminary pipe structural analyses, we never know the reducer manufacturer and we never know the cone angle. You never can tell what the purchasing agents will buy. Most of us just used the maximum 2.0 SIF.

And for as long as there have been computer programs for structural analysis, I have been using the average diameter and the averave wall thickness with the B16.9 Standard length. But always was sure to put the SIF on the end of the previous (real) pipe length and the beginning of the next (real) pipe length. I really look forward to using the new reducer thingie - way to go COADE!!

Regards, John.

Kind of makes one wonder of the SIFs validity?

Another question to ponder is why B31.1 calls for a SIF on only concentric reducers?

And the final question that I always like to ask... which is, Why does B31.1 use a SIF for a component that is non-stadrdized SIF wise and B31.3 on the other hand uses no SIF on the same component.

Is a reducer used in a Chem plant stronger? But if the same component is used in a power plant its weaker I guess!!!

Gotta love those code inconsistencies!

When the user the pipe correkt modeled, then he
place at the point where the stiffness changed a node.
Ok, you have then two short zylinder between weld
and at the beginn reducer an end recucer.
Sorry i have allways a sketch or a draw for the recucer in nuclear power plants.
Sure when you take sif = 2 then this is conservativ, but i think more realistic stress
calculation is better.