UG 44 ASME Sec VIII Div. I

[Mech2325]

Can someone please explain the statement of clause UG 44 for an ASME B16.9 fitting which states "the thickness of a pipe fitting can be assumed to be that of a straight seamless pipe"
If, for example i have a shell made out of pipe material, say, sa 106 Gr.B (NPS 18,SCH 40) and the pipe fitting of sa 234 Wpb is being used with it. Now the fitting will also be of NPS 18, SCH 40. Now how will the required thickness be calculated based on this statement of UG 44??

 

[SnTMan]

thread794-408382

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[Mech2325]

sntman: I did not get my answer there too.
, so i have posted it as a separate thread. I ran the calculation on software and it was taking the shell formula for calculating the required thickness. PR/SE-0.6P. I have read it on 16.9 too but still isn't clear. As for the above example, SCH40 would be the nominal thickness and the thickness calculated from this formula will be required?? Is my understanding correct??

 

[SnTMan]

Mech2325, question: "Now how will the required thickness be calculated based on this statement of UG 44??"

Answer: "the thickness of a pipe fitting can be assumed to be that of a straight seamless pipe".

In other words, treat the fitting as a section of straight seamless pipe.

Regards,

Mike



The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[Mech2325]

sntman: Thank you for your explanation. The dia of this pipe will be taken as that of fitting and the formula of PR/SE-0.6P will be applied to calculate the thickness??

 

[SnTMan]

Mech2325, yes, or, if the allowables for the pipe and fitting materials are the same, as with SA-106B and SA-234WPB, and you have already done the calculation for the cylinder, just match the wall schedules.

BTW, OD formula are often more convenient for pipe, see Apx 1.

Regards,

Mike

 

[Mech2325]

Thank you sntman for the help