UG 44 ASME Sec VIII Div. I
UG 44 ASME Sec VIII Div. I
(OP)
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??
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??





RE: UG 44 ASME Sec VIII Div. I
The problem with sloppy work is that the supply FAR EXCEEDS the demand
RE: UG 44 ASME Sec VIII Div. I
, 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??
RE: UG 44 ASME Sec VIII Div. I
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
RE: UG 44 ASME Sec VIII Div. I
RE: UG 44 ASME Sec VIII Div. I
BTW, OD formula are often more convenient for pipe, see Apx 1.
Regards,
Mike
RE: UG 44 ASME Sec VIII Div. I