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symphony123 (Mechanical) (OP)
13 May 10 17:07
I have a pipe of A333 Gr 6 material- Iam trying to make minimum wall thickness calculation per B31.8 and B31.3 and trying to compare the difference. In B31.3 there are welding factor and coeefficient factor from Table 304.1.1 to be considered. For the above maerial what would be these two values.

Also for A350 LF2 material also can anyone tell me what could be the welding factor and coeefficient factor from Table 304.1.1 of B31.3

Another question that i have is when does anyone need to comply with B31.3 or B31.8. If a pipe complies for B31.8 can it be considered that its ok for B31.3 also. It doesnt look like that though since my calculations for thickness are different.

I would be very glad and thankful if anyone can guide me through this code calculation process.
I appreciate all your help
Thank you
LV
Helpful Member!  rneill (Mechanical)
13 May 10 18:18
Comments on B31.3 Calculation:

W = weld joint strength reduction factor: see Table 302.3.5. For temperatures below 800 F, this factor can be ignored.

E = Ec - casting quality factor or Ej - quality factor for longitudinally seam welded pipe: (see Tables A-1A and A-1B)

Y = coefficient from Table 304.1.1

A333 Grade 6 material can be seamless or longitudinally welded.  I doubt your design temperature is greater than 800 F so you can ignore W.  Since you are not cast steel you can ignore Ec.  If this is seamless pipe then you can ignore Ej but if it is welded pipe then you need to consider Ej per Table A-1B of B31.3.  For design temperatures less than 900 F, Y = 0.4.  The allowable stress value used in the calculation comes from Table A-1.

You asked about A350 LF2 material but this is a forging material used to make threaded and socket welded fittings or flanges.  Pipe is not made from this material so you would not do the same calculations for A350 LF2 materials.

In regards to your question about B31.3 vs. B31.8.  You can not assume that pipe which complies with B31.8 for a particular service is acceptable for the same service conditions under B31.3. These are different design codes and they have completely different approaches to how they calculate required wall thicknesses and to the stress values that they use in these calculations (even for the same material).  Consequently, you must check the material properties and the calculations for each code separately.
symphony123 (Mechanical) (OP)
14 May 10 11:48
Thanks Neil for the response.
When u say the particular values can be ignored, is it i can consider that value to be equal to 1?

Now with A350 LF2 how would i calculate if it has got enough thickness material-The design with this material is actuall different. Its looks like a cylinder with some bolt holes on it but looks like a boss or a small flange like
The boss is actually welded to the pipe so i use the reinforcement calculations per B31.3 for reinforcement check. But how would i make sure that it has got enough wall thickness
Please suggest
Thank you
rneill (Mechanical)
14 May 10 14:09
When I say a factor can be ignored, it is because the factor would be 1 in that case.

With regards to your A350 LF2 component, it sounds like this is a custom component that would then be considered an "Unlisted Component" in ASME B31.3 and the rules for unlisted components would apply.  It is difficult to comment not having seen the specific component, but I'm guessing that the standard design rules for standard components are not applicable to this custom item.  You should consult the manufacturer and ask them how they validated the design rating for this item.  You may require bust testing and/or FEA.

You should attach a picture of the item ... it may be something like a "weldoflange", "nipoflange" or other similar and somewhat proprietary component (Promat makes these).  These would be one piece forged components that combine an O-Let, a nipple, and a flange so that you have one piece with one weld instead of 3 pieces with 3 welds.

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