UG-39
UG-39
(OP)
What does this mean from UG-39(b)(1):
"...shall have a total cross-sectional area of reinforcement for all planes through the center of the opening...".
I can only imagine what it might mean. Does anyone know what it means?
"...shall have a total cross-sectional area of reinforcement for all planes through the center of the opening...".
I can only imagine what it might mean. Does anyone know what it means?





RE: UG-39
Simple explanation; Any time you have an opening in a vessel, this must be evaluated to assure that the surrounding wall of the vessel is not weakened. Think of it this way, the opening in the vessel needs to be evaluated to determine if the wall thickness of the vessel needs to be increased, or if installing a nozzle in the opening a heavier wall nozzle is required or possibly the shell wall needs to be reinforced with an external plate, etc. The total area relates to A in the equation stated in UG-39 (b) (1), and is related to area of reinforcement provided by the shell or head wall, nozzle wall and weld size (if applicable).
RE: UG-39
"...shall have a total cross-sectional area of reinforcement for all planes through the center of the opening...".
RE: UG-39
RE: UG-39
If you're confused by the "all planes" part, take a look at Fig. UG-37 (which doesn't apply to flat heads) and do a quick search in this forum for Fig. UG-37. On a cylindrical shell, the area of reinforcement may be reduced in certain cases depending on the orientation of the section cut you are considering. In the case of a flat head, the orientation of the section cut is irrelevant, and no reduction in reinforcement is logical.
jt
RE: UG-39
RE: UG-39
If head is of a non-circular shape or nozzle config. is non-circular, more than one plane may need looked at.
RE: UG-39
"...shall have a total cross-sectional area of reinforcement for all planes through the center of the opening...".
Here is what we are doing:
We are taking an ANSI blind flange and drilling holes into the blind flange and pushing heater rods through the holes and fusing the heater rods to the blind flange. For instance, we may put (72) 0.838" holes into a 10" 150# ANSI SA-105 RF blind flange. We will put (36) hairpin heater rods that are 0.475" OD into the holes and fuse the heater rods to the flange. We will make a special flange that is thicker than an ANSI flange for this application. We need to prove via ASME how thick to make this special blind flange.
Our current calculations are making our flanges too thick and we are breaking drill bits especially for the 900#, 1500#, and 2500# applications. We are looking for a new way to calculate the flange thickness.
RE: UG-39
Stand the blind flange on its edge and look at the hole(s) you have cut thru it. Those holes weaken the flange, so you have to reinforce the flange. The procedure used by ASME is based on a simplistic approach wherein one "replaces" the area removed by the hole. Allowing for the fact that the stresses may vary in different directions, the designer must satisfy the requirements "in all planes". The easiest example would be case of a cylindrical shell wherein the longitudinal stresses are about one-half of the circumferential stresses. Back to your case, one plane to consider would be the plane that cuts across the hole from 0 to 180 degrees. Another plane to consider would be the plane that cuts across 90 to 270 degrees. If your case is symmetric, then they are all the same. That would not be true of the flange was obround or the opening is off-center. Hope this helps
Steve Braune
Tank Industry Consultants
www.tankindustry.com
RE: UG-39
RE: UG-39
Someone with a later model flange program might get a result that you can live with.
In reality you have a tubesheet in the Class 150-300 flanges.
Do you want your finished flange to have the allowables of each Class?
Can you comeback with your material and design conditions?
RE: UG-39
If you are using an ASNI flange as a tube sheet, there was an interpretation issued several yeas ago that stated that this was not premitted. In any case if there are tube sheets they are not calculated per UG-37 thru 39. Use that provisions of UHX.
RE: UG-39
you are leaving an empty space so the surrounding has to be stronger to compensate.
ER
RE: UG-39
About 5 years ago an electric heater manufacture asked ASME if the blind flanges should be subjected to code calculations. There are now two recommended practices that say yes. ASME has also said that electric process heaters do NOT fall into the UHX category so we are left with UG-37 thru 39.
We don't mind using UG-37 thru 39 thicknesses for 150#, 300# and 600# construction, but the larger 600#, 900#, 1500# and 2500# flanges are too thick. So we are revisting UG-37 thru 39 to see if we can make the flanges thinner. We are an ASME code shop, but we don't claim to be experts with every detail of ASME.
RE: UG-39
to install their heater hairpins.
It is my understanding that the Code does not even allow drilling of blind flanges because by doing that looses its ANSI values.
I have no idea what the industry will do but there are many questions.
I like to know more.
(AClark, the heaters are very thin U tubes " copper, steel, stainless steel, incoloy), filled with magnesium oxide as an insulator and in the center the electric coil/resistor,
they do not offer any strength/stiffening consideration as they are marely floating, the ends are brazed inside or outside or tig welded on the outside)
ER