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UG-39 (d)(1)

UG-39 (d)(1)

UG-39 (d)(1)

(OP)
Suppose that you have a flat head (welded to a vessel) which complies with condition UG-39 (d)(1). The head has an opening with diameter < (head diameter/2), as prescribed in the condition. Following the ASME, I have to use formula UG-34 (c)(2)(1), replacing the value of C with 2C or 0.5, whichever is less (the case is sketch (b-2), figure UG-34). But in this formula, does not appear the inside diameter of the opening. How is it possible? Does this mean that all the openings with inside diameter < (Head diam./2), according to these conditions, shall require all the same thickness for the flat head? I mean, if I had a small hole, the reinforcement should be lesser than the case of bigger holes. Am I wrong? Thanks

RE: UG-39 (d)(1)

UG-39(d)(1) is a simple method to avoid the need for reinforcement calculations. Essentially the flat cover is thickened significantly, and it is accepted that the additional thickness will be suitable for any opening less than (head diameter/2).

It's true that for smaller openings, this will be excessive reinforcement and the head shouldn't need to be this thick, but that's the price you pay for simplified calculations. If you want to use a thinner head, you'll have to perform for detailed calculations per UG-39(b)(1).

Hope that helps.

RE: UG-39 (d)(1)

(OP)
Thank you! Very helpful

Consider a flat head with a central opening, welded to an outer vessel with an external hub, and welded also to an inner vessel with hubs that branch off from the border of opening, in direction 45° from the vertical axis. For some reason, this head doesn't reach minimum thickness prescribed by UG-39 (d)(1) for the opening. Question: could I use as area for reinforcement the area of protuberances that will be welded to the inner vessel? Part of this area is inside the limit of reinforcement prescribed by UG-40; however, this strange geometrical shape doesn't appear in any case shown in figure UG-40. Can I do that? Or should I demonstrate to respect the U-2 (g) "as safe as"? Thanks



RE: UG-39 (d)(1)

(OP)
Another question.
Looking at UG-39 (b)(1), there's 0.5 that multiplies the Area. What does this mean? When I compare this value with the available area of the reinforcerment, does this mean that I have to consider only half of the area available? Is it only a semplification by simmetry or is there another interpretation? Thanks

RE: UG-39 (d)(1)

ElCidCampeador, I can't really answer your question, but comparing the equations for required area of reinforcement from UG-39(b)(1) and UG-37 they are identical, except for two things: 1) F factor, which would not apply in a flat head, 2) 0.5 factor. So I conclude that Code only requires half the area of reinforcement for an opening in a flat head than it requires for an opening in a cylinder, formed head, cone, etc.

Why? Dunno.

Regards,

Mike

RE: UG-39 (d)(1)

The factor 0.5 is due to the fact that only bending stresses are present in the flat head, so that the average stress (absolute value) is 50% of the maximum.

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: UG-39 (d)(1)

Thanks prex

RE: UG-39 (d)(1)

(OP)
Thanks prex!

Do you know where I could find online some notes that explain in detail this fact of 50%?

RE: UG-39 (d)(1)

Another perspective on this is that the flat plate is not likely to fail on the compressive side - failure will be expected on the tensile side. Given that area replacement focuses on replacing area which is in tension (so 100% for circular shells under external pressure only), only 50% of the flat plate is in tension. Thus, the tensile area to be replaced is only half the required thickness.

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