"50%" meaning in UG-39(b)(2)? 4

[pdnachocivicos]

Hi,

I have two openings in a flat head with spacing below 1 1/4 their average diameter, thus the statement "the required reinforcement for each opening in the pair, as determined by (1) above, shall be summed toguether and then distributed such that 50% of the sum is located between the two openings" in UG-39(b)(2) would apply.

But I have read this statement I think a hundred times and still I do not understand it: How should that "50%" distribution be applied? Could someone give some practical example of this calculus?

Thanks to you all,

J.I.G.C.
Chemical Engineer

 

[unclesyd]

Take a look at this link.

http://www.pdhonline.org/courses/m205/m205content.pdf

 

[pdnachocivicos]

Dear unclesyd, thank you for your fast reponse but, unfortunately, after having carefully read this document, I did not find any example of such case in it. Could anyone provide a detailed example?

J.I.G.C.
Chemical Engineer

 

[Duwe6]

Assume two nozz's on a horizontal line -- a left one & a right one.
Assume that the required reinforcement is 40 square-inches of repad [for this explanation, thickness does not matter, just that the reinforcement calc's require 40 sq-inches]
The "50%" clause means that 25% [10 sq-inches] of the repad will be to the left of the centerline of the Left Nozz, 25% of the repad will be right of the Right Nozz c-l, and 50% [20 sq-inches] of the repad will be between the nozz centerlines.
This clause is to force you to keep the nozzles each in their half of the combined repad area. No crowding of nozz's togather. Crowding makes the effective areas of the nozz + repad overlap, making the strength of the overlap area too low.

 

[prex]

If the spacing is really below 1 1/4 the average diameter, then you are out of UG-39(b)(2), and fall into U-2(g). You possibly meant equal to 1 1/4?
Anyway that statement is somewhat the same as the other one "Not less than half the required reinforcement shall be on each side of the center line of single openings.". This one, strictly speaking, is only applicable to shells and formed heads, but should be equally valid for flat heads.
Now if, also for opening pairs, you put, for each opening, 50% of the required reinforcement on each side, you'll immediately see that this is equivalent to the statement of UG-39(b)(2)

prex

http://www.xcalcs.com

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http://www.megamag.it

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http://www.levitans.com

: Air bearing pads

 

[pdnachocivicos]

Thank to you all!

Prex, sorry for my mistake: yes, I meant "2 > p/dave >= 1 1/4". Following the orientatin from prex and Duwe6, thus I expect that the situation for my flat head depicted in the pdf attached would work:

t = 18 mm, trepad=10 mm, p/dave = 1.9
Areq1 = Areinf1 = 432 mm²
Areq2 = Areinf2 = 180 mm²
a/b = d1/d2 = dan1/dan2 = Areq1/Areq2

Could you please, finally, confirm? It would be of great help for me in my calculation checking. Thanks in advance,

J.I.G.C.
Chemical Engineer

 

[Duwe6]

Spacing is correct,for a minimum case. However, when using repads, this will be required to be one oval, combined repad.