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Hey guys, when using SW6, what does the maximum value of the sum of the hole diameters d1+d2 in the same cross-section refer to? For example, if I make holes in a flat plate, with three connection ports labeled a, b, and c, do I need to add the outer diameters of the three ADC connection ports as well? Please be more detailed in your answers, dear ones
They need to be added up on the same diameter line. If the three holes are evenly spaced at 120°, then it’s one hole
If the outer diameters of the nozzles with three holes are different, then is d1+d2 equal to the outer diameter of the nozzle with the largest diameter?
Yes. Take the maximum value of the sum of the opening diameters on the same diameter cross-section
Should the size of the opening be considered?
Of course, when it gets larger, it functions as a flange. :lol
I encountered this problem too and it’s really frustrating; I had no experience, so I simply used the diameter of the hole as a guide
Dear seniors, can we eliminate the reinforcing ring on circular flat-head end caps and use reinforcements instead? What are the effects of removing it on the stress distribution?
It’s been so many years since the original post was made – do you have any updated insights on this issue? I, as a beginner, still don’t quite understand it. Please help clarify it for me. Thank you
Firstly, the same cross-section mentioned by the poster is the one passing through the center of the flat cover; d1... represents the inner diameter (chord length) of the pipe at the center of a cross-section passing through the center of the flat cover, plus 2 corrosion allowances. The so-called sum is the larger of the sums obtained from the openings in the cross-sections passing through the center of the flat cover. Secondly, it is necessary to meet 6.4.3 of GB150.3. If the distance between any two pipe openings is greater than the sum of the two openings, it can be calculated for reinforcement using the normal flat-cover opening method.