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Minimum assumed radius on a blind drill hole

Minimum assumed radius on a blind drill hole

Minimum assumed radius on a blind drill hole

(OP)
Does anyone have any input on the minimum radius on a blind drill hole at the edge between the cone and cylindrical portions of the hole? Without any radius, the stress concentration is huge. In reality, there will always be some radius even though none is specified in the manufacturing drawings. Since this is a very difficult feature to measure, I have been unable to answer my question this way. Currently I am using a 0.1mm radius for a 8mm hole but I would appreciate any feedback I could get on this fairly fundamental modeling question.

Thank you

RE: Minimum assumed radius on a blind drill hole

Interesting question.
Since no one has come up with a reply yet, I will proffer some thoughts.
1. If this is just a typical part with a typical drilled (and perhaps tapped) hole, it is not usual to bother about the radius in FE modelling, because the average stress is normally low deep within the material.
2. If you are really going to model such a small radius, you would need a tremendous number of elements - this is only something you can afford to do in special cases, which I presume this is. I expect you realize this.
3. You say that it is difficult to measure the radius - implying that you have a physical example on hand. But you can measure the radius by using a metrology type casting resin, which you then measure using an optical projector.
4. If you are asking what the radius actually is in a typical drilled hole, I believe it would have to be determined experimentally, using the method in 3 above. It will not be the same as the (very small) radius on the tool used to create it - it will be larger, and the size will depend on material, yield point, spring back effect etc. In practice, tool wear will be a factor if you are producing a lot of holes - that sharp edge is one of the first things to go on a drill. If you have special drills made with a known radius, regrinding will be very expensive.

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