Auto-definition of a diameter for a given surface

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1

Hello 

I'm trying to find a formula but I don't know if it's fatigue or lack of handling but I'm blocking...:)

Let me explain:

I have an M10 screw whose resistant section I know that does not move from 58 mm² of the original (without drilling).

I would like to find the formula that would give me or get closer to a current diameter, Ø2 in this case on the plan knowing the remaining resistant surface (40.97 mm²) in order to minimize the iterations to reach this surface of 40.97, but I have the impression of having too many unknowns, Does anyone have an answer?

Thank you in advance.


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Can it help you?

According to this site: http://b.urbani.free.fr/pagevisserie/visserie.htm 


Section of an 8mm screw  
(8mm being its external diameter we subtract 1mm 
for threads that do not contribute to the resistance of the screw)
its radius is therefore 3.5
3.14159 x 3.5 x 3.5 = 38.50


if the screw is marked 8.8 (hard steel) 
Its tensile strength is 80 kgs/mm² 
and therefore 56 kgs/mm² in shear;  
It will therefore resist :
38.50 x 80 = 3.08 Tons in traction
38.50 x 56 = 2.50 Tons in shear 


if the screw is marked 12.9 (extra hard HR HR steel for High Strength ) 
its tensile strength is 120 kgs/mm² 
and therefore 84 kgs/mm² in shear;  
It will therefore resist :
38.50 x 120 = 4.62 Tons in traction
38.50 x 84 = 3.23 Tons in Shear 



Section of an 8mm screw
(8mm being its outer diameter we subtract 1.25mm
for threads that do not contribute to the resistance of the screw)
its radius is therefore 3.38
3.14159 x 3.38 x 3.38 = 35.88


if the screw is marked 12.9 (extra hard HR HR steel for High Strength )
its tensile strength is 120 kgs/mm²
and therefore 84 kgs/mm² in shear;
It will therefore resist :
35.88 x 120 = 4.3 Tons in traction
35.88 x 84 = 3.0 Tons in Shear

@+

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I think it's this formula that you're missing:

 

I think you'll be able to connect alpha to the diameter of the hole...


surface_segment_circulaire.jpg
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Unless I'm mistaken, I think R.Cos(alpha/2)=diatrou/2

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and don't forget that your residual section is made up of 2 circular segments

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Thank you for your answers, I'll look into it this weekend or next week, I'm a little too busy with work today;)

Have a good day to you.

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Hello

So, with r the radius of the small hole and R the radius of the large hole I arrive at:

Edit: The whole thing *2...

Basically, I made the ratio of the sector to the complete disk to which I subtract a rectangle of width r and length tone L/2.

When will there be an equation editor on Lynkoa. (:

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Hello, sorry to answer you late, I'm looking back at it currently not having really had time before ...

I think I expressed myself badly in fact, badly made myself understood because I remain on too much of an unknown in fact, at least I think, unless I missed something. I will try to be clearer.

I have a screw in which I make a hole to weaken it in traction. This diameter of the bore is unknown.

A kind of "fusible" screw.

This screw will still be in M10 so a resistant section of 58mm².

I know my value of course of the final section of the screw with drilling that I have to reach, for example 48 mm² (the 2 circular segments)

Until now, I worked by trying each drilling diameter (Ø1, 1.2,1.5,2 etc ...)  until it reaches the desired value, or close to it (48mm²).

So if it's feasible, I'd like to make a table with all the tests I've done with all the various drills and I'll consult the table), that when I enter my final cross-section value, it automatically gives me the drill size of my drill...

Thank you for your help in any case. Kind regards.

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haaa I have connections that are made ... I review your messages and I keep you informed lol

 

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Well, no, actually... I thought I could get out of it by extracting the alpha angle of the formula from the air on Steph Beneo's screen... and then the angle use it in the other formulas ... Of course not

I'm drying it ...