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ASTM E10 section 7.6.4 varbiage 1

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SivanG

Materials
May 25, 2023
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Hello,

I am trying to create a verification formula based on ASTM E10 section 7.6.4:
"When indentations are made on a curved surface, the
minimum radius of curvature[sup]1[/sup] of the surface shall be two and a
half times the diameter of the ball
[sup]2[/sup]. Indentations made on
curved surfaces may be slightly elliptical rather than circular in
shape. The measurements of the indentation shall be taken as
the mean of the major and minor axes."

So here's the issue

[sup]1[/sup]R[sub]min[/sub]=b[sup]2[/sup]/a
2a = minor axis = d[sub]0[/sub]​
2b = major axis = d[sub]90[/sub]​
=> R[sub]min[/sub]=d[sub]90[/sub][sup]2[/sup]/2d[sub]0[/sub]​

[sup]2[/sup]R[sub]min[/sub]=2.5D

"Shall be" implies equals to. I was expecting some sort of greater than or less than situation. it seems this scenario is highly unlikely!

Therefore
d[sub]90[/sub][sup]2[/sup]/d[sub]0[/sub]=5D

So unless I'm not thinking through this correctly.. in a case where a 10mm ball is used, the only scenario where this equation is satisfied is when the ratio of d[sub]90[sup]2[/sup][/sub]/d[sub]0[/sub] comes out to exactly 50mm. for example d[sub]90[/sub]= 10mm and d[sub]0[/sub]=2mm.

This seems like a very strict requirement...

 
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What is wrong with the word "minimum"?
That means that "shall be" is the minimum, doesn't it?

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed
 
No. The word minimum is referring to the minimum radius of curvature, which is the calculated value R[sub]min[/sub]. The sentence would need to be structured as "the minimum radius of curvature of the surface shall be at a minimum two and a half times the diameter of the ball" to make sense.

Unless, again, I'm not looking at this correctly.
 
R=b^2/a is for curvature of an ellipse. The min radius of curvature in ASTM refers to the surface of test piece. for example, flat surface has an infinite radius, the ideal situation to calculate the BHN numbers since the indentation will be symmetric if the material is isotropic.

if the radius of curvature is too small, the test surface is very non-smooth, the diameter of indentation will be too small, yielding a higher BHN than would-be for a convex surface.

in addition, avoid testing a concave surface (if not flat). No idea if ASTM specifies to use a convex surface.
 
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