Hardness, thinness and edge deformation

[theonew]

As counterintuitive as it is, I can get my brain around the idea that given two pieces of the same steel with the exact same geometry but at different hardnesses the force required to flex the steel will be the same. But accepting that leads me to question why people are reluctant to put a very acute edge angle on any steel that is softer than say 60 HRC. Here's what I don't understand:

Let's say I have two of the exact same knives in CPM 3V. One is at 62 HRC and the other at 58 HRC and I've thinned the edge of both down to 10 degrees per side. Now let's say I go to forcefully cut some thick plastic or maybe baton through some knotty wood. The thin edge on both should presumably be flexing to the same degree. If the edge on the knife at 58 rolls because the force exerted has exceeded the elastic range of deformation shouldn't the edge on the harder steel snap or chip? Boiling it down, my question is does a harder steel of the exact same type have a greater elastic deformation range than if it is run softer? Seems counterintuitive to me. If it doesn't than why is it so often claimed here that only harder steels (60+ HRC) can be taken down to thin edges?

 

[theamazingdrew]

I never put thin edges on any knives over 60rc its just asking for chips I thin my machetes and large choppers BECAUSE the edge will roll and not chip like harder steels.
so I agree with you 100% doesnt make sense to me either.

 

[hardheart]

The harder the sample, the less the diamond indenter penetrates, because the harder steel has a higher resistance to deformation.

here's a good topic on it, http://forums.swordforum.com/showthread.php?s=&threadid=55349

 

[fujita yuji]

Hardness and strength are correlated to some extent.

http://www.efunda.com/units/hardness/convert_hardness.cfm?cat=Steel&HD=Approx. TS

Which means more force is needed to warp harder steel.
In other words, harder edge is less likely to be deformed.

 

[theonew]

Ahhh, makes sense now, thanks for the links :thumbup:

 

[me2]

Boiling it down, my question is does a harder steel of the exact same type have a greater elastic deformation range than if it is run softer?

Yes. As the yeld strength goes up, the elastic range also increases.

 

[eKretz]

These guys have it exactly right. Some examples I use to explain it to people are: Take two identical springs made of the same steel. One is hardened to higher hardness than the other. Both springs will compress the same amount with the same load applied. The softer spring will take a "set" when compressed to a certain height, and will not return to its original height. The harder spring, when compressed to the same height, will return to its original height.

Another example, take two steel bars of 1/4" diameter and a foot long. Same as before, identical steel, same dimensions, but differing hardness. Set the bars up so they are supported at the ends and hang a weight in the middle of the bars. With identical weights and supports the same distance apart, both bars will sag the same amount in the middle, as long as the elastic limit isn't reached. Again, when the elastic limit is reached with a certain weight on the softer bar, it will remain permanently bent. At the same weight, the harder bar will return to straight when the weight is removed.

 

[Broos]

Thanks for the interesting read and the links, guys.

I am having trouble with taking the relationship that governs blade flex and trying to equate it to edge failures. Are edge failures caused by the edge exceeding the maximum range of plastic deformation from a single cut? I suspect failures of this nature are rare. Maybe rolling is caused even more by exceeding the compressive strength of the very edge causing plastic deformation, and chips are caused by elastic deformations over some period of cutting that cause microcracks and brittle type failures (chipping) at the edge.

Nevertheless here's a good article by Phil Wilson applicable to this discussion:

http://www.seamountknifeworks.com/article 3.htm