[TheForge] Hardening steel

Sat Sep 7 04:46:01 2002

Hardness is another easily misunderstood term. Actually there are a number
of "hardness" measures to us metallurgical types, but I'm sure you are
referring to an indentation hardness measure such as Rockwell or Brinell. A
simplified explanation of a Rockwell C-scale test is as follows:

- First push a diamond point into the surface of the material to be tested,
using a light load (known as the "minor load"). The purpose of this load is
simply to minimize the effect of surface finish, superficial blemishes, etc.
Set the dial to zero.
- Next push the diamond point into the surface using a much higher load
(known as the "major load").
- Release the major load and measure how much deeper into the surface the
diamond point is sitting, compared to before applying the major load. The
difference is a direct measure of hardness.

A Rockwell hardness tester can be viewed as a dial indicator graduated in
hardness numbers, instead of inches. You apply the minor load and set it to
zero. You then simply apply the major load, release it and read the dial.

If you think about the above, you'll see that hardness is really a measure
of strength, not stiffness. What is being measured is resistance to taking a
permanent set. The hardness machine doesn't care how much the material
temporarily deflected when the major load was applied (i.e. how stiff it
was), it only cares by how much the indenter failed to return to its
original position once the load is released (how much the material
plastically/permanently deformed). If you look in a reference book, you will
find tables to convert directly between Rockwell (or Brinell, Vickers,
Knoop, etc.) hardness numbers and Ultimate Tensile Strength.

Hope this helps.

Dan

> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]]On Behalf Of Peter Fels and
> Phoebe Palmer
> Sent: Saturday, September 07, 2002 4:02 AM
> To: [email protected]
> Subject: RE: [TheForge] Hardening steel
>
>
> At 01:08 PM 9/7/02, you wrote:
>
>
> Thanks Dan;
> Understood what was said and accept it and have no doubt that it
> holds true
> as calculated..can't argue with empirical data. Just don't understand
> WHY  the modulus of elasticity would be the same for hard and soft steel.
> If it is harder, why isn't it stiffer?
>
>
>
> >Mike Schermerhorn's explanation is correct. A material's modulus of
> >elasticity is a direct measure of its stiffness (resistance to deflection
> >or, more properly, strain). The value for all steels is around 30,000,000
> >psi. It cannot be altered by heat treatment. It is the same for annealed
> >material as it is for quenched and tempered. Mathematically, it
> is used as
> >follows: stress (psi) = modulus of elasticity (psi) x strain (in/in). In
> >other words, if you put a hook on each end of a 1" square bar and hang a
> >weight on it (say 100 lbs.), that bar will stretch a predicable amount:
> >strain = stress / modulus of elasticity = [(100 lbs.)/(1 square
> >inch)]/30,000,000 psi = .0000033 in/in. That is the amount by which each
> >inch of length stretches when the 100 lb. weight is placed on the hook
> >(which results in a stress of 100 psi since the cross sectional
> area of the
> >bar is one square inch)). If the bar was 24" long unloaded, it
> would be 24"
> >+ (.0000033 x 24") or 24.0008" long with the 100 lb weight
> hanging on it. If
> >you anneal the bar, the elastic elongation will be the same as
> if you quench
> >the bar.
> >
> >The confusion seems to be over the yield strength part. Take
> 1045 steel for
> >example. Annealed, it has a yield strength around 50,000 psi.
> Water quenched
> >and then tempered at 1000 degrees F, it has a yield strength
> around 73,000
> >psi. Both conditions will have a modulus of elasticity of very nearly
> >29,700,000 psi. Continuing with the above example, consider two identical
> >bars as described above. Each bar has a cross section of one
> square inch, so
> >the stress (psi) is numerically equal to the weight hung on the
> bar. If one
> >were to hang heavier and heavier weights on each of the the two
> bars, they
> >would continue to stretch the same amount, and then return to
> their original
> >condition when the weight was removed, up to a point. That point
> is when the
> >yield strength of one of the bars is reached. In this example,
> it will occur
> >around 50,000 lbs. Anything heavier, and the annealed bar will
> stretch and
> >not return to its original length. It will have yielded, and it will
> >permanently elongate more and more as heavier weights are added. The
> >quenched and tempered bar will continue to return to its
> original condition
> >until the applied weight exceeds 73,000 lbs. After that, it too
> will take a
> >permanent set.
> >
> >What it comes down to is this. If you have two identical pieces of steel,
> >differing only in their heat treat condition, they will be
> equally stiff. If
> >you load them identically, they will deflect identically. The only
> >difference will be in how heavily you can load them, before one of them
> >yields. This thread started out discussing a fireplace piece.
> That item will
> >not be made any stiffer by heat treating it, if you understand
> stiffness to
> >mean how much the item resists bending when loaded below its
> yield strength.
> >It would however be made stronger by heat treatment, allowing it
> to be used
> >more heavily and less susceptible taking a set.
> >
> >When it comes to running an experiment, keep in mind what Mike said about
> >limiting the applied loads. You'll have to run the tests such
> that the loads
> >remain below yield strength of the softer material. In that
> case, there will
> >be now sensible differences in stiffness between annealed and hardened
> >steel. If you run the tests such that one of the test piece's
> yield strength
> >is exceeded, you are no longer testing stiffness, you are
> testing strength.
> >
> >Dan Hayes
> >(For what its worth, a practicing professional/licensed
> Materials Engineer
> >with over thirty years experience. That said, you might think I
> should have
> >been able to say it more succinctly)
> >
> > > -----Original Message-----
> > > From: [email protected]
> > > [mailto:[email protected]]On Behalf Of Peter Fels and
> > > Phoebe Palmer
> > > Sent: Saturday, September 07, 2002 1:17 AM
> > > To: [email protected]
> > > Subject: Re: [TheForge] Hardening steel
> > >
> > >
> > > At 08:40 PM 9/6/02, you wrote:
> > >
> > > My old cooked brain is having a lot of trouble with this
> one...not a good
> > > week in that regard. First i get outsmarted by a large piece of
> > > 1" plate ,
> > > now this...It's the problem with being self-taught by a fool.
> > > OK....when you take a rod of carbon steel and you flex it,
> the outside of
> > > the curve has to stretch and the inside compresses. If you harden
> > > the rod,
> > > then it becomes more resistant to stretching and compressing.
> > > Therefore it
> > > should become stiffer....NO?..apparently not . I don't get it.
> > > Would someone be merciful and explain this  to us igorantses...Pete F
> > >
> > >
> > > >Mike,
> > > >
> > > >I tried to find this information in the metals handbook but the
> > > only thing I
> > > >found on Modulus of Elasticity was in relation to different
> temperatures.
> > > >I'm not trained in this but got the books from a friend when he
> > > got the new
> > > >edition. The index said that it would be K1 on the charts but I
> > > didn't see
> > > >that on any of them.
> > > >
> > > >We are off to a sale this weekend but I'll try to run some tests
> > > next week.
> > > >
> > > >Bob Ehrenberger
> > > >Shelbyville, Mo
> > > >
> > > > > I will have to test this myself
> > > >
> > > >Hi Bob,
> > > >   It is one of those things that don't seem to make sense at
> > > first look, but
> > > >is the case.  If you look at a data sheet on most any grade
> of steel, you
> > > >will see the Modulus is shown as a single number, not like
> > > tensile and yield
> > > >strengths, which show the properties with differing tempering
> > > temperatures.
> > > >I agree it still seems like it shouldn't be that way, but if
> you want to
> > > >change the amount of bend or deflection under a given load,
> you need to
> > > >change the cross section of the material, or change to a
> > > different material.
> > > >The average modulus for steel is 29,000,000 which you get by
> dividing the
> > > >stress (pounds per square inch) by the strain (inches per inch).
> > > >Aluminum for example is only 10,000,000 (same as glass), Titanium
> > > >17,000,000,
> > > >etc.
> > > >Good luck with your testing,
> > > >Mike Schermerhorn
> > > >
> > > >
> > > >
> > > >
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