[TheForge] Hardening steel

Sun Sep 8 01:41:00 2002

At 02:36 PM 9/7/02, you wrote:


Thank you Dan;
It is sinking in, but my intuition is still quacking......Pete


>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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