[TheForge] Hardening steel

Sat Sep 7 03:18:00 2002

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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