9.2.2 Designing Low Alloy Steels
|Let's Design a High-Strength Steel|
|Now let's replace carbon as the major
"hardening" agent and design a
high-strength low-alloy steel without much carbon in earnest. I'm going just
through a few principles that illustrate the salient points I'm trying to make.
For more information use this
We want to make relatively cheap steel that can be conventionally welded (liquid welding) but is quite "strong", i.e. hard and ductile. That means it must be low in carbon since we do not want martensite formation, which is bad for welding. For the same reason we do not want a lot of pearlite. If we think in terms of carbon steel the only choice left is relatively soft low-carbon steel.
|We want, however, rather large
hardness (plus a large number of other good properties) that we simply cannot
get from a low-carbon steel. We do not want high-alloy steels either because
they are simply too expensive, not to mention hard-to-weld. What we want to
make is a "high-strength
low-alloy" (HSLA) steel, one of the
major "designer" steels in circulation today. This kind of steel has
a market share of about 12 %, which is quite a lot - roughly 170 million tons
per year. Its development was driven by oil and gas extraction (meaning, for
example, steel for pipelines where welding is a must), construction, and
transportation (meaning cars).
HSLA steels are also known as "micro-alloyed steels". This needs some explaining. When a steel person talks about "alloying", she is referring to elements different from carbon and the commonly accepted amounts of manganese, silicon, etc. Bearing this in mind, we now can distinguish three alloy groups:
|Our HSLA steel to be designed thus
still contains the ubiquitous Manganese (up to 2%) and carbon - but not much.
Around 0.07 % - 0.12 % carbon will suffice. The trick is that we do not use the
carbon for producing cementite or martensite but for making
carbides. Following the
principles of hardening, as discussed in
chapter 8, but
without using martensite, you know now what
you must do:
|For the first mechanisms we need to look at the relevant figure again. We also need to consider all the affordable elements not shown in this figure but here.|
|What we find is that atomically
dissolved carbon and nitrogen have by far the biggest direct effect on the
yield strength (owing to their being interstitials), and that phosphorous kept
in solution is very good, too, (but, remember,
very bad in cold weather.
Then we have silicon (Si), manganese (Mn) titanium (Ti) and copper (Cu, not shown in the figure) and some other elements as pretty good solution hardeners. Copper, however, has drawbacks (including its prize), and silicon causes problems here too. This leaves, in a first cursory evaluation, manganese (Mn), titanium (Ti), and to some extent nickel (N) and vanadium (V) as eligible alloying elements for solid solution hardening.
|Manganese is the clear winner since it also neutralizes spurious sulfur (S) and is quite cheap.|
|The second mechanism calls for alloy elements that form
precipitates either with the iron (of which there is plenty), with the little
bit of carbon left, with iron and carbon,
or with the few other elements around. But not with the manganese (Mn), please, because we
need that dissolved as single atoms for hardness, and to round up and
precipitate residual sulfur (S).
You also want your precipitates to have a melting point higher than iron to avoid hort shortness, and you want them to be hard so neither dislocations nor advancing microcracks can cut right through them.
|Now look up manganese (Mn) in the
periodic table. It
sits right next to iron (Fe) and thus can be expected to be chemically similar.
That makes you suspect that elements that like to form compounds with iron
might also like to form compounds with manganese, which is not good. That rules
out elements for micro alloying that like to form precipitates with iron.
So what kind of precipitates should form? Carbides, of course. That also solves the other problems, because carbides typically have a very high melting point and are extremely hard. We now need alloy elements that are known to be strong "carbide formers". That are all elements that form carbides (surprise!) and win against the iron in the competition for the carbon in there.
All you have to do is to go through the periodic table, looking (experimentally) for good carbide formers. It's a feast for grad students once more. Many Ph.D. theses' and scientific papers will result.
Elements like niobium (Nb), vanadium (V), titanium (Ti), chromium (Cr), zirconium, (Zr) or molybdenum (Mo), but also silicon (Si) or boron (B), emerge as possible contenders. All of them like to form very hard carbides with very high melting points. So we now go and try them out. Each one causes a hell of a lot more work than that little example I gave you with Copper (Cu) in aluminum (Al). What they do if several different ones are there at the same time will provide work for generations to come.
|We still have the third mechanism, keeping the grain size small.|
|That involves two major points:
Of course, the iron-carbon phase diagram is not decisive for that any more. More work for grad students.
|I'm already discussing the processing or in other words the temperature profiling. This is crucial. Whatever you put into your HSLA steel to form carbides, it will only work if you cool down quickly. Otherwise all those carbides have time to grow into large sizes and you want them to remain very small, remember? Just one example: Niobium (Nb) or vanadium (V) works already at low concentrations around 0,1 % - provided you know what you are doing. Tiny niobium carbide particles of about 1 nm size (containing less than 100 alloy atoms) will increase the yield stress from about 20 MPa to 200 MPa, at a concentration of about 0.1 wt % Nb. "Huge" particles with diameters of 10 nm or larger have practically no effect anymore! Look at the simple Cu in Al example. It's all there except that the effects are more pronounced in the iron - niobium system.|
|The problem, of course, is that the
outside always cools down faster than the inside. The "inside" thus
stays "soft". All we can do is
hardening - only the "case", the outer layer is really hard. The
word to introduce and recognize now is "hardening depth". The question simply
is: To what depth is the cooling rate fast enough to allow hardening? And how
can we measure it? For the second question the answer is easy and given in
Whatever happens during quenching, major things only happen when the temperature goes below the transition temperature where austenite intends to change to ferrite. The formation of metal carbides, we might assume, also starts below that temperature. In plain carbon steel this "A1" temperature is around 720 oC (1328 oF). If we could lower it, there wouldn't be so much time for carbide formation and the carbides stay smaller.
Can we? Of courseby alloying the right element(s).
|Even if you overdid your "good stuff break", you should now have gotten the central message:|
|In normal human life, finding out
what is right or wrong gives jobs to a lot of humans. And to lawyers. A hell of
a lot of people working inside some religious organization are concerned about
this topic, and they can tell you unambiguously if what you do is right or
wrong with regard to what you eat, how you deal with your girl friend / spouse,
how to raise your children, and so on. Then there is that parent-in-law who
knows exactly what's right or wrong for her son or daughter, and so on. Just
don't ask them about alloying elements.
In contrast, as far as technical human life is concerned, the number of the scientists and engineers who know the right alloy elements from the wrong ones, and the right way to deal with them, is rather small.
Which group has achieved more I leave open at that point.
|If you are into alloying, you should
be aware of where you could succeed, and where not. The
of applied science comes into its own. Using it, you will know that making
big changes in
modulus by alloying just a little bit of something will not work -
period! Or, to give another example, that alloying will always increase the electrical resistivity.
Nevertheless, alloying plus proper processing does the trick in many cases. The fact that we have more than 1000 different steel alloys, with properties tailored to some particular needs, bears witness to that.
© H. Föll (Iron, Steel and Swords script)