Perhaps I am too naive, but I am having a hard time visualizing how even 'high-carbon' iron and steel alloys are maybe, at most, about four percent carbon by mass, which still means only one (smaller) atom in 120 or so is carbon rather than iron...

How? Is there a diagram that could help me visualize this? Doesn't this mean that most iron atoms are not even remotely close to, or even indirectly affected by, a carbon atom?

  • 3
    $\begingroup$ Maybe think of a stack of dinner plates and same thing with a little sand between the plates. Which stack better resists a side shove? $\endgroup$
    – Ed V
    Sep 13, 2021 at 17:37
  • 2
    $\begingroup$ What is your criterion of calling steel "different" from iron? If you are just talking about steel, it has other metals (nickel, chromium etc.) besides some carbon. $\endgroup$
    – AChem
    Sep 13, 2021 at 18:06
  • 5
    $\begingroup$ You should really brush up on chemical calculation - molar percent here is much higher then mass percent, not lower. $\endgroup$
    – Mithoron
    Sep 13, 2021 at 18:07
  • 2
    $\begingroup$ Have you had a look at iron / steel phase diagrams? $\endgroup$
    – Karl
    Sep 14, 2021 at 10:43
  • 3
    $\begingroup$ Rebar is only a few percent of the mass of reinforced concrete, for an example that's maybe more intuitive. That few percent makes a huge difference. $\endgroup$
    – J...
    Sep 14, 2021 at 13:03

3 Answers 3


One of the key considerations is that much of the interesting mechanical behaviour doesn't occur within the bulk of the material - it occurs at the interfaces between crystals (known as grains). Grain boundary slip, rotation and growth often defines the mechanics - carbon has a tendency to block the motion of grain boundaries, which are one of the key deformation modes.

There are a multitude of reasons, but carbides in the material cause faults in the crystal (which are the key thing that makes metals deform plastically) to slide less easily (reduces dislocation "glide"). This is because carbon forms carbides that cannot be "split" by the dislocation as easily, and have to deform themselves to climb over the carbides, before re-straigtening via dislocation line-tension (energy penalty for being longer than a straight line).

So its a complicated picture already, and this is only scratching the surface of the issue!

I'd highly recommend Dieter's Mechanical metallurgy if you want a more in-depth understanding - its probably mid-undergraduate to post-graduate level though.

Edit: in case it isn't clear, carbon preferentially segregates to grain boundaries, arguably due to the higher open-volume at most boundaries (this is a simplification). Though even in a random solid solution, carbon still has a mild strengthening effect (the same can be seen in Aluminium-copper random solution alloys, vs pure Al)

  • $\begingroup$ Do alloying metals also preferentially segregate to grain boundaries, as you describe carbon doing? $\endgroup$
    – theorist
    Sep 14, 2021 at 16:15
  • 2
    $\begingroup$ Here is a visual tour of steel grains, and where carbon is depending on the temperature history of the material: youtu.be/uG35D_euM-0?t=562 $\endgroup$
    – Karsten
    Sep 14, 2021 at 19:35
  • $\begingroup$ And here is a picture of pearlite grains (containing carbon) between ferrite grains (pure iron): upload.wikimedia.org/wikipedia/commons/7/77/Pearlite.jpg $\endgroup$
    – Karsten
    Sep 14, 2021 at 19:52
  • $\begingroup$ @theorist Yes - the classic example is Gallium in Aluminium, but in that case it causes the Aluminium grains to separate! $\endgroup$ Sep 16, 2021 at 12:24
  • 1
    $\begingroup$ @theorist It depends on what the alloy elements are. Generally intentional alloying elements have good chemical affinity for the parent metal so they will be soluble within the grains of it or else form separate intermetallic phases between parent metal phase grains. But unintended elements (impurity traces) elements like As, Sb, etc that do not have solubility in parent metal phases often migrate to grain boundaries where they form separate phases in either discrete globs or grain surrounding films. $\endgroup$
    – Trunk
    Sep 16, 2021 at 15:06

Elastic deformation properties like stiffness (Young's Modulus) vary very little with alloying element concentration since we are only working on the pre-yield bond strength. A pure iron, a carbon steel and a high alloy steel will only vary by about 15% in stiffness. So elastic properties will, as you suggest, be fairly invariant under small changes in alloy concentration.

But the other mechanical properties - those depending on plastic deformation like yield strength, hardness, toughness - are different in how they are affected by alloy element concentration. Plastic deformation in metals at ambient temperatures, and below 1/3 of the absolute melting point, is mostly about moving dislocations within the crystal structure. Strengthening therefore means finding effective methods of impeding normal dislocation flow.

Let's look into why alloying helps any metal's plastic yield behaviour.

The word "alloying" may embrace situations where two or more metals exist as a mixture of separate elements. Such mixtures will show modifications of properties roughly representable by a rule of mixtures equation, i.e.

$$ S = v_1 S_1 + v_2 S_2$$ where $v_1$, $v_2$ are the volume fractions of the respective elements.

But alloying in the usual meaningful sense will only occur when the chemical equilibrium prevailing favors either a solid solution or intermetallic phase (e.g. $\ce{Ni3Al}$, $\ce{Fe3C}$, etc) in preference to existing as a mixture of two separate elements. Solid solution alloying can occur in two ways. It can occur when an atom of the minor component element substitutes for one in the crystal lattice of the major component element. This usually happens when the minor component has comparable atomic mass, e.g. nickel in iron. Solution alloying can also occur if the minor component has a much smaller atomic mass than that of the major component, e.g. carbon in iron. In this case the minor element atoms fit inside the existing iron lattice, a type of solution known as an interstitial solid solution. Of course, a carbon atom within the body-centered cubic form of iron will expand the lattice around it somewhat. So we have an ambient temperature limit of 0.002% by weight (or about 1 carbon atom per 10,000 iron atoms) on how much carbon is soluble within solid b.c.c. iron, known as ferrite or $\alpha$-iron.

A dislocation within a metallic crystal is a mismatch in the usual ordered 3D-repeating arrangement of ions. It is virtually impossible to crystallize any substance without having lots of dislocations within each crystal. More importantly, since a perfect crystal will have all its bonds yield at once, it is the very existence of dislocations that allow crystalline materials to deform plastically under moderate stress rather than break in a brittle mode at a much higher stress. So the yield strength of any crystalline metallic material is effectively the stress under which its dislocations begin to move. If we want to improve the strength of some element, e.g. iron, we need to find a way to impede the movement of dislocations in it.

Dislocations Schematically

In the vicinity of a dislocation, the lattice is stretched on the side of the missing half-plane of ions and contracted on the side of the extra half-plane. Interstitial solute atoms within the crystal would therefore - given adequate temperature - tend to diffuse and settle on the expanded side of a dislocation zone since here they create minimal lattice strain energy increase. This is what happens with carbon atoms in iron.

Since the chemical interaction between a carbon atom and surrounding iron ions will be of partly covalent nature there will be significantly greater resistance to dislocation flow under stress and thus a higher yield stress. Concentrations of carbon ~ 1 per 10,000 iron atoms would equate to one carbon atom per cube of iron crystal with an edge length of 22 interatomic spacings (~ 60 Angstroms). So a single carbon atom alone can significantly restrain dislocation loop movement within that volume and that is more than adequate for normal dislocation densities observed in carbon steel.

Another factor affecting dislocation impedance is the other carbon-containing phase in carbon steels, i.e. $\ce{Fe3C}$ or cementite. This phase precipitates simultaneously from $\gamma{}$-iron (austenite) or carbon supersaturated ferrite as it cools. This type of simultaneous solid-phase precipitation is referred to as eutectoid decomposition. Eutectoid pair phases are very finely mixed and often show a fixed relationship between their crystal orientations. In the case of carbon steels, the eutectoid mix consists of very fine alternating lamellae of ferrite and cementite. Being so close any dislocation movement towards the edge of the ferrite will be mechanically blocked by the neighboring cementite lamellae.

So the nominally "small" additions of carbon to pure iron will have two important effects in restraining dislocation advance and therefore strengthening the material:

  1. The interstitial solubility of carbon in the ferrite b.c.c. lattice which locates carbon atoms in the "pipes" of dislocations where their partly covalent bonding with neighboring iron atoms on the extra half-planes requires a greater shear stress to move the dislocations within ferrite grains.

  2. The close lamellar morphology of the ferrite/cementite eutectoid mix resists dislocation movement out of ferrite grains.

You have asked a seemingly simple question. But the proper answer is long, involved and demands close study of a number of aspects of physical metallurgy. If your work requires a better understanding of this topic, please consult texts like those by Reed-Hill & Abbaschian and Cahn & Haasen at your local university library.


Let's consider that steel may contain $\pu{1\%}$ carbon. I know it is a bit much. Let us express this mass concentration as molar concentration for a sample of $\pu{100 g}$ of steel. Since this sample contains $\pu{99 g}$ of iron, this amount equates to

$$n(\ce{Fe}) = \frac{\pu{99 g}} { \pu{56 g mol^{-1}} } = \pu{1.768 mol}$$


$$n(\ce{C}) = \frac{\pu{1 g}} { \pu{12 g mol^{-1}} } = \pu{0.083 mol}$$

In iron-carbon alloys, these two elements often combine to form cementite, $\ce{Fe3C}$. It means that $\pu{0.083 mol}$ of $\ce{C}$ is combined with $3 \cdot \pu{0.083 mol} \approx \pu{0.25 mol}$ of $\ce{Fe}$. This $\pu{0.25 mol}$ of iron is not available to act as solvent of carbon containing molecules. So $\pu{100 g}$ of steel with $1$% C contains only $(1.768 - \pu{0.25) mol} = \pu{1.518 mol}$ uncombined iron. The molar proportion of $\ce{Fe3C}$ in steel is $\frac{0.25}{1.768} = \pu{14\%}$. Then $\pu{14\%}$ is a relatively concentrated solution of $\ce{Fe3C}$ in iron taken as the solvent.

This shows that steel is not a diluted solid solution. It is even a moderately concentrated solid solution of carbon containing solute in the solvent iron.

  • 3
    $\begingroup$ Fe3C isn't molecular en.wikipedia.org/wiki/Cementite $\endgroup$
    – Mithoron
    Sep 13, 2021 at 23:02
  • 1
    $\begingroup$ This assumes that all the carbon in a 1 wt% C steel will go preferentially into the cementite phase. This is not so. The amount of cementite phase in a steel depends a lot on its carbon content. Steels with low carbon wt% will have less cementite than those with higher wt%. See the phase diagram tf.uni-kiel.de/matwis/amat/iss/kap_6/illustr/s6_1_2.html $\endgroup$
    – Trunk
    Sep 16, 2021 at 14:47
  • $\begingroup$ Besides, cementite is a hard brittle phase which would be of little use for a material requiring both tensile strength and enough ductility to make it tough. Cementite's main benefit is to provide a measure of wear resistance in the final steel. Its existence in a fine eutectoid mixture with $ \alpha $-iron greatly helps this. $\endgroup$
    – Trunk
    Sep 18, 2021 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.