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BAC2 tech · Technology and Engineering II
Materials and manufacturing · Internal structure of materials

Equilibrium diagrams and the iron-carbon diagram

Steel is not a single material: depending on its carbon percentage and the temperature at which it is observed, its atoms form different phases with very different properties. The iron-carbon diagram is the map that predicts those phases, and the lever rule lets you calculate precisely how much of each one there is.

Before you start, make sure you know: Classification of materials (1st Bachillerato)
Before you start, make sure you know: Percentages (2nd ESO)

Equilibrium diagrams

Before getting into iron and carbon, it is worth understanding what a phase diagram represents in general.

Binary equilibrium diagram with the liquidus and solidus curves, which delimit the liquid, liquid + solid and solid (α phase) regions T (°C) % component B Pure A liquidus solidus liquid liquid + solid solid (α phase)
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Explain what an equilibrium diagram (or phase diagram) is: what variables are represented on its axes (temperature and percentage of each component), what each region…

Explain what an equilibrium diagram (or phase diagram) is: what variables are represented on its axes (temperature and percentage of each component), what each region indicates, and why it is called "equilibrium" (cooling must be very slow so that atoms have time to rearrange). Use water as an analogy: to each temperature and pressure corresponds a stable state.

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Explain what it means for two metals to be completely soluble in the solid state (such as copper and nickel) as opposed to having partial solubility (such as iron and…

Explain what it means for two metals to be completely soluble in the solid state (such as copper and nickel) as opposed to having partial solubility (such as iron and carbon). What shape does the phase diagram of each pair have, and what practical consequence does this have for the alloys that can be manufactured?

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The lever rule

Within a two-phase region, the lever rule lets you calculate what percentage of each phase is present.

Lever rule: horizontal isotherm with phase 1 (C₁) on the left, phase 2 (C₂) on the right and the overall alloy (C₀) as the pivot; each arm determines the percentage of the opposite phase phase 1 (C₁) alloy (C₀) phase 2 (C₂) arm C₀ − C₁ → determines % phase 2 arm C₂ − C₀ → determines % phase 1
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State the lever rule in your own words: what it is used for within a two-phase region, which segments are measured along the tie line (isotherm), and how the percentage…

State the lever rule in your own words: what it is used for within a two-phase region, which segments are measured along the tie line (isotherm), and how the percentage of each phase is obtained. Illustrate it with the analogy of a see-saw: the fulcrum is the alloy composition and the two arms "weigh" the phases at the ends.

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An alloy with $45\,\%$ of component B is at a temperature where an α phase with $20\,\%$ of B and a liquid phase with $60\,\%$ of B coexist, as shown in the figure. a)…

An alloy with $45\,\%$ of component B is at a temperature where an α phase with $20\,\%$ of B and a liquid phase with $60\,\%$ of B coexist, as shown in the figure. a) Read the length of the two lever arms from the isotherm. b) Apply the lever rule to calculate the percentage of α phase and liquid phase. c) Check that both percentages add up to $100\,\%$.

α phase (20% B) alloy (45% B) L phase (60% B) 45 − 20 = 25 60 − 45 = 15 % α = ? % L = ?
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An isomorphous alloy (complete solubility) contains $25\,\%$ of component B. At a certain temperature, the α phase present has $15\,\%$ of B and the liquid has $45\,\%$.…

An isomorphous alloy (complete solubility) contains $25\,\%$ of component B. At a certain temperature, the α phase present has $15\,\%$ of B and the liquid has $45\,\%$. a) Calculate using the lever rule the percentage of each phase present. b) Reason what happens to these percentages if the temperature drops towards the solidus line (which phase gains ground?).

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The iron-carbon diagram and its constituents

The iron-carbon diagram is the most widely used equilibrium diagram in industry: all steels and cast irons rely on it.

Eutectoid zone of the iron-carbon diagram: the point at 727 °C and 0,77 % carbon separates the fields of austenite, ferrite + pearlite and pearlite + cementite T (°C) % C 0 0,77 2,11 6,67 727 austenite (γ) ferrite + pearlite pearlite + cementite eutectoid: 727 °C, 0,77 % C
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Match each constituent of the iron-carbon diagram with its description: (a) ferrite, (b) cementite, (c) pearlite, (d) austenite; with: solid solution of carbon in…

Match each constituent of the iron-carbon diagram with its description: (a) ferrite, (b) cementite, (c) pearlite, (d) austenite; with: solid solution of carbon in α-iron, soft and ductile; iron carbide Fe₃C, very hard and brittle; lamellar mixture of ferrite and cementite in alternating layers; solid solution of carbon in γ-iron, stable only at high temperature.

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State whether true or false, justifying your answer: "Pearlite is a chemical element added to iron to harden it." Explain what pearlite actually is, from which…

State whether true or false, justifying your answer: "Pearlite is a chemical element added to iron to harden it." Explain what pearlite actually is, from which constituent it forms upon cooling, and what appearance it has under the microscope (alternating light and dark lamellae).

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Describe the eutectoid transformation of steel: at what temperature it occurs ($727\ ^\circ\text{C}$), with what composition ($0{,}77\,\%$ C), and what is produced upon…

Describe the eutectoid transformation of steel: at what temperature it occurs ($727\ ^\circ\text{C}$), with what composition ($0{,}77\,\%$ C), and what is produced upon cooling through it (austenite transforms into pearlite). Compare it with the eutectic point of the diagram ($4{,}3\,\%$ C, $1\,147\ ^\circ\text{C}$), where a liquid transforms into two solids.

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Hypoeutectoid and hypereutectoid steels

The carbon percentage of a steel decides which "early" constituent accompanies the pearlite when it cools slowly.

Carbon composition axis between 0 % and 2,11 % that classifies steels into hypoeutectoid, eutectoid and hypereutectoid 0 % 0,77 % 2,11 % hypoeutectoid ferrite + pearlite hypereutectoid pearlite + cementite eutectoid: 100 % pearlite
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Classify these iron-carbon alloys as steel or cast iron and order them from lowest to highest expected hardness: (a) $0{,}10\,\%$ C, (b) $0{,}45\,\%$ C, (c) $0{,}90\,\%$…

Classify these iron-carbon alloys as steel or cast iron and order them from lowest to highest expected hardness: (a) $0{,}10\,\%$ C, (b) $0{,}45\,\%$ C, (c) $0{,}90\,\%$ C, (d) $3{,}5\,\%$ C. Remember that the boundary between steels and cast irons is around $2\,\%$ carbon ($2{,}11\,\%$ in the actual diagram).

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A steel containing $0{,}4\,\%$ carbon is cooled very slowly to room temperature. a) State which constituents its microstructure exhibits. b) Applying the lever rule…

A steel containing $0{,}4\,\%$ carbon is cooled very slowly to room temperature. a) State which constituents its microstructure exhibits. b) Applying the lever rule between ferrite ($0{,}02\,\%$ C) and pearlite ($0{,}77\,\%$ C), calculate the percentage of pearlite in the steel. c) Express what approximate fraction of the microstructure will be ferrite.

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Classify these steels as hypoeutectoid, eutectoid, or hypereutectoid: (a) $0{,}25\,\%$ C, (b) $0{,}77\,\%$ C, (c) $1{,}10\,\%$ C. For each, state which primary…

Classify these steels as hypoeutectoid, eutectoid, or hypereutectoid: (a) $0{,}25\,\%$ C, (b) $0{,}77\,\%$ C, (c) $1{,}10\,\%$ C. For each, state which primary constituent appears alongside pearlite upon slow cooling (proeutectoid ferrite or proeutectoid cementite) and cite an application consistent with their increasing hardness.

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Total cementite and applications of the diagram

The lever rule also lets you calculate the total cementite of a steel (that in the pearlite plus the proeutectoid cementite, if there is any), a figure directly related to its hardness.

Microstructure of metastable cementite (Fe₃C lamellae on ferrite) which, with silicon and time, decomposes into ferrite and graphite flakes of grey cast iron Fe₃C (cementite) Fe + graphite + Si, time Fe₃C graphite metastable hard and brittle grey cast iron better machinability
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Calculate the total percentage of cementite contained in a steel with $1{,}0\,\%$ carbon at room temperature, applying the lever rule between practically pure iron…

Calculate the total percentage of cementite contained in a steel with $1{,}0\,\%$ carbon at room temperature, applying the lever rule between practically pure iron ($0{,}02\,\%$ C) and cementite ($6{,}67\,\%$ C). Comment on whether this result explains why tool steels, with carbon around $1\,\%$, are much harder than structural steels.

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In the expert report on the failure of a transmission shaft manufactured from carbon steel ($0{,}6\,\%$ C) slowly cooled under equilibrium conditions, a technician…

In the expert report on the failure of a transmission shaft manufactured from carbon steel ($0{,}6\,\%$ C) slowly cooled under equilibrium conditions, a technician states the following: "Since pearlite is composed mostly of ferrite ($88\,\%$) and proeutectoid ferrite is also present, nearly $90\,\%$ of the final microstructure is pure alpha ferrite; therefore, the shaft should have behaved as a ductile and soft metal, making its brittle fracture inexplicable." a) Quantitatively determine the microstructure at room temperature by calculating the percentage of microconstituents (proeutectoid ferrite and pearlite) and the total percentage of phases (total alpha ferrite and total cementite), assuming a solubility of $0{,}02\,\%$ and a eutectoid at $0{,}77\,\%$ C. b) Rebut the technician's statement using metallurgical arguments: explain why steel with that proportion of total ferrite does not behave like pure ferrite at all and possesses much higher strength and hardness. c) If it is necessary to increase the toughness of the shaft to prevent its brittle fracture without reducing the carbon percentage, what heat treatment would you propose and what microstructural change does it aim for compared to this slow cooling?

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In the manufacture of machine tool beds, a material capable of effectively damping mechanical vibrations and exhibiting excellent machinability for grinding its…

In the manufacture of machine tool beds, a material capable of effectively damping mechanical vibrations and exhibiting excellent machinability for grinding its slideways is required. During a casting with $3{,}2\,\%$ C, due to a technical failure, the addition of silicon (typically $2\,\%$) is omitted and the component cools excessively fast in a metal mold.

  1. a) After demolding, the bed turns out to be impossible to machine and fractures in a brittle manner under the slightest impact. Justify what occurred from a metallurgical standpoint by comparing the metastable phase diagram (iron-cementite) with the stable one (iron-graphite): what microstructure was intended, and which one was produced instead?
  2. b) Analyze the thermodynamic and kinetic role played by silicon content (graphitizing element) and cooling rate in the decomposition of metastable cementite into ferrite and graphite.
  3. c) Explain at a micromechanical level why the presence of graphite flakes in gray cast iron provides exceptional vibration damping capacity and facilitates chip breaking during machining, compared to a steel with the same pearlitic matrix.
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Choosing the right carbon for each application

The iron-carbon diagram is not just a calculation exercise: it is the tool used to choose the composition of a steel according to its use.

Scale of increasing carbon content in steel with three applications: car body sheet (very ductile), suspension spring (medium-to-high carbon) and knife blade (very hard) increasing carbon high ductility high hardness sheet ~0,1 % C almost all ferrite spring ~0,6-0,8 % C with heat treatment knife >0,77 % C with cementite
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The micrograph of a steel shows white regions and regions with thin alternating light and dark bands, the latter occupying approximately half the surface area. a)…

The micrograph of a steel shows white regions and regions with thin alternating light and dark bands, the latter occupying approximately half the surface area. a) Identify both constituents. b) Estimate the carbon percentage of the steel from the observed fraction of pearlite. c) Reason whether this is a typical steel for car bodywork sheet (very low carbon) or for a machine axle (medium carbon).

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To manufacture rails for a railway line, a hypoeutectoid steel is required that exhibits an equilibrium microstructure containing $75\,\%$ pearlite and $25\,\%$…

To manufacture rails for a railway line, a hypoeutectoid steel is required that exhibits an equilibrium microstructure containing $75\,\%$ pearlite and $25\,\%$ proeutectoid ferrite, with an allowable pearlite range between $70\,\%$ and $80\,\%$.

  1. a) Deduce the required nominal carbon percentage and the allowable carbon tolerance range in the steelmaking process, assuming the eutectoid point at $0{,}77\,\%$ C and the maximum solubility of ferrite at room temperature to be $0{,}008\,\%$ C.
  2. b) A quality control manager argues that, since the steel contains $75\,\%$ pearlite, the total percentage of the ferrite phase in the steel will be approximately $25\,\%$. Analytically demonstrate their error by calculating the actual percentage of total ferrite and total cementite at room temperature.
  3. c) Technically justify why this steel with $75\,\%$ pearlite is selected for a rail instead of a mild steel with $20\,\%$ pearlite or a hypereutectoid steel with grain boundary cementita, assessing the trade-off between friction wear and resistance to brittle impact fracture.
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A materials engineering team must design the steel specification and heat treatment route for three mechanical components subjected to critical and opposing service…

A materials engineering team must design the steel specification and heat treatment route for three mechanical components subjected to critical and opposing service requirements:

1) Deep drawing sheet metal for a vehicle fender (requires maximum cold formability and high energy absorption capacity upon impact).

2) A helical spring for a train suspension system (requires a very high yield strength and fatigue resistance without permanent deformation).

3) The industrial blade of a shear for cutting metal sections (requires extraordinary surface hardness and abrasive wear resistance without chipping).

  1. a) Design the steel selection for each component: specify the recommended carbon range (% C) and justify the required microstructure prior to its forming or machining.
  2. b) Design and justify the final heat treatment route for each part: indicate whether it requires treatment or not, the indicative temperatures with respect to the critical lines of the equilibrium diagram ($A_1$, $A_3$, or $A_{\text{cm}}$), the cooling medium, and subsequent stress relief or tempering treatments to optimize the hardness-toughness trade-off.
  3. c) Predict the technical consequences if an operator mistakenly swaps the materials and manufactures the suspension springs using the fender steel, or manufactures the fender using the blade steel.
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Summary

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