Metals: diagrams and microstructure

1. 1Find the temperature on the vertical axis and rule a horizontal tie line across the two-phase region.
2. 2Read where the tie line meets the liquidus and where it meets the solidus. Write both readings on the diagram with both metals, such as 83% Pb, 17% Sn.
3. 3Rule a vertical line at the alloy composition and mark where it crosses the tie line.
4. 4Each phase takes the arm on the far side from it. Divide that arm by the whole tie line.
5. 5Multiply by 100 for a percentage, or by the total mass for a mass. Check that the two arms add to the whole tie line.

Worked on the 2024 readings. At 200 °C the liquidus reads 45% Pb and the solidus reads 83% Pb. For a 70% Pb alloy:

Percentage solid:

$$\frac{70-45}{83-45}=\frac{25}{38}=65.8\%$$

Percentage liquid:

$$\frac{83-70}{83-45}=\frac{13}{38}=34.2\%$$

The two arms are 25 and 13, and they add to 38, the whole tie line.

1. 1Read the axis label. % Lead and % Tin run in opposite directions, so 38.1% Pb is the same point as 61.9% Sn.
2. 2Find the eutectic at 38.1% Pb and 183 °C, where the two liquidus branches meet. Pure lead melts at 327 °C.
3. 3Name the curves. The liquidus is the upper boundary, with everything liquid above it. The solidus is the lower boundary of each solid-plus-liquid field. The solvus is the boundary below 183 °C between each single-phase solid field and the α + β field.
4. 4For hypo or hyper, decide which solid forms first. Between 38.1% and 81% Pb, primary α forms, so the alloy is hypoeutectic. Between 2.5% and 38.1% Pb, primary β forms, so it is hypereutectic.

Below 2.5% Pb or above 81% Pb the alloy solidifies as a single solid solution with no eutectic, so it is neither hypoeutectic nor hypereutectic.

1. 1Place the carbon percentage in its band. Low carbon is 0.07% to 0.30%, medium carbon is 0.30% to 0.60% and high carbon is 0.60% to 2.0%.
2. 2Compare it with 0.83% C, the eutectoid. Below 0.83% the steel is hypoeutectoid and cools to ferrite and pearlite. Above it the steel is hypereutectoid and cools to pearlite and cementite.
3. 3Name the microstructure and the properties it gives, from the tables below.
4. 4For an application, match it to the syllabus list for that band.

Austenite is the high-temperature form of iron, stable above 723 °C in eutectoid (0.83% C) steel and at higher temperatures for other carbon contents, and it dissolves far more carbon than the room-temperature form. Ferrite is nearly pure iron, soft and ductile with low strength. Cementite is iron carbide, $\text{Fe}_3\text{C}$, hard and brittle. Pearlite is a layered mixture of ferrite and cementite formed at the eutectoid, with properties between the two.

Steel stops at about 2.1% carbon. Above that it is cast iron, which is outside the course. The high-carbon band ends at 2.0%, which is the top of the band and not the steel limit.

1. 1Count what the question asks for: the number of steels, properties and applications. One mark usually sits on each.
2. 2Name the microstructure of the steel’s band.
3. 3Link each constituent to a property, and the property to what it lets the part do.
4. 4For compare or contrast, make every statement comparative, with “than” in it.
5. 5If a decision is asked for, give it its own sentence, with the property that settles it as the reason.

For the full compare and contrast template, see Game plan 7.

1. 1Draw grains as closed cells with visible boundaries, not a block of shading.
2. 2Draw ferrite as plain grains and pearlite as banded grains, so the two kinds are clearly different.
3. 3Label both on the diagram.
4. 4Match the proportions to the carbon content. Low carbon is mostly plain grains and high carbon is mostly banded grains.

Pearlite is alternating ferrite and cementite, so ferrite shows up twice on your sketch, as whole plain grains and as the pale bands inside the pearlite. Label the plain grains as ferrite.