Iron-carbon alloys, containing up to 2.06% of carbon, are called steels. Alloys, containing from 2.06 to 6.67% of carbon, experience eutectic transformation at 2097 ºF (1147 ºC). The eutectic concentration of carbon is 4.3%. In practice only hypoeutectic alloys are used. These alloys (carbon content from 2.06% to 4.3%) are called cast irons. When temperature of an alloy from this range reaches 2097 ºF (1147 ºC), it contains primary austenite crystals and some amount of the liquid phase. The latter decomposes by eutectic mechanism to a fine mixture of austenite and cementite, called ledeburite.
Figure above shows the equilibrium diagram for combinations of carbon in a solid solution of iron. The diagram shows iron and carbons combined to form Fe-Fe3C at the 6.67%C end of the diagram. The left side of the diagram is pure iron combined with carbon, resulting in steel alloys. Three significant regions can be made relative to the steel portion of the diagram. They are the eutectoid E, the hypoeutectoid A, and the hypereutectoid B. The right side of the pure iron line is carbon in combination with various forms of iron called alpha iron (ferrite), gamma iron (austenite), and delta iron. The black dots mark clickable sections of the diagram.
Allotropic changes take place when there is a change in crystal lattice structure. From 2802º-2552ºF the delta iron has a body-centered cubic lattice structure. At 2552ºF, the lattice changes from a body-centered cubic to a face-centered cubic lattice type. At 1400ºF, the curve shows a plateau but this does not signify an allotropic change. It is called the Curie temperature, where the metal changes its magnetic properties.
Two very important phase changes take place at 0.83%C and at 4.3% C. At 0.83%C, the transformation is eutectoid, called pearlite.
gamma (austenite) –> alpha + Fe3C (cementite)
At 4.3% C and 2066ºF, the transformation is eutectic, called ledeburite.
L(liquid) –> gamma (austenite) + Fe3C (cementite)
Problem Solving :
1. Given the Fe-Fe3C phase diagram above, calculate the phases present at the eutectoid composition line at:
a. T = 3000ºF
b. T = 2200ºF
c. T = 1333ºF
d. T = 410ºF
2. Calculate the phases in the cast-iron portion of the diagram at the eutectic composition of 4.3% C in combination with 95.7% ferrite at:
a. T = 3000ºF
b. T = 1670ºF
c. T = 1333ºF
3. A eutectoid steel (about 0.8%C) is heated to 800ºC (1472ºF) and cooled slowly through the eutectoid temperature. Calculate the number of grams of carbide that form per 100g of steel.
4. Determine the amount of pearlite in a 99.5% Fe-0.5%C alloy that is cooled slowly from 870ºC given a basis of 100g of alloy.
1. a. T = 3000ºF. Since the composition E is eutectoid, the carbon content is 0.83%.
b. T = 2200ºF. At this temperature, austenite exists as a single-phase solid.
c. T = 1333ºF. Two phases exist, ferrite and austenite.
The percentages are determined by the lever rule: X ÷ (X+Y) = (Cy-C) ÷ (Cy-Cx).
2. a. T = 3000ºF. At this temperature, the eutectic is all liquid.
b. T = 1670ºF. Since there is a change of the solubility line, there will be a change in the austenite composition, that will transform to the eutectoid at 1333ºF. The overall austenite and cementite composition will be:
A metal matrix composite (MMC) is composite material with at least two constituent parts, one being a metal. [[ This is a content summary only. Visit my website for full links, other content, and more! ]] […]
This resistance to attack is due to the naturally occurring chromium-rich oxide film formed on the surface of the steel. [[ This is a content summary only. Visit my website for full links, other content, and more! ]] […]