Question

a) Sketch the Fe-C carbon phase diagram, denoting the key phases present in this material. Comment on the solubility limit of a-ferrite and the outcome of excess carbon in a-ferrite as a function of temperature. b) Prove that the ideal strength of a material is: ģ 15 where is the Young's modulus of the material. c) Sketch the face centre cubic (FCC) crystal structure and hence, show that the lattice parameter is: a=2RV2 where R is the atomic radius and that the atomic packing factor for an FCC structure is 0.74

Answer 1

Solution a)

Fe-C phase diagram with phases are shown below :

key phases are also shown above and they are

0.1 0.16 0.51 Liquid (L) 1493°C As (Ferrite) L+Y L+FeC 1147°C Y (Austenite) 910°C 768°C A, 143 Ir(Austenite) + FezC(Cementite)] =Ledeburite Temperature - - 723°C a (Ferrite) 0.025%C [a (Ferrite) + Fe C (Cementite)] = Pearlite 210°C ,A 0.8 0.008 Fe % Carbon

1) $\alpha$ ferrite

2)$\delta$ ferrite

3)$\gamma$ austenite

4) Fe​​​​​3​​​​C cementite

5) Pearlite ( $\alpha$ + Fe​​​​​3​​​​C )

6)Ledeuburite ( $\gamma$ + Fe​​​​​3​​​​C)

Solubility limit of carbon in  $\alpha$ ferrite is 0.025 % at the eutectic temperature 723 °C and at room temperature carbon solubility in $\alpha$ ferrite is 0.008% .

If excess carbon is added in $\alpha$ ferrite then we can see cementite phase will be insoluble in it and another phase of cementite will form parallel with $\alpha$ ferrite .

As the temperature increases onset of formation of second phase cementite decreases because temperature increases causes increase in solubility of carbon in $\alpha$ ferrite .Therefore effect of temperature increase is that solubility limit increases of $\alpha$ ferrite and more excess carbon is required to form second phase i.e , cementite .