Often, the properties of multiphase alloys may be approximated by the relationshipE (alloy) = EaVa + EBVBwhere E represents a specific property (modulus of elasticity, hardness, etc.) and Vis the volume fraction. The subscripts a and pdenote the existing phases or microconstituents. Employ the relationship above to determine the approximate Brinell hardness of a99.8 wt% Fe-0.20 wt% C alloy. Assume Brinell hardnesses of 80 and 280 for ferrite and pearlite, respectively, and that volumefractions may be approximated by mass fractions.See Animated Figure 9.24(9.24)i Temperature (°C)1600140012001000800600400(Fe)1538 °C51493 °C1394 °C912 °Ca+yy, Auster0.760.022d., Feritte085 °C1V +L2Composition (at% C)10152.141147 °Ca + Fe CLComposition = 0 wt% CComposition = 0 at% CTemperature = 400 °CTemperature = 752 °F34Composition (wt% C)4.30y + Fe3C727 °C20Cementite (Fe,C)15062525002000150010006.70Temperature (°F)

Given relationship for multiphase alloys E(alloy)=EαVα+EβVβ Where E represents a specific property and V is the volume fraction Given alloy composition is 99.80wt% Fe and 0.20 wt% C Also given hardness of pearlite is 280 and hardness of ferrite is 80The mass fraction of eutectoid ferrite, which is just the difference between total ferrite and pro-eutectoid ferrite mass fraction as shown below: Ws=6.7-0.456.7-0.022-0.76-0.450.76-0.022=0.935-0.420=0.515 The mass fraction is taken from the given phase diagram. Now by using lever arm rule to calculate the amount of ferrite and pearlite, Amount of ferrite will be, Vα=0.76-0.20.76-0.022=0.560.738=0.76 And amount of pearlite, Vβ=0.2-0.0220.76-0.022=0.1780.738=0.24 Then Brinell hardness will be given as shown E(alloy)=EαVα+EβVβHB(alloy)=HBαVα+HBβVβHB(alloy)=(80×0.76)+(280×0.24)HB(alloy)=128 Here HBα and Hβ are the hardness of ferrite and pearlite Answer: Brinell hardness of the given alloy composition is 128.