Question

please show any calculations

(16 points) Label the following microstructures of 10140 (1.4 wt% Carbon) steel given the stated treatments. Be sure to label ALL of the phases AND constituents a. Fully Annealed Condition 4" b. As Quenche Conditiorn c. Quenched and Tempered Condition: 5. (6 points) Calculate how much of each phase is present in condition "a": 6. (6 points) Calculate how much of each constituent is present in condition "a":

Answer 1

The alloy is hyper eutectoid steel beacause the carbon content greater than eutectoid composition and less than 2.1 wt%C.

a) Annealing means heating above critical temperature and cooling in a furnace. So the grains formed are coarse in nature as seen in figure. The white region is the peoeutectoid cementite before crossing the eutectoid temperature.

Also alternate dark and white

black layers of ferrite and cementite is visible. This is known as pearlite.

b) On quenching ie rapid cooling of austenite to ambient condition results in formation of martensite. It is having a body centred tetragonal shape and is hardest transformation product in steel.

c) Martensite is tempere to enhance toughness and Ductility and to reduce hardness. It is heated between 300 to 500 degree celsius and held for sufficient time.

Lever rule is used to find mass composition of the given alloy.

Weight fraction of peoeutectoid cementite= (1.4-0.76)/(6.7-0.76)= 0.107

Weight fraction of pearlite= 1-0.107= 0.893

Weight fraction of ferrite= (6.7-1.4)/(7.7-0.022)=0.794

Weight fraction of cementite in pearlite= 0.794-0.107= 0.686