Question

A steel part made from AISI 1050 is to be cooled from austenite temperature to the room temperature;

(a) draw the micro-structures step by step during cooling to room temperature;

calculate the total amount of (b) ferrite, (c) cementite, (d) pearlite

this is urgent please

Answer 1

Micro structures step by step during cooling temperature.

Hypo Eutectoid steel

Eutectoid steel ( pearlite )

Hyper Eutectoid steel

Microstructures of steel

Nomenclature of critical points

To facilitate reference to the various changes and temperature at which occur, temperature points are referred to by the letter 'A' with various subscript numericals to identify the changes as follows.   

Temperature point	Description
A1 ( lower critical point )	Pearlite austenite ( upper arrow indication heating, lower arrow indication cooling )
A2 ( Curie temperature )	Loss ( due to heating ) or gain ( due to cooling ) of ferromegnetism.
A3 ( Upper critical point )	Perrite absorbed in ( heating ), or nucleated from ( cooling ) austenite.
Acm ( Upper critical point)	Cementite absorbed in ( heating ), or nucleated from ( cooling ) austenite
A4 ( A4 - point )	Austenite - iron ( upper arrow indication heating, lower arrow indication cooling)

The structure of hypo eutectoid steel ( C< 0.8 % ) will consists of pearlite and ferrite. The percentage of these phase can be determined by using lever rule

Steel AISI 1050 Carbon percentage range between 0.470 to 0.55

X	Z	Y
100 % ferrite	C = 0.5 %	C = 0.8 % 100 % pearlite

  

% of pearlite = ( XZ /  XY ) 100 = ( 0.5 / 0.8 ) 100 = 62.5 %

% of ferrite = ( ZY / XY ) 100

= ( 0.8 - 0.5 ) / 0.8 100 = 37.5 %

Neglecting the carbon percent in ferrite, the relative amount of the coexisting phases in hypo eutectoid steel may be obtained using the relations

% of pearlite = ( C / 0.8 ) 100

and % of ferrite = ( 0.8 - C / 0.8 )   100

The structure of eutectoid steel ( C = 0.8 % ) will consists of pearlite only. The relative amounts of ferrite and cementite in pearlite can be computed by applying the level rule.

X	Z	Y
100 % ferrite	C = 0.8 %	C = 6.67 % 100 % Cementite

% of cementite = ( XZ / XY ) 100

= ( 0.8 / 6.67 ) 100 = 12 %

% of ferrite = ( ZY / XY ) 100

= ( 6.67 - 0.8 ) / 6.67   100 = 88 %

The structure of hypo eutectoid steel ( C = 0.8 to 2 % ) will consists of pearlite and cementite. The percentage of pearlite and cementite is determine by using lever rule

X	Z	Y
C = 0.8 % 100 % pearlite	C = 1.2 %	C = 6.67 % 100% Cementite

  

% of pearlite = ( ZY / XY ) 100

= ( 6.67 - 1.2 / 6.67 - 0.8 ) 100 = 93.1 %

% of cementite = ( XZ / XY ) 100

= ( 1.2 - 0.8 / 6.67 - 0.8 ) 100 = 6.8 %

Thus the relative amount of the phases in hypo eutectoid steel may be obtained by using the relations

% of pearlite = ( 6.67 - C / 6.67 - 0.8 ) 100

% of cementite = ( C - 0.8 / 6.67- 0.8 ) 100

In terms of ferrite and cementite

X	Z	Y
100 % ferrite	C = 1.2 %	C = 6.67 % 100 % Cementite

% of pearlite = ( 6.67 - 1.2 / 6.67 ) 100 = 82 %

% of cementite = ( 1.2 / 6.67 ) 100 = 18 %

Thus the relative amount of ferrite are cememtite in hyper eutectoid steel may be obtained by using the relations

% of ferrite = ( 6.67 - C / 6.67 ) 100

% of cementite = ( C / 6.67 ) 100