Find the weight percent of austenite present in the steel, Mechanical Engineering

Find the weight percent of austenite present in the steel:

A 0.4% C hypoeutectoid plain carbon steel is slowly cooled from 1540oC to

(i) slightly above 723oC and (ii) slightly below 723oC.

Calculate the weight percent

1.      austenite present in the steel,

2.      ferrite present in the steel in case (i),

3.      proeutectoid ferrite prevent in the steel, and

4.      eutectoid ferrite and eutectoid cementite % present in the steel in case (ii).


Refer to Figure 8. Point 1 above the liquidus represents the state of liquid steel. The cooling occurs along the line xx and an equilibrium cooling is assumed. Freezing begins at point 2 which is intersection of liquidus and line xx. Temperature at 2 is 1510oC. The steel solidifies completely at point 3 where temperature is 1471oC. Now the entire alloy is composed of austenite (γ-phase) as indicated by first of Figure 19. No change occurs until point 4 on line A3 is reached. At this point the precipitation of ferrite begins out of solid austenite. Further cooling increases the amount of ferrite and austenite decreases. The amount of austenite varies along with IK. The composition of the ferrite varies along with the line IL.

Calculation of % content will be made by lever rule.

The amount of austenite slightly above 723oC is calculated from the line LK itself. i.e. taking LK as tie line.

 (a)       Weight % of austenite =   L5/ LK= ( 0.4 - 0.0250)/(0.8 - 0.025)

                                                       =  0.375/0.775  = 0.484 or 48.4%         ---------- (i)

                     Weight % of ferrite    = 5K / LK       =     0.8 - 0.4/0.8 - 0.025

                                                               =     0.4/0.775 = 0.516 or 51.6%        -------- (ii)

 (b)       Weight % of proeutectoid ferrite slightly below 723oC is same as that slightly above, that means 48.4%. ----------- (iii)

For calculating eutectoid ferrite, the weight of carbide will have to be subtracted form total mass of ferrite and cementite. Just below isothermal line LKM ferrite and pearlite are present and lever arm will extend up to ordinate representing 6.67% C.

Weight % of total (ferrite + cementite) just below 723oC

=     6.67 - 0.4/6.67 - 0.025

 =   6.37/6.645

  = 0.96 or 96%

Weight % of Fe3C just below 723oC

=   0.4 - 0.025/6.67 - 0.025

=  0.375/6.645

= 0.0564 or 5.64%      

Weight % of eutectoid cementite = total ferrite - proeutectoid ferrite

                                                              = 96 - 51.6 = 44.4%

Weight % of eutectoid cementite (by difference)

= 100 - 48.4 - 5.64 - 44.4 = 1.56%                                         --------- (iv)

Posted Date: 1/22/2013 1:31:10 AM | Location : United States

Related Discussions:- Find the weight percent of austenite present in the steel, Assignment Help, Ask Question on Find the weight percent of austenite present in the steel, Get Answer, Expert's Help, Find the weight percent of austenite present in the steel Discussions

Write discussion on Find the weight percent of austenite present in the steel
Your posts are moderated
Related Questions
Q. Explain Double Bolting? Where low strength bolts must be used, initial bolting with higher strength low temperature bolts may be necessary to minimize accumulative strain an

Find the tension in the thread: A thread is wound round a heavy homogeneous cylinder of mass m and radius. The cylinder is permitted to fall from rest and unwinds the thread.

a) Round off the numbers 865250 amd 37.46235 to four significant figures and compute Ea, Er, Ep in each case. b) IF u= 4 x 2 y 3 /z 4 and errors in x, y, z be 0.001, determine

Smallest value of constant horizontal force: A particle containing a mass of 8 kg starts from rest and attains a speed of 1.5 m/sec in a horizontal distance of 10 m. imagine a

Problem - Newton's Third Law Newton's Third Law of Motion states that if two objects interact, the force F 12 exerted by object 1 on object 2 is equal in magnitude and oppos

calculate the head required

an arrangement of the wilson epicyclic gearbox giving four forward and one reverse speed of shaft D for an input speed of shaft E is shown in fig 1st gear Ii is 2 ge

1) Factor of Safety is calculated by using which of the following set of values? [Select the one best answer from the choices given below.] Answer a. Yield Stress and Allowable St

difference between stress intensity factor and stress concentration factor

Q. Describe the Selection of elctrodes? The selection of a proper electrode for electric arc welding is very important to obtain a sound welded joint of proper strength. The va