MSE 230 Solutions for Assignment 10 Fall 2013 Problems with an by them will not be graded but solutions will be provided Problem 1 First we cool the alloys rapidly to 0 C such that no diffusion can occur At this point the microstructure just consists of a polycrystalline single phase microstructure The phase is supersaturated with copper The copper would like to diffuse to form the phase but the temperature is too low to accomplish this Heating to 50 C and holding for 1 hour won t change things because there hasn t been enough of a temperature increase to get diffusion going The microstructure would look the same as when the material was quenched Heating to 200 C for one hour will result in a fine dispersion of precipitates in an matrix while heating to 350 C will result in a smaller number of larger precipitates The ordering in increasing strength will be 50 C 350 C 200 C The alloy treated at 50 C is the lowest because the particles did not form although you should expect some solid solution strengthening The strength of the alloy at 200 C will be the largest because it contains the finest particles and therefore the largest number of particles per unit volume of alloy Many small particles are more effective at inhibiting dislocation motion Problem 2 a After holding for 100 seconds at 600 C all of the austenite has transformed to pearlite Quenching to room temperature does not change the microstructure b After holding for 100 seconds at 700 C none of the austenite has transformed to pearlite So we still have 100 austenite Upon quenching to room temperature virtually all of the austenite transforms to martensite via a displacive or athermal transformation Usually not all of the austenite is able to transform Fig 11 23 so we are left with 90 martensite and 10 austenite c After holding for 100 seconds at 350 C approximately half of the austenite has transformed to bainite Upon quenching to room temperature most of the remaining austenite will be transformed to martensite The bainite formed at 350 C will not be affected by the quench So we are left with 50 bainite 45 martensite and 5 retained austenite Problem 3 The key here is to recognize that if you heat your iron carbon alloy of eutectoid composition above the eutectoid temperature 727 C the alloy will consist of single phase austenite Then you can make a variety of microstructures following the same approach as problem 1 a Making martensite from coarse pearlite heat the coarse pearlite above 727 C say to 800 C and hold it there for sufficient time to transform the pearlite to austenite Then quench the austenite to room temperature This will result in a microstructure that is 90 martensite with the remainder retained austenite b Coarse pearlite from fine pearlite again heat to 800 C until the alloy is all austenite Quickly cool to 650 C and hold it there until the austenite transform completely to coarse pearlite approximately 2 minutes Holding above 650 C but below the eutectoid temperature will produce pearlite that is more coarse but will also take longer c Same thing as part b Problem 4 We are called upon to name the microstructural products that form for specimens of an iron carbon alloy of eutectoid composition that are continuously cooled to room temperature at a variety of rates Figure 10 27 is used in these determinations a At a rate of 200 C s only martensite forms b At a rate of 100 C s both martensite and pearlite form c At a rate of 20 C s only fine pearlite forms Problem 5 Callister 10 D4 This problem asks us to consider the tempering of a water quenched 1080 steel to achieve a hardness of 50 HRC It is necessary to use Figure 10 35 a The time necessary at 425 C is about 650 s b At 315 C the time required by extrapolation is approximately 4 x 106 s about 50 days Problem 6 a Extended heating of pearlite near the eutectoid temperature will cause the plates of ferrite and cementite to break up into spheres of cementite in a ferrite matrix see Fig 10 19 The formation of spheroidite from pearlite is driven by the reduction in interfacial energy between the ferrite and cementite phases This process results in relatively coarse 1 3 m diameter cementite particles Tempered martensite is formed when martensite is heated to a temperature sufficiently large to allow carbon to diffuse If we stay in the Fe3C two phase region this will result in the formation of cementite particles in a ferrite matrix In contrast to spheroidite the cementite particles in tempered martensite will be much smaller see Fig 10 33 in this case 10 times smaller This is because they arise from the nucleation and growth of new cementite particles from carbon that was in solid solution rather than the breakup of existing cementite plates b Tempered martensite will have a greater strength because of the finer dispersion of cementite particles Referring to Table B 4 on page A11 you can see that the strength of oil quenched and tempered 315 C steel is significantly larger than the strength of annealed steel for a number of different steel alloys Problem 7 In this problem we will make use of Figures 10 32 and 10 35 Forming a fine pearlite structure in 1080 steel 0 8 wt C will result in a Brinell hardness of no greater than 300 which is too low unfortunately there is no data for bainite here The data for martensite tempered at 371 C has a value of approximately 475 also too low The data for martensite indicates a Brinell hardness of over 600 too large However we can make martensite and then temper the martensite to reduce the hardness Figure 11 36 indicates that tempering the martensite at 315 C for approximately one hour will result in a steel with a Brinell hardness of approximately 550 To make the martensite in the first place one would likely need to heat above the eutectoid temperature to allow the original microstructure to convert to all austenite before quenching to form martensite Problem 8 We are asked to specify a practical heat treatment for a 2014 aluminum alloy that will produce a minimum tensile strength of 380 MPa and a minimum ductility of 15 EL From Figure 11 27 a the following heat treating temperatures and time ranges are possible to the give the required tensile strength Temperature C Time Range h 204 0 12 h 149 0 1000 h 121 0 With regard to temperatures and times to give the desired ductility Temperature C Time Range h 204 0 2 h 149 10 h 121 500 h From these tabulations the following may be concluded At 204 C the time would be roughly 10
View Full Document
Unlocking...