Chapter 10: Problem 30
Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite.
Chapter 10: Problem 30
Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite.
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Get started for freeFigure \(10.40\) shows the continuous cooling transformation diagram for a \(1.13 \mathrm{wt} \%\) C iron-carbon alloy. Make a copy of this figure and then sketch and label continuous cooling curves to yield the following microstructures: (a) Fine pearlite and proeutectoid cementite (b) Martensite (c) Martensite and proeutectoid cementite (d) Coarse pearlite and proeutectoid cementite (e) Martensite, fine pearlite, and proeutectoid cementite
Briefly explain why there is no bainite transformation region on the continuous cooling transformation diagram for an iron-carbon alloy of eutectoid composition.
Make a copy of the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition (Figure 10.22) and then sketch and label time- temperature paths on this diagram to produce the following microstructures: (a) \(100 \%\) fine pearlite (b) \(100 \%\) tempered martensite (c) \(50 \%\) coarse pearlite, \(25 \%\) bainite, and \(25 \%\) martensite
On the basis of diffusion considerations, explain why fine pearlite forms for the moderate cooling of austenite through the eutectoid temperature, whereas coarse pearlite is the product for relatively slow cooling rates.
Briefly describe the simplest continuous cooling heat treatment procedure that would be used in converting a 4340 steel from one microstructure to another. (a) (Martensite + bainite) to (ferrite + pearlite) (b) (Martensite + bainite) to spheroidite (c) (Martensite + bainite) to (martensite + bainite \(+\) ferrite)
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