HW7 solution - MechEng 382 Winter 2010 Homework #7...

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MechEng 382 Winter 2010 Homework #7 Solutions 1. Ashby & Jones 2: 11.1 From the TTT diagram for a plain carbon steel of eutectoid composition: (a) 100% martensite (quenched below M f where transformation of unstable austenite to martensite is complete) (b) ~75% martensite, 25% austenite ( γ ) (c) After 50% austenite transforms to lower bainite, 75% of the remaining austenite transforms to martensite 50% lower bainite 37.5% martensite 12.5% austenite (d) From (c), martensite transformation of remaining austenite is complete 50% lower bainite 50% martensite (e) 100% lower bainite (f) 50% coarse pearlite (g) 37.5 % martensite (h) 12.5% pearlite
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MechEng 382 Winter 2010 Homework#7 Solutions 2 2. Ashby & Jones 2: 11.2 Room temperature equilibrium microstructure of: (a) Pure iron (b) 0.3 wt% carbon steel (hypoeutectoid) (c) 0.8 wt% carbon steel (eutectoid) (d) 1.2 wt% carbon steel (hypereutectoid) Fe Pearlite ( α +Fe 3 C) α (primary) Pearlite Pearlite Fe 3 C
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MechEng 382 Winter 2010 Homework#7 Solutions 3 3. Ashby & Jones 2: 12.1 (a) Fine-grained steel contains more grain boundaries; i.e. more favorable sites for nucleation than a coarse-grained steel. Diffusive transformation occurs more rapidly; therefore a higher critical cooling rate is needed to avoid transformation (b) Addition of Ni to the 18% Cr steel stabilizes austenite. The alloying elements do the following: Ni lowers the FCC BCC transition temperature. Cr and Ni help to slow the FCC BCC transition. It is more difficult for austenite to transform to ferrite by diffusion.
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HW7 solution - MechEng 382 Winter 2010 Homework #7...

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