Lecture Problem 10'52

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Lecture Problem 10.52

• Jowene Wong
• April 11, 2007

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Problem Statement

• Consider 1.0 kg of austenite containing 1.15 wt
  C, cooled to below 727oC (1341oF).
• What is the proeutectoid phase?
• How many kilograms each of total ferrite and
  cementite form?
• How many kilograms each of pearlite and the
  proeuctectoid phase form?
• Schematically sketch and label the resulting
  microstructure.

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Background

• Microstructures structural features of an alloy
  (e.g. grain and phase structure) that are subject
  to observation under a microscope
• Big picture
• Develop due to carbon content and heat treatment
  (slow heating in this case)
• Metals phases present, proportions, and
  arrangement
• Affect mechanical properties

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Background

• Microconstituent element of the microstructure
  that has an identifiable and characteristic
  structure, may consist of more than one phase
  (i.e. pearlite)
• Stuff you see under the microscope
• Depends on thermal history
• Phase homogeneous portion of a system (material
  under consideration) that has uniform physical
  and chemical characteristics
• Solid, liquid, gas
• Stuff you see in the phase diagram
• For any slow-cooling thermal history that begins
  in a 1-phase region, microconstituents are the
  same as phases unless you cross a 3-phase
  invariant

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Background

• Eutectic (phase/reaction/structure) when a
  liquid transforms isothermally and reversibly
  into two intimately mixed solid phases by cooling
• Eutectoid (reaction) when a solid phase
  transforms isothermally and reversibly into two
  new intimately mixed solid phases
• Proeutectoid forms before eutectoid

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Background Phase Diagram
invariant point
eutectic isotherm (three-phase invariant)
invariant point
eutectoid isotherm (three-phase invariant)
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Background

• Carbon is an interstitial impurity in iron
• Diffused into iron
• Austenite ? iron, 912oC, FCC
• Ferrite a iron, room temp, BCC
• Soft and ductile
• Cementite Fe3C
• Hard and brittle

austenite
ferrite
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Background

• Upon cooling into ? Fe3C phase, cementite
  starts to form along grain boundaries
• Tie line (phase compositions found using
  endpoints)
• Composition of austenite moves along PO boundary
• Composition of cementite stays same (6.7 wt C)
• Austenite ? pearlite
• Proeutectoid cementite

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Background

• Pearlite two-phase microstructure, when
  austenite of eutectoid composition transforms,
  consists of alternating layers of ferrite and
  cementite
• Exists as grains (colonies)
• Layers in same orientation in each colony
• Light ferrite, dark cementite
• Layers form because parent phase (austenite)
  different from product phases (ferrite,
  cementite) and carbon needs to be redistributed
  by diffusion

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Background

• Hypoeutectoid alloy iron-carbon alloy between
  0.022 and 0.76 wt C
• Hypereutectoid alloy iron-carbon alloy between
  0.76 and 2.14 wt C

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Solution (a)

• This is a hypereutectoid alloy, so the
  proeutectoid phase is
• proeutectoid cementite

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Solution (b)

• Lever rule, tie line

Ca
CFe3C
Co
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Solution (b)

• Find the phase compositions.

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Solution (b)

• Lever rule, tie line

Cp
CFe3C
Co
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Solution (c)

• Find the microconstituent compositions.

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Solution (d)

• Figure 10.35 1.4 wt C steel
• Actual schematic has proeutectoid cementite
  because its 1.15 wt C

pearlite
proeutectoid cementite
ferrite
cementite
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Applications

• Form a material with the desired properties
• Steels and cast irons are the primary structural
  materials
• 1.15 wt C steel mainly used for cutting tools
• Can physically check you obtained the desired
  composition

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Questions?!