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    EN380 Homework #7 Solution

    Uploaded by

    Mafer Reyes
    This document provides solutions to homework problems about steel phase diagrams and microstructures. Key points include: 1) Criteria for metals to form a substitutional alloy include similar atomic radii (<15% difference), same crystalline structure, similar electronegativity, and same valence state. 2) A 1095 steel with 0.95% carbon would have 86.02% ferrite, 13.98% cementite, and 97.44% pearlite at just below the eutectoid temperature. 3) Rolling a steel plate from 2" to 1.5/16" thickness results in 3.125% cold work which would increase the yield strength as shown by shifting the stress-

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    EN380 Homework #7 Solution

    Uploaded by

    Mafer Reyes
    31 views6 pages
    This document provides solutions to homework problems about steel phase diagrams and microstructures. Key points include: 1) Criteria for metals to form a substitutional alloy include similar atomic radii (<15% difference), same crystalline structure, similar electronegativity, and same valence state. 2) A 1095 steel with 0.95% carbon would have 86.02% ferrite, 13.98% cementite, and 97.44% pearlite at just below the eutectoid temperature. 3) Rolling a steel plate from 2" to 1.5/16" thickness results in 3.125% cold work which would increase the yield strength as shown by shifting the stress-

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    This document provides solutions to homework problems about steel phase diagrams and microstructures. Key points include: 1) Criteria for metals to form a substitutional alloy include similar atomic radii (<15% difference), same crystalline structure, similar electronegativity, and same valence state. 2) A 1095 steel with 0.95% carbon would have 86.02% ferrite, 13.98% cementite, and 97.44% pearlite at just below the eutectoid temperature. 3) Rolling a steel plate from 2" to 1.5/16" thickness results in 3.125% cold work which would increase the yield strength as shown by shifting the stress-

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    31 views6 pages

    EN380 Homework #7 Solution

    Uploaded by

    Mafer Reyes
    This document provides solutions to homework problems about steel phase diagrams and microstructures. Key points include: 1) Criteria for metals to form a substitutional alloy include similar atomic radii (<15% difference), same crystalline structure, similar electronegativity, and same valence state. 2) A 1095 steel with 0.95% carbon would have 86.02% ferrite, 13.98% cementite, and 97.44% pearlite at just below the eutectoid temperature. 3) Rolling a steel plate from 2" to 1.5/16" thickness results in 3.125% cold work which would increase the yield strength as shown by shifting the stress-

    Copyright:

    © All Rights Reserved
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    Download as pdf or txt
    of 6

    EN380 Homework #7 Solution

    1. What are the criteria for two metals to form a perfect substitutional alloy (that is, these two will form
    the same crystal phase at all potential concentrations of each species)?

    • Atomic radii must not differ by more than 15%.

    • Crystalline Structure must be the same (BCC and BCC, FCC and FCC, etc.).

    • No appreciable difference in electronegativity.

    • Elements should have the same valence state.

    2. Sketch a qualitative plot of σY vs % elongation for the following steel microstructures:

    (a) Ferrite (α)

    (b) Bainite

    (c) Martinsite

    (d) Coarse Pearlite

    (e) Fine Pearlite

    1
    3. The phase diagram for iron-iron carbide (F e - F e3 C) is shown below. For a 0.95% C steel (i.e. 1095
    series) at a temperature just below the eutectoid temperature (727◦ C) determine:

    (a) % C present in the ferrite (α). 0.02%

    (b) % C present in the cementite (F e3 C). 6.67%


    orange CF e3 C −C0 6.67%−0.95%
    (c) total % of the steel that is ferrite (α). Wα = purple+orange = CF e3 C −Cα = 6.67%−0.02% = 86.02%

    (d) total % of the steel that is cementite (F e3 C).


    purple C0 −Cα 0.95%−0.02%
    WF e3 C = purple+orange CF e3 C −Cα = 6.67%−0.02% = 13.98%

    2
    (e) % of steel that is pearlite.
    This is the same as finding how much austenite, γ, was present just above the eutectoid temper-
    ature. Draw the tie line just above 727◦ C.

    Cα = 0.02% Cγ = 0.8%
    green CF e3 C − C0 6.67% − 0.95%
    Wγ = = = = 97.44%
    red + green CF e3 C − Cγ 6.67% − 0.8%

    (f) % of steel that is proeutectoid ferrite.


    As C0 is greater than the eutectoid composition of 0.8%, there cannot be any α present above/before
    the eutectoid transformation temperature. → 0%

    (g) % of steel that is eutectoid ferrite.


    As C0 is greater than the eutectoid composition of 0.8%, all of the α present must have formed
    in the pearlite as eutectoid α ⇒ Wα, Eutectoid = Wα, T otal = 86.02% .

    (h) % of steel that is proeutectoid cementite.


    This is the same as finding how much cementite, F e3 C, was present just above the eutectoid
    temperature
    Draw the tie line just above 727◦ C.

    Cα = 0.02% Cγ = 0.8%
    red C0 − Cγ 0.95% − 0.8%
    Wpro−eutectoid F e3 C = = = = 2.56%
    red + green CF e3 C − Cγ 6.67% − 0.8%

    (i) % of steel that is eutectoid cementite.


    We already know how much cementite is present from part (d) and how much of this F e3 C was
    proeutectoid from (h). Thus:

    Weutectoid F e3 C = Wtotal F e3 C − Wproeutectoid F e3 C = 13.98% − 2.56% = 11.43%

    (j) Compare this material qualitatively with steel with the eutectoid composition of 0.8%C (assuming
    similar thermal history) in terms of:

    i. Cementite content
    This material has more Cementite than 1080 steel

    3
    ii. Strength (σY and σU T S )
    With a higher cementite content (including in the grain boundaries) it would be expected to
    be stronger (higher σY and σU T S ) than 1080 steel.

    iii. Ductility
    With a higher cementite content (including in the grain boundaries) it would be expected to
    be less ductile/more brittle than 1080 steel.

    iv. Toughness
    Similarly it would be expected to be less tough (similar to ductility) than 1080 steel.

    v. Hardness
    Hardness should follow strength → harder than 1080 steel.

    4
    15
    4. A steel plate is passed through rollers which reduce the thickness from 2 in to 1 16 in in a single pass.
    What is the % C.W.? Qualitatively describe what happens to the yield strength of the plate as a result
    and show this using a sketch of the σ −  curve before and after rolling.
    15 1
    t0 − tf 2 in − 1 16 in in
    % C.W. = · 100% = · 100% = 16 · 100% = 3.125% C.W.
    t0 2 in 2 in

    5. An aluminum rod of Diameter D = 2 in is extruded into the T-slotted below. What is the % C.W. if
    the final profile area is A = 1.216 in2 .

    π
    A0 − Af 4 · D02 − Af
    % C.W. = · 100% = π 2 · 100%
    A0 4 · D0

    π
    4 · (2 in)2 − 1.216 in2
    % C.W. = π 2
    · 100%
    4 · (2 in)

    = 61.29% C.W.

    5
    6. Label the Eutectoid steel T T T diagram below to show:

    (a) Regions where the stable phase is: (b) Cooling curves that result in:

    i. Austenite i. 100% Coarse Pearlite


    ii. Coarse Pearlite (α + F e3 C) ii. 50% Fine Pearlite + 50% Martinsite
    iii. Fine Pearlite (α + F e3 C)
    iii. 100% Martinsite
    iv. Upper Bainite

    v. Lower Bainite (For each case, label the quencing medium (or me-
    vi. Martensite dia) used to obtain the final microstructure)

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