Question 10.11

Suppose a steel of eutectoid composition is cooled to 550 C from 760 C in less than 0.5 s and held at this temperature.

  1. How long will it take for the austenite-to-pearlite reaction to go to 50% completion? To 100% completion?
  2. Estimate the hardness of the alloy that has completely transformed to pearlite.

Answer 10.11

  1. Figure 10.5 (in the 6th edition) is helpful here, though you could also refer to Figure 10.13 or Figure 10.15. After dropping perpendiculars from the transformation diagram, we interpolate on the logarithmic x-axis and find that the reaction is 50% complete in 2-3 seconds, 100% complete in 7-9 seconds.
  2. 550 C puts us just above the `knee' of the transformation curve, so we are getting fine pearlite, bordering on bainite. Figure 10.21(a) tells us that fine pearlite at the eutectoid composition has a Brinnell hardness of about 270 HB.

Question 10.18

Make a copy of the isothermal transformation diagram for a 1.13% wt C iron-carbon alloy (Figure 10.28 in 6th edition), then on this diagram sketch and label time-temperature paths to produce the following microstructures:

  1. 6.2% pro-euctectoid cementite and 93.8% coarse pearlite
  2. 50% fine pearlite and 50% bainite
  3. 100% martensite
  4. 100% tempered martensite

Answer 10.18

There are several possibilities for these paths. We assume that all the paths start at time zero, and at 750 C, and they're all going to finish at room temperature, extending to the right-hand side of the figure.

  1. If we cool the alloy through the eutectoid temperature, pro-eutectoid cementite will form; then hold the alloy just below the eutectoid temperature until we're past the 100% line on the transformation diagram.
  2. Cool to just above the knee of the curve, hold it there till we reach the 50% line, then cool to just below the knee and hold till we have 100% transformation.
  3. Quench rapidly to room temperature
  4. Quench rapidly to room temperature, then temper at a higher temperature -- this can be anywhere in the range 200 to 650 C.
So the figure should look something like this:

q1018.jpg

(Using coloured pencils for the four different paths makes answers to this question a lot easier to mark.)

Question 10.35

Estimate the Brinell hardnesses for specimens of a 1.13%wt C iron-carbon alloy that have been subjected to the following heat treatments:

  1. Hold at 920 C until everything turns to austenite; then rapidly cool to 250 C, hold for 1000 seconds, quench to room temperature.
  2. Hold at 920 C until everything turns to austenite; rapidly cool to 700 C, hold at this temperature for 100,000 seconds, then quench to room temperature.
  3. Hold at 920 C until everything turns to austenite; rapidly cool to 600C, hold at this temperature for 7 seconds, rapidly cool to 450 C, hold at this temperature for 4 seconds, then quench to room temperature.

Answer 10.35

  1. This is going to produce martensite, with a Brinell hardness of about 700 HB.
  2. If we just use the isothermal transformation diagram, Fig 10.28, for example, we will conclude that this produces coarse pearlite and some pro-eutectoid cementite, with a hardness of about 200 HB. Callister asks you to answer the question on the basis of the transformation diagram, so I'd accept this answer.

    However, if we read the text on `spheroidite', we see that holding the pearlite at 700 C for 100,000 seconds (28 hours) will allow the carbon to re-arrange itself into spherical lumps, giving the softer and more ductile spheroidite. From Figure 10.21(a), the Brinell hardness of spheroidite with 1.13% carbon is about 180 HB.

  3. 7 seconds at 600 C should be enough for everything to turn to fine pearlite, apart from the initial deposit of pro-eutectoid cementite. (So taking it to 450 C and holding it there for 4 seconds doesn't accomplish anything, as far as I can see.) If we extrapolate Figure 10.21(a) a bit, the hardness of the alloy should be in the range 300-320 HB.

Question 10.38

The room temperature tensile strengths of pure copper and pure silver are 209 MPa and 125 MPa respectively.

  1. Make a schematic graph of the room-temperature tensile strength versus composition for all compositions between pure copper and pure silver.
  2. On the same graph schematically plot tensile strength versus composition at 600 C
  3. Explain the shapes of these two curves as well as any differences between them.

    Answer 10.35

    We can mark two points on the room temperature graph right away -- 209 MPa for 100% Cu, and 125 MPa for 100% Ag. To get any further, we need to consult the CuAg phase diagram (Figure 9.6 in the 6th edition).

    The phase diagram shows that at room temperature, the alpha-phase region of the diagram is very narrow, that is, only very small amounts of silver will form a solid solution in copper. (Some of the answers to this question said that no silver would dissolve in the copper, and vice-versa. This is an overstatement; even when the solvus line gets so close to the vertical axes that you can't see the gap, it's still possible for a few atoms of one metal to dissolve in the other, with measurable effects on strength.) Similarly for the beta-phase. So there will be narrow regions adjacent to the initial two points where the tensile strength goes up, due to solution-hardening. These narrow regions will be wider at 600 C.

    Once we get past the solvus line, we will have a two-phase alloy composed of alpha and beta phases. In general, two-phase alloys are stronger than single-phase alloys, due to precipitation hardening: compare Figure 9.6 with Figure 11.19 and read Section 11.9. How much stronger depends on how the alloy is heat-treated. So we expect a further increase in strength as we cross the solvus lines, and a maximum strength somewhere around the middle of the figure. For the 600 C case, the solvus lines are further in, and the overall strength is less because dislocations are more mobile at the higher temperature. So the graph should look something like this:

    q1035.jpg

    Further Thoughts on Answer 10.35 (added on Monday Feb 16)

    Having thought about it over the weekend, I now think this was a bad question, and I won't set it in subsequent offerings of the course. As far as I can see, Callister doesn't provide a basis for determining the behaviour of the two-phase alloy before Chapter 11; and after we've read Chapter 11, we still can't predict the behaviour of the alloy without knowing its heat treatment history. If it's been treated so as to cause precipitation hardening, then the two-phase alloy will be a lot harder than the single-phase. But it might be possible to come up with a treatment schedule that would make it softer -- holding it at high temperature for a long period until the two phases separated into large spheroidal blobs, for example. So I'm going to give full marks (2 out of 2) to anyone who got the right results for the single-phase parts of the graph.