# ENGR322: Homework Set Six

### Prof. W. H. Warnes Office: Dearborn Hall 303E Phone and Voice Mail: (541) 737-7016 FAX (541) 737-2600

email: warnesw@engr.orst.edu

### Callister, Sixth Edition, Problems:

1) Using the isothermal transformation diagrams shown in figure 10.7 and 10.9, specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at 900 deg C and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (Assume the martensite finish temperature is above room temperature for both alloys.)

a) rapidly cool to 700 deg C, hold for 10 sec and rapidly cool to room temperature;
b) rapidly cool to 575 deg C, hold for 3 sec and rapidly cool to room temperature;
c) rapidly cool to 600 deg C, hold for 1 sec, rapidly cool to room temperature:

2) 11.35 parts b) and d) ONLY

3) 11D5

4) 8.5

5) 8.14

### SOLUTIONS

1) Using the isothermal transformation diagrams shown in figure 10.7 and 10.9, specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at 900 deg C and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (Assume the martensite finish temperature is above room temperature for both alloys.)

a) rapidly cool to 700 deg C, hold for 10 sec and rapidly cool to room temperature;
b) rapidly cool to 575 deg C, hold for 3 sec and rapidly cool to room temperature;
c) rapidly cool to 600 deg C, hold for 1 sec, rapidly cool to room temperature:

There are TWO alloy compositions, two IT curves, and two sets of microstructure to discuss (compare) for each of the three parts of the problem. I have put them in parallel below, starting with the marked up versions of the IT curves.

2) Just like in the example in the book.

First step-- find the effective cooling rate as a function of diameter of the test sample using figure 11.16(b):

Equivalent distance from quenched end in a Jominy test:

 5140 (75mm diam) 8660 (100 mm diam) Surface (1.0 R) 12mm 19mm 0.75 R 17mm 25mm 0.50 R 21.5 mm 32mm center (0 R) 25mm --

Next, find the hardness, using figures 11.13 and 11.14:

 Hardness (HRC) Surface 42 0.75R 37.5 0.5R 34 Center 32

 Hardness (HRC) Surface 57.5 0.75R 53.5 0.5R 50 Center --

Make a plot and you're done!

3) The trick here is to glean what you can from the information in chapter 11, hidden in tables and text, and infer what remains using what you know about strengthening mechanisms. It's a lot like the DaVinci code in that respect... Looking at the ductility can give you an indication of how easily the material could be cold worked for strain hardening.

Alloy Comments Reference Heat Treatable? Cold Work?

R50500 Ti

Commercially pure Ti. Since there are no alloying additions, it can't be precipitation hardened.
Table 11.9
No
Yes
AZ31B Mg A wrought alloy with solid solution strengthening additives.
Table 11.8
No
Yes
6061 Al Good ductility, but usually used only as precipitation hardened.
Table 11.7
Yes
No
C51000 bronze Loads of ductility (64%) allows for substantial cold work strengthening.
Table 11.6
No
Yes
Lead A low melting temperature means that room temperature is essentially a hot working temperature for Pb. It can be strain hardened, but only at lower temperatures than room temperature.
Page 353
No
Sort of
6150 Steel High strength alloy steel can be heatteated by the quench and temper process, and has 20% ductility so it can be cold worked as well.
Table 11.2b
Yes
Yes
304 Stainless steel This is an austenitic steel, so it retains the austenite phase down to room temperature. Because of this, it cannot be quenched and tempered, but can only be cold worked for hardening.
Table 11.4
No
Yes
C17200 Be-Cu

A precipitation hardenable alloy, the ductility is limited to between 4-10%, which limits the amount of cold work hardening you could get.

Table 11.6
Yes
No

4) This problem refers to the stress concentration around a crack of given dimensions, and is discussed in section 8.5. Since this is a ceramic material, and therefore brittle, we expect that the sample will not show much ductility, but will fail at the maximum theoretical strength. From class discussion (at the very beginning of the discussion about deformation and dislocations), we worked out that the theoretical strength of a crystalline solid should be about E/10. For this ceramic, then, the theoretical failure strength is about 25GPa.

Based on the applied stress and the shape and size of the "most severe flaw" in the ceramic, we can estimate the stress concentration around the crack tip from equation 8.1 to be:

Since 15GPa is less than the theoretical strength of 25GPa, the sample WILL NOT FAIL.

5) Here we are using the fracture mechanics approach described by equation 8.4. Based on the geometry and the plane strain fracture toughness, the critical stress for failure is:

Since this stress value is larger than the applied stress of 1030MPa, we do not expect failure to occur.

End of File.