Download MSE/AOE3094 Test 2 - Material Science and Engineering - Prof. Thomas Staley and more Exams Materials science in PDF only on Docsity!
MSE/AOE3094 (Staley) Test#2 02/28/
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This test consists of four questions, each with multiple parts, plus an extra credit problem. Please read all questions carefully before beginning and note point values corresponding to each question. Partial credit is available for all questions, but you must show your reasoning/work on all questions to receive any credit for the right answer. Please answer in the space provided or on the back of each page if necessary. This test is open book & notes , but do not consult anyone about the test material. If you have questions regarding the test, e-mail me ([email protected]) before 5 PM on Thursday, 3/4/04. The test is due back by 10:30 AM on Friday, 3/5/04. It can either be turned in to me in the normal class time, or it can be placed in a box provided for this purpose outside my office door (335 Lane Hall).
Note: By signing this exam, you are acknowledging your compliance with the Virginia Tech Honor Code. In particular, you are noting that you have not discussed the test with anyone else.
- Boron nitride (formula unit = ‘BN’) is a commonly used ceramic material, with applications in armor-plating and composite fibers (among many other things). Boron and nitrogen radii for different bonding scenarios are given in the table below. Using this table, and other relevant information from your book, answer the following [9 pts per section] : Table of Radii: Boron: Atomic radius = 0.083 nm Ionic radius (+3 valence) = 0.023 nm Covalent radius = 0.088 nm
Nitrogen: Atomic radius = 0.071 nm Ionic radius (-3) = 0.132 nm Ionic radius (+3) = 0.030 nm Ionic radius (+5) = 0.027 nm Covalent radius = 0.070 nm
(a) Predict an ionic coordination number and (if possible) a crystal structure for this compound.
(OVER)
(b) Determine the percentage of ionic character the compound, BN, will have, and suggest how this might influence the answers you predicted for part (a).
(c) Do you think this compound might easily form glass (non-crystalline) structures? Why or why not?
- Refer to Figure 15.11 in your textbook (page 491), which shows fatigue data for a variety of common polymer materials. A flexible polymer rod with cross-sectional area of 80 cm 2 is required for use in a bellows system that will experience continuous cyclical forces of ±30 kN at a frequency of 1 cycle per minute.
(a) If the desired use life of the part is 10 years, which of the materials shown in Figure 15.11 would be acceptable choices? Incorporate a factor of safety, N=2, into the expected stress amplitude. [12 pts]
(b) For the materials that you found acceptable, which would you expect to exhibit hydrogen bonding between chains? [4 pts]
(c) For the materials that you found acceptable, which would you expect to crystallize most easily? [4 pts]
(d) For the materials that you found acceptable, which would be able to form distinguishable stereoisomers? [4 pts]
(e) For the materials that you found to be acceptable, which would be able to form distinguishable geometric isomers? [4 pts]
- Pure titanium at room temperature (RT) exists in a hexagonal close packed (HCP) structure, and has an atomic radius of 0.145 nm and atomic mass of 47.88 amu. Alloy addition of about 15 wt% molybdenum metal to titanium will convert the RT structure to a body-centered cubic (BCC) form with average atomic radius of 0.144 nm and average atomic mass of 55.09 amu. Which of the following properties would you expect necessarily to differ between pure Ti and the 85%Ti-15%Mo alloy? Show work as needed to demonstrate your reasoning [3 pts per section] :
(a) Density
(b) Atomic packing factor
(c) Coordination number
(d) Average metallic valence
(e) Grain size
(f) Most favored slip system
(g) Average electronegativity
EXTRA CREDIT (15 points), 3 per section) : On the back of this page, or a separate sheet, classify and designate, as fully as possible, the following ferrous and aluminum alloys. Refer to Handout #1 and Chapter 11 as needed:
(a) 0.5% Carbon, 1.8% Manganese, remainder Iron. (b) 99.5% pure Aluminum cold-worked to a value of ∆=0. (c) 4.0% Carbon, 1.5% Silicon, remainder Iron. (d) 0.15% C, 11.5% Cr, 1.0% Mo, 1.0 % Mn, remainder Iron (e) 1.1% C, 1.1%Ni, 0.25% Mo, remainder Iron.
