MSE 440/540 Test 3 Fall Points Total

Transcription

1 MSE 440/540 Test 3 Fall Points Total Name(print) Undergraduate Or Graduate (circle one) ID number No notes, books, or information stored in calculator memories may be used. The NCSU academic integrity policies apply to this exam. As such, by taking this exam you are implicitly agreeing with the statement: I have neither given nor received unauthorized aid on this test. All work must be written on these pages and turned in. To receive full or partial credit on numerical problems, you must show your calculations in step-by-step fashion. Units must be shown when applicable and plots must have labeled axes. Be sure that you read and answer all parts of each question. Constants, equations, and data are given at the end of the exam. 1. (3 points) Comparing with the homogeneous nucleation, the heterogeneous nucleation has a) lower energy barrier b) lower critical volume for the nucleus c) smaller critical radius for the nucleus d) higher critical undercooling e) both A and B. 2. (3 points) During the cooling of an alloy liquid, a thermal arrest where the temperature does not change with time a) does not happen b) always happens c) sometimes happens d) never happens 3. (3 points) Crystals can grow in which two ways? a) planar and screw growth b) dendritic and planar growth c) laminar and transition growth d) dendritic and straight growth 4. (5 points) Explain why no super heating is needed for melting of a metal. Because γ SL + γ LV < γ SV 5. (3 points) The grain sizes in the heat affected zone after welding are a) larger than those in the base metal b) smaller than those in the base metal c) larger than those in the fusion zone d) smaller than those in the fusion zone e) both a) and c) 1

2 6. (8 points) Determine the value of the strain-hardening exponent for a metal that will cause the average flow stress to be 3/4 of the final flow stress after deformation. Y f = 0.75 Y f Kε n /(1+n) = 0.75 Kε n 1/(1+n) = = 0.75(1+n) = n 0.25 = 0.75n n = (4 points) Name the four basic bulk deformation processes. The four basic bulk deformation processes are (a) rolling, (2) forging, (3) extrusion, and (4) wire and bar drawing. 8. (3 points) What is the temperature rage for warm working? 0.3T m < T warm working < T recrystallization 9. (3 points) The statement that softer materials are always easier to machine is (circle the correct one) (A) True, (B) False 10. (5 points) What is the difference between the shaping and planning? Shaping: The cutting tool moves linearly to cut the workpiece. Planing: The workpiece moves linearly to be cut. 2

3 11. (6 points) What are the stages of tool wear as a function of time? Break-in period: Rapid initial wear Steady-state wear region: Uniform wear Failure region: accelerated wear 12. (3 points) Broaching is (A) a rotational machining, (B) linear machining 13. (3 points) An oil-based cutting fluid is used primarily for: a) cooling, b) lubrication. 14. (4 points) Identify the mechanisms by which cutting tools wear during machining. The important tool wear mechanisms are (1) abrasion, (2) adhesion, (3) diffusion, and (4) plastic deformation of the cutting edge. 15. (3 points) Which of the following part geometry is preferred for a powder metallurgy? B 3

4 16. (5 points) What is the technical difference between blending and mixing in powder metallurgy? Blending refers to combining particles of the same chemistry but different sizes, whereas mixing means combining metal powders of different chemistries. 17. (8 points) In a certain pressing operation, the metallic powder fed into the open die has a packing factor of 0.5. The pressing operation reduces the powders to 2/3 of their starting volume. In the subsequent sintering operation, shrinkage amounts to 10% on a volume basis. Given that these are the only factors that affect the structure of the finished part, determine its final porosity. Packing factor = bulk density / true density Density = (specific volume) -1 Packing factor = true specific volume / bulk specific volume Pressing reduces bulk specific volume to 2/3 = Sintering further reduces the bulk specific volume to 0.90 of value after pressing. Let true specific volume = 1.0 Thus for a packing factor of 0.5, bulk specific volume = 2.0. Packing factor after pressing and sintering = 1.0/(2.0 x.667 x.90) = 1.0/1.2 = By Eq. (18.7), porosity = = (10 points) Tool life tests on a lathe have resulted in the following data: (1) at a cutting speed of 375 ft/min, the tool life was 5.5 min; (2) at a cutting speed of 275 ft/min, the tool life was 53 min. (a) Determine the parameters n and C in the Taylor tool life equation. (b) Using your equation, compute the tool life that corresponds to a cutting speed of 300 ft/min. (c) Compute the cutting speed that corresponds to a tool life T = 10 min. (a) VT n = C; Two equations: (1) 375(5.5) n = C and (2) 275(53) n = C 375(5.5) n = 275(53) n 375/275 = (53/5.5) n = (9.636) n ln = n ln = n n = C = 375(5.5) = 375(1.2629) C = 474 Check: C = 275(53) = 275(1.7221) = 474 (b) At v = 300 ft/min, T = (C/v) 1/n = (474/300) 1/0.137 = (1.579) = 28.1 min (c) For T = 10 min, v = C/T n = 474/ = 474/1.371 = 346 ft/min 19. (3 points) True or False: Ion implantation needs to be performed at relatively high temperature to accommodate the diffusion of alloy element. 4

5 20. (4 points) What is chemical vapor deposition? Chemical vapor deposition involves chemical reaction during the deposition process. 21. (4 points) A researcher plans to make an nanocrystalline alloy using thermal evaporation. What might be the problem that needs to be addressed? The different partial pressures of different alloy element make it hard to maintain the designed alloy composition. 22. (8 points) List 4 metastable phases/microstructures that can be produced by mechanical milling (ball milling). - Amorphous - Metastable crystalline compounds - Supersaturated solid solutions - Quasicrystalline phases (2011 Nobel Prize in Chemistry Schectman) - Nanocrystalline microstructures 5

6 Constants and Equations N A = 6.02 x mole -1 k (or R) = 8.62 x 10-5 ev/atom-k (or 8.31 J/mol K) K = C nm = 1 m 10 2 cm = 1 m γ SL + γ LV < γ SV G = H TS H=U+PV Δ ΔH T m S = = m L T m ΔG LΔT T m TS V = m n T C A C.E.(%) = C% + Si% + P% 3 n σ = Kε F =! wl, L = T = 0.5FL R( t 0! t f ) P = 2! NFL r = d t o c = 1.1 t, SB =! "! b!! b c = at, F = S t L, α Ab = 2π (R + K 360 ba t ) D DR = D K F = b p bf TSwt D 2! =!!!!,! = 1 +!.!!",!!! =!!!!!! =! h!, to 2 r cosα r = R MR = v f d, t R f, vt n = C, V = C I t,!! =!"# c i = tanφ = 32NR 1 r sinα!" 6