ENB231-T1-S1-2008

ENB231-T1-S1-2008 - GUT Surname Given Namels III-I..-...

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Unformatted text preview: GUT Surname Given Namels III-I..- Examination Paper SEMESTER: FIRST SEMESTER EXAMINATIONS 2008 UNIT: ENI3231 MATERIALS AND MANUFACTURING 1 - THEORY 1 DURATION OF EXAMINATION: PERUSAL: 10 MINUTES WORKING: ZHOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS - TWO (2) PER STUDENT CRITERIA SHEET - TWO (2) PAGES ATTACHED EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS CALCULATORS — NON PROGRAMMABLE INSTRUCTIONS TO STUDENTS: Students are prohibited from having mobile phones or any other device capable of communicating information (either verbal or written) in their possession during the examination NOTES MAY BE MADE QJSLI ON THE EXAMINATION PAPER DURING PERUSAL TIME SECTION A — ALL FOUR (4) QUESTIONS ARE TO BE ATTEMPTED SECTION - ALL FOUR (4) QUESTIONS ARE TO BE ATTEMPTED ATTEMPT EACH SECTION IN A SEPARATE EXAMINATION BOOKLET PLEASE ENTER YOUR NAME AND STUDENT NUMBER IN THE SPACES PROVIDED ON PAGE (i) ALL QUESTIONS ARE OF EQUAL VALUE GUT GUT Gardens Point Carseldine 1 SECTION A —- MATERIALS Question 1 A slowly cooled steel shows a microstructure containing 30% pearlite and 70% ferrite. Using the Iron-Carbon Phase diagram (Figure 1): (a) Estimate the carbon content in this alloy. (Show details of how you arrive at your answer.) (b) Describe the microstructure that would be obtained and the phase proportions if this alloy is heated to 730°C (HINT: just above the eutectoid temperature). (c) Describe the process of quenching and tempering of this alloy and explain the phase(s) present at every stage. 1000 Austenite, 7 Temperature 800 Ferrite, or 727C 600 400 0.00 0.40 0.80 1.20 1.60 2.00 Figure 1: Iron—Carbon Phase Diagram. Composition of Fe3C is 6.7%C ENB231T1.081 contl. .. Question 2 Figure 2 shows a typical TTT curve for an alloy. I .T—r" __ 1080- I C"O.?9 i’fn 500 \ mom “(a fiustenitized 01' 900 "C in Size: 6 I ; temperature PC] (21) Using the TTT curve, discuss the austempering process. Why would this process be used? What is the resultant microstructure? (b) What would be the effect on the TTT curve above if alloying had been added to the steel? ENBZ31T1.081 contl. .. Question 3 Referring to Figure 3(a) and Figure 3(b), answer the following: (a) What is hardenability? In order to improve the hardenability of a steel, what are the alloying additions that you would suggest and why? (b) A bar of diameter 90mm is intended to be used in a situation requiring a hardness of 42 HRC. The bar may be made from one of 1045, 1345 or 4145 steel. If the steel is to be quenched in water (H=1), determine which, if any, of these materials is appropriate for this application, giving working and reasoning. Relation between centre of a bar and the position along the Jominy bar. Bar diameter (mm) 0 5 10 15 20 25 30 35 40 45 50 55 Distance from quenched end (mm) Figure 3(a): Correlation between the hardness at the centre of a quenched bar and the distance from the end of a J ominy end quench test. Question 3 continued overleaf ENBZ31T1.081 cont/... Question 3 continued 4 7o 60 r 55 50 a ., 3' E E i c , = AISI 4145 Hardness. HRC. i 3 z AISI 1345 15 . . i . . , i i , 0 5 10 15 20 25 30 35 40 45 50 Distance from quenched end, mm. Figure 3(b): Representative J ominy end quench data for three steels. Question 4 Solid free \ energy \ Q \ \ ===Hnnauua Tnuc T951 T(°C) Figure 4: Free Energy vs. temperature diagram for a pure metal. (a) Discuss, using the energy curve in Figure 4 above, how energy considerations affect the solidification from a melt. How does the presence of second phase particles influence the grain size of the solid formed when the liquid is undercooled? (b) What are the differences between carbiding and nitriding? In your answer include a description of each process, the requirements for the steel used in each process and where each process is applicable. W ENBZ31T1.081 cont/... 5 SECTION B — MANUFACTURING Question 5 In casting experiments performed using a certain alloy and type of sand mold, it took 155 sec for a cube-shaped casting to solidify. The cube was 50 mm on a side. (a) Determine the value of the mold constant the mold constant in Chvorinov‘s Rule. (b) If the same alloy and mold type were used, find the total solidification time for a cylindrical casting in which the diameter x 30 mm and length = 50 mm. Question 6 (a) What is rolling in the context of the bulk deformation processes? (b) A single-pass rolling operatiOn reduces a 20 mm thick plate to 18 mm. The starting plate is 200 mm wide. Roll radius = 250 mm and rotational speed = 12 rev/min. The work material has a strength coefficient = 600 MPa and a strength coefficient “—" 0.22. Determine (i) roll force (ii) roll torque and (iii) power required 27: FLN Use: = p = _....__ F LWYW (60,000)kW Question 7 (a) What is springback in sheet metal bending? (b) A sheet metal with a thickness of 3 mm has a limiting drawing ratio of 2.4. Calculate the largest circular blank that could be used successfully in deep drawing a cylindrical cup with an internal diameter of 75 mm. Also calculate the punch force required when the sheet metal has an ultimate tensile strength of 530 MPa. USE: LDR = Do /D,., F = wito(UTS)[LDR — 0.7] Question 8 (3) Explain the different ways in which changing the die angle affects the extrusion process. (b) Assuming an ideal drawing process, what is the small diameter to which a lOO-mm diameter rod can be drawn? (0) Briefly comment on the grain flow patterns in the extrusion process with respect to dead metal zone and the possible defects the phenomenon can induce in an extruded part. END OF PAPER ENBZ31T1.081 9&9 Emcw €920 wEEmmfi mEow m.me ucm 920m”, 26* m E. 2&5ng mctmmEmcw Scam 90E _m._m:mm 332a mEEmflu 8:39 mEOm m:_v_mE new "2993 Eatoac: 9: ,5 =2 9%: “02.60 2th 0.6 “m5 wcozmcmaxm 5.65 20E...ch .mcozatommn .m:o_mm:ow_n mEom @5255 H3 mmuoSSE O wmm >0 :N .5235 3 Emucfim wEEmmE 8.62m. 269:8 Em 039:6 mEow 9.:me ucm “290$ E999 ucm Emtoafi m5 3 2:3 95w: 89:8 260E 9m 55 mcozmcmfixm BEE mcozEcou .wcofizummu .m:o_mm:om_u 96305 :3 mum—“.22,ch mambmcoEmu 30> . sh gun a m 225% m Emucfim h Emucfim "23355 3:35. utmE 22:55 :33 mem as". mEEmmfi um__mnm_ £02.00 95 wfiflim mEme ucm 536$ “5.329 Em Eatoaé 05 .6 “moE 95m: 50:8 .5ch Em $5 mcozmcmaxw 8.6% mcoEccou .wcozazommu .wco_mm:om_u 020:8 9630.5 :3 039265 mEmhmmfi 3:93. 28950 .25 2.520 633:5 9:me 26 “230mg, Em>m_mc new Emtoaé =m 9%: wcozmcmaxo 5.9.6 macarowa .wcofiwzoflu 3:29.: .mcoEczwu Howtoo mcfisoa I “E @323ch w>_m:m:wanu memcoEmu 20> . actfluflacmE 23 23$me *0 mung—383...: can wag—26:: manhumcofiwn 5:25 Eon—Es: #5th 6:52 Emu—am F 2:38.55: a 2.2.3“: Sumzm 3:5 can—Ea: Emuzum «Em «Em: 50> 5238 30am mtouto mi“ :0 3:3 «Ea—u ammo—n. "whOz 081 u ENB231T1 EEO. “000w 0.00000 00.00 000 09 0000 0.09.00 000 0000005 000 000 >0E 000 >=00000 0000 0.00000 0000 00.00 0.00000 00000.00 000.009 00000.00 05 00 0.00 5.5 20.00000 00000.00 0.00000 5.5 500.000 5.5 >000000 96 50209 05 I 2000909000 20.00.00 I 0E00 00050.00 00000.00 0009 20.00.00 I 00000.00 00050.00 I 00000.00 05 00 0>E0 00> 260 00000.00 05 #0 00000.00 2.000 9 000005200 0200 00> 2.00 5000 00“ 00 0200 00> 500 50> 0. 00000 00000000 0“ 000.00.00.00 50> 000 000005200 0030.00.00 5000 00. 000.00.00.00 0E 50 0.00 250.0 0. 00000 30000000 .500 0 009000 I 50> 0. 00000 50 “00 I 50> 0. 0000.0 000 0.00 .000 0.009 05 09 05 :0 50 000 >000_0 I 0023000000.: 000 0.000009 “9 00.00000 0000 no 09059.0 0000 0>00 >0E 00> o “000 00.099 000 00> . “000 00.00000 000 00> . “000 00.099 000 00> . 00.0902 000 00> . .00_._0E00 0>.om >000000 3.0000000 3:500:000 0.00.~ 000000 >009: 000 .005 00590 000 00590 >000000 30.002300 0000000_ >089 005 55000000 .005 000.000 >0000.Eo00._0 000 .00000000 0.0 .__05 00590 9 000 0000. 5:500:000 0000. 00590 000 00000000 000 00E; 00.000000 00.00000 >000000 0.00 0000 0.00 0>00 000 00000000 0.0.0x0 I 000 00 0.000 50.98 I 05 0000000 300.0000 I 000 2000905 I 000000000. 000 00009095 000 0000 05 00>.000 20000000. 00000095 0000 000.0 000 :0 000 000000000 9 00009095 000 0000 00009090. 000 0.00 000 0000 05 00 “0000000. 20005000 0.00 0000 “00 0000 I 0500.009 00000000. I 0535 0900 00000005 I 0009 “00900.0. 20005000 I 000 >000_0 I “000 3x00 0. “000 00 00m0§ 00 000000 “000 9x0“ “000 9x00 5 0&0“ 0_ ._0 00009. 00.050.030.09 00 00.000. 0_ .000 0. 00 000000 00 0000.6 00 00 00005 00 000000 ._0 “000 9x0“ 0. 00 000.000. 00 00 000000 00 00.000 000 0.0.0309 000000000 000000.600. 00.00“ 0. .00. 000000000 00.00“ 0. .00“ 000000000 00006 ._0 00.00“ 0_ .000 0. 0.00. 000000000 00 00:00:29 000 00.00 05 0002000 000090005 000 0000 0E 000090000. 000 000000000 000000.605 000000.500. 000 0:0 0000 00>.000 00.0.. 0000. 00.09 00> .. 00>.000 9 0.09000 00> . E00 000. 00>.000 00> . 000 0000 000 00>.000 00> . E00 05 00>.000 00> . 0:0 “0.00000. :0 0.00.0000 0.0 0.00030 0 0.00500 0 0.00030 0 E0050 ENB231T1.081 ...
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This note was uploaded on 11/03/2011 for the course EN 40 taught by Professor Mcgregor during the Three '10 term at Queensland Tech.

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ENB231-T1-S1-2008 - GUT Surname Given Namels III-I..-...

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