ENB104-T1-S1-2007 - GUT Student Number Surname Given Name/s...

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Unformatted text preview: GUT Student Number Surname Given Name/s Examination Paper SEMESTER: FIRST SEMESTER EXAMINATIONS 2007 UNIT: ENB‘IO4 ENGINEERING MATERIALS — THEORY 1 DURATION OF EXAMINATION: PERUSAL: 10 MINUTES WORKING: 2 HOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS — THREE (3) PER STUDENT FORMULA AND CONSTANTS SHEET - THREE (3) PAGES ATTACHED CRITERIA SHEET - TWO (2) PAGES ATTACHED EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS CALCULATORS — ANY TYPE 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 QNLX ON THE EXAMINATION PAPER DURING PERUSAL TIME SECTION A - ALL FIVE (5) QUESTIONS ARE TO BE ATTEMPTED SECTION B - ALL THREE (3) QUESTIONS ARE TO BE ATTEMPTED , SECTION C — ALL FOUR (4) QUESTIONS ARE TO BE ATTEMPTED PLEASE ENSURE YOUR NAME AND STUDENT NUMBER ARE INSERTED IN THE SPACES PROVIDED ON THE CRITERIA SHEET ‘ ATTEMPT EACH SECTION IN A SEPARATE EXAMINATION BOOKLET MARKS FOR EACH QUESTION ARE AS INDICATED Queensland University of Technology GUT GUT GUT Gardens Point Kelvin Grove Carseldine QUESTION 1 a. QUESTION 2 a. b. QUESTION 3 a. ' QUESTION 4 a. b. QUESTION 5 a. 1 All Questions are to be Attempted Use a Separate Answer Book for each Section SECTION A (KNOWLEDGE AND APPICATIONS, 25%) Selection of material List four important properties in selecting material for turbine blades and briefly explain each property. Polycrystalline, columnar crystalline and single crystalline nickel based super alloys have been used as the primary structure for the turbine blades. Explain briefly two advantages and two disadvantages from each structure. Dislocation in metals With the aid of diagram, explain the role of dislocations to produce slip. Explain briefly the implication of many slip system and few slip system in materials. Corrosion of material Two pieces of steel are joined mechanically by crimping the edges. Why is this a bad idea if the steel is then exposed to water? . If the water contains salt, would corrosion be affected? Explain why or why not? Polymer Properties List four general characteristics of polymers and explain each briefly. Discuss briefly the effect of molecular weight, degree of cystallinity and heat treatments on polymeric strength. Ceramic Properties Discuss briefly the effect of porosity on the mechanical and optical properties of ceramic materials. Explain how the Young’s Modulus of materials can be measured accurately using piezoelectric materials. ENB104T1.071 .../cont 2 SECTION B (INTERPRETATION AND ANALYSE, 10%) QUESTION 6 Analyse the copper-phosphorous equilibrium phase diagram (Figure l) and answer the following questions: a. What composition would you suggest for the alloy to be used as a filler in metal braising. Explain why. b. What micro-constituents do you find for a 10 Wt% P at room temperature? CugP Cu a) L L— :1 4—0 I'd L- a) o. E a: +4 Cu3P ACU 2.24 ‘ 8.4 12.4 Wt% P Figure 1 Copper-Phosphorous Equilibrium Phase Diagram QUESTION 7 The micrographs in Figure 2, show fracture surfaces of two materials. Examine the modes of fracture of the materials and cite three main differences between the fracture surfaces. Figure 2 micrographs of fractured surfaces of materials ENB104T1.071 .../cont QUESTION 8 a. b. QUESTION 9 QUESTION 10 a. b. QUESTION 11 QUESTION 12 3 Sketch a typical Time-Temperature-Transformation curve (TTT curve) for eutectoid (0.76 wt% C) steel. On the diagram label the phases and micro-constituents present at each position. Using this diagram, show the formation of a eutectoid steel of 50% pearlite and 50% martensite. SECTION C (CALCULATIONS, 15%) Metals such as copper and aluminium are soft and ductile and have the same crystal structure. Show that the atomic packing factor (APF) of these metals is 0.74. A lift car weighs 24,500 N. The lift cable is made of steel of cross— sectional area of 1.13x10'4 m2 and cable length of 24 m. The elastic limit of this type of steel is 250 MPa and has elastic modulus of 207 GPa. Compute, The extension in the cable. The number of people that can get in the lift before the cable plastically deforms (assume one person = 65 kg). The temperature independent diffusion coefficient, D0 of magnesium in aluminium is 1.2x10' mZ/s and the activation energy for diffusion is 131 kJ/mol. Compute the diffusion coefficient of magnesium in aluminium at temperature of 550 °C. A composite material used in modern bicycles is made of carbon fibre with a Young's Modulus of 200 GPa and epoxy with a Young's Modulus of 5 GPa. The volume fraction of fibres is 67%. Calculate the Young's modulus of the composite material if it was tested parallel to the fibres. END OF PAPER ENB104T1.071 (i) FORMULA AND CONS TAN TS SHEET (ENB104 — Engineering Materials) Crystallography A0 -Ad %CW = x 100 (15) A0 HA ,0 = (1) VCNA dn _d0n = Kt (16) Diffusion Failure Mechanisms dC J=—D—— 2 dx ( ) Kc = Y owl/m (17) ac 52C ——=D 2 (3) . at 6x 0. : fl = My; (18) u 2 2 C -C x £=A(AK)m (19) X ° =1—erf[ J (4) “W C5-C0 NE AK : Kmax _ Kmin (20) D = D0 exp[——Q‘1—] (5) RT 91—19— = 5' 2 K20" exp(—& (21) dt S RT Elastic and Plastic Properties 0- : E5 (6) Ceramics n'( A + A ) r 2 Gy (7) p = $921.62; (22) v——§‘——~f~y* (8) 2 _ 82 _ .92 E=E0(1—1.9P+0.9P) (23) E = 20(1+ v) (9) 1. a, = ln[—‘) (10) 10 0' T = “(I + 8) (11) CsCl structure gT=1n(l+£) (12) Strengthening Mechanisms 7: : crss 13 0” (cos mos/l) ( ) “13X 1 0y = 0.0 + kydfli (14) Rocksalt structure ENB104T1 .071 Geometrical constrains for stable coordination of ions. Coordination Cation-Anion number ratio 2 < 0.155 3 0.155-0.225 4 0225-0414 6 0.414—0.732 8 O.732—1.0 Polymers Mn = inMi (24) fihzzmm (a) nu = A1" (26) m nw = A1” (27) m E=ZLW a& Composites EC = EmVin + Epr (29) ENE EC 2 “—p— (30) VmEp + VpEm EC =15”,an +15fo (31) ENB104T1.071 00 gg Q=————~ nq+na EC = KEfo + 15me Corrosion AV=(V2“—K°)——¥m[% I’M CPRzfl pAt Eloctronic Materials J = 0E ptatal = :0: + pi + pd ,0: 2 p0 +aT ,0i = Aci(1—ci) C=2= gogRA V d _ 1».24 Eg(meV) = B: floH+lqu Mmmfi J (32) (33) (34) (35) (36) C37) (38) (39) (40) (41) (42) (43) (44) (45) (iii) Values of Selected Physical Constants Avogadro’s number, N A = 6.023x1023 molecules per mole. Boltzmann’s constant, k = 1.3 8x1023 J/atom—K. Gas constant, R = 8.31 J/mole-K. Permittivity of Free Space; 80 =8.85x10'12 F/m. Permeability of Free Space; H0 =4nx10‘7 H/m. Faraday’s constant; 3 =96 500 C/mole. Planck’s constant, h= 6.63x10'34 J -s or 4.41x10‘15 eV-s. Speed of light, c= 3.0X108 m/s. 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