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- The Sheady Ceap data of stainless cheel tkon at Stress level Fompa ů qiven in te Table Compute he steady-ctate Cleep dets for this alloy at 1a5ok and Shee level of 5omPaf Hhe shees erpinen t fox the alloy ū 7.0. fempersiuse 977 1089 2.0X 183For a Fe-C alloy with a eutectoid composition, an austenite was rapidly cooled down to 450°C, held for 10 sec, and quenched to room temperature. What is the final microstructure based on its TTT curve? Amount Temperature of Transformation, C 700 600 500 400 300 200 Ae₂! Ae₁ Austenite unstable Ms. 10 10² 103 104 105 seconds Austenite stable 50%- 100 MF TELEP P 100% completed F+B Coarse Pearlite - Fine Pearlite -Upper Bainite (Feathery) Lower Bainite → Martensite + Bainite Martensite 1 1sec 1 min 1hr Transformation Time (Log. Scale) O a. 50% bainite and 50% martensite O b. 50% austenite and 50% bainite O c. 100% bainite O d. 100% martensite Finer Lamellac Bainitic Pearlitic Austenite Fine more Acicular Rapid Etching Martensitic cicular, Slow Etching V.P.H. 210 320 450 700 750In this Sn-Pb alloy system, what is the relative percentage of proeutectic alpha phase after alloy 2 is cooled down from 300 °C to room temperature? Alloy 2 Alloy I 327 Solidus Liquidus 300 Liquid Liquidus 250 a + liquid Solidus 200 B+ liquid 183 Eutectic point 150 Solvus +B 100 Solvus 50 19.2 40.0 61.9 97.5 li 60 10 20 30 40 50 70 80 90 100% 100% Weight percent tin Sn Pb O1. 73% O II. 51% O II. 49% O IV. 27% Temperature (C),
- Carbonitriding is a case-hardening process in which a steel is heated in a gaseous atmosphere of such composition that carbon and nitrogen are absorbed simultaneously. Carbonitriding is a modification of the nitriding process. True False Based upon the percent elongation of a tensile sample, a ductile engineering material is defined by a value equal to or greater than 5% and a brittle material is defined by a value less than 5%. True Falseal 72% Í 22:39 365 3bed sleimen 10 minutes ago Question 1 Determine the percentage of ductility of a metal alloy having the following tensile stress-strain diagram. Not yer answered Stress-Strain Diagram Marked out of 15.00 P lag question 025 Strain jem/em Select one: O 25% O 0.26Copper and copper alloys as specified by the Copper Development Association (CDA) and are designated as C00001 thru C99999 in the Unified Numbering System (UNS). True False An alloy steel designated as AISI 4130 WQT 700 means: chromium-molybdenum alloy steel containing .30% carbon in solution, that is through heat treated and water quenched. True False An AISI 4340 alloy steel contains .43 percent or 43 points of the element carbon in solution. True False Chemicals from which the commercially important plastic resins are derived are themselves derived from natural products that are quite plentiful; these raw materials include coal, water, limestone, sulfur, salt, and petroleum. True False There are two (2) types of polymerization; addition or linear and cross-linked or condensation. Addition polymers are thermoplastics and cross-linked polymers are thermosetting materials. True False
- Question One (Compulsory) Composition (atk Sn) 100 20 40 60 80 327C B00 300 Liquid 500 232C 400 200 183C 07.8 18.3 61.9 300 100- H200 100 BO 100 20 40 60 Pb) Compesition (wt% Sn) Figure 1: Lead -tin phase diagram Figure I above shows a lead-tin (alloy) phase diagram. Consider an alloy composition of 55 wt% Sn - 45 wt% Pb. Answer the following questions: (i) Name the phases present and their compositions in the alloy at 150°C. (ii) Describe the structural and composition changes in the alloy when it is cooled from 190°C to 20°C. (iii) What are the compositions of phases at the eutectic point and why is this point important for the alloy? (iv) Given that the average density of a phase is determined as Pa = 100 ,calculate the PA PB volume fractions of the phases at the given compositions and at 150°C. Take the densities of lead and tin as 11.34 g/cm' and 7.36 g/cm' while the mass fraction of a is 0.21. (v) Name and describe four industrial applications of lead-tin alloys. ta aroundu (3.)…800 A Eutectoid temperature H 1400 700 A H 1200 600 1000 500 B 800 400 A 300 600 M(start) 200 M + A 50% 400 M(50%) M(90%) 100 200 10-1 1 10 102 103 104 105 Time (s) Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition, specify the final microstructure and approximate amount of each. Assume a small specimen has been held long enough to have achieved a complete and homogeneous austenitic structure prior to treatment. Sample (1): Quickly cool specimen from 800°C to 575°C, hold for 10 s, then quench to room temperature. Sample (2): Quickly cool specimen from 800°C to 500°C, hold for 100 s, then quench to room temperature. O after treatment, sample 1 is 50% pearlite, 50% austenite O after treatment, sample 2 is bauxite O after treatment, sample 1 is 25% pearlite O after treatment, sample 2 is bainite O after treatment, sample 2 is austenite O None of the answers is correct. O after treatment, sample 2 is coarse pearlite O after treatment, sample 1 is…Question 9 An Ag-Cu alloy contains 25% a and 75% B at 600"C. The composition of the overall alloy is approximately 1000 L+B 800 a /8.87. 28-12 9224 a +8 500 972 400 Ag 10 20 30 40 50 60 70 80 90 Cu Cu (by weight) O A. 73.5% Cu OB. 65.8% Cu OC. 89.6% Cu D 80.2% Cu
- Composition lat% N 40 60 100 1600 2800 1500 1453 C 2600 Liquid 1400 Solidus line Liquidua line 2400 1300 1200 2200 1100 1085 C 2000 1000 20 40 100 Composition (wt N (NO What is the amount for each phase in 1kg of a 50% Ni-50% Cu alloy at temperatures of 1400, 1300, or 1200 degrees C for point A and point B? Temperature ('C) Temperature ("F)O The following engineering stress-strain data were obtained for a 0.2% C plain-carbon steel. (i) Plot the engineering stress-strain curve. (ii) Determine the ultimate tensile strength of the alloy. (iii) Determine the percent elongation at fracture. Engineering Engineering Engineering Engineering Stress Strain Stress Strain (ksi) (in./in.) (ksi) (in./in.) 76 0.08 30 0.001 75 0.10 55 0.002 73 0.12 60 0.005 69 0.14 68 0.010 65 0.16 72 0.020 56 0.18 74 0.040 51 0.19 75 0.060 (Fracture)Question 5: Following is tabulated data that were gathered from a series of Charpy impact tests on a tempered 4140 steel alloy. Temperature (°C) 100 75 50 25 0 -25 -50 -65 -75 -85 -100 -125 -150 -175 Impact Energy (J) 89.3 88.6 87.6 85.4 82.9 78.9 73.1 66.0 59.3 47.9 34.3 29.3 27.1 25.0 (a) Plot the data as impact energy versus temperature. (you can use Excel or other software). (b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (c) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is 70 J.