Microsoft Word - ENGR 0022 Exam 3 Spring 2021 key.docx.pdf

ENGR 0022 Structure and Properties of Materials Exam 3 Spring 2021 Name (print): ______________________________________________ • 60 minutes • Open book • You can use a calculator I have neither given nor received unauthorized assistance on this test: Signature: ________________________________________________

Section 1: True/False (2 points each, 20 points total) __ F __ 1. The first step in age hardening of aluminum alloys is a solution anneal, which requires heating the alloy above the solidus temperature to form a single solid phase. Heating above the solidus will begin melting of the alloy. __ T __ 2. The lamellar spacing of pearlite in plain carbon steels is controlled by changing the cooling rate. __ T __ 3. The martensite start temperature in an iron-carbon alloy is independent of time in a TTT diagram because diffusion is not required to form martensite. __ F __ 4. The components of phase diagrams for ceramic systems are intermetallic compounds rather than elements. The components are ceramic compounds, not intermetallic compounds. Intermetallic compounds appear in metallic binary phase diagrams as intermediate phases. __ T __ 5. If a plain carbon steel with the eutectoid composition is cooled from the austenite single-phase field without rapid quenching, it will transform entirely to pearlite. __ T __ 6. A solid phase transforming to two other solid phases on heating is called a peritectoid reaction. __ T __ 7. The liquidus temperature is the highest temperature at which a solid phase can form during the cooling of an alloy. __ F __ 8. For solid-solid phase transformations, the growth rate of the new phase is low for small undercooling (∆T) and high for large undercooling. Growth rate is high for small undercooling (∆T) and low for large undercooling. __ T __ 9. Proeutectic phases are formed in alloys above the eutectic temperature. __ F __ 10. Rapid quenching of a eutectoid steel to room temperature after holding in the austenite single-phase field produces a supersaturated solid solution of carbon in austenite, which is not at equilibrium. It is then aged to precipitate iron carbide. Rapid quenching of eutectoid steel may produce martensite, but supersaturated austenite will not remain. The step after quenching for steels is called tempering.

Section 2: Problems and short answer questions (60 points total). Phase Diagrams (10 points total) Use the Pb-Sn phase diagram below to answer the following questions. 11. Pb-20wt%Sn alloy is slowly cooled as a liquid from 400°C. At what temperature will solidification begin for this alloy? (2 points) Solidification will begin at the liquidus temperature. (a) 183°C (b) 200°C (c) 240°C (d) 280°C (e) 300°C

12. What are the equilibrium phases present in Pb-20wt%Sn alloy at 100°C? (2 points) (a) Only alpha (b) Alpha and beta (c) Only beta (d) Alpha and liquid (e) Beta and liquid

13. What are the equilibrium phase compositions for Pb-20wt%Sn alloy at 100°C? (3 points) (a) Pb-2%Sn for alpha and Sn-0.5%Pb for beta (b) Pb-6%Sn for alpha and Sn-2%Pb for beta (c) Pb-19%Sn for alpha and Sn-2.5%Pb for beta (d) Pb-18%Sn for alpha and Pb-55%Pb for Liquid (e) Sn-20%Pb for Liquid and Sn-2%Pb for beta

14. What are the equilibrium phase fractions for Pb-20wt%Sn alloy at 100°C? Be sure to show the calculations. (3 points) Equilibrium phase fraction for a is (98 - 20)/(98 - 6) = 0.85 . Equilibrium phase fraction for b is (20 - 6)/(98 - 6) = 0.15 .

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