ts2201s06

# ts2201s06 - EMSE 201 Introduction to Ma teria ls Science...

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EMSE 201 — Introduction to Materials Science & Engineering 29 March 2006 Name: Department of Materials Science and Engineering 1 of 9 Case Western Reserve University Test #2 — 75 minutes; 150 points; 6 questions; 9 pages; 15% of course grade Partial credit will be given for correct set-ups and reasoning. Give units on numerical answers where appropriate. Please write all answers on these pages; use the backs if needed. 1) The interatomic force and potential energy of some material are shown at right. a) (10 points) How could one determine the value of the elastic modulus of this material from one or both of these graphs (if the axes had numerical scales)? Justify your answer using the formula that defines the elastic modulus. To avoid confusion with E representing potential energy in the figure at right, denote the elastic modulus as Y: Y = d σ /d ε = d( F/A )/d( L / ) (4 pts ) In a material under uniaxial stress, will be proportional to the change in interatomic spacing r . (2 pts ) It follows that the elastic modulus Y (at zero strain) will be proportional to the slope of the graph of F vs. at r 0 . (4 pts ) Alternatively, the elastic modulus is proportional to the second derivative of the graph of potential energy vs. at 0 . (4 pts .; 10 pts. max) b) (6 points) The graph of potential energy for a second material has the same values of E 0 and r 0 but is much narrower in the vicinity of r 0 . How would the value of elastic modulus of the second material differ from that of the first material? Justify your answer. Given that the elastic modulus is proportional to the second derivative of the graph of potential energy vs. at 0 (3 pts ) , the narrower curve would have the higher curvature, higher second derivative, and higher elastic modulus (3 pts ) .

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EMSE 201 — Introduction to Materials Science & Engineering 29 March 2006 Department of Materials Science and Engineering 2 of 9 Case Western Reserve University 2) Use the data below (all taken at room temperature) to answer the following questions. alloy elastic modulus, GPa 1040 steel 207 brass 110 copper 115 a) (10 points) Compute the room-temperature modulus of resilience and the modulus of toughness for one of the three alloys (1040 steel, brass, or copper) at 0% cold work, 20% cold work, and at the maximum allowed cold work. Show your work. continued on next page
EMSE 201 — Introduction to Materials Science & Engineering 29 March 2006 Department of Materials Science and Engineering 3 of 9 Case Western Reserve University 2) a) Modulus of resilience, U R = σ y 2 /2E, (2 pts.) and E remains unchanged by cold work. Modulus of toughness, U T ut ε f (used below) or ( ut + y ) ε f /2 (either formula: 2 pts.) . The results are tabulated below

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ts2201s06 - EMSE 201 Introduction to Ma teria ls Science...

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