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Unformatted text preview: EMSE 201 Introduction to Materials Science & Engineering 5 November 2008 Name: SOLUTION Department of Materials Science and Engineering 1 of 5 Case Western Reserve University Test #2 75 minutes; 100 points; 5 questions; 5 pages; 10% of course grade Partial credit will be given for correct setups and reasoning. Give units on numerical answers where ap propriate. Please write all answers on these pages; use the backs if needed. 1) A rod, 10.0 cm long, must be able to support a load of 18.0 kN in uniaxial tension along its length. For each of the four candidate materials listed below, a) (6 points) compute the minimum crosssectional area of the rod to ensure that its yield strength (or in the case of silicon nitride, its fracture strength) will not be exceeded under the specified load; b) (7 points) rank the mass of the resulting rods from 1 (lowest) to 4 (highest); c) (7 points) rank the elongation of the resulting rods under the specified load from 1 (lowest) to 4 (highest); Show your work in the space below. (Use the back if necessary.) material yield strength, MPa density, g cm 3 modulus of elasticity, GPa cross sectional area, cm 2 rank by mass (1 lowest, 4 highest) rank by elonga tion (1 lowest, 4 highest) annealed 316 stainless steel 205 7.80 193 0.878 4 (68.5 g) 1 (0.106%) annealed Ti6Al4V 830 4.43 114 0.217 2 (9.61 g) 3 (0.728%) hotpressed silicon nitride 850* 3.30 304 0.212 1 (6.99 g) 2 (0.280%) polyetheretherketone 91 1.31 1.10 1.98 3 (25.9 g) 4 (8.27%) *) fracture strength For a uniaxial tensile force F on a specimen of uniform initial crosssectional area A , the en gineering stress = F / A ( 2 pts ). Setting the stress equal to the materials strength, A = 18.0 kN/ y where y is the applicable strength parameter ( 1 pt for each correct answer). The mass M of the rod will be given by AL , where L is the length (specified as 10.0 cm) and is the mass density. Substituting A = F / y gives M = FL / y ( 3 pts ). The masses are listed in parentheses in the table above. As a ranking was requested, it was necessary only to compute the ratio / y and rank the resulting values ( 1 pt for each correct ranking or mass). The elongation L of the rod will be given by L = L / E , where is the strain and E is the modulus of elasticity. Substituting = y gives L = L y / E ( 3 pts ). As a ranking was re quested, and L is the same for all of the rods, it was necessary only to compute the strain...
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 Spring '09

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