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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 set-ups 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 cross-sectional 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 Ti-6Al-4V 830 4.43 114 0.217 2 (9.61 g) 3 (0.728%) hot-pressed 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 cross-sectional 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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This document was uploaded on 11/08/2011.
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