12_8

Materials Science and Engineering: An Introduction

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264 12.D2 This problem asks us to select from four alloys (brass, steel, titanium, and aluminum), the one that will support a 50,000 N (11,250 lb f ) load without plastically deforming, and having the minimum weight. From Equation (6.1), the cross-sectional area ( A o ) must necessarily carry the load ( F ) without exceeding the yield strength ( σ y ), as A o = F σ y Now, given the length l , the volume of material required ( V ) is just V = lA o = lF σ y Finally, the mass of the member ( m ) is m = V ρ = ρ lF σ y in which ρ is the density. Using the values given for these alloys m(brass) =
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Unformatted text preview: (8.5 g/cm 3 )(10 cm)(50,000 N) (415 x 10 6 N/m 2 )       1 m 10 2 cm 2 = 102 g m(steel) = (7.9 g/cm 3 )(10 cm)(50,000 N) (860 x 10 6 N/m 2 )       1 m 10 2 cm 2 = 46 g m(aluminum) = (2.7 g/cm 3 )(10 cm)(50,000 N) (310 x 10 6 N/m 2 )       1 m 10 2 cm 2 = 43.5 g m(titanium) = (4.5 g/cm 3 )(10 cm)(50,000 N) (550 x 10 6 N/m 2 )       1 m 10 2 cm 2 = 40.9 g Thus, titanium would have the minimum weight (or mass), followed by aluminum, steel, and brass....
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This document was uploaded on 05/04/2010.

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