Solution10 - without plastic deformation ( F y ). Taking...

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EGN3365-Structure and Properties of Materials TTR 8:30-9:45 Instructor: Dr. S. Seal TA: Mr. Himesh Bhatt Solutions: Home-Work 10 1. We are asked to compute the maximum length of a cylindrical titanium alloy specimen that is deformed elastically in tension. For a cylindrical specimen A o = π 2 where d o is the original diameter. Combining Equations (6.1), (6.2), and (6.5) and solving for l o leads to l o = = = 0.25 m = 250 mm (10 in.) 2. This portion of the problem calls for a determination of the maximum load that can be applied
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Unformatted text preview: without plastic deformation ( F y ). Taking the yield strength to be 275 MPa, and employment of Equation (6.1) leads to F y = σ y A o = (345 x 10 6 N/m 2 )(130 x 10-6 m 2 ) = 44,850 N (b) The maximum length to which the sample may be deformed without plastic deformation is determined from Equations (6.2) and (6.5) as l i = l o = (76 mm) + MPa x MPa 3 10 103 345 1 = 76.254 mm Or ∆ l = l i- l o = 76.254 mm - 76.00 mm = 0.254 mm...
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This note was uploaded on 01/13/2011 for the course EGN 3365 taught by Professor Yong-hosohn during the Spring '08 term at University of Central Florida.

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Solution10 - without plastic deformation ( F y ). Taking...

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