HW02 Solutions2011

# HW02 Solutions2011 - ME 382 Fall 2011 HW02 Solutions...

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Unformatted text preview: ME 382 Fall 2011 HW02 Solutions Problem 1: Dowling 5.16 = 400 × 10 −6 = 1000 × 10 −6 = 800 × 10 −6 = = = 0 From Table 5.2: E = 70.3 GPa and ν =0.345 . 3D Hooke’s Law: = 1 ¡ ¢ + £¤ = 1 ¡ ( + ) ¤ Plugging in known values: 0.004(70,300 ) = ¡ 0.345 0.0010(70,300 ) = ¡ 0.345 Solving simultaneously (2 equations, 2 unknowns) gives: = . and = . For an isotropic material: = = 2(1 ¡ ) Plugging in known values: = 70.3 2(1 ¡ 0.345) = 53.67 = (53,670 )(0.0008) = . To find , use 3D Hooke’s Law again: = 1 ¡ ¢ + £¤ Plugging in known values: = 1 70,300 [0 ¡ 0.345(59.45 + 90.81 )] = × − ME 382 Fall 2011 HW02 Solutions Problem 2: Dowling 5.25 Given: = − 60 = − 100 Block of brass is confined by a die in the z direction: = 0 From Table 5.2: E = 101 GPa and ν =0.35 . = 1 − ¡ + ¢£ = 0 = ¡ + ¢ = − = 1 − ¡ + ¢£ = − . × ¤ = 1 − ( + ) £ = − × ¤ ME 382 Fall 2011 HW02 Solutions Problem 3: A&J:I 4.1 ( ) = − + , ¡ = 2, ¢ = 10 stable molecule: r = 0.3 nm, U = -4 eV ( − 4 )(1.6 × 10 £19 / ) = − 0.3 2 + 0.3 10 [Eq. 1] at r...
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## This note was uploaded on 11/22/2011 for the course ME ME382 taught by Professor Dowling during the Fall '11 term at University of Michigan.

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HW02 Solutions2011 - ME 382 Fall 2011 HW02 Solutions...

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