Homework 2 Solutions

# Homework 2 Solutions - Problem 2.8 E = 210 GPa t = 0.002 l...

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Problem 2.8 E = 210 GPa ε t = 0.002 ε l = 0.005 " = # t l = 0.002 0.005 = 0.4 Generalized Hooke’s Law with considerations to Poisson’s ratio t = 1 E t \$ %# l [ ] l = 1 E l \$ t [ ] Solving the equation for σ l l = E l + \$# t ( ) 1 % \$ 2 = 210 0.005 + 0.4 0.002 ( ) ( ) 1 % 0.4 ( ) 2 = 1.45 GPa t = E t + \$" l = 210 0.002 ( ) + 0.4 1.45 ( ) = 1 GPa Without consideration to Poisson's ratio l = E 1 = 210 0.005 = 1.05 GPa t = E t = 210 0.002 = 0.42 GPa Error in stress calculation if Poisson's ratio is not taken into consideration: l = 1.45 % 1.05 = 400 MPa t = 1 % 0.42 = 580 MPa or l = 1.05 % 1.45 1.45 = % 27.6% l = 0.42 % 1.0 1.0 = % 58%

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Problem 2.9
Problem 2.11 However if the material is assumed to be cubic instead of isotropic

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## This note was uploaded on 11/16/2011 for the course EMA 4223 taught by Professor Mecholsky during the Spring '08 term at University of Florida.

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Homework 2 Solutions - Problem 2.8 E = 210 GPa t = 0.002 l...

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