4020-F2011-HW4Solutions - MSE 4020 HW4-SOLUTION 09/27/2011...

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1 MSE 4020 HW4-SOLUTION 09/2 7 /2011 Problem 1 (a) ! yy = " 0 E # = $ ! xx ! yy = $ ! zz ! yy % ! xx = $#! yy = $# " 0 E ! zz = $#! yy = $# " 0 E ' ( ( ) ( ( where ν , E, and σ 0 are Poisson’s ratio, Young’s modulus, and the tensile stress along y direction, respectively. ! yy = " u y " y # u y = ! yy $ y = % 0 E y Similarly, u x = % 0 E x u z = % 0 E z (b) No shear strain in this problem, thus, the strain tensor is pretty straightforward, i.e. just fill in the tensile strains obtained above. Thus, ! = ! xx 0 0 0 ! yy 0 0 0 ! zz " # $ $ $ $ % ' ' ' ' = () * 0 E 0 0 0 * 0 E 0 0 0 () * 0 E " # $ $ $ $ $ $ $ $ % ' ' ' ' ' ' ' '
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2 Problem 2 (a) By direct observation, the deformation in this problem is simple shear. Only x component of the displacement is non-zero and linearly dependent on z. From given information, we have the following displacement vector, ! u = u x u y u z ! " # # # # $ % = 0.2 ' z 0 0 ! " # # # $ % (b) From above ! ij = 1 2 " u i " x j + " u j " x i # $ % % ' ( ( ) ! xx = ! yy = ! zz
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4020-F2011-HW4Solutions - MSE 4020 HW4-SOLUTION 09/27/2011...

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