The problem makes use of the sparm command to

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Unformatted text preview: t = 0.1 0.25 0.25 0.989 1.40 1.35 0.966 Force at Displacement = 0.1 0.25 0.25 0.997 Force at Displacement = 0.2 HYPER158 0.25 Force at Displacement = 0.2 HYPER58 Force at Displacement = 0.1 Force at Displacement = 0.2 HYPER74 Ratio Force at Displacement = 0.2 HYPER56 ANSYS 1.4 1.38 0.986 With KEYOPT (2) = 3 of CONTA171 HYPER56 Force at Displacement = 0.1 0.25 0.27 1.069 Force at Displacement = 0.2 1.40 1.41 1.004 With KEYOPT (2) = 3 of CONTA172 HYPER74 Force at Displacement = 0.1 0.25 0.25 1.012 Force at Displacement = 0.2 1.40 1.42 1.012 With KEYOPT (2) = 3 of CONTA173 HYPER58 Force at Displacement = 0.1 0.25 0.27 1.065 Force at Displacement = 0.2 1.40 1.40 1.001 With KEYOPT (2) = 3 of CONTA174 ANSYS Verification Manual . ANSYS Release 9.0 . 002114 . © SAS IP, Inc. 1–483 VM211 Target[1] 1. Ratio Force at Displacement = 0.1 0.25 0.25 1.001 Force at Displacement = 0.2 HYPER158 ANSYS 1.4 1.41 1.007 Determined from graphical results. See T. Tussman, K-J Bathe, “A Finite Element Formulation for Nonlinear Incompress...
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This note was uploaded on 12/09/2010 for the course DEPARTMENT E301 taught by Professor Kulasinghe during the Spring '09 term at University of Peradeniya.

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