1402054378

# 032 after three iterations the fr computed 1032 by

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Unformatted text preview: The kinematic torsor for (FE1, FE2) is considered null because the contact between the two planes is assumed to be perfect and form tolerances are not specified in this instance. We must therefore determine the remaining four Small Displacement Torsor Intervals: T1/0 for FE1 relative to FE0, T3/2 relating FE3 to FE2, T5/4 relating FE5 to FE4, as well as T4/3 for the inclusion of FR2 in the analysis. From this, using the initial tolerance values from table I (first section), the final expression of the Jacobian-Torsor formulation with intervals is then represented by equation 5. [ 0, 0 ] [ 0, 0 ] [ − 0.2, + 0.2 ] [ − 0.0013, + 0.0013 ] [ − 0.0025, + 0.0025 ] [ 0, 0 ] FE1 [ − 0.1, + 0.1] [ 0, 0 ] [ − 0.1, + 0.1] [ − 0.002, + 0.002 ] 0 [ 0, 0 ] 0 [ − 0.002, + 0.002 ] FE2 0 • [ − 0.0265, + 0.0265 ] 0 [ 0, 0 ] 0 [ − 0.0265, + 0.0265 ] 1 FE 4 [ − 0.0011, + 0.0011] [ 0, 0 ] [ − 0.0011, + 0.0011] FE3 [ − 0.05, + 0.05 ] [ 0, 0 ] [ − 0.05, + 0.05 ] − 0.003 3, + 0.003 3 [ 0, 0 ] − 0.003 3, + 0.003 3 FE4 [u , u ] 1 [v, v ] 0 [ w, w ] 0 = [α , α ] 0 0 β , β 0 δ , δ FR 00 10 0 0 0 0 0 50 1 − 10...
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