1402054378

# The way to analyze functional tolerances and the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5lry1S2 - 35.35lry 2S2 - ltz1S2 + 1.414ltz 2S2 ⎫ -25rx1S2 - 45ry1S2 - 35.35ry 2S2 - tz1S2 + 1.414tz 2S2 ⎪ ⎪ 25lrx1S2 + 40lry1S2 + ltz1S2 + 25rx1S2 + 40ry1S2 + tz1S2 ⎬ ⎪ -10lrx1S2 + 56.57lry1S2 - ltz1S2 + 1.414ltz 2S2 ⎪ -10rx1S2 + 56.57ry 2S2 - tz1S2 + 1.414tz3S2 ⎭ The last step consists in determining the constraints applied on the positioning. Assuming that the 3 connections are constituted of slipping contact implies: For the primary connection LHP [1]={lrx1S2, lry1S2, ltz1S2}={0,0,0} For the secondary connection, it is necessary to apply a positioning function such as the displacement of the workpiece in direction of the second plane is maximum respecting a non-penetration constraint for each potential point of contact. So to maximize −ltz 2S2 − tz 2S2 Simulation of the Manufacturing Process respecting: 187 -3.54lry1S2 - 25lry 2S2 + ltz 2S2 + 3.54rx 2S2 - 3.54ry1S2 ≥ 0 -3.54lry1S2 + 25lry 2S2 + ltz 2S2 + 3.54rx 2S2 - 3.54ry1S2 ≥ 0 3.54lry1S2 - 25lry 2S2 + ltz 2S2 - 3.54rx 2S2 + 3.54ry1S2 ≥ 0 3.54lry1S2 + 25lry 2S2 + ltz 2S2 - 3.54rx 2S2 + 3.54ry1S2 ≥ 0 For the tertiary connection, the same principle is applied. All of the positioning de...
View Full Document

Ask a homework question - tutors are online