Lecture_6_-_Posit_Slider___Inv_Slider_Cr

Lecture_6_-_Posit_Slider___Inv_Slider_Cr - ME 371 Lecture 6...

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Unformatted text preview: ME 371 Lecture 6 1 ME 371 ME 371 Chapter 4 Chapter 4 Positional Analysis Positional Analysis Slider Slider – – Cranks & Inverted Slider Cranks & Inverted Slider ME 371 Lecture 6 2 Slider – Cranks − For position analysis we will use Vector Loop Approach − We know – Lengths of links 2 & 3 – Offset height – Input angle θ 2 − Calculate – Length of link 1 – Angle θ 3 ME 371 Lecture 6 3 Slider – Cranks Vector loop equation R 2 – R 3 – R 4 – R 1 = 0 Projections: (x and y axis) a cos θ 2 – b¡cos θ 3 – c¡cos θ 4 – d¡=¡0 a sin θ 2 – b¡sin θ 3 – c¡sin θ 4 = 0 here a, b and c are known θ 2 is given θ 4 = 90 ° Solution d = a cos θ 2 – b¡cos θ 3 ME 371 Lecture 6 4 Slider – Cranks Solve for θ 3 using the equation (projection on y axis) a sin θ 2 – b¡sin θ 3 – c¡sin θ 4 = 0 Solution (± 90 of sin-1 ) 1 st configuration 2 nd configuration d = a cos θ 2 – b¡cos θ 3 − = − b c a 2 1 3 sin sin 1 θ θ π θ θ + − − = − b c a 2 1 3 sin sin 2 ME 371 Lecture 6 5 Inverted Slider...
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This document was uploaded on 10/27/2011 for the course MECHANICAL 371 at Michigan State University.

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Lecture_6_-_Posit_Slider___Inv_Slider_Cr - ME 371 Lecture 6...

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