226-Lect 3 - Note: Lab 1 (this Friday) is in 2.G06 Week 3...

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1 Week 3 Week 3 (1) Cam-follower design example (2) Position and velocity analysis of 4 bar mechanism Note: Lab 1 (this Friday) is in 2.G06 Week 3 Cam-follower example Problem 8-43: (use Kloomok & Muffley Eqs) Design a cam to move a follower at a constant velocity of 100 mm/sec for 2 sec then return to its starting position with a total cycle time of 3 sec (Note: The solution is available in eLearning) Week 3 8-43 solution Week 3 8-43 solution
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2 Week 3 8-43 solution Week 3 8-43 solution Week 3 4 bar mechanism analysis • Position analysis (ch 4, Norton) • Velocity analysis – graphical method (ch 6 Norton, ch 8 M & R) • Velocity polygon, velocity image (section 6.2) Week 3 Position Analysis (section 4.5, Norton) Application of basic geometry principles Use of COSINE RULE (can use SINE Rule to complete the solution) B C a c b A () BC A C B a 2 cos 2 2 2 + = c C b B a A sin sin sin = =
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3 Week 3 Position Analysis For a 4 bar mechanism: Given the lengths of each of the links, and the angle between 2 of them (typically between link 2 and the “ground” or link 1) It is ALWAYS POSSIBLE to solve for the angular position of each of the other links Week 3 Position Analysis Find the length “z” then from the cosine rule: γ θ2 r1 z r2 r3 r4 = 4 3 2 4 2 3 2 1 2 cos r r r r z γ Week 3 Position Analysis Note: cos is a “double-valued function” (this relates to the 2 possible branches of assembly), hence you need to be careful in applying this method to select the required “branch”. γ ” is the transmission angle Week 3 Transmission Angle It is preferable that “ γ ” remain in the range 40° to 140° (or 220 ° to 320°, or within 50° of a right angle) throughout the cycle of a mechanism – particularly in higher loaded regions of the cycle. Otherwise the mechanism will tend to bind due to friction at the joints. If “ γ ” reaches 180, a dead point in the mechanism occurs – can cause trouble!
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4 Week 3 Loop Closure This method of determining the position of links in a mechanism is based on the simple observation that, providing the links remain connected, if you move from one link to the next around a mechanism – you end up where you started. Week 3 Loop Closure More formally, considering the following figure, 2 equations can be generated by setting the “sum” of the X components, and the “sum” of the Y components, of each link to zero. r r r r 1 2 3 4 θ2 θ3 θ4 Y X Week 3 Loop Closure X components: r 1 + r 4 cos θ 4 –r 2 cos θ 2 –r 3 cos θ 3 = 0 Y components: r 4 sin θ 4 –r 2 sin θ 2 3 sin θ 3 = 0 NB Norton uses complex number representation of the vectors in this area of analysis Week 3 Vector forms The vectors that are the basis for developing equations 4.6a and 4.6b can be expressed in many forms. In mechanism analysis common methods include: Unit Vector Complex Numbers X components: r 1 + r 4 cos θ 4 –r 2 cos θ 2 3 cos θ 3 = 0 Y components: r 4 sin θ 4 –r 2 sin θ 2 3 sin θ 3 = 0
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5 Week 3 Unit Vector Unit Vector Method R = R cos θ i + R sin θ j or
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226-Lect 3 - Note: Lab 1 (this Friday) is in 2.G06 Week 3...

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