Homework3asol.fall11

# Homework3asol.fall11 - ME 352 Machine Design I Fall...

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1 ME 352 - Machine Design I Name of Student ___________________________ Fall Semester 2011 Lab Section Number ________________________ Homework No. 3 (30 points). Due at the beginning of lecture on Wednesday, September 14th. Consider the variation of the Scotch-yoke mechanism shown in Problem 3.26 on pages 161 and 162. For the given position of the input angle, that is, o 2 45 , θ = perform a position analysis of the mechanism using trigonometry (that is, the law of sines and the law of cosines). If the input link 2 of the mechanism is rotating clockwise with a constant angular velocity of 36 rad/s, that is, 2 36 k ω = − rad/s then for the given position of the mechanism, perform a velocity analysis using: (i) The method of kinematic coefficients. (ii) The method of instantaneous centers of velocity. Draw the Kennedy circle and list the primary instant centers and the secondary instant centers. Clearly write the scale that you used for your scaled drawing of the mechanism. Compare the answers that you obtained from Part (i) with the answers that you obtained from Part (ii).

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2 Solution to Homework 3 (30 Points). Position analysis of the mechanism. The vectors for the variation of the Scotch-yoke mechanism are as shown in Figure 1. Note that 4 R is the vector from the ground pivot 2 O to point B, fixed in link 4, that is, the time rate of change of this vector is the velocity of link 4. Also, note that 34 R is the vector from point B (fixed in link 4) to point A (which is fixed in link 3). The time rate of change of this vector, that is, 34 R , ± is the relative velocity between these two points (which are fixed on different links). Figure 1. The vectors for a kinematic analysis of the mechanism. The known constants are: 2 R 250 mm, = o 4 0 , θ = and o 34 105 . θ = For the given input position o 2 45 θ = , the unknown position variables 4 R and 34 R can be obtained from trigonometry. For example, using the law of sines gives 34 2 4 2 2 2 R R R sin ABO sin AO B sin O AB = = (1) Therefore, the distance between point B and point A can be written as 2 34 2 2 R 250 R sin AO B sin 45 sin ABO sin 75 = = ° ° (2a) Therefore, the distance between point B (fixed in link 4) and point A (fixed in link 3) is 34 R 183.01mm = (2b) Similarly, the distance between the ground pivot 2 O and point B can be written as 2 4 2 2 R 250 R sin O AB sin 60 sin ABO sin 75 = = ° ° (3a) Therefore, the distance between the ground pivot 2 O and point B is 4 R 224.14 mm = (3b)