1ME 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, o245 ,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, 236 kω= −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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2Solution 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 4Ris the vector from the ground pivot 2Oto 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 34Ris 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, 34R,±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: 2R250 mm,=o40 ,θ=and o34105 .θ=For the given input position o245θ=, the unknown position variables 4Rand 34Rcan be obtained from trigonometry. For example, using the law of sines gives 3424222RRRsinABOsinAO BsinO AB==∠∠∠(1) Therefore, the distance between point B and point A can be written as 23422R250RsinAO Bsin 45sinABOsin 75=∠=°∠°(2a) Therefore, the distance between point B (fixed in link 4) and point A (fixed in link 3) is 34R183.01mm=(2b) Similarly, the distance between the ground pivot 2Oand point B can be written as 2422R250RsinO ABsin 60sinABOsin 75=∠=°∠°(3a) Therefore, the distance between the ground pivot 2Oand point B is 4R224.14 mm=(3b)