homework6csol.spring11 - ME 352 - Machine Design I Spring...

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- 1 - ME 352 - Machine Design I Name of Student ____________________________________ Spring Semester 2011 Lab Section Number ________________________________ Homework No. 6. Cam Design. Chapter 6. Part I (30 points). Due at the beginning of lecture on Friday, March 4th. Consider the cam-follower system presented in Problem 6.28 on pages 328 and 329. At the cam angle , 225 o θ = determine numerical values for: (i) The first, second, and third-order kinematic coefficients of the displacement diagram. (ii) The radius of the curvature of the cam surface. (iii) The unit tangent and normal vectors to the cam at the point of contact with the follower. (iv) The coordinates of the point of contact between the cam and the follower. Express your answers in the moving Cartesian coordinate reference frame attached to the cam. (v) The pressure angle of the cam. Is your answer acceptable for this cam-follower system? Part II (30 points). Due at the end of lab on Wednesday, March 9th, and Thursday, March 10th. Solve Problem 6.28 on pages 328 and 329. Templates to draw the lift curve and the cam profile will be provided by your TA.
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- 2 - Solution to Homework 6. Part I. 30 Points. The base circle diameter of the cam is D = 3 inches and the diameter of the reciprocating roller follower is d = 1 inch. Therefore, the radius of the prime circle of the cam is 0 31 2 inches 22 Dd R + + == = (1) Since the problem states that there is a radial follower then the follower offset (or eccentricity) is 0. ε = Therefore, the coordinates of the follower center on the prime circle (that is, when the roller follower is touching the base circle of the cam) in the fixed reference frame are 0 0 X = = (2a) and 00 202 i n c h e s YR =− = = (2b) The displacement of the follower center for full-rise simple harmonic motion, see Equation (6.12a), page 291 in the text book, can be written as * 1c o s 2 base L yy πθ β ⎛⎞ =+− ⎜⎟ ⎝⎠ (3) where base y is the base lift (that is, the lift at the beginning of the full-rise simple harmonic motion). Therefore , from Table P6.28 in the problem statement, the base lift is 0 base y = (4) Also, the total lift during the full-rise simple harmonic motion is 2 inches L = (5) The range of the full-rise simple harmonic motion is oo o 270 210 60 3 rad π =−= = (6a) and 0 * θ θθ = (6b) where 0 is the starting cam angle for the full-rise simple harmonic motion. The cam angle is , 225 o = and the starting cam angle for the full-rise simple harmonic motion is 0 , 210 o = therefore o * 225 210 15 12 rad = (6c) and * 3 12 4 rad βπ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (7)
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This note was uploaded on 04/23/2011 for the course ME 352 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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homework6csol.spring11 - ME 352 - Machine Design I Spring...

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