chpt5soln

# chpt5soln - ME 314 Chapter 5 HW Chapter 5 HW Solution...

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Unformatted text preview: ME 314 Chapter 5 HW March 26, 2010 Chapter 5 HW Solution Problem 5.2: The reciprocating flat-face follower motion is a rise of 2 in with SHM in 180 ◦ of cam rotation, followed by a return with SHM in the remaining 180 ◦ . The prime (base) circle radius R o = 2 in, and there is NO OFFSET . (a) Find the displacement functions y ( θ ) for the full motion and plot the displacement diagram using MATLAB . Use units of DEG for the plot. For the rise (segment 1), from text Figure 5.14 and equation (5.18a), where L 1 = 2 and β 1 = π , we have y 1 = L 1 2 1- cos πθ β 1 = (1- cos θ ) , ≤ θ ≤ π (1) For the return (segment 2), from text Figure 5.17 and equation (5.21a), where L 2 = 2 and β 2 = π , we have y 1 = L 1 2 1 + cos πθ β 1 = (1 + cos θ ) , ≤ θ ≤ π (2) To get the complete displacement function one simply evaluates (1) and (2) for 0 ≤ θ ≤ π , then concatenates both segments together (removing the common point at the segment boundary). The plot is shown below: 30 60 90 120 150 180 210 240 270 300 330 360 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Cam Angle (DEG) Follower Displacement (in) Follower Displacement for Problem 2 (b) Assuming the follower has a circular cross-section, how large must this follower radius be to accommodate the contact point? The distance s from the follower axis to the contact point is simply equal to y . Therefore all we need to is to find max( | y | ). Since both rise and return are “equal,” we can simply examine the rise. From text equation (5.18b), the follower “speed” y is given by y = πL 2 β sin πθ β = ⇒ max | y | = πL 2 β = π (2) 2( π ) = 1 (3) Therefore the follower radius must be r = 1 in (the corresponding diameter is 2 in) 1 ME 314 Chapter 5 HW March 26, 2010 (c) Find and plot the profile that will accomplish this motion. Use 1 ◦ steps for cam angle θ . I used the camprofile.m functions I wrote for this purpose. The format for this functions is: FUNCTION [xc,yc] = camprofile(Ro,y,yp,theta,dir_flag) where Ro is base circle radius, y and yp are displacement y and yprime, theta is cam angle ( RAD ), and dir flag is ’CCW’ for this cam (the only change from CW to CCW is that the x component of cam profile coordinates must be negated—this produces a “mirror image” cam). Ironically, for the circular cam here it doesn’t matter! The displacement is shown below: it is a circle. That is because of the Simple Harmonic Motion. The “cross” indicates the axis of rotation. The vertical follower axis is at the top of the cam. − 3 − 2 − 1 1 2 3 − 4 − 3 − 2 − 1 1 2 Length (in) Length (in) Problem 5.7: This problem is similar to text Example 5.2. The motion is composed of: 1. an initial dwell 2. a full-rise motion 3. a half-return to a uniform velocity 4. a uniform velocity segment 5. a half-return back to rest (a) Plot the position y (in), velocity ˙ y (in/s), and acceleration ¨ y (ft/s 2 ) vs cam angle θ ( DEG ) for the complete motion....
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chpt5soln - ME 314 Chapter 5 HW Chapter 5 HW Solution...

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