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CSU MECH 324 - DESIGN OF MACHINERY

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DESIGN OF MACHINERY SOLUTION MANUAL 15-12-1! PROBLEM 15-12Statement:The cam in Figure P15-2 is a pure eccentric with eccentricity = 20 mm and turns at 200 rpm. The mass of the follower is 1 kg. The spring has a rate of 10 N/m and a preload of 0.2 N. Find the follower force over one revolution. Assume a damping ratio of 0.10. If there is follower jump, respecify the spring rate and preload to eliminate it.Units:rpm 2π.rad.min1.Given:Cam eccentricity and speed: a 20 mm. ω200 rpm.Follower mass: mf1 kg.Damping ratio:ζ0.10Spring: k 10 N.m1.Fpl0.2 N.Solution:See Figure P15-2 and Mathcad file P1512.1. Using the equation given in the figure, write functions for the displacement, velocity, and acceleration of the follower. Note that the displacement function is written such that at it is zero at t = 0.Displacement:sθ() a 1 cosθ()().Velocity: vθ() aω.sinθ().Acceleration: aθ() aω2.cosθ().2. Calculate the damping coefficient using equations 15.2i and 15.3a.c 2ζ.mfk..c 0.632N sec.m=3. Substitute the expressions for displacement, velocity, acceleration, and spring preload into equation 15.9 and solve for the force on the follower as a function of cam angle.Fcθ() mfaθ().cvθ().ksθ().Fpl4. Plot the force on the follower for one revolution of the cam. θ0 deg.2 deg.,360 deg...0 60 120 180 240 300 3601050510FORCE ON FOLLOWER (ORIGINAL SPRING)Cam Rotation Angle, degForce, NFcθ()Nθdeg2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 15-12-25. The cam force becomes negative, which indicates that the follower will jump or lose contact with the cam. New values of spring rate and/or preload will be tried iteratively to make the force always positive.Let the new spring parameters be:k 50 N.m1.Fpl7.5 N.c 2ζ.mfk..c 1.414N sec.m=Fcnθ() mfaθ().cvθ().ksθ().Fpl0 60 120 180 240 300 36005101520FORCE ON FOLLOWER (NEW SPRING)Cam Rotation Angle, degForce, NFcnθ()Nθdeg2nd Edition,


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CSU MECH 324 - DESIGN OF MACHINERY

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