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Fsh = torque on shaft, lb/in (Nm) Fy = compressive contact force in the radial direction between the cam surface and the follower, lb (N) Ic = mass moment of inertia of the cam, lb/in2/RAD (N/m/RAD) kr = follower return spring rate, lb/in/RAD (N/m) ks = drive system exibility lb/in/RAD (N/m/RAD) m = follower mass, lb (kg) Q = variable re ected follower inertia ratio Qm = maximum re ected inertia ratio R = radius from cam center to follower roller center, in (m) T = drive system torque at cam speed, lb/in (N-m) t = time, sec x = follower displacement, in (m) x = follower velocity (dx/dt) in/sec (m/s) = follower acceleration (d2x/dt2) in/sec2 (m/s2) x x = cam slope (dx/dqc), in (m) x = rate of change of cam slope (d2x/dqc2), in (m) xf = nal or total displacement of rise, in (m) y = normalized cam position, in (m) b = variable cam drive system windup ratio bm = maximum cam drive system windup ratio g = variable follower radial force ratio gm = maximum follower radial force ratio h = limiting linear frequency ratio qc = cam angular position, rad q c = cam angular velocity, rad/sec qc = cam angular acceleration, rad/sec2 qf = nal or total duration of rise, rad qi = input power source position, rad q i = input power source speed, rad/sec f = cam surface pressure angle, rad wb = base natural frequency rad/sec wl = limiting linear natural frequency, rad/sec wn = variable natural frequency, rad/sec The system takes account of camshaft exibility and leads to a set of nonlinear equations owing to the fact that the driving-cam function is no longer a known function of time In Fig 1210, where qi(t), the input shaft rotating at a known angular rate, is not equal to the cam angle, qc, or qi qc This is due to camshaft exibility; ks is the camshaft spring rate and the cam-follower rise is x A convenient method of formulation is derivable from Lagrange s equations as follows: T = 1 Mx 2 + 1 Iq c2 2 2 V = 1 ks (q c - q i ) + 1 ks x 2 2 2 L = T - V The follower is constrained by the cam function x = x(q c ) where qi = qi(t), by the Lagrange multiplier method.

10-6

Fn dx - Fn x dq c = 0 x = dx (q c ) dq c Fsh dq1 - Fshq1 dt = 0 q = dq i dt d dL - dL dx + F dx = 0 mx + kx + F = 0 n n dt dx dx d dL dL dq - Fn x dq c = 0 Iq c + ks (q c - q i ) - Fn x = 0 dt dq c dq c c d dL dL dq + Fsh = 0 ks (q i - q c ) + Fsh = 0 dt dq i dq i i (1212) (1213)

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We may use a computer algebra utility like Mathematica or Maple to calculate the integral exactly (to six decimal places) to equal 0746824 We thus see that the answer we obtained with the Trapezoid Rule is certainly accurate to two decimal places It is not accurate to three decimal places

Eliminating Fn between Eqs (1212) and (1213) and taking two time derivatives of x(qc) yields mxx + k f xx + Icq c + ks (q c - q i ) = 0 x = x q 2 + q x

whence ( Ic + mx 2 )q c + mx x q c2 + k f xx + ksq c = ksq i (t ) (1214)

It should be noted that Maple and Mathematica both use numerical techniques, like the ones being developed in this section, to calculate integrals So our calculations merely emulate what these computer algebra utilities do so swiftly and so well

yields the basic second degree vibration equation characterized as the spring-windup equation The equation is consistent with those reported in cam windup studies by Bloom and Radcliffe (1964) and Koster (1975a and b) The problem being studied is one of self-induced torsional vibration in the cam system The transient response is of interest with the angular velocity of the power source being a constant Also the torque response for the open-track system uctuates about the same rotational speed as the input system Four dimensionless parameters are used to characterize the dynamic response of the system These parameters, according to Szakallas and Savage (1980), are the frequency ratio h, the re ected inertia ratio Qm (measure of nonlinearity of the system), the maximum drive windup ratio bm, and the maximum radial force ratio gm h= Qm = wl q f w f 2pq i m( xmax ) Ic

Solution: The worst looking part of this integral is the l n x so make a substitution 1 v = In x so that dv = - dx This transforms the integral into a standard form

four decimal places of accuracy in the last example If you have some facility with a computer, use the Trapezoid Rule with that partition and confirm that your answer agrees with Mathematica's answer to four decimal places EXAMPLE

(1215)

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