CAM SYSTEM MODELING in Software

Generate Data Matrix ECC200 in Software CAM SYSTEM MODELING

EA (30 10 6 psi )( 0075 in 2 ) = = 47 10 5 lb in L 475 in EA (30 10 6 psi )( 0075 in 2 ) = = 30 10 5 lb in L 74 in

All of the compliances act like springs in series, and hence are dominated by the spring with the smallest spring rate Since the lowest rate belongs to the coil spring, this spring becomes the most important spring element to retain in any single degree of freedom model The other two stiffnesses are so large relative to the coil spring that the pushrod and valve stem may be reasonably modeled as rigid bodies With the valve stem, the rocker

Inertial Parameters of Valve-gear System Mass or Inertia ml = 0270 lbm mpr = 0114 lbm Jra = 0132 lbm-in2 mvs = 0251 lbm mcs = 0153 lbm

arm, the pushrod and the lifter now assumed rigid, the system takes on the schematic form of Fig 1124a To obtain the equivalent system of Fig 1124b re ected to the lifter, it is necessary to equate the kinetic energy of the original system to that of the equivalent system to account for the mechanical advantage of the rocker arm Recalling that the mass contribution of a spring xed at one end is a third of its mass at the free end, equating the kinetic energy of the equivalent system and the original system gives 1 1 1 1 1 11 2 2 2 2 meq vl2 = m f vl2 + m pr v pr + Jraw ra + mvs vvs + mcs vvs 2 2 2 2 2 23 Kinematically, there are several relationships between velocities that follow from the modeling assumptions made earlier Since the lifter and pushrod are assumed to be rigid, their velocities are the same, so v pr = vl Assuming small angles of rotation on the rocker arm, the rocker arm rotational velocity can be obtained from w ra = vl v = vs lra , pr lra ,vs

this also gives the relationship between the linear velocities of the pushrod and valve as l vl 1 = ra , pr = vvs lra ,vs rra where rra is the rocker arm ratio For this rocker arm, lra,pr = 0875 in, lra,vs = 14875 in and rra = 17 Putting this all together gives: meq = ml + m pr + Jra m 2 2 + mvs rra + cs rra 2 3 l ra , pr 0132 lbm in 2 0153 lbm 2 (17)2 + 0251 lbm(17) + 3 (0875 in )2

Since the system is referred to the lifter, the spring rate of the coil spring must be modi ed to account for the pushrod ratio Equating the potential energy of the equivalent model and the original model 1 1 1 2 2 keq xl2 = kcs xvs = kcs rra xl2 2 2 2 so the equivalent stiffness is

Although the pushrod and valve stem were considered rigid, it is reasonable to consider adding their stiffnesses in series to that of the coil spring as Chen (1982) suggests in reducing a similar model In this case, it makes very little difference since there is such a vast discrepancy in the stiffnesses of the pushrod and valve stem and that of the coil spring If these springs are added in series with proper accounting for the rocker arm ratio, 1 1 1 1 = 2 + 2 + Keq rra Kcs rra Kvs K pr 1 1 1 = + + (17)2 (230 lb in ) (17)2 ( 47 10 5 lb in ) 30 10 5 lb in so Keq = 660 lb/in, which is exactly the same as Kvs to two signi cant gures (the values differ by about 03 percent) The fact that the equivalent stiffness resembles that of the coil spring provides additional justi cation for assuming these elements to be rigid Thus a simpli ed model of this system can be obtained by using these equivalent mass and stiffness values Over what frequency range is this model likely to be valid To get a rough idea, consider that the natural frequency of the system with this mass and stiffness is w n = 322 12 Keq Meq = 420 rad s = 67 Hz

This makes sense since redline for the engine corresponds to a camshaft rotational speed on the order of 3000 rpm (or 50 Hz) and resonance in this basic operating region should be avoided To get a rough idea of the next resonance in the system, consider treating the