Page 364 - Cam Design Handbook

THB11 9/19/03 7:34 PM Page 352

352 CAM DESIGN HANDBOOK

this also gives the relationship between the linear velocities of the pushrod and valve as

v l 1
l = ra pr =
,
v vs l ra vs r ra
,
where r ra is the rocker arm ratio. For this rocker arm, l ra,pr = 0.875in., l ra,vs = 1.4875in and
r ra = 1.7. Putting this all together gives:
J m
m = m + m + ra + mr + cs r ra 2
2
l
pr
eq
vs ra
l 2 3
,
ra pr
0.132 lbm ◊ in 2 . 0153 lbm
2
= . 0270 lbm + . 0114 lbm + + . 0251 lbm (1.7 ) + ( . 17 ) 2
(0.875 in ) 2 3
= . 143 lbm
Since the system is referred to the lifter, the spring rate of the coil spring must be mod-
ified to account for the pushrod ratio. Equating the potential energy of the equivalent model
and the original model
1 1 1
2
2
kx = k x = k r x 2
2
2 eq l 2 cs vs 2 cs ra l
so the equivalent stiffness is
2
)(
2
in
K = K r = (2301 7 ) = 660 lb in.
lb
.
eq
cs ra
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 Ê 1 1 1 ˆ
= Á + + ˜
2
2
K eq Ë r K cs r K vs K ¯
ra
pr
ra
Ê 1 1 1 ˆ
= Á + + ˜
1 7) (
¥
5
.
¥
5
( Ë . 2 230 lb in ) ( . 2 47 10 lb in ) 3010 lb in ¯
17) ( .
so K eq = 660lb/in, which is exactly the same as K vs to two significant figures (the
values differ by about 0.3 percent). The fact that the equivalent stiffness resembles
that of the coil spring provides additional justification for assuming these elements to be
rigid.
Thus a simplified 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
K eq
s
=
w = 32 2 12. ¥ ¥ = 42067 Hz
rad
n
M eq
This makes sense since redline for the engine corresponds to a camshaft rotational speed
on the order of 3000rpm (or 50Hz) 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