Page 238 - Cam Design Handbook

THB8 9/19/03 7:25 PM Page 226

226 CAM DESIGN HANDBOOK

Solution In Fig. 8.7a we see a schematic picture of the mechanism. The time is
1
60/1200 = 1/20sec/rev; the time for total rise of 1 / 4 in. is 160/360 ¥ 1/20 = 1/45sec.
In Fig. 8.7b the displacement diagram divides into two parts, denoted by 1 and 2. We
are given q 1 + q 2 = 160 degrees. Since the ratio of acceleration is 3:1, we see that
q = 40 degrees
1
q = 120 degrees
2
The time

40 Ê 1 ˆ 1
t = = sec
1 Ë ¯
160 45 180
120 Ê 1 ˆ 1
t = = sec.
Ë
2
160 45 ¯ 60
Also, we know that the displacements
1
y + y = 1 .
2
1
4
For the parabolic curve from Eq. (2.24) the follower displacement
1
y = t Ain
2
2
where
A = follower accceleration, in sec 2
The displacement

1 Ê 1 ˆ 2 5
20200)
y = ( , = in
1
2 Ë 180 ¯ 16
1 Ê 1 2 15
y = ( 6750) ˆ = in.
2 Ë ¯
2 60 16
The displacement diagram is shown plotted in Fig. 8.7b. The inertia loads from Eq. (8.1)
w
F = y ˙˙
i
g
where w is the equivalent weight:
weight of spring
w = + weight of linkage
3
= assumed negligible + = lb.2 2
The inertia force which increases the cam surface load
2
F = (20 200, ) =+104lb.
i
386
The critical inertia force tending to remove follower from cam