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CAM DESIGN HANDBOOK
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In Fig 87a we see a schematic picture of the mechanism The time is 60/1200 = 1/20 sec/rev; the time for total rise of 11/4 in is 160/360 1/20 = 1/45 sec
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In Fig 87b the displacement diagram divides into two parts, denoted by 1 and 2 We are given q1 + q2 = 160 degrees Since the ratio of acceleration is 3 : 1, we see that q1 = 40 degrees q 2 = 120 degrees The time t1 = t2 = 40 1 1 = sec 45 180 160 120 1 1 = sec 160 45 60
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Also, we know that the displacements 1 y1 + y2 = 1 4 For the parabolic curve from Eq (224) the follower displacement y= where A = follower accceleration, in sec 2 The displacement y1 = y2 = 1 1 5 (20, 200) = in 180 2 16 1 1 15 (6750) = in 60 2 16
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The displacement diagram is shown plotted in Fig 87b The inertia loads from Eq (81) Fi = where w is the equivalent weight: w= weight of spring + weight of linkage 3 = assumed negligible + 2 = 2 lb w y g
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The inertia force which increases the cam surface load Fi = 2 (20, 200) = +104 lb 386
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The critical inertia force tending to remove follower from cam
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CAM MECHANISM FORCES
Displacement y 1 " 14
2 15 " 16
5 " 16
Transition point 40
20 dwell 120 160 160 (b) Displacement diagram Cam angle q, degrees +20,200 in/sec2 +20,200 in/sec2 Dwell
(a) Cam-follower schematic
(c) Acceleration, in/sec2
+104 lb
+104 lb 6750 in/sec2
(d ) Inertia force, lb 35 lb
+
(e) External load, lb
+10 lb + +2 lb + +16 lb +16 lb
(f ) w, weight of follower, lb
Combined loads + Critical design point +12 lb +12 lb +14 lb +14 lb 23 lb +30 lb +30 lb Spring load +146 lb +146 lb +78 lb +130 lb +130 lb +90 lb (h) Total load +55 lb +55 lb on cam, lb +7 lb +7 lb +26 lb (g) Combined above + loads for spring design, lb
FIGURE 87 Cam surface load example
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Fi =
2 ( -6750) = -35lb 386
The acceleration curves and inertia are shown in Figs 87c and 87d The external load on the follower is +10 lb in Fig 87c The weight of follower linkage is +2 lb, Fig 87f Now, we shall combine these forces Fig 87g shows the superimposed values to give a combined force diagram We see a uctuating load from +116 lb to -23 lb, etc It was stated that the spring must exceed the net negative loads by a margin of 7 lb In Fig 87g, we shall plot the spring load curve At the transiting point, the spring force must equal 23 lb + 7 lb = 30 lb At the lowest point, let us assume an initial spring preload of 14 lb The spring has a spring index ks = where DS = spring force, lb d = de ection, in and the spring force at maximum rise S2 - 14 = 512 1 14 S2 = 780 lb Superimposing the spring force in Fig 87g, we see that the spring force curve rises appreciably as the cam rises Note that the spring force is +( ) although it is now plotted in reverse for better understanding of margin safety Also, this spring may have to be redesigned to t the mechanism or to eliminate any spring surge Last, it must be remembered that the highest cam speed requires the largest spring since at any lower speed the inertia forces are reduced Let us now combine all the forces This is shown in Fig 87h, in which we see a uctuating load curve The critical load point is the transition point where the minimum difference of 7 lb exists between the spring load and the external load A slightly stronger spring could have been chosen with a negligible effect on the system It should be noted that the foregoing problem serves primarily as a guide since recalculations are necessary One of the factors for redesign would be to include the spring weight of the follower Note that the use of a spring-loaded system produces a load on the cam regardless of the speed and accordingly affects the size and wear of the designed parts DS 30 - 14 += = + 512 lb in 5 Dd 16
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