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CHAPTER 40 CAM MECHANISMS Andrzej A. Ol?dzki, D.Sc. f Warsaw Technical University, Poland SUMMARY / 40.1 40.1 CAM MECHANISM TYPES, CHARACTERISTICS, AND MOTIONS / 40.1 40.2 BASIC CAM MOTIONS / 40.6 40.3 LAYOUT AND DESIGN; MANUFACTURING CONSIDERATIONS / 40.17 40.4 FORCE AND TORQUE ANALYSIS / 40.22 40.5 CONTACT STRESS AND WEAR: PROGRAMMING / 40.25 REFERENCES / 40.28 SUMMARY This chapter addresses the design of cam systems in which flexibility is not a consid- eration. Flexible, high-speed cam systems are too involved for handbook presenta- tion. Therefore only two generic families of motion, trigonometric and polynomial, are discussed. This covers most of the practical problems. The rules concerning the reciprocating motion of a follower can be adapted to angular motion as well as to three-dimensional cams. Some material concerns circu- lar-arc cams, which are still used in some fine mechanisms. In Sec. 40.3 the equations necessary in establishing basic parameters of the cam are given, and the important problem of accuracy is discussed. Force and torque analysis, return springs, and con- tact stresses are briefly presented in Sees. 40.4 and 40.5, respectively. The chapter closes with the logic associated with cam design to assist in creating a computer-aided cam design program. 40.7 CAM MECHANISM TYPES, CHARACTERISTICS, AND MOTIONS Cam-and-follower mechanisms, as linkages, can be divided into two basic groups: 1. Planar cam mechanisms 2. Spatial cam mechanisms In a planar cam mechanism, all the points of the moving links describe paths in par- allel planes. In a spatial mechanism, that requirement is not fulfilled. The design of mechanisms in the two groups has much in common. Thus the fundamentals of pla- nar cam mechanism design can be easily applied to spatial cam mechanisms, which f Prepared while the author was Visiting Professor of Mechanical Engineering, Iowa State University, Ames, Iowa. FIGURE 40.1 (a) Planar cam mechanism of the internal-combustion-engine D-R-D-R type; (b) spatial cam mechanism of the 16-mm film projector R-D-R type. is not the case in linkages. Examples of planar and spatial mechanisms are depicted in Fig. 40.1. Planar cam systems may be classified in four ways: (1) according to the motion of the follower—reciprocating or oscillating; (2) in terms of the kind of follower sur- face in contact—for example, knife-edged, flat-faced, curved-shoe, or roller; (3) in terms of the follower motion—such as dwell-rise-dwell-return (D-R-D-R), dwell- rise-return (D-R-R), rise-return-rise (R-R-R), or rise-dwell-rise (R-D-R); and (4) in terms of the constraining of the follower—spring loading (Fig. 40.1«) or positive drive (Fig. 40.16). Plate cams acting with four different reciprocating followers are depicted in Fig. 40.2 and with oscillating followers in Fig. 40.3. Further classification of reciprocating followers distinguishes whether the cen- terline of the follower stem is radial, as in Fig. 40.2, or offset, as in Fig. 40.4. Flexibility of the actual cam systems requires, in addition to the operating speed, some data concerning the dynamic properties of components in order to find dis- crepancies between rigid and deformable systems. Such data can be obtained from dynamic models. Almost every actual cam system can, with certain simplifications, be modeled by a one-degree-of-freedom system, shown in Fig. 40.5, where m e FIGURE 40.2 Plate cams with reciprocating followers. FIGURE 40.3 Plate cams with oscillating followers. denotes an equivalent mass of the system, k e equals equivalent stiffness, and s and y denote, respectively, the input (coming from the shape of the cam profile) and the output of the system. The equivalent mass m e of the system can be calculated from the following equation, based on the assumption that the kinetic energy of that mass equals the kinetic energy of all the links of the mechanism: 1 ^? YHjV] 1 ^ I 1 W] m e = L—^-^L-^T Z = I A i = 1 * where ra, = mass of link i VI = linear velocity of center of mass of z'th link Ii = moment of inertia about center of mass for /th link co, = angular velocity of ith link 5 = input velocity The equivalent stiffness k e can be found from direct measurements of the actual system (after a known force is applied to the last link in the kinematic chain and the displacement of that link is measured), and/or by assuming that k e equals the actual stiffness of the most flexible link in the chain. In the latter case, k e can usually be cal- culated from data from the drawing, since the most flexible links usually have a sim- ple form (for example, a push rod in the automotive cam of Fig. 40.16c). In such a FIGURE 40.4 Plate cam with an offset recip- rocating roller follower. FIGURE 40.5 The one-degree-of-freedom cam system model. model, the natural frequency of the mass m e is co e = \/k e lm e and should be equal to the fundamental frequency co n of the actual system. The motion of the equivalent mass can be described by the differential equation m e y + k e (y -s) = Q (40.1) where y denotes acceleration of the mass m e . Velocity s and acceleration s at the input to the system are ds ds J0 , /A ^^ s = — = — — = s'co (40.2) dt dQ dt v ' and d , d?' , Jco d«' JG J=-T 5 CO= (0 + 5 -—- = a5 —- CO+ S OC Jr Jt Jf (40.3) = s"co 2 + s'cc where 6 = angular displacement of cam a = angular acceleration of cam s '= ds/ JG, the geometric velocity s "= ds 7 JG = J 2 ^JG 2 , the geometric acceleration When the cam operates at constant nominal speed co = CO 0 , Jco/Jf = oc = O and Eq. (40.3) simplifies to s = s"(*i (40.4) The same expressions can be used for the actual velocity y and for the actual accel- eration y at the output of the system. Therefore y = y'(Q (40.5) y = y"<i? + y'a (40.6) or y=y"a?o co = CO 0 = constant (40.7) Substituting Eq. (40.7) into Eq. (40.1) and dividing by k e gives ^ d y " +y = s (40.8) where [i d = (m e /A: e )cOo, the dynamic factor of the system. Tesar and Matthew [40.10] classify cam systems by values of (i rf , and their recom- mendations for the cam designers, depending on the value of JLI^, are as follows: [i d = 10~ 6 (for low-speed systems; assume s = y) [i d = 10" 4 (for medium-speed systems; use trigonometric, trapezoidal motion specifi- cations, and/or similar ones; synthesize cam at design speed co = CO 0 , use good manu- facturing practices and investigate distortion due to off-speed operations) (i rf = 10~ 2 (for high-speed systems; use polynomial motion specification and best available manufacturing techniques) FIGURE 40.6 Types of follower motion. In all the cases, increasing k e and reducing m e are recommended, because it reduces ji rf . There are two basic phases of the follower motion, rise and return. They can be combined in different ways, giving types of cams classifiable in terms of the type of follower motion, as in Fig. 40.6. For positive drives, the symmetric acceleration curves are to be recommended. For cam systems with spring restraint, it is advisable to use unsymmetric curves because they allow smaller springs. Acceleration curves of both the symmetric and unsymmetric types are depicted in Fig. 40.7. FIGURE 40.7 Acceleration diagrams: (a), (b) spring loading; (c), (d) positive drive. 40.2 BASICCAMMOTIONS Basic cam motions consist of two families: the trigonometric and the polynomial. 40.2.1 Trigonometric Family This family is of the form s "= C 0 + C 1 sin 00 + C 2 cos bQ (40.9) where C 0 , C 1 , a, and b are constants. For the low-speed systems where \i d < 10" 4 , we can construct all the necessary dia- grams, symmetric and unsymmetric, from just two curves: a sine curve and a cosine curve. Assuming that the total rise or return motion S 0 occurs for an angular displace- ment of the cam 0 = p 0 , we can partition acceleration curves into i separate segments, where / = 1,2,3, with subtended angles P 1 , p 2 , P 3 , so that P 1 + P 2 + P 3 + - = Po- The sum of partial lifts S 1 , S 2 , S 3 , in the separate segments should be equal to the total rise or return S 0 : ^i + S 2 + S 3 + — = SQ. If a dimensionless description 0/p of cam rotation is introduced into a segment, we will have the value of ratio 0/p equal to zero at the beginning of each segment and equal to unity at the end of each segment. All the separate segments of the acceleration curves can be described by equa- tions of the kind s"=Asin^ /1 = ^,1,2 (40.10) P 2 or s"= A cos^ (40.11) where A is the maximum or minimum value of the acceleration in the individual segment. The simplest case is when we have a positive drive with a symmetric acceleration curve (Fig. 40.7d). The complete rise motion can be described by a set of equations /0 1 2710 \ „ 27W 0 27C0 J=ffo fe - & sm "T J 5= ~F sm ~T (40.12) , S 0 ( - 2710 \ ,„ 4n 2 s 0 2710 ^Tl 1 - 008 !-) s = -p- cos T The last term is called geometric jerk (s' = coY"). Traditionally, this motion is called cycloidal The same equations can be used for the return motion of the follower. It is easy to prove that ^return ~ ^O ~~ ^rise $ return ~ ~$ rise (40.13) v' — -v' ?'"——?'" 1 ^ return 3 rise J return 1 ^ rise FIGURE 40.8 Trigonometric standard follower motions (according to the equation of Table 40.1, for c = d = O). All the other acceleration curves, symmetric and unsymmetric, can be constructed from just four trigonometric standard follower motions. They are denoted further by the numbers 1 through 4 (Fig. 40.8). These are displayed in Table 40.1. Equations in Table 40.1 can be used to represent the different segments of a fol- lower's displacement diagram. Derivatives of displacement diagrams for the adja- cent segments should match each other; thus several requirements must be met in order to splice them together to form the motion specification for a complete cam. Motions 1 through 4 have the following applications: Motion 1 is for the initial part of a rise motion. Motion 2 is for the end and/or the middle part of a rise motion and the initial part of a return motion. The value c is a constant, equal to zero only in application to the end part of a rise motion. Motion 3 is for the end part of a rise motion and/or the initial or middle part of a return motion. The value d is a constant, equal to zero only in application to the initial part of a return motion. Motion 4 is for the end part of a return motion. The procedure of matching the adjacent segments is best understood through examples. Example 1. This is an extended version of Example 5-2 from Shigley and Uicker [40.8], p. 229. Determine the motion specifications of a plate cam with reciprocating TABLE 40.1 Standard Trigonometric Follower Motions Parameter Motion 1 Motion 2 Motion 3 Motion 4 5 j,/V0 . 7r0\ . vo e re (e i\ f e i *e\ T~ sm T S2 * m ™"* c ~n 53008 TF + ^U"? M 1 T-^ 11 T If \P\ PlJ 2/3 2 P2 2^ 3 \0 3 2/ V 0 4 TT #4/ 5' Sj/. 7T^\ 5 2 1T *0 C SjT . IfB d S 4 L *6\ A I 1 -""ft) 2fe C ° S 2ft + ^ -2^ Wn 2ft + ^ -id 1+C ° S £J 5* tr5, . IT^ S 2 ir 2 . *0 5 3 ir 2 TT^ TTS 4 . ir0 W- 11 A' ~^ sm % -^'^ ^r sin ^ J^" TT 2 ^ 1 TT^ J 2 * 3 TO Si** • *0 T 2 S* V0 1T COS ^ -8^ C ° S % 8l Sm 2^ 7T C ° S £ S ' /1 - O^ &• - - ^ 4 "Ur } 2^ A ft ^. (L-t\ 25, £ _M + ^ *"*U- / /J 1 ft 2ft + ft **«.;**. .» -2*1 __a! .» _af ,"-2^ ^ max ^j S min - ^ * ™" 4j g2 m " ~ /Sj FIGURE 40.9 Example 1: (a) displacement diagram, in; (b) geometric velocity diagram, in/rad; (c) geomet- ric acceleration diagram, in/rad 2 . follower and return spring for the following requirements: The speed of the cam is con- stant and equal to 150 r/min. Motion of the follower consists of six segments (Fig. 40.9): 1. Accelerated motion to s^ end = 25 in/s (0.635 m/s) 2. Motion with constant velocity 25 in/s, lasting for 1.25 in (0.03175 m) of rise 3. Decelerated motion (segments 1 to 3 describe rise of the follower) 4. Return motion 5. Return motion 6. Dwell, lasting for t > 0.085 s The total lift of the follower is 3 in (0.0762 m). Solution. Angular velocity CG = 15071730 = 15.708 radians per second (rad/s). The cam rotation for 1.25 in of rise is equal to p 2 = 1.25 mlS 2 = 1.25 in/1.592 in/rad = 0.785 rad - 45°, where si = 25/15.708 - 1.592 in/rad. The following decisions are quite arbitrary and depend on the designer: 1. Use motion 1; then S 1 = 0.5 in, < ax - 0.057C/P 2 ! - 0.5jc/(0.628) 2 - 4 in/rad 2 (0.1016 m/rad 2 ). s"^ d = 2(0.5)/pi; so P 1 - 1/1.592 = 0.628 rad, or 36°. 2. For the motion with constant velocity, S 2 -1.592 in/rad (0.4044 m/rad); S 2 = 1.25 in. 3. Motion type 2: S 3 = S 2 = 1.25 in, $3'^ = s 3 7i/(2p 3 ) = 1.592 in/rad; therefore p 3 = 1.257C/[2(1.592)] -1.233 rad = 71°, ^n = -(1.257i 2 )/[4(1.233) 2 ] = -2 in/rad 2 . (Points 1 through 3 describe the rise motion of the follower.) 4. Motion type 3:s4' init =s 4 7r 2 /(4p 2 ) = -2 in/rad 2 (the same value as that of s^), s£ end = -7tt4/(2p 4 ), S 4 + S 5 = 3 in. 5. Motion type 4: s 5 " max = 7W 5 /P 2 , s 5 'i nit = - s( end = -2s 5 /fi 5 . We have here the four unknowns p 4 , S 4 , p 5 , and S 5 . Assuming time I 6 = 0.85 s for the sixth segment (a dwell), we can find (3 6 = COf 6 = 15.708(0.08) = 1.2566 rad, or 72°. Therefore P 4 + P 5 = 136°, or 2.374 rad (Fig. 40.9). Three other equations are S 4 + S 5 = 3,s 4 n 2 /(4$) = 2, and 7cs 4 /(2p 4 ) = 2s 5 /p 5 . From these we can derive the quadratic equation in p 4 . 0.696Pi + 6.044p 4 -12 = Q Solving it, we find p 4 = 1.665 848 rad = 95.5° and p 5 = 40.5°. Since S 4 Is 5 = 4p 4 /(7ip 5 ) = 3.000 76, it is easy to find that S 5 = 0.75 in (0.019 05 m) and S 4 = 2.25 in (0.057 15 m). Maximum geometric acceleration for the fifth segment s 5 ' max = 4.7 in/rad 2 (0.0254 m/rad 2 ), and the border (matching) geometric velocity s 4>end = s^ = 2.12 in/rad (0.253 m/rad). Example 2. Now let us consider a cam mechanism with spring loading of the type D-R-D-R (Fig. 40.70). The rise part of the follower motion might be constructed of three segments (1,2, and 3) described by standard follower motions 1,2, and 3 (Fig. 40.8). The values of constants c and d in Table 40.1 are no longer zero and should be found from the boundary conditions. (They are zero only in the motion case R-R-D, shown in Fig. 40.Jb, where there is no dwell between the rise and return motions.) For a given motion specification for the rise motion, the total follower stroke S 0 , and the total angular displacement of the cam p 0 , we have eight unknowns: P 1 , Si, P 2 , $2, Ps> S 3 , and constants c and d. The requirements of matching the displacement derivatives will give us only six equations; thus two more must be added to get a unique solution. Two additional equations can be written on the basis of two arbi- trary decisions: 1. The maximum value of the acceleration in segment l,s" tmsa should be greater than that in segment 2 because of spring loading. So s" max = -as" min where s^'mm is the minimum value of the second-segment acceleration and a is any assumed num- ber, usually greater than 2. 2. The end part of the rise (segment 3), the purpose of which is to avoid a sudden drop in a negative accelerative curve, should have a smaller duration than the basic negative part (segment 2). Therefore we can assume any number b (greater than 5) and write p 2 = &p 3 .The following formulas were found after all eight equa- tions for the eight unknowns were solved simultaneously: R Po Q _« a ^~l + a + alb ^' 1 S(I + *) + * TC^ ,_ 4a Sl ~ S ° b 2 (n + 4a) + 4a(2a + l) " 2 ~ Sl K 4a Sa 2 53 " " 1 TtZ) 2 C ~ Sl nb 2 d = 2s 3 SQ = Si+ 52 +c + S 3 We can assume practical values for a and b (say a = 2, b = 10) and find from the above equations the set of all the parameters (as functions of S 0 and p 0 ) necessary to form