.1. i L. l. K4 I II ”W t \ l i W W at I \ fl 5% m DETERMINATION OF THE DYNAMiC CHARACTERISTECS OF MECHANSSMS USING ENERGY METHODS Thesis far fho Dagree cf M. S. MECHIGAN STATE COLLEGE Ivan E. Morse, Jr. 1954 -t- L" 4 A. ' ‘ ‘ 4..- .Qfi ,. . . d \ ‘ I; c :. ‘ h 1‘ 1 J' ‘ ‘ . '9. — ~ . ..-.,, ~. - ' h I . ,- 1.. - r: ' - o.- . m2" ‘~\ . ‘ ‘H w F1:_ ‘ . {1‘ f“ on '- um” . um .\ "7.71..“ "3-".qu 1;“ ~ y'" 4 ’1‘4 ~ . I .1 I". 3‘ ‘ ‘ ' ’l I I ‘ ; '. . _. I‘ 1‘. A . ‘ . .l ' V. . _' ‘duWi‘J d‘J&\d"I Lin-.1 hh‘ii‘i.n‘ui‘.égudh~;h"‘ 54 ‘1‘ .- l4 7 9.5.4.1} d r‘ V".."“'".‘ ‘ " " \ ‘ I '3 ‘7’" - at _ ‘n‘dil‘llr‘q‘ ' 1c \7 DI‘JI UKUILidILTL )3 '-Jf 1.11.: .J; uJLniq hall-liul 41...".I- Di 1‘4.) x . ' \ - W11 u -~.. - ~ ~ . \"S 4 . - .. ’1 V H. V ~“ ’ .v 4J‘If1‘3.‘ Duh.) !.4'-.) L . ! a—J“ LJA‘.‘\4 l .14 LJI‘Iti -’AJ' Even 5, horse Jr. Submitted to the School of ”undue .53 Studies of ..':'Lc‘:1.1.;;an State College of Agriculture and Apglied Science in partial fulfillment of the reqqirements for the degree of .7. “rY‘H ' [\I: S“‘i‘l“,C ‘\ . . L-‘JLLJJ. .4 b. .2 L44. 4.4 Department of mechanical Engineering ar l95h v‘ (D TABLE til“ CfliITailTS Acknowledgments Symbols and Sign Convention I. Introduction . . . . . . . . . II. Development of the finergy Lethod Equations A, Equivalent Moment of Inertia. . . m . filUch’ilBllt filass o o o o o o 0 III. Applications of the Energy nethod bquations A, Parallel Crank Mechanism , . . . B. Four-Ear mechanism . . . . . . C. Four-Bar mechanism Used as 3 Switch D. Slider Crank Mechanism. . . . . IV. Summary . . . . . . . . . . . Li St ‘Df liner'QflCeS . o o o o o o o o C’E‘H n.r"/\.r."\." 4' l-(n 1x14 I\.Vl.!'lu.{fl1’._lu‘n ‘15 The antrnr wishes tJ exoress his sincere it:l€Ci3thfl to Dr. nolland T. linfile, under whose supervision this thesis was undertaken, for extending his time and Knowledge toward the preparation of this thesis. The author is also indebted to rrofessor China-U Ip for his valuable suggestions lending to the development of this thesis. SYMFULS Kinetic Energg. (ft-Its) ,2, L“. o o 0 Input or Jutyut inergy. {ft—lbs) (I; T . . . Torque. (ft~lbs) F , . . Force. (lbs) I . . . Moment of Inertia. (slugoft2) A . . . mass. (slugs) CL) . . . Instantaneous Angular VelocitV. (rad/sec) a . . . Instantaneous Angular Acceleration. (rad/sec?) V . . . Instantaneous Linear Velocity. (ft/sec) A . . . Instantaneous Linear Acceleration. (ft/secz) 9, Q . . Angular Displacement. (radians) s . . . Linear D'sylacement.' (feet) t . . . Time. (seconds) G . . . Center of Gravity. SIGN CONVENTION Torques, angular diSplacements, angular velocities and angular accelerations are considered positive in a counterclockwise direction. Forces, linear displacements, linear velocities and linear ac- celerations are considered positive if they act in the direction of the positive X or Y axis. INTRJJJCTlJN The study of dynamic characteristics of mechanisms and machines has become more important with the advent of increased speeds in high- speed machinery. The designer of these mechanisms requires a method for accurately analyzing their dynamic characteristics. The purpose of this thesis is to develop and apply energy methods to the analysis of the dynamic characteristics of mechanisms. Two general types of problems which are involved in the design of a mechanism are: A. Detennination of tne dynamic forces resulting in a mechanism which has specified velocities and accelerations. B. Determination of the velocities and accelerations resulting from the application of Known forces. The problem of determining the forces required to produce Specified dynamic characteristics in a mechanism can be solved using vector polygon methods.1 This method, however, does not provide a direct approach to the determination of velocities and accelerations resulting from the application of known forces. The dynamic characteristics usually are analyzed for one motion cycle of the mechanism and the vector polygon method becomes lengthy and tedious. The required forces are determined for a sufficient number of successive phases of the lHam, C. W. and E. J. Crane. Mechanics of Machinery, 3rd ed., McGrawa Hill Book Company, Inc., New lorE,'I9h8,'§§8fipp. 2. mechanism to yield a curve of the driving force versus the mechanism phase. The mechanism phase is either designated by an angular dis— placement of one of the linxs of the mechanism or by a linear displace- ment of some point in toe mechanism which has translation, A driving force, which will vary in the manner determined by the vector polygon method, may be impractical to produce. n designer has at his disposal certain devices which can be adapted to driving mechanisms. Springs, air and hydraulic cylinders, and solenoids are three of these devices. All of these devices have known force-displacement relations. Placing these limitations on the available driving sources, it becomes necessary to analyze the dynamic characteristics resulting from the application of known forces. This indicates tnat a method for solving type B problems would be desirable. Several attempts have been made toward this goal.2’3 However, the methods developed were not general in their application. Studying a mechanism from an energy viewpoint limits the analysis to scalar quantities and allows for easier analytical solutions. The energy in a mechanism is a function of the velocities. Therefore, a dynamic analysis of a mechanism would only require a velocity analysis to determine certain characteristics whi h can be applied in the energy method equations. The energy method solutions indicate two possible 'methods of representing and analyzing a mechanism. They are an equiva— lent moment of inertia or an equivalent mass system. Either of these 2Quinn, B. B. Energy Method for Determining Dynamic Characteristics of Mechanisms, igurnal of Applied Mechanics (ASME Trans.) Vol. 71, l9h9, 283-288 _. “' 3VanSickle, R. C. and T. P. Goodman. Spring Actuated Linkage Analysis to Increase Speed, Product Engineering, Vol. 2h, No. 7, 1953, 152-157 - ‘ ' " .' . ‘.;l-3 7 systems 15 adaptable to tne solution 0. either t,, viously mentioned. r0 DE‘JBLOrAiififlT OF TVS 5215M}! ..IL“TH.)U EQUATLJTIS In the analysis that follows, strain energy, potential energy and bearing friction will be neglected. Therefore, the only energy that will be considered is the input, output and kinetic energy of the mechanism. Equivalent Moment of Inertia The kinetic energy of a rigid body having plane motion is equal to the kinetic energy due to rotation about its center of gravity plus the kinetic energy due to translation of its center of gravity. Referring to Figure l, the kinetic energy of each of the links is: . .fl 2 . , . 2 , 2 11m 3 Kb. = 1/2 1g3w3 + 1/2 a, v (1) 3 £53 link h K81; 9- 1/2 lbw: .1 I2 and I are the moments of inertia of links 2 and b about their h respective centers of rotation. IE3 is the moment of inertia of link 3 about its center of gravity (G3). The subscript on the angular velocity symbols (6&1) indicate the respective links. Vg3 represents the instantaneous linear velocity of the center of gravity of link 3. The total kinetic energy of a system of rigid bodies is equal to the algebraic sum of the kinetic energy of each body in the system. Therefore, the total kinetic energy (Kt) of the mechanism for the con- figuration shown is: FIGURE I. A FOUR-BAR MECHANISM. g.- 5 3 3H "\ | ,a I 2 / \ l \ S I \ / \ r-t- \—/ \ zo: \ A '3'; l \ n 3!:- . l mo 1 | ‘ l 2‘“ 411- ANGULAR DISPLACEMENT- 6 FIGURE 2. AN EOUWALENT MOMENT OF "ERTIA CURVE. O\ at = K52 + K83 + nth or, “rid 1/21w2+l/2I w2+l’2“ V2 +l/216i)2 (2) a 1.1 t 2 2 w 3 / 333 ' u n This instantaneous kinetic energy is attributed to a variable mass which is rotating about the fixed point 02 at an instantaneous angular velocity equal to that of link 2; link 2 is the input link of he mechanism. It is apparent that the kinetic energy of this variable mass will be: 2 Kbeq = 1/2 quw2 (3) qu is the instantaneous moment of inertia of tne variable mass. Equating the two kinetic energy expressions and solving for the equiva— lent moment of inertia, the following expressions result: ‘I - v Kbeq KEt “m2 =Is+1 (w 7) +1£> +m2i d) qu g3 Equation h gives a relation between the moments of inertia of each link in the mechanism and the equivalent moment of inertia, when the kinetic energy of the mechanism is referred to link 2. A similar ex- pression will result if a different reference is chosen. A more general expression for the equivalent moment of inertia for any mechanism, regardless of its complexity is: K - n 2 I - (ms). (t ) eq wz :5 a r K=2 where (6&2r) is tne angular velocity of the reference link and (n) is the number of links in tne mechanism. In Equation b there are ratios of tie angular velocities of the links of the mechanism which can be determined by a velocity analysis. The angular velocity ratios are a function of the lengths of the links and the angular positions of the links. Therefore, it is not necessary to know a Specific value for the angular velocity of a link. The analysis is usually carried out assuming a constant angular velocity for the input or reference link. After this velocity ratio analysis has been completed it is possible to construct a curve of the equivalent moment of inertia versus the angular position of the reference link, using Equation h. A typical equivalent moment of inertia curve is shown in Figure 2. Some of the simple mechanisms will be adaptable to complete anal- ytical solution by writing the angular velocity ratios as functions of the crank angle (9), then substituting in Equation h to obtain an anal- ytical expression for the equivalent moment of inertia. The mechanism is supplied with energy in the form of an input torque (T1), at link 2, and energy is removed in the form of an output torque (T0) at link h. The instantaneous angular velocity of link 2, after link 2 has moved through an angular displacement ((39), can be found by applying the principle of dynamics that is stated below: "The work done on a system of particles by all of the external and internal forces in any displacement of the system is equal to the change in the kinetic energy of the system in the same displacement." ESeely, M. S. and N. E. Ensign. Analytical Mechanics for Engineers, 3rd ed., John Wiley and Sons, Inc., New York;*I9HB, p 299 E) J. The total work done, on the mechanism, will be equal to tne dif- ference between the input and output energies. neferrint to Fiqure 3, the input energy (Bi) will be: 5. = Ti d9 (5) From Figure h, the output energy (do) will be: a. so a To up (6) ”a t The limits used to evaluate the output energy from Equation 6 must be compatible with the limits used to evaluate the input energy from Equation 5. Each set of limits should represent the same change in position of the links of the mechanism. To avoid errors in the choice of limits it is possible to reconstruct the output torque curve of Figure h to a curve of output torque versus the reference crank angle (9), since the angle (fl) is a function of the angle (9). However, in many cases the values of the limits, pa and flb, will be detennined which are compatible with the limits, Ga and 9b. To evaluate the energy quantities given by the Equations 5 and 6, it will be necessary to know the relation between the torques and their respective angles. if the torque relations are not adaptable to analytical expressions it will be necessary to construct the two torque curves and by graphical or num— erical means evaluate the energy. The difference between the input and output energy, or the net input energy, can be determined by constructing curves similar to those in Figure 5. The area, cdef, between the two .W I.» ' 5 :3 : I % I 7% 'a a :m—“l 91 Assault msPuoeuan- e 5 FIGURE 3. AN INPUT TORQUE CURVE. .: I gW g I .: 3 . 2:3, I °1 / ’a $7 qt ANGULAR DISPLAOEIENT - FIGURE 4. AN OUTPUT TORQUE CURVE. I O t‘ L) INPUT TORQUE-T. EQUIVALENT OUTPUT TORQUE-T 93 .5 ANGULAR DISPLACEMENT- 9 FIGURE 5. INPUT AND EQUIVALENT OUTPUT TORQUE CURVES. INPUT TORQUE - Ta EQUIVALENT OUTPUT TORQUE-T ANGULAR OISPLAOENENT- 9 FIGURE 6. INPUT AND EQUIVALENT OUTPUT TORQUE CURVES. ll. torque curves within the diSplacement interval will represent the net input energy during that interval. This area can represent a negative input or loss in net input energy if more than one half of the total area indicated lies above the input torque curve. This is illustrated in Figure 6. The area, abc, represents a net gain of input energy while the area, cde, represents a net loss in input energy to the mechanism. I To construct the curve in Figure 5 for an equivalent output torque (To) versus angular displacement (9) it is necessary to use the following relations: 95 lab 1 To d9 . To d¢ 93 75a Differentiation with respect to 9 yields: l T = T 92 O 0 d9 OJ hilt, g2 3......9. d9 w, 1. f o T, - T wh ( ) .here ore. o - 0-127— 7 Equation 7 indicates that the value of the output torque (T0) at any position can be converted to an equivalent output torque with ref- erence to the input link, if it is multiplied by the angular velocity ratio of the output to the input link at that position. Subtraction of Equation 6 from Equation 5 and equating this result to the change in the kinetic energy of the equivalent system during the same displacement interval yields the following expressions: 12. 2 2 . _ = / _ / 1 ’2 T1 d9 TO d?) 1,2(qu)bb)b l, 2( eq)awa to) 93 la The subscripts on qu and C0’ indicate the angular positions of the equivalent system and the reference link of the actual mechanism. The equivalent system is rotating at the same angular velocity as the ref- erence link, link 2. Therefore,¢k) b will equal the instantaneous angular velocity of link 2 at the angular position G~. Solving Equation 8 for the angular velocity yields the following equation: 9b ¢b ' 2 3 (qu)b 9 £75 3. a If the net input energy supplied to the mechanism during a particular angular displacement of the reference link is equal to zero, Cr. ( i 2 EC), then the instantaneous angular velocity after this displace- ment will be given by the following relation: cub = wa _._SL(I€‘ )e (8b) (qu)b The terms (leq)a and (qu)b denote the value of the equivalent moment of inertia of the mechanism at positions Ga and 9b, respectively. If the reference link of the mechanism is rotating at an instantaneous angular velocity'(€u)a), at the angular position 9 and is allowed to a, rotate to position 9b without a change in the total energy of the 3. mechanism, the instantaneous angular velocity of the reference link will change in accordance with Equation 3b. This change in angular velocity of the reference link would indicate a tron fer of energy be- tween the links of the mechanism. To determine the angular acceleration (ab) of fhe reference link at the angular position 9%, equation Ea is differentiated once with re- spect to 9, and the relation, (1 =cud w, is used to obtain: d9 “I. _ Wit - b (Ughfl (lb “ " die a. A q u .3 , A ( d9 )b[§ TidQ - % Todd/5 + 1/2I10q)aw;] - a a (9) / 'T' 2 \ieq)b Substitution of Equation 8a in dquation 9 yields: “II mob - (Tub (mu (“2% d1 (Lb = - -—-~-—‘ (—33 )b (9a) (qu)b 2(qu)b (19 When performing the differentiation, it must be remembered that the terms (qu)b, Ti, TO and (Lib are functions of the angular position 9, while (qu)a and C~Ia have specific values as determined at position 93. In Equation 9a, tne tenns (Ti)b, (To)b’ (leq)b’ (Egga)b and (423%)b are determined for the angular position, 9b, of the reference link. The torques (Ti)b and (To)b are obtained from the input and output torque GU 1‘) CLJ2 b curves or analytical expressions. The angular velocity ratio ( 114. can be determined by graphical methods or from the data used to cor- struct the equivalent moment of inertia curve. The value of the equiv- HT 1 - - 1 ”‘ * E’C) . -. alent moment 01 inertia (leq)b and the slcpe (‘55’)b are obtained irom dIQC) the equivalent moment of inertia curve. The slope (—-;3) can be deter- d9 mined graphicall; for each particular position being analyzed or the equivalent moment of inertia curve can be graphically or numerically differentiated and a slope curve constructed to permit a complete analysis of the mechanism. The angular acceleration can also be obtained by graphical differ— entiation of the angular velocity versus the angular displacement curve, d<2J which is obtained from Equation 8a, and using the relationship, a =GU':;". 0 Both methods require one graphical differentiation and would introduce approximately the same error in the analysis. However, the equivalent moment of inertia curve has been constructed for use in obtaining the angular velocity from Equation 8a. it would therefore seem more desir- able to differentiate this curve anl use Equation 9a to obtain the angular acceleration. Rewriting Equation 9a yields: ‘01: (602) di b< dSQ)b + (aqua.b <9b> ‘ * [(1'in " (To)b(w2)b] a This equation is more convenient for finding the torques required to meet specified angular velocities and accelerations. To determine the time required for the linkage to move through a particular angular displacement, it is necessary to construct a curve . . . 1 . of the reciprocal oi the angular velOCity (227) versus the angular dis— placement. Equation 8a is used to compute the angular velocity. The 15:. area under this curve, within the displacement limits, will then rep— resent the time. This can be shown by the following equations: 00:93 dt 1 , or, dt=a7d9 l tb-ta=S-6b-'dg (10) Ga The curve of (zng versus 9 can be graphically integrated to obtain a time versus angular displacement curve. In some cases it would only be necessary to measure the total area under the curve to determine the time for a complete motion cycle of the reference link. Equivalent Mass The preceding outline offers a method for the complete dynamic analysis of a mechanism when the mechanism is reduced to a single rota- ting mass with a variable moment f inertia. A similar analysis results for an equivalent mass system if a reference point which has translation is chosen instead of a reference link which is rotating about a fixed point. Referring to the mechanism in Figure 7, the Kinetic energy for any configuration of the mechanism can be determined from the following expression: .. 2 2 . 2 .. ,2 at = 1/2 12602 + 1/2 ig3w3 + 1/2 u3vg3 + 1/2 mth (11) LINK 2 9 CEDO L— STROKE FIGURE 7. A SLIDER CRANK ME OHANISM. I ll 1 q Mao 3 EQUIVALENT MASS" II D O ww— LINEAR DISPLACEMENT- 8 FIGURE 8. AN EQUIVALENT MASS CURVE. 17. The instantaneous kinetic energy of a variable mass, Which is translating at an instantaneous linear velocity equal to that of the point ”P" in the actual mechanism, will be: ME = eq (1?) If the mechanism and the variable mass arc assumed to possess the same Kinetic energy at all times, then the mass of the equivalent system m‘st vary in order to maintain this energy balance. The ex- pression used to find the value of the equivalent mass of the variable mass body, for any configuration of tne mechanism, can be determined h; equating the two Kinetic energy eXpressions and solving for j .eq . ‘fi’2 1 («I3 a V 3 2 ... I""EBQ [I2 (-7];- + 133(_V-p-) + "13 (fl) + “hi (13) For a mechanism with "n" links and a point "P" chosen as the trans- lation reference point, the equivalent mass equation becomes: 4 = n M - 2 (mg) eq v2 K (13a) p 2<=2 The velocity ratios in Equation 13 vary figuration of the mechanism. Therefore, the several configurations which Will completely of the reference point. Then a curve of the linear displacement of the reference point is constructed. with each particular con- ratios are determined for represent one motion cycle equivalent mass versus the Figure 8 in— dicates a method for representing the data obtained from Equation 13 or 13a. 13. Because the reference point in this analysis has reciprocating motion, the velocity becomes zero at each end of its path. Similarly, the angular velocity for an oscillating cran< would become zero at the end of its path. heferrin: to Ehuation l2, this would indicate that the kinetic ene‘gy of the equivalent system becomes equal to zero, However, the kinetic energy of the actual mechanism is not nac;sserily equal to zero. Therefore, the equivalent mass must become infinite y large. The value of this equivalent mass will he indeterminate from hquation 13. It will therefore be necessary to construct the equivalent mass curve as accurately as possible up to the head end and crana end dead center CD positions, but not including these positions. As the curve in Figure indicates, there would be consideraole error introduced at the two end d Meg ds alent mass curve will be required for the analysis. This slope, in the positions. It will be shown later that the slope ( ) of the equiv- region at the ends of the path of motion, will be very susceptible to error and this error may be objectionable in the analysis, however, some mechanisms of this type will only be analyzed through a portion of their entire motion cycle and if this motion is restricted to the ac- curate portion of the equivalent mass curve, the equivalent mass analysis can be used. If a complete analysis of this mechanism is required it will be necessary to use the equivalent moment of inertia method; in which case, the input or output torque curve can be replaced by a force- displacemcnt relation similar to the one shown in Figure 9. This will not introduce any new problems to the analysis because the energy, either input or output, will be determinate from the force-displacement curve. ~ STROKE FORGE - F {gamm‘x J- a: HEDG g; g b LINEAR DISPLACEMENT - 5 FIGURE 9. A FORGE CURVE. 0 l9. 23. The linear velocity (Up) of the reference point in the mechanism of Figure 7 can be determined in a manner similar to that previously outlined for finding the angular velocity of tne reference link in the mechanism of Figure 1. Tue input energy supplied to the mechanism during a change in the configuration of the meghanism, assuming an in- Y" 4 put torque versus angular displacement relation as shown in rigure 3 is known, can be determined by the following equation: 9 c.) The output energy during the same configuration change can be de- termined if the output force-displacement relation is known. The curve in Figure 9 represents a general relation between the force acting on the piston and the linear displacement of the piston. The output energy is obtained from the following equation: E = F ds (15) ‘where the limits, 3a and sb, are compatible with the limits, 9a and 9b, respectively. The force-displacement curve of Figure 9 can be changed to an equivalent output torque (Tg) versus angular diSplacement curve if the value of the force at a particular configuration is multiplied by the ratio of the linear velocity of the piston to the angular velocity of the input link at that position. This relation can be shown by the follow— ing equations: 9b 5b I 1 d’ = h 1 o ~.9 (S s 9a a re H V t . . . . . . ‘ wnere (To) is toe equivalent output torque function acting on lan 2. Differentiation witn respect to 9 yields: rds p__ O (14-) but, 25 = ’4 J : therefore, T' = F (5%.) 2 The net input (16) nergy supplied to the mechanism during a particular displacement interval can be detentined by subtracting Equation 15 from Equation lb. The following exyressions result if the net input energy supplied to the mecoanism is equatel to the change in the Kinetic energy of the equivalent system during the some disolacement interval, d. - f = ' — ’5 l uo Ktob A a 9b Sb . 2 . . 2 Ti d9 - F ds = l/2(;eq)b(vp)b - l/2(Aeq)a(\lp)a 9a 8a Solving Equation 17 for the instantaneous linear velocity of the 3ields the following equation: F % Ti d9- _ 9a 58 _J ‘Sb W )2 2 (V = . p b (Ineq)b 2 F ds + 1/2(:.-:req)a(vp)a (1?) piston (17a) 22. Equation 173 can be usci to compute the instantaneous linear velocity of the reference point in the mechanism after a particular clim We in tEEB confi,niratinri of tanznectmniism, The irstantaneous linear velocity of the reference point, after particular dis lacemont durin_; which the not input energy suppliei to the mechanism is equal to zero (ii = 60), will be oLtained from tie following equation: (*"eq )3 up)b = (v (m) (L.qu)b Equation 17b inricates that if the mechanism was supplied with a certain amount of energy then allowed to coast, without a chanbe in the total energy of t e mechmnism, the velocity of the reference point will vary, This indicates that the kinetic energy of each link varies in order to maintain the overall energy level and conse uently energy must be transferred between the links of the mechanism. Differentiation of Equation 17a with respect to 5 ani use of the avg) 9%? relation, AP: V ), yields the following eXpression for the instan— taneous linear acceleration of the reference ‘HOllt: a) 9b Sb [ohm-ft, - (bub) (35%)}, é‘ri d9 45:1? as + 1/2<:..eq>a: A = (p)b (221 )b 2 eq (uéqh) (18) Substitution of Equation 17a in equation 18 yields. [(T > (v- p?) (F) 1 )2 b b b ('p b _ dl: n \ (Ap)b= .. - 2": ) (d e: Sjjb (183) (22-:q>b \weq b 23. ‘ 2 r: ,....~ a w \r‘ - (-7 .+‘ . , ~ ‘~.', fne terms (Heq)a and (Vp)a3 in blHdJlOfl l7a, axe not a function of the linear disglacement but have Specific values as deternined at (ea). All other tenns in eyuation l7a are a function of tne diSplace- ment. The subscript, outside the parentheses, on the terns in Equation mo l 18a indicate the position at which the functions are evaluated. ne ”“1 fbrme(F) is a function of the linear diSplacemcnt (5). inc torque (Ti) is a function of the cran: angle (9), and the equivalent mass (seq) is a function of the linear displacement. All of these functions are rep- resented hy their resyective graphs or curves. The instantaneous linear acceleration of the reference point can also be obtained by graphical differentiation of the linear velocity versus linear displacement curve. The velocity—displacenent curve can be constructed using fiquation 17a, The time required for the reference point to move through a particular linear displacement can be determined by constructing a 1 curve of the reelprocal of the linear veloc1ty (7-) versus the linear displacement. The area under this curve, within the diaplacement limits, will represent the time for that displacement. Tne equations for ob— taining the time are as follows: v =15. dt Sb 1 or, dt = '3 ds Sa Integration yields: Sb ds (1?) tb ' ta 3 a) is equal to zero and the angle 9 is equal to the angle fl, r-3 .I l) ft—lbs ......... T- d9 10 9 id ll \JI F“. (+- I P-S O c. (1.) ll \,1 (I) Substitution of the above relations in Jquation 83 yields the following general equation for the instantaneous angular veloci g of the reference link as a function of the angular displacement. w = (13.78 1/;- m. - rs — . 11" VI *; (‘4 1 i fie did—fil- {1.1 d J L-JA‘) A x . . .. -acenent \4) is measured in rallans. , r ‘V N I \ .' -I r- ‘rs " . ' r. -. 1r .r‘. . ‘- . ,7 ' l""’~' y‘ angulal VBlLClLQ Vferc twe angular disolncenent is olixn (- o o I Q ' -., ,‘ . ‘ v-‘i - f‘ .yir ’T“ I‘..,-y-< ‘ ., ,— lor :iré IQVJI.HLLNL'JL tie IKLJBTUHCC la n.. la: dLuMJ; silatiill for more than one revulution. (3.) The angular acceleration curve is constructel next equation 2. fiuation is simplified by the coalition tnat constant. (4)1, Ti " Tofu-5) a: qu 10-5 -—02 0.2178 23 rad/sec2 (any-- it) up in Figuro ll ‘will agpl; ‘15}ng l is (Dq " d The angular acceleration for this problem is constant. This can also be determined by differentiation of the equation for tne angular velocity. ca 6.78 “/5- (OZ C~> II E. U (D 2003—— =u6 d9 a L6/2 = 23 rad/sec2 (14.) cycle can be detennined in two ways. a. problem is substituted in Squation 10. The angular velocity equation determined for this The time required for the mechanism to complete one motion . ll’v» ‘ ° ‘. P ¥ : t I II ‘ I! l-‘ I ulli I I: . lll.‘ I a" . cud I1. "‘0!! . A u j \\ . 31. c+ II in. (l) M (I) t = m. = 0,7ll seconds cx . :3 b. Since qu is constant, the .echanism can be treated as a rigid body rotating about a fixed point. Therefore, the equations of motion for a rigid body apply. The rigid body starts from res and is subjected to a con— stant acceleration. (of w0+at (of 23t The angular velocity after one motion cycle is equal to l? rad/sec (from Figure ll). Therefore, t = 17/23 = 0.7h seconds A check on the results of this problem can be made using the vector polygon methods. 7.1" | ’3', V. . rr'. "" b.J.lI—,dl “seabed-bu; The four-bar mechanism shown in Fifure 2a has the following physical properties: Weight of linK 2 . . . . . . . . . . . . . . . . . . . . . 13 lbs 'Jeight of link 3 . . . . . . . . . . . . . . . . . . . . 3.22 lbs Weight of linK h . . . . . . . . . . . . . . . . . . . . . lo lbs Loments of inertia of linxs 2 and b about their respective centers of gravity . . . . . . . . . . . . . . . . . .0.0l2 slug—ft2 Moment of inertia of link 3 about its center of gravity . . . . . , . , , 0.320 slug-ft2 The mechanism starts from rest at a position corresponding to a = 30 and is subjected to a constant input torque of 13 ft—lbs. The torque acts in the direction indicated on the mechanism. The angular velocity and the annular acceleration curves, for link 2, are constructed in the following manner: (1.) Equation h is used to construct the equivalent moment of hwfifiacmwe, wk 2 3 2 n {ll-:3 2 la; ‘ [12 * lid-‘33) * IgB‘W‘g) + ““30? 1 The velocity ratios, in Equation h, are determined from a velocity vector polygon. The velocity polygon is drawn for a sufficient number of phases to construct the equivalent moment of inertia curve. A samp e irelocity polygon is shown in Figure 12b. The data obtained from the Paolygons and the information required to construct the curve are tab- Lllated in Table I. The equivalent moment of inertia versus the angular clisplacement curve is shown in Figure 13. (2.) The angular velocity curve is constructed using aquation 8a. 53filnce the mechanism starts from rest at Q = 0° and is subjected to a <3<>nstant input torque of 13 ft—lbs, the following conditions apply: LINK 2 _&___\_- FIGURE I2. 33. 6: LINK 3 LINK 4 A T; G. G: 9 4b _ - _ - ___°2n ___l SCALEu- 4"sl' (a). FOUR-BAR MECHANISM °v (b), VELOCITY POLYGON F OUR - BAR MECHANISM AND VELOCITY POLYGON. O m .1 .-~ r ‘1 .' {‘ng L’AHJ I ,1« v. ‘\. ' j . 1‘ 'Jbfii'b 31kt AuJA‘I‘IE F. 111 d.djll\;1l .‘J‘thlrr 1": 311.1571 "1‘ CF I‘: All? 1 1:). Dl‘tvl‘llk E? Jll. f-IlSII. _- 2 'F'"‘" * I.) w 11 £03 £2 IGQ 2 degrees a.) 2 2 2 slug—ft 0 1.000 1.000 0.308 0.0923 7 1/2 0.610 0.825 0.199 0.0608 15 0.238 0.633 0.192 0.0009 22 1/2 0.0S< 0.070 0.263 0.0029 30 .281 0.300 0.333 0.0070 37 1/2 0.030 0.250 0.389 0.0538 05 0.532 0.187 0.020 0.0590 60 0.660 0.130 0.067 0.0672 75 0.725 0.050 0.092 0.0722 90 0.750 0.008 0.500 0.0700 105 0.750 0.021 0.500 0.0700 120 0.725 0.059 0.090 0.0720 135 0.682 0.100 0.075 0.0687 150 0.600 0.162 0.005 0.0631 165 0.082 0.201 0.000 0.0559 133 0.338 0.330 0.309 0.0090 195 0.185 0.020 0.299 0.0009 210 0.050 0.080 0.258 0.0028 225 0.055 0.510 0.206 0.0028 200 0.150 0.513 0.250 0.0037 255 0.209 0.078 0.253 0.0060 270 0.350 0.008 0.333 0.0095 285 0.075 0.292 0.391 0.0560 300 0.662 0.100 0.068 0.0670 307 1/2 0.782 0.038 0.508 0.0760 315 0.913 0.200 0.550 0.0888 322 1/2 1.075 0.396 0.588 0.1050 330 1.220 0.600 0.600 0.1213 337 1/2 1.355 0.830 0.600 0.1388 305 1.390 1.000 0.550 0.1020 352 1/2 1.265 1.060 0.005 0.1201 360 1.000 1.000 0.308 0.0923 in; u) * . . 2 V qu --= {.0311 + .0310 (—J’i) + .020(w--;&3—)2 + 06.3% 30. foul. ‘: -I ‘ o. . . l 7 .1: [little I» Ill. lII’J‘ Ti = l) ft—los ......... Ti d9 = 10 Q o T = O o wa=o Therefore, (4) = 20 0 qu Table II contains the results of computations based on the above equation. These datr are plotted in Figure lh to graphically represent the angular velocity af linx 2 versus the aujular diSplacement. (3.) The angular acceleration curve is constructed using Equation 9a. This equation is simplified by the conditions of the problem to obtain: 2 Ti - wzfiifzay a. = __ di“ 2 qu The construction of the angular acceleration curve, using the above equation, requires three intermediate steps. First, the value of the angular velocity must be read from the angular velocity curve for the number of positions required to adequately represent a complete motion cycle. Secondly, the equivalent moment of inertia must be tab- ulated for the same positions. Thirdly, the slope of the equivalent moment of inertia curve must be detenuined at each of these positions. 'Graphical methods for the de,ermination of the slepe are sufficiently accurate for use in most problems. The values of the slope in Table III Inere determined for the required positions. Table III also contains the 'information necessary to compute the angular acceleration. The angular nvv, ‘ P n ' .. r‘fi 11-.l'J‘.JL.’i JL‘AwJ‘Vll FIT" TAB L11) 1 I c) ‘(IllT‘fl -\ J yfi—h‘ 'J r); -4 ‘ “ ‘\ I 1 1.111111 FJi- 1‘ uu.n—L11.Ll ....J,. .‘L; 1 u 37. 9 9 2 ’3 9 1 eq (.0 2 CU degrees radians ft—lh—rad slug—ft2 (rad/sec)2 rad/sec 0 0 0 0.0923 0 0 15 0.262 5.20 0.0009 116.3 10.8 30 0.520 10.08 0.0070 221.0 10.9 60 1.005 20.90 3.0672 311.3 17.6 90 1.571 31.00 0.0700 025.0 20.6 120 2.095 01.90 0.0720 530.0 20.- 150 2.620 52.00 0.0631 830.0 28.8 180 3.102 62.80 0.0090 1273.0 35.6 210 3.670 73.00 0.0028 1718.0 01.5 200 0.190 83.80 0.0037 1918.0 03.7 270 0.710 90.20 0.0095 1910.0 03.6 300 5.200 100.80 0.0670 1550.0 39.0 315 5.500 110.00 0.0888 1200.0 35.2 330 5.760 115.2' 0.1210 950.0 30.8 305 6.020 120.00 0.1020 805.0 29.0 360 6.283 125.60 0.0923 1360.0 36.9 330 6.800 136.00 0.0070 2875.0 53.6 .201001.00~0.1.1>‘.111 20: F0 0-01; 1 11.15;. 9 (1)2 01 qu a degrees (rad/sec)2 slu*—ft2 slug-ft,2 rad/3e02 d‘1 10:18 O 0 —O.202 0.0923 108 15 117 -0. 05 3 0.1009 290 30 221 0.003 0.0070 100 60 311 0.020 0.0672 90 90 025 0.000 0.0700 135 120 580 —0.010 0.0720 179 150 830 —0.028 0.0631 302 180 1273 -0.020 0.0090 510 210 1718 -0.005 0.0020 330 200 1918 0.005 ‘0.0037 120 270 1910 0.020 0.0095 -130 300 1550 0.030 0.0670 -505 315 1200 0.110 0.0838 -685 330 950 0.130 0.1210 -000 305 805 -0.019 0.1020 127 360 1360 -0.202 0.0923 1890 curve is plotted in Figure 15. zero at tw0 different values 250 degrees and 305 degree n i ‘ Figure la, idesr 1 .11 anrular displacement values carresrfie 00 . The anfular acceleration is equal to 1. A r- ,. ,-‘ I . .‘ —, ‘ ~ \ rq‘r‘y 1‘ , .00 0030101 dlsyluLQmaflt. 13‘ are ‘ferrin; to the angular velocitg curve, ,01 t7 points on the curve where tne 51000 is equal t0 zero, which is eXpec*ed. V I '1 y—+—4 «Pf—+- +— L—o-J' ;—+—¢‘~-Q— IO .+ T 2+ + 1 f 1 -. 1,._? .1 v a‘ ' \J l- '0‘, I Four—ear fiechnnism deed as 0 Switch \ The four-bar mechanism Shown in Figure 15 has the same pnysical properties as the mechanism in the yreceding problem. Tee mecnanism is considered to be an opening electrical switcn. The mechanism starts from rest at a position where e = 22 l/2°. it is subjected to an input torque that is represented graphically in Figure 17 and an output torque that is represented in Figure 13. The equivalent minent of inertia curve is the same curve as constructed for the preceding problem and is shown in Figure 13. The angular velocity curve is constructed for the displacement interval represented by 9 = 22 l/2° to 9 = 180°. Squation 8a is used to compute the angular velocity of link 2 at a series of angular dis- placements. Table IV contains the computed angular velocities. it also contains the data obtained from tne equivalent moment of inertia curve and that data obtained from the two torque curves. These data are plotted in Figure 19. FIGURE l6. FOUR - BAR SCALE=- 3"- I' MECHANISM USED AS A SWITCH. 03. INPUT TORQUE-Ti (FT-LBS) OUTPUT TORQUE-To (FT-Les) —--- O 22 l/Z 90 ANGULAR DISPLACEMENT- 6 (DEGREES) FIGURE l7. INPUT TORQUE CURVE. W2 I80 ANGULAR DISPLAcEMENT-43 (DEGREES) FIGURE 51’ ll! OUTPUT TO ROUE CURVE . I48 5000011 VJLQCITI 01;. 0.» 0301-030 0' “010. 0390 15 “ qvch. 0.-.... “ . 9 fii d9 fio dfl ieq Cd" 0) degrees ft-lbs ft-lbs slug-ft2 (rad/sec)2 rad/sec 0 0 0 __ __ __ 22 1/2 0 0 0.0029 0 0 30 1.20 0.175 0.0070 05.0 6.70 37 1/2 2.33 0.393 0.0538 71.8 8.06 05 3.26 0.707 0.0590 86.7 9.30 52 1/2 0.07 1.090 0.0632 90.5 9.70 60 0.72 1.080 0.0672 96.0 3.80 67 1/2 3.25 1.965 0.0700 90.0 9.69 75 5.62 2.000 0.0722 89.3 9.05 82 1/2 5.83 2.830 0.0738 80.0 8.95 90 5.90 3.360 0.0703 60.3 8.39 105 5.90 3.363 0.0700 68.8 8.39 120 5.90 3.360 0.0720 70.5 8.00 135 5.90 3.360 0.06:7 70.0 8.60 150 5.90 3.360 0.0631 80.5 8.95 155 3.90 3.363 0.0559 91.0 9.55 180 5.90 3.360 0.0090 103.0 10.10 xuz. 'I-I o. x o- Slider Crank Mechanism Hr- The slider crank mec1anism shown in Figure 2D has the {allowing phjsical praperties: Weight of 110K 2 . . . . . . . . . . . . . . . . . . 3,2” lee 'I‘N'eitrzht Oi‘ link 3 O D l O n o o o o a o o o a I I O a 0 30:3,“ :1 0 Weight of the piston . . . . . . . . . . . . . . . . . 3.22 lbs Moment of inertia of link 2 about its center of gravity . . . . . . . . . . . 0.3280 slug-ft2 Moment of inertia of linx 3 about 2 its center of gravity . . . . . . . . . . . 0.0203 slugqft The mechanism starts from rest at a position corresoonding to 9 = 0°. The piston is subjected to an input force which decreases linearly from the head-end dead center position. The force-displacement re— lation is expressed analytically as: Fi = 103 - 200 s where (s) is the linear diSplacement, in feet, of the giston. It is measured positively from the head-end dead center position. force is wupal to zero at the crank-end dead center position. The mechanism complc,es the retaining one—half cycle without energy being supplied or removed. The angular velocity curve for link 2 is constructed in the follow— ing manner: (1.) Equation 0 is used to construct the equivalent moment of inertia curve. For this mecnanism, the equation is rewritten in the following fonn: CL) 1 3 V. 2 _ V 2 1eq = [12 + Ig3(02)2 + M3953“) + RIM—2.9 l 2 ‘02 Table V contains the data obtainei.from a velocity analysis and also the computed data necessary to construct the equivalent moment of SCALE =- 4": l' FIGURE 20. SLIDER CRANK MECHANISM. T x J; Ll“. V EQUIVALENT ninsui 02 IflihTiP DATA Fan Stinrnztuuxng..00958550 V 37 9 ‘P '03 “”3 leg degrees a.) 2 w 2 co 2 slug-£152 0 0 0.125 0.250 3.0328 15 0.080 0.101 0.202 0.0338 30 0.150 0.175 0.217 0.0362 05 0.208 0.213 0.179 0.0392 60 0.206 0.238 0.127 0.0018 75 0.258 0.252 0.067 0.0031 90 0.250 0.250 3 0.0025 105 0.225 0.238 0.067 0.0008 120 0.190 0.213 0.127 0.0383 135 0.100 0.180 0.179 0.0357 150 0.098 0.155 0.217 0.0303 165 0.050 3.130 3.202 0.0332 180 0 0.125 0.250 0.0328 95 0.050 0.130 0.202 0.0332 210 0.093 3.155 0.217 0.0303 225 0.100 0.180 0.179 0.0357 200 0.190 0.213 0.127 0.0383 55 0.225 0.238 0.067 0.0008 270 0.250 0.250 0 0.0025 285 0.258 0.252 0.067 0.0031 300 0.206 0.238 0.127 0.0018 315 0.208 0.213 0.179 0.0392 330 0.150 0.175 0.217 0.0362 1 305 0.080 0.101 0.202 0.0338 360 0 0.125 0.250 0.0328 .3! w 2 'T v ‘1 = [.030 + .02ot-—2) + .l(182)2 + .1(l£LJ inertia curve. Tues: data are platted in Figure 21. For slider crank mechanisms, the eqiivalent moment of 'ne Lia curve is very nearly sinusoidal in nature and the portion of tie curve from 189 degrees to 369 degrees '5 a mirror image of the yoliion tr); 0 degrees to lBJ de- G.) .. \L‘ rn A gr (2.) For this mechanism, equation Ca is rewritten and singlified to obtain: 2 F ds where, F ds = E. Q i O and, 100 s (l - s) [13 II The above equation is used to compile the data in Table VI. These data are plotted in Figure 22 to represent the angular velocity of link 2 for one motion cycle. In this analysis, it is assumed that the input force on the piston can cause turning of the crank when the mechanism is at the head—end dead center phase. ( I .. '0. Ifu‘cl xuz. l" o. x o- Tal'r } -i ’ II I I- I” I II‘F. ill I Ell! Ahouuhtvstaina'aiTAIHHTSLnnaicaix -— TixlfiL-E Vi . .. . . ”1.". ll ”.2131“: 11.3.31 9 5i 16C,1 2 CO degrees ft lbs slug-ft2 (rail/sec)2 rad/sec 0 0 0.0328 0 o 15 0.99 0.0338 59 7.7 30 3.92 0.0362 217 1L,7 05 8.06 0.0392 012 20.3 60 12.65 0.0018 . 606 20.6 75 16.70 3.3031 775 27.8 90 20.18 (L0025 952 30.9 105 22.50 0.0008 1100 33.2 120 23.90 0.0383 1250 35.3 135 20.60 0.0357 1330 37.2 150 20.89 0.0303 1050 38.1 165 20.91 0.0332 1530 38.7 180 25.00 0.0328 1525 39.0 210 25.00 0.0303 1060 38.2 200 25.00 0.0383 1305 36.1 270 25.00 0.0025 1175 30.2 300 25.00 0.0018 1195 30.6 330 25.00 0.0362 1380 37.2 360 25.00 0.0328 1525 39.0 S tilifnril' The equations and metnods derived from the energy equations are general. They can he used forcietermining the dynamic characteristics of any mechanism. This is evident from the fact that eituer of the two general types of problems can be solved using these equations. The velocity polygon will suygly the information necessary to perform a complete analysis of a mecnanism. it will not be necessary to deter- mine accelerations which may be difficult in the case of complex mechanisms. The proposed methods consist of the following steps which lead to the comolete analysis of a mechanism subjected to Known forces. A. Construction of an equivalent moment of inertia or equivalent mass curve. 8. Construction of an angular velocity curve. . Construction of an angular acceleration curve. D. Construction of a time curve. Step A requires a preliminary operation involving a velocity—vector polygon. The remaining steps are gerfonnel using the equations and methods preposed. The energy method solutions indicate two methods of representing a mechanism for analysis. They are the equivalent moment of inertia and the equivalent mass systems. The equivalent moment :f inertia is the more general of the two systems. However, both systems are applic- able for the solution of the two types of general problems. 22. Trna erréryy inctriii :ni a;§.li:¥i t >‘noc uniis m; hin: cexigoir1riiznxi- vantages which limit the accuracy or the analysis. The effect of friction is cumulative, onfi the errwr may increase during the motion cycle. The determination of the velicit; ratios from the velocity polygon and the determination of the slope of the e11ivalent moment of inertia or anfiular velocity curves will introfixce errors. Such errors, however, are difficult to deternine analytically, Some of these errors can be minimized by careful apolication of the :raghical methods re- quired. LIST OF ittFl‘QiitiNCbS Ham, C. N. and a, J. Crane. Mechanics of Aachinery, 3rd ed., McGraw—Hill Book Company, lnc., New lork, I900, 538 pp, hinkle, nollcnd T. Kinematics of Lachines, lst ed., Brentice- Hall, Inc., New Bork, 1353, 231 pp, Quinn, 3 C, Energy netnod for Determining Dynamic Character- istics of iechanisms, Journal of Applied Mechanics, (ASMB Transactions) Vol. 71, 1909, 253-288. Seelr, h. S. and N. j. dnsign. Analytical Mechanics for Engineers, 3 d ed., John Wiley and Sons, Inc., New Tori, 1905, p 299, VanSickle, n. C. and T. P. Goodman. Spring Actuated ninkage Analysis to Increase Speed, Product Engineering, Vol. 20, No. 7, 1953, 152-157. a ROOM USE ONLY . HI. I" m ix) 1:, L4.) G :x 7" 1|