α. Figure 1(iii) shows the inertia force and


 Jeffrey Lloyd
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1 CHPTER DYNMIC ORCE NLYSIS Whe the ierti fores re osiere i the lysis of the mehism, the lysis is kow s ymi fore lysis. Now pplyig D lemert priiple oe my reue ymi system ito equivlet stti system use the tehiques use i stti fore lysis to stuy the system. Ierti fore ouple i = m m G α m G Ι g α igure 1: Illustrtio of ierti fore (i) trsltig oy (ii) ompou peulum, (iii) ierti fore ouple o ompou peulum. Cosier oy of mss m movig with elertio s show i figure 1(i). orig to D lemert Priiple, the oy e rought to equilirium positio y pplyig fore equl to i = m i iretio opposite to the iretio of elertio. igure 1 (ii) shows ompou peulum of mss m, momet of ierti I g out eter of mss G while rottig t its eter of mss hs lier elertio of ouple tig o the peulum. Equivlet offset Ierti fore gulr elertio of α. igure 1(iii) shows the ierti fore G I i i H G i i H G i m G H (i) (ii) igure : (i) Illustrtio of equivlet offset ierti fore igure (i) shows oy with ierti fore i ierti ouple I. The ouple e reple y two prllel fores (equl i mgitue opposite i iretio) tig t G H respetively s show i igure (ii). If we osier their mgitue of these fores sme s tht of ierti fore, the the equl (iii) (iv) 30
2 opposite fores t poit G will el eh other the resultig fore will e fore t H whih is i the sme iretio s ierti fore. If h is the miimum iste etwee the fore t G H, the where I I h = i re mgitue of I i i respetively. This fore tig t H is kow s equivlet offset ierti fore. or the ompou peulum show i igure 1(iii), the equivlet offset ierti fore is show i igure (iii). Dymi fore lysis of four r mehism Let us stuy the four r mehism where m, m m 3 4 re mss of lik,3 4 respetively. We hve to fi the torque require t lik for ymi equilirium whe exterl fore 4 ts o lik 4 s show i figure 3. Now for ymi fore lysis the followig steps my e followe. T 4 Exterl fore igure 1: our r mehism showig exterl ostrit fores Drw the elertio igrm or use y lytil metho to etermie elertio Determie gulr elertio of lik,3 4. Determie lier elertio of eter of mss ( gi i =,3, 4 ) of lik 3 4. The mgitue of ierti fore of lik i ( i =,3 or 4) e etermie y multiplyig mss of lik i with the orrespoig elertio of the eter of mss. The iretio of the ierti fore is opposite to the iretio of the elertio. Determie the mgitue of ierti ouple whih is equl to Iiα i The iretio of the ierti ouple is opposite to tht of gulr elertio. Reple the ierti fore ouple y the equivlet offset ierti fore for eh lik. Tret these offset ierti fore s exterl fore follow the proeure for stti fore lysis. Oe my use either superpositio priiple or priiple of virtul work to fi the require = m gi i gi torque for equilirium. 31
3 Dymi ore lysis of our r Mehism usig Mtrix Metho I the four r mehism show i igure 1, Lik 1 is the grou lik (sometimes lle the frme or fixe lik), is ssume to e motioless. Liks 4 eh rotte reltive to the grou lik out fixe pivots ( D). Lik 3 is lle the oupler lik, is the oly lik tht tre pths of ritrry shpe (euse it is ot rottig out fixe pivot). Usully oe of the "groue liks" (lik or 4) serves s the iput lik, whih is the lik whih my either e ture y h, or perhps rive y eletri motor or hyruli or peumti ylier. If lik is the iput lik, the lik 4 is lle the follower lik, euse its rottio merely follows the motio s etermie y the iput oupler lik motio. If lik is the iput lik its possile rge of motio is ulimite, it is lle rk, the likge is lle rkroker. Crkrokers re very useful euse the iput lik e rotte otiuously while poit o its oupler tres lose omplex urve. ig.1. Simple fourr likge. The ymi fore lysis prolem ws solve usig the mtrix metho y reuig it to oe requirig stti lysis. or this purpose, D lemert s Priiple whih sttes tht the ierti fores ouples, the exterl fores torques o the oy together give sttilly equilirium, ws osiere. The ierti fores g i s ierti momets Tg i s re give y, gi = (1) m i gi = α () Tgi Ii gi 3
4 where, m i is the mss of the lik i, I i is the momet of ierti out xis pssig through the etre of mss g i perpeiulr to ple of rottio of the lik i, gi α re the elertio gulr elertio of the etre of mss of the i th lik respetively. gi ig.. The reeoy igrms of () Lik (rk/iput lik) () Lik 3(oupler) Lik 4(follower lik) () Give positio, veloity, elertio, ierti properties suh s mss mss momet of ierti for eh movig lik of fourr likge, fore lysis for the likge e performe. rom the free oy igrms (ig..) three stti equilirium equtios, i terms of fores i the X Y iretios momet out the eter of grvity of the lik, e writte for eh lik. or lik, we get + + = (3) 1x 3x gx 0 m = (4) g 1y 3y gy 0 T r + ( r r ) + T =0 s g 1 g 3 g where, r = r exp ( i) is the positio vetor from joit to the eter of grvity of lik g g 1 3 g. re the joit fores tig o lik. (5) T re the ierti fore ierti momet of lik. m is the mss of lik Ts is the rivig torque. Similrly for lik 3, we get g 33
5 + + = (6) 3x 43x g3x 0 m = (7) 3g 3y 43y g3y 0 r + ( r r ) + T =0 g3 3 3 g3 43 g3 where, r = r exp ( i) is the positio vetor from joit to the eter of grvity of lik g3 g g3 m 3 3. re the joit fores tig o lik 3. momet of lik 3. is the mss of lik 3. Similrly for lik 4, we get (8) T 3 re the ierti fore ierti g + + = (9) 34x 14x g4x 0 m = (10) 4g 34y 14y g4y 0 r + ( r r ) + T + T =0 g g4 34 g4 1 where, r = r exp ( i) is the positio vetor from joit D to the eter of grvity of lik g4 g g 4 m4 T1 4. re the joit fores tig o lik 4. T g 4 re the ierti fore ierti momet of lik 4. is the mss of lik 4 is the torque of exterl lo. The equtios (5), (8), (11) e expresse s, (11) ( θ θ δ ) ( ) T r os + r si + r osθ r os s g 1y g 1x g 3y r si r os( + ) + T = 0 g 3x g ( θ θ δ ) ( ) r os + r si + r osθ r os g y g x 3 3 g y r si r os( + ) + T = g x g3 ( θ θ δ ) ( ) r os + r si + r osθ r os g y g x 4 4 g y r si r os( + ) + T = g x g4 (1) (13) (14) Here, it ws tke ito out tht ijx = jix ijy = jiy. Thus the equtios (311) e writte s ie lier equtios i terms of ie ukows. They e expresse i symoli form x = (15) x 1, 1, 3, 3, 34, 34, 14, 14, where, = the trspose of ( x y x y x y x yt s ) is vetor osistig of the ukow fores iput torque, = the trspose of (, m, T,, m, T, m, T + T ) is vetor tht otis gx gy g g g3x g3y 3g g3 g4x 4g g4 1 exterl lo plus ierti fores ierti torques. the mtrix whih is 9x9 mtrix, is fou to e 34
6 r g si r g os r siθ r g os r osθ + r g os r g 3 si 3 3 r g 3 os 3 3 r siθ 3 3 r g3 os 3 3 r osθ + r 3 3 g3 os r siθ + r 4 4 g 4 os 4 4 r osθ r g 3 os 4 4 r g 4 si 4 4 r g 4 os 4 4 Solutio proeure lrge umer of iputs re require from the user, viz. lik legth hs of the four liks, their msses, rius of gyrtio eprtures of the etre of mss from the lik positios (i.e. gleδ ), iput gle, iitil gulr veloity, gulr elertio the exterl lo torque. The proeure followe for solvig the ymi fore lysis usig the ove formulte mtrix metho is s follows: 1. Oe the iputs were tke, the lik legths were heke for fesiility.. Whe fou to e fesile, the other gles (θ ) were ompute. It ws fou tht two sets of solutios/oriettios re possile. 3. gulr veloities elertios were lulte for oth the possile oriettios, whih were use i the lultio of the ierti fores ouples. 4. fter this the mtries were ompute. Similrly, mtries C were ompute for the seo oriettio. 5. The solutio sets X Y { = the trspose of ( 1 x, 1 y, 3 x, 3 y, 34 x, 34 y, 14 x, 14 y, T s )}, vetors osistig of the ukow fores iput torque were otie y the followig formule, 1 X = (16) 1 Y = C (17) The MTL oe geerte for the simultio of the ove prolem is show elow. (Coe evelope y Pritish Rj Pri( ), s prt of ssigmet i ME308) %DYNMIC ORCE NLYSIS O OURR MECHNISM %TKING INPUTS ROM THE USER OR THE OURR MECHNISM = iput('eter the legth of the lik : '); = iput('eter the legth of the lik C : '); = iput('eter the legth of the lik CD : '); = iput('eter the legth of the lik D(fixe lik) : '); 35
7 m = iput('eter the mss of lik : '); m = iput('eter the mss of lik C : '); m = iput('eter the mss of lik CD : '); k = iput('eter the rius of gyrtio of lik : '); k = iput('eter the rius of gyrtio of lik C : '); k = iput('eter the rius of gyrtio of lik CD : '); rg = iput('eter the mgitue of the p.v. of the.g. of lik from the fixe pivot : '); rg = iput('eter the mgitue of the p.v. of the.g. of lik C from the pivot : '); rg = iput('eter the mgitue of the p.v. of the.g. of lik CD from the fixe pivot D : '); elt = iput('eter the evitio gle of the p.v. of the.g. of from the p.v. of pivot : '); elt = iput('eter the evitio gle of the p.v. of the.g. of C from the p.v. of pivot C(wrt ) : '); elt = iput('eter the evitio gle of the p.v. of the.g. of CD from the p.v. of pivot C(wrt D) : '); thet = iput('eter the iput gle (gle etwee D) i egrees : '); omeg = iput('eter the gulr veloity of lik : '); lph = iput('eter the gulr elertio of the lik : '); Tl = iput('eter the lo torque : '); %CONVERTING DEGREES TO RDINS ND CHECKING OR ESIILITY thet = *thet/180; K = ((*)  (*) + (*) + (*))/; P = K  (*()*os(thet))  (*) ; Q = ***si(thet); R = K  (*(+)*os(thet))  (*); flg=0; if ((Q*Q  4*P*R)<0) isp('wrog vlues of the lik legths'); flg=1; e %CLCULTION O OTHER NGLES while(flg==0) thet1 = *t( ((1*Q) + sqrt(q*q  4*P*R))/(*P)); thet = *t( ((1*Q)  sqrt(q*q  4*P*R))/(*P)); if(thet1<=0) thet1 = *t( ((1*Q) + sqrt(q*q  4*P*R))/(*P)) ; e if(thet<=0) thet = *t( ((1*Q)  sqrt(q*q  4*P*R))/(*P)) ; e thet1 = si( ((*si(thet1))  (*si(thet)))/) ; thet = si( ((*si(thet))  (*si(thet)))/) ; %CLCULTION O NGULR VELOCITIES omeg1 = (1**omeg*si(thet1  thet))/(*si(thet1  thet1)); omeg = (1**omeg*si(thet  thet))/(*si(thet  thet)); omeg1 = (1**omeg*si(thet1  thet))/(*si(thet1  thet1)); omeg = (1**omeg*si(thet  thet))/(*si(thet  thet)); %CLCULTION O NGULR CCELERTIONS 36
8 lph1 = ((1**lph*si(thet1  thet)) + (*omeg*omeg*os(thet1  thet)) + (*omeg1*omeg1*os(thet1  thet1))  (*omeg1*omeg1))/(*si(thet1  thet1)); lph = ((1**lph*si(thet  thet)) + (*omeg*omeg*os(thet  thet)) + (*omeg*omeg*os(thet  thet))  (*omeg*omeg))/(*si(thet  thet)); lph1 = ((1**lph*si(thet1  thet)) + (*omeg*omeg*os(thet1  thet)) + (*omeg1*omeg1)  (*omeg1*omeg1*os(thet1  thet1)))/(*si(thet1  thet1)); lph = ((1**lph*si(thet  thet)) + (*omeg*omeg*os(thet  thet)) + (*omeg*omeg)  (*omeg*omeg*os(thet  thet)))/(*si(thet  thet)); %CLCULTION O THE ELEMENTS O THE '' MTRIX 1 = 1*m*rg*(lph*os(thet + elt  ( /)) + omeg*omeg*os(thet + elt)); = m* m*rg*(lph*si(thet + elt  ( /)) + omeg*omeg*si(thet + elt)); 3 = m*k*k*lph; 4 = 1*m*rg*(lph1*os(thet1 + elt  ( /)) + omeg1*omeg1*os(thet1 + elt)); 5 = m* m*rg*(lph1*si(thet1 + elt  ( /)) + omeg1*omeg1*si(thet1 + elt)); 6 = m*k*k*lph1; 7 = 1*m*rg*(lph1*os(thet1 + elt  ( /)) + omeg1*omeg1*os(thet1 + elt)); 8 = m* m*rg*(lph1*si(thet1 + elt  ( /)) + omeg1*omeg1*si(thet1 + elt)); 9 = m*k*k*lph1  Tl; %CLCULTIONS O THE ELEMENTS O THE '' MTRIX 31 = rg*si(thet + elt); 3 = 1*rg*os(thet + elt); 33 = *si(thet)  rg*os(thet + elt); 34 = rg*os(thet + elt)  *os(thet); 39 = 1; 63 = rg*si(thet1 + elt); 64 = 1*rg*os(thet1 + elt); 65 = *si(thet1)  rg*os(thet1 + elt); 66 = rg*os(thet1 + elt)  *os(thet1); 95 = rg*os(thet1 + elt)  *si(thet1); 96 = *os(thet1)  rg*os(thet1 + elt); 97 = rg*si(thet1 + elt); 98 = 1*rg*os(thet1 + elt); = [ ]; = [ ; ; ; ; ; ; ; ; ]; %CLCULTION O THE IRST SOLUTION X = (iv())*'; thet11 = (t(x(,1)/x(1,1)))*180/ ; if(x(1,1)<0) thet11 = (t(x(,1)/x(1,1)) )*180/ ; e thet13 = (t(x(4,1)/x(3,1)))*180/ ; if(x(3,1)<0) thet13 = (t(x(4,1)/x(3,1)) )*180/ ; e 37
9 thet134 = (t(x(6,1)/x(5,1)))*180/ ; if(x(5,1)<0) thet134 = (t(x(6,1)/x(5,1)) )*180/ ; e thet114 = (t(x(8,1)/x(7,1)))*180/ ; if(x(7,1)<0) thet114 = (t(x(8,1)/x(7,1)) )*180/ ; e %CLCULTION O THE ELEMENTS O THE '' MTRIX 1 = 1*m*rg*(lph*os(thet + elt  ( /)) + omeg*omeg*os(thet + elt)); = m* m*rg*(lph*si(thet + elt  ( /)) + omeg*omeg*si(thet + elt)); 3 = m*k*k*lph; 4 = 1*m*rg*(lph*os(thet + elt  ( /)) + omeg*omeg*os(thet + elt)); 5 = m* m*rg*(lph*si(thet + elt  ( /)) + omeg*omeg*si(thet + elt)); 6 = m*k*k*lph; 7 = 1*m*rg*(lph*os(thet + elt  ( /)) + omeg*omeg*os(thet + elt)); 8 = m* m*rg*(lph*si(thet + elt  ( /)) + omeg*omeg*si(thet + elt)); 9 = m*k*k*lph  Tl; %CLCULTIONS O THE ELEMENTS O THE 'C' MTRIX C31 = rg*si(thet + elt); C3 = 1*rg*os(thet + elt); C33 = *si(thet)  rg*os(thet + elt); C34 = rg*os(thet + elt)  *os(thet); C39 = 1; C63 = rg*si(thet + elt); C64 = 1*rg*os(thet + elt); C65 = *si(thet)  rg*os(thet + elt); C66 = rg*os(thet + elt)  *os(thet); C95 = rg*os(thet + elt)  *si(thet); C96 = *os(thet)  rg*os(thet + elt); C97 = rg*si(thet + elt); C98 = 1*rg*os(thet + elt); D = [ ]; C = [ ; ;C31 C3 C33 C ; ; ;0 0 C63 C64 C65 C ; ; ; C95 C96 C97 C98 0]; %CLCULTION O THE SECOND SOLUTION Y = (iv(c))*d'; thet1 = (t(y(,1)/y(1,1)))*180/ ; if(y(1,1)<0) thet1 = (t(y(,1)/y(1,1)) )*180/ ; e thet3 = (t(y(4,1)/y(3,1)))*180/ ; if(y(3,1)<0) thet3 = (t(y(4,1)/y(3,1)) )*180/ ; e thet34 = (t(y(6,1)/y(5,1)))*180/ ; if(y(5,1)<0) thet34 = (t(y(6,1)/y(5,1)) )*180/ ; e thet14 = (t(y(8,1)/y(7,1)))*180/ ; if(y(7,1)<0) thet14 = (t(y(8,1)/y(7,1)) )*180/ ; e %DISPLY O RESULTS isp('there re two sets of solutios possile : '); 38
10 isp('set I : '); isp('x = [1x 1y 3x 3y 34x 34y 14x 14y Ts] ');isp(x); isp('1 = ');isp(sqrt(x(1,1)^ + X(,1)^)); isp('thet_1 = ');isp(thet11); isp('3 = ');isp(sqrt(x(3,1)^ + X(4,1)^)); isp('thet_3 = ');isp(thet13); isp('34 = ');isp(sqrt(x(5,1)^ + X(6,1)^)); isp('thet_34 = ');isp(thet134); isp('14 = ');isp(sqrt(x(7,1)^ + X(8,1)^)); isp('thet_14 = ');isp(thet114); isp('set II : '); isp('y = [1x 1y 3x 3y 34x 34y 14x 14y Ts] ');isp(y); isp('1 = ');isp(sqrt(y(1,1)^ + Y(,1)^)); isp('thet_1 = ');isp(thet1); isp('1 = ');isp(sqrt(y(3,1)^ + Y(4,1)^)); isp('thet_3 = ');isp(thet3); isp('1 = ');isp(sqrt(y(5,1)^ + Y(6,1)^)); isp('thet_34 = ');isp(thet34); isp('1 = ');isp(sqrt(y(7,1)^ + Y(8,1)^)); isp('thet_14 = ');isp(thet14); flg=flg+; e %END O CODE Dymi lysis of Slier Crk Mehism C D x β θ O x =isplemet of pisto from ier e eter = r[( + 1) ( os β + os θ )] = r + ( 1 osθ ) ( ) si θ If the oetig ro is very lrge The equtio overts to x = r(1 os θ ) is very lrge, hee si θ will pproh. 39
11 This is the expressio for SHM. Thus the pisto exeutes SHM whe oetig ro is lrge. Veloity of Pisto: x x θ v = =. t θ t si θ = rω siθ + si θ if is lrge ompre to si θ the si θ v = rω siθ + If si θ e eglete (whe is quite less) the v= rω siθ elertio of Pisto: v v θ = =. t θ t si θ os θ = rω siθ ω rω osθ θ + = + If is very lrge 0 Whe θ = 0 Whe 0 θ = 180 = rω osθ whih is SHM 1 i.e. t IDC = rω 1+ 1 i.e. t ODC = rω 1+ 0 t θ = 180 whe the iretio of motio is reverse = rω 1 1 gulr Veloity gulr elertio of Coetig ro: siθ si β = o 40
12 β osθ Differetitig = ω t os β osθ ω = ω = ω si θ osθ 1 si α = gulr elertio of the oetig ro ω ω θ ω θ t θ t = =. = si 1 ( si θ ) (3/) The egtive sig iites tht the sese of gulr elertio of the ro is suh tht it tes to reue the gle β. Net or Effetive fore o the Pisto: 1= re of the over e = re of the pisto e 1 P = pressure of the over e P = pressure of the pisto e m = mss of the reiprotig prts ore o the pisto p = P 1 1 P os θ Ierti fore = m= mrω osθ + Net or Effetive fore o the pisto = p Crk Effort: It is the et effort (fore) pplie t the rk pi perpeiulr to the rk, whih gives the require turig momet o the rkshft. = rk effort t = fore o the oetig ro s r = rsi( θ + β ) t t = si( θ + β) = si( θ +β) osβ θ 41
13 T = r t = si( θ + β) r os β r = (siθ os β + osθsi β ) os β 1 = r siθ + osθsi β os β si θ = r siθ + si θ lso The T m + m = m m= m = r t = r si( θ + β ) os β = (OD os β ) os β = (OD) r si( θ + β ) = OD os β Poit msses t two poits, if it is esure tht the two msses together hve the sme ymi properties. l G The two memer will e ymilly similr if (i) The sum of two msses is equl to the totl mss (ii) The omie etre of mss oiies with tht of the ro (iii) The momet of ierti of the two poit msses out perpeiulr xis through their omie etre of mss is equl to tht of the ro. 4
14 m + m = m m= m m = m + = m m + Let m e the totl mss of the ro oe of the msses e lote t the smll e. Let the seo mss e ple t D = mss t m m = mss t D l G G D Tke G = DG = The m + m = m m= m m = m m = m + + I = m + m = m + m + + = m = mk K = Iste of keepig the mss t D if we keep t the first two oitios e stisfie s follows. 43
15 The m + m = m m= m m = m m = m + + ut I = m, ssumig >, I > I So y osierig the two msses t iste of D, the ierti torque is irese from the tul vlue of T = Iα = mα ( + ) ( + ) [ ] = mα ( l L) This muh torque must e pplie to the two mss system i the opposite iretio to tht of gulr elertio to mke the system ymilly equivlet to tht of the tul ro. The orretio ouple will e proue y two equl, prllel, opposite fores y tig t the gugeo pi rk pi es perpeiulr to the lie of stroke. ore t is tke y the retio of guies. y l r y O Torque t the rk shft ue to fore t or orretio torque, 44
16 T = rosθ y T mα ( l L) osθ mα ( l L)osθ osθ = r os θ = = = m α ( l L) os os r si θ si β l β 1 θ lso ue to weight of mss t, torque is exerte o the rkshft whih is give y T = ( m g ) r os θ I se of vertil egies, torque is lso exerte o the rkshft ue to weight t e give y, si θ ( mgr ) siθ + si θ The et torque o the rk shft will e vetoril sum of the torquest, T, T T. Exmple1: The oetig ro of IC egie is 450 mm log hs mss of kg. The eter of the mss of the ro is 300 mm from the smll e its rius of gyrtio out xis through this eter is 175 mm. The mss of the pisto the gugeo pi is.5 kg the stroke is 300 mm. The ylier imeter is 115 mm. Determie the mgitue iretio of the torque pplie o the rkshft whe the rk is t 40 egree the pisto is movig wy from the iere eter uer effetive gs pressure of N/mm. The egie spee is 1000 rpm. Solutio: = l/r =450/150 = 3. l=300 mm G ω l=450 mm 0 40 r=150 mm O p Ierti fores ue to reiprotig msses: Divie the mss of the ro ito two ymilly equivlet prts 45
17 Mss of the rk pi, m = (m ) / l = 300/450 where m is mss of the ro = kg. Mss t the gugeo pi, m = = kg. Totl mss of the reiprotig prts, m= = kg osθ + os θ Ierti fore ue to reiprotig msses, = mrω = (π 1000/60) (os40+(os80)/3) = N. ore o the pisto, p = 10 6 π/4 (0.115) = N Net fore, = p Torque ue to this fore, T = r siθ + T =004.7 Nm Torque to osier the orretio ouple: α ω siθ ( 1 si θ ) = 3 / = r/s. L = + = +k / = mm. si θ si θ osθ T = mα ( l L) where m is mss of the ro si θ = Nm Torque ue to mss t : T = m gr osθ = Nm. Totl torque o the rkshft: Net torque o the rk shft =T T + T = = 01.1 Nm. (swer) Exmple : The oetig ro of vertil reiprotig egie is m log hs weight of 300 kg. The mss eter is 85 mm from ig e erig. Whe suspee s peulum from 46
18 the gugeo pi xis, it mkes 10 omplete osilltios i 5 se. Clulte rius of gyrtio of the oetig ro. The rk is 400 mm log rottes t 00 rpm. Whe the rk hs ture through 45 egree from the top e eter the pisto is movig owwrs, lytilly grphilly fi the torque tig t the rkshft. Solutio: = l/r =000/400 = 5. Ierti fores ue to reiprotig mss: Divie the mss of the ro ito two ymilly equivlet prts Mss of the rk pi, m = m l = 300 (0.85)/ (where m is mss of the ro) = kg. Mss t the gugeo pi, m = = kg. Ierti fore ue to reiprotig msses, = m osθ + os θ rω = (π 00/60) (os45+(os90)/5) = N. o 45 O l=m ω r=0.4 m Torque ue to this fore, T = r siθ + T = Nm si θ si θ Torque to osier the orretio ouple: We kow tht + = +k / =L = = m L t = π = 5/10 g L = m Now, k /1.175 =
19 α k = m. ω siθ ( 1 si θ ) = 3 / = r/s. osθ T = mα ( l L) where m is mss of the ro. si θ = Nm Torque ue to weight of mss t : T = m gr siθ = 489 Nm. Torque ue to weight of mss t : = m T gr siθ + = 39 Nm si θ si θ Totl ierti torque o the rkshft: Totl ierti torque o the rkshft = (1381.) = Nm. Grphil metho: Klei s ostrutio is show i the followig igrm. rom the igrm, go = 0.38 m, IY = m, IX =.095 m, IP =.611 m, IC =.7977 mm. We kow tht, C = m ω (go) = 300 (0.94) 0.38 = N W C = = 943 N Now, tke the momet of the fores out the poit I. T (IC) = C (IY) W C (IX) rom this we lulte the ukow T T = N Ierti torque, T = T r = = 5487 Nm. 48
20 Y N P x I W C C R Z G 0.4 D L E N Q C K T O g D L E N Q C K O g 49
21 Summery I this hpter the followig oepts re lere Ierti fores ouples Offset ierti fore Dymilly equivlet system Determitio of rius of gyrtio of rkshft y metho of osilltio Pisto effort, rk effort Dymi fore lysis of mehisms usig grphil metho, vetor metho virtul work priiple Exerise Prolems 1. Determie the torque require to rive lik of the four r mehism i the positio show i figure elow uer the tio of fores 3 with mgitues of 50 N 75 N, respetively. ore ts i the horizotl iretio 3 ts s show i the figure. Weight of liks, C CD re 5 N, 7.5 N 8 N respetively. =30 m, C=40 m, CD=50 m the fixe lik D=75 m CE=15 C= 0m. Crk is rottig t 100rpm i lokwise iretio, Use oth grphil lytil methos E C D. ssumig = 3 = 0, etermie the torque require to overome the ierti fores of the liks of the four r mehism i prolem 1 if () lik elerte t the rte of 5 r/se, () eelerte t the rte of 5 r/se. Use virtul work priiple, grphil vetor metho to solve the prolem. 3 The oetig ro of reiprotig ompressor is m log hs weight of 300 kg. The mss eter is 85 mm from ig e erig. The rius of gyrtio of the oetig ro is 0.7 m. The rk is 40 mm log rottes t 50 rpm. Whe the rk hs ture through 5 egree from the ier e eter the pisto is movig towrs left, lytilly grphilly fi the torque tig t the rkshft. Cosier mss of rk, oetig ro pisto to e 5 kg, 4 kg kg respetively. 4. igure elow shows mehism use to rush roks. Cosierig mss of lik, C CDE to e 0.8, respetively, i the positio show, etermie the torque require to rive the iput lik whe the rushig fore tig i the horizotl iretio is 5000N. Here, = 50 m, C=100 m, CD=10 m the fixe lik D=150 m CE=5 m 50
22 the gle CED of the terry lik CED is ssumig lik is rottig with 500 rpm i lokwise iretio, use () grphil metho, () lytil metho () virtul work priiple to etermie the erig fores power require y the motor. C E 5000 N 45 0 D 51
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