Computer Aided Analysis and design

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  1002 1002 Theory of Machines Theory of Machines 25 F F F F Fea ea ea ea eatur tur tur tur tures (Main) es (Main) es (Main) es (Main) es (Main) Computer Computer Computer Computer Computer 1. Introduction. 2. Computer Aided Analysis for Four Bar Mechanism Aided Aided Aided Aided Aided (Freudenstein’s Equation). 3. Programme for Four Bar Mechanism. Analysis and Analysis and Analysis and Analysis and Analysis and 4. Computer Aided Analysis for Slider Crank Mechanism. 6. Coupler Curves. 7. Synthesis of Mechanisms. Synthesis of Synthesis of Synthesis of Synthesis of Synthesis of 8. Classifications of Synthesis Problem. 9. Precision Points for Function Mechanisms Mechanisms Mechanisms Mechanisms Mechanisms Generation. 10. Angle Relationship for function Generation. 25.1. Introduction 11. Graphical Synthesis of Four Bar Mechanism. We have already discussed in chapters 7 and 8, the 12. Graphical Synthesis of Slider graphical methods to determine velocity and acceleration Crank Mechanism. analysis of a mechanism. It may be noted that graphical 13. Computer Aided (Analytical) method is only suitable for determining the velocity and Synthesis of Four Bar acceleration of the links in a mechanism for a single position Mechanism. of the crank. In order to determine the velocity and 15. Least square Technique. acceleration of the links in a mechanism for different 16. Programme using Least positions of the crank, we have to draw the velocity and Square Technique. acceleration diagrams for each position of the crank which 17. Computer Aided Synthesis of is inconvenient. In this chapter, we shall discuss the analytical Four Bar Mechanism With Coupler Point. expressions for the displacement, velocity and acceleration 18. Synthesis of Four Bar in terms of general parameters of a mechanism and Mechanism for Body calculations may be performed either by a desk calculator Guidance or digital computer. 19. Analytical Synthesis for slider Crank Mechanism 1002 Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1003 25.2. Computer Aided Analysis for Four Bar Mechanism (Freudenstein’s Equation) Consider a four bar mechanism ABCD, as shown in Fig. 25.1 (a), in which AB = a, BC = b, CD = c, and DA = d. The link AD is fixed and lies along X-axis. Let the links AB (input link), BC (coupler) and DC (output link) make angles θβ,andφ respectively along the X-axis or fixed link AD. (a) Four bar mechanism. (b) Components along X-axis and Y-axis. Fig. 25.1 The relation between the angles and link lengths may be developed by considering the links as vectors. The expressions for displacement, velocity and acceleration analysis are derived as discussed below : 1. Displacement analysis For equilibrium of the mechanism, the sum of the components along X-axis and along Y-axis must be equal to zero. First of all, taking the sum of the components along X-axis as shown in Fig. 25.1 (b), we have ab cosθ+ cosβ−c cosφ−d = 0 . . . (i) or bc cosβ= cosφ+d −a cosθ Squaring both sides 22 2 cos ( cos cos ) bc β= φ+d −a θ 22 2 22 cd cos φ+ + 2cd cosφ+a cos θ = −φ 2ac cos cosθ− 2a d cosθ . . . (ii) Now taking the sum of the components along Y-axis, we have ab sinθ+ sinβ−csinφ = 0 . . . (iii) or bc sinβ= sinφ−a sinθ Squaring both sides, 22 2 sin ( sin sin ) bc β= φ−a θ 22 22 sin sin 2 sin sin . . . (iv) =φca+θ−acφθ Adding equations (ii) and (iv), 2 22 2 22 2 2 22 (cos sin ) (cos sin ) 2 cos (cos sin ) bc β+ β = φ+ φ +d +cd φ+a θ+ θ −φ 2( accoscosθ+sinφsinθ)−2a dcosθ 1004 Theory of Machines 222 2 or bc=+d+ 2cd cosφ+−a 2ac(cosφcosθ+ sinφsinθ)− 2ad cosθ 22 2 2 or 2ac(cosφcosθ+φ sin sinθ)=− a b+c++ d 2c d cosφ− 2a d cosθ 22 2 2 ab−+c+d d d cosφcosθ+ sinφsinθ= +φ cos−θ cos . . . (v) 2ac a c 222 2 −++ dd abcd Let . . . (vi) ==kk;;and =k 12 3 ac 2ac Equation (v) may be written as cos cos sin sin cos cos φθ+ φ θ=kk φ− θ+k . . . (vii) 12 3 or cos (φ – θ) or cos(θ−φ) =kk cosφ− cosθ+k 12 3 The equation (vii) is known as Freudenstein’s equation. Since it is very difficult to determine the value of φ for the given value of , from θ equation (vii), therefore it is necessary to simplify this equation. From trigonometrical ratios, we know that 2 1t−φ an ( /2) 2tan(φ/2) cos and φ= sinφ= 2 2 1t+φ an ( /2) 1tan ( /2) +φ Substituting these values of sin φ and cosφ in equation (vii), 2 1−φ tan ( / 2) 2 tan(φ / 2) cos sin ×θ+ × θ 22 1t+φ an ( /2) 1t+φ an ( /2) 2 1tan ( /2) −φ =×kk− cosθ+k 123 2 1tan ( /2) +φ 2 cosθ− 1 tan (φ / 2)+ 2sinθ tan (φ / 2) 22 2 =kk 1− tan (φ− / 2) cosθ 1+ tan (φ+ / 2)k 1+ tan (φ / 2) 12 3 2 cosθ− cosθ tan (φ / 2)+ 2sinθ tan (φ / 2) 22 2 =− kk tan (φ / 2)−k cosθ−k cosθ tan (φ / 2)+k+k tan (φ/ 2) 11 2 2 3 3 Rearranging this equation, 22 2 2 −θ cos tan (φ / 2)+kk tan (φ+ / 2) cosθ tan (φ / 2)−k tan (φ / 2)+ 2sinθ tan (φ / 2) 12 3 =− cosθ+kk − cosθ+k 12 3 2 −φ tan ( / 2) cosθ−kk− cosθ+k + 2sinθtan (φ/ 2)−k−k+ cosθ(1+k )= 0 12 3 1 3 2 2 (1)cos tan /2(2sin)tan/2 (1)cos0 −kk θ+ −k φ+− θ φ+k+k− +k θ= 231 13 2 (By changing the sign) 2 or tan ( / 2) tan( / 2) 0 . . . (viii) AB φ+ φ +C= Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1005 Ak =− (1 ) cosθ+k+k where  . . . (ix) 231,  B =−2sinθ,and   Ck=+k− (1+k ) cosθ 13 2  Inner view of an aircraft engine. Note : This picture is given as additional information and is not a direct example of the current chapter. The equation (viii) is a quadratic equation in tan(φ / 2). Its two roots are 2 −±BB − 4AC tan (φ= / 2) 2A  2 −±BB − 4AC −1  2tan φ= or . . . (x) 2A   From this equation (x), we can find the position of output link CD (i.e. angle φ ) if the length of the links (i.e. a, b, c and d) and position of the input link AB (i.e. angle θ ) is known. If the relation between the position of input link AB (i.e. angle ) and the position of θ coupler link BC (i.e. angle β ) is required, then eliminate angle φ from the equations (i) and (iii). The equation (i) may be written as ca cosφ= cosθ+b cosβ−d ... (xi) Squaring both sides, 22 22 22 ca cos φ= cos θ+b cos β+ 2ab cosθ cosβ 2 + 2 cos 2 cos . . . (xii) da−θ d −bdβ Now equation (iii) may be written as ca sinφ= sinθ+bsinβ . . . (xiii) 1006 Theory of Machines Squaring both sides, 2 2 22 22 ... (xiv) ca sin φ= sin θ+b sin β+ 2ab sinθsinβ Adding equations (xii) and (xiv), 22 2 22 2 22 2 ca (cos φ+ sin θ) = (cos θ+ sin θ)+b (cos β+ sin β) 2 2 (cos cos sin sin ) 2 cos 2 cos +θ ab β+θβ+ d− a dθ− b dβ 22 2 or ca=+b+ 2ab(cosθcosβ+ sinθsinβ) 2 +−da 2d cosθ− 2bd cosβ 22 2 2 or 2ab(cosθcosβ+θ sin sinβ)=c−a−b−+ d 2ad cosθ+ 2bd cosβ 22 2 2 ca−−b−d d d cosθβ cos+ sinθsinβ= + cosθ+ cosβ . . . (xv) 2 ab b a 22 2 2 c−− abd− dd Let and = k . . . (xvi) ==kk;; 14 5 2ab ab Equation (xvi) may be written as ∴ cosθβ cos+ sinθsinβ=kk cosβ+ cosθ+k . . . (xvii) 14 5 From trigonometrical ratios, we know that 2 2tan( /2) 1t−β an ( /2) β cos sinβ= , and β= 2 2 1t+β an ( /2) 1t+β an ( /2) Substituting these values of sin β and cosβ in equation (xvii), 2  1−β tan ( / 2) 2 tan(β / 2) cosθ+ sinθ  22 1t+β an ( /2) 1t+an (β/2)   2  1t−β an ( /2) =+ kk cosθ+k  145 2  1t+β an ( /2)  2 cos 1 tan ( / 2) 2sin tan( / 2) θ− β + θ β 22 2   =−kk 1 tan (β / 2)+ cosθ 1+ tan (β / 2) ++ k 1tan (β/2) 14 5   2 cosθ− cosθ tan (β/ 2)+ 2sinθ tan(β/ 2) 22 = kk− tan (β / 2)+θ k cos+θ k cos tan (β / 2) 11 4 4 2 tan ( / 2) ++ kk β 55 22 2 2 cos tan ( / 2) tan ( / 2) cos tan ( / 2) tan ( / 2) −θ β +kk β− θ β−k β 14 5 +θ 2sin tan(β / 2)−kk− cosθ−k+ cosθ= 0 14 5 2 − tan (β/ 2)(kk +1)cosθ+−k + 2sinθtan(β/ 2)−(kk −1)cosθ++k = 0 451 451 Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1007 2 or (kk +θ 1) cos+−k tan (β / 2)+ (−2sinθ) tan(β/ 2)+(kk −θ 1) cos++k = 0 451 451 (By changing the sign) 2 or . . . (xviii) DE tan (β+ / 2) tan(β / 2)+F= 0 Dk=+(1)cosθ+k−k,  451  where E =−2sinθ, and ... (xix)   F= (k−θ 1)cos+k+k 451  The equation (xviii) is a quadratic equation in tan(β/ 2) . Its two roots are 2 −±EE − 4DF tan(β= / 2) 2D  2 4 −±EE −DF −1 or  . . . (xx) β= 2tan  2 D  From this equation (xx), we can find the position of coupler link BC (i.e. angle β ). Note: The angle may be obtained directly from equation (i) or (iii) after determining the angle φ. α 2. Velocity analysis Let ω = Angular velocity of the link AB=θ d / dt , 1 BC=β d / dt ω = Angular velocity of the link , and 2 ω = Angular velocity of the link CD=φ d / dt . 3 Differentiating equation (i) with respect to time, dd θβ dφ −θabc sin× −β sin× + sinφ× = 0 dt dt dt or −ωab sinθ− ω sinβ+ωc sinφ= 0 ... (xxi) 12 3 Again, differentiating equation (iii) with respect to time, dd θβ dφ ab cosθ× + cosβ× −c cosφ× = 0 dt dt dt or ab ωθ cos+ω cosβ−cω cosφ= 0 ... (xxii) 12 3 cosβ sinβ Multiplying the equation (xxi) by and equation (xxii) by , −ab ω sinθβ cos−ω sinββ cos+cω sinφβ cos = 0 ... (xxiii) 12 3 and ab ω cosθsinβ+ωβ cos sinβ−cωφ cos sinβ= 0 ... (xxiv) 12 3 Adding equations (xxiii) and (xxiv), ac ωβ sin(−θ)+ω sin (φ−β)= 0 13 −ω a sin(β−θ) 1 ... (xxv) ∴ ω= 3 csin(φ−β) 1008 Theory of Machines Again, multiplying the equation (xxi) by cosφ and equation (xxii) by sinφ , −ωab sinθcosφ− ω sinβcosφ+cω sinφcosφ= 0 ... (xxvi) 12 3 and ab ω cosθsinφ+ ω cosβsinφ−cω cosφsinφ = 0 ... (xxvii) 123 Adding equations (xxvi) and (xxvii), ab ωφ−θ+ωφ−β= sin( ) sin( ) 0 12 −ω a sin(φ−θ) 1 ... (xxviii) ∴ ω= 2 bsin(φ−β) From equations (xxv) and (xxviii), we can find ω and ω , if ab ,,c,θφ , ,β and ω are 3 2 1 known. 3. Acceleration analysis Let α = Angular acceleration of the link AB=ω d dt , / 1 1 α = Angular acceleration of the link BC=ω d / dt , and 2 2 α = Angular acceleration of the link CD=ω d / dt . 3 3 Differentiating equation (xxi) with respect to time, dd θβdd ωω  12 −ωab cosθ× + sinθ× − ω cosβ× + sinβ× 12   dt dt dt dt    dφ dω 3 +ω c cosφ× + sinφ× = 0 3  dt dt  dd  vdu ∵ ()uuvv =× +× . . .  dx dx dx  22 or −ωaa cosθ− sinθα −bb ω cosβ− sinβα 112 2 2 ... (xxix) +ωcc φ+ φα = cos sin 0 33 Again, differentiating equation (xxii) with respect to time, dd θβdd ωω   12 ab ω×−sinθ× + cosθ× + ω ×−sinβ× + cosβ× 12   dt dt dt dt   dφ dω  3 −ω c ×−sinφ× + cosφ× = 0 3  dt dt  22 or −ωaa sinθ+ cosθα −bω sinβ+b cosβα 112 2 2 sin cos 0 ... (xxx) +ωcc φ− φα = 33 cosφ Multiplying equation (xxix) by , and equation (xxx) by sin φ , 22 −aa ω cosθφ cos−α sinθφ cos−bω cosβφ cos 11 2 22 ... (xxxi) −αbc sinβcosφ+ ω cos φ+cα sinφcosφ= 0 23 3 Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1009 22 and −aa ω sinθφ sin+ α cosθφ sin −bω sinβφ sin 112 22 ... (xxxii) +αbc cosβsinφ+ω sin φ−cα cosφsinφ= 0 23 3 Adding equations (xxxi) and (xxxii), 2 −ωaa (cosφcosθ+ sinφsinθ)+ α (sinφcosθ− cosφsinθ) 11 2 −ωbb (cosφcosβ+ sinφsinβ)+ α (sinφcosβ− cosφsinβ) 22 22 2 +ω c (cos φ+ sin φ)= 0 3 22 2 −aω cos(φ−θ)+aα sin (φ−θ)−bbc ω cos(φ−β)+ α sin (φ−β) + ω = 0 11 2 2 3 22 2 −αaa sin (φ−θ)+ ω cos(φ−θ)+ωb cos(φ−β)−cω 11 2 3 ... (xxxiii) ∴ α= 2 bsin (φ−β) cosβ sin β Again multiplying equation (xxix) by and equation (xxx) by , 222 −aa ω cosθβ cos−α sinθβ cos−bω cosβ−bα sinβcosβ 11 2 2 2 ... (xxxiv) +ωcc cosφcosβ+ α sinφcosβ= 0 33 222 and −ωaa sinθsinβ+ α cosθsinβ−bω sin β+bα cosβsinβ 11 2 2 2 ... (xxxv) +cc ω sinφβ sin−α cosφβ sin = 0 33 Adding equations (xxxiv) and (xxxv), 2222 −ωaa (cosβcosθ+ sinβsinθ)+ α (sinβcosθ− cosβsinθ)−bω (cos β+ sin β) 112 2 +ωcc (cosφcosβ+sinφsinβ)+ α (sinφcosβ− cosφsinβ)= 0 33 222 −ωaa cos (β−θ)+ α sin (β−θ)−bω +cω cos(φ−β)+cα sin (φ−β)= 0 11 23 3 222 −αaa sin (β−θ)+ ω cos (β−θ)+bω −cω cos(φ−β) 11 23 ... (xxxvi) ∴ a = 3 sin ( ) c φ−β From equations (xxxiii) and (xxxvi), the angular acceleration of the links BC and CD (i.e. α and α ) may be determined. 2 3 25.3. Programme for Four Bar Mechanism The following is a programme in Fortran for determining the velocity and acceleration of the links in a four bar mechanism for different position of the crank. C PROGRAM TO FIND THE VELOCITY AND ACCELERATION IN A FOUR-BAR C MECHANISM DIMENSION PH (2), PHI (2), PP (2), BET (2), BT (2), VELC (2), VELB (2), ACCC (2), ACCB (2), C1 (2), C2 (2), C3 (2), C4 (2), B1 (2), B2 (2), B3 (2), B4 (2) READ (, ) A, B, C, D, VELA, ACCA, THETA PI = 4.0 ATAN (1.0) THET = 0 IHT = 180/THETA DTHET = PI/IHT 1010 Theory of Machines DO 10 J = 1, 2 IHT THET = (J – 1) DTHET AK = (A A – B B + C C + D D) 0.5) TH = THET 180/PI AA = AK – A (D – C) COS (THET) – (C D) BB = – 2.0 A C SIN (THET) CC = AK – A (D + C) COS (THET) + (C D) AB = BB 2 – 4 AA CC IF (AB . LT . 0) GO TO 10 PHH = SQRT (AB) PH (1) = – BB + PHH PH (2) = – BB – PHH DO 9 I = 1, 2 PHI (I) = ATAN (PH (I) 0.5/AA) 2 PP (I) = PHI (I) 180/PI BET (I) = ASIN ((C SIN (PHI (I)) – A SIN (THET)) / B) BT (I) = BET (I) 180/PI VELC (I) = A VELA SIN (BET (I) – THET) / (C SIN (BET (I) – PHI (I))) VELB (I) = (A VELA SIN (PHI (I) – THET) ) / (B SIN (BET (I) – PHI (I)))) C1 (I) = A ACCA SIN (BET (I) – THET) C2 (I) = A VELA 2 COS (BET (I) – THET) + B VELB (I) 2 C3 (I) = C VELC (I) 2 COS (PHI (I) – BET (I) ) C4 (I) = C SIN (BET (I) – PHI (I)) ACCC (I) = (C1 (I) – C2 (I) + C3 (I) ) / C4 (I) B1 (I) = A ACCA SIN (PHI (I) – THET ) B2 (I) = A VELA 2 COS (PHI (I) – THET ) B3 (I) = B VELB (I) 2 COS (PHI (I) – BET (I) ) – C VELC (I) 2 B4 (I) = B (SIN (BET (I) – PHI (I)))) 9 ACCB (I) = (B1 (I) – B2 (I) – B3 (I)) / B4 (I) IF (J . NE . 1) GO TO 8 WRITE (, 7) 7 FORMAT (4X,’ THET’, 4X,’ PHI’, 4X,’ BETA’, 4X,’ VELC’, 4X,’ VELB’, 4X,’ ACCC’, 4X,’ ACCB’) 8 WRITE (, 6) TH, PP (1), BT (1), VELC (1), VELB (1), ACCC (1), ACCB (1) 6 FORMAT (8F8 . 2) WRITE (, 5) PP (2), BT (2), VELC (2), VELB (2), ACCC (2), ACCB (2) 5 FORMAT (8X, 8F8 . 2) 10 CONTINUE STOP END The various input variables are A, B, C, D = Lengths of the links AB, BC, CD, and DA respectively in mm, THETA = Interval of the input angle in degrees, VELA = Angular Velocity of the input link AB in rad/s, and 2 ACCA = Angular acceleration of the input link in rad/s . The output variables are : THET = Angular displacement of the input link AB in degrees, PHI = Angular displacement of the output link DC in degrees, BETA = Angular displacement of the coupler link BC in degrees, VELC = Angular velocity of the output link DC in rad/s, VELB = Angular velocity of the coupler link BC in rad/s, 2 ACCC = Angular acceleration of the output link DC in rad/s , 2 ACCB = Angular acceleration of the coupler link BC in rad/s . Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1011 Example 25.1. ABCD is a four bar mechanism, with link AD fixed. The lengths of the links are AB = 300 mm; BC = 360 mm; CD = 360 mm and AD = 600 mm. 2 The crank AB has an angular velocity of 10 rad/s and an angular retardation of 30 rad/s , both anticlockwise. Find the angular displacements, velocities and accelerations of the links BC and CD, for an interval of 30° of the crank AB. Solution. Given input : A = 300, B = 360, C = 360, D = 600, VA = 10, ACCA = –30, THETA = 30 OUTPUT : THET PHI BETA VELC VELB ACCC ACCB .00 – 114.62 – 65.38 – 10.00 – 10.00 – 61.67 121.67 114.62 65.38 – 10.00 – 10.00 121.67 – 61.67 30.00 – 144.88 – 82.70 – 8.69 – .84 101.52 181.43 97.30 35.12 – .84 – 8.69 181.43 101.52 60.00 – 166.19 – 73.81 – 6.02 6.02 38.02 77.45 106.19 13.81 6.02 – 6.02 77.45 38.02 90.00 174.73 – 47.86 – 8.26 12.26 – 180.18 216.18 132.14 – 5.27 12.26 – 8.26 216.18 – 180.18 270.00 – 132.14 5.27 12.26 – 8.26 – 289.73 229.73 – 174.73 47.86 – 8.26 12.26 229.73 – 289.73 300.00 – 106.19 – 13.81 6.02 – 6.02 – 113.57 – 1.90 166.19 73.81 – 6.02 6.02 – 1.90 – 113.57 330.00 – 97.30 – 35.12 – .84 – 8.69 – 170.39 – 49.36 144.88 82.70 – 8.69 – .84 – 49.36 – 176.39 25.4. Computer Aided Analysis For Slider Crank Mechanism A slider crank mechanism is shown in Fig. 25.2 (a). The slider is attached to the connecting rod BC of length b. Let the crank AB of radius a rotates in anticlockwise direction with uniform (a) (b) Fig. 25.2 Slider crank mechanism. 2 angular velocity ω rad/s and an angular acceleration α rad/s . Let the crank makes an angle θ 1 1 with the X-axis and the slider reciprocates along a path parallel to the X-axis, i.e. at an eccentricity CD = e, as shown in Fig. 25.2 (a). 1012 Theory of Machines The expressions for displacement, velocity and acceleration analysis are derived as discussed below : 1. Displacement analysis For equilibrium of the mechanism, the sum of the components along X-axis and along Y- axis must be equal to zero. First of all, taking the sum of the components along X-axis, as shown in Fig. 25.2 (b), we have ab cosθ+ cos(−β)−x = 0 ... ( β in clockwise direction from X-axis is taken – ve) or bx cosβ= −a cosθ ... (i) Squaring both sides, 22 2 22 ... (ii) bx cos β= +a cos θ− 2xa cosθ Now taking the sum of components along Y-axis, we have be sin(−β)+ +a sinθ = 0 or −βbe sin+=a sinθ be sinβ= −a sinθ ... (iii) ∴ Squaring both sides, 22 2 22 sin sin 2 sin ... (iv) be β= +a θ−ea θ Adding equations (ii) and (iv), 22 2 2 2 22 2 (cos sin ) (cos sin ) 2 cos 2 sin b β+ β = x +e +a θ+ θ − xa θ− ea θ 22 2 2 bx=+e+a−2cxaosθ−2easinθ 2222 or xa +−(2 cosθ)x+a −b +e −2easinθ=0 2 or ... (v) xk++x k= 0 12 22 2 where ka =−2cosθ , and 2sin ... (vi) ka= −+− b e ea θ 1 2 The equation (v) is a quadratic equation in x. Its two roots are 2 −±kk − 4k 11 2 ... (vii) x = 2 From this expression, the output displacement x may be determined if the values of a, b, e and are known. The position of the connecting rod BC (i.e. angle β) is given by θ sinθ− ae sin (−β) = b ea−θ sin or sinβ= b ea−θ sin  −1 ... (viii) ∴ β= sin  b  Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1013 Note : When the slider lies on the X-axis, i.e. the line of stroke of the slider passes through the axis of rotation of the crank, then eccentricity, e = 0. In such a case, equations (vi) and (viii) may be written as 22 ka =−2cosθ , and ka=−b 1 2 −θ asin  −1 β= sin and  b  2. Velocity analysis Let ω = Angular velocity of the crank AB=θ d /, dt 1 ω = Angular velocity of the connecting rod BC=β d / dt , and 2 dx/. dt v = Linear velocity of the slider = S Differentiating equation (i) with respect to time, ddβθ x d ba ×−sinβ× = − ×−sinθ× dt dt dt dx or ... (ix) −ωab sinθ− ω sinβ− = 0 12 dt Again, differentiating equation (iii) with respect to time, dd βθ ba cosβ× = − cosθ× dt dt or ... (x) ab ωθ cos+ω cosβ= 0 12 cosβ Multiplying equation (ix) by and equation (x) by sinβ , dx ... (xi) −ωab sinθcosβ−ω sinβcosβ− × cosβ= 0 12 dt and ab ωθ cos sinβ+ω cosβsinβ= 0 ... (xii) 12 Adding equations (xi) and (xii), dx aωβ (sin cosθ− cosβsinθ)− × cosβ= 0 1 dt dx aωβ sin(−θ)= × cosβ 1 dt dx aωβ sin(−θ) 1 = ... (xiii) ∴ dt cosβ From this equation, the linear velocity of the slider (v ) may be determined. S The angular velocity of the connecting rod BC (i.e. ω ) may be determined from equa- 2 tion (x) and it is given by −ω a cosθ 1 ω= 2 ... (xiv) cos b β 1014 Theory of Machines 3. Acceleration analysis Let α = Angular acceleration of the crank AB=ω d / dt , 1 1 α = Angular acceleration of the connecting rod = dd ω/,t and 2 2 22 dx /dt a = Linear acceleration of the slider = S Differentiating equation (ix) with respect to time, 2 dd θβdd ωωdx   12 −ωab cosθ× + sinθ× − ω cosβ× + sinβ× − = 0 12   2 dt dt dt dt   dt 2 dx 22   −αab sinθ+ω cosθ− α sinβ+ω cosβ− = 0 ... (xv) 11 2 2 2   dt The chain-belt at the bottom of a bulldozer provides powerful grip, spreads weight and force on the ground, and allows to exert high force on the objects to be moved. Note : This picture is given as additional information and is not a direct example of the current chapter. Differentiating equation (x) with respect to time, dd θβdd ωω   12 ab ω×−sinθ× + cosθ× + ω ×−sinβ× + cosβ× = 0 12   dt dt dt dt   22   ab α cosθ−ω sinθ + α cosβ−ω sinβ = 0 ... (xvi) 11 2 2   cos β Multiplying equation (xv) by and equation (xvi) by sin β , 222   −αab sinθcosβ+ω cosθcosβ−α sinβcosβ+ω cosβ 11 2 2   2 dx −× cosβ= 0 ... (xvii) 2 dt 222   ab α cosθsinβ−ω sinθsinβ + α cosβsinβ−ω sin β = 0 and ... (xviii) 11 2 2   Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1015 Adding equations (xvii) and (xviii), 2  aα (sinβcosθ− cosβsinθ)−ω (cosβcosθ+βθ sin sin ) 11  2 dx 22 2 −ω b (cos β+ sin β)− ×cosβ= 0 2 2 dt 2 dx 22 sin ( ) cos ( ) cos 0 aa α β−θ − ω β−θ −bω − × β = 11 2 2 dt 2 22 dxaa α sin (β−θ)− ω cos (β−θ)−bω 11 2 ... (xix) ∴ = 2 cosβ dt From this equation, the linear acceleration of the slider (a ) may be determined. S The angular acceleration of the connecting rod BC (i.e. α ) may be determined from 2 equation (xvi) and it is given by, 22 ab (αθ cos−ω sinθ)−ω sinβ 11 2 α= ... (xx) 2 b cosβ 25.5. Programme for a Slider Crank Mechanism The following is a programme in Fortran to find the velocity and acceleration in a slider crank mechanism. c PROGRAM TO FIND THE VELOCITY AND ACCELERATION IN A SLIDER c CRANK MECHANISM READ (, ) A, B, E, VA, ACC, THA PI = 4 ATAN (1.) TH = 0 IH = 180/THA DTH = PI / IH DO 10 I = 1, 2 I H TH = (I – 1) DTH BET = ASIN (E – A SIN (TH) ) / B) VS = – A VA SIN (TH – BET) / (COS (BET) 1000) VB = – A VA COS (TH) / B COS (BET) AC1 = A ACC SIN (BET – TH) – B VB 2 AC2 = A VA 2 COS (BET – TH) ACS = (AC1 – AC2) / (COS (BET) 1000) AC3 = A ACC COS (TH) – A VA 2 SIN (TH) AC4 = B VB 2 SIN (BET) ACB = – (AC3 – AC4) / (B COS (BET) ) I F (i . EQ . 1) WRITE (, 9) 9 FORMAT (3X,’ TH’, 5X,’ BET’, 4X,’ VS,’ 4X,’ VB,’ 4X,’ ACS’, 4X,’ ACB’) 10 WRITE (, 8) TH 180 / P I , BET 180 / P I, VS, VB, ACS, ACB 8 FORMAT (6 F 8 . 2) STOP END 1016 Theory of Machines The input variables are : A, B, E = Length of crank AB (a), connecting rod BC (b) and offset (e) in mm, VA = Angular velocity of crank AB (input link) in rad/s, 2 ACC = Angular acceleration of the crank AB (input link) in rad/s , and THA = Interval of the input angle in degrees. The output variables are : THA = Angular displacement of the crank or input link AB in degrees, BET = Angular displacement of the connecting rod BC in degrees, VS = Linear velocity of the slider in m/s, VB = Angular velocity of the crank or input link AB in rad/s, 2 ACS = Linear acceleration of the slider in m/s , and 2 ACB = Angular acceleration of the crank or input link AB in rad/s . Example 25.2. In a slider crank mechanism, the crank AB = 200 mm and the connecting rod BC = 750 mm. The line of stroke of the slider is offset by a perpendicular distance of 50 mm. 2 If the crank rotates at an angular speed of 20 rad/s and angular acceleration of 10 rad/s , find at an interval of 30° of the crank, 1. the linear velocity and acceleration of the slider, and 2. the angular velocity and acceleration of the connecting rod. Solution. Given input : A = 200, B = 750, E = 50, VA = 20, ACC = 10, THA = 30 OUTPUT : T H B E T V S V B ACS AC B .00 3.82 .27 – 5.32 – 101.15 – .78 30.00 – 3.82 – 2.23 – 4.61 – 83.69 49.72 60.00 – 9.46 – 3.80 – 2.63 – 35.62 91.14 90.00 – 11.54 – 4.00 .00 14.33 108.87 120.00 – 9.46 – 3.13 2.63 44.71 93.85 150.00 – 3.82 – 1.77 4.61 55.11 54.35 180.00 3.82 – .27 5.32 58.58 4.56 210.00 11.54 1.29 4.53 62.42 – 47.90 240.00 17.31 2.84 2.55 57.93 – 93.34 270.00 19.47 4.00 .00 30.28 – 113.14 300.00 17.31 4.09 – 2.55 – 21.45 – 96.14 330.00 11.54 2.71 – 4.53 – 75.44 – 52.61 25.6. Coupler Curves It is often desired to have a mechanism to guide a point along a specified path. The path generated by a point on the coupler link is known as a coupler curve and the generating point is called a coupler point (also known as tracer point). The straight line mechanisms as discussed in chapter 9 (Art. 9.3) are the examples of the use of coupler curves. In this article, we shall discuss Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1017 the method of determining the co-ordinates of the cou- pler point in case of a four bar mechanism and a slider crank mechanism. 1. Four bar mechanism Consider a four bar mechanism ABCD with an offset coupler point E on the coupler link BC, as shown in Fig. 25.3. Let the point E makes an angle with BC α in the anticlockwise direction and its co-ordinates are E (x , y ). E E γ First of all, let us find the value of BD, and β . From right angled triangle BB D, 1 Fig. 25.3. Four bar mechainsm BB BB asin θ 11 tan γ= = = with a coupler point. B D AD−− AB d a cosθ 11  sin a θ − 1 or γ= tan  da−θ cos  22 22 2 and ( ) () ( ) () ( ) BD=+ BB B D=+ BB AD− AB 11 1 1 22 (sin ) ( cos ) =θad+−a θ 22 2 2 2 =θad sin++a cosθ− 2ad cosθ 22 2 2 (sin cos ) 2 cos =θad + θ+−adθ 22 =+ ad−2c adosθ Now in triangle DBC, 222 ()()( BD+− BC CD) cos(γ+β) = ... (cosine law of triangle) 2 BC × BD 22 2 fb+−c = 2bf 22 2  fb+−c −1 or γ+β= cos  2bf  22 2  fb+−c −1 ... (i) ∴ β= cos−γ  bf 2  Let us now find the co-ordinates x and y . From Fig. 25.3, we find that E E x== AEAB+BEA= B+BE ... ( BE = BE ) ∵ 12 1 E2 1 12 1 1 =θae cos+ cos(α+β) ... (ii) and y== EEEE+EEB= B+EE ... ( EE =BB ) ∵ 21 1 E2 21 1 1 1 =θae sin+ sin (α+β) ... (iii) From the above equations, the co-ordinates of the point E may be determined if a, e, , α θ and β are known. 1018 Theory of Machines 2. Slider crank mechanism Consider a slider crank mechanism with an offset coupler point E, as shown in Fig. 25.4. Let the point E makes an angle α with BC in the anticlockwise direction and its co-ordinates are E (x , y ). E E β First of all, let us find the angle . From right angled triangle BC C, 1 BC BB−θ B C a sin− e 11 11 1 sinβ= = = BC BC b ae sinθ−  −1 1 ... (iv) ∴ β= sin  b  Now xA==E AB+BE=AB+BB E1 1 11 1 2 =θae cos+ cos(α−β) ... (v) and y== EEEB+BEB= B+BE E1 12 2 1 2 =θae sin+ sin (α−β) ... (vi) Fig. 25.4 Slider crank mechanism with From the above equations, the co-ordinates of the coupler point. β point E may be determined, if a, b, e, e , θ , α and 1 are known. Note : When the slider lies on the X-axis, i.e. the line of stroke of the slider passes through the axis of rotation of the crank, then eccentricity e = 0. In such a case equation (iv) may be written as 1 asin θ  −1 β= sin  b  25.7. Synthesis of Mechanisms In the previous articles, we have discussed the computer-aided analysis of mechanisms, i.e. the determination of displacement, velocity and acceleration for the given proportions of the mechanism. The synthesis is the opposite of analysis. The synthesis of mechanism is the design or creation of a mechanism to produce a desired output motion for a given input motion. In other words, the synthesis of mechanism deals with the determination of proportions of a mechanism for the given input and output motion. We have already discussed the application of synthesis in designing a cam (Chapter 20) to give follower a known motion from the displacement diagram and in the determination of number of teeth on the members in a gear train (Chapter 13) to produce a desired velocity ratio. Roller conveyor. In the application of synthesis, to Note : This picture is given as additional information and is the design of a mechanism, the problem not a direct example of the current chapter. divides itself into the following three parts: Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1019 1. Type synthesis, i.e. the type of mechanism to be used, 2. Number synthesis, i.e. the number of links and the number of joints needed to produce the required motion, and 3. Dimensional synthesis, i.e. the proportions or lengths of the links necessary to satisfy the required motion characteristics. In designing a mechanism, one factor that must be kept in mind is that of the accuracy required of the mechanism. Sometimes, it is possible to design a mechanism that will theoretically generate a given motion. The difference between the desired motion and the actual motion produced is known as structural error. In addition to this, there are errors due to manufacture. The error resulting from tolerances in the length of links and bearing clearances is known as mechanical error. 25.8. Classifications of Synthesis Problem The problems in synthesis can be placed in one of the following three categories : 1. Function generation ; 2. Path generation ; and 3. Body guidance. These are discussed as follows : 1. Function generation. The major classification of the synthesis problems that arises in the design of links in a mechanism is a function generation. In designing a mechanism, the frequent requirement is that the output link should either rotate, oscillate or reciprocate according to a specified function of time or function of the motion of input link. This is known as function genera- tion. A simple example is that of designing a four bar mechanism to generate the function y = f (x). In this case, x represents the motion of the input link and the mechanism is to be designed so that the motion of the output link approximates the function y. Note : The common mechanism used for function generation is that of a cam and a follower in which the angular displacement of the follower is specified as a function of the angle of rotation of the cam. The synthesis problem is to find the shape of the cam surface for the given follower displacements. 2. Path generation. In a path generation, the mechanism is required to guide a point (called a tracer point or coupler point) along a path having a prescribed shape. The common requirements are that a portion of the path be a circular arc, elliptical or a straight line. 3. Body guidance. In body guidance, both the position of a point within a moving body and the angular displacement of the body are specified. The problem may be a simple translation or a combination of translation and rotation. 25.9. Precision Points for Function Generation In designing a mechanism to generate a particular function, it is usually impossible to accurately produce the function at more than a few points. The points at which the generated and desired functions agree are known as precision points or accuracy points and must be located so as to minimise the error generated between these points. The best spacing of the precision points, for the first trial, is called Chebychev spacing. According to Freudenstein and Sandor, the Chebychev spacing for n points in the range xx ≤≤x SF (i.e. when x varies between x and x ) is given by S F 11 π−(2 j1)  xx=+()x−(x−x)cos j SF F S ... (i)  22 2n  11 π−(2 j1)  =+() xx−×∆×xcos SF  22 2 n  1020 Theory of Machines where x = Precision points j ∆ x = Range in xx=−x , and FS j = 1, 2, ... n The subscripts and indicate start and finish positions respectively. S F The precision or accuracy points may be easily obtained by using the graphical method as discussed below. 1. Draw a circle of diameter equal to the range ∆=xx− x . FS 2. Inscribe a regular polygon having the number of sides equal to twice the number of precision points required, i.e. for three precision points, draw a regular hexagon inside the circle, as shown in Fig. 25.5. 3. Draw perpendiculars from each corner which intersect the diagonal of a circle at preci- sion points x , x , x . 1 2 3 Now for the range 13 ≤≤xx, =1;x =3, and SF ∆=xx− x= 31 − = 2 ∴ FS or radius of circle, rx =∆ /2= 2/2= 1 ∆ x 2 xx=+r=x+ =12 + = ∴ 2S S 22 ∆ x xx=−r cos30°=−x cos30° 12 2 2 2 =− 2 cos30°= 1.134 2 Fig. 25.5. Graphical method for ∆ x determining three precision xx=+r cos30°=x+ cos30° and 32 2 2 points. 2 =+ 2 cos30°= 2.866 2 25.10. Angle Relationships for Function Generation (a) Four bar mechanism. (b) Linear relationship between x and θ. Fig. 25.6 Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1021 Consider a four bar mechanism, as shown in Fig. 25.6 (a) arranged to generate a function y = f (x) over a limited range. Let the range in x is (x – x ) and the corresponding range in is θ F S () θ−θ () yy − φ() φ−φ . Similarly, let the range in y is ( and the corresponding range in is . FS FS FS The linear relationship between x and is shown in Fig. 25.6 (b). From the figure, we find θ that θ−θ FS θ=θ + () xx − ... (i) SS xx − FS Similarly, the linear relationship between y and φ may be written as φ−φ FS φ=φ + () yy − SS ... (ii) yy − FS An automatic filling and sealing machine. Note : This picture is given as additional information and is not a direct example of the current chapter. For n points in the range, the equation (i) and (ii) may be written as θ−θ ∆θ FS θ=θ + () − =θ + () − xx xx jjSSSS j − ∆ xx x FS φ−φ ∆φ FS φ=φ + () yy − =φ + () yy − jjSSSS j and yy − ∆ y FS where j = 1, 2, ... n, ∆=xx− x; ∆θ = θ − θ , FS FS ∆=yy− y ; and ∆φ = φ −φ FS FS

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