ME 5 - Machine Design I Spring Seester 009 Nae Lab. Div. FINAL EXAM. OPEN BOOK AND LOSED NOTES. Friday, May 8th, 009 Please use the blank paper for your solutions. Write on one side of the paper only. Any work that can not be followed will assued to be in error. Proble (5 Points). For the echanis in the position shown in Figure, the velocity acceleration of the input link are V = 5 j / s A 5 j s, = / respectively. The first second-order kineatic coefficients of link are θ =+ 0.577 rad / θ = 0.85 rad /, respectively. The geoetry, the free length spring rate of the spring, the daping constant of the daper, the asses second oents of ass about the ass centers of links are: OG GA AG GD R O K 0.5.0 0.5 0. 0.6 50 N/ 5 Ns/ 5kg 7kg kg- I G I G 9 kg- Gravity is acting vertically downward the effects of friction can be neglected. (i) Deterine the first-order kineatic coefficients for the linear spring the viscous daper. (ii) Deterine the equivalent ass of the echanis. (iii) Write the equation of otion for this echanis in sybolic for. (iv) Deterine the torque T acting on the crankshaft at O. Specify the agnitude the direction. Figure. A Planar Mechanis.
ME 5 - Machine Design I FINAL EXAM. Spring Seester 009. Nae Lab. Div. Proble (5 Points). Part I. The steel shaft shown in Figure is siply supported by two bearings at A D. Two gears are rigidly attached to the shaft at locations B. Gear at location B weighs 50 N, gear at location weighs 90 N, the weight of the shaft can be neglected. The known stiffness coefficients are 4 4 4 k =.5 0 N/, k = 5.0 0 N/, k = 4.0 0 N/. (i) Deterine the first critical speed of the shaft using: (a) the Dunkerley approxiation; (b) the Rayleigh-Ritz ethod. (ii) Deterine the first second critical speeds of the shaft using the exact solution. (iii) If gear is now placed at location B gear is placed at location, then deterine the first critical speed of the new syste using the Dunkerley approxiation. Figure. A Steel Shaft Siply Supported by Two Rolling Eleent Bearings. Part II. The first critical speeds of a rotating shaft with two ass disks, obtained fro three different atheatical techniques, are presented in the following table. ω ω ω 50 rad/s 5 rad/s 5 rad/s (i) Specify the values in this table which correspond to the first critical speed fro the exact solution, the Rayleigh-Ritz ethod, the Dunkerley approxiation. (ii) If the influence coefficients a 4 a 0 /N = = the asses of the two disks are specified as = = 0.0 kg, then use the Rayleigh-Ritz ethod to calculate the influence coefficient a. 4 (iii) If the influence coefficients a = a = 0 /N the asses of the two disks are identical then use the Dunkerley approxiation to calculate the ass of each disk.
ME 5 - Machine Design I FINAL EXAM. Spring Seester 009. Nae Lab. Div. Proble (5 Points). Part I. The shaft shown in Figure (a) is rotating counterclockwise with a constant angular velocity ω= 0 rad / s. Three ass particles with asses =. 5kg, = 7. 5kg, = 5. 0kg, are rigidly attached to the shaft as shown in the figure. The distances fro the shaft axis to the ass particles are R = 5. 0, R = 7. 5, R =. 5. The distances a = b = 50, c = e = f = 00, d = 00. The ass of the shaft can be neglected. (i) Deterine the inertial forces caused by the three rotating ass particles. (ii) Deterine the agnitudes of the bearing reaction forces at locations A B their angular orientations easured fro a rotating reference line (assued to be coincident with the X axis). (iii) Deterine the agnitude of the bearing reaction force at location A using the graphical technique. Figure (a). A Rotating Shaft with Three Mass Particles. Part II. The distributed ass syste shown in Figure (b) is rotating counterclockwise with a constant angular velocity ω= 50 rad / s. The forces acting fro the distributed ass syste (denoted as ) on the foundation (denoted as ) at bearings A B are (F ) A = 40 N (F ) B = 90 N, respectively. Deterine the agnitudes the locations of the correction asses to be reoved in the correction planes () () at the radius R = R = 5c. Figure (b). A Rotating Shaft with a Distributed Mass.
ME 5 - Machine Design I FINAL EXAM. Spring Seester 009. Nae Lab. Div. Proble 4 (5 Points). The crankshaft of the three-cylinder engine shown in Figure 4 is rotating counterclockwise with a constant angular velocity ω = 00 rad/s. The effective asses of the three pistons are = 5 kg, the lengths of the connecting rods are L= 75c, the length of the throw of the crank is R = 7.5c. (i) Deterine the X Y coponents of the priary shaking force on the crankshaft bearing. (ii) Deterine the agnitude(s) the location(s); i.e., the angle(s), of the correcting force(s) created by the correcting ass(es) which will balance the priary shaking force. (iii) Deterine the agnitude(s) the location(s) of the correcting ass(es) which will balance the priary shaking force if the radius of the correcting ass(es) is R = R = ρ = c. (iv) Show the agnitude direction of the priary shaking force vector on the figure below to the o right, which corresponds to the crank angle θ = 45. Also, show the location(s) of the correcting ass(es) on this figure. Figure 4. A Three-ylinder Engine. 4
Solution to Proble. Kineatic Analysis. The vectors for the echanis are shown in Figure. Figure. The vectors for the echanis. Fro the vectors shown in Figure, the vector loop equation (VLE) for the echanis is The X Y coponents of Equation () are?? I R R R 0 () = R cosθ R cosθ R cosθ = 0 (a) R sinθ R sinθ R sinθ = 0 (b) Differentiating Equations () with respect to the input position R gives R sinθ θ + R cosθ cosθ = 0 (a) + R cosθ θ + R sinθ sinθ = 0 (b) 5
Equations () can be written in atrix for as R sin θ + cos θ θ + cos θ Rcos sin R = sin + θ + θ + θ (4) The deterinant of the coefficient atrix in Equation (4) is DET = R sin θ R cos θ = R =. 5 (5) Using raer s rule, the first-order kineatic coefficients can be written as R cos θ + cos θ sin θ + sin θ θ = = DET R sin θ + cos θ + R cos θ + sin θ = = DET sin ( θ θ ) R R cos ( θ θ ) R (6a) (6b) Substituting the given data into these equations, the first-order kineatic coefficients are ( o o 0 90 ) sin θ = =+ 0. 577 rad / 5. o o ( ) (7a) R = cos 0 90 =+ 0. 500 / (7b) Differentiating Equation () with respect to the input position R gives R sinθθ R cosθθ R sinθθ + R cosθ = 0 (8a) + R cosθθ R sinθθ + R cosθθ + R sinθ = 0 (8b) Equations (8) can be written in atrix for as R sin θ + cos θ θ + Rcos θθ + R sin θθ Rcos sin R = + θ + θ + R sin θθ R cos θθ (9) Using raer s rule, the second-order kineatic coefficients can be written as R cos θθ + R sinθθ + cos θ sin θθ cosθθ + sin θ + θ R R R θ = = (0a) DET R 6
R sin θ R cos θ θ + R sinθ θ θ R θθ R θ θ R θ +R cos sin cos R = = (0b) DET R Substituting the known data into these equations, the second-order kineatic coefficients are + ( + 0. 500)( + 0. 577) θ = = 0. 85 rad / 5. (a) R =+ R θ =+ (. 50)( + 0.577 ) =+ 0. 500 / (b) The coordinates of the center of gravity of link, in the fixed X-Y reference frae, are X G = Rcos θ + RGcosθ =.99 YG = Rsinθ + RGsinθ = 0.60 (a) Differentiating Equation (a) with respect to the input position R, the first-order kineatic coefficients for the center of gravity of link are X G = R Gcosθ = 0 Y G = R Gsinθ = (b) Differentiating Equation (b) with respect to the input position R, the second-order kineatic coefficients for the center of gravity of link are X = Y = 0 (c) G 0 The coordinates of the center of gravity of link are XG = OG cosθ YG = OG sinθ (a) G that is X G = 0.5cos 0 = 0.4 YG = 0.5sin 0 = 0.5 (b) Differentiating Equation (a) with respect to the input position R, the first-order kineatic coefficients for the center of gravity of link are X G = OG sinθθ Y G = OG cosθθ (4a) that is X G = 0.5sin 0 (0.577) = 0.44 / Y G = 0.5cos0 (0.577) = 0.50 / (4b) Differentiating Equation (4a) with respect to the input position R, the second-order kineatic coefficients for the center of gravity of link are X = OG sinθ θ OG cosθ θ G Y = OG cosθ θ OG sinθθ (5a) G that is X G = 0.5sin 0( 0.85) 0.5cos 0 (0.577) Y G = 0.5cos 0 ( 0.85) 0.5sin 0 (0.577) (5b) 7
that is = X G 0.048 / (i) 4 Points. The length of the spring can be written as = (5c) YG 0.50 / that is RS = Rcosθ AG GD (a) R = 0. 75 0. 5 0. 0 = 0. 5 (b) S Therefore, the first-order kineatic coefficient of the spring is The length of the viscous daper is R = R = / (c) S Therefore, the first-order kineatic coefficient of the viscous daper is R = R R (a) R = R = / (b) (ii) 5 points. The equivalent ass of the echanis can be written as where the coefficient for link j is EQ = A (a) j= j j Gj Gj Gj j j A = ( x + y ) + I θ (b) Substituting the known values into Equation (a), the equivalent ass of the echanis is that is that is = + ( x + y ) + I θ (4a) EQ G G G 5 7[( 0.44) ( 0.50) ] 9( 0.577) EQ = + + + + + (4b) EQ = 5 + 7(0.007 + 0.065) + 9(0.9) =.908 kg (4c) (iii) 0 Points. The power equation for the echanis can be written as + ω = EQ + j + j Gj + S 0 S + j= j= P. V T. [ R B R ]R gy R K(R R )R R R R (5) This is a scalar equation in one unknown variable, that is, the agnitude the direction of the torque T acting on link. The external force acting at point P on link is given as P=+ P j the velocity of link is given as V = 5 j / s. 8
Therefore, the dot product is P. V =+ PR =+ 75R (6a) The dot product of the torque T the angular velocity of link can be written as T. ω =+ T ( θ R ) (6b) where the sign is positive because the unknown torque T is assued to be acting in the sae direction as the angular velocity ω. Therefore, Equation (6b) can be written as T. ω =+ 0577T. R (6c) Substituting Equations (6a) (6c) into Equation (5), cancelling the input velocity R, the equation of otion for the echanis can be written as EQ j j Gj S 0 S j= j= P + 0. 577T = R + B R + g y + K(R R )R + R R (7) (iv) 0 Points. The coefficient for link j can be written as B = ( x x + y y ) + I θ θ (8) j j Gj Gj Gj Gj G j j Substituting the known values into Equation (8), the coefficient for links is j that is B + B = 7[( 0.44)( 0.048) + (0.50)( 0.50)] + 9(0.577)( 0.85) (9a) B + B = 7[(0.0069) + ( 0.065)] + 9( 0.) = 4.609 kg/ (9b) The tie rate of change of the potential energy due to gravity is which can be written as du = j gy Gj R (0a) dt = j du = ( g y G + g y )R G (0b) dt Substituting the known values into Equation (0a) gives that is du = [5 (9. 8)( + ) + 7(9. 8)( + 0.50)]R [49 05 7.68]R =. + dt (a) du =+ 66. 8R dt (b) The tie rate of change in the potential energy stored in the spring can be written as du dt S = K(R R )R R (a) S 0 S 9
Substituting known values Equation (a) into Equation (9a) gives du dt S = 50(05. 06)R. = 5R (b) The energy dissipated over tie by the viscous daper can be written as dw dt = R R (a) Substituting the known values into Equation (a), the power dissipated by the daper is dw = (5)( ) R 5 R =+ dt (b) Substituting Equations (4c), (9b), (b), (b), (b) into Equation (7), the equation of otion for the echanis can be written as + 75 + 0. 577 T =. 908 R 4. 609R + 66. 8 5+ 5R (4) The change in the input length is positive which eans that the relationship between the tie rate of change of the length of vector R the velocity acceleration of the input link can be written as: R = V i j =+ 5/ s R = A i j = 5/ s (5) Substituting Equations (5) into Equation (4) gives + +. =.. + +. + + (6a) 75 0 577 T 908 ( 5) 4 609( 5) 66 8 5 5( 5) that is + 75 + 0. 577 T = 78. 60 5. 5 + 66. 8 5 + 75 (6b) Rearranging this equation, the torque can be written as T Therefore, the torque acting on link is 78. 60 5. 5 + 66. 8 5 + 75 75. 67 = = N (7a) 0577. 0577. T = 40. 7 N (7b) The negative sign indicates that the direction of the torque is in the opposite direction to the angular velocity of link. The angular velocity of link is ω =θ R = ( + 0. 577)(5) =+.885rad / s (8) The positive sign indicates that link is rotating counterclockwise (as link oves upward). Since the torque is opposing the otion then the torque on link ust be acting in the clockwise direction. 0
Solution to Proble. Part I. (ii) 6 Points. (a) Using the Dunkerley approxiation, the first critical speed of the shaft with the two gears can be written as ω = a + a The influence coefficients are the reciprocal of the spring coefficients; i.e., () a ii = /N k ii a jk = akj = /N k jk () Therefore, the influence coefficients are.5 x0 4x0 /N 5 a = =, 4 5.0 x0 5 a = =.5 x0 /N (a) 4 4.0 x0 x0 /N 5 a = a = = (b) 4 The asses of the two gears are 50 N 9.8 /s = = 5.9 kg Substituting Equations () (4) into Equation () gives 90 N = = 9.7 kg (4) 9.8 /s Therefore, the first critical speed of the shaft is = 4 0 5( 5.9) +.5 0 5( 9.7) = 84.085 0 5 ω (5a) ω = 4.49 rad / s (5b) Note that the Dunkerley approxiation is a lower bound to the first critical speed of the shaft. (b) Using the Rayleigh-Ritz approxiation, the first critical speed of the shaft with the two gears can be written as Wx + Wx ω = g [ ] Wx + Wx (6) The total deflections of the shaft at the locations of the two gears can be written as x = aw + aw x = aw + aw (7) Substituting the weights of the gears Equations () into Equations (7), the total deflections are x 7.8 x 0 = x 5.5 x 0 = (8)
Substituting Equations (8) the known values into Equation (6), the first critical speed of the shaft with the two gears can be written as or as ω [(50)(7.8 x 0 ) + (90)(5.5 x 0 )] = 9.8 [(50)(7.8 x 0 ) + (90)(5.5 x 0 ) ] (9a) 6. / s ω = (9b) 5 60.66 x 0 Therefore, the first critical speed of the shaft is ω = 7.6 rad/s (0) Note that the Rayleigh-Ritz approxiation is an upper bound to the first critical speed of the shaft. (iii) 5 Points. The exact solution for the first second critical speeds of the shaft can be written as, ω ω ( + ) ± ( + ) 4( ) = a a a a a a a a () Substituting Equations () (4) into Equation () gives or, ω ω 5 0 + 84.0850 x 0 ± (84.0850 65.0) x 0 = (a), 4.045 x 0 0.454 x 0 ω 5 5 =+ ± (b) ω Use the positive sign for the first critical speed the negative sign for the second critical speed. Therefore, the first second critical speeds of the shaft are ω = 7.4 rad/s ω = 9.8 rad/s () (iii) 5 Points. When the two gears are interchanged then the Dunkerley approxiation, see Equation (), can be written as new new = a + a (4) ω Note that the influence coefficients of the shaft do not change (even though the two gears were interchanged). Therefore, substituting these values into Equation (4) gives ω ( )( ) ( )( ) = 4 0 5 9.7 +.5 0 5 5.9 = 74.9 0 5 (5a) Therefore, the first critical speed of the shaft is ω = 6.5 rad / s (5b) Note that this answer is greater than the answer obtained in Equation (5b).
Part II. (i) 5 Points. The first critical speed fro the Rayleigh-Ritz approxiation is an upper bound, therefore, ω= 5 rad / s corresponds to the answer fro the Rayleigh-Ritz approxiation. The Dunkerley approxiation gives a lower liit to the first critical speed, therefore, ω= 50 rad / s corresponds to the answer fro the Dunkerley approxiation. The value of the first critical speed fro the exact ethod is ω= 5 rad / s. The answers are suarized in Table. ω ω ω 5 rad/s Exact 5 rad/s Rayleigh-Ritz 50 rad/s Dunkerley (ii) Points. The Rayleigh-Ritz equation can be written as Wx + Wx ω = g [ ] Wx + Wx Since the two ass disks have the sae then this equation can be written as x + x ω = g [ ] x + x The total deflections of the shaft at locations can be written as x = aw + aw x = aw + aw () Since the influence coefficients a = a a = a the ass disks = then the deflections x = x = x. Therefore, Equation () can be written as g ω = (4) x Substituting the known data Equation () into Equation (4) gives 5 Solving for the influence coefficient gives = 9.8 4 (0.0)(9.8)( 0 + a) 4 a.7 x 0 / N = (6) (iii) Points. Since the first critical speed the influence coefficients are given then the Dunkerley approxiation can be used to calculate the two ass disks; i.e., () () (5) = a + a ω Substituting the given inforation the first critical speed fro the table above gives Therefore, the ass is 50 0 0 4 4 = + (8) 4 0 = = 0.074 kg (9) 50 (7)
Solution to Proble. Part I. (i) 5 Points. The inertial forces caused by the three rotating ass particles are F F ( )( )( ) ( )( )( ) = = = (a) Rω.5 kg 0.05 0 rad/s 75 N F Rω 7.5 kg 0.075 0 rad/s 5.5 N = = = (b) ( )( )( ) Rω 5 kg 0.05 0 rad/s 5 N = = = (c) Equations () can be written in vector for as F = 75 N 60 = + 7.5 iˆ+ 64.95 ˆj N F = 5.5 N 0 = + 45.47 iˆ 6.5 ˆj N F = 5 N 0 =.65 iˆ.5 ˆj N (a) (b) (c) (ii) 4 Points. The su of the oents about the left h bearing at A is Fro the given figure, this equation can be written as M A = 0 () ˆ ˆ ˆ ( ) ( ) + ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) ( ) ( ) ( B ) 0.k.65i.50 j 0.k + 45.47i 6.5 j + 0.4 k + 7.5i + 64.95 j + 0.6k F = 0 (4) Solving this equation for the reaction force at bearing B gives F = 6.55iˆ.47 ˆj N (5a) Therefore, the agnitude the direction of the reaction force at bearing B is B F B = 48.89 N.6 (5b) The su of the oents about the right h bearing at bearing B is This equation can be written as M B = 0 (6) ( ˆ ) ( ) ( ˆ ) ( ˆ ˆ ) ( ˆ A ) ( ˆ ˆ ) ( ˆ ) ( ˆ ˆ ) 0.6 k F + 0.5 k.65i.50 j + 0.4k + 45.47i 6.5 j + 0.k 7.5i + 64.95 j = 0 (7) Solving this equation, the reaction force at bearing A is F = 4.77 iˆ+ 6.7 ˆj N (8a) A 4
Therefore, the agnitude the direction of the reaction force at bearing A is F = 5.55 N 65.80 (8b) A (iii) 5 Points. The inertial oents necessary to draw a oent polygon to solve for the force at bearing A are M M M ( )( )( )( ) ( )( )( )( ) ZRω 0..5 kg 0.05 0 rad/s 5 N = = = (9a) ZRω 0.4 7.5 kg 0.075 0 rad/s N = = = (9b) ( )( )( )( ) ZRω 0.5 5 kg 0.05 0 rad/s.5 N = = = (9c) These inertial oents can be drawn at the sae angle as the inertial forces as shown in Figure.. Figure.. Moent Polygon for the oent at bearing A. Scale c = N. Fro the oent polygon, shown in Figure., the inertial oent M A is easured as 7.665 c. Therefore, fro the scale c = N, the oent about bearing A is Therefore, the reaction force at bearing A is The angle is easured as 66. M A = 5. N (0) M 5. N F A A = = = 5.55 N () ZA 0.6 These results are in good agreeent with the analytical approach, see Eq. (8b). 5
heck: The force polygon can be drawn to solve for the force at bearing B. Figure.. The Force polygon for the reaction force at bearing B. Scale c = 0 N. The reaction force at bearing B is easured as F B = 48.88 N at an angle of () This answer is in good agreeent with the analytical solution, see Eq. (5b). Part II. (i) Points. The X Y coponents of the forces acting on the frae (or the ground) at bearing A are ( F ) = 40 cos 45 = 8.84 N Ax (a) ( F ) = 40 sin 45 = 8.84 N Ay (b) The X Y coponents of the forces acting on the frae at bearing B are ( F ) 90 cos (40 ) 45 N Bx = = (a) ( F ) 90 sin (40 ) 77.94 N By = = (b) For static balancing, see Equations (9-8), page 6, the equations can be written as ( F ) + ( F ) + [ cos + cos ] = 0 Bx ω r φ r Ax φ (a) ( F ) + ( F ) + [ sin + sin ] = 0 By ω r φ r Ay φ (b) For dynaic balancing, see Equations (9-9), page 6, the oents about the right-h bearing B can be written as Z ( ) + [ cos + cos ] = 0 A F Ax ω Z r φ Z r φ (4a) 6
Z ( F ) + Z r sin + Z r sin = 0 A where the distances fro the right-h bearing B are [ ] Ay ω φ φ (4b) Z A =+ 90 c, Z = + 70 c, Z = + 0 c (5) The four equations, Equations (a), (b), (4a), (4b), contain four unknowns; naely, the two correction asses the two angles; i.e.,,, φ φ. Substituting Equations (5) the known data into Equations (a), (b), (4a), (4b) gives ( )( ) φ 6.76 N + 5 c 50 rad/s ( cos + cos φ ) = 0 (6a) ( )( ) φ φ 49.658 N + 5 c 50 rad/s ( sin + sin ) = 0 (6b) ( )( ) φ 545.56 N c + 5 c 50 rad/s [ (70 c) cos + (0 c) cos φ ] = 0 (6c) ( )( ) φ 545.56 N c + 5 c 50 rad/s [(70 c) sin + (0 c) sin φ ] = 0 (6d) Equations (6) can be written as ( )( ) φ φ 5 c 50 rad/s ( cos + cos ) =+ 6.76 N (7a) ( )( ) φ φ 5 c 50 rad/s ( sin + sin ) =+ 49.658 N (7b) ( )( ) φ φ 5 c 50 rad/s [(70c) cos + (0c) cos )] = 545.56 N c (7c) ( )( ) φ φ 5 c 50 rad/s [(70c) sin + (0c) sin ] = 545.56 N c (7d) Rearranging Equations (7a) (7b) gives cosφ + 6.76 = cosφ (8a) (5 c)(50 rad / s) sinφ + 49.658 = sinφ (8b) (5 c)(50 rad / s) Then substituting Equations (8a) (8b) into Equations (7c) (7d) gives ( )( ) φ 5 c 50 rad/s [ +.0 (50c) cos ) = 545.56 N c (9a) ( )( ) φ 5 c 50 rad/s [ + 9.695 (50c) sin ] = 545.56 N c (9b) Rearranging Equations (9a) Equation (9b) gives 7
cosφ = + 0.98 kg (9c) sinφ = + 0. kg (9d) Dividing Equation (9d) by Equation (9c) gives the angle for the second correction ass; i.e., φ + 0. = tan = 58. + 0.98 Substituting Equation (0a) into Equation (9c) or Equation (9d) gives the second correction ass ( + 0.98 kg) = = 0.77 kg cos 58. (0a) (0b) Substituting Equations (0) into Equations (8), then rearranging dividing Equation (8b) by Equation (8a) gives the angle for the first correction ass; i.e., φ 0.888 = tan =+ 0.87 0.56 Substituting Equation (8a) into Equation (6a) or Equation (6b) gives the first correction ass ( 0.56) = = 0.4 kg cos 0.87 (a) (b) Therefore, the distributed ass syste can be dynaically balanced by perforing the following: At the correcting Plane. The correction ass is The correction ass is to be reoved at the angle = 0.4 kg (a) which is ( φ ) ( ) 80 REMOVE = φ ADD (b) ( ) 0.87 80 40.87 50.87 φ REMOVE = + = =+ (c) At the correcting Plane. The correction ass is = 0.77 kg (a) The correction ass is to be reoved at the angle which is ( φ ) = ( φ ) + 80 (b) REMOVE ADD ( φ ) = 58. + 80 = 8. (c) REMOVE The locations of the two asses to be reoved at the two correcting planes are as shown in the figure below. 8
Figure. The locations of the two correcting asses to be reoved. 9
Solution to Proble 4. (i) 0 Points. The two correcting forces, for the first haronic, can be written as F A D B = ( + ) + ( ) F ( A D ) ( B ) Also, the location angles of the correcting forces are = + + () B tan γ = ( A + D) tan γ = B+ D A () To deterine the coefficients A, B,, D. The X Y coponents of the first haronic force for cylinder can be written as S = Pcos( θ ψ )cosψ (a) X S = Pcos( θ ψ )sinψ (b) Y The X Y coponents of the first haronic force for cylinder can be written as S = P cos( θ ψ + φ )cosψ (4a) X S = P cos( θ ψ + φ )sinψ (4b) Y The X Y coponents of the first haronic force for cylinder can be written as S = P cos( θ ψ + φ ) cosψ (5a) X S = P cos( θ ψ + φ )sinψ (5b) Y The X-coponent of the resultant of the first haronic forces can be written as S = S + S + S (6a) PX X X X the Y-coponent of the resultant of the first haronic forces can be written as S = S + S + S (6b) PY Y Y Y The angles ψ = 45, ψ = 5, ψ = 70, φ = 0, φ = 0, φ = 0 Substituting these angles P = P = Rω, P = P = Rω, P = P = Rω into Equations (a) (4a) (5a), the results into Equation (6a), the X-coponent of the resultant of the first haronic forces can be written as S = Pcos( θ 45 )cos 45 + Pcos( θ 5 + 0 )cos(5 ) + Pcos( θ 70 + 0 )cos(70 ) (7a) PX or as S = + Pcosθ + 0Psinθ (7b) PX 0
Substituting the angles into Equations (b) (4b) (5b) the results into Equation (6b), the Y- coponent of the resultant of the first haronic forces can be written as or as S = Pcos( θ 45 ) sin 45 + Pcos( θ 5 + 0 ) sin(5 ) + P cos( θ 70 + 0 ) sin(70 ) (8a) PX S = + 0Pcosθ + Psinθ (8b) PY The agnitude of the resultant of the first haronic forces is S = S + S (9) P PX PY Substituting Equations (7b) (8b) into this equation, the agnitude of the resultant of the first haronic forces is or as S P P P P P = ( cosθ + 0 sin θ) + (0 cosθ + sin θ) (0a) SP P sin θ = + (0b) For the crank position θ = 0 o, the agnitude of the resultant of the first haronic forces is The direction of the resultant of the first haronic forces is S = P = P () P S = tan PY τ (a) S Substituting Equations (7b) (8b) into Equation (a), the direction of the resultant of the first haronic forces is PX + 0 Pcosθ + Psinθ τ = tan + Pcosθ + 0Psinθ (b) For the crank position θ = 0 o, the X Y coponents of the priary shaking force are S =+ Pcosθ + 0Psinθ =+ P S = + 0Pcosθ + Psinθ = 0 () PX Substituting Equations () into Equation (a), the direction of the resultant of the first haronic forces is PY + 0 τ = = + P tan tan 0 (4a) Since the nuerator is positive the denoinator is positive then the direction of the resultant of the first haronic forces is τ = + 0 o (4b) The priary shaking force (or the first haronic force) is shown in Figure 4..
(ii) 0 Points. oparing Equation (4b) with Equation (7a) coparing Equation (5b) with Equation (7b), the coefficients are A =+ P, B = 0, = 0, D =+ P (5) Substituting Equations (5) into Equation (), the first correcting force is which can be written as F ( ) ( ) = P+ P + 0 0 (6a) F = P 9+ 0 =.5P (6b) Substituting Equations (5) into Equation (), the second correcting force is which can be written as The force P can be written as F ( ) ( ) = P P + 0+ 0 (7a) F = P + 0 = 0.5P (7b) P = = 5 0.075 00 = 8,750 N (8) Rω Therefore, the correcting forces are F =.5 x 8,750 = 8,5 N (9a) F = 0.5 x 8,750 = 9,75 N (9b) Substituting Equations (5) into Equation (), the location angle for the first correcting ass is 0 0 + 0 tan γ = = ( P + P) P (0a) Since the denoinator (or the cosine of the angle) is negative then the angle is γ = 80 (0b) Substituting Equations (5) into Equation (), the location angle for the second correcting ass is 0+ 0 + 0 tan γ = = P P + P (a) Since the denoinator is negative then the angle is γ = 0 (b) (iii) 5 Points. The first correcting force can be written as
Therefore, the correcting ass can be written as The second correcting force can be written as Therefore, the correcting ass can be written as F R ω 8,5 N = = (a) 8,5 N =.44kg ( c)(00 rad / s) = (b) F = = (a) R ω 9,75N 9,75 N = 7.8kg ( c)(00 rad / s) = (b) Substituting Equations (7b) (8b) into Equation (a), the direction of the resultant of the first haronic forces is + P sin45 τ = = = + P cos 45 o o tan tan 6.4 o (4) (iv) 5 Points. For the crank position 45, o θ = the priary shaking force (or the first haronic force) the locations of the two correcting asses are as shown on Figure 4..
o Figure 4.. Magnitudes Locations of the Two orrecting Masses. For the crank angle θ = 45. 4