ME Machine Design I

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1 ME 35 - Machine Design I Summer Semester 008 Name Lab. Div. FINAL EXAM. OPEN BOOK AND CLOSED NOTES. Wednesday, July 30th, 008 Please show all your work for your solutions on the blank white paper. Write on one side of the paper only. Credit will be given for clear presentation. Problem (5 Points). For the mechanism in the position shown in Figure, the kinematic coefficients are 3 = m θ, θ 3 = 6.98 m, R =.73 ND, R = 8 m. The velocity the acceleration of the input link are V = 5 j m/s A 0 j m/s = the force acting on link is F = 00 j N. The length of link 3 is AB = m the distance GG 3 = 0.5 m. A linear spring is attached between points O A with a free length L= 0.5m a spring constant K = 500 N/m. A viscous damper with a damping coefficient C= 5N.s/m is connected between the ground link. The masses mass moments of inertia of the links are m = 0.75 kg, m 3 =.0 kg, m =.5 kg, I G = 0.5 N.m.s, G3 =.0 N.m.s I G = 0.35 N.m.s. I Assume that gravity acts in the negative Y-direction (as shown in the figure) the effects of friction in the mechanism can be neglected. (i) Determine the first-order kinematic coefficients of the linear spring the viscous damper. (ii) Determine the equivalent mass of the mechanism. (iii) Determine the magnitude direction of the horizontal force P that is acting on link. Figure. A Planar Mechanism.

2 ME 35 - Machine Design I Summer Semester 008 Name Lab. Div. Problem (5 Points). Part A. A rotating steel shaft is simply supported by two rolling element bearings at A B as shown in Figure. Gears,, 3 are rigidly attached to the shaft at locations C, D, E, respectively. The gear at location C weighs 50 N, the gear at location D weighs 0 N, the gear 3 at location E weighs 30 N, the weight of the shaft can be neglected. The first critical speeds of the shaft with each gear attached separately are ω = 50 rad/s, ω = 650 rad /s, ω 33 = 00 rad/s, respectively. (i) Determine the first critical speed of the shaft with all three gears using the Dunkerley approximation. (ii) Determine the influence coefficients a, a a 33. (iii) If gear is now placed at location D gear is placed at location C, then determine the first critical speed of the new system using the Dunkerley approximation. Figure. A Rotating Shaft with Three Gears. Part B. The weights of two flywheels rigidly attached to a rotating shaft are W = 55N W = 95N. From the exact equation, the first second critical speeds of the shaft with the two flywheels are ω = 50 rad/s ω = 75 rad/s. The known influence coefficients of the shaft are 5 5 a = 9.60x0 cm/n a = 9.5x0 cm/n. Determine the influence coefficient a. Recall that the exact equation for the first second critical speeds of a rotating shaft with two mass disks was derived in lecture can be written as (a m + a m ) ± (a m + a m ) (a a a a )m m, = ω ω

3 ME 35 - Machine Design I Summer Semester 008 Name Lab. Div. Problem 3 (5 Points). Part A. The angular speed of three mass particles rigidly attached to the shaft shown in Figure 3(a) is a constant 350 rev/min. The masses of the three particles are m = 3 kg, m = kg, m3 =.5 kg. The radial distances of the three mass particles, from the shaft axis, are R = 5 mm, R = 65 mm, R3 = 50 mm. (i) Determine the magnitudes of the bearing reaction forces at locations A B. (ii) Determine the magnitudes of the masses that must be removed in the correcting planes () (). The radial distances of the masses to be removed in the correcting planes, from the shaft axis, are RC = RC = 70 mm. (iii) Determine the angular orientations, relative to the X-axis, of the masses to be removed in the correcting planes () (). Figure 3(a). A Rotating Shaft with Three Mass Particles. 3

4 ME 35 - Machine Design I Summer Semester 008 Name Lab. Div. Problem 3 (Continued). Part B. The three mass particles in Figure 3(a) are now replaced by a continuous mass system (denote as ) as shown in Figure 3(b). The angular speed of the rotating shaft is a constant 80 rev/min. The magnitudes of the bearing reaction forces at locations A B are specified as (F ) A = 75 N at 75 (F ) B = 5 N at 50, respectively. The radial distances of the two correcting masses in the correcting planes () () are RC = RC = 30 mm. Determine the two correcting masses the angular orientations of these two masses that must be added in the correcting planes () (). Figure 3(b). A Rotating Shaft with a Continuous Mass System.

5 ME 35 - Machine Design I Summer Semester 008 Name Lab. Div. Problem (5 Points). The two cranks of the two-cylinder engine shown in Figure are perpendicular. The effective mass of each piston is m = m = 0 kg m = m= 0kg, the length of each connecting rod is L= 75cm, the length of the throw of each crank is R = 5cm. The angular velocity of the crankshaft is a constant ω= θ = 50 rad/s counterclockwise. (i) Determine the X Y components of the resultant of the first harmonic forces (i.e., the primary shaking forces). (ii) Determine the magnitudes the locations (i.e., the angles) of the two correcting forces created by the two correcting masses. (iii) Draw your answers on the figure to the right below where the reference line is shown at the crank angle θ = 0 o. Figure. A Two Cylinder Engine. 5

6 Solution to Problem. (i) 8 points. The vectors for the linear spring are shown in Figure.. Fig... The Vectors for the Linear Spring. The vector loop for the linear spring can be written as the scalar equation? V IV RS R = 0 (a) RS = R (b) Differentiating Equation (b) with respect to the input position R, the first-order kinematic coefficient of the spring is R S = + ND () Note that the sign is positive because for the negative input, the length of the linear spring is decreasing. Also, note that the first-order kinematic coefficient of the mass center of the input link is Y R ND (3) G = S =+ The vectors for the viscous damper are shown in Figure.. Fig... The Vectors for the Viscous Damper. The vector loop for the damper can be written as VV? V VV R9 RC R 0 = () 6

7 The X Y components of Equation () are R cosθ R cosθ R cosθ = 0 (5a) 9 9 C C R sinθ R sinθ R sinθ = 0 (5b) 9 9 C C Differentiating Equations (5) with respect to the input position R gives R cos θ R cos θ = 0 (6a) C C R sinθ R sinθ = 0 (6b) C C 0 Substituting θ C = 0 damper is 0 θ = 0 into Equation (6a), the first-order kinematic coefficient of the viscous R C = R =+.73 ND (7) The sign agrees with our intuition, for a negative input the length of the viscous damper is decreasing. (ii) 8 Points. The equivalent mass of the mechanism can be written as For Link : ( ) meq = A j (8) j= A = m X G + Y G + I θ G (9) The vector loop for the center of mass of the input link can be written as The X Y components of Equation (0) are R?? IV = R (0) G G = 0 X (a) Y = R (b) G Differentiating Equations () with respect to the input position R gives X = ND (a) G 0 Y = R =+ ND (b) G Differentiating Equations () with respect to the input position R gives = = m (3a) XG 0 0 Y = R = m (3b) 0 G 7

8 Substituting Equations () (3) into Equation (9) gives ( ) ( ) A = = kg () 3 = m3 X θ G YG I G 3 For Link 3: ( ) The vector loop for the center of mass of link 3 can be written as The X Y components of Equation (6) are A (5)?? IV V? R = R + R (6) G3 3G X = R cosθ + R cosθ = 0.5 m (7a) G3 3G 3 Y = R sinθ + R sinθ = 0.33 m (7b) G3 3G 3 Differentiating Equations (7) with respect to the input position R gives X = cosθ R sinθθ = ND (8a) G3 3G 3 3 Y = sinθ + R cosθθ =+ 0.5 ND (8b) G3 3G 3 3 Then differentiating Equations (8) with respect to the input position R gives X = R sinθθ R cosθθ =.0 m (9a) G3 3G 3 3 3G 3 3 Y = R cosθθ R sinθθ = 0 m (9b) G3 3G 3 3 3G 3 3 Substituting the known data Equations (8) (9) into Equation (5) gives A =.0 ( 0.866) ( ) =+ 6 kg (0) For Link : ( ) 3 = m X + + θ G YG I G A (a) Note that X R.73 m, = = G therefore, Equation (a) can be written as ( ) ( ) A =.5[ ] =+.5kg (b) Therefore, the equivalent mass of the mechanism from Equation (8) is meq = =+.5kg () (iii) 9 points. The power equation for the mechanism can be written as F.V dt du dw + P.V = + + f dt dt dt (3a) 8

9 Substituting the time rate of change of energy terms into the right-h side of Equation (3a) gives F.V 3 + P.V = ( 0) + j j j G j s S S S C j= j= j= ARR BR m gy R K R R RR CR R (b) The linear velocity of link the force acting on link are both in the same direction (i.e., both downward). Assuming that the force P is in the same direction as the velocity of link, that is, to the right, then Eq. (b) can be written as FV 3 + PV = ( 0) + j j j G j s S S S C j= j= j= ARR BR m gy R K R R RR CR R (a) The velocity of the input link is V =+ R the velocity of link is V =+ R, therefore, Equation (a) can be written as + FR + = ARR + BR + m gy R + K R R RR + CR R (b) 3 PR ( 0) j j j G j s S S S C j= j= j= Canceling the input velocity is R in Equation (b) gives the equation of motion for the mechanism, that ( 0) j j j G j s S S S C j= j= j= + F+ PR = A R + B R + m gy + K R R R + C R R (5) where the first-order coefficient of link is given as R =.73 ND. The sum of the B terms can be written as daj B j = dθ (6) j= j= For Link : B = m( X X + Y Y ) + I θ θ (7a) G G G G G For Link 3: ( ) B = 0.75( 0 + 0) + 0.5( 0) = 0 (7b) B = m X X + Y Y + I θ θ (8a) 3 3 G3 G3 G3 G3 G3 3 3 = + + =+ kg m (8b) B3 [(.866)( ) (0.5)(0)] ( )( 6.98) 0.78 B θ (9a) For Link : m ( X X + Y Y ) + I θ = G G G G G ( )( ) B =.5[(.73)( 8) + (0)(0)] = kg m (9b) 9

10 Adding Equations (7b), (8b), (9b) gives The change in potential energy due to gravity is Bj = B+ B + B3 = =+.56 kg m (30) j= mj gy G j (3) j= For link : m gy G (3a) Which can be written as m g = (0.75)(9.8) = N (3b) For link 3: m3 gy G 3 (33a) Which can be written as m3 gy 3 = ()(9.8)( ) = 39. N (33b) G For Link : m gy G (3a) Which can be written as m gy = (.5)(9.8)(0) = 0 N (3b) Substituting Equations (3b), (33b) (3b) into Equation (3) gives j j= G m gy = = 3.80 N (35) G j Then substituting Equations (5), (0), (), (30) (35) into Equation (5) gives ( ) + F + PR =+.5R +.56 R K R R ( + ) + C ( +.73) R (36a) S S S0 Rearranging this equation, the force acting on link can be written as P = F.5 R.56 R 3.80 ( 0) (.73) R R K + + S RS RS C (36b) The input velocity R = 5m/s, the input acceleration R = 0m/s the force F = 00 N. Substituting these values the known data into this equation, the force acting on link can be written as P = 00.5( 0).56( 5) ( ) 5 (.73) ( 5) (37a) or as P = [ ] (37b).73 or as 0

11 Therefore, the force acting on link is +. P = (37c).73 P = N (38) The negative sign indicates that the force P (acting on link ) is acting to the left; that is, in the opposite direction to the velocity of the output link. Recall that the force P was assumed to be acting to the right.

12 Solution to Problem. Part A. (i) 5 Points. From the Dunkereley approximation, the first critical speed of the shaft can be written as = + + ω ω ω ω () 33 where the double numbered subscripts correspond to the critical speeds of the shaft with each gear attached searately. Substituting the known critical speeds corresponding to each gear into Eq. (), the first critical speed of the shaft is = + + =. 0 ω () ω = Therefore, the first critical speed of the shaft is. 0 5 rad / s (3a) ω = 7.7 rad / s (3b) (ii) 5 Points. The influence coefficients for each location can be determined using the mass at that location the corresponding critical speed, that is ma = ω () Rearranging Equation () solving for the influence coefficient gives 6 a = = = m/ N mω (50 / 9.8)(50) (5a) Similarly, the influence coefficient a = = = m N mω (0 / 9.8)(650) / (5b) Similarly, the influence coefficient a = = = m N 33 m3ω33 (30 / 9.8)(00) / (5c) (iii) 5 Points. When gears are interchanged, the first critical speed of the new system is = a m + a m + ω ω33 (6)

13 Equation (6) can be written as = ( )(0 / 9.8) + (.60 0 )(50 / 9.8) + = ω (7a) or as ω = rad / s (7b) Therefore, the first critical speed of the news ystem is ω = rad / s (8) Part B. (i) 5 Points. The exact solution for the first second critical speeds of the shaft can be written as, ω ω ( am + am ) ± ( am + am ) mm ( aa aa) = () The sum of the first second critical speeds of the shaft can be written as + = a m + a m ω ω () Substituting the known values into Equation () gives = 9.60 x 0 + a (3) Rearranging this equation gives a = x 0 () Therefore, the influence coefficient is 0.97 a = x (5) a = cm N (6) / 3

14 Solution to Problem 3(a). (i) 5 Points. The constant angular velocity of the rotating shaft is The inertial forces of the three rotating mass particles are F (π rad/rev)(350 rev/min) ω = = rad/s () (60 s/min) mrω (3 kg)(0.05 m)(36.65 rad/s) 8.3 N = = = (a) F F mr ω ( kg)(0.065 m)(36.65 rad/s) 7.6 N = = = (b) 3 m3rω 3 (.5 kg)(0.050 m)(36.65 rad/s) 00.7 N = = = (c) Therefore, the inertial forces of the three rotating mass particles can be written as F = 8.3 N 60 = i j N F = 7.6 N 50 = 5.3 i j N F3 = 00.7 N 300 = i 87. j N (3a) (3b) (3c) The sum of the inertial forces of the three rotating mass particles can be written as F = F + F + F 3 = 0.9 i j N = 57.5 N 93.7 () Therefore, the reaction forces at the two bearings A B can be written as FA + FB = F+ F + F3 = 0.9 i 57. j N ( ) The sum of the two correcting forces can be written as FC + FC = F+ F + F3 = 0.9 i 57. j N ( ) (5) (6) The sum of the moments about the correcting plane () is M () = 0 (7a) 0.65 k F k F k F k F C = 0 (7b) or as F i F j = 7.8 i j N CY C X (7c) Therefore, the inertial force in the correcting plane () can be written as FC = i 7.8 j = N 3.5 (8)

15 The correcting mass in the correcting plane () can be written as m F C C = ω RC (9a) m N = = 0.86 kg (9b) (0.070 m)(36.65 rad / s) C The orientation of the correcting mass in the correcting plane () is θ C = 3.5 (0a) Since the mass is to be subtracted then the angle must be rotated 80. The orientation of the mass, from the X-axis, that is to be removed in correcting plane () is θ C = =.5 (0b) Substituting Equation (8) into Equation (6a), the force in the second correcting plane is FC = 0.9 i 57. j i j = 9.8 i 09.8 j N (a) FC = 0. N 5.5 (b) The correcting mass in correcting plane () can be written as m F C C = ω RC (a) m 0. N = =.77 kg (b) (0.070 m)(36.65 rad / s) C The orientation of the correcting mass in the correcting plane () is θ C = 5.5 (3a) Since the mass is to be subtracted then the angle must be rotated 80. The orientation of the mass, from the X-axis, that is to be removed in correcting plane () is (iii) 5 Points. The sum of the moments about bearing A can be written as θ C = = 65.5 (3b) M A = 0 (a) M = 0 = 0.0 k F k F k F k F (b) A 3 B 5

16 Therefore, the reaction force at bearing B is FBYi FBX j = i j F = 0.5 i j N = N 5.5 B Substituting Equation (5b) into Equation (5), the reaction force at bearing A is F = i 68.3 j N = 85.0 N A (5a) (5b) (6) Solution to Problem 3(b). (i) 5 Points. The sum of the bearing reaction forces must be equal to the sum of the correcting forces; i.e., or as F = 0 (a) F F F F (b) A + B C C = 0 F + F = F + F A B C C (c) The sum of the moments about the correcting plane () is M () = 0 (a) 0.55 k F k F 0.05 k F = 0 A C B (b) Substituting the known data into Equation (b), the first correcting force can be written as 0 = 0. k (.65 i +.50 j) + 0. k F k (9. i 7. j) C (3a) F i F j = i j CY C X (3b) Therefore, the correcting force in the correcting plane () is FC = 9.68 i j N = 98.7 N 98. (3c) Substituting F F Equation (3c) into Equation (b) gives A B (.65 i +.50 j) + (9. i 7. j) ( 9.68 i j) F = 0 () C Therefore, the correcting mass in the correcting plane () is FC =+ 7. i 53.6 j N = N 80. (5) Therefore, the correcting mass in the correcting plane () is m F C C = RC ω (6a) 6

17 Substituting the known date into Eq. (6a), the correction mass is m 98.7 N = = 9. kg (0.030 m)(8.85 rad/ s) C (6b) The angular location of the correcting mass from the X-axis is The correcting mass in the correcting plane () is θ C = 98. (7) m m F C C = RC ω N = =.6 kg (0.030 m)(8.85 rad / s) C (8a) (8b) The angular location of the correcting mass from the X-axis is θ C = 80. (9) 7

18 Solution to Problem. (i) 5 Points. The X Y components of the first harmonic force for cylinder can be written as S = Pcos( θ ψ )cosψ (a) X S = Pcos( θ ψ )sinψ (b) Y The X Y components of the first harmonic force for cylinder can be written as S = P cos( θ ψ + φ )cosψ (a) X S = P cos( θ ψ + φ )sinψ (b) Y From Figure, the angles ψ =+ 0, ψ = 60, φ = 70 = 90. Substituting these angles P = P = mrω P = P = mrω into Equations (a) (a), the X-component of the resultant of the first harmonic forces can be written as S = S + S = Pcos( θ 60 )cos60 + Pcos( θ )cos ( 60 ) (3a) X X X Therefore, the X-component of the resultant of the first harmonic forces is S = +.6 Pcosθ Psinθ (3b) X Also, substituting the angles into Equations (b) (b), the Y-component of the resultant of the first harmonic forces can be written as S = S + S = Pcos( θ 60 )sin 60 + Pcos( θ )sin ( 60 ) (a) Y Y Y Therefore, the Y-component of the resultant of the first harmonic forces is The magnitude of the resultant of the first harmonic force is S =.067 Pcosθ 0.6 Psinθ (b) Y S = S + S = P cos θ +.33cosθsinθ (5a) P X Y the direction of the resultant of the first harmonic force is SY.067 cosθ 0.6sinθ tan = tan S +.6 cosθ sinθ X (5b) The primarry shaking force (or the first harmonic force) is shown in Figure. Recall that the X Y components of the resultant of the first harmonic forces for any multicylinder reciprocating engine (see page 65 in the Uicker, et al., text book) can be written in the form S = Acos θ + B sin θ (6a) X S = C cos θ + D sin θ (6b) Y 8

19 The first harmonic forces can always be balanced by a pair of rotating masses, or in some special cases a single mass, as shown in Figure 9.8, page 65, in the Uicker, et al., text book. The correcting mass m creates a correcting force F at a location angle of ( θ + γ ) from the X-axis the correcting mass m creates a correcting force F at a location angle of ( θ + γ ) from the X-axis. For balance, these two correcting forces plus the resultant of the first harmonic forces (or the primary shaking force) must be equal to zero; i.e., Acos θ + Bsin θ + F cos ( θ + γ ) + F cos [ ( θ + γ ) ] = 0 (7a) C cos θ + Dsin θ + F sin ( θ + γ ) + F sin [ ( θ + γ ) ] = 0 (7b) Exping these two equations, in terms of the angles of θ, γ,γ, rearranging, gives ( F cos γ + F cos γ ) cos θ ( F sin γ + F sin γ ) sin θ = Acos θ Bsin θ (8a) ( F sin γ F sin γ ) cos θ + ( F cos γ F cos γ ) sin θ = C cos θ D sin θ (8b) To satisfy Equations (8), for all values of the crank angle θ, the necessary conditions are F cos γ + F cos γ = A (9a) F sin γ + F sin γ =+ B (9b) F sin γ F sin γ = C (9c) F cos γ F cos γ = D (9d) Solving Equations (9), the correcting forces are F = ( A+ D ) + ( B C ) (0a) F = ( A D ) + ( B+ C ) (0b) Also, from Equations (9), the location angles of the correcting forces are B C tan γ = ( A + D) (a) tan γ B + C = ( A D) (b) Comparing Equation (3b) with Equation (6a) comparing Equation (b) with Equation (6b), the coefficients are A =+.6 P, B = P, C =.067 P, D = 0.6 P () 9

20 Substituting Equation () into Equations (0), the two correcting forces are F = (.6 P+ 0.6 P ) + ( P.067 P ) =.36 P (3a) F = (.6 P 0.6 P ) + ( 0.933P+.067 P ) =.39 P (3b) The magnitude of P is P= m Rω = = 3,5 N () Therefore, the two correcting forces are F = m R = P= N ω.36 6,987.5 F = m R = P= N (5) ω.39 3,87.9 Substituting Equation () into Equation (a), the location angle for the first correcting mass is [( P) (.067 P)] tan γ = =.000 [( +.6 P) + ( 0.6 P)] (6a) Since the denominator (or the cosine of the angle) is negative then the location angle is γ = 96.6 o (6b) Substituting Equation () into Equation (b), the location angle for the second correcting mass is [( P) + (.067 P)] tan γ = = [( +.6 P) ( 0.6 P)] (7a) Since the denominator (or the cosine of the angle) the numerator (or the sine of the angle) are negative then the location angle is o For the crank position θ = 0, the answers are shown on Figure. γ = 86. o (7b) 0

21 Figure. The Magnitudes the Locations of the Two Correcting Masses. o For the position θ = 0.

of the four-bar linkage shown in Figure 1 is T 12

of the four-bar linkage shown in Figure 1 is T 12 ME 5 - Machine Design I Fall Semester 0 Name of Student Lab Section Number FINL EM. OPEN BOOK ND CLOSED NOTES Wednesday, December th, 0 Use the blank paper provided for your solutions write on one side