20k rad/s and 2 10k rad/s,

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1 ME 35 - Machine Design I Summer Semester 0 Name of Student: Lab Section Number: FINAL EXAM. OPEN BOOK AND CLOSED NOTES. Thursday, August nd, 0 Please show all your work for your solutions on the blank paper. Write on one side of the paper only, not on the exam. Credit will be given for clear presentation good figures diagrams. Problem (5 Points). For the mechanism in the position shown in Figure, O A = m, O B =.6 m, BE = 0. m, EO C = 0.7 m, G 3 D =.3 m, X G3 Y G3 X G3 ' Y G3 ' X G3 " Y G3 " X G X G ' X G " (m) (m) (m/rad) (m/rad) (m/rad ) (m/rad ) (m) (m/rad) (m/rad ) The angular velocity acceleration of the input link are 0k rad/s 0k rad/s, respectively. The free length of the spring R SO = 0.5 m the spring stiffness K S = 0 N/m. The damping constant C = 0 N-s/m. The masses the mass moments of inertia of links, 3, are: m (kg) m 3 (kg) m (kg) I G (kg-m ) I G3 (kg-m ) I G (kg-m ) Gravity acts in the negative Y-direction friction in the mechanism can be neglected, determine: (i) The first-order kinematic coefficients of the spring the damper. (ii) The kinetic energy of the mechanism. (iii) Using the equation of motion determine the magnitude the direction of the torque T. Figure. A planar mechanism.

2 ME 35 - Machine Design I Summer Semester 0 Name of Student: Lab Section Number: Problem (5 points). For the linkage shown in Figure, the vertical link is rigidly fixed to the ground (at point A) pinned to the horizontal link 3 at point B where AB =.5 m. The length of the horizontal link 3 is DH = 3 m where DB = BH =.5 m. Link is pinned to link 3 at point H the mass of link is 00 kg. The vertical link 5 is pinned to link 3 at D to the ground at point O 5. Link has a solid circular cross-section with diameter d 5 mm, a compressive yield strength Syc 370 MPa, a modulus of elasticity E 05GPa. The factor of safety guarding against buckling, for link, is N =.5 the end-condition constant C =. Note that gravity is acting vertically downward links, 3, 5 are assumed to be massless compared to link. (i) For link determine: (a) the slenderness ratio; (b) the slenderness ratio at the point of tangency between the Euler column formula the Johnson parabolic formula. (ii) Determine the critical unit load acting on link. (iii) If the diameter of link is increased to be d 90mm then is link an Euler column or a Johnson column? Clearly show all your calculations to justify your answer. Figure. A planar linkage.

3 ME 35 - Machine Design I Summer Semester 0 Name of Student: Lab Section Number: Problem 3 (5 points). Part I. The 80 cm long shaft shown in Figure 3(a) is simply supported by the bearings at B D. Flywheel at location A weighs 0 N flywheel at location C weighs 70 N. The weight of the shaft 6 6 can be neglected. The stiffness coefficients of the shaft are k.0 0 N/m, k N/m, 6 k.0 0 N/m. Determine the first second critical speeds of the shaft using the exact equation. Then determine the first critical speed of the shaft using: (i) the Rayleigh-Ritz method; (ii) the Dunkerley approximation. Figure 3(a). A shaft simply supported at B D. Part II. Figure 3(b) shows three mass particles fixed to a simply supported shaft rotating counterclockwise with a constant angular velocity ω 5 rad/s. The masses are m 3kg, m kg, m3 kg the distances are R = 5 mm, R = 35 mm, R 3 = 0 mm. The axial distances are a = 50 mm, b = 5 mm, c = 75 mm, d = 00 mm, e = 50 mm, f = 5 mm. (i) Determine the magnitudes the directions of the bearing reaction forces at A B. (ii) Use the analytical method to determine the magnitudes locations of the correcting masses that must be added in the two correcting planes () () to dynamically balance the system. The correcting masses are to be placed at a radial distance R C = 5 mm from the shaft axis. Figure 3(b). Three mass particles attached to a rotating shaft. 3

4 ME 35 - Machine Design I Summer Semester 0 Name of Student: Lab Section Number: Problem (5 Points). Consider the two-cylinder engine shown in Figure. The effective masses of the pistons are m = m = 5 kg, the lengths of the connecting rods are L = L = 6 cm the lengths of the throws of the cranks are R = R = 8 cm. The angular velocity of the crankshaft is a constant ω 00 rad/s counterclockwise. To balance the primary shaking force on the crankshaft bearing, correcting masses are to be added at a radial distance of R C = cm from the crankshaft. Determine: (i) The X Y components of the primary shaking force in terms of the crank angle θ. (ii) The magnitudes the angular locations of the correcting masses. (iii) Show on the figure to the right, which corresponds to θ = 30, the magnitude direction of the primary shaking force the locations of the correcting masses. Figure. A two-cylinder engine.

5 Solution to Problem. (i) 8 points. The first order kinematic coefficient of the spring can be obtained from an inspection of the mechanism, that is R Y (a) S G3 Substituting the given value into Equation (a), the first order kinematic coefficient is R S Y G m/rad (b) The positive sign indicates that the spring becomes shorter as the input link rotates clockwise. Similarly, the first-order kinematic coefficient for the damper can be obtained from an inspection of the mechanism, that is R X (a) C G Substituting the given value into Equation (a), the first order kinematic coefficient for the damper is R ( 0.500) m/rad m/rad (b) C The positive sign indicates that the damper becomes shorter as the input link rotates clockwise. (ii) 8 Points. The kinetic energy of the mechanism can be written as j EQ j that is the equivalent mass moment of inertia of the mechanism is T A I (3a) I EQ A (3b) j j For Link : which can be written as For Link 3: which can be written as For Link : which can be written as G G G A m X Y I (a) A (b) kg-m 3 3 G3 G3 G3 3 A m X Y I (5a) A3 3[ ] kg-m (5b) G G G A m X Y I (6a) A 0[ ] kg-m (6b) Therefore, the equivalent mass moment of inertia of the mechanism is EQ j 3 3 kg-m (7) j I A A A A 5

6 Substituting Equation (7) into Equation (3a), the kinetic energy of the mechanism can be written as x 3x Nm (iii) 9 points. The power equation for the mechanism can be written as T (8) dt du dw f T (8) dt dt dt The positive sign on the left h side of Equation (8) indicates that the input torque is assumed to be acting in the same direction as the angular velocity of the input link, that is, clockwise. Cancelling the input angular velocity in the power equation gives the equation of motion for the mechanism. Therefore, the equation of motion for this mechanism can be written as j j j Gj s S S0 S C j j j T A B m gy K R R R C R (9) The B terms in Equation (9) can be written as For Link : which can be written as For Link 3: which can be written as For Link : which can be written as B da j B j j j dr X X G G YY G G IGR m R B B m X X Y Y I 3 3 G3 G3 G3 G3 G3 3 3 B3 3[ ] B X G X G YG YG IG m B 0[ ] kg-m (0) (a) (b) (a) (b) (3a) (3b) Substituting Equations (), (), (3) into Equation (0) gives The gravitational term can be written as Bj B B3B kg-m () j m j gy Gj mgy G m3gy G3mgY G m3gy G3 (5a) j 6

7 Substituting the known data into Equation (5a), the gravitational term is mgy j Gj N-m (5b) j The spring term can be written as Ks RS RS0 R S N-m (6) The damping term can be written as CR N-m C (7) Substituting Equations (), (5b), (6) (7) into Equation (9), the input torque can be written as that is T N-m T N-m (8a) (8b) Therefore, the input torque is T 79 N-m 7.9 kn-m (9) The positive sign for the input torque indicates that it is acting in the same direction as the angular velocity of the input link, that is, clockwise. 7

8 Solution to Problem. (i) 8 Points. (a) Points. The slenderness ratio of link is defined as where the radius of gyration is S r L (a) k I k (b) A The second moment of area for a solid circular cross-section is the cross-sectional area of link is I m d m (a) d x5 A mm (b) Substituting Equations (a) (b) into Equation (b), the radius of gyration of link is k m m m (3a) Check: The radius of gyration of a solid circular cross-section, see Table, page 87, is d k m (3b) Substituting the length L =.5 m Equation (3) into Equation (a), the slenderness ratio of link is.5 Sr 00 () (b) Points. The slenderness ratio at the point of tangency between the Euler column formula the Johnson parabolic formula can be written as CE Sr (5) D S where the end condition constant for link is given as yc C (6) Substituting the given data into Equation (5), the slenderness ratio at the point of tangency is S r D 05GPa 7.90 (7) 370MPa 8

9 Comparing Equation (7) with Equation () indicates that S S (8) r r D Therefore, link is an Euler column. (ii) Points. The free body diagram for link is shown in Figure. Figure. The free body diagram for link. The sum of the vertical forces acting on link can be written as or as F F Y 0 (9a) 3Y m g 0 (9b) Substituting the given data into Equation (9b), the vertical reaction force at pin H is 3Y / 96 N The free body diagram for link 3 is shown in Figure. F kg m s (9c) Figure. The free body diagram for link 3. The sum of the moments about pin B is M B 0 (0a) 9

10 Therefore, the sum of the moments about pin B is F R F R (0b) 3Y BC 53Y BD 0 Since R BC = R BD then Equation (0b) can be written as The sum of the vertical forces acting on link 3 can be written as F53 F3 96 N () Y Y or as F 3Y 0 (a) F F F (b) 3Y 53Y 3Y 0 Substituting the given data into Equation (b), the vertical component of the internal reaction force acting on link 3 from link at pin B is F3 Y 96 N 96 N 39 N (3a) Therefore, the vertical component of the reaction force acting on link from link 3 at pin B is In other words, the reaction force acting on link from link 3 at pin B is Therefore, the compressive load acting on link at pin B is The critical unit load acting on link is defined as F3 F3 39 N (3b) Y Y F3 0 i 39 j N () P 39 N (5) P CR A (6a) where the critical load on link can be written as PCR N P (6b) Substituting Equation (5) the given factor of safety N =.5 into Equation (6b), the critical load is P.5 x N (7) CR Then substituting Equations (b) (7) into Equation (6a), the critical unit load on link is P CR 980 N 9.98 N/mm A (8) mm 0

11 (iii) 5 Points. If the diameter of link is d 90mm, then is link an Euler or a Johnson column? In this case, the new cross-sectional area of link is d x 0.09 A The new radius of gyration of a solid circular cross-section is Therefore, the new slenderness ratio is m (9) d 0.09 k 0.05 m (0).5 Sr. () 0.05 Recall that the slenderness ratio at the point of tangency is given by Equation (7), that is 7.90 Comparing Equation () with Equation () indicates that S () r S D S (3) r r D Therefore, link is a Johnson column. Recall that in Part (ii) the answer is that link is an Euler column.

12 Solution to Problem 3. PART I. (i) 5 Points. The masses of the flywheels are that is W 0N m g 9.8 m/ s m W 70N g 9.8 m/ s (a) m.077kg m 7.36kg (b) The influence coefficients are the reciprocal of the spring stiffness coefficients, that is or a, a, a.0 0 N/m N/m.0 0 N/m k k k a.00 m/ N, a.6670 m/ N, a.5 0 m/ N (b) (a) From Maxwell s reciprocity theorem, the influence coefficients a 7 a m / N (3) The exact solution for the first second critical speeds of the shaft can be written as (a m a m ) (a m a m ) (a a a a )m m, ω ω (a) Rearranging Equation (a), the first second critical speeds of the shaft can be written as ω, ω (a m a m ) (a m a m ) (a a a a )m m (b) Substituting the given data into Equation (b), the exact solutions for the first second critical speeds of the shaft are 77. rad/s 83.8 rad/s (5) (ii) 3 Points. The total deflections of the shaft at locations can be written as that is x awaw x awaw x x (6) (7a) (7b) Using the Rayleigh-Ritz method, the approximation to the first critical speed is Wx Wx g Wx (8a) Wx

13 Therefore, the approximation to the first critical speed from Equation (8a) is that is (8b) rad/s (8c) Comparing Equations (6) (8c) we note that the Rayleigh-Ritz equation predicts a slightly greater first critical speed than the exact solution. The Rayleigh-Ritz equation is an upper bound. (iii) 3 Points. The Dunkerley approximation for the first critical speed is given by which may be written as or as a m a m (9a) a m a m (9b) 3.05 rad/s (9c) 6 7 Note that the Dunkerley approximation underestimates the first critical speed. PART II. (i) 6 Points. The magnitudes of the inertial forces caused by the rotating mass particles are Equations () can be written in vector form as F mr (a) (3 kg)(0.05 m)(5 rad/s) 0.5 N F mr ( kg)(0.035 m)(5 rad/s) N (b) F m R (c) ( kg)(0.00 m)(5 rad/s) 9 N F 0.5 N i j N (a) F N i j N (b) F3 9 N i.500 j N (c) The sum of moments about the bearing A can be written as M A 0 (3a) 3

14 Therefore, the sum of moments about the bearing A are 0.5 k5.063i j0.59 k 6.80i j 0.9 k7.79i.500 j0.7 kfb 0 Rearranging Equation (3b) writing out the separate components gives (0.7) F ( 0.5)(5.063) ( 0.59)( 6.80) ( 0.9)( 7.79) N (a) BX (0.7) F ( 0.5)(8.769) ( 0.59)( 3.938) ( 0.9)(.500).69N (b) BY Therefore, the bearing reaction force at point B is (3b) F.9 i j N N 08.5 (5) B The sum of moments about the bearing B can be written as M B 0 that is 0.55 k 5.063i j +0.5 k 6.80i j 0.5 k7.79i.500 j 0.7 kfa 0 (6a) (6b) Rearranging Equation (6b) writing out the separate components gives ( 0.7) F (0.55)(5.063) (0.5)( 6.80) (0.5)(7.79) 3.583N (7a) AX ( 0.7) F (0.55)(8.769) (0.5)( 3.938) (0.5)(.500).8878N (7b) AY Therefore, the bearing reaction force at point A is F.8i j N 6.9 N 8.9 (8) A Check. The sum of the inertial forces the bearing reaction forces must sum to be zero, that is Fi FA FB 0 (9) i Substituting the known values the results from Equations (5) (8) into Equation (9) gives that is 5.063i j 6.80i j 7.79i.500 j.9i 3.57 j(.8i j) 0 (0a) 0 0 (0b) (ii) 8 Points. For dynamic balance, the sum of the moments must be zero. The sum of the moments about the correcting plane () can be written as M () 0 ()

15 The sum of the moments about the correcting plane () can be written as the two scalar equations ZmR i i i cosi ZCmC RC cosc 0 (a) i ZmR i i i sini ZCmCRC sinc 0 (b) i Dividing Equations () by gives ZmR i i icosi ZCmCRC cosc 0 (3a) i ZmR i i isini ZC mcrcsinc 0 (3b) i Substituting the known values into Equations (3) gives ZmR i i i cos i i ( 0.65)(0.05)(3) cos 60 ( 0.60)(0.035)() cos 0 ( 0.50)(0.00)() cos kg-m i i i i ZmRsin i ( 0.65)(0.05)(3) sin 60 ( 0.60)(0.035)() sin 0 ( 0.50)(0.00)() sin kg-m (a) (b) Rearranging Equations (3), the angle of the correction force can be written as tan C i i ZmRsin i i i i ZmRcos i i i i (5) Substituting Equations () into Equation (5) gives tan C (6) Since the numerator the denominator are both positive then the angle is.30 (7) Rearranging Equation (3a), the correcting mass in correction plane () can be written as C m C Z i ZmRcos C i i i R C cos C i (8) 5

16 Then substituting Equation (7) into Equation (8), the correcting mass is m C 0.56 kg (9) ( 0.95)(0.05) cos(83.6) Similarly, to find the correcting forces the mass in the correction plane (), take moments about the correction plane (), that is M () 0 (0) The sum of the moments about the correcting plane () can be written as the two scalar equations ZmR i i i cosi ZCmCRC cosc 0 (a) i ZmR i i i sini ZCmCRC sinc 0 (b) i Dividing Equations () by gives ZmR i i icosi ZCmCRC cosc 0 (a) i ZmR i i isini ZCmCRCsinC 0 (b) i Substituting the known values into Equations () gives i i (0.375)(0.00)() cos kg-m ZmRcos (0.650)(0.05)(3) cos 60 (0.75)(0.035)() cos 0 i i i i (0.375)(0.00)()sin kg-m ZmRsin (0.650)(0.05)(3) sin 60 (0.75)(0.035)()sin 0 i i i i (3a) (3b) Rearranging Equations (), the angle of the correction force can be written as tan C i i ZmRsin i i i i ZmRcos i i i i () Substituting Equations (3) into Equation () gives tan C (5)

17 Since the numerator is positive the denominator is negative in Equation (5) then the angle.03 (6) Rearranging Equation (a), the correcting mass in correction plane () can be written as C m C Z i ZmRcos C i i i R C cos C i (7) Substituting Equation (6) into Equation (8), the correcting mass is m C.07 kg (8) (0.95)(0.05)cos(.03 ) Check. The sum of the inertial forces the correcting forces must sum to zero for static balance, i.e., Fi FA FB 0 (9) i or as scalar equations Dividing Equations (30) by gives mr i i cosi mcrc cosc mc RC cosc 0 (30a) i mr i i sini mcrc sinc mc RC sinc 0 (30b) i mr i icosi mcrc cosc mc RCcosC 0 (3a) i mr i isini mcrcsin C mcrcsinc 0 (3b) i Inserting the known values, the masses angles from (7), (9), (6), (8), gives (0.05)(3) cos 60 (0.035)() cos 0 (0.00)() cos330 (0.05)(0.5599) cos.30 (0.05)(.07) cos.03 0 (0.05)(3) sin 60 (0.035)()sin 0 (0.00)()sin 330 (0.05)(0.5599)sin.30 (0.05)(.07) sin.03 0 (3a) (3b) which gives (33) This confirms that the correcting masses do, in fact, balance the system of three mass particles. 7

18 Solution to Problem. (i) Points. The X Y components of the primary shaking force from the first cylinder are S Pcos( )cos (a) X S Pcos( )sin (b) Y The X Y components of the primary shaking force from the second cylinder are S P cos( )cos (a) X S P cos( )sin (b) Y The total X Y components of the primary shaking force is then given by S S S (3a) PX X X S S S (3b) PY Y Y The X Y components of the resultant primary shaking force can be written in the form S Acos Bsin (a) PX S Ccos Dsin (b) PY where the coefficients are A Pcos ( ) cos (5a) i i i i i i B Psin ( ) cos (5b) i i i i C Pcos ( ) sin (5c) i i i i i i D Psin ( ) sin (5d) i i i i From Figure, the given angles are 5, 300, 0, 0 (6) Substituting these values in Equation (6) into Equations (5), the coefficients are A Pcos 5cos 5Pcos 60cos3000.5P 0.5P 0.75P (7a) B Psin 5cos 5Psin 60cos P 0.33P 0.933P (7b) C Pcos 5sin 5Pcos60sin P 0.33P 0.067P (7c) 8

19 where D Psin 5sin 5Psin 60sin P 0.75P 0.5P (7d) P P P mr (8) (5 kg)(0.08 m)(00 rad/s) 8, 000 N Substituting Equation (8) into Equations (7), the numerical values of the four coefficients are A N, B 785 N, C 35. N, D 000 N (9) Substituting Equations (6) into Equations (5), the total X Y primary shaking forces can be written as S 0.75Pcos 0.933Psin S 0.067Pcos 0.5Psin (0a) PX Substituting Equations (8) into Equations (5), the total X Y primary shaking forces are S 36000cos 785sin N S 35. cos 000sin N (0b) PX (ii) 8 Points. The magnitudes of the correcting forces can be written as F A D B C (a) F A D B C (b) The angles of the correcting forces can be written as PY PY tan tan sin B C cos ( A D) (c) sin B C cos ( A D) (d) The correcting forces on the crankshaft bearing are shown in Figure.. Substituting Equations (9) into Equations (), the correcting forces are ( ) ( ) F 0.75P0.5P 0.933P P P (a) ( ) ( ) (b) F P P P P P Substituting Equation (8) into Equations (), the correcting forces are F,000 N F 33, 9 N (c) To determine the correcting masses. The inertial force for a rotating mass i can be written as F i mr i i (3) 9

20 Therefore, the two correcting masses are m m F,000 N 5.0 kg (a) C RC (0.0 m)(00 rad/s) F 33,9 N. kg R (b) C C (0.0 m)(00 rad/s) The location of the correction masses for an arbitrary crank angle. Figure.. The location of the correction masses for an arbitrary crank angle. Substituting Equations (8) into Equations (), the angular location of the first correcting mass can be written as 0.933P0.067P tan (0.75P0.5 P) 0. 5 (5a) Since the numerator (or the sine of the angle) is positive the denominator (or the cosine of the angle) is negative then the location angle is 0 (5b) Similarly, the angular location of the second correcting mass can be written as 0.933P0.067P.0 tan (0.75P0.5 P).0 (5c) 0

21 Since the numerator (or the sine of the angle) is positive the denominator (or the cosine of the angle) is negative then the location angle is 35 (5d) (iii) 5 points. Substituting 30 into Equation (9b), the X Y components of the resultant primary shaking force are SPX N S 35. N (5a) Therefore, the primary shaking force on the crankshaft bearing is PY S 53569i 35. j N N 3. (5b) P The primary shaking force on the crankshaft bearing is shown in Figure.. The locations of the two correcting masses are also shown on the figure. Figure.. The location of the correction masses for 30.

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