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 of the paper only. Where necessary, you can use the figures provided on the exam to show vectors. ny work that cannot be followed is assumed to be in error. t the completion of your exam, please staple each problem separately staple your crib sheet to the end of Problem. Problem (5 Points). The torque acting on link at the crankshaft O of the four-bar linkage shown in Figure is T 000 k Nm. lso, there is a torque T acting on link at the crankshaft O to hold the linkage in static equilibrium. The length of the coupler link is B 00 mm the crosssection is rectangular with width t thickness t (into the paper). The coupler link is a steel alloy with compressive yield strength Syc 05 MPa a modulus of elasticity E 07 GPa. If the critical load for the coupler link is PCR 75kN, the effects of gravity are ignored, then determine: (i) The factor of safety guarding against buckling. (ii) The slenderness ratio at the point of tangency between the Euler equation the Johnson equation. (iii) The numerical value of the parameter t. (iv) If the rectangular cross-section is replaced by a circular cross-section of the same material diameter d 8mm then determine the factor of safety guarding against buckling of the coupler link. t t t 60 o O T = 000 Nm T 60 o 0 o O B Figure. four-bar linkage in static equilibrium.
ME 5 - Machine Design I Fall Semester 0 Name of Student Lab Section Number Problem (5 Points). For the mechanism in the position shown in Figure, the kinematic coefficients of link are 0.5 rad / rad 0.60 rad/rad. The angular velocity acceleration of the input link are 5krad/s krad/s. The input torque T 0k Nm there is a horizontal force F of unknown magnitude direction acting at the mass center of link. The lengths are O 50mm, OS 5 mm, D 00mm, G 60mm. The free length stiffness of the spring are Ro 0mm K 000 N / m the damping coefficient of the damper is C 50 Ns / m. The masses mass moments of inertia of the links are m m 6 kg, m kg, I I 5kg-m, I kg-m. Neglect the effects of friction in the mechanism. G G G (i) Determine the first-order kinematic coefficients of the horizontal spring viscous damper. (ii) Determine the kinetic energy of this mechanism. (iii) Using the equation of motion, determine the magnitude direction of the horizontal force F. K = 000 N/m O s g = 9.8 m/s LO F G T = 0 Nm O,G O,G C = 50 Ns/m D Figure. planar mechanism.
ME 5 - Machine Design I Fall Semester 0 Name of Student Lab Section Number Problem (5 Points). The effective mass of each piston in the two-cylinder engine shown in Figure is mm 6kg. The length of each connecting rod is L L 600 mm the length of the throws are RR 75mm. The crankshaft is rotating with a constant angular velocity ω 55 k rad/s. (i) Determine the magnitude direction of the primary shaking force in terms of the crank angle θ. (ii) If correcting masses are required to balance the primary shaking force then determine: (a) The magnitudes directions of the inertia forces created by these correcting masses. (b) The magnitudes angular locations of the correcting masses. The radii of the correcting masses are specified as R C R C 50 mm. (iii) Determine the answers for Parts (i) (ii) when the reference line (attached to crank ) is specified o at the crank angle θ 0. Show your answers on the left-h figure of Figure. θ = 0 o ω R Ref. Line θ 50 o -ω 5o R 60 o Crank ngle θ = 0 o Reference Line L L m m Figure. two-cylinder engine.
ME 5 - Machine Design I Fall Semester 0 Name of Student Lab Section Number Problem (5 Points). Part. The shaft shown in Figure is simply supported in the bearings at B is rotating clockwise with a constant angular velocity ω 50 rad/s. The masses of the two particles attached to the shaft are m kg m 5kg the radial distances are R 60mm R 5 mm. (i) Determine the magnitudes angular locations of the correcting masses that must be removed in the correcting planes () () to dynamically balance the system. The radial distances from the shaft axis are specified as RC R C 70 mm. (ii) Show the angular orientations of the correcting masses on the right-h figure of Figure. B m Z ω m m R m R 0 o 75 o 00 mm 50 mm 00 mm Figure. rotating shaft with two mass particles. Part B. The weights of two gears rigidly attached to a shaft are W 00 N W 50N. The shaft is rotating with a constant operating speed ω 00 rad/s. From a deflection analysis, the influence 6 6 6 coefficients for the shaft are a 5 0 mm/n, a 90 0 mm/n, a 50 0 mm/n. ssuming that the mass of the shaft can be neglected then determine: (i) The first second critical speeds of the shaft using the exact equation. Is the operating speed of the shaft acceptable? (ii) The first critical speed of the shaft using the Rayleigh-Ritz method. (iii) The first critical speed of the shaft using the Dunkerley approximation.
Solution to Problem. (i) 5 Points. The geometry of the four-bar linkage is shown in Figure. γ T α = 60 o β = 0 o α = 60 o O O B Figure. The geometry of the four-bar linkage. The angles α β are given as 60 o 0 o, respectively, therefore, the angle 90 o () For static equilibrium, link is a two-force member. Therefore, the internal reaction forces F F must be equal, opposite, collinear. The FBD of the coupler link is shown in Figure. LO F Figure. The free body diagram of link. B LO F 5
Since link has two forces a torque then the internal reaction forces F F must be equal, opposite, parallel. The FBD of link is shown in Figure. LO F d T = 000 Nm O LO F Figure. The free body diagram of link. The sum of the moments about the ground pivot O can be written as or as MO 0 (a) ET T (b) d F 0 The perpendicular distance between the LO of the force F the ground pin O is measured as Check: The length of the coupler link is B is given as d 7.0 mm () B 00 mm, therefore, the length of link is o B sin 00sin 0 O d 7.05 mm () o sin sin 60 Substituting Eq. () the torque T = 000 Nm clockwise into Eq. (b), rearranging, the internal reaction force between links is F T d 000 0.705 57.00 N (5) This internal reaction force is acting upward at point on link as shown in Figure. 6
F = 57 N T = 000 Nm F = 57 N O Figure. The free body diagram of link. Therefore, the free body diagram of the coupler link is as shown in Figure 5. F = 57 N Figure 5. The free body diagram of link. The coupler link is in compression the compressive load is B F = 57 N F F 57 N (6) The factor of safety guarding against buckling of the coupler link is defined as Substituting Eq. (6) PCR N P CR (7a) F 75KN into Eq. (7a), the factor of safety guarding against buckling is 7
75000 N 6.5 (7b) 57 Since the width h is less than the thickness b then link will buckle in the plane (referred to as in-plane buckling), see Figure 6. h = t B Figure 6. In-plane buckling of link. (ii) Points. The slenderness ratio at the point of tangency for link can be written as CE (8) Sr D Syc Since link is buckling in the - plane then the end-conditions are pinned-pinned the endcondition constant is C (9) Substituting Eq. (9) the given compressive yield strength modulus of elasticity into Eq. (8), the slenderness ratio at the point of tangency for link is 070 Sr.8 D 6 050 9 (0) (iii) 0 Points. The slenderness ratio of the coupler link is defined as The radius of gyration is defined as the cross-sectional area is Sr L () k k I (a) t x t t m (b) The area moment of inertia of link (with a rectangular cross-section) can be written as 8
I b h (a) Substituting the thickness b = t the width h = t into Eq. (a), the area moment of inertia of link is (t) (t) I 9t m (b) where the parameter t is in meters. Note that if the thickness b = t the width h = t then the area moment of inertia of link is (t) (t) I 6t m (c) The smaller of the two values in Eqs. (b) (c) must be used, see Example., page 597 (the member will begin to buckle in the weakest plane), that is, Eq. (b) must indeed be used to determine the slenderness ratio of link the critical unit load. lso, note that if link is assumed to be buckling out of the - plane then the end-conditions of link are not known. It is common to assume that for out of plane buckling that the ends of link are fixed-fixed. In such a case, the end-condition constant for a fixed-fixed link (see page 57) is C. Substituting Eqs. (b) (b) into Eq. (a), the radius of gyration of link can be written as 9t k 0.866t m () t Substituting Eq. () the length L B 00 mm 0.00 m into Eq. (), the slenderness ratio of link can be written in terms of the unknown parameter t (expressed in meters) as Sr 0.00 0.5 0.87 t t (5) Case I. ssume that link is an Euler column. ccording to the Euler column formula, the critical load for link can be written as C E PCR (6a) S r Substituting Eqs. (b) (5) the given data into Eq. (6a) gives or 9 x x070 75000 t 0.5 ( ) t (7a) 75000 0 t N (7b) Rearranging this equation, the parameter t is 9
t. mm 0.00 m (7c) Substituting Eq. (7b) into Eq. (5), the slenderness ratio of the coupler link is 0.5 Sr 79.9 0.00 (8) Recall that the Euler column formula is only valid when the slenderness ratio Sr Sr D (9) Comparing Eq. (8) with Eq. (0) shows that link the coupler link cannot be an Euler column, that is, link must be a Johnson column. Therefore, the value of the parameter t can be obtained as follows. Using the Johnson parabolic equation, the critical load for link can be written as PCR SycSr Syc CE (0a) Substituting Eqs. (0b) () the given data into Eq. (0a) gives or 6 0. (05 0 ) 6 75000 t 05 0 t 9 x070 6 59.8 75000 t 050 t (0b) (0c) Rearranging this equation, the parameter is t 5.78 0 m 5.78mm (iv) 6 Points. The area moment of inertia for a circular cross-section is () I d (a) 6 Substituting the diameter d 8mm into Eq. (a), the area moment of inertia of link is x 0.008 0 (b) I.0 0 m 6 The cross-sectional area of link is d x 0.008 6 50.7 0 m () Substituting Eqs. (b) () into Eq. (a), the radius of gyration of link is 0
0.00 k 0.00m 6 50.70 () Substituting Eq. () the length L B 0.m into Eq. (), the slenderness ratio is 0.00 Sr 50 0.00 (5) Comparing Eq. (5) with Eq. (0), the conclusion is that the slenderness ratio Sr Sr D (6) Therefore, the circular cross-section link is an Euler column. Substituting Eqs. () (5) the given data into Eq. (6a), the critical load for link can be written as 9 6 x x 070 CR P 50.70 N 50 (7a) Therefore, the critical load for link is PCR 56. N (7b) Substituting Eqs. (6) (7b) into Eq. (7a), the factor of safety guarding against buckling is 56. N 0.0 (8) 57 Since the factor of safety is less than one then buckling is predicted to occur in the circular cross-section link.
Solution to Problem. The vectors for a kinematic analysis of the mechanism are shown in Figure. k O s R G R R O D O C β Figure. The vectors for a kinematic analysis of the mechanism. The vector loop equation (VLE) for the mechanism can be written as I?? R R R 0 () The components of Equation () are RcosRcosRcos 0 (a) RsinRsinRsin 0 (b) Differentiating Equations () with respect to the input position gives R sin R cos R sin 0 (a) R cos R sin R cos 0 (b) Then writing Equations () in matrix form gives R sin cos Rsin R cos sin R Rcos () o where the input angle 90, o o O 60 60 sin ( ) (5a) D
that is o 50 o o o 60 sin ( ) 60 0 0 (5b) 00 Substituting Equation (5b) the given dimensions into Equation () gives 50 mm 0.866 rad 50 mm 86.60 mm 0.50 rad R 0 (6a) The determinant of the coefficient matrix in Equation (6a) is DET ( 50 mm)( 0.5 rad) ( 86.6 mm)( 0.866 rad) 00 mm (6b) Using Cramer s rule, the first-order kinematic coefficients of link are 0.5 mm / mm (7a) R.0 mm (7b) Differentiating Equations () with respect to the input position gives R cos R cos R sin R cos R sin 0 (8a) R sin R sin R cos R sin R cos 0 (8b) Then writing Equations (8) in matrix form gives R sin cos R cos R cos R sin R cos sin R Rsin Rsin Rcos (9) Substituting the known data into Equation (9) gives 50 mm 0.866 rad.55 mm 86.60 mm 0.50 rad R 68. mm (0a) Check: The coefficient matrix in Equation (0a) is the same as the coefficient matrix in Equation (6a). Therefore, the determinant of the coefficient matrix in Equation (0a) is given by Equation (6b), that is DET Using Cramer s rule, the second-order kinematic coefficients of link are 00 mm (0b) 0.60 mm / mm (a) R.96mm (b)
(i) 6 Points. The vectors for the kinematic analysis of the spring are shown in Figure. R = R R s k O s R G O D O C Figure. The vectors for the spring. The first order kinematic coefficients for the spring can be obtained from the vector equation for point, that is?? I R R () The components of point are Rcos (a) Rsin (b) The first-order kinematic coefficients of point is Rsin 50 mm/rad (a) Rcos 0 (b) Note that the first-order kinematic coefficient for the spring is Rs 50 mm/rad () The positive sign indicates that the spring is increasing in length as the input link rotates in the counterclockwise direction or the spring is decreasing in length as the input link rotates in the clockwise direction.
The vectors for the kinematic analysis of the viscous damper are shown in Figure. k O s R G R O c D O R D Figure. The vectors for the damper. The first order kinematic coefficients for the damper can be obtained from the vector equation for point D, that is?? I RD R R (5) The components of point D are R cos R cos (6a) D R sin R sin (6b) D The first-order kinematic coefficients of point D are R sin R sin (7a) D R cos R cos (7b) D Substituting the given data into Eqs. (7), the first-order kinematic coefficients of point D are o o 50sin 90 00sin 0 ( 0.5) 7.50 mm/rad (8a) D o o 50cos90 00cos0 ( 0.5).65 mm/rad 8b) D Therefore, the first-order kinematic coefficient for the damper is RD D 7.50 mm/rad (9) 5
The negative sign indicates that the damper is decreasing in lenth as the input link rotates in counterclockwise or the damper is increasing in length as the input link rotates terclockwise. (ii) 8 Points. The kinetic energy of the mechanism can be written as T I (0a) EQ where the equivalent mass moment of inertia of the mechanism is EQ j j ( ) Gj Gj G j j j j I m I (0b) For link : Given that the center of mass of link is coincident with the ground pin O, Eq. (0b) gives 0 5() 5 kg-m () The vectors for a kinematic analysis of the mass center of link are as shown in Figure. k O s R R G R G O D O Figure. The vectors for the mass center of link. C The vector equation for the mass center of link can be written as?? I R = R + R () G The coordinates of the mass center of link are R cosθ R cosθ 5.96 mm (a) G =R G sinθ Rsinθ = 0.00 mm (b) 6
Differentiating Eqs. () with respect to the input position, the first-order kinematic coefficients of the mass center of link are = R sinθ R sin θθ =.50 mm/rad (a) G = R cosθ R cosθθ =.99 mm/rad (b) G Differentiating Eqs. () with respect to the input position, the second-order kinematic coefficients of link are G = R cosθ R cosθθ R sinθθ =.75 mm/rad (5a) G (5b) = R sinθ R sinθθ R cosθθ = 6.95 mm/rad Substituting Eqs. () the given data into Eq. (0b) gives 6 [( 0.050) ( 0.099) ] 5( 0.5) 0. kg-m (6a) For link : Since the center of mass of link is coincident with the ground pin O, then Eq. (0b) gives 0 ( 0.5) 0.5 kg-m (6b) Substituting Eqs. (), () () into Eq. (0b), the equivalent mass moment of inertia is I EQ j 5 0. 0.5 5.57 kg-m (7) j Substituting Eq. (7) the input angular velocity into Eq. (0a), the kinetic energy is (5.57) ( 5) 69.675 Nm (iii) points. The power equation for the mechanism can be written as T (8) or as dt du dw dt dt dt f. FV. G (9a) T G i i i ( S S0) i S C T F B m gy K R R R CR The sum of the B terms in Eq. (7c) can be written as 7 (9b) Note that the external force F is assumed to pointing in the same direction as the -component of the velocity of the mass center of link, that is, in the positive -direction. Cancelling the input angular velocity ω on both sides of Eq. (7b) gives the equation of motion for the mechanism, that is (0) G i i i ( S S0) i S C T F B m gy K R R R CR
dj Bj mj G j G j G j G I j G j j j d j j j () For link. Since the center of mass of link is coincident with the ground pin O, then Eq. (8) gives B 600 50 0 For link. Substituting Eqs. (7a), (a), (), (5) into Eq. (8) gives (a) B 6[( 0.050)(0.075) (0.099)( 0.0695)] 50.50.6 0.75 kg-m (b) For link. Since the center of mass of link is at O, then Eq. (8) gives Substituting Eqs. () into Eq. (8) gives The gravitational term can be written as B 0 00.50.6 0.6 kg-m (c) B j 0 0.75 0.6.5 kg-m () j m jggj mgg mggm gg j (a) Substituting Eq. () the given data into Eq. (a), the gravitational term is mg j Gj 0 69.80.0990 0.765 N-m (b) j Substituting Eq. () into the spring term gives Ks RS RS0 R S 000(0.05 0.00)( 0.050). N-m (5) Substituting Eq. (9) into the damping term gives D CR 50 ( 0.0750) ( 5).05 N-m (6) Substituting Eqs. (5) ()-(5) the input torque into Eq. (7c) gives 0 F( 0.05) (5.57)() (.5)( 5) 0.765..05 (7) Rearranging Equation (7), the external force can be written as 6.7.65 0.765..05 0 F 65 N 0.05 (8a) 8
The negative sign indicates that the assumption that the external force is pointing in the positive - direction is incorrect. The external force is pointing to the left, therefore, the external force can be written as F 65 i N (8b) 9
Solution to Problem. (i) Points. The -components of the resultant of the primary shaking force can be written as S S S (a) S S S (b) The components of the primary shaking force for cylinder can be written as S Pcos( )cos (a) S Pcos( )sin (b) The components of the primary shaking force for cylinder can be written as S P cos( )cos (a) S P cos( )sin (b) From Figure, the angles are 5, 0, 00, 50 Substituting these angles P P P mr into Eqs. (a) (a), the results into Eq. (a), the -component of the resultant of the primary shaking force is that is S P cos( 5 ) cos 5P cos( 0050 ) cos 00 (a) S 0.067 Pcos 0.75 Psin (b) Similarly, substituting the angles into Eqs. (b) (b), the results into Eq. (b), the -component of the resultant of the primary shaking force is S Pcos( 5 ) sin 5P cos( 0050 ) sin 00 (5a) Therefore, the -component of the resultant of the primary shaking force is S.5 Pcos 0.067 Psin (5b) The magnitude of the resultant of the primary shaking force can be written as S S S (6a) P Substituting Eqs. (b) (5b) into Eq. (6a), the magnitude is SP P.567 cos 0.567sin 0.68sincos (6b) The direction of the resultant of the primary shaking force can be written as 0
S S tan (7a) Substituting Eqs. (b) (5b) into Eq. (7a), the direction of the resultant of the primary shaking force is tan.5 P cos 0.067 P sin.5 cos 0.067 sin tan 0.067 Pcos 0.75 Psin 0.067cos 0.75sin (7b) The force magnitude is P m R 6 0.075 55 6.5 N (8) Substituting Eq. (8) the crank shaft position 0 o into Eqs. (6b) (7b) gives SP 6.5.567 cos (0) 0.567sin (0) 0.68sin(0)cos(0) 69.55 N (9a).5 cos 0 0.067 sin 0.60 0.067 cos 0 0.75 sin 0 0.0 o tan tan 8.79 (9b) Since the numerator the denominator are both negative then the angle is in the third quadrant. (ii) 0 Points. The components of the resultant of the primary shaking force for a multicylinder reciprocating engine can be written in the form S cos B sin (0a) S C cos D sin (0b) The primary shaking force can be balanced by a pair of rotating masses. These two correcting forces plus the resultant of the primary shaking force must be equal to zero; i.e., cos Bsin F cos ( ) F cos [ ( ) ] 0 (a) C cos Dsin F sin ( ) F sin [ ( ) ] 0 (b) Exping these two equations, in terms of the angles of,,, rearranging, gives ( F cos F cos ) cos ( F sin F sin ) sin cos Bsin (a) ( F sin F sin ) cos ( F cos F cos ) sin C cos D sin (b) To satisfy Eqs. (), for all values of the crank angle, the necessary conditions are F cos F cos (a) F sin F sin B (b) F sin F sin C (c) F cos F cos D (d)
Solving Eqs. (), the correcting forces are F D B C (a) F D B C (b) lso, from Eqs. (), the location angles of the correcting forces are B C tan ( D) (5a) tan B C ( D) (5b) Note that these equations are similar to Eqs. (7.5) (7.6) of the text book page 807. Comparing Eq. (b) with Eq. (0a) Eq. (b) with Eq. (0b), the coefficients are 0.067 P, B 0.75 P, C.5 P, D 0.067P (6) Substituting Eq. (6) into Eqs. (), the two correcting forces are The magnitude is F 0.067 P 0.067 P 0.75 P.5 P 0.59P (7a) F 0.067 P 0.067 P.5P 0.75 P P (7b) P m R 6 0.075 55 6.5 N (8) Substituting Eq. (8) into Eq. (7), the two correcting forces are Recall that the correcting forces are defined as F 5.56N F 6.5 N (9) F m C R C (0a) F m C R C (0b) Substituting RC R C 0.5m into Eqs. (0), rearranging, the correcting masses are F 5.56 N mc 0.777 kg RC 0.5 x 55 m/s F 6.5 N mc kg R 0.5 x 55 m/s C (a) (b)
Substituting Eq. (6) into Eq. (5a), the location angle for the first correcting mass is 0.75.5 0.5 tan P P.7 (0.067P0.067 P) 0. (a) Since the numerator denominator are both negative, the angle for the second correcting mass is 55 o (b) Substituting Eq. (6) into Eq. (5b), the location angle for the second correcting mass is 0.75P.5P tan (a) (0.067 P 0.067 P) 0 Since the numerator is positive, the angle for the second correcting mass is 90 o (b) (iii) Points. The answers for parts (i) (ii), that is, Eqs. (9a), (9b), (9), (), (), (), are o shown on the figure below where the reference line is specified at the crank angle θ 0. F θ+γ = 05 o τ = 68.79 o θ = 0 o F ω Reference Line -(θ+γ ) = - 00 o -ω S P Figure.. The correcting masses where the reference line is for the crank angle o θ 0.
Solution to Problem. Part. Points. The inertial forces of the two rotating mass particles are F (a) mr ( kg)(0.06 m)(50 rad/s) 800 N F mr (5 kg)(0.05 m)(50 rad/s).5 N (b) Therefore, the inertial forces of the two rotating mass particles can be written as F 800 N 7565.87 i 78.67 j N F.5 N 0656.5 i 6.66 j N (a) (b) The sum of the inertial forces of the two rotating mass particles can be written as F F F 90.8 i 60.0 j N 6.9 N 07.55 () Therefore, the reaction forces at bearings B can be written as F FB FF 90.8 i60.0 j N The sum of the moments about bearing can be written as 0.kF 0.55 k( 65.87 i78.67 j) 0.7 k( 656.5 i6.66 j) 0 B () (5) The components of Eq. (5) can be written as 0. FB ( 0.55)(78.67) ( 0.7)(6.66) 0 (6a) 0. F B ( 0.55)(65.87) ( 0.7)( 656.5) 0 (6b) Therefore, the components of the force at bearing B are F 507.87 N F 79.8 N (7a) B B Therefore, the force at bearing B can be written as F 507.87 i 79.8 j N B Substituting Eq. (7b) into Eq. () gives F 507.87i 79.8 j N 90.8 i 60.0 j N Therefore, the force at bearing can be written as F 7.9 i777.8 j N (7b) (8a) (8b) The sum of the two correcting forces can be written from Eq. () as
90.8 F F F F i 60.0 j N C C The sum of the moments about correcting plane () can be written as 0.5 k F0.5 kf C 0 or as 0.5F i 0.5F j 0.5(78.67) i 0.5(65.87) j 0 C C Rearranging Eq. (0b), the inertial force in correcting plane () can be written as FC 65.87i 78.67 j 800 N 55 (9) (0a) (0b) () The correcting mass in correcting plane () can be written as which can be written as m F C C RC (a) 800 N mc 0.86 kg (b) (50 rad/s) (0.07 m) The orientation of the correcting mass in correcting plane () is C 55 (a) Therefore, the orientation of the mass from the -axis, that is to be removed in correcting plane (), is C 8580 75 (b) Substituting Eq. () into Eq. (9), the force in the second correcting plane is FC 90.8 i 60.0 j + 65.87 i 78.67 j 656.5 i 6.66 j N (a) which can be written as FC.50 N 60 (b) The correcting mass in correcting plane () can be written as which can be written as m F C C RC (5a).50 N mc 7.5 kg (5b) (50 rad/s) (0.07 m) The orientation of the correcting mass in the correcting plane () is C 60 (6a) Therefore, the orientation of the mass from the -axis, that is to be removed in correcting plane (), is 5
C 6080 0 (6b) These answers are shown in Figure.. Z ω B m m R m R 0 o m C m 75 o 00 mm 50 mm 00 mm m C Figure.. The locations of the correction masses. Part B. (i) 6 Points. The exact equation for the first second critical speeds can be written as (a W a W) (a W a W ) (a a a a )W W, ω ω g () Substituting the influence coefficients the masses into Eq. () gives 9 9 9 9 9 9 8 8 (50 ) (00) ( 900 )(50) [(5 0 ) (00) ( 900 )(50)] [ (50 )( 900 ) (50 )( 50 )] (00)(50), ω x9.8 ω () Simplifying this equation gives Therefore, the first critical speed is,.80 0 s,. 0 s ω 6 7 () ω ω 5.98 rad/s () the second critical speed is ω 85.0 rad/s (5) The operating speed of the shaft is much less than the first critical speed of the shaft. Therefore, the operating speed of the shaft is acceptable. 6
(ii) 5 Points. The Rayleigh-Ritz equation can be written as where the deflections are Wx Wx ω g Wx Wx (6) 9 9 5 x a W a W 5 0 00 50 0 50. x 0 m (7a) 9 9 5 x a W a W 50 0 00 90 0 50.5x 0 m (7b) Substituting Eqs. (7) into Eq. (6), the Rayleigh-Ritz equation can be written as (00)(. x0 ) (50)(.5 x0 ) ω 9.8[ ] rad /s (00)(. x0 ) (50)(.5 x0 ) 5 5 5 5 (8) Therefore, the first critical speed of the shaft is ω 5. rad / sec (9) Note that the Rayleigh-Ritz approximation to the first critical speed of the shaft is greater than the exact answer, see Eq. (). This is consistent with the fact that the Rayleigh-Ritz approximation is an upper bound to the first critical speed. (iii) Points. The Dunkerley approximation to the first critical speed of the shaft can be written as a m a m ω (0) Substituting the numerical values into Equation (0), the Dunkerley approximation to the first critical speed of the shaft can be written as 00 50 ω 9.8 9.8 9 9 5 0 90 0 sec (a) Therefore, the Dunkerley approximation to the first critical speed of the shaft is ω 50.78 rad / sec (b) Note that the Dunkerley approximation to the first critical speed of the shaft is less than the exact answer, see Eq. (). This is consistent with the fact that the Dunkerley approximation is a lower bound to the first critical speed. 7