205MPa and a modulus of elasticity E 207 GPa. The critical load 75kN. Gravity is vertically downward and the weight of link 3 is W3

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1 ME 5 - Machine Design I Fall Semester 06 Name f Student: Lab Sectin Number: Final Exam. Open bk clsed ntes. Friday, December 6th, 06 ur name lab sectin number must be included in the spaces prvided at the tp right h crner f each prblem. Write yur slutin n the blank pages at the end f each prblem write n ne side f the paper nly. Any wrk that cannt be fllwed will be assumed t be wrng. Please make sure that each prblem is stapled separately befre submitting yur exam. Prblem (5 pints). The mechanism shwn in Figure is in static equilibrium due t frces F B =.5 i kn F P = 5 jkn, acting n link, trque T 4 =500 k Nm frm the shaft at B. The distances OA.75 m, AP 0.75 m, AG GB.5 m. The rectangular crss-sectin f link has the width t the thickness h 5t (int the paper). Link is a steel ally with a cmpressive yield strength S yc fr link is PCR 05MPa a mdulus f elasticity E 07 GPa. The critical lad 75kN. Gravity is vertically dwnward the weight f link is W 5kN. If the weights f links 4 frictin in the mechanism can be neglected then determine fr link : (i) The factr f safety guarding against buckling. (ii) The slenderness rati at the pint f tangency between the Euler frmula the Jhnsn equatin. (iii) The numerical value f the width t. (iv) If the rectangular crss-sectin f link is replaced by a circular crss-sectin f the same material a diameter d 60 mm, then determine the new factr f safety guarding against buckling. P A g = 9.8 m/s F P = 5 kn G T 4 = 500 Nm t F B =.5 kn 0 B 4 O Figure. A planar mechanism in static equilibrium.

2 ME 5 - Machine Design I Fall Semester 06 Name f Student: Lab Sectin Number: Prblem (5 Pints). Fr the mechanism in the psitin shwn in Figure, the input link is rlling withut slipping n grund link at pint D with an angular velcity angular acceleratin f 5krad. / s 5krad. / s. The knwn dimensins are BG GP.5 m, OCA.5m, DB.5 m. The frce acting n link at pin P is F P 50 i N a trque T 4 is acting at the shaft O 4. The free length stiffness f the spring are m 5 kn / m, the damping cefficient is 00 Ns / m. The mass f link is m 5kg the mass mment f inertia f the link abut the mass center is I 9kg-m. The masses f links 4 frictin in the mechanism can G be neglected. The first-rder secnd-rder kinematic cefficients f links 4 the secndrder kinematic cefficients f the mass center f link are: (rad/rad) (rad/rad ) 4 (rad/rad) 4 (rad/rad ) x G (m/rad ) y G (m/rad ) (i) Determine the first-rder kinematic cefficients f the spring. (ii) Determine the first-rder kinematic cefficients f the hrizntal viscus damper. (iii) Using the equatin f mtin, determine the magnitude directin f the trque T 4. P F P = 50 N g = 9.8 m/s A,G C = 00 Ns/m O C k = 5 kn/m T B 60 O 4 D Figure. A planar mechanism.

3 ME 5 - Machine Design I Name f Student: Fall Semester 06 Lab Sectin Number: Prblem (5 Pints). Part I. The shaft shwn in Figure (a) is rtating clckwise with a cnstant angular velcity 0 rad / s. The mass particles m 5kg., m 75kg., m 5kg, are rigidly attached t the shaft at the distances R 00mm, R 75 mm, R 5 mm frm the shaft axis. The distances a = f = 75 mm, b = c = 75 mm, d = e = 50 mm. The crrectin planes are dented as n the figure. If the mass f the shaft can be neglected then determine: (i) The inertia frces created by the rtating mass particles. (ii) The magnitudes f the reactin frces at the bearings A B their angular rientatins measured frm a rtating reference line (assumed t be cincident with the axis). (iii) The magnitudes rientatins f the crrecting frces t dynamically balance the system. a b c d e f m R m 45 ω 0 0 R m R m Z A B ω m ` m Figure (a). A rtating shaft with three mass particles. Part II. The rtr (dented as ) in Figure (b) is rtating cunterclckwise with a cnstant angular velcity 50 rad / s. The frces acting n the bearings A B are measured as (F ) A 80 N (F ) 0 N, respectively. The distance a = b = 75 mm. Determine the magnitudes lcatins f B the crrecting masses t be remved in crrecting planes () () at the radius R R 50 mm. C C a 750 mm b ω A B Z 60 (F ) B = 0 N 0 (F ) A = 80 N Figure (b). A rtr supprted by tw bearings.

4 ME 5 - Machine Design I Fall Semester 06 Name f Student: Lab Sectin Number: Prblem 4 (5 Pints). The crankshaft f the three-cylinder engine shwn in Figure 4 is rtating cunterclckwise with a cnstant angular velcity ω 50 rad/s. The effective masses f the three pistns are m 75 kg, the lengths f the cnnecting rds are L 500 mm, the length f the thrw f the crank is R 60mm. Part I. In terms f the symblic crank angle, determine: (i) The cmpnents f the primary shaking frce n the crankshaft bearing. (ii) The magnitude(s) the lcatin(s); i.e., the angle(s), f the crrecting frce(s) created by the crrecting mass(es) which will balance the primary shaking frce. (iii) The magnitude(s) the lcatin(s) f the crrecting mass(es) which will balance the primary shaking frce if the radius f the crrecting mass(es) is R R ρ 80 mm. C C Part II. Fr the crank angle 60, determine the magnitude directin f the primary shaking frce vectr. Shw yur answers n the figure belw t the right. Als, shw the lcatin(s) f the crrecting mass(es) n this figure. ` Pistn mass = m L Reference Line Crank L Pistn mass = m Reference Line 60 R 0 60 L ω ρ -ω Pistn mass = m Crank angle = 60 Figure 4. A three-cylinder engine. 4

5 Slutin t Prblem. (i) 5 Pints. Link is a tw frce member link has fur frces a trque. The free bdy diagram f link is shwn in Figure. P A g = 9.8 m/s F P = 5 kn LOA F 4 G LOA F W = 5 kn 0 B T 4 = 500 Nm F B =.5 kn Figure. The free bdy diagram f link. Fr static equilibrium, the sum f the mments abut the shaft at B can be written as that is M B 0 (a) ET T (BG cs0 )W (BAcs0 ) F (BPcs0 ) F 0 (b) 4 P Nte that the internal reactin frce at pin A is assumed t be acting in the psitive -directin. Rearranging Eq. (b), the internal reactin frce at pin A can be written as F T 4 (BG cs 0 ) W (BP cs 0 ) FP (c) BA cs 0 Substituting the given trque T4 500 Nm the knwn values int Eq. (c), the internal reactin frce acting n link frm link at pin A is F 500 (.990)(5000) (.476)(5000) 97 N 9. kn (a).598 The psitive sign indicates that the internal reactin frce at pin A is acting in the psitive directin. 5

6 Therefre, the internal reactin frce at pin A acting n link frm link is F F 9. kn (b) The sum f the frces acting n link in the -directin can be written as that is F 0 (a) F ET F (b) 0 Rearranging Eq. (b) substituting Eq. (b) int the resulting equatin, the frces acting n link are F F 9. kn (c) Nte that link is in cmpressin, therefre, there is the pssibility that the link culd buckle. The free bdy diagram f link is shwn in Figure. Since the width t is less than the thickness h then the deflectin f link is in the plane (referred t as the in-plane buckling prblem). F = 9. kn A t O F = 9. kn Figure. In-plane buckling f link. The factr f safety guarding against buckling f link is defined as N P CR (4a) F 6

7 Substituting Eq. (c) the critical lad P CR against buckling f link is 75kN int Eq. (4a), the factr f safety guarding 75 N.9 (4b) 9. Therefre, link is safe against buckling. (ii) 4 Pints. The slenderness rati at the pint f tangency fr link can be written as CE (5) Sr D Syc Since the deflectin f link is in the - plane then the end-cnditins are pinned-pinned the endcnditin cnstant is C (6) Substituting Eq. (6) the given cmpressive yield strength mdulus f elasticity int Eq. (5), the slenderness rati at the pint f tangency fr link is Sr 4.8 D (7) (iii) 0 Pints. The slenderness rati f link is defined as The radius f gyratin f link is defined as Sr k L (8a) k I A (8b) If the width t is in meters then the crss-sectinal area f link is 7 A hxt5txt 5t m (8c) The area mment f inertia f link with the rectangular crss-sectin can be written as I b h (9a) Substituting the thickness b = t the width h = t int Eq. (9a), the area mment f inertia is (5t) (t) 4 4 I t m (9b) Nte that if the thickness b = t the width h = t then the area mment f inertia f link is (t) (5t) 4 4 I t m (9c)

8 Nte that the smaller f the tw values in Eqs. (9b) (9c) must be used t determine the slenderness rati f link the critical unit lad. Als, nte that fr ut f the plane buckling (that is, ut f the - plane), the end-cnditins f link are nt knwn. It is cmmn t assume that the ends are fixedfixed fr ut f plane buckling, in such a case, the end-cnditin cnstant is C 4. Substituting Eqs. (8c) (9b) int Eq. (8b), the radius f gyratin f link is t k t m (0) 5t Substituting Eq. (0) the length L OA.75m int Eq. (8a), the slenderness rati f link in terms f the width t (which is expressed in meters) is Sr () t t Case I. Assume that link is an Euler clumn. Accrding t the Euler clumn frmula, the critical lad fr link can be written as C E PCR A (a) S r Substituting Eqs. (8c) () the given data int Eq. (a) gives 9 x x t 0.780t N 6.06 ( ) t (b) Rearranging this equatin, the width f link is Substituting Eq. (c) int Eq. (), the slenderness rati f link is t m (c) 6.06 Sr () Recall that the Euler clumn frmula is nly valid when the slenderness rati is greater than the slenderness rati at the pint f tangency, that is 8 Sr Sr (4) D Cmparing Eq. (4) with Eq. (5) shws that link is an Euler clumn, that is, ur initial assumptin is crrect. (iv) 6 Pints. The area mment f inertia fr a circular crss-sectin is I 4 d 64 (5a)

9 Substituting the diameter d 60 mm = 0.06 m int Eq. (5a), the area mment f inertia f link is 4 x I m 64 (5b) The crss-sectinal area f link is d x A m (6) 4 4 Substituting Eqs. (5b) (6) int Eq. (7a), the radius f gyratin f link is k m (7) Substituting Eq. (7) the length L AO.75m int Eq. (6), the slenderness rati is.750 Sr 6.67 (8) Cmparing Eq. (8) with Eq. (5), the cnclusin is that the slenderness rati Sr Sr (9) D Therefre, the circular crss-sectin link is a Jhnsn clumn. The critical lad fr this clumn is calculated using S C S r PCR A S C CE (0a) S r Substituting Eqs. (6) (8) the given data int Eq. (0a), the critical lad fr link can be written as (050 ) 6.67 PCR (0b) 9 x 070 Therefre, the critical lad fr link is PCR N 8.76 kn (0c) Substituting Eqs. (6) (0c) int Eq. (7a), the factr f safety guarding against buckling is 8.76 N 9.7 () 9. Since the factr f safety is greater than ne, therefre the clumn is nt expected t buckle in the circular crss-sectin link. 9

10 Slutin t Prblem. (i) 5 Pints. The vectrs fr the spring are shwn in Figure. P R F P = 50 N g = 9.8 m/s R S A,G C k T 4 4 R 4 B,G 60 O 4,G 4 D Figure. The vectrs fr the spring. The vectr lp equatin fr the spring can written as The cmpnents f Eq. () can be written as?? R4 R RS 0 () R4 4 R R S S cs cs cs 0 (a) R4 4 R R S S sin sin sin 0 (b) Frm the right-angled triangle O4AP, the length f the spring the distance O4A, respectively, are RS.5 m R4.65 m (a) Check: Substituting S 90 knwn values int Eq. (b), the length f the spring can be written as RS R4 sin 60.5 sin50 (b) Then substituting Eq. (a) int Eq. (b), the length f the spring fr the given input psitin is R.65(0.866).5(0.5).5 m (c) S 0

11 Differentiating Eqs. () with respect t the input psitin gives R sin R sin R cs R sin 0 (4a) S S S S S R cs R cs R sin R cs 0 (4b) S S S S S Substituting S 90 the knwn values int Eq. (4b), the first-rder kinematic cefficient f the spring RS R4 cs44 Rcs.085 m/rad (5a) The psitive sign indicates that the spring is increasing in length as the input link rtates cunterclckwise. Similarly frm Eq. (4a), the first-rder kinematic cefficient f the spring R4sin44 Rsin RS cs S S 0.5 rad/rad (5b) R S The psitive sign indicates that the spring is rtating cunterclckwise as the input link rtates cunterclckwise. (ii) 5 Pints. The vectrs fr the viscus damper are shwn in Figure. P F P = 50 N g = 9.8 m/s k R 4 A,G R C C R T B,G O 4,G 4 D Figure. The vectrs fr the viscus damper. The vectr lp equatin fr the viscus damper can be written as

12 ?? R R R 4 C 0 (6) The cmpnents f Eq. (6) can be written as R R R 4 cs 4 C cs C cs 0 (7a) R R R 4 sin 4 Csin C sin 0 (7b) Differentiating Eqs. (7) with respect t the input psitin gives Substituting damper R sin R cs R sin 0 (8a) C C C C C R R R 4cs 4 4 C sin C C cs C C 0 (8b) C 80 the knwn values int Eq. (8a), the first-rder kinematic cefficient f the R C.65(sin 60 )( 0.5) m/rad (9a) The psitive sign indicates that the length f the damper is increasing as the input link rtates cunterclckwise. Similarly, frm Eq. (8b), the first-rder kinematic cefficient f the damper R4 cs (cs 60 )( 0.5) C rad/rad (9b) R.5 C The negative sign indicates that the damper is rtating clckwise as the input link rtates cunterclckwise. (iii) pints. The pwer equatin fr the mechanism can be written as In terms f kinematic cefficients, Eq. (0a) can be written as dt du dw f T4. 4 FP. VP (0a) dt dt dt (0b) 4 ( 4 ) P( P ) i i i ( S S0) i S C T F A B m gy K R R R CR Nte that the trque T 4 is assumed t act in the same directin as the angular velcity f link 4. Cancelling the input angular velcity (which is psitive) in Eq. (0b) gives the equatin f mtin, that is T4 ( 4) FP( P ) A Bi migyi K( RS RS0) R i S CRC () Cnsider the terms n the right h side f Eq. (). The cefficients G G G 0 (a) A m I

13 A m G G IG (b) 4 4 G 4 G 4 G4 4 0 (c) A m I The vectrs fr a kinematic analysis f the mass center f link are shwn in Figure. P F P = 50 N g = 9.8 m/s A,G C k R 4 4 T 4 B,G 60 R 7 O 4,G 4 R 9 D Figure. The vectrs fr the mass center f link. The vectr equatin fr the mass center f link can be written as The crdinates f the mass center f link are?? R G = R 4 (a) G 4 4 (4a) R cs θ.65(cs 60 ).085 m G 4 4 (4b) = R sin θ.65(sin 60 ) =.8750 m Differentiating Eqs. (4) with respect t the input psitin, the first-rder kinematic cefficients f the mass center f link are G = R sin θθ.65(sin 60 )( 0.5) m/rad (5a)

14 G The crdinates f pint P are = R csθθ =.65(cs 60 )( 0.5) 0.54 m/rad (5b) P RScsθS 0 (6a) =R sinθ.5 m P S S (6b) Differentiating Eqs. (6) with respect t the input psitin, the first-rder kinematic cefficients f pint P are P = RSsinθθ S S RScsθ S = RS θs 0.65 m / rad (7a) P = RScsθθ S S RSsin θs RS.085 m/rad (7b) Substituting the knwn values Eqs. (5) int Eq. (b) gives A 5[( 0.975) ( 0.54) ] 9( 0.5) 9.88 kg-m (8) Nte that Eq. (8) is the equivalent mass mment f inertia f the mechanism. Fr links,, 4, the cefficients are B mg G G G IG 0 (9a) G G G G G B m I B m I Substituting the knwn values int Eq. (9b) gives (9b) 4 4 G4 G4 G4 G4 G (9c) B 5[( 0.975)( 0.706) ( 0.54)( 0.78)] kg-m (0) The gravitatinal term can be written as mg j Gj mg G j Substituting the knwn data int Eq. (a), the gravitatinal term is (a) mg j Gj N-m (b) j Substituting the knwn values int the spring term gives Ks RS RS0 R S 5000 (.5 )(.085) N-m () Substituting the knwn values int the damping term gives 4

15 C CR 00 ( 0.975) (.5) N-m () Substituting the knwn values int Eq. (), the equatin f mtin can be written as ( 0.5)T 50( 0.65) [(9.88)(.5) ( 0.487)(.5) ] (4a) 4 that is 0.5 T ( ) (4b) Rearranging Eq. (4b), the trque can be written as T Nm (5a) The psitive sign indicates that the assumptin that the trque is in the same directin as the angular velcity f link 4, that is, cunterclckwise, is crrect. Therefre, the trque frm the shaft acting n link 4 can be written as T k Nm 6.58 k knm (5b) 5

16 Slutin t Prblem. Part I. (i) Pints. The inertia frces f the three rtating mass particles are F (a) mr (.5 kg)(0.00 m)(0 rad/s) 000 N F F mr (7.5 kg)(0.75 m)(0 rad/s) 55 N (b) mr (5 kg)(0.5 m)(0 rad/s) 450 N (c) Therefre, the inertia frces f the three rtating mass particles can be written as F 000 N 5707.i707. j N F 55 N i 6.50 j N F 450 N 089.7i5.00 j N (a) (b) (c) (ii) 5 pints. The sum f the inertia frces f the tw rtating mass particles can be written as F F F F i 9.6 j N N 64. () Therefre, the sum f the reactin frces at bearings A B can be written as FA FB FF F i9.6 j N The sum f the mments abut bearing A can be written as.00kf k( 707.i 707. j) 0.45 k( i 6.50 j) B 0.5 k( 89.7i 5.00 j) 0 (4) (5) The cmpnents f Eq. (5) can be written as.00 FB ( 0.675)(707.) ( 0.45)( 6.50) ( 0.5)( 55.00) 0 (6a).00 F B ( 0.675)( 707.) ( 0.45)( ) ( 0.5)( 89.7) 0 (6b) Therefre, the cmpnents f the reactin frce at bearing B are F 45.4 N F 6.05 N (7a) B B Therefre, the frce at bearing B acting n the shaft can be written as F 45.4i 6.05 j N = N.9 B Substituting Eq. (7b) int Eq. (4) rearranging, the frce at bearing A acting n the shaft is F 45.4i 6.05 j i9.6 j N A (7b) (8a) 6

17 Therefre, the frce at bearing A acting n the shaft can be written as F 46.7 i96.44 j N = N 5.54 A (iii) 6 Pints. The sum f the tw crrecting frces can be written frm Eq. (4) as F F F F F i 9.6 j N C C The sum f the mments abut crrecting plane () can be written as 0.5 kf0.5 kf0.675 kf 0.85 kf C 0 (8b) (9) (0) The cmpnents f Eq. (0) can be written as (0.5)( 707.) (0.5)( 6.50) (0.675)( 55.00) (0.85) 0 (a) F C F C (0.5)( 707.) (0.5)( ) (0.6)(89.7) (0.85) 0 (b) Rearranging Eqs. (), the inertia frce in crrecting plane () can be written as FC 9.79 i.84 j 64.8 N.5 Substituting Eq. () int Eq. (9), the inertia frce in the secnd crrecting plane is FC 9.79 i.84 j i9.6 j N which can be written as FC 4.7 i 5.44 j N 0.8 () (a) (b) Part II. Pints. The sum f the mments abut the crrecting plane is M 0 (a) () which can be written as [ZA FA ZB F B] ZC FC 0 (b) where ZA 95mm, ZB 75mm, ZC t as the crrecting frce) created by adding a crrecting mass C (b), the frces frm the rtr acting n the grund are 750 mm. Nte that F C is the inertia frce (referred m in crrecting plane (). Frm Figure F 69.8 i 40 j N (a) A FB 60 i 0.9 j N (b) Substituting Eqs. () int Eq. (b), the sum f the mments abut crrecting plane can be written as 0.95 k (69.8 i 40 j) 0.75 k ( 60 i 0.9 j) k F 0 () C 7

18 Rearranging Eq. () gives 8.8 i j 0.75( F i F j) 0 (7) C C Rearranging Eq. (7), the crrecting frce in plane () can be written as FC i FC j 5.08i j N (8) Therefre, the magnitude f the crrecting frce in plane () is the angle is C FC 0.56 N (9a) 5.08 tan (9b) Nte that Eq. (9b) is valid fr adding crrecting mass, that is, the crrecting frce in plane () is FC 0.56 N i 5.08 j N (0) Since the prblem states that mass must be remved then the angular rientatin f the crrecting mass t be remved in the crrecting plane () is btained frm the cnstraint ( ) ( ) 80 (7a) C remval C adding Substituting Eq. (5b) int Eq. (7a), the crrecting mass remved in crrecting in plane () is ( ) (7b) C remval The crrecting mass that must be remved in the crrecting plane () can be written as m F 0.56 N 0.75 kg (8) C C RC (0.5 m)(50 rad/s) The sum f the mments abut the crrecting plane is M 0 (0a) () which can be written as [ZA FA ZB F B] ZC FC 0 (0b) where ZA 75mm, ZB 95 mm, ZC 750mm. Nte that F C is the inertia frce (referred m in the crrecting plane (). t as the crrecting frce) created by adding a crrecting mass C Substituting Eqs. () int Eq. (0b), the crrecting frce in plane () can be written as r as 0.75 k ( 69.8 i 40 j) 0.95 k ( 60 i 0.9 j) 0.75 k F C 0 (a) FCi FCj 8.84 i 90.7 j (b) Therefre, the magnitude f the crrecting frce in plane () is 8

19 the angle is C FC 49.7 N (a) 8.84 tan (b) Nte that Eqs. () are valid fr adding crrecting mass, that is, the crrecting frce in plane () is FC 49.7 N i 8.84 j N () Since the prblem states that mass must be remved then the angular rientatin f the crrecting mass that must be remved in the crrecting plane () is ( ) ( ) 80 (4) C remval C adding Substituting Eq. (b) int Eq. (4), the angular rientatin f the crrecting mass that must be remved in crrecting plane () is ( ) (5) C remval The crrecting mass in crrecting plane () can be written as which can be written as m F (6a) C C RC 49.7 N mc kg (6b) (0.5 m)(50 rad/s) Check: Fr frce balanced, the sum f the frces can be written as that is F 0 (8a) [ F F ] F F 0 (8b) A B C C Substituting Eqs. (), (6), () int Eq. (8b) gives ( 69.8 i 40 j) ( 60 i 0.9 j) ( i 5.08 j) ( 90.7 i 8.84 j) 0 (9a) that is 0 0 (9b) This prves that the tw crrectin frces prduced by the remval f the tw crrecting masses causes the system t be dynamically balanced. The tw crrecting masses their angular lcatins are shwn in Figure.. 9

20 F C = 49.7 N F C = 0.56 N 4.6 ω (F ) B = 0 N (F ) A = 80 N (m C ) remve (m C ) remve Figure.. The tw crrecting masses t be remved. 0

21 Slutin t Prblem 4. (i) 8 pints. The -cmpnents f the resultant f the primary shaking frce can be written as (a) S S S S S S S S (b) The cmpnents f the primary shaking frce fr cylinder can be written as S Pcs( )cs (a) S Pcs( )sin (b) The cmpnents f the primary shaking frce fr cylinder can be written as S Pcs( )cs (a) S Pcs( )sin (b) The cmpnents f the primary shaking frce fr cylinder can be written as S P cs( )cs (4a) S P cs( )sin (4b) Frm Figure 4, the angles are 0, 0, 0, 70, 0 Substituting these angles P P P P mr int Eqs. (a), (a) (4a), the results int Eq. (a), the -cmpnent f the resultant f the primary shaking frce is S Pcs( 0 )cs0 Pcs( 700 )cs 70Pcs( 00 )cs0 (5a) that is S.0 Pcs (5b) Similarly, substituting the angles int Eqs. (b) (b), the results int Eq. (b), the -cmpnent f the resultant f the primary shaking frce is S Pcs( 0 )sin 0 Pcs( 700 )sin 70Pcs( 00 )sin0 (6a) Therefre, the -cmpnent f the resultant f the primary shaking frce is S.0 Psin (6b) The magnitude f the resultant f the primary shaking frce can be written as S S S (7a) P

22 Substituting Eqs. (5b) (6b) int Eq. (6a), the magnitude is SP P sin The directin f the resultant f the primary shaking frce can be written as (7b) S S Substituting Eqs. (5b) (6b) int Eq. (8a), the directin f the resultant f the primary shaking frce is tan (8a) The magnitude f the frce P can be calculated as Psin sin P cs cs tan tan (8b) Substituting Eq. (9) 60 int Eqs. (5b) (6b) gives P m R kn (9) S 8.50 cs kN (0a) S 56.50sin kN (0b) Substituting Eq. (9) 60 int Eqs. (7b) (8b) gives S 8.50 sin 60 P 507.0kN (a) sin 60.7 cs tan tan 7.90 (b) Equatins (0a) (0b) are the x y cmpnents f the primary shaking frces. (ii) 8 Pints. The cmpnents f the resultant f the primary shaking frce fr the engine, that is, Eqs. (), can be written as S Acs Bsin (a) S C cs Dsin (b) The primary shaking frce can be balanced by a pair f rtating masses. These tw crrecting frces plus the resultant f the primary shaking frce must be equal t zer; i.e., Acs Bsin F cs ( ) F cs [ ( ) ] 0 (a) C cs Dsin F sin ( ) F sin [ ( ) ] 0 (b) Exping these tw equatins, in terms f the angles f,,, rearranging, gives ( F cs F cs ) cs ( F sin F sin ) sin Acs Bsin (4a)

23 ( F sin F sin ) cs ( F cs F cs ) sin C cs Dsin (4b) T satisfy Eqs. (4), fr all values f the crank angle, the necessary cnditins are F cs F cs A (5a) F sin F sin B (5b) F sin F sin C (5c) F cs F cs D (5d) Slving Eqs. (5), the crrecting frces are F A D B C (6a) F A D B C (6b) Als, frm Eqs. (5), the lcatin angles f the crrecting frces are B C tan ( A D) (7a) tan B C ( A D) (7b) Cmparing Eq. (5b) with Eq. (a) Eq. (6b) with Eq. (b), the cefficients are A.0 P, B0, C 0, D.0 P (8) Substituting Eq. (8) int Eqs. (6), the tw crrecting frces are F.0 P.0 P P (9a) F.0 P.0 P P (9b) Substituting Eq. (9) int Eqs. (9), the tw crrecting frces are F kn F kn (0) (iii) 5 pints. Recall that the crrecting frces are defined as F m C R C F m C R C () Substituting RC R C 0.08m int Eqs. (), rearranging, the crrecting masses are

24 m m F 4875 N 84.75kg C RC m/s F 4065 N 8.5 kg C RC m/s (a) (b) Substituting Eq. (8) int Eq. (7a), the lcatin angle fr the first crrecting mass is 0P 0P 0 tan ( P P) (a) Since the numeratr is 0 the denminatr is negative then the angle fr the first crrecting mass is 80 (b) Substituting Eq. (8) int Eq. (7b), the lcatin angle fr the secnd crrecting mass is 0P 0P 0 tan ( P P) (4a) Since the numeratr is 0 the denminatr is psitive, then the angle fr the secnd crrecting is 0 (4b) (iv) 6 Pints. The answers fr parts (i) (ii) (iii), that is, Eqs. (a), (b), (0), (), (), (4), are shwn n Figure 4. where the reference line is specified at the crank angle θ 60. 4

25 S p = kn Reference Line ω 60 τ = 7.9 θ+γ = 40 m C = kg ρ -ω θ + γ ) = 60 m C = 8.5 kg F = kn F = kn Figure 4.. The reference line is at the crank angle θ 60. 5

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