Figure 1a. A planar mechanism.

Transcription

1 ME 5 - Machine Design I Fall Semester 0 Name f Student Lab Sectin Number EXAM. OPEN BOOK AND CLOSED NOTES. Mnday, September rd, 0 Write n ne side nly f the paper prvided fr yur slutins. Where necessary, use the figures prvided n the exam t shw vectrs instant centers. Any wrk that cannt be fllwed is assumed t be in errr. Staple each prblem separately staple yur crib sheet t the end f Prblem. Prblem (5 Pints). Part I. (5 Pints). (i) Determine the mbility f the mechanism shwn in Fig. a (number the links label the lwer higher pairs n the figure). Des the Kutzbach criterin give the crrect answer? (ii) Define suitable vectrs fr a kinematic analysis. Label shw each vectr n the figure. (iii) Write the vectr lp equatins that are required fr the kinematic analysis. Clearly identify suitable input(s) list: (a) the knwn quantities; (b) the unknwn variables; (c) any cnstraints. If yu identified cnstraints in part (c) then write the cnstraint equatins. Figure a. A planar mechanism. Part II. (0 Pints). Fr the fur-bar linkage shwn in Fig. b, the input link OA 60mm, the cupler link AB 0 mm, the utput link BO 00 mm, the grund link O O 80 mm. Determine the input angles the transmissin angles when the mechanical advantage is infinite. O Y A O Figure b. The fur-bar linkage. B X

2 ME 5 - Machine Design I Fall Semester 0 Name f Student Lab Sectin Number Prblem (5 Pints). Fr the mechanism in the psitin shwn in Fig., the distance frm the grund pivt O t pin A (cnnecting links ) is 75 mm. The length f link is OC 5 mm. The velcity f the input link is V 5 i m/s. Using the analytical methd, determine: (i) The first-rder kinematic cefficients f the mechanism. (ii) The angular velcity f link. Give the magnitude directin f this vectr. (iii) The velcity f link alng link. Give the magnitude directin f this vectr. (iv) The magnitude directin f the velcity f pint C. Shw the directin f this vectr n Fig.. Y 60 O X V = 5 m/s A C Figure. A planar mechanism.

3 ME 5 - Machine Design I Fall Semester 0 Name f Student Lab Sectin Number Prblem (5 Pints). Fr the mechanism in the psitin shwn in Fig., the angular velcity f the input link is ω 50 rad / s cunterclckwise. The mechanism is drawn full size; i.e., mm = mm. (i) List the primary instant centers list the secndary instant centers. (ii) Using the given Kennedy circle, shw the lcatin f all the instant centers n Fig.. Using the lcatin f the instant centers, determine: (iii) The first-rder kinematic cefficients f links,, 5. (iv) The angular velcities f links. Give the magnitude directin f each vectr. (v) The velcity f the slider, link 5. Give the magnitude directin f this vectr. Kennedy Circle B. C. 5 ω = 50 rad/s. O. A. O Figure. A planar mechanism. Scale: full size mm = mm.

4 ME 5 - Machine Design I Fall Semester 0 Name f Student Lab Sectin Number Prblem (5 Pints). Fr the mechanism in the psitin shwn in Fig., pin B cnnecting links is lcated vertically abve the grund pint O. The link lengths are OA 500mm, AB 00 mm, the radius f the circular grund link is 50 mm. The radius f the circular wheel which is rlling withut slipping at the pint f cntact C n the grund link is 50 mm. The cnstant angular velcity f the input link is ω 5 rad / s cunterclckwise. (i) Write a suitable vectr lp equatin fr this mechanism. Draw yur vectrs clearly n Fig.. (ii) Using yur vectr lp equatin determine the first-rder kinematic cefficients fr the mechanism. (iii) Determine the angular velcities f links. Give the magnitude directin f each vectr. (iv) Determine the velcity f pin B. Give the magnitude directin f this vectr. Y A ω = 5 rad/s ρ = 50 mm 60. C B ρ = 50 mm O O X Figure. A planar mechanism.

5 Slutin t Prblem. Part I. (i) pints. The links, the lwer pairs, the higher pairs f the mechanism are as shwn in Figure (a). j j 5 j j j j 6 j 7 j j Figure (a). The links the jints f the mechanism. The Kutzbach mbility criterin can be written as M (n ) j j () The number f links, the number f lwer pairs (r j jints), the number f higher pairs (r j jints), respectively, are n 7, j 8, j () Substituting Equatin () int Equatin (), the mbility f the mechanism can be written as Therefre, the mbility f the mechanism is M (7 ) (8) () () M () This is the crrect answer fr this mechanism, that is, fr a single input there is a unique utput. 5

6 (ii) Pints. Suitable vectrs fr a kinematic analysis f the mechanism are shwn in Figure (b). R R R R 5 R φ R R 5 R 6 R R 6 R 6 7 R Figure (b). Vectrs fr the mechanism. (iii) 7 Pints. (a) The input link is chsen here t be link (see Figure a) the input variable is the angle. Nte that link wuld be suitable as an input link since it pinned t the grund link the input variable wuld be the angle. Als, the slider (link 5) wuld be suitable as an input link the input variable wuld be the length R. 5 Anther ptin fr the input is the psitin f link 6 (the pistn) relative t link 7 (the cylinder ). Anther ptin fr the input is link 7 since it pinned t the grund link the input variable wuld be the angle. 7 Other chices fr the input link are nt really practical. (b) The six unknwn variables are the fur angles,, R 5, 6 the tw distances R R 6. (c) There are three cnstraints here, the three cnstraint equatins can be written as (a) 6 (b) (c) Since there are 6 unknwn variables then three independent vectr lp equatins are required fr a kinematic analysis f the mechanism. The three vectr lp equatins can be written as: Vectr lp is Vectr lp is I?? R RR5R 0 (5a) I C? C? R R R R R R 0 (5b) 6

7 Vectr lp is I C?? R R6 R6 R 0 (5c) Part II. 0 pints. The mechanical advantage is infinite when links are fully extended r flded n tp f each ther (tggle psitins), that is, when the angle 0 r 80. Y A β O O X Figure (c). The fur-bar linkage. B Grashf s law fr a planar fur-bar linkage, see Sectin.9, that is, Equatin (.6), page 6, states that in rder fr the fur-bar linkage t be a crank-rcker fur-bar linkage then The lengths f the fur links f the fur-bar linkage are specified as s l p q () s 60 mm (a) l 0 mm (b) p 80 mm (c) q 00 mm (d) Substituting these dimensins int Equatin () gives r 60 mm 0 mm 80 mm 00 mm (a) 80 mm 80 mm (b) Since the inequality hlds then the fur-bar linkage is a crank-rcker fur-bar linkage. Als, since the shrtest link s is adjacent t the grund link then the shrtest link is a crank, see Sectin.9. Fr the first tggle psitin, that is, when the angle 0, then Using the law f csines gives OB ABOA mm () 7

8 BO OO OB OO OBcs(80 ) (5) Substituting Equatin () BO 00 mm OO 80 mm int Equatin (5) gives Rearranging this equatin gives cs(80 ) (6a) cs(80 ) (6b) Therefre, the angular psitin f the input link is Using the law f csines, the transmissin angle can be written as (7) OO OB BO OBBO cs (8) Substituting Equatin () BO 00 mm OO 80 mm int Equatin (8) gives Rearranging this equatin gives cs (9a) cs (9b) Therefre, the transmissin angle is (0) The law f sines can als be used t determine the transmissin angle. Hwever, nte that the transmissin angle is greater than 90, that is, in the secnd quadrant. This is especially imprtant when the sine f the angle is used t determine the transmissin angle. This will give 89. which is the wrng answer. Fr the secnd tggle psitin, that is, when the angle Using the law f csines gives 8 80 then OB ABOA 80 mm () BO OO OB OO OBcs(60 ) () Substituting Equatin (7) BO 00 mm OO 80 mm int Equatin () gives cs(60 ) (a)

9 Rearranging this equatin gives cs(60 ) 8080 (a) Therefre, the angular psitin f the input link is Using the law f sines gives () 0 OB BO sin sin (60 ) (5a) Rearranging Equatin (5a) gives Substituting Equatin () OO Therefre, the transmissin angle is OB sin (60 ) sin (5b) BO 80 mm BO 00 mm int Equatin (5b) gives sin 0 (6a) 0 r 80 Fr this fur-bar linkage, the crrect answer fr the transmissin angle is (7a) 0 (7b) 9

10 Slutin t Prblem. (i) Pints. A suitable chice f vectrs fr the mechanism are as shwn in Fig. a. Y O R X V = 5 m/s C A R R Figure a. A suitable chice f vectrs fr the mechanism. The vectr lp equatin (VLE) fr the mechanism can be written as I?? R R R 0 (a) The input is the psitin f link, that is, R, the cnstraints are (b) The X Y cmpnents f the VLE, see Equatin (a), are Rcs RcsR cs 0 (a) Rsin Rsin R sin 0 (b) Differentiating Equatins () with respect t the input psitin R gives cs R cs R sin 0 (a) sin R sin R cs 0 (b) 0

11 Equatins () can be written in matrix frm as cs Rsin R cs sin Rcs sin () Using Cramer s rule, the first-rder kinematic cefficients are where the determinant is R R cs( ) DET (5a) sin ( ) DET (5b) DET R 75 mm (6) Substituting Equatins (b) (6) int Equatins (5), the first-rder kinematic cefficients are R 0.5rad/rad (7a) where the psitive sign indicates that the vectr R is getting shrter as the input link mves t the right (that is, the vectr R is getting shrter).55 rad / m (7b) where the negative sign indicates that link is rtating cunterclckwise as the input link mves t the right (that is, the vectr R is getting shrter). (ii) pints. The angular velcity f link can be written as Nte that the input velcity is R (8a) (8b) R V 5m/s that is, the vectr R is getting shrter as link mves t the right. Substituting Equatins (7b) (8b) int Equatin (8a), the angular velcity f link is The psitive sign indicates that link is rtating cunterclckwise. (iii) Pints. The velcity f link alng link can be written as (.55) ( 5) rad/ s (9) V R R (0a) Substituting Equatins (7a) (8b) int Equatin (0a), the velcity f link alng link is V ( 0.5)( 5).5 m / s (0b)

12 The negative sign indicates that link is mving upward alng link, that is, the vectr R is getting shrter fr the given input psitin velcity. (iv) Pints. The magnitude f the velcity f pint C can be written as where R the velcity vectr V C R (a) V c are shwn in Fig. b. The directin f the velcity f pint C is 0 (b) Y O X A R C V C Substituting Equatin (9) R C is Figure b. The vectr fr pint C. Alternatively. The directin f the velcity f pint C is 5mm int Equatin (a), the magnitude f the velcity f pint VC (0.5 m) 6.09 m / s (a) V c (6.09cs0 i 6.09sin 0 j).5 i 8.05 j m / s (b) Check. The X Y crdinates f pint C can be written as X Y c Rcs (a) c Rsin (b) Differentiating Equatins () with respect t the input R gives

13 X Y c Rsin (a) c Rcs (b) Substituting Equatin (8b) int Equatins () gives X c ( 0.5)(sin 0 )(.55).5 m / rad (5a) Y c ( 0.5)(cs 0 )(.55) 0.7 m / rad (5b) The velcity f pint C can be written as V c (Xci Yc j)r (6) Substituting Equatins (5) the velcity f the input link int Equatin (6), the velcity f pint C is V c (.5 i 0.7 j)( 5.00).5 i 8.00 j m / s (7a) The magnitude f the velcity f pint C is Vc m / s Nte that Equatins (7) are in gd agreement with Equatins (). (7b) Alternative vectr lp. Vectrs fr the mechanism can be chsen as shwn in Figure. Y R O X R V = 5 m/s R R A C Figure. The vectrs fr the mechanism.

14 The vectr lp equatin (VLE) fr the mechanism can be written as I?? C R R R R 0 (a) Nte that the input is the linear psitin f link, that is, R the cnstraints are 0, 80, 70 (b) (i) pints. The X Y cmpnents f the VLE, see Equatin (), are R cs R cs R cs R cs 0 (a) R sin R sin R sin R sin 0 (b) Substituting Equatins (b) int Equatins (), gives R R R cs 0 (a) R R sin 0 (b) Differentiating Equatins () with respect t the input psitin R gives R cs R sin 0 (a) R sin R cs 0 (b) Equatins () can be expressed in matrix frm as cs Rsin R sin R cs 0 (5) Using Cramer s rule, the first-rder kinematic cefficients are where R R cs DET (6a) sin DET (6b) DET R 75 mm (7) Finally, using the Equatins (6) t (7), the first-rder kinematic cefficients are R cs 0.5 rad / rad (8a)

15 sin R.55 rad / m (8b) The negative sign f R indicates that the vectr R is getting shrter as the input vectr R is getting lnger (that is, link is mving t the right) the psitive sign f indicates that link is rtating cunterclckwise as link mves t the right (that is, the vectr R is getting lnger). Nte that these answers are in agreement with the first vectr lp equatin. The angular velcity f link can be written as R (9a) Nte that R V 5m/s, that is, the vectr R is getting shrter as link mves t the right. Substituting Equatin (7b) the velcity f the input link int Equatin (8a) gives The psitive sign indicates that link is rtating cunterclckwise. The velcity f link alng link can be written as (.55) ( 5) rad/ s (9b) V R R (0a) Substituting Equatin (8a) the velcity f the input link int Equatin (0a) gives V ( 0.5)( 5).5 m / s (0b) The negative sign indicates that link is mving upward alng link, that is, the vectr R is getting shrter fr the given input psitin velcity. 5

16 Slutin t Prblem. (i) 5 Pints. The number f links is five, therefre, the ttal number f instant centers is n( n) 5 N 0 () There are 5 primary instant centers; namely, I, I, I, I 5, I 5. Therefre, there are 5 secndary instant centers; namely, I, I, I 5, I 5 I. The prcedure t lcate the secndary instant centers is: (i) The pint f intersectin f the line thrugh I I the perpendicular line t the link (where center I is lcated) is the instant center I. (ii) The pint f intersectin f the line thrugh I 5 I 5 the line thrugh I I is the instant center I. (iii) The pint f intersectin f the line thrugh I I 5 the line thrugh I I 5 is the instant center I 5. (iv) The pint f intersectin f the line thrugh I I 5 the line thrugh I I 5 is the instant center I 5. (v) The pint f intersectin f the line thrugh I I the line thrugh I I is the instant center I. I 5 at I 5 at I I 5 I I I I 5 I 5 I 5 at I I Figure. The lcatins f the instant centers. 6

17 (ii) 0 Pints. The first rder kinematic cefficient f link can be written as I I (a) II Frm Figure, the distances are measured as I I 7.80 cm I I.57 cm. Therefre, the first rder kinematic cefficient f link is I I rad/rad (b) II.57 The relative instant center I des nt lie between the tw abslute relative centers I I, therefre, the first-rder kinematic cefficient f link is psitive. The first rder kinematic cefficient f link can be written as I I (a) II Frm Figure, the distances are measured as I I 9.85 cm I I 6.89 cm. Substituting these values int Equatin (a), the first-rder kinematic cefficient f link is rad/rad (b) 6.89 The relative instant center I lies between the abslute relative centers I I, therefre, the firstrder kinematic cefficient f link is negative. The first rder kinematic cefficient f link 5 can be written as R I I (a) 5 5 Frm Figure, the distance is measured as I I 5.6 cm. Therefre, the first-rder kinematic cefficient f link 5 is R 5.6 cm/rad (b) Nte that the crrect sign f the first-rder kinematic cefficient f link 5 depends n hw the vectr R 5 is defined. If the vectr is t the left f link 5 then the first-rder kinematic cefficient f link 5 is negative if the vectr is t the right f link 5 then the first-rder kinematic cefficient f link 5 is psitive (link 5 is mving t the left as the input angle rtates cunterclckwise). (iv) 6 Pints. The angular velcity f link can be written as (5a) Substituting Equatin () the input angular velcity int Equatin (5a) gives ( 0.6)( 50.00).00 rad s (5b) The psitive sign implies that the directin f the angular velcity f link is cunterclckwise. The angular velcity f link can be written as 7

18 (6a) Substituting Equatin () the given input angular velcity int Equatin (6a) gives (.)( 50) 7.50 rad/s (6b) The negative sign implies that the directin f the angular velcity f link is clckwise (fr the given cunterclckwise angular velcity f the input link). The velcity f link 5 can be written as V R (7a) 5 5 Substituting Equatin () the given input angular velcity int Equatin (7a) gives V5 (.6)( 50).00 cm/s (7b) The negative sign implies that link 5 is mving t the left as the input link rtates cunterclckwise. (v) Pints. The velcity f pint C n link 5 is equal t the velcity f pint C n link, that is V ( I C ) ( I I ) (8a) C 5 The distance is measured as I I 5.98 cm. Substituting this measurement Equatin (6b) int Equatin (8a), the velcity f pint C is V (.98)( 7.50 rad/s).07 cm/s (8b) C The negative sign indicates that the velcity f pint C is directed t the left as the input link rtates cunterclckwise. 8

19 Slutin t Prblem. (i) 8 Pints. A set f vectrs fr a kinematic analysis f the mechanism are shwn in Fig. (a). Y A ω = 5 rad/s R R ρ = 50 mm B C 60 R 7 ρ = 50 mm O R O X Figure (a). Suitable vectrs fr the mechanism. The vectr lp equatin (VLE) can be written as I?? R RR7 R 0 () where the vectr R 7 is the arm. Nte that link des nt appear in the (VLE). Therefre, the angular velcity f link must be btained later frm the rlling cntact cnstraint. Psitin Analysis. T determine the angular psitin f link using trignmetry, see Fig. (b). 9

20 A D β B O O O Figure (b). The trignmetry f the mechanism. Using the right-angled triangle OAO, the vertical distance AO is Therefre, the distance AD is AO R sin 500sin 60.0 mm (a) AD AO DO AO BO mm (b) Using the right-angled triangle ADB, the angle ABD can be btained frm the relatin Therefre, the angle ABD is AD.0 mm sin AB 00 mm (a) (b) Therefre, the angular psitin f link is (c) (ii) 0 Pints. The X Y cmpnents f the VLE, see Equatin (), are Rcs Rcs R7cs 7 Rcs 0 (a) Rsin Rsin R7sin 7 Rsin 0 (b) Nte that the angles fr the given psitin f the mechanism are 0, 60, 7 90 (5) 0

21 Differentiating Equatins () with respect t the input psitin gives R sin R sin R sin 0 (6a) R cs R cs R cs 0 (6b) Writing Equatins (6) in matrix frm gives Rsin R7sin 7 Rsin Rcs R7cs 7 7 Rcs (7a) Substituting Equatins () (5) int Equatin (7a) gives 00sin sin sin cs cs90 500cs 60 (7b) The determinant f the cefficient matrix in Equatin (7b) is DET mm (8) Using Cramer s rule, the first-rder kinematic cefficient f link is rad / rad (9) The negative sign indicates that link is rtating clckwise fr a cunterclckwise rtatin f the input link. Similarly, the first-rder kinematic cefficient f link 7 (the arm) is rad / rad (0) The psitive sign indicates that link 7 is rtating cunterclckwise fr a cunterclckwise rtatin f the input link. (ii) 5 Pints. The angular velcity f link can be written as (a) Substituting Equatin (7) the input angular velcity int Equatin (9a) gives (.)( 5).00 rad/ s (b) The negative sign indicates that link is rtating clckwise as the input link rtates cunterclckwise. The angular velcity f link 7 (the arm) can be written as 7 7 (a) Substituting Equatin (0) the input angular velcity int Equatin (a), the angular velcity f link 7 is 7 (.7)( 5) 9.75 rad/ s (b)

22 The psitive sign indicates that the arm (link 7) is rtating cunterclckwise. The rlling cntact cnstraint between link the grund link can be written as 7 7 (a) The crrect sign is negative because there is external cntact between link link. Nting that 0 substituting Equatin (9) int Equatin (a) gives 50 (.7) 50 0 (.7) (b) Therefre, the first-rder kinematic cefficient fr link is The angular velcity f link can be written as.8 rad/rad () (5a) Substituting Equatin () int Equatin (a), the angular velcity f link is (.80 rad / rad)( 5 rad/s) 7.00 rad / s (b) The psitive sign indicates that link is rtating cunterclckwise as the input link rtates cunterclckwise. (iv) Pints. The velcity f pint B can be written as VB (a) Substituting Equatin (0b) the given input angular velcity int Equatin (a), the velcity f pint B is VB ( 9.75 rad/s)( 00 mm) 8.55 m/s (b) The psitive sign implies that pint B is mving t the left as the input link rtates cunterclckwise. Check. The psitin f pint B can be written as The X Y crdinates f pint B can be written as 7 7 I R R R () B XB Rcs Rcs (5a) YB Rsin Rsin (5b) Differentiating Equatins (5) with respect t the input gives X R sin R sin (6a) B

23 Y R cs R cs (6b) B Substituting Equatins (), (8) the input psitin int Equatins (6) gives B X ( )(sin 60 ) ( 00.00)(sin 09.0 )(.) mm / rad (7a) B Y ( )(cs 60 ) ( 00.00)(cs09.0 )(.) 0.57 mm / rad (7b) Nte here that Y B is nt zer due t the rund ff errr prpagated in the slutin. If the accurate values f the angle the first-rder kinematic cefficient are used then the first-rder kinematic cefficients fr pint B are The velcity f pint B can be written as X B 7.9 mm / rad YB V B (XBi YB j) 0mm/rad (8) Substituting Equatin (9) the velcity f the input link int Equatin (8), the velcity f pint B is V B ( i 0.57 j)( 5) i. j mm / s (9) The magnitude f the velcity f pint B is Pint B is mving t the left as the input link rtates cunterclckwise. VB 8.5m/s (0)