Figure 1. A planar mechanism. 1

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1 ME Machine Design I Summer Semester 201 Name of Student Lab Section Number EXAM 1. OPEN BOOK AND CLOSED NOTES. Wednesday, July 2nd, 201 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 and instant centers. Any work that cannot be followed is assumed to be in error. At the completion of your exam, please staple each problem separately. Staple your crib sheet to the end of Problem 1. Problem 1 (25 Points). For the planar mechanism in the position shown in Figure 1: (i) Determine the mobility using the Kutzbach criterion. Clearly number each link and label the lower pairs and the higher pairs on the given figure. (ii) Define suitable vectors for a kinematic analysis of the mechanism. Label and show the direction of each vector clearly on the given figure. (iii) Write the vector loop equation(s) that are required for a kinematic analysis of the mechanism. Clearly identify suitable input(s) for the mechanism. List: (a) known quantities; (b) unknown variables; and (c) any constraints. If you identified constraints in part (b) then write the constraint equations. Figure 1. A planar mechanism. 1

2 ME Machine Design I Summer Semester 201 Name of Student Lab Section Number Problem 2 (25 Points). For the mechanism in the position shown in Figure 2, gear 3 is rolling without slipping on the ground link 1 at the point of contact C. The radius of gear 3 is 3 75 mm, the radius of the ground link 1 is mm, and the length of link is AB 90 mm. The constant angular velocity of the input link 2 is 2 25 rad / s counterclockwise. (i) Write the vector loop equation(s) that are suitable for a kinematic analysis of this mechanism. Draw your vectors clearly on the figure. (ii) Using the vector loop equation(s) determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and, and the linear velocity of the slider (link 5). Give the magnitudes and directions of these three vectors. Figure 2. A planar mechanism. 2

3 ME Machine Design I Summer Semester 201 Name of Student Lab Section Number Problem 3 (25 Points). For the mechanism in the position shown in Figure 3, the input link 2 is rotating counterclockwise with a constant angular velocity 2 30 rad / s. (i) List the primary instant centers and the secondary instant centers for the mechanism. (ii) Using the given Kennedy circle, show the locations of all the instant centers on Figure 3. Using the locations of the instant centers, determine: (iii) The first-order kinematic coefficients of links 3,, and 5. (iv) The angular velocities of links 3,, and 5. Give the magnitudes and directions of each vector. Figure 3. A planar mechanism. (The figure is drawn full size.) 3

4 ME Machine Design I Summer Semester 201 Name of Student Lab Section Number Problem (25 Points). For the mechanism in the position shown in Figure, link 3 is rolling without slipping on the input link 2 at the point of contact C. The link lengths are OA 3 90mm, AB 150 mm, and the radius of link 3 is 3 85 mm. The constant velocity of the input link 2 is V2 15 i m/s. (i) Write the vector loop equation(s) that are suitable for a kinematic analysis of the mechanism. Draw your vectors clearly on the given figure. (ii) Using the vector loop equation(s) determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and. Give the magnitudes and directions of each vector. (iv) Determine the velocity of pin B connecting link to the slider (link 5). Give the magnitude and direction of this vector. Figure. A planar mechanism.

5 Solution to Problem 1. (i) 6 Points. The links, the lower pairs, and the higher pairs of the planar mechanism are as shown in Figure 1a. Figure 1a. The links and kinematic pairs of the planar mechanism. The Kutzbach mobility criterion can be written as M 3(n 1) 2j1 1j2 (1) The number of links, the number of lower pairs (or j 1 joints), and the number of higher pairs (or j 2 joints), respectively, are n 6, j1 6, and j2 2 (2) Substituting Equation (2) into Equation (1), the mobility of the mechanism can be written as Therefore, the mobility of the mechanism is M 3(61) 2x6 1x2 (3a) M (3b) This is the correct answer for the mobility of the mechanism in this position, that is, for a single input there is a unique output. 5

6 (ii) 5 Points. A suitable set of vectors for a kinematic analysis of the mechanism are shown in Figure 1b. Figure 1b. The vectors for the mechanism. (iii) 1 Points. Three independent vector loop equations are required for a kinematic analysis of the mechanism. The three vector loop equations can be written as Loop 1: I?? () Loop 2: C 1?? (5) Loop 3:? C 2? (6) The input link is chosen to be link 2 (see Figure 1b) and the input variable is the angle θ 2. Other suitable inputs are link and link 5 since they are pinned to the ground. Link 6 would not be the best option for the input despite being pinned to ground. The reason is that a clockwise rotation of link 6 could cause joint separation, so the remaining links would not be driven by link 6 for all positions of the mechanism. Link 3 is not a practical input because it does not have a connection to the ground link. 6

7 (a) The seven known quantities of the mechanism are O E, AE, O A, O B, O C, CD, and O D (7) Additionally, the magnitudes and the directions of the ground vectors 1, 11, and 111 (b) The six unknown variables are are known. 3,, 5, 6, 7, and 5 (8) (c) There are two constraints. The two constraint equations are identified as follows. Constraint 1: The angle between the vectors 2 and 22 is constant, that is Constraint 2: The directions for vectors and 5 are always equal, that is 22 2 (9) (10) 5 7

8 Solution to Problem 2. (i) 6 Points. A suitable set of vectors for a kinematic analysis of the mechanism is shown in Figure 2. Figure 2. The vectors for a kinematic analysis of the mechanism. The vector loop equation (VLE) for the mechanism can be written as I?? (1) The input is the angular position of link 2 (connecting O 2 to A) which is the arm of the gear train (links 1, 2, and 3). The length of the arm is 2 = 200 mm, that is, the sum of the radii of the ground link 1 and link 3. Also, in the given position θ 2 = 0, θ = 90, θ 5 = 30, θ 11 = 0, and θ 1 = 90. (ii) 13 Points. The X and Y components of Equation (1) are and cos cos cos cos cos 0 (2a) sin sin sin sin sin 0 (2b) Differentiating Equations (2) with respect to the input position θ 2 gives and sin sin cos 0 (3a) cos cos sin 0 (3b) Writing Equations (3) in matrix form gives sincos52sin2 cos sin 5 5 2cos 2 8 ()

9 The determinant of the coefficient matrix in Equation () is DET sin sin cos cos (5a) 5 5 which simplifies by the sum-difference trigonometric identity to DET cos( ) (5b) 5 Substituting θ = 90 and θ 5 = 30 into Equation (5b) gives Therefore, the determinant of the coefficient matrix is DET cos (90 30 ) cos 60 (6a) DET (90mm) cos 60 5 mm (6b) Note that the determinant, see Equation (5b), is zero when θ = 120 or 300, that is, when link is perpendicular to the line of sliding of link 5. Substituting the known data into Equation () gives which simplifies to 90sin(90 ) mm cos(30 ) 200sin(0 ) mm 90cos(90 ) mm sin(30 ) 5 200cos(0 ) mm 90 mm 3 / 2 0mm 0mm 1/2 200 mm 5 (7a) (7b) Using Cramer s rule, the first-order kinematic coefficient for link can be written as 0mm 3/ mm 1/ mm 2 5 mm 5 mm (8a) Therefore, the first-order kinematic coefficient for link is 20 3 mm mm 3.89 (8b) 9 mm mm The positive sign indicates that link is rotating counterclockwise for a counterclockwise rotation of the input link. The first-order kinematic coefficient for link 5 is 5 90 mm 0 mm 2 0 mm 200 mm mm 00 mm 5 mm 5 mm (9) The positive sign indicates that link 5 is moving up the inclined plane for a counterclockwise rotation of the input link. 9

10 The rotation of gear 3 is constrained by the rolling contact equation between gear 3 and the ground link. The rolling contact equation can be written as Differentiating Equation (10a) with respect to the input position θ 2 gives (10a) (10b) The correct sign here is negative because there is external rolling between links 3 and 1. Since the arm (link 2) is the input then, by definition, the first-order kinematic coefficient 2 1. Also, since link 1 denotes the ground link then the angle θ 1 does not change. Therefore, by definition, the first-order kinematic coefficient 1 0. Substituting the known kinematic coefficients into Equation (10b) gives 01 3 (11a) 1 31 earranging Equation (11a), the first-order kinematic coefficient for gear 3 can be written as (11b) 3 Substituting the known data into Equation (11b), the first-order kinematic coefficient for gear 3 is 125 mm 8mm mm (12) 75 mm 3mm mm The positive sign indicates that link 3 is rotating counterclockwise for a counterclockwise rotation of the input link 2. (iii) 6 Points. The angular velocity of gear 3 can be written from the chain rule as 3 2 (13a) 3 Substituting Eq. (12) into Equation (13a), the angular velocity of gear 3 is x 25 k rad/s k rad/s The angular velocity of link can be written from the chain rule as 2 (13b) (1a) Substituting Equation (8) into Equation (1a), the angular velocity of link is 20 3 mm k rad/s k rad/s 9 mm 9 (1b) 10

11 Therefore, the angular velocity of link is k rad/s (1c) The velocity of link 5 can be written from the chain rule as Substituting Equation (9) into Equation (15a), the velocity of link 5 is 5 00 mm x 25 rad/s 10, 000 mm/s The velocity of link 5 is in the direction of the line of sliding, up the inclined plane, that is V 5 10,000 30mm/s (15a) (15b) (16) 11

12 Solution to Problem 3. (i) 5 Points. The number of links in the mechanism is five, therefore the total number of instant centers for this mechanism is nn ( 1) 5 N 10 (1) 2 2 There are six primary instant centers, namely: I 12, I 23, I 3, I 5, I 15, and I 13. The instant centers I 12, I 23, I 3, I 5, and I 15 are primary because they are the given pin joints. The instant center I 13 is also regarded as a primary instant center because this instant center must lie somewhere on the line that passes through the pin of link 3 and be perpendicular to the ground slot. From Kennedy s theorem, the primary instant center I 13 must also lie on the line connecting the instant centers connecting I 12 I 23. Note that these two lines are parallel in this position, therefore, I 13 is at infinity. There are four secondary instant centers, namely: I 1, I 35, I 2, and I 25. (ii) 12 Points. The locations of the ten instant centers are as shown in Figure 3. Figure 3. The locations of the ten instant centers. 12

13 The procedure to find the four secondary instant centers is outlined by the following steps: (i) The point of intersection of the line through I 3 I 13 and the line through I 15 I 5 is the instant center I 1. (ii) The point of intersection of the line through I 13 I 15 and the line through I 3 I 5 is the instant center I 35. (iii) The point of intersection of the line through I 23 I 3 and the line through I 12 I 1 is the instant center I 2. (iv) The point of intersection of the line through I 23 I 35 and the line through I 12 I 15 is the instant center I 25. (iii) 5 Points. The general notation of the first-order kinematic coefficient of link j can be written as (2) IijI1i j IijI 1 j where the correct sign is determined from the location of the relative instant center relative to the locations to the two absolute instant centers. (efer to Sections 3.17 and 3.18 of the text book for more details). The first-order kinematic coefficient of link 3 can be written as I23I12 3 (3) From the scaled drawing, see Figure 3, the distance I 13 I 23 is infinite and the distance I23I13 I 12 I 23 = in = mm () Substituting these values into Equation (3), the first-order kinematic coefficient of link 3 is in 3 in 0 in/in (5) Therefore, the motion of link 3 is pure translation (in this position). The first-order kinematic coefficient of link can be written as I2I12 (6) The correct sign for the first-order kinematic coefficient of link is negative because the relative instant center I 2 lies between the two absolute instant centers I 12 and I 1. Check: By inspection, since the input link is rotating counterclockwise then link must rotate clockwise. From the scaled drawing, the distances are measured as I2I1 I 12 I 2 = in = mm and I 1 I 2 = in = mm (7) Substituting Equation (7) into Equation (6), the first-order kinematic coefficient of link is in 0.28 in/ in (8) in The negative sign indicates that link is rotating clockwise for a counterclockwise rotation of the input link. 13

14 The first-order kinematic coefficient of link 5 can be written as I25I12 5 (9) The correct sign for the first-order kinematic coefficient of link 5 is negative because the relative instant center I 25 lies between the two absolute instant centers I 12 and I 15. Check: By inspection, since the input link is rotating counterclockwise then link 5 must rotate clockwise. From the scale drawing, the distances are measured as I25I15 I 12 I 25 = 2.19 in = mm and I 15 I 25 = in = mm (10) Substituting Equation (10) into Equation (9), the first-order kinematic coefficient of link 5 is 2.19 in in/in (11) in The negative sign indicates that link 5 is rotating clockwise for a counterclockwise rotation of the input link. (iv) 3 Points. The angular velocity of link 3 can be written as (12a) Substituting Equation (5) and the input angular velocity into Equation (12a), the angular velocity of link 3 is 3 (0 in/in)(30 k rad/s) 0 (12b) The angular velocity of link can be written as 2 (13a) Substituting Equation (8) and the input angular velocity into Equation (13a), the angular velocity of link is ( 0.28 in/in)(30 k rad/s) 12.8 k rad/s (13b) The angular velocity of link 5 can be written as (1a) Substituting Equation (11) and the input angular velocity into Equation (1a), the angular velocity of link 5 is 5 ( in/in)(30 k rad/s) k rad/s (1b) 1

15 Solution to Problem. (i) 3 Points. A suitable set of vectors for a kinematic analysis of the mechanism is shown in Figure. Figure. A suitable set of vectors for a kinematic analysis of the mechanism. The vector loop equation (VLE) for this mechanism can be written as IC?? (1) where θ 2 = 180, θ 7 = 90, θ 3 = 90, and θ 5 = 150 in this position. The input is link 2 (the rack), and 2 is the vector from point E on the ground to point C (that is, the point of contact between link 2 and link 3). The arm (link 7), that is, the vector 7, is the vector from the point of contact C to O3 and is always perpendicular to the rack. The vector 3 is the vector from O3 to A, is the vector from B to A, and 5 is the vector from point E to pin B along the line of sliding of link 5. (ii) 15 Points. The angle θ can be obtained from the triangle O 3 AB. From the sine rule, the angle Then the supplement to the angle α is sin sin 1 3 (2) (3a) 15

16 Substituting the known dimensions into Equation (3a) gives 1 90 mm 180sin mm (3b) The rolling contact constraint between the rack (link 2) and the pinion (gear 3) can be written as ( ) (a) The correct sign here is negative because a decrease in the length of the vector 2 (when the rack moves to the right) causes a counterclockwise rotation of gear 3. Differentiating Equation (a) with respect to the input position, the first-order kinematic coefficient of link 2 can be written as ( ) (b) Since link 2 is the input then by definition the kinematic coefficient of the input link is 2 1 (5a) Also, the arm (link 7) is always perpendicular to the rack, that is, the angle θ 7 does not change. Therefore, the kinematic coefficient of the arm is 7 0 (5b) Substituting Equations (5) into Equation (b), and rearranging, the first-order kinematic coefficient of link 3 can be written as 1 3 (6a) Substituting the radius of gear 3 into Equation (6a), the first-order kinematic coefficient of link 3 is (6b) 85 mm mm The negative sign indicates that link 3 is rotating counterclockwise for the given velocity of the input link (that is, the input vector is becoming shorter as the input link moves to the right). The VLE in Equation (1) can be written in terms of the X and Y components as and cos cos cos cos cos 0 (7a) sin sin sin sin sin 0 (7b) Differentiating Equations (7) with respect to the input position gives and cos sin sin cos 0 (8a) sin cos cos sin 0 (8b)

17 Writing Equations (8) in matrix form and substituting in Equation (6a) gives 1 2cos2 3 sin3 sincos5 3 cos sin sin2 3 cos 3 3 (9) The determinant of the coefficient matrix in Equation (9) is which can be written as DET sin sin cos cos (10a) 5 5 DET cos( ) (10b) 5 The determinant is zero when θ = 60 or 20, that is, when link is perpendicular to the line of sliding for link 5. Substituting the known data into Equation (10b), the determinant of the coefficient matrix is DET 150cos( ) mm mm (11) Substituting the known information into Equation (9) gives 1 mm 1cos(180 ) 90 sin(90 ) 150sin(13.13 ) mm cos(150 ) 85 mm 150 cos(13.13 ) mm sin(150 ) 5 1 mm 1sin(180 ) 90cos(90 ) 85 mm (12a) Simplifying this equation gives 3 90 mm 90 mm mm 1 mm 120 mm mm (12b) Using Cramer s ule, the first-order kinematic coefficient of link can be written as 90 mm mm 2 mm 1 0 mm mm (13a) Therefore, the first-order kinematic coefficient of link is 1 90 mm mm mm mm (13b) 17

18 The negative sign indicates that link is rotating counterclockwise as the rack is moving to the right, that is, as the vector 2 is decreasing in length. Using Cramer s ule, the first-order kinematic coefficient of the slider link 5 is 90 mm 90 mm 1 85 mm 5 mm 120 mm 0 mm mm (1a) Therefore, the first-order kinematic coefficient of the slider link 5 is mm mm mm mm mm (1b) The negative sign here indicates that link 5 is sliding upward as the rack is moving to the right, that is, as the vector 2 is decreasing in length. (iii) 5 Points. The angular velocity of link 3 can be written as (15) where 2 has equal magnitude and opposite direction of the input velocity V 2, that is, the vector 2 decreases in length if the input velocity V 2 is to the right (and the vector 2 increases in length if the input velocity V 2 is to the left). Substituting Equation (5b) into Equation (15), the angular velocity of the pinion link 3 is mm ( mm/s) rad/s rad/s The positive sign indicates that link 3 is rotating counterclockwise, that is k rad/s (16a) (16b) The angular velocity of link can be written as 2 (17) Substituting Equation (13) into Equation (17), the angular velocity of link is 1 ( mm )( mm/s) rad/s (18a) The positive sign indicates that link is rotating counterclockwise, that is k rad/s (18b) (iv) 2 Points. The velocity of pin B is equal to the velocity of link 5 because the slider is in pure translation. The velocity of pin B and the velocity of link 5 can be written as 18

19 V V B (19) Substituting Equation (1) into Equation (19), the velocity of pin B and link 5 are VB V5 5 ( 0.07 mm/mm)( mm/s) 711 mm/s The positive sign indicates that the slider is moving upward along the line of sliding, that is V V mm/s B (20a) (20b) 19