Namee of Student. link and I R 2 ?? = 0. and Δ θ. Calculate the. in the next. iteration. 1. = 6cm and θ * 9 = 135.

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1 ME 52 - Machine Design I Fall Semester 2010 EXAM 1. OPEN BOOK AND CLOSED NOTES. Namee of Student Lab. Div. Number Wednesday, September 29th, 2010 Use the blank paper provided for your solutions. Write on one side of the paper only. Where necessary, you can use the given figures to show vectors instantt centers. Attach your one page crib sheet to the end of problem 1. Any work that cannot be followed will assume to be wrong. Problem 1 (25 Points). Part I. (i) Determine the mobility of the mechanism shownn in Figure 1. Clearly number each link label the lower pairs the higher pairs on the figure. (ii) Define vectors that are suitable for a complete kinematicc analysis off the mechanism. Label show the direction of each vector on Figure 1. (iii) Write the vector loop equation(s) for this mechanismm identify: (a) suitable input(s) for the mechanism; (b) the known quantities, the unknown variables, any constraints. If you identified constraints then write the constraint equation(s). Figure 1. A Planar Mechanism. Part II. For a particular planar mechanism, the vector loop equation can be written as I R 2?? R9 R1 = 0 where R 2 = 5 cm, 2 = 45, R 1 = 8cm, 1 = 0. The initial estimatess of the position variables R 9 9 for the Newton-Raphson technique are R * 9 = 6cm * 9 = 15. Calculate the numerical values of the corrections Δ R * 9 Δ * 9 (to four decimal places) that are to be used in the next iteration. 1

2 ME 52 - Machine Design I Fall Semester 2010 Namee of Student Lab. Div. Number Problem 2 (25 Points). For the mechanism in the position shown in Figure 2, the angular velocity of the input link 2 is ω 2 = 12 k rad / s. The length of the ground length is O2 O4 = 10 cm. (i) On Figure 2, draw clearly label all the vectors that are required for a complete kinematic analysis of the mechanism. (ii) Write the vector loop equations. Clearly indicate the input, the known quantities, the unknown variables, any constraints. If you identified constraints then write the constraint equations. (iii) Calculate the numerical values of the first-order kinematic coefficients 4 R 4 (where R 4 is the vector from the ground pivot O 4 to point A on link ). (iv) Determine the angular velocity of link 4 the velocityy of point A relative to the ground pivot O. 4 Give the magnitude the direction of each vector. Figure 2. A planar mechanism. 2

3 ME 52 - Machine Design I Fall Semester 2010 Namee of Student Lab. Div. Number Problem (25 Points). For the mechanism in the position shown in Figure, the input link 2 is rotating counterclockwise with an angular velocity ω 2 = 9 rad / s. Thee mechanismm is drawn full scale in the figure. (i) List the primary instant centers the secondary instant centers for the mechanism. (ii) Show the locations of all instant centers on Figure. Please use the Kennedy circle shown below. Using the locations of the instant centers, determine: (iii) The first-order The magnitudes directions of the angular velocities of links kinematic coefficients of links, 4, link 7. (iv) 4. Figure. The Planar Mechanism. Drawn full scale.

4 ME 52 - Machine Design I Fall Semester 2010 Namee of Student Lab. Div. Number Problem 4 (25 Points). For the mechanism in the position shown in Figure 4, the input link 2 is rotating counterclockwise with an angular velocity ω 2 = 15rad/s. The wheel, link 4, is in rolling contact with the circular ground surface at point C. The radius of the ground d link is ρ 1 = 10 cm the lengths of links 2 are O2A = AB = 8 cm. (i) On Figure 4, draw clearly label all the vectors that are required for a complete kinematic analysis of the mechanism. (ii) Write the vector loop equations. Clearly indicate the input, the known quantities, the unknown variables, any constraints. If you identified constraints then write the constraint equations. (iii) Determine the first-order kinematic coefficients of links 4. (iv) The angular velocity of link. Give the magnitude the direction of this vector. (v) The angular velocity of link 4. Give the magnitude the direction of this vector. Figure 4. The Planar Mechanism. 4

5 SOLUTION TO PROBLEM 1. (i) The labels on the link numbers the joint types in the mechanism are shown on Figure 1. Figure 1. Number of links joint types in the mechanism. The Kutzbach mobility criterion can be written as m= (n 1) 2( j ) 1( j ) 1 2 (1) For the wheel rolling without slipping on the ground link thenn n = 7, j = 8, Substituting Equation (2) into Equation (1) gives 1 j2 = 1 (2) m = (7 1) 2(8) 1(1) = (a) Therefore, the mobility of the mechanism is m = 1 Note that if the wheel is rolling slipping on the ground mobility of the mechanismm is link then j 1 = 7 j 2 (b) = 2. In this case, the m = (7 1) 2(7) 1(2) = = 2 (4) 5

6 (ii) The vectors for a kinematic analysis of the mechanism aree shown in Figure 2. Figure 2. The vectors for a kinematic analysis. (iii) Three vector loop equations are necessary for the kinematic analysis of this mechanism, namely I?? C R2 + R R14 R9 = 0 C Ι? C?? R R + R + R R = I C C? R2 R + R55 R 7 = 0 (5a) (5b) (5c) (a) The input for the mechanism is the angular position of link 2, that is, 2. (b) The known quantities are the seven lengths R 2, R, R, R 6, R 5, R 55 5, R 9, the two angles 14, 7. The unknownn variables are the three lengths R 14, R 26, R 7, the three angles, 5, 6. There are four constraint equations, namely = , 26 = 2, =, 55 = +γ 5 (6) 6

7 Since there is rolling contact between link 4 (the pinion) the ground link 1 (the rack) then the rolling constraint equation can be written as ± Δ R 14 =ρ4( Δ4 Δ 14) (7a) The positive sign must be used for the mechanism in the given position, that is, the vector R 14 is getting longer for a counterclockwise rotation of link 4, or the vector is getting shorter for a clockwise rotation of link 4. Therefore, Equation (7a) can be written as + Δ R 14 =ρ4( Δ4 Δ 14) (7b) (iv) From the given VLE, the X Y components of the error vector ε can be written as * * R2 cos 2 R2 cos 2 R1 cos 1 =ε X (1a) * * R2 sin 2 R2 sin 2 R1 sin 1 =ε Y (1b) Taking the partial derivates of Equation (1a) with respect to the estimates of the of the position variables gives εx * ε = cos * 2 X * * = + R * 2 sin2 (2a) R 2 2 Also, taking the partial derivates of Equation (8b) with respect to the estimates of the position variables gives εy * = sin * 2 R 2 εy * * = R * 2 cos2 2 (2b) From Cramers rule, the coefficient matrix can be written as εx εx * * R2 2 εy ε Y * * R2 2 (a) Substituting Equations (2) into Equation (a), the coefficient matrix is * * * cos 2 + R2 sin 2 * * * sin 2 R2 cos 2 (b) The matrix equation for the corrections to the position variables can be written as * * * * cos 2 + R2 sin 2 ΔR 2 εx = * * * * sin 2 R2 cos ε 2 Δ2 Y (4) 7

8 Using Cramer s rule, the corrections can be written from Equation (11) as * * * * * εxr2 cos 2 +εyr2 sin R 2 2 DET Δ = (5a) * * * cos 2εY sin 2εX 2 DET Δ = (5b) The determinant of the coefficient matrix, see Equation (b), is * * * * * * * DET = ( cos 2)( R2 cos 2) ( sin 2)( + R2 sin 2) = R2 (6) The initial estimates of the position variables are R * 9 = 6cm * 9 = 15. Substituting this information the given data into Equations (1), the X Y components of the error vector are ε = 5cos 45 6cos15 8cos0 = cm (7a) X ε = 5sin 45 6sin15 8sin 0 = cm (7b) Y Also, substituting the initial estimates of the position variables into Equation (6), the determinant is * DET = R 2 = 6 cm (8) Substituting Equations (7) (8) into Equations (5), the corrections can be written as * ( cm)(6 cm)cos15 + ( cm)(6 cm)sin15 Δ R 2 = 6cm (9a) * cos15 ( cm) sin15 ( cm) Δ 2 = 6cm (9b) Therefore, the corrections that are required for the next iteration are 2 2 * cm.0000 cm Δ R 2 = = 0.42cm 6cm (10a) * cm cm Δ 2 = = rad 6cm (10b) 8

9 SOLUTION TO PROBLEM 2 (i) The vectors that are suitable for a velocity analysis of the mechanism are as shown in Figure 1. Figure 1. The vectors for a velocity analysis of the mechanism. There are two independent vector loops for this mechanism. The firstt vector loop equation is the second vector loop equation is I R 2 I R 2?? R4 R 1? C? + R45 R 5 = 0 = 0 (1a) (1b) The input is the angular position of the input link 2, that is, 2 2. The four unknown variables are the lengths R 4, R 45, R 5 the angular position of link 4, namely, 4 4. The constraint equation is = 4 = 45 = 4 From the geometry of the mechanism, the length of the input link is (2) 5 10 R 4 = R45 = R2 = = cm = 5.77 cm cos0 () (ii) The X Y components of Equation (1a) are R2cos2 R 4 cos4 R1cos 1 = 0 R2sin2 R 4 sin4 R 1 sin 1 = 0 9 (4a) (4b)

10 Differentiating Equations (4) with respect to the input position gives R2sin2 R4 cos4 + R4sin44 = 0 (5a) R2cos2 R4 sin4 R4sin44 0 = (5b) Writing Equations (5) in matrix form gives cos4 + R4sin4 R4 + R2 sin2 sin4 R4cos 4 = 4 R2cos 2 (6) Solving Equation (6) using Cramer s rule, the first-order kinematic coefficients can be written as where the determinant is RR 2 4sin( 4 2) R4 = (7a) DET R2cos( 4 2) 4 = (7b) DET DET = R 4 (8) Substituting Equation (8) into Equations (7), the first-order kinematic coefficients can be written as RR 2 4sin( 4 2) R4 = = R2 sin( 4 2) (9a) R4 R2cos( 4 2) 4 = = cos( 4 2) (9b) R4 Substituting the given data into Equations (9), the first-order kinematic coefficients for the mechanism are 10 R 4 = cm sin(150 0 ) =+ 5 cm/rad (10a) The positive sign indicates that the vector R 4 is getting shorter for a clockwise rotation of the input link 4 = cos(150 0 ) = 0.5 rad/rad (11) The negative sign indicates that link 4 is rotating counterclockwise (for the input rotating clockwise). (iv) 5 Points. The angular velocity of link 4 is The angular velocity of link 4 is counterclockwise. The velocity of point A relative to the ground pivot O 4 is This velocity is acting towards the ground pivot O. 4 ω = ω 4 2 = 0.5 ( 12 rad/s) =+ 6 rad/s (12) V4 = R 4ω2 =+ 5 cm / rad ( 12 rad/s) = 60 cm/s (1) 10

11 SOLUTION TO PROBLEM The number of instantt centers for this mechanism is n( n 1) 7(7 1) N = = = (1) (i) 5 points. There are 111 primary instant centers, namely, I12, I 1, I 14, I 1 16, I 17, I 2, I 24, I 25, I 4, I 45, I 57 ; 10 secondary instant centers, namely, I 15, I 56, I 26, I 27, I 67, I 47, I 46, I 6, I 7, I 5. (ii) 5 Points. The procedure to locate the secondary instant centers (see the Kennedy circle) is: 1. Connect instant centers I 12 I 2 5. Connect instant centers I 17 I 57. The intersection of these two lines gives the instant center I Connect instant centers I 15 I16. Draw a line perpendicular to the slot through the contact. The intersection of these two lines gives the instant center I 56.. Connect instant centers I 12 I 1 6. Connect instant centers I 25 I 56. The intersection of these two lines gives the instant center I Connect instant centers I 12 I 1 7. Connect instant centers I 25 I 57. The intersection of these two lines gives the instant center I Connect instant centers I 16 I 1 7. Connect instant centers I 56 I 57. The intersection of these two lines gives the instant center I Connect instant centers I 14 I 1 7. Connect instant centers I 45 I 57. The intersection of these two lines gives the instant center I Connect instant centers I 14 I 1 6. Connect instant centers I 47 I 67. The intersection of these two lines gives the instant center I Connect instant centers I 1 I 1 6. Connect instant centers I 4 I 46. The intersection of these two lines gives the instant center I Connect instant centers I 1 I 1 7. Connect instant centers I 4 I 47. The intersection of these two lines gives the instant center I Connect instant centers I 1 I15. Connect instant centers I 7 I 57. The intersection of these two lines gives the instant center I 5. Figure 1. The Kennedyy Circle. 11

12 The construction of the secondary instant centers is shown inn Figure 2. Figure 2. The location of the primary instant centers. (iii) 5 Points. The rolling contact equation between gear 4 the fixed gear 1 can be written as ρ1 ± 4 2 = ρ (2a) where the positive sign must be used here because of the internal rolling contact. Substituting the measurements the known kinematic coefficients into Equation (2a) gives 4cm + = 1cm (2b) Therefore, the first-order kinematic coefficient of gear 4 is = rad/rad 4 (2c) The negative sign indicates that gear 4 is rotating in the opposite direction to the input link 2 (that is, the arm). Therefore, gear 4 is rotating in the clockwise direction. The rolling contact equation between gear gear 4 can be written as ρ4 ± 2 = ρ 4 2 (a) where the negative sign must be used here because of external rolling contact. 12

13 Substituting the measurements the known kinematic coefficients into Equation (a) gives 1cm 1 = (b) 2cm 1 Substituting Equation (2c) into Equation (b), the first-order kinematic coefficient of gear is 4 = + rad/rad (c) The positive sign indicates that gear is rotating in the same direction to the input link 2 (that is, the arm). Therefore, gear is rotating in the counterclockwise direction. Alternative approach. The first-order kinematic coefficient for link j can be written as I I 12 2 j j = (4a) I1jI2 j Therefore, the first-order kinematic coefficient for gear 4 can be written as I I = (4b) I14I24 The distances required to obtain the first-order kinematic coefficient of link 4 are measured as I12I 24 = cm I14I 24 = 1cm (5a) Substituting Equation (5a) into Equation (4b), the first-order kinematic coefficient of link 4 is cm 4 = = rad/rad (5b) 1cm Note that the relative instant center I 24 is between the two absolute instant centers I 12 I 14. Therefore, the first-order kinematic coefficient for gear 4 must be negative, that is Similarly, the first-order kinematic coefficient for gear can be written as 4 = rad/rad (5c) I I 12 2 = (6) I1I2 Note in this case that the instant centers I 12, I 2 I 1 are coincident, see Figure 2. Therefore, the firstorder kinematic coefficient of link cannot be obtained from this equation. Consider the point of contact between gears 4. Denote this point as A (note that point A on gear is coincident with point A on gear 4 (that is, point A is the instant center I 4 ). Therefore, the velocity of point A can be written as V = V (7a) which can be written as A A 4 ω ( I I ) = ω ( I I ) (7b)

14 Dividing both sides of Equation (7b) by the input angular velocity ω 2 gives The positive sign indicates that gear is rotating counterclockwise. 14 ( I I ) = ( I I ) (7c) Then rearranging this equation, the first-order kinematic coefficient of link can be written as I I = ( ) (7d) I1I4 The lengths necessary to solve for the first-order kinematic coefficient of link are measured to be I1I 4 = 2cm I14I 4 = 2cm (8a) Substituting Equations (5c) (8a) into Equation (7d) gives 2 = ( ± )( ) = ± rad/rad (8b) 2 1 Note that the relative instant center I 4 is between the absolute instant centers I 1, I 14. Therefore, the first-order kinematic coefficient for gear has the opposite sign as the first-order kinematic coefficient of gear 4. Therefore, the first-order kinematic coefficient of link must be positive, that is = + rad/rad (9) The first-order kinematic coefficient for a slider, say link k, can be written as R = I I (10a) j 12 2 j Therefore, the first-order kinematic coefficient for link 7 can be written as R = I I (10b) The distance required for the first-order kinematic coefficient of link 7 is measured as I12I 27 = 4.75 cm (11) Substituting Equation (11) into Equation (10b), the first-order kinematic coefficient for link 7 is R 7 = 4.75 cm (12) The negative sign must be used here because the vector R 7 is getting shorter for a positive change in the position of the input link 2, see Figure 4. In other words, link 7 is moving upward for a positive change in the rotation of the input link 2. (iv) 5 points. The angular velocity of gear can be written as ω = ω (1a) 2 Substituting Equation (9) the input angular velocity into Equation (1) gives ω = ( + )( + 9) = + 27 rad/s (1b)

15 The angular velocity of gear 4 can be written as ω = ω (14a) Substituting Equation (6) the input angular velocity into Equation (15) gives ω = ( )( + 9) = 27 rad/s 4 ( (14b) The negative sign indicates that gear 4 is rotating clockwise. Figure 4. A suitable choice of vectors for the mechanism. 15

16 SOLUTION TO PROBLEM 4. Note thatt the lengths of link 2 triangle, that is link are equal, therefore, the triangle O 2 AB is an AO 2 B = O2BA = 60 isosceles (1a) Using the sum of the interior angles of a triangle, O2 AB = 180 AO2B OBA 2 = 60 Therefore, the triangle O2AB 2 is an equilateral triangle O2 A= = AB = O2B = 8cm (1b) (1c) Subtracting the length of O 2 B from the radius of the ground link then the radius of link 4 is ρ4 = ρ 1 OB 2 = 2 cm (1d) (i) 5 points. A suitable choice of vectors for the mechanism are shown in Figure 1. Figure 1. The vectors for thee mechanism. The vector loop equation can be written as C?? R2 + R R9 = 0 (1) The X Y components of Equation (1) can be written as R2cos 2 + R cos R9cos 9 = 0 R2sin 2 + R sin R9sin 9 = 0 (2a) (2b) Differentiting Equation (2) with respect to the input position 2 gives R2sin2 R + R = 0 sin 9sin R2cos2 + Rcos R9cos 9 9 = 0 (a) (b) Then writing Equations () in matrix form gives Rsin + R9sin9 + R2 sinn 2 + Rcos R9cos = 9 9 R2 coss 2 16 (4)

17 Solving Equation (4) using Cramer s rule, the first-order kinematic coefficients can be written as RR 2 9sin( 2 9) = DET (5a) RR 2 sin( 2 ) 9 = DET (5b) where the determinant of the coefficient matrix can be written as DET = R R (6) 9sin( 9) Substituting Equation (6) into Equations (5), the first-order kinematic coefficients can be written as R2sin( 2 9) = Rsin( 9) R2sin( 2 ) 9 = R9sin( 9) (7a) (7b) Substituting the position data into Equations (7), the first-order kinematic coefficients are 8cmsin( 60 0 ) = =+ 1 rad/rad 8cmsin(60 0 ) 8cmsin( ) 9 = =+ 1 rad/rad 8cmsin(60 0 ) (8a) (8b) The positive sign for the first-order kinematic coefficients in Equations (8) indicate that links 9 are rotating in the same angular velocity as the input link 2. The answers in Equations (8) also imply that links 9 have the same angular velocities as the input link 2. Note. For the given input position link dimensions the answers in Equations (8) should be intuitively obvious. Since the lengths of O 2 A, AB, O 2 B are equal then O 2 AB is a rigid equilateral triangle rotating about the ground pivot O 2. Since the triangle experiences rigid body motion then the angular velocity of this triangle must be same as the angular velocity of link 2. Therefore, the first-order kinematic coefficients of links 9 must be 1. (ii) 5 points. To determine the angular velocity of link 4, consider the rolling contact equation between links 1 4. Recall that the general form of the rolling contact equation between a gear a pinion can be written as ρgear Δpinion Δarm ± = (9) ρ Δ Δ pinion gear arm where the positive sign is used for internal contact the negative sign is used for external contact. For the given mechanism the gear is link 1, the pinion is link 4, the arm is link 2. Substituting this notation into Equation (9) gives ρ1 Δ4 Δ + = 2 (10) ρ4 Δ 1 Δ 2 17

18 Differentiating Equation (10) with respect to the input position 2 gives ρ = ρ4 0 1 (11) Solving Equation (11) for the first-order kinematic coefficient of link 4 gives ρ 1 4 = 1 (12) ρ4 Then substituting the radius of the ground the radius of link 4 into Equation (12) gives (iii) 5 points. The angular velocity of link can be written as 10 cm 4 = 1 = 4rad/rad (1) 2cm ω 2 Substituting the known values into Equation (14a) gives ω = (14a) ω = ( + 1rad/rad)( + 15rad/s) =+ 15rad/s (14b) The positive sign indicates that the angular velocity of link is counterclockwise. (iv) 5 points. The angular velocity of link 4 can be written as 4 ω 4 2 Substituting the known values into Equation (15a) gives ω = (15a) ω 4 = ( 4 rad/rad)( + 15 rad/s) = 60 rad/s (15b) The negative sign indicates that the angular velocity of link 4 is clockwise. 18