ME 35 - Machine Design I Spring Semester 010 Name of Student Lab. Div. Number EXAM 1. OPEN BOOK AND CLOSED NOTES Thursday, February 18th, 010 Please use the blank paper provided for your solutions. Write on one side of the paper only. Where necessary, you can use the figures that are provided on the exam to show vectors and instantaneous centers of velocity. Problem 1 (5 Points). Part I (13 Points). (i) Clearly number each link and label the lower pairs and the higher pairs of the mechanism shown in Figure 1(a). Then determine the mobility of this mechanism. (ii) Define vectors that would be suitable for a kinematic analysis of the mechanism. Label and show the direction of each vector on Figure 1(a). (iii) Write the vector loop equation(s) for the mechanism and: (a) Identify suitable input(s) for the mechanism. (b) Identify the known quantities, the unknown variables, and write any constraint equations. (c) Write the rolling constraint equation in terms of the position variables. Figure 1(a). A Planar Mechanism. 1
ME 35 - Machine Design I Spring Semester 010 Name of Student Lab. Div. Number Problem 1 (continued). Part II (1 Points). Consider the four-bar linkage in the position shown in Figure 1(b). The angular o position of the input link is given by the angle θ = 10, measured counterclockwise from the ground link which is chosen to be coincident with the fixed X-axis. The lengths of the four links are R1= OO = 15cm, R = OA= 6cm, R3 = AB= 10cm, and R = OB= 10cm. Use Freudenstein's equation to determine the angular position of the output link θ as shown in the figure. Note that the mechanism is not drawn to scale in Figure 1(b). Figure 1(b). A Planar Four-Bar Linkage.
ME 35 - Machine Design I Spring Semester 010 Name of Student Lab. Div. Number Problem (5 Points). For the mechanism in the position shown in Figure, the input link has a constant angular velocity ω = 5rad/s clockwise. The link lengths and dimensions are OB = 5cm, AB = BC = 5 cm, CO = 10 cm, and OD = 5cm. (i) Write a vector loop equation that would be suitable for a kinematic analysis of the mechanism. Clearly indicate the input, the knowns, and any constraints. Draw your vectors clearly on Figure. (ii) Determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and. Give the magnitude and direction of each vector. (iv) Determine the relative velocity of point C fixed in link 3 with respect to point B fixed in link. Give the magnitude and the direction of this vector. Figure. A Planar Mechanism. 3
ME 35 - Machine Design I Spring Semester 010 Name of Student Lab. Div. Number Problem 3 (5 Points). For the mechanism in the position shown in Figure 3, the angular velocity of the input link is a constant ω = 7rad/s counterclockwise. Link 5 is rolling without slipping on link 3 at the point of contact C. The mechanism is drawn full scale in Figure 3, that is, 1 cm = 1 cm. (i) List the primary instant centers and the secondary instant centers for the mechanism. (ii) Using the Kennedy circle shown below, 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 magnitudes and directions of the angular velocities of links 3 and 5. (v) The magnitude and direction of the velocity of point C fixed in link 5. Figure 3. A Planar Mechanism. The scale is 1 cm = 1 cm.
ME 35 - Machine Design I Spring Semester 010 Name of Student Lab. Div. Number Problem (5 Points). For the mechanism in the position shown in Figure, the input link has a velocity ˆ V = icm/s and an acceleration A ˆ = 5i cm/s. The roller, link, is rolling without slipping on the ground link 1. The known link lengths are AB = 15 cm, ρ 1 = 5cm, and ρ = cm. (i) Write a suitable vector loop equation for a kinematic analysis of this mechanism. Indicate the input, the knowns, and the unknown variables. Draw your vectors clearly on Figure. (ii) Write symbolic equations for the rolling contact equation between link and the ground link 1 in terms of: (a) the position variables, and (b) the first-order kinematic coefficients of the mechanism. (iii) Determine numerical values for the first-order kinematic coefficients of the mechanism. (iv) Determine the angular velocities of links 3 and. Give the magnitudes and directions of each vector. Figure. A Planar Mechanism. 5
Solution to Problem 1. Part I. (i) points. There are 6 links in the mechanism and the joint types are as shown in Figure 1(a). Figure 1(a). Joint Types of the Mechanism. The Kutzbach mobility criterion for a planar mechanism can be written as M = 3(n 1) J1 1J (1) For this mechanism, the number of links, the number of lower pairs (or J 1 joints), and the number of higher pairs (or J joints), respectively, are n = 6, J1 = 7, and J = 0 () Substituting Equation () into Equation (1), the mobility of the mechanism is that is M = 3(6 1) (7) 0 (3a) M = 15 1 0= 1 (3b) This is the correct answer for this mechanism, that is, for a single input there is a unique output. 6
(ii) 5 points. Suitable vectors for a kinematic analysis of the mechanism are shown in Figure 1(b). Figure 1(b). Vectors for the Mechanism. (iii) points. There are 5 unknown variables, therefore, independent vector loop equations are required and one rolling contact equation. The two independent vector loop equations can be written as + + = 0 (a) Ι C? C? Loop 1: R R R3 R R R1 + + + = 0 (b)? C C?? Loop : R R R33 R5 R9 R11 (a) The input link is chosen to be the disk, denoted as link, and the input variable is the angle θ. (b) The five unknown variables are the angles θ, θ, θ, θ, and θ. 3 5 6 9 There are three constraint equations, namely θ = θ α (5a) and θ33 = θ3 β (5b) θ = (5c) θ 90 7
(c) The rolling contact equation between the wheel (denoted as link 6) and the ground link 1 can be written as ρ Δθ Δθ 1 6 9 ± = (6a) ρ6 Δ θ1 Δ θ9 The correct sign in Equation (6a) is the negative sign because there is external rolling contact, that is ρ ρ 1 = Δθ6 Δθ9 0 Δθ 6 9 (6b) Note that Equations (6) can also be written in terms of the first-order kinematic coefficients as that is ρ θ θ 1 6 9 ± = ρ6 θ1 θ9 ρ ρ 1 = θ 6 θ 9 0 θ 6 9 (7a) (7b) Part II (1 points). The vectors for the four-bar linkage are shown in Figure 1(b). Figure 1(b). The vector loop for the four-bar linkage. The vector loop equation (VLE) can be written as I?? R + R3 R R1= 0 (1) The X and Y components of Eq. (1) are and R cosθ + R cosθ R cosθ R cosθ = 0 (a) 3 3 1 1 8
Freudenstein's Equation can be written as R sinθ + R sinθ R sinθ R sinθ = 0 (b) 3 3 1 1 where and Acosθ + Bsinθ = C (3) A = R R cos θ R R cos θ (a) 1 1 B = R R sin θ R R sin θ 1 1 (b) C= R R R R 3 1 + R 1 R cos( θ 1 θ ) (c) Substituting the known data into Equations () gives A = 15 10 cos 0 6 10 cos 10 =+ 360 cm (5a) and B = 15 10 sin 0 6 10 sin 10 = 103.9 cm (5b) C = 10 15 6 10 + x 15 x 6 cos (0 10 ) = 351 cm (5c) Substituting Equations (5) into Equation (3) gives + 360 cos θ 103.9 sin θ = 351 cm (6) To determine the output angle we can write this transcendental equation as an algebraic equation, (namely, a quadratic equation). The procedure is to use the tangent of the half-angle relationship; i.e., which gives Z sin 1 Z θ = (7a) Z tan ( ) θ = + and 1 Z cos θ = (7b) 1 + Z Substituting Equations (7b) into Equation (3), and rearranging, gives The solution to this quadratic equation can be written as (A + C) Z ( B) Z + (C A) = 0 (8) + B± B (A + C)(C A) Z = A + C (9) Substituting Equations (5) into Equation (9) gives 103.9 ± ( 103.9) (360 351)( 351 360) Z = (360 351) (10a) 9
The two roots to this quadratic equation are Writing Equation (7a) as Z =+ 3.05 and Z = 6.118 (10b) 1 θ = and θ II = tan 1 Z (11) I tan 1 Z 1 Then substituting Equations (10b) into Equations (11) gives I tan 13.05 θ = and θ II = tan 1 6.118 (1) Therefore, the two possible answers for the angular position of link are θ = 13.1 and θ = 175.61 (13) The correct answer for the angular position of link, for the given open configuration shown in Figure 1(b), is θ = 13.1 (1) 10
Solution to Problem. (i) 8 Points. A suitable set of vectors for a kinematic analysis of the mechanism are shown in Fig. (a). Figure (a). Suitable vectors for the planar mechanism. The vector loop equation (VLE) can be written as I? C? C R + R3+ R3 R1 = 0 (1) The input is the angular position of link, that is, θ. For the given input position θ = 315. The knowns are the length of link, the length of the ground link 1, and the angle of the ground link. The angular position of link 3 is constrained to be perpendicular to link, that is θ3 = θ + 90 = 5 () Also, the angular position of link is constrained to be perpendicular to link 3, that is θ = θ + 90 = θ + 180 = 135 (3) 3 3 11
(ii) 6 points. The X and Y components of the VLE, see Eq. (1), are and R cosθ + R3 cos θ 3 + R3 cos θ3 R1 cosθ 1 = 0 (a) R sin θ + R3 sin θ 3 + R3 sin θ3 R1 sin θ 1 = 0 (b) Differentiating Equations () with respect to the input position θ gives and R sin θ R sin θ + cos θ R R sin θ + cos θ R = 0 (5a) 3 3 3 3 3 3 3 3 R cosθ + R cos θ + sin θ R + R cos θ + sin θ R = 0 (5b) 3 3 3 3 3 3 3 3 Then writing Equations (5) in matrix form gives cos θ3 cos θ3 R 3 R sin θ + R3 sin θ 3 + R3 sin θ3 = sin θ sin θ R R cos θ R cos θ R sin θ 3 3 3 3 3 3 3 (6) Substituting the known data into Equation (6) gives + 0.7071 radians 0.7071 radians R 3 + 7.071 cms = + 0.7071 radians + 0.7071 radians R 3 0 (7a) The determinant of the coefficient matrix in Equation (7a) can be written as DET = ( + 0.7071 radians)( + 0.7071 radians) ( + 0.7071 radians)( 0.7071 radians) (7b) Therefore, the determinant of the coefficient matrix is DET = 1.000 radians (7c) Using Cramer s rule, the first-order kinematic coefficient for link 3 with respect to link is + 5cm R 3 = =+ 5 cm / rad + 1radians (8) The positive sign indicates that the distance from point B to point C, that is, the vector R 3 decreasing in length as link rotates clockwise. From Cramer s rule, the first-order kinematic coefficient for link 3 with respect to link is, is 5cm R 3 = = 5 cm / rad + 1radians (9) The negative sign indicates that the distance from point C to the ground pivot O, that is the vector R 3, is increasing in length as link rotates clockwise. Differentiating Equation () with respect to the input position, the third kinematic coefficient of the mechanism is θ = θ = θ =+ (10) 3 3 1 1
(iii) 7 Points. The angular velocity of link 3 can be written as ω 3 =θ 3 ω (11) Substituting Equation (10) into Equation (11), the angular velocity of link 3 is The negative sign indicates that link 3 is rotating clockwise. The angular velocity of link can be written as ω 3 =+ 1ω = 5rad/s (1) ω =θ ω (13) Differentiating Equation (3) with respect to the input position, and using Equation (11) gives θ = θ = θ =+ (1) 3 3 1 Then substituting Equation (1) into Equation (13) gives ω =+ 1ω = 5rad/s (15) The negative sign indicates that link is rotating clockwise. (iv) 7 Points. The relative velocity between point C fixed in link 3 and point B fixed in link can be written as VCB = R 3ω (16) The relative velocity between point C fixed in link 3 and point B fixed in link is directed along the line BC in the direction from C to B. Substituting Equation (8) and the input angular velocity into Equation (16) gives V CB = ( + 5cm/rad)( 5rad/s) = 5cm/s (17) The positive sign indicates that point C is moving towards point B, that is, the distance BC is decreasing (the angle of the velocity vector is inclined at 135 degrees clockwise to the horizontal axis). 13
Solution to Problem 3. (i) and (ii) 15 Points. This mechanism has six links, therefore, the total number of instant centers is N n( n 1) 6x5 = = = 15 (1) There are 7 primary instant centers; namely, Therefore, there are 8 secondary instant centers; namely, I 1, I 3, I 3, I 1, I 35, I 56, and I 16 () I 13, I 15, I, I 5, I 6, I 36, I 5, and I 6 (3) The 8 secondary instant centers can be obtained as follows: (i) The point of intersection of the line through I 1 I 3 and the line through I 1 I 3 is the instant center I 13. (ii) The point of intersection of the line through I 13 I 35 and the line through I 16 I 56 is the instant center I 15. (iii) The point of intersection of the line through I 1 I 1 and the line through I 3 I 3 is the instant center I. (iv) The point of intersection of the line through I 1 I 15 and the line through I 3 I 35 is the instant center I 5. (v) The point of intersection of the line through I 5 I 56 and the line through I 1 I 16 is the instant center I 6. (vi) The point of intersection of the line through I 35 I 56 and the line through I 13 I 16 is the instant center I 36. (vii) The point of intersection of the line through I 1 I 15 and the line through I 3 I 35 is the instant center I 5. (viii) The point of intersection of the line through I 1 I 16 and the line through I 5 I 56 is the instant center I 6. The procedure to locate these 8 secondary instant centers is indicated on the Kennedy circle below, see Figure 3(a). Figure 3(a). The Kennedy Circle. The locations of all 15 instant centers for this mechanism are shown on Figure 3(b). 1
Figure 3(b). The locations of the fifteen instant centers. (iii) 6 Points. The first-order kinematic coefficient of link 3 can be written as I I 1 3 θ 3 = (a) I13I3 The distance I 1 I 3 is measured as I 1 I 3 =.55 cm and the distance I 13 I 3 is measured as I13I 3 = 7.88 cm. Therefore, the first-order kinematic coefficient of link 3 is.55 cm θ 3 = = 0.3 rad/rad (b) 7.88 cm Note that the correct sign the first-order kinematic coefficient of link 3 is negative because the relative instant center I 3 lies between the two absolute instant centers I 1 and I 13, that is 15
The first-order kinematic coefficient of link can be written as θ 3 = 0.3 rad/rad (c) R = I I (5a) 1 From measuring figure, I1I = 1.70 cm, therefore, the first-order kinematic coefficient of link is R = 1.70 cm (5b) Note that the correct sign is negative because link will move to the left for a positive change in the input (that is, the distance from O to point B is decreasing for a positive change in the input). Therefore, the first-order kinematic coefficient of link is The first-order kinematic coefficient of link 5 can be written as R = 1.70 cm (5c) I I 1 5 θ 5 = (6a) I15I5 The distance I 1 I 5 is measured as I 1 I 5 = 3.38 cm and the distance I 15 I 5 is measured as I15I 5 =.0 cm. Therefore, the first-order kinematic coefficient of link 5 is 3.38 cm θ 5 = = 0.805 rad/rad (6b).0 cm Note that the correct sign is negative because the relative instant center I 5 lies between the two absolute instant centers I 1 and I 15, that is, the first-order kinematic coefficient of link 5 is The first-order kinematic coefficient of link 6 can be written as θ 5 = 0.805 rad/rad (6c) I I 1 6 θ 6 = (7a) I16I6 The distance I 1 I 6 is measured as I 1 I 6 = 1.65 cm and the distance I 16 I 6 is measured as I16I 6 = 8.00 cm. Therefore, the first-order kinematic coefficient of link 6 is 1.65 cm θ 6 = = 0.06 rad/rad (7b) 8.00 cm Note that the correct sign is negative because the relative instant center I 6 lies between the two absolute instant centers I 1 and I 16, that is, the first-order kinematic coefficient of link 6 is (iv) Points. The angular velocity of link 3 can be written as θ 6 = 0.06 rad/rad (7c) 16
ω3 = θ 3ω = ( 0.3 rad rad)( + 7 rad sec) =.68 rad sec (8) The negative sign implies that the direction of the angular velocity of link 3 is clockwise, see Fig. 3(c). The angular velocity of link 5 can be written as ω5 = θ 5 ω = ( 0.805 rad rad)( + 7 rad sec) = 5.635 rad sec (9) The negative sign implies that the direction of the angular velocity of link 5 is clockwise, see Fig. 3(c). Check. The relative instant center I 35 does not lie between the two absolute instant centers I 13 and I 15, therefore, the angular velocity of link 5 is in the same direction as the angular velocity of link 3. The angular velocity of link 6 can be written as ω6 = θ 6 ω = ( 0.06 rad rad)( + 7 rad sec) = 1. rad sec (10) The negative sign implies that the direction of the angular velocity of link 6 is clockwise, see Fig. 3(c). Figure 3(c). The angular velocities of links 3, 5 and 6 and the velocities of points B and C. 17
Check. The relative instant center I 6 lies between the two absolute instant centers I 1 and I 16. Therefore, the angular velocity of link 6 is in the opposite direction to the angular velocity of link, that is, clockwise. (v) Points. The velocity of point C, fixed in link 3, can be written as V C 3(I13C) 3 = ω (11) The distance I13C is measured as I13C = 5.06 cm. Therefore, the velocity of point C fixed in link 3 is V C 3 = (.68 rad / s)(5.06 cm) = 11.8 cm / s (1) The direction of the velocity of point C is perpendicular to the line I 13 C, that is, 130 degrees to the horizontal as shown in Figure 3(c). Check: The velocity of point C, fixed in link 5, can be written as VC V C 5(I15C) = =ω (13) 5 3 The distance I15C is measured as I15C =.0 cm. Therefore, the velocity of point C, fixed in link 5, is V C = (5.635 rad / s)(.0 cm) = 11.50 cm / s (1) The direction of the velocity of point C is perpendicular to the line I15C as shown in Figure 3(c). The velocity of point B, fixed in link, can be written as VB V B 3(I13B) = =ω (15) 3 The distance I13B is measured as I13B= 5.5 cm. Therefore, the velocity of point B fixed in link is V B = (.68 rad / s)(5.5 cm) = 11.91 cm / s (16) The direction of the velocity of point B is along the X-axis as shown in Figure 3(c). Check: The velocity of point B, fixed in link can be written as VB R = ω (17) Substituting Equation (5b) into Equation (17), the velocity of point B fixed in link is This answer is in good agreement with Equation (16). V B = (1.70 cm)(7 rad / s) = 11.90 cm / s (18) 18
Solution to Problem. (i) 5 Points. A suitable set of vectors for a kinematic analysis of the mechanism are shown in Fig. (a). The vector loop equation (VLE) can be written as Figure (a). Suitable vectors for the mechanism. I?? R R1+ R9 R3 = 0 (1) where the input is the length R which defines the position of link, and the two unknown variables θ 9 and θ 3 are the angular positions of the arm (link 9) and link 3. (ii) 6 points. The rolling contact constraint between link and the ground link 1 of the mechanism can be written as ρ Δθ Δθ 1 9 ± = (a) ρ Δ θ1 Δ θ9 The correct sign is negative because there is external contact between links and 1, that is 19
ρ Δθ Δθ ρ θ θ 1 9 = (b) Δ Δ 1 9 Equation (b) can be written in terms of the first-order kinematic coefficients of the mechanism as ρ θ θ (3) 1 9 = ρ θ1 θ9 (iii) 9 points. The X and Y components of the VLE, see Eq. (1), are and Rcos θ R1cos θ 1+ R9cos θ9 R3cos θ 3 = 0 (a) Rsin θ R1sin θ 1+ R9sin θ9 R3sin θ 3 = 0 (b) Differentiating Equations () with respect to the input position R gives and + 1cosθ R sin θ θ + R sin θ θ = 0 (5a) 9 9 9 3 3 3 + 1sin θ + R cos θ θ R cos θ θ = 0 (5b) 9 9 9 3 3 3 Then writing Equations (5) in matrix form gives R9sin θ 9 + R3sin θ3 θ 9 1cosθ R9cos 9 R3cos = 3 3 1sin + θ θ θ θ (6) Substituting the known data into Equation (6) gives.95 cm 1.99 cm θ 9 1 rad = +.95 cm 7.5 cm θ 0 3 (7a) The determinant of the coefficient matrix in Equation (7a) is DET = (.95 cm)( 7.5 cm) ( +.95 cm)( 1.99 cm) =+ 101.3 cm (7b) Using Cramer s rule, the first-order kinematic coefficient for link 9 is + 7.5 cm θ 9 = =+ 0.07 rad / cm + 101.3 cm (8) The positive sign indicates that link 9 is rotating clockwise as the change in the input position R decreases in length. From Cramer s rule, the first-order kinematic coefficient for link 3 is +.95 cm θ 3 = =+ 0.09 rad / cm + 101.3 cm (9) The positive sign indicates that link 3 is rotating clockwise as the change in the input position R decreases in length. 0
The first-order kinematic coefficient of link in the mechanism can be obtained from Equation (3). The equation can be written as ρ ρ 1 = θ θ 9 0 θ 9 (10a) Rearranging this equation, the first-order kinematic coefficient of link can be written as 1 θ θ9 1 ρ = + ρ (10b) Substituting known values into Equation (10b), the first-order kinematic coefficient for link is 5cm θ =+ 0.07 rad/cm 1 + =+ 0.59 rad/cm cm (11) (iv) 5 Points. The angular velocity of link 3 can be written as ω =θ R (1) 3 3 Note that the length of the vector R is decreasing as the input link is moving to the right (the velocity of link is given to the right). Therefore, the time rate of change in the length of this vector is R = V = cm/s (13) Substituting Eq. (9) and Eq. (13) into Eq. (1), the angular velocity of link 3 is ω 3 = ( + 0.09 rad/cm)( cm/s) = 0.098 rad / s (1) The negative sign indicates that link 3 is indeed rotating clockwise, that is, the angular velocity of link 3 is clockwise. The angular velocity of link can be written as ω =θ R (15) Substituting Eq. (11) and Eq. (13) into Eq. (15), the angular velocity of link is The negative sign means that link is rotating clockwise. Aside. The angular velocity of link 9 can be written as ω = ( + 0.59 rad/cm)( cm/s) = 0.518 rad / s (16) ω =θ R (17) 9 9 Substituting Eq. (8) and Eq. (13) into Eq. (17), the angular velocity of link 9 is ω 9 = ( + 0.07 rad/cm)( cm/s) = 0.18 rad / s (18) The negative sign means that link 9 (that is, the arm) is rotating clockwise, that is, the angular velocity of link 9 is clockwise 1