ME Machine Design I. EXAM 1. OPEN BOOK AND CLOSED NOTES. Wednesday, September 30th, 2009

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1 ME - Machine Design I Fall Semester 009 Name Lab. Div. EXAM. OPEN BOOK AND CLOSED NOTES. Wednesday, September 0th, 009 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. Any work that cannot be followed will assume to be wrong. Problem ( Points). Part I. ( Points). (i) Clearly number each link and label the lower pairs and the higher pairs on the mechanism shown in Figure (a). Then determine the mobility of this mechanism. (ii) Define vectors that are suitable for a complete kinematic analysis of the mechanism. Label and show the direction of each vector on Figure (a). (iii) Write the vector loop equation(s) for the mechanism and clearly identify: (a) suitable input(s) for the mechanism; and (b) the known variables, the unknown variables, and any constraints. (c) If you identified constraints in part (b) then write the constraint equation(s). Figure (a). A Planar Mechanism.

2 ME - Machine Design I Fall Semester 009 Name Lab. Div. Problem (continued). Part II. ( Points). Consider the four-bar linkage in the position shown in Figure (b). The angle of o the input link is θ = 60, measured counterclockwise from the ground link which is coincident with the fixed X-axis. The lengths of the four links are R= OO = 7cm, R = OA= 8cm, R = AB= cm, and R = OB= 0cm. Use Freudenstein's equation to determine the angular position of the output link θ. Figure (b). A Planar Four-Bar Linkage.

3 ME - Machine Design I Fall Semester 009 Name Lab. Div. Problem ( Points). For the mechanism in the position shown in Figure, the input link is rotating counterclockwise with a constant angular velocity ω = 0 rad / s. Also for this position, the coupler link is horizontal and is sliding along the vertical output link. The known link lengths are OO 0cm, OA= cm, and AB =. cm. = (i) Write a vector loop equation that would be suitable for a complete kinematic analysis of this mechanism. Indicate the input, the known variables, the unknown variables, and any constraints. Draw your vectors clearly on Figure. (ii) Determine the first-order kinematic coefficients for the mechanism from your vector loop equation. (iii) Determine the angular velocities of links and. Give the magnitudes and the directions. (iv) Determine the velocity of point B fixed in link relative to the velocity of the coincident point B fixed in link. Give the magnitude and the direction of this vector. Figure. A Planar Mechanism.

4 ME - Machine Design I Fall Semester 009 Name Lab. Div. Problem ( points). For the mechanism in the position shown in Figure, the input link is rotating clockwise with a constant angular velocity ω = rad / s. Link is in contact with link at point C and link is in contact with the slot in link at point E. The figure is drawn full scale, i.e., in = in. (i) List the primary instant centers and the secondary instant centers for the mechanism. (ii) Using the Kennedy circle, show the location of all the instant centers on Figure. Using the location of the instant centers, determine: (iii) The first-order kinematic coefficients of links and. (iv) The magnitudes and directions of the angular velocities of links and. (v) The magnitude and direction of the velocity of point B, and the magnitude and direction of the slipping velocity at point E. Kennedy Circle. Figure. A Planar Mechanism. (Drawn Full Scale: in = in).

5 ME - Machine Design I Fall Semester 009 Name Lab. Div. Problem ( points). For the gear mechanism in the position shown in Figure, the input link is rotating with an angular velocity ω = 0 rad / s clockwise and an angular acceleration α = rad / s clockwise. Link is pinned to the ground at O and is pinned to the center of gear at point A. The center of gear is also pinned to the ground at O and gear is pinned to the ground at O. Gears, and are all in rolling contact at point B. The radii of the fixed gear and the moving gears, and, are ρ = 0 cm, ρ = 0 cm, ρ = 0 cm, and ρ = 0 cm, respectively. Determine: (i) The first-order kinematic coefficients for gears,, and. (ii) The angular velocities of gears,, and. Specify the magnitudes and directions. (iii) The angular accelerations of gears,, and. Specify the magnitudes and directions. (iv) The velocity of point B fixed in gear. Figure. A Gear Mechanism.

6 Solution to Problem. Part I. (i) points. There are five links in this mechanism and the joint types connecting these five links are as shown in Figure (a). Figure (a). Joint Types of the Mechanism. For this mechanism, the number of links, number of lower pairs (or J joints), and number of higher pairs (or J joints), respectively, are n =, J =, and J = () The Kutzbach mobility criterion for a planar mechanism can be written as M = (n ) J J () Substituting Equation () into Equation (), the mobility of the mechanism is M = ( ) () = () This is the correct answer for this mechanism, that is, for a single input there is a unique output. 6

7 (ii) points. Suitable vectors for a kinematic analysis of the mechanism are shown in Figure (b). Figure (b). Vectors for the Mechanism. (iii) points. There are unknown variables, therefore, independent vector loop equations are required and one rolling contact equation. If the input link is chosen to be the slider, that is, link then the two independent vector loops can be written as Ι C? C? Loop : R R R R R + + = 0 (a) = (b)? C?? Loop : R 0 R R R7 R9 R (a) Since the input link is the slider, that is, link, then the input variable is the length R. (b) The four unknown variables in Equations () are the angular displacements θ, θ, and the linear distances R and R 9. (c) There are four constraint equations, namely: θ = + (a) θ 90 and θ = + (b) θ 90 7

8 θ = (c) θ 90 Note that the angular displacement of the wheel θ is constrained to the linear distance R 9 by rolling contact. The rolling contact equation between link and the ground link can be written as ( ) ±Δ R = ρ Δθ Δ θ = ρ Δ θ (6a) 9 9 Note that this equation can also be written in terms of the first-order kinematic coefficients as ± R = ρ (6b) 9 The correct sign in Equations (6) is positive because for a positive rotation of the wheel (that is, counterclockwise) the length of the vector R 9 is increasing or for a negative rotation of the wheel (that is, clockwise) the length of the vector R 9 is decreasing. Part II. points. The vectors for the four-bar linkage are shown in Figure (b). Figure (b). The vector loop for the four-bar linkage. The vector loop equation (VLE) can be written as I?? R + R R R= 0 (a) The X and Y components of Eq. () are R cosθ + R cosθ R cosθ R cosθ = 0 (a) 8

9 and R sinθ + R sinθ R sinθ R sinθ = 0 (b) Freudenstein's Equation can be written as where and Acosθ + Bsinθ = C () A = R R cos θ R R cos θ (a) B = R R sin θ R R sin θ (b) C= R R R R + R R cos( θ θ ) (c) Substituting the known data into Equations () gives A = 7 0 cos cos 60 =+ 60 cm (a) and B = 7 0 sin sin 60 = 8.6 cm (b) C = x 7 x 8 cos (0 60 ) = cm (c) Substituting Equations () into Equation () gives + 60 cos θ 8.6 sin θ = 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 Z θ = (7a) Z tan ( ) θ = + and Z cos θ = (7b) + Z Substituting Equations (7b) into Equation (), and rearranging, gives The solution to this 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 () into Equation (9) gives 8.6 ± ( 8.6) (60 )( 60) Z = 60 (0a) 9

10 The two roots to this quadratic equation are Equation (7a) can be written as Z = and Z =+.969 (0b) I tan Z θ = and θ II = tan Z () Substituting Equations (0b) into Equations () gives I tan θ = and θ II = tan.969 () Therefore, the two possible answers for the angular position of link are θ = 8.7 and θ =.6 () The answer for the angular position of link, for the given open configuration shown in Figure (b), is θ = 8.7 0

11 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 Mechanism. The vector loop equation (VLE) can be written as I?? C R + R R R = 0 (a) where the constraint is θ = θ = θ + 90 (b) The X and Y components of the VLE, see Eq. (), are and Rcos θ + Rcos θ Rcos θ Rcos θ = 0 (a) Rsin θ + Rsin θ Rsin θ Rsin θ = 0 (b) (ii) 7 Points. Differentiating Equations () with respect to the input position θ gives and R sin θ R sin θ R cos θ + R sin θ θ = 0 (a) R cos θ + R cos θ R sin θ R cos θ θ = 0 (b) Then writing Equations () in matrix form gives R sinθ + R sinθ cosθ θ + Rsinθ R cos R cos sin R = Rcos + θ θ θ θ ()

12 Substituting the known data into Equation () gives +.99 cm 0 rad +.99 cm = +. cm rad R 7. cm (a) The determinant of the coefficient matrix in Equation (a) is DET = ( +.99 cm)( rad) ( +. cm)(0 rad) =.99 cm (b) Using Cramer s rule, then from Eq. (a), the first-order kinematic coefficient for link is.99 cm θ = =+ cm/cm.99 cm (6a) The positive sign indicates that link is rotating in the same direction as the input link, i.e., counterclockwise. Also this answer indicates that link is rotating with same angular velocity as the input link; i.e., by definition θ =θ = rad / rad (6b) Also, using Cramer s rule, the first-order kinematic coefficient for links and is ( +.99 cm) ( 7. cm) ( +. cm)( +.99 cm) R = =+ 0cm /cm.99 cm (7a) The positive sign indicates that the vector R is increasing in length for a positive input, that is, point B on link is moving away from the ground pivot of link. (iii) 7 Points. The angular velocity of link can be written as Note that the first-order kinematic coefficient for link is ω = ω (8) θ =θ =θ =+ rad / rad (9) Substituting Eq. (9) and the input angular velocity into Eq. (8), the angular velocity of link is ω = ( + rad/rad)( + 0rad/s) =+ 0rad/s (0) The positive sign means that link is, indeed, rotating counterclockwise. (iv) Points. The velocity of point B fixed in link relative to the velocity of point B fixed in link can be written as R = R ω () B Substituting Equation (7a) and the input angular velocity into Equation (), the velocity is B = + + =+ () R ( 0 cm / cm)( 0 rad / s) 900 cm / s The magnitude and the direction of the velocity of point B fixed in link relative to the velocity of point B fixed in link is V =+ R j =+ 900 j cm / s () B B

13 The velocity of point B fixed in link relative to the velocity of point B fixed in link is directed vertically upwards. This can be verified by finding the instantaneous centers of velocity, see Figure (c). Figure (c). The location of the instant centers. Check. Using instant centers, the velocity of point B fixed in link relative to the velocity of the coincident point B fixed in link can be written as V = V V () B B B Using instant centers, the magnitude of the velocity of point B fixed in link can be written as V = (I B) ω (a) B The distance I B is measured as I B =.69 cm. Therefore, the velocity of point B fixed in link is V B = (.69)(0 rad / s) = cm / s (b) Using instant centers, the velocity of point B fixed in link can be written as V = (I B) ω i (6a) B The distance I B is measured as I B =.990 cm. Therefore, the velocity of point B fixed in link is V B = (.990)(0 rad / s) = 89.7 i cm / s (6b) Substituting Eqs. (b) and (6b) into Eq. (), the velocity of point B fixed in link relative to the velocity of the coincident point B fixed in link is V = V V = (89.7 i j) 89.7 i = 900 j cm / s (7) B B B The direction of this vector is indeed vertically upward as shown in Figure (c).

14 Solution to Problem. (i) 0 Points. The number of links in the mechanism is four, therefore, the total number of instant centers for this mechanism is six; i.e., n( n ) x N = = = 6 () There are three primary instant centers; namely, I, I, and I. The three secondary instant centers; namely, I, I, and I can be obtained as follows: (i) It is important to note that the secondary instant center I is not located at point C because there is slip between links and. However, the instant center I must lie on the line that is perpendicular to the line BC through the point of contact C. The point of intersection of this line with the line connecting the instant centers I I is the instant center I. (ii) The point of intersection of the line through I I and the perpendicular to the slot in link that passes through point E is the instant center I. (iii) The point of intersection of the line through I I and the line through I I is the instant center I. The procedure to locate the three secondary instant centers are marked on the Kennedy circle. Figure (a). The Kennedy Circle. The location of the six instant centers for this mechanism are shown on Figure (b). (ii) Points. The first-order kinematic coefficient of link can be written as I I = (a) II The distance I I is measured as I I =.87 in and the distance I I is measured as II =. in. Therefore, the first-order kinematic coefficient of link is.87 in = = 0.6 in/in (b). in Note that the correct sign is negative because the relative instant center I lies between the absolute instant centers, that is = 0.6 rad/rad (c) The first-order kinematic coefficient of link can be written as I I = (a) II

15 The distance I I is measured as I I =.78 in and the distance I I is measured as II =.07 in. Therefore, the first-order kinematic coefficient of link is.78 in = =.07 in / in (b).07 in Note that the correct sign is negative because the relative instant center I lies between the absolute instant centers I and I, that is =.07 rad/rad (c) Figure (b). The location of the instant centers. (iii) Points. The magnitude of the angular velocity of link can be written as ω = ω = ( 0.6 rad rad)( rad sec) =+ 9 rad sec (6) The positive sign indicates that the direction of the angular velocity of link is counterclockwise, as shown in Figure (c). The magnitude of the angular velocity of link can be written as ω = ω = (.07 rad rad)( rad sec) =+ 8.0 rad sec (7) The positive sign indicates that the direction of the angular velocity of link is counterclockwise, as shown in Figure (c).

16 Figure (c). Angular velocities of links and. (iv) Points. The velocity of point B can be written as VB ( ) = I B ω (8a) The distance IB is measured as IB =.90 in, therefore, the velocity of point B is VB ( I B) ω ( ) = = (.9 in) 9 rad sec =.6 in sec (8b) The direction of the velocity of point B is perpendicular to the line connecting the instant center I to point B as shown in Figure (d). The velocity of point B is directed degrees below the X-axis. The slipping velocity at point E can be written as where the slipping velocity at point E is defined here as V = ( I E) ω = ( I E)( ω ω ) (9) Slip V = V = V V (0) Slip E /E E E The direction of the slipping velocity. The angular velocity of link is counterclockwise and the angular velocity of link is counterclockwise. Since the angular velocity of link is greater than the angular velocity of link then the angular velocity of link relative to the angular velocity of link is 6

17 counterclockwise about the instant center I. This means that the slipping velocity of point E along the slot is pointed downward as shown in Figure (d). The distance IE is measured as IE =.0 in () Substituting Eqwuation () into Equation (0), the slipping velocity at point E is VSlip If the slipping velocity at point E is defined as then Equation () would be written as VSlip ( I E) ω ( ) = = =.7 in sec () V = V = V V () Slip E /E E E ( I E) ω ( ) = = =.7 in sec () Since the angular velocity of link relative to the angular velocity of link is clockwise about the instant center I. This means that the slipping velocity of point E along the slot (based on this definition) is pointed upward as shown in Figure (d). Figure (d). Velocity of point B and the slipping velocity at point E. 7

18 Solution to Problem. (i) 6 Points. The rolling contact equation for gear rolling on the ground link can be written as ρ ± = ρ The correct sign on the left hand side is negative since there is external contact, that is ρ = ρ 0 and ρ 0 = ρ 0 0 The rolling contact equation between gear and gear can be written as ρ ± = ρ The correct sign on the left hand side is positive since there is internal contact, that is and ρ + = ρ ρ 0 + = ρ 0 The rolling contact equation between gear and gear can be written as ρ ± = ρ The correct sign on the left hand side is positive since there is internal contact, that is (a) (b) (c) (a) (b) (c) (a) ρ 0 + = (b) ρ 0 and ρ 0 + = (c) ρ 0 (ii) 7 Points. Substituting the known radius for gear and the radius of gear into Equation (b), and rearranging, gives = (a) Therefore, the first-order kinematic coefficient for gear is = + rad/rad (b) 8

19 Equation (a) can also be written as ± ρ( ) = ρ( ) (a) or as ± ρ( ) = ρ( ) (b) Substituting the known second-order kinematic coefficients into Equation (b) gives ± ρ (0 0) = ρ ( 0) (6a) Therefore, the second-order kinematic coefficient for gear is Substituting the known radius for gears and into Equation (b) gives = + 0 rad/rad (6b) 0 + = (7a) 0 Then rearranging this equation gives 0 = ( ) + (7b) 0 Substituting Equation (b) into this equation gives 0 0 = ( ) + =+ rad/rad (8a) 0 0 Therefore, the first-order kinematic coefficient for gear is = +. rad/rad (8b) Substituting the known radius for gears and into Equation (c) the second-order kinematic coefficient for gear is =+ 0 rad/rad (8c) Substituting the known radius for gears and into Equation (b) gives 0 + = (9a) 0 θ Substituting Equation (6b) into this equation gives = + + =+ rad/rad (9b) Therefore, the first-order kinematic coefficient for gear is = rad/rad (0a) Substituting the known radius for gears and into Equation (c), the second-order kinematic coefficient for gear is 9

20 (iii) 8 Points. The angular velocity of gear can be written as =+ 0rad/rad (0b) ω = θω (a) Substituting Equation (a) into Equation (a), the angular velocity of gear is The negative sign indicates that gear is indeed rotating clockwise. The angular velocity of gear can be written as ω = ( + )( 0) = 00 rad/s (b) ω = θω (a) Substituting Equation (6b) into Equation (0a), the angular velocity of gear is The negative sign indicates that gear is indeed rotating clockwise. The angular velocity of gear can be written as ω = ( +.)( 0) = rad/s (b) ω = θω (a) Substituting Equation (8a) into Equation (a), the angular velocity of gear is The negative sign indicates that gear is indeed rotating clockwise. The angular acceleration of gear can be written as ω = ( + 0.8)( 0) = 0 rad/s (b) α = θα + θω (a) Substituting Equation (c) into Equation (a), the angular acceleration of gear is The negative sign indicates that gear is accelerating clockwise. The angular acceleration of gear can be written as α = ( + )( ) + (0)( 0) = 0 rad/s (b) α = θα + θω (a) Substituting Equation (6c) into Equation (a), the angular acceleration of gear is α = ( +.)( ) + (0)( 0) = 0 rad/s (b) The negative sign indicates that gear is accelerating clockwise. The angular acceleration of gear can be written as α = θα + θω (6a) Substituting Equation (8b) into Equation (6a), the angular acceleration of gear is α = ( + 0.8)( ) + (0)( 0) = rad/s (6b) 0

21 The negative sign indicates that gear is accelerating clockwise. (iv) Points. The velocity of point B fixed in gear can be written as ( ) V B = V I B B = ω (7a) Since point B is coincident with the instant center I. Substituting the values into Equation (7a) gives V B 00 = V B = 0 = 000 cm/s (7b) The velocity of point B fixed in gear is directed vertically downward.