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Unformatted text preview: Solutions to Chapter 6 Exercise Problems Problem 6.1: Design a double rocker, fourbar linkage so that the base link is 2in and the output rocker is 1in long. The input link turns counterclockwise 60˚ when the output link turns clockwise through 90˚. The initial angle for the input link is 30˚ counterclockwise from the horizontal, and the initial angle for the output link is 45˚. The geometry is indicated in the figure. A2 60˚ O2 30˚ A1 O4 45˚ B2 Solution: Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it clockwise 60˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate the intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first figure. Measure the lengths os 02 A1 and A1 B1 . This gives the linkage shown in the second figure. The link lengths are: r1 = 6 " r2 = 3.72" r3 = 7.29" r4 = 4 " 90˚ B1  222  A2 60˚ A1 30˚ O2 O4 45˚ 90˚ B2 B1 B'2 Basic Construction A2 A1 r2 30˚ O2 r1 45˚ 90˚ B2 B1 r4 r3 O4 B'2 Final Linkage  223  Problem 6.2: Design a double rocker, fourbar linkage so that the base link is 4in and the output rocker is 2in long. The input link turns counterclockwise 40˚ when the output link turns counterclockwise through 80˚. The initial angle for the input link is 20˚ counterclockwise from the horizontal, and the initial angle for the output link is 25˚. The geometry is indicated in the figure. A2 60˚ O2 20˚ A1 O4 B2 80˚ 25˚ B1 Solution Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it clockwise 60˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate the intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first figure. Measure the lengths os 02 A1 and A1 B1 . This gives the linkage shown in the second figure. The link lengths are: r1 = 4 " r2 = 2.77" r3 = 3.21" r4 = 2 "  224  Basic Construction Final Linkage Problem 6.3: In a back hoe, a fourbar linkage is added at the bucket in part to amplify the motion that can be achieved by the hydraulic cylinder attached to the link that rotates the bucket as shown in the figure. Design the link attached to the bucket and the coupler if the frame link is 13in and the input link is 12in long. The input link driven by the hydraulic cylinder rotates through an angle of 80˚ and the output link rotates through an angle of 120˚. From the figure, determine reasonable angles for the starting angles ( 0 and 0 ) for both of the rockers.  225  Solution Determine reasonable starting angles for each crank relative to the beam member. Use 40˚ for the input rocker and 70˚ for the output rocker. This is shown in the figure. The basic problem can then be redrawn as shown in the following figure.  226  Solution Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it clockwise 120˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate the intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first figure below. Measure the lengths os 02 A1 and A1 B1 . After unscaling, the linkage is shown in the second figure that follows. The link lengths are: r1 = 13" r2 = 8.27" r3 = 19.34" r4 = 12"  227  Basic Construction Final Linkage  228  Problem 6.4 In the drawing, AB = 1.25 cm. Use A and B as circle points, and design a fourbar linkage to move its coupler through the three positions shown. Use Grashof’s equation to identify the type of fourbar linkage designed.
Y B3 θ 3 = 60˚ A 3(2, 3) B2 θ2 = 45˚
X A 2 (2, 1) B1 A 1 (0, 0), θ1 = 0 Solution Find the center points A* and B* and measure the link lengths. Then, AA* = 2.027" AB = 1.25" A*B* = 0.670" BB* = 2.903" l + s = 2.903 + 0.670= 3.573" p + q = 2.027 + 1.25 = 2.277 Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker.  229  Y B3 A3 B* A* B2 A2 A1 Problem 6.5 B1 X Using points A and B as circle points, design a fourbar linkage that will position the body defined by AB in the three positions shown. Draw the linkage in position 1, and use Grashof’s equation to identify the type of fourbar linkage designed. Position A1 B1 is horizontal, and position A2 B2 is vertical. AB = 1.25 in.
Y A 1 (0, 1.75) B1 (1.63, 1.25) A2 B3 35˚ A3 (2.13, 0.63) B2 X Solution Find the center points A* and B* and measure the link lengths. Then, AA* = 2.15" AB = 1.25"  230  A*B* = 2.32" BB* = 1.11" l + s = 2.32 + 1.11= 3.43" p + q = 2.15 + 1.25 = 3.40 Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker. Y A1 B1 A2 B* B3 A3 X B2 A* Problem 6.6 Design a fourbar linkage to move its coupler through the three positions shown below using points A and B as moving pivots. AB = 4 cm. What is the Grashof type of the linkage generated?
Y B3 B2 60˚ A3(2, 2.4) 50˚ A 2 (2, 0.85) A1 (0, 0) X B1 Solution Find the center points A* and B* and measure the link lengths. Then, AA* = 1.705 cm AB = 4.000 cm  231  A*B* = 3.104 cm BB* = 6.954 cm l + s = 6.954 + 1.705= 8.659" p + q = 3.104 + 4.000 = 7.104 Therefore, l+s > p+q and the linkage is a nonGrashof linkage and a double rocker. Y B3 B* A3 A* A2 A1 B1 B2 X  232  Problem 6.7 A fourbar linkage is to be designed to move its coupler plane through the three positions shown. The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of the other crank is at C*. Draw the linkage in position 1 and use Grashof’s equation to identify the type of fourbar linkage designed. Also determine whether the linkage changes branch in traversing the design positions. Positions A1 B1 and A2 B2 are horizontal, and position A3 B3 is vertical. AB = 3 in.
Y A2 B 2 (0.0, 2.88) A1 B 1 (1.94, 0.94) A3 C* X (1.0, 3.0) B3 Solution Find the center points C1 and A* and measure the link lengths. Note that the linkage is to be drawn in position 1 so the motion must be referred to position 1 when locating C1 . Then, CC* = 2.67 in AC = 2.38 in A*C* = 3.00 in AA* = 2.06 in l + s = 3.00 + 2.06= 5.06 p + q = 2.67 + 2.38 = 5.05 Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker.  233  Y A2 B2 * C3 A1 A* C 1 B1 C* A3 X * C2 B3 To determine if the linkage changes branch, draw the linkage in the three positions, and measure the sign of . From the drawing below, the sign of is different in the three positions, and the mechanism changes branch.  234  Y A2 B2 C2 ψ2
A* A1 C1 B1 C* X A3 ψ1 ψ3 C3 B3 Problem 6.8 Design a fourbar linkage to move a coupler containing the line AB through the three positions shown. The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of the other crank is at C*. Draw the linkage in position 1 and use Grashof’s equation to identify the type of fourbar linkage designed. Position A1 B1 is horizontal, and positions A2 B2 and A3 B3 are vertical. AB = 4 in. Y A2 A3 A1 (0, 2) B1 C1 * (0, 0) (2, 0) B2 (4, 0) B3 X  235  Solution The solution is shown in the figure. The link lengths are: r = 3.22 = p 1 A2 A3 A1 r2 r3 r1 B2 r4 C1 A* B3 B1 C * C* 2 C* 3 r2 = 3.32 = q r3 = 3.62 = l r4 = 2.24 = s Grashof l+ s < p+ q For this mechanism, and l + s = 3.62 + 2.24 = 5.86 p + q = 3.22 + 3.32 = 6.54 Therefore, l + s < p + q , and the mechanism is a crankrocker or rockercrank.  236  Problem 6.9 A mechanism must be designed to move a computer terminal from under the desk to top level. The system will be guided by a linkage, and the use of a fourbar linkage will be tried first. As a first attempt at the design, do the following: a) Use C* as a center point and find the corresponding circle point C in position 1. b) Use A as a circle point and find the corresponding center point A*. c) Draw the linkage in position 1. d) Determine the type of linkage (crank rocker, double rocker, etc.) resulting. e) Evaluate the linkage to determine whether you would recommend that it be manufactured.
Desk Y Position 3 (Horizontal) A3 (3.3, 2.6) C* A1 (2.2, 0.3) Position 1 Position 2 135˚ A2 (2.1, 0.1) 160˚ X  237  Desk Pos'n 3
* A A3 C* 2 C* 3 A1 C* Pos'n 2 A2
Pos'n 1 A1C 1 = 2.6864 C1C* = 3.0684 A * C* = 1.0301 A1A* = 1.395 A * C * +C 1C* = 4.0985 A1C1 + A1A* = 4.0814 Therefore, A * C * +C 1C* > A1C 1 + A1A * and the linkage is a nonGrashof linkage and a double rocker. C1 This is a poor solution because C 1is below the floor plane. Problem 6.10 Design a fourbar linkage to move the coupler containing line segment A B through the three positions shown. The moving pivot for one crank is to be at A, and the fixed pivot for the other crank is to be at C*. Draw the linkage in position 1 and determine the classification of the resulting linkage (e.g., crank rocker, double crank). Positions A2 B2 and A3 B3 are horizontal, and position A1 B1 is vertical. AB = 3.5 in.
B1 (1.0, 2.5) A3 B3 (0.0, 2.0) Y A2 B2 (0.0, 1.0) C* 1 X A1  238  B1 C* 2 C* 2 A3 A* A2 B2 B3 C1 C* A1 From the figure, r1 = A*C* = 1.8228 in r2 = A*A1 = 2.4974 in r3 = A1 C1 = 2.6966 in r4 = C*C1 = 1.3275 in Grashof calculation: l +s?p+q [2.6966 + 1.3275=(4.0241)] < [1.8228 + 2.4974=(4.3202)] The linkage is a crank rocker.  239  Problem 6.11 Design a fourbar linkage to move a coupler containing the line AB through the three positions shown. The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of the other crank is at C*. Draw the linkage in position 1, and use Grashof’s equation to identify the type of fourbar linkage designed. Position A1 B1 is horizontal, and positions A2 B2 and A3 B3 are vertical. AB = 6 cm.
Y B1 A3 (2, 3) A2 (2, 3) A1 (6, 4.3) C *(0, 0) X B3 B2  240  A* C* 2 Y C1 A3 B1 A1 A2 C*
3 C* X B3 B2 Find A*. Then find C'2 and C'3 and find the circle point C1 . Draw the linkage. Then, AA* = 13.0 AC = 6.84 CC* = 4.41 C*A* = 15.8 l + s = 4.41 + 15.8 = 20.2 p + q = 6.84 + 13.0 = 19.8 Therefore, l+s > p+q and the linkage is a Type II double rocker.  241  Problem 6.12 Design a fourbar linkage to move the coupler containing line segment A B through the three positions shown. The moving pivot for one crank is to be at A, and the fixed pivot for the other crank is to be at C*. Draw the linkage in position 1 and determine the classification of the resulting linkage (e.g., crank rocker, double crank). Also check to determine whether the linkage will change branch as it moves from one position to another. Position A1 B1 is horizontal, and position A3 B3 is vertical. AB = 5.1 cm.
Y (1.5, 3.7) A2
45 ˚ B2 A3 X (7.8, 1.0) A1 (5.0, 1.0) C* ( 5.0, 5.0) B1 B3 Solution Find the center points C1 and A* and measure the link lengths. Note that the linkage is to be drawn in position 1 so the motion must be referred to position 1 when locating C1 . Then, CC* = 7.81 cm AC = 9.73 cm A*C* = 6.68 cm AA* = 6.82 cm l + s = 9.73 + 6.68 = 16.41 cm p + q = 7.81 + 6.82 = 14.63 cm Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker.  242  Y A2 X A1 B1 B2 A3 A* C* B3 C1 * C2 * C3 To determine if the linkage changes branch, redraw the linkage in the three positions, and determine if the transmission angle changes. This is shown below. The linkage does not change branch.  243  Y A2 X A1 B1 B2 A3 A* C* C2 C1 B3 C3 Problem 6.13 Synthesize a fourbar mechanism in position 2 that moves its coupler through the three positions shown below if points C* and D* are center points. Position A1 B1 and position A3 B3 are horizontal. AB = 4 cm.
Y D* (3.0, 2.6) B2 A 3 (3.4, 1.6) C* 45˚ A2 (2.7,  0.7) A1 (0.7,  1.8) B1 B3 X Solution Find the circle points C2 and D2 and measure the link lengths. Notice that the linkage is to be drawn in position 2. Therefore, position 2 for the coupler is the position to which the positions of D* and C* are referred for finding the circle points. Then,  244  DD* = 2.767 cm CD = 4.045 cm C*D* = 3.985 cm CC* = 2.300 cm l + s = 4.045 + 2.300= 6.345 p + q = 2.767 + 3.985 = 6.752 Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker. D* 1 Y D* D2 C* 1 C* D* 3 A1 C2 A2 B1 A3 X B2 B3 C* 3  245  Problem 6.14 Synthesize a fourbar mechanism in position 2 that moves its coupler through the three positions shown below. Point A is a circle point, and point C* is a center point. Position A1 B1 and position A3 B3 are horizontal. AB = 4 cm.
Y C* (2.3, 4.5) B2 A3 (2.7, 3.5) 45˚ A2 (2.0, 1.0) X B1 B3 Solution A1 Find the center points C2 and A* and measure the link lengths. Note that the linkage is to be drawn in position 2 so the motion must be referred to position 2 when locating C2 . Then, CC* = 2.257 cm AC = 3.146 cm A*C* = 2.989 cm AA* = 3.065 cm l + s = 3.146 + 2.257= 5.403 p + q = 3.065 + 2.989 = 6.054 Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker. C* 1 Y C2 A* A3 C3 * A2 A1 B1 X C* B2 B3  246  Problem 6.15 A hardware designer wants to use a fourbar linkage to guide a door through the three positions shown. Position 1 is horizontal, and position 3 is vertical. As a tentative design, she selects point B* as a center point and A as a circle point. For the three positions shown, determine the location of the circle point B corresponding to the center point B* and the center point A* corresponding to the circle point A. Draw the linkage in position 1 and determine the Grashof type for the linkage. Indicate whether you think that this linkage should be put into production.
Y Position 3 135˚ Position 2 A3 (0, 8.3) A2 (3.1, 6.6) Position 1 A1(4.0, 2.8) B* X Solution Position 2 A3 A2 Position 3 Position 1 A1 A* B1 B* 3 B* B* 2  247  AA* = 1.7552" BB* = 1.2903" AB = 2.0069" A*B* = 1.5576" l + s = 2.0069 +1.2903 = 3.2972 < p + q = 1.7552 +1.5576 = 5.0680 crank rocker The mechanism would not be acceptable because of the location of the fixed and moving pivots inside the wall. Problem 6.16 Design a slidercrank mechanism to move the coupler containing line segment AB through the three positions shown. The moving pivot for the crank is to be at A. Determine the slider point, and draw the linkage in position 1. Also check to determine whether the linkage will move from one position to another without being disassembled. Position A1 B1 is horizontal, and position A3 B3 is vertical. AB = 2.0 in.
Y A2 (2.69, 1.44) 135˚ A3 (5.06, 0.0) A1(0, 0) B1 B2 X B3 Solution Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next select a slider point (C1 ), and find the location of that point in the other positions. This establishes the slider line. Note that a difference linkage results for each choice of C1 . From the figure, r2 = A*A1 = 2.9393 in r3 = A1 C1 = 4.3606 in Find the transmission angle in the three positions. The linkage changes mode because the sign of the transmission angle changes.  248  A2 A1 r2 = 2.9393 B1 B2 A3 r3 = 4.3606 P2
3 A* B3 P' 23 P1
3 C3 C2 P1
2 C1  249  Problem 6.17 Design a slidercrank mechanism to move a coupler containing the line A B through the three positions shown. The line AB is 1.25" long. The moving pivot (circle point) of the crank is at A. The approximate locations of the three poles (p 1 2, p 1 3, p 2 3) are shown, but these should be determined accurately after the positions are redrawn. Find A*, the slider point that lies above B1 on a vertical line through B1 , and draw the linkage in position 1.
Y P13 A1 (1.25, 1.70) B1 (1.65, 1.25) A 2 P12 P23 B3 (3.15 , 1.32 A3 (2.15, 0.58) B2 X Solution Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next select the slider point (C1 ) that lies above B1 . Then find the location of that point in the other positions. This establishes the slider line. Draw the linkage in the three positions. By inspection of the positions of C, the mechanism changes mode and goes through the positions in the wrong order.  250   251  Problem 6.18 Design a slidercrank linkage to move a coupler containing the line AB through the three positions shown. The fixed pivot (center point) of the other crank is at C*. Draw the linkage (including the slider line) in Position 1. Position A1 B1 is horizontal, and positions A2 B2 and A3 B3 are vertical. AB = 4 in. Y A2 A3 A1 (0, 2) B1 C1 * (0, 0) (2, 0) B2 (4, 0) B3 X Solution This problem illustrates the type of rigid body guidance problem that cannot be solved directly using the elementary techniques developed in the text book. Therefore, to correct answer is that there is no solution. However, a partial solution can be developed, and the students should work the problem far enough to illustrate that the solution procedure breaks down. With the information give, it is possible to find C1 . This is illustrated in the following construction.  252  Finding C1 Next find the poles as shown in the following construction. This is where a problem occurs. Note that P1 2 and P1 3 are coincident and P2 3 is at infinity on the line shown. Therefore, the circle of sliders appears to be on a straight line, but the orientation of the line cannot be determined from the elementary theory provided.  253  Finding the poles  254  Problem 6.19 Design a slidercrank mechanism to move a coupler containing the line with A through the three positions shown. The moving pivot (circle point) of the crank is at A. Find the slider point which lies on Line BC and draw the linkage (including the slider line) in position 1. Note that Line BC is NOT the line on which the slider moves. 60˚ (0.8, 0.8) A 3 A2 (0.8, 0) B 0.4 30˚ C A1 (0, 0) Solution Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next select the slider point (C1 ) that lies on the line BC. There are two solutions, but the one on the right is implied by the problem statement. Then find the location of that point in the other positions. This establishes the slider line. Draw the linkage in the three positions. By inspection of the positions of C, the mechanism changes mode and goes through the positions in the wrong order. The construction steps are shown in the following. Finding the center point A*  255  Finding the poles Finding the circle of sliders  256  Finding the slider point  257  Final solution Problem 6.20 A device characterized by the inputoutput relationship = a1 + a2 cos is to be used to generate / 4. (approximately) the function = 2 ( and both in radians) over the range 0 a) Determine the number of precision points required to compute a1 and a2 . b) Choose the best precision point values for from among 0, 0.17, 0.35, and 0.52, and determine the values of a1 and a2 that will allow the device to approximate the function. c) Find the error when = /8.  258  Solution Two precision points can be used because there are two design variables. For this, first determine 1 and 2 . Look at Chebychev spacing to see what values of are reasonable. From the figure, 135˚ 0 θ1 π 8 45˚ θ2 π 4 1 2 = = 8 8 (1 + cos135°) = 0.115 (1 + cos45°) = 0.670 The angles which are closest to these values are 0.17 and 0.52. Then,
1 = ( 1 )2 = (0.17)2 = 0.0289 2 2 2 = ( 2 ) = (0.52) = 0.270 Substituting into the equation for the system model,
1 = a1 + a2 cos 2 = a1 + a2 cos 1 2 or, 0.0289 = a1 + a2 cos(0.17) = a1 + a2 (0.985) 0.270 = a1 + a2 cos(0.52) = a1 + a2 (0.868) Subtract the first equation from the second, and solve for a2 0.241 = a2 ( 0.117) and a2 = 2.060 Now back substituting into the first equation, a1 = 0.0289 a2 (0.985) = 0.0289 + 2.060(0.985) = 2.058 . The error is given by  259  e= ideal act = 2 (2.058 2.060 cos )
= /8 = 22.5˚, Substituting in the given value e = (0.3926 )2 (2.058 2.060 cos 22.5 ) = 0.000673 ˚ Problem 6.21
2 A mechanical device characterized by the inputoutput relationship = 2a1 + 3a2 sin + a3 is to be 2 over the range 0 / 4 . Exterior used to generate (approximately) the function = 2 constraints on the design require that the parameter a3 = 1. a) Determine the number of precision points required to complete the design of the system. b) Use Chebyshev spacing, and determine the values for the unknown design variables which will allow the device to approximate the function. c) Find the error when = /6. Solution Two precision points can be used because there are two design variables. For this, first determine 1 and 2 . Look at Chebychev spacing to see what values of are reasonable. From the figure, 135˚ 0 θ1 π 8 45˚ θ2 π 4 1 2 = = 8 8 (1 + cos135°) = 0.115 (1 + cos45°) = 0.670 Then,  260  1 2 = 2( 1 )2 = 2( 0.115)2 = 0.02645 = 2(
2) 2 = 2( 0.670)2 = 0.8978 Substituting into the equation for the system model,
1 2 = 2a1 + 3a2 sin = 2 a1 + 3a2 sin 2 1 + a3 2 = 2 a1 + 3a2 sin 1 +1 2 2 + a3 = 2 a1 + 3a2 sin +1 or, 0.02645 = 2 a1 + 3a2 sin(0.115) + 1 = 2a1 + 0.3442a2 + 1 0.8978 = 2 a1 + 3a2 sin(0.670 ) +1 = 2a1 + 1.8629a2 +1 Subtract the first equation from the second, and solve for a2 and 0.8713 = 1.5187 a2 a2 = 0.5737 Now back substituting into the first equation, 2 a1 = 0.02645 1 0.3442 a2 = 0.02645 1 0.3442(0.5737) = 1.1710 or a1 = 1.1710 / 2 = 0.5855 The error is given by e= =
ideal 22 (2a1 + 3a2 sin + 1) = 2 [ 0.1710 + 1.7214 sin ]
act =2 2 2 [2( 0.5855) + 3( 0.5737)sin +1] Substituting in the given value = /6 = 30˚, e = 2( 0.5235)2 [ 0.1710 + 1.7214 sin 30˚ ] = 0.1416 Problem 6.22
2 A mechanical device characterized by the inputoutput relationship = 2a1 + a2 tan + a3 is to be used to generate (approximately) the function = 3 3 ( and both in radians) over the range / 3 . Exterior constraints on the design require that the parameter a3 = 1. 0 a) Determine the number of precision points required to complete the design of the system. b) Use Chebyshev spacing, and determine the values for the unknown design variables that will allow the device to approximate the function. c) Find the error when = /6.  261  Solution: There are two unknowns so the number of precision points is two. The precision points according to Chebyshev spacing are:
1= ( max + min) 2 +( max min) 2 cos135 = 6 + cos135 = 0.1534 6 and
2 =( max + min) 2 +( max min ) 2 cos45 = 6 + 6 cos45 = 0.8938 The corresponding values of are:
3 1 = 2 1 = 2(0.1534)3 = 0.0072 2 =2 3 = 2(0.8938)3 2 = 1.428 We can now solve for a1 and a2 using the desired inputoutput relationship.
2 = 2a1 + a2 tan + a3 = 2a1 + a2 tan + 1 Then, 0.0072 = 2a1 + a2 tan(0.1534) + 1 = 2a1 + 0.1546a2 + 1 and 1.428 = 2a1 + a2 tan(0.8938) + 1 = 2a1 + 1.2442a2 + 1 Subtracting the two equations gives, 1.4209 = 1.0902a2 a2 = 1.3033 Backsubstituting to determine a1 gives 2a1 = 1.4281 1.2442a2 1 = 0.4281 1.2442(1.3033) = 1.1935 The error is given by e= =
ideal 32 a1 = 0.5968 (2 a1 + a2 tan + 1) = 3 [ 0.1936 + 1.3033 tan ]
act =3 2 2 [2 ( 0.5968) + (1.3033)tan + 1] Substituting in the given value = /6 = 30˚, e = 3( 0.5235)2 [ 0.1936 + 1.3033 tan 30˚ ] = 0.2632  262  Problem 6.23 A mechanical device characterized by the inputoutput relationship = 2a1 + a2 sin is to be used to generate (approximately) the function y = 2x 2 over the range 0 x / 2 where x, y, , and are all in radians. Assume that the use of the device will be such that the starting point and range for x can be the same as those for , and the range and starting point for y can be the same as those for . a) Determine the number of precision points required to complete the design of the system. b) Use Chebyshev spacing, and determine the values for the unknown design variables that will allow the device to approximate the function. c) Compute the error generated by the device for x = /4. Solution: There are two unknowns so the number of precision points is two. The precision points according to Chebyshev spacing are: x1 = and x2 = (x max + x min) (x max x min) + cos 45 = + cos 45 = 1.3407 44 2 2 (xmax + xmin) (xmax x min) + cos 135 = + cos135 = 0.2300 44 2 2 The corresponding values of y are:
2 y1 = 2 x1 = 2( 0.2300)2 = 0.1058 y2 = 2 x 2 = 2 (1.3407)2 = 3.5949 2 We must now relate to x and to y. Since and x have the same starting value and same range, we can interchange them exactly. The same applies to and y. Therefore, we can use the following pairs of numbers to solve the problem. = 0.2300 rad = 13.178˚ 1 = 0.1058
1 and = 1.3407rad = 76.816˚ 2 = 3.5949
2 We can now solve for a1 and a2 using the desired inputoutput relationship. = 2a1 + a2 sin Then,  263  1 = 2a1 + a2 sin 2 = 2 a1 + a2 sin 1 = 0.1058 = 2 a1 + a2 sin 13.178 = 2a1 + a2 (.2280) ˚ ˚ 2 = 3.5949 = 2 a1 + a2 sin 76.816 = 2 a1 + a2 (.9736) Subtracting the two equations gives, 3.4891 = 0.7456a2 a2 = 3.4891 / 0.7456 = 4.6795 Backsubstituting to determine a1 gives a1 = [0.1058 a2 (.2280 )] / 2 = [0.1058 4.6795(.2280)] / 2 = 0.4806 The error is given by e= =
ideal 22 (2a1 + a2 sin ) = 2 [ 0.9611 + 4.6795 sin ]
act =2 2 2 [2( 0.4806) + (4.6795)sin ] Substituting in the given value e =2
2 = /4 = 45˚, [ 0.9611 + 4.6795 sin ] = 2(.7854 )2 [ 0.9611 + 4.6795 sin 45 ] = 1.1141 ˚ Problem 6.24 Determine the link lengths and draw a fourbar linkage that will generate the function = 2 ( and both in radians) for values of between 0.5 and 1.0 radians. Use Chebyshev spacing with three position points. The base length of the linkage must be 2 cm. Solution Determine the precision points using Chebychev spacing. Then
0 f = 0.5 = 1.0 =
f 0 0 cos 30˚ = 1.5 0.5 cos 30˚ = 0.53349364905389 2 2 2 0 1.5 = = 0.75 2 1.5 0.5 0 0 +f cos 30˚ = + cos 30˚ = 0.96650635094611 2 2 2 f 1 + 2 f+ 2= 2 f+ 3= 2 Compute the 's
1 = 2= 3= 2 1 = 0.28461547358084 2 2 = 0.56250000000000 2 3 = 0.93413452641916  264  1 cos A = 1 cos 1 cos cos( cos( B= cos( Then
1 2 3 1 2 3 1) cos cos cos 1 2 3 1 0.95976969417465 = 1 0.84592449923107 1 0.59451456313420 0.86103565250177 0.73168886887382 0.56817793658800 0.96918935579862 ) = 0.98247331310126 2 0.99947607824379 3) x1 1 cos x2 = 1 cos x3 1 cos and 1 2 3 cos cos cos 1 2 3 1 cos( cos( cos( 1 2 3 1) 1.05801273775729 ) = 0.00195486778387 2 0.10097974323242 3) r2 = 1 = 511.5 x2 r4 = 1 = 9.903 x3 r3 = 1 + r22 + r22 2r2r4 x1 = 522.0 Now unscale the values by multiplying each by 2. Then R1 = 2.0; R2 = 2(511.5) = 1023; R3 = 2(522.0) = 1044; R4 = 2(9.903) = 19.80; Check for linkage type: l + s = 1044 + 2.0 = 1046 p + q = 19.80 + 1023 = 1042.8 Then 1046> 1042 nonGrashof, double rocker. Note that the linklength ratios vary greatly. This design would be considered very undesirable. Problem 6.25 Determine the link lengths and draw a fourbar linkage that will generate the function = sin( ) for values of between 0 and 90 degrees. Use Chebyshev spacing with three position points. The base length of the linkage must be 2 cm.  265  Solution Determine the precision points using Chebychev spacing. Then
0 =0 ˚ f = 90 =
f 0 0 0 f 0 1 + 2 f+ 2= 2 f+ 3= 2 2 = + 4
f cos 30˚ = 4 4 cos 30˚ = 0.10522340180962 = 0.78539816339745
0 2 cos 30˚ = 4 + 4 cos 30˚ = 1.46557292498528 Compute the 's
1 = sin( 1 ) = 0.10502933764983 2 = sin( 2 ) = 0.70710678118655 3 = sin( 3 ) = 0.99446912382076
1 2 3 1 cos A = 1 cos 1 cos cos( B = cos( cos( Then
1 2 3 cos cos cos
1) 1 2 3 1 0.99448948752450 0.99446912382076 = 1 0.76024459707563 0.70710678118655 1 0.54494808990266 0.10502933764983 0.99999998116955 2 ) = 0.99693679488540 0.89106784875455 3) 1 cos x1 x2 = 1 cos x3 1 cos and 1 2 3 cos cos cos 1 2 3 1 cos( cos( cos( 1 2 3 1) 1.05576595368848 2 ) = 0.36099675791050 0.30492802741671 3) r2 = 1 = 2.77010798043768 x2 1 r4 = = 3.27946239796913 x3
2 r3 = 1 + r22 + r2 2r2 r4 x1 = 0.49621992921794 Now unscale the values by multiplying each by 2. Then  266  R1 = 2.0; R2 = 2(2.7701) = 5.5402; R3 = 2( 0.4962) = 0.9924; R4 = 2(3.2794) = 6.5588; A scaled drawing of the linkage is given in the following: Problem 6.26 Design a fourbar linkage that generates the function y = x x + 3 for values of x between 1 and 4. Use the Chebyshev spacing for three position points. The base length of the linkage must be 2 in. Use the following angle information:
0 0 = 45˚ = 30˚ = 50˚ = 70˚ Compute the error at x = 2 Solution The solution is given in the following. From the given information,  267  x0 = 1; xf = 4. Using Chebychev spacing for the precision points, x1 = xf + x0 2 xf x0 cos30° = 4 +1 2 2 4 1 cos30° = 1.20096189 2 Similarly, x2 = 2.5 and x3 = 3.799038105. Then, the corresponding values for y are: yf = y0 = y1 = y2 = y3 = xf x0 x1 x2 x3 xf + 3 = 1 x0 + 3 = 3 x1 + 3 = 2.894922175 x2 + 3 = 2.081138830 x3 + 3 = 1.150074026 Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. For the range of linkage angles, we have:
0 = 45°; = 50° and
0 = 30°; = 70° The precision points in terms of are:
1 = = = x1 x 0 x f x0 x2 x0 x f x0 x3 xf x0 x0 + + + 0 = = = 1.20096189 1 50°+ 45° = 48.34936490538903° 3 2.5 1 50° + 45° = 70° 3 3.799038105 1 50°+ 45° = 91.65063509461096° 3 2 0 3 0 Similarly,
1 y1 y0 y f y0 y2 y0 2= y f y0 y3 y0 3= y f y0 = + + + 0 = 0 2.894922175 3 70° + 30 = 33.67772386020241° 2 2.081138830 3 = 70°+ 30 = 62.16014094705335° 2 0 = 1.150074026 3 70° + 30 = 94.74740905563471° 2 Using the matrix solution procedure,  268  z1 1 cos z2 = 1 cos z3 1 cos 1 2 3 cos cos cos 1 2 3 1 cos( cos( cos( 1 2 3 1) 2) 3) 1 1 0.8321697783 0.6645868192 = 1 0.4670019053 0.3420201433 1 0.0827631422 0.0288050322 and r2 = r4 = and
2 r3 = 1 + r22 + r4 0.9673932335 1.0037381398 0.9906531872 = 0.1450642022 0.9985397136 0.2363317277 1 1 = = 6.8934994632 z2 0.1450642022 1 1 = = 4.2313404537 z3 0.2363317277 2r2 r4 z1 2(6.893499)( 4.231340)(1.003738) = 2.8051768 = 1 + (6.893499) 2 + (4.231340)2 For the overall size of the linkage, use a base link length of 2 in. Then the lengths of the other links become R1 = 1( 2) = 2 in R2 = 6.8934994632(2) = 13.7869989 in R4 = 4.2313404537(2 ) = 8.46268090 in R3 = 2.8051768(2 ) = 5.610353611 in Note that the linklength ratios would make this linkage undesirable and possibly unusable. To compute the error at x = 2, compute the ideal y and the generated y. yideal = x x + 3 = 2 2 + 3 = 2.4142 To compute the generated value, first compute the value of corresponding to x = 2. Then,
e = xe xf x0 x0 + 0 = 21 50°+ 45° = 61.666° 3 Using this value of and the link lengths given above, find the output value for . This can be done graphically, or by using the routine fourbar_cr. The value of is
e = 51.02815161 The corresponding value for y is,  269  yact = The error is e 0 ( yf y0 ) + y0 = 51.02815161 30 ( 2 ) + 3 = 2.3992 70 error = yideal Problem 6.27 yact = 2.4142 2.3992 = 0.0150 Design a fourbar linkage to generate the function y=x2 1 for values of x between 1 and 5. Use Chebyshev spacing with three position points. The base length of the linkage must be 2 cm. Use the following angle information:
0 0 = 30˚ = 45˚ = 60˚ = 90˚ Compute the error at x = 3. Solution The solution is given in the following. From the given information, x0 = 1; xf.= 5. Using Chebychev spacing for the precision points, x1 = xf + x0 2 xf 2 x0 cos30° = 1 + 5 2 5 1 cos30° = 1.26794919243112 2 Similarly, x2 = 3 and x3 = 4.73205080756888. Then, the corresponding values for y are: yf = x f2 y0 = x0 2 y1 = x12 y2 = x22 y3 = x32 1 = 24 1= 0 1 = 0.60769515458674 1= 8 1 = 21.39230484541327 Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. The linkage angles are:
0 = 30°; = 60° and
0 = 45°; = 90° The precision points in terms of are:  270  1= x1 x0 xf x0 + + + 0 = 1.26794919243112 1 60° + 30° = 34.01923788646684° 4 = 3 1 60°+ 30° = 60° 4 = 4.73205080756888 1 60° + 30° = 85.98076211353316° 4 2 = x2 x0 xf x0 x3 x0 xf x0 0 3= 0 Similarly,
1 y1 y0 y f y0 y2 y0 2= y f y0 y3 y0 3= y f y0 = + + + 0 = 0 0.60769515458674 0 90°+ 45 = 47.278856829° 24 80 = 90°+ 45 = 75° 24 0 = 21.39230484541327 0 90° + 45 = 125.221143170° 24 Using the matrix solution procedure, 1 cos z1 z2 = 1 cos z3 1 cos
1 2 3 cos cos cos 1 2 3 1 cos( cos( cos( 1 2 3 1) 2) 3) 1 1 0.6784308203 0.8288497687 = 1 0.2588190451 0.5000000000 1 0.5767338180 0.0700914163 and r2 = r4 = and
2 r3 = 1 + r22 + r4 0.973340766 1.1947633778 0.965925826 = 0.6332388583 0.774498853 0.7854636563 1 1 = = 1.57918293 z2 0.6332388583 1 1 = = 1.27313338 z3 0.7854636563 2r2 r4 z1 = 1 + (1.579182)2 + 1.2731332 2(1.579182)(1.273133)(1.1947633) = 0.557242 For the overall size of the linkage, use a base link length of 2 cm. Then the lengths of the other links become  271  R1 = 1( 2) = 2 cm R2 = 1.579182(2 ) = 3.158364 cm R4 = 1.273133( 2) = 2.546266 cm R3 = 0.557242( 2) = 1.114484 cm Since x=3 is a precision point, the error will be zero at that point. To prove this, compute the ideal y and the generated y. yideal = x 2 1 = 32 1 = 8 To compute the generated value, first compute the value of corresponding to x = 3. Then,
e = xe xf x0 x0 + 0 = 31 60°+ 30° = 60° 3 Using this value of and the link lengths given above, find the output value for . This can be done graphically, or by using the routine fourbar_cr. The value of is
e = 75 ˚ The corresponding value for y is, yact = The error is error = yideal yact = 8 8 = 0
e 0 ( yf y0 ) + y0 = 75 45 (24) + 0 = 8 90 Problem 6.28 The output arm of a lawn sprinkler is to rotate through an angle of 90˚, and the ratio of the times for the forward and reverse rotations is to be 1 to 1. Design the crankrocker mechanism for the sprinkler. If the crank is to be 1 inch long, give the lengths of the other links. Solution = 180 Q 1 = 180 1 1 = 0° Q +1 1+ 1 This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at the angle shown. Then, the center location for O2 is located on a line through B1 and B2 . Any point will work as long as it is selected to the left of B2 .  272  4.314 1.500 B2 O2 3.232" 90˚ B1 O4 For the location selected, r2 + r3 = 4.314 r3 r2 = 1.500 Then 2(r3 ) = 5.814 or and r3 = 2.907" r2 = r3 1.500 = 2.907 1.500 = 1.407" From the figure, r1 = 3.232". If the crank is 1 in long, then each dimension must be scaled. To do this multiply each length by K where K= Then, r1 = 0.711*(3.232) = 2.298 in r2 =0.711*(1.407) = 1.00 in, r3 = 0.711*(2.907) = 2.067 in r4 = 0.711*(2.0) = 1.422 in, 1 = 0.711 1.407  273  Problem 6.29 Design a crankrocker mechanism such that with the crank turning at constant speed, the oscillating lever will have a time ratio of advance to return of 3:2. The lever is to oscillate through an angle of 80o , and the length of the base link is to be 2 in. Solution = 180 Q 1 = 180 1.5 1 = 36° Q +1 1.5 + 1 This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at the angle shown. Then, the construction of the center location for O2 is shown in the following figure. From the figure, r2 + r3 = 3.731 r3 r2 = 1.687 Then 2(r3 ) = 5.418 or and r3 = 2.709" r2 = r3 1.687 = 2.709 1.687 = 1.022" From the figure, r1 = 2.155". If the distance between the fixed pivots is really 2 in, then each dimension must be scaled. To do this multiply each length by K where K= 2 = 0.928 2.155 B2 B1 3.731" 36˚ 1.687" O2 80˚ 2.155" O4 Then, r1 = 0.928*2.155 = 2 in  274  r2 = 0.928*(1.022) = 0.948 in, r3 = 0.928*(2.709) = 2.514 in, r4 = 0.928*(2.0) = 1.856 in. Note that a different linkage is generated for each line drawn through B1 . Problem 6.30 A packing mechanism requires that the crank (r2 ) rotate at a constant velocity. The advance part of the cycle is to take twice as long as the return to give a quickreturn mechanism. The distance between fixed pivots must be 0.5 m. Determine the lengths for r2 , r3 , and r4 .
B r3 A
1 ω2 ∆θ = 80˚ r4 O4 S chematic Drawing r2 O2 Solution = 180 Q 1 = 180 2 1 = 60° Q +1 2 +1 This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at the angle shown. Then, the construction of the center location for O2 is: B2 B1 2.9365" 1.1482" O2 1.4728" O4 60˚ 80˚ r2 + r3 = 2.9365 r3 r2 = 1.1482  275  Then 2(r3 ) = 4.0847 or and r3 = 2.0424" r2 = r3 1.1482 = 2.0424 1.1482 = 0.8942" From the figure, r1 = 1.4728". If the distance between the fixed pivots is 0.5 meters, then each dimension must be scaled. To do this multiply each length by K where K = 0.5 = 0.3395 1.4728 Then, r1 = 0.3395)*(1.4728) = 0.5 meters r2 = 0.3395)*(0.8942) = 0.3036 meters, r3 = 0.3395)*(2.0424) = 0.6934 meters, r4 = 0.3395)*(2.0) = 0.6789 meters, Note that a different linkage is generated for each line drawn through B1 . Problem 6.31 The rocker O4 B of a crankrocker linkage swings symmetrically about the vertical through a total angle of 70˚. The return motion should take 0.75 the time that the forward motion takes. Assuming that the two pivots are 2.5 in apart, find the length of each of the links. Solution For this problem, the time ratio of forward stroke to return is 1.33. Therefore, = 180 Q 1 = 180 1.33 1 = 25.7° Q +1 1.33 + 1 This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at the angle shown. Then, the construction of the center location for O2 is: r2 + r3 = 4.2014 r3 r2 = 2.4112 Then 2(r3 ) = 6.6126" or and r3 = 3.3063"  276  r2 = r3 2.4112 = 3.3063 2.4112 = 0.8951" B2 B1 70˚ 25.7˚ 4.2014" 2.4112" O2 2.5984" O4 From the figure, r1 = 2.5984". If the distance between the fixed pivots is really 2.5", then each dimension must be scaled. To do this multiply each length by K where K= Then, r1 = 0.9621*(2.5984) = 2.5" r2 = 0.9621*(0.8951) = 0.9612", r3 = 0.9621*(3.3063) = 3.1810", r4 = 0.9621*(2.0) = 1.9243", Note that a different linkage is generated for each line drawn through B1 . Problem 6.32 A crank rocker is to be designed such that with the crank turning at a constant speed CCW, the rocker will have a time ratio of advance to return of 1.25. The rocking angle is to be 40˚, and it rocks symmetrically about a vertical line through O4 . Assume that the two pivots are on the same horizontal line, 3 in apart. Solution: This problem does not have a unique solution. To solve this problem, the procedure given in Section 4.4.3.1 can be used. First draw the line beween the pivot points as shown. Next compute the angle using 2.5 = 0.9621 2.5984  277  = 180 Q 1 = 180 1.25 1 = 20° Q +1 1.25 + 1 B2 40˚ 2.3127" 4.4885" O4 20˚ 40˚ G Locate G as shown in Fig. 4.46, and draw the locus of B2 . To locate the oscillation angle symetrically about the vertical, locate B2 at an angle of 20˚ from the vertical. We can then locate B1 at an angle of 80˚ from the line O2 B2 . Next compute r2 and r3 from r2 + r3 = 4.4885 r3 r2 = 2.3127 Then 2(r3 ) = 6.8012 or and r3 = 3.4006 r2 = r3 2.3127 = 3.4006 2.3127 = 1.0879" Also from the drawing, r1 = 3.0000" r4 = 1.9238" The transmission angles do not approach either 0 or 180; therefore, the linkage is reasonably efficient. Problem 6.33 Design a crankrocker mechanism that has a base length of 2.0, a time ratio of 1.3, and a rocker oscillation angle of 100˚. The oscillation is to be symmetric about a vertical line through O4 . Specify the length of each of the links. Solution: 1.9238" 80˚ B1 O2 3.00"  278  This problem does not have a unique solution. To solve this problem, the procedure given in Section 4.4.3.1 can be used. First draw the line beween the pivot points as shown. Next compute the angle using = 180 Q 1 = 180 1.3 1 = 23.48° 1.3 + 1 Q +1 and 2 = 50 23.48 = 26.52 Locate G as shown in Fig. 4.46, and draw the locus of B2 . To locate the oscillation angle symetrically about the vertical, locate B2 at an angle of 50˚ from the vertical. We can then locate B1 at an angle of 100˚ from the line O2 B2 . Next compute r2 and r3 from r2 + r3 = 2.9890 r3 r2 = 1.3229 Then 2(r3 ) = 4.3119 or r3 = 2.1559 100˚ 1.3229" B2 50˚ 2.9890" B1 O2 2.00" 26.52˚ 50˚ G O4 1.2794" and r2 = r3 1.3229 = 2.1559 1.3229 = 0.8330" Also from the drawing, r1 = 2.0000" r4 = 1.2794" Problem 6.34 A crankrocker mechanism with a time ratio of 2 1 and a rocker oscillation angle of 72˚ is to be 3 designed. The oscillation is to be symmetric about a vertical line through O4 . Draw the mechanism  279  in any position. If the length of the base link is 2 in, give the lengths of the other three links. Also show the transmission angle in the position in which the linkage is drawn. Solution: This problem does not have a unique solution. To solve this problem, the procedure given in Section 4.4.3.1 can be used. First draw the line beween the pivot points as shown. Next compute the angle using 21 1 = 72˚ = 180 Q 1 = 180 3 Q +1 2 1 +1 3 2 = 36 72 = 36˚ and This means that the point G is at infinity, and the B2 locus is a straight line through O2 at an angle of 36˚ to the horizontal. The result is shown in the following: 2.8617" B2 1.6251" 36˚ 72˚ B1 1.1629" O2 36˚ 2.00" 36˚ G O4 To locate the oscillation angle symetrically about the vertical, locate B2 at an angle of 36˚ from the vertical. We can then locate B1 at an angle of 72˚ from the line O2 B2 . Next compute r2 and r3 from r2 + r3 = 2.8619 r3 r2 = 1.6251 Then 2(r3 ) = 4.4870 or and r3 = 2.2435" r2 = r3 1.6251 = 2.2435 1.6251 = 0.6184" Also from the drawing, r1 = 1.0000" and  280  r4 = 1.1629" The linkage is shown in an arbitrary position below. The transmission angle is shown. B 69.44˚ O2 O4 Problem 6.35 The mechanism shown is used to drive an oscillating sanding drum. The drum is rotated by a splined shaft that is cycled vertically. The vertical motion is driven by a fourbar linkage through a rackandpinion gear set (model as a rolling contact joint). The total vertical travel for the sander drum is 3 in, and the pinion has a 2 in radius. The sander mechanism requires that the crank (r2 ) rotate at a constant velocity, and the advance part of the cycle is to take the same amount of time as the return part. The distance between fixed pivots must be 4 in. Determine the lengths for r2 , r3 , and r4 . Sander drum Total travel = 3" A
1ω 2 r3 B r2
O2 r4
O4 2" ∆θ Rack Schematic Drawing Pinion Belt drive Splined Shaft Motor Solution: This problem does not have a unique solution. To solve this problem, we must determine the oscillation angle. If the drum rotates 3 inches, we can find the oscillation angle form and r4 = d = d = 3 = 1.5 rad = 85.94˚ r4 2  281  To start the construction, first draw the line beween the pivot points as shown. Next compute the angle using = 180 Q 1 = 180 1 1 = 0˚ 1+ 1 Q +1 Arbitrarily pick r4 as 2" at the angle shown. Then, the center location for O2 is located on a line through B1 and B2 . Any point will work as long as it is selected to the left of B2 . 4.250 1.500 B2 O2 3.2253" 85.94˚ B1 O4 For the location selected, r2 + r3 = 4.250 r3 r2 = 1.500 Then 2(r3 ) = 5.750 or and r3 = 2.875" r2 = r3 1.500 = 2.875 1.500 = 1.375" From the figure, r1 = 3.2253". If the crank is really 1 in long, then each dimension must be scaled. To do this multiply each length by K where K= Then, r1 = 1.2413*(3.2253) = 4.000 in r2 =1.2413*(1.375) = 1.706 in, 4 = 1.2413 3.2253  282  r3 = 1.2413*(2.875) = 3.569 in r4 = 1.2413*(2.0) = 2.482 in, Problem 6.36 The mechanism shown is proposed for a rock crusher. The crusher hammer rotates through an angle of 20˚, and the gear ratio R G/RP is 4:1, that is, the radius r G is four times the radius r p . Contact between the two gears can be treated as rolling contact. The crusher mechanism requires that the crank (r2 ) rotate at a constant velocity, and the advance part of the cycle is to take 1.5 times as much as the return part. The distance between fixed pivots O2 and O4 must be 4 ft. Determine the lengths for r2 , r3 , and r4 . A
1ω 2 r3 r4
O4 B ∆θ = 20˚ r2
O2 Schematic Drawing rP rG
5 Solution = 180 Q 1 = 180 1.5 1 = 36° Q +1 1.5 + 1 = 4(20) = 80° This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at the angle shown. Then, the construction of the center location for O 2 is: B2 B1 1.6602" 36˚ O2 3.7087" 80˚ O4  283  r2 + r3 = 3.7087 r3 r2 = 1.6602 Then 2(r3 ) = 5.3689 or and r3 = 2.6845" r2 = r3 1.6602 = 2.6845 1.6602 = 1.0242" From the figure, r1 = 2.1418". If the distance between the fixed pivots is really 4 feet, then each dimension must be scaled. To do this multiply each length by K where K= Then, r1 = 4.0 feet, r2 = 1.9126 feet, r3 = 5.0133 feet, r4 = 3.7350 feet, Note that a different linkage is generated for each line drawn through B1 . Problem 6.37 The mechanism shown is proposed for a shaper mechanism. The shaper cutter moves back and forth such that the forward (cutting) stroke takes twice as much time as the return stroke. The crank (r2 ) rotates at a constant velocity. The follower link (r4 ) is to be 4 in and to oscillate through an angle of 80˚. Determine the lengths for r1 , r2 , and r3 . 4 = 1.8675 2.1418 A
1ω 2 r3 r4
O4 C B r2
O2 Schematic Drawing r5
D 6 Cutter Part Being Machined Solution: This problem does not have a unique solution. To start the procedure, determine the angle the time ratio.  284 from = 180 Q 1 = 180 (2 1) = 60˚ Q +1 (2 +1) B2 0.90" 80˚ B1 2.90" 60˚ O2 2.00" 1.64" O4 From the diagram: r1 = 1.64" r2 = O2 B1 O2 B2 = 2.90 0.90 = 1.00" 2 2 r3 = O2B1 + O2B2 = 2.90 + 0.90 = 1.90" 2 2 r4 = 2" Scalling the results, R4 = 4 in R2 = r2 R 4 = 1.00 4 = 1.00(2) = 2.00 in r4 2 R 4 = 1.90(2) = 3.80 in R3 = r3 r4 R4 = 1.64(2) = 3.28 in R1 = r1 r4 The linkage is shown to scale in a general position in the following: B A O2 O4  285  Problem 6.38 A crank rocker is to be used in a doorclosing mechanism. The door must open 100˚. The crank motor is controlled by a timer mechanism such that it pauses when the door is fully open. Because of this, the mechanism can open and close the door in the same amount of time. If the crank (r2 ) of the mechanism is to be 10 cm long, determine the lengths of the other links (r1 , r3 , and r4 ). Sketch the mechanism to scale. Schematic Drawing Door B A
1ω 2 r3 r2
O2 Wall r4
O4 Solution The time ration is 1 so is 0. This problem does not have a unique solution. Initially pick the output link length to be 2. Then the following scalled values are determined as shown in the figure below. B1O2 = 5.0"= r3 + r2 B2O2 = 1.953"= r3 r2 Then r3 = 3.476" r2 = 5.0 3.476 = 1.523" r4 = 2" r1 = 3.713"  286  B1 B2 O2
100˚ O4 Now unscale the output. K = R2 = 10 = 6.566 r2 1.532 Then, R1 = 6.566(3.713) = 24.37 cm R2 = 6.566(3.713) = 10 cm R3 = 6.566(3.713) = 22.82 cm R 4 = 6.566(2) = 13.13 cm A scaled version of the linkage if given in the following. B A R2 O2 R3 Scaled Linkage R1 R4 O4  287  Problem 6.39 A crank rocker is to be used for the rock crusher mechanism shown. The oscillation angle for the rocker is to be 80˚, and the working (crushing) stroke for the rocker is to be 1.1 times the return stroke. If the frame link (r1 ) of the mechanism is to be 10 ft long, determine the lengths of the other links (r2 , r3 , and r4 ). Sketch the mechanism to scale.
B 3 A 2 O1 O2 4 Roc k Solution: = 180 Q 1 = 180 (1.1 1) = 8.6˚ Q +1 (1.1 +1) This problem does not have a unique solution. Initially pick the output link length to be 2. Then from the diagram, B2 3.24" O1 5.74" 8.6˚ 2.00" 80˚ B1 4.39" O2 r1 = 4.39" r2 = O1B1 O1B2 = 5.74 3.24 = 1.25" 2 2 O1B1 + O1B2 = 5.74 + 3.24 = 4.49" r3 = 2 2 r4 = 2" Scalling the results, R1 = 10 ft  288  R2 = r2 R1 = 1.25 10 = 1.25(2.33) = 2.91 ft r1 4.39 R1 = 4.49(2.33) = 10.46 ft R3 = r3 r1 R 4 = r4 R1 = 2(2.33) = 4.66 ft r1 The linkage is shown to scale in a general position in the following: B A O1 Problem 6.40 O2 A crank rocker is to be used in a windshieldwiping mechanism. The wiper must oscillate 80˚. The time for the forward and return stroke for the wiper is the same. If the base link (r1 ) of the mechanism is to be 10 cm long, determine the lengths of the other links (r2 , r3 , and r4 ). Sketch the mechanism to scale. Wiper Schematic Drawing 1ω 2 r1 Frame O4 r4 r3 B O2 r2 A Solution The time ration is 1 so  289  = 180 Q 1 = 180 0 = 0 2 Q+1 This problem does not have a unique solution. Initially pick the output link length to be 2. Then the linkage can be constructed as shown O4 R1 R4 B2 A2 B1 O2 A1 R2 From the figure, R4 = 2 in R1 = 3.512 in R2 + R3 = O2 B2 R3 R2 = O2 B1 or R2 = (O2 B2 O2 B1) / 2 = (4.5078 2.0018) / 2 = 1.253 in R3 = (O2B2 + O2B1) / 2 = (4.5078 + 2.0018) / 2 = 3.881in Determine the scaling factor, r1 = 10 = 2.7732 = r2 = r3 = r4 R1 3.6059 R2 R3 R4 Using the unscalled lengths from the figure, r2 = 2.7732 R2 = 2.7732(1.253) = 3.4749cm r3 = 2.7732 R3 = 2.7732(3.881) = 10.7629cm r4 = 2.7732 R4 = 2.7732(2) = 5.5465cm The mechanism is drawn to scale as:  290  O4 R4 B O2 R3 R2 A Problem 6.41 Design a sixbar linkage like that shown in Fig. 6.62 such that the output link will do the following for one complete revolution of the input crank: 1. Rotate clockwise by 30˚ for a clockwise rotation of 210˚ of the input crank. 2. Rotate counterclockwise by 30˚ for a clockwise rotation of 150˚ of the input crank. Solution Virtually any curve that has a general oval shape can be made to work for this problem. After selecting such a curve, pick a starting place on the curve as one of the two extreme locations for point F. Draw a line perpendicular to the curve and select a length for link 5. The value for the link length is somewhat arbitrary although it needs to be long enough to permit the mechanism to operate for the whole cycle. Next count around the curve 42 dashes corresponding to 210˚ of crank rotation. This will be the other extreme location for link 5, and the link will be perpendicular to the curve at this location also. Locate the link perpendicular to the curve, and this will locate the second extreme location for Point F. Connect the two extreme locations of F by a cord line, and locate the perpendicular bisector of the cord line. Draw a line through one of the extreme locations of F at an angle of 15˚ to the perpendicular bisector. This will locate point G and the length of link 6. Draw the dyad GRE to complete the 6bar linkage. For the mechanism shown, the following values apply: AB = 0.762”; BE = 1.313”; BC = 2.191”; CD = 1.193”; AD = 2.488”;
1 = 11.83˚ YG = 0.532” = 34.03˚; EF = 1.00”; FG = 1.564”; XG = 2.644”; The linkage was analyzed for the values given above using the program sixbar.m, and the results are shown in following the solution drawing. The results are fairly accurate; however, the accuracy could be improved by moving the location of G slightly.  291   292  Problem 6.42 Design a sixbar linkage like that shown in Fig. 6.62 such that the output link will make two complete 35˚ oscillations for each revolution of the driving link. (Hint: Select a coupler curve that is shaped like a "figure 8".) Solution Locate a curve that is roughly in the shape of a symmetric figure 8. Draw a circle which is tangent to the top of the curve at the two points b and d. Here, b and d should be approximately in the center of the loops of the figure 8. Label the center of the circle as point F', which is one extreme location for the point F. Next draw a circle of the same radius as that of the first circle but tangent to two ponits on the bottom of the curve. The tangent points are a and c. The center of this circle is point F", which is the other extreme location for the point F. Points a, b, c, d will be the extreme locations of the coupler point E. Bisect the line F'F". The pivot G must lie on this line. Locate G such that the included angle between GF' and GF" is 35˚. Draw the dyad GFE to complete the 6bar linkage. The linkage was analyzed for the values given above using the program sixbar.m, and the results are shown in following the solution drawing. The results are fairly accurate; however, the accuracy could be improved by moving the location of G slightly. Y B 2 3 β C X 4 b E a D c 5 d A AD = 2.1429" AB = 1.0994" BC = 2.1806" CD = 1.6212" BE = 2.1005" EF = 2.9" FG = 1.0614" β = 35.5˚ θ = 26.46˚ x G = 0.7222" yG = 2.95" G
35 ˚ F' 6 F"  293 F Problem 6.43 Design a sixbar linkage like that shown in Fig. 6.62 such that the output link will do the following for one complete revolution of the input crank: 1. Rotate clockwise by 40˚ 2. Rotate counterclockwise by 35˚ 3. Rotate clockwise by 30˚ 4. Rotate counterclockwise by 35˚ (Hint: Select a figure 8 or kidney bean shaped coupler curve.) Solution: Figure 4.51 is repeated here for simplicity.  294  G 6 F E 5 3 C B 2 A 4 1 Fig. PB6.43.1: Six bar linkage D For this problem, the displacement of link 6 can be represented schematically as shown in Fig. PB6.43.2, and the oscillation of link 6 would then appear as shown in Fig. 6.43.3. When link 6 is in an extreme position, link 5 is perpendicular to the coupler curve. Therefore, we need only select different coupler curves and locate candidate points for extreme positions. For this problem, select a figure 8 coupler curve. The problem can be solved using the following steps.
40 30 20 10 0a c Rota tion of cr a nk e b d Fig. PB6.43.2: Motion of link 6 1) First draw two identical circles, one tangent to two locations at the top of the curve and the other tangent to two locations at the bottom. 2) Find the centers of the two circles and call these points Fa and Fb . Next bisect the line between Fa and Fb . On this line, locate the point G such that the angle FaGFb is 40˚. Draw an arc centered at G and of radius GFb . This is shown in Fig. PB6.43.4. 3) Locate the lines GFc and GFd at an angle of 5˚ from the first two lines as shown in Fig. PB6.43.5. Using Fc and Fd as centers, draw two more circles of the same radius as for the circles in Step 1. Rota tion of link 6  295  b d 35˚ 40˚ 30˚ 35˚ G 6 a e
Fig. PB6.43.3: Positions of link 6 c F 4) Use the geometry in Fig. PB6.43.5 as a single entity and rotate and scale it to fit it to the coupler curve such that each of the circles is tangent to the coupler curve at one location. 5) Draw the dyad GFE to complete the linkage. The result is shown in Fig. PB6.43.6. The coordinates have been rotated to facilitate the analysis of the mechanism. 6) Analyze the linkage to ensure that it meets the requirements closely enough. If the requirements are not met adequately, select a other coupler curve and repeat the procedure. The mechanism n has been analyzed using the program sixbar.m, and the results are given in Fig. PB6.43.7. The results are reasonably accurate; however, the results can be improved somewhat by changing the location of point G slightly. Fb 6 G Fa 40˚ Fig. PB6.43.4: Finding equaldiameter circles each tangent to coupler curve at two locations.  296  Fb 6 G Fa Fd Fc
40 ˚ 5˚ 5˚ Fig. PB6.43.5: Setting extremes of motion for link 6 Y 3 B 2 A β 4 AD = 2.1429" AB = 1.0994" BC = 2.1806" CD = 1.6212" BE = 2.1005" EF = 1.60" FG = 1.00" β = 35.5˚ x G = 1.70" y G = 1.60" C E 5 Fb 6 G Fd Fc Fa D X Fig. PB6.43.6: Final solution  297  Fig. PB6.43.7: Results from s i x b a r . m Problem 6.44 Design a sixbar linkage like that shown in Fig. 6.62 such that the displacement of the output link (link 6) is the given function of the input link rotation. The output displacement reaches maximum values of 30˚ and 60˚ at input rotations of 60˚ and 240˚, respectively. The rotation of the output link is zero when the input rotation angle is 0, 120˚, and 360˚.
Displacement of link 6, degrees 60˚ 50˚ 40˚ 30˚ 20˚ 10˚ 0˚ 60˚ 120˚ 180˚ 240˚ 300˚ 360˚ Rotation of crank 2, degrees Solution Because of the double oscillation, either a figure 8 or kidney bean shaped curved can be used. The procedure will be illustrated with the figure 8 curve used in Example 6.8. Locate a curve that is roughly symmetrical as shown in the following figure. Draw a circle of radius r5 which is tangent to the top of the curve at two locations. The center of this circle is at point F’ which is one extreme location of F. The radius of the circle gives the length of link 5. The value for this radius is a  298  design decision. The points a and b in the figure correspond to the locations where the oscillation angle is zero. Starting at b, count 24 dashes (120˚) and draw a second circle of radius r5 tangent to the curve at point c. The center of this circle will be F”, the other extreme location of F. Locate point G by bisecting the cord line between F’ and F” and finding the point on the perpendicular bisector that will give a 60˚ oscillation angle between F’ and F”. The distance from G to F’ or to F” is the length of link 6.  299  Locate the center of the arc between F’ and F”. Draw a third circle of radius r5 centered at this point. This circle should be tangent to the coupler curve at a location that is approximately 12 dashes from point a. If the error is too large, select another curve. Once the linkage is found to be acceptable, draw the dyad GFE to complete the 6bar linkage. In the figure, Gx = 3.543; G y = 3.848; r5 = 3.324”; r 6 = 1.329”. The original fourbar linkage values are given in Example 6.8. The solution given is only approximate. It is difficult to solve problems like this exactly. Therefore, it is usually necessary to approximate either the timing or the oscillation angle. Also, a 60˚ oscillation angle is relatively large for a mechanism such as this. Once a viable linkage is designed, it can be analyzed using the sixbar analysis program. Simple adjustments can then be done directly with the linkage in that program. Problem 6.45 Design a sixbar linkage like that shown in Fig. 6.62 such that the displacement of the output link (link 6) is the given function of the input link rotation. The output link dwells for 90˚ of input rotation starting at 0 and 180 degrees. The maximum rotation angle for link 6 is 15˚.
Displacement of link 6, degrees 15˚ 10˚ 5˚ 0˚ 180˚ 270˚ 90˚ Rotation of crank 2, degrees 360˚ Solution For this problem, locate a curve that is a kidneybean shaped curve and isroughly symmetrical as shown in the following figure. The “top” and “bottom” of the kidney bean must have approximately the same radius of curvature for the two dwell periods. The radius of curvature corresponds to r5 . Draw a circle of radius r5 which is tangent to the top of the curve. The center of this circle is at point F’ which is one extreme location of F. The radius of the circle gives the length of link 5. Draw a second circle of radius r5 tangent to bottom part of the curve. The center of this circle will be F”, the other extreme location of F. Locate point G by bisecting the cord line between F’ and F” and finding the point on the perpendicular bisector that will give a 15˚ oscillation angle between F’ and F”. The distance from G to F’ or to F” is the length of link 6. Once the linkage is found to be acceptable, draw the dyad GFE to complete the 6bar linkage. For the solution shown, the following values apply:  300  AB = 0.619; BC = 1.562; CD = 1.564; AD = 0.938 BE = 2.519; EF = 2.95; FG = 1.573; Gx = 2.377; Gy = 0.711 = 35.65˚ It is apparent from the figure that a relatively long dwell can be achieved for the top region of the curve, but the dwell on the bottom region is shorter than the problem statement specifies. If the approximation is ultimately unacceptable, a different curve can be used. Also, the problem can be analyzed using the sixbar routine, and minor adjustments can be made directly with the program. Problem 6.46 Design an eightbar linkage like that shown in Fig. 6.70 such the coupler remains horizontal while the given point on the coupler moves approximately along the path given.  301  700 mm 60 mm Coupler Po 500 mm 450 mm 100 mm Solution Search for a coupler curve that is roughly the shape of the figure shown. It is not possible to find a curve that exactly duplicates the figure; however, it is possible to find a curve that generally scans the same area. A solution is shown in the following. Once the coupler curve is found, the 8bar linkage is completed using the procedure given in Section 6.6.2 Problem 6.47 Resolve Problem 6.46 if the coupler is inclined at an angle of 45˚.  302  700 mm Coupler Po 500 mm 60 mm 45˚ 450 mm 100 mm Solution Search for a coupler curve that is roughly the shape of the figure shown. It is not possible to find a curve that exactly duplicates the figure; however, it is possible to find a curve that generally scans the same area. A solution is shown in the following. Once the coupler curve is found, the 8bar linkage is completed using the procedure given in Section 6.6.2 Problem 6.48 Design an eightbar linkage like that shown in Fig. 6.70 such the coupler remains horizontal while the given point on the coupler moves approximately along the path from A to B to C. The coupler can return either by retracing the path from C to B to A or by going directly from C to A . This means that the basic 4bar linkage need not be a crank rocker.  303  B Equilateral Triangle Coupler Point 1" 5" C 8" A Solution Search for a coupler curve that is roughly the shape of the figure shown. It is not possible to find a curve that exactly duplicates the figure; however, it is possible to find a curve that generally scans the same area. A solution is shown in the following. Once the coupler curve is found, the 8bar linkage is completed using the procedure given in Section 6.6.2  304  Problem 6.49 Determine the two 4bar linkages cognate to the one shown below. The dimensions are MA = 10 cm, AB = 16 cm, AC = 32 cm, QB = 21 cm, and MQ = 24 cm. Draw the cognates in the position for = 90˚.
C A
θ M B Q Solution Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by recognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated in Figs. 6.76 and 6.77. The cognates are shown in the following figure. F C B G A Q M D O E This figure can be checked using the program cognates.m. The coupler curve and cognates are given in the following.  305  Problem 6.50 Determine the two 4bar linkages cognate to the one shown below. The dimensions are MQ = 1.5 in, AB = BC = BQ = AC= 1 in, and AM = 0.5 in. Draw the cognates in the position for = 90˚. C B A θ M Solution Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by recognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated in Figs. 6.76 and 6.77. The cognates are shown in the following figure. Q  306  C F G A M D O B E Q This figure can be checked using the program cognates.m. The coupler curve and cognates are given in the following.  307  Problem 6.51 Determine the two 4bar linkages cognate to the one shown below. The dimensions are MQ = 2 in, AB = BC = BQ =1 in, and AM = 1.5 in. AC = 0.75 in. Draw the cognates in the position for = 45˚.
B C A M Q Solution Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by recognizing that triangle MQO is similar to ABC. Then complete the parallagrams indicated in Figs. 6.76 and 6.77. The cognates are shown in the following figure. This figure can be checked using the program cognates.m. The coupler curve and cognates are given in the following.  308  Problem 6.52 Determine the two 4bar linkages cognate for the draglink mechanism shown. The dimensions are MQ = 1 m, AM = BQ = 4 m, AB =2 m, and angles CAB and CBA both equal 45˚. Notice that the cognates will also be draglink mechanisms. Draw the cognates in the position for = 180˚. θ B Q C 45˚ M A Solution  309  Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by recognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated in Figs. 6.76 and 6.77. The cognates are shown in the following figure.
F E B Q G O C D M A This figure can be checked using the program cognates.m. The coupler curve and cognates are given in the following.  310   311  ...
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This note was uploaded on 02/20/2011 for the course MEC 411 taught by Professor Shudong during the Winter '11 term at Ryerson.
 Winter '11
 Shudong

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