Solutions to Chapter 6 Exercise Problems A 1 O 4 B 2

Transcription

1 Solutions to Chapter 6 Exercise Problems Problem 6.1: Design a double rocker, four-bar linkage so that the base link is 2-in and the output rocker is 1-in 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. A 2 60 A 1 O 2 30 O 4 45 B 2 90 B 1 Solution: Invert the motion and solve the problem graphically. Draw a line from O 2 to B 2 and rotate it clockwise 60 to locate the position of B' 2. Draw the perpendicular bisector of B' 2 B 1 and locate the intersection of that line with the line O 2 A 1. This will locate A 1. The solution is given in the first figure. Measure the lengths os 0 2 A 1 and A 1 B 1. This gives the linkage shown in the second figure. The link lengths are: r 1 = 6" r 2 = 3.72" r 3 = 7.29" r 4 = 4"

2 A 2 60 A 1 O 2 30 O B 2 B 1 B' 2 Basic Construction A 2 A 1 O 2 r 2 r 3 30 r 1 O r 4 B 2 B 1 B' 2 Final Linkage

3 Problem 6.2: Design a double rocker, four-bar linkage so that the base link is 4-in and the output rocker is 2-in 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. A 2 B 2 O B 1 A 1 80 O 4 25 Solution Invert the motion and solve the problem graphically. Draw a line from O 2 to B 2 and rotate it clockwise 60 to locate the position of B' 2. Draw the perpendicular bisector of B' 2 B 1 and locate the intersection of that line with the line O 2 A 1. This will locate A 1. The solution is given in the first figure. Measure the lengths os 0 2 A 1 and A 1 B 1. This gives the linkage shown in the second figure. The link lengths are: r 1 = 4" r 2 = 2.77" r 3 = 3.21" r 4 = 2"

4 Basic Construction Final Linkage Problem 6.3: In a back hoe, a four-bar 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 13-in and the input link is 12-in 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

5 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

6 Solution Invert the motion and solve the problem graphically. Draw a line from O 2 to B 2 and rotate it clockwise 120 to locate the position of B' 2. Draw the perpendicular bisector of B' 2 B 1 and locate the intersection of that line with the line O 2 A 1. This will locate A 1. The solution is given in the first figure below. Measure the lengths os 0 2 A 1 and A 1 B 1. After unscaling, the linkage is shown in the second figure that follows. The link lengths are: r 1 = 13" r 2 = 8.27" r 3 = 19.34" r 4 = 12"

7 Basic Construction Final Linkage

8 Problem 6.4 In the drawing, AB = 1.25 cm. Use A and B as circle points, and design a four-bar linkage to move its coupler through the three positions shown. Use Grashof s equation to identify the type of fourbar linkage designed. Y B 3 A 3 θ 3 = 60 (2, 3) B2 A 2 θ 2 = 45 (2, 1) Solution A 1 (0, 0),θ 1 =0 B 1 X 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 = = 3.573" p + q = = Therefore, l +s > p+q and the linkage is a nongrashof linkage and a double rocker

9 Y B 3 B* A 3 A* B 2 A 2 A 1 B 1 X Problem 6.5 Using points A and B as circle points, design a four-bar 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 four-bar linkage designed. Position A 1 B 1 is horizontal, and position A 2 B 2 is vertical. AB = 1.25 in. Y A 1 (0, 1.75) B 1 (1.63, 1.25) A2 B 3 35 A 3 (2.13, 0.63) X Solution B 2 Find the center points A* and B* and measure the link lengths. Then, AA* = 2.15" AB = 1.25"

10 A*B* = 2.32" BB* = 1.11" l + s = = 3.43" p + q = = 3.40 Therefore, l +s > p+q and the linkage is a nongrashof linkage and a double rocker. Y A 1 B 1 A 2 B* B 3 A 3 X A* B 2 Problem 6.6 Design a four-bar 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 B 3 60 B2 Solution A 3 (2, 2.4) 50 A 2 (2, 0.85) A 1 (0, 0) B 1 X Find the center points A* and B* and measure the link lengths. Then, AA* = cm AB = cm

11 A*B* = cm BB* = cm l + s = = 8.659" p + q = = Therefore, l+s > p+q and the linkage is a nongrashof linkage and a double rocker. Y B 3 B* A 3 B2 A* A 1 A 2 B 1 X

12 Problem 6.7 A four-bar 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 four-bar linkage designed. Also determine whether the linkage changes branch in traversing the design positions. Positions A 1 B 1 and A 2 B 2 are horizontal, and position A 3 B 3 is vertical. AB = 3 in. Y A 2 B 2 (0.0, 2.88) A 1 B 1(-1.94, 0.94) A 3 C* X B 3 (-1.0, -3.0) Solution Find the center points C 1 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 C 1. Then, CC* = 2.67 in AC = 2.38 in A*C* = 3.00 in AA* = 2.06 in l + s = = 5.06 p + q = = 5.05 Therefore, l +s > p+q and the linkage is a nongrashof linkage and a double rocker

13 Y A2 B2 C* 3 A 1 A* C 1 B 1 C* X A 3 C* 2 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. B

14 A 2 C 2 Y B 2 ψ 2 A 1 A* C 1 B 1 ψ 1 C* X A 3 ψ 3 C 3 B 3 Problem 6.8 Design a four-bar 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 four-bar linkage designed. Position A 1 B 1 is horizontal, and positions A 2 B 2 and A 3 B 3 are vertical. AB = 4 in. Y A 2 A 3 (0, 2) A 1 B 1 C 1 * (0, 0) (2, 0) (4, 0) B 2 B 3 X

15 Solution The solution is shown in the figure. The link lengths are: r 1 = 3.22 = p A 2 A 3 A 1 B 1 r 2 r 3 r 1 A* C * B 2 B 3 C 2 * r 4 C 1 C 3 * r2 = 3.32 = q r3 = 3.62 = l r 4 = 2.24 = s Grashof l + s < p + q For this mechanism, and l + s = = 5.86 p + q = = 6.54 Therefore, l + s < p + q, and the mechanism is a crank-rocker or rocker-crank

16 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 four-bar 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) A 3 (3.3, 2.6) C* Position X A1(-2.2, -0.3) Position A2 (2.1, -0.1)

17 Desk Pos'n 3 C * 3 C * 2 C 1 C * A * A 3 Pos'n 2 A 1 C 1 = A 1 A 2 C 1 C* = Pos'n 1 A*C*= A 1 A* = A*C*+C 1 C* = A 1 C 1 + A 1 A* = Therefore, A*C*+C 1 C* > A 1 C 1 + A 1 A* and the linkage is a nongrashof linkage and a double rocker. This is a poor solution because C is below the floor plane. 1 Problem 6.10 Design a four-bar linkage to move the coupler containing line segment AB 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 A 2 B 2 and A 3 B 3 are horizontal, and position A 1 B 1 is vertical. AB = 3.5 in. A 3 (-1.0, 2.5) B1 Y B3 (0.0, 2.0) A 2 B2 (0.0, 1.0) C 1 * X A

18 B 1 C* 2 C* 2 A 3 B 3 A * A 2 B 2 C 1 C * A 1 From the figure, r 1 = A*C* = in r 2 = A*A 1 = in r 3 = A 1 C 1 = in r 4 = C*C 1 = in Grashof calculation: l + s? p + q [ =(4.0241)] < [ =(4.3202)] The linkage is a crank rocker

19 Problem 6.11 Design a four-bar 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 four-bar linkage designed. Position A 1 B 1 is horizontal, and positions A 2 B 2 and A 3 B 3 are vertical. AB = 6 cm. Y A3 (-2, 3) B 1 A 2 (2, 3) A 1 (6, 4.3) C* (0, 0) X B3 B

20 A* C* 2 C 1 Y B1 A 1 A3 A 2 C* 3 C* X B3 B 2 Find A*. Then find C' 2 and C' 3 and find the circle point C 1. Draw the linkage. Then, AA* = 13.0 AC = 6.84 CC* = 4.41 C*A* = 15.8 l + s = = 20.2 p + q = = 19.8 Therefore, l+s > p+q and the linkage is a Type II double rocker

21 Problem 6.12 Design a four-bar linkage to move the coupler containing line segment AB 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 A 1 B 1 is horizontal, and position A 3 B 3 is vertical. AB = 5.1 cm. Y A2 (1.5, 3.7) 45 B2 X A 1 (-5.0, -1.0) B 1 A 3 (7.8, -1.0) C* (-5.0, -5.0) B 3 Solution Find the center points C 1 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 C 1. Then, CC* = 7.81 cm AC = 9.73 cm A*C* = 6.68 cm AA* = 6.82 cm l + s = = cm p + q = = cm Therefore, l +s > p+q and the linkage is a nongrashof linkage and a double rocker

22 Y A 2 X A 1 B2 A 3 B 1 A* C* C 1 B 3 C 2 * C 3 * 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

23 Y A 2 X A 1 B2 A 3 B 1 A* C* C 1 C 2 B 3 C3 Problem 6.13 Synthesize a four-bar mechanism in position 2 that moves its coupler through the three positions shown below if points C* and D* are center points. Position A 1 B 1 and position A 3 B 3 are horizontal. AB = 4 cm. Y D* (3.0, 2.6) B2 C* B3 A 3 (3.4, 1.6) X 45 A 2 (2.7, - 0.7) Solution B1 A 1 (0.7, - 1.8) Find the circle points C 2 and D 2 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,

24 DD* = cm CD = cm C*D* = cm CC* = cm l + s = = p + q = = Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker. D* 1 D2 Y D* B2 B 3 C* C* 1 A 3 X D* 3 A 2 A 1 B 1 C 2 C*

25 Problem 6.14 Synthesize a four-bar 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 A 1 B 1 and position A 3 B 3 are horizontal. AB = 4 cm. Y C* (2.3, 4.5) B2 B3 A 3 (2.7, 3.5) 45 A 2 (2.0, 1.0) X Solution A 1 B1 Find the center points C 2 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 C 2. Then, CC* = cm AC = cm A*C* = cm AA* = cm l + s = = p + q = = Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker. C* 1 Y C* A* C 2 A 3 B 2 B 3 C* 3 A 1 A 2 B 1 X

26 Problem 6.15 A hardware designer wants to use a four-bar 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 135 Position 3 Position 2 A 3 (0, 8.3) A 2 (-3.1, 6.6) Position 1 A 1 (-4.0, 2.8) B* X Solution Position 2 A 3 Position 3 A2 Position 1 A* B1 B* 3 A1 B* B*

27 AA* = " BB* = " AB = " A*B* = " l + s = = < p + q = = 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 slider-crank 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 A 1 B 1 is horizontal, and position A 3 B 3 is vertical. AB = 2.0 in. Y A2 (2.69, 1.44) 135 A 3 (5.06, 0.0) A 1 (0, 0) B 1 B2 X Solution B3 Find the center ponit A*. Then find the poles, the image pole I' 23, and the circle of sliders. Next select a slider point (C 1 ), 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 C 1. From the figure, r 2 = A*A 1 = in r 3 = A 1 C 1 = in Find the transmission angle in the three positions. The linkage changes mode because the sign of the transmission angle changes

28 A 2 A 1 B 1 A 3 r = B 2 r = A * P 2 3 B 3 C 3 C2 P 1 3 P 23 ' P 1 2 C

29 Problem 6.17 Design a slider-crank mechanism to move a coupler containing the line AB 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 12, p 13, p 23 ) are shown, but these should be determined accurately after the positions are redrawn. Find A*, the slider point that lies above B 1 on a vertical line through B 1, and draw the linkage in position 1. Y P 13 A 1 (1.25, 1.70) B 1 B A 3 (1.65, 1.25) 2 P 23 (3.15, 1.32 P 12 B 2 A 3 (2.15, 0.58) X Solution Find the center ponit A*. Then find the poles, the image pole I' 23, and the circle of sliders. Next select the slider point (C 1 ) that lies above B 1. 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

30

31 Problem 6.18 Design a slider-crank 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 A 1 B 1 is horizontal, and positions A 2 B 2 and A 3 B 3 are vertical. AB = 4 in. Y A 2 A 3 (0, 2) A 1 B 1 C 1 * (0, 0) B 2 (2, 0) (4, 0) B 3 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 C 1. This is illustrated in the following construction

32 Finding C 1 Next find the poles as shown in the following construction. This is where a problem occurs. Note that P 12 and P 13 are coincident and P 23 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

33 Finding the poles

34 Problem 6.19 Design a slider-crank 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 B C A 1 A 2 (0, 0) (0.8, 0) Solution Find the center ponit A*. Then find the poles, the image pole I' 23, and the circle of sliders. Next select the slider point (C 1 ) 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*

35 Finding the poles Finding the circle of sliders

36 Finding the slider point

37 Final solution Problem 6.20 A device characterized by the input-output relationship = a1 + a2 cos is to be used to generate (approximately) the function = 2 ( and both in radians) over the range 0 /4. a) Determine the number of precision points required to compute a 1 and a 2. b) Choose the best precision point values for from among 0, 0.17, 0.35, and 0.52, and determine the values of a 1 and a 2 that will allow the device to approximate the function. c) Find the error when = /

38 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, θ 1 π θ 2 8 π 4 1 = (1+ cos135 ) = = (1+ cos45 ) = The angles which are closest to these values are 0.17 and Then, 1 = (1) 2 = (0.17) 2 = = ( 2 ) 2 = (0.52) 2 = Substituting into the equation for the system model, or, 1 = a 1 + a 2 cos 1 2 = a 1 + a 2 cos = a1 + a2 cos(0.17) = a1 + a2(0.985) = a1 + a2cos(0.52) = a1 + a2(0.868) Subtract the first equation from the second, and solve for a = a2(0.117) and a2 = Now back substituting into the first equation, a1 = a2(0.985) = (0.985) = The error is given by

39 e = ideal act = 2 ( cos) Substituting in the given value = /8 = 22.5, e = (0.3926) 2 ( cos 22.5 )= Problem 6.21 A mechanical device characterized by the input-output relationship = 2a1 + 3a2sin + a 2 3 is to be used to generate (approximately) the function = 2 2 over the range 0 /4. Exterior constraints on the design require that the parameter a 3 = 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, θ 1 π θ 2 8 π 4 1 = (1+ cos135 ) = = (1+ cos45 ) = Then,

40 1 = 2(1) 2 = 2(0.115) 2 = = 2( 2 ) 2 = 2(0.670) 2 = Substituting into the equation for the system model, or, 1 = 2a 1 + 3a 2 sin 1 + a 3 2 = 2a 1 +3a 2 sin = 2a 1 +3a 2 sin 2 + a 3 2 = 2a 1 +3a 2 sin = 2a 1 +3a 2 sin(0.115) +1= 2a a = 2a 1 +3a 2 sin(0.670) +1 = 2a a 2 +1 Subtract the first equation from the second, and solve for a = a2 and a2 = Now back substituting into the first equation, or 2a1 = a2 = (0.5737) = a1 = / 2 = The error is given by e = ideal act = 2 2 ( 2a 1 + 3a 2 sin +1)= 2 2 [ 2(0.5855) + 3(0.5737)sin +1] = [ sin] Substituting in the given value = /6 = 30, e = 2(0.5235) 2 [ sin30 ]= Problem 6.22 A mechanical device characterized by the input-output relationship = 2a1 + a2tan +a2 3 is to be used to generate (approximately) the function = 3 3 ( and both in radians) over the range 0 /3. Exterior constraints on the design require that the parameter a 3 = 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 that will allow the device to approximate the function. c) Find the error when = /

41 Solution: There are two unknowns so the number of precision points is two. The precision points according to Chebyshev spacing are: and 1 = ( max +min) 2 2 = ( max + min) 2 The corresponding values of are: + ( max min) cos135 = cos135 = ( max min) cos45 = cos45 = = = 2(0.1534) 3 = = = 2(0.8938) 3 =1.428 We can now solve for a 1 and a 2 using the desired input-output relationship. Then, and = 2a1 + a2tan +a 3 2 = 2a1 + a2 tan = 2a 1 + a2 tan(0.1534) +1= 2a a = 2a 1 + a2 tan(0.8938) +1= 2a a2 +1 Subtracting the two equations gives, =1.0902a 2 a2 = Backsubstituting to determine a 1 gives 2a1 = a2 1 = (1.3033) = a1 = The error is given by e = ideal act = 3 2 ( 2a 1 + a 2 tan +1)= 3 2 [ 2(0.5968)+ (1.3033)tan +1] = 3 2 [ tan ] Substituting in the given value = /6 = 30, e = 3(0.5235) [ tan30 ]=

42 Problem 6.23 A mechanical device characterized by the input-output relationship = 2a1 + a2sin 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: and x1 = (x max + x min ) 2 x2 = (x max+ x min ) 2 The corresponding values of y are: y 1 = 2x 2 1 = 2(0.2300) 2 = y 2 = 2x 2 = 2(1.3407) 2 = (x max x min ) cos135 = cos135 = (x max x min ) cos45 = cos45 = 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. and 1 = rad = = =1.3407rad = = We can now solve for a 1 and a 2 using the desired input-output relationship. Then, = 2a1 + a2sin

43 1 = 2a1 + a2 sin1 = = 2a1 +a2 sin = 2a1 + a2(.2280) 2 = 2a 1 + a 2 sin 2 = = 2a 1 +a 2 sin = 2a 1 +a 2 (.9736) Subtracting the two equations gives, = a2 a2 = / = Backsubstituting to determine a 1 gives a1 = [ a2(.2280) ]/ 2 = [ (.2280) ]/ 2 = The error is given by e = ideal act = 2 2 ( 2a 1 + a 2 sin )= 2 2 [ 2(0.4806) + (4.6795)sin] = [ sin ] Substituting in the given value = /4 = 45, e = [ sin]= 2(.7854) [ sin 45 ]= Problem 6.24 Determine the link lengths and draw a four-bar 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 = 0.5 f =1.0 1 = f = f = f Compute the 's f 0 2 = = f 0 2 cos30 = = 12 = = 2 2 = = 32 = cos30 = cos30 = cos30 =

44 Then and 1 cos 1 cos A = 1 cos 2 cos 2 1 cos 3 cos 3 = cos(1 1) B = cos( 2 2 ) cos( 3 3 ) = x1 x 2 x 3 1 cos1 cos1 1 cos(1 1) = cos 2 cos 2 cos( 2 2 ) 1 cos 3 cos 3 cos( 3 3 ) = r 2 = 1 x 2 = r4 = 1 x 3 = r 3 = 1+ r 22 +r 22 2r 2 r 4 x 1 = Now unscale the values by multiplying each by 2. Then R1 = 2.0; R2 = 2(-511.5) = -1023; R 3 = 2(522.0) =1044; R 4 = 2(9.903) =19.80; Check for linkage type: Then l + s = =1046 p + q = = > 1042 nongrashof, double rocker. Note that the link-length ratios vary greatly. This design would be considered very undesirable. Problem 6.25 Determine the link lengths and draw a four-bar 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

45 Solution Determine the precision points using Chebychev spacing. Then 0 = 0 f = 90 1 = f = f = f Compute the 's Then and f 0 2 cos30 = 4 cos30 = = 4 = f = sin( 1 ) = = sin(2) = = sin(3) = cos30 = 4 + cos30 = cos 1 cos A = 1 cos 2 cos 2 = cos 3 cos cos( 1 1 ) B = cos( 2 2 ) = cos( 3 3 ) x1 x 2 x 3 1 cos 1 cos 1 1 cos( 1 1 ) = 1 cos 2 cos 2 cos( 2 2 ) 1 cos 3 cos 3 cos( 3 3 ) = r 2 = 1 x 2 = r4 = 1 x 3 = r 3 = 1+ r r 2 2 2r 2 r 4 x 1 = Now unscale the values by multiplying each by 2. Then

46 R 1 = 2.0; R 2 = 2( ) = ; R 3 = 2(0.4962) = ; R4 = 2( ) = ; A scaled drawing of the linkage is given in the following: Problem 6.26 Design a four-bar 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 = 45 0 = 30 = 50 = 70 Compute the error at x = 2 Solution The solution is given in the following. From the given information,

47 x 0 = 1; x f = 4. Using Chebychev spacing for the precision points, x1= x f + x 0 2 x f x 0 cos30 = cos30 = Similarly, x 2 = 2.5 and x 3 = Then, the corresponding values for y are: y f = x f x f + 3 =1 y 0 = x 0 x 0 +3 = 3 y1= x1 x1 +3 = y2= x2 x2 + 3 = y3= x3 x3 +3 = Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. For the range of linkage angles, we have: and 0 = 45 ; = 50 0 = 30 ; = 70 The precision points in terms of are: Similarly, 1 = x 1 x 0 x f x 0 +0 = 2 = x 2 x 0 x f x 0 +0 = = = = x 3 x 0 +0 = = x f x = y 1 y 0 y f y = 2 = y2 y0 y f y 0 +0 = 3 = y 3 y 0 y f y = Using the matrix solution procedure, = = =

48 z 1 z2 z cos 1 cos 1 cos(1 1 ) = 1 cos2 cos2 cos(2 2) 1 cos 3 cos 3 cos( 3 3 ) = = and r2 = 1 z 2 = = r4 = 1 1 = z = and r 3 = 1+ r r 4 2 2r 2 r 4 z 1 = 1+( ) 2 + ( ) 2 2( )( )( ) = For the overall size of the linkage, use a base link length of 2 in. Then the lengths of the other links become R 1 =1(2) = 2 in R 2 = (2) = in R4 = (2) = in R3 = (2) = in Note that the link-length 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 = = To compute the generated value, first compute the value of corresponding to x = 2. Then, e = x e x 0 +0 = = x f x 0 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 = The corresponding value for y is,

49 The error is yact = e 0 (y f y0) + y0 = (2)+ 3 = error = yideal yact = = Problem 6.27 Design a four-bar linkage to generate the function y=x 2-1 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 = 30 0 = 45 = 60 = 90 Compute the error at x = 3. Solution The solution is given in the following. From the given information, x 0 = 1; x f.= 5. Using Chebychev spacing for the precision points, x1= x f + x 0 2 x f x 0 cos30 = cos30 = Similarly, x 2 = 3 and x 3 = Then, the corresponding values for y are: yf = xf 2 1 = 24 y0 = x0 2 1= 0 y1= x1 2 1 = y2= x2 2 1= 8 y3= x3 2 1= Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. The linkage angles are: and 0 = 30 ; = 60 0 = 45 ; = 90 The precision points in terms of are:

50 Similarly, 1 = x 1 x 0 x f x = 2 = x 2 x 0 +0 = = 60 x f x = x 3 x 0 x f x 0 +0 = 1 = y 1 y 0 y f y = y2 y0 2 = = = = = = 75 y f y = y 3 y 0 y f y = Using the matrix solution procedure, and and z 1 z2 z = cos 1 cos 1 cos(1 1 ) = 1 cos2 cos2 cos(2 2) 1 cos 3 cos 3 cos( 3 3 ) = r2 = 1 z 2 = = r4 = 1 1 = z = = r 3 = 1+ r r 4 2 2r 2 r 4 z 1 = 1+( ) ( )( )( ) = For the overall size of the linkage, use a base link length of 2 cm. Then the lengths of the other links become

51 R 1 =1(2) = 2cm R 2 = (2) = cm R4 = (2) = cm R3 = (2) = 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= = 8 To compute the generated value, first compute the value of corresponding to x = 3. Then, e = x e x 0 +0 = = 60 x f x 0 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, The error is yact = e 0 (y f y0) + y0 = error = yideal yact =8 8= (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 crank-rocker mechanism for the sprinkler. If the crank is to be 1 inch long, give the lengths of the other links. Solution =180 Q 1 Q +1 = = 0 This problem does not have a unique solution. To start the construction, arbitrarily pick r 4 as 2" at the angle shown. Then, the center location for O 2 is located on a line through B 1 and B 2. Any point will work as long as it is selected to the left of B

52 B 2 B 1 O " 90 O 4 For the location selected, r2 + r3 = r3 r2 =1.500 Then 2(r3) = or r3 = 2.907" and r2 = r = =1.407" From the figure, r 1 = 3.232". If the crank is 1 in long, then each dimension must be scaled. To do this multiply each length by K where Then, K = = r 1 = 0.711*(3.232) = in r 2 =0.711*(1.407) = 1.00 in, r 3 = 0.711*(2.907) = in r 4 = 0.711*(2.0) = in,

53 Problem 6.29 Design a crank-rocker 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 80 o, and the length of the base link is to be 2 in. Solution =180 Q 1 = Q = 36 This problem does not have a unique solution. To start the construction, arbitrarily pick r 4 as 2" at the angle shown. Then, the construction of the center location for O 2 is shown in the following figure. From the figure, r2 + r3 = r3 r2 =1.687 Then 2(r3) = or r3 = 2.709" and r2 = r = =1.022" From the figure, r 1 = 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 = = B 2 B " " 80 O " O 4 Then, r 1 = 0.928*2.155 = 2 in

54 r 2 = 0.928*(1.022) = in, r 3 = 0.928*(2.709) = in, r 4 = 0.928*(2.0) = in. Note that a different linkage is generated for each line drawn through B 1. Problem 6.30 A packing mechanism requires that the crank (r 2 ) rotate at a constant velocity. The advance part of the cycle is to take twice as long as the return to give a quick-return mechanism. The distance between fixed pivots must be 0.5 m. Determine the lengths for r 2, r 3, and r 4. B 1ω 2 A r 2 r 3 r 4 θ = 80 Solution O 2 O 4 Schematic Drawing =180 Q 1 = Q = 60 This problem does not have a unique solution. To start the construction, arbitrarily pick r 4 as 2" at the angle shown. Then, the construction of the center location for O 2 is: B 2 B " O " " 80 O 4 r2 + r3 = r3 r2 =

55 Then 2(r3) = or r3 = " and r2 = r = = " From the figure, r 1 = ". 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 Then, K = = r 1 = )*(1.4728) = 0.5 meters r 2 = )*(0.8942) = meters, r 3 = )*(2.0424) = meters, r 4 = )*(2.0) = meters, Note that a different linkage is generated for each line drawn through B 1. Problem 6.31 The rocker O 4 B of a crank-rocker 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 Therefore, =180 Q 1 = Q = 25.7 This problem does not have a unique solution. To start the construction, arbitrarily pick r 4 as 2" at the angle shown. Then, the construction of the center location for O 2 is: r2 + r3 = r3 r2 = Then 2(r3) = " or r3 = " and

56 r2 = r = = " B 2 B " " O " O 4 From the figure, r 1 = ". 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 Then, K = = r 1 = *(2.5984) = 2.5" r 2 = *(0.8951) = ", r 3 = *(3.3063) = ", r 4 = *(2.0) = ", Note that a different linkage is generated for each line drawn through B 1. 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 The rocking angle is to be 40, and it rocks symmetrically about a vertical line through O 4. 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 can be used. First draw the line beween the pivot points as shown. Next compute the angle using

57 =180 Q 1 = Q = 20 B 2 80 B " " " O " O 4 Locate G as shown in Fig. 4.46, and draw the locus of B 2. To locate the oscillation angle symetrically about the vertical, locate B 2 at an angle of 20 from the vertical. We can then locate B 1 at an angle of 80 from the line O 2 B 2. Next compute r 2 and r 3 from r2 + r3 = r3 r2 = Then 2(r3) = or r3 = and r2 = r = =1.0879" Also from the drawing, r1= " r4 =1.9238" The transmission angles do not approach either 0 or 180; therefore, the linkage is reasonably efficient. Problem 6.33 Design a crank-rocker 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 O 4. Specify the length of each of the links. Solution: G

58 This problem does not have a unique solution. To solve this problem, the procedure given in Section can be used. First draw the line beween the pivot points as shown. Next compute the angle using =180 Q 1 = Q = and 2 = = Locate G as shown in Fig. 4.46, and draw the locus of B 2. To locate the oscillation angle symetrically about the vertical, locate B 2 at an angle of 50 from the vertical. We can then locate B 1 at an angle of 100 from the line O 2 B 2. Next compute r 2 and r 3 from r2 + r3 = r3 r2 = Then 2(r3) = or r3 = " B " B 1 O " 50 O " and r2 = r = = " G Also from the drawing, r1= " r4 =1.2794" Problem 6.34 A crank-rocker mechanism with a time ratio of and a rocker oscillation angle of 72 is to be designed. The oscillation is to be symmetric about a vertical line through O 4. Draw the mechanism

59 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 can be used. First draw the line beween the pivot points as shown. Next compute the angle using and =180 Q 1 Q +1 = = = = 36 This means that the point G is at infinity, and the B2 locus is a straight line through O 2 at an angle of 36 to the horizontal. The result is shown in the following: " " B B " O " G 2 O 4 36 To locate the oscillation angle symetrically about the vertical, locate B 2 at an angle of 36 from the vertical. We can then locate B 1 at an angle of 72 from the line O 2 B 2. Next compute r 2 and r 3 from r2 + r3 = r3 r2 = Then 2(r3) = or r3 = " and r2 = r = = " Also from the drawing, and r1= "

60 r4 =1.1629" The linkage is shown in an arbitrary position below. The transmission angle is shown. B O 2 O 4 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 four-bar linkage through a rack-and-pinion 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 (r 2 ) 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 r 2, r 3, and r 4. Solution: A Sander drum B 1ω 2 r 2 O O 4 2 2" Schematic Drawing Pinion r 3 r 4 Splined Shaft θ Rack Belt drive Total travel = 3" Motor 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 = r d = 3 =1.5 rad =

61 To start the construction, first draw the line beween the pivot points as shown. Next compute the angle using =180 Q 1 Q +1 = = 0 Arbitrarily pick r 4 as 2" at the angle shown. Then, the center location for O 2 is located on a line through B 1 and B 2. Any point will work as long as it is selected to the left of B B 2 B 1 O " O 4 For the location selected, r2 + r3 = r3 r2 =1.500 Then 2(r3) = or r3 = 2.875" and r2 = r = =1.375" From the figure, r 1 = ". If the crank is really 1 in long, then each dimension must be scaled. To do this multiply each length by K where Then, K = = r 1 = *(3.2253) = in r 2 =1.2413*(1.375) = in,

62 r 3 = *(2.875) = in r 4 = *(2.0) = 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 /R P 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 (r 2 ) 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 O 2 and O 4 must be 4 ft. Determine the lengths for r 2, r 3, and r 4. A r 3 r 4 B θ = 20 1ω 2 r 2 O2 O 4 r P r G 5 Schematic Drawing Solution =180 Q 1 = Q = 36 = 4(20) = 80 This problem does not have a unique solution. To start the construction, arbitrarily pick r 4 as 2" at the angle shown. Then, the construction of the center location for O 2 is: B 2 B " " 80 O 2 O

63 r2 + r3 = r3 r2 = Then 2(r3) = or r3 = " and r2 = r = =1.0242" From the figure, r 1 = ". 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 Then, K = = r 1 = 4.0 feet, r 2 = feet, r 3 = feet, r 4 = feet, Note that a different linkage is generated for each line drawn through B 1. 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 (r 2 ) rotates at a constant velocity. The follower link (r 4 ) is to be 4 in and to oscillate through an angle of 80. Determine the lengths for r 1, r 2, and r 3. 1 ω2 A r 3 r 2 r 4 O 2 O 4 Schematic Drawing C B r 5 D 6 Cutter Part Being Machined Solution: This problem does not have a unique solution. To start the procedure, determine the angle from the time ratio

64 =180 Q 1 =180 (2 1) Q +1 (2 +1) = 60 B 2 80 B " " O " 2.00" O 4 From the diagram: r1= 1.64" r2 = O 2B1 O2B2 2 = =1.00" r3 = O 2B1 +O2B2 = 2 r4 = 2" =1.90" Scalling the results, R 4 = 4 in R2 = r2 R 4 = =1.00(2) = 2.00 in r4 2 R3 = r3 R 4 =1.90(2) = 3.80 in r 4 R1 = r1 R 4 =1.64(2) =3.28 in r4 The linkage is shown to scale in a general position in the following: B A O 2 O

65 Problem 6.38 A crank rocker is to be used in a door-closing 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 (r 2 ) of the mechanism is to be 10 cm long, determine the lengths of the other links (r 1, r 3, and r 4 ). Sketch the mechanism to scale. Schematic Drawing Door 1ω 2 B r A 3 r r 4 2 O 4 O 2 Wall 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. Then B1O2 = 5.0"= r3 + r2 B2O2 =1.953"= r3 r2 r3 = 3.476" r2 = =1.523" r4 = 2" r1= 3.713"

66 B 2 B 1 O O 4 Now unscale the output. Then, K = R 2 r 2 = = R 1 = 6.566(3.713) = cm R 2 = 6.566(3.713) = 10 cm R3 = 6.566(3.713) = cm R4 = 6.566(2) =13.13 cm A scaled version of the linkage if given in the following. B A R 2 R 3 Scaled Linkage R 4 O 2 R 1 O

67 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 (r 1 ) of the mechanism is to be 10 ft long, determine the lengths of the other links (r 2, r 3, and r 4 ). Sketch the mechanism to scale. Solution: 3 4 A 2 Roc k O1 O 2 B =180 Q 1 =180 (1.11) Q +1 (1.1+1) = 8.6 This problem does not have a unique solution. Initially pick the output link length to be 2. Then from the diagram, B 2 80 B 1 O " 5.74" " 4.39" O 2 r1= 4.39" r2 = O 1B1 O1B2 = =1.25" 2 2 r3 = O 1B1+ O1B2 = = 4.49" 2 2 r4 = 2" Scalling the results, R 1 = 10 ft

68 R2 = r2 R1 r1 = =1.25(2.33) = 2.91 ft 4.39 R3 = r3 R 1 r 1 = 4.49(2.33) =10.46 ft R4 = r4 R1 r1 = 2(2.33) = 4.66 ft The linkage is shown to scale in a general position in the following: B A O 1 O 2 Problem 6.40 A crank rocker is to be used in a windshield-wiping mechanism. The wiper must oscillate 80. The time for the forward and return stroke for the wiper is the same. If the base link (r 1 ) of the mechanism is to be 10 cm long, determine the lengths of the other links (r 2, r 3, and r 4 ). Sketch the mechanism to scale. Wiper Schematic Drawing 1ω 2 r 1 O 2 Frame O 4 r 2 A r 4 r 3 B Solution The time ration is 1 so

69 =180 Q1 Q+1 = [ ] = 0 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 O 4 R 1 R 4 O 2 B 2 A 1 R 2 A 2 B1 From the figure, R 4 = 2 in R 1 = in R2 +R3 = O2B2 R3 R2 = O2B1 or R 2 = (O 2 B 2 O 2 B 1 )/2= ( ) / 2 =1.253 in R3 =(O2B2 +O2B1)/2= ( ) / 2 = 3.881in Determine the scaling factor, r 1 = 10 R = = r 2 = r 3 = r 4 R 2 R 3 R 4 Using the unscalled lengths from the figure, r2 = R2 = (1.253) = cm r3 = R3 = (3.881) = cm r4 = R4 = (2) = cm The mechanism is drawn to scale as:

70 O 4 R 4 O 2 B R 3 R 2 A Problem 6.41 Design a six-bar linkage like that shown in Fig 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 6-bar linkage. For the mechanism shown, the following values apply: AB = ; BC = ; CD = ; AD = ; 1 = BE = ; = ; EF = 1.00 ; FG = ; XG = ; YG = 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