solutions to chapter 6 exercise problems -...

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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-inlong. 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 anglefor the output link is -45˚. The geometry is indicated in the figure.

A1

B1

O 2

A2

B2

O 4

90˚

45˚

30˚

60˚

Solution:

Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate itclockwise 60˚ to locate the position of B'2. Draw the perpendicular bisector of B'2B1 and locate theintersection of that line with the line O2A1. This will locate A1. The solution is given in the firstfigure. Measure the lengths os 02A1 and A1B1. This gives the linkage shown in the second figure.

The link lengths are:

r1 = 6"

r2 = 3.72"

r3 = 7.29"

r4 = 4"

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A1

B1

O 2

A2

B2

O4

90˚

45˚

30˚

60˚

B'2

Basic Construction

A1

B1

O 2

B2

O4

90˚

45˚

30˚

B'2

r1

r2r3

r4

A2

Final Linkage

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Problem 6.2:

Design a double rocker, four-bar linkage so that the base link is 4-in and the output rocker is 2-inlong. The input link turns counterclockwise 40˚ when the output link turns counterclockwisethrough 80˚. The initial angle for the input link is 20˚ counterclockwise from the horizontal, and theinitial angle for the output link is 25˚. The geometry is indicated in the figure.

A1B1

O 2

A2B2

O 4

80˚

25˚20˚

60˚

Solution

Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate itclockwise 60˚ to locate the position of B'2. Draw the perpendicular bisector of B'2B1 and locate theintersection of that line with the line O2A1. This will locate A1. The solution is given in the firstfigure. Measure the lengths os 02A1 and A1B1. This gives the linkage shown in the second figure.


r1 = 4"

r2 = 2.77"

r3 = 3.21"

r4 = 2"

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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 beachieved 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 is12-in long. The input link driven by the hydraulic cylinder rotates through an angle of 80˚ and theoutput link rotates through an angle of 120˚. From the figure, determine reasonable angles for thestarting angles ( 0 and 0) for both of the rockers.

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Solution

Determine reasonable starting angles for each crank relative to the beam member. Use 40˚ for theinput 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.

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Solution

Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate itclockwise 120˚ to locate the position of B'2. Draw the perpendicular bisector of B'2B1 and locatethe intersection of that line with the line O2A1. This will locate A1. The solution is given in the firstfigure below. Measure the lengths os 02A1 and A1B1. After unscaling, the linkage is shown in thesecond figure that follows.


r1 = 13"

r2 = 8.27"

r3 = 19.34"

r4 = 12"

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Basic Construction

Final Linkage

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Problem 6.4

In the drawing, AB = 1.25 cm. Use A and B as circle points, and design a four-bar linkage to moveits coupler through the three positions shown. Use Grashof’s equation to identify the type of four-bar linkage designed.

θ3 = 60˚

θ2 = 45˚

A3

A2

A1 ,θ1 =0

B1

B2

B3

X

Y

(2, 1)

(0, 0)

(2, 3)

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.

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A1B1

B3

A3

B2

X

Y

A*

B*

A2

Problem 6.5

Using points A and B as circle points, design a four-bar linkage that will position the body definedby AB in the three positions shown. Draw the linkage in position 1, and use Grashof’s equation toidentify the type of four-bar linkage designed. Position A1B1 is horizontal, and position A2B2 isvertical. AB = 1.25 in.

B1

B2

A2

B3

A3X

Y

35˚

A1 (0, 1.75)

(1.63, 1.25)

(2.13, 0.63)

Solution


AA* = 2.15"AB = 1.25"

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A*B* = 2.32"BB* = 1.11"

l + s = 2.32 + 1.11= 3.43"p + q = 2.15 + 1.25 = 3.40


A1B1

B3

A3

B2

X

Y

A*

B*A2

Problem 6.6

Design a four-bar linkage to move its coupler through the three positions shown below using pointsA and B as moving pivots. AB = 4 cm. What is the Grashof type of the linkage generated?

B1

B2

B3

X

Y

60˚

50˚

A1(0, 0)

A2(2, 0.85)

A3(2, 2.4)

Solution


AA* = 1.705 cmAB = 4.000 cm

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A*B* = 3.104 cmBB* = 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.

B1

B3

A3

B2

X

Y

A*

B*

A2

A1

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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 othercrank is at C*. Draw the linkage in position 1 and use Grashof’s equation to identify the type offour-bar linkage designed. Also determine whether the linkage changes branch in traversing thedesign positions. Positions A1B1 and A2B2 are horizontal, and position A3B3 is vertical. AB = 3 in.

A1

B3

A3

C*

A2

B1(-1.94, 0.94)

(-1.0, -3.0)

B2 (0.0, 2.88)

X

Y

Solution

Find the center points C1 and A* and measure the link lengths. Note that the linkage is to be drawnin position 1 so the motion must be referred to position 1 when locating C1. Then,

CC* = 2.67 inAC = 2.38 inA*C* = 3.00 inAA* = 2.06 in

l + s = 3.00 + 2.06= 5.06p + q = 2.67 + 2.38 = 5.05


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A1

B3

A3

A2

B1

B2

X

Y

A*

C*2

3C*

C1

C*

To determine if the linkage changes branch, draw the linkage in the three positions, and measure thesign of . From the drawing below, the sign of is different in the three positions, and themechanism changes branch.

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A1

B3

A3

C*

A2

B1

B2

X

Y

C1

A*

C2

C3

ψ1

ψ3

ψ2

Problem 6.8

Design a four-bar linkage to move a coupler containing the line AB through the three positionsshown. The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of theother crank is at C*. Draw the linkage in position 1 and use Grashof’s equation to identify the typeof four-bar linkage designed. Position A1B1 is horizontal, and positions A2B2 and A3B3 are vertical.AB = 4 in.

A1 B1

B 2

A2

B3

A3

X

Y

(2, 0) (4, 0)

(0, 2)

C1* (0, 0)

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Solution

The solution is shown in the figure. The link lengths are:

r1 = 3.22 = p

C3*

A1 B1

B 2

A2

B3

A 3

C*

A*

C2*

C 1

r1

r2

r3

r4

r2 = 3.32 = q

r3 = 3.62 = l

r4 = 2.24 = s

Grashof l+ s < p + q

For this mechanism,

l+ s = 3.62 + 2.24 = 5.86and

p + q = 3.22 +3.32 = 6.54

Therefore, l+ s < p + q , and the mechanism is a crank-rocker or rocker-crank.

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Problem 6.9

A mechanism must be designed to move a computer terminal from under the desk to top level. Thesystem will be guided by a linkage, and the use of a four-bar linkage will be tried first. As a firstattempt 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.

C*Position 2

Position 1

Desk

A3 (3.3, 2.6)

A2 (2.1, -0.1)A1 (-2.2, -0.3)

X

Y

160˚

135˚

Position 3 (Horizontal)

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C*

A1 A2

A3

Pos'n 3

Pos'n 2

Pos'n 1

Desk

A*

C*2

C*3

C1

A1C1 = 2.6864

C1C* = 3.0684

A * C* = 1.0301

A1A* = 1.395

A * C * +C1C* = 4.0985

A1C1 + A1A* = 4.0814

Therefore,

A * C * +C1C* > A1C1 + A1A *

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 threepositions shown. The moving pivot for one crank is to be at A, and the fixed pivot for the othercrank is to be at C*. Draw the linkage in position 1 and determine the classification of the resultinglinkage (e.g., crank rocker, double crank). Positions A2B2 and A3B3 are horizontal, and positionA1B1 is vertical. AB = 3.5 in.

A 3 B3

A1

B1

C1*

A2 B2

X

Y

(0.0, 2.0)

(0.0, 1.0)

(-1.0, 2.5)

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A3 B3

A1

B1

A*

A2B2

C*

C*2 C*2

C1

From the figure,

r1 = A*C* = 1.8228 in

r2 = A*A1 = 2.4974 in

r3 = A1C1 = 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.

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Problem 6.11

Design a four-bar linkage to move a coupler containing the line AB through the three positionsshown. The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of theother crank is at C*. Draw the linkage in position 1, and use Grashof’s equation to identify thetype of four-bar linkage designed. Position A1B1 is horizontal, and positions A2B2 and A3B3 arevertical. AB = 6 cm.

A1B1

B3

C*

A3 (-2, 3)(6, 4.3)

X

Y

(0, 0)

B2

A2 (2, 3)

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A1B1

B3

C*

A3

X

Y

B2

A2

A*

C*2

C*3

C1

Find A*. Then find C'2 and C'3 and find the circle point C1. Draw the linkage. Then,

AA* = 13.0AC = 6.84CC* = 4.41C*A* = 15.8

l + s = 4.41 + 15.8 = 20.2p + q = 6.84 + 13.0 = 19.8

Therefore, l+s > p+q and the linkage is a Type II double rocker.

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Problem 6.12

Design a four-bar linkage to move the coupler containing line segment AB through the threepositions shown. The moving pivot for one crank is to be at A, and the fixed pivot for the othercrank is to be at C*. Draw the linkage in position 1 and determine the classification of the resultinglinkage (e.g., crank rocker, double crank). Also check to determine whether the linkage will changebranch as it moves from one position to another. Position A1B1 is horizontal, and position A3B3 isvertical. AB = 5.1 cm.

A1

B3

A3

C*

A2

B1(-5.0, -1.0)

(7.8, -1.0)

B2 X

Y

(-5.0, -5.0)

(1.5, 3.7)

45˚

Solution


CC* = 7.81 cmAC = 9.73 cmA*C* = 6.68 cmAA* = 6.82 cm

l + s = 9.73 + 6.68 = 16.41 cmp + q = 7.81 + 6.82 = 14.63 cm


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B3

A1

A2

A3

B1

X

Y

B2

C*

A*

C3*

C 2*

C1

To determine if the linkage changes branch, redraw the linkage in the three positions, and determineif the transmission angle changes. This is shown below. The linkage does not change branch.

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B3

A1

A2

A3

B1

X

Y

B2

C*

A*

C1

C 2

C3

Problem 6.13

Synthesize a four-bar mechanism in position 2 that moves its coupler through the three positionsshown below if points C* and D* are center points. Position A1B1 and position A3B3 arehorizontal. AB = 4 cm.

B

B

B

A11

A2

2

A33

C*

D*

X

Y

(0.7, - 1.8)

(3.0, 2.6)

(2.7, - 0.7)

(3.4, 1.6)

45˚

Solution

Find the circle points C2 and D2 and measure the link lengths. Notice that the linkage is to bedrawn in position 2. Therefore, position 2 for the coupler is the position to which the positions ofD* and C* are referred for finding the circle points. Then,

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DD* = 2.767 cmCD = 4.045 cmC*D* = 3.985 cmCC* = 2.300 cm

l + s = 4.045 + 2.300= 6.345p + q = 2.767 + 3.985 = 6.752

Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker.

B1

B3

A3

X

YD*

C*

A1

B2

A2

D*1

D2

D*3

C*1

C*3

C2

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Problem 6.14

Synthesize a four-bar mechanism in position 2 that moves its coupler through the three positionsshown below. Point A is a circle point, and point C* is a center point. Position A1B1 and positionA3B3 are horizontal. AB = 4 cm.

B

B

A1 B1

A2

2

A33

C*

(2.0, 1.0)

(2.7, 3.5)

(2.3, 4.5)

X

Y

45˚

Solution


CC* = 2.257 cmAC = 3.146 cmA*C* = 2.989 cmAA* = 3.065 cm

l + s = 3.146 + 2.257= 5.403p + q = 3.065 + 2.989 = 6.054

Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker.

B1

B3

A3

X

Y C*

A1

B2

A2

C*1

C*3

C2

A*

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Problem 6.15

A hardware designer wants to use a four-bar linkage to guide a door through the three positionsshown. Position 1 is horizontal, and position 3 is vertical. As a tentative design, she selects pointB* as a center point and A as a circle point. For the three positions shown, determine the location ofthe circle point B corresponding to the center point B* and the center point A* corresponding to thecircle 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.

B*

Position 1

Position 2

Position 3

A3 (0, 8.3)

A2 (-3.1, 6.6)

A1(-4.0, 2.8)

Y

X

135˚

Solution

B*

Position 1

Position 2

Position 3

A1

A2

A3

A*

2B*

3B*B1

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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 pivotsinside the wall.

Problem 6.16

Design a slider-crank mechanism to move the coupler containing line segment AB through the threepositions shown. The moving pivot for the crank is to be at A. Determine the slider point, and drawthe linkage in position 1. Also check to determine whether the linkage will move from one positionto another without being disassembled. Position A1B1 is horizontal, and position A3B3 is vertical.AB = 2.0 in.

B3

B1

A2 (2.69, 1.44)

B2

A3 (5.06, 0.0)

Y

X

135˚

A1(0, 0)

Solution

Find the center ponit A*. Then find the poles, the image pole I'23, and the circle of sliders. Nextselect a slider point (C1), and find the location of that point in the other positions. This establishesthe slider line. Note that a difference linkage results for each choice of C1.

From the figure,

r2 = A*A1 = 2.9393 in

r3 = A1C1 = 4.3606 in

Find the transmission angle in the three positions. The linkage changes mode because the sign ofthe transmission angle changes.

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A*

A3

B3

A1 B1

A2

B2

P12

P23

P13P23'

C1

C2

C3

r = 2.93932

r = 4.36063

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Problem 6.17

Design a slider-crank mechanism to move a coupler containing the line AB through the threepositions 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 (p12, p13, p23) are shown, but these should bedetermined accurately after the positions are redrawn. Find A*, the slider point that lies above B1 ona vertical line through B1, and draw the linkage in position 1.

A1B1

B2

A 2B 3

A3P12

P13

P23

X

Y

(1.25, 1.70)

(1.65, 1.25)

(2.15, 0.58)

(3.15 , 1.32

Solution

Find the center ponit A*. Then find the poles, the image pole I'23, and the circle of sliders. Nextselect the slider point (C1) that lies above B1. Then find the location of that point in the otherpositions. This establishes the slider line.

Draw the linkage in the three positions. By inspection of the positions of C, the mechanismchanges mode and goes through the positions in the wrong order.

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Problem 6.18

Design a slider-crank linkage to move a coupler containing the line AB through the three positionsshown. The fixed pivot (center point) of the other crank is at C*. Draw the linkage (including theslider line) in Position 1. Position A1B1 is horizontal, and positions A2B2 and A3B3 are vertical.AB = 4 in.

A1 B1

B 2

A2

B3

A 3

X

Y

(2, 0) (4, 0)

(0, 2)

C1* (0, 0)

Solution

This problem illustrates the type of rigid body guidance problem that cannot be solved directlyusing the elementary techniques developed in the text book. Therefore, to correct answer is thatthere is no solution. However, a partial solution can be developed, and the students should work theproblem 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.

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Finding C1

Next find the poles as shown in the following construction. This is where a problem occurs. Notethat P1 2 and P1 3 are coincident and P2 3 is at infinity on the line shown. Therefore, the circle ofsliders appears to be on a straight line, but the orientation of the line cannot be determined from theelementary theory provided.

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Finding the poles

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Problem 6.19

Design a slider-crank mechanism to move a coupler containing the line with A through the threepositions shown. The moving pivot (circle point) of the crank is at A. Find the slider point whichlies on Line BC and draw the linkage (including the slider line) in position 1. Note that Line BC isNOT the line on which the slider moves.

A

A

1

3B C

A 2

(0, 0) (0.8, 0)

(0.8, 0.8)

60˚

30˚

0.4

Solution

Find the center ponit A*. Then find the poles, the image pole I'23, and the circle of sliders. Nextselect the slider point (C1) that lies on the line BC. There are two solutions, but the one on the rightis implied by the problem statement. Then find the location of that point in the other positions. Thisestablishes the slider line.

Draw the linkage in the three positions. By inspection of the positions of C, the mechanismchanges mode and goes through the positions in the wrong order. The construction steps areshown in the following.

Finding the center point A*

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Finding the poles

Finding the circle of sliders

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Finding the slider point

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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 a1 and a2.

b) Choose the best precision point values for from among 0, 0.17, 0.35, and 0.52, and determinethe values of a1 and a2 that will allow the device to approximate the function.

c) Find the error when = /8.

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Solution

Two precision points can be used because there are two design variables. For this, first determine1 and 2. Look at Chebychev spacing to see what values of are reasonable. From the figure,

045˚

θ1 θ2

135˚

π4

π8

1 = 8(1+ cos135°) = 0.115

2 = 8(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 = (0.52)2 = 0.270

Substituting into the equation for the system model,

1 = a1 + a2cos 1

2 = a1 + a2cos 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

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e = ideal act =2 2.058 2.060cos( )

Substituting in the given value = /8 = 22.5˚,

e = (0.3926)2 2.058 2.060cos 22.5̊( ) = 0.000673

Problem 6.21

A mechanical device characterized by the input-output relationship = 2a1 + 3a2 sin + a32 is to be

used to generate (approximately) the function = 2 2 over the range 0 / 4 . Exteriorconstraints 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 willallow the device to approximate the function.


Solution

Two precision points can be used because there are two design variables. For this, first determine1 and 2. Look at Chebychev spacing to see what values of are reasonable. From the figure,

045˚

θ1 θ2

135˚

π4

π8

1 = 8(1+ cos135°) = 0.115

2 = 8(1+ cos45°) = 0.670

Then,

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1 = 2( 1)2 = 2(0.115)2 = 0.02645

2 = 2( 2 )2 = 2(0.670)2 = 0.8978

Substituting into the equation for the system model,

1 = 2a1 + 3a2 sin 1 + a32 = 2a1 +3a2 sin 1 +1

2 = 2a1 +3a2 sin 2 + a32 = 2a1 +3a2 sin 2 +1

or,

0.02645 = 2a1 +3a2 sin(0.115) +1= 2a1 + 0.3442a2 +1

0.8978 = 2a1 +3a2 sin(0.670) +1 = 2a1 +1.8629a2 +1

Subtract the first equation from the second, and solve for a2

0.8713 =1.5187a2and

a2 = 0.5737

Now back substituting into the first equation,

2a1 = 0.02645 1 0.3442a2 = 0.02645 1 0.3442(0.5737) = 1.1710or

a1 = 1.1710 /2 = 0.5855


e = ideal act = 2 2 2a1 + 3a2 sin +1( ) = 2 2 2( 0.5855) + 3(0.5737)sin +1[ ]

= 2 2 0.1710 +1.7214sin[ ]

Substituting in the given value = /6 = 30˚,

e= 2(0.5235)2 0.1710 +1.7214sin30̊[ ] = 0.1416

Problem 6.22

A mechanical device characterized by the input-output relationship = 2a1 + a2tan +a32 is to beused to generate (approximately) the function = 3 3 ( and both in radians) over the range0 / 3 . Exterior constraints on the design require that the parameter a3 = 1.


b) Use Chebyshev spacing, and determine the values for the unknown design variables that willallow the device to approximate the function.


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Solution:

There are two unknowns so the number of precision points is two. The precision points accordingto Chebyshev spacing are:

1 =( max + min)

2+( max min)

2cos135 =

6+6cos135 = 0.1534

and

2 =( max + min)

2+( max min)

2cos45 =

6+6cos45 = 0.8938

The corresponding values of are:

1 = 2 13 = 2(0.1534)3 = 0.0072 2 = 2 2

3 = 2(0.8938)3 =1.428

We can now solve for a1 and a2 using the desired input-output relationship.

= 2a1 + a2tan +a32 = 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 a1 = 0.5968


e = ideal act = 3 2 2a1 + a2 tan +1( ) = 3 2 2( 0.5968)+ (1.3033)tan +1[ ]

= 3 2 0.1936 +1.3033tan[ ]


e= 3(0.5235)2 0.1936 +1.3033tan30̊[ ] = 0.2632

- 263 -

Problem 6.23

A mechanical device characterized by the input-output relationship = 2a1 + a2 sin is to be used togenerate (approximately) the function y = 2x 2 over the range 0 x / 2 where x, y, , and areall in radians. Assume that the use of the device will be such that the starting point and range for xcan be the same as those for , and the range and starting point for y can be the same as those for .


b) Use Chebyshev spacing, and determine the values for the unknown design variables that willallow 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 accordingto Chebyshev spacing are:

x1 =(xmax + xmin)

2+(xmax xmin)

2cos135 =

4+4cos135 = 0.2300

and

x2 =(xmax + xmin)

2+(xmax xmin)

2cos45 =

4+4cos45 =1.3407

The corresponding values of y are:

y1 = 2x12 = 2(0.2300)2 = 0.1058

y2 = 2x22 = 2(1.3407)2 = 3.5949

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 followingpairs of numbers to solve the problem.

1 = 0.2300 rad =13.178̊

1 = 0.1058

and

2 =1.3407rad = 76.816̊

2 = 3.5949

We can now solve for a1 and a2 using the desired input-output relationship.

= 2a1 + a2 sin

Then,

- 264 -

1 = 2a1 + a2 sin 1 = 0.1058 = 2a1 +a2 sin13.178̊ = 2a1 + a2 (.2280)

2 = 2a1 + a2 sin 2 = 3.5949 = 2a1 +a2 sin76.816̊ = 2a1 +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


e = ideal act = 2 2 2a1 + a2 sin( ) = 2 2 2( 0.4806) + (4.6795)sin[ ]

= 2 2 0.9611+ 4.6795sin[ ]


e =2 2 0.9611+4.6795 sin[ ] = 2(.7854)2 0.9611+ 4.6795sin 45̊[ ] = 1.1141

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 = 0.5f =1.0

1 =f + 0

2f 0

2cos30̊ =1.5

20.52cos30̊ = 0.53349364905389

2 =f + 0

2=1.52= 0.75

3 =f + 0

2+

f 0

2cos30̊ =

1.52+0.52cos30̊ = 0.96650635094611

Compute the 's

1 = 12 = 0.28461547358084

2 = 22 = 0.56250000000000

3 = 32 = 0.93413452641916

- 265 -

A =

1 cos 1 cos 1

1 cos 2 cos 2

1 cos 3 cos 3

=

1 0.95976969417465 -0.861035652501771 0.84592449923107 -0.731688868873821 0.59451456313420 -0.56817793658800

B=

cos( 1 1)cos( 2 2)cos( 3 3)

=

0.969189355798620.982473313101260.99947607824379

Then

x1x2x3

=

1 cos 1 cos 1

1 cos 2 cos 2

1 cos 3 cos 3

1 cos( 1 1)cos( 2 2)cos( 3 3)

=

1.05801273775729-0.001954867783870.10097974323242

and

r2 = 1x2= -511.5

r4 = 1x3= 9.903

r3 = 1+ r22 + r22 2r2r4x1 = 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 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( ) forvalues of between 0 and 90 degrees. Use Chebyshev spacing with three position points. Thebase length of the linkage must be 2 cm.

- 266 -

Solution

Determine the precision points using Chebychev spacing. Then

0 = 0

f = 90̊

1 =f + 0

2f 0

2cos30̊ =

4 4cos30̊ = 0.10522340180962

2 =f + 0

2=4= 0.78539816339745

3 =f + 0

2+

f 0

2cos30̊ =

4+4cos30̊ =1.46557292498528

Compute the 's

1 = sin( 1) = 0.10502933764983

2 = sin( 2 ) = 0.70710678118655

3 = sin( 3 ) = 0.99446912382076

A =

1 cos 1 cos 1

1 cos 2 cos 2

1 cos 3 cos 3

=

1 0.99448948752450 -0.994469123820761 0.76024459707563 -0.70710678118655

1 0.54494808990266 -0.10502933764983

B=

cos( 1 1)

cos( 2 2 )

cos( 3 3)

=

0.999999981169550.99693679488540

0.89106784875455

Then

x1x2x3

=

1 cos 1 cos 1

1 cos 2 cos 2

1 cos 3 cos 3

1 cos( 1 1)

cos( 2 2 )

cos( 3 3 )

=

1.05576595368848-0.36099675791050

-0.30492802741671

and

r2 =1x2= -2.77010798043768

r4 =1x3= -3.27946239796913

r3 = 1+ r22 + r2

2 2r2r4x1 = 0.49621992921794

Now unscale the values by multiplying each by 2. Then

- 267 -

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 four-bar linkage that generates the function y = x x +3 for values of x between 1 and4. Use the Chebyshev spacing for three position points. The base length of the linkage must be 2in. Use the following angle information:

0 = 45˚ = 50˚

0 = 30˚ = 70˚

Compute the error at x = 2

Solution

The solution is given in the following. From the given information,

- 268 -

x0 = 1; xf = 4.

Using Chebychev spacing for the precision points,

x1=xf + x02

xf x02

cos30° = 4 +12

4 12cos30° = 1.20096189

Similarly, x2 = 2.5 and x3 = 3.799038105. Then, the corresponding values for y are:

yf = xf xf + 3 =1y0 = x0 x0 +3 = 3y1= x1 x1 +3 = 2.894922175y2= x2 x2 + 3 = 2.081138830y3= x3 x3 +3 =1.150074026

Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. Forthe range of linkage angles, we have:

0 = 45°; = 50°

and

0 = 30°; = 70°

The precision points in terms of are:

1 =x1 x0x f x0

+ 0 =1.20096189 1

350°+ 45° = 48.34936490538903°

2 =x2 x0x f x0

+ 0 =2.5 1

350° +45° = 70°

3 =x3 x0x f x0

+ 0 =3.799038105 1

350°+ 45° = 91.65063509461096°

Similarly,

1 =y1 y0y f y0

+ 0 =2.894922175 3

270° +30 = 33.67772386020241°

2 =y2 y0yf y0

+ 0 =2.081138830 3

270°+ 30 = 62.16014094705335°

3 =y3 y0y f y0

+ 0 =1.150074026 3

270° +30 = 94.74740905563471°

Using the matrix solution procedure,

- 269 -

z1z2z3

=

1 cos 1 cos 1

1 cos 2 cos 2

1 cos 3 cos 3

1 cos( 1 1)

cos( 2 2 )cos( 3 3)

=

1 0.8321697783 -0.6645868192

1 0.4670019053 -0.3420201433

1 -0.0827631422 0.0288050322

1 0.9673932335

0.9906531872

0.9985397136

=

1.0037381398

0.1450642022

0.2363317277

and

r2 =1z2=

10.1450642022

= 6.8934994632

r4 =1z3=

10.2363317277

= 4.2313404537

and

r3 = 1+ r22 + r4

2 2r2r4 z1

= 1+ (6.893499)2 + (4.231340)2 2(6.893499)(4.231340)(1.003738) = 2.8051768

For the overall size of the linkage, use a base link length of 2 in. Then the lengths of the other linksbecome

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 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 = 2 2 +3 = 2.4142

To compute the generated value, first compute the value of corresponding to x = 2. Then,

e =xe x0x f x0

+ 0 =2 1

350°+ 45° = 61.666°

Using this value of and the link lengths given above, find the output value for . This can be donegraphically, or by using the routine fourbar_cr. The value of is

e = 51.02815161

The corresponding value for y is,

- 270 -

yact =e 0 (yf y0 ) + y0 =

51.02815161 3070

( 2)+ 3 = 2.3992

The error is

error = yideal yact = 2.4142 2.3992 = 0.0150

Problem 6.27

Design a four-bar linkage to generate the function y=x2-1 for values of x between 1 and 5. UseChebyshev spacing with three position points. The base length of the linkage must be 2 cm. Usethe following angle information:

0 = 30˚ = 60˚

0 = 45˚ = 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 + x02

xf x02

cos30° = 1+52

5 12cos30° =1.26794919243112

Similarly, x2 = 3 and x3 = 4.73205080756888. Then, the corresponding values for y are:

yf = xf2 1 = 24y0 = x02 1= 0

y1= x12 1 = 0.60769515458674y2= x22 1= 8y3= x32 1= 21.39230484541327

Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. Thelinkage angles are:

0 = 30°; = 60°

and

0 = 45°; = 90°

The precision points in terms of are:

- 271 -

1 =x1 x0xf x0

+ 0 =1.26794919243112 1

460°+30° = 34.01923788646684°

2 =x2 x0xf x0

+ 0 =3 1

460°+ 30° = 60°

3 =x3 x0xf x0

+ 0 =4.73205080756888 1

460°+30° = 85.98076211353316°

Similarly,

1 =y1 y0y f y0

+ 0 =0.60769515458674 0

2490°+ 45= 47.278856829°

2 =y2 y0yf y0

+ 0 =8 024

90°+ 45 = 75°

3 =y3 y0y f y0

+ 0 =21.39230484541327 0

2490° + 45=125.221143170°

Using the matrix solution procedure,

z1z2z3

=

1 cos 1 cos 1

1 cos 2 cos 2

1 cos 3 cos 3

1 cos( 1 1)

cos( 2 2 )cos( 3 3)

=

1 0.6784308203 -0.8288497687

1 0.2588190451 -0.5000000000

1 -0.5767338180 -0.0700914163

1 0.973340766

0.965925826

0.774498853

=

1.1947633778

0.6332388583

0.7854636563

and

r2 =1z2=

10.6332388583

=1.57918293

r4 =1z3=

10.7854636563

=1.27313338

and

r3 = 1+ r22 + r4

2 2r2r4 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 otherlinks become

- 272 -

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 yand 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 x0x f x0

+ 0 =3 1

360°+30° = 60°

Using this value of and the link lengths given above, find the output value for . This can be donegraphically, or by using the routine fourbar_cr. The value of is

e = 75̊

The corresponding value for y is,

yact =e 0 (yf y0 ) + y0 =

75 4590

(24) + 0 = 8

The error is

error = yideal yact =8 8 = 0

Problem 6.28

The output arm of a lawn sprinkler is to rotate through an angle of 90˚, and the ratio of the times forthe forward and reverse rotations is to be 1 to 1. Design the crank-rocker mechanism for thesprinkler. If the crank is to be 1 inch long, give the lengths of the other links.

Solution

=180Q 1Q +1

=1801 11+1

= 0°

This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" atthe angle shown. Then, the center location for O2 is located on a line through B1 and B2. Anypoint will work as long as it is selected to the left of B2.

- 273 -

B1B2

O4

O2

90˚3.232"

1.500

4.314

For the location selected,

r2 + r3 = 4.314r3 r2 =1.500

Then2(r3) = 5.814

orr3 = 2.907"

andr2 = 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 Kwhere

K = 11.407

= 0.711

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,

- 274 -

Problem 6.29

Design a crank-rocker mechanism such that with the crank turning at constant speed, the oscillatinglever will have a time ratio of advance to return of 3:2. The lever is to oscillate through an angle of80o, and the length of the base link is to be 2 in.

Solution

=180Q 1Q +1

=1801.5 11.5 +1

= 36°

This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" atthe angle shown. Then, the construction of the center location for O2 is shown in the followingfigure. From the figure,

r2 + r3 = 3.731r3 r2 =1.687

Then2(r3) = 5.418

orr3 = 2.709"

andr2 = r3 1.687 = 2.709 1.687 =1.022"


If the distance between the fixed pivots is really 2 in, then each dimension must be scaled. To dothis multiply each length by K where

K = 22.155

= 0.928

B1

B2

3.731"

O4O2

80˚

2.155"

1.687"

36˚

Then,

r1 = 0.928*2.155 = 2 in

- 275 -

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 ofthe cycle is to take twice as long as the return to give a quick-return mechanism. The distancebetween fixed pivots must be 0.5 m. Determine the lengths for r2, r3, and r4.

Schematic Drawing

∆θ = 80˚

��

A

B

r2

r3

r4

O2 O4

ω21

Solution

=180Q 1Q +1

=180 2 12 +1

= 60°

This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" atthe angle shown. Then, the construction of the center location for O2 is:

B1

B2

2.9365"

O4

O2

80˚60˚

1.1482"

1.4728"

r2 + r3 = 2.9365r3 r2 =1.1482

- 276 -

Then2(r3) = 4.0847

orr3 = 2.0424"

andr2 = r3 1.1482 = 2.0424 1.1482 = 0.8942"


If the distance between the fixed pivots is 0.5 meters, then each dimension must be scaled. To dothis multiply each length by K where

K = 0.51.4728

= 0.3395

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,


Problem 6.31

The rocker O4B of a crank-rocker linkage swings symmetrically about the vertical through a totalangle of 70˚. The return motion should take 0.75 the time that the forward motion takes. Assumingthat 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,

=180Q 1Q +1

=1801.33 11.33 +1

= 25.7°

This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" atthe angle shown. Then, the construction of the center location for O2 is:

r2 + r3 = 4.2014r3 r2 = 2.4112

Then2(r3) = 6.6126"

orr3 = 3.3063"

and

- 277 -

r2 = r3 2.4112 = 3.3063 2.4112 = 0.8951"

B1

B2

O4O2

70˚

25.7˚2.4112"

4.2014"

2.5984"


If the distance between the fixed pivots is really 2.5", then each dimension must be scaled. To dothis multiply each length by K where

K = 2.52.5984

= 0.9621

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",


Problem 6.32

A crank rocker is to be designed such that with the crank turning at a constant speed CCW, therocker will have a time ratio of advance to return of 1.25. The rocking angle is to be 40˚, and itrocks symmetrically about a vertical line through O4. Assume that the two pivots are on the samehorizontal line, 3 in apart.

Solution:

This problem does not have a unique solution. To solve this problem, the procedure given inSection 4.4.3.1 can be used. First draw the line beween the pivot points as shown. Next computethe angle using

- 278 -

=180Q 1Q +1

=1801.25 11.25 +1

= 20°

B1B2

O4O2

20˚ 40˚

G

80˚

3.00"

4.4885"2.3127"

40˚

1.9238"

Locate G as shown in Fig. 4.46, and draw the locus of B2. To locate the oscillation anglesymetrically about the vertical, locate B2 at an angle of 20˚ from the vertical. We can then locate B1at an angle of 80˚ from the line O2B2. Next compute r2 and r3 from

r2 + r3 = 4.4885r3 r2 = 2.3127

Then2(r3) = 6.8012

orr3 = 3.4006

andr2 = 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 reasonablyefficient.

Problem 6.33

Design a crank-rocker mechanism that has a base length of 2.0, a time ratio of 1.3, and a rockeroscillation angle of 100˚. The oscillation is to be symmetric about a vertical line through O4.Specify the length of each of the links.

Solution:

- 279 -


=180Q 1Q +1

=1801.3 11.3 +1

= 23.48°

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 anglesymetrically about the vertical, locate B2 at an angle of 50˚ from the vertical. We can then locate B1at an angle of 100˚ from the line O2B2. Next compute r2 and r3 from

r2 + r3 = 2.9890r3 r2 =1.3229

Then2(r3) = 4.3119

orr3 = 2.1559

B1

B2

O4O2

50˚

G

2.00"

1.3229"

1.2794"

2.9890"

26.52˚

100˚

50˚

andr2 = r3 1.3229 = 2.1559 1.3229 = 0.8330"


r1= 2.0000"

r4 =1.2794"

Problem 6.34

A crank-rocker mechanism with a time ratio of 2 13 and a rocker oscillation angle of 72˚ is to bedesigned. The oscillation is to be symmetric about a vertical line through O4. Draw the mechanism

- 280 -

in any position. If the length of the base link is 2 in, give the lengths of the other three links. Alsoshow the transmission angle in the position in which the linkage is drawn.

Solution:


=180Q 1Q +1

=1802 13 12 13 +1

= 72̊

and

2= 36 72 = 36˚

This means that the point G is at infinity, and the B2 locus is a straight line through O2 at an angleof 36˚ to the horizontal. The result is shown in the following:

B1

B2

O4O2

36˚ G2.00"

2.8617"

1.1629"

1.6251"

36˚

72˚

36˚

To locate the oscillation angle symetrically about the vertical, locate B2 at an angle of 36˚ from thevertical. We can then locate B1 at an angle of 72˚ from the line O2B2. Next compute r2 and r3from

r2 + r3 = 2.8619r3 r2 =1.6251

Then2(r3) = 4.4870

orr3 = 2.2435"

andr2 = r3 1.6251 = 2.2435 1.6251 = 0.6184"


r1= 1.0000"

and

- 281 -

r4 =1.1629"

The linkage is shown in an arbitrary position below. The transmission angle is shown.

B

O4O2

69.44˚

Problem 6.35

The mechanism shown is used to drive an oscillating sanding drum. The drum is rotated by asplined shaft that is cycled vertically. The vertical motion is driven by a four-bar linkage through arack-and-pinion gear set (model as a rolling contact joint). The total vertical travel for the sanderdrum 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 asthe return part. The distance between fixed pivots must be 4 in. Determine the lengths for r2, r3,and r4.

Schematic Drawing

A

B

r2

r3

O2O4

��

��

��

��Splined Shaft

Pinion

Rack

Motor

Belt drive

Sander drum Total travel = 3"

2"

ω 21r4

∆θ

Solution:

This problem does not have a unique solution. To solve this problem, we must determine theoscillation angle. If the drum rotates 3 inches, we can find the oscillation angle form

r4 = dand

= dr4= 32=1.5 rad =85.94˚

- 282 -

To start the construction, first draw the line beween the pivot points as shown. Next compute theangle using

=180Q 1Q +1

=1801 11+1

= 0̊

Arbitrarily pick r4 as 2" at the angle shown. Then, the center location for O2 is located on a linethrough B1 and B2. Any point will work as long as it is selected to the left of B2.

B1B2

O4

O2

85.94˚3.2253"

1.500

4.250

For the location selected,

r2 + r3 = 4.250r3 r2 =1.500

Then2(r3) = 5.750

orr3 = 2.875"

andr2 = r3 1.500 = 2.875 1.500 =1.375"


If the crank is really 1 in long, then each dimension must be scaled. To do this multiply each lengthby K where

K = 43.2253

=1.2413

Then,

r1 = 1.2413*(3.2253) = 4.000 in

r2 =1.2413*(1.375) = 1.706 in,

- 283 -

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 anangle of 20˚, and the gear ratio RG/RP is 4:1, that is, the radius rG is four times the radius rp.Contact between the two gears can be treated as rolling contact. The crusher mechanism requiresthat the crank (r2) rotate at a constant velocity, and the advance part of the cycle is to take 1.5 timesas much as the return part. The distance between fixed pivots O2 and O4 must be 4 ft. Determinethe lengths for r2, r3, and r4.

Schematic Drawing

∆θ = 20˚

��

AB

r2

r3

r4

O25

rP

rGO4

ω21

Solution

=180Q 1Q +1

=1801.5 11.5 +1

= 36°

= 4(20) = 80°

This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" atthe angle shown. Then, the construction of the center location for O 2 is:

B1

B2

3.7087"1.6602"

O4O2

36˚ 80˚

- 284 -

r2 + r3 = 3.7087r3 r2 =1.6602

Then2(r3) = 5.3689

orr3 = 2.6845"

andr2 = r3 1.6602 = 2.6845 1.6602 =1.0242"


If the distance between the fixed pivots is really 4 feet, then each dimension must be scaled. To dothis multiply each length by K where

K = 42.1418

=1.8675

Then,

r1 = 4.0 feet,

r2 = 1.9126 feet,

r3 = 5.0133 feet,

r4 = 3.7350 feet,


Problem 6.37

The mechanism shown is proposed for a shaper mechanism. The shaper cutter moves back andforth 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 anangle of 80˚. Determine the lengths for r1, r2, and r3.

Schematic Drawing

A B

r2

r3

r412ω

O2 O4

C

D

r5

6

Cutter

Part Being Machined

Solution:

This problem does not have a unique solution. To start the procedure, determine the angle fromthe time ratio.

- 285 -

=180Q 1Q +1

=180 (2 1)(2 +1)

= 60˚

O4

80˚

2.90"0.90"

B2B1

O21.64"

2.00"

60˚

From the diagram:

r1= 1.64"

r2 = O2B1 O2B22

= 2.90 0.902

=1.00"

r3 = O2B1 +O2B22

= 2.90 + 0.902

=1.90"

r4 = 2"

Scalling the results,

R4 = 4 in

R2 = r2 R4r4=1.00 4

2=1.00(2) = 2.00 in

R3 = r3R4r4=1.90(2) = 3.80 in

R1 = r1 R4r4=1.64(2) =3.28 in

The linkage is shown to scale in a general position in the following:

O4O2

B

A

- 286 -

Problem 6.38

A crank rocker is to be used in a door-closing mechanism. The door must open 100˚. The crankmotor is controlled by a timer mechanism such that it pauses when the door is fully open. Becauseof this, the mechanism can open and close the door in the same amount of time. If the crank (r2) ofthe mechanism is to be 10 cm long, determine the lengths of the other links (r1, r3, and r4). Sketchthe mechanism to scale.

Schematic Drawing

A

r2

r3

r41 2ω

O2

O4

B

Wall

Door

Solution

The time ration is 1 so is 0. This problem does not have a unique solution. Initially pick theoutput link length to be 2. Then the following scalled values are determined as shown in the figurebelow.

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"

- 287 -

O4

B1B2

O2 100˚

Now unscale the output.

K =R2r2= 101.532

= 6.566

Then,

R1 = 6.566(3.713) = 24.37 cmR2 = 6.566(3.713) = 10 cmR3 = 6.566(3.713) = 22.82 cmR4 = 6.566(2) =13.13 cm

A scaled version of the linkage if given in the following.

B

A

R4

R3

R2

R1

Scaled Linkage

O4O2

- 288 -

Problem 6.39

A crank rocker is to be used for the rock crusher mechanism shown. The oscillation angle for therocker is to be 80˚, and the working (crushing) stroke for the rocker is to be 1.1 times the returnstroke. If the frame link (r1) of the mechanism is to be 10 ft long, determine the lengths of the otherlinks (r2, r3, and r4). Sketch the mechanism to scale.

Rock

A

B

O

2

3

4

1 O 2

��

��

Solution:

=180Q 1Q +1

=180 (1.1 1)(1.1+1)

= 8.6̊

This problem does not have a unique solution. Initially pick the output link length to be 2. Thenfrom the diagram,

O2

80˚

8.6˚3.24"

5.74"

B2B1

O1

4.39"

2.00"

r1= 4.39"

r2 = O1B1 O1B22

= 5.74 3.242

=1.25"

r3 = O1B1+O1B22

= 5.74 + 3.242

= 4.49"

r4 = 2"

Scalling the results,

R1 = 10 ft

- 289 -

R2 = r2 R1r1=1.25 10

4.39=1.25(2.33) = 2.91 ft

R3 = r3R1r1= 4.49(2.33) =10.46 ft

R4 = r4 R1r1= 2(2.33) = 4.66 ft

The linkage is shown to scale in a general position in the following:

O2O1

B

A

Problem 6.40

A crank rocker is to be used in a windshield-wiping mechanism. The wiper must oscillate 80˚. Thetime for the forward and return stroke for the wiper is the same. If the base link (r1) of themechanism is to be 10 cm long, determine the lengths of the other links (r2, r3, and r4). Sketch themechanism to scale.

Schematic Drawing

A

r2

r3

r4

1 2ω

O2 O4

B

Frame

Wiper

r1

Solution

The time ration is 1 so

- 290 -

=180 Q 1Q+1

=180 0

2[ ] = 0

This problem does not have a unique solution. Initially pick the output link length to be 2. Thenthe linkage can be constructed as shown

O4

O2

B1

B2

A2A1

R2

R1 R4

From the figure,

R4 = 2 in

R1 = 3.512 in

R2 +R3 =O2B2R3 R2 =O2B1

or

R2 = (O2B2 O2B1) / 2 = (4.5078 2.0018) / 2 =1.253 inR3 = (O2B2 +O2B1) / 2 = (4.5078 +2.0018) / 2 = 3.881in

Determine the scaling factor,

r1R1

= 103.6059

= 2.7732 = r2R2

=r3R3

=r4R4

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:

- 291 -

O4

O2

B

A

R4

R3

R2

Problem 6.41

Design a six-bar linkage like that shown in Fig. 6.62 such that the output link will do the followingfor 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. Afterselecting such a curve, pick a starting place on the curve as one of the two extreme locations forpoint F. Draw a line perpendicular to the curve and select a length for link 5. The value for the linklength is somewhat arbitrary although it needs to be long enough to permit the mechanism tooperate for the whole cycle. Next count around the curve 42 dashes corresponding to 210˚ of crankrotation. This will be the other extreme location for link 5, and the link will be perpendicular to thecurve at this location also. Locate the link perpendicular to the curve, and this will locate the secondextreme location for Point F.

Connect the two extreme locations of F by a cord line, and locate the perpendicular bisector of thecord line. Draw a line through one of the extreme locations of F at an angle of 15˚ to theperpendicular bisector. This will locate point G and the length of link 6. Draw the dyad GRE tocomplete the 6-bar linkage.

For the mechanism shown, the following values apply:

AB = 0.762”; BC = 2.191”; CD = 1.193”; AD = 2.488”; 1 = -11.83˚

BE = 1.313”; = -34.03˚; EF = 1.00”; FG = 1.564”; XG = 2.644”; YG = -0.532”

The linkage was analyzed for the values given above using the program sixbar.m, and the results areshown in following the solution drawing. The results are fairly accurate; however, the accuracycould be improved by moving the location of G slightly.

- 292 -

- 293 -

Problem 6.42

Design a six-bar linkage like that shown in Fig. 6.62 such that the output link will make twocomplete 35˚ oscillations for each revolution of the driving link. (Hint: Select a coupler curve that isshaped like a "figure 8".)

Solution

Locate a curve that is roughly in the shape of a symmetric figure 8. Draw a circle which is tangentto the top of the curve at the two points b and d. Here, b and d should be approximately in thecenter of the loops of the figure 8. Label the center of the circle as point F', which is one extremelocation for the point F. Next draw a circle of the same radius as that of the first circle but tangentto two ponits on the bottom of the curve. The tangent points are a and c. The center of this circle ispoint F", which is the other extreme location for the point F. Points a, b, c, d will be the extremelocations of the coupler point E. Bisect the line F'F". The pivot G must lie on this line. Locate Gsuch that the included angle between GF' and GF" is 35˚.

Draw the dyad GFE to complete the 6-bar linkage.

The linkage was analyzed for the values given above using the program sixbar.m, and the results areshown in following the solution drawing. The results are fairly accurate; however, the accuracycould be improved by moving the location of G slightly.

b d

a

c

AB

C

DE

F'

F"

G

2

3

4

5

F

X

Y

β

635˚

AD = 2.1429" AB = 1.0994" BC = 2.1806"

BE = 2.1005" CD = 1.6212"

EF = 2.9"FG = 1.0614" β = -35.5˚

x = 0.7222" Gy = -2.95" G

θ = -26.46˚

- 294 -

Problem 6.43

Design a six-bar linkage like that shown in Fig. 6.62 such that the output link will do the followingfor 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.

- 295 -

A

B

C

D

EF

1

2

4

5

6

G

��

3

Fig. PB6.43.1: Six bar linkage

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, weneed only select different coupler curves and locate candidate points for extreme positions. For thisproblem, select a figure 8 coupler curve. The problem can be solved using the following steps.

Rota tion of crank0

10

20

30

40

Rot

atio

n of

link

6

a

b

c

d

e

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 othertangent 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 betweenFa and Fb. On this line, locate the point G such that the angle FaGFb is 40˚. Draw an arccentered 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 thecircles in Step 1.

- 296 -

G

6a

b

c

d

eF

40˚

35˚

30˚

35˚

Fig. PB6.43.3: Positions of link 6

4) Use the geometry in Fig. PB6.43.5 as a single entity and rotate and scale it to fit it to thecoupler 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. Thecoordinates 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 requirementsare not met adequately, select another coupler curve and repeat the procedure. The mechanismhas been analyzed using the program sixbar.m, and the results are given in Fig. PB6.43.7. Theresults are reasonably accurate; however, the results can be improved somewhat by changing thelocation of point G slightly.

G6

Fb

aF

40˚

Fig. PB6.43.4: Finding equal-diameter circles each tangent to coupler curve at two locations.

- 297 -

G

640˚

5˚

5˚bF

dF

cF

aF

Fig. PB6.43.5: Setting extremes of motion for link 6

AB

C

DE

G

2

3

4

5

X

Y

β

6

bF

dF

cF

aF

AD = 2.1429" AB = 1.0994" BC = 2.1806"

BE = 2.1005" CD = 1.6212"

EF = 1.60"FG = 1.00" β = -35.5˚x = 1.70" Gy = -1.60" G

Fig. PB6.43.6: Final solution

- 298 -

Fig. PB6.43.7: Results from s ixbar .m

Problem 6.44

Design a six-bar 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 maximumvalues of 30˚ and 60˚ at input rotations of 60˚ and 240˚, respectively. The rotation of the output linkis zero when the input rotation angle is 0, 120˚, and 360˚.

10˚

20˚

30˚

40˚

50˚

60˚

0˚ 60˚ 120˚ 300˚ 360˚Rotation of crank 2, degreesD

ispl

acem

ent o

f lin

k 6,

deg

rees

180˚ 240˚

Solution

Because of the double oscillation, either a figure 8 or kidney bean shaped curved can be used. Theprocedure will be illustrated with the figure 8 curve used in Example 6.8. Locate a curve that isroughly symmetrical as shown in the following figure. Draw a circle of radius r5 which is tangentto the top of the curve at two locations. The center of this circle is at point F’ which is one extremelocation of F. The radius of the circle gives the length of link 5. The value for this radius is a

- 299 -

design decision. The points a and b in the figure correspond to the locations where the oscillationangle is zero. Starting at b, count 24 dashes (120˚) and draw a second circle of radius r5 tangent tothe curve at point c. The center of this circle will be F”, the other extreme location of F. Locatepoint G by bisecting the cord line between F’ and F” and finding the point on the perpendicularbisector that will give a 60˚ oscillation angle between F’ and F”. The distance from G to F’ or toF” is the length of link 6.

- 300 -

Locate the center of the arc between F’ and F”. Draw a third circle of radius r5 centered at thispoint. This circle should be tangent to the coupler curve at a location that is approximately 12dashes 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 6-bar linkage.

In the figure, Gx = 3.543; Gy = -3.848; r5 = 3.324”; r6 = 1.329”. The original four-bar linkagevalues 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, a60˚ oscillation angle is relatively large for a mechanism such as this. Once a viable linkage isdesigned, it can be analyzed using the sixbar analysis program. Simple adjustments can then bedone directly with the linkage in that program.

Problem 6.45

Design a six-bar 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 inputrotation starting at 0 and 180 degrees. The maximum rotation angle for link 6 is 15˚.

5˚

10˚

15˚

0˚ 90˚ 270˚ 360˚Rotation of crank 2, degreesD

ispl

acem

ent o

f lin

k 6,

deg

rees

180˚

Solution

For this problem, locate a curve that is a kidney-bean shaped curve and isroughly symmetrical asshown in the following figure. The “top” and “bottom” of the kidney bean must haveapproximately the same radius of curvature for the two dwell periods. The radius of curvaturecorresponds to r5. Draw a circle of radius r5 which is tangent to the top of the curve. The center ofthis circle is at point F’ which is one extreme location of F. The radius of the circle gives the lengthof link 5.

Draw a second circle of radius r5 tangent to bottom part of the curve. The center of this circle willbe F”, the other extreme location of F. Locate point G by bisecting the cord line between F’ andF” and finding the point on the perpendicular bisector that will give a 15˚ oscillation angle betweenF’ and F”. The distance from G to F’ or to F” is the length of link 6. Once the linkage is foundto be acceptable, draw the dyad GFE to complete the 6-bar linkage.

For the solution shown, the following values apply:

- 301 -

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 thecurve, but the dwell on the bottom region is shorter than the problem statement specifies. If theapproximation is ultimately unacceptable, a different curve can be used. Also, the problem can beanalyzed using the sixbar routine, and minor adjustments can be made directly with the program.

Problem 6.46Design an eight-bar linkage like that shown in Fig. 6.70 such the coupler remains horizontal whilethe given point on the coupler moves approximately along the path given.

- 302 -

100 mm

500 mm

700 mm

60 mm

450 mm

Coupler Po

Solution

Search for a coupler curve that is roughly the shape of the figure shown. It is not possible to find acurve that exactly duplicates the figure; however, it is possible to find a curve that generally scansthe same area. A solution is shown in the following. Once the coupler curve is found, the 8-barlinkage is completed using the procedure given in Section 6.6.2

Problem 6.47Re-solve Problem 6.46 if the coupler is inclined at an angle of 45˚.

- 303 -

100 mm

500 mm

700 mm

60 mm450 mm

Coupler Po

45˚

Solution


Problem 6.48Design an eight-bar linkage like that shown in Fig. 6.70 such the coupler remains horizontal whilethe given point on the coupler moves approximately along the path from A to B to C. The couplercan return either by retracing the path from C to B to A or by going directly from C to A. Thismeans that the basic 4-bar linkage need not be a crank rocker.

- 304 -

Coupler Point

8"

1"

5"

Equilateral Triangle

A

B

C

Solution


- 305 -

Problem 6.49

Determine the two 4-bar linkages cognate to the one shown below. The dimensions are MA = 10cm, AB = 16 cm, AC = 32 cm, QB = 21 cm, and MQ = 24 cm. Draw the cognates in the positionfor = 90˚.

A B

M Q

C

θ

Solution

Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O byrecognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated inFigs. 6.76 and 6.77.

The cognates are shown in the following figure.

M

A

B

QO

G

D

C

F

E

This figure can be checked using the program cognates.m. The coupler curve and cognates aregiven in the following.

- 306 -

Problem 6.50

Determine the two 4-bar linkages cognate to the one shown below. The dimensions are MQ = 1.5in, AB = BC = BQ = AC= 1 in, and AM = 0.5 in. Draw the cognates in the position for = 90˚.

M

AB

Q

C

θ

Solution



- 307 -

G

D

F O

M

AB

Q

C E


- 308 -

Problem 6.51

Determine the two 4-bar 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˚.

A

B

M Q

C

Solution

Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O byrecognizing that triangle MQO is similar to ABC. Then complete the parallagrams indicated inFigs. 6.76 and 6.77.



- 309 -

Problem 6.52

Determine the two 4-bar linkages cognate for the drag-link mechanism shown. The dimensions areMQ = 1 m, AM = BQ = 4 m, AB =2 m, and angles CAB and CBA both equal 45˚. Notice that thecognates will also be drag-link mechanisms. Draw the cognates in the position for = 180˚.

M

Q

A

B

C 45˚

θ

Solution

- 310 -



B

C

QG

D

E

M

O

A

F


- 311 -

solutions to chapter 6 exercise problems -...

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