# Mécanismes à barres trajectoires et conception .pdf

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**Mécanismes à barres - trajectoires et conception.pdf**

**Microsoft Word - Solns, Ch6 (12-28-03).doc**

**Gary Kinzel**

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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-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.

A2

60˚

O2

A1

O4

30˚

45˚

B2

90˚

B1

Solution:

Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it

clockwise 60˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate the

intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first

figure. Measure the lengths os 02 A1 and A1 B1 . This gives the linkage shown in the second figure.

The link lengths are:

r1 = 6"

r2 = 3.72"

r3 = 7.29"

r4 = 4"

- 222 -

A2

60˚

A1

30˚

O4

O2

45˚

90˚

B2

B1

B'2

Basic Construction

A2

A1

r3

r2

30˚

O2

O4

r1

45˚

90˚

r4

B2

B1

B'2

Final Linkage

- 223 -

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.

B2

A2

60˚

O2

B1

80˚

A1

O4

20˚

25˚

Solution

Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it

clockwise 60˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate the

intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first

figure. Measure the lengths os 02 A1 and A1 B1 . This gives the linkage shown in the second figure.

The link lengths are:

r1 = 4"

r2 = 2.77"

r3 = 3.21"

r4 = 2"

- 224 -

Basic Construction

Final Linkage

Problem 6.3:

In a back hoe, a 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.

- 225 -

Solution

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

input rocker and 70˚ for the output rocker. This is shown in the figure.

The basic problem can then be redrawn as shown in the following figure.

- 226 -

Solution

Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it

clockwise 120˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate

the intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first

figure below. Measure the lengths os 02 A1 and A1 B1 . After unscaling, the linkage is shown in the

second figure that follows.

The link lengths are:

r1 = 13"

r2 = 8.27"

r3 = 19.34"

r4 = 12"

- 227 -

Basic Construction

Final Linkage

- 228 -

Problem 6.4

In the drawing, AB = 1.25 cm. Use A and B as circle points, and design a 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

B3

θ 3 = 60˚

A 3(2, 3)

B2

A 2 (2, 1)

θ2 = 45˚

X

B1

A 1 (0, 0), θ1 = 0

Solution

Find the center points A* and B* and measure the link lengths. Then,

AA* = 2.027"

AB = 1.25"

A*B* = 0.670"

BB* = 2.903"

l + s = 2.903 + 0.670= 3.573"

p + q = 2.027 + 1.25 = 2.277

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

- 229 -

B3

Y

A3

B*

A*

B2

A2

X

B1

A1

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 A1 B1 is horizontal, and position A2 B2 is

vertical. AB = 1.25 in.

Y

A 1 (0, 1.75)

B1

(1.63, 1.25)

A2

B3

35˚

A3 (2.13, 0.63)

B2

Solution

Find the center points A* and B* and measure the link lengths. Then,

AA* = 2.15"

AB = 1.25"

- 230 -

X

A*B* = 2.32"

BB* = 1.11"

l + s = 2.32 + 1.11= 3.43"

p + q = 2.15 + 1.25 = 3.40

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

Y

B1

A1

A2

B3

B*

A3

X

B2

A*

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

B3

B2

60˚

A3(2, 2.4)

50˚

X

A 2 (2, 0.85)

A1 (0, 0)

Solution

B1

Find the center points A* and B* and measure the link lengths. Then,

AA* = 1.705 cm

AB = 4.000 cm

- 231 -

A*B* = 3.104 cm

BB* = 6.954 cm

l + s = 6.954 + 1.705= 8.659"

p + q = 3.104 + 4.000 = 7.104

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

B3

Y

B2

B*

A3

A*

A2

B1

A1

- 232 -

X

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 A1 B1 and A2 B2 are horizontal, and position A3 B3 is vertical. AB = 3 in.

Y

B 2 (0.0, 2.88)

A2

A1

B 1 (-1.94, 0.94)

C*

A3

X

(-1.0, -3.0)

B3

Solution

Find the center points C1 and A* and measure the link lengths. Note that the linkage is to be drawn

in position 1 so the motion must be referred to position 1 when locating C1 . Then,

CC* = 2.67 in

AC = 2.38 in

A*C* = 3.00 in

AA* = 2.06 in

l + s = 3.00 + 2.06= 5.06

p + q = 2.67 + 2.38 = 5.05

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

- 233 -

Y

B2

A2

C 3*

A1

B1

A* C

1

C*

X

A3

C 2*

B3

To determine if the linkage changes branch, draw the linkage in the three positions, and measure the

sign of . From the drawing below, the sign of is different in the three positions, and the

mechanism changes branch.

- 234 -

Y

B2

A2

C2

ψ2

A*

A1

B1

C1

C*

ψ1

X

A3

ψ3

C3

B3

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 A1 B1 is horizontal, and positions A2 B2 and A3 B3 are vertical.

AB = 4 in.

Y

A1

*

C1

A2

(0, 2)

(0, 0)

A3

B1

(2, 0)

B2

- 235 -

(4, 0)

B3 X

Solution

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

r1 = 3.22 = p

A3

A2

A1

B1

r2

r3

A*

r1

*

C

B2

r4

B3

C*2

C1

C*3

r2 = 3.32 = q

r3 = 3.62 = l

r4 = 2.24 = s

Grashof l + s < p + q

For this mechanism,

and

l + s = 3.62 + 2.24 = 5.86

p + q = 3.22 + 3.32 = 6.54

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

- 236 -

Problem 6.9

A mechanism must be designed to move a computer terminal from under the desk to top level. The

system will be guided by a linkage, and the use of a 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.

Y

Desk

Position 3 (Horizontal)

A3 (3.3, 2.6)

Position 2

C*

A1 (-2.2, -0.3)

135˚ A2 (2.1, -0.1)

Position 1

- 237 -

160˚

X

Desk

Pos'n 3

A3

A*

C*2

C*3

C*

Pos'n 2

A1

A1C 1 = 2.6864

C1C* = 3.0684

A * C* = 1.0301

A1A* = 1.395

A2

Pos'n 1

A * C * +C 1C* = 4.0985

A1C1 + A1A* = 4.0814

Therefore,

A * C * +C 1C* > A1C 1 + A1A *

and the linkage is a nonGrashof linkage and a double rocker.

C1

This is a poor solution because C 1is below the floor plane.

Problem 6.10

Design a 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 A2 B2 and A3 B3 are horizontal, and position

A1 B1 is vertical. AB = 3.5 in.

Y

B1

(-1.0, 2.5)

A3

B3 (0.0, 2.0)

A2

B2 (0.0, 1.0)

C*1

A1

- 238 -

X

C*2

B1

A3

C*2

B3

A*

B2

A2

C1

C*

A1

From the figure,

r1 = A*C* = 1.8228 in

r2 = A*A1 = 2.4974 in

r3 = A1 C1 = 2.6966 in

r4 = C*C1 = 1.3275 in

Grashof calculation:

l +s?p+q

[2.6966 + 1.3275=(4.0241)] < [1.8228 + 2.4974=(4.3202)]

The linkage is a crank rocker.

- 239 -

Problem 6.11

Design a 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 A1 B1 is horizontal, and positions A2 B2 and A3 B3 are

vertical. AB = 6 cm.

Y

B1

A3 (-2, 3)

A1

(6, 4.3)

A2 (2, 3)

X

C *(0, 0)

B3

B2

- 240 -

A*

C*

2

Y

C1

B1

A3

A1

A2

C*

3

X

C*

B3

B2

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

AA* = 13.0

AC = 6.84

CC* = 4.41

C*A* = 15.8

l + s = 4.41 + 15.8 = 20.2

p + q = 6.84 + 13.0 = 19.8

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

- 241 -

Problem 6.12

Design a 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 A1 B1 is horizontal, and position A3 B3 is

vertical. AB = 5.1 cm.

Y

(1.5, 3.7)

A2

45

˚

A1

(-5.0, -1.0)

B1

B2

X

A3

(7.8, -1.0)

C*

(-5.0, -5.0)

B3

Solution

Find the center points C1 and A* and measure the link lengths. Note that the linkage is to be drawn

in position 1 so the motion must be referred to position 1 when locating C1 . Then,

CC* = 7.81 cm

AC = 9.73 cm

A*C* = 6.68 cm

AA* = 6.82 cm

l + s = 9.73 + 6.68 = 16.41 cm

p + q = 7.81 + 6.82 = 14.63 cm

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

- 242 -

A2

Y

X

B2

A1

A3

B1

A*

C*

B3

C1

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.

- 243 -

A2

Y

X

B2

A1

A3

B1

A*

C*

C2

B3

C1

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 A1 B1 and position A3 B3 are

horizontal. AB = 4 cm.

Y

D* (3.0, 2.6)

B2

B3

A 3 (3.4, 1.6)

X

C*

45˚

A2 (2.7, - 0.7)

Solution

B1

A1 (0.7, - 1.8)

Find the circle points C2 and D2 and measure the link lengths. Notice that the linkage is to be

drawn in position 2. Therefore, position 2 for the coupler is the position to which the positions of

D* and C* are referred for finding the circle points. Then,

- 244 -

DD* = 2.767 cm

CD = 4.045 cm

C*D* = 3.985 cm

CC* = 2.300 cm

l + s = 4.045 + 2.300= 6.345

p + q = 2.767 + 3.985 = 6.752

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

D*1

Y

D*

D2

B2

B3

A3

C*1

X

C*

D*3

A2

B1

A1

C2

C*

3

- 245 -

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 A1 B1 and position

A3 B3 are horizontal. AB = 4 cm.

Y

C* (2.3, 4.5)

B2

B3

A3 (2.7, 3.5)

45˚

A2 (2.0, 1.0)

B1

A1

Solution

X

Find the center points C2 and A* and measure the link lengths. Note that the linkage is to be drawn

in position 2 so the motion must be referred to position 2 when locating C2 . Then,

CC* = 2.257 cm

AC = 3.146 cm

A*C* = 2.989 cm

AA* = 3.065 cm

l + s = 3.146 + 2.257= 5.403

p + q = 3.065 + 2.989 = 6.054

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

C*1

C*

Y

B2

C2

A*

B3

A3

C*

3

A2

B1

A1

- 246 -

X

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

Position 3

135˚

Position 2

A3 (0, 8.3)

A2 (-3.1, 6.6)

Position 1

A1(-4.0, 2.8)

B*

X

Solution

Position 2

A3

Position 3

A2

A*

B1

Position 1

B*

3

A1

B*

- 247 -

B*2

AA* = 1.7552"

BB* = 1.2903"

AB = 2.0069"

A*B* = 1.5576"

l + s = 2.0069 +1.2903 = 3.2972 < p + q = 1.7552 +1.5576 = 5.0680 crank rocker

The mechanism would not be acceptable because of the location of the fixed and moving pivots

inside the wall.

Problem 6.16

Design a 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 A1 B1 is horizontal, and position A3 B3 is vertical.

AB = 2.0 in.

Y

A2 (2.69, 1.44)

135˚

A3 (5.06, 0.0)

A1(0, 0)

B2

B1

X

B3

Solution

Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next

select a slider point (C1 ), and find the location of that point in the other positions. This establishes

the slider line. Note that a difference linkage results for each choice of C1 .

From the figure,

r2 = A*A1 = 2.9393 in

r3 = A1 C1 = 4.3606 in

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

the transmission angle changes.

- 248 -

A2

A1

A3

B1

B2

r2 = 2.9393

r3 = 4.3606

A*

P2

B3

3

C3

C2

P23'

P1

P1

2

3

C1

- 249 -

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 1 2, p 1 3, p 2 3) are shown, but these should be

determined accurately after the positions are redrawn. Find A*, the slider point that lies above B1 on

a vertical line through B1 , and draw the linkage in position 1.

Y

P13

(1.25, 1.70)

B1

A1

(1.65, 1.25) A 2

P12

B3

(3.15 , 1.32

P23

A3

(2.15, 0.58)

X

B2

Solution

Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next

select the slider point (C1 ) that lies above B1 . Then find the location of that point in the other

positions. This establishes the slider line.

Draw the linkage in the three positions. By inspection of the positions of C, the mechanism

changes mode and goes through the positions in the wrong order.

- 250 -

- 251 -

Problem 6.18

Design a 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 A1 B1 is horizontal, and positions A2 B2 and A3 B3 are vertical.

AB = 4 in.

Y

A1

*

C1

A2

(0, 2)

(0, 0)

A3

B1

(2, 0)

B2

(4, 0)

B3

X

Solution

This problem illustrates the type of rigid body guidance problem that cannot be solved directly

using the elementary techniques developed in the text book. Therefore, to correct answer is that

there is no solution. However, a partial solution can be developed, and the students should work the

problem far enough to illustrate that the solution procedure breaks down.

With the information give, it is possible to find C1 . This is illustrated in the following construction.

- 252 -

Finding C1

Next find the poles as shown in the following construction. This is where a problem occurs. Note

that P1 2 and P1 3 are coincident and P2 3 is at infinity on the line shown. Therefore, the circle of

sliders appears to be on a straight line, but the orientation of the line cannot be determined from the

elementary theory provided.

- 253 -

Finding the poles

- 254 -

Problem 6.19

Design a 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

A1

(0, 0)

B

C

0.4

A2

(0.8, 0)

30˚

Solution

Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next

select the slider point (C1 ) that lies on the line BC. There are two solutions, but the one on the right

is implied by the problem statement. Then find the location of that point in the other positions. This

establishes the slider line.

Draw the linkage in the three positions. By inspection of the positions of C, the mechanism

changes mode and goes through the positions in the wrong order. The construction steps are

shown in the following.

Finding the center point A*

- 255 -

Finding the poles

Finding the circle of sliders

- 256 -

Finding the slider point

- 257 -

Final solution

Problem 6.20

A device characterized by the 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 determine

the values of a1 and a2 that will allow the device to approximate the function.

c) Find the error when = /8.

- 258 -

Solution

Two precision points can be used because there are two design variables. For this, first determine

1 and 2 . Look at Chebychev spacing to see what values of are reasonable. From the figure,

135˚

π

4

45˚

0

π

8

θ1

θ2

1 = (1+ cos135°) = 0.115

8

2 = (1+ cos45°) = 0.670

8

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 + a2 cos1

2 = a1 + a2 cos2

or,

0.0289 = a1 + a2 cos(0.17) = a1 + a2 (0.985)

0.270 = a1 + a2 cos(0.52) = a1 + a2 (0.868)

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

0.241 = a2 (0.117)

and

a2 = 2.060

Now back substituting into the first equation,

a1 = 0.0289 a2 (0.985) = 0.0289 + 2.060(0.985) = 2.058.

The error is given by

- 259 -

e = ideal act = 2 (2.058 2.060cos )

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

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

Problem 6.21

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

2

used to generate (approximately) the function = 2 over the range 0 / 4. Exterior

constraints on the design require that the parameter a3 = 1.

a) Determine the number of precision points required to complete the design of the system.

b) Use Chebyshev spacing, and determine the values for the unknown design variables which will

allow the device to approximate the function.

c) Find the error when = /6.

Solution

Two precision points can be used because there are two design variables. For this, first determine

1 and 2 . Look at Chebychev spacing to see what values of are reasonable. From the figure,

135˚

π

4

45˚

0

θ1

π

8

1 = (1+ cos135°) = 0.115

8

2 = (1+ cos45°) = 0.670

8

Then,

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θ2

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 sin1 + a32 = 2a1 + 3a2 sin1 + 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

and

0.8713 = 1.5187a2

a2 = 0.5737

Now back substituting into the first equation,

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

or

a1 = 1.1710 / 2 = 0.5855

The error is given by

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 + 1.7214sin30˚ ] = 0.1416

Problem 6.22

A mechanical device characterized by the input-output relationship = 2a1 + a2 tan + a32 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 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 that will

allow the device to approximate the function.

c) Find the error when = /6.

- 261 -

Solution:

There are two unknowns so the number of precision points is two. The precision points according

to Chebyshev spacing are:

1 = (max + min) + (max min) cos135 = + cos135 = 0.1534

2

2

6 6

and

2 = (max + min) + (max min ) cos45 = + cos45 = 0.8938

2

2

6 6

The corresponding values of are:

1 = 213 = 2(0.1534)3 = 0.0072

2 = 232 = 2(0.8938)3 = 1.428

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

= 2a1 + a2 tan + 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

The error is given by

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

= 3 2 [ 0.1936 + 1.3033tan ]

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

e = 3(0.5235)2 [ 0.1936 + 1.3033tan30˚ ] = 0.2632

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Problem 6.23

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

generate (approximately) the function y = 2x 2 over the range 0 x / 2 where x, y, , and are

all in radians. Assume that the use of the device will be such that the starting point and range for x

can be the same as those for , and the range and starting point for y can be the same as those for .

a) Determine the number of precision points required to complete the design of the system.

b) Use Chebyshev spacing, and determine the values for the unknown design variables that will

allow the device to approximate the function.

c) Compute the error generated by the device for x = /4.

Solution:

There are two unknowns so the number of precision points is two. The precision points according

to Chebyshev spacing are:

x1 =

(xmax + xmin) (xmax x min)

+

cos135 = + cos135 = 0.2300

4 4

2

2

x2 =

(x max + x min) (x max x min)

+

cos45 = + cos45 = 1.3407

4 4

2

2

and

The corresponding values of y are:

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

y2 = 2x 22 = 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 following

pairs of numbers to solve the problem.

1 = 0.2300 rad = 13.178˚

1 = 0.1058

and

2 = 1.3407rad = 76.816˚

2 = 3.5949

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

= 2a1 + a2 sin

Then,

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1 = 2a1 + a2 sin1 = 0.1058 = 2a1 + a2 sin13.178˚ = 2a1 + a2 (.2280)

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

The error is given by

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

= 2 2 [0.9611+ 4.6795sin ]

Substituting in the given value = /4 = 45˚,

e = 2 2 [0.9611+ 4.6795 sin ] = 2(.7854)2 [ 0.9611+ 4.6795sin 45˚ ] = 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 = 0.5

f = 1.0

f + 0 f 0

cos30˚ = 1.5 0.5 cos30˚ = 0.53349364905389

2

2

2

2

f + 0 1.5

2 =

=

= 0.75

2

2

+ 0 f 0

1.5 0.5

3 = f

+

cos30˚ =

+

cos30˚ = 0.96650635094611

2

2

2

2

1 =

Compute the 's

1 = 12 = 0.28461547358084

2 = 22 = 0.56250000000000

3 = 32 = 0.93413452641916

- 264 -

cos1 1 0.95976969417465

cos2 = 1 0.84592449923107

cos3 1 0.59451456313420

1 cos1

A = 1 cos2

1 cos

3

-0.86103565250177

-0.73168886887382

-0.56817793658800

cos(1 1) 0.96918935579862

B = cos(2 2) = 0.98247331310126

cos( ) 0.99947607824379

3

3

Then

x1 1 cos 1

x2 = 1 cos2

x3 1 cos3

cos1 1cos(1 1) 1.05801273775729

cos 2 cos(2 2) = -0.00195486778387

cos3

cos( 3 3)

0.10097974323242

and

r2 = 1 = -511.5

x2

r4 = 1 = 9.903

x3

r3 = 1+ r22 + r22 2r2r4 x1 = 522.0

Now unscale the values by multiplying each by 2. Then

R1 = 2.0;

R2 = 2(-511.5) = -1023;

R3 = 2(522.0) = 1044;

R4 = 2(9.903) = 19.80;

Check for linkage type:

l + s = 1044 + 2.0 = 1046

p + q = 19.80 + 1023 = 1042.8

Then

1046> 1042 nonGrashof, double rocker.

Note that the 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.

- 265 -

Solution

Determine the precision points using Chebychev spacing. Then

0 = 0

f = 90˚

f + 0 f 0

cos30˚ = cos30˚ = 0.10522340180962

2

2

4 4

f + 0

2 =

= = 0.78539816339745

2

4

+ 0 f 0

3 = f

+

cos30˚ = + cos30˚ = 1.46557292498528

2

2

4 4

1 =

Compute the 's

1 = sin(1 ) = 0.10502933764983

2 = sin( 2 ) = 0.70710678118655

3 = sin( 3 ) = 0.99446912382076

1 cos1

A = 1 cos 2

1 cos 3

cos1 1 0.99448948752450 -0.99446912382076

cos 2 = 1 0.76024459707563 -0.70710678118655

cos 3 1 0.54494808990266 -0.10502933764983

cos(1 1 ) 0.99999998116955

B = cos( 2 2 ) = 0.99693679488540

cos( 3 3 ) 0.89106784875455

Then

x1 1 cos 1

x2 = 1 cos 2

x3 1 cos 3

cos 1 1 cos(1 1 ) 1.05576595368848

cos 2 cos(2 2 ) = -0.36099675791050

cos 3

cos( 3 3 )

-0.30492802741671

and

r2 = 1 = -2.77010798043768

x2

1

r4 = = -3.27946239796913

x3

r3 = 1+ r22 + r22 2r2 r4 x1 = 0.49621992921794

Now unscale the values by multiplying each by 2. Then

- 266 -

R1 = 2.0;

R2 = 2(-2.7701) = -5.5402;

R3 = 2(0.4962) = 0.9924;

R4 = 2(-3.2794) = -6.5588;

A scaled drawing of the linkage is given in the following:

Problem 6.26

Design a 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˚

= 50˚

0 = 30˚

= 70˚

Compute the error at x = 2

Solution

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

- 267 -

x0 = 1; xf = 4.

Using Chebychev spacing for the precision points,

x1 =

xf + x0 xf x0 cos30° = 4 +1 4 1 cos30° = 1.20096189

2

2

2

2

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

yf =

y0 =

y1 =

y2 =

y3 =

xf xf + 3 = 1

x0 x0 + 3 = 3

x1 x1 + 3 = 2.894922175

x2 x2 + 3 = 2.081138830

x3 x3 + 3 = 1.150074026

Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. For

the range of linkage angles, we have:

0 = 45°; = 50°

and

0 = 30°; = 70°

The precision points in terms of are:

1 =

x1 x 0

1.20096189 1

+ 0 =

50°+ 45° = 48.34936490538903°

x f x0

3

2 =

x2 x0

2.5 1

+ 0 =

50° + 45° = 70°

x f x0

3

3 =

x3 x0

3.799038105 1

+ 0 =

50°+ 45° = 91.65063509461096°

x f x0

3

Similarly,

y1 y0

2.894922175 3

+ 0 =

70° + 30 = 33.67772386020241°

y f y0

2

y y

2.081138830 3

2 = 2 0 + 0 =

70°+ 30 = 62.16014094705335°

y f y0

2

y y

3 = 3 0 + 0 = 1.150074026 3 70° + 30 = 94.74740905563471°

y f y0

2

1 =

Using the matrix solution procedure,

- 268 -

1

z1 1 cos 1 cos1 cos(1 1 )

z2 = 1 cos 2 cos 2 cos( 2 2 )

z3 1 cos 3 cos 3 cos( 3 3 )

1

1 0.8321697783 -0.6645868192 0.9673932335 1.0037381398

= 1 0.4670019053 -0.3420201433 0.9906531872 = 0.1450642022

1 -0.0827631422 0.0288050322

0.9985397136

0.2363317277

and

r2 =

1

1

=

= 6.8934994632

z2 0.1450642022

r4 =

1

1

=

= 4.2313404537

z3 0.2363317277

and

r3 = 1+ r22 + r42 2r2 r4 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 links

become

R1 = 1(2) = 2 in

R2 = 6.8934994632(2) = 13.7869989 in

R4 = 4.2313404537(2) = 8.46268090 in

R3 = 2.8051768(2) = 5.610353611 in

Note that the 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 x0

2 1

+ 0 =

50°+ 45° = 61.666°

x f x0

3

Using this value of and the link lengths given above, find the output value for . This can be done

graphically, or by using the routine fourbar_cr. The value of is

e = 51.02815161

The corresponding value for y is,

- 269 -

yact =

e 0

51.02815161 30

(y f y0 ) + y0 =

(2)+ 3 = 2.3992

70

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. Use

Chebyshev spacing with three position points. The base length of the linkage must be 2 cm. Use

the 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 + x0 xf x0 cos30° = 1+ 5 5 1 cos30° = 1.26794919243112

2

2

2

2

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

yf = x f2 1 = 24

y0 = x0 2 1= 0

y1 = x12 1 = 0.60769515458674

y2 = x22 1= 8

y3 = x32 1= 21.39230484541327

Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. The

linkage angles are:

0 = 30°; = 60°

and

0 = 45°; = 90°

The precision points in terms of are:

- 270 -

1 = x1 x0 + 0 = 1.26794919243112 1 60° + 30° = 34.01923788646684°

4

xf x0

2 =

x2 x0 + = 3 1 60°+ 30° = 60°

0

4

xf x0

3 =

x3 x0 + = 4.73205080756888 1 60° + 30° = 85.98076211353316°

0

4

xf x0

Similarly,

y1 y0

0.60769515458674 0

+ 0 =

90°+ 45= 47.278856829°

y f y0

24

y y

8 0

2 = 2 0 + 0 =

90°+ 45 = 75°

y f y0

24

y y

3 = 3 0 + 0 = 21.39230484541327 0 90° + 45= 125.221143170°

y f y0

24

1 =

Using the matrix solution procedure,

1

z1 1 cos 1 cos1 cos(1 1 )

z2 = 1 cos 2 cos 2 cos( 2 2 )

z3 1 cos 3 cos 3 cos( 3 3 )

1

1 0.6784308203 -0.8288497687 0.973340766 1.1947633778

= 1 0.2588190451 -0.5000000000 0.965925826 = 0.6332388583

1 -0.5767338180 -0.0700914163

0.774498853

0.7854636563

and

r2 =

1

1

=

= 1.57918293

z2 0.6332388583

r4 =

1

1

=

= 1.27313338

z3 0.7854636563

and

r3 = 1+ r22 + r42 2r2 r4 z1

= 1+ (1.579182)2 + 1.2731332 2(1.579182)(1.273133)(1.1947633) = 0.557242

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

links become

- 271 -