Mécanismes à barres trajectoires et conception .pdf



Nom original: Mécanismes à barres - trajectoires et conception.pdfTitre: Microsoft Word - Solns, Ch6 (12-28-03).docAuteur: 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 cos 1
2 = a1 + a2 cos 2
or,
0.0289 = a1 + a2 cos(0.17) = a1 + a2 (0.985)
0.270 = a1 + a2 cos(0.52) = a1 + a2 (0.868)
Subtract the first equation from the second, and solve for a2
0.241 = a2 ( 0.117)
and
a2 = 2.060
Now back substituting into the first equation,
a1 = 0.0289 a2 (0.985) = 0.0289 + 2.060(0.985) = 2.058.
The error is given by

- 259 -

e = ideal act = 2 (2.058 2.060cos )
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,

- 260 -

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 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
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 = 2 13 = 2(0.1534)3 = 0.0072

2 = 2 32 = 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

- 262 -

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,

- 263 -

1 = 2a1 + a2 sin 1 = 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 -

cos 1 1 0.95976969417465
cos 2 = 1 0.84592449923107
cos 3 1 0.59451456313420

1 cos 1
A = 1 cos 2
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 cos 2

x3 1 cos 3

cos 1 1 cos( 1 1) 1.05801273775729


cos 2 cos( 2 2) = -0.00195486778387
cos 3
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 cos 1
A = 1 cos 2

1 cos 3

cos 1 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 cos 1 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 cos 1 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 -


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