2 limit position

7/30/2019 2 Limit Position

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2

Minimum Transmission Angle at 2=00

Maximum Transmission Angle at 2=1800

The limits of link lengths are theoretical. However, additional considerations

must be given regarding the quality of the transmission of motion between

the links for any practical design. One common consideration is the

transmission angle of the mechanism.

The ideal value of trans-

mission angle is 90. In

this instance, the line of

action of the interactiveforce between links 3and 4 matches the line of action of

motion of the kinematic pair between

the two links. This ideal value cannot

be maintained in a moving mechanism.However, it is generally acceptable if

transmission angles fall within the

range, 45 < < 135

Values outside this range result in inefficient transmission of motion.


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Maximum Transmission Angle at 2=2700

Minimum Transmission Angle at 2=900


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Figure 2.18 Configuration of a four-bar mechanism with a small transmission angle.

The figure shows the configuration of a

four-bar mechanism having a small

transmission angle, outside theacceptable range.

Driving torque M12 is applied to link 2. As shown in Figure (b), the direction of

force transmitted from link 3 on link 4 results in a small torqueM14 about point

O4 , but a high bearing force at the same point. For transmission angles close to

0 or 180, there is a tendency for a mechanism to bind.


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Applying the cosine law for the triangle

formed by points O2O4B,

Using the cosine law again for the triangle

formed by points DO4B,

22122212 cos24 rrrrrBO

cos2 432

4

2

3

2

4rrrrrBO

43

22

4

2

31

2cos 4

rr

rrr BO

The transmission angle for a particularangular displacement of link 2.

For a four-bar mechanism, extreme values

of the transmission angle occur when

max

min

0

2

0

2 180,0 and


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Mechanics of MachinesCleghorn

6

Figure 2.19 Transmission angle of a four-bar mechanism.


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Figure 2.12 Limit positions of a slider crank mechanism: (a) mechanism, (b) geometry of first limit position, (c) geometry of second limit position.

Limit Position

Limit Position


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Limit PositionLimit Position

122 180

32

11

1sin

rr

r

23

11

2sin

rr

r

2

1

2

321 rrrs

2

1

2

232

rrrs

21

2

13

2

1

2

3221 rrrrrrsssstroke

The time ratio as the time for the

slider to move in one direction

between the limit positions, divided

by the time it takes to move in the

opposite direction between the

same limit positions. The time ratiois dimensionless.


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The time ratio of the mechanism is

The average velocity of the slider,2

2

2

1

2

t

t

TR

2

2

21

1

,4

ss

t

strokev

leftavgTo the left,

To the right, 2

2

21,4 22

ss

t

strokev

leftavg

Between the limit positions, The time taken for the slider to move

To the left To the right

2

2

1

t

2

2

2

2

t


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Example : For the slider crank mechanism with dimensions and crank angular speed;

r1=2.0 cm, r2=3.5cm and r3=10cm ;the angular velocity, 20.00 rads-1 CCW. Determine

the average slider velocity to the left and right.

01

32

11

152.8

105.3

2sinsin

rr

r

01

23

11

292.17

5.310

2sinsin

rr

r

rad31.34.18952.892.17180180 00122

cmrrrrrrsssstroke 17.7212132123221

1

2

2

21

1

,4 35.43

2031.3

17.7

cms

ss

t

strokev

leftavg

1

2

2

21

1

,4 14.482

cmssst

strokevleftavg


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11

1321

2

32

2

1

2

4cos2 rrrrrrr

141

2

4

2

1

2

32cos2 rrrrrr

From triangle (b)

321

2

4

2

32

2

111

2cos

rrr

rrrr

41

2

4

2

32

2

111

2cos

rr

rrrr

The amplitude of motion of the rocker may be expressed as, 214

(b)(c)


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22312

23

2

1

2

4cos2 rrrrrrr

241

2

4

2

1

2

23cos2 rrrrrr

From triangle (c)

231

2

4

2

23

2

11

22cos rrr

rrrr

41

2

4

2

23

2

11

2 2cos rr

rrrr

The amplitude of motion of the rocker maybe expressed as, 214

(c)

2

2

4

,4

CCWavg

To the left,To the right,

2

2

4

,4 2

CWavg


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Figure 2.15 Limit position of a quick-return mechanism.

1

21cosrr

421 OO

rr 22 2PO

rr Where,

It shows a quick-return mechanism in

one of its two limit positions. We willassume that link 2 is driven at a

constant rate in the counterclockwise

direction. From Figure points O2, P2,

and O4 form a right triangle. Therefore


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Figure 2.15 Limit position of a quick-return mechanism.

Due to symmetry about the vertical axis

,the rotation of link 2 required to move link

6 to the right, from the second limit

position to the first, is 2.

The time required to execute this motion is,

2

1

21

2

1

cos22

rr

t

the time required to move link 6 to the left, from

the first limit position to the second, is

2

1

21

2

2

cos2222

rr

t

The time ratio of the motions of link 6 to the

right and to the left is

2

1

t

tTR

2 limit position

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