Solution manual for mechanisms and machines kinematics dynamics and synthesis 1st edition by stanisic

Problem 1.2 If you fix freeze the P1 joint between bodies 2 and 1 then 3 and 4 cannot move... To see this consider the four bar mechanism of links 1, 2, 3 and 4 which moves with 1 degree

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Chapter 1

Introduction - Solutions

Problem 1.1

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drawing 015.soltn

00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 0000000

00000000 00000000 11111111 11111111

1

1 2

3

4

5

6

P1

N = 6

P1= 7

P2= 0

F = 3(N − 1) − 2P1− P2= 3(6 − 1) − 2(7) − 0 = 1

If you fix (freeze) the P1 joint between bodies 2 and 1 then 3 cannot roll on 4, so 3 and 4 cannot move and

M = F in this mechanism If 4 cannot move, neither 5 or 6 will move Since only one P1 was eliminated to

make this mechanism into a structure we have M = 1 M and F in this mechanism

Problem 1.2

If you fix (freeze) the P1 joint between bodies 2 and 1 then 3 and 4 cannot move If 4 cannot move, neither 5

or 6 will move Since only one P1was eliminated to make this mechanism into a structure we have M = 1 M

and F in this mechanism

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fig001.soltn 000000000 000000000 111111111 111111111

00 00 00 00

11 11 11 11

1

1 2

3

4

5 6

P1, P1

P1

P1= 7

P2= 0

F = 3(N − 1) − 2P1− P2= 3(6 − 1) − 2(7) − 0 = 1

Problem 1.3

fig 005.soltn

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0

1

0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000

1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111

1

2

3 4

5

P1

P2

N = 5

P1= 5

P2= 1

F = 3(N − 1) − 2P1− P2= 3(5 − 1) − 2(5) − 1 = 1

If you freeze the P1joint between 5 and 1, 2 cannot rotate without breaking the slipping contact between 2

and 5 If 2 and 5 do not move, then 3 and 4 are fixed in place Thus the mechanism has become a structure

and M = 1 M = F in this mechanism

Problem 1.4

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

fig008a.soltn 000 1

1

2

3

P1

P2

N = 3

P1= 2

P2= 1

F = 3(N − 1) − 2P1− P2= 3(3 − 1) − 2(2) − 1 = 1

If you freeze the P1 joint between 1 and 3, then 2 can no longer move and maintain a rolling contact with 1 So

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00000000000 00000000000 00000000000 00000000000 00000000000 00000000000 00000000000 00000000000 00000000000 00000000000

11111111111 11111111111 11111111111 11111111111 11111111111 11111111111 11111111111 11111111111 11111111111 11111111111

fig151.soltn

00000

11111

1

1 2

3

4

P1

P2

N = 4

P1= 3

P2= 2

F = 3(N − 1) − 2P1− P2

= 3(4 − 1) − 2(3) − 2 = 1

If we freeze the P1 joint between 1 and 2, then 3 cannot move without breaking (violating) one of the two

slipping contacts between 4 and 1 Hence this makes the mechanism into a structure and M = 1 M = F in

this mechanism

Problem 1.6

00000000000000000

11111111111111111 11111111111111111

0000

1111 1111

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fig093a.soltn

1

1 2

3

4

P1

P2

N = 4

P1= 3

P2= 1

F = 3(N − 1) − 2P1− P2= 3(4 − 1) − 2(3) − 1 = 2

If we freeze the P1 joint between 4 and 1, 3 can still spin and roll against 2, which slips against 1 If we then

freeze the P1 joint between 3 and 4, 2 can no longer roll on 3 and maintain contact with 1 so the system is a

structure and M = 2 M = F for this mechanism

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000 000

111 111

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

fig107a.soltn

1 1

2

3

P1

P2

N = 3

P1= 2

P2= 1

F = 3(N − 1) − 2P1− P2= 3(3 − 1) − 2(2) − 1 = 1

If you freeze the P1 joint between 3 and 2, 2 can no longer move and the system is a structure so M = 1 and

M = F for this mechanism

Problem 1.8

000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000 000000000000000000000

111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111 111111111111111111111

fig025.soltn

0000 0000 1111 1111

1

1 2

3

4

P1

N = 4

P1= 4

P2= 0

F = 3(N − 1) − 2P1− P2= 3(4 − 1) − 2(4) − 0 = 1

If you freeze the P1 joint between 1 and 2, 3 can no longer roll on 4 and we have a structure, so M = 1 and

M = F in this case

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fig273.soltn 000

000

111 111

1

2 3

4

5

P1

P2

N = 5

P1= 5

P2= 1

F = 3(N − 1) − 2P1− P2= 3(5 − 1) − 2(5) − 1 = 1

If we freeze the P1 joint between 1 and 2, then the four bar formed by links 1, 2, 3 and 4 does not move so 5

can no longer move and M = 1 M = F for this mechanism

Problem 1.10

000

+

fig274.soltn 0000

0000

1111 1111

00000000000 00000000000

11111111111

1

1 2

3

4

5

6

P1

P1, P1

P2

N = 6

P1= 7

P2= 1

F = 3(N − 1) − 2P1− P2= 3(6 − 1) − 2(7) − 1 = 0

This mechanism is already a structure so M = 0 To see this consider the four bar mechanism of links 1, 2, 3

and 4 which moves with 1 degree-of-freedom We see that as this four bar moves, 6 gets get pulled and it slips

(drags) along 1 The problem is that as 6 makes this motion 5 cannot maintain a sliding contact with it So

there is no relative movement between 6 and 5 Consequently the four bar mechanism does not move and the

mechanism does not move M = F for this mechanism

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000000000 000000000 000000000 000000000 000000000 000000000 000000000 000000000

111111111 111111111 111111111 111111111 111111111 111111111 111111111 111111111

fig275.soltn

0000

000 000

111 111

1 1

1

2

P1

N = 4

P1= 5

P2= 0

F = 3(N − 1) − 2P1− P2= 3(4 − 1) − 2(5) − 0 = −1 Link 3 cannot be rotated without 4 breaking its contact with 1, therefore 1, 3 and 4 are a structure, so 2

cannot move either In fact, 2 will either be too long or too short, making this system a statically indeterminate

structure Since no P1 joint needs to be frozen to make this a structure, M = 0 In this mechanism M 6= F

and M > F because the structure is statically indeterminate

Problem 1.12

000000000 000000000 000000000 000000000 000000000 000000000 000000000 000000000

111111111 111111111 111111111 111111111 111111111 111111111 111111111 111111111

fig276.soltn

4

P1

P1= 4

P2= 0

F = 3(N − 1) − 2P1− P2= 3(4 − 1) − 2(4) − 0 = 1 Freezing the P1 joint between 1 and 2 makes the mechanism a structure and so M = 1 M = F

Problem 1.13

0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000

1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111

000

111

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fig277.soltn

000

111

1

2

3

4 P1

P1

P2

N = 4

P1= 3

P2= 2

F = 3(N − 1) − 2P1− P2= 3(4 − 1) − 2(3) − 2 = 1

If you freeze the P1joint between 1 and 2, 4 can no longer move and 3 cannot roll on 2 and maintain a slipping

contact with 1 So M = 1 and M = F

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0000 0000 1111 1111

000

111

000 000 000 111 111 111

fig011a.soltn

1

1 2 3

4

P1

N = 4

P1= 4

P2= 0

F = 3(N − 1) − 2P1− P2= 3(4 − 1) − 2(4) − 0 = 1

If you freeze the P1joint between 4 and 1, then 2 and 3 cannot move without 2 breaking its rolling contact with

4, so the system becomes a structure and M = 1 M = F

Problem 1.15

0000 0000 1111 1111

000 000 111 111

fig011b.soltn1 1

2 3

4

P1

N = 4

P1= 4

P2= 0

F = 3(N − 1) − 2P1− P2= 3(4 − 1) − 2(4) − 0 = 1

If you freeze the P1joint between 4 and 1, then 2 and 3 cannot move without 2 breaking its rolling contact with

4, so the system becomes a structure and M = 1 M = F

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000

111

000 000 000 000

111 111 111 111

0000 0000 1111 1111

fig011c.soltn 1

1 2

3

P1

P2

N = 3

P1= 2

P2= 1

F = 3(N − 1) − 2P1− P2= 3(3 − 1) − 2(2) − 1 = 1

If you freeze the P1joint between 1 and 2, 3 can no longer rotate without violating its slipping contact with 2,

so the mechanism becomes a structure and M = 1 M = F

Problem 1.17

0000 0000 1111 1111

000 000 111 111

fig011d.soltn

1 1

2

3

P1

P2

N = 3

P1= 2

P2= 1

F = 3(N − 1) − 2P1− P2= 3(3 − 1) − 2(2) − 1 = 1

If you freeze the P1joint between 1 and 2, 3 can no longer rotate without violating its slipping contact with 2,

so the mechanism becomes a structure and M = 1 M = F

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fig528a.soltn

1 2

5

N = 5

P1= 5

P2= 0

F = 3(N − 1) − 2P1− P2= 3(5 − 1) − 2(5) − 0 = 2

If you freeze the P1 joint between 1 and 2, then 5 can still roll against 1 while the ”dyad” formed by 3 and 4

will extend or contract to reach the connection point between 4 and 5, so it is still moveable If you now freeze

the P1 joint between 5 and 1, then the system becomes a structure, so M = 2 M = F

Problem 1.19

000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000

111111 111111 111111 111111 111111 111111 111111 111111 111111 111111 111111 111111 111111 111111 111111 111111 111111

000000000000000000000000

111111111111111111111111

fig154.soltn

1

2

3

P1

N = 3

P1= 3

P2= 0

F = 3(N − 1) − 2P1− P2= 3(3 − 1) − 2(3) − 0 = 0

The system is moveable If you push 3 to the left, you see that 2 will move up If you freeze the P1 joint

between 3 and 1, then 2 is now longer moveable, so M = 1 M 6= F , and M > F The explanation for this is

as follows The development of Gruebler’s Criterion assumes that every body in the system has the potential

of a rotational degree-of-freedom But, when the system consists entirely of sliding joints, as in this case, this

rotational degree-of-freedom is not possible Gruebler’s Criterion as developed is not applicable to this system

Problem 1.20

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00000

11111

00000000000000000000000

11111111111111111111111

fig153.soltn

1

2

3

P1

P2

N = 3

P1= 2

P2= 1

F = 3(N − 1) − 2P1− P2= 3(3 − 1) − 2(2) − 1 = 1

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00 00 00 00 00 00 00 00

11 11 11 11 11 11 11 11

0000000000000000 0000000000000000 1111111111111111 1111111111111111

000000000000000 000000000000000 111111111111111 111111111111111

+ +

fig002.soltn 1

1

2

3

4

5

6

?

P1

P1, P1

The questioned joints are point contacts, so they can only of rolling (P1) or slipping (P2) Not counting those

two joints we have, N = 6, P1= 5 and P2= 0, so

F = 3(N − 1) − 2P1− P2= 3(6 − 1) − 2(5) − 0 = 5

In order that the mechanism have F = 1, 4 more degrees of freedom must be removed, which means that both

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Give all the kinematic inversions of the mechanism shown in figure 1.31 a.).

279a.soltn

00 00 00 00 00

11 11 11 11 11

0 0 0 0 0

1 1 1 1 1

0 0 0 0 0

1 1 1 1 1

0 0 0 0 0

1 1 1 1 1

0000

000 000

111 111

00 00 00 00 00

11 11 11 11 11

0 0 0 0

1 1 1 1

1

1 1

1

1 2

2

2 2

2

3

3 3

3

4

4 4

5 5

The fifth kinematic inversion is the mechanism as shown in figure 1.31 No need to redraw it

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(F) and mobility (M) of that mechanism.

fig278.soltn

0000

0000 0000

1111 1111

1

2

3

4 5

5 6

P1

N = 6

P1= 6

P2= 0

F = 3(N − 1) − 2P1− P2= 3(6 − 1) − 2(6) − 0 = 3

If we freeze the P1 joint between 1 and 6, 6 is immobilized, but the rest of the mechanism still moves If we

continue and freeze the P1 joint between 1 and 2, 2 is immobilized, but 1, 3 and 4 continue to be moveable

If we then freeze the P1 joint between 1 and 2, then 3 and 4 are also immobilized and the system becomes a

structure Thus M = 3 and M = F

Problem 1.24

The figure below shows a pair of gears in mesh The smaller gear is referred to as the “pinion” and the larger

gear is referred to as the “gear.” Typically, the pinion is the driver and the gear is driven Since the pinion is

smaller than the gear, the gear rotates slower than the pinion, but the torque is increased Generally speaking,

that is the purpose of a gear pair, to reduce speed and to increase torque This is because most prime movers

(motors, engines etc.) run at high speeds but produce relatively low torques You will see all this later when

we study gears

What type of joint must exist between these two gear teeth so that the system is movable, with one dof?

A B

fig120a.soltn

000 000 000 111 111 111 0000

0000 0000 1111 1111 1111

?

P1

If we neglect the contact in question, we have N = 3, P1= 2, P2= 0 and then

F = 3(N − 1) − 2P1− P2= 3(3 − 1) − 2(2) − 0 = 2

so in order that F = 1, the contact between a pair of gear teeth must be a slipping contact

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