solutions instructor manual chapter 12 dynamic force analysis planar

8/11/2019 Solutions Instructor Manual Chapter 12 Dynamic Force Analysis Planar

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Chapter 12Dynamic Force Analysis (Planar)

1

12.1 The steel bell crank shown in the figure is used as an oscillating cam follower. Using

7.83 Mg/ 3m for the density of steel, find the mass moment of inertia of the lever aboutan axis through O.

For the vertical arm, using Appendix Table 5,

( ) ( ) ( )( )318.75 mm 93.75 mm 9.375 mm 7.83 Mg/m 129 g= = =m wht

( ) ( ) ( ) ( )2 22 2 2212 129 g/9810 mm/s 18.75 mm 93.75 mm 12 10.02 g mm s+ = + = = G m a cI

( ) ( )22 22 210.02 g mm s 129 g/98.10 mm/s 37.5 mm 28.51 g mm sO GI I md = + = = + For the horizontal arm, using Appendix Table 5,

( ) ( ) ( )( )39.375 mm 18.75 mm 150 mm 7.83 Mg/m 206 g= = =m wht

( ) ( ) ( ) ( )2 22 2 2212 206 g/9810 mm/s 18.75 mm 150 mm 12 39.99 g mm s=+ + = = G m a cI

( ) ( )22 22 239.99 g mm s 206 g/9810 mm/s 84.375 mm 189.48 g mm s= + = = +O GI I md

For the roller, using Appendix Table 5,( ) ( )( )

22 312.5 mm 12.5 mm 7.83 Mg/m 48 g = = =m r t

( )( )22 2 22 48 g 9810 mm/s 12.5 mm 2 0.382 g mm s= = = GI mr

( ) ( )22 2 22 0.382 g mm s 48 g 9810 mm/s 150 mm 110.47 g mm s= + = = +O GI I md For the compositie lever

( ) ( ) ( )129 g 206 g 48 g 383 g= + + =m

( ) ( ) ( )2 2 2 228.51 g mm s 189.48 g mm s 110.47 g mm s 328.46 g mm s= + + = OI 2

0.33 kg mm s=

Ans.

1Unless stated otherwise, solve all problems without friction and without gravitational loads.


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12.2 A 5- by 50- by 300-mm steel bar has two round steel disks, each 50 mm in diameter and20 mm long, welded to one end as shown. A small hole is drilled 25 mm from the other

end. The density of steel is 7.80 Mg/ 3m . Find the mass moment of inertia of thisweldment about an axis through the hole.

For the rectangular bar, using Appendix Table 5,( )( )( )( )30.050 m 0.300 m 0.005 m 7.80 Mg/m 0.585 kgm wht = = =

( ) ( ) ( ) ( )2 22 2 212 0.585 kg 0.050 m 0.300 m 12 0.004 509 kg mGI m a c = + = + =

( ) ( )22 2 20.004 509 kg m 0.585 kg 0.125 m 0.013 650 kg m

O GI I md= + = + =

For the two circular disks, using Appendix Table 5,

( ) ( ) ( )22 32 2 0.025 m 0.020 m 7.80 Mg/m 0.613 kgm r t = = =

( ) ( )22 22 0.613 kg 0.025 m 2 0.000 191 kg mGI mr= = =

( ) ( )

22 2 20.000 191 kg m 0.613 kg 0.250 m 0.038 480 kg mO G

I I md= + = + =

For the compositie lever

( ) ( )0.585 kg 0.613 kg 1.198 kgm= + =

( ) ( )2 2 20.013 650 kg m 0.038 480 kg m 0.052 130 kg mOI = + = Ans.

12.3 Find the external torque which must be applied to link 2 of the four-bar linkage shown todrive it at the given velocity.

The DAlembert inertia forces and offsets are:

22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( ) ( )3

2 2

3 3

0.3214 kg 9.81 m/s 1896 +225 m/s

62.1 7.37 kg 63.56 kg 186.77

=

=

= =

Gm

i j

f A

i j

( ) ( )4

2 2

4 4

0.3214 kg 9.81 m/s 684 +225 m/s

22.4 7.37 kg 26.33 kg 198.21

=

=

= =

Gm

i j

f A

i j

( )( )33 3

2 217.47 g cm s 4 950 rad/s

86.48 kg cm 0.8648 kg m

=

=

= =

GIt

k

k k

( )( )44 4

2 212.91 g cm s 8 900 rad/s

114.94 kg cm

=

=

=

GIt

k

k

( ) ( ) ( ) ( )3 3 3 4 4 4

0.8648 kg m 63.56 kg 13.6 mm, 1.149 kg m 26.33 kg 43= = = = = =h t f h t f

Next, the free body diagrams are drawn. Since the lines of action for the forces on thefree body diagrams can not be discovered from two- and three-force member concepts,

the force 34F is divided into radial and transverse components. (Notice that it is totally

coincidental that the reaction components are also exactly aligned with the radial andtransverse axes of link 3.)


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4 4 4 44 4 34

t

O G O BO= + + =M R f t R F 0 ( ) ( ) ( ) ( )( )

34

45 60 mm 22.4 7.37 kg 1.149 kg m 90 120 mm 0.800 0.600 + + + + + =t

Fi j i j k i j i j 0

( ) ( ) ( ) 34 1.676 kg m 1.149 kg m 0.15 m = + +tF 0k k k

34 19.98 kg=t

F

3 3 3 43

r

A G A BA= + + =M R f t R F 0

( ) ( ) ( ) ( ) ( ) 43 80 60 mm 62.1 7.37 kg 0.8648 kg m 160 120 mm 0.600 0.800+ + + + =r


( ) ( ) ( ) 43 3.1364 kg m 0.8648 kg m 0.2 m = + +rF 0k k k

43 11.36 kg= r

F 34 9 21.34 kg 23.15 kg 67.26= + = F i j

43 3 23= + + =F F f F 0 23 72.18 29 kg 77.63 kg 21.82= + = F i j

2 2 32 120

O AO= + =M R F M 12 21.47 N m or 2.167 kg m= M k Ans.

12.4 Crank 2 of the four-bar linkage shown in the figure is balanced. For the given angularvelocity of link 2, find the forces acting at each joint and the external torque that must beapplied to link 2.


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2

0h t f= =

( ) ( )3

2 2

3 3

1.2 kg 9.81 m/s 948 +78.6 m/s

115.96 9.61 kg 116.36 kg 4.74

=

=

= =

Gm

i j

f A

i j

( ) ( )4

2 2

4 4

3 kg 9.81 m/s 240 633 m/s

73.39 193.58 kg 207 kg 69.24

=

=

= + =

Gm

i j

f A

i j


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184

( )( )33 3

2 20.0066 N m s 6 500 rad/s

42.9 N m

GI=

=

=

t

k

k

( )( )44 4

2 20.059 N m s 240 rad/s

14.16 N m

GI=

=

=

t

k

k

( ) ( )3 3 3 42.9 N m 116.36 9.81 N 37.58 mm= = =h t f ,

( ) ( )4 4 4 14.16 N m 207 9.81 N 6.97 mm= = =h t f


the force 34F is divided into radial and transverse components.

4 4 4 44 4 34

t

O G O BO= + + =M R f t R F 0

( ) ( ) ( ) ( )( )34

36.52 93.1 mm 720 1899 N 14.16 N m 73 186 mm 0.931 0.365+ + + + + =t


( ) ( ) ( ) 34 2.32 N m 14.16 N m 0.2 m = + + t

F 0k k k

34 82.4 N=t

F

3 3 3 43 43

t r

A G A BA BA= + + + =M R f t R F R F 0

( ) ( ) ( ) ( ) (

( ) ( ) 43

181.35 110.77 mm 1157 97.9 N 44.52 N m 362.7 221.55 mm 75.65 31.1

362.7 221.55 mm 0.365 0.931

+ + + + +

+ + =rF

i j i j k i j i

i j i j 0

( ) ( ) ( ) ( ) 43 148.2 N m 44.52 N m 18.66 N m 297.175 mm+ = + + rF 0k k k k

43 280.35 N= r

F 34 26.7 293.7 N 293.7 N 95.06= = F i j Ans.

34 4 14= + + =F F f F 0 14 720.9 1668.75 N 1815.6 N 113.27= = F i j Ans.

43 3 23= + + =F F f F 0 23 1183.7 195.8 N 1201.5 N 170.57= = F i j Ans.

32 12= + =F F F 0 12 1183.7 195.8 N 1201.5 N 9.43= + = F i j Ans.

2 2 32 120O AO= + =M R F M 12 49.6 N m= M k Ans.


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12.5 For the angular velocity of crank 2 given in the figure, find the reactions at each joint andthe external torque to be applied to the crank.


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( )( )2 233 3

1.54 kg 9.81 m/s 2385 1487 m/s

374.4 233.4 kg 441.7 kg 31.95

Gm

=

=

= + =

i j

f A

i j

( ) ( )4

2 2

4 4

1.298 kg 9.81 m/s 2394 m/s

316.75 kg 316.75 kg 0.00

=

=

= =

Gm

i

f A

i

43,330.7 N 31.95= 3107.3 N 0.00=

( ) ( )33 3

2 20.012 N m s 7 670 rad/s

92 N m

=

=

=

GIt

k

k

44 4G

I= =t 0

( ) ( )3 33 92 N m 43,330.7 N 21.23 mm= == t fh , 4 4 4 0h t f= =

Next, the free body diagrams are drawn and the solution proceeds.

3 3 3 4 14A G A BA BA= + + + =M R f t R f R F 0

( ) ( ) ( )

( ) ( ) ( ) ( ) 14

110.725 19.875 mm 3675.7 2291.75 N 92 N m

295.275 53.025 mm 316.75 9.81 N 295.275 53.025 mm

+ +

+ + =F

i j i j k

i j i i j j 0

( ) ( ) ( ) ( ) 14 331.99 N m 92 N m 167 N m 0.295 m+ = + + F 0k k k k

14 1975.8 N= F 141975.8 N 1975.8 N 90= = F j Ans.

4 14 34= + + =F f F F 0 34 3107.3 1975.8 lb 3675.7 N 147.50= + = F i j Ans.

43 3 23= + + =F F f F 0 23 6777.3 315.95 N 6788.25 N 177.33= = F i j Ans.32 12

= + =F F F 0 12 6777.3 315.95 N 6788.25 N 177.33= = F i j Ans.

2 2 32 120O AO= + =M R F M 12 348 N m= M k Ans.


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12.6 The figure shows a slider-crank mechanism with an external force BF applied to the

piston. For the given crank velocity, find all the reaction forces in the joints and thecrank torque.


( )( )2 222 2

0.43 kg 9.81 m/s 685.8 396 m/s

30.4 17.7 kg 35.4 kg 30.00

+= =

= =

Gm

i j

f A

i j

22 2G

I= =t 0

2 2 2 0h t f= =

( )( )2 233 3

1.59 kg 9.81 m/s 1708.8 680.1 m/s

276.9 110.2 kg 298 kg 21.70

+

=

=

= =

Gm

i j

f A

i j

( ) ( )4

2 2

4 4

1.13 kg 9.81 m/s 1884 m/s

217 kg 217 kg 0.00

=

=

= =

Gm

i

f A

i

( )( )33 3

2 20.012 Nms 3 090 rad/s

37 N m

GI=

=

=

t

k

k

44 4G

I= =t 0

( ) ( )3 33 37 N m 298 9.81 N 12.65 mm= == t fh , 4 4 4 0h t f= =

Next, the free body diagrams are drawn and the solution proceeds.

3 3 3 4 14A G A BA BA B BA= + + + + =M R f t R f R F R F 0

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 14

0.087 0.011 m 2716.4 1081 N 37 N m 0.298 0.0375 m 2128.7 N

0.298 0.0375 m 3560 N 0.298 0.0375 m

+ + + +

+ + + + =F

i j i j k i j i

i j i i j j 0

( ) ( ) ( ) ( ) ( ) 14 123.93 N m 37 N m 79.83 N m 133.5 N m 0.298 m+ = + + + F 0k k k k k

14 111.6 N=F 14 111.6 N 111.6 N 90= = F j Ans.

4 14 34B= + + + =F f F F F 0 34 1388 111.6 N 1393 N 4.95= = F i j Ans.

43 3 23= + + =F F f F 0 23 1366 975 N 1678 N 144.50 + = = i jF Ans.

32 12= + =F F F 0 12 1664 1148 N 2020 N 145.40 + = = i jF Ans.

2 2 32 120O AO= + =M R F M 12 12.2 N m= M k Ans.


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188

( ) ( ) ( ) 34 831 N m 88 N m 0.800 mtF = + + 0k k k

34 1 149 Nt

F =

3 3 3 43 43

t r


( ) ( ) ( ) ( ) ( )( ) ( ) 43

0.631 0.154 m 1 210 6 227 N 360 N m 1.498 0.059 m 259 1 119 N

1.498 0.059 m 0.974 0.226 rF

+ + + +

+ =

i j i j k i j i j

i j i j 0

( ) ( ) ( ) ( ) 43 3 745 N m 360 N m 1 661 N m 0.395 inr

F+ = + + 0k k k k

43 4 365 Nr

F = 34

3 993 2 104 N 4 513 N 152.21= = i jF Ans.

34 4 14= + + =F F f F 0 14 3 993 28 N 3 993 N 0.40= = F i j Ans.

43 3 23= + + =F F f F 0 23 2 783 4 123 N 4 974 N 124.02 + = = i jF Ans.

32 12= + =F F F 0 12 2 783 4 123 N 4 974 N 124.02 + = = i jF Ans.

2 2 32 12 0O AO= + =

M R F M 12

1 411 N m

= M k Ans.

12.8 Solve Problem 12.7 with an additional external force 12 0 kND = F acting at pointD.

The method of superposition is used for the solution. The force components of ProblemP12.7 are denoted with primes and the additional force increments with double primes.The figure here shows the incremental forces only.

4 4 4 34O DO D BO = + =M R F R F 0

( ) ( ) ( )( ) 34 0.162 0.366 m 12 000 N 0.779 0.180 m 0.999 0.040 F+ + + + =i j i i j i j 0

( ) ( )34 4 386 N m 0.211 m F= + 0k k

34 20 772 NF =

34 20 756 822 N 20 772 N 177.73= + = i jF

34 24 749 1 282 N 24 782 N 182.97= = i jF Ans.

14 8 756 822 N 8 794 N 5.36 = = F i j 14

12 749 850 N 12 777 N 3.81= = i jF Ans.

23 20 756 822 N 20 772 N 177.73= + = i jF 23

23 539 4 945 N 24 053 N 168.14 + = = i jF Ans.

12 20 756 822 N 20 772 N 177.73= + = i jF 12

23 539 4 945 N 24 053 N 168.14 + = = i jF Ans.

125 121 N m = M k 12

6 532 N m= M k Ans.


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189

12.9 Make a complete kinematic and dynamic analysis of the four-bar linkage of Problem 12.7

using the same data, but with 2 2 2170 , 12 rad/s ccw, 0, = = = and an external

force D 8.94 64.3 kN= F acting at pointD.

Kinematic Analysis:

( ) ( )22 12 rad/s 0.295 0.052 m 0.625 3.545 m/s 3.600 m/s 100

A AO= = +

= =

V R k i j

i j

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

4

3 4

3 3 4 4

3 4

0.625 3.545 m/s rad/s 1.304 0.740 m rad/s 0.109 0.793 m

0.625 3.545 m/s 0.740 1.304 m 0.793 0.109 m

B A BA BO

+ + = +

= + + = +

= + =

= i j k i j k i j

i j i j i j

V V R R

33.020 rad/s= k 4

3.607 rad/s= k Ans.

2.859 0.393 m/s 2.886 m/s 172.16B = + = V i j

( ) ( )2 2 222

2

2 2

12 rad/s 0.295 0.052 m

42.543 7.502 m/s 43.200 m/s 10

A AO AO= + = +

= =

A R R i j

i j

( ) ( ) ( ) ( )

( ) ( ) ( )

( )

4 4

2 23 3 4 4

2 2 23

2 24

23

42.543 7.502 m/s 11.893 6.749 m/s rad/s 1.304 0.740 m

1.419 10.311 m/s rad/s 0.109 0.793 m

32.069 3.940 m/s 0.740 1.

B A BA BA BO BO

+

= + = +

= + + +

= +

= + +

A A R R R R

i j i j k i j

i j k i j

i j i( ) ( )3 4 4 304 m 0.793 0.109 m = +j i j

2

30.390 rad/s= k 24

40.816 rad/s= k Ans.

( ) ( ) ( ) ( )3 3 3

23 3

2 2 2 42.543 7.502 m/s 4.149 4.234 m/s 0.390 rad/s 0.455 0.464 m

G A G A G A= +

= + + +

A A R R

i j i j k i j

3

2 2 38.575 11.913 m/s 40.373 m/s 17.16G = = A i j Ans.

( ) ( ) ( )4 4 4 4 4

24 4

2 2 0.932 5.780 m/s 40.816 rad/s 0.072 0.444 m

O OG G G= +

= + +

A R R

i j k i j

4

2 2 19.054 2.841 m/s 19.265 m/s 8.48G = = A i j Ans.

Dynamic Analysis:


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =


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190

( ) ( )3

2

3 3

65.8 kg 38.575 11.913 m/s

2 538 784 N 2 657 N 162.84

Gm

=

=

= + =

i j

f A

i j

( )( )4

2

4 4

21.8 kg 19.054 2.841 m/s

415 62 N 420 N 171.52

Gm

=

=

= + =

i j

f A

i j

( ) ( )33 3

2 24.200 kg m 0.390 rad/s

1.638 N m

GI=

=

=

t

k

k

( )( )44 4

2 20.51 kg m 40.816 rad/s

21 N m

GI=

=

=

t

k

k

( ) ( )3 33 1.638 N m 2 657 N 0.001 mt fh = == , ( ) ( )4 4 4 21 N m 420 N 0.050 mh t f= = =



4 4 4 4 44 4 34

t

O G O DO D BO= + + + =M R f t R F R F 0

( ) ( ) ( ) ( ) ( )

( ) ( )34

0.072 0.444 m 415 62 N 21 N m 0.284 0.282 m 3 877 8 056 N

0.109 0.793 m 0.991 0.136 tF

+ + + + +

+ + =

+

i j i j k i j i j

i j i j 0

( ) ( ) ( ) ( ) 34 180 N m 21 N m 3 379 N m 0.800 mt

F = + + + 0k k k k

34 3 972 Nt

F =


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191

3 3 3 43 43

t r


( ) ( ) ( ) ( ) ( )

( ) ( ) 43

0.455 0.464 m 2 538 784 N 1.638 N m 1.304 0.740 m 3 935 542 N

1.304 0.740 m 0.136 0.991r

F

+ + + + +

+ + =

i j i j k i j i j

i j i j 0

( ) ( ) ( ) ( ) 43 1 534 N m 2 N m 3 618 N m 1.192 in rF+ = + + 0k k k k

43 1 747 Nr

F = 34

4 173 1 189 N 4 339 N 164.10= = i jF Ans.

34 4 14D= + + + =F F f F F 0 14 711 6 929 N 6 966 N 84.14= = F i j Ans.

43 3 23= + + =F F f F 0 23 1 635 1 972 N 2 562 N 129.65 = = iF Ans.

32 12= + =F F F 0 12 1 635 1 972 N 2 562 N 129.65 = = iF Ans.

2 2 32 120O AO= + =M R F M 12 668 N m= M k Ans.

12.10 Repeat Problem 12.9 using 2 2 2200 , 12 rad/s ccw, 0, = = = and an external force

8.49 45 kNC

= F acting at point C.

Kinematic Analysis:

( ) ( )22

12 rad/s 0.282 0.103 m

1.231 3.383 m/s 3.600 m/s 70

A AO= =

= =

V R k i j

i j

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

4

3 4

3 3 4 4

3 4

1.231 3.383 m/s rad/s 1.198 0.902 m rad/s 0.016 0.799 m

1.231 3.383 m/s 0.902 1.198 m 0.799 0.016 m

B A BA BO

+ + = +

= + + = +

= + =

= i j k i j k i j

i j i j i j

V V R R

3

2.846 rad/s=

k 4

1.671 rad/s=

k Ans. 1.336 0.027 m/s 1.337 m/s 178.84

B= + = V i j

( ) ( )2 2 2

22

2

2 2

12 rad/s 0.282 0.103 m

40.595 14.775 m/s 43.200 m/s 20

A AO AO= + =

= + =

A R R i j

i j

( ) ( ) ( ) ( )

( ) ( ) ( )

( )

4 4

2 23 3 4 4

2 2 23

2 24

2

3

40.595 14.775 m/s 9.703 7.306 m/s rad/s 1.198 0.902 m

0.045 2.233 m/s rad/s 0.016 0.799 m

30.937 9.702 m/s 0.902 1.1

B A BA BA BO BO

+

= + = +

= + + + +

= +

= + + +

A A R R R R

i j i j k i j

i j k i j

i j i

( ) ( )3 4 4

98 m 0.799 0.016 m = +j i j

2

38.760 rad/s= k 24


( ) ( ) ( ) ( )3 3 3

2

3 3

2 2 2 40.595 14.775 m/s 3.169 4.204 m/s 8.760 rad/s 0.391 0.519 m

G A G A G A= +

= + + + +

A A R R

i j i j k i j

3

2 2 41.972 7.146 m/s 42.576 m/s 9.66G = + = A i j Ans.


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192

( ) ( ) ( )4 4 4 4 4

2

4 4

2 2 0.343 1.209 m/s 48.558 rad/s 0.123 0.433 m

G G O G O= +

= + +

A R R

i j k i j

4

2 2 21.365 4.754 m/s 21.887 m/s 12.54G = + = A i j Ans.

Dynamic Analysis:


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( ) ( )3

2

3 3

65.8 kg 41.972 7.146 m/s

2 762 470 N 2 802 N 170.34

Gm

+

=

=

= =

i j

f A

i j

( )( )4

2

4 4

21.8 kg 21.365 4.754 m/s

466 104 N 477 N 167.46

Gm

+

=

=

= =

i j

f A

i j

( ) ( )33 3

2 24.200 kg m 8.760 rad/s

37 N m

GI=

=

=

t

k

k

( )( )44 4

2 20.51 kg m 48.558 rad/s

25 N m

GI=

=

=

t

k

k

( ) ( )3 33 37 N m 2 802 N 0.013 mt fh = == , ( ) ( )4 4 4 25 N m 477 N 0.052 mh t f= = =




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193

4 4 4 44 4 34

t

O G O BO= + + =M R f t R F 0

( ) ( ) ( ) ( ) ( )34

0.123 0.433 m 466 104 N 25 N m 0.016 0.799 m 0.999 0.020t

F + + + + =i j i j k i j i j 0

( ) ( ) ( ) 34 214 N m 25 N m 0.800 mtF = + + 0k k k

34 299 NtF =

3 3 3 43 43

t r

A G A CA C BA BA= + + + + =M R f t R F R F R F 0 ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 43

0.391 0.519 m 2 762 470 N 37 N m 0.291 0.799 m 6 003 6 003 N

1.198 0.902 m 299 6 N 1.198 0.902 m 0.020 0.999r

F

+ + + + +

+ + + + + =

i j i j k i j i j

i j i j i j i j 0

( ) ( ) ( ) ( ) ( ) 43 1 250 N m 37 N m 3 050 N m 277 N m 1.179 inrF+ + = + + 0k k k k k

43 1 260 Nr

F = 34 274 1 266 N 1 295 N 77.80= = F i j Ans.

34 4 14= + + =F F f F 0 14 192 1 370 N 1 383 N 82.01= + = F i j Ans.

43 3 23C= + + + =F F f F F 0 23 2 968 6 799 N 7 419 N 113.58 = = iF Ans.

32 12= + =F F F 0 12 2 968 6 799 N 7 419 N 113.58 = = iF Ans.

2 2 32 120

O AO= + =M R F M 12 1 612 N m= M k Ans.

12.11 At 2 270 = , 2 18 rad/s ccw = , 2 0 = , a kinematic analysis of the linkage whose

geometry is given in Problem 12.7 gives 3 46.6 = , 4 80.5 = ,2

3 178 rad/s cw = ,

24 256 rad/s cw = ,

3

2112 22.7 m/sG = A ,4

2119 352.5 m/sG = A . An external

force 8.94 64.3 kND = F acts at point D. Make a complete dynamic analysis of the

linkage.


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( ) ( )3

2

3 3

65.8 kg 103.324 43.221 m/s

6 799 2 844 N 7 370 N 157.30

Gm

+

= =

=

= i j

i j

f A

( )( )4

2

4 4

21.8 kg 117.982 15.533 m/s

2 572 339 N 2 594 N 172.50

Gm

=

=

= + =

i j

f A

i j

( ) ( )33 3

2 24.200 kg m 178 rad/s

748 N m

GI=

=

=

t

k

k

( )( )44 4

2 20.51 kg m 256 rad/s

131 N m

GI=

=

=

t

k

k

( ) ( )3 33 748 N m 7 370 N 0.101 mt fh = == , ( ) ( )4 4 4 131 N m 2 594 N 0.050 mh t f= = =

Next, the free body diagrams are drawn. Since the lines of action for the forces on the


14/36

194

free body diagrams can not be discovered from two- and three-force member concepts,


4 4 4 4 44 4 34

t

O G O DO D BO= + + + =M R f t R F R F 0

( ) ( ) ( ) ( ) ( )

( ) ( ) 34

0.059 0.446 m 2 572 339 N 131 N m 0.276 0.290 m 3 877 8 056 N

0.131 0.789 m 0.986 0.164 tF

+ + + + + +

+ + =

i j i j k i j i j

i j i j 0

( ) ( ) ( ) ( )34

1 127 N m 131 N m 3 348 N m 0.800 m tF = + + + 0k k k k

34 2 613 Nt

F =

3 3 3 43 43

t r

A G A BA BA= + + + =M R f t R F R F 0 ( ) ( ) ( ) ( ) ( )

( ) ( ) 43

0.300 0.577 m 6 799 2 844 N 748 N m 1.031 1.089 m 2 578 429 N

1.031 1.089 m 0.164 0.986 rF

+ + +

+ + =

+i j i j k i j i j

i j i j 0

( ) ( ) ( ) ( ) 43 3 070 N m 748 N m 3 250 N m 0.839 inr

F+ = + + 0k k k k

43 677 Nr

F = 34 2 466 1 097 N 2 699 N 156.01= + = i jF Ans.

34 4 14D= + + + =F F f F F 0 14 1 411 9 153 N 9 261 N 98.76= = i jF Ans.

43 3 23= + + =F F f F 0 239 265 1 747 lb 9 428 N 10.68+ = = iF Ans.

32 12= + =F F F 0 12 9 265 1 747 lb 9 428 N 10.68+ = = iF Ans.

2 2 32 120

O AO= + =M R F M 12 2 780 N m= M k Ans.


15/36


16/36

196

4 4 4 44 4 34

t

O G O BO= + + =M R f t R F 0

( ) ( ) ( ) ( ) ( )34

0.004 0.125 m 21 44 N 0.281 N m 0.008 0.249 m 0.999 0.030t

F + + + + + + =i j i j k i j i j 0

( ) ( ) ( ) 34 2.844 N m 0.281 N m 0.250 mt

F = + + 0k k k

34

13 NtF =

3 3 3 43 43

t r

A G A CA C BA BA= + + + + =M R f t R F R F R F 0

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 43

0.170 0.106 m 92 342 N 2.431 N m 0.280 0.226 m 601 195 N

0.292 0.130 m 12.5 N 0.292 0.130 m 0.030 0.999 rF

+ + + +

+ + + =

+

+

i j i j k i j i j

i j i i j i j 0

( ) ( ) ( ) ( ) ( ) 43 48 N m 2.431 N m 191 N m 1.623 N m 0.296 inr

F+ + = + + 0k k k k k

43 496 Nr

F = 34 2 496 N 496 N 89.71= = i jF Ans.

34 4 14= + + =F F f F 0 14 24 452 N 453 N 93.02= + = i jF Ans.

43 3 23C= + + + =F F f F F 0 23 690 643 N 944 N 137 = = iF Ans.

32 12= + =F F F 0 12690 643 N 944 N 137 = = iF Ans.

2 2 32 120

O AO= + =M R F M 12 82.83 N m= M k Ans.

12.13 Repeat Problem 12.12 at 2 260 = . Analyze both the kinematics and the dynamics of

the system at this position.

Kinematic Analysis:

( ) ( )22

32 rad/s 0.021 0.118 m

3.782 0.667 m/s 3.840 m/s 10

A AO= =

= =

V R k i j

i j

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

4

3 4

3 3 4 4

3 4

3.782 0.667 m/s rad/s 0.138 0.289 m rad/s 0.183 0.171 m

3.782 0.667 m/s 0.289 0.138 m 0.171 0.183 m

B A BA BO

+ + = +

= + + =

= + =

= i j k i j k i j

i j i j i j

V V R R

310.554 rad/s= k 4

4.314 rad/s= k Ans.

0.736 0.789 m/s 1.079 m/s 46.99B

= + = V i j

( ) ( )2 2 2

22

2

2 2

32 rad/s 0.021 0.118 m

21.338 121.013 m/s 122.880 m/s 80

A AO AO= + =

= + =

A R R i j

i j

( ) ( ) ( ) ( )

( ) ( ) ( )

4 4

2 23 3 4 4

2 2 23

2 24

21.338 121.013 m/s 15.374 32.158 m/s rad/s 0.138 0.289 m

3.402 3.174 m/s rad/s 0.183 0.171 m

B A BA BA BO BO

+

= + = +

= + + + +

= +

A A R R R R

i j i j k i j

i j k i j

( ) ( ) ( )2 3 3 4 4 2.562 92.029 m/s 0.289 0.138 m 0.171 0.183 m + + + = i j i j i j


17/36

197

2

3199.622 rad/s= k 24

352.356 rad/s= k Ans.

( ) ( ) ( ) ( )3 3 3

2

3 3

2 2 2 21.338 121.013 m/s 6.718 21.240 m/s 199.622 rad/s 0.060 0.191 m

G A G A G A= +

= + + + +

A A R R

i j i j k i j

3

2 2 52.748 87.796 m/s 102.423 m/s 59.00G = + = A i j Ans.

( ) ( ) ( )4 4 4 4 4

2

4 4

2 2 1.701 1.587 m/s 352.356 rad/s 0.091 0.085 m

G G O G O= +

= + +

A R R

i j k i j

4

2 2 31.651 30.477 m/s 43.939 m/s 43.92G = + = A i j Ans.

Dynamic Analysis:


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( ) ( )3

2

3 3

4.0 kg 52.748 87.796 m/s

211 351 N 410 N 121.00

Gm

+

=

=

= =

i j

f A

i j

( ) ( )4

2

4 4

1.5 kg 31.651 30.477 m/s

47 46 N 66 N 136.08

Gm

+

=

=

= =

i j

f A

i j

( ) ( )33 3

2 20.011 N m s 199.622 rad/s

2.196 N m

GI=

=

=

t

k

k

( )( )44 4

2 20.002 3 N m s 352.356 rad/s

0.810 N m

GI=

=

=

t

k

k

( ) ( )3 3 3 2.196 N m 410 N 0.005 mh t f= = = , ( ) ( )4 4 4 0.810 N m 66 N 0.012 mh t f= = =




18/36

198

4 4 4 44 4 34

t

O G O BO= + + =M R f t R F 0

( ) ( ) ( ) ( ) ( )34

0.091 0.085 m 47 46 N 0.810 N m 0.183 0.170 m 0.682 0.731t

F + + + + + =i j i j k i j i j 0

( ) ( ) ( ) 34 8.212 N m 0.810 N m 0.250 mt

F = + + 0k k k

34

36 NtF =

3 3 3 43 43

t r


( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 43

0.060 0.191 m 211 351 N 2.196 N m 0.066 0.354 m 601 195 N

0.138 0.289 m 25 26 N 0.138 0.289 m 0.731 0.682 rF

+ + +

+ + + =

+

+

i j i j k i j i j

i j i j i j i j 0

( ) ( ) ( ) ( ) ( ) 43 19 N m 2.196 N m 226 N m 3.629 N m 0.305 inr

F+ + = + + 0k k k k k

43 658 Nr

F = 34 505 422 N 659 N 39.88= = i jF Ans.

34 4 14= + + =F F f F 0 14 458 468 N 655 N 134.39= + = i jF Ans.

43 3 23C= + + + =F F f F F 0 23 115 124 lb 169 N 46.97+ = = iF Ans.

32 12= + =F F F 0 12115 124 lb 169 N 46.97+ = = iF Ans.

2 2 32 120

O AO= + =M R F M 12 11.03 N m= M k Ans.

12.14 Repeat Problem 12.13 at 2 300 = .

Kinematic Analysis:

( ) ( )22

32 rad/s 0.060 0.104 m

3.326 1.920 m/s 3.840 m/s 30

A AO= =

= + =

V R k i j

i j

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

4

3 4

3 3 4 4

3 4

3.326 1.920 m/s rad/s 0.093 0.306 m rad/s 0.147 0.202 m

3.326 1.920 m/s 0.306 0.093 m 0.202 0.147 m

B A BA BO

+ + + = +

= + + + =

= + =

= i j k i j k i j

i j i j i j

V V R R

31.597 rad/s= k 4


2.844 2.066 m/s 3.515 m/s 36.04B = + = V i j

( ) ( )2 2 2

22

2

2 2

32 rad/s 0.060 0.104 m

61.440 106.417 m/s 122.880 m/s 120

A AO AO= + =

= + =

A R R i j

i j

( ) ( ) ( ) ( )

( ) ( ) ( )

4 4

2 2

3 3 4 4

2 2 23

2 24

61.440 106.417 m/s 0.237 0.780 m/s rad/s 0.093 0.306 m

29.105 39.995 m/s rad/s 0.147 0.202 m

B A BA BA BO BO

+

= + = += + + + +

= +

A A R

R R

R

i j i j k i j

i j k i j

( ) ( ) ( )2 3 3 4 4 90.782 145.632 m/s 0.306 0.093 m 0.202 0.147 m + + + = i j i j i j 2

3671.302 rad/s= k 24

567.507 rad/s= k Ans.


19/36

199

( ) ( ) ( ) ( )3 3 3

2

3 3

2 2 2 61.440 106.417 m/s 0.079 0.504 m/s 671.302 rad/s 0.031 0.198 m

G A G A G A= +

= + + + +

A A R R

i j i j k i j

3

2 2 71.399 85.103 m/s 111.087 m/s 50.00G = + = A i j Ans.

( ) ( ) ( )4 4 4 4 4

2

4 4

2 2 14.547 20.022 m/s 567.507 rad/s 0.073 0.101 m

G G O G O= +

= + +

A R R

i j k i j

4

2 2 71.865 21.406 m/s 74.986 m/s 16.59G = + = A i j Ans.

Dynamic Analysis:


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( ) ( )

3

2

3 3

4.0 kg 71.399 85.103 m/s 286 340 N 444 N 130.00

Gm

+

=

== =

i j

f A

i j

( ) ( )

4

2

4 4

1.5 kg 71.865 21.406 m/s 108 32 N 112 N 163.41

Gm

+

=

== =

i j

f A

i j

( ) ( )33 3

2 20.011 N m s 671.302 rad/s

7.384 N m

GI=

=

=

t

k

k

( )( )44 4

2 20.002 3 N m s 567.507 rad/s

1.305 N m

GI=

=

=

t

k

k

( ) ( )3 3 3 7.384 N m 444 N 0.017 mh t f= = = , ( ) ( )4 4 4 1.305 N m 112 N 0.012 mh t f= = =




20/36

200

4 4 4 44 4 34

t

O G O BO= + + =M R f t R F 0

( ) ( ) ( ) ( ) ( )34

0.073 0.101 m 108 32 N 1.305 N m 0.147 0.202 m 0.809 0.588t

F + + + + + =i j i j k i j i j 0

( ) ( ) ( ) 34 13.273 N m 1.305 N m 0.250 mt

F = + + 0k k k

34

58 NtF =

3 3 3 43 43

t r


( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 43

0.032 0.197 m 286 340 N 7.384 N m 0.013 0.360 m 601 195 N

0.094 0.306 m 47 34 N 0.094 0.306 m 0.588 0.809 rF

+ + +

+ + + =

+

+

i j i j k i j i j

i j i j i j i j 0

( ) ( ) ( ) ( ) ( ) 43 46 N m 7.384 N m 219 N m 11.170 N m 0.256 inr

F+ + = + + 0k k k k k

43 603 Nr

F = 34 401 453 N 606 N 48.50= = i jF Ans.

34 4 14= + + =F F f F 0 14 293 486 N 567 N 121.13= + = i jF Ans.

43 3 23C= + + + =F F f F F 0 23 86 81 N 119 N 43.26+ = = iF Ans.

32 12= + =F F F 0 1286 81 N 119 N 43.26+ = = iF Ans.

2 2 32 120

O AO= + =M R F M 12 13.85 N m= M k Ans.

12.15 Analyze the dynamics of the offset slider-crank mechanism shown in the figure using the

following data: 0.06 ma= ,2

0.1 mAOR = , 0.38 mABR = , 0.4 mCAR = , 2 32 = ,

30.26 mG AR = , 22= , 2 2.5 kgm = , 3 7.4 kgm = , 4 2.5 kgm = , 2

20.005 N m sGI = ,

3

20.013 6 N m sGI = , 2 120 = , and 2 18 rad/s cw = with 2 0 = , B2 000 N= F i ,

C1 000 N= F i . Assume a balanced crank and no friction forces.

Kinematic Analysis:

( ) ( )22

18 rad/s 0.050 0.087 m

1.559 0.900 m/s 1.800 m/s 30

A AO= = +

= + =

V R k i j

i j

( ) ( ) ( )

( ) ( )3

3 3

3

1.559 0.900 m/s rad/s 0.351 0.147 m

1.559 0.900 m/s 0.147 0.351 m

B A BA

BV

+ +

= + + +

= +

= i j k i j

i j i j

V V R

i

32.567 rad/s= k 1.182 m/sB =V i Ans.

( ) ( )2 2 2

22

2

2 2

18 rad/s 0.050 0.087 m

16.200 28.059 m/s 32.400 m/s 60

A AO AO= + = +

= =

A R R i j

i j

( ) ( ) ( ) ( )

23 3

2 2 23

16.200 28.059 m/s 2.310 0.966 m/s rad/s 0.351 0.147 mB

B A BA BA

A

= +

= + + + i

A A R R

i j i j k i j


21/36

201

( ) ( )2 3 3 13.890 27.093 m/s 0.147 0.351 mBA = + +i i j i j 2

377.188 rad/s= k 225.156 m/s

B=A i Ans.

( ) ( ) ( ) ( )3 3 3

2

3 3

2 2 2 16.200 28.059 m/s 1.713 0.021 m/s 77.188 rad/s 0.260 0.003 m

G A G A G A= +

= + + +

A A R R

i j i j k i j

3

2 2 14.466 7.969 m/s 16.516 m/s 28.85G = = A i j Ans.

Dynamic Analysis:


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( )( )3

2

3 3

7.4 kg 14.466 7.969 m/s

107 59 N 122 N 151.15

Gm

=

=

= + =

i j

f A

i j

( ) ( )24 4

2.5 kg 25.156 m/s

63 N 63 N 180

Bm

=

=

= =

i

f A

i

( ) ( )33 3

2 20.013 6 N m s 77.188 rad/s

1.050 N m

GI=

=

=

t

k

k

44 4G

I= =t 0

( ) ( )3 3 3 1.050 N m 122 N 0.009 mh t f= = = , 4 4 4 0h t f= =

3 3 3 4 14A G A CA C BA BA B BA= + + + + + =M R f t R F R f R F R F 0

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )14

0.260 0.003 m 107 59 N 1.050 N m 0.395 0.065 m 1 000 N

0.351 0.147 m 63 N 0.351 0.147 m 2 000 N 0.351 0.147 m 1.000 F

+ + + +

+ + + =

i j i j k i j i

i j i i j i i j j 0


22/36

202

( ) ( ) ( ) ( ) ( ) ( ) 14 15 N m 1.050 N m 65 N m 9.261 N m 294 N m 0.351 in F + + + + + =k k k k k k 0

14 639 NF = 14639 N 639 N 90.00= = jF Ans.

14 4 34B= + + + =F F f F F 0 34 2 063 639 N 2 160 N 17.21= = i jF Ans.43 3 23C= + + + =F F f F F 0 23 3 170 698 N 3 246 N 12.42 = = iF Ans.

32 12= + =F F F 0 12 3 170 698 N 3 246 N 12.42 = = iF Ans.

2 2 32 120

O AO= + =M R F M 12 241 N m= M k Ans.

12.16 Analyze the system of Problem 12.15 for a complete rotation of the crank. Use C=F 0

and B1 000 N= F i when BV is toward the right, but use B =F 0 when BV is toward

the left. Plot a graph of 12M versus 2 .

32

1 2 3 4

jjjR R e R e R

+ + = 32

0.060 0.100 0.380

jj

Bj e e R

+ + =

2 30.060 0.100sin 0.380sin 0 + + = ( )1

3 2sin 0.158 0.263sin = +

2 30.100 cos 0.380 cosBR = + ( )32

3 31 2

jj

G G AjR R e R e

+= + +R ( )323

220.060 0.100 0.260

jj

Gj e e

+ = + +R

( ) ( )3 2 3 2 3

0.100cos 0.260cos 22 0.060 0.100sin 0.260sin 22G j = + + + + + + R

The first-order kinematic coefficients are found as follows:32

3 40.100 0.380jj

j e j e R + =

2 3 30.100cos 0.380cos 0 + = 2 3 3 40.100 sin 0.380 sin R =

3 2 30.263cos cos = ( )2 2 30.100 sin cos tanBR = ( )32

3

22

30.100 0.260jj

Gj e j e

+ = +R

( ) ( )3 2 3 3 2 3 3

0.100sin 0.260sin 22 0.100cos 0.260cos 22G j = + + + + R

Similarly, the second-order kinematic coefficients are as follows:3 32 2

3 30.100 0.380 0.380j jj

Be j e e R

+ = 2

2 3 3 3 30.100sin 0.380cos 0.380sin 0 + = 2

2 3 3 3 30.100cos 0.380sin 0.380cos BR = 2

3 2 3 3 30.263sin cos tan = + , ( )2

3 2 3 30.100cos 0.380 cosBR = +

( ) ( )3 323

22 22 2

3 30.100 0.260 0.260j jj

Ge j e e

+ + = + R

( ) ( )

( ) ( )

3

2

2 3 3 3 3

2

2 3 3 3 3

0.100cos 0.260sin 22 0.260cos 22

0.100sin 0.260cos 22 0.260sin 22

G

j

= + +

+ + + +

R

By virtual work we can formulate the dynamic input torque requirement as:

312 3 3 3 4

G B B B = + + +M f R t k f R F R i i i i


23/36

203

The individual elements of this equation are:

( )3 3 3

2 2

3 3 3 2 2 397.6 kg/sG G Gm m = = = f A R R

( ) ( )

( ) ( )

2

3 2 3 3 3 3

2

2 3 3 3 3

240cos 623sin 22 623cos 22

240sin 623cos 22 623sin 22j

= + + + +

+ + + +

f

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]

3

2

3 2 3 3 3 3 2 3 3

2

2 3 3 3 3 2 3 3

240 cos 623sin 22 623cos 22 0.100sin 0.260 sin 22

240sin 623cos 22 623sin 22 0.100 cos 0.260 cos 22

G

= + + + + +

+ + + + + +

f Ri

( ) ( ) ( )33 3 2 3 3 3 2 3 3 3

62sin 22 1 62cos 22 162G

= + + +f Ri

( )3 3

2

3 3 3 2 3 4.406 N mG GI I = = = t k k

( )3 3 3 3 4.406 N m = t ki

( )2 24 4 4 2 810 kg/sB B Bm m = = = f A R R

( ) 24 3 2 3 381cos 308 N cos = + f i

( ) ( ) 2 24 3 2 3 2 3 3sin 8.1cos 30.8 N m cosB = + f Ri

( ) ( ){ }2 3 2 3 2 3 500 1 sgn cos tan sin N=500 1+sgn sin cos NB = + F i i

( ){ } ( )3 2 3 3 2 350 1+sgn sin cos sin cos N mB B = F Ri

Finally, putting these pieces together, we obtain:

( ) ( ) ( )

( ) ( )

( ){ } ( )

12 3 2 3 3 3 2 3 3 3

2 2

3 2 3 2 3 3

3 2 3 3 2 3

62sin 22 1 62cos 22 158

sin 8.1cos 30.8 cos

50 1+sgn sin cos sin cos N m

M

= + + +

+ +

+

The plot of this torque requirement is shown below. The sinusoidal curve in the first halfof the cycle is caused primarily by the mass of the connecting rod; the mass of the pistonis included also. The applied force FB causes the rise in the second half of the cycle.Note that the mass of link 3 causes dynamic torque which helps to overcome up to onethird of the applied force effect.


24/36

204

12.17 A slider-crank mechanism similar to that of Problem 12.15 has 0a= ,2

0.1 mAOR = ,

0.45 mBAR = , 0CAR = , 0C = , 3 0.2 mG AR = , 0= , 2 1.5 kgm = , 3 3.5 kgm = ,

4 1.2 kgm = , 220.01 N m sGI = , 3

20.060 N m sGI = , and 12 60 N mM = .

Corresponding to2

120 = and2

24 rad/s cw = with2

0 = , a kinematic analysis

gave 3 9 = , 0.374 mBR = ,2

3 89.3 rad/s ccw = ,240.6 m/sB=A i , and

3

2 40.6 22.6 m /sG = A i j . Assume link 2 is balanced and find 14F and 23F .


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( )( )3

2

3 3

3.5 kg 40.6 22.6 m / s

142 79 N 163 N 150.90

Gm

=

=

= + =

i j

f A

i j

( ) ( )24 4

1.2 kg 40.6 m/s

49 N 49 N 180

Bm

=

=

= =

i

f A

i

( )( )33 3

2 20.060 N m s 89.3 rad/s

5.358 N m

GI=

=

=

t

k

k

44 4G

I= =t 0

( ) ( )3 3 3 5.358 N m 163 N 0.033 mh t f= = = , 4 4 4 0h t f= =

3 3 3 4 14A G A BA BA= + + + =M R f t R f R F 0

( ) ( ) ( ) ( ) ( )

( ) ( ) 14

0.196 0.038 m 142 79 N 5.358 N m 0.442 0.087 m 49 N

0.442 0.087 m 1.000 F

+ + +

+ =

i j i j k i j i

i j j 0

( ) ( ) ( ) ( ) 14 10.039 N m 5.358 N m 4.243 N m 0.442 m F + + + =k k k k 0

14 1 NF = 141 N 1 N 90.00= = jF Ans.

14 4 3 23= + + + =F F f f F 0 23 191 78 N 206 N 22.21 = = iF Ans.


25/36

205

12.18 Repeat Problem 12.17 for 2 240 = . The results of a kinematic analysis are 3 11.1 = ,

0.392 mBR = ,2

3 112 rad/s cw = ,235.2 m/B s=A i , and 3

2 31.6 27.7 m /sG = +A i j .


22 2 Gm= =f A 0 22 2GI= =t

0 2 2 2 0h t f= =

( )( )3

2

3 3

3.5 kg 31.6 27.7 m / s

111 97 N 147 N 138.76

Gm

+

=

=

= =

i j

f A

i j

( ) ( )24 4

1.2 kg 35.2 m/s

42 N 42 N 180

Bm

=

=

= =

i

f A

i

( )( )33 3

2 20.060 N m s 112 rad/s

6.720 N m

GI=

=

=

t

k

k

44 4G

I= =t 0

( ) ( )3 3 3 6.720 N m 147 N 0.046 mh t f= = = , 4 4 4 0h t f= =

3 3 3 4 14A G A BA BA= + + + =M R f t R f R F 0 ( ) ( ) ( ) ( ) ( )

( ) ( )14

0.196 0.038 m 111 97 N 6.720 N m 0.442 0.087 m 42 N

0.442 0.087 m 1.000 F

+ + + +

+ + =

i j i j k i j i

i j j 0

( ) ( ) ( ) ( ) 14 22.961 N m 6.720 N m 3.637 N m 0.442 in F + + + =k k k k 0

14 29 NF = 1429 N 29 N 90.00= = jF Ans.

14 4 3 23= + + + =F F f f F 0 23 153 68 N 168 N 24.11+ = = iF Ans.


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206

12.19 A slider-crank mechanism, as in Problem 12.15, has 0.008 ma= ,2

0.25 mAOR = ,

1.25 mBAR = , 1.0 mCAR = , 38C = , 3 0.75 mG AR = , 18= , 2 10 kgm = ,

3 140 kgm = , 4 50 kgm = , 222.0 N m sGI = , and 3

28.42 N m sGI = , and has a

balanced crank. Make a complete kinematic and dynamic analysis of this system at

2 120 = with2

2 6 rad/s ccw = and 2 0 = , using B =50 180 kN F and

C =80 60 kN F .

Kinematic Analysis:

( ) ( )22

6 rad/s 0.125 0.217 m

1.299 0.750 m/s 1.500 m/s 150

A AO= = +

= =

V R k i j

i j

( ) ( ) ( )

( ) ( )3

3 3

3

1.299 0.750 m/s rad/s 1.227 0.225 m

1.299 0.750 m/s 0.225 1.227 m

B A BA

BV

+

= + +

= +

= i j k i j

i j i j

V V R

i

30.611 rad/s= k 1.162 m/sB = V i Ans.

( ) ( )2 2 2

22

2

2 2

6 rad/s 0.125 0.217 m

4.500 7.794 m/s 9.000 m/s 60

A AO AO= + = +

= =

A R R i j

i j

( ) ( ) ( ) ( )

23 3

2 2 23

4.500 7.794 m/s 0.459 0.084 m/s rad/s 1.227 0.225 mB

B A BA BA

A

= +

= + + + i

A A R R

i j i j k i j

( ) ( )2 3 3 4.041 7.710 m/s 0.225 1.227 mBA = + +i i j i j 2

36.284 rad/s= k 25.455 m/sB =A i Ans.

( ) ( ) ( ) ( )3 3 3

2

3 3

2 2 2 4.500 7.794 m/s 0.247 0.133 m/s 6.284 rad/s 0.660 0.357 m

G A G A G A= +

= + + +

A A R R

i j i j k i j

3

2 2 6.496 3.514 m/s 16.516 m/s 28.41G

= = A i j Ans.

Dynamic Analysis:


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( )( )3

2

3 3

140 kg 6.496 3.514 m/s

909 492 N 1 034 N 151.59

Gm

=

=

= + =

i j

f A

i j

( ) ( )24 4

50 kg 5.455 m/s

273 N 273 N 180

Bm

=

=

= =

i

f A

i


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207

( )( )33 3

2 28.42 N m s 6.284 rad/s

52.911 N m

GI=

=

=

t

k

k

44 4G

I= =t 0

( ) ( )3 3 3 52.911 N m 1 034 N 0.051 mh t f= = = , 4 4 4 0h t f= =

3 3 3 4 14A G A CA C BA BA B BA= + + + + + =M R f t R F R f R F R F 0

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) 14

0.660 0.357 m 909 492 N 52.911 N m 0.263 0.301 m 40 000 69 282 N

1.227 0.225 m 273 N 1.227 0.225 m 50 000 N 1.227 0.225 m 1.000 F

+ + +

+ + + =

i j i j k i j i j

i j i i j i i j j 0

( ) ( ) ( ) ( ) ( ) ( )14

0.207 N m 52.911 N m 6 157 N m 61.425 N m 11 250 N m 1.227 in F + + + + + =k k k k k k 0

14 14 280 NF = 14 14 280 N 14 280 N 90.00= = jF Ans.

14 4 34B= + + + =F F f F F 0 34 50 273 14280 N 52 262 N 15.86= = F i j Ans.

43 3 23C= + + + =F F f F F 0 23 11 182 54 510 N 55 645 N 78.41+ = = i jF Ans.

32 12= + =F F F 0 12 11 182 54 510 N 55 645 N 78.41+ = = i jF Ans.

2 2 32 120

O AO= + =M R F M 12 9 240 N m= M k Ans.


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208

12.20 Cranks 2 and 4 of the cross-linkage shown in the figure are balanced. The dimensions of

the linkage are2

150 mm=AOR , 4 2 450 mm=O OR , 450 mm=BAR , 4 150 mm=BOR ,

600 mm=CAR , and 3 300 mm=G AR . Also, 3 1.8 kg=m , 2 420.007 N m s= = G GI I ,

and3

20.055 N m s= GI . Corresponding to the position shown, and with

2 10 rad/s ccw = and 2 0 = , a kinematic analysis gave as results 3 1.43 rad/s cw = ,

4 11.43 rad/s cw = ,2

3 4 84.7 rad/s cw = = , and 32 14.28 21.1 m/s= +GA i j . Find

the driving torque and the pin reactions if 133.5 N= CF j .


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( ) ( )33 3

2

25.7 37.98 N 45.84 N 124.10

1.8 kg 14.28 21.1 m/s=

= =

=

+

Gm

i j

f A

i j 44 4 G

m= =f A 0

( ) ( )33 3

2 20.007 N m s 84.7 rad/s

0.593 N m

=

=

=

GIt

k

k

( ) ( )44 4

2 20.007 N m s 84.7 rad/s

0.593 N m

=

=

=

GIt

k

k

( ) ( )3 3 3 0.593 N m 45.84 N 12.93 mm= = =h t f , 4 0h =



4 44 34

t

O BO= + =M t R F 0


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209

( ) ( )( )34

0.593 N m 21.42 148.45 mm 0.990 0.143t

F + =k i j i j 0

34 3.96 N= t

F

3 3 3 43 43

t r


( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 43

0.23 0.18 m 25.7 37.98 N 0.593 N m 0.47 0.37 m 133.5 N

0.35 0.28 m 3.92 0.56 N 0.35 0.28 m 0.143 0.990 rF

+ +

+ + + =

i j i j k i j j

i j i j i j i j 0

( ) ( ) ( ) ( ) ( ) 43 14.24 N m 0.6 N m 63.96 N m 0.9 N m 0.39 mrF = + + + + 0k k k k k

43 195.8 N=r

F 34 31.18 188.8 N 191.35 N 99.38= = F i j

43 3 23C= + + + =F F f F F 0 23 5.28 18.82 N 19.55 N 105.67= = F i j

2 2 32 120

O AO= + =M R F M 12 0.74 N m= M k Ans.

12.21 Find the driving torque and the pin reactions for the mechanism of Problem 12.20 underthe same dynamic conditions, but with crank 4 as the driver.

Given the same dynamic conditions, the DAlembert forces and torques are the same asin Problem P12.20. However, crank 2 is now a two-force member with no appliedmoment. Therefore the free body diagrams appear as:

The solution now proceeds as follows:

3 3 3 23B G B CB C AB= + + + =M R f t R F R F 0 ( ) ( ) ( ) ( ) ( )( ) ( ) 23

0.117 0.093 m 25.7 37.98 N 0.593 N m 0.117 0.093 m 133.5 N

0.35 0.28 m 0.500 0.866

+ + +

+ + =F

i j i j k i j j

i j i j 0

( ) ( ) ( ) ( ) 23 7.119 N m 0.603 N m 15.93 N m 0.445 m = + + + F 0k k k k

23 18.15 N=F 23 9.075 15.72 N 18.15 N 120= = F i j Ans.


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210

23 3 43C= + + + =F F f F F 0 43 35.4 188.08 N 191.38 N 79.34= + = F i j Ans.

4 4 34 4 140O BO= + + =M R F t M 14 0.639 N m= M k Ans.

12.22 A kinematic analysis of the mechanism of Problem 12.20 at 2 210 = with

2 10 rad/s = ccw and 2 0 = gave 3 14.7 = , 4 164.7 = , 3 4.73 rad/s ccw = ,

4 5.27 rad/s cw = , 3 4 10.39 rad/s cw = = , and 327.8 20.85 m / s= GA . Compute

the crank torque and the pin reactions for this phase of the motion using the same force

CF as in Problem 12.20.


22 2 Gm= =f A 0

22 2GI= =t 0

2 2 2 0h t f= =

( ) ( )33 3

2

13.42 5.11 N 14.37 N 159.15

1.8 kg 7.29 2.78 m/s=

= =

=

+

Gm

i j

f A

i j 44 4 G

m= =f A 0

( )( )33 3

2 20.055 N m s 10.39 rad/s

0.571 N m

=

=

=

GIt

k

k

( ) ( )44 4

2 20.007 N m s 10.39 rad/s

0.0727 N m

=

=

=

GIt

k

k

( ) ( )3 3 3 0.571 N m 14.37 N 39.73 mm= = =h t f , 4 0h =

Next, the free body diagrams are drawn. Since the lines of action for the forces on the

free body diagrams can not be discovered from two- and three-force member concepts,the force 34F is divided into radial and transverse components.


31/36

211

4 44 34

t

O BO= + =M t R F 0

( ) ( )( )34

0.0727 N m 0.144 0.039 m 0.263 0.965 + + + =t

Fk i j i j 0

34 0.485 N=t

F

3 3 3 43 43

t r


( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 43

0.29 0.076 m 13.42 5.11 N 0.0727 N m 0.58 0.153 m 133.5 N

0.435 0.114 m 0.129 0.467 N 0.435 0.114 m 0.965 0.263

+ + + +

+ + + + =rF

i j i j k i j j

i j i j i j i j 0

( ) ( ) ( ) ( ) ( ) 0.128 N m 0.074 N m 78.65 N m 0.191 N m 0.225 m43

+ + + + =rFk k k k k 0

43 78 lbr

F = 34 334.86 89 N 347 N 164.80= + = F i j Ans.

43 3 23C= + + + =F F f F F 0 23 322.27 230.13 N 396 N 144.47= + = F i j Ans.

2 2 32 120O AO= + =M R F M 12 54.8 N m= M k Ans.

12.23

The figure shows a linkage with an extended coupler having an external force of CF acting during a portion of the cycle. The dimensions of the linkage are

20.14 mAOR = ,

4 21 m= =O O BAR R , 4 1.4 m=BOR , 3 0.8 m=G AR , and 4 4 0.5 m=G OR . Also

3 100.78 kg=m , 4 94.4 kg=m , 3225.14 N m s= GI , and 4

229.37 N m s= GI , and the

crank is balanced. Make a kinematic and dynamic analysis for a complete rotation of the

crank using 2 10 rad/s ccw = , 2.225 3.94 kN= +CF i j for 290 300 and

C =F 0 for other angles.

For position analysis, the loop closure equation is:

32 4

2 3 1 4

jj jR e R e R R e

+ = + 32 40.4 1 1 1.4 + = +jj je e e

2 3 40.4cos 1cos 1 1.4cos + = + 2 3 40.4sin 1sin 1.4sin + =

Eliminating 3 we find 4 from the roots of the quadratic

( ) ( ) ( )22 4 2 4 20.425 0.2cos tan 2 1.4sin tan 2 1.2cos 3.075 0 + + = Ans.

Then ( ) ( )13 4 2 4 2tan 0.175sin 0.05sin 0.425cos 0.125 0.05cos = + Ans.

32

30.4 0.8 = + jjG e eR

4

41 0.5 = + jG eR Ans.

( )32 4.086

0.4 1.403

= + jjC e eR Ans.

The first-order kinematic coefficients are found by differentiating the above:32 4

3 40.4 1 1.4 + =jj jj e j e j e

2 3 3 4 40.4sin 1sin 1.4sin = 2 3 3 4 40.4cos 1cos 1.4cos + =


32/36


33/36

213

44 4 42436.3 N m = i

Gf R

( )4 4

2

4 4 4 2 4 2983 N m = = = G GI It k k

4 4 4 4 2983 N m = it k

( ) ( )

( ) ( )

2 3 3

2 3 3

4450 N cos120 0.4sin 1.403sin 4.086 m

4450 N sin120 0.4cos 1.403cos 4.086 m

=

+ +

iC CF R

( ) ( )2 3 31808sin 120 6344sin 124.086 N m = iC CF R

Reassembling the elements we must remember that force is only nonzero for a portion ofthe cycle. Therefore,

( ) ( ) ( ) ( )[ ] ( )

( ) ( ) ( )3 2 3 3 3 2 3 3 3 3 3 4 4 4 4

3 3 4 43 2 3 3 2 3 3

12 3327 sin 1 cos 2 2554 21 560 26 400 N m

N m54193327 cos 3327sin 1 9209=

= +

M

for the entire cycle and, for 290 300 , an additional increment is added:

( ) ( )2 3 312 1808sin 120 6344sin 124.086 N m =M Ans.

This input torque requirement is shown in the following plot. Notice the very smalldiscontinuities in the curve when force FCbegins and ends its effect.

12.24 The figure shows a motor geared to a shaft on which a flywheel is mounted. The mass

moments of inertia of the parts are as follows: flywheel, 20.308 N m s= I ; flywheel

shaft, 20.1724 N cm s= I ; gear, 21.9 N cm s= I ; pinion, 20.0388 N cm s= I ;

motor, 20.9612 N cm s= I . If the motor has a starting torque of 8.34 N m , what is theangular acceleration of the flywheel shaft at the instant the motor is turned on ?

If we identify the motor shaft as 2 and the flywheel shaft as 3 then2 2 2

2 0.009612 0.000388 0.01= + = I N m s N m s N m s 2 2 2 2

3 0.308 0.001724 0.019 0.3287= + + = I N m s N m s N m s N m s

( )3 2 3 2R R = ( )3 2 3 2R R =

3 3 23 3 3M R F I= = ( )223 2 3 3 2F R R I =


34/36

214

2 12 2 32 2 2M M R F I= = ( )2

12 2 2 2 32 2 2 3 3 2M I R F I R R I = + = +

Now, substituting the numeric values,

( ) ( )22 2 2 28.475 0.01 25 112.5 0.3287 0.02623 = + = 2N m N m s mm mm N m s N m s

2

8.475 0.02623 323.1032 2N m N m s rad/s = =

( ) ( )3 2 3 2 25 112.5 323.103 71.80 = = = 2 2

R R mm mm rad/s rad/s Ans.

12.25 The disk cam of Problem 14.31 is driven at a constant input shaft speed of

2 20 rad/s ccw. = Both the cam and the follower have been balanced so that the

centers of mass of each are located at their respective fixed pivots. The mass of the camis 0.075 kg with radius of gyration of 30 mm, and for the follower the mass is 0.030 kg

with radius of gyration of 35 mm. Determine the moment 12M required on the camshaft

at the instant shown in the figure to produce this motion.

( ) ( )3

22

3 3 0.030 0.035 0.000 036 75

2

GI m k kg m kg m= = =

For full-rise cycloidal cam motion, Eq. (5.19),

2 30 112.51 cos 1 cos 2 0.200

150 150

Ly

= = =

( )( )

( )2 2

360 302 2 112.5sin sin 2 0.480

150150

Ly

= = =

( )22

3 2 0.480 20 1922

y rad/s rad/s = = =

( )( )33 3

0.000 036 75 192 0.007 0562 2Gt I kg m rad/s N m= = =

By virtual work,

( )

( ) ( )

312 3

0.200 0.007 056 0.150 8 sin 45 0.168

CO CM y t

N m m N N m ccw

= +

= + =

R F k i

Ans.

12.26 Repeat Problem 12.25 with a shaftspeed of 2 40 rad/s ccw. =

( ) ( )3

22

3 3 0.030 0.035 0.000 036 752

GI m k kg m kg m= = =

For full-rise cycloidal cam motion, Eq. (5.19),

2 30 112.5

1 cos 1 cos 2 0.200150 150

L

y

= = =

( )( )

( )2 2

360 302 2 112.5sin sin 2 0.480

150150

Ly

= = =

( )22

3 2 0.480 40 7682

y rad/s rad/s = = =

( )( )33 3

0.000 036 75 768 0.028 2242 2Gt I kg m rad/s N m= = =


35/36

215

By virtual work,

( )

( ) ( )

312 3

0.200 0.028 224 0.150 8 sin 45 0.164

CO CM y t

N m m N N m ccw

= +

= + =

R F k i

Ans.

12.27

A rotating drum is pivoted at 2O and is decelerated by the double-shoe brake mechanismshown in the figure. The mass of the drum is 105 kg and its radius of gyration is

142 mm. The brake is actuated by force 445 N= P j , and it is assumed that the contact

between the two shoes and the drum act at points C and D, where the coefficients ofcoulomb friction are 0.300.= Determine the angular acceleration of the drum and the

reaction force at the fixed pivot 12 .F

( ) ( ) ( ) ( )5 65 300 75= + = mm P mm F M i j j i 0 65 4 1780= =F P N

5 65 35= + + =F P F F 0 35 1778.63 445 1833.4 14.04= + = N NF i j

The friction angle is ( )1

tan 0.300 16.70

= = .

( ) ( ) ( )3 53 23 550 125 225 cos sin = + + = mm mm F M j F i j i j 0

23 3408.7=F N 32 3264.9 979.5 3408.7 16.70F N N= + = i j


36/36

216

( ) ( ) ( )4 64 24 625 125 225 cos sinmm mm F = + + + =M j F i j i j 0

24 6194.4=F N 42 5933 1780 6194.4 163.30F N N= = i j

( )( )2

22

2 2 105 0.1415 2.156= = = 2

GI m k kg m N m s

( ) ( ) ( ) ( )2 2 2

0.2 3261.85 979 0.2 5931.85 1780= + + =

O Gm N m N I M i i j i i j

2261 2rad/s= k Ans.

2 12 32 42= + + =F F F F 0 12 2668.2 800.5 2785.7 16.70= + = F lb N i j Ans.

Note that gravitational effects are not yet included. If standard gravity acts in the

direction then the j component is 410 lb. Since the main bearing at O2 supports this

weight, it does not affect the friction forces and can be added by superposition. Ifweights of the other parts were known, however, these weight might have some smalleffect on the friction forces and the braking forces, and would have to be includedsimultaneously. Superposition could not be applied.

solutions instructor manual chapter 12 dynamic force analysis planar

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