solutions instructor manual chapter 11 static force analysis

7/27/2019 Solutions Instructor Manual Chapter 11 Static Force Analysis

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PART III

DYNAMICS OF MACHINES


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Chapter 11

Static Force Analysis

11.1 The figure shows four mechanisms and the external forces and torques exerted on or bythe mechanisms. Sketch the free-body diagram of each part of each mechanism. Do notattempt to show the magnitudes of the forces, except roughly, but do sketch them in theirproper locations and directions.


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11.2 What moment 12M must be applied to the crank of the mechanism shown if

0.9 kNP= ?

Kinematic analysis:

( ) ( )1 1sin sin sin 75 mmsin105 350 mm 11.95r = = =


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Force analysis:900 N= P i

( ) ( )14 34 14 34 cos sin 900 N 0.978 0.207 0F F F F = + + = + + =F P j i j i j i j 34900 N 0.978 0F + = 34 900 N 0.978 920 NF = =

14 340.207 0F F = ( )14 0.207 920 N 190 NF = = ( )32 34 920 N 0.978 0.207 900 190 N= = + = +F F i j i j

( ) ( )12 2 32 12 75 mm cos105 sin105 900 190 N= + = + + + =M M r F M i j i j 0

1261.5 N m+ =M k 0 12 61.5 N m= M k Ans.

11.3 If 12 100 N mM = for the mechanism shown, what force Pis required to maintain staticequilibrium?

Kinematic analysis:

Recall 11.95= from Prob. P11.2.cos cos 75 mmcos105 350 mmcos11.95 323 mmx r = + = + =

Force analysis:

14 12 0z

O xF M= =M 14 12 100N m 0.323 m 310 NF M x= = = Recall the force polygon on link 4 from Prob. P11.2.

14 tan 310 N tan11.95 1 463 NP F = = = Ans.

11.4 Find the frame reactions and torque 12M necessary to maintain equilibrium of the four-

bar linkage shown in the figure.

Kinematic analysis:

2 88 mm 210 75.775 43.75 mm= = AOR i j 4 150 mm 135.53 107.04 105.07= = +BOR i j 150 mm 82.83 18.725 148.825 mm= = +

BAR i j 4 100 mm 135.53 71.375 70.05 = = +COR i j


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Force analysis:

( ) ( )

( ) ( )( )

4 4 4 34

34

34

71.375 70.05 mm 311.72 317.57 N

107.04 105.07 mm cos82.83 sin 82.83

45.2 N m 119.325 mm

= + =

+ +

+ + + =

=

O CO BO

F

F

M R P R F 0

i j i j

i j i j 0

k 0

34 9.47 N m= F 34 372.94 N 82.83 46.55 370.02 N= = F i j

( ) ( )2 212 32

12 75.775 43.75 mm 46.55 370.02 N

= + =

+ =

O AOM

M M R F 0

k i j i j 0

12 26.408 N m= M 12 26.408 N m= M k Ans.

11.5

What torque must be applied to link 2 of the linkage shown to maintain staticequilibrium?

Kinematic analysis:

2

87.5 mm 240 43.75 75.775 mm= = AOR i j 4150 mm 152.64 133.225 68.92= = +BOR i

150 mm 105.26 39.475 144.7 mm= = +BAR i j 4

175 mm 152.64 155.425 80.42= = +DOR i


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Force analysis:

( ) ( )

( ) ( )( )

4 4 4 34

34

34

155.425 80.425 mm 222.5 N

133.225 68.925 mm cos105.26 sin105.26

18.176 N m 110.375 mm

= + =

+

+ + + =

=

O DO BO

F

F

M R P R F 0

i j i

i j i j 0

k 0

34 4.116 N m= F 34 162.109 N 105.26 42.666 156.39 N= = F i j

( ) ( )2 212 32

12 43.75 75.775 mm 42.666 156.39 N

= + =

+ + =

O AOM

M M R F 0

k i j i j 0

12 10.23 N m= M 12 10.23 N m= M k Ans.

11.6 Sketch a complete free-body diagram of each link of the linkage shown. What force Pisnecessary for equilibrium?

Kinematic analysis:

2

100 mm 90 100 mmAO = =R j 4 125 mm 44.3 89 87 mmBO = = +R i j

150 mm 4.86 149 13 mmBA

= = R i j 4

200 mm 44.3 143 140 mmCO = = +R i j

400 mm 20.44 375 140 mmDC

= = R i j

Force analysis:

( ) ( )( )

2 212 32

32

32 32

90 N m 100 mm cos 4.86 sin 4.86

90 N m 99.640 mm 903 N 4.86

O AO

F

F

= + =

+ + =

= =

M M R F 0k j i j 0

k 0 F


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( ) ( )

( ) ( )( )

4 4 434 54

54

54 54

89 87 mm cos175.14 sin175.14 903 N

143 140 mm cos 20.44 sin 20.44

86 N m 181 mm 472 N 20.44 =44 165 N

O BO CO

F

F

= + =

+ +

+ + + =

= =

M R F R F 0i j i j

i j i j 0

k 0 F i j

16 443 N 165 N =P F= + +F i i j j 0 443 N=P i Ans.

11.7 Determine the torque 12M required to drive slider 6 of the figure against a load of

445 N=P at a crank angle of 30= , or as specified by your instructor.

Kinematic analysis:

2

63 mm 30 54.125 31.25 mm= = +AOR i j 4 186.65 mm 73.37 53.4 178.85 m= = +AOR i j

4

400 mm 73.37 114.45 383.275 mm= = +BO

R i j 200 mm 175.20 199.3 16.725 m= = +CB

R i j

Force analysis:

( )16 56 cos175.20 sin175.20P F F= + + + =F i j i j 0

56 cos175.20 445 N 0.996 446.56 N= = =F P ; 56 446.56 N 175.20 445 37.34 N= = +F i j

( ) ( )( ) ( )

( )

4 4 454 34

34

34 34

114.45 383.275 mm 445 37.34 N

53.4 178.85 mm cos163.37 sin163.37

177.58 N m 186.65 mm 936.67 N 163.37 = 897.49

= + =

+

+ + + =

+ = = +

O BO AO

F

F

M R F R F 0

i j i j

i j i j 0

k 0 F i 268.07 N


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( ) ( )2 212 32

12 54.125 31.25 mm 897.49 268.07 N

= + =

+ + =

O AOM

M M R F 0

k i j i j 0

12 43.22 N m= M 12 43.22 N m= M k Ans.

11.8

Sketch complete free-body diagrams for the illustrated four-bar linkage. What torque

12M must be applied to 2 to maintain static equilibrium at the position shown?

Kinematic analysis:

2

200 mm 60 100 173 mmAO = = +R i j 4 350 mm 109.05 114 331 mmCO = = R i j

400 mm 46.06 278 288 mmBA= = R i j , 700 mm 46.06 486 504 mmCA= = R i j

Force analysis:Since the lines of action of all constraint forces can not be found from two- and three-

force members, the force 34F is resolved into radial and transverse components,

34 34andr

F F . Then

( ) ( )4 414 34 34

45 N m+ 114 331 mm cos 19.05 sin 19.05O CO F = + = + =M M R F k i j i j

34 45 N m+350 mmF =k k 0 34 129 N 19.05 =122 42 N

= F i j

( ) ( ) ( ) ( )

( ) ( )

43 43

43

278 288 mm 350 N + 486 504 mm 122 42 N

+ 486 504 mm cos70.95 sin 70.95

rA BA CA CA

rF

= + + =

+

+ =

M R P R F R F 0i j i i j i j

i j i j 0

43 101 N m 41 N m+624 mm rF =k k k 0, 43 228 N 70.95 =74 215 N

r = +F i j


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43 43 43 48 257 N 261 N 100.58r = + = + = F F F i j

Now the lines of action for other forces may be found as shown.

43 43 23

r = + + + =F F F P F 0 23

74 215 N 122 42 N 350 N+ + + =i j i j i F 0, 23 =398 257 N 474 N 32.85 = F i j

( ) ( )2 212 32

12 100 173 mm 398 257 N

O AO

M

= + =+ + + =

M M R F 0k i j i j 0

12 94.55 N mM = 12 94.55 N m= M k Ans.

11.9 Sketch free-body diagrams of each link and show all the forces acting. Find themagnitude and direction of the moment that must be applied to link 2 to drive the linkageagainst the forces shown.

Kinematic analysis:

2

100 mm 30 86.6 50 mm= = +AO

R i j 4

250 mm 84.34 24.675 248.775 m= = +CO

R i j

350 mm 67.81 132.2 324.075 mm= = +BAR i j 350 mm 34.61 288.075 198.775 m= = +CAR i j

4

175 mm 84.34 17.275 174.15 mm= = +DOR i j



34 34andr F F . Then

4 4 4 34

= + = O DO D COM R P R F 0 ( ) ( ) ( )

( ) 34

17.275 174.15 mm 858.85 231.4 N + 24.675 248.775 mm

cos 5.66 sin 5.66 F

+ + +

+ =

i j i j i j

i j 0


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34 155.94 N m 249.975 mm =Fk k 0 34 614.1 N 5.66 =609.65 62.3 N

= F i j

( ) ( ) ( ) ( )

( ) ( )

43 43

43

132.2 324.075 mm 445 N + 288.075 198.775 mm 609.65 62.3 N

+ 288.075 198.775 mm cos 95.66 sin 95.66

= + + =

+ + +

+ + =

rA BA B CA CA

rF

M R P R F R F 0

i j i i j i j

i j i j 0

43 146.448 N m 141.363 N m 267.075 mm + =rFk k k 0,

43 1059.1 N 95.66 = 102.35 1054.65 N= rF i j

43 43 43 712 992.35 N 1219.3 125.66= + = = rF F F i j


43 43 23

r

B

= + + + =F F F P F 0 23

102.35 1054.65 N 609.65 62.3 N 445 N + + =i j i j i F 0,

23 =1157 992.35 N 1526.35 N 40.62+ = F i j

( ) ( )2 212 32

12 86.6 50 mm 1157 992.35 N

= + =+ + =

O AO

M

M M R F 0

k i j i j 0

12 28.476 N m= M 12 28.476 N m= M k Ans.

11.10 The figure shows a four-bar linkage with external forces applied at points B and C. Drawa free-body diagram of each link and show all the forces acting on each. Find the torquethat must be applied to link 2 to maintain equilibrium.

Kinematic analysis:

2

75 mm 30 65 38 mmAO = = R i j 4

200 mm 124.56 113 165 mmCO = = +R i j

200 mm 16.00 192 55 mmBA= = +R i j 300 mm 42.38 222 202 mmCA= = +R i j


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Force analysis:

( ) ( ) ( ) ( )

( ) ( )

43

43

192 55 mm 354 354 N + 222 202 mm 1 800 N

+ 222 202 mm cos124.56 sin124.56

A BA B CA C CA

F

= + + =

+ + +

+ + =

M R P R P R F 0i j i j i j i

i j i j 0

43 87.438 N m 363.000 N m 297 mm rF + =k k k 0,

43 927 124.56 N= 526 763 N= +F i j

43 23B C= + + + =F F P P F 0 23

526 763 N 354 354 N+1 800 N + + + =i j i j i F 0,

23 = 920 1 117 N 1 447 N 129.48 = F i j

( ) ( )2 212 32

12 65 38 mm 920 1 117 N

O AO

M

= + =

+ + =


12 107.57 N mM = 12 107.57 N m= M k Ans.

11.11 Draw a free-body diagram of each of the members of the mechanism shown in the figure,and find the magnitude and the direction of all the forces and moments. Compute themagnitude and direction of the torque that must be applied to link 2 to maintain staticequilibrium.

Kinematic analysis:

2

100 mm 180 100 mm= = AOR i 4 200 mm 124.23 112.5 165.35 m= = +COR i j

350 mm 55.98 195.85 290.075 mm= = +BAR i j 250 mm 41.41 187.5 165.35 mm= = +CAR i j

4 150 mm 95.27 13.775 149.375 mm= = +DOR i j


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34 34andr

F F . Then

( ) ( ) ( )

( )

4 4 4 34

34

13.775 149.375 mm 694.2 400.5 N + 112.5 165.35 mm

cos34.23 sin34.23

O DO D CO

F

= + =

+ + +

+ =

M R P R F 0

i j i j i j

i j 0

34 99.779 N m 200 mm =Fk k 0 34 489.5 N 34.23 =404.95 275.9 N

= +F i j

( ) ( ) ( ) ( )

( ) ( )

43 43

43

195.85 290.075 mm 534 N + 187.5 165.35 mm 404.95 275.9 N

+ 187.5 165.35 mm cos 55.77 sin 55.77

= + + =

+ +

+ + =

rA BA B CA CA

rF

M R P R F R F 0

i j i i j i j

i j i j 0

43 157.296 N m 15.481 N m 248.025 mm + =r

Fk k k 0

43 685.3 N 55.77 =387.15 565.15 N= rF i j

43 43 43 17.8 841.05 N 841.05 N 91.21= + = = rF F F i j Ans.


43 23B= + + =F F P F 0

23 17.8 841.05 N 534 N + =i j i F 0, 23 =551.8 841.05 N 1005.7 N 56.73+ = F i j Ans.

34 43 17.8 841.05 N 841.05 N 88.79= = + = F F i j Ans.

34 14D= + + =F F P F 0

14 17.8 841.05 N 694.2 400.5 N+ + + =i j i j F 0,

14 =676.4 1241.55 N 1415.1 N 61.42 = F i j Ans.

32 23 551.8 841.05 N 1005.7 N 123.27= = = F F i j Ans.

12 32 551.8 841.05 N 1005.7 N 56.73= = + = F F i j Ans.

( ) ( )2 212 32

12 100 mm 551.8 841.05 N

= + =

+ =

O AOM

M M R F 0

k i i j 0

12 85.428 N m= M 12 85.428 N m= M k Ans.

11.12

Determine the magnitude and direction of the forces that must be applied to link 2 tomaintain static equilibrium.

Kinematic analysis:

2

75 mm 90 75 mm= =AOR j 350 mm 12.37 341.875 75 mm= = CAR i j

175 mm 34.93 143.475 100.225 mm= = BAR i j


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Force analysis:

( ) ( ) ( ) ( )

( ) ( )

14

14

143.475 100.225 mm 222.5 N + 341.875 75 mm 445 N

+ 341.875 75 mm

= + + =

=

A BA B CA C CA

F

M R P R P R F 0

i j j i j i

i j j 0

14 32.431 N m 33.9 N m 350 mm + =Fk k k 0 14 4.45 N 90 =4.45 N= F j

14 23B C= + + + =F P P F F 0

23 222.5 N 445 N 4.45 N + + =i j F 0, 23 =445 226.95 N 498.5 N 27.02 = F i j

2 212 32O AO= + =M M R F 0 ( ) ( )12 75 mm 445 226.95 N+ + =M k j i j 0

12 33.9 N m= M 12 33.9 N m= M k Ans.

11.13 The photograph shows the Figee floating crane with leminscate boom configuration. Also

shown is a schematic diagram of the crane. The lifting capacity is 16 T (with 1 T = 1metric ton =1 000 kg) including the grab which is about 10 T. The maximum outreach is

30 m, which corresponds to 2 49 = . Minimum outreach is 10.5 m at 2 49 . = Other

dimensions are given in the figure caption. For the maximum outreach position and a

grab load of 10 T, find the bearing reactions at A,B, 2 4, and ,O O and the moment 12M

required. Notice that the photograph shows a counterweight on link 2; neglect thisweight and also the weights of the members.

Kinematic analysis:

2

14.700 m 49.00 9.644 11.094 mAO = = +R i j , 4 19.300 m 59.70 9.739 16.663 mBO = = +R i j

6.500 m 2.37 6.494 0.269 mBA= = +R i j 22.300 m 14.39 21.600 5.543 mCA= = +R i j


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Force analysis:Note that a metric ton is a unit of mass; the weight of a metric ton in standard gravity is

( )( )21 000 kg 9.81 m/s 9.810 kNW mg= = = . Therefore the rated load of the crane is98.100 kNF= .

( ) ( ) ( ) ( )43

43 6.494 0.269 m cos59.70 sin59.70 + 21.600 5.543 m 98.100 kN

A BA CA

F

= + =

+ + + =

M R F R F 0i j i j i j j 0

43 5.471 m 2 119 kN mF =k k 0, 43 387 59.70 kN=195 334 kN= +F i j

14 34 387 kN 59.70 =195 334 kN= = +F F i j Ans.

43 23= + + =F F F F 0

23 195 334 kN 98.1 kN+ + =i j j F 0

23 = 195 236 kN 307 kN 129.59 = F i j

12 32 = = 195 236 kN 307 kN 129.59 = F F i j Ans.

( ) ( )2 212 32

12 9.644 11.094 m 195 236 kN

O AO

M

= + =

+ + =


1247 kN mM =

12

47 kN m= M k Ans.

11.14 Repeat Problem 11.13 forthe minimum outreach position.

Kinematic analysis:

2

14.700 m 132.00 9.836 10.924 mAO = = +R i j , 4 19.300 m 120.35 9.751 16.656BO = = +R i j

6.500 m 3.81 6.486 0.432 mBA

= = +R i j 22.300 m 15.83 21.454 6.083 mCA= = +R i j


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Force analysis:Note that a metric ton is a unit of mass; the weight of a metric ton in standard gravity is

( )( )21 000 kg 9.81 m/s 9.810 kNW mg= = = . Therefore the rated load of the crane is98.100 kNF= .

( ) ( ) ( ) ( )43

43 6.486 0.432 m cos120.35 sin120.35 + 21.454 6.083 m 98.1 kN

A BA CA

F

= + =

+ + + =

M R F R F 0i j i j i j j 0

43 5.815 m 2 105 kN mF =k k 0 43 362 kN 120.35 = 183 312 kN= +F i j

14 34 362 kN 120.35 = 183 312 kN= = +F F i j Ans.

43 23= + + =F F F F 0 23 183 312 kN 98.1 kN + + =i j j F 0

23

=183 214 kN 282 kN 49.51 = F i j

12 32 = =183 214 kN 282 kN 49.51 = F F i j Ans.

( ) ( )2 212 32

12 9.836 10.924 m 183 214 kN

O AO

M

= + =

+ + + =


12 109 kN mM = 12 109 kN m= M k Ans.

11.15 Repeat Problem 11.7 assuming coefficients of Coulomb friction 0.20c = between links

1 and 6 and 0.10c = between links 3 and 4. Determine the torque 12M necessary to

drive the system, including friction, against the load P.

See the figure and solution for Problem P11.7 for the kinematic and frictionless solution.

For friction between links 1 and 6, the friction angle is ( )1tan 0.20 11.31 = = . Since

the impending motion6 /1C

V is to the left the friction force 16 16n

cf F= is toward the right.

Also, since the non-friction normal force 16n

F is downward (from the solution for Prob.

P11.7), the total force 16F acts at the angle 90 11.31 78.69 + = . Therefore,


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( ) ( )16 56 cos 78.69 sin 78.69 cos175.20 sin175.20P F F= + + + + =F i i j j i j 0 16 56445 N cos 78.69 cos175.20 0+ + =F F , and 16 56sin 78.69 sin175.20 0F F + =

16 38.76 N=F 56 454.19 N=F

56 454.19 N 175.20 452.6 38 N= = +F i j

For friction between links 3 and 4, the friction angle is ( )1tan 0.10 5.71 = = . Since

the impending motion3 / 4A

V is upward the friction force 34 34n

cf F= is upward. Also,

since the non-friction normal force 34n

F is toward the left (from the solution for Prob.

P11.7), the total force 34F acts at the angle 163.37 5.71 157.66 = . Therefore,

( ) ( )

( ) ( )

4 4 454 34

34

114.45 383.275 mm 452.6 38 N

53.4 178.85 mm cos157.66 sin157.66

= + =

+

+ + + =

O BO AO

F

M R F R F 0

i j i j

i j i j 0

( )34 34 180.176 N m 185.725 mm 957.45 N 157.66 = 885.58 363.94 N + = = +F k 0 F i j

( ) ( )2 212 32

12 54.125 31.25 mm 885.58 363.94 N

= + =

+ + =

O AOM

M M R F 0

k i j i j 0

12 48.117 N m= M 12 48.117 N m= M k Ans.

11.16 Repeat Problem 11.12 assuming a coefficient of static friction 0.15 = between links 1and 4. Determine the torque 12M necessary to overcome friction.

See the figure and solution for Problem P11.12 for the kinematic and frictionless

solution. For friction between links 1 and 4, the friction angle is ( )1tan 0.15 8.53 = = .

Since the impending motion4 /1C

V is to the right the friction force 14 14n

cf F= is toward

the left. Also, since the non-friction normal force 14n

F is upward (from the solution for

Prob. P11.12), the total force 14F acts at the angle 90 8.53 98.53 + = . Therefore,

( ) ( ) ( ) ( )

( ) ( )

14

14

143.475 100.225 mm 222.5 N + 341.875 75 mm 445 N

+ 341.875 75 mm cos98.53 sin 98.53

= + + =

+ =

A BA B CA C CA

F

M R P R P R F 0

i j j i j i

i j i j 0

14 32.431 N m 33.9 N m 326.95 mm + =Fk k k 0, 14 4.42 N 98.53 = 0.654 4.374 N= +F i j

14 23B C= + + + =F P P F F 0 23

222.5 N 445 N 0.654 4.374 N + + =i i j F 0, 23 =445.445 226.95 N 500.18 N 26 = F i j

2 212 32O AO= + =M M R F 0 ( ) ( )12 75 mm 445.445 226.95 N+ + M k j i j

12 33.94 N m= M 12 33.94 N m= M k Ans.


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11.17 In each case shown, pinion 2 is the driver, gear 3 is an idler, the gears have module of

4 mm/tooth and 20 pressure angle. For each case, sketch the free-body diagram ofgear 3 and show all forces acting. For (a) pinion 2 rotates at 600 rev/min and transmits13.43 kw to the gearset. For (b) and (c), pinion 2 rotates at 900 rev/min and transmits18.65 kw to the gearset.

a) 22m 18 teeth 4 mm/tooth

36 mm2 2

= = =

NR 33

m 34 teeth 4 mm/tooth68 mm

2 2

= = =

NR

( )( )2 600 rev/min 2 62.832 rad/s cw60 s/min

= = , 23 23

36 mm 62.832 rad/s 33.264 rad/s ccw68 mm

= = =RR

( ) ( )23 3 3

13.43 1000 w (1 N m) (1000 mm/m)5.937 kN

68 mm 33.264 rad/s

= = =t

PF

R

23 23 cos 5937 N cos20 6318 Nt

F F = = = , 43 23 6318 N= =F F

23 43 13= + + =F F F F 0 13 23 43 5937 5937 11.874 kN= + = + =t tF F F Ans.

b) 22m 18 teeth 4 mm/tooth

36 mm2 2

= = =

NR 33


2 2

= = =

NR

( ) ( )2

900 rev/min 294.248 rad/s ccw

60 s/min

= = , 23 2

3

36 mm94.248 rad/s 47.124 rad/s cw

72 mm = = =

R

R

( ) ( )23 3 3(18.65 kw) (1000 w/kw) (1 N m/w) (1000 mm/m) 5497 N

72 mm 47.124 rad/s = = =t PF

R

23 23 cos 5497 N cos20 5849 N= = =t

F F , 43 23 5849 N= =F F


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23 43 13= + + =F F F F 0 ( )

( )

13 23 43

5849 N 20 5849 N 110

4945 N 225

= +

= +

=

F F F

Ans.

c) 22m 18 teeth 4 mm/tooth

36 mm2 2

= = =

NR 33


2 2

= = =

NR

( ) ( )2 900 rev/min 2 94.248 rad/s ccw60 s/min

= = , 23 23

36 mm 94.248 rad/s 47.124 rad/s cw72 mm

= = =RR

( )( ) ( )

( ) ( )23

3 3

18.65 kw 1000 w/kw 1 N m/w 1000 mm/m5497 N

75 mm 47.124 rad/s

= = =t

PF

R

23 23 cos 5497 N cos20 5849 N= = =t

F F , 43 23 5849 N= =F F

23 43 13= + + =F F F F 0

( )

( )13 23 43

5849 N 20 5849 N 70

10.6 N 45

= +

= +

=

F F F

Ans.

11.18 A 15-tooth spur pinion has a module of 5 mm/tooth and 20 pressure angle, rotates at600 rev/min, and drives a 60-tooth gear. The drive transmits 18.65 kw. Construct a free-body diagram of each gear showing upon it the tangential and radial components of theforces and their proper directions.

22

m 15 teeth 5 mm/tooth37.5 mm

2 2

= = =

NR 33


2 2

= = =

NR

( )( )2

600 rev/min 262.832 rad/s

60 s/min

= = 23 2

3

37.5 mm62.832 rad/s 15.708 rad/

150 mm = = =

R

R


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( )( ) ( ) ( )( )( )32 32 323 3

18.65 kw 1000 w/kw 1 N m/w 1000 mm/m7915 N, tan 288

37.5 mm 62.832 rad/s

= = = = =t r t

PF F F

R

23 32 7915 N= =t t

F F 23 32 2880.8 N= =r r

F F

11.19 A 16-tooth pinion on shaft 2 rotates at 1 720 rev/min and transmits 3.73 kw to the

double-reduction gear train. All gears have 20 pressure angle. The distances betweencenters of the bearings and gears for shaft 3 are shown in the figure. Find the magnitudeand direction of the radial force that each bearing exerts against the shaft.

22


2 2

= = =

NR


2 2

= = =AA

NR


2 2

= = =BB

NR 4


2 2

= = =A

NR

( ) ( )2

1 720 rev/min 2180.118 rad/s

60 s/min

= = 23 2

24 mm180.118 rad/s 45.029 rad/s

96 mm

= = =A

R

R

4 3

4

48 mm45.029 rad/s 30.020 rad/s

72 mm = = =B

R

R

( )( ) ( ) ( )

( ) ( )23

3

3.73 kw 1000 w/kw 1 N m/w 1000 mm/m862.8 N

96 mm 45.029 rad/s

= = =t

A

PF

R,

23 23 cos 918.17 N= =t

F F

43 23

96 mm862.8 N 1725.6 N

48 mm= = =t tA

B

RF F

R 43 43 cos 1836.3 N= =

tF F


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169

Choosing a coordinate system with origin at Cas shown we have

23 918.17 N 20 862.8 314.03 lb= = = +AF F i j 96 48 mm= +AR j k

43 1836.3 N 20 1725.56 628 N= = = BF F i j 48 240 mm= +BR j k

x yC C CF F= +F i j C=R 0

x yD D DF F= +F i j 288 mm=DR k

C A A B B D D= + + =M R F R F R F 0

( ) ( ) ( ) ( )

( ) ( )

96 48 mm 862.8 314.03 N 48 240 mm 1725.56 628 N

288 mm

+ + + +

+ + =x yD D

F F

j k i j j k i j

k i j 0

( ) ( ) ( ) 15 41.35 82.72 N m 150.7 414 82.72 N m 288 288 mm + + + + + + =y xD DF Fi j k i j k i j 01495 N, 445 N= =x yD DF F 1557.5 N 163.42 1495 445 N= = +DF i j Ans.

A B C D= + + + =F F F F F 0 ( ) ( ) ( ) 862.8 314.03 N 1495 445 N 1495 445 N+ + + + + =Ci j i j F i j 0

961.2 N 188.86 952.3 146.85 N= = CF i j Ans.

11.20 Solve Problem 11.17 if each pinion has right-hand helical teeth with a 30 helix angleand a 20 pressure angle. All gears in the train are helical, and, of course, the normalmodule is 4 mm/tooth for each case.


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170

a) 22m 18 teeth 4 mm/tooth

36 mm2 2

= = =

NR 33


2 2

= = =

NR

( )( )2

600 rev/min 262.832 rad/s cw

60 s/min

= = , 23 2

3

36 mm62.832 rad/s 33.264 rad/s ccw

68 mm = = =

R

R

( )( ) ( )( )( )

23

3 3

13.42 kw 1000 w/kw 1 N m/w 1000 mm/m 550 ft lb/s/hp 12 in/ft593268 mm 33.268 rad/s

= = =

t PF R

Since the pressure angles and the helix angle are related by cos tan tann t = ,

( ) ( )1 1tan tan cos tan tan 20 cos30 22.80t n = = =

( )23 23 tan 5932 N tan 22.80 2493.5 N= = =r t

tF F , ( )23 23 tan 5932 N tan 30 3424.8 N= = =

a tF F

23 5932 2493.5 3424.8 N= +F i j k 43 5923 2493.5 3424.8 N= + F i j k

23 43 13= + + =F F F F 0 13 11.864 N= F i Ans.3 23 3 43 13 R R= + =M j F j F M 0

( ) ( ) ( ) ( )13 68 mm 5932 2493.5 3424.8 N 68 mm 5932 2493.5 3424.8 N + + + =i j k j i j k M 0

( ) ( ) ( ) ( ) 13 68 mm 5932 2493.5 3424.8 N 68 mm 5932 2493.5 3424.8 N + + + =i j k j i j k M 0

13465.45 N m= M i This moment must be supplied by the shaft bearings. Ans.

b) 22m 18 teeth 4 mm/tooth

36 mm2 2

= = =

NR 33


2 2

= = =

NR

( ) ( )2


60 s/min

= = , 23 2

3

36 mm94.248 rad/s 47.124 rad/s cw

72 mm = = =

R

R

( )( ) ( )( ) ( )23 3 3

18.65 kw 1000 w/kw 1000 w/kw 1000 mm/m5496.7 N

72 mm 47.124 rad/s

= = =t

PF

R

Since the pressure angles and the helix angle are related by cos tan tann t

= ,

( ) ( )1 1

tan tan cos tan tan 20 cos30 22.80t n

= = = ( )23 23 tan 5496.7 N tan 22.80 2310.59 N= = =

r t

tF F , ( )23 23 tan 5496.7 N tan 30 3173= = =a t

F F


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171

23 5496.7 2310.59 3173.5 N= F i j k 43 2310.59 5496.7 3173.5 N= + +F i j k

23 43 13= + + =F F F F 0 13 3186.11 3186.11 N= F i j Ans.3 23 3 43 13 R R= + + =M j F i F M 0

( ) ( ) ( )

( ) 13

72 mm 5496.7 2310.59 3173.5 N 72 mm

2310.59 5496.7 3173.5 N

+

+ + + =

j i j k i

i j k M 0

13 228.49 228.49 N m= + M i j This moment must be supplied by the shaft bearings. Ans.

c) 22m 18 teeth 4 mm/tooth

36 mm2 2

= = =

NR 33


2 2

= = =

NR

( )( )2


60 s/min

= = , 23 2

3

36 mm94.248 rad/s 47.124 rad/s cw

72 mm = = =

R

R

( ) ( )23 3 3

(18.65 kw) (1000 w/kw) (1 N m/w) (1000 mm/m)5496.7 N

72 mm 47.124 rad/s

= = =t

PF

R

Since the pressure angles and the helix angle are related by cos tan tann t = ,

( ) ( )1 1tan tan cos tan tan 20 cos30 22.80t n = = =

( )23 23 tan 5496.7 N tan 22.80 2310.59 N= = =r t

tF F ,

( )23 23 tan 5496.7 N tan 30 3173.5 N= = =a tF F

23 5496.7 2310.59 3173.5 N= F i j k 43 2310.59 5496.7 3173.5 N= +F i j k

23 43 13= + + =F F F F 0 13 7807.29 7807.29 N= +F i j Ans.3 23 3 43 13 R R= + + =M j F i F M 0

( ) ( ) ( )

( ) 13

72 mm 5496.7 2310.59 3173.5 N 72 mm

2310.59 5496.7 3173.5 N

+

+ + =

j i j k i

i j k M 0

13 228.49 228.49 N m= M i j This moment must be supplied by the shaft bearings. Ans.


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172

11.21 Analyze the gear shaft of Example 11.8 and find the bearing reactions andC DF F .

The solution is shown in Fig. 11.20c.

11.22 In each of the bevel gear drives shown in the figure, bearing Atakes both thrust load andradial load, while bearingBtakes only radial load. The teeth are cut with a 20 pressureangle. For (a) 2

20 N m= T i and for (b) 2 26.7 N m= T k . Compute the bearing

loads for each case.

a) ( )1tan 32 teeth 16 teeth 63.43 = = 32 2 2 20 N m/0.018 m 1111 N= = =t

F T R

32 32 tan cos 180.87 N= =r t

F F 32 32 tan sin 361.66 N= =a t

F F

23 180.87 1111 361.66 N= + +F i j k

23 3A BA B PA= + + =M R F R F T 0

( ) ( ) ( ) ( )3

50 mm 35 59 mm 180.87 1111 361.66 N + + + + + =

x y

B BF F Tk i j i k i j k k 0

( ) ( ) 3 50 mm 50 mm 70 1.9 40.68 N m + + + =y xB BF F Ti j i j k k 0

340.68 N m= T k 37.825 1370.6 N= BF i j Ans.

23A B= + + =F F F F 0 218.695 259.0 361.66 N= + AF i j k Ans.

b) ( )1tan 18 teeth 24 teeth 36.87 = = 32 2 2 26.7 N m 32 mm 834.375 Nt

F T R= = =

32 32 tan cos 242.95 N = =r t

F F 32 32 tan sin 182.21 N = =a t

F F

32 242.95 834.375 182.21 N= +F i j k

32 2A BA B PA= + + =M R F R F T 0 ( ) ( ) ( ) ( ) ( ) 50 mm 32 20 mm 242.95 834.375 182.21 N 26.7 N m + + + + x yB BF Fk i j i k i j k k

( ) ( ) ( ) 50 mm 50 mm 16.95 0.9 26.7 N m 26.7 N m + + + + + =y xB BF Fi j i j k k 0 17.8 333.75 N= BF i j Ans.

23A B= + + =F F F F 0 222.5 1170.35 182.45 N= + AF i j k Ans.

11.23 The figure shows a gear train composed of a pair of helical gears and a pair of straightbevel gears. Shaft 4 is the output of the train and delivers 4.5 kw to the load at a speed of

370 rev/min. All gears have pressure angles of 20 . If bearing Eis to take both thrustload and radial load, while bearing Fis to take only radial load, determine the forces thateach bearing exert against shaft 4.

The diameters of the bevel gears at their large ends are

( )4 4m 2 3 mm/tooth 40 teeth 2 60 mmR N= = =

( )3 3mN 2 3 mm/tooth 20 teeth 2 30 mmR = = =


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173

( )1 4 3tan 63.43R R = = ( )1 3 4tan 26.57R R

= =

The average pitch radii are

4,avg 4 0.500 sin 59.55 mm= =R R 3,avg 3 0.500 sin 29.78 mm= =R R

( )( )4

370 rev/min 238.746 rad/s

60 s/min

= =

( )( )34 4,avg 4

4.5 kw 1000 w/kw 1 N m/w 1000 mm/m1950 N

59.55 mm 38.746 rad/s

= = =t

PF

R

34 34 tan cos 317.46 N= =r t

F F 34 34 tan sin 634.78 N= =a t

F F

34 634.78 317.46 1950 N= + F i j k

34 4E FE F PE= + + =M R F R F T 0

( ) ( ) ( ) ( ) ( ) 60 mm 18 59.56 mm 634.78 317.46 1950 N 115.5 N m + + + + =y zF FF Fi j k i j i j k i

( ) ( ) ( ) 60 mm 60 mm 115.5 40.7 31 N m 115.5 N m + + + + =z yF FF Fk i j k i 0

430.98 564.192 N= +FF j k Ans.

34E F= + + =F F F F 0 634.78 748.34 1386.97 N= +EF i j k Ans.

11.24 Using the data of Problem 11.23, find the forces exerted by bearings CandDonto shaft3. Which of these bearings should take the thrust load if the shaft is to be loaded incompression?

The pitch radius of the helical gear S is

m 2 2 mm/tooth 35 teeth 2 35 mm= = =S SR N

( )23 43 3,avg 1950 N 29.78 mm/35 mm 1659.17 N= = =t t

SF F R R

( )23 23 tan 1659.17 N tan 20 603.89 N= = =r tF F ( )23 23 tan 1659.17 N tan 30 957.92= = =

a tF F

23 603.89 957.92 1659.17 N= + +F i j k

43 23C DC D PC RC = + + =M R F R F R F 0 ( ) ( ) ( ) ( )

( ) ( )

44 mm 29.78 78.21 mm 634.78 317.96 1950 N

35 22 mm 603.89 957.92 1659.17 N

+ + + +

+ + + + =

x z

D DF Fj i k i j i j k

i j i j k 0

( ) ( ) ( ) 44 mm 44 mm 152.5 58 40 N m 36.5 58 46.81 N m + + + + =z xD DF Fi k i j k i j k 0 133.5 4192 N=

DF i k Ans.

23 43C D= + + + =F F F F F 0 284.8 538.45 418.3 N= +CF i j k Ans.Since the thrust force is in the direction, Cshould be a thrust bearing.


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174

11.25 Use the method of virtual work to solve the slider-crank mechanism of Problem 11.2.

( ) ( )1 1sin sin sin 75 mmsin105 350 mm 11.95r = = =

cos cos 75 mmcos105 350 mmcos11.95 323 mmx r = + = + = The first-order kinematic coefficient is

24tan 323 mm tan11.95 68.36 mm

Px dx d y x = = = = =

( )12 900 N 68.36 mm 61.5 N m cwM Px= = = Ans.

11.26 Use the method of virtual-work to solve the four-bar linkage of Problem 11.5.

24 264.425 mm=P OR , 24 4 114.425 mm=P OR , 4 175 mm 152.64= DOR

The first-order kinematic coefficient is

24 2 24 44 4 264.425 mm/114.425 mm 0.563 = = = =P O P Od d R R

( )414

sin152.64 175 mm 222.5 N sin152.64 17.89 N m cw= = = DOM R P

( )( )12 14 4 2 14 4 17.89 N m cw 0.563 10.07 N m ccw = = = = M M d d M Ans.

11.27 Use the method of virtual work to analyze the crank-shaper linkage of Problem 11.7.

Given that the load remains constant at 445 N,=P i find and plot a graph of the cranktorque 12M for all positions in the cycle using increments of 30 for the input crank.

4 4 4 2cos 63cos = =AO AOx R 4 4 4 2sin 150 63sin = = +AO AOy R

1 24

2

150 63sintan

63cos

+=

4 2

1056.25 750sin= +AOR


26/30

175

24 4 4 4sinP O AOy R = 24 4

4

24 4

4

4 42

150

1 150 sin

= = =

P O

AOP O

yd

Rd y

5 4200sin 400 400sin = = BCy ( )1

5 4sin 2 2sin =

( )

( ) ( )

( )

46

5 5 4 5

5 4 5

200sin 200cos 400cos 200cos

200 sin 2cos tan

=

= =

P C BC C CBy y x x

46 44 5 4 5400 200sin 400cos tan = = + C P Odx d y

( ) ( )12 4 4 2CM dx d d d P =

Values for one cycle are shown in the following table.

2 (deg.) 4 (deg.) 4 (mm)AOR 4 2d d 5 (deg.) 4 (mm)Cdx d 12 (N m)M

0 67.38 162.5 0.14793 -8.85 393.18 26.2930 73.37 189.15 0.24015 -4.80 392.88 42.6560 81.30 206.5 0.28197 -1.32 396.79 50.5790 90.00 212.5 0.29412 0.00 400 53.18120 98.70 206.5 0.28197 -1.32 393.99 50.21150 106.63 189.15 0.24015 -4.80 373.65 40.56180 112.62 162.5 0.14793 -8.85 395.48 23.09210 114.50 130.5 -0.04593 -10.37 333.64 -6.93240 108.05 100.85 -0.41416 -5.65 368.04 -68.89270 90.00 87.5 -0.71429 0.00 400 -129.14300 71.95 100.85 -0.41416 -5.65 392.57 -73.49330 65.50 130.5 -0.04593 -10.37 394.35 -8.19360 67.38 162.5 0.14793 -8.85 393.18 26.29


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176

The values of 12M from this table are graphed as follows:

11.28 Use the method of virtual work to solve the four-bar linkage of Problem 11.10.

23 12 23 133 3 275 mm 688 mm 0.1089P P P Pd d R R = = = =

( )132 3

0.1089 568 mm 135.33 61.881 mm 45.33B B BPd d = = = = R R k R k

( )132 3

0.1089 662 mm 124.56 72.153 mm 34.56C C CPd d = = = = R R k R k

( ) ( ) ( ) ( )12

500 N 135 61.881 mm 45.33 1 800 N 0 72.153 mm 34.56

30.940cos89.67 N m+129.876cos34.56 N m

B B C CM = +

= +

=

P R P Ri i

i i

12 107 N m cwM = Ans.


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177

11.29 A car (link 2) which weighs 8900 N is slowly backing a 4450 N trailer (link 3) up a 30 inclined ramp as shown in the figure. The car wheels are of 325 mm radius, and thetrailer wheels have 250 mm radius; the center of the hitch ball is also 325 mm above the

roadway. The centers of mass of the car and trailer are located at 2G and 3,G

respectively, and gravity acts vertically downward in the figure. The weights of thewheels and friction in the bearings are considered negligible. Assume that there are nobrakes applied on the car or on the trailer, and that the car has front-wheel drive.Determine the loads on each of the wheels and the minimum coefficient of static frictionbetween the driving wheels and the road to avoid slipping.

For the trailer:

3

945 863.15 mm 1280 mm 42.41= + = G BR i j

313 31250 mm= + = B G BFM k R W 0

13 3364.2 N 120 1682.1 2914.75 N= = +F i j Ans.

13 23 3= + + =F F F W 0 23 2278.4 N 42.41 1682.1 1535.25 N= = +F i j

For the car:

( ) ( )12 32 2000 mm 800 mm 8900 N 2900 325 mm= + + =R

P FM k k i j F 0

125513.5 N=RF j Ans.

12 12 3212

F Rf= + + + =F F i F F 0 12 4921.7 N=FF j Ans.12

1682.1 N=f i Ans.

12 12 1682.1 N 4921.7 NF

f F = 0.342 Ans.

11.30 Repeat Problem 11.29 assuming that the car has rear-wheel drive rather than front-wheeldrive.

The entire solution is identical with that of Problem P11.29 except that friction force 12f acts on the rear wheel of the car instead of on the front wheel. The solution process andall values are the same until the final step. Then

12 12 1682.1 N 5513.5 NR

f F = 0.305 Ans.


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178

11.31 The low-speed disk cam with oscillating flat-faced follower shown in the figure is drivenat a constant shaft speed. The displacement curve for the cam has a full-rise cycloidalmotion, defined by Eq. (5.19) with parameters 30L= , 30= , and a prime circleradius 30 mmoR = ; the instant pictured is at 2 112.5 = . A force of 8 NCF = is

applied at point Cand remains at 45 from the face of the follower as shown. Use thevirtual work approach to determine the moment 12M required on the crankshaft at theinstant shown to produce this motion.

The moment on link 3 caused by the output load is

( )( )313

150 mm 8 N sin 135 0.849 N mCO C= = = M R F k k

From Eq. (5.19b),

2 30 112.51 cos 1 cos 2 0.200

150 150

Ly

= = =

From virtual work

( )12 3 2 13 13 0.200 0.849 N m 0.170 N md d y = = = = M M M k k Ans.

11.32 Repeat Problem 11.31 for the entire lift portion of the cycle, finding 12M as a function of

2.

From Problem P11.31

( )( )313

150 mm 8 N sin 135 0.849 N mCO C= = = M R F k k

( )2 23602 30

1 cos 1 cos 0.200 1 cos 2.4150 150

Ly

= = =

( )( ) ( )12 13 2 2 0.200 1 cos 2.4 0.849 N m 0.170 1 cos 2.4 N my = = = M M k k Ans.

11.33 A disk 3 of radius R is being slowly rolled under a pivoted bar 2 driven by an appliedtorque Tas shown in the figure. Assume a coefficient of static friction of between the

disk and ground and that all other joints are frictionless. A force F is acting vertically

downward on the bar at a distance dfrom the pivot 2.O Assume that the weights of the

links are negligible in comparison to F. Find an equation for the torque Trequired as a

function of the distance2CO

X R= , and an equation for the final distance X that is

reached when friction no longer allows further movement.


30/30

179

20

O =M 32

cos

cosB B

d dF F F

X X

= =

0C=M 23 sin sinB

RdT F R F

X = =

But, for geometric compatibility, sinR X = . Therefore,

( )22sinB BT F d F d R X = = Ans.

Also, 2

2

sec 1 tan 11

sin tan tan tan

RX R R R

+= = = = +

Motion is still possible as long as tan , or as long as21 1X R + Ans.

solutions instructor manual chapter 11 static force analysis

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