PART III
DYNAMICS OF MACHINES
149
151
Chapter 11
Static Force Analysis 11.1
The figure shows four mechanisms and the external forces and torques exerted on or by the mechanisms. Sketch the free-body diagram of each part of each mechanism. Do not attempt to show the magnitudes of the forces, except roughly, but do sketch them in their proper locations locations and directions. directions.
151
Chapter 11
Static Force Analysis 11.1
The figure shows four mechanisms and the external forces and torques exerted on or by the mechanisms. Sketch the free-body diagram of each part of each mechanism. Do not attempt to show the magnitudes of the forces, except roughly, but do sketch them in their proper locations locations and directions. directions.
11.2
152
What moment M12 must be applied to the crank of the mechanism shown if P = 0.9 kN ? Kinematic analysis: φ = sin −1 ⎡⎣( r sin θ ) ⎤⎦ = sin −1 ⎡⎣( 75 mm sin 105 ° ) 350 mm ⎤⎦ = 11.95 °
153
Force analysis: P = −900ˆi N F = P + F ˆj + F
∑
14
34
ˆj) = 0 ( cos φ ˆi − sin φ ˆj) = −900 Nˆi + F ˆj + F (0.978ˆi − 0.207 14
−900 N + 0.978 F 34 = 0 F14 − 0.207 F 34 = 0 F32
= 900 N 0.978 = 920 N F 14 = 0.207 ( 920 N ) = 190 N F 34
= −F34 = 920 N ( −0.978ˆi + 0.207 ˆj) = −900ˆi + 190 ˆj N
∑M = M
12
+ r2 × F32 = M12 + 75 mm ( cos105ˆi + sin105ˆj) × ( −900 ˆi + 190 ˆj N ) = 0
M12 + 61.5kˆ N ⋅ m = 0 11.3
34
M12
= −61.5kˆ N ⋅ m
Ans.
If M 12 = 100 N ⋅ m for the mechanism shown, what force P is required to maintain static equilibrium? Kinematic analysis: Recall φ = 11.95° from Prob. P11.2. x = r cos θ + cos φ = 75 mm cos105 ° + 350 mm cos11.95 ° = 323 mm
Force analysis: z M O = xF14 − M 12
∑
=0
F14
= M 12
Recall the force polygon on link 4 from Prob. P11.2. P = F 14 tan φ = 310 N tan11.95 ° = 1 463 N
11.4
x = 100N ⋅ m 0.323 m
= 310 N Ans.
Find the frame reactions and torque M12 necessary to maintain equilibrium of the four bar linkage shown in the figure. Kinematic analysis: R AO2 = 88 mm∠210° = −75.775ˆi − 43.75 ˆj mm R BA
= 150 mm∠82.83° = 18.725ˆi + 148.825ˆj mm
R BO4
= 150 mm∠135.53° = −107.04 ˆi +105.07 ˆj mm
R CO4
= 100 mm∠135.53° = −71.375ˆi + 70.05 ˆj mm
154
Force analysis: M O4 = R CO4 × P + R BO4
×F = 0 ∑ ( −71.375ˆi + 70.05ˆj mm ) × (311.72ˆi + 317.57 ˆj N ) + ( −107.04ˆi + 105.07 ˆj mm ) × (cos82.83 °ˆi + sin 82.83 °ˆj) F 34
34
( −45.2 N − m − 119.325 mmF 34 ) kˆ = 0 F 34 = −9.47 N − m ∑ MO = M12 + R AO × F32 = 0 2
M 12kˆ + M 12
11.5
F34
=0
= −372.94 N∠82.83° = −46.55ˆi − 370.02ˆj N
2
( −75.775ˆi − 43.75ˆj mm ) × ( −46.55ˆi − 370.02 ˆj N ) = 0
= 26.408 N − m
M12
= 26.408kˆ N − m
Ans.
What torque must be applied to link 2 of the linkage shown to maintain static equilibrium? Kinematic analysis: R AO2 = 87.5 mm∠240° = −43.75ˆi − 75.775 ˆj mm R BA
= 150 mm∠105.26° = −39.475ˆi + 144.7ˆj mm
R BO4
= 150 mm∠152.64° = −133.225ˆi + 68.925 ˆj mm
R DO4
= 175 mm∠152.64° = −155.425ˆi + 80.425 ˆj mm
155
Force analysis: M O4 = R DO4 × P + R BO4
×F = 0 ∑ ( −155.425ˆi + 80.425ˆj mm ) × (222.5ˆi N ) + ( −133.225ˆi + 68.925ˆj mm ) × (cos105.26 °ˆi + sin105.26 °ˆj) F 34
34
=0
( −18.176 N − m −110.375 mmF 34 ) kˆ = 0 F 34 = −4.116 N − m F34 = −162.109 N∠105.26° = 42.666ˆi − 156.39ˆj N ∑ MO = M12 + R AO × F32 = 0 2
M 12kˆ + M 12
11.6
2
( −43.75iˆ − 75.775ˆj mm ) × ( −42.666ˆi +156.39 ˆj N ) = 0
= 10.23 N − m
M12
= 10.23kˆ N − m
Ans.
Sketch a complete free-body diagram of each link of the linkage shown. What force P is necessary for equilibrium? Kinematic analysis: R AO2 = 100 mm∠90° = 100 ˆj mm R BA
= 150 mm∠ − 4.86° = 149ˆi − 13ˆj mm
R BO4
= 125 mm∠44.3° = 89ˆi + 87 ˆj mm
R CO4
= 200 mm∠44.3° = 143ˆi +140 ˆj mm
R DC = 400 mm∠ − 20.44° = 375ˆi − 140ˆj mm
Force analysis: M O2 = M12 + R AO2 × F32
∑
(
90kˆ N ⋅ m + 100ˆj mm
=0
) × (cos − 4.86 °ˆi + sin − 4.86 °ˆj) F
( 90 N ⋅ m − 99.640 mmF 32 ) kˆ = 0
32
F32
=0
= 903 N∠ − 4.86°
156
∑M = R ×F + R ×F = 0 (89ˆi + 87ˆj mm ) × (cos175.14°ˆi + sin175.14 °ˆj)903 N + (143ˆi + 140ˆj mm ) × ( cos − 20.44 °ˆi + sin − 20.44 °ˆj) F O4
BO4
34
CO4
54
54
=0
(86 N ⋅ m −181 mmF 54 ) kˆ = 0 F54 = 472 N ∠ − 20.44 °=44 ˆi −165 ˆj N P = 443ˆi N ∑ F = Pˆi − 443 Nˆi +165 Nˆj + F 16ˆj=0 11.7
Ans.
Determine the torque M12 required to drive slider 6 of the figure against a load of P = 445 N at a crank angle of θ = 30° , or as specified by your instructor. Kinematic analysis: R AO2 = 63 mm∠30° = 54.125ˆi + 31.25ˆj mm R BO4
= 400 mm∠73.37° = 114.45ˆi + 383.275 ˆj mm
Force analysis: F = Pˆi + F16 ˆj + cos175.20°ˆi + sin175.20 ˆj F 56
∑
(
)
R AO4
= 186.65 mm∠73.37° = 53.4ˆi + 178.85ˆj mm
R CB
= 200 mm∠175.20° = −199.3ˆi + 16.725 ˆj mm
=0
= − P cos175.20 = −445 N −0.996 = 446.56 N ; F56 = 446.56 N∠175.20° = −445ˆi + 37.34 ˆj N ∑ MO = R BO × F54 + R AO × F34 = 0 F56
4
4
4
(114.45ˆi + 383.275ˆj mm ) × (445ˆi − 37.34 ˆj N ) + ( 53.4ˆi + 178.85ˆj mm ) × ( cos163.37 °ˆi + sin163.37 °ˆj) F
34
( −177.58 N − m + 186.65 mmF 34 ) kˆ = 0
F34
=0
= 936.67 N ∠163.37 °= − 897.49 ˆi + 268.07 ˆ N
157
∑M
O2
= M12 + R AO × F32 = 0 2
(
M 12kˆ + 54.125ˆi + 31.25ˆj mm M 12
11.8
= 43.22 N − m
) × (897.49ˆi − 268.07 ˆj N ) = 0 M12
= 43.22kˆ N − m
Ans.
Sketch complete free-body diagrams for the illustrated four-bar linkage. What torque M12 must be applied to 2 to maintain static equilibrium at the position shown? Kinematic analysis: R AO2 = 200 mm∠60° = 100ˆi + 173ˆj mm R BA
R CO4
= 350 mm∠ − 109.05° = −114ˆi − 331ˆj mm
= 400 mm∠ − 46.06° = 278ˆi − 288ˆj mm , R CA = 700 mm∠ − 46.06° = 486ˆi − 504 ˆj mm
Force analysis: Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components, θ
r
F34 and F34 . Then
∑M
O4
= M14 + R CO × F34θ = −45kˆ N ⋅ m+ ( −114ˆi − 331ˆj mm ) × (cos −19.05 °ˆi + sin −19.05 °ˆj) F 34θ = 0 4
−45kˆ N ⋅ m+350kˆ mmF 34θ = 0 ∑ M A = R BA × P + RCA × F43θ + RCA × F43r = 0
θ F34
= 129 N∠ − 19.05°=122ˆi − 42 ˆj N
( 278ˆi − 288ˆj mm ) × ( −350ˆi N) + ( 486ˆi − 504ˆj mm ) × ( −122ˆi + 42ˆj N) + ( 486ˆi − 504ˆj mm ) × ( cos 70.95°ˆi + sin 70.95° ˆj) F = 0 r 43
−101kˆ N ⋅ m − 41kˆ N ⋅ m+624kˆ mmF 43r = 0 ,
r
F43
= 228 N∠70.95°=74ˆi + 215ˆj N
158
F43
= F43r + F43θ = −48ˆi + 257 ˆj N = 261 N∠100.58°
Now the lines of action for other forces may be found as shown. θ + P + F23 = 0 F = F43r + F43
∑
74ˆi + 215ˆj N − 122ˆi + 42 ˆj N − 350 ˆi N + F23 = 0 , F23 =398ˆi − 257ˆj N = 474 N∠ − 32.85°
∑M
O2
= M12 + R AO × F32 = 0 2
(
M 12kˆ + 100ˆi + 173ˆj mm M 12
11.9
) × ( −398ˆi + 257 ˆj N ) = 0
= −94.55 N ⋅ m
M12
= −94.55kˆ N ⋅ m
Ans.
Sketch free-body diagrams of each link and show all the forces acting. Find the magnitude and direction of the moment that must be applied to link 2 to drive the linkage against the forces shown. Kinematic analysis: R AO2 = 100 mm∠30° = 86.6ˆi + 50 ˆj mm
R CO4
= 250 mm∠84.34° = 24.675ˆi + 248.775 ˆj mm
= 350 mm∠67.81° = 132.2ˆi + 324.075 ˆj mm R CA = 350 mm∠34.61° = 288.075ˆi + 198.775 ˆj mm R DO = 175 mm∠84.34° = 17.275ˆi + 174.15 ˆj mm R BA
4
Force analysis: Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components, θ F34 and F34 . Then r
∑M = R × P + R × F = 0 (17.275ˆi + 174.15ˆj mm ) × ( −858.85ˆi + 231.4 ˆj N ) + (24.675ˆi + 248.775 ˆj mm ) × ( cos− 5.66°ˆi + sin − 5.66°ˆj) F = 0 O4
DO4
D
CO4
θ 34
θ 34
159
155.94kˆ N − m − 249.975 kˆ mmF 34θ
=0 ∑ M A = R BA × PB + RCA × F43θ + RCA × F43r = 0
θ F34
= 614.1 N∠ − 5.66°=609.65ˆi − 62.3ˆj N
(132.2ˆi + 324.075ˆj mm ) × ( −445ˆi N ) + ( 288.075ˆi + 198.775ˆj mm ) × ( −609.65ˆi + 62.3ˆj N ) + ( 288.075ˆi + 198.775ˆj mm ) × ( cos− 95.66°ˆi + sin − 95.66°ˆj) F = 0 r 43
146.448kˆ N − m + 141.363kˆ N − m − 267.075 kˆ mm F 43r r F = 1059.1 N∠ − 95.66°= − 102.35ˆi − 1054.65 ˆj N
= 0 ,
43
F43
= F43r + F43θ = −712ˆi − 992.35ˆj N = 1219.3∠ − 125.66°
Now the lines of action for other forces may be found as shown. θ r F = F43 + F43 + P B + F23 = 0
∑
−102.35ˆi −1054.65ˆj N − 609.65ˆi + 62.3ˆj N − 445 ˆi N + F23 = 0 , F23 =1157ˆi + 992.35ˆj N = 1526.35 N ∠40.62 ° ∑ MO = M12 + R AO × F32 = 0 2
2
(
M 12kˆ + 86.6ˆi + 50ˆj mm M 12
) × ( −1157ˆi − 992.35ˆj N ) = 0
= 28.476 N − m
M12
= 28.476kˆ N − m
Ans.
11.10 The figure shows a four-bar linkage with external forces applied at points B and C. Draw a free-body diagram of each link and show all the forces acting on each. Find the torque that must be applied to link 2 to maintain equilibrium.
Kinematic analysis: R AO2 = 75 mm∠ − 30° = 65ˆi − 38ˆj mm R BA
= 200 mm∠16.00° = 192ˆi + 55ˆj mm
R CO4 R CA
= 200 mm∠124.56° = −113ˆi +165 ˆj mm = 300 mm∠42.38° = 222ˆi + 202 ˆj mm
160
Force analysis: M A = R BA × PB
+ R ×P + R ×F = 0 ∑ (192ˆi + 55ˆj mm ) × ( −354ˆi + 354ˆj N ) + ( 222ˆi + 202ˆj mm ) × (1 800ˆi N ) + ( 222ˆi + 202ˆj mm ) × ( cos124.56°ˆi + sin124.56°ˆj) F = 0 CA
C
CA
43
43
87.438kˆ N ⋅ m − 363.000kˆ N ⋅ m + 297 kˆ mm F 43r F = 927∠124.56° N= − 526ˆi + 763ˆj N
= 0 ,
43
∑F = F
43
+ P B + PC + F23 = 0
−526ˆi + 763ˆj N − 354ˆi + 354ˆj N+1 800 ˆi N + F23 = 0 , F23 = − 920ˆi − 1 117 ˆj N = 1 447 N∠ − 129.48° ∑ MO = M12 + R AO × F32 = 0 2
2
(
M 12kˆ + 65ˆi − 38ˆj mm M 12
) × (920ˆi +1 117 ˆj N ) = 0
= −107.57 N ⋅ m
M12
= −107.57kˆ N ⋅ m
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 the magnitude and direction of the torque that must be applied to link 2 to maintain static equilibrium.
Kinematic analysis: R AO2 = 100 mm∠180 ° = −100 ˆi mm
R CO4
= 200 mm∠124.23° = −112.5ˆi +165.35 ˆj mm
= 350 mm∠55.98° = 195.85ˆi + 290.075 ˆj mm R CA = 250 mm∠41.41° = 187.5ˆi + 165.35ˆj mm R DO = 150 mm∠95.27° = −13.775ˆi +149.375 ˆj mm R BA
4
161
Force analysis: Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components, θ
r F34 and F34 . Then
∑M = R ×P + R ×F = 0 ( −13.775ˆi + 149.375ˆj mm ) × ( −694.2ˆi + 400.5ˆj N ) + ( −112.5ˆi +165.35 ˆj mm ) × ( cos 34.23°ˆi + sin 34.23 °ˆj) F = 0 O4
DO4
D
θ 34
CO4
θ 34
99.779kˆ N − m − 200kˆ mmF 34θ
=0
θ
F34
= 489.5 N∠34.23°=404.95ˆi + 275.9 ˆj N
∑M = R ×P + R ×F + R ×F = 0 (195.85ˆi + 290.075ˆj mm ) × ( −534ˆi N ) + (187.5ˆi + 165.35ˆj mm) × ( −404.95ˆi − 275.9ˆj N ) + (187.5ˆi + 165.35ˆj mm ) × ( cos− 55.77°ˆi + sin− 55.77°ˆj) F = 0 A
BA
B
CA
θ 43
CA
r 43
r 43
157.296kˆ N − m + 15.481kˆ N − m − 248.025 kˆ mm F 43r = 0 r F = 685.3 N∠ − 55.77°=387.15ˆi − 565.15 ˆj N 43
F43
= F + F43θ = −17.8ˆi − 841.05ˆj N = 841.05 N∠ − 91.21° r 43
Ans.
Now the lines of action for other forces may be found as shown. F = F43 + P B + F23 = 0
∑
−17.8ˆi − 841.05ˆj N − 534ˆi N + F23 = 0 , F23 =551.8ˆi + 841.05ˆj N = 1005.7 N∠56.73 ° Ans. Ans. F34 = − F43 = 17.8ˆi + 841.05ˆj N = 841.05 N∠88.79° ∑ F = F34 + P D + F14 = 0 17.8ˆi + 841.05 ˆj N − 694.2ˆi + 400.5 ˆj N + F14 = 0 , F =676.4ˆi − 1241.55ˆj N = 1415.1 N ∠ − 61.42 ° 14
= −F23 = −551.8ˆi − 841.05ˆj N = 1005.7 N∠ − 123.27° F12 = − F32 = 551.8ˆi + 841.05ˆj N = 1005.7 N∠56.73 ° ∑ MO = M12 + R AO × F32 = 0 F32
2
M 12kˆ + M 12
Ans. Ans. Ans.
2
( −100ˆi mm ) × ( −551.8ˆi − 841.05ˆj N ) = 0
= −85.428 N − m
M12
= −85.428kˆ N − m
Ans.
11.12 Determine the magnitude and direction of the forces that must be applied to link 2 to maintain static equilibrium.
Kinematic analysis: R AO2 = 75 mm∠90° = 75ˆj mm R BA
R CA
= 350 mm∠ − 12.37° = 341.875ˆi − 75ˆj mm
= 175 mm∠ − 34.93° = 143.475ˆi −100.225 ˆj mm
162
Force analysis: M A = R BA × PB
+ R ×P + R ×F = 0 ∑ (143.475ˆi − 100.225ˆj mm) × ( 222.5ˆj N ) + ( 341.875ˆi − 75ˆj mm ) × ( −445ˆi N ) + ( 341.875ˆi − 75ˆj mm ) × ( ˆj) F = 0 CA
C
CA
14
14
32.431kˆ N − m − 33.9kˆ N − m + 350kˆ mm F 14
∑F = P
B
M 12
O2
F14
= 4.45 N∠90°=4.45ˆj N
+ PC + F14 + F23 = 0
222.5ˆ N − 445ˆi N + 4.45ˆj N + F23
∑M
=0
= 0 ,
= M12 + R AO × F32 = 0 2
= −33.9 N − m
F23 =445ˆi − 226.95ˆj N = 498.5 N∠ − 27.02 °
(
M 12kˆ + 75ˆj mm
M12
) × ( −445ˆi + 226.95ˆj N ) = 0
= −33.9kˆ N − m
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 = 1 metric ton =1 000 kg) including the grab which is about 10 T. The maximum outreach is 30 m, which corresponds to θ 2 = 49° . Minimum outreach is 10.5 m at θ 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, O2 , and O4 , and the moment M12 required. Notice that the photograph shows a counterweight on link 2; neglect this weight and also the weights of the members. Kinematic analysis: R AO2 = 14.700 m∠49.00° = 9.644 ˆi + 11.094 ˆj m , R BO4 R BA
= 6.500 m∠2.37° = 6.494ˆi + 0.269 ˆj m
R CA
= 19.300 m∠59.70° = 9.739ˆi +16.663ˆj m = 22.300 m∠14.39° = 21.600 ˆi + 5.543ˆj m
163
Force analysis: Note that a metric ton is a unit of mass; the weight of a metric ton in standard gravity is 2 W = mg = (1 000 kg ) ( 9.81 m/s ) = 9.810 kN . Therefore the rated load of the crane is F = 98.100 kN . M A = R BA × F43 + R CA × F = 0
∑ ( 6.494ˆi + 0.269ˆj m ) × ( cos 59.70°ˆi + sin 59.70° ˆj) F + ( 21.600ˆi + 5.543ˆj m) × ( −98.100ˆj kN ) = 0 43
5.471kˆ mF 43 − 2 119kˆ kN ⋅ m = 0 , F43 F = − F = 387 kN∠59.70° =195ˆi + 334ˆj kN 14
= 387∠59.70° kN=195ˆi + 334ˆj kN Ans.
34
∑F = F
43
+ F + F23 = 0
195ˆi + 334ˆj kN − 98.1ˆj kN + F23 = 0 F = − 195ˆi − 236ˆj kN = 307 kN∠ − 129.59° 23
F12 = − F32 = − 195ˆi − 236ˆj kN = 307 kN∠ − 129.59°
∑M
O2
= M12 + R AO × F32 = 0 2
(
M 12kˆ + 9.644ˆi + 11.094ˆj m M 12
Ans.
) × ( −195ˆi − 236ˆj kN ) = 0
= −47 kN ⋅ m
M12
= −47kˆ kN ⋅ m
Ans.
11.14 Repeat Problem 11.13 for the minimum outreach position.
Kinematic analysis: R AO2 = 14.700 m∠132.00 ° = −9.836 ˆi +10.924 ˆj m , R BO4 R BA
= 6.500 m∠3.81° = 6.486ˆi + 0.432ˆj m
R CA
= 19.300 m∠120.35° = −9.751ˆi +16.656 ˆj m = 22.300 m∠15.83° = 21.454ˆi + 6.083ˆj m
164
Force analysis: Note that a metric ton is a unit of mass; the weight of a metric ton in standard gravity is 2 W = mg = (1 000 kg ) ( 9.81 m/s ) = 9.810 kN . Therefore the rated load of the crane is F = 98.100 kN . M A = R BA × F43 + R CA × F = 0
∑ ( 6.486ˆi + 0.432ˆj m ) × ( cos120.35°ˆi + sin120.35°ˆj) F + ( 21.454ˆi + 6.083ˆj m) × ( −98.1ˆj kN ) = 0 43
5.815kˆ mF 43 − 2 105kˆ kN ⋅ m = 0
F43
= 362 kN∠120.35°= − 183ˆi + 312ˆj kN
= −F34 = 362 kN∠120.35°= − 183ˆi + 312ˆj kN −183ˆi + 312ˆj kN − 98.1ˆj kN + F23 = 0 ∑ F = F43 + F + F23 = 0
F14
Ans.
F23 =183ˆi − 214ˆj kN = 282 kN∠ − 49.51° F12 = − F32 =183ˆi − 214ˆj kN = 282 kN∠ − 49.51°
∑M
O2
M 12kˆ +
Ans.
= M12 + R AO × F32 = 0 2
( −9.836ˆi + 10.924ˆj m ) × ( −183ˆi + 214ˆj kN ) = 0 = 109kˆ kN ⋅ m
Ans.
11.15 Repeat Problem 11.7 assuming coefficients of Coulomb friction
μc = 0.20 between links
M 12
= 109 kN ⋅ m
1 and 6 and
M12
μc = 0.10 between links 3 and 4. Determine the torque
M12 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 φ = tan −1 ( 0.20 ) = 11.31° . Since the impending motion VC 6 /1 is to the left the friction force f16
= μ c F 16n is toward the right.
Also, since the non-friction normal force F 16n is downward (from the solution for Prob. P11.7), the total force F16 acts at the angle
−90° + 11.31° = −78.69° . Therefore,
165
∑ F = Pˆi + ( cos− 78.69°ˆi + sin − 78.69°ˆj) F ˆj + (cos175.20 °ˆi + sin175.20 °ˆj)F 16
445 N + cos − 78.69 °F16 + cos175.20 °F 56
56
=0
= 0 , and sin − 78.69°F16 + sin175.20 °F 56 = 0 F 56 = 454.19 N
= 38.76 N F56 = 454.19 N∠175.20° = −452.6ˆi + 38ˆj N F 16
For friction between links 3 and 4, the friction angle is φ = tan −1 ( 0.10 ) = 5.71° . Since the impending motion V A3 / 4 is upward the friction force f34
= μ c F 34n is upward. Also,
since the non-friction normal force F 34n is toward the left (from the solution for Prob. P11.7), the total force F34 acts at the angle 163.37° − 5.71° = 157.66° . Therefore,
∑M = R ×F + R ×F = 0 (114.45ˆi + 383.275ˆj mm ) × (452.6ˆi − 38ˆj N ) + ( 53.4ˆi + 178.85ˆj mm ) × ( cos157.66 °ˆi + sin157.66 °ˆj) F O4
54
BO4
34
AO4
34
( −180.176 N − m + 185.725 mmF 34 ) kˆ = 0 ∑ MO = M12 + R AO × F32 = 0 2
= 957.45 N ∠157.66 °= − 885.58 ˆi + 363.94 ˆj N
2
(
M 12kˆ + 54.125ˆi + 31.25ˆj mm M 12
F34
=0
) × (885.58ˆi − 363.94 ˆj N ) = 0
= 48.117 N − m
M12
= 48.117kˆ N − m
11.16 Repeat Problem 11.12 assuming a coefficient of static friction
Ans.
μ = 0.15 between links 1
and 4. Determine the torque M12 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 φ = tan −1 ( 0.15 ) = 8.53° . Since the impending motion VC 4 /1 is to the right the friction force f14
= μ c F 14n is toward
the left. Also, since the non-friction normal force F 14n is upward (from the solution for Prob. P11.12), the total force F14 acts at the angle 90° + 8.53° = 98.53° . Therefore,
∑M = R ×P + R ×P + R ×F = 0 (143.475ˆi − 100.225ˆj mm ) × ( 222.5ˆj N ) + ( 341.875ˆi − 75ˆj mm ) × ( −445ˆi N ) + ( 341.875ˆi − 75ˆj mm ) × ( cos 98.53°ˆi + sin 98.53°ˆj) F = 0 A
BA
B
CA
C
CA
14
14
32.431kˆ N − m − 33.9kˆ N − m + 326.95kˆ mmF 14
∑F = P
B
M 12
O2
F14
= 4.42 N∠98.53°= − 0.654ˆi + 4.374ˆj N
+ PC + F14 + F23 = 0
222.5ˆ N − 445ˆi N − 0.654ˆi + 4.374ˆj N + F23
∑M
= 0 ,
= M12 + R AO × F32 = 0 2
= −33.94 N − m
= 0 ,
F23 =445.445ˆi − 226.95ˆj N = 500.18 N∠ − 26.98°
(
M 12kˆ + 75ˆj mm M12
) × ( −445.445ˆi + 226.95ˆj N ) = 0
= −33.94kˆ N − m
Ans.
166
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 of gear 3 and show all forces acting. For ( a) pinion 2 rotates at 600 rev/min and transmits 13.43 kw to the gearset. For (b) and (c), pinion 2 rotates at 900 rev/min and transmits 18.65 kw to the gearset.
a) R2
ω 2
=
=
N 2 m
=
18 teeth × 4 mm/tooth
2 ( 600 rev/min )( 2π ) 60 s/min P
R3ω 3
=
= 62.832 rad/s cw ,
ω3
R3
=
=
R2 R3
N 3m
2
ω 2
=
=
34 teeth × 4 mm/tooth 2
36 mm 68 mm
= 68 mm
62.832 rad/s = 33.264 rad/s ccw
13.43 × 1000 w × (1 N − m) × (1000 mm/m)
= 5.937 kN ( 68 mm )( 33.264 rad/s ) t F23 = F 23 cos φ = 5937 N cos20 ° = 6318 N , F43 = F 23 = 6318 N t t F13 = F23 + F 43 = 5937 + 5937 = 11.874 kN ∑ F = F23 + F43 + F13 = 0 t
F 23
=
2
= 36 mm
Ans.
b) R2
ω 2
=
=
N 2 m
=
18 teeth × 4 mm/tooth
2 ( 900 rev/min )( 2π ) 60 s/min P
= 94.248 rad/s ccw , ω3 =
=
F23
= F 23t cos φ = 5497 N
R3ω 3
R2 R3
N 3m
ω 2
2
=
=
36 teeth × 4 mm/tooth
36 mm 72 mm
2
( 72 mm )( 47.124 rad/s ) cos20 ° = 5849 N , F43 = F 23 = 5849 N
= 72 mm
94.248 rad/s = 47.124 rad/s cw
(18.65 kw) × (1000 w/kw) × (1 N − m/w) × (1000 mm/m)
F 23
t
=
2
= 36 mm R3 =
= 5497 N
167
F13
∑F = F
23
+ F43 + F13 = 0
= − ( F23 + F43 ) = − ( 5849 N∠ − 20° + 5849 N∠110 ° ) = 4945 N∠225°
Ans.
c) R2
ω 2
=
=
N 2 m
=
18 teeth × 4 mm/tooth
2 ( 900 rev/min )( 2π ) 60 s/min
2
= 36 mm R3 =
= 94.248 rad/s ccw , ω3 =
R2 R3
N 3m
2 ω 2
=
36 teeth × 4 mm/tooth 2
= 72 mm
36 mm 94.248 rad/s = 47.124 rad/s cw 72 mm
=
(18.65 kw )(1000 w/kw )(1 N − m/w ×1000 mm/m ) = 5497 N R3ω 3 ( 75 mm )( 47.124 rad/s ) t F23 = F 23 cos φ = 5497 N cos20 ° = 5849 N , F43 = F 23 = 5849 N F13 = − ( F23 + F43 ) = − ( 5849 N∠ − 20° + 5849 N∠ − 70 ° ) ∑ F = F23 + F43 + F13 = 0 = 10.6 N∠ − 45° t
F 23
=
P
=
Ans.
11.18 A 15-tooth spur pinion has a module of 5 mm/tooth and 20° pressure angle, rotates at 600 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 the forces and their proper directions. R2
ω 2
=
=
N 2 m
=
15 teeth × 5 mm/tooth
2 2 ( 600 rev/min )( 2π ) 60 s/min
= 37.5 mm
= 62.832 rad/s
N 3m
R3
=
ω3
=
2 R2 R3
=
ω 2
60 teeth × 5 mm/tooth
=
2 37.5 mm 150 mm
= 150 mm
62.832 rad/s = 15.708 rad/s
168
(18.65 kw )(1000 w/kw )(1 N − m/w ) × (1000 mm/m ) = 7915 N, F32r = F 32t tan φ = 2880.8 N R3ω 3 ( 37.5 mm )( 62.832 rad/s ) t t r r F23 = F 32 = 7915 N F23 = F 32 = 2880.8 N t
F32
P
=
=
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 between centers of the bearings and gears for shaft 3 are shown in the figure. Find the magnitude and direction of the radial force that each bearing exerts against the shaft. R2
=
N 2 m
2
=
16 teeth × 3 mm/tooth 2
= 24 mm R A =
N A m
2
=
64 teeth × 3 mm/tooth 2
= 96 mm
N m 36 teeth × 4 mm/tooth 24 teeth × 4 mm/tooth = 48 mm R4 = A = = 72 mm 2 2 2 2 (1 720 rev/min )( 2π ) R 24 mm 180.118 rad/s = 45.029 rad/s ω 2 = = 180.118 rad/s ω3 = 2 ω 2 = 60 s/min 96 mm R A
R B
=
N B m
=
ω4
=
R B
=
R4
ω 3
48 mm 72 mm
45.029 rad/s = 30.020 rad/s
( 3.73 kw )(1000 w/kw )(1 N − m/w ) × (1000 mm/m ) = 862.8 N , R Aω 3 ( 96 mm )( 45.029 rad/s ) t F23 = F 23 cos φ = 918.17 N t
=
t
=
F 23
F43
P
R A R B
=
t
F 23
=
96 mm 48 mm
862.8 N = 1725.6 N
F43
= F 43t cos φ = 1836.3 N
169
Choosing a coordinate system with origin at C as shown we have F A = F23 = 918.17 N∠20° = 862.8ˆi + 314.03ˆj lb R A = −96ˆj + 48kˆ mm F = F = 1836.3 N∠ − 20° = 1725.56 ˆi − 628 ˆj N R = 48ˆj + 240kˆ mm B
43
B
= FC x ˆi + F C y ˆj x y F D = FD ˆi + F D ˆj ∑ MC = R A × FA + R B × FB + RD × FD = 0
R C = 0
FC
R D
= 288kˆ mm
( −96 jˆ + 48kˆ mm ) × (862.8ˆi + 314.03ˆj N ) + (48ˆj + 240kˆ mm ) × (1725.56 ˆi − 628 ˆj N ) + ( 288kˆ mm ) × ( F ˆi + F ˆj) = 0 ( −15ˆi + 41.35ˆj + 82.72kˆ N − m ) + (150.7 ˆi + 414 ˆj −82.72kˆ N − m ) + ( −288 F ˆi + 288 F ˆj mm ) = 0 x D
y D
y D
= −1495 N, F Dy = 445 N ∑ F = F A + FB + FC + FD = 0 x
F D
F D
x D
= 1557.5 N∠163.42° = −1495ˆi + 445 ˆj N Ans.
(862.8ˆi + 314.03ˆj N ) + (1495ˆi − 445ˆj N ) + F + ( −1495ˆi + 445ˆj N ) = 0 C
FC = 961.2 N∠188.86° = −952.3ˆi − 146.85 ˆj N Ans. 11.20 Solve Problem 11.17 if each pinion has right-hand helical teeth with a 30 ° helix angle and a 20° pressure angle. All gears in the train are helical, and, of course, the normal module is 4 mm/tooth for each case.
170
a) R2
ω 2
=
=
N 2 m
=
18 teeth × 4 mm/tooth
2 ( 600 rev/min )( 2π ) 60 s/min
2
= 36 mm R3 =
= 62.832 rad/s cw ,
ω3
=
R2 R3
N 3m
2
ω 2
=
=
34 teeth × 4 mm/tooth 2
36 mm 68 mm
= 68 mm
62.832 rad/s = 33.264 rad/s ccw
(13.42 kw ×1000 w/kw ×1 N − m/w ×1000 mm/m )(550 ft ⋅lb/s/hp )(12 in/ft ) = 5932 N R3ω 3 ( 68 mm )( 33.268 rad/s ) Since the pressure angles and the helix angle are related by cos ψ = tan φn tan φ t , φt = tan −1 ( tan φn cos ψ ) = tan −1 ( tan 20 ° cos 30 ° ) = 22.80 ° r t a t F23 = F 23 tan φ t = ( 5932 N ) tan 22.80 ° = 2493.5 N , F23 = F 23 tanψ = ( 5932 N ) tan 30° = 3424.8 N F23 = 5932ˆi − 2493.5ˆj + 3424.8kˆ N F43 = 5923ˆi + 2493.5ˆj − 3424.8kˆ N Ans. F13 = −11.864ˆi N ∑ F = F23 + F43 + F13 = 0 ∑ M = R3ˆj × F23 − R3ˆj × F43 + M13 = 0 t
F 23
=
P
=
( 68ˆ mm ) × (5932ˆi − 2493.5ˆj + 3424.8kˆ N ) − (68ˆj mm ) × (5932 ˆi + 2493.5 ˆj −3424.8 kˆ N ) + M ( 68ˆ mm ) × (5932ˆi − 2493.5ˆj + 3424.8kˆ N ) − (68ˆj mm ) × (5932 ˆi + 2493.5 ˆj −3424.8 kˆ N ) + M M13
= −465.45ˆi N − m This moment must be supplied by the shaft bearings.
b) R2
ω 2
=
=
N 2 m
=
18 teeth × 4 mm/tooth
2 ( 900 rev/min )( 2π ) 60 s/min
2
= 36 mm R3 =
= 94.248 rad/s ccw , ω3 =
R2 R3
N 3m
ω 2
2
=
=
36 teeth × 4 mm/tooth
36 mm 72 mm
2
13
=0
13
=0
Ans.
= 72 mm
94.248 rad/s = 47.124 rad/s cw
(18.65 kw )(1000 w/kw ) × (1000 w/kw ×1000 mm/m ) = 5496.7 N R3ω 3 ( 72 mm )( 47.124 rad/s ) Since the pressure angles and the helix angle are related by cos ψ = tan φn tan φ t , φt = tan −1 ( tan φn cos ψ ) = tan −1 ( tan 20 ° cos 30 ° ) = 22.80 ° r t a t F23 = F 23 tan φ t = ( 5496.7 N ) tan 22.80 ° = 2310.59 N , F23 = F 23 tanψ = ( 5496.7 N ) tan 30° = 3173.5 N t
F 23
=
P
=
171
= 5496.7ˆi − 2310.59ˆj − 3173.5kˆ N ∑ F = F23 + F43 + F13 = 0
= −2310.59ˆi + 5496.7 ˆj + 3173.5kˆ N F13 = −3186.11ˆi − 3186.11ˆj N
F23
F43
Ans.
∑ M = R ˆj × F + R ˆi × F + M = 0 ( 72 jˆ mm ) × ( 5496.7ˆi − 2310.59ˆj − 3173.5kˆ N ) + ( 72iˆ mm ) × ( −2310.59ˆi + 5496.7ˆj + 3173.5kˆ N ) + M = 0 3
23
3
43
13
13
M13
= 228.49ˆi + 228.49ˆj N − m This moment must be supplied by the shaft bearings.
c) R2
=
N 2 m
=
18 teeth × 4 mm/tooth
= 36 mm R3 =
2 2 ( 900 rev/min )( 2π ) ω 2 = = 94.248 rad/s ccw , ω3 60 s/min t
F 23
=
P R3ω 3
=
=
R2 R3
N 3m
2 ω 2
=
=
36 teeth × 4 mm/tooth 2
36 mm 72 mm
Since the pressure angles and the helix angle are related by cos ψ
= 72 mm
94.248 rad/s = 47.124 rad/s cw
(18.65 kw) × (1000 w/kw) × (1 N − m/w) × (1000 mm/m)
( 72 mm )( 47.124 rad/s )
Ans.
= 5496.7 N
= tan φn tan φ t ,
= tan −1 ( tan φn cosψ ) = tan −1 ( tan 20° cos 30° ) = 22.80° r t F23 = F 23 tan φ t = ( 5496.7 N ) tan 22.80° = 2310.59 N , F23a = F 23t tanψ = ( 5496.7 N ) tan 30° = 3173.5 N
φt
= 5496.7ˆi − 2310.59ˆj − 3173.5kˆ N ∑ F = F23 + F43 + F13 = 0
= 2310.59ˆi − 5496.7ˆj + 3173.5kˆ N F13 = −7807.29ˆi + 7807.29ˆj N
F23
F43
Ans.
∑ M = R ˆj × F + R ˆi × F + M = 0 ( 72 jˆ mm ) × ( 5496.7ˆi − 2310.59ˆj − 3173.5kˆ N ) + ( −72iˆ mm ) × ( 2310.59ˆi − 5496.7ˆj + 3173.5kˆ N ) + M = 0 3
23
3
43
13
13
M13
= 228.49ˆi − 228.49ˆj N − m This moment must be supplied by the shaft bearings.
Ans.
172
11.21 Analyze the gear shaft of Example 11.8 and find the bearing reactions FC and FD .
The solution is shown in Fig. 11.20c. 11.22 In each of the bevel gear drives shown in the figure, bearing A takes both thrust load and radial load, while bearing B takes only radial load. The teeth are cut with a 20° pressure angle. For (a) T = −20ˆi N − m and for ( b) T = −26.7kˆ N − m . Compute the bearing 2
2
loads for each case.
Γ = tan −1 ( 32 teeth 16 teeth ) = 63.43° r t F32 = F 32 tan φ cos Γ = 180.87 N F23 = −180.87ˆi + 1111ˆj + 361.66kˆ N ∑ M A = R BA × FB + RPA × F23 + T3 = 0
a)
= T2 R2 = 20 N − m/0.018 m = 1111 N a t F32 = F 32 tan φ sin Γ = 361.66 N t
F32
( −50kˆ mm ) × ( F ˆi + F ˆj) + (35ˆi − 59kˆ mm ) × ( −180.87ˆi +1111ˆj + 361.66 kˆ N ) + T kˆ = 0 ( 50 mmF ˆi − 50 mmF ˆj) + (70ˆi −1.9ˆj + 40.68kˆ N − m ) + T kˆ = 0 x B
y B
y B
3
x B
= −40.68kˆ N − m ∑ F = F A + FB + F23 = 0
T3
b) γ = tan −1 (18 teeth 24 teeth ) = 36.87°
= F 32t tan φ cos γ = 242.95 N F32 = 242.95ˆi − 834.375ˆj + 182.21kˆ N ∑ M A = R BA × FB + RPA × F32 + T2 = 0 r
F32
3
= −37.825ˆi − 1370.6ˆj N F A = 218.695ˆi + 259.0ˆj − 361.66kˆ N F B
Ans. Ans.
= T2 R2 = 26.7 N − m 32 mm = 834.375 N a t F32 = F 32 tan φ sin γ = 182.21 N
t
F32
( 50kˆ mm ) × ( F ˆi + F ˆj) + ( −32ˆi − 20kˆ mm ) × (242.95ˆi − 834.375ˆj +182.21kˆ N ) + ( −26.7 kˆ N −m ) = 0 ( −50 mmF ˆi + 50 mmF ˆj) + ( −16.95ˆi + 0.9ˆj + 26.7kˆ N − m ) + ( −26.7 kˆ N − m ) = 0 x B
y B
y B
x B
= −17.8ˆi − 333.75ˆj N F A = −222.5ˆi + 1170.35ˆj − 182.45kˆ N F B
∑F = F
A
+ FB + F23 = 0
Ans. Ans.
11.23 The figure shows a gear train composed of a pair of helical gears and a pair of straight bevel 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 E is to take both thrust load and radial load, while bearing F is to take only radial load, determine the forces that each bearing exert against shaft 4.
The diameters of the bevel gears at their large ends are R4 = mN 4 2 = 3 mm/tooth × ( 40 teeth ) 2 = 60 mm R3
= mN 3 2 = 3 mm/tooth× ( 20 teeth ) 2 = 30 mm
173
Γ = tan −1 ( R4
γ = tan −1 ( R3 R4 ) = 26.57°
R3 ) = 63.43°
The average pitch radii are R4,avg = R4 − 0.500 sin Γ = 59.55 mm
= R3 − 0.500 sin γ = 29.78 mm
( 370 rev/min )( 2π )
= 38.746 rad/s 60 s/min P 4.5 kw × 1000 w/kw × 1 N − m/w ×1000 mm/m t F 34 = = R4,avgω 4 ( 59.55 mm )( 38.746 rad/s)
ω 4
=
R3,avg
= F 34t tan φ cos Γ = 317.46 N F34 = −634.78ˆi + 317.46ˆj − 1950kˆ N ∑ M E = R FE × FF + RPE × F34 + T4 = 0 r
a
F34
F34
= 1950 N
= F 34t tan φ sin Γ = 634.78 N
( 60ˆi mm ) × ( F ˆj + F kˆ ) + (18ˆi − 59.56ˆj mm ) × ( −634.78ˆi + 317.46 ˆj −1950 kˆ N ) + ( −115.5 ˆi N −m ) = 0 ( −60 mmF ˆ + 60 mmF kˆ ) + (115.5ˆi + 40.7 ˆj − 31kˆ N − m ) + ( −115.5ˆi N − m ) = 0 y F
z F
z F
∑F = F
E
y F
+ FF + F34 = 0
FF = 430.98ˆj + 564.192kˆ N
Ans.
F E = 634.78ˆi − 748.34ˆj + 1386.97kˆ N
Ans.
11.24 Using the data of Problem 11.23, find the forces exerted by bearings C and D onto shaft 3. Which of these bearings should take the thrust load if the shaft is to be loaded in compression?
The pitch radius of the helical gear S is RS = mN S 2 = 2 mm/tooth × 35 teeth 2
= 35 mm t t F23 = F43 R3,avg RS = 1950 N ( 29.78 mm/35 mm ) = 1659.17 N r t a t F23 = F 23 tan φ = (1659.17 N ) tan 20° = 603.89 N F23 = F 23 tanψ = (1659.17 N ) tan 30° = 957.92 N
= −603.89ˆi + 957.92ˆj + 1659.17kˆ N ∑ MC = R DC × FD + R PC × F43 + R RC × F 23 = 0
F23
( 44 jˆ mm ) × ( F ˆi + F kˆ ) + ( −29.78ˆi + 78.21ˆj mm ) × ( 634.78ˆi − 317.96ˆj + 1950kˆ N ) + ( 35ˆi + 22ˆj mm ) × ( −603.89ˆi + 957.92ˆj + 1659.17kˆ N ) = 0 ( 44 mmF ˆi − 44 mmF kˆ ) + (152.5ˆi + 58ˆj − 40kˆ N − m ) + (36.5ˆi − 58ˆj + 46.81kˆ N − m ) = 0 x D
z D
z D
x D
= 133.5ˆi − 4192kˆ N FC = −284.8ˆi − 538.45ˆj + 418.3kˆ N F D
∑F = F
C
+ FD + F23 + F43 = 0
Since the thrust force is in the
−ˆ direction, C should be a thrust bearing.
Ans. Ans.
174
11.25 Use the method of virtual work to solve the slider-crank mechanism of Problem 11.2.
= sin −1 ⎡⎣( r sin θ ) ⎤⎦ = sin −1 ⎡⎣( 75 mm sin105 ° ) 350 mm ⎤⎦ =11.95 ° x = r cos θ + cos φ = 75 mm cos105 ° + 350 mm cos11.95 ° = 323 mm φ
The first-order kinematic coefficient is x′ = dx dθ = yP24 = x tan φ = 323 mm tan11.95 ° = 68.36 mm M 12
= Px′ = 900 N ( 68.36 mm ) = 61.5 N ⋅ m cw
Ans.
11.26 Use the method of virtual-work to solve the four-bar linkage of Problem 11.5.
RP24O2
= 64.425 mm ,
RP24O4
= 114.425 mm , R DO = 175 mm∠152.64° 4
The first-order kinematic coefficient is θ 4′ = dθ 4 dθ 2 = RP24O2 RP24O4 = 64.425 mm/114.425 mm
= 0.563 M 14 = R DO P sin 152.64° = 175 mm ( 222.5 N ) sin 152.64° = 17.89 N − m cw M 12 = − M 14 dθ 4 dθ 2 = − M 14θ 4′ = − (17.89 N − m cw )( 0.563 ) = 10.07 N − m ccw 4
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 P = 445ˆi N, find and plot a graph of the crank
torque M 12 for all positions in the cycle using increments of 30° for the input crank.
= RAO cos θ 4 = 63cos θ 2 ⎛ 150 + 63sin θ 2 ⎞ θ 4 = tan −1 ⎜ ⎟ ⎝ 63cos θ 2 ⎠
x AO4
4
y AO4
= RAO sin θ 4 = 150 + 63sin θ 2
R AO4
= 1056.25 + 750sin θ 2
4
175
yP24O4
= RAO sin θ 4 4
y BC = 200sin θ 5 yP46C
= yBC ( xC
= 400 − 400sin θ 4
d θ 4
y P24O4
=
θ5
= sin −1 ( 2 − 2 sin θ 4 )
dθ 2
=
yP24O4
= 1 − 150 sin θ 4
R AO4
xCB )
= 200sin θ5 ⎡⎣( 200 cos θ5 − 400 cos θ4 ) (200 cos θ5 ) ⎤⎦ = 200 ( sin θ5 − 2 cos θ4 tan θ 5 ) dxC dθ 4 = yP O = 400 + 200sin θ5 − 400 cos θ4 tan θ5 M 12 = ( dxC dθ 4 )( dθ 4 dθ 2 ) P 46
− 150
θ 4′
4
Values for one cycle are shown in the following table. θ 2 (deg.)
θ 4 (deg.)
R AO4 (mm)
dθ 4 d θ 2
θ 5 (deg.)
dxC dθ 4 (mm)
M 12 (N − m)
0 30 60 90 120 150 180 210 240 270 300 330 360
67.38 73.37 81.30 90.00 98.70 106.63 112.62 114.50 108.05 90.00 71.95 65.50 67.38
162.5 189.15 206.5 212.5 206.5 189.15 162.5 130.5 100.85 87.5 100.85 130.5 162.5
0.14793 0.24015 0.28197 0.29412 0.28197 0.24015 0.14793 -0.04593 -0.41416 -0.71429 -0.41416 -0.04593 0.14793
-8.85 -4.80 -1.32 0.00 -1.32 -4.80 -8.85 -10.37 -5.65 0.00 -5.65 -10.37 -8.85
393.18 392.88 396.79 400 393.99 373.65 395.48 333.64 368.04 400 392.57 394.35 393.18
26.29 42.65 50.57 53.18 50.21 40.56 23.09 -6.93 -68.89 -129.14 -73.49 -8.19 26.29
176
The values of M 12 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.
θ 3′ = dθ 3 dθ 2 R B′
= RP
23 P12
RP23P13
= 75 mm 688 mm = −0.1089
= θ 3′kˆ × R BP = −0.1089kˆ × ( 568 mm∠135.33 °) = 61.881 mm ∠45.33 ° R′C = dR C d θ 2 = θ 3′kˆ × R CP = −0.1089kˆ × ( 662 mm∠124.56° ) = 72.153 mm ∠34.56 ° M 12 = P B i R′B + PC i R′C = ( 500 N∠135° )i( 61.881 mm∠45.33° ) + (1 800 N∠0° )i( 72.153 mm∠34.56° ) = 30.940cos89.67° N ⋅ m+129.876cos34.56° N ⋅ m M 12 = 107 N ⋅ m cw Ans. = dR B
d θ 2
13
13
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 the trailer 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 G2 and G3 ,
respectively, and gravity acts vertically downward in the figure. The weights of the wheels and friction in the bearings are considered negligible. Assume that there are no brakes 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 friction between the driving wheels and the road to avoid slipping.
For the trailer: R G3 B = 945ˆi + 863.15ˆj mm = 1280 mm ∠42.41 °
∑M
B
= 1250 mmF 13kˆ + RG B × W3 = 0 3
= 3364.2 N∠120° = −1682.1ˆi + 2914.75ˆj N Ans. F23 = 2278.4 N∠42.41° = 1682.1ˆi + 1535.25ˆj N ∑ F = F13 + F23 + W3 = 0
F13
For the car:
∑M
P
= 2000 mmF 12 Rkˆ − 800 mm (8900 N ) kˆ + (2900 ˆi + 325 ˆj mm ) × F32 = 0 = 5513.5ˆj N F F12 = 4921.7ˆj N
Ans.
= 1682.1ˆi N μ ≥ 0.342
Ans.
R
F12
∑F = F
F 12
+ f 12ˆi + F12R + F32 = 0
f12
μ ≥ f12 F 12F = 1682.1 N 4921.7 N
Ans.
Ans.
11.30 Repeat Problem 11.29 assuming that the car has rear-wheel drive rather than front-wheel drive.
The entire solution is identical with that of Problem P11.29 except that friction force f 12 acts on the rear wheel of the car instead of on the front wheel. The solution process and all values are the same until the final step. Then μ ≥ f12 F 12 R = 1682.1 N 5513.5 N μ ≥ 0.305 Ans.
178
11.31 The low-speed disk cam with oscillating flat-faced follower shown in the figure is driven at a constant shaft speed. The displacement curve for the cam has a full-rise cycloidal motion, defined by Eq. (5.19) with parameters L = 30° , β = 30° , and a prime circle
= 30 mm ; the instant pictured is at θ 2 = 112.5° . A force of F C = 8 N is applied at point C and remains at 45° from the face of the follower as shown. Use the radius Ro
virtual work approach to determine the moment M12 required on the crankshaft at the instant shown to produce this motion. The moment on link 3 caused by the output load is M13 = R CO3 × FC = (150 mm )(8 N ) sin −135 °kˆ = −0.849 kˆ N ⋅ m From Eq. (5.19b), 2πθ ⎞ 30° ⎛ 112.5° ⎞ L ⎛ = − 1 cos 2 y ′ = ⎜ 1 − cos π ⎟ ⎜ ⎟ = 0.200 150° ⎠ β ⎝ β ⎠ 150° ⎝ From virtual work M12
= − dθ 3
dθ 2 M13
= − y′M13 = −0.200 ( −0.849kˆ N ⋅ m ) = 0.170kˆ N ⋅ m
Ans.
11.32 Repeat Problem 11.31 for the entire lift portion of the cycle, finding M12 as a function of
θ 2 .
From Problem P11.31 M13 = R CO3 × FC = (150 mm )(8 N ) sin −135 °kˆ = −0.849 kˆ N ⋅ m 360°θ 2 ⎞ 2πθ ⎞ 30° ⎛ = 1 − cos 1 − cos ⎜ ⎟ ⎜ ⎟ = 0.200 (1 − cos 2.4θ 2 ) β ⎝ β ⎠ 150° ⎝ 150° ⎠ M12 = − y′M13 = −0.200 (1 − cos 2.4θ 2 ) ( −0.849 kˆ N ⋅ m ) = 0.170 (1 − cos 2.4θ 2 ) kˆ N ⋅m
y ′ =
L ⎛
Ans.
11.33 A disk 3 of radius R is being slowly rolled under a pivoted bar 2 driven by an applied torque T as shown in the figure. Assume a coefficient of static friction of μ 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 d from the pivot O2 . Assume that the weights of the links are negligible in comparison to F . Find an equation for the torque T required as a function of the distance X = RCO2 , and an equation for the final distance X that is reached when friction no longer allows further movement.
